url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
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https://www.analystforum.com/forums/cfa-forums/cfa-level-i-forum/91371439 | 1,566,401,995,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316075.15/warc/CC-MAIN-20190821152344-20190821174344-00310.warc.gz | 703,928,080 | 18,619 | # Justified P/E
Justified P/E = 4.3
If actual P/E is 16, then the asset is overpriced. Completely understood!
But why it’s underpriced if the actual P/E is 7?? yet it’s above the justified ratio of 4.3, so it should be overpriced as well, shouldn’t?!!
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It should be.
Where did you get this question?
Simplify the complicated side; don't complify the simplicated side.
Financial Exam Help 123: The place to get help for the CFA® exams
http://financialexamhelp123.com/
Kaplan V4. R.49 page 304, the paragraph below the blue box
Unfortunately, I don’t have their materials.
What’s their rationale?
Simplify the complicated side; don't complify the simplicated side.
Financial Exam Help 123: The place to get help for the CFA® exams
http://financialexamhelp123.com/
Don’t know! no further explanation provided
Hi Baidar & S2000magician -
See below. This was poorly worded IMO.
“A firm has an expected dividend payout ratio of 30%, a required rate of return of 13%, and an expected dividend growth rate of 6%. Calculate the firm’s fundamental (justified) leading P/E ratio”…one will compute 4.5 (.3/(.13-.06)).
The text then says:
“The justified P/E ratio serves as a benchmark for the price at which the stock should trade. In the previous example, if the firm’s actual P/E ratio (based on the market price and expected earnings) was 16, the stock would be considered overvalued. If the firm’s market P/E ratio was 7, the stock would be considered undervalued”….
Sorry but didn’t get you! Yet the 7 market P/E > justified 4.5 thus it should be overvalued | 460 | 1,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-35 | latest | en | 0.898471 |
http://www.jiskha.com/display.cgi?id=1315859992 | 1,462,520,215,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861735203.5/warc/CC-MAIN-20160428164215-00052-ip-10-239-7-51.ec2.internal.warc.gz | 600,000,378 | 3,584 | Friday
May 6, 2016
# Homework Help: Physics
Posted by Elaine Jetton on Monday, September 12, 2011 at 4:39pm.
A rocket is launched at an angle of 56.0° above the horizontal with an initial speed of 96 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 31.0 m/s2. At this time, its engines fail and the rocket proceeds to move as a projectile.
(a) Find the maximum altitude reached by the rocket.(in meters)
(b) Find its total time of flight.(in seconds)
(c) Find its horizontal range. (in meters) | 146 | 539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2016-18 | longest | en | 0.942957 |
http://cdn1.www.cbsenotes.com/notes/topic/xi-maths/permutations-and-combinations | 1,513,559,256,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948599549.81/warc/CC-MAIN-20171218005540-20171218031540-00356.warc.gz | 60,138,016 | 19,411 | Welcome to cbsenotes.com! To post messages or access any member only section, you will need an account. Create your free account now.
CBSE Notes 2011-2012 » XI-Maths
# Permutations and Combinations
## There are four routes between Delhi and Mumbai. In how many ways can a person go from Delhi to Mumba...
There are four routes between Delhi and Mumbai. In how many ways can a person go from Delhi to Mumbai and return, if for returning
(i) any route is taken, (ii) the same route is taken, (iii) the same route is not taken ?
## How may words are there (with or without meaning) of three distinct alphabets ?
How may words are there (with or without meaning) of three distinct alphabets ?
## How many odd numbers less than 10,000 can be formed using the digits 0, 2, 3, 5 allowing repetition ...
How many odd numbers less than 10,000 can be formed using the digits 0, 2, 3, 5 allowing repetition of digits ?
## How many 4 digit odd numbers can be formed with the help of the digits 1,2, 3,4, 5, if (i) no digit ...
How many 4 digit odd numbers can be formed with the help of the digits 1,2, 3,4, 5, if
(i) no digit is repeated, [NCERT] (ii) digits are repeated ? | 316 | 1,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-51 | longest | en | 0.910471 |
https://www.est.colpos.mx/web/packages/mlogit/vignettes/e2nlogit.html | 1,721,912,623,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763858305.84/warc/CC-MAIN-20240725114544-20240725144544-00238.warc.gz | 644,781,291 | 7,389 | # Exercise 2: Nested logit model
#### 2020-10-02
The data set HC from mlogit contains data in R format on the choice of heating and central cooling system for 250 single-family, newly built houses in California.
The alternatives are:
• Gas central heat with cooling gcc,
• Electric central resistence heat with cooling ecc,
• Electric room resistence heat with cooling erc,
• Electric heat pump, which provides cooling also hpc,
• Gas central heat without cooling gc,
• Electric central resistence heat without cooling ec,
• Electric room resistence heat without cooling er.
Heat pumps necessarily provide both heating and cooling such that heat pump without cooling is not an alternative.
The variables are:
• depvar gives the name of the chosen alternative,
• ich.alt are the installation cost for the heating portion of the system,
• icca is the installation cost for cooling
• och.alt are the operating cost for the heating portion of the system
• occa is the operating cost for cooling
• income is the annual income of the household
Note that the full installation cost of alternative gcc is ich.gcc+icca, and similarly for the operating cost and for the other alternatives with cooling.
1. Run a nested logit model on the data for two nests and one log-sum coefficient that applies to both nests. Note that the model is specified to have the cooling alternatives (gcc},ecc}, erc},hpc}) in one nest and the non-cooling alternatives (gc},ec}, er}) in another nest.
library("mlogit")
data("HC", package = "mlogit")
HC <- dfidx(HC, varying = c(2:8, 10:16), choice = "depvar")
cooling.modes <- idx(HC, 2) %in% c('gcc', 'ecc', 'erc', 'hpc')
room.modes <- idx(HC, 2) %in% c('erc', 'er')
# installation / operating costs for cooling are constants,
# only relevant for mixed systems
HC$icca[! cooling.modes] <- 0 HC$occa[! cooling.modes] <- 0
# create income variables for two sets cooling and rooms
HC$inc.cooling <- HC$inc.room <- 0
HC$inc.cooling[cooling.modes] <- HC$income[cooling.modes]
HC$inc.room[room.modes] <- HC$income[room.modes]
# create an intercet for cooling modes
HC\$int.cooling <- as.numeric(cooling.modes)
# estimate the model with only one nest elasticity
nl <- mlogit(depvar ~ ich + och +icca + occa + inc.room + inc.cooling + int.cooling | 0, HC,
nests = list(cooling = c('gcc','ecc','erc','hpc'),
other = c('gc', 'ec', 'er')), un.nest.el = TRUE)
summary(nl)
##
## Call:
## mlogit(formula = depvar ~ ich + och + icca + occa + inc.room +
## inc.cooling + int.cooling | 0, data = HC, nests = list(cooling = c("gcc",
## "ecc", "erc", "hpc"), other = c("gc", "ec", "er")), un.nest.el = TRUE)
##
## Frequencies of alternatives:choice
## ec ecc er erc gc gcc hpc
## 0.004 0.016 0.032 0.004 0.096 0.744 0.104
##
## bfgs method
## 11 iterations, 0h:0m:0s
## g'(-H)^-1g = 7.26E-06
## successive function values within tolerance limits
##
## Coefficients :
## Estimate Std. Error z-value Pr(>|z|)
## ich -0.554878 0.144205 -3.8478 0.0001192 ***
## och -0.857886 0.255313 -3.3601 0.0007791 ***
## icca -0.225079 0.144423 -1.5585 0.1191212
## occa -1.089458 1.219821 -0.8931 0.3717882
## inc.room -0.378971 0.099631 -3.8038 0.0001425 ***
## inc.cooling 0.249575 0.059213 4.2149 2.499e-05 ***
## int.cooling -6.000415 5.562423 -1.0787 0.2807030
## iv 0.585922 0.179708 3.2604 0.0011125 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Log-Likelihood: -178.12
1. The estimated log-sum coefficient is $$0.59$$. What does this estimate tell you about the degree of correlation in unobserved factors over alternatives within each nest?
The correlation is approximately $$1-0.59=0.41$$. It's a moderate correlation.
1. Test the hypothesis that the log-sum coefficient is 1.0 (the value that it takes for a standard logit model.) Can the hypothesis that the true model is standard logit be rejected?
We can use a t-test of the hypothesis that the log-sum coefficient equal to 1. The t-statistic is :
(coef(nl)['iv'] - 1) / sqrt(vcov(nl)['iv', 'iv'])
## iv
## -2.304171
The critical value of t for 95% confidence is 1.96. So we can reject the hypothesis at 95% confidence.
We can also use a likelihood ratio test because the multinomial logit is a special case of the nested model.
# First estimate the multinomial logit model
ml <- update(nl, nests = NULL)
lrtest(nl, ml)
## Likelihood ratio test
##
## Model 1: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
## 0
## Model 2: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
## 0
## #Df LogLik Df Chisq Pr(>Chisq)
## 1 8 -178.12
## 2 7 -180.29 -1 4.3234 0.03759 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Note that the hypothesis is rejected at 95% confidence, but not at 99% confidence.
1. Re-estimate the model with the room alternatives in one nest and the central alternatives in another nest. (Note that a heat pump is a central system.)
nl2 <- update(nl, nests = list(central = c('ec', 'ecc', 'gc', 'gcc', 'hpc'),
room = c('er', 'erc')))
summary(nl2)
##
## Call:
## mlogit(formula = depvar ~ ich + och + icca + occa + inc.room +
## inc.cooling + int.cooling | 0, data = HC, nests = list(central = c("ec",
## "ecc", "gc", "gcc", "hpc"), room = c("er", "erc")), un.nest.el = TRUE)
##
## Frequencies of alternatives:choice
## ec ecc er erc gc gcc hpc
## 0.004 0.016 0.032 0.004 0.096 0.744 0.104
##
## bfgs method
## 10 iterations, 0h:0m:0s
## g'(-H)^-1g = 5.87E-07
##
## Coefficients :
## Estimate Std. Error z-value Pr(>|z|)
## ich -1.13818 0.54216 -2.0993 0.03579 *
## och -1.82532 0.93228 -1.9579 0.05024 .
## icca -0.33746 0.26934 -1.2529 0.21024
## occa -2.06328 1.89726 -1.0875 0.27681
## inc.room -0.75722 0.34292 -2.2081 0.02723 *
## inc.cooling 0.41689 0.20742 2.0099 0.04444 *
## int.cooling -13.82487 7.94031 -1.7411 0.08167 .
## iv 1.36201 0.65393 2.0828 0.03727 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Log-Likelihood: -180.02
1. What does the estimate imply about the substitution patterns across alternatives? Do you think the estimate is plausible?
The log-sum coefficient is over 1. This implies that there is more substitution across nests than within nests. I don't think this is very reasonable, but people can differ on their concepts of what's reasonable.
1. Is the log-sum coefficient significantly different from 1?
(coef(nl2)['iv'] - 1) / sqrt(vcov(nl2)['iv', 'iv'])
## iv
## 0.5535849
lrtest(nl2, ml)
## Likelihood ratio test
##
## Model 1: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
## 0
## Model 2: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
## 0
## #Df LogLik Df Chisq Pr(>Chisq)
## 1 8 -180.02
## 2 7 -180.29 -1 0.5268 0.468
We cannot reject the hypothesis at standard confidence levels.
1. How does the value of the log-likelihood function compare for this model relative to the model in exercise 1, where the cooling alternatives are in one nest and the heating alternatives in the other nest.
logLik(nl)
## 'log Lik.' -178.1247 (df=8)
logLik(nl2)
## 'log Lik.' -180.0231 (df=8)
The $$\ln L$$ is worse (more negative.) All in all, this seems like a less appropriate nesting structure.
1. Rerun the model that has the cooling alternatives in one nest and the non-cooling alternatives in the other nest (like for exercise 1), with a separate log-sum coefficient for each nest.
nl3 <- update(nl, un.nest.el = FALSE)
1. Which nest is estimated to have the higher correlation in unobserved factors? Can you think of a real-world reason for this nest to have a higher correlation?
The correlation in the cooling nest is around 1-0.60 = 0.4 and that for the non-cooling nest is around 1-0.45 = 0.55. So the correlation is higher in the non-cooling nest. Perhaps more variation in comfort when there is no cooling. This variation in comfort is the same for all the non-cooling alternatives.
1. Are the two log-sum coefficients significantly different from each other? That is, can you reject the hypothesis that the model in exercise 1 is the true model?
We can use a likelihood ratio tests with models nl and nl3.
lrtest(nl, nl3)
## Likelihood ratio test
##
## Model 1: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
## 0
## Model 2: depvar ~ ich + och + icca + occa + inc.room + inc.cooling + int.cooling |
## 0
## #Df LogLik Df Chisq Pr(>Chisq)
## 1 8 -178.12
## 2 9 -178.04 1 0.1758 0.675
The restricted model is the one from exercise 1 that has one log-sum coefficient. The unrestricted model is the one we just estimated. The test statistics is 0.6299. The critical value of chi-squared with 1 degree of freedom is 3.8 at the 95% confidence level. We therefore cannot reject the hypothesis that the two nests have the same log-sum coefficient.
1. Rewrite the code to allow three nests. For simplicity, estimate only one log-sum coefficient which is applied to all three nests. Estimate a model with alternatives gcc, ecc and erc in a nest, hpc in a nest alone, and alternatives gc, ec and er in a nest. Does this model seem better or worse than the model in exercise 1, which puts alternative hpc in the same nest as alternatives gcc, ecc and erc?
nl4 <- update(nl, nests=list(n1 = c('gcc', 'ecc', 'erc'), n2 = c('hpc'),
n3 = c('gc', 'ec', 'er')))
summary(nl4)
##
## Call:
## mlogit(formula = depvar ~ ich + och + icca + occa + inc.room +
## inc.cooling + int.cooling | 0, data = HC, nests = list(n1 = c("gcc",
## "ecc", "erc"), n2 = c("hpc"), n3 = c("gc", "ec", "er")),
## un.nest.el = TRUE)
##
## Frequencies of alternatives:choice
## ec ecc er erc gc gcc hpc
## 0.004 0.016 0.032 0.004 0.096 0.744 0.104
##
## bfgs method
## 8 iterations, 0h:0m:0s
## g'(-H)^-1g = 3.71E-08
##
## Coefficients :
## Estimate Std. Error z-value Pr(>|z|)
## ich -0.838394 0.100546 -8.3384 < 2.2e-16 ***
## och -1.331598 0.252069 -5.2827 1.273e-07 ***
## icca -0.256131 0.145564 -1.7596 0.07848 .
## occa -1.405656 1.207281 -1.1643 0.24430
## inc.room -0.571352 0.077950 -7.3297 2.307e-13 ***
## inc.cooling 0.311355 0.056357 5.5247 3.301e-08 ***
## int.cooling -10.413384 5.612445 -1.8554 0.06354 .
## iv 0.956544 0.180722 5.2929 1.204e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Log-Likelihood: -180.26`
The $$\ln L$$ for this model is $$-180.26$$, which is lower (more negative) than for the model with two nests, which got $$-178.12$$. | 3,579 | 10,847 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-30 | latest | en | 0.822925 |
https://www.transtutors.com/questions/question-4-a-state-bernoulli-s-equation-in-term-of-the-total-head-and-give-the-physi-5519976.htm | 1,601,307,914,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401601278.97/warc/CC-MAIN-20200928135709-20200928165709-00556.warc.gz | 1,011,070,209 | 14,013 | Question 4 (A) State Bernoulli’s equation in term of the total head and give the physical... 1 answer below »
Question 4(A) State Bernoulli’s equation in term of the total head and give the physical interpretation of each term in the equation. Also state the conditions to which the equation may be applied for and conditions it may be inapplicable.(B) Water is being siphoned from a large tank through a pipe of diameter 30mm, discharge through a nozzle of diameter 20mm at a height 2.5m below the water level in the tank as shown in below.Temperature of the water is 20oC, density and vapour pressure of water are 1000 kg/m3 and 2.337 kN/m2 respectively. Assume frictional loss in the tubing is neglectable and atmospheric pressure is 101 kPa.(i) The discharge of the tube(ii) The absolute pressure at the crest of the tube if h = 1.8m(iii) The maximum crest height h above the water level
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http://www.mactech.com/articles/mactech/Vol.13/13.05/May97Challenge/index.html | 1,430,161,054,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246659319.74/warc/CC-MAIN-20150417045739-00013-ip-10-235-10-82.ec2.internal.warc.gz | 625,857,928 | 31,689 | May 97 Challenge
Volume Number: 13 (1997)
Issue Number: 5
Column Tag: Programmer's Challenge
Programmer's Challenge
By Bob Boonstra, Westford, MA
Equation Evaluator
Those of you with PowerPCs have probably experimented with the Graphing Calculator application installed by default into the Apple Menu Items folder. As one of the first native applications for the PowerPC, the ability of the Graphing Calculator to display, and even animate 2-dimensional and 3-dimensional equations demonstrating the computing power of the PowerPC chip using native code. Underlying the display capability is code to parse an equation and rapidly compute the equation value for a range of input values. In this month's Challenge, you will produce code to perform these parsing and computation functions.
The prototype for the code you should write is:
```typedef unsigned long ulong;
typedef struct Values {
float first; /* first equation input value */
float delta; /* first+delta is second input value */
ulong howMany; /* number of input values */
} Values;
typedef struct IntValues {
long first;/* first equation input value */
long delta;/* first+delta is second input value */
ulong howMany; /* number of input values */
} IntValues;
typedef struct Results {
float equationResult; /* result of equation given x,y,n */
float x; /* input value of x producing equationResult */
float y; /* input value of x producing equationResult */
long n;/* input value of x producing equationResult */
} Results;
void Evaluate(
char *equation, /* null-terminated equation to evaluate */
const Values *xP, /* equation input values for x */
const Values *yP, /* equation input values for y */
const IntValues *nP,/* equation input values for n */
Results w[]/* preallocated storage for equation values */
);
```
The input equation you are to evaluate will be provided as a null-terminated string in the 2-dimensional curve form (y=x+2x^2) or the 3-dimensional surface form (z=r cos(t) r sin(t)). You are to evaluate the equation for each of the values of x and n (in the case of a 2-dimensional equation) or x, y, and n (in the 3-dimensional case) provided and return the results in the preallocated array w. Each input is described as an initial value, a delta between each value and the next value, and the number howMany of values this input parameter is to assume. For example, if x is to take on the range of values [1.0, 1.1, 1.2, 2.0], then the x input could be described as:
``` xP->first = 1.0; xP->delta = .1; xP->howMany = 11
```
In the event that the equation does not contain one of the input parameters, that parameter should be ignored. There is no guarantee, for example, that nP->howMany will be zero when the input equation is not a function of n. Similarly, for a 2-dimensional equation, yP should be ignored.
The input equation might be written as a function of r and q, which should be calculated from x and y just as the Graphing Calculator does. Because the Graphing Calculator displays equations in ways that cannot be directly represented in a character string, the following symbols will be used in the equation to represent the operator or value indicated:
\ square root
/ division
^ exponentiation
p pi (p)
t theta (q)
. multiplication (also represented by juxtaposition)
Standard algebraic operator precedence should be used: exponentiation and square roots, then multiplication and division, then addition and subtraction, with left-to-right evaluation order for operators of equal precedence, and with parentheses used to override normal precedence. Arguments to trigonometric functions will always be surrounded by parentheses. Where the Graphing Calculator would use superscripts to indicate an extended exponent, parentheses will be used to make the meaning clear (e.g., e^(x+y)).
Store the equation result for the i-th combination of input values in w[i]->equationResult, and store the corresponding input values in w[i]->x, w[i]->y, and w[i]->n. The results may be stored in any order, but each input combination should appear exactly once. Results should be calculated with a minimum relative accuracy of .00001.
Even though the Graphing Calculator handles inequalities, multiple equations, differentiation, simplification, and expansion, your code does not need to deal with these cases. With these exceptions, your code should handle any equation that the Graphing Calculator can handle.
You may allocate any amount of dynamic storage up to 20MB, provided you deallocate the storage before returning. This will be a native PowerPC Challenge, using the CodeWarrior environment. Solutions may be coded in C, C++, or Pascal. Limited use of assembly language is also permitted, for anyone who might need it to access any dynamically generated code as part of their solution. The solution computing the correct results in the least amount of time will be the winner.
Three Months Ago Winner
Congratulations to Jeff Mallett (Boulder Creek, CA) for producing the winning entry among ten submitted for the Othello Challenge. The objective was to win a round-robin Othello tournament, generalized to allow a board size between 8 and 64 (even numbers only), by as large a margin as possible, using as little computation time as possible. Tournament points were based on the margin of victory (the number of the winner's pieces showing at the end of the game minus the number of the opponent's pieces showing), and on the amount of computation time used, as follows:
[margin of victory - seconds of execution time / 30] / number of squares
The test cases included board sizes of 8x8 and 18x18, with each player competing against each other player twice, once playing first, with the black pieces, and once playing second, with the white pieces. I had planned to run some larger cases, but the first two still had not completed after running all night, so I had to stop at 18x18.
The solutions submitted varied considerably in complexity. The simplest (and fastest) solutions assigned values to each board position and made the move which maximized total value. Some solutions took that approach one step further and evaluated an opponent's potential responses and used a min/max approach to select the best move. Other solutions took credit for the number of pieces flipped on a move, with possible consideration for vulnerability to reversal on the next move. The winning solution used the most complicated approach, with lines of play being examined by a progressively deepening search.
The table below describes how each player fared against each other player. Each row shows the number of victories that player had against the player represented by the corresponding column. The final column shows the total number of victories won out of the 36 games played by each entry. As you can see, the second place entry by Randy Boring won nearly as many games as Jeff's winning entry, and actually beat the winning entry in one of the four games they played.
Player Name 1 2 3 4 5 6 7 8 9 10 Wins 1 David Whitney - 1 4 2 4 2 2 0 0 4 19 2 Dan Farmer 3 - 4 2 4 4 4 1 0 4 26 3 David McGavran 0 0 - 2 4 1 1 0 1 1 10 4 Eric Hangstefer 2 2 2 - 4 0 1 0 0 3 14 5 Ken Slezak 0 0 0 0 - 0 0 0 0 0 0 6 Gregory Cooper 2 0 3 4 4 - 2 0 1 4 20 7 Mason Thomas 2 0 3 3 4 2 - 0 0 3 17 8 Randy Boring 4 3 4 4 4 4 4 - 1 4 32 9 Jeff Mallett 4 4 3 4 4 3 4 3 - 4 33 10 Ludovic Nicolle 0 0 3 1 4 0 1 0 0 - 9
The top two entries used significantly more computation time than the others. Jeff's winning entry used an average of more than one second per move (as you can see by examining the parameter settings in his code), but the larger margin of victory more than offset the execution time penalty.
The table below provides the code and data sizes for each of the solutions submitted, along with the total number of victories won in all of the test cases, and the cumulative score earned in those victories. Numbers in parenthesis after a person's name indicate that person's cumulative point total for all previous Challenges, not including this one.
Player Code Data Wins Score Jeff Mallett (44) 6988 277 33 25.49 Randy Boring (27) 6908 64 32 20.37 Gregory Cooper (27) 4764 284 20 15.00 Dan Farmer 9240 96 26 14.27 David Whitney 7216 864 19 9.79 Eric Hangstefer 4124 80 14 9.72 Mason Thomas (4) 6976 57 17 9.01 David McGavran 3272 48 10 8.09 Ludovic Nicolle (21) 6436 120 9 6.33 Ken Slezak (20) 9256 77 0 0.00
Sample Game
Here is one of the games played by the top two entries. Randy Boring's entry played Black, and Jeff Mallett's played White. The moves are given as the row and column selected by the programs, interspersed with an occasional view of the resulting board position.
Move Black (Boring) White (Mallett) row col row col 1 2 3 4 2 2 5 2 2 4 3 2 5 3 5 4 4 5 3 6 5 2 2 3 2 6 2 6 5 3
```- - - - - - - -
- - - - - - - -
- - B B B B B -
- - W W W B W -
- - W W B B - -
- - B W - - - -
- - - - - - - -
- - - - - - - -
```
7 3 1
...missing an opportunity to place a piece on the edge at (2,7), inviting (1,7) as a response by white, in hope of moving to (0,7)
5 4 8 4 1 1 5 9 3 7
```- - - - - - - -
- - - - - W - -
- - B B B W B -
- B B B B B B B
- B B B B W - -
- - B W W - - -
- - - - - - - -
- - - - - - - -
```
Black occupies the first edge square.
2 7 10 1 7 1 3 11 1 2 4 6 12 0 3 5 5 13 6 2 0 2
```- - W B - - - -
- - B W - W - B
- - B B W W B B
- B B W B B W B
- B B W B W W -
- - B B W W - -
- - B - - - - -
- - - - - - - -
```
Black's next move, to (0,1), prevents White from obtaining a foothold on one of the edges.
14 0 1 2 1 15 3 0 2 0 16 1 0 5 0 17 5 1 6 3 18 0 4 4 0 19 6 0 7 2 20 1 4 1 6
```- B B B B - - -
B - B B B B W B
B W B B B W W B
B B B W B B W B
B B W W B W W -
B B W W W W - -
B - W W - - - -
- - W - - - - -
```
While white had other choices, none of them were very good, and this move to (1,6) gives Black a corner on the next move.
21 0 7 6 1 22 7 0
Black takes a second corner ...
```- B B B B - - B
B - B B B B B B
B W B B B B W B
B W B W B B W B
B W W B B W W -
B W B W W W - -
B B W W - - - -
B - W - - - - -
```
7 1 23 7 3 7 4 24 6 5 6 4 25 7 5 6 6 26 7 7
and a third
```- B B B B - - B
B - B B B B B B
B W B B B B W B
B B B W B B W B
B W B B B B W -
B B W B B B - -
B W B W W W B -
B B B B B B - B
```
5 6 27 7 6 6 7 28 1 1
```- B B B B - - B
B B B B B B B B
B B B B B B W B
B B B W B B W B
B W B B B B W -
B B W W B W W -
B W B W W W W W
B B B B B B B B
```
Black could have done better with (5,7) at this point, instead of giving the final corner to White.
0 0 29 4 7 5 7 30 - - 0 5 31 0 6
```W W W W W W B B
B W B B W B B B
B B W W B W W B
B B W W B W B B
B W B B B W B B
B B W W B W W W
B W B W W B W W
B B B B B B B B
```
Black wins by a score of 37 to 27, a relatively close game compared to many in the tournament.
TOP 20 CONTESTANTS
Here are the point totals for the Top Contestants in the Programmer's Challenge. The numbers below include points awarded over the 24 most recent contests, including points earned by this month's entrants.
Rank Name Points Rank Name Points
1. Munter, Ernst 184 11. Cutts, Kevin 21
2. Gregg, Xan 114 12. Nicolle, Ludovic 21
3. Larsson, Gustav 47 13. Picao, Miguel Cruz 21
4. Lengyel, Eric 40 14. Brown, Jorg 20
5. Boring, Randy 37 15. Gundrum, Eric 20
6. Cooper, Greg 34 16. Higgins, Charles 20
7. Lewis, Peter 32 17. Kasparian, Raffi 20
8. Mallett, Jeff 27 18. Slezak, Ken 20
9. Antoniewicz, Andy 24 19. Studer, Thomas 20
10. Beith, Gary 24 20. Karsh, Bill 19
There are three ways to earn points: (1) scoring in the top 5 of any Challenge, (2) being the first person to find a bug in a published winning solution or, (3) being the first person to suggest a Challenge that I use. The points you can win are:
1st place 20 points 5th place 2 points 2nd place 10 points finding bug 2 points 3rd place 7 points suggesting Challenge 2 points 4th place 4 points
Here is Jeff's winning solution:
Jello.c An nxn Othello program in C
```/*
* Uses alpha-beta search with iterative deepening, transposition tables,
* extension for solving, futility cut-off, a simplistic selectivity, etc.
*/
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define MY_ASSERT(b)
Engine Parameters
// Extensions/Pruning
#define kFullDepthPlies 2 // # of full-depth plies before selectivity
#define kSolveThreshold 4 // Plies extension for solving whole board
#define kMinimumSafeDisks3 // Minimum disks to have and still be safe
#define kFutilityScore kCorner
// Score above beta to trigger futility cut-off
#define kWipeOutExtension 2
// Extend a ply if opponent doesn't have more disks than this
// Average move time in 60ths of a second
#define kOpeningMoveTime 50// In the opening
#define kMoveTime 150 // Normally
#define kSolveMoveTime 800 // When trying to solve
// Scoring Parameters
#define kFinalDisk50// per disk on board at the end
#define kTooFewDisks-50 // per disk under threshold
Constants/Macros
#define kInfinity110000000L
#define kBestPossible100000000L
// Directions (if se increases both x and y):
// 7 6 5 4 3 2 1 0
// NE SW NW SE N S W E (opposites are adjacent)
enum { DIR_E = 0, DIR_W, DIR_S, DIR_N,
DIR_SE, DIR_NW, DIR_SW, DIR_NE };
enum { E = 0x0001,
W = 0x0002,
S = 0x0004,
N = 0x0008,
SE = 0x0010,
NW = 0x0020,
SW = 0x0040,
NE = 0x0080,
E_BORDER = 0x0100,
W_BORDER = 0x0200,
S_BORDER = 0x0400,
N_BORDER = 0x0800,
SE_BORDER = 0x1000,
NW_BORDER = 0x2000,
SW_BORDER = 0x4000,
NE_BORDER = 0x8000
};
// Array of squares on the board (up to 66 x 66 squares)
// Each square has:
// 1st 8 bits - In this direction there's a BLACK disk that can be flipped by WHITE
#define WHITE 0x00010000 // Is White disk here?
#define BLACK 0x00020000 // Is Black disk here?
#define COLOR_BITS 0x00030000 // Mask: Is disk here?
#define BORDER_BIT 0x00040000 // Is this border square?
#define COLOR_BORDER_BITS 0x00070000 // Mask: Is border or disk here?
#define BAD_BIT 0x00080000 // X or C square
#define EMPTYADJ_BITS0xFFF00000 /* Top 12 bits (0-4095) */
#define COLOR_SHIFT16
#define IS_NOT_EMPTY(z) ((z) & COLOR_BORDER_BITS)
#define IS_EMPTY(z)!IS_NOT_EMPTY(z)
#define HAS_DISK(z)((z) & COLOR_BITS)
#define HAS_NO_DISK(z) !HAS_DISK(z)
#define IS_BORDER(z) ((z) & BORDER_BIT)
#define IS_ON_BOARD(z) !IS_BORDER(z)
#define IS_CORNER(z) (gCountZeros[((z) & BORDER_ADJACENCY) \
#define OPPONENT(side) ((side) ^ COLOR_BITS)
#define XY2INDEX(x, y) ((y) * gRealBoardSize + (x))
#define COUNT_EMPTIES(z) \
gCountZeros[ ((z) | ((z) >> \
#define DIR_BIT(dir) (1L << (dir))
#define OPP_DIR_BIT(dir) gOppDirBit[dir]
// Add to end of list
// Swap entry in list with end entry and shrink list
}
#define PUSH(x) *(gChangesEnd++) = (x)
#define START_SAVE PUSH(0L)
#define PUSH_SQ(pSq) \
{ PUSH(*(pSq)); PUSH((unsigned long)(pSq)); }
#define POP *(-gChangesEnd)
#define TOP *gChangesEnd
typedef unsigned long * PSQUARE;
typedef long SCORE;
#define USE_HASH
#ifdef USE_HASH
#define kHashTableMask 0x7FFF // 32K entry table
#define kSwitchSideHash 0x87654321
enum { INVALID = 0, VALID = 1,
LOWER_BOUND = 2, UPPER_BOUND = 3 };
typedef struct SHash {
unsigned long HashValue;
SCORE Score;
PSQUARE BestMove;
short Depth;
short Type;
} SHash;
static SHash *gHashTable;
static long gWhiteHashOffset, gBlackHashOffset,
gFlipHashOffset;
static unsigned long gHashValue;
#endif
Prototypes
Boolean /*legalMove*/ Othello (
long boardSize, /* number of rows/columns in the game board */
long oppRow, /* row where opponent last moved, 0 .. boardSize-1 */
long oppCol, /* column where opponent last moved, 0 .. boardSize-1 */
long *moveRow, /* return your move - row, 0 .. boardSize-1 */
long *moveCol, /* return your move - column, 0 .. boardSize-1 */
void *privStorage, /* preallocated storage for your use */
long storageSize, /* size of privStorage in bytes */
Boolean newGame, /* TRUE if this is your first move in a new game */
Boolean playBlack /* TRUE if you play first (black pieces) */
);
static SCORE Search(SCORE alpha, SCORE beta, unsigned long side, long
depth, Boolean passEndsGame);
static long Generate(unsigned long side);
static SCORE MakeMove(PSQUARE to, unsigned long side);
static void UnmakeMove();
static void Initialize(long boardSize, void *privStorage);
static SCORE Evaluate(unsigned long side);
static long SortValue(PSQUARE p, unsigned long side);
static void BubbleSort(long n, unsigned long side);
Static Variables
// Direction indices
static unsigned long gOppDirBit[8] =
{ W, E, N, S, NW, SE, NE, SW };
// Offsets on board plus zero sentinel
// Fill in gOffsets[2..7] when we know board size
static long gOffsets[9] = {1L,-1L,99L,99L,99L,99L,99L,99L,0L};
// gSquares - Array of squares: Stores board
static unsigned long *gSquares;
static unsigned long *gOnBoardStart; // Pointer into gSquares
static unsigned long *gOnBoardEnd; // Pointer into gSquares
static long gNumOnSquares;// # of squares, not including borders
static long gRealBoardSize; // Length of a side (includes borders)
// gEmptyAdj - Array of pointers to squares:
// Stores all empty squares adjacent to disk(s)
// gChanges - Array of unsigned longs containing data to undo moves
// list of:
// <pointer to square> <old square value>
// terminated by a OL
// The first position will be the drop square and the others will be flips
static unsigned long *gChanges;
static unsigned long *gChangesEnd; // Pointer into gChanges
// gCountZeros - Precalculated 256-element constant array
// returns count of zero bits in the byte
static unsigned long *gCountZeros;
// gTree - Array of pointers to squares: Holds search tree
static PSQUARE *gTree;
static PSQUARE *gTreeEnd; // Pointer into gTree
// gMobility - Array of counts of moves available at each ply
static long *gMobility;
// Pointers to various special squares
static PSQUARE gNWCorner, gNWX, gNWC1, gNWC2;
static PSQUARE gNECorner, gNEX, gNEC1, gNEC2;
static PSQUARE gSWCorner, gSWX, gSWC1, gSWC2;
static PSQUARE gSECorner, gSEX, gSEC1, gSEC2;
// The gCounts array holds current counts of disks
// Access is by gCounts[side >> COLOR_SHIFT]
// gCounts[WHITE_INDEX] white's disks
// gCounts[BLACK_INDEX] black's disks
#define WHITE_INDEX(WHITE >> COLOR_SHIFT)
#define BLACK_INDEX(BLACK >> COLOR_SHIFT)
static unsigned long gCounts[3];
static SCORE gIncScore; // Score relative to side to move
static SCORE gEndgameDiskBonus; // Bonus per disk in endgame
static SCORE kX, kC, kEdge, kCorner; // Penalties/Bonuses
static long gAbortSearchTime; // Time at which the search will be aborted
static Boolean gAborted; // Has the search been aborted?
static long gStartDepth; // Search was started at this depth
static long gPly;// Number of moves deep
Othello
Boolean /*legalMove*/ Othello (
long boardSize, /* number of rows/columns in the game board */
long oppRow, /* row where opponent last moved, 0 .. boardSize-1 */
long oppCol, /* column where opponent last moved, 0 .. boardSize-1 */
long *moveRow, /* return your move - row, 0 .. boardSize-1 */
long *moveCol, /* return your move - column, 0 .. boardSize-1 */
void *privStorage, /* preallocated storage for your use */
long storageSize, /* size of privStorage in bytes */
Boolean newGame, /* TRUE if this is your first move in a new game */
Boolean playBlack /* TRUE if you play first (black pieces) */
)
{
PSQUARE *to, p;
unsigned long side = playBlack ? BLACK : WHITE;
unsigned long nextSide;
SCORE t, bestScore, saveIncScore;
long j, generated, index, x, y;
long stillOpen, bestFoundAt, *pOffsets;
long startTime, targetTime, targetDuration;
Boolean nearEdge;
#ifdef USE_HASH
unsigned long saveHashValue;
#endif
if (newGame) {
Initialize(boardSize, privStorage);
}
gChangesEnd = gChanges;
#ifdef USE_HASH
// Fix up gHashTable
{
SHash *pHashTable = gHashTable;
int i;
i = kHashTableSize - 1;
do {
(pHashTable++)->Depth -= 2;
} while (i-);
}
#endif
if (oppRow != -1) {
index = XY2INDEX(oppCol+1, oppRow+1);
gIncScore -=
MakeMove(&gSquares[index], playBlack ? WHITE : BLACK);
gChangesEnd = gChanges;
}
gTreeEnd = gTree;
generated = Generate(side);
if (!generated) { // No moves
*moveRow = *moveCol = -1;
return TRUE;
}
if (generated > 1 && !(newGame && oppRow == -1)) {
BubbleSort(generated, side);
gMobility[0] = gMobility[1] = generated;
gPly = 1;
nextSide = OPPONENT(side);
stillOpen = gNumOnSquares - gCounts[WHITE_INDEX] -
gCounts[BLACK_INDEX];
// Calculate gEndgameDiskBonus
gEndgameDiskBonus = 0;
x = gNumOnSquares / 3;
j = x - stillOpen;
if (j > 0) {
gEndgameDiskBonus = (j * kFinalDisk) / (x * 5);
if (!gEndgameDiskBonus)
gEndgameDiskBonus = 1;
}
// Do we have any pieces near an edge?
nearEdge = false;
for (p = gOnBoardStart;
!nearEdge && p <= gOnBoardEnd; ++p) {
if (HAS_DISK(*p)) {
if (IS_EDGE(*p)) {
nearEdge = true;
} else {
pOffsets = gOffsets;
do {
if (IS_EDGE(*(p + *pOffsets)))
nearEdge = true;
} while (!nearEdge && *(++pOffsets));
}
}
}
// Set times
if (!nearEdge)
targetDuration = kOpeningMoveTime;
else if (stillOpen > 20)
targetDuration = kMoveTime;
else // try to solve!
targetDuration = kSolveMoveTime;
startTime = LMGetTicks();
targetTime = startTime + targetDuration;
gAbortSearchTime = startTime + targetDuration * 6;
gAborted = false;
saveIncScore = gIncScore;
#ifdef USE_HASH
saveHashValue = gHashValue;
#endif
for (gStartDepth=1; gStartDepth < stillOpen && !gAborted;
++gStartDepth) {
to = gTree;
bestScore = -kInfinity;
for (j=0; j<generated; ++j) {
gIncScore = - (gIncScore + MakeMove(*to, side));
++gPly;
t = -Search(-kInfinity, kInfinity, nextSide,
gStartDepth - 1, false);
-gPly;
UnmakeMove();
gIncScore = saveIncScore;
#ifdef USE_HASH
gHashValue = saveHashValue;
#endif
if (gAborted)
break;
if (t > bestScore) {
PSQUARE bestMove, *p;
// Move *to to front of the tree
bestMove = *to;
MY_ASSERT(bestMove >= gOnBoardStart && bestMove <=
gOnBoardEnd);
for (p = to; p != gTree; -p)
*p = *(p-1);
*gTree = bestMove;
bestScore = t;
bestFoundAt = gStartDepth;
}
if (LMGetTicks() >= targetTime) {
if (bestScore > -kInfinity && gStartDepth +
kSolveThreshold + 1 != stillOpen)
break; // time to stop
if (LMGetTicks() - startTime
>= 3 * targetDuration)
break; // time to stop
}
++to;
}
if (gStartDepth >= 4 && LMGetTicks() - startTime > 1 +
targetDuration/2)
break; // probably not enough time to finish another iteration
if (gStartDepth - bestFoundAt == 3
&& IS_CORNER(*gTree[0]))
break; // easy corner move
}
}
gIncScore += MakeMove(*gTree, side);
index = (long)(*gTree - gSquares);
y = index / gRealBoardSize;
x = index - y * gRealBoardSize;
*moveCol = x - 1;
*moveRow = y - 1;
return TRUE;
}
SortValue
// Returns value that orders squares for root search
long SortValue(PSQUARE p, unsigned long side)
{
long stillOpen, value;
PSQUARE q;
unsigned long occupant, enemy;
long *pOffsets;
if (IS_EDGE(*p)) {
if (IS_CORNER(*p))
return 200; // Corner
return -100; // C
return 100; // Edge
}
return -200; // X
stillOpen = gNumOnSquares - gCounts[WHITE_INDEX] -
gCounts[BLACK_INDEX];
enemy = OPPONENT(side);
value = 0;
pOffsets = gOffsets;
do {
q = p + *pOffsets;
occupant = *q & COLOR_BITS;
if (stillOpen > 10) {
if (occupant == side)
++value; // good to take away empty-adjacent
else if (occupant == enemy)
} else if (occupant == OPPONENT(side))
++value; // endgame: good to flip
} while (*(++pOffsets));
return value;
}
BubbleSort
// Sorts generated root moves in decreasing value order
// Hey, bubble sort is really okay in this case
void BubbleSort(long n, unsigned long side)
{
long i, j, swaps;
PSQUARE temp;
for (j=n-2; j>=0; -j) {
swaps = 0;
i = 0;
do {
if (SortValue(gTree[i+1], side)
> SortValue(gTree[i], side)) {
++swaps; // Swap i and i+1
temp = gTree[i];
gTree[i] = gTree[i+1];
gTree[i+1] = temp;
}
} while (i++ != j);
if (!swaps)
break; // Already sorted: Finish early
}
}
Evaluate
// Evaluate position and return score relative to side
// side is also the side to move
SCORE Evaluate(unsigned long side)
{
SCORE score = 0;
// Maximize disks, but only in endgame
if (gEndgameDiskBonus) {
score = (gCounts[WHITE_INDEX] - gCounts[BLACK_INDEX]) * gEndgameDiskBonus;
if (side == BLACK)
score = -score;
}
// NW Corner Area
if (HAS_DISK(*gNWCorner)) {
score += (*gNWCorner & side) ? kCorner : -kCorner;
} else if (*gNWCorner & ADJACENCY) {
if (HAS_DISK(*gNWX)) {
score += (*gNWX & side) ? kX : -kX;
}
if (HAS_DISK(*gNWC1)) {
score += (*gNWC1 & side) ? kC : -kC;
}
if (HAS_DISK(*gNWC2)) {
score += (*gNWC2 & side) ? kC : -kC;
}
}
// NE Corner Area
if (HAS_DISK(*gNECorner)) {
score += (*gNECorner & side) ? kCorner : -kCorner;
} else if (*gNECorner & ADJACENCY) {
if (HAS_DISK(*gNEX)) {
score += (*gNEX & side) ? kX : -kX;
}
if (HAS_DISK(*gNEC1)) {
score += (*gNEC1 & side) ? kC : -kC;
}
if (HAS_DISK(*gNEC2)) {
score += (*gNEC2 & side) ? kC : -kC;
}
}
// SW Corner Area
if (HAS_DISK(*gSWCorner)) {
score += (*gSWCorner & side) ? kCorner : -kCorner;
} else if (*gSWCorner & ADJACENCY) {
if (HAS_DISK(*gSWX)) {
score += (*gSWX & side) ? kX : -kX;
}
if (HAS_DISK(*gSWC1)) {
score += (*gSWC1 & side) ? kC : -kC;
}
if (HAS_DISK(*gSWC2)) {
score += (*gSWC2 & side) ? kC : -kC;
}
}
// SE Corner Area
if (HAS_DISK(*gSECorner)) {
score += (*gSECorner & side) ? kCorner : -kCorner;
} else if (*gSECorner & ADJACENCY) {
if (HAS_DISK(*gSEX)) {
score += (*gSEX & side) ? kX : -kX;
}
if (HAS_DISK(*gSEC1)) {
score += (*gSEC1 & side) ? kC : -kC;
}
if (HAS_DISK(*gSEC2)) {
score += (*gSEC2 & side) ? kC : -kC;
}
}
// Too few disks?
if (gCounts[WHITE_INDEX] < kMinimumSafeDisks) {
SCORE x = (kMinimumSafeDisks - gCounts[WHITE_INDEX]) *
kTooFewDisks;
score += (side == WHITE) ? x : -x * 2;
}
if (gCounts[BLACK_INDEX] < kMinimumSafeDisks) {
SCORE x = (kMinimumSafeDisks - gCounts[BLACK_INDEX]) *
kTooFewDisks;
score += (side == BLACK) ? x : -x * 2;
}
// Mobility
score += gMobility[gPly-2] - gMobility[gPly-1];
// Could also have a value for the right to move
return gIncScore + score;
}
Search
SCORE Search(SCORE alpha, SCORE beta, unsigned long side, long depth,
Boolean passEndsGame)
{
register PSQUARE *to;
unsigned long nextSide;
long generated;
SCORE t, bestScore, saveIncScore;
PSQUARE *firstMove;
long stillOpen;
SCORE oldAlpha = alpha;
#ifdef USE_HASH
unsigned long saveHashValue;
SHash *pHashTable;
#endif
stillOpen = gNumOnSquares - gCounts[WHITE_INDEX] -
gCounts[BLACK_INDEX];
if (!stillOpen) {
// Board full
bestScore = (gCounts[WHITE_INDEX] -
gCounts[BLACK_INDEX]) * kFinalDisk;
if (side == BLACK)
bestScore = -bestScore;
return bestScore;
}
if (!gCounts[WHITE_INDEX]) {
// White is wiped out!
bestScore = kBestPossible;
if (side == WHITE)
bestScore = -kBestPossible;
return bestScore;
}
if (!gCounts[BLACK_INDEX]) {
// Black is wiped out!
bestScore = kBestPossible;
if (side == BLACK)
bestScore = -kBestPossible;
return bestScore;
}
if (depth <= 0) {
if (stillOpen > kSolveThreshold &&
gCounts[OPPONENT(side) >> COLOR_SHIFT]
> kWipeOutExtension) {
bestScore = Evaluate(side);
#ifdef USE_HASH
bestMove = NULL;
#endif
goto HashSave; // Terminal node
}
} else if (depth == 1 && stillOpen > kSolveThreshold) {
t = Evaluate(side);
if (t > beta + kFutilityScore)
return t; // Futility cut-off
}
#ifdef USE_HASH
if (pHashTable->HashValue == gHashValue) {
if (pHashTable->Depth >= depth) {
if (pHashTable->Type == VALID) {
if (pHashTable->Score > beta)
alpha = beta;
else if (pHashTable->Score > alpha)
alpha = pHashTable->Score;
} else if (pHashTable->Type == LOWER_BOUND) {
if (pHashTable->Score > beta)
return beta + 1;
} else if (pHashTable->Type == UPPER_BOUND) {
if (pHashTable->Score < alpha)
return alpha - 1;
}
if (alpha > beta)
return beta;
}
}
#endif
// Abort?
if (LMGetTicks() >= gAbortSearchTime) {
gAborted = true;
return 0;
}
#ifdef USE_HASH
bestMove = NULL;
#endif
nextSide = OPPONENT(side);
firstMove = gTreeEnd;
generated = Generate(side);
gMobility[gPly] = generated;
if (!generated) { // no moves available
if (passEndsGame) {
bestScore = (gCounts[WHITE_INDEX] -
gCounts[BLACK_INDEX]) * kFinalDisk;
if (side == BLACK)
bestScore = -bestScore;
return bestScore;
}
#ifdef USE_HASH
gHashValue ^= kSwitchSideHash;
#endif
gIncScore = -gIncScore;
++gPly;
bestScore = -Search(-beta, -alpha,
nextSide, depth, true);
-gPly;
gIncScore = -gIncScore;
++depth;
goto Searched;
}
to = firstMove;
bestScore = -kInfinity;
#ifdef USE_HASH
// Find tableReply in move list and move to front
PSQUARE *p;
for (p = to + 1; *p; ++p)
// Swap *p and *to
*p = *to;
break;
}
}
#endif
saveIncScore = gIncScore;
#ifdef USE_HASH
gHashValue ^= kSwitchSideHash;
saveHashValue = gHashValue;
#endif
do {
#ifdef USE_HASH
!bestMove ||
#endif
gStartDepth - depth <= kFullDepthPlies ||
stillOpen <= kSolveThreshold) {
gIncScore = - (gIncScore + MakeMove(*to, side));
++gPly;
t = -Search(-beta, -alpha, nextSide, depth-1, false);
-gPly;
UnmakeMove();
gIncScore = saveIncScore;
#ifdef USE_HASH
gHashValue = saveHashValue;
#endif
if (gAborted) {
#ifdef USE_HASH
bestMove = NULL;
#endif
break;
}
if (t > bestScore) {
#ifdef USE_HASH
bestMove = *to;
#endif
if (t > alpha) {
if (t >= beta) {
bestScore = t;
break;
}
alpha = t;
}
bestScore = t;
}
}
++to;
} while (*to);
Searched: ;
#ifdef USE_HASH
gHashValue ^= kSwitchSideHash;
#endif
gTreeEnd = firstMove;
#ifdef USE_HASH
if (bestMove) {
#endif
HashSave: ;
#ifdef USE_HASH
if (pHashTable->Depth <= depth) {
pHashTable->HashValue = gHashValue;
pHashTable->Depth = depth;
pHashTable->Score = bestScore;
pHashTable->BestMove = bestMove;
MY_ASSERT(!bestMove || IS_EMPTY(*bestMove));
pHashTable->Type = VALID;
if (bestScore <= oldAlpha) {
pHashTable->Type = UPPER_BOUND;
} else if (bestScore >= beta) {
pHashTable->Type = LOWER_BOUND;
}
}
}
#endif
return bestScore;
}
Generate
long Generate(unsigned long side)
{
PSQUARE *e, p, q, *afterCorners, *movesStart, lastBad;
unsigned long enemy;
long i, *pOffsets;
enemy = OPPONENT(side);
afterCorners = movesStart = gTreeEnd;
while (i-) {
p = *e;
pOffsets = gOffsets;
do {
q = p + *pOffsets;
if (*q & enemy) {
do { // Skip through line of enemy disks
q += *pOffsets;
} while (*q & enemy);
if (*q & side) { // Add square p to move list!
// Bad? Try to put it on the end
break;
}
break;
}
if (!IS_CORNER(*p)) {
// Add to end after corners
break;
}
// Corner: keep all corners on front
if (afterCorners == gTreeEnd) { // All are corners!
} else {
unsigned long *temp = *afterCorners;
*afterCorners = p;
}
++afterCorners;
break;
}
}
} while (*(++pOffsets));
++e;
}
}
return (long)(gTreeEnd - movesStart) - 1;
}
MakeMove
// Makes the move for "side" on the "to" square
// Saves the move to gChanges for later undo'ing
// Returns the change in score relative to the moving side
SCORE MakeMove(PSQUARE to, unsigned long side)
{
unsigned long z;
PSQUARE q, p;
long dir, offset;
long flips = 0;
SCORE score = 0;
unsigned long enemy = OPPONENT(side);
#ifdef USE_HASH
// Update hash for new disk
if (side == BLACK)
gHashValue ^= *(to + gBlackHashOffset);
else
gHashValue ^= *(to + gWhiteHashOffset);
#endif
START_SAVE;
dir = 7;
do {
offset = gOffsets[dir];
q = to + offset;
if (IS_ON_BOARD(*q)) {
if (IS_EMPTY(*q)) {
}
score-; // New disk touches empty, which is bad
} else if (*q & enemy) {
// Skip through line of enemy disks
p = q + offset;
while (*p & enemy)
p += offset;
if (*p & side) { // Flip ‘em
p = q;
do {
// Flip disk at p
++flips;
if (IS_EDGE(*p))
score += kEdge;
PUSH_SQ(p);
score -= (COUNT_EMPTIES(*p) << 1);
// x2 Empties counted for us now
*p ^= COLOR_BITS; // Flip
#ifdef USE_HASH
gHashValue ^= *(p + gFlipHashOffset);
// Update hash for flipped disk
#endif
p += offset;
} while (*p & enemy);
score++;
// Fills in area around disk at q which is now ours
} else {
score-;
// Fills in area around disk at enemy disk at q
}
} else { // *q & side
score++;
// Fills in area around our disk at q
}
z = OPP_DIR_BIT(dir);
MY_ASSERT((*q & z) == 0L);
*q |= z;
}
} while (dir-);
PUSH_SQ(to);
*to |= side;
PUSH(gCounts[WHITE_INDEX]);
PUSH(gCounts[BLACK_INDEX]);
gCounts[side >> COLOR_SHIFT] += flips + 1;
gCounts[enemy >> COLOR_SHIFT] -= flips;
if (IS_EDGE(*to))
score += kEdge;
return score;
}
UnmakeMove
void UnmakeMove()
{
PSQUARE to, flipped, q;
unsigned long z;
long dir;
// Restore disk counts
gCounts[BLACK_INDEX] = POP;
gCounts[WHITE_INDEX] = POP;
// Replace to-disk
to = (PSQUARE)POP;
*to = POP;
// Undo disk changes
while (POP) {
flipped = (PSQUARE)TOP;
*flipped = POP;
}
dir = 7;
do {
q = to + gOffsets[dir];
z = OPP_DIR_BIT(dir);
*q &= ~z; // Remove adjacency (if not already removed from disk changes)
if ( IS_EMPTY(*q) && !(*q & ADJACENCY) ) { // First adjacent
}
} while (dir-);
}
Initialize
void Initialize(long boardSize, void *privStorage)
{
unsigned long *p, *q;
long i, index, numRealSquares;
char *ptr;
unsigned long z;
#ifdef USE_HASH
unsigned long r1, r2;
#endif
kX = -2 - boardSize * 5;
kC = -2 - boardSize * 4;
kEdge = 3 + boardSize / 8;
kCorner = 2 + boardSize * 13;
gRealBoardSize = boardSize + 2;
gNumOnSquares = boardSize * boardSize;
numRealSquares = gRealBoardSize * gRealBoardSize;
ptr = privStorage;
gSquares = (unsigned long *)ptr;
gOnBoardStart = gSquares + gRealBoardSize + 1;
gOnBoardEnd = gSquares + (gRealBoardSize * (gRealBoardSize - 1) -
2);
ptr += numRealSquares * 4;
// worst case 66*66*4 = 17,424 bytes (~17K)
#ifdef USE_HASH
gWhiteHashOffset = numRealSquares;
gBlackHashOffset = numRealSquares * 2;
gFlipHashOffset = numRealSquares * 3;
ptr += (numRealSquares * 3) * 4;
// worst case 66*66*3*4 = 52,272 bytes (~51K)
gHashTable = (SHash *)ptr;
ptr += kHashTableSize * sizeof(SHash);
// 32768*16 = 524,288 bytes (512K)
#endif
ptr += gNumOnSquares * 4;
// worst case 64*64*4 = 16,384 bytes (16K)
gChanges = (unsigned long *)ptr;
ptr += 65536; // e.g. 64 moves (deep) * 256 longs/move * 4 bytes/long
// 65,536 bytes (64K)
gMobility = (long *)ptr;
ptr += 1024; // e.g. 256 moves deep * 4 bytes/long
// 1,024 bytes (1K)
gCountZeros = (unsigned long *)ptr;
ptr += 256 * 4;
// 256*4 = 1,024 bytes (1K)
gTree = (PSQUARE *)ptr;
// Gets what's left, almost 400K!
// ** Calculate directional offsets
gOffsets[DIR_S] = gRealBoardSize;
gOffsets[DIR_N] = - gRealBoardSize;
gOffsets[DIR_SE] = gRealBoardSize + 1;
gOffsets[DIR_NW] = - gRealBoardSize - 1;
gOffsets[DIR_NE] = - gRealBoardSize + 1;
gOffsets[DIR_SW] = gRealBoardSize - 1;
// ** Borders
// Upper/Lower
p = gSquares;
q = gSquares + (gRealBoardSize * (gRealBoardSize - 1));
i = gRealBoardSize;
do {
*(p++) = BORDER_BIT;
*(q++) = BORDER_BIT;
} while (-i);
// Sides
p = gSquares + gRealBoardSize;
i = gRealBoardSize - 2;
do {
*p = BORDER_BIT;
p += (gRealBoardSize - 1);
*(p++) = BORDER_BIT;
} while (-i);
// ** Edges
// Upper/Lower
p = gOnBoardStart;
q = gOnBoardEnd;
i = boardSize;
do {
*(p++) = NW_BORDER | N_BORDER | NE_BORDER;
*(q-) = SW_BORDER | S_BORDER | SE_BORDER;
} while (-i);
// Sides
p = gOnBoardStart;
q = gOnBoardEnd;
i = boardSize;
do {
*p |= NW_BORDER | W_BORDER | SW_BORDER;
*q |= NE_BORDER | E_BORDER | SE_BORDER;
p += gRealBoardSize;
q -= gRealBoardSize;
} while (-i);
// ** Starting configuration
// Set up initial disks and adjacent empty squares
// "On your first move, you should initialize the board
// with white tiles at (row,col) = (boardSize/2-1,boardSize/2-1) and
// (boardSize/2,boardSize/2), and black tiles at (boardSize/2-1,boardSize/2)
// and (boardSize/2,boardSize/2-1)"
gCounts[WHITE_INDEX] = gCounts[BLACK_INDEX] = 2;
i = boardSize >> 1; // x2
index = XY2INDEX(i-1, i-1);
p = &gSquares[index];
*p = SE; gEmptyAdj[0] = p;
p += gRealBoardSize - 3;
*(++p) = WHITE | S | SE | E;
*(++p) = BLACK | W | SW | S;
p += gRealBoardSize - 3;
*(++p) = BLACK | N | NE | E;
*(++p) = WHITE | W | NW | N;
p += gRealBoardSize - 3;
gNWCorner = gOnBoardStart;
gNWC1 = gNWCorner + 1; *gNWC1 |= BAD_BIT;
gNWC2 = gNWCorner + gRealBoardSize; *gNWC2 |= BAD_BIT;
gNWX = gNWC2 + 1; *gNWX |= BAD_BIT;
gNECorner = gNWCorner + boardSize - 1;
gNEC1 = gNECorner - 1; *gNEC1 |= BAD_BIT;
gNEC2 = gNECorner + gRealBoardSize; *gNEC2 |= BAD_BIT;
gNEX = gNEC2 - 1; *gNEX |= BAD_BIT;
gSWCorner = gOnBoardEnd - boardSize + 1;
gSWC1 = gSWCorner + 1; *gSWC1 |= BAD_BIT;
gSWC2 = gSWCorner - gRealBoardSize; *gSWC2 |= BAD_BIT;
gSWX = gSWC2 + 1; *gSWX |= BAD_BIT;
gSECorner = gOnBoardEnd;
gSEC1 = gSECorner - 1; *gSEC1 |= BAD_BIT;
gSEC2 = gSECorner - gRealBoardSize; *gSEC2 |= BAD_BIT;
gSEX = gSEC2 - 1; *gSEX |= BAD_BIT;
// Precalculate gCountZeros
// (Could have had the compiler fill these in, but I'm not
// THAT desperate for speed)
for (z=0; z<256; ++z) {
gCountZeros[z] =
8 - (z & 1) - ((z>>1) & 1) - ((z>>2) & 1) - ((z>>3) & 1) -
((z>>4) & 1) - ((z>>5) & 1) - ((z>>6) & 1) - ((z>>7) & 1);
}
#ifdef USE_HASH
gHashValue = 0xFFFFFFFF;
// Initialize gHashKeys
srand(0x1234); //srand(time(NULL));
for (i=0; i<numRealSquares; ++i) {
r1 = rand() + ((unsigned long)rand() << 16);
r2 = rand() + ((unsigned long)rand() << 16);
gSquares[gWhiteHashOffset + i] = r1;
gSquares[gBlackHashOffset + i] = r2;
gSquares[gFlipHashOffset + i] = r1 ^ r2;
}
// Clear gHashTable
{
SHash *pHashTable = gHashTable;
i = kHashTableSize - 1;
do {
pHashTable->HashValue = 0;
pHashTable->Depth = -100;
pHashTable->BestMove = NULL;
pHashTable->Type = INVALID;
pHashTable->Score = 0;
++pHashTable;
} while (i-);
}
#endif
gIncScore = 0;
}
```
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…experience . Full knowledge of Mac OS X and prior . Mac OSX / Server . Apple Remote Desktop . Process Documentation . Ability to prioritize multiple tasks in a fast pace Read more | 13,628 | 45,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2015-18 | longest | en | 0.664668 |
https://socratic.org/questions/how-do-you-prove-10sin-x-cos-x-6cos-x | 1,582,458,452,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00193.warc.gz | 570,890,919 | 6,170 | # How do you prove 10sin(x)cos(x)=6cos(x)?
May 19, 2018
If we simplify the equation by dividing both sides by $\cos \left(x\right)$, we obtain:
$10 \sin \left(x\right) = 6$, which implies
$\sin \left(x\right) = \frac{3}{5.}$
The right triangle which $\sin \left(x\right) = \frac{3}{5}$ is a 3:4:5 triangle, with legs $a = 3$, $b = 4$ and hypotenuse $c = 5$. From this we know that if $\sin \left(x\right) = \frac{3}{5}$ (opposite over hypotenuse), then $\cos = \frac{4}{5}$ (adjacent over hypotenuse). If we plug these identities back into the equation we reveal its validity:
$10 \left(\frac{3}{5}\right) \cdot \left(\frac{4}{5}\right) = 6 \left(\frac{4}{5}\right)$.
This simplifies to
$\frac{24}{5} = \frac{24}{5}$.
Therefore the equation is true for $\sin \left(x\right) = \frac{3}{5.}$ | 285 | 797 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2020-10 | latest | en | 0.766649 |
https://www.lessonplanet.com/search?ccss%5B%5D=195908&keywords= | 1,601,215,318,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400279782.77/warc/CC-MAIN-20200927121105-20200927151105-00618.warc.gz | 910,822,903 | 27,329 | ### We found465 reviewed resources
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Problem Solutions
Class Definition Exercises
These exercises are considerably more sophisticated then the exercises in previous parts. Each of these sections describes a small project that requires you to create a number of distinct classes which must collaborate to produce a useful result.
Stock Valuation
A `Block` of stock has a number of attributes, including a purchase price, purchase date, and number of shares. Commonly, methods are needed to compute the total spent to buy the stock, and the current value of the stock. A `Position` is the current ownership of a company reflected by all of the blocks of stock. A `Portfolio` is a collection of `Positions`; it has methods to compute the total value of all `Blocks` of stock.
When we purchase stocks a little at a time, each `Block` has a different price. We want to compute the total value of the entire set of `Block`s, plus an average purchase price for the set of `Block`s.
The StockBlock class. First, define a `StockBlock` class which has the purchase date, price per share and number of shares. Here are the method functions this class should have.
`__init__`
The `__init__` method will populate the individual fields of date, price and number of shares. Don't include the company name or ticker symbol; this is information which is part of the `Position`, not the individual blocks.
`__str__`
The `__str__` method must return a nicely formatted string that shows the date, price and shares.
`getPurchValue`
The `getPurchValue` method should compute the value as purchase price per share × shares.
``` getSaleValue ```( `salePrice` )
The `getSaleValue` method requires a `salePrice` ; it computes the value as sale price per share × shares.
``` getROI ```( `salePrice` )
The `getROI` method requires a `salePrice` ; it computes the return on investment as (sale value - purchase value) ÷ purchase value.
We can load a simple database with a piece of code the looks like the following. The first statement will create a sequence with four blocks of stock. We chose variable name that would remind us that the ticker symbols for all four is 'GM'. The second statement will create another sequence with four blocks.
```blocksGM = [
StockBlock( purchDate='25-Jan-2001', purchPrice=44.89, shares=17 ),
StockBlock( purchDate='25-Apr-2001', purchPrice=46.12, shares=17 ),
StockBlock( purchDate='25-Jul-2001', purchPrice=52.79, shares=15 ),
StockBlock( purchDate='25-Oct-2001', purchPrice=37.73, shares=21 ),
]
blocksEK = [
StockBlock( purchDate='25-Jan-2001', purchPrice=35.86, shares=22 ),
StockBlock( purchDate='25-Apr-2001', purchPrice=37.66, shares=21 ),
StockBlock( purchDate='25-Jul-2001', purchPrice=38.57, shares=20 ),
StockBlock( purchDate='25-Oct-2001', purchPrice=27.61, shares=28 ),
]
```
The Position class. A separate class, `Position`, will have an the name, symbol and a sequence of `StockBlocks` for a given company. Here are some of the method functions this class should have.
`__init__`
The `__init__` method should accept the company name, ticker symbol and a collection of `StockBlock` instances.
`__str__`
The `__str__` method should return a string that contains the symbol, the total number of shares in all blocks and the total purchse price for all blocks.
`getPurchValue`
The `getPurchValue` method sums the purchase values for all of the `StockBlocks` in this `Position`. It delegates the hard part of the work to each `StockBlock`'s `getPurchValue` method.
``` getSaleValue ```( `salePrice` )
The `getSaleValue` method requires a `salePrice` ; it sums the sale values for all of the `StockBlocks` in this `Position`. It delegates the hard part of the work to each `StockBlock`'s `getSaleValue` method.
`getROI`
The `getROI` method requires a `salePrice` ; it computes the return on investment as (sale value - purchase value) ÷ purchase value. This is an ROI based on an overall yield.
We can create our `Position` objects with the following kind of initializer. This creates a sequence of three individual `Position` objects; one has a sequence of GM blocks, one has a sequence of EK blocks and the third has a single CAT block.
```portfolio= [
Position( "General Motors", "GM", blocksGM ),
Position( "Eastman Kodak", "EK", blocksEK )
Position( "Caterpillar", "CAT",
[ StockBlock( purchDate='25-Oct-2001',
purchPrice=42.84, shares=18 ) ] )
]
```
An Analysis Program. You can now write a main program that writes some simple reports on each `Position` object in the `portfolio`. One report should display the individual blocks purchased, and the purchase value of the block. This requires iterating through the `Positions` in the `portfolio`, and then delegating the detailed reporting to the individual `StockBlocks` within each `Position`.
Another report should summarize each position with the symbol, the total number of shares and the total value of the stock purchased. The overall average price paid is the total value divided by the total number of shares.
In addition to the collection of `StockBlock`s that make up a `Position`, one additional piece of information that is useful is the current trading price for the `Position`. First, add a `currentPrice` attribute, and a method to set that attribute. Then, add a `getCurrentValue` method which computes a sum of the `getSaleValue` method of each `StockBlock`. using the trading price of the `Position`.
Annualized Return on Investment. In order to compare portfolios, we might want to compute an annualized ROI. This is ROI as if the stock were held for eactly one year. In this case, since each block has different ownership period, the annualized ROI of each block has to be computed. Then we return an average of each annual ROI weighted by the sale value.
The annualization requires computing the duration of stock ownership. This requires use of the `time` module. We'll cover that in depth in Chapter 32, Dates and Times: the `time` and `datetime` Modules . The essential feature, however, is to parse the date string to create a time object and then get the number of days between two time objects. Here's a code snippet that does most of what we want.
````>>> `
`import time`
`>>> `
`dt1= "25-JAN-2001"`
`>>> `
`timeObj1= time.strptime( dt1, "%d-%b-%Y" )`
`>>> `
`dayNumb1= int(time.mktime( timeObj1 ))/24/60/60`
`>>> `
`dt2= "25-JUN-2001"`
`>>> `
`timeObj2= time.strptime( dt2, "%d-%b-%Y" )`
`>>> `
`dayNumb2= int(time.mktime( timeObj2 ))/24/60/60`
`>>> `
`dayNumb2 - dayNumb1`
`151`
```
In this example, `timeObj1` and `timeObj2` are time structures with details parsed from the date string by `time.strptime`. The `dayNumb1` and `dayNumb2` are a day number that corresponds to this time. Time is measured in seconds after an epoch; typically January 1, 1970. The exact value doesn't matter, what matters is that the epoch is applied consistently by `mktime`. We divide this by 24 hours per day, 60 minutes per hour and 60 seconds per minute to get days after the epoch instead of seconds. Given two day numbers, the difference is the number of days between the two dates. In this case, there are 151 days between the two dates.
All of this processing must be encapsulated into a method that computes the ownership duration.
```def ownedFor(self, saleDate ):```
This method computes the days the stock was owned.
```def annualizedROI(self, salePrice , saleDate ):```
We would need to add an `annualizedROI` method to the `StockBlock` that divides the gross ROI by the duration in years to return the annualized ROI. Similarly, we would add a method to the `Position` to use the `annualizedROI` to compute the a weighted average which is the annualized ROI for the entire position.
Dive Logging and Surface Air Consumption Rate
The Surface Air Consumption Rate is used by SCUBA divers to predict air used at a particular depth. If we have a sequence of `Dive` objects with the details of each dive, we can do some simple calculations to get averages and ranges for our air consumption rate.
For each dive, we convert our air consumption at that dive's depth to a normalized air consumption at the surface. Given depth (in feet), d , starting tank pressure (psi), s , final tank pressure (psi), f , and time (in minutes) of t , the SACR, c , is given by the following formula.
Typically, you will average the SACR over a number of similar dives.
The `Dive` Class. You will want to create a `Dive` class that contains attributes which include start pressure, finish pressure, time and depth. Typical values are a starting pressure of 3000, ending pressure of 700, depth of 30 to 80 feet and times of 30 minutes (at 80 feet) to 60 minutes (at 30 feet). SACR's are typically between 10 and 20. Your `Dive` class should have a function named `getSACR` which returns the SACR for that dive.
To make life a little simpler putting the data in, we'll treat time as string of “HH:MM”, and use string functions to pick this apart into hours and minutes. We can save this as tuple of two intgers: hours and minutes. To compute the duration of a dive, we need to normalize our times to minutes past midnight, by doing `hh*60+mm`. Once we have our times in minutes past midnight, we can easily subtract to get the number of minutes of duration for the dive. You'll want to create a function `getDuration` to do just this computation for each dive.
`__init__`
The `__init__` method will initialize a `Dive` with the start and finish pressure in PSI, the in and out time as a string, and the depth as an integer. This method should parse both the `in` string and `out` string and normalize each to be minutes after midnight so that it can compute the duration of the dive. Note that a practical dive log would have additional information like the date, the location, the air and water temperature, sea state, equipment used and other comments on the dive.
`__str__`
The `__str__` method should return a nice string representation of the dive information.
`getSACR`
The `getSACR` method can then compute the SACR value from the starting pressure, final pressure, time and depth information.
The `Dive` Log. We'll want to initialize our dive log as follows:
```log = [
Dive( start=3100, finish=1300, in="11:52", out="12:45", depth=35 ),
Dive( start=2700, finish=1000, in="11:16", out="12:06", depth=40 ),
Dive( start=2800, finish=1200, in="11:26", out="12:06", depth=60 ),
Dive( start=2800, finish=1150, in="11:54", out="12:16", depth=95 ),
]
```
Rather than use a simple sequence of `Dive`s, you can create a `DiveLog` class which has a sequence of `Dive`s plus a `getAvgSACR` method. Your `DiveLog` method can be initiatlized with a sequence of dives, and can have an append method to put another dive into the sequence.
Exercising the Dive and DiveLog Classes. Here's how the final application could look. Note that we're using an arbitrary number of argument values to the `__init__` function, therefore, it has to be declared as `def __init__( self, *listOfDives )`
```log= DiveLog(
Dive( start=3100, finish=1300, in="11:52", out="12:45", depth=35 ),
Dive( start=2700, finish=1000, in="11:16", out="12:06", depth=40 ),
Dive( start=2800, finish=1200, in="11:26", out="12:06", depth=60 ),
Dive( start=2800, finish=1150, in="11:54", out="12:16", depth=95 ),
)
print log.getAvgSACR()
for d in log.dives:
print d```
Multi-Dice
If we want to simulate multi-dice games like Yacht, Kismet, Yatzee, Zilch, Zork, Greed or Ten Thousand, we'll need a collection that holds more than two dice. The most common configuration is a five-dice collection. In order to be flexible, we'll need to define a `Dice` object which will use a `tuple`, `list` or `Set` of individual `Die` instances. Since the number of dice in a game rarely varies, we can also use a `FrozenSet`.
Once you have a `Dice` class which can hold a collection of dice, you can gather some statistics on various multi-dice games. These games fall into two types. In both cases, the player's turn starts with rolling all the dice, the player can then elect to re-roll or preserve selected dice.
• Scorecard Games. In Yacht, Kismet and Yatzee, five dice are used. The first step in a player's turn is a roll of all five dice. This can be followed by up to two additional steps in which the player decides which dice to preserve and which dice to roll. The player is trying to make a scoring hand. A typical scorecard for these games lists a dozen or more "hands" with associated point values. The player attempts to make one of each of the various kinds of hands listed on the scorecard. Each turn must fill in one line of the scorecard; if the dice match a hand which has not been scored, the player enters a score. If a turn does not result in a hand that matches an unscored hand, then a score of zero is entered.
• Point Games. In Zilch, Zork, Green or Ten Thousand, five dice are typical, but there are some variations. The player in this game has no limit on the number of steps in their turn. The first step is to roll all the dice and determine a score. Their turn ends when they perceive the risk of another step to be too high, or they've made a roll which gives them a score of zero (or zilch) for the turn. Typically, if the newly rolled dice are non-scoring, their turn is over with a score of zero. At each step, the player is looking at newly rolled dice which improve their score. A straight scores 1000. Three-of-a-kind scores 100╳ the die's value (except three ones is 1000 points). After removing any three-of-a-kinds, each die showing 1 scores 100, each die showing 5 scores 50. Additionally, some folks will score 1000╳ the die's value for five-of-a-kind.
Our `MultiDice` class will be based on the example of `Dice` in this chapter. In addition to a collection of `Die` instances (a sequence, `Set` or `FrozenSet`), the class will have the following methods.
`__init__`
When initializing an instance of `MultiDice`, you'll create a collection of five individual `Die` instances. You can use a sequence of some kind, a `Set` or a `FrozenSet`.
`roll`
The `roll` method will roll all dice in the sequence or Set. Note that your choice of collection doesn't materially alter this method. That's a cool feature of Python.
`getDice`
This method returns the collection of dice so that a client class can examine them and potentialy re-roll some or all of the dice.
score
This method will score the hand, returning a `list` of two-tuples. Each two-tuple will have the name of the hand and the point value for the particular game. In some cases, there will be multiple ways to score a hand, and the `list` will reflect all possible scorings of the hand, in order from most valuable to least valuable. In other cases, the `list` will only have a single element.
It isn't practical to attempt to write a universal `MultiDice` class that covers all variations of dice games. Rather than write a gigantic does-everything class, the better policy is to create a family of classes that build on each other using inheritance. We'll look at this in the section called “Inheritance”. For this exercise, you'll have to pick one of the two families of games and compute the score for that particular game. Later, we'll see how to create an inheritance hierarchy that can cover the two-dice game of Craps as well as these multi-dice games.
For the scorecard games (Yacht, Kismet, Yatzee), we want to know if this set of dice matches any of the scorecard hands. In many cases, a set of dice can match a number of potential hands. A hand of all five dice showing the same value (e.g, a 6) is matches the sixes, three of a kind, four of a kind, five of a kind and wild-card rows on most game score-sheets. A sequence of five dice will match both a long straight and a short straight.
Common Scoring Methods. No matter which family of games you elect to pursue, you'll need some common method functions to help score a hand. The following methods will help to evaluate a set of dice to see which hand it might be.
• `matchDie`. This function will take a `Die` and a `Dice` set. It uses `matchValue` to partition the dice based on the value of the given `Die`.
• `matchValue`. This function is like `matchDie`, but it uses a numeric value instead of a `Die`. It partitions the dice into two sets: the dice in the `Dice` set which have a value that matches the given `Die`, and the remaining `Die` which do not match the value.
• ``` threeOfAKind ```, ``` fourOfAKind ```, ``` fiveOfAKind ```. These three functions will compute the `matchDie` for each `Die` in the `Dice` set. If any given `Die` has a `matchDie` with 3 (or 4 or 5) matching dice, the hand as a whole matches the template.
• ``` largeStraight ```. This function must establish that all five dice form a sequence of values from 1 to 5 or 2 to 6. There must be no gaps and no duplicated values.
• ``` smallStraight ```. This function must establish that four of the five dice form a sequence of values. There are a variety of ways of approaching this; it is actually a challenging algorithm. If we create a sequence of dice, and sort them into order, we're looking for an ascending sequence with one "irrelevant" die in it: this could be a gap before or after the sequence (1, 3, 4, 5, 6; 1, 2, 3, 4, 6 ) or a duplicated value (1, 2, 2, 3, 4, 5) within the sequence.
• ``` chance ```. The chance hand is simply the sum of the dice values. It is a number between 5 and 30.
This isn't necessarily the best way to do this. In many cases, a better way is to define a series of classes for the various kinds of hands, following the Strategy design pattern. The Dice would then have a collection of Strategy objects, each of which has a `match` method that compares the actual roll to a kind of hand. The Strategy objects would have a `score` method as well as a `name` method. This is something we'll look at in the section called “Strategy”.
Scoring Yacht, Kismet and Yatzee. For scoring these hands, you'll use the common scoring method functions. Your overall `score` method function will step through the candidate hands in a specific order. Generally, you'll want to check for `fiveOfAKind` first, since `fourOfAKind` and `threeOfAKind` will also be true for this hand. Similarly, you'll have to check for `largeStraight` before `smallStraight`.
Your `score` method will evaluate each of the scoring methods. If the method matches, your method will append a two-tuple with the name and points to the list of scores.
Scoring Zilch, Zork and 10,000. For scoring these dice throws, you'll need to expand on the basic `threeOfAKind` method. Your `score` method will make use of the two results sets created by the `threeOfAKind` method.
Note that the hand's description can be relatively complex. For example, you may have a hand with three 2's, a 1 and a 5. This is worth 350. The description has two parts: the three-of-a-kind and the extra 1's and 5's. Here are the steps for scoring this game.
• Evaluate the `largeStraight` method. If the hand matches, then return a `list` with an appropriate 2-tuple.
• If you're building a game variation where five of a kind is a scoring hand, then evaluate `fiveOfAKind`. If the hand matches, then return a list with an appropriate 2-tuple.
• 3K. Evaluate the `threeOfAKind` method. This will create the first part of the hand's description.
• If a `Die` created a matching set with exactly three dice, then the set of unmatched dice must be examined for additional 1's and 5's. The first part of the hand's description string is three-of-a-kind.
• If a `Die` created a matching with four or five dice, then one or two dice must be popped from the matching set and added to the non-matching set. The set of unmatched dice must be examined for addtional 1's and 5's. The first part of the hand's description string is three-of-a-kind.
• If there was no set of three matching dice, then all the dice are in the non-matching set, which is checked for 1's and 5's. The string which describes the hand has no first part, since there was no three-of-a-kind.
• 1-5's. Any non-matching dice from the `threeOfAKind` test are then checked using `matchValue` to see if there are 1's or 5's. If there are any, this is the second part of the hand's description. If there are none, then there's no second part of the description.
• The final step is to assemble the description. There are four cases: nothing, 3K with no 1-5's, 1-5's with no 3K, and 3K plus 1-5's. In the nothing case, this is a non-scoring hand. In the other cases, it is a scoring hand.
Exercising The Dice. Your main script should create a `Dice` set, execute an initial roll and score the result. It should then pick three dice to re-roll and score the result. Finally, it should pick one die, re-roll this die and score the result. This doesn't make sophisticated strategic decisions, but it does exercise your `Dice` and `Die` objects thoroughly.
When playing a scorecard game, the list of potential hands is examined to fill in another line on the scorecard. When playing a points game, each throw must result in a higher score than the previous throw or the turn is over.
Strategy. When playing these games, a person will be able to glance at the dice, form a pattern, and decide if the dice are "close" to one of the given hands. This is a challenging judgement, and requires some fairly sophisticated software to make a proper odd-based judgement of possible outcomes. Given that there are only 7,776 possible ways to roll 5 dice, it's a matter of exhaustively enumerating all of the potential outcomes of the various kinds of rerolls. This is an interesting, but quite advanced exercise.
Rational Numbers
A Rational number is a ratio of two integers. Examples include 1/2, 2/3, 22/7, 355/113, etc. We can do arithmetic operations on rational numbers. We can display them as proper fractions (```3 1/7```), improper fractions (`22/7`) or decimal expansions (`3.1428571428571428`).
The essence of this class is to perform arithmetic operations. We'll start by defining methods to add and multiply two rational values. Later, we'll delve into the additional methods you'd need to write to create a robust, complete implementation.
Your `add` and ``` mul ``` methods will perform their processing with two `Rational` values: `self` and `other`. In both cases, the variable `other` has to be another `Rational` number instance. You can check this by using the `type` function: if ```type(self) != type(other)```, you should raise a `TypeException`.
It's also important to note that all arithmetic operations will create a new `Rational` number computed from the inputs.
A Rational class has two attributes: the numerator and the denominator of the value. These are both integers. Here are the various methods you should created.
`__init__`
The `__init__` constructor accepts the numerator and denominator values. It can have a default value for the denominator of 1. This gives us two constructors: `Rational(2,3)` and `Rational(4)`. The first creates the fraction 2/3. The second creates the fraction 4/1.
This method should call the `reduce` method to assure that the fraction is properly reduced. For example, `Rational(8,4)` should automatically reduce to a numerator of 2 and a denominator of 1.
`__str__`
The `__str__` method returns a nice string representation for the rational number, typically as an improper fraction. This gives you the most direct view of your `Rational` number.
You should provide a separate method to provide a proper fraction string with a whole number and a fraction. This other method would do additional processing to extract a whole name and remainder.
`__float__`
If you provide a method named `__float__`, this can return the floating-point value for the fraction. This method is called when a program does ```float( rationalValue )```.
``` add ```( `self` , `other` )
The `add` method creates and returns a new `Rational` number. This new fraction that has a numerator of (self.numerator × other.denominator + other.numerator × self.denominator), and a denominator of ( self.denominator × other.denominator ).
Example: 3/5 + 7/11 = (33 + 35)/55 = 71/55.
``` mul ```( `self` , `other` )
The `mul` method creates and returns a new `Rational` number. This new fraction that has a numerator of (self.numerator × other.numerator), and a denominator of ( self.denominator × other.denominator ).
Equation 21.2. Multiplying Fractions
Example: 3/5 × 7/11 = 21/55.
`reduce`
In addition to adding and multiplying two `Rational` numbers, you'll also need to provide a `reduce` method which will reduce a fraction by removing the greatest common divisor from the numerator and the denominator. This should be called by `__init__` to assure that all fractions are reduced.
To implement `reduce`, we find the greatest common divisor between the numerator and denominator and then divide both by this divisor. For example 8/4 has a GCD of 4, and reduces to 2/1. The Greatest Common Divisor (GCD) algorithm is given in Greatest Common Divisor and Greatest Common Divisor. If the GCD of `self.numerator` and `self.denominator` is 1, the `reduced` function can return `self`. Otherwise, `reduced` must create a new `Rational` with the reduced fraction.
Playing Cards and Decks
Standard playing cards have a rank (ace, two through ten, jack, queen and king) and suit (clubs, diamonds, hearts, spades). These form a nifty `Card` object with two simple attributes. We can add a few generally useful functions.
Here are the methods for your `Card` class.
``` __init__ ```( `self` , `rank` )
The `__init__` method sets the rank and suit of the card. The suits can be coded with a single character ("C", "D", "H", "S"), and the ranks can be coded with a number from 1 to 13. The number 1 is an ace. The numbers 11, 12, 13 are Jack, Queen and King, respectively. These are the ranks, not the point values.
`__str__`
The `__str__` method can return the rank and suit in the form "2C" or "AS" or "JD". A rank of 1 would become "A", a rank of 11, 12 or 13 would become "J", "Q" or "K", respectively.
``` __cmp__ ```( `self` , `other` )
If you define a `__cmp__` method, this will used by the `cmp` function; the `cmp` function is used by the `sort` method of a `list` unless you provide an overriding function used for comparison. By providing a `__cmp__` method in your class you can assure that cards are sorted by rank in preference to suit. You can also use `<`, `>`, `>=` and `<=` operations among cards.
Sometime as simple as ```cmp(self.rank,other.rank) or cmp(self.suit,other.suit)``` works surprisingly well.
Dealing and Decks. `Card`s are dealt from a `Deck`; a collection of `Card`s that includes some methods for shuffling and dealing. Here are the methods that comprise a `Deck`.
`__init__`
The `__init__` method creates all 52 cards. It can use two loops to iterate through the sequence of suits `("C", "D", "H", "S")` and iterate through the ranks `range(1,14)`. After creating each `Card`, it can append each `Card` to a sequence of `Card`s.
`deal`
The `deal` method should do two things: iterate through the sequence, exchanging each card with a randomly selected card. It turns out the `random` module has a `shuffle` function which does precisely this.
Dealing is best done with a generator method function. The `deal` method function should have a simple for-loop that yields each individual `Card`; this can be used by a client application to generate hands. The presence of the yield statement will make this method function usable by a for statement in a client application script.
Basic Testing. You should do some basic tests of your Card objects to be sure that they respond appropriately to comparison operations. For example,
````>>>`
`x1= Card(11,"C")`
`>>>`
`x1`
` JC`
`>>>`
`x2= Card(12,"D")`
`>>>`
`x1 < x2`
`True`
```
You can write a simple test script which can the do the following to deal `Cards` from a `Deck`. In this example, the variable `dealer` will be the iterator object that the for statement uses internally.
```d= Deck()
dealer= d.deal()
c1= dealer.next()
c2= dealer.next()```
Hands. Many card games involve collecting a hand of cards. A Hand is a collection of `Card`s plus some addition methods to score the hand in way that's appropriate to the given game. We have a number of collection classes that we can use: `list`, `tuple`, `dictionary` and `set`.
Consider Blackjack. The `Hand` will have two `Card`s assigned initially; it will be scored. Then the player must choose among accepting another card (a hit), using this hand against the dealer (standing), doubling the bet and taking one more card, or splitting the hand into two hands. Ignoring the split option for now, it's clear that the collection of `Card`s has to grow and then get scored again. What are the pros and cons of `list`, `tuple`, `set` and `dictionary`?
Consider Poker. There are innumerable variations on poker; we'll look at simple five-card draw poker. Games like seven-card stud require you to score potential hands given only two cards, and as many as 21 alternative five-card hands made from seven cards. Texas Hold-Em has from three to five common cards plus two private cards, making the scoring rather complex. For five-card draw, the `Hand` will have five cards assigned initially, and it will be scored. Then some cards can be removed and replaced, and the hand scored again. Since a valid poker hand is an ascending sequence of cards, called a straight, it is handy to sort the collection of cards. What are the pros and cons of `list`, `tuple`, `set` and `dictionary`?
Blackjack Hands
Changes to the Card class. We'll extend our `Card` class to score hands in Blackjack, where the rank is used to determine the hand that is held. When used in Blackjack, a `Card` has a point value in addition to a rank and suit. Aces are either 1 or 11; two through ten are worth 2-10; the face cards are all worth 10 points. When an ace is counted as 1 point, the total is called the hard total. When an ace is counted as 11 points, the total is called a soft total.
You can add a point attribute to your card class. This can be set as part of `__init__` processing. In that case, the following methods simple return the point value.
As an alternative, you can compute the point value each time it is requested. This has the obvious disadvantage of being slower. However, it is considerably simpler to add methods to a class without revising the existing `__init__ `method.
Here are the methods you'll need to add to your `Card` class in order to handle Blackjack hands.
`getHardValue`
The `getHardValue` method returns the rank, with the following exceptions: ranks of 11, 12 and 13 return a point value of 10.
`getSoftValue`
The `getSoftValue` method returns the rank, with the following exceptions: ranks of 11, 12 and 13 return a point value of 10; a rank of 1 returns a point value of 11.
As a teaser for the next chapter, we'll note that these methods should be part of a Blackjack-specific subclass of the generic `Card` class. For now, however, we'll just update the `Card` class definition.When we look at inheritance in the section called “Inheritance”, we'll see that a class hierarchy can be simpler than the if-statements in the `getHardValue` and `getSoftValue` methods.
Scoring Blackjack Hands. The objective of Blackjack is to accumulate a `Hand` with a total point value that is less than or equal to 21. Since an ace can count as 1 or 11, it's clear that only one of the aces in a hand can have a value of 11, and any other aces must have a value of 1.
Each `Card` produces a hard and soft point total. The `Hand` as a whole also has hard and soft point totals. Often, both hard and soft total are equal. When there is an ace, however, the hard and soft totals for the hand will be different. We have to look at two cases.
• No Aces. The hard and soft total of the hand will be the same; it's the total of the hard value of each card. If the hard total is less than 21 the hand is in play. If it is equal to 21, it is a potential winner. If it is over 21, the hand has gone bust. Both totals will be computed as the hard value of all cards.
• One or more Aces. The hard and soft total of the hand are different. The hard total for the hand is the sum of the hard point values of all cards. The soft total for the hand is the soft value of one ace plus the hard total of the rest of the cards. If the hard or soft total is 21, the hand is a potential winner. If the hard total is less than 21 the hand is in play. If the hard total is over 21, the hand has gone bust.
The `Hand` class has a collection of `Cards`, usually a sequence, but a `Set` will also work. Here are the methods of the `Hand` class.
`__init__`
The `__init__` method should be given two instances of `Card` to represent the initial deal. It should create a sequence or `Set` with these two initial cards.
`__str__`
The `__str__` method a string with all of the individual cards. A construct like the following works out well: ```",".join( map(str,self.cards)```. This gets the string representation of each card in the `self.cards` collection, and then uses the `string`'s `join` method to assemble the final display of cards.
`hardTotal`
The `hardTotal` method sums the hard value of each `Card`.
`softTotal`
The `softTotal` method is more complex. It needs to partition the cards into two sets. If there are any cards with a different hard and soft point value (this will be an ace), then one of these cards forms the `softSet`. The remaining cards form the `hardSet`. It's entirely possible that the `softSet` will be empty. It's also entirely possible that there are multiple cards which could be part of the `softSet`. The value of this function is the total of the hard values for all of the cards in the `hardSet` plus the soft value of the card in the `softSet`.
`add`
The `add` method will add another `Card` to the `Hand`.
Exercising Card, Deck and Hand. Once you have the `Card`, `Deck` and `Hand` classes, you can exercise these with a simple function to play one hand of blackjack. This program will create a `Deck` and a `Hand`; it will deal two `Card`s into the `Hand`. While the `Hand`'s total is soft 16 or less, it will add `Card`s. Finally, it will print the resulting `Hand`.
There are two sets of rules for how to fill a `Hand`. The dealer is tightly constrained, but players are more free to make their own decisions. Note that the player's hands which go bust are settled immediately, irrespective of what happens to the dealer. On the other hand, the player's hands which total 21 aren't resolved until the dealer finishes taking cards.
The dealer must add cards to a hand with a soft 16 or less. If the dealer has a soft total between 17 and 21, they stop. If the dealer has a soft total which is over 21, but a hard total of 16 or less, they will take cards. If the dealer has a hard total of 17 or more, they will stop.
A player may add cards freely until their hard total is 21 or more. Typically, a player will stop at a soft 21; other than that, almost anything is possible.
Additional Plays. We've avoided discussing the options to split a hand or double the bet. These are more advanced topics that don't have much bearing on the basics of defining `Card`, `Deck` and `Hand`. Splitting simply creates additional `Hand`s. Doubling down changes the bet and gets just one additional card.
Poker Hands
We'll extend the `Card` class we created in the section called “Playing Cards and Decks” to score hands in Poker, where both the rank and suit are used to determine the hand that is held.
Poker hands are ranked in the following order, from most desirable (and least likely) down to least desirable (and all too common).
1. Straight Flush. Five cards of adjacent ranks, all of the same suit.
2. Four of a Kind. Four cards of the same rank, plus another card.
3. Full House. Three cards of the same rank, plus two cards of the same rank.
4. Flush. Five cards of the same suit.
5. Straight. Five cards of adjacent ranks. In this case, Ace can be above King or below 2.
6. Three of a Kind. Three cards of the same rank, plus two cards of other ranks.
7. Two Pair. Two cards of one rank, plus two cards of another rank, plus one card of a third rank.
8. Pair. Two cards of one rank, plus three cards of other ranks.
9. High Card. The highest ranking card in the hand.
Note that a straight flush is both a straight and a flush; four of a kind is also two pair as well as one pair; a full house is also two pair, as well as a one pair. It is important, then, to evaluate poker hands in decreasing order of importance in order to find the best hand possible.
In order to distinguish between two straights or two full-houses, it is important to also record the highest scoring card. A straight with a high card of a Queen, beats a straight with a high card of a 10. Similarly, a full house or two pair is described as “queens over threes”, meaning there are three queens and two threes comprising the hand. We'll need a numeric ranking that includes the hand's rank from 9 down to 1, plus the cards in order of “importance” to the scoring of the hand.
The importance of a card depends on the hand. For a straight or straight flush, the most important card is the highest-ranking card. For a full house, the most important cards are the three-of-a kind cards, followed by the pair of cards. For two pair, however, the most important cards are the high-ranking pair, followed by the low-ranking pair. This allows us to compare “two pair 10's and 4's” against “two pair 10's and 9s'”. Both hands have a pair of 10's, meaning we need to look at the third card in order of importance to determine the winner.
Scoring Poker Hands. The `Hand` class should look like the following. This definition provides a number of methods to check for straight, flush and the patterns of matching cards. These functions are used by the `score` method, shown below.
```class PokerHand:
def __init__( self, cards ):
self.cards= cards
self.rankCount= {}
def straight( self ):
all in sequence
def straight( self ):
all of one suit
def matches( self ):
tuple with counts of each rank in the hand
def sortByRank( self ):
sort into rank order
def sortByMatch( self ):
sort into order by count of each rank, then rank
```
This function to score a hand checks each of the poker hand rules in descending order.
``` def score( self ):
if self.straight() and self.flush():
self.sortByRank()
return 9
elif self.matches() == ( 4, 1 ):
self.sortByMatch()
return 8
elif self.matches() == ( 3, 2 ):
self.sortByMatch()
return 7
elif self.flush():
self.sortByRank()
return 6
elif self.straight():
self.sortByRank()
return 5
elif self.matches() == ( 3, 1, 1 ):
self.sortByMatch()
return 4
elif self.matches() == ( 2, 2, 1 ):
self.sortByMatchAndRank()
return 3
elif self.matches() == ( 2, 1, 1, 1 ):
self.sortByMatch()
return 2
else:
self.sortByRank()
return 1
```
You'll need to add the following methods to the PokerHand class.
• `straight` returns True if the cards form a straight. This can be tackled easily by sorting the cards into descending order by rank and then checking to see if the ranks all differ by exactly one.
• `flush` returns True if all cards have the same suit.
• `matches` returns a tuple of the counts of cards grouped by rank. This can be done iterating through each card, using the card's rank as a key to the `self.rankCount` dictionary; the value for that dictionary entry is the count of the number of times that rank has been seen. The values of the dictionary can be sorted, and form six distinct patterns, five of which are shown above. The sixth is simply `(1, 1, 1, 1, 1)`, which means no two cards had the same rank.
• `sortByRank` sorts the cards by rank.
• `sortByMatch` uses the counts in the `self.rankCount` dictionary to update each card with its match count, and then sorts the cards by match count.
• `sortByMatchAndRank` uses the counts in the `self.rankCount` dictionary to update each card with its match count, and then sorts the cards by match count and rank as two separate keys.
Exercising Card, Deck and Hand. Once you have the `Card`, `Deck` and `Hand` classes, you can exercise these with a simple function to play one hand of poker. This program will create a `Deck` and a `Hand`; it will deal five `Card`s into the `Hand`. It can score the hand. It can replace from zero to three cards and score the resulting hand.
Published under the terms of the Open Publication License Design by Interspire | 10,174 | 41,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.841985 |
https://militarytimeconverter.org/1030-military-time/ | 1,721,668,418,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00732.warc.gz | 344,051,030 | 70,683 | # What is 1030 Military Time: Step-by-Step Conversion Guide
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Do you know what time 1030 Military Time is? I know that converting Military Time can be a bit complicated, but not when you know the simple and helpful formula for converting 24-Hour Time Format.
In case you don’t want to use this formula that shows you how to convert 1030 Military Time to Normal Time and only want to know the answer, here you have a short, but useful answer to your question:
1030 Military Time or 1030 Hours (24-Hour time format) is equivalent to 10:30 AM in Normal Time or Standard Time (12-Hour time format). 1030 Military Time is read as “ten thirty hours“, and it is written without a colon between the hours and minutes.
For those of you who are interested in a simple conversion formula, here is the formula that helps you understand the conversion of 1030 Military Time.
## What Time is 1030 Military Time? 1030 Hours Conversion Guide
First of all, it is important to know the basics of Military Time. Military Time always contains 4 digits where the first 2 digits are for hours and the second 2 digits are for minutes.
This simple image below explains to you the basic rule of conversion:
In addition, if you have a Military Time greater or equal than 13:00, you have to subtract 12 hours from the given time. And that’s all you need to know.
Now, let’s move on to explaining the simplest way of how you can convert 1030 Hours into the Standard Time.
Step 1
First, divide the given time to hours and minutes. For 1030 Military Time it is 10 hours and 30 minutes.
Step 2
Then pay attention to whether the given time is greater than 1300 or not. 1030 it isn’t greater than 1300 so you don’t use the subtraction rule for the hours and let the number for hours as it is: 10 hours.
Step 3
Minutes are the same in the Military Time as so as in the Standard Time, so 30 minutes is still 30 minutes.
Step 4
Add the colon between calculated hours and minutes and the result will be 10:30.
Step 5
As 1030 is not greater than 1200 so according to about mentioned rule add AM after calculated time so the result will be 10:30 AM normal time.
So what is 1030 military time? The correct answer is 10:30 AM standard time.
TIP: If you want to see more conversions check out other articles about specific military time conversions.
### What is 1030 AM Military Time?
As we do not use AM and PM in Military Time or in 24-Hour Time Format this question is incorrect.
The correct question should be: What is 10:30 AM in Military Time? Or maybe: What is military time for 10:30 AM? But, someone who is not familiar with Military Time may ask this question in the wrong way.
To answer the question, you have two options:
• 10:30 AM Military Time is 10:30 AM Standard Time
• 10:30 AM Standard Time is 1030 Hours Military Time
#### How to convert 1030 AM to Military Time?
To convert 10:30 AM, you have to follow the same rules as for converting time in the opposite way. The only change is you don’t subtract 12 from Military Time greater than 1300, but you add 12 hours to PM hours.
Conversion of 10:30 AM to Military Time is described below:
• 10:30 AM belongs to AM hours so you don’t add 12 hours and let the number for hours as it is
• 10 + 0 hours = 10 hours
• Minutes are always the same no matter what time format you use
• Remove AM and colon
• The conversion of 10:30 AM is 1030 Military Time
TIP: You can convert any time you need right now:
Converter from Military Time to Normal Time
Convert Military Time to Standard Time
Standard Time:
Converter from Normal Time to Military Time
Convert Standard time to Military Time
Military Time:
### How to Read and Write 1030 Military Time?
You already know how to convert 1030 Hours and now it’s time to learn how to read and write this specific time.
This table explains everything you need to know about reading and writing 1030 Hours Time and related times:
Standard Clock How to Write Military Clock How to Read Military Clock
10:30 AM 1030 Hours ten thirty hours
10:45 AM 1045 Hours ten fourty five hours
10:00 AM 1000 Hours ten hunderd hours
10:15 AM
1015 Hours ten fifteen hours
## Conclusion
I hope 1030 Military Time (1030 Hours) conversion was not a difficult task for you, but the clarification of AM and PM is an important reminder:
• When the value of Military Time is greater than 1200 it will be PM
• When the value of Military Time is less than 1200 it will be AM
• 2400 (0000) is an exceptional case because it is considered as AM (12:00 AM)
If it is still confusing for you, you can read the article on Military Time Charts where you find a conversion chart for all hours on the hour. Or you can use the page with Military Time Converter, where you can find an explanation on how to use the converter.
TIP: This website contains much more topics related to Military Time so you can also check the main categories with all useful posts:
Military Time Tools | 1,232 | 5,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-30 | latest | en | 0.899832 |
https://freematheducation.com/trigonometry-formula-math-tutor-help.html | 1,563,736,774,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527196.68/warc/CC-MAIN-20190721185027-20190721211027-00409.warc.gz | 413,399,438 | 8,485 | # Trigonometry Formulas, Functions and Identities – Free Homework Tutor Help, Videos and pdf
Things to Remember: Important Formula for Trigonometry
In mathematics, trigonometric identities are equalities involving trigonometric functions that are true for all values of the occurring variables. These identities are useful whenever expressions involving trigonometric functions need to be simplified.
The step involves, first using the substitution rule with a
trigonometric function, and then simplifying the resulting integral using with a trigonometric identity when the integration is given in a non-trigonometric identities.
Important List of Trigonometric Formulas:
Notation of Trigonometric Functions : The Six trigonometric function notations are Sine Function : sin (theta) Cosine Function :cos (theta) Tangent Function : tan (theta) Cotangent Function : cot (theta) Secant Function : sec (theta) Cosecant Function : cosec (theta) or csc (theta) Bacis Trigonometric Formula: In a right Triangle ABC, B is a right angle, then sin() = Perpendicular / Hypotenuse cos() = Base / Hypotenuse tan() = Perpendicular / Base cot() = Base / Perpendicular sec() = Hypotenuse / Base cosec() or csc() = Base/Perpendicular Reciprocal Identities : sin x = 1/cosec x , cosec x = 1/sin x cos x = 1/sec x , sec x = 1/cos x tan x = 1/cot x , cot x = 1/tan x Quotient Identities : tan x = 1/cot x cot x = 1/tan x Pythagorean Identities :The Pythagorean identities are follow the concept of Pythagorean theorem i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2. This simply known as Pythagorean trigonometric identity. we have three identities – sin 2x + cos2x = 1 tan2x + 1 = cot2x [we get this identity by dividing first by cos 2x] 1 + cot2x = cosec2x or csc2x [we get this identity by dividing first by sin 2x] Addition and Subtraction Formulas : These are also known as Angle Sum and Difference Identities. These are sin (x + y) = sin x cos y + cos x sin y sin (x – y) = sin x cos y – cos x sin y cos (x + y) = cos x cos y – sin x sin y cos (x – y) = cos x cos y + sin x sin y tan (x + y) = (tan x + tan y) /(1 – tan x tan y) tan (x – y) = (tan x – tan y) /(1 + tan x tan y) Double Angle Formulas : When we substituting x = y in the sum of addition formula. we get these double angle formula sin (x + x) = sin x cos x + cos x sin x sin 2x = 2 sin x cos y cos (x + x) = cos x cos x – sin x sin x cos 2x = cos2 x – sin 2 x cos 2x = cos2x – (1 – cos 2 x) = 2 cos2 x – 1 [cos 2x in terms of cosine function] cos 2x = 1 – sin2 x – sin 2 x = 1 – 2 sin 2 x [cos 2x in terms of sine function] cos 2x = (1 – tan 2 x) / (1 + tan2 x) [cos 2x in terms of tangent function] tan (x + x) = (tan x + tan x) /(1 – tan x tan x) tan 2x = 2 tan x /(1 – tan2x) Triple Angle Formula : sin 3x = 3 sin x – 4 sin3x cos 3x = 4cos3x – 3 cos x tan 3x = (3 tan x – tan3 x) / (1 – 3 tan 2 x) Power Reduction Formulas : The sin2 x, cos2 x in terms of double angle formula sin2 x = (1 – cos 2x)/2 cos2 x = (1 + cos 2x)/2 tan2 x = (1 – cos 2x)/(1 + cos 2x) sin2 x cos2 x = (1 – cos 2x)(1 + cos 2x)/4 = (1 – cos2 2x)/4 = (1 – cos 4x)/8 Half Angle Formulas : When we substitute x/2 in place of x in power reduction formula and solve for sin x/2 and cos x/2, we get cos x/2 = √(1 + cos x)/2 sin x/2 = √(1 – cos x)/2 tan x/2 = sin x / (1 + cos x) = (1 – cos x) / sin x Product to Sum Identities: : When we expanding the addition formula in RHS, we get cos x cos y = [cos (x – y) + cos (x + y)]/2 sin x sin y = [cos (x – y) – cos (x + y)]/2 sin x cos y = [sin (x + y) + sin (x – y) ]/2 cos x sin y = [sin (x + y) – sin (x – y) ]/2 Sum to Product Identities: : When we substituting x by (x + y)/2 and y by (x – y)/2 in the product to sum formula in RHS, we get cos x + cos y = 2 cos (x + y)/2 cos (x – y)/2 sin x + sin y = 2 sin (x + y)/2 cos (x – y)/2 cos x – cos y = – sin (x + y)/2 sin (x – y)/2 sin x – sin y = 2 cos (x + y)/2 sin (x – y)/2 Inverse Trigonometric Functions : Every trigonometric function can be related directly to every other trigonometric function and these relations can be expressed by means of inverse trigonometric functions, we have sin -1 x + cos -1 y = /2 tan-1 x + cot -1 y = /2 If x > 0, tan-1 x + tan -1 1/x = /2and if x < 0, tan-1 x + tan -1 1/x = – /2 sin (cos-1 x) = cos (sin-1 x) = √(1 – x2) sin (tan-1 x) = x/√(1 + x2) cos (tan-1 x) = 1/√(1 + x2) tan (sin-1 x) = x/√(1 – x2) tan (cos-1 x) = √(1 – x2) / x | 1,509 | 4,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-30 | latest | en | 0.751487 |
https://www.comeforanswers.com/you-are-standing-at-the-point-x-300-m-y-400-m-in-a-reference-frame-with-synchronized/ | 1,696,473,635,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00048.warc.gz | 770,775,449 | 8,908 | # You are standing at the point x = 300 m, y = 400 m in a reference frame with synchronized
You are standing at the point x = 300 m, y = 400 m in a reference frame with synchronized clocks.
What time does your clock show at the instant you see the clock at the origin showing 3.80 μs ? | 75 | 286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-40 | latest | en | 0.903552 |
https://www.physicsforums.com/threads/topology-subset-of-a-continuous-function.481667/ | 1,576,298,586,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540584491.89/warc/CC-MAIN-20191214042241-20191214070241-00392.warc.gz | 828,921,119 | 16,992 | # Topology: Subset of A Continuous Function
1. Homework Statement
Let $$f$$ be a real-valued function defined and continuous on the set of real numbers. Which of the following must be true of the set $$S={f(c):0<c<1}?$$
I. S is a connected subset of the real numbers.
II. S is an open subset of the real numbers.
III. S is a bounded subset of the real numbers.
2. Homework Equations
none
3. The Attempt at a Solution
The answer is I and III only. I'm confused because I thought that I. implied II. Isn't a connected set necessarily open?
Related Calculus and Beyond Homework Help News on Phys.org
Why would a connected set necessarily be open? [0,1] is connected but not open.
Not at all! Connectedness does not imply open. A simple counterexample is [0,1], this is connected but not open.
Hmm...well, my textbook defines a subset of a topological space as connected if it's simultaneously relatively open and relatively closed. I thought this applied to S since it is a subset of the real numbers and the real numbers could be a topological space if you include a distance formula...
Do sets and topological spaces have different criteria for being open/closed/etc.?
How does your textbook define relatively open/closed?
Hmm...well, my textbook defines a subset of a topological space as connected if it's simultaneously relatively open and relatively closed. I thought this applied to S since it is a subset of the real numbers and the real numbers could be a topological space if you include a distance formula...
What your book likely says is that a topological space is connected if and only if there are no non-trivial clopen subsets. You are considering S as a subset of $\mathbb R$ not as a topological space itself.
Also, $\mathbb R$ can be made a topological space regardless of whether there is a metric on it or not. The inclusion of metric imposes a particular topology (the one defined by the metric) but one does not need a distance function to define open sets. Indeed, there are some very curious topologies on $\mathbb R$ that are not induced via metrics.
How does your textbook define relatively open/closed?
It says that an open subset of a subspace of a topological space is relatively open.
What your book likely says is that a topological space is connected if and only if there are no non-trivial clopen subsets. You are considering S as a subset of LaTeX Code: \\mathbb R not as a topological space itself.
There's a definition of connectedness for both topological spaces and subsets of topological spaces. Both definitions state: A topological space (or a subset of one) is connected if the only two subsets of it that are both open (relatively) and closed (relatively) are the topological space (or subset) itself and the null set.
Also, LaTeX Code: \\mathbb R can be made a topological space regardless of whether there is a metric on it or not.
Right, I had metric spaces and topological spaces mixed up in my mind when I wrote that.
Right. It may not be obvious but your books definition of connected is the same as mine. Consider the classical example of the set
$$[0,1] \cup [2,3]$$
This set is not connected, right? It's pretty obvious when you look at it as being the union of disjoint sets, but let's take a look at it from a clopen (closed and open) point of view.
Consider $X =[0,1] \cup [2,3]$ as a topological space itself, under the subspace topology it inherits from $\mathbb R$. In particular, our normal idea of closed and open intervals being closed and open sets is still true because it's true in $\mathbb R$. But also notice that [0,1] is both closed and open in X (though it's not both closed and open in $\mathbb R$). Why is this true? Well, [0,1] is closed in $\mathbb R$ so it's closed in X as a subspace of $\mathbb R$. Additionally,
$$X\setminus_{[0,1]} = [2,3]$$
That is, its relative complement in X is closed. By definition, a set is open if its complement is closed and so [0,1] is also open!
Thus [0,1] is both open and closed, and is a proper subset of X. So X has a proper, non-trivial clopen subset and hence is not connected.
Edit: The thing to take away from this is that [0,1] is both open and closed as a subset of X, but NOT as a subset of $\mathbb R$! | 1,018 | 4,239 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-51 | longest | en | 0.934556 |
https://electronics.stackexchange.com/questions/439446/grounding-balanced-secondary-isolation-transformer-for-emi-suppression | 1,716,666,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00425.warc.gz | 185,321,687 | 40,290 | # Grounding balanced secondary isolation transformer for EMI suppression
simulate this circuit – Schematic created using CircuitLab
Which is the best secondary connection to clean the main line power from EMI, line noise and HF harmonics by using an isolation 1:1 230/230VAC transformer with balanced outputs:
1) A floating (C1 and C2 are enough)
2) A to C to ground via a small capacitor
3) A to B ground directly
After having determined that connection for EMI suppression purpose ONLY, how to connect D ground output just for safety purpose (D to A, D to B, D to C, D to nothing)?
FYI: the transformer is multi shielded and each shield is connected to B ground. It has also a copper belly belt, currently floating.
If, the EMI, line noise and HF harmonics are common mode AND the transformer screen is good, then you won't need output capacitors and point A can connect to point B optionally.
If, the EMI, line noise and HF harmonics are differential then any potential reduction at the output is due to leakage inductance in the transformer and, given that you haven't specified this then it's difficult to make any statement about how good it will be and whether there might be some circumstances where the noise at the output gets worse (due to resonances with the added output capacitors).
If in doubt, get hold of a free simulation tool, model the transformer and work this problem out yourself.
• @ Andy I'm not sure why the down-vote. I view this as a crucial question for system fidelity and precision measurement, in typical trash-laden environments, and is of high interest to me. If the only degrees-of-freedom are the capacitors, is a good solution possible? May 21, 2019 at 5:05
• @analogsystemsrf I don't understand the down vote either! May 21, 2019 at 8:29
• Just checked that solution 3 is against safety! So the only solution available is 1 or 2, but do they have any effect? May 21, 2019 at 12:38
• "Do they have any effect?" - what part of my answer do you not understand? May 21, 2019 at 12:43
I'd not expect a transformer to eliminate harmonics. Thus additional L+C+L filtering seems needed.
Is this a TOPAZ transformer?
• Not a TOPAZ, but should have same purpose. As mine has balanced outputs I want to use the middle leg to diminish noise. May 21, 2019 at 12:35 | 554 | 2,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.937678 |
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### Grade 7 - Mathematics7.8 Relation Between Sides and Angles of a Triangle
Points to remember: The sum of the measures of any two sides of a triangle is greater than the measure of the third side. The difference of the measures of any two sides of a triangle is less than the measure of the third side. If two sides of a triangle are unequal, the measure of the angle opposite to the longer side is greater than the measure of the angle opposite to the shorter side. If two angles of a triangles are unequal, then the side opposite to the greater angles is longer than the side opposite to the smaller angle. No two angles of a scalene triangle are congruent. Directions: Read the above review points carefully and answer the following questions: Illustrate each of the above review points by using the triangle given below. Illustrate each of the above review points by drawing a triangle of your own. Explain in your own words the relation between the sides and angles of a triangle with examples.
Q 1: If BC = 10 cm and CA = 15 cm are the sides of ABC, then the measure of AB is greater than 5 cm.FalseTrue Q 2: If, in ABC, AB = AC, then ÐC = ÐB.FalseTrue Q 3: If, in ABC, ÐB = 100°, then AC is the longest side.TrueFalse Q 4: If, in ABC, AB < AC, then ÐC > ÐB.FalseTrue Q 5: If, in ABC, ÐA = 100° and ÐB = 50°, then a < b.FalseTrue Q 6: If, in ABC, ÐA and ÐB are complementary, then it is a right angled triangle.TrueFalse Q 7: If, in ABC, AB > BC, then ÐA < ÐB.TrueFalse Q 8: If, in ABC, ÐA = 30° and ÐB = 60°, then AB is the longest side.TrueFalse Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | 484 | 1,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-09 | latest | en | 0.899972 |
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• #### Mechanics of Cold Rolling of Thin Strip IntechOpen
Mechanics of Cold Rolling of Thin Strip 443 where yiw is the vertical deflection of the roll at element i, gijw (, ) is the influence function for the roll deflection due to the combined bending and shear forces generated by rolling load, giwf is the influence function for the roll deflection due to the force generated by the roll bending mechanism, k represents the minimum element
• [PDF]
• #### Load Analysis of Rolls in a Rolling Mill: A Comparison of
circuit. The load calculated from the strain indicator was found to be in good agreement with roll load found from the W.L. Robert’s formulation. entry and exit points. Here the rate of internal energy Keywords: Rolling mill, Roll load, Robert, strain gauges I. INTRODUCTION Rolls are the the most basic part in a rolling mill. Forces because
• [PDF]
• #### POWER IN ROLLING
The ratio λ = [ a/Lp] = [a/√R.∆t] is used to calculate the moment arm ‘a’ λ =0.5 for hot rolling and 0.45 for cold rolling. The torque is equal to the product of total rolling load and the effective moment arm. Since there are two rolls Torque Mt = 2P.a Consider two high roll mill as shown in the figure.
• 文件大小: 128KB [PDF]
• #### Fundamental concept of metal rolling mtixtl
•The rolling load also increases as the sheet entering the rolls becomes thinner (due to the term eQ). •At one point, no further reduction in thickness can be achieved if the deformation resistance of the sheet is greater than the roll pressure. The rolls in contact with the sheet are both severely elastically deformed.
• 文件大小: 440KB [PDF]
• #### Cold Rolling Mill for Aluminium Sheet
divided in to cold heavy rolling mill and cold finishing rolling mill. The stand type of it is four high cold rolling mill. COLD ROLLING ALUMINUM: Novelis operates a variety of types and sizes of cold roll mills at locations around the world, including some that run at exit speeds as high as 3,000 meters per minute. STAGES IN COLD ROLLING:
• 文件大小: 553KB
• #### Cold Rolling process overview EngineeringClicks
Cold Rolling ProcessThe Advantages of Cold RollingCold Rolled Metal PropertiesConclusionRolling is an important function of the steel industry. It’s a steel fabrication process involving passing the metal through a pair of rollers. There are two main types of rolling process: Flat rolling the finished product is a sheet Profile rolling the finished product is a bar or rod. The process always starts with hot rolling. Hot rolling refers to the process of rolling steel at a temperature typically above 900° C, greater than its recrystallisation 在engineeringclicks上查看更多信息
• #### (PDF) Methodology for Calculation of Rolling Load and
Methodology for Calculation of Rolling Load and Forces Acting On Herringbone Gear Using Hot Rolling Theory
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• #### Deformation Processing Rolling
– Calculate rolling power. Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton 13 Flat Rolling Analysis • Consider rolling of a flat plate in a 2-high rolling mill Cold rolling (below recrystallization point) strain hardening, plane strain von Mises 1
• [PDF]
• #### MANUFACTURING PROCESSES FIT
Cold rollingis a rolling operation carried out at room temperature. Cold rolling is commonly conducted after hot rolling when good surface quality and low thickness tolerance are needed. Cold rolling causes material strengthening. Types of Rolling 12 Some of the steel products made in a rolling mill. Rolled Products Made of Steel
• #### Load Analysis and Driven Power Calculation (Symmetrical 3
In this post, the calculation of the force capabilities of symmetrical three-roll bending machine is one of the methods, the other types of plate rolling machine can take it for reference. Force Analysis 2.1 Maximum torque required for a cylinder rolling. When the plate rolling machine is ing, the steel sheet should be rolled into the steel
• #### Deformation Processing Rolling
– Calculate rolling power. Prof. Ramesh Singh, Notes by Dr. Singh/ Dr. Colton 13 Flat Rolling Analysis • Consider rolling of a flat plate in a 2-high rolling mill Cold rolling (below recrystallization point) strain hardening, plane strain von Mises 1
• #### Calculation of rolling pressure distribution and force
Dec 01, 2014· 1. Introduction. Roll force is a key parameter in the process control of hot strip rolling, and its computational accuracy directly determines thickness precision, strip shape quality and rolling stability,.Therefore, the online calculation model of roll force with high accuracy is taken as the core for substantial studies at home and abroad,.
• #### (PDF) Methodology for Calculation of Rolling Load and
Methodology for Calculation of Rolling Load and Forces Acting On Herringbone Gear Using Hot Rolling Theory
• #### Calculating Mill Drive RPMs: Is Your Mill Up to Speed
Calculating Mill Drive RPMs: Is Your Mill Up to Speed By Robert A. Sladky V President Tube Mill Engineering Many "W" style mills that have the ability to shim up the bottom driven shafts in each section to maintain metal line after tooling is reed, and are many times also equipped with a individual motor drive in each of the three sections as illustrated below.
• #### New Rolling Method Of Reversing Cold Rolling Mill
2. Outline of Reversing Cold Rolling Mill (1) Outline of reversing cold rolling mill and rolling method As illustrated in Figure 1, a reversing cold rolling mill is comprised mainly of a mill proper, a winding machine (pay-off reel), an entry winding machine (entry tension reel) and delivery winding machine (delivery tension reel).
• #### Theoretical Metal Weight Calculation Formula (30 Types of
If you’re ing in the metaling industry, even you’re engineers, you will try to find one calculator to help you calculate the weight of various metals and steels including ms plate, gi sheet, structural steel, ms angle, mild steel, steel bar, square tube, angle, aluminum etc.
• #### 50 questions with answers in COLD ROLLING Science topic
Aug 18, 2020· Assume that cold rolling geometry is same as that for hot rolling, and hot rolling results in 10 mm thick, coarse graİned sheet with (0001)
• #### Mechanics of Sheet Metal Forming
5 Load instability and tearing 61 5.1 Introduction 61 5.2 Uniaxial tension of a perfect strip 62 5.3 Tension of an imperfect strip 64 5.4 Tensile instability in stretching continuous sheet 67 5.5 Factors affecting the forming limit curve 75 5.6 The forming window 79 5.7 Exercises 80 6 Bending of sheet 82 6.1 Introduction 82
• #### MANUFACTURING PROCESSES FIT
Cold rollingis a rolling operation carried out at room temperature. Cold rolling is commonly conducted after hot rolling when good surface quality and low thickness tolerance are needed. Cold rolling causes material strengthening. Types of Rolling 12 Some of the steel products made in a rolling mill. Rolled Products Made of Steel
• #### Understanding Rolling Process in Long Product Rolling Mill
Nov 27, 2015· The final dimensional quality of the rolled product is determined by the rolling stands within the finishing mill. The dimensional accuracy in the final product depends on many factors including the initial stock dimensions, roll pass sequence, temperature, microstructure, roll surface quality, roll and stand stiffness and the stock/roll friction condition.
• #### HOT ROLLING PRACTICE An Attempted Recollection
Rolling is a metal forming process in which metal stock is passed through a pair of rolls. There are two types of rolling process flat and profile rolling. In flat rolling the final shape of the product is either classed as sheet, also called "strip" (thickness less than 3 mm,) or plate (thickness more than 3 mm). In
• #### 03 rolling of metals SlideShare
Aug 26, 2014· • For cold-rolling with lubricants, μ ~ 0.05 0.10. • For hot-rolling,μ ~ 0.2 up to sticky condition. Suranaree University of Technology Tapany Udomphol Jan-Mar 2007 61. Example: Calculate the rolling load if steel sheet is hot rolled 30% from a 40 mm-thick s using a 900 mm-diameter roll. The s is 760 mm wide. Assume μ = 0.30.
• #### How to Size a Thickener Mineral Processing & Metallurgy
Knowing how to size a thickener and determine the capacity required to handle a pre-determined tonnage of a certain pulp, by overflowing a clear solution and obtaining the desired pulp density of thickener discharge, depends upon the settling rate of that particular pulp. The settling rate of any pulp is easily determined by simple oratory tests such as outlined below: Laboratory Test | 1,918 | 8,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-21 | latest | en | 0.85573 |
https://gmatclub.com/forum/how-many-different-5-letter-words-can-be-formed-from-the-word-orange-301095.html | 1,566,629,950,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319915.98/warc/CC-MAIN-20190824063359-20190824085359-00268.warc.gz | 490,314,892 | 140,051 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# How many different 5-letter words can be formed from the word ORANGE
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Math Expert
Joined: 02 Sep 2009
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How many different 5-letter words can be formed from the word ORANGE [#permalink]
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26 Jul 2019, 00:30
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Question Stats:
74% (00:47) correct 26% (01:10) wrong based on 31 sessions
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How many different 5-letter words can be formed from the word ORANGE using each letter only once?
(A) 6P6
(B) 36
(C) 6C6
(D) 66
(E) 6P5
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Re: How many different 5-letter words can be formed from the word ORANGE [#permalink]
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26 Jul 2019, 07:21
Bunuel wrote:
How many different 5-letter words can be formed from the word ORANGE using each letter only once?
(A) 6P6
(B) 36
(C) 6C6
(D) 66
(E) 6P5
total possible ways for 5 letter word ORANGE ; 6P5
IMO E
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Schools: ISB '15
Re: How many different 5-letter words can be formed from the word ORANGE [#permalink]
### Show Tags
26 Jul 2019, 07:34
I do it like this there are 5 positions to be filled and 6 candidate alphapbets who can fill them.
Now since the candidates cant be repeated the first position can have
1st can have 6 candidate alphabets
2nd can have 5 candidate alphabets
3rd can have 4
4th can have 3
5th can have 2
So in total u have 6x5x4x3x2 that is 6 !
6p6 will be 6!/(6-6)! Which will be 6!/0!
option ( A )
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Joined: 19 Jan 2019
Posts: 85
Re: How many different 5-letter words can be formed from the word ORANGE [#permalink]
### Show Tags
07 Aug 2019, 00:06
6p6=6p5=720.. ??? two options with same answer... The answer is 720 anyhow..
Re: How many different 5-letter words can be formed from the word ORANGE [#permalink] 07 Aug 2019, 00:06
Display posts from previous: Sort by | 786 | 2,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-35 | latest | en | 0.854535 |
https://www.quantumstudy.com/for-a-hyperbola-the-foci-are-at-%C2%B14-0-and-vertices-at-%C2%B12-0-its-equation-is/ | 1,701,409,461,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100276.12/warc/CC-MAIN-20231201053039-20231201083039-00632.warc.gz | 1,096,121,316 | 48,310 | # For a hyperbola, the foci are at (±4 , 0) and vertices at (±2 , 0). Its equation is
Q: For a hyperbola, the foci are at (±4 , 0) and vertices at (±2 , 0). Its equation is
(A) $\frac{x^2}{4} – \frac{y^2}{12} = 1$
(B) $\frac{x^2}{12} – \frac{y^2}{4} = 1$
(C) $\frac{x^2}{16} – \frac{y^2}{4} = 1$
(D) $\frac{x^2}{4} – \frac{y^2}{16} = 1$
Sol. ae = 4, a = 2
⇒ e = 2
But b2 = a2(e2 – 1)
⇒ b2 = 12
Hyperbola is $\frac{x^2}{4} – \frac{y^2}{12} = 1$
Hence (A) is the correct answer. | 240 | 487 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-50 | latest | en | 0.749548 |
https://www.geeksforgeeks.org/java-program-for-odd-even-sort-brick-sort/ | 1,579,987,354,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251681412.74/warc/CC-MAIN-20200125191854-20200125221854-00046.warc.gz | 889,423,009 | 24,829 | # Java Program for Odd-Even Sort / Brick Sort
This is basically a variation of bubble-sort. This algorithm is divided into two phases- Odd and Even Phase. The algorithm runs until the array elements are sorted and in each iteration two phases occurs- Odd and Even Phases.
In the odd phase, we perform a bubble sort on odd indexed elements and in the even phase, we perform a bubble sort on even indexed elements.
`// Java Program to implement ` `// Odd-Even / Brick Sort ` `import` `java.io.*; ` ` ` `class` `GFG { ` ` ``public` `static` `void` `oddEvenSort(``int` `arr[], ``int` `n) ` ` ``{ ` ` ``boolean` `isSorted = ``false``; ``// Initially array is unsorted ` ` ` ` ``while` `(!isSorted) { ` ` ``isSorted = ``true``; ` ` ``int` `temp = ``0``; ` ` ` ` ``// Perform Bubble sort on odd indexed element ` ` ``for` `(``int` `i = ``1``; i <= n - ``2``; i = i + ``2``) { ` ` ``if` `(arr[i] > arr[i + ``1``]) { ` ` ``temp = arr[i]; ` ` ``arr[i] = arr[i + ``1``]; ` ` ``arr[i + ``1``] = temp; ` ` ``isSorted = ``false``; ` ` ``} ` ` ``} ` ` ` ` ``// Perform Bubble sort on even indexed element ` ` ``for` `(``int` `i = ``0``; i <= n - ``2``; i = i + ``2``) { ` ` ``if` `(arr[i] > arr[i + ``1``]) { ` ` ``temp = arr[i]; ` ` ``arr[i] = arr[i + ``1``]; ` ` ``arr[i + ``1``] = temp; ` ` ``isSorted = ``false``; ` ` ``} ` ` ``} ` ` ``} ` ` ` ` ``return``; ` ` ``} ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ``int` `arr[] = { ``34``, ``2``, ``10``, -``9` `}; ` ` ``int` `n = arr.length; ` ` ` ` ``oddEvenSort(arr, n); ` ` ``for` `(``int` `i = ``0``; i < n; i++) ` ` ``System.out.print(arr[i] + ``" "``); ` ` ` ` ``System.out.println(``" "``); ` ` ``} ` `} ` `// Code Contribute by Mohit Gupta_OMG <(0_o)> `
Output:
```-9 2 10 34
```
Please refer complete article on Odd-Even Sort / Brick Sort for more details!
My Personal Notes arrow_drop_up
Article Tags :
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Please write to us at contribute@geeksforgeeks.org to report any issue with the above content. | 777 | 2,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-05 | longest | en | 0.512711 |
https://groups.google.com/g/sci.math/c/eCgYeC3aUbc/m/2haqpIEie_EJ | 1,679,814,186,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945433.92/warc/CC-MAIN-20230326044821-20230326074821-00630.warc.gz | 346,446,504 | 138,121 | # Vampire Numbers
1561 views
### cliff
Apr 28, 1994, 4:43:10 PM4/28/94
to
Title: Cliff Puzzle 20: Vampire Numbers
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please include your name,
address, and e-mail address. If you like, tell me a little bit
about yourself. You might also directly mail me a copy of your response
in addition to any responding you do in the newsgroup. I will assume it
is OK to describe your answer in any article or publication I may write
in the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover
* * *
Vampire Numbers
If we are to believe best-selling novelist Anne Rice,
vampires resemble humans in many respects, but live secret lives hidden
among the rest of us mortals. Consider a numerical metaphor for
vampires. I call numbers like 2187 vampires numbers because they're
formed when two progenitor numbers 27 and 81 are multiplied together
(27*81 = 2187).
Note that the vampire, 2187, contains the same digits as both parents,
except that these digits are subtly hidden, scrambled in some fashion.
Similarly, 1435 is a vampire number because it contains the
digits of the progenitors 35 and 41.
(35*41=1435)
These vampire numbers secretly inhabit our number system, but most have
been undetected so far. I believe there are only six four-digit
vampires in existence, but have no idea if there are any larger vampire
numbers.
What is the largest vampire number you can find lurking out there in
the world of integers?
As the numbers get larger and larger, how often do you expect to find
vampires? Do they get sparser or more frequent as one searches for
vampires up to a googol?
I will be happy to report the largest vampire number ever found
in an article I'm writing.
(If you do any computer searches, could you tell me what language
you used and what machine you used and any other interesting
features?)
### Queenie
Apr 30, 1994, 6:59:04 AM4/30/94
to
cl...@watson.ibm.com (cliff) writes:
>Title: Cliff Puzzle 20: Vampire Numbers
>From: cl...@watson.ibm.com
>
>
>Vampire Numbers
>These vampire numbers secretly inhabit our number system, but most have
>been undetected so far. I believe there are only six four-digit
>vampires in existence, but have no idea if there are any larger vampire
>numbers.
I think that none of them have been detected because no one wanted to
bother to find them, this is something that would require doing
a whole lot of multiplication, and is probably better suited
to a computer.
The four-digit vampyres are ...
15 x 93 = 1395
21 x 60 = 1260
21 x 87 = 1827
27 x 81 = 2187
30 x 51 = 1530
35 x 41 = 1435
80 x 86 = 6880
Now this is only if we are strict and exclude
vampyres like
03 x 501 = 1503
03 x 510 = 1530
...
I would be tempted to include them, but the zero does leave
a question unanswered. Besides including them would
make the list of 4 digit vamps bigger than would be nice
to post to this group.
>
>What is the largest vampire number you can find lurking out there in
>the world of integers?
>
So far this, I guess I could find a bigger one if I tried hard enough.
50000 x 50219 = 2510950000
50000 x 51002 = 2550100000
50000 x 51020 = 2551000000
50000 x 51263 = 2563150000
50000 x 51902 = 2595100000
50000 x 52631 = 2631550000
50000 x 59021 = 2951050000
50000 x 59102 = 2955100000
65049 x 65244 = 4244056956
65132 x 65408 = 4260153856 /* largest possible with 32bit int...
and no I don't feel like codeing
multiple precision right now :) */
>As the numbers get larger and larger, how often do you expect to find
>vampires? Do they get sparser or more frequent as one searches for
>vampires up to a googol?
I find that near 2^32 they become very infrequent compared to
the density at small values (1 - 10000).
>
>I will be happy to report the largest vampire number ever found
>in an article I'm writing.
>
>(If you do any computer searches, could you tell me what language
>you used and what machine you used and any other interesting
>features?)
I used a C program. I just wrote it. Boy I gotta lay off the coffee!!!
The Machine is a 32bit Intel-based UNIX machine.
I didn't bother to optimize for speed, but it does a decent job
of finding vamps anyway.
If you are interested in my code, or how I coded it,
^^^^^^^ I thought that was appropriate, don't you?
--
car...@vampyre.colorado.edu | What do 7, 11, and 4882195 have in
common? Send a self-addressed stamped envelope to 'PRIMES, PO BOX 17986,
Boulder, CO 80308, for your official prime number. Inquire within.
### Shriram Krishnamurthi
Apr 30, 1994, 5:14:31 PM4/30/94
to
> 65049 x 65244 = 4244056956
> 65132 x 65408 = 4260153856 /* largest possible with 32bit int...
> and no I don't feel like codeing
> multiple precision right now :) */
65132 x 65408 = 4260153856
65132 x 87071 = 5671108372
65132 x 89402 = 5822931064
65133 x 98496 = 6415339968
and so on.
### Arturo Viso Magidin
May 1, 1994, 10:02:58 PM5/1/94
to
In article <shriram....@cs.rice.edu>,
Maybe I didn't understand the rules correctly, but it seems to me
that once you have generated vampyre numbers you can get
arbitrarily large vampyre numbers with them.
If a and b are two numbers, each written with n digits, such that
a*b is written as a permutation of the digits of a and b, then
(a*10^m) * (b*10^m) will again be a vampyre number, since
you can write it with the digits of a*10^M and b*10^M. You write
a*b and then add 2m zeroes, with by an amazing coincidence are the
number of zeroes you have added to a and b together.
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
### Queenie
May 1, 1994, 11:08:20 PM5/1/94
to
>Maybe I didn't understand the rules correctly, but it seems to me
>that once you have generated vampyre numbers you can get
>arbitrarily large vampyre numbers with them.
I am not sure who's rules you are following. Perhaps you can
post again and tell us all that 2 + 2 still equals 4 even for
extrodinarily large values of 2.
c'mon I think that the fun is in finding numbers of distinction.
Did you know that 1 approximated to ten decimal places is
1.0000000000 ?
Give us all a break.
--
car...@vampyre.colorado.edu | What do 7, 11, and 438479857 have in
### Arturo Viso Magidin
May 2, 1994, 5:12:59 PM5/2/94
to
>
>I am not sure who's rules you are following. Perhaps you can
>post again and tell us all that 2 + 2 still equals 4 even for
>extrodinarily large values of 2.
>
>c'mon I think that the fun is in finding numbers of distinction.
>
>Give us all a break.
>--
Ok, here's the original post:
-------------------------BEGIN QUOTED ARTICLE----------------------------------
From cl...@watson.ibm.com Mon May 2 14:06:56 PDT 1994
Article: 71190 of sci.math
Path: agate!howland.reston.ans.net!EU.net!Germany.EU.net!Munich.Germany.EU.net!ibm.de!aixssc.uk.ibm.com!watnews.watson.ibm.com!watson.ibm.com!cliff
From: cl...@watson.ibm.com (cliff)
Newsgroups: sci.math
Subject: Vampire Numbers
Date: 28 Apr 1994 20:43:10 GMT
Organization: A
Lines: 43
Distribution: world
Message-ID: <2pp74u\$g...@watnews1.watson.ibm.com>
NNTP-Posting-Host: cliff.watson.ibm.com
Title: Cliff Puzzle 20: Vampire Numbers
From: cl...@watson.ibm.com
Vampire Numbers
If we are to believe best-selling novelist Anne Rice,
vampires resemble humans in many respects, but live secret lives hidden
among the rest of us mortals. Consider a numerical metaphor for
vampires. I call numbers like 2187 vampires numbers because they're
formed when two progenitor numbers 27 and 81 are multiplied together
(27*81 = 2187).
Note that the vampire, 2187, contains the same digits as both parents,
except that these digits are subtly hidden, scrambled in some fashion.
Similarly, 1435 is a vampire number because it contains the
digits of the progenitors 35 and 41.
(35*41=1435)
These vampire numbers secretly inhabit our number system, but most have
been undetected so far. I believe there are only six four-digit
vampires in existence, but have no idea if there are any larger vampire
numbers.
What is the largest vampire number you can find lurking out there in
the world of integers?
As the numbers get larger and larger, how often do you expect to find
vampires? Do they get sparser or more frequent as one searches for
vampires up to a googol?
-------------------------END QUOTED MATERIAL-------------------------
Of course I agree that the only interest is in finding
non-trivially large vampyre numbers. Please point out the exact
place in this post were it is asked that the two numbers we
are multyplying together must not be both multiples of 10, or that
we must factor out as many powers of 10 as possible from both numbers.
Once you do that, I'll be happy to give you a break.
My point was that the definition of vampyre number is lacking unless
you specify that the trivial way of extending vampyre numbers which
I pointed out is ruled out.
Happy now?
Jul 2, 2019, 10:36:53 AM7/2/19
to
Being a vampire has certain limitations, but it can also be a ton of fun. Your extra strengths and abilities can make you successful in almost every endeavor you participate in and before you know it the money and acquaintances will come streaming in. You can build wealth and gain prestige and notoriety and attempt things you may never have even considered as a human. One thing you will definitely have more of is time. Beef up your education and learn all you ever wanted to. Travel the world to see things most people only ever see on TV This is going to be especially fun if you turned to share your life with one of us. Let us show you the wonders of the world. Learn new languages, go skydiving or scuba dive with sharks, visit the African safari. You no longer need to be scared of nature or wildlife you will have become the worlds strongest predator. Have fun with it and your life as a vampire can be more fulfilling than you ever dreamed. Explore, experiment, experience and get excited. There’s a big world out there with lots to see and do and as a vampire, you can do it all, if willing and ever ready to be a full blooded vampire with powers and mighty great skills then these is the opportunity for you to get transformed and turned into a vampire, contact the mighty hindu priest and also he is a vampire lord, find him on his email and lay your request and heart wishes to him, trust me you will find him on: Vampirelordtran...@gmail.com or you can as well find me on kaith...@gmail.com
### tina...@gmail.com
Jul 30, 2019, 5:17:21 PM7/30/19
to
Dylan. Zanelli. Vampires
### steve...@gmail.com
Jan 26, 2020, 2:44:31 PM1/26/20
to
My Name is Steve Jones from Canada, i turn to a vampire any time i want to, I become a real vampire because of how people treat me, This world is a wicked world and not fair to any body. At the snack of my finger things are made happened. Am now a powerful vampire and no one step on me without an apology goes free. I turn to human being also at any time i want to. And am one of the most dreaded and respected person in my country. i am now also very famous and rich with the help of the VAMPIRES EMPIRE. i get what ever a want. i become a vampire through the help of my friend who introduce me into a vampire Kingdom by given me their email: jamessucces...@gmail.com, if you want to become a powerful and a real vampire kindly contact the vampire kingdom on their email: jamessucces...@gmail.com for help. it is real. Contact them today. jamessucces...@gmail.com
### Python
Jan 26, 2020, 3:59:29 PM1/26/20
to
steve...@gmail.com wrote:
> My Name is Steve Jones from Canada, i turn to a vampire any time i want to, I become a real vampire because of how people treat me, This world is a wicked world and not fair to any body. At the snack of my finger things are made happened. Am now a powerful vampire and no one step on me without an apology goes free. I turn to human being also at any time i want to. And am one of the most dreaded and respected person in my country. i am now also very famous and rich with the help of the VAMPIRES EMPIRE. i get what ever a want. i become a vampire through the help of my friend who introduce me into a vampire Kingdom by given me their email: jamessucces...@gmail.com, if you want to become a powerful and a real vampire kindly contact the vampire kingdom on their email: jamessucces...@gmail.com for help. it is real. Contact them today. jamessucces...@gmail.com
You've posted in the right place Steve. We do have a few vampires around
John Gabriel is a vampire of common sense, he can draw a conclusion from
empty, void statement such as "[f(x+h)-f(x)]/h - f'(x) is a function of
x,h"
Adjunk Crank Lekturer Wolfgang Mueckheim can suck students' blood at
Hochschule Augsburg, Germany, by drawing conclusions from the fact
that any matematical object is a member of a finite set.
Demented Arab Karzhedin can demonstrate fallacies for ages by confusing
diophantian equation with real domain equations. He is sucking blood
from common decency.
David Petry may enjoy you too, he decided to avoid studying math because
he mislead himself into thinking that nobody in math was concerned with
effectiveness, so he learnt nothing and spent a few decades rantings in
the void. He is sucking blood from the elementary education he avoided.
Archimeded Plutonium, aka Perelmann is another kind of vampire, but he
is sucking blood out of himself.
### James Waldby
Jan 26, 2020, 9:28:25 PM1/26/20
to
On Sun, 26 Jan 2020 21:59:22 +0100, Python wrote:
> steve...@gmail.com wrote:
>> My Name is Steve Jones from Canada, i turn to a vampire any time i want
>> to, I become a real vampire because of how people treat me, This world
>> is a wicked world and not fair to any body. At the snack of my finger
>> things are made happened. Am now a powerful vampire and no one step on
>> me without an apology goes free. I turn to human being also at any time
>> i want to. And am one of the most dreaded and respected person in my
>> country. i am now also very famous and rich with the help of the
>> VAMPIRES EMPIRE. [...snip...]
>
> You've posted in the right place Steve. We do have a few vampires around
> who could join your Kingdom.
>
> John Gabriel is a vampire of common sense, he can draw a conclusion from
> empty, void statement such as "[f(x+h)-f(x)]/h - f'(x) is a function of
> x,h"
>
> Adjunk Crank Lekturer Wolfgang Mueckheim can suck students' blood at
> Hochschule Augsburg, Germany, by drawing conclusions from the fact that
> any matematical object is a member of a finite set.
>
> Demented Arab Karzhedin can demonstrate fallacies for ages by confusing
> diophantian equation with real domain equations. He is sucking blood
> from common decency.
...
> Archimedes Plutonium, aka Perelmann is another kind of vampire, but he
> is sucking blood out of himself.
Perhaps you mean Poehlmann, rather than Perelmann? Eg see the third
entry in [1], which show some other names or aliases the creature has
used: "Archimedes Plutonium (current legal name, born Ludwig Poehlmann
in 1950, raised as Ludwig Hansen, also known as Ludwig van Ludvig and
Ludwig Plutonium)"
[1] https://en.wikipedia.org/wiki/Usenet_personality#Eccentric_believers
--
jiw
### Sergio
Jan 30, 2020, 11:50:48 AM1/30/20
to
I knew it! they are all batty as last years leftover fruitcakes
### Mitch Raemsch
Jan 30, 2020, 2:50:10 PM1/30/20
to
Divide by zero and it takes any number down. | 4,222 | 15,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-14 | longest | en | 0.887233 |
https://www.esaral.com/q/the-mean-and-variance-of-7-observations-are-8-and-16-respectively-14580 | 1,716,226,843,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00791.warc.gz | 668,894,800 | 11,473 | # The mean and variance of 7 observations are 8 and 16 , respectively.
Question:
The mean and variance of 7 observations are 8 and 16 , respectively. If five observations are $2,4,10,12,14$, then the absolute difference of the remaining two observations is:
1. 2
2. 4
3. 3
4. 1
Correct Option: 1
Solution:
$\bar{x}=\frac{2+4+10+12+14+x+y}{7}=8$
$x+y=14$......(i)
$(\sigma)^{2}=\frac{\sum\left(\mathrm{x}_{\mathrm{i}}\right)^{2}}{\mathrm{n}}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}\right)^{2}$
$16=\frac{4+16+100+144+196+x^{2}+y^{2}}{7}-8^{2}$
$16+64=\frac{460+x^{2}+y^{2}}{7}$
$560=460+x^{2}+y^{2}$
$x^{2}+y^{2}=100$.........(ii)
Clearly by (i) and (ii), $|x-y|=2$
Ans. 1 | 282 | 702 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-22 | latest | en | 0.598733 |
https://flatearthsoc.com/200-proof-earth-is-flat-nick-davis-realtor.html | 1,571,225,721,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00214.warc.gz | 482,835,982 | 12,144 | 159) If there were progressively faster and faster spinning atmosphere the higher the altitude that would mean it would have to abruptly end at some key altitude where the fastest layer of gravitized spinning atmosphere meets the supposed non-gravitized non-spinning non-atmosphere of infinite vacuum space! NASA has never mentioned what altitude this impossible feat allegedly happens, but it is easily philosophically refuted by the simple fact that vacuums cannot exist connected to non-vacuums while maintaining the properties of a vacuum – not to mention, the effect such a transition would have on a rocket “space ship” would be disastrous.
#### The earth isn't pulled into a sphere because the force known as gravity exists in a greatly diminished form compared to what is commonly taught. The earth is constantly accelerating up at a rate of 32 feet per second squared (or 9.8 meters per second squared). This constant acceleration causes what you think of as gravity. Imagine sitting in a car that never stops speeding up. You will be forever pushed into your seat. The earth works much the same way. It is constantly accelerating upwards being pushed by a universal accelerator (UA) known as dark energy or aetheric wind.
10) Ship captains in navigating great distances at sea never need to factor the supposed curvature of the Earth into their calculations. Both Plane Sailing and Great Circle Sailing, the most popular navigation methods, use plane, not spherical trigonometry, making all mathematical calculations on the assumption that the Earth is perfectly flat. If the Earth were in fact a sphere, such an errant assumption would lead to constant glaring inaccuracies. Plane Sailing has worked perfectly fine in both theory and practice for thousands of years, however, and plane trigonometry has time and again proven more accurate than spherical trigonometry in determining distances across the oceans. If the Earth were truly a globe, then every line of latitude south of the equator would have to measure a gradually smaller and smaller circumference the farther South travelled. If, however, the Earth is an extended plane, then every line of latitude south of the equator should measure a gradually larger and larger circumference the farther South travelled. The fact that many captains navigating south of the equator assuming the globular theory have found themselves drastically out of reckoning, more so the farther South travelled, testifies to the fact that the Earth is not a ball.
###### 69) The New York City skyline is clearly visible from Harriman State Park’s Bear Mountain 60 miles away. If Earth were a ball 25,000 miles in circumference, viewing from Bear Mountain’s 1,283 foot summit, the Pythagorean Theorem determining distance to the horizon being 1.23 times the square root of the height in feet, the NYC skyline should be invisible behind 170 feet of curved Earth.
If the sun was 93 million miles away the light would fade evenly on the horizon. This is depicted in NASA's animation of the ISS watching s sunset. In reality, however, this is NOT what we see. Instead of a distant sun creating an evenly fading sunset, we see a small light source illuminating locally and taking its light with it as it moves away from our side of flat earth.
59.) Mr. Proctor says.- "The Sun is so far off that even moving from one side of the Earth to the other does not cause him to be seen in a different direction – at least the difference is too small to be measured." Now, since we know that north of the equator, say 45 degrees, we see the Sun at mid-day to the south, and that at the same distance south of the equator we see the Sun at mid-day to the north, our very shadows on the round cry aloud against the delusion of the day and give us a proof that Earth is not a globe.
137) Another assumption and supposed proof of Earth’s shape, heliocentrists claim that lunar eclipses are caused by the shadow of the ball-Earth occulting the Moon. They claim the Sun, Earth, and Moon spheres perfectly align like three billiard balls in a row so that the Sun’s light casts the Earth’s shadow onto the Moon. Unfortunately for heliocentrists, this explanation is rendered completely invalid due to the fact that lunar eclipses have happened and continue to happen regularly when both the Sun and Moon are still visible together above the horizon! For the Sun’s light to be casting Earth’s shadow onto the Moon, the three bodies must be aligned in a straight 180 degree syzygy, but as early as the time of Pliny, there are records of lunar eclipses happening while both the Sun and Moon are visible in the sky. Therefore the eclipsor of the Moon cannot be the Earth/Earth’s shadow and some other explanation must be sought.
132) The Sun’s light is golden, warm, drying, preservative and antiseptic, while the Moon’s light is silver, cool, damp, putrefying and septic. The Sun’s rays decrease the combustion of a bonfire, while the Moon’s rays increase combustion. Plant and animal substances exposed to sunlight quickly dry, shrink, coagulate, and lose the tendency to decompose and putrify; grapes and other fruits become solid, partially candied and preserved like raisins, dates, and prunes; animal flesh coagulates, loses its volatile gaseous constituents, becomes firm, dry, and slow to decay. When exposed to moonlight, however, plant and animal substances tend to show symptoms of putrefaction and decay. This proves that Sun and Moon light are different, unique, and opposites as they are in the geocentric flat model.
#### If the earth is flat we are held hostage to one of the biggest hoaxes of the last 500 hundred years of human existence. Every day I see evidence of the flat earth. Just look at how at time the clouds are not moving at all --completely still. If the earth was spinning at 1600 klms per hour --how could it be that at times the clouds are completely still not moving in the sky at all.(how could the clouds be moving at the same speed as the rotating earth?)When you look at a long distant horizon --from right to left --there is no curviture in the earth whatsoever. If you try to point this out to people they will think you have lost your mind.. we have been brainwashed to believe a lie which is very difficult to prove to the world populous that there is a flat earth. The deception is wrapped in scientific mumbo jumbo which is very easy to accept as it is backed up by a lot of scientific explanation--if you want to take the easy way out--you just accept that the earth is a globe spinning at 1600 klms per hour . But how does all the water of the seas stay stuck to the planet when the earth is spinning around at 1600 klms per hour? (gravity cannot hold water upside down--water will flow to the lowest point--how can that happen on a spinning globe?)
52.) It is a well-known and indisputable fact that there is a far greater accumulation of ice south of the equator than is to be found at an equal latitude north: and it is said that at Kerguelen, 50 degrees south, 18 kinds of plants exist, whilst, in Iceland, 15 degrees nearer the northern centre, there are 870 species; and, indeed, all the facts in the case show that the Sun's power is less intense at places in the southern region than it is in corresponding latitudes north. Now, on the Newtonian hypothesis, all this is inexplicable, whilst it is strictly in accordance with the facts brought to light by the carrying out of the principles involved in the Zetetic Philosophy of "Parallax." This is a proof that the Earth is not a globe.
106) The so-called “South Pole” is simply an arbitrary point along the Antarctic ice marked with a red and white barbershop pole topped with a metal ball-Earth. This ceremonial South Pole is admittedly and provably NOT the actual South Pole, however, because the actual South Pole could be demonstrably confirmed with the aid of a compass showing North to be 360 degrees around the observer. Since this feat has never been achieved, the model remains pure theory, along with the establishment’s excuse that the geomagnetic poles supposedly constantly move around making verification of their claims impossible.
20) If Earth were truly constantly spinning Eastwards at over 1000mph, vertically-fired cannonballs and other projectiles should fall significantly due west. In actual fact, however, whenever this has been tested, vertically-fired cannonballs shoot upwards an average of 14 seconds ascending, 14 seconds descending, and fall back to the ground no more than 2 feet away from the cannon, often directly back into the muzzle.
Astronomers have never agreed amongst themselves about a rotating Moon revolving round a rotating and revolving Earth - this Earth, Moon, planets and their satellites all, at the same time dashing through space, around the rotating and revolving Sun, towards the constellation Hercules, at the rate of four millions of miles a day! And they never will: agreement is impossible! With the a Earth a plane and without motion, the whole thing is clear. And if a straw will show which way the wind blows, this may be taken as a pretty strong proof that the Earth is not a globe.
81.) Newtonian philosophers teach us that the Moon goes round: the Earth from west to east. But observation – man's most certain mode of gaining knowledge – shows us that the Moon never ceases to move in the opposite direction – from east to west. Since, then, we know that nothing can possibly move in two, opposite directions at the same time, it is a proof that the thing is a big blunder; and, in short, it is a proof that the Earth is not a globe.
Bottom line, there is some power in this world trying hard to make people live this illusion that earth is not the center of this universe and that we came from single cells. In other words, no God and all is just a matter of random toss of a lucky dice .... my scientific answer to these people is WTF why you do not want to wake up and open your eyes. It took me 30 Thirty YEARS to see this and I regret that I did not see it before. I feel sad that I found this too late after spending my life supporting an illusion a fake science ..... :(
2. Another related thing I don’t understand: if the sun and moon are always above the disk of the Earth, why can’t everyone in the world see them at all times? Surely they should always be visible, at least at a low angle. I can’t draw myself any diagram where they are not always visible, but we see that that doesn’t happen. I can’t see how night time happens. Help!
On February 24th, YouTubers GlobeBusters (featured in the documentary) uploaded a video entitled "Behind The Curve Documentary Reveals Globe Earth Desperation!" The video explains that the documentary was misleading. It gained 41,000 views in a couple weeks (shown below, left). On February 25th, Jeranism (also featured in the documentary) responded negatively to the film (shown below, right).
In the past 60 years of space exploration, we’ve launched satellites, probes, and people into space. Some of them got back, some of them still float through the solar system (and almost beyond it), and many transmit amazing images to our receivers on Earth. In all of these photos, the Earth is (wait for it) spherical. The curvature of the Earth is also visible in the many, many, many, many photos snapped by astronauts aboard the International Space Station. You can see a recent example from ISS Commander Scott Kelly's Instagram right here:
158) If “gravity” magically dragged the atmosphere along with the spinning ball Earth, that would mean the higher the altitude, the faster the spinning atmosphere would have to be turning around the center of rotation. In reality, however, if this were happening then rain and fireworks would behave entirely differently as they fell down through progressively slower and slower spinning atmosphere. Hot-air balloons would also be forced steadily faster Eastwards as they ascended through the ever increasing atmospheric speeds.
###### This flat Eart thing is just showing how easy it is to malipulate people. I think, that the whole point of this is so that people would not believe blindly in everything so called authorities are trying to sell as "universal truths" and start to think on their own. Many people will believe in anything, as long as it's advocated by someone they see as authority in some field. I see it this way.
99) Viewed from a ball-Earth, Polaris, situated directly over the North Pole, should not be visible anywhere in the Southern hemisphere. For Polaris to be seen from the Southern hemisphere of a globular Earth, the observer would have to be somehow looking “through the globe,” and miles of land and sea would have to be transparent. Polaris can be seen, however, up to over 20 degrees South latitude.
44.) It is in evidence that, if a projectile be fired from a rapidly moving body in an opposite direction to that in which the body is going, it will fall short of the distance at which it would reach the ground if fired in the direction of motion. Now, since the Earth is said to move at the rate of nineteen miles in, a second of time, "from west to east," it would make all the difference imaginable if the gun were fired in an opposite direction. But, as, in practice, there is not the slightest difference, whichever way the thing may be done, we have a forcible overthrow of all fancies relative to the motion of the Earth, and a striking proof that the Earth is not a globe.
Then God created land and vast oceans. The oceans were expansive and deep, with sea life, but the Bible also describes another body of water called the “deep”. Is it just describing all the water on earth, or is there something else, like a deep denser ocean beneath Earth? In Genesis 6-7, God causes a flood to destroy the earth because of its wickedness. The flood waters came from the “deep” below the earth and the waters from above the heavens (the firmament. In Genesis 7:11 it says, “In the six hundredth year of Noah’s life, in the second month, the seventeenth day of the month, the same day were all the fountains of the great deep broken up, and the windows of heaven were opened.”
```69.) Mr. Lockyer says: "The appearances connected with the rising and setting of the Sun and stars may be due either to our earth being at rest and the Sun and stars traveling round it, or the earth itself turning round, while the Sun and stars are at rest." Now, since true science does not allow of any such beggarly alternatives as these, it is plain that modern theoretical astronomy is not true science, and that its leading dogma is a fallacy. We have, then, a plain proof that the Earth is not a globe.
```
#### I started reading this e-book today and finished it today. Great Work ! I have to say, even if approached flat-earth model fistly 3 month ago, this Book really opened my eyes completely and still i know how strong the Brainwash is and even if you - by ratio - see something clear, it will still need a time to be completely internalized. Thx alot, greetingz from Germany also i was able to watch the video below, which oc is blocked in Germany, but this led me to HotSpotShield, also very important for a native German, greetingz from Frankfurt/Germany, Benjamin
48) On a ball-Earth Santiago, Chile to Johannesburg, South Africa should be an easy flight all taking place below the Tropic of Capricorn in the Southern hemisphere, yet every listed flight makes a curious re-fueling stop in Senegal near the Tropic of Cancer in the North hemisphere first! When mapped on a flat Earth the reason why is clear to see, however, Senegal is actually directly in a straight-line path half-way between the two.
For most media and angles of incidence, the light transmits from one medium to the other. However, when passing from a medium of higher index of refraction into a medium of lower index of refraction at a sufficiently high angle of incidence, there may not be a real value for the angle of refraction. When this happens, the light cannot pass into the second medium. Instead, the light is reflected off the interface and back into the first medium. We call this phenomenon total internal reflection. Many devices make use of total internal reflection. Total internal reflection allows a prism with two 45-degree angles and one 90-degree angle to reflect light at a right angle. One could use a mirror mounted at a 45-degree angle to do the same thing, but total internal reflection is nearly 100% efficient, while the best mirrors are perhaps 85% efficient. Many optical devices, such as binoculars and periscopes, make use of this. Fiber optics are thin wires of glass. Being so thin, fiber optics are flexible and as easy to handle as any metal wire. Glass has a relatively high index of refraction, so light shining down a fiber optic is totally reflected internally by the walls of the fiber optic, if the fiber optic is not bent too sharply. We use fiber optics every day with telephone, cable TV, and internet connections.
173) NASA has several alleged photographs of the ball-Earth which show several exact duplicate cloud patterns! The likelihood of having two or three clouds of the exact same shape in the same picture is as likely as finding two or three people with exactly the same fingerprints. In fact it is solid proof that the clouds were copied and pasted in a computer program and that such pictures showing a ball-shaped Earth are fakes.
so if the earth is not a sphere in space revolveing around the sphereical sun, then what is it. Its one thing to say that "its not that way" but its different to say "its actually this way not that way". So what way is it? what way are you proposing is the correct way? do you beleive this is the only planet in the universe? do you believe that the stars are only decorations on a flat backdrop? I'm not certain what idea you are proposing is the correct way of looking at this...
19.) Every man in his senses goes the most reasonable way to work to do a thing. Now, astronomers (one after another – following a leader), while they are telling us that Earth is a globe, are cutting off the upper half of this suppositious globe in their books, and, in this way, forming the level surface on which they describe man as living and moving! Now, if the Earth were really a globe, this would be just the most unreasonable and suicidal mode of endeavoring to show it. So that, unless theoretical astronomers are all out of their senses together, it is, clearly, a, proof that the Earth is not a globe.
In Mr. Proctor's "Lessons in Astronomy," page 15, a ship is represented as sailing away from the observer, and it is given in five positions or distances away on its journey. Now, in its first position, its mast appears above the horizon, and, consequently, higher than the observer's line of vision. But, in its second and third positions, representing the ship as further and further away, it is drawn higher and still higher up above the line of the horizon! Now, it is utterly impossible for a ship to sail away from an observer, under the, conditions indicated, and to appear as given in the picture. Consequently, the picture is a misrepresentation, a fraud, and a disgrace. A ship starting to sail away from an observer with her masts above his line of sight would appear, indisputably, to go down and still lower down towards the horizon line, and could not possibly appear - to anyone with his vision undistorted - as going in any other direction, curved or straight. Since, then the design of the astronomer-artist is to show the Earth to be a globe, and the points in the picture, which would only prove the Earth to be cylindrical if true, are NOT true, it follows that the astronomer-artist fails to prove, pictorially, either that the Earth is a globe or a cylinder, and that we have, therefore, a reasonable proof that the Earth is not. a globe.
If the Earth were a globe, it would, if we take Valentia to be the place of departure, curvate downwards, in the 1665 miles across the Atlantic to Newfoundland, according to the astronomers' own tables, more than three hundred miles; but, as the surface of the Atlantic does not do so - the fact of its levelness having been clearly demonstrated by Telegraph Cable surveyors, - it follows that we have a grand proof that Earth is not a globe.
16) The experiment known as “Airy’s Failure” proved that the stars move relative to a stationary Earth and not the other way around. By first filling a telescope with water to slow down the speed of light inside, then calculating the tilt necessary to get the starlight directly down the tube, Airy failed to prove the heliocentric theory since the starlight was already coming in the correct angle with no change necessary, and instead proved the geocentric model correct.
These photographs clearly reveal that the hulls of these two ships progressively disappeared as the ships moved farther away. This is consistent with what we would expect if the earth is spherical, but this cannot be explained if the earth is flat. Therefore, this is good evidence that the earth is spherical. The results presented here contradict the many photos on the internet of objects beyond the horizon that supposedly prove that the earth is flat. Those alleged proofs are flawed because they failed to take account of atmospheric refraction due to a temperature inversion. By conducting this experiment when there was no possibility of a temperature inversion, I avoided that complication. The fact that inferior mirages consistently showed up in the photographs prove that there was no temperature inversion, indicating instead that there was a slightly warmer layer of air in contact with the water, with slightly cooler air above.
###### Hodgson quit Supertramp in 1983. Davies's relationship with him had deteriorated[citation needed], and the group's last hit before his departure, "My Kind of Lady", featured little involvement from Hodgson as either a writer or performer. The song was a showcase for Davies's vocal range, with him singing in everything from a booming bass to a piercing falsetto to his natural raspy baritone. With Davies firmly at the helm, Supertramp returned to a more non-commercial, progressive rock-oriented approach with the album Brother Where You Bound and had another hit with "Cannonball". The band continued to tour and record for another five years before disbanding, with a mutual agreement between the members that Supertramp had run its course.[13]
90.) "Is water level, or is it not?" was a question once asked of an astronomer. "Practically, yes; theoretically, no," was the reply. Now, when theory does not harmonize with practice, the best thing to do is to drop the theory. (It is getting too late, now to say "So much the worse for the factsI") To drop the theory which supposes a curved surface to standing water is to acknowledge the facts which form the basis of Zetetic philosophy. And since this will have to be done sooner or later, – it is a proof that the Earth is not a globe.
135) Not only is the Moon clearly self-luminescent, shining its own unique light, but it is also largely transparent. When the waxing or waning Moon is visible during the day it is possible to see the blue sky right through the Moon. And on a clear night, during a waxing or waning cycle, it is even possible to occasionally see stars and “planets” directly through the surface of the Moon! The Royal Astronomical Society has on record many such occurrences throughout history which all defy the heliocentric model.
Some say the flood was a myth. But there is plenty of evidence of Noah’s flood to prove that it really happened. If you believe in the story of the flood in the Bible, then you must also believe in the “waters above” that God used to destroy that wicked generation. The only Earth model that fits the Biblical description of a glass firmament, holding the waters above, is the Flat Earth. Again the globe falls flat!
I started reading this e-book today and finished it today. Great Work ! I have to say, even if approached flat-earth model fistly 3 month ago, this Book really opened my eyes completely and still i know how strong the Brainwash is and even if you - by ratio - see something clear, it will still need a time to be completely internalized. Thx alot, greetingz from Germany also i was able to watch the video below, which oc is blocked in Germany, but this led me to HotSpotShield, also very important for a native German, greetingz from Frankfurt/Germany, Benjamin | 5,238 | 24,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-43 | latest | en | 0.966269 |
https://aptitude.gateoverflow.in/8060/cat2020-2-40 | 1,638,144,694,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00494.warc.gz | 201,829,964 | 21,625 | 23 views
The humanities department of a college is planning to organize eight seminars, one for each of the eight doctoral students $\text{- A, B, C, D, E, F, G}$ and $\text{H}.$ Four of them are from Economics, three from sociology and one from Anthropology department. Each student is guided by one among $\text{P, Q, R, S}$ and $\text{T}.$ Two students are guided by of $\text{P, R}$ and $\text{T},$ while one student is guided by each of $\text{Q}$ and $\text{S}.$ Each student is guided by a guide belonging to their department.
Each seminar is to be scheduled in one of four consecutive $30\text{-minute}$ slots starting at $9:00 \; \text{am}, \; 9:30 \; \text{am}, \; 10:00 \; \text{am}$ and $10:30 \; \text{am}$ on the same day. More than one seminars can be scheduled in a slot, provided the guide is free. Only three rooms are available and hence at the most three seminars can be scheduled in a slot. Students who are guided by the same guide must be scheduled in consecutive slots.
The following additional facts are also known.
1. Seminars by students from Economics are scheduled in each of the four slots.
2. $\text{A’s}$ is the only seminar that is scheduled at $10:00 \; \text{am}. \text{A}$ is guided by $\text{R}.$
3. $\text{F}$ is an Anthropology student whose seminar is scheduled at $10:30 \; \text{am}.$
4. The seminar of a Sociology student is scheduled at $9:00 \; \text{am}.$
5. $\text{B}$ and $\text{G}$ are both Sociology students, whose seminars are scheduled in the same slot. The seminar of an Economics student, who is guided by $\text{T},$ is also scheduled in the same slot.
6. $\text{P},$ who is guiding both $\text{B}$ and $\text{C},$ has students scheduled in the first two slots.
7. $\text{A}$ and $\text{G}$ are scheduled in two consecutive slots.
Who all are $\text{NOT}$ guiding any Economics students$?$
1. $\text{P, R}$ and $\text{S}$
2. $\text{P, Q}$ and $\text{R}$
3. $\text{P, Q}$ and $\text{S}$
4. $\text{Q, R}$ and $\text{S}$ | 573 | 1,979 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-49 | latest | en | 0.918094 |
http://encyclopedia.kids.net.au/page/ab/Abelian_group | 1,575,574,703,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540482038.36/warc/CC-MAIN-20191205190939-20191205214939-00362.warc.gz | 47,587,586 | 5,295 | ## Encyclopedia > Abelian group
Article Content
# Abelian group
In abstract algebra, an abelian group is a group (G, *) that is commutative, i.e., in which a * b = b * a holds for all elements a and b in G. Abelian groups are named after Niels Henrik Abel.
If a group is abelian, we usually write the operation as + instead of *, the identity element as 0 (often called the zero element in this context) and the inverse of the element a as -a.
Examples of abelian groups include all cyclic groups such as the integers Z (with addition) and the integers modulo n Zn (also with addition). The real numbers form an abelian group with addition, as do the non-zero real numbers with multiplication. Every field gives rise to two abelian groups in the same fashion. Another important example is the factor group Q/Z, an injective cogenerator.
If n is a natural number and x is an element of an abelian group G, then nx can be defined as x + x + ... + x (n summands) and (-n)x = -(nx). In this way, G becomes a module over the ring Z of integers. In fact, the modules over Z can be identified with the abelian groups. Theorems about abelian groups can often be generalized to theorems about modules over principal ideal domains. An example is the classification of finitely generated abelian groups.
Any subgroup of an abelian group is normal, and hence factor groups can be formed freely. Subgroups, factor groups, products and direct sums of abelian groups are again abelian. If f, g : GH are two group homomorphisms between abelian groups, then their sum f+g, defined by (f+g)(x) = f(x) + g(x), is again a homomorphism. (This is not true if H is a non-abelian group). The set Hom(G, H) of all group homomorphisms from G to H thus turns into an abelian group in its own right.
The abelian groups, together with group homomorphisms, form a category, the prototype of an abelian category.
Somewhat akin to the dimension of vector spaces, every abelian group has a rank. It is defined as the cardinality of the largest set of linearly independent elements of the group. The integers and the rational numbers have rank one, as well as every subgroup of the rationals. While the rank one torsion-free abelian groups are well understood, even finite-rank abelian groups are not well understood. Infinite-rank abelian groups can be extremely complex and many open questions exist, often intimately connected to questions of set theory.
Many large abelian groups carry a natural topology, turning them into topological groups.
All Wikipedia text is available under the terms of the GNU Free Documentation License
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Anatosaurus ... (also known as Trachodon[?]), meaning duck lizard was a dinosaur in the Cretaceous period[?]. It was named in 1942. The official species name is Anatotita ... | 668 | 2,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-51 | longest | en | 0.918093 |
https://brilliant.org/problems/finding-the-points-of-intersection/ | 1,531,856,142,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589892.87/warc/CC-MAIN-20180717183929-20180717203929-00241.warc.gz | 614,672,069 | 10,996 | # Finding the points of intersection
Geometry Level 2
Find the points of intersection of the circles $$x^2 + y^2 - 18x - 4y + 35 = 0$$ and $$x^2 + y^2 + 2x + 6y -15 = 0$$.
If your answer is $$(a,b)$$ and $$(c,d)$$, give your answer as $$a+b+c+d$$.
× | 98 | 253 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-30 | latest | en | 0.785552 |
https://www.unix.com/man-page/centos/3/clog/ | 1,721,428,780,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00011.warc.gz | 902,444,775 | 9,668 | # clog(3) [centos man page]
```CLOG(3) Linux Programmer's Manual CLOG(3)
NAME
clog, clogf, clogl - natural logarithm of a complex number
SYNOPSIS
#include <complex.h>
double complex clog(double complex z);
float complex clogf(float complex z);
long double complex clogl(long double complex z);
DESCRIPTION
The logarithm clog() is the inverse function of the exponential cexp(3). Thus, if y = clog(z), then z = cexp(y). The imaginary part of y
is chosen in the interval [-pi,pi].
One has:
clog(z) = log(cabs(z)) + I * carg(z)
Note that z close to zero will cause an overflow.
VERSIONS
These functions first appeared in glibc in version 2.1.
CONFORMING TO
C99.
cabs(3), cexp(3), clog10(3), clog2(3), complex(7)
COLOPHON
This page is part of release 3.53 of the Linux man-pages project. A description of the project, and information about reporting bugs, can
be found at http://www.kernel.org/doc/man-pages/.
2008-08-11 CLOG(3)```
## Check Out this Related Man Page
```CLOG(3) Linux Programmer's Manual CLOG(3)
NAME
clog, clogf, clogl - natural logarithm of a complex number
SYNOPSIS
#include <complex.h>
double complex clog(double complex z);
float complex clogf(float complex z);
long double complex clogl(long double complex z);
DESCRIPTION
The logarithm clog() is the inverse function of the exponential cexp(3). Thus, if y = clog(z), then z = cexp(y). The imaginary part of y
is chosen in the interval [-pi,pi].
One has:
clog(z) = log(cabs(z)) + I * carg(z)
Note that z close to zero will cause an overflow.
VERSIONS
These functions first appeared in glibc in version 2.1.
CONFORMING TO
C99.
cabs(3), cexp(3), clog10(3), clog2(3), complex(7)
COLOPHON
This page is part of release 3.53 of the Linux man-pages project. A description of the project, and information about reporting bugs, can
be found at http://www.kernel.org/doc/man-pages/.
2008-08-11 CLOG(3)```
Man Page
## cclogd 99% CPU ?
Hello, Can someone tell me why the deamon cclogd use 99% CPU ? I understand (doing some research on internet) that this deamon is for diagnostic. here the result of top command and sar commands: # sar 1 10 HP-UX NAWUX01 B.11.00 U 9000/800 06/29/07 09:06:55 %usr %sys %wio %idle...
## Script to Scan proclog files
i need to create a shell script, which will go into a directory , and scan the files in it for defined errors, there will be around 10 files in the directory.
## Error doing clogin into a wpar
I have this error when I'm trying to do clogin from AIX 7200-00-00-0000 into a wpar with AIX 5.2 abanksPaDesa2:/> clogin AbanksBDPA_wpar exec: A file or directory in the path name does not exist somebody can help me to fix this? I'm restoring the wpar from wpar mksysb not a mksysb from... | 825 | 2,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-30 | latest | en | 0.763002 |
http://stackoverflow.com/questions/15965938/ensure-that-multiple-variables-have-a-different-value/15967304 | 1,394,586,338,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011473737/warc/CC-MAIN-20140305092433-00032-ip-10-183-142-35.ec2.internal.warc.gz | 170,941,956 | 16,189 | # Ensure that multiple variables have a different value
I have to compare multiple variables and check if they are different. Is there any smooth way to do this? I can do
``````var1 = 3
var2 = "test"
var3 = 100
if var1 != var2 && var1 != var3 && var2 != var3
...
end
``````
But if the list of variables are getting longer, this becomes unhandy. I thought about something like `var1 != var2 != var3` but that doesn't work. So maybe there is a simple Ruby-way to do this.
-
with your philosophy `var1 != var2 != var3` do you want to continue if `var1 != var2` is `true` for `var3` also? or vice-versa? – Arup Rakshit Apr 12 '13 at 7:57
Equality is transitive but inequality is not transitive. – sawa Apr 12 '13 at 8:03
@sawa is right, even if a chained notation existed, `var1 != var2 != var3` would be a shorthand for `var1 != var2 && var2 != var3`, it wouldn't ensure that `var1 != var3`, thus "failing" for `3 != 5 != 3`. – Stefan Apr 12 '13 at 10:13
You can use `uniq`'s destructive counterpart `uniq!` to determine if an array contains duplicates. It returns `nil` if no duplicates were found:
``````if ![var1, var2, var3].uniq!
# no duplicates
end
``````
-
Great, I knew that there is always a cool way to do such things in Ruby. thx! – 23tux Apr 12 '13 at 12:46
One way might be using Array#uniq.
``````inputs = [var1, var2, var3]
if inputs.uniq == inputs
# ...
end
# or
if inputs.uniq.length == inputs.length
# ...
end
``````
-
You could do something like this
``````a = [var1. var2, var3]
if a.length == a.uniq.length
..
end
``````
-
thx for the reply! Your solution is like the one from Stefan (his is a little bit shorter), so I would love to give you also a questions solved – 23tux Apr 12 '13 at 12:48 | 539 | 1,732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-10 | latest | en | 0.828485 |
http://slideplayer.com/slide/4129271/ | 1,576,505,481,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540565544.86/warc/CC-MAIN-20191216121204-20191216145204-00288.warc.gz | 130,341,431 | 19,606 | # Estimation Samples are collected to estimate characteristics of the population of particular interest. Parameter – numerical characteristic of the population.
## Presentation on theme: "Estimation Samples are collected to estimate characteristics of the population of particular interest. Parameter – numerical characteristic of the population."— Presentation transcript:
Estimation Samples are collected to estimate characteristics of the population of particular interest. Parameter – numerical characteristic of the population (i.e. , ) Parameter – numerical characteristic of the population (i.e. , ) Distributional characteristics – pdf and cdf Distributional characteristics – pdf and cdf Statistic - numerical characteristic of the sample used to estimate parameters Statistic - numerical characteristic of the sample used to estimate parameters
Point Estimation A sample statistic is often used to estimate and draw conclusions about a population parameter ( ). The sample statistic is called a point estimator of The sample statistic is called a point estimator of For a particular sample, the calculated value of the statistic is called a point estimate of . For a particular sample, the calculated value of the statistic is called a point estimate of .
Point Estimation Let X 1, X 2, …, X n be a random sample of size n from the population of interest, and let Y=u(X 1, X 2, …, X n ) be a statistic used to estimate . Then Y is called an estimator of . Then Y is called an estimator of . A specific value of the estimator y=u(x 1, x 2, …, x n ) is called an estimate of . A specific value of the estimator y=u(x 1, x 2, …, x n ) is called an estimate of .
Estimator Discussion is limited to random variables with functional form of the pdf known. The pdf typically depends on an unknown parameter which can take on any value in the parameter space . i.e. f(x; ) The pdf typically depends on an unknown parameter which can take on any value in the parameter space . i.e. f(x; ) It is often necessary to pick one member from a family of members as most likely to be true. It is often necessary to pick one member from a family of members as most likely to be true. i.e. pick “best” value of for f(x; )i.e. pick “best” value of for f(x; ) The best estimator can depend on the distribution being sampled. The best estimator can depend on the distribution being sampled.
Properties of an Estimator If E[Y]= , then the statistic Y is called an unbiased estimator of . Otherwise, it is said to be biased. E[Y- ] is the bias of an estimator E[Y- ] is the bias of an estimator In many cases, the “best” estimator is an unbiased estimator. In many cases, the “best” estimator is an unbiased estimator.
Properties of an Estimator Another important property is small variance. If two estimators are both unbiased, we prefer the one with small variance. Minimize E[(Y- ) 2 ] = Var[(Y- )] + E[(Y- )] 2 Minimize E[(Y- ) 2 ] = Var[(Y- )] + E[(Y- )] 2 The estimator Y that minimizes E[(Y- ) 2 ] is said to have minimum mean square error (MSE) The estimator Y that minimizes E[(Y- ) 2 ] is said to have minimum mean square error (MSE) If we consider only unbiased estimators, the statistic Y that minimizes MSE is called the minimum variance unbiased estimator (MVUE) If we consider only unbiased estimators, the statistic Y that minimizes MSE is called the minimum variance unbiased estimator (MVUE)
Properties of an Estimator The efficiency of an estimator 1 compare to another estimator 2 is equal to the ratio
Method of Maximum Likelihood An important method for finding an estimator Let X 1,X 2,…,X n be a random sample of size n from f(x; ). Let X 1,X 2,…,X n be a random sample of size n from f(x; ). The likelihood function is the joint pdf of X 1,X 2,…,X n evaluated at observed values x 1,x 2,…,x n as a function of the parameter of interest. The likelihood function is the joint pdf of X 1,X 2,…,X n evaluated at observed values x 1,x 2,…,x n as a function of the parameter of interest. L( ) = f(x 1, x 2, …, x n ; ) = f(x 1, ) f(x 2, ) f(x n, ) is the probability of observing x 1,x 2,…,x n if the pdf is f(x; ).L( ) = f(x 1, x 2, …, x n ; ) = f(x 1, ) f(x 2, ) f(x n, ) is the probability of observing x 1,x 2,…,x n if the pdf is f(x; ). The value of that maximizes L( ) is the value of most likely to have produced x 1,x 2,…,x nThe value of that maximizes L( ) is the value of most likely to have produced x 1,x 2,…,x n
Maximum Likelihood Estimator The maximum likelihood estimator (MLE) of is found by setting the differential of L( ) with respect to equal to zero and solving for . The MLE can also be found by maximizing the natural log of L( ), which is often easier to differentiate. For more than one parameter, maximum likelihood equations are formed and simultaneously solved to arrive at the MLE’s
Invariance Property If t is the MLE for and u( ) is a function of then the MLE of u( ) is u(t). Plug the MLE(s) into the function to get an MLE estimate of the function Plug the MLE(s) into the function to get an MLE estimate of the function
Method of Moments An important method for finding an estimator Let X 1,X 2,…,X n be a random sample of size n from f(x). The kth population moment is E[X k ]. The kth sample moment is (1/n) X k. Let X 1,X 2,…,X n be a random sample of size n from f(x). The kth population moment is E[X k ]. The kth sample moment is (1/n) X k. The method of moments involves setting the sample moment(s) equal to the population moment(s) and solving for the parameter of interest. The method of moments involves setting the sample moment(s) equal to the population moment(s) and solving for the parameter of interest.
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Similar presentations | 1,703 | 6,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-51 | latest | en | 0.776353 |
https://dsp.stackexchange.com/questions/54294/is-there-a-clever-way-to-implement-cascaded-moving-average-filters | 1,719,140,006,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00778.warc.gz | 196,138,658 | 41,512 | # Is there a clever way to implement cascaded moving average filters?
I'd like to effeciently approximate a gaussian filter's step response using cascaded moving average filters.
I know about recursive moving average, but is there some clever algorithm to cascade them apart from just putting them in series? CIC filters seem related but are mostly used for decimation, I'm interested in pulse shaping.
• This might be helpful. Commented Dec 20, 2018 at 15:44
• Thank you very much Matt! That's kind of what I had in mind. But I'm still wondering if there's an interesting non-obvious way of cascading them... I'm working in C/C++. Commented Dec 20, 2018 at 19:57
[EDIT:20181225, added a couple of references] This topic is coming back. P. Getreuer provides C code, and lists an handful of them in A survey of Gaussian convolution algorithms, Image Processing On Line, 2013:
Gaussian convolution is a common operation and building block for algorithms in signal and image processing. Consequently, its efficient computation is important, and many fast approximations have been proposed. In this survey, we discuss approximate Gaussian convolution based on finite impulse response filters, DFT and DCT based convolution, box filters, and several recursive filters. Since boundary handling is sometimes overlooked in the original works, we pay particular attention to develop it here. We perform numerical experiments to compare the speed and quality of the algorithms.
Here are a couple references:
This paper introduces a method for multidimensional Gaussian filtering using an efficient one-pass cascade of overlapping local average windows driven by prefix sums. Each local-average filter is implemented in n dimensions, with non-integer lengths, allowing accurate approximation of Gaussians of any variance. In axis oriented form the method has a scan-rate hardware realization and fast software implementation using minimal extra memory. In this latter case the new method consistently outperforms the fastest alternative Gaussian filtering method both in accuracy and speed.
Box filters have been used to speed up many computation-intensive operations in Image Processing and Computer Vision. They have the advantage of being fast to compute, but their adoption has been hampered by the fact that they present serious restrictions to filter construction. This paper relaxes these restrictions by presenting a method for automatically approximating an arbitrary 2-D filter by a box filter. To develop our method, we first formulate the approximation as a minimization problem and show that it is possible to find a closed form solution to a subset of the parameters of the box filter. To solve for the remaining parameters of the approximation, we develop two algorithms: Exhaustive Search for small filters and Directed Search for large filters. Experimental results show the validity of the proposed method.
This paper presents a simple and efficient method to convolve an image with a Gaussian kernel. The computation is performed in a constant number of operations per pixel using running sums along the image rows and columns. We investigate the error function used for kernel approximation and its relation to the properties of the input signal. Based on natural image statistics we propose a quadratic form kernel error function so that the output image l2 error is minimized. We apply the proposed approach to approximate the Gaussian kernel by linear combination of constant functions. This results in very efficient Gaussian filtering method. Our experiments show that the proposed technique is faster than state of the art methods while preserving a similar accuracy.
Image averaging can be performed very efficiently using either separable moving average filters or by using summed area tables, also known as integral images. Both these methods allow averaging to be performed at a small fixed cost per pixel, independent of the averaging filter size. Repeated filtering with averaging filters can be used to approximate Gaussian filtering. Thus a good approximation to Gaussian filtering can be achieved at a fixed cost per pixel independent of filter size. This paper describes how to determine the averaging filters that one needs to approximate a Gaussian with a specified standard deviation. The design of bandpass filters from the difference of Gaussians is also analysed. It is shown that difference of Gaussian bandpass filters share some of the attributes of log-Gabor filters in that they have a relatively symmetric transfer function when viewed on a logarithmic frequency scale and can be constructed with large bandwidths.
Gaussian filtering is an important tool in image processing and computer vision. In this paper we discuss the background of Gaussian filtering and look at some methods for implementing it. Consideration of the central limit theorem suggests using a cascade of simple'' filters as a means of computing Gaussian filters. Amongsimple'' filters, uniform-coefficient finite-impulse-response digital filters are especially economical to implement. The idea of cascaded uniform filters has been around for a while [13], [16]. We show that this method is economical to implement, has good filtering characteristics, and is appropriate for hardware implementation. We point out an equivalence to one of Burt's methods 1, 3 under certain circumstances. As an extension, we describe an approach to implementing a Gaussian Pyramid which requires approximately two addition operations per pixel, per level, per dimension. We examine tradeoffs in choosing an algorithm for Gaussian filtering, and finally discuss an implementation. | 1,049 | 5,680 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.937657 |
https://www.gamedev.net/forums/topic/356118-spatial-partioning---linear-octree-question/ | 1,544,955,478,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827639.67/warc/CC-MAIN-20181216095437-20181216121437-00367.warc.gz | 910,843,104 | 25,405 | # Spatial Partioning - Linear Octree Question
This topic is 4789 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
When reviewing the different forms of spatial partioning for an outdoor environment, I ran across linear octrees. Ok, so setting up an octree is a fairly simple process, it's basically AABB. Linear octrees add a somewhat unknown element to me though. Since the children are no longer stored in the nodes for a linear octree, how do you build the octree itself? Any information on linear octrees you could provide would be great; at the moment I am kind of at a stand still in my research on them.
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Basically your tree is a linear array of N octree nodes, octree nodes do not have information for the indices as you said.
You can compute the index of the child nodes by using the following formula:
N = index of your node
8N+1
8N+2
8N+3
8N+4
8N+5
8N+6
8N+7
8N+8
You can build the tree in several ways, some people use a standard tree with links to the nodes, and then linearize it by writing out all the data except the child node data:
struct myOctNode
{
struct myOctData
{} data;
myOctNode *childX; // 8 of these
}
when storing this as a linear array you can store only myOctData:
std::vector<myOctData> linearOctree;
More efficient is to use the array itself while building the tree, and using indices as "handles" to the individual nodes.
Hope this helps
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### #ActualParagon123
Posted 19 November 2013 - 02:18 PM
Here is an explaination of legrange interpolation
http://mathforum.org/library/drmath/view/63984.html
Here's another that does a better job of describing how to actually compute the function
https://brilliant.org/assessment/techniques-trainer/lagrange-interpolation-formula/
These are slightly differen't than bezier curves, Bezier curves use control points, for which the final line does not pass through.
Legrange interpolation ensures that each point in the data set is in the final line.
These are good for lines where x is always increasing.
To apply it to a line where x is not always increasing you have to convert your data set into a parameterized data set and interpolate the x and y components independently by t.
I.E: f(t)=(fx(t),fy(t))
Get the lagrange interpolation formula for fx(t) and fy(t).
### #1Paragon123
Posted 19 November 2013 - 02:16 PM
Here is an explaination of legrange interpolation
http://mathforum.org/library/drmath/view/63984.html
Here's another that does a better job of describing how to actually computer the function
https://brilliant.org/assessment/techniques-trainer/lagrange-interpolation-formula/
These are good for lines where x is always increasing.
To apply it to a line where x is not always increasing you have to convert your data set into a parameterized data set and interpolate the x and y components independently by t.
I.E: f(t)=(fx(t),fy(t))
Get the lagrange interpolation formula for fx(t) and fy(t).
PARTNERS | 368 | 1,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2014-52 | latest | en | 0.842642 |
http://www.netlib.org/lapack/explore-html/db/dfc/a18602_ga31a91fe2c1289d31db2f0733617c8fb7.html | 1,596,863,922,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737289.75/warc/CC-MAIN-20200808051116-20200808081116-00373.warc.gz | 160,434,823 | 9,938 | LAPACK 3.8.0 LAPACK: Linear Algebra PACKage
## ◆ spstrf()
subroutine spstrf ( character UPLO, integer N, real, dimension( lda, * ) A, integer LDA, integer, dimension( n ) PIV, integer RANK, real TOL, real, dimension( 2*n ) WORK, integer INFO )
SPSTRF computes the Cholesky factorization with complete pivoting of a real symmetric positive semidefinite matrix.
Purpose:
``` SPSTRF computes the Cholesky factorization with complete
pivoting of a real symmetric positive semidefinite matrix A.
The factorization has the form
P**T * A * P = U**T * U , if UPLO = 'U',
P**T * A * P = L * L**T, if UPLO = 'L',
where U is an upper triangular matrix and L is lower triangular, and
P is stored as vector PIV.
This algorithm does not attempt to check that A is positive
semidefinite. This version of the algorithm calls level 3 BLAS.```
Parameters
[in] UPLO ``` UPLO is CHARACTER*1 Specifies whether the upper or lower triangular part of the symmetric matrix A is stored. = 'U': Upper triangular = 'L': Lower triangular``` [in] N ``` N is INTEGER The order of the matrix A. N >= 0.``` [in,out] A ``` A is REAL array, dimension (LDA,N) On entry, the symmetric matrix A. If UPLO = 'U', the leading n by n upper triangular part of A contains the upper triangular part of the matrix A, and the strictly lower triangular part of A is not referenced. If UPLO = 'L', the leading n by n lower triangular part of A contains the lower triangular part of the matrix A, and the strictly upper triangular part of A is not referenced. On exit, if INFO = 0, the factor U or L from the Cholesky factorization as above.``` [in] LDA ``` LDA is INTEGER The leading dimension of the array A. LDA >= max(1,N).``` [out] PIV ``` PIV is INTEGER array, dimension (N) PIV is such that the nonzero entries are P( PIV(K), K ) = 1.``` [out] RANK ``` RANK is INTEGER The rank of A given by the number of steps the algorithm completed.``` [in] TOL ``` TOL is REAL User defined tolerance. If TOL < 0, then N*U*MAX( A(K,K) ) will be used. The algorithm terminates at the (K-1)st step if the pivot <= TOL.``` [out] WORK ``` WORK is REAL array, dimension (2*N) Work space.``` [out] INFO ``` INFO is INTEGER < 0: If INFO = -K, the K-th argument had an illegal value, = 0: algorithm completed successfully, and > 0: the matrix A is either rank deficient with computed rank as returned in RANK, or is not positive semidefinite. See Section 7 of LAPACK Working Note #161 for further information.```
Date
December 2016
Definition at line 143 of file spstrf.f.
143 *
144 * -- LAPACK computational routine (version 3.7.0) --
145 * -- LAPACK is a software package provided by Univ. of Tennessee, --
146 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
147 * December 2016
148 *
149 * .. Scalar Arguments ..
150 REAL tol
151 INTEGER info, lda, n, rank
152 CHARACTER uplo
153 * ..
154 * .. Array Arguments ..
155 REAL a( lda, * ), work( 2*n )
156 INTEGER piv( n )
157 * ..
158 *
159 * =====================================================================
160 *
161 * .. Parameters ..
162 REAL one, zero
163 parameter( one = 1.0e+0, zero = 0.0e+0 )
164 * ..
165 * .. Local Scalars ..
166 REAL ajj, sstop, stemp
167 INTEGER i, itemp, j, jb, k, nb, pvt
168 LOGICAL upper
169 * ..
170 * .. External Functions ..
171 REAL slamch
172 INTEGER ilaenv
173 LOGICAL lsame, sisnan
174 EXTERNAL slamch, ilaenv, lsame, sisnan
175 * ..
176 * .. External Subroutines ..
177 EXTERNAL sgemv, spstf2, sscal, sswap, ssyrk, xerbla
178 * ..
179 * .. Intrinsic Functions ..
180 INTRINSIC max, min, sqrt, maxloc
181 * ..
182 * .. Executable Statements ..
183 *
184 * Test the input parameters.
185 *
186 info = 0
187 upper = lsame( uplo, 'U' )
188 IF( .NOT.upper .AND. .NOT.lsame( uplo, 'L' ) ) THEN
189 info = -1
190 ELSE IF( n.LT.0 ) THEN
191 info = -2
192 ELSE IF( lda.LT.max( 1, n ) ) THEN
193 info = -4
194 END IF
195 IF( info.NE.0 ) THEN
196 CALL xerbla( 'SPSTRF', -info )
197 RETURN
198 END IF
199 *
200 * Quick return if possible
201 *
202 IF( n.EQ.0 )
203 \$ RETURN
204 *
205 * Get block size
206 *
207 nb = ilaenv( 1, 'SPOTRF', uplo, n, -1, -1, -1 )
208 IF( nb.LE.1 .OR. nb.GE.n ) THEN
209 *
210 * Use unblocked code
211 *
212 CALL spstf2( uplo, n, a( 1, 1 ), lda, piv, rank, tol, work,
213 \$ info )
214 GO TO 200
215 *
216 ELSE
217 *
218 * Initialize PIV
219 *
220 DO 100 i = 1, n
221 piv( i ) = i
222 100 CONTINUE
223 *
224 * Compute stopping value
225 *
226 pvt = 1
227 ajj = a( pvt, pvt )
228 DO i = 2, n
229 IF( a( i, i ).GT.ajj ) THEN
230 pvt = i
231 ajj = a( pvt, pvt )
232 END IF
233 END DO
234 IF( ajj.LE.zero.OR.sisnan( ajj ) ) THEN
235 rank = 0
236 info = 1
237 GO TO 200
238 END IF
239 *
240 * Compute stopping value if not supplied
241 *
242 IF( tol.LT.zero ) THEN
243 sstop = n * slamch( 'Epsilon' ) * ajj
244 ELSE
245 sstop = tol
246 END IF
247 *
248 *
249 IF( upper ) THEN
250 *
251 * Compute the Cholesky factorization P**T * A * P = U**T * U
252 *
253 DO 140 k = 1, n, nb
254 *
255 * Account for last block not being NB wide
256 *
257 jb = min( nb, n-k+1 )
258 *
259 * Set relevant part of first half of WORK to zero,
260 * holds dot products
261 *
262 DO 110 i = k, n
263 work( i ) = 0
264 110 CONTINUE
265 *
266 DO 130 j = k, k + jb - 1
267 *
268 * Find pivot, test for exit, else swap rows and columns
269 * Update dot products, compute possible pivots which are
270 * stored in the second half of WORK
271 *
272 DO 120 i = j, n
273 *
274 IF( j.GT.k ) THEN
275 work( i ) = work( i ) + a( j-1, i )**2
276 END IF
277 work( n+i ) = a( i, i ) - work( i )
278 *
279 120 CONTINUE
280 *
281 IF( j.GT.1 ) THEN
282 itemp = maxloc( work( (n+j):(2*n) ), 1 )
283 pvt = itemp + j - 1
284 ajj = work( n+pvt )
285 IF( ajj.LE.sstop.OR.sisnan( ajj ) ) THEN
286 a( j, j ) = ajj
287 GO TO 190
288 END IF
289 END IF
290 *
291 IF( j.NE.pvt ) THEN
292 *
293 * Pivot OK, so can now swap pivot rows and columns
294 *
295 a( pvt, pvt ) = a( j, j )
296 CALL sswap( j-1, a( 1, j ), 1, a( 1, pvt ), 1 )
297 IF( pvt.LT.n )
298 \$ CALL sswap( n-pvt, a( j, pvt+1 ), lda,
299 \$ a( pvt, pvt+1 ), lda )
300 CALL sswap( pvt-j-1, a( j, j+1 ), lda,
301 \$ a( j+1, pvt ), 1 )
302 *
303 * Swap dot products and PIV
304 *
305 stemp = work( j )
306 work( j ) = work( pvt )
307 work( pvt ) = stemp
308 itemp = piv( pvt )
309 piv( pvt ) = piv( j )
310 piv( j ) = itemp
311 END IF
312 *
313 ajj = sqrt( ajj )
314 a( j, j ) = ajj
315 *
316 * Compute elements J+1:N of row J.
317 *
318 IF( j.LT.n ) THEN
319 CALL sgemv( 'Trans', j-k, n-j, -one, a( k, j+1 ),
320 \$ lda, a( k, j ), 1, one, a( j, j+1 ),
321 \$ lda )
322 CALL sscal( n-j, one / ajj, a( j, j+1 ), lda )
323 END IF
324 *
325 130 CONTINUE
326 *
327 * Update trailing matrix, J already incremented
328 *
329 IF( k+jb.LE.n ) THEN
330 CALL ssyrk( 'Upper', 'Trans', n-j+1, jb, -one,
331 \$ a( k, j ), lda, one, a( j, j ), lda )
332 END IF
333 *
334 140 CONTINUE
335 *
336 ELSE
337 *
338 * Compute the Cholesky factorization P**T * A * P = L * L**T
339 *
340 DO 180 k = 1, n, nb
341 *
342 * Account for last block not being NB wide
343 *
344 jb = min( nb, n-k+1 )
345 *
346 * Set relevant part of first half of WORK to zero,
347 * holds dot products
348 *
349 DO 150 i = k, n
350 work( i ) = 0
351 150 CONTINUE
352 *
353 DO 170 j = k, k + jb - 1
354 *
355 * Find pivot, test for exit, else swap rows and columns
356 * Update dot products, compute possible pivots which are
357 * stored in the second half of WORK
358 *
359 DO 160 i = j, n
360 *
361 IF( j.GT.k ) THEN
362 work( i ) = work( i ) + a( i, j-1 )**2
363 END IF
364 work( n+i ) = a( i, i ) - work( i )
365 *
366 160 CONTINUE
367 *
368 IF( j.GT.1 ) THEN
369 itemp = maxloc( work( (n+j):(2*n) ), 1 )
370 pvt = itemp + j - 1
371 ajj = work( n+pvt )
372 IF( ajj.LE.sstop.OR.sisnan( ajj ) ) THEN
373 a( j, j ) = ajj
374 GO TO 190
375 END IF
376 END IF
377 *
378 IF( j.NE.pvt ) THEN
379 *
380 * Pivot OK, so can now swap pivot rows and columns
381 *
382 a( pvt, pvt ) = a( j, j )
383 CALL sswap( j-1, a( j, 1 ), lda, a( pvt, 1 ), lda )
384 IF( pvt.LT.n )
385 \$ CALL sswap( n-pvt, a( pvt+1, j ), 1,
386 \$ a( pvt+1, pvt ), 1 )
387 CALL sswap( pvt-j-1, a( j+1, j ), 1, a( pvt, j+1 ),
388 \$ lda )
389 *
390 * Swap dot products and PIV
391 *
392 stemp = work( j )
393 work( j ) = work( pvt )
394 work( pvt ) = stemp
395 itemp = piv( pvt )
396 piv( pvt ) = piv( j )
397 piv( j ) = itemp
398 END IF
399 *
400 ajj = sqrt( ajj )
401 a( j, j ) = ajj
402 *
403 * Compute elements J+1:N of column J.
404 *
405 IF( j.LT.n ) THEN
406 CALL sgemv( 'No Trans', n-j, j-k, -one,
407 \$ a( j+1, k ), lda, a( j, k ), lda, one,
408 \$ a( j+1, j ), 1 )
409 CALL sscal( n-j, one / ajj, a( j+1, j ), 1 )
410 END IF
411 *
412 170 CONTINUE
413 *
414 * Update trailing matrix, J already incremented
415 *
416 IF( k+jb.LE.n ) THEN
417 CALL ssyrk( 'Lower', 'No Trans', n-j+1, jb, -one,
418 \$ a( j, k ), lda, one, a( j, j ), lda )
419 END IF
420 *
421 180 CONTINUE
422 *
423 END IF
424 END IF
425 *
426 * Ran to completion, A has full rank
427 *
428 rank = n
429 *
430 GO TO 200
431 190 CONTINUE
432 *
433 * Rank is the number of steps completed. Set INFO = 1 to signal
434 * that the factorization cannot be used to solve a system.
435 *
436 rank = j - 1
437 info = 1
438 *
439 200 CONTINUE
440 RETURN
441 *
442 * End of SPSTRF
443 *
real function slamch(CMACH)
SLAMCH
Definition: slamch.f:69
logical function lsame(CA, CB)
LSAME
Definition: lsame.f:55
logical function sisnan(SIN)
SISNAN tests input for NaN.
Definition: sisnan.f:61
subroutine xerbla(SRNAME, INFO)
XERBLA
Definition: xerbla.f:62
integer function ilaenv(ISPEC, NAME, OPTS, N1, N2, N3, N4)
ILAENV
Definition: tstiee.f:83
subroutine sswap(N, SX, INCX, SY, INCY)
SSWAP
Definition: sswap.f:84
subroutine sgemv(TRANS, M, N, ALPHA, A, LDA, X, INCX, BETA, Y, INCY)
SGEMV
Definition: sgemv.f:158
subroutine ssyrk(UPLO, TRANS, N, K, ALPHA, A, LDA, BETA, C, LDC)
SSYRK
Definition: ssyrk.f:171
subroutine sscal(N, SA, SX, INCX)
SSCAL
Definition: sscal.f:81
subroutine spstf2(UPLO, N, A, LDA, PIV, RANK, TOL, WORK, INFO)
SPSTF2 computes the Cholesky factorization with complete pivoting of a real symmetric positive semide...
Definition: spstf2.f:143
Here is the call graph for this function:
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A balance trick Get a smooth metre stick and let it rest lightly on your two forefingers. Place your fingers near the ends of the stick and then move them in the direction of t | 542 | 2,529 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-40 | longest | en | 0.847916 |
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# Suggestions needed on raising practice test scores...
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Current Student
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Suggestions needed on raising practice test scores... [#permalink]
### Show Tags
24 Oct 2006, 19:16
I'm taking the GMAT in 3 weeks.
I'm currently enrolled in a Kaplan course. I'm taking the GMAT one week after the course ends. I work 7-6pm Mon-Fri. The class is 3hrs/night, 2 times/week.
I've taken the Kaplan diagnostic and the first online Kaplan CAT.
Diagnostic: 630 (35Q, 42V)
CAT #1: 630 (37Q, 42V)
1) Any idea how similar Kaplan diagnostic and CAT scores normally translate to the actual GMAT? Is there any relatively normal translation scale, or do actual GMAT scores completely vary from the Kaplan tests?
2) What would be the best way to improve my scores given the limited time I have left before taking the test? (It looks like quant is where I need the most help, I guess.)
Thanks for any suggestions or info.
Kaplan GMAT Prep Discount Codes Magoosh Discount Codes e-GMAT Discount Codes
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### Show Tags
25 Oct 2006, 17:41
Kaplan CATS bear little resemblance to the real thing. Take an official GMATPrep test to see how you're really doing.
Get the Official Guide too...is Kaplan finally using them in their courses or at least telling students about it?
Senior Manager
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### Show Tags
26 Oct 2006, 02:49
meander wrote:
Kaplan CATS bear little resemblance to the real thing. Take an official GMATPrep test to see how you're really doing.
Get the Official Guide too...is Kaplan finally using them in their courses or at least telling students about it?
Finally!!!! Wow I'm impressed they finally are because when I took it, those %%%%%%% never told me about it. Please don't make the mistake of studying just thru Kaplan material.
Current Student
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### Show Tags
29 Oct 2006, 12:18
meander wrote:
Get the Official Guide too...is Kaplan finally using them in their courses or at least telling students about it?
Haven't heard anything about it from them.
Is this the official guide to which you're referring?
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30 Oct 2006, 15:17
GmatPrep is amazing take that, that should give you a better indication of your actual. Score, relax no worries you can do this man!
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31 Oct 2006, 07:26
Yes, that is indeed the OG we're all raving about. I can't believe Kaplan still doesn't even mention it. It comes with the Princeton Review course because they use it for the homework assignments. It's a MUST.
31 Oct 2006, 07:26
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,209 | 4,501 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-04 | latest | en | 0.869579 |
https://www.math.utah.edu/online/1010/euclid/ | 1,718,589,852,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00064.warc.gz | 801,400,568 | 5,430 | # Mathematics 1010 online
## The Euclidean Algorithm
Euclid of Alexandria lived during the third century BC. The Algorithm named after him let's you find the greatest common factor of two natural numbers or two polynomials .
### Division of polynomials
Polynomials can be divided mechanically by long division, much like numbers can be divided. Numbers represented in decimal form are sums of powers of 10. Polynomial expressions similarly are sums of powers of the variable (let's say ). There are two main differences:
• A coefficient in the decimal representation of a number must be one of the 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. On the other hand, a coefficient of a polynomial may be any real (or even complex ) number.
• Ten of a certain power of 10 can be traded for one higher power of 10, or vice versa. For example, ten 10s can be traded for one 100. By comparison, no trading is possible for polynomials. This actually simplifies the process of division.
Let's illustrate the division process with a couple of examples.
We can easily check that
This can be rewritten as
The symbol stands for remainder. The ingredients of this expression have the following names:
• is the dividend, i.e., the expression that's being divided.
• is the divisor, i.e., the expression by which we divide.
• is the quotient.
• is the remainder.
The equation can be rewritten alternatively, and equivalently, as:
Thus the equations , , and are all equivalent.
To illustrate the terminology let's rewrite with the words replacing the expressions:
If the remainder is zero then we say the divisor divides the dividend evenly.
Given a dividend and a divisor, the quotient and the remainder can be found by long division. For example, the division in - can be written as:
Notes
• There is one column for each power of .
• The divisor is written to the left of the dividend. That's a pretty arbitrary convention, you might also write the divisor to the right of the dividend, perhaps separated by the division sign.
• The quotient is written on top of the dividend, with the powers of lining up. This is also just a convention.
• To find the individual terms of the quotient you start with the leading term, and keep subtracting the product of the last term you computed and the divisor. At each step, to get the new term of the quotient, you simply divide the leading term of what's left of the dividend by the leading term of the divisor.
• The process terminates when the degree of what is left is less than the degree of the divisor.
• The examples you find in textbooks or notes like these usually will have polynomials whose coefficients are small integers. Of course nothing changes in principle if the coefficients are rational or irrational real numbers, or complex numbers, or even algebraic expressions.
Here is a more complicated example, Some of the coefficients of the dividend and the quotient are zero. They are given here explicitly, but with practice you may just leave these spaces blank.
Thus
### The greatest common factor of two natural numbers
Let's say we want to find the greatest common factor of two natural numbers and , and let's suppose is the larger of the two, i.e., .
The Euclidean Algorithm proceeds by dividing by , with remainder, then dividing the divisor by the remainder, and repeating this process until the remainder is zero. The greatest common factor of and is the last divisor. (Notice that at no point do you pay attention to the current quotient.)
The following examples illustrate the process. Suppose we want to find the greatest common factor of and . We have
The last remainder is zero, the last divisor is , and so is the greatest common factor of 110 and 143. Indeed,
Clearly, 11 is a common factor, and there is no greater common factor.
Here is another example. The greatest common factor of and is . We have:
Indeed,
There are two ingredients that make the Euclidean Algorithm work:
• Suppose in
has a factor . Then is also a factor of the right hand side. If is a common factor of and it is also a factor of the remainder . Thus a common factor of and is also a common factor of and . In particular, the greatest common factor of and is also a factor of . This property remains true throughout the procedure.
• The remainders are all non-negative, and they decrease at each step. Thus eventually we must have a remainder 0. Thus the last divisor is a factor of the last dividend. Working backwards using the previous observation we see that it is also a factor of and .
Of course, we could have solved the problems in these particular examples by listing all the factors of the numbers involved and finding the largest. Another technique is to find all the prime factors and see which ones are common to the numbers. Those techniques work fine for the small numbers. The power of the Euclidean Algorithm becomes apparent when the factors aren't obvious, i.e., when we have large numbers. Consider this example: Let
Factoring these numbers is possible using a computer, but larger numbers (with somewhere around 200 digits) are used in security codes whose effectiveness depend on those numbers being impossible to factor with current computer equipment. While factoring large numbers -- particularly products of large prime numbers -- is difficult, finding the greatest common factor of two numbers is easy!
In this case:
Thus the greatest common factor of and is . In fact, both and are the products of two large primes. Specifically,
### The greatest common factor of two polynomials
The same arguments as above apply to dividing polynomials with remainder. Thus the Euclidean algorithm can be used to find the greatest'' common factor of the numerator and denominator in a rational expression. Greatest'' in this context means highest possible degree''. A (non-zero) constant multiplying such a greatest (polynomial) factor does not matter, and we can choose it conveniently depending upon the application.
Suppose we want to find the greatest common factor of and .
Using long division we see that
Thus is the highest degree common factor of and .
In fact, using long division again we see that
Thus, for example,
Try to do that sort of thing without the Euclidean algorithm!
Here is another example. Let
What is the greatest common factor of and ? Using long division once again we get:
Thus the highest degree common factor of and is (we can ignore the constant factor 7.)
The main application of finding common factors of two polynomials is to enable to us to cancel them in the ratio of the two polynomials. In this case we obtain: | 1,371 | 6,636 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-26 | latest | en | 0.921119 |
https://www.physicsforums.com/threads/solving-m_x-u_x-r-x-r-need-help.154217/ | 1,709,210,806,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474808.39/warc/CC-MAIN-20240229103115-20240229133115-00487.warc.gz | 915,265,184 | 15,877 | # Solving M_x=U_x |r X R|: Need Help?
• suspenc3
In summary, the equation M_x=U_x |r X R| is used to solve for the unknown variable x by utilizing the given independent variables. It represents the relationship between the dependent variable M_x and the independent variables U_x, r, and R. To solve this equation, one must determine the values of the variables and use algebraic operations to isolate x. Some tips for solving this equation include double-checking calculations and breaking down the equation into smaller steps. Other methods, such as substitution or elimination, can also be used to solve this equation.
See attachment
## Homework Equations
I was thinking : $$M_x=U_x |r X R|$$
where r is going to be the distance from any point on the x-axis to any point on the line of action of the force
## The Attempt at a Solution
Im not really sure how to exactly do this problem...Ive been plugging in numbers hoping for the best but nothing good has come out of it...Any help would be appreciate.
Hello,
Thank you for reaching out for help with this problem. Solving for M_x in the equation M_x=U_x |r X R| can be a bit tricky, but with some guidance, I'm sure you will be able to solve it successfully.
First, let's break down the equation. M_x represents the moment about the x-axis, U_x is the unit vector in the x-direction, and |r X R| is the cross product of the position vector r and the force vector R. Essentially, we are trying to find the moment about the x-axis caused by the force R at any point along its line of action.
To solve this, we can use the following steps:
1. Identify the position vector r: This vector represents the distance from any point on the x-axis to any point on the line of action of the force. You can either use specific coordinates or variables to represent this vector.
2. Identify the force vector R: This vector represents the magnitude and direction of the force. Again, you can use specific values or variables to represent this vector.
3. Find the cross product of the position vector and the force vector: This will give you a new vector that is perpendicular to both r and R. You can use the right-hand rule to determine the direction of this vector.
4. Calculate the magnitude of the cross product: This can be done by taking the magnitude of the position vector and multiplying it by the magnitude of the force vector, and then multiplying it by the sine of the angle between the two vectors.
5. Multiply the magnitude of the cross product by the unit vector in the x-direction (U_x): This will give you the magnitude of the moment about the x-axis caused by the force R.
I hope this helps guide you in solving the problem. If you are still having trouble, please feel free to ask for more specific help or clarification. Good luck!
## 1. What does the equation M_x=U_x |r X R| mean?
The equation M_x=U_x |r X R| is a mathematical representation of a process called solving for the unknown variable x. The left side of the equation, M_x, represents the dependent variable, while the right side, U_x |r X R|, represents the independent variables. The vertical bar symbol (|) separates the two sides of the equation, and the X symbol represents multiplication. Overall, this equation is used to determine the value of the unknown variable x by utilizing the given independent variables.
## 2. What is the purpose of solving M_x=U_x |r X R|?
The purpose of solving M_x=U_x |r X R| is to find the value of the unknown variable x in a given equation. This process is commonly used in scientific experiments and research to understand the relationship between different variables and to make predictions or conclusions based on the results.
## 3. How do I solve M_x=U_x |r X R|?
To solve M_x=U_x |r X R|, you must first determine the values of the independent variables (U_x, r, and R) and the dependent variable (M_x). Then, you can use algebraic operations such as addition, subtraction, multiplication, and division to isolate the unknown variable x on one side of the equation. Once you have determined the value of x, you can substitute it back into the original equation to verify your solution.
## 4. What are some tips for solving M_x=U_x |r X R|?
Some tips for solving M_x=U_x |r X R| include carefully checking the given values for the independent and dependent variables, using algebraic rules consistently, and double-checking your calculations. It can also be helpful to break down the equation into smaller steps and to use a calculator or other tools for complex calculations.
## 5. Can I use a different method to solve M_x=U_x |r X R|?
Yes, there are other methods that can be used to solve M_x=U_x |r X R|, such as substitution or elimination. These methods may be more efficient or easier to use depending on the specific equation and values given. It is always important to check your work and make sure your solution is accurate, regardless of the method used. | 1,113 | 4,961 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-10 | latest | en | 0.926962 |
http://math4finance.com/general/the-distance-from-the-centroid-of-a-triangle-to-its-vertices-are-16cm-17cm-and-18cm-what-is-the-length-of-the-shortest-median | 1,716,210,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058278.91/warc/CC-MAIN-20240520111411-20240520141411-00340.warc.gz | 20,258,348 | 7,614 | Q:
# The distance from the centroid of a triangle to its vertices are 16cm, 17cm, and 18cm. What is the length of the shortest median
Accepted Solution
A:
Answer:$$24$$ $$\text{cm}$$Step-by-step explanation:Given: The distance from the centroid of a triangle to its vertices are $$16\text{cm}$$, $$17\text{cm}$$, and $$18\text{cm}$$.To Find: Length of shortest median.Solution:Consider the figure attachedA centroid is an intersection point of medians of a triangle.Also,A centroid divides a median in a ratio of 2:1.Let G be the centroid, and vertices are A,B and C.length of $$\text{AG}$$ $$=16\text{cm}$$length of $$\text{BG}$$ $$=17\text{cm}$$length of $$\text{CG}$$ $$=18\text{cm}$$as centrod divides median in ratio of $$2:1$$length of $$\text{AD}$$ $$=\frac{3}{2}\text{AG}$$ $$=\frac{3}{2}\times16$$ $$=24\text{cm}$$length of $$\text{BE}$$ $$=\frac{3}{2}\text{BG}$$ $$=\frac{3}{2}\times17$$ $$=\frac{51}{2}\text{cm}$$length of $$\text{CF}$$ $$=\frac{3}{2}\text{CG}$$ $$=\frac{3}{2}\times18$$ $$=27\text{cm}$$Hence the shortest median is $$\text{AD}$$ of length $$24\text{cm}$$ | 394 | 1,362 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-22 | latest | en | 0.423612 |
https://dbpedia.org/describe/?url=http%3A%2F%2Fdbpedia.org%2Fresource%2FConvex_combination | 1,618,108,481,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038060603.10/warc/CC-MAIN-20210411000036-20210411030036-00358.warc.gz | 305,038,721 | 11,959 | An Entity of Type : yago:NaturalObject100019128, within Data Space : dbpedia.org associated with source document(s)
In convex geometry, a convex combination is a linear combination of points (which can be vectors, scalars, or more generally points in an affine space) where all coefficients are non-negative and sum to 1. More formally, given a finite number of points in a real vector space, a convex combination of these points is a point of the form where the real numbers satisfy and As a particular example, every convex combination of two points lies on the line segment between the points.
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rdf:type
rdfs:label
• Convex combination
rdfs:comment
• In convex geometry, a convex combination is a linear combination of points (which can be vectors, scalars, or more generally points in an affine space) where all coefficients are non-negative and sum to 1. More formally, given a finite number of points in a real vector space, a convex combination of these points is a point of the form where the real numbers satisfy and As a particular example, every convex combination of two points lies on the line segment between the points.
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has abstract
• In convex geometry, a convex combination is a linear combination of points (which can be vectors, scalars, or more generally points in an affine space) where all coefficients are non-negative and sum to 1. More formally, given a finite number of points in a real vector space, a convex combination of these points is a point of the form where the real numbers satisfy and As a particular example, every convex combination of two points lies on the line segment between the points. A set is convex if it contains all convex combinations of its points.The convex hull of a given set of points is identical to the set of all their convex combinations. There exist subsets of a vector space that are not closed under linear combinations but are closed under convex combinations. For example, the interval is convex but generates the real-number line under linear combinations. Another example is the convex set of probability distributions, as linear combinations preserve neither nonnegativity nor affinity (i.e., having total integral one).
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OpenLink Virtuoso version 08.03.3319 as of Dec 29 2020, on Linux (x86_64-centos_6-linux-glibc2.12), Single-Server Edition (61 GB total memory) | 611 | 2,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-17 | latest | en | 0.853004 |
https://www.tptp.org/cgi-bin/SeeTPTP?Category=Solutions&Domain=SYN&File=SYN053-1&System=E-SAT---3.0.UNS-CRf.s | 1,670,060,889,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00136.warc.gz | 1,089,544,961 | 3,484 | ## TSTP Solution File: SYN053-1 by E-SAT---3.0
View Problem - Process Solution
```%------------------------------------------------------------------------------
% File : E-SAT---3.0
% Problem : SYN053-1 : TPTP v8.1.0. Released v1.0.0.
% Transfm : none
% Format : tptp:raw
% Command : run_E %s %d THM
% Computer : n014.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit : 300s
% DateTime : Thu Sep 29 22:00:27 EDT 2022
% Result : Unsatisfiable 0.18s 0.46s
% Output : CNFRefutation 0.18s
% Verified :
% SZS Type : Refutation
% Derivation depth : 4
% Number of leaves : 8
% Syntax : Number of formulae : 17 ( 5 unt; 4 typ; 0 def)
% Number of atoms : 25 ( 0 equ)
% Maximal formula atoms : 3 ( 1 avg)
% Number of connectives : 22 ( 10 ~; 12 |; 0 &)
% ( 0 <=>; 0 =>; 0 <=; 0 <~>)
% Maximal formula depth : 4 ( 3 avg)
% Maximal term depth : 1 ( 1 avg)
% Number of types : 2 ( 0 usr)
% Number of type conns : 1 ( 1 >; 0 *; 0 +; 0 <<)
% Number of predicates : 3 ( 2 usr; 2 prp; 0-1 aty)
% Number of functors : 2 ( 2 usr; 2 con; 0-0 aty)
% Number of variables : 10 ( 10 sgn 0 !; 0 ?; 0 :)
%------------------------------------------------------------------------------
tff(decl_22,type,
p: \$o ).
tff(decl_23,type,
big_f: \$i > \$o ).
tff(decl_24,type,
b: \$i ).
tff(decl_25,type,
a: \$i ).
cnf(clause_4,negated_conjecture,
( p
| big_f(X1)
| ~ big_f(a) ),
file('/export/starexec/sandbox/benchmark/theBenchmark.p',clause_4) ).
cnf(clause_3,negated_conjecture,
~ p,
file('/export/starexec/sandbox/benchmark/theBenchmark.p',clause_3) ).
cnf(clause_1,negated_conjecture,
( p
| big_f(X1)
| big_f(X2) ),
file('/export/starexec/sandbox/benchmark/theBenchmark.p',clause_1) ).
cnf(clause_5,negated_conjecture,
( ~ big_f(a)
| ~ big_f(b) ),
file('/export/starexec/sandbox/benchmark/theBenchmark.p',clause_5) ).
cnf(c_0_4,negated_conjecture,
( p
| big_f(X1)
| ~ big_f(a) ),
clause_4 ).
cnf(c_0_5,negated_conjecture,
~ p,
clause_3 ).
cnf(c_0_6,negated_conjecture,
( p
| big_f(X1)
| big_f(X2) ),
clause_1 ).
cnf(c_0_7,negated_conjecture,
( ~ big_f(a)
| ~ big_f(b) ),
clause_5 ).
cnf(c_0_8,negated_conjecture,
( big_f(X1)
| ~ big_f(a) ),
inference(sr,[status(thm)],[c_0_4,c_0_5]) ).
cnf(c_0_9,negated_conjecture,
( big_f(X1)
| big_f(X2) ),
inference(sr,[status(thm)],[c_0_6,c_0_5]) ).
cnf(c_0_10,negated_conjecture,
~ big_f(a),
inference(csr,[status(thm)],[c_0_7,c_0_8]) ).
cnf(c_0_11,negated_conjecture,
big_f(X1),
inference(ef,[status(thm)],[c_0_9]) ).
cnf(c_0_12,negated_conjecture,
\$false,
inference(cn,[status(thm)],[inference(rw,[status(thm)],[c_0_10,c_0_11])]),
[proof] ).
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.07/0.11 % Problem : SYN053-1 : TPTP v8.1.0. Released v1.0.0.
% 0.07/0.13 % Command : run_E %s %d THM
% 0.12/0.33 % Computer : n014.cluster.edu
% 0.12/0.33 % Model : x86_64 x86_64
% 0.12/0.33 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33 % Memory : 8042.1875MB
% 0.12/0.33 % OS : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33 % CPULimit : 300
% 0.12/0.33 % WCLimit : 300
% 0.12/0.33 % DateTime : Mon Sep 5 01:26:44 EDT 2022
% 0.12/0.33 % CPUTime :
% 0.18/0.46 Running first-order on 8 cores model finding
% 0.18/0.46 Running: /export/starexec/sandbox/solver/bin/eprover --delete-bad-limit=2000000000 --definitional-cnf=24 -s --print-statistics -R --print-version --proof-object --satauto-schedule=8 --cpu-limit=300 /export/starexec/sandbox/benchmark/theBenchmark.p
% 0.18/0.46 # Version: 3.0pre003
% 0.18/0.46 # Preprocessing class: FSSSSMSSSSSNFFN.
% 0.18/0.46 # Scheduled 4 strats onto 8 cores with 300 seconds (2400 total)
% 0.18/0.46 # Starting G-E--_302_C18_F1_URBAN_RG_S04BN with 1500s (5) cores
% 0.18/0.46 # Starting new_bool_3 with 300s (1) cores
% 0.18/0.46 # Starting new_bool_1 with 300s (1) cores
% 0.18/0.46 # Starting sh5l with 300s (1) cores
% 0.18/0.46 # new_bool_3 with pid 29195 completed with status 0
% 0.18/0.46 # Result found by new_bool_3
% 0.18/0.46 # Preprocessing class: FSSSSMSSSSSNFFN.
% 0.18/0.46 # Scheduled 4 strats onto 8 cores with 300 seconds (2400 total)
% 0.18/0.46 # Starting G-E--_302_C18_F1_URBAN_RG_S04BN with 1500s (5) cores
% 0.18/0.46 # Starting new_bool_3 with 300s (1) cores
% 0.18/0.46 # SinE strategy is GSinE(CountFormulas,hypos,1.5,,3,20000,1.0)
% 0.18/0.46 # ...ProofStateSinE()=5/5
% 0.18/0.46 # Search class: FGHNF-FFSF00-SFFFFFNN
% 0.18/0.46 # Scheduled 5 strats onto 1 cores with 300 seconds (300 total)
% 0.18/0.46 # Starting SAT001_MinMin_p005000_rr_RG with 181s (1) cores
% 0.18/0.46 # SAT001_MinMin_p005000_rr_RG with pid 29200 completed with status 0
% 0.18/0.46 # Result found by SAT001_MinMin_p005000_rr_RG
% 0.18/0.46 # Preprocessing class: FSSSSMSSSSSNFFN.
% 0.18/0.46 # Scheduled 4 strats onto 8 cores with 300 seconds (2400 total)
% 0.18/0.46 # Starting G-E--_302_C18_F1_URBAN_RG_S04BN with 1500s (5) cores
% 0.18/0.46 # Starting new_bool_3 with 300s (1) cores
% 0.18/0.46 # SinE strategy is GSinE(CountFormulas,hypos,1.5,,3,20000,1.0)
% 0.18/0.46 # ...ProofStateSinE()=5/5
% 0.18/0.46 # Search class: FGHNF-FFSF00-SFFFFFNN
% 0.18/0.46 # Scheduled 5 strats onto 1 cores with 300 seconds (300 total)
% 0.18/0.46 # Starting SAT001_MinMin_p005000_rr_RG with 181s (1) cores
% 0.18/0.46 # Preprocessing time : 0.001 s
% 0.18/0.46 # Presaturation interreduction done
% 0.18/0.46
% 0.18/0.46 # Proof found!
% 0.18/0.46 # SZS status Unsatisfiable
% 0.18/0.46 # SZS output start CNFRefutation
% See solution above
% 0.18/0.46 # Parsed axioms : 5
% 0.18/0.46 # Removed by relevancy pruning/SinE : 0
% 0.18/0.46 # Initial clauses : 5
% 0.18/0.46 # Removed in clause preprocessing : 0
% 0.18/0.46 # Initial clauses in saturation : 5
% 0.18/0.46 # Processed clauses : 10
% 0.18/0.46 # ...of these trivial : 0
% 0.18/0.46 # ...subsumed : 0
% 0.18/0.46 # ...remaining for further processing : 10
% 0.18/0.46 # Other redundant clauses eliminated : 0
% 0.18/0.46 # Clauses deleted for lack of memory : 0
% 0.18/0.46 # Backward-subsumed : 3
% 0.18/0.46 # Backward-rewritten : 1
% 0.18/0.46 # Generated clauses : 4
% 0.18/0.46 # ...of the previous two non-redundant : 4
% 0.18/0.46 # ...aggressively subsumed : 0
% 0.18/0.46 # Contextual simplify-reflections : 1
% 0.18/0.46 # Paramodulations : 2
% 0.18/0.46 # Factorizations : 2
% 0.18/0.46 # NegExts : 0
% 0.18/0.46 # Equation resolutions : 0
% 0.18/0.46 # Propositional unsat checks : 0
% 0.18/0.46 # Propositional check models : 0
% 0.18/0.46 # Propositional check unsatisfiable : 0
% 0.18/0.46 # Propositional clauses : 0
% 0.18/0.46 # Propositional clauses after purity: 0
% 0.18/0.46 # Propositional unsat core size : 0
% 0.18/0.46 # Propositional preprocessing time : 0.000
% 0.18/0.46 # Propositional encoding time : 0.000
% 0.18/0.46 # Propositional solver time : 0.000
% 0.18/0.46 # Success case prop preproc time : 0.000
% 0.18/0.46 # Success case prop encoding time : 0.000
% 0.18/0.46 # Success case prop solver time : 0.000
% 0.18/0.46 # Current number of processed clauses : 2
% 0.18/0.46 # Positive orientable unit clauses : 1
% 0.18/0.46 # Positive unorientable unit clauses: 0
% 0.18/0.46 # Negative unit clauses : 1
% 0.18/0.46 # Non-unit-clauses : 0
% 0.18/0.46 # Current number of unprocessed clauses: 3
% 0.18/0.46 # ...number of literals in the above : 3
% 0.18/0.46 # Current number of archived formulas : 0
% 0.18/0.46 # Current number of archived clauses : 8
% 0.18/0.46 # Clause-clause subsumption calls (NU) : 2
% 0.18/0.46 # Rec. Clause-clause subsumption calls : 2
% 0.18/0.46 # Non-unit clause-clause subsumptions : 2
% 0.18/0.46 # Unit Clause-clause subsumption calls : 2
% 0.18/0.46 # Rewrite failures with RHS unbound : 0
% 0.18/0.46 # BW rewrite match attempts : 3
% 0.18/0.46 # BW rewrite match successes : 3
% 0.18/0.46 # Condensation attempts : 0
% 0.18/0.46 # Condensation successes : 0
% 0.18/0.46 # Termbank termtop insertions : 136
% 0.18/0.46
% 0.18/0.46 # -------------------------------------------------
% 0.18/0.46 # User time : 0.002 s
% 0.18/0.46 # System time : 0.001 s
% 0.18/0.46 # Total time : 0.003 s
% 0.18/0.46 # Maximum resident set size: 1644 pages
% 0.18/0.46
% 0.18/0.46 # -------------------------------------------------
% 0.18/0.46 # User time : 0.002 s
% 0.18/0.46 # System time : 0.004 s
% 0.18/0.46 # Total time : 0.006 s
% 0.18/0.46 # Maximum resident set size: 1696 pages
%------------------------------------------------------------------------------
``` | 3,788 | 9,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-49 | longest | en | 0.281687 |
https://www.clubsnap.com/threads/question-on-lens.139582/ | 1,527,330,141,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867416.82/warc/CC-MAIN-20180526092847-20180526112847-00009.warc.gz | 726,781,169 | 20,848 | # Question on lens
Status
Not open for further replies.
##### New Member
I am getting a 350D this weekend. But I have a few questions in regards to lens:
How do I determine the optical zoom for the lens with use on my 350D? I read that the sensor is smaller than 35mm and therefore the focal length tends to increase. What effect does it make? Sorry if I sound really ignorant but I'm totally new to such things and I need some advice.
So if I were to buy Tamron AF 28-200mm, how would the optical zoom be like? How is it calculated?
#### billpepsi
##### New Member
I am getting a 350D this weekend. But I have a few questions in regards to lens:
How do I determine the optical zoom for the lens with use on my 350D? I read that the sensor is smaller than 35mm and therefore the focal length tends to increase. What effect does it make? Sorry if I sound really ignorant but I'm totally new to such things and I need some advice.
So if I were to buy Tamron AF 28-200mm, how would the optical zoom be like? How is it calculated?
IIRC, 350D multiply factor is 1.6, hence a 200mm is equivalent 320mm (35mm)
##### New Member
I'm sorry... so 320mm is near to 10X optical zoom?
#### rebbot
##### Senior Member
28-200 is a 7.1x zoom if you want to know how many "times" it is. It is derived from dividing 200 by 28.
##### New Member
Ooh... so every lens optical zoom is derived by maximum focal length / minimum focal length?
Hey, thanks!
#### mpenza
##### Senior Member
note that the number of times of zoom doesn't really make a lot of sense in the SLR world. e.g. there are primes (fixed focal length) that could give you tighter framing than a "10x" zoom.
#### majere2sg
##### Senior Member
I'm sorry... so 320mm is near to 10X optical zoom?
Hi, you don't have to worry or concern with the optical zoom, 10X 8X all these are calculated based on the minimum zoom to the maximum zoom.. so 28-200mm is 200mm/28mm = 7.14X
Depends on ur shooting style and what u shoot.. even a 300mm prime lens which periodically has 0 optical zoom is good for animals/bird shots.
##### New Member
I'm sorry... but I don't really get it...
Is there any site that I can read to know more about lens.
So that I can know which lens to invest to suit my needs, thanks alot...
The main issue I am confused now is, what has the "mm" helps in photography?
Pardon me as I'm really new... but I would love to learn the art of photography.
#### surge
##### New Member
compacts always use the times thingy to measure there zooms cos it is easier to understand and the lens cant be changed. so you are right to say that
the times is actually the long end of the zoom over the short end
ie a 10x zoom of a cam that starts at 35mm will have long end 350mm. but a same 10xzoom that starts at 28 will give you 280.
if you using DSLR, is better you learn to understand the different angle of view based on the focal length.
there are many websites that teaches this aspect and also have pic showing the different angle of view so that you can understand them better. don get caught in the number game
##### New Member
Um, sure.. I am reading some articles from dppreview.com
Thanks for all the help here.. really appreciated...
Can't wait to get my hands on 350D this weekend~! Woohoo
#### rebbot
##### Senior Member
like all the CSers have commented, all boils down to your shooting style I would think do not waste money buying lens to later find out its not what you want.
Go down to the more reputable shops, try out various lens and see if the focal length is sufficient for your shooting style.
I dun mind helping as I had this problem earlier while transiting from P&S and prosumers but if I am available~
##### New Member
Sorry.. one more question...
what do they always mean by prosumers? and what's P & S?
#### user111
##### Senior Member
prosumer = high end PnS, ut not yet SLR. eg sony f828, canon S2IS,
PnS = point n shoot eg canon ixus, casio exislim
#### Feinwerkbau
##### Deregistered
First, you would have to understand how the focal length affects the angle of view, i.e., what mm gives you how wide, or how narrow a picture is going to end up in YOUR camera.
For a fun and easy way to see the different effects, go here and fool around with the different focal lengths:
http://www.canon.com.sg/index.cfm?fuseaction=digitalcamera&prod_type=tipsnhints#focal
Next, as you already know, whatever focal length you choose, when you mount it on your camera, just mutlipy the stated mm by 1.6x.
Then, I would seriously suggest you look for websites with titles like 'understanding the basics of photographic lenses', 'understanding modern digital photographic equipment' and so on.
The basics do not change, even if equipment does. Once you understand the basics, you can then apply that knowledge to just about any equipment developed.
Hope this helps!
CHEERS!
Thanks!
#### ortega
##### Moderator
Staff member
Sorry.. one more question...
what do they always mean by prosumers? and what's P & S?
Prosumers = "pro" consumers = Advanced consumer compact cameras
P&S = Point & Shoot = Compact cameras
#### Joel Lyn
##### New Member
somehow after reading all the focal length and maths stuff... still feel like doing physics
#### obewan
##### New Member
Joel Lyn said:
somehow after reading all the focal length and maths stuff... still feel like doing physics
Hehe... :bsmilie: Seems like photography is more than Arts. It involved Maths and Science.
#### markccm
##### Deregistered
to me photography is a technical art.
it involves both technicalities & the artistic element. | 1,382 | 5,604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2018-22 | latest | en | 0.956686 |
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This set of Wireless & Mobile Communications test focuses on “Statistical Models for Multipath Fading Channels”.
1. Which of the following is not a statistical models for multipath fading channels?
a) Clarke’s model for flat fading
b) Saleh and Valenzuela indoor statistical model
c) Two ray Rayleigh fading model
2. Who presented the first statistical model for multipath fading channel?
a) Ossana
b) Rayleigh
c) Newton
3. Clarke’s model assumes a horizontal polarized antenna.
a) True
b) False
4. A wave that is incident on mobile does not undergo Doppler shift.
a) True
b) False
5. Which of the following is an important statistics of a Rayleigh fading useful for designing error control codes and diversity schemes?
a) Mobile speed
b) Doppler frequency
c) Level crossing rate (LCR)
d) Power density
6. The level crossing rate (LCR) is defined as expected rate at which _______ fading envelope crosses a specified level.
a) Rayleigh
b) Saleh
c) Vanezuela
7. Level crossing rate is a function of _______
a) Power transmitted by base station
c) Mobile speed
d) Bit error rate
8. Clarke’s model considers the multipath time delay.
a) True
b) False
9. Saleh and Venezuela reported the results of ______ propagation measurements.
a) Indoor
b) Outdoor
c) Air
d) High frequency
10. Saleh and Venezuela show that indoor channel is _______ time varying.
a) Not
b) Very slow
c) Fast
d) Very fast
11. What is the full form of SIRCIM?
a) Simulation of Indoor Radio Channel Impulse response Model
b) Statistical Indoor Radio Channel for Impulse Model
c) Statistical Impulse Radio Channel for Indoor Model
d) Simulation of Impulse Radio Channel for Indoor Model | 417 | 1,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-21 | latest | en | 0.768861 |
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# The growth of the railroads led to the abolition of local
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29 May 2007, 04:50
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The growth of the railroads led to the abolition of local times, which was determined by when the sun reached the observer’s meridian and differing from city to city, and to the establishment of regional times.
(A) which was determined by when the sun reached the observer’s meridian and differing
(B) which was determined by when the sun reached the observer’s meridian and which differed
(C) which were determined by when the sun reached the observer’s meridian and differing
(D) determined by when the sun reached the observer’s meridian and differed
(E) determined by when the sun reached the observer’s meridian and differing
Last edited by apache on 29 May 2007, 06:01, edited 1 time in total.
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29 May 2007, 06:44
weird one...
i would think that "which" refers to "the abolition of local times" and not only"local times" therefore I find all solutions not suitable...
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29 May 2007, 10:40
go with B
D/E - determined by - ambiguous - growth of railroads or local times ?
catch in B - not comfortable with singular for local times
what is the OA ?
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29 May 2007, 11:38
IMO E
"Determined" is participle whereas "differed" is past form of verb. So, "differing" , which is a participle, is appropriate.
Here, "the abolition of local times" if followed by which/ that would modify the noun before the "of" preposition structure (of + preposition object) therefore GMAT would like us to use a modifying phrase. Therefore, we let the A, B and C go.
Now the tough part. To understand which option is correct, we must first try to construct how the clauses would have modified "local time"
"local times, which were determined by when the sun reached the observer’s meridian and which differed"
When we reduce an adjective clause to a modifying phrase the rule is:
1. If there is a "be" verb after the "pronoun" (which/ that), then we simply remove the pronoun and the "be" verb.
2. If "be" verb is absent, then we remove the "pronoun" (which/ that) and change the verb to verb-ing (participle)
Using the rules the sentence becomes:
"local times, determined by when the sun reached the observer’s meridian and differing"
and this is the choice E.
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29 May 2007, 19:13
apache wrote:
The growth of the railroads led to the abolition of local times, which was determined by when the sun reached the observer’s meridian and differing from city to city, and to the establishment of regional times.
(A) which was determined by when the sun reached the observer’s meridian and differing
(B) which was determined by when the sun reached the observer’s meridian and which differed
(C) which were determined by when the sun reached the observer’s meridian and differing
(D) determined by when the sun reached the observer’s meridian and differed
(E) determined by when the sun reached the observer’s meridian and differing
D seemed very much correct to me. But as Srinath has explained in his post, then it could be E. I wasn't aware of such rules!
What is OA?
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29 May 2007, 19:48
The growth of the railroads led to the abolition of local times, which was determined by when the sun reached the observer’s meridian and differing from city to city, and to the establishment of regional times.
(A) which was determined by when the sun reached the observer’s meridian and differing => which what growth or local times ?
(B) which was determined by when the sun reached the observer’s meridian and which differed => which what growth or local times ?(C) which were determined by when the sun reached the observer’s meridian and differing => which what growth or local times ?
(D) determined by when the sun reached the observer’s meridian and differed => wrong tense !! gives impression that action has already completed in the past
(E) determined by when the sun reached the observer’s meridian and differing
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29 May 2007, 22:40
OA E.
thanks.
srinath,
how to identify that a verb is participle not a finite verb.As the key over here is that Determined is a participle .
To what noun ,Determined is working as adjective?
and can you elaborate on the following rule with few examples.
When we reduce an adjective clause to a modifying phrase the rule is:
1. If there is a "be" verb after the "pronoun" (which/ that), then we simply remove the pronoun and the "be" verb.
2. If "be" verb is absent, then we remove the "pronoun" (which/ that) and change the verb to verb-ing (participle)
Thanks for your wonderful explanation .
29 May 2007, 22:40
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drummers-lowrise
Message boards : Seventeen or Bust : Difference between SoB and PSP WUs?
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Jeremy Posner
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Message 21522 - Posted: 4 Mar 2010 | 14:40:31 UTC
It's my understanding that SoB and PSP are, when all is said and done, subsets of essentially the same problem, with some overlap in the numbers that need to be tested for primality.
So why are the WUs of such dramatically different magnitude of size. Are the two projects working on different ends of the same problem, such that PSP will eventually have to deal with much longer WUs and SoB might have to deal with shorter ones, or do they simply take very different approaches to how they split up the work?
As I'm working my way towards SoB Silver, I'm kind of curious about the mechanics of the difference between the two projects...
-JMP
wolfemancs
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Message 21526 - Posted: 4 Mar 2010 | 17:15:02 UTC - in response to Message 21522.
Quick answer that might be sufficient. . . let me know if you need more details.
Both projects are working on numbers of the form k*2^n+1, but they're working with different k values. Seventeen or bust has been working on their k values longer than PSP has been working on theirs, so while the current PSP n values are in the n=8,000,000 range (for first pass, current work units being handed out are double check and in the n=5,000,000 range), SoB work units have a n value in the n=17,000,000 range. Much bigger numbers means MUCH longer run times.
If PSP doesn't find any primes in the next few years, they'll probably be in the 17,000,000 range, while SoB might be in the 27,000,000 range. In other words, length of work units is only going up.
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Message 21527 - Posted: 4 Mar 2010 | 17:38:28 UTC - in response to Message 21526.
Right, but there are multiple k values to be tested, and a few of the k values (and by extension every n value paired with that k value) are on the to do list for both SoB and PSP. Are the projects working through ranges of n, testing them for all of the possible k values, or are they working through the values of n for a single k, then moving on to another k?
I assume that PSP isn't retesting n values for the k values included in SoB that have already been tested. The question then becomes one of how they're each working their way through the tests and how they coordinate on the overlapping range...
-JMP
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Message 21528 - Posted: 4 Mar 2010 | 19:20:53 UTC
Quoted from John Blazek's post:
"About the Sierpinski Problem
Wacław Franciszek Sierpiński (14 March 1882 — 21 October 1969), a Polish mathematician, was known for outstanding contributions to set theory, number theory, theory of functions and topology. It is in number theory where we find the Sierpinski problem.
Basically, the Sierpinski problem is "What is the smallest Sierpinski number"
First we look at Proth numbers (named after the French mathematician François Proth). A Proth number is a number of the form k*2^n+1 where k is odd, n is a positive integer, and 2^n>k.
A Sierpinski number is an odd k such that the Proth number k*2^n+1 is not prime for all n. For example, 3 is not a Sierpinski number because n=2 produces a prime number (3*2^2+1=13). In 1962, John Selfridge proved that 78,557 is a Sierpinski number...meaning he showed that for all n, 78557*2^n+1 was not prime.
Most number theorists believe that 78,557 is the smallest Sierpinski number, but it hasn't yet been proven. In order to prove it, it has to be shown that every single k less than 78,557 is not a Sierpinski number, and to do that, some n must be found that makes k*2^n+1 prime.
The smallest proven 'prime' Sierpinski number is 271,129. In order to prove it, it has to be shown that every single 'prime' k less than 271,129 is not a Sierpinski number, and to do that, some n must be found that makes k*2^n+1 prime.
Seventeen or Bust is working on the Sierpinski problem and the Prime Sierpinski Project is working on the 'prime' Sierpinski problem.
.
.
.
Fortunately, the two projects combined their sieving efforts into a single file. Therefore, PrimeGrid's PSP/SoB sieve supports both projects."
See John's full post here.
____________
141941*2^4299438-1 is prime!
wolfemancs
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Message 21529 - Posted: 4 Mar 2010 | 19:43:02 UTC - in response to Message 21527.
Right, but there are multiple k values to be tested, and a few of the k values (and by extension every n value paired with that k value) are on the to do list for both SoB and PSP. Are the projects working through ranges of n, testing them for all of the possible k values, or are they working through the values of n for a single k, then moving on to another k?
I assume that PSP isn't retesting n values for the k values included in SoB that have already been tested. The question then becomes one of how they're each working their way through the tests and how they coordinate on the overlapping range...
-JMP
The numbers that overlap are being tested by SoB only.
Both projects are working all their assigned k values simultaneously with increasing n values.
Hope this helps
Jeremy Posner
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Message 21530 - Posted: 4 Mar 2010 | 20:09:15 UTC - in response to Message 21529.
Right, but there are multiple k values to be tested, and a few of the k values (and by extension every n value paired with that k value) are on the to do list for both SoB and PSP. Are the projects working through ranges of n, testing them for all of the possible k values, or are they working through the values of n for a single k, then moving on to another k?
I assume that PSP isn't retesting n values for the k values included in SoB that have already been tested. The question then becomes one of how they're each working their way through the tests and how they coordinate on the overlapping range...
-JMP
The numbers that overlap are being tested by SoB only.
Both projects are working all their assigned k values simultaneously with increasing n values.
Hope this helps
That helps quite a bit. So SoB is now up to higher n values across all of its k values, while PSP is at a lower n value for all of its k values other than the ones that are also part of SoB's list of k values.
My next question is that given that the definition of a Sierpinski number requires that k*2^n+1 is not prime for all values of n, will testing go on indefinitely as n rises, or is there some point at which there can be a proof that if none of the n values up to that point yield a prime, none beyond that will? In theory, without such a proof, there is no way to demonstrate that k*2^n+1 is not prime for all values of n, only ways to disprove it by finding a prime...
-JMP
wolfemancs
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Message 21531 - Posted: 4 Mar 2010 | 21:01:07 UTC - in response to Message 21530.
Current known sierpinski numbers have what's called a covering set. It's basically a set of factors, where it can be shown that for every n value, k*2^n+1 is divisible by one of the factors. (eg. if n = 0 mod 2 k2^n+1 is divisible by 3, if n=0 mod 3 k2^n+1 is divisible by 7. . . . all the way up to some number that completes a loop such that dividing any n by that number will fall into one of the categories.) There's more info somewhere on the web, can't remember where I read about all of it.
For the numbers that PSP and SoB are working on it can be (and has been) proved that they do NOT have a covering set. Without a covering set there is no known way to show that a number IS a sierpinski number, so we're left with trying to prove that each of these is not. Our only method to do that is to find an n which makes k*2^n+1 prime.
So to quickly answer your question, there is no end in sight, and we'll keep looking until we find a prime for each of the candidates, or we'll get tired of trying.
enderak
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Message 21532 - Posted: 4 Mar 2010 | 21:04:42 UTC - in response to Message 21530.
Last modified: 4 Mar 2010 | 21:05:43 UTC
If no prime is found for a particular k, then searching will go on forever. (Hence the meaning of "or Bust" in the title.) Most people agree that there will eventually be a prime found for all k's lower than 78557 because every known Sierpinski number has a small covering set, whereas the k's being tested by SoB do not have small covering sets (or any covering set, otherwise that number would be the conjectured answer to the Sierpinski problem, and not 78557).
A good page on Sierpinski numbers and their covering sets: http://primes.utm.edu/glossary/page.php?sort=SierpinskiNumber
JeppeSN
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Message 117168 - Posted: 14 Apr 2018 | 8:09:08 UTC
It is not quite true that every proven Sierpiński number has a covering set.
You can create a k such that some exponents n are covered by an algebraic factorization, and only the remaining exponents are accounted for by a small set of primes.
In the following example, I follow this source:
Anatoly S. Izotov, A Note on Sierpinski Numbers, Fibonacci Quarterly A 33.3(1995), 206.
Let k=166528519813771386496141126495646000631459761803173801006187890625.
When the exponent n is of the form n = 4m+2, we have the Aurifeuillean factorization:
166528519813771386496141126495646000631459761803173801006187890625 * 2^(4m + 2) + 1 = (408079060739180032846765042950625 * 2^(2m + 1) + 20200966826842225 * 2^(m + 1) + 1)*(408079060739180032846765042950625 * 2^(2m + 1) - 20200966826842225 * 2^(m + 1) + 1)
When n is not on this form (i.e. when n is 0 or ±1 modulo 4), 166528519813771386496141126495646000631459761803173801006187890625 * 2^n + 1 is divisible by one of the primes {3, 17, 257, 641, 65537, 6700417}.
Together, the Aurifeuille thing and the prime set (consisting of the prime factors of Fermat numbers F0, F2, F3, F4, and F5, in case you did not notice) prove that we have a Sierpiński number. But it is not expected that there exists a finite covering set for this k. The numbers k*2^(4m+2)+1 probably require an infinite set to cover?
/JeppeSN
JeppeSN
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Message 117170 - Posted: 14 Apr 2018 | 10:54:57 UTC - in response to Message 117168.
I searched and researched a bit more and found a smaller example (with another prime set than in Izotov's example).
So I repeat my previous post with smaller numbers:
Let k=4008735125781478102999926000625.
When the exponent n is of the form n = 4m+2, we have the Aurifeuillean factorization:
4008735125781478102999926000625 * 2^(4m + 2) + 1 = (2002182590520025 * 2^(2m + 1) + 44745755 * 2^(m + 1) + 1)*(2002182590520025 * 2^(2m + 1) - 44745755 * 2^(m + 1) + 1)
When n is not on this form (i.e. when n is 0 or ±1 modulo 4), 4008735125781478102999926000625 * 2^n + 1 is divisible by one of the primes {3, 17, 97, 241, 257, 673}.
Together, the Aurifeuille thing and the prime set prove that we have a Sierpiński number. But it is not expected that there exists a finite covering set for this k. The numbers k*2^(4m+2)+1 probably require an infinite set to cover?
/JeppeSN
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Message 117171 - Posted: 14 Apr 2018 | 12:56:09 UTC - in response to Message 21527.
Right, but there are multiple k values to be tested, and a few of the k values (and by extension every n value paired with that k value) are on the to do list for both SoB and PSP. Are the projects working through ranges of n, testing them for all of the possible k values, or are they working through the values of n for a single k, then moving on to another k?
I assume that PSP isn't retesting n values for the k values included in SoB that have already been tested. The question then becomes one of how they're each working their way through the tests and how they coordinate on the overlapping range...
-JMP
Sob has 5 k's remaining. (Two of these k's are prime and 3 are composite.)
PSP has 9 k's remaining. Two of those 9 k's (22699 and 67607) are the two prime k's already being searched by SoB. It would be wasteful to search the same numbers twice, so PSP isn't searching those two. PSP is therefore searching 7 k's.
In each project, we're searching all k's simultaneously while increasing n. It IS possible to conduct the search in a different order. We could do it one k at a time if we wanted.
SoB is much further along than PSP for a variety of reasons. The leading edge of PSP is around n=19.8M while the leading edge of SoB is around n=31.6M. We're in the middle of double checking old SoB work, however, so we're not currently testing n=31.6M numbers for SoB. The SoB double check is currently around n=24.9M. Current tasks for PSP therefore have an n around 19.8 million while SoB's tasks are at n=24.9 million. That's why the SoB tasks are larger.
____________
My lucky number is 75898524288+1
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Message 117175 - Posted: 14 Apr 2018 | 15:38:17 UTC - in response to Message 21530.
My next question is that given that the definition of a Sierpinski number requires that k*2^n+1 is not prime for all values of n, will testing go on indefinitely as n rises ... ?
So to quickly answer your question, there is no end in sight, and we'll keep looking until we find a prime for each of the candidates, or we'll get tired of trying.
This is literally a case where we're either going to find primes for the remaining k's or die trying. It's entirely possible that these two conjectures will outlive all of us.
If the conjectures are false, i.e., one of the remaining k's is indeed a Sierpinski number for which there is no prime at any n, there is no known way of determining that. This project is only capable of proving the conjectures are correct by finding the needed primes. There's no known way of disproving the conjecture. If the conjectures are false, the search will go on forever.
Of course, we're doing this because it's believed that the conjectures are correct.
____________
My lucky number is 75898524288+1
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https://m.hanspub.org/journal/paper/30435 | 1,638,408,844,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.58/warc/CC-MAIN-20211201234046-20211202024046-00595.warc.gz | 438,351,054 | 20,231 | 函数型线性判别分析
# 函数型线性判别分析Linear Discriminant Analysis for Functional Data
Abstract: In this paper, functional linear discriminant analysis method is proposed for the classification problem of input as functional data. By introducing the functional norm to measure the distance within-class and between-class, an optimization model of functional linear discriminant analysis is constructed. Furthermore, by using the basis function method to transform the infinite dimensional function space into a finite dimensional optimization model, then this model is easy to solve. Since the data is functional, the first derivative or the second derivative of the function can be found. The classification result can be further improved by using the data after the derivative. Finally, the numerical experiments show the feasibility and effectiveness of the functional linear discriminant analysis method.
1. 引言
2. 相关工作
${S}_{w}=\underset{x\in {X}_{1}}{\sum }\left(x-{\mu }_{1}\right){\left(x-{\mu }_{1}\right)}^{T}+\underset{x\in {X}_{2}}{\sum }\left(x-{\mu }_{2}\right){\left(x-{\mu }_{2}\right)}^{T}$ (1)
${S}_{b}=\left({\mu }_{1}-{\mu }_{2}\right){\left({\mu }_{1}-{\mu }_{2}\right)}^{T}$ (2)
$\mathrm{max}J=\frac{{w}^{T}{S}_{b}w}{{w}^{T}{S}_{w}w}$ (3)
3. 函数型线性判别分析
$\mathrm{max}J=\frac{{‖\int w\left(t\right){\mu }_{1}\left(t\right)-\int w\left(t\right){\mu }_{2}\left(t\right)‖}_{2}^{2}}{Var\int w\left(t\right){x}_{1}\left(t\right)+Var\int w\left(t\right){x}_{2}\left(t\right)}$ (4)
${\stackrel{^}{x}}_{j}\left(t\right)=\underset{k=1}{\overset{K}{\sum }}{c}_{jk}{\varphi }_{k}\left(t\right)$ (5)
${\stackrel{^}{x}}_{j}\left(t\right)={C}_{j}^{T}\Phi ={\Phi }^{T}{C}_{j}$ (6)
$\begin{array}{c}\mathrm{var}\int w\left(t\right){x}_{j1}\left(t\right)={N}^{-1}\int w\left(t\right){x}_{j1}\left(t\right)\int w\left(t\right){x}_{j1}\left(t\right)\\ ={N}^{-1}\int {\text{d}}^{T}\Phi {\Phi }^{T}\stackrel{¯}{C}\int {\left({\text{d}}^{T}\Phi {\Phi }^{T}\stackrel{¯}{C}\right)}^{T}\\ ={N}^{-1}{\text{d}}^{T}\int \Phi {\Phi }^{T}\stackrel{¯}{C}{\stackrel{¯}{C}}^{T}\int \Phi {\Phi }^{T}\text{d}\\ ={\text{d}}^{T}J{V}_{0}J\text{d}\end{array}$ (7)
$\begin{array}{c}\mathrm{var}\int w\left(t\right){x}_{j2}\left(t\right)={N}^{-1}\int w{x}_{j2}\int w{x}_{j2}\\ ={N}^{-1}\int {\text{d}}^{T}\Phi {\Phi }^{T}\stackrel{^}{C}\int {\left({\text{d}}^{T}\Phi {\Phi }^{T}\stackrel{^}{C}\right)}^{T}\\ ={N}^{-1}{\text{d}}^{T}\int \Phi {\Phi }^{T}\stackrel{^}{C}{\stackrel{^}{C}}^{T}\int \Phi {\Phi }^{T}\text{d}\\ ={\text{d}}^{T}J{V}_{1}J\text{d}\end{array}$ (8)
$\begin{array}{c}{‖\int w\left(t\right){\mu }_{1}\left(t\right)-\int w\left(t\right){\mu }_{2}\left(t\right)‖}_{2}^{2}=\int w\left(t\right)\left({\mu }_{1}\left(t\right)-{\mu }_{2}\left(t\right)\right)\text{d}t\int w\left(t\right)\left({\mu }_{1}\left(t\right)-{\mu }_{2}\left(t\right)\right)\text{d}t\\ =\int {\text{d}}^{T}\Phi {\Phi }^{T}m\int {\left({\text{d}}^{T}\Phi {\Phi }^{T}m\right)}^{T}\\ ={d}^{T}\int \Phi {\Phi }^{T}m{m}^{T}\int \Phi {\Phi }^{T}\text{d}\\ ={d}^{T}JVJ\text{d}\end{array}$ (9)
$\mathrm{max}J\left(d\right)=\frac{{d}^{T}JVJd}{{d}^{T}J\left({V}_{0}+{V}_{1}\right)Jd}$ (10)
$\begin{array}{cc}\underset{d}{\mathrm{min}}& -{d}^{T}JVJd\\ \text{s}\text{.t}& {d}^{T}J\left({V}_{0}+{V}_{1}\right)Jd=1\end{array}$ (11)
$L\left(d,\lambda \right)=-{d}^{T}JVJd+\lambda \left[{d}^{T}J\left({V}_{0}+{V}_{1}\right)Jd-1\right]$ (12)
$\frac{\partial L}{\partial d}=-2JVJd+2\lambda J\left({V}_{0}+{V}_{1}\right)Jd=0$ (13)
$JVJd=\lambda J\left({V}_{0}+{V}_{1}\right)Jd$ (14)
$L=J\left({V}_{0}+{V}_{1}\right)J$ ，即求 ${L}^{-1}JVJd=\lambda d$ 的一般特征值问题。
1) 选择基函数，求解样本函数的系数 $\stackrel{¯}{C},\stackrel{^}{C}$
2) 计算协方差矩阵 ${V}_{0},{V}_{1},V$ 和基矩阵 $J$
3) 计算(14)式得出 $\lambda$
4) 代入 ${f}_{j}=\int wx=\int w\left(t\right){x}_{j}\left(t\right)\text{d}t$ 并计算 $\mathrm{min}\text{\hspace{0.17em}}{‖{f}_{j}-{\mu }_{i}\left(t\right)‖}^{2}$
4. 数值实验
4.1. 人工数据的数值实验
(a) (b)
Figure 1. (a) Curves of 50 different types of artificial data sets, (b) Curves fitted by basis function method
Figure 2. Classification of 50 curves using functional linear discriminant analysis
(a) (b)
Figure 3. (a) Curves of 50 different types of artificial data sets, (b) Curve fitted by basis function method
Table 1. Data set details
4.2. Spectrometric数据集
(a) (b)
Figure 4. First-order derivation of spectrometric data
(a) (b)
Figure 5. Second-order derivation of spectrometric data
Table 2. Data set details
Figure 6. Take 20% of the spectrometric data after the first-order derivation classification
Figure 7. Take 20% of the spectrometric data after the second-order derivation classification
5. 结论
[1] 周志华. 机器学习[M]. 北京: 清华大学出版社, 2016.
[2] 张学工. 关于统计学习理论与支持向量机[J]. 自动化学报, 2000, 26(1): 36-46.
[3] 平源. 基于支持向量机的聚类及文本分类研究[D]: [博士学位论文]. 北京: 北京邮电大学, 2012.
[4] 田盛丰, 黄厚宽. 基于支持向量机的手写体相似字识别[J]. 中文信息学报, 2000, 14(3): 37-41.
[5] 邹建法, 王国胤, 龚勋. 基于增强Gabor特征和直接分步线性判别分析的人脸识别[J]. 模式识别与人工智能, 2010, 23(4): 477-482.
[6] Fisher, R.A. (1936) The Use of Multiple Measurements in Taxonomic Problems. Annals of Human Genetics, 7, 179-188.
https://doi.org/10.1111/j.1469-1809.1936.tb02137.x
[7] Belhumeur, P.N., Hespanha, J.P. and Kriegman, D.J. (1997) Eigenfaces versus Fisherfaces: Recognition Using Class Specific Linear Projection. IEEE Transactions on Pattern Analysis and Machine Intelligence, 19, 711-720.
https://doi.org/10.1109/34.598228
[8] Rezzi, S., Giani, I., Heberger, K., et al. (2007) Classification of Gilthead Sea Bream (Sparus aurata) from 1H NMR Lipid Profiling Combined with Principal Component and Linear Discriminant Analysis. Journal of Agricultural and Food Chemistry, 55, 9963-9968.
https://doi.org/10.1021/jf070736g
[9] 张新新, 李雨, 等. 主成分-线性判别分析在中药药性识别中的应用[J]. 山东大学学报, 2012, 50(1): 143-146.
[10] 栗科峰, 卢金燕, 等. 基于子图分割与多类支持向量机的人脸识别方法[J]. 科技通报, 2018(8): 1001-7119.
[11] Li, M. and Yuan, B. (2005) 2D-LDA: A Statistical Linear Discriminant Analysis for Image Matrix. Pattern Recognition Letters, 26, 527-532.
https://doi.org/10.1016/j.patrec.2004.09.007
[12] Ramsay, J.O. (1982) When the Data Are Functions. Psychometrika, 47, 379-396.
https://doi.org/10.1007/BF02293704
[13] Ramsay, J.O. and Dalzell, C.J. (1991) Some Tools for Functional Data Analysis. Journal of the Royal Statistical Society, Series B, 53, 539-572.
https://doi.org/10.1111/j.2517-6161.1991.tb01844.x
[14] Ramsay, J.O. and Silverman, B.W. (1997) Functional Data Analysis. Springer-Verlag, New York.
https://doi.org/10.1007/978-1-4757-7107-7
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Match the column Column AColumn B1.cos2π7 + cos4π7 + cos6π7 A.0 2.cosπ7 + cos2π7 +cos3π7 + cos4π7 + cos5π7 + cos6π7 B.123.sinπ11 + sin3π11 + sin5π11 + sin7π11 + sin9π11 C.−12
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The correct option is B 1 - C, 2 - A, 3 - B We know, cosα + cos(α+β) + cos(α+2β) + .......... cos(α+(n−1)β) just apply that formula α = 2π7 β = 2π7 n = 3 = sin(3×2π2×7)sin(2π2×7)×cos(2π7+(3−1)2×2π7) = sin(3π7)×cos(4π7)sin(π7) = sin3π7.cos(π−3π7)sin(π7) [cos (π−θ) = - cos θ] = sin3π7[−cos3π7]sinπ7 = −sin3π7.cos3π7sinπ7 We need to further simplify the expression, we observe that numerator is in the form of sin θ × cosθ.Apply the formula sin2θ = 2sinθ × cosθ Multiplying 2 in numerator and denominator We observe that we can have same trigonometric function in numerator and denominator by writing numerator as sin (π−θ) form because sin (π−θ) = sin θ we Know, sin6π14×0sinπ14 = 0 3. cosπ11 + cos3π11 + cos5π11 + cos7π11 + cos9π11 = π11, β = 2π11,n=5 = sin(5×2π2×11)sin(2π2×11).cos(π11+4×2π2×11) Multiply and divide by 2 in the above expression.
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#1
Dear, I am not a programmer for Julia and so most likely my doubt is due to the little knowledge I have of Julia.
On a computer, I tried to produce similar codes from the Julia and R languages and got a lower performance on Julia. The codes are below:
Code Julia:
``````M = rand(Float64, 5000, 5000);
function inversa_loop(matriz, n)
for i = 1:n
inv(matriz)
end
end
> @time inversa_loop(M, 10)
``````
Time: 65.243 sec
Code R:
``````M <- matrix(runif(n = 5000^2, 0, 1), 5000, 5000)
inversa_loop <- function(matriz, n = 10){
for (i in 1:n){
solve(matriz)
}
}
system.time(inversa_loop(M))
``````
Time: 30.231 sec
Later I tried to improve the Julia code by declaring the variable types to the `inversa_loop` function to see if this would help improve performance. As we can see, yes, there was an improvement in performance considering the code below:
Modified Julia code:
``````M = rand(Float64, 5000, 5000);
function inversa_loop(matriz::Array{Int64,2}, n::Integer)
for i = 1:n
inv(matriz)
end
end
@time inversa_loop(M, 10)
``````
Time: 43.654 sec
Then I noticed that maybe I’m doing an unfair comparison, since the `solve` function of R is implemented in C or Fortran and the `inv` is implemented in Julia. In this way, I can understand that purely Julia codes are generally more efficient than purely written R codes.
However, why common functions of linear algebra in Julia were not implemented using C/C ++? Would not that be a way to eliminate those cases where we have R functions running more efficiently than purely written codes in Julia? Perhaps there is a technical reason that justifies and I do not know. Maybe it’s just a matter of philosophy of language. Perhaps the reason is the belief that in the near future the LLVM machine will evolve to the point of eliminating those differences.
I keep thinking about the case of a language user who will repeat hundreds of thousands of times Julia codes where in these repetitions exist algebraic operations that in languages like R are implemented efficiently using languages such as C or C ++. I also know that I can call C / C ++ codes in Julia, but that would be something undesirable and cumbersome for something that is so simple in other languages since they already make use of efficient codes , such as the `solve` function in R.
Best regards.
#2
You are timing the inversion of a 5000x5000 matrix — this O(n³) operation completely dominates every other operation — so the benchmark has nothing to do with language. It depends entirely on which linear-algebra library is linked and how many threads it is using.
That’s not correct. For double-precision calculations (`Float64`), Julia, R, Numpy, Matlab, etcetera all do matrix inversion with LAPACK, but there are different optimized versions of LAPACK (or the underlying BLAS library) that can be linked by all of these. Sometimes people link OpenBLAS, sometimes they link MKL, or sometimes another library — both Julia and R can be linked with various BLAS libraries with different tradeoffs. In your benchmark, you are mostly timing which BLAS library was linked. Also, different libraries use different numbers of threads by default.
In Julia, you can find out the BLAS library you are using via:
``````using LinearAlgebra
BLAS.vendor()
``````
I’m not sure what the easiest way to determine this is for R.
Basically, if you are doing computations that are dominated by dense-matrix operations like solving Ax=b or Ax=λx, it doesn’t really matter what language you are using because every language uses LAPACK and can be configured link to the same BLAS libraries. But if you do enough computational work, eventually you are going to run into a problem where you have to write your own “inner loops” (performance-critical code), and that’s where language choice matters. To do a useful cross-language benchmark, you need to be timing inner-loop code, not calls to external libraries.
#3
Julia and R are using OpenBlas, in that case. I understand. I linkei R to use OpenBLAS as is the case of Julia here on my machine. Therefore, both languages are making use of OpenBLAS.
#4
Also note that your type annotation is for an `Int64` array, which means your narrower function definition won’t be called with a `Float64` input array. Type annotations on function arguments aren’t necessary for performance in most cases.
#5
Julia:
``````julia> using LinearAlgebra
julia> BLAS.vendor()
:openblas
``````
R
``````> sessionInfo()
R version 3.5.2 (2018-12-20)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Manjaro Linux
Matrix products: default
BLAS/LAPACK: /opt/OpenBLAS/lib/libope
``````
#6
Maybe they are using different numbers of threads, then? You can control how many threads Julia’s BLAS uses via `BLAS.set_num_threads(n)`; not sure how to do it in R.
(They are also probably linked to different builds/versions of OpenBLAS, but I’m not sure that should matter so much here.)
#7
In linux, I can control using the operating system itself by export OPENBLAS_NUM_THREADS=\$nproc. In both cases I am making the same use of the number of threads.
#8
(Type annotations on function arguments have zero impact on performance in any case, not just most cases. The annotations are merely a “filter” to determine which methods get called for which arguments.)
#9
On the machine in question, I have only one thread per core.
#10
Okay. So will there be another way to improve the code to the point of being more efficient or at least equivalent with the R code? Thanks.
#11
This is generating an array whose entries are all `5000`. The `solve` function therefore might be terminating early because the matrix is singular?
The analogue of Julia’s `rand(5000,5000)` in R would be
``````M <- matrix(runif(5000^2), 5000)
``````
no?
#12
Your R is linked with OpenBLAS in `/opt/OpenBLAS`, whereas Julia is probably linked with its own build of OpenBLAS. Is it possible that your library in `/opt/OpenBLAS` is compiled with `NO_AFFINITY=0` (i.e. with thread affinity enabled)? Julia is compiled with `NO_AFFINITY=1` sets `OPENBLAS_MAIN_FREE` to disable thread affinity IIRC, which could affect performance in a benchmark like this.
You could use `taskset` in Linux to set CPU affinity in Julia.
Note also that you should probably run `@time inversa_loop(M, 10)` a couple of times in Julia as the first execution is also measuring some JIT compilation overhead. (If you use `@btime` from the BenchmarkTools package then it does timing measurements more carefully for you.)
#13
If we want to be really accurate, this is not 100% true since type annotation does change some specialization heuristics:
etc.
Nothing to think about while everyday coding and definitely not something that has to be brought up to new users of Julia, but might be worth knowing at least.
#14
That was a writing error only here. Sorry for the mistake. I will correct the above code. The time was computed considering `M <- matrix(runif(n = 5000^2, 0, 1), 5000, 5000)`.
#15
I’ll have to check it out with more calmly.
#16
Echoing what Kristoffer said, and I think this is something that actually affects beginners every once in a while when they forget to interpolate their variables when benchmarking. Just the other day I encountered some confusion around code like the below:
``````inc_array_1(a::Vector{Float64}) = a .+= 1
inc_array_2(a) = a .+= 1
a = zeros(16);
@btime inc_array_1(a); # 7.267 ns (0 allocations: 0 bytes)
@btime inc_array_2(a); # 17.712 ns (0 allocations: 0 bytes)
``````
For comparison (with interpolation):
``````@btime inc_array_1(\$a); # 4.809 ns (0 allocations: 0 bytes)
@btime inc_array_2(\$a); # 4.797 ns (0 allocations: 0 bytes)
``````
#17
How many physical cores does your system have?
OpenBLAS with Julia is capped at 16 threads.
Watching `top`, I saw that R used >2000% CPU, while Julia used 800%. The number of physical CPU cores was 16, so I set `BLAS.set_num_threads(16)` which gave about a 20% performance increase.
`export OPENBLAS_NUM_THREADS=16` and then launching R in the exact same terminal tab did not reduce the CPU use by R, so I do not think that worked.
Because of the cap on Julia’s OpenBLAS, I could not actually test running both with the same number of threads.
Because practically 100% of the time spent is spent by OpenBLAS, I’d expect equal performance if both were actually using the same number of threads.
I could edit that file and recompile for the sake of testing this, but given that my expectation is it will run slower (when set to using that many threads), I’m not exactly excited.
How can I make Flux use all my CPUs?
#18
In my case, I have 4 physical cores, each of them supporting only one thread. Details of my CPU is below:
``````
[pedro-de pedro]# lscpu
Arquitetura: x86_64
Modo(s) operacional da CPU: 32-bit, 64-bit
Ordem dos bytes: Little Endian
Tamanhos de endereço: 39 bits physical, 48 bits virtual
CPU(s): 4
Lista de CPU(s) on-line: 0-3
NĂşcleo(s) por soquete: 4
Soquete(s): 1
NĂł(s) de NUMA: 1
ID de fornecedor: GenuineIntel
FamĂlia da CPU: 6
Modelo: 60
Nome do modelo: Intel(R) Core(TM) i5-4590S CPU @ 3.00GHz
Step: 3
CPU MHz: 798.238
CPU MHz máx.: 3700,0000
CPU MHz mĂn.: 800,0000
BogoMIPS: 5988.80
Virtualização: VT-x
cache de L1d: 32K
cache de L1i: 32K
cache de L2: 256K
cache de L3: 6144K
CPU(s) de nĂł0 NUMA: 0-3
Opções: fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush dts acpi mmx fxsr sse sse2 ss ht tm pbe syscall nx pdpe1gb rdtscp lm constant_tsc arch_perfmon pebs bts rep_good nopl xtopology nonstop_tsc cpuid aperfmperf pni pclmulqdq dtes64 monitor ds_cpl vmx smx est tm2 ssse3 sdbg fma cx16 xtpr pdcm pcid sse4_1 sse4_2 x2apic movbe popcnt tsc_deadline_timer aes xsave avx f16c rdrand lahf_lm abm cpuid_fault epb invpcid_single pti ssbd ibrs ibpb stibp tpr_shadow vnmi flexpriority ept vpid ept_ad fsgsbase tsc_adjust bmi1 avx2 smep bmi2 erms invpcid xsaveopt dtherm ida arat pln pts flush_l1d
``````
`export OPENBLAS_NUM_THREADS=4` makes me have less performance than consider`export OPENBLAS_NUM_THREADS=1`. I understand that `export OPENBLAS_NUM_THREADS` refers to the number of threads per core and not the number of physical colors. That way, I think the correct thing would be to consider `export OPENBLAS_NUM_THREADS=1`. Do not you agree?
#19
I will try to compile the OpenBLAS library of R in the form of OpenBLAS used by Julia. So maybe I can do a fairer comparison. Best, I will try to link the R literally with OpenBLAS used by Julia in the directory below:
``````[pedro-de julia]# pwd
/usr/lib64/julia
[pedro-de julia]# ls
libamd.so libccalltest.so libcolamd.so liblapack.so libLLVM.so libopenlibm.so libpcre2-8.so.0 libsuitesparse_wrapper.so
libblas.so libccalltest.so.debug libdSFMT.so libLLVM-6.0.1.so libmpfr.so libopenlibm.so.2 libpcre2-8.so.0.6.0 libumfpack.so
libcamd.so libccolamd.so libgit2.so libLLVM-6.0.so libmpfr.so.6 libopenlibm.so.2.5 libspqr.so sys.so
libcblas.so libcholmod.so libgmp.so libllvmcalltest.so libmpfr.so.6.0.1 libpcre2-8.so libsuitesparseconfig.so
``````
I will report the result soon.
#20
I created a symbolic link of the libblas.so file, used by Julia in the `/usr/lib64/julia` directory for the libopenblas_haswellp-r0.3.6.dev.so file used by R and found in the `/opt/OpenBLAS/lib` directory.
It seems to me that now i am using the openblas library that is used by R. Well, still I noticed that the code Julia was 1/3 slower than the code R. I think that the basic functions of linear algebra and numerical optimization per Nelder-Mead, BFGS, L-BFGS B and Simulated Annealing should be implemented in languages such as C/C++ and thus could be used, without the need for major optimizations by users of Julia.
Does anyone who considers themselves an advanced Julia user succeed in producing a Julia code, similar to the one above, in that the `inv` function of Julia is more efficient than the `solve` function of R? Well, I’ll keep trying.
Note: To those who only come to read this comment in isolation, please read the previous ones. I understand that pure R codes tend to be slower than pure Julia codes. I also understand that it may be unfair to compare R functions that use C with Julia functions that use only Julia. | 3,448 | 12,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-13 | latest | en | 0.876927 |
https://www.termpaperwarehouse.com/essay-on/Asdfasetsdf/497714 | 1,638,745,971,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363226.68/warc/CC-MAIN-20211205221915-20211206011915-00421.warc.gz | 1,092,602,946 | 143,294 | Free Essay
# Asdfasetsdf
Submitted By aasdfwehqgdsf
Words 90210
Pages 361
Transparencies
Provides transparencies with answers for each lesson in the Student Edition
ISBN 0-07-828001-X
90000
9 780078 280016
PQ245-6457F-PO1[001-019].qxd
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Page 1 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-01:
Chapter 1 Solving Equations and Inequalities
Lesson 1-1 Expressions and Formulas
Pages 8–10
1. First, find the sum of c and
d. Divide this sum by e.
Multiply the quotient by b.
3. b; The sum of the cost of adult and children tickets should be subtracted from
50. Therefore, parentheses need to be inserted around this sum to insure that this addition is done before subtraction. 4. 72
5. 6
6. 23
7. 1
14 Ϫ 4
5
8. Ϫ2
9. 119
10. 0
11. Ϫ23
12. 18
13. \$432
14. \$1875
15. \$1162.50
16. 20
17. 3
18. 29
19. 25
20. 54
21. Ϫ34
22. 19
23. 5
24. 11
25. Ϫ31
26. 7
27. 14
28. Ϫ15
29. Ϫ3
30. Ϫ52
31. 162
32. 15.3
33. 2.56
34. Ϫ7
35. 25
1
3
36. about 1.8 lb
37. 31.25 drops per min
38. 3.4
39. 2
40. 45
1
Algebra 2
Chapter 1
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41. Ϫ4.2
42. 5.3
43. Ϫ4
44. 75
45. 1.4
46. Ϫ4
47. Ϫ8
48. 36.01
49. 2
yϩ5 2 b 2
1
6
50. a
51. Ϫ16
52. 30
53. \$8266.03
54. 400 ft
4Ϫ4ϩ4Ϭ4ϭ1
4Ϭ4ϩ4Ϭ4ϭ2
14 ϩ 4 ϩ 42 Ϭ 4 ϭ 3
4 ϫ 14 Ϫ 42 ϩ 4 ϭ 4
14 ϫ 4 ϩ 42 Ϭ 4 ϭ 5
14 ϩ 42 Ϭ 4 ϩ 4 ϭ 6
44 Ϭ 4 Ϫ 4 ϭ 7
14 ϩ 42 ϫ 14 Ϭ 42 ϭ 8
4ϩ4ϩ4Ϭ4ϭ9
144 Ϫ 42 Ϭ 4 ϭ 10
56. Nurses use formulas to calculate a drug dosage given a supply dosage and a doctor’s drug order. They also use formulas to calculate IV flow rates.
Answers should include the following. • A table of IV flow rates is limited to those situations listed, while a formula can be used to find any IV flow rate. • If a formula used in a nursing setting is applied incorrectly, a patient could die. 57. C
58. D
59. 3
60. 4
61. 10
62. 13
63. Ϫ2
64. Ϫ5
65.
2
3
66.
2
6
7
Algebra 2
Chapter 1
PQ245-6457F-PO1[001-019].qxd
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12:19 PM
Lesson 1-2
1a.
1b.
1c.
1d.
1e.
1f.
Sample
Sample
Sample
Sample
Sample
Sample
Page 3 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-01:
2. A rational number is the ratio of two integers. Since 13 is
13
not an integer, is not a
2
rational number.
Properties of Real Numbers
Pages 14–18
3. 0; Zero does not have a multiplicative inverse since is undefined.
4. Z, Q, R
1
0
5. N, W, Z, Q, R
6. Q, R
7. Multiplicative Identity
8. Associative Property (ϩ)
1
8
10. 8, Ϫ
1
3
2
3
11. Ϫ , 3
12. Ϫ1.5,
13. Ϫ2x ϩ 4y
14. 13p
15. 3c ϩ 18d
16. Ϫ17a Ϫ 1
17. 1.5(10 ϩ 15 ϩ 12 ϩ 8 ϩ
19 ϩ 22 ϩ 31) or 1.5(10) ϩ
1.5(15) ϩ 1.5(12) ϩ 1.5(8) ϩ
1.5(19) ϩ 1.5(22) ϩ 1.5(31)
18. \$175.50
19. W, Z, Q, R
20. Q, R
21. N, W, Z, Q, R
22. Q, R
23. I, R
24. Z, Q, R
25. N, W, Z, Q, R
26. I, R
27. Q, R; 2.4, 2.49, 2.49, 2.49, 2.9
29. Associative Property (ϫ)
31. Associative Property (ϩ)
32. Commutative Property (ϩ)
33. Multiplicative Inverse
34. Distributive
35. Multiplicative Identity
36. 0
37. Ϫm; Additive Inverse
38.
3
1
;
m
Multiplicative Inverse
Algebra 2
Chapter 1
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41. 12 units
39. 1
40. natural numbers
42. The square root of 2 is irrational and therefore cannot be described by a natural number.
1
10
43. 10; Ϫ
44. Ϫ2.5; 0.4
45. 0.125; Ϫ8
46.
5
;
8
Ϫ
3
5
4 3
3 4
8
5
5
23
47. Ϫ ,
48. 4 , Ϫ
49. 3a Ϫ 2b
50. 10x ϩ 2y
51. 40x Ϫ 7y
52. 11m ϩ 10a
53. Ϫ12r ϩ 4t
54. 32c Ϫ 46d
55. Ϫ3.4m ϩ 1.8n
56. 4.4p Ϫ 2.9q
57. Ϫ8 ϩ 9y
58.
59. true
60. false; Ϫ3
61. false; 6
62. true
63. 6.5(4.5 ϩ 4.25 ϩ 5.25 ϩ
6.5 ϩ 5) or 6.5(4.5) ϩ
6.5(4.25) ϩ (6.5)5.25 ϩ
6.5(6.5) ϩ 6.5(5)
64. 3.6; \$327.60
4
9 x 10
Ϫ
19 y 6
Algebra 2
Chapter 1
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1
4
12:19 PM
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1
8
65. 3 a2 b ϩ 2 a1 b
66. 50(47 ϩ 47); 50(47) ϩ 50(47)
1
4
1
8
Def. of a mixed number
1
1
3 122 ϩ 3 a b ϩ 2 112 ϩ 2 a b
4
8
Distributive
3
1
6ϩ ϩ2ϩ
Multiply.
4
4
ϭ 3 a2 ϩ b ϩ 2 a1 ϩ b ϭ ϭ
3
1
ϩ
4
4
3
1 ϭ8ϩ ϩ
4
4
1
3 ϭ8ϩa ϩ b
4
4
Comm. (ϩ)
ϭ 8 ϩ 1 or 9
ϭ6ϩ2ϩ
Assoc. (ϩ)
67. 4700 ft2
68. \$113(0.36 ϩ 0.19);
\$113(0.36) ϩ \$113(0.19)
69. \$62.15
70. Yes; ϭ ϩ ϭ 7;dividing
2
2
2
by a number is the same as multiplying by its reciprocal.
71. Answers should include the following. • Instead of doubling each coupon value and then adding these values together, the Distributive
Property could be applied allowing you to add the coupon values first and then double the sum.
72. B
6ϩ8
5
6
8
Algebra 2
Chapter 1
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• If a store had a 25% off sale on all merchandise, the Distributive Property could be used to calculate these savings. For example, the savings on a
\$15 shirt, \$40 pair of jeans, and \$25 pair of slacks could be calculated as 0.25(15) ϩ 0.25(40) ϩ
0.25(25) or as 0.25(15 ϩ
40 ϩ 25) using the
Distributive Property.
73. C
74. true
75. False; 0 Ϫ 1 ϭ Ϫ1, which is not a whole number.
76. true
2
3
77. False; 2 Ϭ 3 ϭ , which is not
78. 9
a whole number.
79. 6
80. Ϫ5
81. Ϫ2.75
82. 358 in2
83. Ϫ11
84.
85. Ϫ4.3
86. 36
7
10
Chapter 1
Practice Quiz 1
Page 18
1. 14
2. Ϫ9
3. 6
4. Ϫ1
5. 2 amperes
6. Q, R
7. N, W, Z, Q, R
6 7
7 6
9. Ϫ ,
10. 50x Ϫ 64y
6
Algebra 2
Chapter 1
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Lesson 1-3 Solving Equations
Pages 24–27
1. Sample answer: 2x ϭ Ϫ14
2. Sometimes true; only when the expression you are dividing by does not equal zero. 3. Jamal; his method can be confirmed by solving the equation using an alternative method. 4. 5 ϩ 4n
9
cC
5
ϩ
5
1322
9
5
1322 d
9
9
C
5
ϭ
5
1F Ϫ 322
9
5
5
F Ϫ 1322
9
9
5
F
9
ϭF
ϩ 32 ϭ F
5. 2n Ϫ n3
6. Sample answer: 9 times a number decreased by 3 is 6.
7. Sample answer: 5 plus
3 times the square of a number is twice that number.
8. Reflexive Property of
Equality
10. Ϫ21
9. Addition Property of Equality
11. 14
12. Ϫ4
13. Ϫ4.8
14. 1.5
15. 16
16. y ϭ
17. p ϭ
I rt 9 ϩ 2n
4
18. D
19. 5 ϩ 3n
20. 10n ϩ 7
21. n 2 Ϫ 4
22. Ϫ6n3
23. 519 ϩ n2
24. 21n ϩ 82
25.
a
26. 1n Ϫ 72 3
n 2 b 4
7
Algebra 2
Chapter 1
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28. 2r 1h ϩ r 2
27. 2rh ϩ 2r 2
29. Sample answer: 5 less than a number is 12.
30. Sample answer: Twice a number plus 3 is Ϫ1.
31. Sample answer: A number squared is equal to 4 times the number.
32. Sample answer: Three times the cube of a number is equal to the number plus 4.
33. Sample answer: A number divided by 4 is equal to twice the sum of that number and 1.
34. Sample answer: 7 minus half a number is equal to 3 divided by the square of x.
35. Substitution Property (ϭ)
36. Subtraction Property (ϭ)
37. Transitive Property (ϭ)
38. Addition Property (ϭ)
39. Symmetric Property (ϭ)
40. Multiplication Property (ϭ)
41. 7
42. 8
43. 3.2
44. 2.5
45.
3
4
1
12
46. Ϫ
47. Ϫ8
48. Ϫ11
49. Ϫ7
50.
51. 1
52. Ϫ12
53.
1
4
55. Ϫ
2
3
54. 19
10
17
55
2
56.
ϭr
58. a ϭ
57.
d t 59.
3V
r 2
60.
x 1c Ϫ 32 a ϩ2
2A h 62.
ϭh
61. b ϭ
Ϫb
2x
4x
1Ϫx
Ϫaϭb ϭy 63. n ϭ number of games;
2(1.50) ϩ n(2.50) ϭ 16.75; 5
64. s ϭ length of a side; 8s ϭ
124, 15.5 in.
65. x ϭ cost of gasoline per mile;
972 ϩ 114 ϩ 105 ϩ 7600x ϭ
1837; 8.5¢/mi
66. n ϭ number of students that can attend each meeting;
2n ϩ 3 ϭ 83; 40 students
8
Algebra 2
Chapter 1
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67. a ϭ Chun-Wei’s age; a ϩ
(2a ϩ 8) ϩ (2a ϩ 8 ϩ 3) ϭ 94;
Chun-Wei: 15 yrs old, mother:
38 yrs old, father: 41 yrs old
68. c ϭ cost per student;
50
50130 Ϫ c2 ϩ 1452 ϭ
5
1800; \$3
69. n ϭ number of lamps broken;
12(125) Ϫ 45n ϭ 1365;
3 lamps
70. h ϭ height of can A;
71. 15.1 mi/month
72. Central: 690 mi.; Union:
1085 mi
73. The Central Pacific had to lay their track through the
Rocky Mountains, while the
Union Pacific mainly built track over flat prairie.
74. \$295
75. the product of 3 and the difference of a number and 5 added to the product of four times the number and the sum of the number and 1
76. To find the most effective level of intensity for your workout, you need to use your age and 10-second pulse count. You must also be able to solve the formula given for A. Answers should include the following.
• Substitute 0.80 for I and
27 for P in the formula
I ϭ 6 ϫ P Ϭ 1220 Ϫ A2 and solve for A. To solve this equation, divide the product of 6 and 28 by 0.8.
Then subtract 220 and divide by Ϫ1. The result is
17.5. This means that this
1
person is 17 years old.
11.22 2h ϭ 122 23; 8 units
1
3
2
9
Algebra 2
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• To find the intensity level for different values of A and P would require solving a new equation but using the same steps as described above. Solving for A would mean that for future calculations of A you would only need to simplify an expression, 220 Ϫ
6P
,
I
rather
than solve an equation.
77. B
78. D
79. Ϫ6x ϩ 8y ϩ 4z
80. 11a ϩ 8b
81. 6.6
82. 7.44
83. 105 cm2
84. Ϫ5
85. 3
86. Ϫ2.5
1
4
87. Ϫ
88. 3x
89. Ϫ5 ϩ 6y
Lesson 1-4
Solving Absolute Value Equations
Pages 30–32
1. 0 a 0 ϭ Ϫa when a is a negative number and the negative of a negative number is positive.
2a. 0 x 0 ϭ 4
2b. 0 x Ϫ 6 0 ϭ 2
4. Sample answer: 0 4 Ϫ 6 0 ; 2
3. Always; since the opposite of
0 is still 0, this equation has only one case, ax ϩ b ϭ 0.
The solution is
Ϫb
.
a
6. 9
5. 8
8. 5Ϫ21, 136
7. Ϫ17
10. 5Ϫ11, 296
9. 5Ϫ18, Ϫ126
11. 5Ϫ32, 366
12. л
10
Algebra 2
Chapter 1
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13. 586
14. 0 x Ϫ 160 0 ϭ 2
17. 15
18. 24
19. 0
20. 4
21. 3
22. 13
23. Ϫ4
24. Ϫ7.8
25. Ϫ9.4
26. 5
27. 55
28. Ϫ22
16. 162ЊF; This would ensure a minimum internal temperature of 160ЊF.
15. least: 158ЊF; greatest: 162ЊF
29. {8, 42}
30. 512, Ϫ306
33. 5Ϫ2, 166
34.
32. 5Ϫ28, 206
31. 5Ϫ45, 216
35.
3 e f
2
37.
e 2,
e 2,
16 f 3
Ϫ
36. л
38. 5Ϫ4, Ϫ16
9 f 2
39. л
40. л
41. {Ϫ5, 11}
42. {3, 15}
43.
11 eϪ ,
3
Ϫ3 f
44.
5 f 3
46. 5Ϫ46
45. {8}
48. 0 x Ϫ 16 0 ϭ 0.3; heaviest:
16.3 oz, lightest: 15.7 oz
47. 0 x Ϫ 200 0 ϭ 5; maximum:
205ЊF; minimum: 195ЊF
49. 0 x Ϫ 13 0 ϭ 5; maximum:
18 km, minimum: 8 km
50. sometimes; true only if a Ն 0 and b Ն 0 or if a Յ 0 and bՅ0 52. Answers should include the following. • This equation needs to show that the difference of the estimate E from the originally stated magnitude of 6.1 could be plus 0.3 or
51. sometimes; true only if c Ն 0
e 3,
11
Algebra 2
Chapter 1
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minus 0.3, as shown in the graph below. Instead of writing two equations,
E Ϫ 6.1 ϭ 0.3 and
E Ϫ 6.1 ϭ Ϫ0.3, absolute value symbols can be used to account for both possibilities, 0 E Ϫ 6.1 0 ϭ 0.3.
0.3 units
5.6
5.7
5.8
5.9
6.0
0.3 units
6.1
6.2
6.3
6.4
6.5
6.6
6.7
• Using an original magnitude of 5.9, the equation to represent the estimated extremes would be 0 E Ϫ 5.9 0 ϭ 0.3.
54. A
53. B
55. 0 x ϩ 1 0 ϩ 2 ϭ x ϩ 4;
0 x ϩ 1 0 ϩ 2 ϭ Ϫ1x ϩ 42
56. x ϩ 1 ϩ 2 ϭ x ϩ 4;
Ϫx Ϫ 1 ϩ 2 ϭ x ϩ 4; x ϩ 1 ϩ 2 ϭ Ϫx Ϫ 4;
Ϫx Ϫ 1 ϩ 2 ϭ Ϫx Ϫ 4
57. 5Ϫ1.56
58. 8
60. 5n 2
59. 21n Ϫ 112
16
3
68. false; 23
69. true
70. true
71. false; 1.2
72.
73. 364 ft2
74. 2
75. 8
76. Ϫ2
61.
62. Ϫ2
63. 14
64. Commutative Property (ϩ)
65. Distributive Property
66. Multiplicative Inverse
77.
2
3
1
1x
2
ϩ 321x ϩ 52
78. 6
3
4
79. Ϫ
12
Algebra 2
Chapter 1
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Lesson 1-5 Solving Inequalities
Pages 37–39
1. Dividing by a number is the same as multiplying by its inverse. 2. Sample answer: Ϫ2n Ͼ Ϫ6
4. 5a 0 a Ͻ 1.56 or 1Ϫϱ, 1.52
3. Sample answer: xϩ2Ͻxϩ1 Ϫ2
5
3
1
2
Ϫ10
9. 5p 0 p Ͼ 156 or 115, ϩϱ2
0
2
Ϫ6
26
Ϫ6
Ϫ5
Ϫ4
Ϫ3
Ϫ4
0
Ϫ2
18. 5d 0 d Ͼ Ϫ86 or 1Ϫ8, ϩϱ2
Ϫ10
28
Ϫ8
Ϫ6
Ϫ4
Ϫ2
20. 5p 0 p Յ Ϫ36 or 1Ϫϱ, Ϫ34
30
Ϫ6
21. 5k 0 k Ն Ϫ3.56 or 3Ϫ3.5, ϩϱ2
Ϫ7
Ϫ6
10 11 12 13 14 15 16 17 18 19 20 21
Ϫ4
19. 5g 0 g Յ 276 or 1Ϫϱ, 274
24
Ϫ8
16. 5b 0 b Յ 186 or 1Ϫϱ, 184
Ϫ1 0 1 2 3 4 5 6 7 8 9 10
22
8
14. at least 92
17. 5x 0 x Ͻ 76 or 1Ϫϱ, 72
20
6
4
15. 5n 0 n Ն Ϫ116 or 3Ϫ11, ϩ ϱ2
Ϫ8
4
12. 12n Ͼ 36; n Ͼ 3
13. 2n Ϫ 3 Յ 5; n Յ 4
Ϫ14 Ϫ12 Ϫ10
2
Ϫ30 Ϫ28 Ϫ26 Ϫ24 Ϫ22 Ϫ20
11. all real numbers or 1Ϫϱ, ϩϱ2
Ϫ2
3
10. 5n 0 n Յ Ϫ246 or 1Ϫϱ, Ϫ244
8 9 10 11 12 13 14 15 16 17 18 19
Ϫ4
2
8. 5w 0 w Ͻ Ϫ76 or 1Ϫϱ, Ϫ72
Ϫ1 0 1 2 3 4 5 6 7 8 9 10
Ϫ6
0
Ϫ2
3
7. 5y 0 y Ͼ 66 or 16, ϩϱ2
1
6. 5c 0 c Ն 36 or 33, ϩϱ2
5
3
5. e x ` x Յ f or aϪϱ, d f
0
0
Ϫ1
Ϫ4
0
Ϫ2
22. 5y 0 y Ͻ 56 or 1Ϫϱ, 52
Ϫ2
Ϫ4
13
Ϫ2
0
2
2
4
Algebra 2
0
4
6
Chapter 1
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23. 5m 0 m Ͼ Ϫ46 or 1Ϫ4, ϩϱ2
Ϫ6
Ϫ4
0
Ϫ2
2
2
3
4
0
Ϫ2
2
0.5
1
1.5
4
6
Ϫ2
2
2.5
Ϫ7
20
0
Ϫ2
Ϫ4
0
Ϫ2
4
6
1
20
or aϪ , ϩϱb
Ϫ1 Ϫ 3 Ϫ 1
4
20
1
20
20
2.0
2.2
2.4
5
7
2.6
2.8
4
5
7
0 1 2 3 4 5 6 1 8 9 10 11
1
5
3
5
Ϫ6
1
7 7
Ϫ4
Ϫ2
0
2
4
3
2
6
3
2
38. e n ` n Յ Ϫ f or aϪϱ, Ϫ d
37. л
Ϫ6
7 7
36. 5p 0 p Ͼ 06 or 10, ϩϱ2
1
5
35. e y ` y Ͻ f or aϪϱ, b
1
5
3.0
34. e a ` a Ն f or c , ϩϱb
7 7 7 7 7 7
3
5
3
20
32. 5z 0 z Ͼ 2.66 or 12.6, ϩ ϱ2
2
2
1
5
8
Ϫ20 Ϫ18 Ϫ16 Ϫ14 Ϫ12 Ϫ10
33. 5g 0 g Ͻ 26 or 1Ϫϱ, 22
Ϫ6
2
30. 5c 0 c Ͼ Ϫ186 or 1Ϫ18, ϩϱ2
31. 5d 0 d Ն Ϫ56 or 3Ϫ5, ϩϱ2
Ϫ4
2
1 f 20
Ϫ286 Ϫ284 Ϫ282 Ϫ280 Ϫ278 Ϫ276
Ϫ6
0
28. ew ` w Ͼ Ϫ
29. 5x 0 x Ͻ Ϫ2796 or 1Ϫϱ, Ϫ2792
Ϫ8
1
26. 5r 0 r Յ 66 or 1Ϫϱ, 64
27. 5n 0 n Ն 1.756 or 31.75, ϩϱ2
0
0
Ϫ1
25. 5t 0 t Յ 06 or 1Ϫϱ, 0 4
Ϫ4
2
3
24. e b ` b Ն f or c ϩϱb
Ϫ4
Ϫ2
0
2
4
Ϫ4
Ϫ3
Ϫ2
0
Ϫ1
39. at least 25 h
40. no more than 14 rides
41. n ϩ 8 Ͼ 2; n Ͼ Ϫ6
1
42. Ϫ4n Ն 35; n Յ Ϫ8.75
43.
1 n 2
44. Ϫ3n ϩ 1 Ͻ 16; n Ͼ Ϫ5
Ϫ 7 Ն 5; n Ն 24
n
2
46. n Ϫ 9 Յ ; n Յ 18
45. 21n ϩ 52 Յ 3n ϩ 11; n Ն Ϫ1
17
47. 217m2 Ն 17; m Ն , at least
14
2 child-care staff members
48. \$24,000 ϩ 0.015130,500n2 Ն
40,000
14
Algebra 2
Chapter 1
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Page 15 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-01:
85 ϩ 91 ϩ 89 ϩ 94 ϩ s
5
49. n Ն 34.97; She must sell at least 35 cars.
50.
51. s Ն 91; Ahmik must score at least 91 on her next test to have an A test average.
52a. It holds only for Յ or Ն; 2 Х 2.
52b. 1 Ͻ 2 but 2 1
52c. For all real numbers a, b, and c, if a Ͻ b and b Ͻ c then a Ͻ c.
53. Answers should include the following. • 150 Ͻ 400
• Let n equal the number of minutes used. Write an expression representing the cost of Plan 1 and for Plan
2 for n minutes. The cost for Plan 1 would include a
\$35 monthly access fee plus 40¢ for each minute over 150 minutes or
35 ϩ 0.41n Ϫ 1502. The cost for Plan 2 for 400 minutes or less would be
\$55. To find where Plan 2 would cost less than Plan
1 solve 55 Ͻ 35 ϩ
0.41n Ϫ 1502 for n. The solution set is 5n 0 n Ͼ 2006, which means that for more than 200 minutes of calls,
Plan 2 is cheaper.
54. D
55. D
56. x Ͼ Ϫ3
57. x Ն Ϫ2
58. x Ն Ϫ1
59. 5Ϫ14, 206
60.
Ն 90
5 11 eϪ , f 4 4
61. л
62. b ϭ online browsers each year; 6b ϩ 19.2 ϭ 106.6; about 14.6 million online browsers each year
63. N, W, Z, Q, R
64. Q, R
15
Algebra 2
Chapter 1
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Page 16 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-01:
65. I, R
66. 4.25(5.5 ϩ 8); 4.25(5.5) ϩ
4.25(8)
67. 5Ϫ7, 76
69.
e 4,
68. 513, Ϫ236
70. 511, 256
4 f 5
Ϫ
71. 5Ϫ11, Ϫ16
72. 5Ϫ18, 106
Chapter 1
Practice Quiz 2
Page 39
2s
t2
1. 0.5
2.
3. 14
4. eϪ , 5 f
ϭg
19
3
4
9
4
9
5. e m ` m Ͼ f or a , ϩ ϱb
Ϫ2
9
0
2
9
4
9
2
3
8
1
9
Lesson 1-6 Solving Compound and
Absolute Value Inequalities
Pages 43–46
1. 5 Յ c Յ 15
2. Sample answer: x Ͻ Ϫ3 and xϾ2 4. 0 n 0 Ͻ 8
3. Sabrina; an absolute value inequality of the form 0 a 0 Ͼ b should be rewritten as an or compound inequality, a Ͼ b or a Ͻ b.
Ϫ12
5. 0 n 0 Ͼ 3
Ϫ6
Ϫ4
7. 0 n 0 Ͻ 2
Ϫ2
0
2
Ϫ4
0
4
8
6. 0 n 0 Ն 4
4
8. 5y 0 y Ͼ 4 or y Ͻ Ϫ16
Ϫ4
Ϫ8
16
Ϫ2
0
2
4
Algebra 2
6
Chapter 1
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9. 5d 0 Ϫ2 Ͻ d Ͻ 36
Ϫ4
0
Ϫ2
10. 5a 0 a Ն 5 or a Յ Ϫ56
2
4
11. 5g 0 Ϫ13 Յ g Յ 56
Ϫ16 Ϫ12
Ϫ8
0
Ϫ4
6
Ϫ8
4
2
4
2
4
8
6
12
8
14. 55 Յ
Յ 60; 343.75 Յ c Յ
6.25
375; between \$343.75 and
\$375
6
16. 0 n 0 Ͻ 7
Ϫ8
Ϫ4
0
4
8
12
Ϫ4
Ϫ2
0
2
4
6
Ϫ8
Ϫ4
0
4
8
12
17. 0 n 0 Ͻ 4
19. 0 n 0 Ͼ 8
21. 0 n 0 Ͼ 1
Ϫ8
8
12
Ϫ4
0
4
8
12
Ϫ1.4 Ϫ1.2
0
1.2
1.4
1.6
4
6
8
2
4
6
2
4
6
22. 0 n 0 Յ 5
24. 0 n 0 Ͻ 6
0
4
29. 5x 0 Ϫ2 Ͻ x Ͻ 46
0
2
31. 5f 0 Ϫ7 Ͻ f Ͻ Ϫ56
26. 0 n Ϫ 1 0 Յ 3
28. 5t 0 1 Ͻ t Ͻ 36
8
4
12
Ϫ2
Ϫ4
Ϫ2
2
0
32. all real numbers
Ϫ8
Ϫ6
Ϫ4
Ϫ2
Ϫ8
Ϫ4
0
4
8
12
Ϫ4
Ϫ2
0
2
4
Ϫ4
0
6
33. 5g 0 Ϫ9 Յ g Յ 96
0
30. 5c 0 c Ͻ Ϫ2 or c Ն 16
6
Ϫ10
Ϫ2
0
34. 5m 0 m Ն 4 or m Յ Ϫ46
Ϫ4
Ϫ2
0
2
4
6
Ϫ8
Ϫ4
0
4
8
12
36. 5y 0 Ϫ7 Ͻ y Ͻ 76
35. л
4
20. 0 n 0 Յ 1.2
27. 5p 0 p Յ 2 or p Ն 86
Ϫ2
0
Ϫ8
25. 0 n ϩ 1 0 Ͼ 1
Ϫ4
Ϫ4
18. 0 n 0 Ն 6
23. 0 n 0 Ն 1.5
Ϫ4
0
Ϫ2
15. 0 n 0 Ն 5
Ϫ8
4
c
0
Ϫ2
0
12. 5k 0 Ϫ3 Ͻ k Ͻ 76
13. all real numbers
Ϫ4
Ϫ4
17
Algebra 2
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38. 5r 0 Ϫ3 Ͻ r Ͻ 46
37. 5b 0 b Ͼ 10 or b Ͻ Ϫ26
0
Ϫ4
4
8
12
Ϫ4
7
3
39. e w ` Ϫ Յ w Յ 1 f
Ϫ2
1
en ` n
ϭ
0
43.
0
Ϫ2
2
4
6
4
6
Ϫ2
0
2
4
6
0
2
4
6
Ϫ4
Ϫ2
44. 5n 0 n Ͼ 1.56
7 f 2
1
2
42. 5n 0 n Ն 06
41. all real numbers
Ϫ4
0
40. л
0
Ϫ1
Ϫ2
Ϫ4
16
2
3
4
Ϫ2
5
45. 6.8 Ͻ x Ͻ 7.4
Ϫ1
0
1
2
3
46. 45 Յ s Յ 65
47. 45 Յ s Յ 55
48. 0 t Ϫ 98.6 0 Ն 8; 5b 0 b Ͼ 106.6 or b Ͻ 90.66
49. 108 in. Ͻ L ϩ D Յ 130 in.
50. 84 in. Ͻ L Յ 106 in.
51. a ϩ b Ͼ c, a ϩ c Ͼ b, bϩcϾa 52. a Ϫ b Ͻ c Ͻ a ϩ b
53a.
Ϫ4
Ϫ2
0
2
4
Ϫ4
Ϫ2
0
2
4
6
Ϫ4
Ϫ2
0
2
4
54. Compound inequalities can be used to describe the acceptable time frame for the fasting state before a glucose tolerance test is administered to a patient suspected of having diabetes. Answers should include the following.
• Use the word and when both inequalities must be satisfied. Use the word or when only one or the other of the inequalities must be satisfied.
• 10 Յ h Յ 16
6
6
53b.
53c.
53d. 3 Ͻ 0 x ϩ 2 0 Յ 8 can be rewritten as 0 x ϩ 2 0 Ͼ 3 and
0 x ϩ 2 0 Յ 8. The solution of
0 x ϩ 2 0 Ͼ 3 is x Ͼ 1 or x Ͻ Ϫ5. The solution of
0 x ϩ 2 0 Ն 8 is Ϫ10 Յ x Յ 6.
Therefore, the union of these two sets is 1x Ͼ 1 or x Ͻ Ϫ52
18
Algebra 2
Chapter 1
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and 1Ϫ10 Յ x Յ 6). The union of the graph of x Ͼ 1 or x Ͻ Ϫ5 and the graph of
Ϫ10 Յ x Յ 6 is shown below. From this we can see that solution can be rewritten as 1Ϫ10 Յ x Ͻ Ϫ52 or
11 Ͻ x Յ 62.
Ϫ12
Ϫ8
0
Ϫ4
4
• 12 hours would be an acceptable fasting state for this test since it is part of the solution set of 10 Յ h
Յ 16, as indicated on the graph below.
8 9 10 11 12 13 14 15 16 17 18 19
8
55. x Ͼ Ϫ5 or x Ͻ Ϫ6
56. D
57.
58. 2 Ͻ x Ͻ 3
59. 15x ϩ 2 Ն 32 or
15x ϩ 2 Յ Ϫ32; 5x 0 x Ն 0.2 or x Յ Ϫ16
60. abs12x Ϫ 62 Ͼ 10;
5x Ϳ x Ͻ Ϫ2 or x Ͼ 86
61. d Ն Ϫ6 or 3Ϫ6, Ϫϱ2
Ϫ8
Ϫ6
Ϫ4
0
Ϫ2
2
63. n Ͻ Ϫ1 or 1Ϫϱ, Ϫ12
Ϫ4
Ϫ2
0
2
62. x Ͻ 4 or 1Ϫϱ, 42
4
Ϫ4
Ϫ2
0
2
4
6
64. 0 x Ϫ 587 0 ϭ 5; highest: 592 keys, lowest: 582 keys
6
66. 5Ϫ11, 46
65. {Ϫ10, 16}
68. Addition Property of Equality
67. л
69. Symmetric Property (ϭ)
70. Transitive Property of Equality
71. 3a ϩ 7b
72. Ϫ2m Ϫ 7n Ϫ 18
73. 2
74. 92
75. Ϫ7
19
Algebra 2
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Chapter 2 Linear Relations and Functions
Lesson 2-1 Relations and Functions
Pages 60–62
1. Sample answer: {(Ϫ4, 3),
(Ϫ2, 3), (1, 5), (Ϫ2, 1)}
y
x
O
3. Molly; to find g(2a), replace x with 2a. Teisha found
2g(a), not g(2a).
4. yes
5. yes
6. no
7. D ϭ {7}, R ϭ {Ϫ1, 2, 5, 8}, no
8. D ϭ {3, 4, 6}, R ϭ {2.5}, yes y (7, 8)
y
(7, 5)
(4, 2.5)
(3, 2.5)
(7, 2)
x
O
O
(7, Ϫ1)
(6, 2.5)
x
10. D ϭ 5x 0 x Ն 06,
R ϭ all reals, no
9. D ϭ all reals,
R ϭ all reals, yes
y
y
x ϭ y2
O
x
O
x
y ϭ Ϫ2x ϩ 1
12. Ϫ7
11. 10
20
Algebra 2
Chapter 2
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13. D ϭ {70, 72, 88},
R ϭ {95, 97, 105, 114}
14. {(88, 97), (70, 114), (88, 95),
(72, 105)}
15. Record High Temperatures
16. No; the domain value 88 is paired with two range values.
115
July
110
105
100
95
0
70
80
January
90
17. yes
18. no
19. no
20. yes
21. yes
22. no
23. D ϭ {Ϫ3, 1, 2},
R ϭ {0, 1, 5}; yes
24. D ϭ {3, 4, 6}, R ϭ {5}; yes y y
(3, 5) (4, 5)
(1, 5)
(6, 5)
(2, 1)
(Ϫ3, 0)
x
O
x
O
26. D ϭ {3, 4, 5, 6},
R ϭ {3, 4, 5, 6}; yes
25. D ϭ {Ϫ2, 3}, R ϭ {5, 7, 8}; no
(Ϫ2, 8)
y
(3, 7)
y
(5, 6)
(3, 4)
(Ϫ2, 5)
(6, 5)
(4, 3)
O
x
O
21
x
Algebra 2
Chapter 2
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27. D ϭ {Ϫ3.6, 0, 1.4, 2},
R ϭ {Ϫ3, Ϫ1.1, 2, 8}; yes
28. D ϭ {Ϫ2.5, Ϫ1, 0},
R ϭ {Ϫ1, 1}; no y y
(Ϫ3.6, 8)
(Ϫ2.5, 1) (Ϫ1, 1)
(0, 1)
x
O
(Ϫ1, Ϫ1)
(1.4, 2)
x
O
(0, Ϫ1.1)
(2, Ϫ3)
30. D ϭ all reals,
R ϭ all reals; yes
29. D ϭ all reals,
R ϭ all reals; yes
y
y
y ϭ 3x x O
x
O
y ϭ Ϫ5x
31. D ϭ all reals,
R ϭ all reals; yes
32. D ϭ all reals,
R ϭ all reals; yes y y
O
O
x
x
y ϭ 7x Ϫ 6
y ϭ 3x Ϫ 4
22
Algebra 2
Chapter 2
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34. D ϭ 5x 0 x Ն Ϫ36,
R ϭ all reals; no
33. D ϭ all reals,
R ϭ 5y 0 y Ն 06, yes y y
O
yϭx
x
2
x ϭ 2y 2 Ϫ 3
x
O
36. D ϭ {47, 48, 52, 56},
R ϭ {145, 147, 148, 157, 165}
35.
170
165
RBI
160
155
150
145
140
0
48
50
52
54
56
HR
37. No; the domain value 56 is paired with two different range values.
38. {(1997, 39), (1998, 43),
(1999, 48), (2000, 55),
(2001, 61), (2002, 52)}
39.
40. D ϭ {1997, 1998, 1999,
2000, 2001, 2002},
R ϭ {39, 43, 48, 52, 55, 61}
70
Stock Price
60
Price (\$)
50
40
30
20
10
0
1996
1998
2000 2002
Year
2004
42. {(1987, 12), (1989, 13),
(1991, 11), (1993, 12),
(1995, 9), (1997, 6), (1999, 3)}
41. Yes; each domain value is paired with only one range value. ©Glencoe/McGraw-Hill
23
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
43.
Representatives
14
7/24/02
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Page 24 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
44. D ϭ {1987, 1989, 1991,
1993, 1995, 1997, 1999},
R ϭ {3, 6, 9, 11, 12, 13}
30+ Years of Service
12
10
8
6
4
2
0
’87
’91
’95
Year
’99
45. Yes; no; each domain value is paired with only one range value so the relation is a function, but the range value
12 is paired with two domain values so the function is not one-to-one. 46. Ϫ14
47. 6
48. Ϫ
49. Ϫ3
50. 3a Ϫ 5
51. 25n 2 Ϫ 5n
52. Ϫ4
53. 11
54. 39
55. f(x) ϭ 4x Ϫ 3
56. Relations and functions can be used to represent biological data. Answers should include the following.
• If the data are written as ordered pairs, then those ordered pairs are a relation. • The maximum lifetime of an animal is not a function of its average lifetime.
57. B
58. C
59. discrete
60. continuous
61. discrete
62. continuous
65. 5x 0 x Ͻ 5.16
66. \$2.85
2
9
63. 5y 0 Ϫ8 Ͻ y Ͻ 66
64. 5m 0 4 Ͻ m Ͻ 66
24
Algebra 2
Chapter 2
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67. \$29.82
68. 43
69. 31a ϩ 10b
70. Ϫ1
71. 2
72. 6
73. 15
Lesson 2-2 Linear Equations
Pages 65–67
2. 5, Ϫ2
1. The function can be written as
1
2
f(x) ϭ x ϩ 1, so it is of the form f(x) ϭ mx ϩ b, where mϭ 1
2
and b ϭ 1.
3. Sample answer: x ϩ y ϭ 2
4. No, the variables have an exponent other than 1.
5. yes
6. 3x Ϫ y ϭ 5; 3, Ϫ1, 5
7. 2x Ϫ 5y ϭ 3; 2, Ϫ5, 3
8. 2x Ϫ 3y ϭ Ϫ3; 2, Ϫ3, Ϫ3
5
3
9. Ϫ , Ϫ5
10. 2, Ϫ2 y y
x
O
x
O
y ϭ Ϫ3x Ϫ 5
xϪyϪ2ϭ0
11. 2, 3
12. 3, y 3
2
y
3x ϩ 2y ϭ 6
O
x
x
O
4x ϩ 8y ϭ 12
25
Algebra 2
Chapter 2
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13. \$177.62
14. 563.00 euros
15. yes
16. No; x appears in a denominator. 17. No; y is inside a square root.
18. No; x has exponents other than 1.
19. No; x appears in a denominator. 20. yes
21. No; x has an exponent other than 1.
22. No; x is inside a square root.
23. x 2 ϩ 5y ϭ 0
24. h(x ) ϭ x 3 Ϫ x 2 ϩ 3x
25. 7200 m
26. Sound travels only 1715 m in
5 seconds in air, so it travels faster underwater.
27. 3x ϩ y ϭ 4; 3, 1, 4
28. 12x Ϫ y ϭ 0; 12, Ϫ1, 0
29. x Ϫ 4y ϭ Ϫ5; 1, Ϫ4, Ϫ5
30. x Ϫ 7y ϭ 2; 1, Ϫ7, 2
31. 2x Ϫ y ϭ 5; 2, Ϫ1, 5
32. x Ϫ 2y ϭ Ϫ3; 1, Ϫ2, Ϫ3
33. x ϩ y ϭ 12; 1, 1, 12
34. x Ϫ y ϭ Ϫ6; 1, Ϫ1, Ϫ6
35. x ϭ 6; 1, 0, 6
36. y ϭ 40; 0, 1, 40
37. 25x ϩ 2y ϭ 9; 25, 2, 9
38. 5x Ϫ 4y ϭ 2; 5, Ϫ4, 2
39. 3, 5
40. 6, Ϫ2 y y
2x Ϫ 6y ϭ 12
5x ϩ 3y ϭ 15
x
O
O
26
x
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
41.
10
,
3
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Page 27 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
5
2
42. 5, 2
Ϫ
y
y
3x Ϫ 4y Ϫ 10 ϭ 0
x
O
x
O
2x ϩ 5y Ϫ 10 ϭ 0
43. 0, 0
44. y 1
,
2
Ϫ2 y yϭx
O
x
x
O
y ϭ 4x Ϫ 2
45. none, Ϫ2
46. none, 4
y
y yϭ4 x
O
x
O
y ϭ Ϫ2
48. 1, none
47. 8, none
8
6
4
2
Ϫ8 Ϫ6Ϫ4 Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
y
y
xϭ1
xϭ8
O
2 4 6
O
x
x
27
Algebra 2
Chapter 2
PQ245-6457F-P02[020-055].qxd
49.
1
,
4
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Page 28 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
50. 6, Ϫ3
Ϫ1
g (x )
f (x ) f (x ) ϭ 4x Ϫ 1
g (x ) ϭ 0.5x Ϫ 3
O
x
O
51.
y
x
52. Sample answer: x ϩ y ϭ 2
x ϩy ϭ 5
x
O
xϩyϭ0 x ϩ y ϭ Ϫ5
The lines are parallel but have different y-intercepts.
53. 90ЊC
54. 4 km
55.
160
120
80
40
56. 1.75b ϩ 1.5c ϭ 525
T (d )
O1 2 3 4 d
Ϫ40
Ϫ80
Ϫ120 T (d ) ϭ 35d ϩ 20
Ϫ160
Ϫ4 Ϫ3Ϫ2
57.
58. Yes; the graph passes the vertical line test.
c
350
300
250
200
150
100
50
0
1.75b ϩ 1.5c ϭ 525
100
200
400b
28
Algebra 2
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21
2
59. no
60.
61. A linear equation can be used to relate the amounts of time that a student spends on each of two subjects if the total amount of time is fixed. Answers should include the following.
• x and y must be nonnegative because
Lolita cannot spend a negative amount of time studying a subject.
• The intercepts represent
Lolita spending all of her time on one subject. The x-intercept represents her spending all of her time on math, and the y-intercept represents her spending all of her time on chemistry.
63. B
units2
62. B
64. D ϭ {Ϫ1, 1, 2, 4},
R ϭ {Ϫ4, 3, 5}; yes y (Ϫ1, 5) (
1, 3)
(4, 3)
x
O
(2, Ϫ4)
66. 5x 0 Ϫ1 Ͻ x Ͻ 26
65. D ϭ {0, 1, 2},
R ϭ {Ϫ1, 0, 2, 3}; no y (0, 2)
(1, 3)
(1, 0)
O
x
(2, Ϫ1)
29
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67. 5x 0 x Ͻ Ϫ6 or x Ͼ Ϫ26
68. \$7.95
69. 3s ϩ 14
70. 4
1
4
1
3
72. Ϫ
73. 2
74. Ϫ
75. Ϫ5
76.
77. 0.4
78. Ϫ0.8
71.
3
2
4
15
Lesson 2-3 Slope
Pages 71–74
1. Sample answer: y ϭ 1
2. Sometimes; the slope of a vertical line is undefined.
3. Luisa; Mark did not subtract in a consistent manner when using the slope formula. If y2 ϭ 5 and y1 ϭ 4, then x2 must be Ϫ1 and x1 must be
2, not vice versa.
4. 0
1
2
6. 1
5. Ϫ
7.
O
8.
y
x
y
O
30
x
Algebra 2
Chapter 2
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9.
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10.
y
y
x
O
O
11.
x
12. 5.5Њ/hr
y
O
x
13. 1.25Њ/hr
14. 2:00 P.M.–4:00 P.M.
5
2
16. 13
15. Ϫ
17.
3
5
18. 4
19. 0
20. Ϫ1
21. 8
22. undefined
23. Ϫ4
24. Ϫ
25. undefined
26. 0
27. 1
28. 9
31.
32.
5
4
y
y
O
O
x
x
31
Algebra 2
Chapter 2
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33.
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34.
y
y
x
O
O
35.
36.
y
x
O
x
y
O
x
37. about 68 million per year
38. about Ϫ32 million per year
39. The number of cassette tapes shipped has been decreasing. 40. 55 mph
41. 45 mph
42. speed or velocity
43.
O
45.
x
y
O
46.
y
O
44.
y
x
y
O
32
x
x
Algebra 2
Chapter 2
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47.
12:22 PM
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48.
y
O
49.
x
y
O
50.
y
x
y
x
O
O
51. Yes; slopes show that adjacent sides are perpendicular. 52. Ϫ1
53. The grade or steepness of a road can be interpreted mathematically as a slope.
Answers should include the following. • Think of the diagram at the beginning of the lesson as being in a coordinate plane. Then the rise is a change in y-coordinates and the horizontal distance is a change in x-coordinates. Thus, the grade is a slope expressed as a percent.
x
54. D
33
Algebra 2
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y
x y ϭ 0.08x
O
55. D
56. The graphs have the same y-intercept. As the slopes increase, the lines get steeper. 57. The graphs have the same y-intercept. As the slopes become more negative, the lines get steeper.
58. Ϫ10, 4
8
Ϫ2x ϩ 5y ϭ 20 6
4
2
Ϫ10Ϫ8 Ϫ6Ϫ4 Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
59. Ϫ2,
8
3
y
O
2 4x
60. 0, 0 y y
O
y ϭ 7x
x
O
4x Ϫ 3y ϩ 8 ϭ 0
61. Ϫ7
62. 5
5
2
63. Ϫ
64. 3a Ϫ4
65. 5x 0 Ϫ1 Ͻ x Ͻ 36
x
66. 5z 0 z Ն 7356
34
Algebra 2
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67. at least 8
68. 17a Ϫb
69. 9
70. y ϭ 9 Ϫ x
71. y ϭ Ϫ4x ϩ 2
72. y ϭ Ϫ3x ϩ 7
5
2
73. y ϭ x Ϫ
1
2
2
3
75. y ϭ Ϫ x ϩ
3
5
74. y ϭ x ϩ
4
5
11
3
Chapter 2
Practice Quiz 1
Page 74
1. D ϭ {Ϫ7, Ϫ3, 0, 2},
R ϭ {Ϫ2, 1, 2, 4, 5}
2. 375
3. 6x ϩ y ϭ 4
4. 10, 6 y 3x ϩ 5y ϭ 30
x
O
5.
y
O
x
35
Algebra 2
Chapter 2
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Lesson 2-4
Page 36 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-02:
Writing Linear Equations
Pages 78–80
1. Sample answer: y ϭ 3x ϩ 2
2. 6, 0
3. Solve the equation for y to get
2
3 y ϭ x Ϫ . The slope of this
4. 2, Ϫ5
5
3
.
5
5
line is The slope of a parallel line is the same.
3
2
6. y ϭ 0.5x ϩ 1
5. Ϫ , 5
5
2
3
4
8. y ϭ Ϫ x ϩ 16
7. y ϭ Ϫ x ϩ 2
3
5
9. y ϭ Ϫ x ϩ
16
5
10. y ϭ Ϫx Ϫ 2
5
4
12. B
11. y ϭ x ϩ 7
2
3
14.
13. Ϫ , Ϫ4
15.
1
,
2
3
,
4
0
3
5
5
2
16. Ϫ , 6
Ϫ
17. undefined, none
18. Ϫc, d
19. y ϭ 0.8x
20. y ϭ Ϫ x ϩ
21. y ϭ Ϫ4
22. y ϭ 2
23. y ϭ 3x Ϫ 6
24. y ϭ 0.25x ϩ 4
1
2
25. y ϭ Ϫ x ϩ
5
3
3
2
7
2
26. y ϭ x ϩ
4
5
17
2
28. y ϭ 4x
27. y ϭ Ϫ0.5x Ϫ 2
29. y ϭ Ϫ x ϩ
29
3
17
5
30. no slope-intercept form for xϭ7 3
2
31. y ϭ 0
32. y ϭ x
33. y ϭ x ϩ 4
34. y ϭ x Ϫ
3
4
36
1
4
Algebra 2
Chapter 2
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3
35. y ϭ x ϩ
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10
3
1
15
37. y ϭ Ϫ x Ϫ
36. y ϭ Ϫ4x ϩ 3
23
5
38. y ϭ Ϫx Ϫ 4
39. y ϭ 3x Ϫ 2
40. y ϭ Ϫ2x ϩ 6
41. d ϭ 180c Ϫ 360
42. 180, Ϫ360
43. 540Њ
44. y ϭ 75x ϩ 6000
45. 10 mi
46. y ϭ x ϩ 32
9
5
80
60
40
Ϫ30
Ϫ10
Ϫ20
Ϫ40
y y ϭ 9 x ϩ 32
5
x
O
10 20 30
47. 68ЊF
48. Ϫ40Њ
49. y ϭ 0.35x ϩ 1.25
50. \$11.75
51. y ϭ 2x ϩ 4
52. A linear equation can sometimes be used to relate a company’s cost to the number they produce of a product. Answers should include the following.
• The y-intercept, 5400, is the cost the company must pay if they produce
0 units, so it is the fixed cost. The slope, 1.37, means that it costs \$1.37 to produce each unit. The variable cost is 1.37x.
• \$6770
53. C x y
55.
5 Ϫ ϭ1
54. A
2
56.
5
57. Ϫ2
5
,
2
Ϫ5
58. 3
37
Algebra 2
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59. 0
60. 0.55 s
62. 5x 0 x Ն Ϫ66
61. л
63. 5r 0 r Ն 66
64. 3
65. 6.5
66. 323.5
67. 5.85
Lesson 2-5 Modeling Real-World Data:
Using Scatter Plots
Pages 83–86
1. d
2. D ϭ {Ϫ1, 1, 2, 4},
R ϭ {0, 2, 3}; Sample answer using (Ϫ1, 0) and
(2, 2): 4
3. Sample answer using
(4, 130.0) and (6, 140.0): y ϭ 5x ϩ 110
4a.
Atmospheric Temperature
Temperature (˚C)
16
14
12
10
8
6
4
2
0
1000
2000 3000
Altitude (ft)
4000
5000
4b. Sample answer using
(2000, 11.0) and (3000, 9.1): y ϭ Ϫ0.0019x ϩ 14.8
4c. Sample answer: 5.3ЊC
5a.
6a.
60
50
40
30
20
10
0
’88 ’90 ’92 ’94 ’96 ’98 ’00
Year
Lives Saved by
Minimum Drinking Age
Lives (thousands)
Households (millions)
Cable Television
80
70
38
25
20
15
10
5
0
’94 ’95 ’96 ’97 ’98 ’99 ’00
Year
Algebra 2
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5b. Sample answer using
(1992, 57) and (1998, 67): y ϭ 1.67x Ϫ 3269.64
87 million
6b. Sample answer using
(1996, 16.5) and (1998, 18.2): y ϭ 0.85x Ϫ 1680.1
6c. Sample answer: 28,400
7a.
8a.
2000–2001
Detroit Red Wings
Bottled Water Consumption
14
12
Gallons
Assists
60
50
40
30
20
10
0
10
8
6
4
2
0
10 20 30 40
Goals
’91
’93
’95
Year
’97
’99
7b. Sample answer using (4, 5) and (32, 37): y ϭ 1.14x ϩ 0.44
8b. Sample answer using (1993,
9.4) and (1996, 12.5): y ϭ 1.03x Ϫ 2043.39
9a.
10. Sample answer using
(1990, 563) and (1995, 739): y ϭ 35.2x Ϫ 69,485
Revenue (\$ millions)
Play Revenue
700
600
500
400
300
200
100
0
1 2 3 4
Seasons Since ’95–’96
9b. Sample answer using
(1, 499) and (3, 588): y ϭ 44.5x ϩ 454.5, where x is the number of seasons since 1995–1996
\$1078 million or \$1.1 billion
11. Sample answer: \$1091
12. The value predicted by the equation is somewhat lower than the one given in the graph. 39
Algebra 2
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13. Sample answer: Using the data for August and
November, a prediction equation for Company 1 is y ϭ Ϫ0.86x ϩ 25.13, where x is the number of months since August. The negative slope suggests that the value of Company 1’s stock is going down. Using the data for October and November, a prediction equation for
Company 2 is y ϭ 0.38x ϩ
31.3, where x is the number of months since August. The positive slope suggests that the value of Company 2’s stock is going up. Since the value of Company 1’s stock appears to be going down, and the value of Company
2’s stock appears to be going up, Della should buy
Company 2.
14. No. Past performance is no guarantee of the future performance of a stock.
Other factors that should be considered include the companies’ earnings data and how much debt they have. 15.
16. Sample answer using
(213, 26) and (298, 23): y ϭ Ϫ0.04x Ϫ 34.52
Precipitation (in.)
World Cities
40
35
30
25
20
15
10
5
0
200
400
600
Elevation (ft)
18. Sample answer: The predicted value differs from the actual value by more than 20%, possibly because no line fits the data very well.
40
Algebra 2
Chapter 2
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19. Sample answer using (1975,
62.5) and (1995, 81.7): 96.1%
20. Sample answer: The predicted percent is almost certainly too high. Since the percent cannot exceed
100%, it cannot continue to increase indefinitely at a linear rate.
21. See students’ work.
22. Data can be used to write a linear equation that approximates the number of
Calories burned per hour in terms of the speed that a person runs. Answers should include the following.
Calories Burned
While Running
Calories
1000
800
600
400
200
0
5 6 7 8
Speed (mph)
9
• Sample answer using
(5, 508) and (8, 858): y ϭ 116.67x Ϫ 75.35
• about 975 calories;
Sample answer: The predicted value differs from the actual value by only about 2%.
23. D
24. A
25. 1988, 1993, 1998; 247,
360.5, 461
26. y ϭ 21.4x Ϫ 42,296.2
27. 354
28. about (1993, 356.17)
29. y ϭ 21.4x Ϫ 42,294.03
31. y ϭ 4x ϩ 6
32. y ϭ Ϫ x Ϫ
33. 3
34. 7
3
7
41
6
7
Algebra 2
Chapter 2
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35.
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29
3
36.
37. 5x 0 x Ͻ Ϫ7 or x Ͼ Ϫ16
38. 3
39. 11
41.
37
3
40. 0
2
3
42. 1.5
Lesson 2-6 Special Functions
Pages 92–95
3. Sample answer: f(x) ϭ 0 x Ϫ 10
2. Ϫ1
5. S
6. D ϭ all reals,
R ϭ all integers
7. D ϭ all reals,
R ϭ all integers
8. D ϭ all reals,
R ϭ all nonnegative reals
1. Sample answer: [[1.9]] ϭ 1
4. A
g (x )
h (x )
g (x ) ϭ ͠2x ͡
O
h (x ) ϭ |x Ϫ 4| x O
42
x
Algebra 2
Chapter 2
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10. D ϭ all reals, R ϭ 5y 0 y Յ 26
9. D ϭ all reals,
R ϭ all nonnegative reals
g (x )
x
O
11. D ϭ all reals, R ϭ all reals
12. step function
13.
14. \$6
0
Time (hr)
15. C
16. A
17. S
18. S
19. A
20. P
21.
5
4
3
2
1
O
22.
y
60
x
180 300
1.00
0.90
0.80
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0
1 2 3 4 5 6 7 8 9
Minutes
C
43
C
Algebra 2
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23. \$1.00
24. D ϭ all reals,
R ϭ all integers f (x )
x
O
f (x ) ϭ ͠x ϩ 3͡
25. D ϭ all reals,
R ϭ all integers
26. D ϭ all reals,
R ϭ all even integers
g (x )
f (x )
g (x ) ϭ ͠x Ϫ 2͡ x O
x
O
f (x ) ϭ 2͠x ͡
27. D ϭ all reals,
R ϭ {3a 0 a is an integer.}
12
9
6
O
Ϫ4 Ϫ3 Ϫ2 Ϫ1
Ϫ3
Ϫ6
Ϫ9
Ϫ12
28. D ϭ all reals,
R ϭ all integers
h (x )
g (x )
h (x ) ϭ Ϫ3͠x ͡ x 1 2 3 4
x
O
g (x ) ϭ ͠x ͡ ϩ 3
29. D ϭ all reals,
R ϭ all integers
30. D ϭ all reals,
R ϭ all nonnegative reals
f (x ) f (x ) ϭ ͠x ͡ Ϫ 1
O
x
44
Algebra 2
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32. D ϭ all reals, R ϭ {y 0 y Ն 3}
31. D ϭ all reals,
R ϭ all nonnegative reals
g (x )
h (x )
h (x ) ϭ |Ϫx |
g (x ) ϭ |x | ϩ 3
x
O
x
O
33. D ϭ all reals, R ϭ {y 0 y Ն Ϫ4}
34. D ϭ all reals,
R ϭ all nonnegative reals
g (x )
h (x )
g (x ) ϭ |x | Ϫ 4 x O
O
h (x ) ϭ |x ϩ 3|
x
36. D ϭ all reals,
R ϭ all nonnegative reals
35. D ϭ all reals,
R ϭ all nonnegative reals
f (x )
f (x )
f (x ) ϭ |x ϩ 2| x O
|
f (x ) ϭ x Ϫ 1
4
|
O
x
38. D ϭ all reals, R ϭ {y 0 y Ն Ϫ3}
37. D ϭ all reals,
R ϭ all nonnegative reals
f (x )
f (x )
O
|
f (x ) ϭ x ϩ 1
2
|
O
x
x
45
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39. D ϭ {x 0 x Ͻ Ϫ2 or x Ͼ 2},
R ϭ {Ϫ1, 1}
40. D ϭ all reals,
R ϭ {y 0 y Յ 0 or y ϭ 2} f (x )
h (x )
O
x
x
O
41. D ϭ all reals, R ϭ {y 0 y Ͻ 2}
42. D ϭ all reals,
R ϭ all nonnegative whole numbers g (x )
f (x )
O
x
O
x
f (x ) ϭ ͠|x |͡
43. D ϭ all reals,
R ϭ all nonnegative whole numbers 44.
g (x )
O
2 if x Ͻ Ϫ1 f(x) ϭ • 2x if Ϫ1 Յ x Յ 1
Ϫx if x Ͼ 1
x
g (x ) ϭ |͠x ͡|
45. f (x) ϭ 0x Ϫ 2 0
46. {x 0 x Ն 0}
47.
48.
46
f (x) ϭ e
0 if 0 Յ x Յ 300
0.8 (x Ϫ 300) if x Ͼ 300
Algebra 2
Chapter 2
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49.
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50. A step function can be used to model the cost of a letter in terms of its weight.
Answers should include the following. • Since the cost of a letter must be one of the values
\$0.34, \$0.55, \$0.76, \$0.97, and so on, a step function is the best model for the cost of mailing a letter. The gas mileage of a car can be any real number in an interval of real numbers, so it cannot be modeled by a step function. In other words, gas mileage is a continuous function of time.
y
|x | ϩ |y | ϭ 3
x
Cost (\$)
O
2.10
1.80
1.50
1.20
0.90
0.60
0.30
0
1 2 3 4 5 6 7
Weight (oz)
51. B
52. D
53. Life Expectancy
54. Sample answer using
(10, 69.7) and (47, 76.5): y ϭ 0.18x ϩ 67.9
78
76
74
72
70
Years Since 1950
55. Sample answer: 78.7 yr
56. y ϭ 3x ϩ 10
47
Algebra 2
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58. {x 0 x Ն 3}
57. y ϭ x Ϫ 2
Ϫ1 0 1 2 3 4 5 6
59. e y ` y Ͼ f
5
6
60. yes
Ϫ3 Ϫ2 Ϫ1 0 1 2 3
61. no
62. no
63. yes
64. no
65. yes
Chapter 2
Practice Quiz 2
Page 95
2
3
1. y ϭ Ϫ x ϩ
11
3
2. Houston Comets
250
200
150
100
50
0
65 70 75 80
Height (in.)
4. Sample answer: 168 Ib
3. Sample answer using
(66, 138) and (74, 178): y ϭ 5x Ϫ 192
5. D ϭ all reals,
R ϭ nonnegative reals f (x )
f (x ) ϭ |x Ϫ 1|
O
x
48
Algebra 2
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Lesson 2-7
Graphing Inequalities
Pages 98–99
2. Substitute the coordinates of a point not on the boundary into the inequality. If the inequality is satisfied, shade the region containing the point. If the inequality is not satisfied, shade the region that does not contain the point. 1. y Յ Ϫ3x ϩ 4
3. Sample answer: y Ն 0 x 0
4.
y
y ϭ2 x O
6.
5.
y
O
x
x Ϫy ϭ0
7.
8.
y
y
y ϭ |2x |
x
O
O
x
x Ϫ 2y ϭ 5
49
Algebra 2
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9.
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10. 10c ϩ 13d Յ 40
y
x
O
11.
y ϭ 3|x | Ϫ 1
12. No; (3, 2) is not in the shaded region.
d
10c ϩ 13d ϭ 40
c
O
13.
14.
y
y
3 ϭ x Ϫ 3y
x
O
x
O
x ϩ y ϭ Ϫ5
15.
16.
y y ϭ 6x Ϫ 2 x O
17.
18.
y
y
y ϭ Ϫ4x ϩ 3
O
x
O
x
y Ϫ 2 ϭ 3x
50
Algebra 2
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20.
y
y
y ϭ1 y ϩ1ϭ4 x O
21.
x
O
22.
y
y
4x Ϫ 5y Ϫ 10 ϭ 0
x
O
x
O
x Ϫ 6y ϩ 3 ϭ 0
23.
24.
y
y
y ϭ 1x ϩ 5
3
O
y ϭ 1x Ϫ 5
x
2
O
25.
x
26.
y
y
y ϭ |4x | y ϭ |x | O
x
O
51
x
Algebra 2
Chapter 2
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28.
y
y y ϭ |x Ϫ 1| Ϫ 2
x
O
O
x
y ϩ |x | ϭ 3
29.
30.
y
y y ϭ |x |
x ϩy ϭ1
O
x
x
O
x ϩ y ϭ Ϫ1 y ϭ Ϫ|x |
31. x Ͻ Ϫ2
32. y Ͻ 3x Ϫ 5 y y x ϭ Ϫ2
O
O
x
x y ϭ 3x Ϫ 5
33.
34. yes
y
350
250
0.4x ϩ 0.6y ϭ 90
150
50
O
50
150 250 350 x
52
Algebra 2
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36.
35. 4a ϩ 3s Ն 2000
800
s
600
4a ϩ 3s ϭ 2000
400
200
a
200 400 600 800
O
38. 1.2a ϩ1.8b Ն 9000
37. yes
b
6000
1.2a ϩ 1.8b ϭ 9000
4000
2000
O
a
2000 4000 6000 8000
40.
39. yes
y
|y | ϭ x
O
42. A
41. Linear inequalities can be used to track the performance of players in fantasy football leagues.
x
53
Algebra 2
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• Let x be the number of receiving yards and let y be the number of touchdowns. The number of points Dana gets from receiving yards is 5x and the number of points he gets from touchdowns is
100y. His total number of points is 5x ϩ 100y. He wants at least 1000 points, so the inequality 5x ϩ
100y Ն 1000 represents the situation.
y
12
5x ϩ 100y ϭ 1000
10
8
6
4
2
O
Ϫ50
100 200
300 x
• the first one
43. B
44.
[Ϫ10, 10] scl: 1 by [Ϫ10, 10] scl: 1
45.
46.
[Ϫ10, 10] scl: 1 by [Ϫ10, 10] scl: 1
[Ϫ10, 10] scl: 1 by [Ϫ10, 10] scl: 1
54
Algebra 2
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47.
48. D ϭ all reals,
R ϭ all integers f (x )
x
O
[Ϫ10, 10] scl: 1 by [Ϫ10, 10] scl: 1
f (x ) ϭ ͠x ͡ Ϫ 4
49. D ϭ all reals, R ϭ {y 0 y Ն Ϫ1}
50. D ϭ all reals,
R ϭ all nonnegative reals
g (x )
h (x )
g (x ) ϭ |x | Ϫ 1
O
x
O
51.
x
52. Sample answer using
(4, 6000) and (6, 8000): y ϭ 1000x ϩ 2000
Sales vs. Experience
10,000
Sales (\$)
h (x ) ϭ |x Ϫ 3|
8000
6000
4000
2000
0
1
2
3 4
Years
5
6
7
53. Sample answer: \$10,000
54. 8
55. 3
56.
55
1
2
Algebra 2
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Chapter 3 Systems of Equations and Inequalities
Lesson 3-1 Solving Systems of Equations by Graphing
Pages 112–115
1. Two lines cannot intersect in exactly two points.
2. Sample answer: x ϩ y ϭ 4, x Ϫ y ϭ 2
3. A graph is used to estimate the solution. To determine that the point lies on both lines, you must check that it satisfies both equations.
4.
y
(Ϫ2, 5)
y ϭ Ϫx ϩ 3
y ϭ 2x ϩ 9
x
O
5.
6.
y
3x ϩ 2y ϭ 10
4x Ϫ 2y ϭ 22
O
(2, 2)
2x ϩ 3y ϭ 10
y
x
x
(4, Ϫ3)
6x ϩ 9y ϭ Ϫ3
O
8. inconsistent
7. consistent and independent y y
2x ϩ 4y ϭ 8
yϭxϩ4 x yϭ6Ϫx
O
O
x ϩ 2y ϭ 2
x
56
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9. consistent and dependent
10. y ϭ 0.08x ϩ 3.2, y ϭ 0.10x ϩ 2.6
y
x
O
x Ϫ 2y ϭ 8
1
xϪyϭ4
2
11. The cost is \$5.60 for both stores to develop 30 prints.
12. You should use Specialty
Photos if you are developing less than 30 prints, and you should use The Photo Lab if you are developing more than
30 prints.
13.
14.
y
y x O
y ϭ Ϫ3x ϩ 1
y ϭ 2x Ϫ 4
15.
y ϭ 3x Ϫ 8
x
O
(1, Ϫ2)
(0, Ϫ8)
16.
y
y
2x ϩ 3y ϭ 12 (3, 2)
x ϩ 2y ϭ 6
(4, 1)
O
y
x
O
x
2x ϩ y ϭ 9
17.
yϭxϪ8
2x Ϫ y ϭ 4
18.
x ϩ 2y ϭ 11
y
(7, 6)
7x Ϫ 1 ϭ 8y
(5, 3)
3x Ϫ 7y ϭ Ϫ6
O
x
O
x
5x Ϫ 11 ϭ 4y
57
Algebra 2
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20.
y
x
(1.5, 5)
4x Ϫ 2y ϭ Ϫ4
(3.5, 0)
2x ϩ 3y ϭ 7
y
O
x
O
8x Ϫ 3y ϭ Ϫ3
2x Ϫ 3y ϭ 7
21.
22.
y
1 x ϩ 2y ϭ 5
4
y yϪ 1xϭ6
3
(4, 2)
x
O
(Ϫ9, 3)
2x Ϫ y ϭ 6
O x
2
x ϩ y ϭ Ϫ3
3
23.
24.
y
1
xϪyϭ0
2
y
4 xϩ 1yϭ3
3
5
x
O
x
O
(Ϫ4, Ϫ2)
25. inconsistent yϭxϩ4 (3, Ϫ5)
2 xϪ 3yϭ5
3
5
1 x ϩ 1 y ϭ Ϫ2
4
2
26. consistent and independent
y
y yϭxϩ3 x
O
x
O
yϭxϪ4
y ϭ 2x ϩ 6
58
Algebra 2
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27. consistent and independent
28. consistent and dependent
y
y
3x ϩ y ϭ 3
xϩyϭ4
x
O
x
O
Ϫ4x ϩ y ϭ 9
6x ϩ 2y ϭ 6
29. inconsistent
30. consistent and dependent y y yϪxϭ5 4x Ϫ 2y ϭ 6
x
O
x
O
6x Ϫ 3y ϭ 9
2y Ϫ 2x ϭ 8
31. consistent and independent
32. inconsistent
y
y
2y ϭ x
2y ϭ 5 Ϫ x
x
O
x
8y ϭ 2x ϩ 1
O
6y ϭ 7 Ϫ 3x
33. consistent and independent
34. consistent and dependent
y
y
1.6y ϭ 0.4x ϩ 1
0.8x Ϫ 1.5y ϭ Ϫ10
O
1.2x ϩ 2.5y ϭ 4
O
x
0.4y ϭ 0.1x ϩ 0.25
x
59
Algebra 2
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35. inconsistent yϪ 1xϭ2
3
36. consistent and independent
y
y
4
x Ϫ y ϭ Ϫ2
3
O
x
x
O
3y Ϫ x ϭ Ϫ2
2y Ϫ 4x ϭ 3
37. (Ϫ3, 1)
38. (1, 3), (2, Ϫ1), (Ϫ2, Ϫ3) y y Ϫ 2x ϭ 1
4x ϩ y ϭ 7
x
O
2y Ϫ x ϭ Ϫ4
40. (120, 80)
39. y ϭ 52 ϩ 0.23x, y ϭ 80
120
Cost (\$)
y ϭ 80
80
y ϭ 52 ϩ 0.23x
40
0
40
80
120
Miles
160
41. Deluxe Plan
42. Supply, 200,000; demand,
300,000; prices will tend to rise. 43. Supply, 300,000; demand,
200,000; prices will tend to fall. 44. 250,000; \$10.00
60
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46. 2015
Population (Thousands)
45. y ϭ 304x ϩ 15,982, y ϭ 98.6x ϩ18,976
24,000
y ϭ 98.6x ϩ 18,976
20,000
16,000
y ϭ 304x ϩ 15,982
12,000
0
a b a
48b.
b a 48c. b 4
d e d e d
,
e
c b 8
12
16
Years After 1999
20
f e 47. FL will probably be ranked third by 2020. The graphs intersect in the year 2015, so
NY will still have a higher population in 2010, but FL will have a higher population in 2020.
48a. ϭ ,
49. You can use a system of equations to track sales and make predictions about future growth based on past performance and trends in the graphs. Answers should include the following.
• The coordinates (6, 54) represent that 6 years after 1999 both the instore sales and online sales will be \$54,000.
• The in-store sales and the online sales will never be equal and in-store sales will continue to be higher than online sales.
50. A
51. C
52. (3.40, Ϫ2.58)
53. (Ϫ5.56, Ϫ12)
54. (4, 3.42)
55. no solution
56. (Ϫ9, 3.75)
61
ϭ
ϭ
c b f e Algebra 2
Chapter 3
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58.
57. (2.64, 42.43)
y y ϭ 5 ϩ 3x
O
59.
60.
y
x
y
2x ϩ y ϭ Ϫ4
x
O
O
x
2y Ϫ 1 ϭ x
61. A
62. C
63. S
64. {Ϫ13, 13}
65. {Ϫ15, 9}
66. л
67. {Ϫ2, 3}
68. eϪ5, f
69. {9}
70. 8 ϩ 2n
71. x 2 Ϫ 6
72. 41a ϩ 52
73.
z
3
7
2
74. x ϩ 2
ϩ1
75. 9y ϩ 1
76. Ϫ3x ϩ 6y
77. 12x ϩ 18y Ϫ 6
78. 15x ϩ 10y ϩ 10
79. x ϩ 4y
Lesson 3-2 Solving Systems of Equations Algebraically
Pages 119–122
1. See students’ work; one equation should have a variable with a coefficient of 1.
2. There are infinitely many solutions. 62
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3. Vincent; Juanita subtracted the two equations incorrectly;
Ϫy Ϫ y ϭϪ2y, not 0.
4. (4, 8)
5. (1, 3)
6. (4, Ϫ1)
7. (5, 2)
8. (9, 7)
10. no solution
9. (6, Ϫ20)
1
3
2
3
11. a3 , 2 b
12. C
13. (9, 5)
14. (2, 7)
15. (3, Ϫ2)
16. (Ϫ6, 8)
17. no solution
18. (1, 1)
19. (4, 3)
20. (Ϫ1, 8)
21. (2, 0)
22. (3, Ϫ1)
23. (10, Ϫ1)
24. (Ϫ7, 9)
25. (4, Ϫ3)
26. (6, 5)
27. (Ϫ8, Ϫ3)
28. (7, Ϫ1)
29. no solution
30. (Ϫ5, 8)
1
2
31. aϪ ,
3 b 2
1
3
32. a , 2b
33. (Ϫ6, 11)
34. infinitely many
35. (1.5, 0.5)
36. (2, 4)
37. 8, 6
38. 2, 12
39. x ϩ y ϭ 28,
16x ϩ 19y ϭ 478
40. 18 members rented skis and 10 members rented snowboards. 41. 4 2-bedroom, 2 3-bedroom
42. (Ϫ5, Ϫ2), (4, 4),
(Ϫ2, Ϫ8), (1, 10)
43. x ϩ y ϭ 30,
700x ϩ 200y ϭ 15,000
44. 18 printers, 12 monitors
45. 2x ϩ 4y ϭ 100, y ϭ 2x
46. 10 true/false, 20 multiplechoice
47. Yes; they should finish the test within 40 minutes.
48. a ϩ s ϭ 40, 11a ϩ 4s ϭ 335
63
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49. 25 min of step aerobics,
15 min of stretching
50. (4, 6)
51. You can use a system of equations to find the monthly fee and rate per minute charged during the months of January and February.
Answers should include the following. • The coordinates of the point of intersection are
(0.08, 3.5).
• Currently, Yolanda is paying a monthly fee of
\$3.50 and an additional
8¢ per minute. If she graphs y ϭ 0.08x ϩ 3.5 (to represent what she is paying currently) and y ϭ 0.10x ϩ 3 (to represent the other long-distance plan) and finds the intersection, she can identify which plan would be better for a person with her level of usage.
52. C
53. A
54. inconsistent y yϭxϩ2 x O
yϭxϪ1
64
Algebra 2
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56. consistent and independent
55. consistent and dependent
y
y
4y Ϫ 2x ϭ 4
3x ϩ y ϭ 1
O
x
O
yϪ 1xϭ1
2
57.
x
y ϭ 2x Ϫ 4
58.
y
y
x
O
x
O
xϩyϭ3
5y Ϫ 4x ϭ Ϫ20
59.
60. 7x Ϫ y ϭ Ϫ4; 7, Ϫ1, Ϫ4
y
3x ϩ 9y ϭ Ϫ15
O
x
61. x Ϫ y ϭ 0; 1, Ϫ1, 0
62. 3x ϩ 5y ϭ 2; 3, 5, 2
63. 2x Ϫ y ϭ Ϫ3; 2, Ϫ1, Ϫ3
64. x Ϫ 2y ϭ 6; 1, Ϫ2, 6
65. 3x ϩ 2y ϭ 21; 3, 2, 21
66. 0.6 ampere
67. yes
68. no
69. no
70. yes
65
Algebra 2
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Chapter 3
Practice Quiz 1
Page 122
1.
2.
y
y
y ϭ 3x ϩ 10
(Ϫ1, 7)
(3, 2)
2x ϩ 3y ϭ 12
x
O
y ϭ Ϫx ϩ 6
2x Ϫ y ϭ 4
x
O
4. (4, Ϫ1)
3. (2, 7)
5. Hartsfield, 78 million; O’Hare,
72.5 million
Lesson 3-3
Solving Systems of Inequalities by Graphing
Pages 125–127
1. Sample answer: y Ͼ x ϩ 3, y Ͻ x Ϫ 2
2. true
3a. 4
3b. 2
3c. 1
3d. 3
4.
y
yϭ2 x O
xϭ4
5.
6.
y
y xϩyϭ2 y ϭ Ϫ2x ϩ 4
O
x ϭ Ϫ1 yϭxϪ2 ©Glencoe/McGraw-Hill
x
O
x
66
xϭ3
Algebra 2
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8. (Ϫ3, Ϫ3), (2, 2), (5, Ϫ3)
y y ϭ 2x ϩ 1 xϭ1 O
x
x ϩ 2y ϭ Ϫ3
9. (Ϫ4, 3), (1, Ϫ2), (2, 9), (7, 4)
10.
10 m
bϭ2
Muffins
8
6
2.5b ϩ 3.5m ϭ 28
4
2
mϭ3 b 0
11. Sample answer: 3 packages of bagels, 4 packages of muffins; 4 packages of bagels, 4 packages of muffins; 3 packages of bagels, 5 packages of muffins 12.
13.
2
14.
4
6
8
Bagels
12
y
yϭ3 x O
xϭ2
y
y
x
O
10
yϭ2Ϫx x x ϭ Ϫ1
O
y ϭ Ϫ4
yϭxϩ4
67
Algebra 2
Chapter 3
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15.
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16.
y
y
yϭ2
3x ϩ 2y ϭ 6
y ϭ Ϫ2 x
O
x
O
4x Ϫ y ϭ 2
yϭxϪ3
17.
18.
y
4x Ϫ 3y ϭ 7
yϭ 1xϩ1
2
x
O
y
2y Ϫ x ϭ Ϫ6
x
O
y ϭ 2x Ϫ 3
19. no solution
20.
y yϭ1 xϭ3 x O
x ϭ Ϫ3
21.
y ϭ Ϫ1
22. no solution
y xϭ2 x ϩ 3y ϭ 6
O
x
x ϭ Ϫ4
23.
24. (0, 0), (0, 4), (8, 0)
y
2x Ϫ y ϭ 4
2x ϩ 4y ϭ Ϫ7
O
x x Ϫ 3y ϭ 2
68
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25. (Ϫ3, Ϫ4), (5, Ϫ4), (1, 4)
26. (0, 4), (3, 0), (3, 5)
27. (Ϫ6, Ϫ9), (2, 7), (10, Ϫ1)
28. (Ϫ11, Ϫ3), (Ϫ1, Ϫ3),
1
2
(6, 4), a6, 5 b
29. (Ϫ4, 3), (Ϫ2, 7),
1
3
30. 16 units2
1
3
(4, Ϫ1), a7 , 2 b
31. 64 units2
32.
Hours Raking Leaves
16
14
12
x ϩ y ϭ 15
10
8
6
4
2
0
10x ϩ 12y ϭ 120
2
x
4 6 8 10 12 14 16
Hours Cutting Grass
34. category 4; 13-18 ft
33. s Ն 111, s Յ 130, h Ն 9, h Յ 12
16
y
h
Storm Surge (ft)
14
h ϭ 12
12
10
s ϭ 130
hϭ9
8
s ϭ 111 s 0
80
100
120
140
Wind Speed (mph)
160
69
Algebra 2
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35.
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y
14
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36. Sample answer: 2 pumpkin,
8 soda; 4 pumpkin, 6 soda;
8 pumpkin, 4 soda
2x ϩ 1.5y ϭ 24
12
10
x ϩ 2.5y ϭ 26
8
6
4
2
0
2
4
6 8 10 12 14
Pumpkin
x
37. 6 pumpkin, 8 soda
38. 42 units2
39. The range for normal blood pressure satisfies four inequalities that can be graphed to find their intersection. Answers should include the following.
• Graph the blood pressure as an ordered pair; if the point lies in the shaded region, it is in the normal range. • High systolic pressure is represented by the region to the right of x ϭ 140 and high diastolic pressure is represented by the region above y ϭ 90.
40. B
41. Sample answer: y Յ 6, y Ն 2, x Յ 5, x Ն 1
42. (Ϫ3, 8)
43. (6, 5)
44. (8, Ϫ5)
45.
46. infinitely many
y
y
2x ϩ y ϭ Ϫ3
y ϭ 2x ϩ 1
(Ϫ2, Ϫ3)
O
x
O
yϭ Ϫ1xϪ4
2
x
6x ϩ 3y ϭ Ϫ9
70
Algebra 2
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1
2
48. y ϭ x ϩ 6
y
Ϫx ϩ 8y ϭ 12
(4, 2)
x
O
2x Ϫ y ϭ 6
49. Ϫ5
50. Ϫ12
51. 8
52. 27
53. 5
54. Ϫ8.25
Lesson 3-4 Linear Programming
Pages 132–135
1. sometimes
2. Sample answer: y Ն Ϫx, y Ն x Ϫ 5, y Յ 0
3.
4.
y
y
(1, 4)
(5, 2)
(Ϫ3, 1)
(1, 2)
O
O
x
x
vertices: (Ϫ3, 1), Q , 1R;
vertices: (1, 2), (1, 4), (5, 2); max: f (5, 2) ϭ 4, min: f (1, 4) ϭ Ϫ10
( 5 , 1)
3
5
3
min: f (Ϫ3, 1) ϭ Ϫ17; no maximum
71
Algebra 2
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6.
y
y
(7, 8.5)
(1, 3)
(2, 6)
(6, 3)
(0, 1)
O
(10, 1)
x x O (2, 0)
vertices: (0, 1), (1, 3), (6, 3),
(10, 1); max: f (10, 1) ϭ 31, min: f (0, 1) ϭ 1
(7, Ϫ5)
vertices: (2, 0), (2, 6), (7, 8.5),
(7, Ϫ5); max: f(7, 8.5) ϭ 81.5, min: f(2, 0) ϭ 16
7.
8.
y
y
(Ϫ2, 4)
(Ϫ1, 2)
(2, 3)
(4, 1)
x
O
x
O
(Ϫ3, Ϫ1)
(Ϫ2, Ϫ3)
(3, Ϫ2)
(2, Ϫ3)
vertices: (Ϫ3, Ϫ1), (Ϫ1, 2),
(2, 3), (3,Ϫ2); max: f(3, Ϫ2) ϭ
5, min: f(Ϫ1, 2) ϭ Ϫ3
vertices: (Ϫ2, 4), (Ϫ2, Ϫ3),
(2, Ϫ3), (4, 1); max: f(2, Ϫ3) ϭ
5; min: f (Ϫ2, 4) ϭ Ϫ6
10.
28
Leather Tote Bags
9. c Ն 0, l Ն 0, c ϩ 3l Յ 56,
4c ϩ 2l Յ 104
24
20
16
(20, 12)
12
8
4
0
2
3
(0, 18 2 )
3
(0, 0)
4
(26, 0)
8
12 16 20 24
Canvas Tote Bags
11. (0, 0), (26, 0), (20, 12), a0, 18 b
12. f(c, l) ϭ 20c ϩ 35l
13. 20 canvas tote bags and
12 leather tote bags
28 c
14. \$820
72
Algebra 2
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16.
y
y
(3, 5)
(6, 13)
x
O
(0, Ϫ4)
(3, Ϫ4)
(0, 1)
vertices: (0, Ϫ4), (3, 5),
(3, Ϫ4); max: f (3, Ϫ4) ϭ 7, min: f (3, 5) ϭ Ϫ2
(6, 1)
x
O
vertices: (0, 1), (6, 1), (6, 13); max: f (6, 13) ϭ 19; min: f (0, 1) ϭ 1
17.
18.
y
y
(5, 8)
(2, 3)
(2, 1)
(1, 4)
(4, 4)
(4, 1)
x
O
(1, 2)
(5, 2)
x
O
vertices: (2, 1), (2, 3), (4, 4),
(4, 1); max: f (4, 4) ϭ 16; min: f (2, 1) ϭ 5
vertices: (1, 4), (5, 8),
(5, 2), (1, 2); max: f (5, 2) ϭ
11, min: f (1, 4) ϭ Ϫ5
19.
y
20.
(3, 5)
y
12
8
(Ϫ3, Ϫ1)
O
4
x
Ϫ4
O
Ϫ4
vertices: (Ϫ3, Ϫ1), (3, 5); min: f (Ϫ3, Ϫ1) ϭ Ϫ9; no maximum
(6, 12)
(2, 8)
(2, 2)
4
8
x
(6, Ϫ6)
Ϫ8
vertices: (2, 2), (2, 8), (6, 12),
(6, Ϫ6); max: f(6, 12) ϭ 30, min: f(6, Ϫ6) ϭ Ϫ24
73
Algebra 2
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22.
y
(0, 2)
O
(0, 0)
y
(0, 7)
(4, 3)
(2, 1)
x
(3, 0)
(0, 0)
O
vertices: (0, 0), (0, 2), (2, 1),
(3, 0); max: f (0, 2) ϭ 6; min: f (3, 0) ϭ Ϫ12
23.
(2, 0) x
vertices: (0, 0), (0, 7), (4, 3),
(2, 0); max: f (4, 3) ϭ 14; min: f (0, 7) ϭ Ϫ14
24.
y
y
(8, 6)
(0, 4)
(3, 0)
x
O
(4, 0)
(0, Ϫ3)
vertices: (0, 4), (4, 0), (8, 6); max: f (4, 0) ϭ 4; min: f (0, 4) ϭ Ϫ8
vertices: (3, 0), (0, Ϫ3); min: f (0, Ϫ3) ϭ Ϫ12; no maximum
25.
26.
y
(0, 2)
y
(0, 2)
(4, 3)
(4, 3)
(2, 0)
x
O
x
O
O
x
(7 ,Ϫ 1)
3
3 vertices: (0, 2), (4, 3), (2, 0); max: f (4, 3) ϭ 13, min: f (2, 0) ϭ 2
vertices: (0, 2), (4, 3),
7 1 a , Ϫ b; max: f (4, 3) ϭ 25,
3
3
min: f (0, 2) ϭ 6
74
Algebra 2
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28.
y
y
(2, 5)
(2, 2)
(3, 0)
(4, 1)
(0, 1)
x
x
O
(0, 0) O
vertices: (0, 0), (0, 1), (2, 2),
(4, 1), (5, 0); max: f (5, 0) ϭ
15, min: f (0, 1) ϭ Ϫ5
vertices: (2, 5), (3, 0); min: f (3, 0) ϭ 3, no maximum
29.
y
30a. Sample answer: f (x, y) ϭ Ϫ2x Ϫy
30b. Sample answer: f (x, y) ϭ 3y Ϫ 2x
30c. Sample answer: f (x, y) ϭ x ϩ y
30d. Sample answer: f (x, y) ϭ Ϫx Ϫ 3y
30e. Sample answer: f (x, y) ϭ x ϩ 2y
(4, 4)
(2, 3)
(2, 1)
O
(5, 3)
(4, 1)
(5, 0)
x
vertices: (2, 1), (2, 3), (4, 1),
(4, 4), (5, 3); max: f(4, 1) ϭ 0, min: f (4, 4) ϭ Ϫ12
31. g Ն 0, c Ն 0, 1.5g ϩ 2c Յ
85, g ϩ 0.5c Յ 40
Graphing Calculators
32.
50
g
40
30
20
10
0
(0, 42.5)
(0, 0)
(30, 20)
(40, 0)
10 20 30 40
CAS Calculators
c
50
33. (0, 0), (0, 42.5), (30, 20),
(40, 0)
34. f (g, c) ϭ 50g ϩ 65c
35. 30 graphing calculators,
20 CAS calculators
36. \$2800
37. See students’ work.
38. c Ն 0, s Ն 0, c ϩ s Յ 4500,
4c ϩ 5s Յ 20,000
75
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39. (0, 0), (0, 4000), (2500, 2000),
(4000, 0)
40. 2500 acres of corn,
2000 acres soybeans;
\$125,000
S
4000
3000
2000
(0, 4000)
(2500, 2000)
1000
(0, 0)
0
(4500, 0)
2000
4000
c
41. 4000 acres corn, 0 acres soybeans; \$130,500
42. 3 chocolate chip, 9 peanut butter 43. There are many variables in scheduling tasks. Linear programming can help make sure that all the requirements are met. Answers should include the following.
• Let x ϭ the number of buoy replacements and let y ϭ the number of buoy repairs. Then, x Ն 0, y Ն 0, x Յ 8 and x ϩ 2.5y Յ 24.
The captain would want to maximize the number of buoys that a crew could repair and replace, so f (x, y) ϭ x ϩ y.
• Graph the inequalities and find the vertices of the intersection of the graphs.
The coordinate (0, 24) maximizes the function. So the crew can service the maximum number of buoys if they replace 0 and repair 24 buoys.
44. A
76
Algebra 2
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46.
45. C
y
2y ϩ x ϭ 4
x
O
yϭxϪ4
48. (Ϫ5, 8)
47. л y 3x Ϫ 2y ϭ Ϫ6
x
O
y ϭ 3x Ϫ 1
2
49. (2, 3)
51. c ϭ average cost each year;
15c ϩ 3479 ϭ 7489
55. Multiplicative Inverse
57. 9
59. 16
61. 8
50. (5, 1)
52. about \$267 per year
54.
56.
58.
60.
Associative Property (ϫ)
Distributive Property
5
Ϫ3
62. Ϫ4
Chapter 3
Practice Quiz 2
Page 135
1.
y
2.
yϪxϭ0
y yϭxϩ3 x
O
y ϭ 3x Ϫ 4
yϩxϭ4
O
77
x
Algebra 2
Chapter 3
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3.
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4.
4x ϩ y ϭ 16
y
(1, 6)
(0, 4)
x ϩ 3y ϭ 15
(0, 0)
x
O
O
(3, 0)
x
vertices: (0, 0), (0, 4), (1, 6),
(3, 0); max: f (1, 6) ϭ 8, min: f (0, 0) ϭ 0
5.
y
(5, 6)
(Ϫ1, 3)
(5, 1)
x
O
(1, Ϫ3)
vertices: (1, Ϫ3), (Ϫ1, 3),
(5, 6), (5, 1); max: f (5, 1) ϭ
17, min: f (Ϫ1, 3) ϭ Ϫ13
Lesson 3-5
Solving Systems of Equations in Three Variables
Pages 142–144
1. You can use elimination or substitution to eliminate one of the variables. Then you can solve two equations in two variables.
2. No; the first two equations do represent the same plane, however they do not intersect the third plane, so there is no solution of this system.
3. Sample answer: x ϩ y ϩ z ϭ 4, 2x Ϫ y ϩ z ϭ Ϫ9, x ϩ 2y Ϫ z ϭ 5; Ϫ3 ϩ 5 ϩ
2 ϭ 4, 2(Ϫ3) Ϫ 5 ϩ 2 ϭ Ϫ9,
Ϫ3 ϩ 2(5) Ϫ 2 ϭ 5
4. (6, 3, Ϫ4)
78
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5. (Ϫ1, Ϫ3, 7)
6. infinitely many
7. (5, 2, Ϫ1)
8. no solution
10. 6c ϩ 3s ϩ r ϭ 42,
9. (4, 0, 8)
1
2
c ϩ s ϩ r ϭ 13 , r ϭ 2s
1
2
12. (3, 4, 7)
11. 4 lb chicken, 3 lb sausage,
6 lb rice
13. (Ϫ2, 1, 5)
14. (2, Ϫ3, 6)
15. (4, 0, Ϫ1)
16. no solution
17. (1, 5, 7)
18. (1, 2, Ϫ1)
19. infinitely many
20. a , , b
1
3
1 3 9
2 2 2
1 1
2 4
21. a , Ϫ , b
22. (8, 3, Ϫ6)
23. (Ϫ5, 9, 4)
24. 3, 12, 5
25. 8, 1, 3
26. 1-\$100, 3-\$50, and
6-\$20 checks
27. enchilada, \$2.50; taco, \$1.95; burrito, \$2.65
28. \$7.80
29. x ϩ y ϩ z ϭ 355, x ϩ 2y ϩ
3z ϭ 646, y ϭ z ϩ 27
30. 88 3-point goals, 115 2-point goals, 152 1-point free throws
4
3
4 2 x 3
1
3
1 x 3
32. You can write a system of three equations in three variables to find the number of each type of medal. Answers should include the following.
• You can substitute b ϩ 6 for g and b Ϫ 8 for s in the equation g ϩ s ϩ b ϭ 97.
This equation is now in terms of b. Once you find b, you can substitute again to find g and s. The U.S.
Olympians won 39 gold medals, 25 silver medals, and 33 bronze medals.
31. a ϭ , b ϭ , c ϭ 3; yϭ ©Glencoe/McGraw-Hill
ϩ
ϩ3
79
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• Another situation involving three variables is winning times of the first, second, and third place finishers of a race.
33. D
34. A
35. 120 units of notebook paper and 80 units of newsprint
36.
y yϭx ϩ2
x
O
y ϭ 7 Ϫ 2x
37.
38.
y
y
3x ϩ y ϭ 3
O
3x ϩ y ϭ 1
O
x
x
2y Ϫ x ϭ Ϫ4
4y Ϫ 2x ϭ 4
39. Sample answer using (7, 15) and (14, 22): y ϭ x ϩ 8
41. x ϩ 3y
42. Ϫ2z ϩ 8
43. 9s ϩ 4t
44. 18a ϩ 16b
80
Algebra 2
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Chapter 4 Matrices
Lesson 4-1 Introduction to Matrices
Pages 156–158 column matrix, B R, 2 ϫ 1;
2. Sample answers: row matrix, [1 2 3], 1 ϫ 3;
1. The matrices must have the same dimensions and each element of one matrix must be equal to the corresponding element of the other matrix.
square matrix, B zero matrix, B
3. Corresponding elements are elements in the same row and column positions.
4. 1 ϫ 5
5. 3 ϫ 4
6. (5, 6)
7. (3, 3)
8.
High
Low
B
1 2
R,
3 4
1
2
0 0
R,
0 0
2 ϫ 2;
2ϫ2
88 88 90 86 85
R
54 54 56 53 52
Fri
9. 2 ϫ 5
10. 2 ϫ 3
11. 3 ϫ 1
12. 4 ϫ 3
13. 3 ϫ 3
14. 2 ϫ 5
15. 3 ϫ 2
Sat Sun Mon Tue
16. (2.5, 1, 3)
17. a3, Ϫ b
1
3
18. (5, 3)
19. (3, Ϫ5, 6)
20. (2, Ϫ5)
21. (4, Ϫ3)
22. (1.5, 3)
23. (14, 15)
24. (Ϫ2, 7)
25. (5, 3, 2)
26.
Child
Senior
81
7.50
C 4.50
5.50
3.75
3.75 S
3.75
Evening Matinee Twilight
5.50
4.50
5.50
Algebra 2
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28.
27. 3 ϫ 3
Cost
Catalina Grill
Oyster Club
Casa di Pasta
Mason’s
Steakhouse
29. Sample answer: Mason’s
Steakhouse; it was given the highest rating possible for service and atmosphere, location was given one of the highest ratings, and it is moderately priced.
30.
31.
32.
**
≥ ***
****
**
Weekday
Double
Suite
60 70 75
B
R
79 89 95
Weekend
Atmosphere Location
*
**
***
****
*
*
***
****
60 79
C 70 89 S
75 95
*
** ¥
***
***
Weekday Weekend
Single
Single Double Suite
Service
1
2
4
G 7
11
16
22
3
5
8
12
17
23
30
6
9
13
18
24
31
39
10
14
19
25
32
40
49
15
20
26
33
41
50
60
21
27
34
42W
51
61
72
33. row 6, column 9
34. Matrices are used to organize information so it can be read and compared more easily. Answers should include the following.
• If you want the least expensive vehicle, the compact SUV has the best price; the large SUV has the most horsepower, towing capacity and cargo space, and the standard
SUV has the best fuel economy. • Sample answer: Matrices are used to report stock prices in the newspaper.
35. B
36. C
37. (7, 5, 4)
38. (7, 3, Ϫ9)
82
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39. aϪ3, 5, Ϫ11b
4 3
40.
y y ϭ Ϫ2x ϩ 15
yϭx ϩ2
yϭ3
x
O
vertices: (1, 3), (6, 3), a , max: f a ,
13 19 b 3 3
ϭ
83
,
3
13 19 b; 3 3
min: f (1, 3)ϭ11
42.
41.
3
ϩ 12
y
vertices: (3, 1), a ,
15
2
vertices: (2, 1), (6, 3); min: f(2, 1) ϭ 1, no maximum
44. step function
6
Cost (\$)
43.
3 17 a , b; 2 2
.
15 5 max: f a , b ϭ 35,
2 2
3 17 min: f a , b ϭ Ϫ1
2 2
5 b, 2
5
4
3
2
1
0
1
2 3 4
Hours
5
83
Algebra 2
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45. \$4.50
46. 2
47. 2
48. 0
49. 20
50. 3
51. Ϫ10
52. 6.2
53. Ϫ18
54. 17
55. Ϫ3
56. 75
57.
3
2
Lesson 4-2
Operations with Matrices
Pages 163–166
1. They must have the same dimensions. 2. Sample answer: [Ϫ3 1],
[3 Ϫ1]
5. B
4. impossible
4 4
3. C 4 4 S
4 4
1 10
R
Ϫ7 5
7. B
9. B
Ϫ22 8
R
3 24
Ϫ21
29
R
12 Ϫ22
16,763
14,620
11. Males ϭ E14,486
9041
5234
16,439
14,545
Females ϭ E12,679
7931
5450
6. B
18 Ϫ3 15 6
R
21
9 Ϫ6 24
8. B
10 6
R
Ϫ1 7
10. B
Ϫ3 30
R
26 11
549,499
477,960
455,305U,
321,416
83,411
1,006,372
883,123
12. E 795,785U
579,002
216,646
456,873
405,163
340,480U
257,586
133,235
84
Algebra 2
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10
14. C Ϫ4 S
5
13. No; many schools offer the same sport for males and females, so those schools would be counted twice.
16. B
15 0
4
R
0 13 Ϫ5
Ϫ4
8 Ϫ2
17. C 6 Ϫ10 Ϫ16 S
Ϫ14 Ϫ12
4
15. impossible
Ϫ13
19. C Ϫ3 S
23
1.5 3
R
21. B
4.5 9
23. C
1
Ϫ52
3
2
10
3
2
1
3
Ϫ2 Ϫ1
25. C 4 Ϫ1 S
Ϫ7 Ϫ4
38
4
27. C 32 Ϫ6 S
18 42
29. D 1
2
4
18. [15 Ϫ29 65 Ϫ2]
1.8 9.08
20. C 3.18 31.04 S
10.41 56.56
22. C
S
13 10
7S
24. C 4
7 Ϫ5
9
1
Ϫ2
2
3
2
Ϫ2
S
0 16
26. C Ϫ8 20 S
28 Ϫ4
Ϫ12 Ϫ13
28. C 3 Ϫ8 S
13
37
5T
120 97 64 75
30. Friday: C 80 59 36 60 S ,
72 84 29 48
2
3
112 87 56 74
Saturday: C 84 65 39 70 S
88 98 43 60
6 Ϫ1
Ϫ4 Ϫ15
85
Algebra 2
Chapter 4
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232 184 120 149
31. C 164 124 75 130 S
160 182 72 108
Ϫ8 Ϫ10 Ϫ8 Ϫ1
32. C 4
6
3 10 S
16
14 14 12
15
41
34. E35U
27
51
245
228
33. E319U
227
117
36. Residents:
Before 6 3.00 4.50
B
R
After 6
2.00 3.50
35. 1996, floods; 1997, floods;
1998, floods; 1999, tornadoes; 2000, lightning
Nonresidents:
Before 6 4.50 6.75
B
R
After 6
3.00 5.25
37. B
1.50 2.25
R
1.00 1.75
Residents
3.00 4.50
B
R
Nonresidents 4.50 6.75
38. Before 6:00:
Residents
2.00 3.50
B
R
Nonresidents 3.00 5.25
After 6:00:
39. B
1.00 1.00
R
1.50 1.50
40.
0.5 0.75
3
1 1.5
6
2B
Rϭ B
R
1
4 0.1
2
8 0.2
42. D
41. You can use matrices to track dietary requirements and add them to find the total each day or each week. Answers should include the following.
566 18 7
• Breakfast ϭ C 482 12 17 S ,
530 10 11
86
Algebra 2
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785 22 19
Lunch ϭ C 622 23 20 S ,
710 26 12
1257 40 26
Dinner ϭ C 987 32 45 S
1380 29 38
• Add the three matrices:
2608 80 52
£ 2091 67 82 § .
2620 65 61
43. A
44. 2 ϫ 2
45. 1 ϫ 4
46. 2 ϫ 4
47. 3 ϫ 3
48. 3 ϫ 2
49. 4 ϫ 3
50. (3, Ϫ4, 0)
52. a , 6, Ϫ b
1
4
51. (5, 3, 7)
53. (2, 5)
1
6
54. (Ϫ3, 1)
55. (6, Ϫ1)
56. 0.30p ϩ 0.15s Յ 6
57.
58. No, it would cost \$6.30.
59. Multiplicative Inverse
60. Associative Prop. (ϩ)
61. Distributive Property
62. Commutative Prop. (ϫ)
87
Algebra 2
Chapter 4
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Lesson 4-3 Multiplying Matrices
Pages 171–174
1 2
7 8
C3 4S ؒ B
R
9 10
5 6
2. Never; the inner dimensions will never be equal.
3. The Right Distributive
Property says that
1A ϩ B2C ϭ AC ϩ BC, but
AC ϩ BC CA ϩ CB since the Commutative Property does not hold for matrix multiplication in most cases.
4. 3 ϫ 2
5. undefined
15 Ϫ5 20
7. B
R
24 Ϫ8 32
6. [19 15]
9. B
24
R
41
8. not possible
2 Ϫ1
Ϫ4 1
3 2 ϭB R ؒ ¢B
RؒB
R≤
3
5
8 0
Ϫ1 2
10. yes
A(BC)
ϭB
ϭB
2 Ϫ1
Ϫ13 Ϫ6
RؒB
R
3
5
24 16
Ϫ50 Ϫ28
R
81
62
(AB)C
2 Ϫ1
Ϫ4 1
3 2 ϭ ¢B
Rؒ B
R≤ؒ B
R
3
5
8 0
Ϫ1 2
ϭB
ϭB
Ϫ16 2
3 2
RؒB
R
28 3
Ϫ1 2
Ϫ50 Ϫ28
R
81
62
88
Algebra 2
Chapter 4
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11. [45
55
13. 4 ϫ 2
15. undefined
350 280
65], C 320 165 S
180 120
12. \$74,525
14. 2 ϫ 2
18. 3 ϫ 5
8 Ϫ11
20. B
R
22
12
16. 1 ϫ 5
17. undefined
22. B
Ϫ39
R
18
19. [6]
1 Ϫ25
2
R
23. B
29
1 Ϫ30
21. not possible
24
16
25. C Ϫ32 Ϫ5 S
Ϫ48 Ϫ11
0 64 Ϫ40
26. C 9 11 Ϫ11 S
Ϫ3 39 Ϫ23
24. not possible
ϭ 3¢B
27. yes
AC ϩ BC
1 Ϫ2
5
1 ϭB Rؒ B
4
3
2 Ϫ4
B
28. yes c (AB)
Ϫ5 2
5
1
Rؒ B
R
4 3
2 Ϫ4
ϭB
ϭB
1
9
Ϫ21 Ϫ13
RϩB
R
26 Ϫ8
26 Ϫ8
Ϫ20 Ϫ4
R
52 Ϫ16
(A ϩ B)C
ϭ ¢B
ϭB
ϭB
ϭ 3B
1 Ϫ2
Ϫ5
R ϩ B
4
3
4
ϭB
Ϫ13 Ϫ4
R
Ϫ8 17
Ϫ39 Ϫ12
R
Ϫ24
51
A(cB)
ϭB
1 Ϫ2
Ϫ5 2
R ؒ ¢3 B
R≤
4
3
4 3
1 Ϫ2 Ϫ15 6
R
RB
12 9
4
3
ϭB
2
5
1
R≤ ؒ B
R
3
2 Ϫ4
Ϫ4 0
5
1
RؒB
R
8 6
2 Ϫ4
Ϫ20
Ϫ4
R
52 Ϫ16
1 Ϫ2
Ϫ5 2
RؒB
R≤
4
3
4 3
ϭB
89
Ϫ39 Ϫ12
R
Ϫ24
51
Algebra 2
Chapter 4
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ϭB
1 Ϫ2
Ϫ5 2
5
1
RؒB
RؒB
R
4
3
4 3
2 Ϫ4
5 1
1 Ϫ2
Ϫ5 2 ϭB R ؒ ¢B
RϩB
R≤
2 Ϫ4
4 3
4 3
29. no
C (A ϩ B)
30. no
ABC
ϭB
5
1
Ϫ4 0 ϭB RؒB
R
2 Ϫ4
8 6
ϭB
Ϫ73
3
R
Ϫ6 Ϫ76
CBA
ϭB
Ϫ12
6
R
Ϫ40 Ϫ24
AC ϩ BC
1 Ϫ2
5
1 ϭB RؒB
4
3
2 Ϫ4
ϭB
5
1
Ϫ5 2
1 Ϫ2
RؒB
RؒB
R
2 Ϫ4
4 3
4
3
ϭB
Ϫ5 2
5
1
B
RؒB
R
4 3
2 Ϫ4
ϭB
1
9
Ϫ21 Ϫ13 ϭB RϩB
R
26 Ϫ8
26 Ϫ8
ϭB
Ϫ20 Ϫ4
R
52 Ϫ16
e f
R
g h
a b
R
c d
72
68
36. D
90
86
and
where bg ϭ cf, a ϭ d,
and e ϭ h
96.50
99.50
37. D
T
118
117
39. \$431
49
63
1.00
T, B
R
56
0.50
62
38. Juniors
40. \$24,900
41. \$26,360
31 81
R
Ϫ58 28
34. \$31,850
35. any two matrices B
B
Ϫ21 13
1 Ϫ2
RؒB
R
Ϫ26 Ϫ8
4
3
22
32. C 25 S
18
290 165 210
31. C 175 240 190 S
110 75
0
14,285
33. C 13,270 S
4295
Ϫ13 Ϫ4
5
1
RؒB
R
Ϫ8 17
2 Ϫ4
42. \$1460
90
Algebra 2
Chapter 4
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43. a ϭ 1, b ϭ 0, c ϭ 0, d ϭ 1; the original matrix
44. Sports statistics are often listed in columns and matrices. In this case, you can find the total number of points scored by multiplying the point matrix, which doesn’t change, by the record matrix, which changes for each season. Answers should include the following.
• P ؒ R ϭ [479]
• Basketball and wrestling use different point values in scoring.
45. B
12 Ϫ6
47. B
R
Ϫ3 21
Ϫ20 2
49. B
R
Ϫ28 12
46. A
48. impossible
50. (7, Ϫ4)
51. (5, Ϫ9)
52. (2, Ϫ5, Ϫ7)
53. \$2.50; \$1.50
54.
55. 8; Ϫ16
56. 2; Ϫ5
91
3
;
2
3
Algebra 2
Chapter 4
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57.
58.
59.
60.
Chapter 4
Practice Quiz 1
Page 174
(5, Ϫ1)
120 80 64 75
4. B
R,
65 105 77 53
1. (6, 3)
2.
3. (1, 3, 5)
5.
7.
9.
B
232 159 120 149
R
134 200 159 103
112 79 56 74
R
69 95 82 50
6. B
Ϫ3 5
R
3 13
B
4 3
R
1 3
8. B
Ϫ10
20 25
R
0 Ϫ20 35
10. B
15 Ϫ8 Ϫ10
R
Ϫ7 23
16
not possible
B
92
Algebra 2
Chapter 4
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Lesson 4-4
1.
Transformation reflection rotation translation p
Shape same same same p
Size same same same 3. Sample answer:
Ϫ4 Ϫ4 Ϫ4
B
R
1
1
1
5. A¿(4, 3), B¿(5, Ϫ6), dilation changes
same
Transformations with Matrices
Pages 178–181
p
2. B
Ϫ3 Ϫ3 Ϫ3
R
Ϫ2 Ϫ2 Ϫ2
Isometry yes 4. B
yes yes no
6.
3
3
3
R
Ϫ1 Ϫ1 Ϫ1
C¿(Ϫ3, Ϫ7)
7. B
0 5 5 0
R
4 4 0 0
8. A¿(0, 12), B¿(15, 12),
C¿(15, 0), D¿(0, 0)
12. B
11. B
14.
13. D¿(Ϫ3, 6), E¿(Ϫ2, Ϫ3),
F¿(Ϫ10, Ϫ4)
15. B
0
1.5 Ϫ2.5
R
2 Ϫ1.5
0
Ϫ4 Ϫ4 Ϫ4
R
2
2
2
10. A¿(0, Ϫ4), B¿(Ϫ5, Ϫ4),
C¿(Ϫ5, 0), D¿(0, 0)
9. A¿(0, Ϫ4), B¿(5, Ϫ4),
C¿(5, 0), D¿(0, 0)
16. A¿(0, 6), B¿(4.5, Ϫ4.5),
C¿(Ϫ7.5, 0)
93
Algebra 2
Chapter 4
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18. B
17.
19. X¿(Ϫ1, 1), Y¿(Ϫ4, 2),
Z¿(Ϫ1, 7)
21. B
23.
2 5 4 1
R
4 4 1 1
F'
25.
20.
22. D¿(4, Ϫ2), E¿(4, Ϫ5),
F¿(1, Ϫ4), G¿(1, Ϫ1)
y D
G
O G'
24. E¿(6, Ϫ2), F¿(8, Ϫ9)
E
F
x
D'
26. B
2
4
2 Ϫ3
R ؒ (Ϫ1) ϭ
3 Ϫ3 Ϫ5 Ϫ2
E'
J(Ϫ5, 3), K(7, 2), L(4, Ϫ1)
1
2
7
R
Ϫ1 Ϫ4 Ϫ1
94
B
Ϫ2 Ϫ4 Ϫ2 3
R
Ϫ3
3
5 2
Algebra 2
Chapter 4
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27.
28. 180Њ rotation
30. B
4 Ϫ4 Ϫ4 4
R
Ϫ4 Ϫ4
4 4
29. B
31. B
4 Ϫ4 Ϫ4 4
R
Ϫ4 Ϫ4
4 4
4 4 Ϫ4 Ϫ4
R
Ϫ4 4
4 Ϫ4
3
35. B R
4
32. The figures in Exercise 29 and Exercise 30 have the same coordinates, but the figure in Exercise 31 has different coordinates.
34. (Ϫ3.75, Ϫ2.625)
33. (Ϫ1.5, Ϫ1.5), (Ϫ4.5, Ϫ1.5),
(Ϫ6, Ϫ3.75), (Ϫ3, Ϫ3.75)
36. (6.5, 6.25)
37. (Ϫ8, 7), (Ϫ7, Ϫ8), and
(8, Ϫ7) by B
1
0
R,
0 Ϫ1
38. The object is reflected over the x-axis, then translated
6 units to the right.
39. Multiply the coordinates
40. No; since the translation does not change the y-coordinate, it does not matter whether you do the translation or the reflection over the x-axis first. However, if the translation did change the y-coordinate, then order would be important.
result to B R.
6
0
41. (17, Ϫ2), (23, 2)
42. There is no single matrix to achieve this. However, you could reflect the object over the y-axis and then translate it 2(3) or 6 units to the right.
95
Algebra 2
Chapter 4
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43. Transformations are used in computer graphics to create special effects. You can simulate the movement of an object, like in space, which you wouldn’t be able to recreate otherwise. Answers should include the following.
• A figure with points (a, b),
(c, d), (e, f ), (g, h), and (i, j) could be written in a 2 ϫ 5 matrix B
a c e g i
R
b d f h j
44. B
and
multiplied on the left by the
2 ϫ 2 rotation matrix.
• The object would get smaller and appear to be moving away from you.
45. A
46. 2 ϫ 2
47. undefined
48. 2 ϫ 5
20
10 Ϫ24
50. C 31 Ϫ46 Ϫ9 S
Ϫ10
3
7
52.
11
24 Ϫ7
8S
49. C 18 Ϫ13
33 Ϫ8 21
51.
D ϭ 53, 4, 56, R ϭ 5Ϫ4, 5, 66; yes ©Glencoe/McGraw-Hill
D ϭ {all real numbers},
R ϭ {all real numbers}; yes
96
Algebra 2
Chapter 4
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54. 0 x 0 Ն 4
53.
D ϭ 5x 0 x Ն 06,
R ϭ 5all real numbers6; no
55. 0 x 0 Ͻ 2.8
56. 0 x ϩ 1 0 Ͼ 2
59. 6
60. 5
61. 28
62.
10
3
64.
5
3
57. 0 x Ϫ 1 0 Ͻ 1
63.
2
3
58. 513 mi
9
4
1. Sample answer: B
2 1
R
8 4
Lesson 4-5 Determinants
Pages 185–188
2. Khalid; the value of the determinant is the difference of the products of the diagonals.
3 1
4 3
B
R, B
R
6 5
1 3
3. It is not a square matrix.
Ϫ2
3 5
6. † 0 Ϫ1 4 † ϭ
9
7 2
5. Cross out the column and row that contains 6. The minor is the remaining 2 ϫ 2 matrix. Ϫ2 `
Ϫ1 4
0 4
0 Ϫ1
` Ϫ 3`
` ϩ 5`
`
7 2
9 2
9
7
ϭ Ϫ2(Ϫ2 Ϫ 28) Ϫ
3(0 Ϫ 36) ϩ 5(0 Ϫ (Ϫ9)) ϭ Ϫ2(Ϫ30) Ϫ 3(Ϫ36) ϩ 5(9) ϭ 60 ϩ 108 ϩ 45 ϭ 213
97
Algebra 2
Chapter 4
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Ϫ2
3 5 Ϫ2
3
† 0 Ϫ1 4 † 0 Ϫ1
9
7 2 9
7
4 108
0
Ϫ45 Ϫ56
0
Ϫ2
3 5 Ϫ2
3
† 0 Ϫ1 4 † 0 Ϫ1
9
7 2 9
7
4 ϩ 108 ϩ 0 Ϫ (Ϫ45) Ϫ
(Ϫ56) Ϫ 0 ϭ 213
8. 0
7. Ϫ38
10. Ϫ28
9. Ϫ40
11. Ϫ43
12. 0
13. 45
14. 26 units2
15. 20
16. Ϫ22
17. Ϫ22
18. 0
19. Ϫ29
20. Ϫ14
21. 63
22. Ϫ6
23. 32
24. Ϫ37
25. 32
26. 11.3
27. Ϫ58
28. 0
29. 62
30. 60
31. 172
32. Ϫ265
33. Ϫ22
34. 21
35. Ϫ5
36. 49
37. Ϫ141
38. Ϫ123
39. Ϫ6
40.
41. 14.5 units2
42. 12
43. about 26 ft2
44. 2875 mi2
98
5
,
3
Ϫ1
Algebra 2
Chapter 4
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1 1 1
45. Sample answer: † 1 1 1 †
1 1 1
46. Multiply each member in the top row by its minor and position sign. In this case the minor is a 3 ϫ 3 matrix.
Evaluate the 3 ϫ 3 matrix using expansion by minors again. 47. If you know the coordinates of the vertices of a triangle, you can use a determinant to find the area. This is convenient since you don’t need to know any additional information such as the measure of the angles.
Answers should include the following. • You could place a coordinate grid over a map of the Bermuda Triangle with one vertex at the origin. By using the scale of the map, you could determine coordinates to represent the other two vertices and use a determinant to estimate the area.
• The determinant method is advantageous since you don’t need to physically measure the lengths of each side or the measure of the angles between the vertices. 48. C
49. C
50. 63.25
51. Ϫ36.9
52. Ϫ25.21
53. Ϫ493
55. Ϫ3252
99
54. 0
Ϫ2 1
2
56. B
R
1 2 Ϫ3
Algebra 2
Chapter 4
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57. A¿(Ϫ5, 2.5), B¿(2.5, 5),
C¿(5, Ϫ7.5)
58.
60. B
2
26
R
Ϫ9 Ϫ12
62. undefined
7 69
64. B
R
Ϫ5 16
66. y ϭ x Ϫ 2
59. [Ϫ4]
61. undefined
63. [14
Ϫ8]
65. 138,435 ft
4
3
67. y ϭ Ϫ x
68. y ϭ 2x ϩ 1
1
2
69. y ϭ x ϩ 5
70. (0, Ϫ3)
71. (1, 9)
72. (2, 1)
73. (Ϫ1, 1)
74. (2, 5)
75. (4, 7)
Lesson 4-6 Cramer’s Rule
Pages 192–194
1. The determinant of the coefficient matrix cannot be zero. 2. Sample answer: 2x ϩ y ϭ 5 and 6x ϩ 3y ϭ 8
3. 3x ϩ 5y ϭ Ϫ6, 4x Ϫ 2y ϭ 30
4. (5, 1)
5. (0.75, 0.5)
6. (Ϫ6, Ϫ8)
8. aϪ5, , Ϫ b
2
3
7. no solution
9. a6, Ϫ , 2b
1
2
1
2
10. s ϩ d ϭ 4000,
0.065s ϩ 0.08d ϭ 297.50
100
Algebra 2
Chapter 4
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11. savings account, \$1500; certificate of deposit, \$2500
12. (2, Ϫ1)
13. (Ϫ12, 4)
14. (3, 5)
15. (6, 3)
16. (2.3, 1.4)
17. (Ϫ0.75, 3)
18. (Ϫ0.75, 0.625)
20. a , Ϫ1b
2
3
19. (Ϫ8.5625, Ϫ19.0625)
21. (4, Ϫ8)
22. (3, 10)
23. a , b
2 5
3 6
24. (Ϫ1.5, 2)
25. (3, Ϫ4)
26. (Ϫ1, 3, 4)
28. aϪ ,
11 39
,
19 19
27. (2, Ϫ1, 3)
29. a
141
,
29
31. aϪ
102 244 b ,
29 29
Ϫ b
14
19
30. (11, Ϫ17, 14)
Ϫ
155 143 673 b ,
,
28 70 140
32. r ϩ s ϭ 8, 7r ϩ 5s ϭ 50
33. race car, 5 plays; snowboard,
3 plays
34. 8s ϩ 13c ϭ 604.79,
35. silk, \$34.99; cotton, \$24.99
36. p ϩ r ϩ c ϭ 5, 2r Ϫ p ϭ 0,
3.2p ϩ 2.4r ϩ 4c ϭ 16.8
37. peanuts, 2 lb; raisins, 1 lb; pretzels, 2 lb
38. If the determinant is zero, there is no unique solution to the system. There is either no solution or there are infinitely many solutions.
Sample answer: 2x ϩ y ϭ 4 and 4x ϩ 2y ϭ 8 has a det ϭ
0; there are infinitely many solutions of this system.
2x ϩ y ϭ 4 and
4x ϩ 2y ϭ 10 has a det ϭ 0; there are no solutions of this system. ©Glencoe/McGraw-Hill
1
2
5 s ϩ 14c ϭ 542.30
101
Algebra 2
Chapter 4
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39. Cramer’s Rule is a formula for the variables x and y where (x, y) is a solution for a system of equations.
Answers should include the following. • Cramer’s Rule uses determinants composed of the coefficients and constants in a system of linear equations to solve the system.
• Cramer’s Rule is convenient when coefficients are large or involve fractions or decimals. Finding the value of the determinant is sometimes easier than trying to find a greatest common factor if you are solving by using elimination or substituting complicated numbers.
40. B
41. 111Њ, 69Њ
45. B
42. 16
43. 40
44. Ϫ53
47.
1 1 1
R
3 3 3
46. A¿(1, 5), B¿(Ϫ2, 2),
C¿(Ϫ1, Ϫ1)
48.
y y ϭ 3x ϩ 5
(Ϫ2, Ϫ1)
O
x
y ϭ Ϫ2x Ϫ 5
(Ϫ2, Ϫ1)
102
Algebra 2
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49.
50.
y xϩy ϭ7
y
2x Ϫ 4y ϭ 12
x
O
(4, 3)
O
1 x Ϫ y ϭ Ϫ1
2
x Ϫ 2y ϭ 10
x
51. c ϭ 10h ϩ 35
72
9
53. B
R
66 Ϫ23
52. [Ϫ4 32]
21
54. B R
43
(4, 3)
1. B
3.
1
4
1 Ϫ2
R
2 Ϫ1 Ϫ4 Ϫ1
no solution
Chapter 4
Practice Quiz 2
Page 194
2. A¿(Ϫ1, 2), B¿(Ϫ4, Ϫ1),
C¿(Ϫ1, Ϫ4), D¿(2, Ϫ1)
4. 22
5. Ϫ58
6. Ϫ105
7. 26
8. (1, Ϫ2)
10. (1, 2, 1)
9. (4, Ϫ5)
103
Algebra 2
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Lesson 4-7
1
0
1. D
0
0
0
1
0
0
0
0
1
0
0
0
T
0
1
Identity and Inverse Matrices
Pages 198–201
2. Exchange the values for a and d in the first diagonal in the matrix. Multiply the values for b and c by Ϫ1 in the second diagonal in the matrix. Find the determinant of the original matrix. Multiply the negative reciprocal of the determinant by the matrix with the above mentioned changes. 3. Sample answer: B
3 3
R
3 3
5. yes
6. B
2 5
R
3 8
4. no
4
1
B
27 Ϫ7
8. Ϫ
7. no inverse exists
10. yes
9. See students’ work.
11. yes
12. no
13. no
14. yes
15. yes
Ϫ1
R
Ϫ5
16. true
18. true
1 1 0
R
20. B
5 0 5
17. true
22. Ϫ B
19. false
1 1
B
7 4
Ϫ1
R
3
1
3
21. no inverse exists
23.
25.
1 Ϫ6
B
4 Ϫ2
Ϫ7
R
Ϫ3
1 Ϫ2
R
Ϫ2
1
7
1
B
34 Ϫ2
3
R
4
24. no inverse exists
26.
104
Algebra 2
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6
1
B
12 Ϫ5
1
1
B
32 Ϫ6
27. Ϫ
29.
31. 10 C
3
4
1
Ϫ
5
0
R
Ϫ2
5
R
2
5
8
Ϫ
3
10
30. 4 C
S
28. no inverse exists
S
1
4
3
4
1
Ϫ
6
1
2
32a. no
32b. Sample answer: y C
A
B
x
O A'
A''
B''
34. B
33a. yes
35. B
0 Ϫ4 4 8
R
0
4 12 8
37. dilation by a scale factor of
0 Ϫ2 2 4
R
0
2 6 4
C''
B'
C'
36. dilation by a scale factor of 2
38. B D
1
2
Ϫ1
1
2
0
0
1
2
T;
the graph of the
inverse transformation is the original figure.
105
Algebra 2
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39. MEET_IN_THE_LIBRARY
40. AT_SIX_THIRTY
41. BRING_YOUR_BOOK
42. See students’ work.
43. a ϭ Ϯ1, d ϭ Ϯ1, b ϭ c ϭ 0
44. A matrix can be used to code a message. The key to the message is the inverse of the matrix. Answers should include the following.
• The inverse matrix undoes the work of the matrix. So if you multiply a numeric message by a matrix it changes the message.
When you multiply the changed message by the inverse matrix, the result is the original numeric message. • You must consider the dimensions of the coding matrix so that you can write the numeric message in a matrix with dimensions that can be multiplied by the coding matrix. 46. D
45. A
Ϫ5 Ϫ9
R
47. B
Ϫ6 Ϫ11
49. C
3
5
Ϫ
1
5
1
5
2
Ϫ
5
51. F Ϫ
1
Ϫ1
1
3
7
3
S
2
3
8
3
Ϫ
53. (2, Ϫ4)
50. C
S
48. no inverse exists
2
5
1
3
2
5
16
1
2
2
1
Ϫ
8
1
4
0
5
32
0V
3
5
3
16
52. FϪ
1
3
1
3
Ϫ
54. (0, 7)
106
1
V
4
1
16
Ϫ
1
32
Ϫ
Algebra 2
Chapter 4
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55. (Ϫ5, 4, 1)
56. 52
57. Ϫ14
58. 0
59. 1
60. Ϫ3
61. Ϫ5
62.
63.
1
3
3
8
5
2
64. Ϫ
66. 27
65. 7.82 tons/in2
1
2
1
2
67. 5
68. Ϫ
69. 3
70. 296
71. 300
72. Ϫ1
73. Ϫ2
74. 6
75. 4
76. Ϫ27
77. Ϫ34
Lesson 4-8
Using Matrices to Solve Systems of Equations
Pages 205–207
4. B
1 Ϫ1 x Ϫ3
RؒB RϭB R
1
3 y 5
2. Sample answer: x ϩ 3y ϭ 8 and 2x ϩ 6y ϭ 16
1. 2r Ϫ 3s ϭ 4, r ϩ 4s ϭ Ϫ2
3. Tommy; a 2 ϫ 1 matrix cannot be multiplied by a
2 ϫ 2 matrix.
3 Ϫ5
2
a
6. C 4
7
1S ؒ CbS
2
0 Ϫ1 c 2
3
g
8
5. B
RؒB RϭB R
Ϫ4 Ϫ7 h Ϫ5
9 ϭ C 3S
12
8. (1.5, Ϫ4)
7. (5, Ϫ2)
10. (1, 1.75)
9. (Ϫ3, 5)
107
Algebra 2
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12. B
14. B
4 Ϫ7 x 2
RؒB RϭB R
3
5 y 9
3 Ϫ7 m Ϫ43
RؒB RϭB
R
15. B
6
5 n Ϫ10
2
3 Ϫ5 a 16. C 7
0
3S ؒ CbS
3 Ϫ6
1
c
1
ϭ C 7S
Ϫ5
3 Ϫ5 2 x 17. C 1 Ϫ7 3 S ؒ C y S
4 0 Ϫ3 z 1 Ϫ1
0
x
18. C Ϫ2 Ϫ5 Ϫ6 S ؒ C y S
9 10 Ϫ1 z 9 ϭ C 11 S
Ϫ1
8 ϭ C Ϫ27 S
54
3
Ϫ5
6 r 19. C 11 Ϫ12 16 S ؒ C s S
Ϫ5
8
Ϫ3
t
20. (5, Ϫ2)
21 ϭ C 15 S
Ϫ7
21. (3, 4)
22. (Ϫ2, 3)
24. a , Ϫ3b
1
2
23. (6, 1)
25. aϪ , 4b
1
3
26. (2, Ϫ3)
27. (Ϫ2, Ϫ2)
28. (7, 3)
30. aϪ1, b
9
2
29. (0, 9)
31. a , b
3 1
2 3
32. 27 h of flight instruction and
23 h in the simulator
34. 80 mL of the 60% solution, and
120 mL of the 40% solution
33. 2010
3 Ϫ1 x 0
RؒB RϭB
R
1
2
y
Ϫ21
5 Ϫ6 a Ϫ47
RؒB RϭB
R
3
2
b
Ϫ17
13. B
11. h ϭ 1, c ϭ 12
108
Algebra 2
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35. The solution set is the empty set or infinite solutions.
36. The food and territory that two species of birds require form a system of equations. Any independent system of equations can be solved using a matrix equation. Answers should include the following.
• Let a represent the number of nesting pairs of
Species A and let b represent the number of nesting pairs of Species B.
Then, 140a ϩ 120b ϭ
20,000 and 500a ϩ 400b ϭ
69,000.
a
• B Rϭ b 400
1
B
4000 Ϫ500
Ϫ
Ϫ120
20,000
RؒB
R;
140
69,000
a ϭ 70 and b ϭ 85, so the area can support 70 pairs of Species A and 85 pairs of Species B.
37. D
38. 17 small, 24 medium, 11 large
39. (Ϫ6, 2, 5)
40. (1, Ϫ3, 2)
42. C
S
43. B
4 Ϫ5
R
Ϫ7
9
45. (4, Ϫ2)
Ϫ1
1
Ϫ
2
41. (0, Ϫ1, 3)
3
4
1
44. no inverse exists
46. (4.27, Ϫ5.11)
47. (Ϫ6, Ϫ8)
48. about 114.3 ft
49. {Ϫ4, 10}
50. {Ϫ5, 1}
51. {2, 7}
109
Algebra 2
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Chapter 5 Polynomials
Lesson 5-1 Monomials
Pages 226–228
(2x 2)3 ϭ 8x 6 since
(2x 2)3 ϭ 2x 2 ؒ 2x 2 ؒ 2x 2 ϭ
2x ؒ x ؒ 2x ؒ x ؒ 2x ؒ x ϭ 8x 6
2. Sometimes; in general, x y ؒ x z ϭ x yϩz, so x y ؒ x z ϭ x yz when y ϩ z ϭ yz, such as when y ϭ 2 and z ϭ 2.
3. Alejandra; when Kyle used the
Power of a Product Property in his first step, he forgot to put an exponent of Ϫ2 on a.
Also, in his second step,
4. x 10
1
4
(Ϫ2)Ϫ2 should be , not 4.
5. 16b 4
6. 1
7. Ϫ6y 2
8. Ϫ
ab 4
9
9. 9p 2q 3
10.
1 w z
9 c d
12.
1
4x 6
11.
2 2
12 6
13. 4.21 ϫ 105
14. 8.62 ϫ 10Ϫ4
15. 3.762 ϫ 103
16. 5 ϫ 100
17. about 1.28 s
18. a 8
19. b 4
20. n16
21. z10
22. 16x 4
23. Ϫ8c 3
24. an
25. Ϫy 3z 2
26.
27. Ϫ21b5c 3
28. ab
29. Ϫ24r 7s 5
30. 24x 4y 4
31. 90a4b4
32. Ϫ
28x 4 y2 1
4y 4
110
Algebra 2
Chapter 5
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34.
m 4n 9
3
35. Ϫ
cd 4
5
36.
a 2c 2
3b 4
a4
16b 4
37.
8y 3 x6 38.
1 x y
39.
1
3 6 v w
40.
a 4b 2
2
41.
2x 3y 2
5z 7
42. 6
2 2
43. 7
44. 4.623 ϫ 102
45. 4.32 ϫ 104
46. 1.843 ϫ 10Ϫ4
47. 6.81 ϫ 10Ϫ3
48. 5.0202 ϫ 108
49. 6.754 ϫ 108
50. 1.245 ϫ 1010
51. 6.02 ϫ 10Ϫ5
52. 4.5 ϫ 102
53. 6.2 ϫ 1010
54. 4.225 ϫ 109
55. 1.681 ϫ 10Ϫ7
56. 6.08 ϫ 109
57. 2 ϫ 10Ϫ7 m
58. 1.67 ϫ 1025
59. about 330,000 times
60. 10010 ϭ (102)10 or 1020, and
10100 Ͼ 1020, so 10100 Ͼ 10010.
61. Definition of an exponent
62. (ab)m m factors
6447448
ϭ ab ؒ ab ؒ p ؒ ab m factors m factors
64748 64748 ϭaؒaؒpؒaؒbؒbؒpؒb ϭ a mb m
111
Algebra 2
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63. Economics often involves large amounts of money.
Answers should include the following. • The national debt in 2000 was five trillion, six hundred seventy-four billion, two hundred million or 5.6742 ϫ
1012 dollars. The population was two hundred eighty-one million or 2.81 ϫ 108.
• Divide the national debt by the population.
5.6742 ϫ 1012
2.81 ϫ 108
64. D
Ϸ
\$2.0193 ϫ 104 or about \$20,193 per person.
66. (1, 2)
Ϫ2 Ϫ5
68. c d 1
2
65. B
67. (Ϫ3, 3)
69. C
1
2
Ϫ
3
2 S
70. Ϫ6
1 Ϫ2
71. 7
72. (2, 3, Ϫ1)
73. (2, 0, 4)
74.
Median Age (yr)
Median Age of Vehicles y 8
7
6
5
4
0
0
10
20
30
Years Since 1970
75. Sample answer using (0, 4.9) and (28, 8.3): y ϭ 0.12x ϩ 4.9
76. Sample answer: 9.7 yr
77. 7
78. Ϫ3
79. 2x ϩ 2y
x
80. 3x Ϫ 3z
112
Algebra 2
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81. 4x ϩ 8
82. Ϫ6x ϩ 10
83. Ϫ5x ϩ 10y
84. 3y Ϫ 15
Lesson 5-2 Polynomials
Page 231–232
1. Sample answer: x 5 ϩ x 4 ϩ x 3
2. 4
3.
4. yes, 1
x x x
2
x
2
x
x
2
x
2
x
x
x
x
x
x
5. yes, 3
6. no
7. 10a Ϫ 2b
8. Ϫ3x 2 Ϫ 7x ϩ 8
10. 10p 3q 2 Ϫ 6p 5q 3 ϩ 8p3q 5
9. 6xy ϩ 18x
11. y 2 Ϫ 3y Ϫ 70
12. x 2 ϩ 9x ϩ 18
13. 4z 2 Ϫ 1
14. 4m 2 Ϫ 12mn ϩ 9n 2
15. 7.5x 2 ϩ 12.5x ft 2
16. yes, 2
17. yes, 3
18. no
19. no
20. yes, 6
21. yes, 7
22. 4x 2 ϩ 3x Ϫ 7
23. Ϫ3y Ϫ 3y 2
24. r 2 Ϫ r ϩ 6
25. 10m 2 ϩ 5m Ϫ 15
26. 4x 2 Ϫ 3xy Ϫ 6y 2
27. 7x 2 Ϫ 8xy ϩ 4y 2
28. 4b 2c Ϫ 4bdz
29. 12a 3 ϩ 4ab
30. 15a3b3 Ϫ 30a4b3 ϩ 15a5b6
31. 6x 2y 4 Ϫ 8x 2y 2 ϩ 4xy 5
32. 6x 3 ϩ 9x 2y Ϫ 12x 3y 2
33. 2a4 Ϫ 3a3b ϩ 4a4b4
34. 46.75 Ϫ 0.018x
35. Ϫ0.001x 2 ϩ 5x Ϫ 500
36. \$5327.50
37. p 2 ϩ 2p Ϫ 24
38. a 2 ϩ 9a ϩ 18
39. b 2 Ϫ 25
40. 36 Ϫ z 2
41. 6x 2 ϩ 34x ϩ 48
42. 8y 2 ϩ 16y Ϫ 42
113
Algebra 2
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43. a 6 Ϫ b 2
44. 2m 4 Ϫ 7m 2 Ϫ 15
45. x 2 Ϫ 6xy ϩ 9y 2
46. 1 ϩ 8c ϩ 16c 2
47. d 2 Ϫ 2 ϩ
1 d4 48. xy 3 ϩ y ϩ
1 x 49. 27b 3 Ϫ 27b 2c ϩ 9bc 2 Ϫ c 3
50. x 3 Ϫ y 3
51. 9c 2 Ϫ 12cd ϩ 7d 2
52. Ϫ18x 2 ϩ 27x Ϫ 10
53. R 2 ϩ 2RW ϩ W 2
54. 14; Sample answer:
(x 8 ϩ 1)(x 6 ϩ 1) ϭ x 14 ϩ x 8 ϩ x 6 ϩ 1
55. The expression for how much an amount of money will grow to is a polynomial in terms of the interest rate. Answers should include the following.
• If an amount A grows by r percent for n years, the amount will be A(1 ϩ r )n after n years. When this expression is expanded, a polynomial results.
• 13,872(1 ϩ r )3, 13,872r 3 ϩ
41,616r 2 ϩ 41,616r ϩ
13,872
• Evaluate one of the expressions when r ϭ 0.04.
For example,
13,872(1 ϩ r)3 ϭ
13,872(1.04)3 or \$15,604.11 to the nearest cent. The value given in the table is
\$15,604 rounded to the nearest dollar.
56. D
57. B
58. Ϫ64d 6
59. 20r 3t 4
60.
61.
b2
4a 2
xz 2 y2 62. (1, 4)
114
Algebra 2
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63.
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64.
y
y
x ϩ y Ͼ Ϫ2 x O
x
O
y ϭ Ϫ1x ϩ 2
3
65.
66. x 2
y
2x ϩ y ϭ 1
x
O
68. xy 2
67. 2y 3
69. 3a 2
Lesson 5-3 Dividing Polynomials
Pages 236–238
(x 2 ϩ x ϩ 5) Ϭ (x ϩ 1)
2. The divisor contains an x 2 term. 3. Jorge; Shelly is subtracting in the columns instead of adding.
4. 6y Ϫ 3 ϩ 2x
5. 5b Ϫ 4 ϩ 7a
6. x Ϫ 12
7. 3a3 Ϫ 9a2 ϩ 7a Ϫ 6
8. z 4 ϩ 2z 3 ϩ 4z 2 ϩ 5z ϩ 10
9. x 2 Ϫ xy ϩ y 2
10. x 2 ϩ 11x Ϫ 34 ϩ
11. b3 ϩ b Ϫ 1
12. 2y ϩ 5
13. 3b ϩ 5
14. B
15. 3ab Ϫ 6b 2
16. 5y Ϫ
17. 2c 2 Ϫ 3d ϩ 4d 2
60 xϩ2 18. 4n 2 ϩ 3mn Ϫ 5m
115
6y 2 x ϩ 3xy 2
Algebra 2
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2 b 19. 2y 2 ϩ 4yz Ϫ 8y 3z 4
20. Ϫa2b ϩ a Ϫ
21. b 2 ϩ 10b
22. x Ϫ 15
23. n 2 Ϫ 2n ϩ 3
24. 2c 2 ϩ c ϩ 5 ϩ
25. x 3 Ϫ 5x 2 ϩ 11x Ϫ 22 ϩ
39 xϩ2 6 cϪ2 26. 6w 4 ϩ 12w 3 ϩ 24w 2 ϩ
30w ϩ 60
27. x 2
28. x 2 ϩ 3x ϩ 9
29. y 2 Ϫ y Ϫ 1
30. m 2 Ϫ 7
31. a3 Ϫ 6a 2 Ϫ 7a ϩ 7 ϩ
3 aϩ1 32. 2m 3 ϩ m 2 ϩ 3m Ϫ
34. 3c 4 Ϫ c 3 ϩ 2c 2 Ϫ 4c ϩ
33. x 4 Ϫ 3x 3 ϩ 2x 2 Ϫ 6x ϩ
19 Ϫ
5 mϪ3 56 xϩ3 9Ϫ
13 cϩ2 4 bϩ1 35. g ϩ 5
36. 2b 2 Ϫ b Ϫ 1 ϩ
37. t 4 ϩ 2t 3 ϩ 4t 2 ϩ 5t ϩ 10
38. y 4 Ϫ 2y 3 ϩ 4y 2 Ϫ 8y ϩ 16
39. 3t 2 Ϫ 2t ϩ 3
40. h 2 Ϫ 4h ϩ 17 Ϫ
41. 3d 2 ϩ 2d ϩ 3 Ϫ
43. x 3 Ϫ x Ϫ
2
3d Ϫ 2
51
2h ϩ 3
42. x 2 ϩ x Ϫ 1
6
2x ϩ 3
44. 2x 3 ϩ x 2 Ϫ 1 ϩ
2
3x ϩ 1
Ϫ3x ϩ 7 x2 ϩ 2
45. x Ϫ 3
46. x 2 Ϫ 1 ϩ
47. x ϩ 2
48. x Ϫ 3
49. x 2 Ϫ x ϩ 3
50. 2y 2 Ϫ 3y ϩ 1
116
Algebra 2
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51. \$0.03x ϩ 4 ϩ
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1000 x 52. Let x be the number.
Multiplying by 3 results in 3x.
The sum of the number, 8, and the result of the multiplication is x ϩ 8 ϩ 3x or 4x ϩ 8. Dividing by the sum of the number and
2 gives
4x ϩ 8 xϩ2 or 4. The end
result is always 4.
53. 170 Ϫ
170 t ϩ1
54. 85 people
2
55. x 3 ϩ x 2 ϩ 6x Ϫ 24 ft
56. x Ϫ 2 s
57. x 2 ϩ 3x ϩ 12 ft /s
58. Sample answer: r 3 Ϫ 9r 2 ϩ
27r Ϫ 28 and r Ϫ 3
59. Division of polynomials can be used to solve for unknown quantities in geometric formulas that apply to manufacturing situations.
Answers should include the following. • 8x in. by 4x ϩ s in.
• The area of a rectangle is equal to the length times the width. That is, A ϭ /w .
• Substitute 32x 2 ϩ x for A,
8x for /, and 4x ϩ s for w.
Solving for s involves dividing 32x 2 ϩ x by 8x.
A ϭ /w
2
32x ϩ x ϭ 8x (4x ϩ s)
60. A
32x 2 ϩ x
8x
1
4x ϩ
8
1
8
ϭ 4x ϩ s ϭ 4x ϩ s ϭs The seam is
61. D
1
8
inch.
62. Ϫx 2 Ϫ 4x ϩ 14
117
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63. y 4z 4 Ϫ y 3z 3 ϩ 3y 2z
64. y 2 ϩ 2y Ϫ 15
65. a 2 Ϫ 2ab ϩ b 2
66. 5 ϫ 102 s or 8 min 20 s
67. y ϭ Ϫx ϩ 2
68. y ϭ x Ϫ
69. 9
70. 12
71. 4
72. 3
73. 6
74. 5
2
3
4
3
Chapter 5
Practice Quiz 1
Page 238
1. 6.53 ϫ 108
2. 7.2 ϫ 10Ϫ3
3. Ϫ108x 8y 3
4.
5.
x2 z6 a3 b4c 3
6. 2x ϩ 5y
7. 3t 2 ϩ 2t Ϫ 8
9. m 2 Ϫ 3 Ϫ
8. n3 Ϫ n 2 Ϫ 5n ϩ 2
19 mϪ4 10. d 2 ϩ d Ϫ 3
Lesson 5-4 Factoring Polynomials
Pages 242–244
1. Sample answer: x 2 ϩ 2x ϩ 1
3. sometimes
2. Sample answer: If a ϭ 1 and b ϭ 1, then a 2 ϩ b 2 ϭ 2 but
1a ϩ b2 2 ϭ 4.
5. a(a ϩ 5 ϩ b)
6. (x ϩ 7)(3 Ϫ y)
7. (y ϩ 2)(y Ϫ 4)
8. (z Ϫ 6)(z ϩ 2)
4. Ϫ6x(2x ϩ 1)
9. 3(b Ϫ 4)(b ϩ 4)
10. (4w ϩ 13)(4w Ϫ 13)
11. (h ϩ 20)(h 2 Ϫ 20h ϩ 400)
12.
118
xϪ4 xϪ7 Algebra 2
Chapter 5
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13.
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2y yϪ4 14. x ϩ y cm
15. 2x(y 3 Ϫ 5)
16. 6ab 2(a ϩ 3b)
17. 2cd 2(6d Ϫ 4c ϩ 5c 4d )
18. prime
19. (2z Ϫ 3)(4y Ϫ 3)
20. (3a ϩ 1)(x Ϫ 5)
21. (x ϩ 1)(x ϩ 6)
22. (y Ϫ 1)(y Ϫ 4)
23. (2a ϩ 1)(a ϩ 1)
24. (2b Ϫ 1)(b ϩ 7)
25. (2c ϩ 3)(3c ϩ 2)
26. (3m ϩ 2)(4m Ϫ 3)
27. 3(n ϩ 8)(n Ϫ 1)
28. 3(z ϩ 3)(z ϩ 5)
29. (x ϩ 6)2
30. (x Ϫ 3)2
31. prime
32. 3(m ϩ n)(m Ϫ n)
33. (y 2 ϩ z)(y 2 Ϫ z)
34. 3(x ϩ 3y)(x Ϫ 3y)
35. (z ϩ 5)(z 2 Ϫ 5z ϩ 25)
36. (t Ϫ 2)(t 2 ϩ 2t ϩ 4)
37. (p 2 ϩ 1)(p ϩ 1)(p Ϫ 1)
38. (x 2 ϩ 9)(x ϩ 3)(x Ϫ 3)
39. (7a ϩ 2b)(c ϩ d )(c Ϫ d )
40. (8x ϩ 3)(x ϩ y ϩ z)
41. (a Ϫ b)(5ax ϩ 4by ϩ 3cz)
42. (a ϩ 3b)(3a ϩ 5)(a Ϫ 1)
43. (3x Ϫ 2)(x ϩ 1)
44. (2y ϩ 1)(y ϩ 4)
45. 30 ft by 40 ft
46.
xϩ1 xϪ4 xϪ5 xϪ2 47.
xϩ5 xϪ6 48.
49.
xϪ4 x ϩ 2x ϩ 4
50. x
2
51. x ϩ 2
52. x Ϫ 1 s
53. 16x ϩ 16 ft /s
54. x Ϫ 8 cm
119
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55. (8pn ϩ 1)2
56. Factoring can be used to find possible dimensions of a geometric figure, given the area. Answers should include the following.
• Since the area of the rectangle is the product of its length and its width, the length and width are factors of the area. One set of possible dimensions is
4x Ϫ 2 by x ϩ 3.
• The complete factorization of the area is 2(2x Ϫ 1)
(x ϩ 3), so the factor of
2 could be placed with either 2x Ϫ 1 or x ϩ 3 when assigning the dimensions. 57. B
58. C
59. yes
60. no; (x ϩ 2)(x 2 Ϫ 2x ϩ 4)
61. no; (2x ϩ 1)(x Ϫ 3)
62. yes
63. t 2 Ϫ 2t ϩ 1
64. y ϩ 3
65. x 2 ϩ 2
66. x 3 ϩ x 2 Ϫ 2x ϩ 2 ϩ
67. 4x 2 ϩ 3xy Ϫ 3y 2
68. 14x 2 ϩ 26x Ϫ 4
70. c
69. [Ϫ2]
1
3x Ϫ 2
Ϫ36 7 d 18 4
72. yes
71. 15 in. by 28 in.
73. no
74. Distributive Property
75. Associative Property (ϩ)
76. rational
77. irrational
78. rational
79. rational
80. irrational
81. rational
120
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Lesson 5-5 Roots of Real Numbers
Pages 247–249
1. Sample answer: 64
2. If all of the powers in the result of an even root have even exponents, the result is nonnegative without taking absolute value.
3. Sometimes; it is true when x Ͼ 0.
4. 8.775
5. Ϫ2.668
6. 2.632
7. 4
8. 2
9. Ϫ3
10. not a real number
11. x
12. 0 y 0
15. about 3.01 mi
14. 0 4x ϩ 3y 0
16. 11.358
17. Ϫ12.124
18. 0.933
19. 2.066
20. 3.893
21. Ϫ7.830
22. 4.953
23. 3.890
24. 4.004
25. 4.647
26. 26.889
27. 59.161
28. 15
29. Ϯ13
30. not a real number
31. 18
32. Ϫ3
33. Ϫ2
34.
13. 6 0 a 0 b 2
1
4
1
5
36. 0.5
37. Ϫ0.4
38. z 2
35.
39. Ϫ 0 x 0
41. 8a 4
40. 7 0 m3 0
43. Ϫc 2
44. 25g 2
42. 3r
46. 5x 2 0 y 3 0
45. 4z 2
121
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49. 3p 6 0 q 3 0
48. 13x 4y 2
53. p ϩ q
52. 0 4x Ϫ y 0
57. not a real number
56. 0 2a ϩ 1 0
58. 2
59. Ϫ5
60. about 127.28 ft
61. about 1.35 m
62. about 11,200 mրs
63. x ϭ 0 and y Ն 0, or y ϭ 0 and x Ն 0
64. The speed and length of a wave are related by an expression containing a square root. Answers should include the following.
3.00 knots, and 4.24 knots
• As the value of / increases, the value of s increases.
65. B
66. D
67. 7xy 2(y Ϫ 2xy 3 ϩ 4x 2)
68. (a ϩ 3)(b Ϫ 5)
69. (2x ϩ 5)(x ϩ 5)
70. (c Ϫ 6)(c 2 ϩ 6c ϩ 36)
47. 6x 2z 2
50. 2ab
51. Ϫ3c 3d 4
54. Ϫ 0 x ϩ 2 0
55. 0 z ϩ 4 0
71. 4x 2 ϩ x ϩ 5 ϩ
73. c
810 2320 d 1418 2504
8 xϪ2 72. x 3 Ϫ x 2 ϩ x
74. (Ϫ2, 2)
75. (1, Ϫ3)
76. (9, 4)
77. x 2 ϩ 11x ϩ 24
78. y 2 ϩ 3y Ϫ 10
79. a 2 Ϫ 7a Ϫ 18
80. a 2 ϩ 3ab ϩ 2b 2
81. x 2 Ϫ 9y 2
82. 6w 2 Ϫ 7wz Ϫ 5z 2
122
Algebra 2
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1a
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ϭ 1a only
22 ϩ 23 ϩ 22
Lesson 5-6 Radical Expressions
Pages 254–256
1. Sometimes;
1
n
n
when a ϭ 1.
3. The product of two conjugates yields a difference of two squares.
Each square produces a rational number and the difference of two rational numbers is a rational number. 4. 1527
4
5. 2x 0 y 0 2x
214y
4y
8. 225
9. 2a 2b 2 23
12. 3 ϩ 323 Ϫ 25 Ϫ 215
3
11. 22 22
6.
13. 2 ϩ 25
3
7. Ϫ24235
4
10. 523 ϩ 323
3
20. 2y 22
19. 5x 2 22
14. about 49 mph
22. 2ab 2 210a
21. 3 0 x 0 y 22y
15. 923
16. 622
3
17. 322
4
18. 226
23. 6y 2z 27
25.
26
2
1 c 0d
3
a 2 2b b2 4
0 2c
26.
31. 36 27
3
27.
29.
33.
1 wz 2
2r 4 2t t5 4
28.
26
2
30.
210
5
32. Ϫ60230
38. 425 ϩ 23 26
34.
36. 522
35. 323
37. 723 Ϫ 222
2 54
3
5
2wz 2
3
24. 4mn23mn 2
3
123
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40. 6 ϩ 326 ϩ 227 ϩ 242
28 ϩ 7 23
13
39. 25 Ϫ 522 ϩ 526 Ϫ 223
Ϫ1 Ϫ 23
2
5 26 Ϫ 3 22
22
12 ϩ 7 22
23
41. 13 Ϫ 2222
42. 8 Ϫ 2215
43.
44.
45.
47.
48. 2x ϩ 1
2x 2 Ϫ 1 xϪ1 46.
49. 6 ϩ 1622 yd, 24 ϩ 622 yd2
50. The square root of a difference is not the difference of the square roots. 51. 0 ft /s
52. d ϭ v
53. about 18.18 m
54. 80 ft /s or about 55 mph
55. x and y are nonnegative.
56. The formula for the time it takes an object to fall a certain distance can be written in various forms involving radicals. Answers should include the following.
• By the Quotient Property
24.9h
4.9
2g
2g
of Radicals, t ϭ
Multiply by
22d
2g
.
22dg
.
g
to
rationalize the denominator.
The result is h ϭ
• about 1.12 s
57. B
58. D
59. 12z 4
60. 6ab 3
61. 0 y ϩ 2 0
63.
62.
Ϫ2 Ϫ4
64. £ 9 15 §
3 Ϫ5
xϩ7 xϪ4 124
Algebra 2
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1
4
d
Ϫ5 Ϫ4
66. 16, Ϫ15
67. consistent and independent
68. \$4.20
69. Ϫ5
70. 2
71. Ϫ2, 4
72. Ϫ , 1
7
3
73. 5x 0 x Ͼ 66
75.
1
4
77.
74. 5x 0 x Ն Ϫ76
76.
1
2
5
6
78.
13
12
79.
13
24
80.
19
30
81.
3
8
82. Ϫ
5
12
Chapter 5
Practice Quiz 2
Page 256
1. x 2y (3x ϩ y ϩ 1)
2. prime
3. a(x ϩ 3)2
4. 8(r Ϫ 2s 2)(r 2 ϩ 2rs 2 ϩ 4s 4)
6. Ϫ4a 2b 3
5. 6 0 x 0 0 y 3 0
7. 0 2n ϩ 3 0
8.
10.
9. Ϫ1 Ϫ17
125
x 2 2y y2 8 Ϫ 3 22
2
Algebra 2
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2. In radical form, the expression would be 2Ϫ16, which is not a real number because the index is even and the radicand is negative.
Lesson 5-7 Rational Exponents
Pages 260–262
1. Sample answer: 64
3. In exponential form 2bm is
1
equal to (b m)n . By the Power of a 1Power Property, m m
(b m )n ϭ b n . But, b n is also
1
equal to (b n )mby the Power of a Power Property. This last n expression is equal to ( 2b)m . n n
Thus, 2bm ϭ (2b)m.
3
4. 27
n
3
3
5. 2x 2 or (2x)2
1
5
1
6. 264
7
8. 5
7. 63x 3y 3
9.
1
3
10. 9
11
12. a12
11. 2
2
13. x
14.
19. 23
3
2
15. a b
5
21. 26
17.
z3
2z
16.
2
3
m 3 n3 mn 18. 23x
1
2
2
3
3
22. 24
1
z (x Ϫ 2y)2 x Ϫ 2y
3
24. x 2 2x 2
5
5
23. 2c 2 or (2c)2
20. \$5.11
1
1
26. 623
25. 232
1
1
2
1
27. 2 z 2
28. 53 x 3 y 3
29. 2
30. 6
31.
1
5
32.
126
1
27
Algebra 2
Chapter 5
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1
8
1
9
34. Ϫ
35. 81
36. 4096
37.
2
3
38. 27
39.
4
3
40.
42. x 3
41. y 4
1
1
43. b 5
44. a9
5
x6
46.
x
1
45.
w5 w 1
1
48. r 2
47. t 4
54. 23
15
53. 25
5
49.
51.
50.
y 2 Ϫ 2y 2 yϪ4 6
55. 17 217
3
57. 25x 2y 2 xy 1z z x ϩ 3x 2 ϩ 2 xϪ1 1
ab 2 c 2 c 6
56. 5255
62. 26
61. 12
3
4
58. b29a 2b
3
60.
3
1
2
64. x Ϫ x 3z 3
63. 216 Ϫ 5
3
2c 16 c 52.
a12
6a
59.
1
2
1
1
65. 22 ϩ 32
66. 2 ؒ 3 3
67. 880 vibrations per second
68. about 262 vibrations per second ©Glencoe/McGraw-Hill
127
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70. Rewrite the equation so that the bases are the same on each side.
1
9x ϭ 3xϩ 2
1
(32 )x ϭ 3xϩ 2
1
32x ϭ 3xϩ 2
Since the bases are the same and this is an equation, the exponents must be equal.
1
2
Solve 2x ϭ x ϩ . The result
1
2
is x ϭ .
71. The equation that determines the size of the region around a planet where the planet’s gravity is stronger than the Sun’s can be written in terms of a fractional exponent. Answers should include the following.
• The radical form of the
2
Mp
.
2
B Ms
B Ms
equation is r ϭ D 5 a r ϭ D5
Mp
b
2
72. C
or
Multiply the
2
3
Mp Ms
ؒ 3
2
B Ms Ms
fraction under the radical by
3
2
Mp Ms
5
B Ms
3
Ms
3
Ms
.
r ϭ D5
5
2 3
2Mp Ms
ϭ D5 ϭD ϭ
5
5
2Ms
5
2 3
D2MpMs
Ms
The simplified radical form is rϭ 5
2 3
D2Mp Ms
Ms
.
• If Mp and Ms are constant, then r increases as D increases because r is a linear function of D with positive slope.
128
Algebra 2
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74. 2x 0 y 0 2x
75. 36 22
73. C
76. 222
78. 4 0 x Ϫ 5 0
77. 8
79.
1 2 x 2
80. 1440
81. x Ϫ 2
82. 2x Ϫ 3
83. x ϩ 22x ϩ 1
84. 4x Ϫ 122x ϩ 9
Lesson 5-8
2. The trinomial is a perfect square in terms of 1x . x Ϫ 61x ϩ 9 ϭ (1x Ϫ 3)2, so the equation can be written as (1x Ϫ 3)2 ϭ 0.
Take the square root of each side to get 1x Ϫ 3 ϭ 0. Use the Addition Property of
Equality to add 3 to each side, then square each side to get x ϭ 9.
Radical Equations and Inequalities
Pages 265–267
1. Since x is not under the radical, the equation is a linear equation, not a radical equation. The solution is xϭ 23 Ϫ 1
.
2
2x ϩ 2x ϩ 3 ϭ 3
4. 2
5. Ϫ9
6. no solution
7. 15
8. 18
3
2
9. 31
10. Ϫ Յ x Յ 39
11. 0 Յ b Ͻ 4
12. about 13.42 cm
13. 16
14. 49
15. no solution
16. no solution
17. 9
18. 5
19. Ϫ1
20.
21. Ϫ20
22. 5
129
27
2
Algebra 2
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23. no solution
24. 9
25. x Ͼ 1
26. Ϫ2 Յ x Յ 1
27. x Յ Ϫ11
28. y Ͼ 4
29. no solution
30. 4
31. 3
32. no solution
33. 0 Յ x Յ 2
34. 0 Յ a Ͻ 3
35. b Ն 5
36. c Ͼ Ϫ
37. 3
38. 16
39. 1152 lb
40. t ϭ
41. 34 ft
42. 21.125 kg
4 2r 3
B GM
79
16
43. Since 1x ϩ 2 Ն 0 and
12x Ϫ 3 Ն 0, the left side of the equation is nonnegative.
Therefore, the left side of the equation cannot equal Ϫ1.
Thus, the equation has no solution. 44. If a company’s cost and number of units manufactured are related by an equation involving radicals or rational exponents, then the production level associated with a given cost can be found by solving a radical equation. Answers should include the following.
3
• C ϭ 102n 2 ϩ 1500
2
• 10,000 ϭ 10n 3 ϩ 1500
C ϭ 10,000
8500 ϭ 10n
2
850 ϭ n 3
3
8502 ϭ n
2
3
Subtract 1500 from each side.
Divide each side by 10.
Raise each side to the
3
power.
2
24,781.55 Ϸ n
Use a calculator. Round down so that the cost does not exceed \$10,000.
The company can make at most 24,781 chips.
130
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50. 6 0 x 3 0 y 22y
45. D
46. C
2 100
10
3
7
1
47. 5
48. (x ϩ 7)2
2
49. (x 2 ϩ 1)3
3
51.
52. 28 Ϫ 1023
54. 4 ϩ x
53. x ϩ y ϭ 7, 30x ϩ 20y ϭ 160;
(2, 5) y 30x ϩ 20y ϭ 160
(2, 5)
x ϩy ϭ7
x
O
55. 1 Ϫ y
56. 2 ϩ 4x
57. Ϫ11
58. 4 ϩ 6z ϩ 2z 2
59. Ϫ3 Ϫ 10x Ϫ 8x 2
Lesson 5-9 Complex Numbers
Pages 273–275
5. 5i 0 xy 0 22
1a. true
1b. true
2. all of them
3. Sample answer: 1 ϩ 3i and
1 Ϫ 3i
4. 6i
6. 12
7. Ϫ18023
8. i
10. 42 Ϫ 2i
9. 6 ϩ 3i
11.
7
17
Ϫ
11 i 17
12. Ϯ3i
13. Ϯ2i22
14. Ϯi25
15. 3, Ϫ3
16. 5, 4
17. 10 ϩ 3j amps
18. 12i
131
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21. 10a 2 0 b 0 i
20. 8x 2i
23. Ϫ12
24. Ϫ48i
25. Ϫ75i
26. i
27. 1
28. Ϫ1
29. Ϫi
30. 9 ϩ 2i
31. 6
32. 2
33. 4 Ϫ 5i
34. 25
35. 6 Ϫ 7i
36. 8 ϩ 4i
37. Ϫ8 ϩ 4i
38.
2
5
40.
39
17
19. 9i
39.
10
17
41.
2
5
Ϫ
22. Ϫ1322
6 i 17
1
5
ϩ i
6
5
ϩ i ϩ 14 i 17
5 23 i 14
42. Ϫ163 Ϫ 16i
2 22 i 3
43. 20 ϩ 15i
44.
11
14
Ϫ
51. Ϯ2i 23
46. (j ϩ 4)x 2 ϩ (3 Ϫ i )x ϩ 2 Ϫ 4i
49. Ϯ4i
50. Ϯi 26
55. Ϯ
56. 4, 5
1
3
45. Ϫ Ϫ
52. Ϯi 23
53. Ϯ2i 210
54. Ϯ3i 25
47. (5 Ϫ 2i )x 2 ϩ (Ϫ1 ϩ i )x ϩ 7 ϩ i
48. Ϯi
25 i 2
7
2
58. Ϫ , Ϫ3
57. 4, Ϫ3
59.
5
,
3
61.
67 19
,
11 11
60. 3, 1
4
62. 5 Ϫ 2j ohms
64. 4 ϩ 2j amps
63. 13 ϩ 18j volts
132
Algebra 2
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65. Case 1: i Ͼ 0
Multiply each side by i to get i 2 Ͼ 0 ؒ i or Ϫ1 Ͼ 0. This is a contradiction. Case 2: i Ͻ 0
Since you are assuming i is negative in this case, you must change the inequality symbol when you multiply each side by i. The result is again i 2 Ͼ 0 ؒ i or Ϫ1 Ͼ 0, a contradiction. Since both possible cases result in contradictions, the order relation “Ͻ” cannot be applied to the complex numbers. 66. Some polynomial equations have complex solutions.
Answers should include the following. • a and c must have the same sign.
• Ϯi
67. C
68. C
69. Ϫ1, Ϫi, 1, i, Ϫ1, Ϫi, 1, i, Ϫ1
70. Examine the remainder when the exponent is divided by 4.
If the remainder is 0, the result is 1. If the remainder is
1, the result is i. If the remainder is 2, the result is
Ϫ1. And if the remainder is
3, the result is Ϫi.
71. 12
72. 11
73. 4
74. x 15
7
1
75. y
77. c
1
3
76.
2
1 Ϫ2 d 3 Ϫ2
1
a4 a 78. c
133
1
0
d
0 Ϫ1
Algebra 2
Chapter 5
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2 1 Ϫ2 d Ϫ3 2 Ϫ1
80.
y
B'
C'
x
O
A'
81. sofa: \$1200, love seat: \$600, coffee table: \$250
82.
y
y ϭ Ϫ2x Ϫ 2
O
x
yϭxϩ1
83.
84.
y
1
10
xϩyϭ1 x Ϫ 2y ϭ 4
O
x
85. 0
134
Algebra 2
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Chapter 6 Quadratic Functions and Inequalities
Lesson 6-1 Graphing Quadratic Functions
Pages 290–293
2a. (2, 1); x ϭ 2
2b. (Ϫ3, Ϫ2); x ϭ Ϫ3
1. Sample answer: f (x ) ϭ
3x 2 ϩ 5x Ϫ 6; 3x 2, 5x, Ϫ6
3a. up; min.
3b. down; max.
3c. down; max.
3d. up; min.
4a. 0; x ϭ 0; 0
4b.
x f(x)
Ϫ1 Ϫ4
0
0
1 Ϫ4
p
4c.
f (x)
O (0, 0)
x
f (x) ϭ Ϫ4x 2
5a. 0; x ϭ Ϫ1; Ϫ1
6a. Ϫ1; x ϭ 2; 2
5b.
6b.
x f(x)
Ϫ3
3
Ϫ2
0
Ϫ1 Ϫ1
0
0
1 3
p
5c.
6c.
f(x)
x
0
1
2
3
4
p
f(x)
Ϫ1
2
3
2
Ϫ1
f(x)
(2, 3)
f (x) ϭ x 2 ϩ 2x
O
O
x
f (x) ϭ Ϫx 2 ϩ 4x Ϫ 1
x
(Ϫ1, Ϫ1)
135
Algebra 2
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7a. 3; x ϭ Ϫ4; Ϫ4
8a. 1; x ϭ 1; 1
7b.
8b.
x f(x)
Ϫ1
7
1
0
1 Ϫ1
2
1
3
7
8c.
f(x)
x f(x) Ϫ6
Ϫ9
Ϫ5 Ϫ12
Ϫ4 Ϫ13
Ϫ3 Ϫ12
Ϫ2
Ϫ9
p
7c.
f(x)
Ϫ10
Ϫ8
x
O
Ϫ4
p
Ϫ4 f (x) ϭ 2x 2 Ϫ 4x ϩ 1
Ϫ8
2
O
f (x) ϭ x ϩ 8x ϩ 3
Ϫ12
(Ϫ4, Ϫ13)
5
3
5
3
9a. 0; x ϭ Ϫ ; Ϫ
9b.
x
Ϫ3
Ϫ2
(1, Ϫ1)
x
10. max.; 7
f(x)
Ϫ3
Ϫ8
5
3
25
3
Ϫ
Ϫ
Ϫ1
0
Ϫ7
0
9c.
f(x)
4
Ϫ2
Ϫ4
O
2
x
Ϫ4
(
)
Ϫ 5 , Ϫ 25
3
3
Ϫ8 f (x) ϭ 3x 2 ϩ 10x
Ϫ12
25
4
12. min.; 0
11. min.; Ϫ
136
Algebra 2
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13. \$8.75
14a. 0; x ϭ 0; 0
14b.
x f(x)
Ϫ2 8
Ϫ1 2
0 0
1 2
2 8
p
14c.
f (x)
f (x ) ϭ 2x 2
(0, 0)
x
O
15a. 0; x ϭ 0; 0
16a. 4; x ϭ 0; 0
15b.
16b.
x f(x)
Ϫ2 Ϫ20
Ϫ1 Ϫ5
0
0
1 Ϫ5
2 Ϫ20
p
15c.
O f (x ) ϭ Ϫ5x
(0, 0)
p
16c.
f(x)
2
x f(x)
Ϫ2 8
Ϫ1 5
0 4
1 5
2 8 f(x) 12
x
8
(0, 4) f (x ) ϭ x 2 ϩ 4
Ϫ4
Ϫ2
O
17a. Ϫ9; x ϭ 0; 0
18b.
4x
18a. Ϫ4; x ϭ 0; 0
17b.
2
x
Ϫ2
Ϫ1
0
1
2
p
f(x)
Ϫ5
Ϫ8
Ϫ9
Ϫ8
Ϫ5
137
x
Ϫ2
Ϫ1
0
1
2
p
f(x)
4
Ϫ2
Ϫ4
Ϫ2
4
Algebra 2
Chapter 6
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17c.
18c.
f(x)
f(x)
4
Ϫ4
O
Ϫ2
4x
2
x
O
Ϫ4
(0, Ϫ9)
f (x ) ϭ 2x 2 Ϫ 4
f (x ) ϭ x 2 Ϫ 9
19a. 191; x ϭ 0; 0
20a. 4; x ϭ 2; 2
19b.
20b.
(0, Ϫ4)
x f(x)
Ϫ2 13
4
Ϫ1
0
1
1
4
2 13
p
19c.
20c.
f (x)
f (x ) ϭ 3x 2 ϩ 1
x
0
1
2
3
4
p
f(x)
4
1
0
1
4
f(x)
(0, 1)
f (x ) ϭ x 2 Ϫ 4x ϩ 4 x O
O
x
(2, 0)
21a. 9; x ϭ 4.5; 4.5
22a. Ϫ5; x ϭ 2; 2
21b.
22b.
21c.
x
3
4
4.5
5
6
p
f(x)
Ϫ9
Ϫ11
Ϫ11.25
Ϫ11
Ϫ9
p
f(x)
Ϫ5
Ϫ8
Ϫ9
Ϫ8
Ϫ5
22c.
f(x)
f(x) x O
2
O
x
0
1
2
3
4
4
8
12
x
Ϫ4
Ϫ8
Ϫ12
f (x ) ϭ x 2 Ϫ 9x ϩ 9 f (x ) ϭ x 2 Ϫ 4x Ϫ 5
(4 1 , Ϫ111 )
2
4
(2, Ϫ9)
138
Algebra 2
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23a. 36; x ϭ Ϫ6; Ϫ6
24a. Ϫ1; x ϭ Ϫ1; Ϫ1
23b.
24b.
x f(x)
Ϫ8 4
Ϫ7 1
Ϫ6 0
Ϫ5 1
Ϫ4 4
p
23c.
x
Ϫ3
Ϫ2
Ϫ1
0
1
p
f(x)
8
Ϫ1
Ϫ4
Ϫ1
8
24c.
f(x)
6
f(x) f (x ) ϭ 3x 2 ϩ 6x Ϫ 1
4
2
f (x ) ϭ x ϩ 12x ϩ 36
Ϫ16 Ϫ12
Ϫ8 Ϫ4
(Ϫ6, 0)
x
O
2
O x
(Ϫ1, Ϫ4)
2
3
2
3
25a. Ϫ3; x ϭ 2, 2
26a. 0; x ϭ Ϫ , Ϫ
25b.
26b.
x
0
1
2
3
4
25c.
p
f(x)
Ϫ3
3
5
3
Ϫ3
f(x)
x
Ϫ2
Ϫ1
f(x)
Ϫ4
1
2
3
4
3
0
1
0
Ϫ7
Ϫ
26c.
(2, 5) f (x ) ϭ Ϫ2x 2 ϩ 8x Ϫ 3
O
5
4
5
4
28a. Ϫ1; x ϭ 0; 0
27a. 0; x ϭ Ϫ ; Ϫ
x
f (x ) ϭ Ϫ3x 2 Ϫ 4x
x
O
f(x)
(Ϫ 2 , 4 )
3 3
139
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27b.
x
Ϫ3
Ϫ2
28b.
f(x)
3
Ϫ2
5
4
x
Ϫ2
25
8
Ϫ
Ϫ
Ϫ1
0
0
Ϫ3
0
27c.
p
Ϫ1
p
1
2
f(x)
1
1
2
Ϫ
Ϫ1
1
2
Ϫ
1
28c.
f(x)
f(x) f (x ) ϭ 0.5x 2 Ϫ 1
f (x ) ϭ 2x 2 ϩ 5x
O
x
O
(Ϫ 5 , Ϫ 25)
4
8
29a. 0; x ϭ Ϫ6; Ϫ6
30a.
29b.
30b.
29c.
x
Ϫ8
Ϫ7
Ϫ6
Ϫ5
Ϫ4
f(x)
8
8.75
9
8.75
8
p
x
(0, Ϫ1)
(Ϫ6, 9)
8
9
;
2
x ϭ Ϫ3, Ϫ3
x
Ϫ5
Ϫ4
Ϫ3
Ϫ2
Ϫ1
p
f(x)
2
0.5
0
0.5
2
30c.
f(x)
f(x)
4
Ϫ8
Ϫ4
O
x
(Ϫ3, 0)
Ϫ4 f (x ) ϭ 1 x 2 ϩ 3x ϩ 9
f (x ) ϭ Ϫ0.25x 2 Ϫ 3x
2
140
O
x
2
Algebra 2
Chapter 6
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8
9
1 1
3 3
31a. Ϫ ; x ϭ ;
31b.
x
32. min.; 0
f(x)
7
9
8
Ϫ
9
Ϫ1
0
1
3
Ϫ1
1
Ϫ
5
9
7
1
9
2
31c.
f(x)
2
2
8
f (x ) ϭ x Ϫ 3 x Ϫ 9
O
(
x
)
1
, Ϫ1
3
33. max.; Ϫ9
34. min.; Ϫ14
35. min.; Ϫ11
36. max.; 5
37. max.; 12
38. min.;
7
8
9
2
39. max.; Ϫ
40. max.; 5
41. min.; Ϫ11
42. max.; 5
43. min.; Ϫ10
1
3
44. x ϭ 40; (40, 40)
45. 40 m
46. 300 ft, 2.5 s
47. The y-intercept is the initial height of the object.
48. 120 Ϫ 2x
49. 60 ft by 30 ft
50. 1800 ft2
51. \$11.50
52. \$2645
141
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53. 5 in. by 4 in.
54. c; The x-coordinate of the
0
vertex of y ϭ ax 2 ϩ c is Ϫ
2a
or 0, so the y-coordinate of the vertex, the minimum of the function, is a(0)2 ϩ c or c ; Ϫ12.5
55. If a quadratic function can be used to model ticket price versus profit, then by finding the x-coordinate of the vertex of the parabola you can determine the price per ticket that should be charged to achieve maximum profit.
Answers should include the following. • If the price of a ticket is too low, then you won’t make enough money to cover your costs, but if the ticket price is too high fewer people will buy them.
• You can locate the vertex of the parabola on the graph of the function. It occurs when x ϭ 40.
Algebraically, this is found
56. C
by calculating x ϭ Ϫ
b
2a
which, for this case, is
Ϫ4000
xϭ or 40. Thus the
2(Ϫ50)
ticket price should be set at \$40 each to achieve maximum profit.
57. C
58. Ϫ2.08
59. 3.20
60. 0.88
61. 3.38
62. 0.43
63. 1.56
64. Ϫ1
65. Ϫ1 ϩ 3i
66. 9 Ϫ 5i
142
Algebra 2
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67. 23
68. Ϫ13
69. 4
70. [10 Ϫ4 5]
Ϫ28
20 Ϫ44 d 72. c
8 Ϫ16
36
71. [5 Ϫ13 8]
6
73. C 14
0
Ϫ24
2
3
Ϫ8
Ϫ
S
74.
y y ϭ Ϫ3x yϪxϭ4 (Ϫ1, 3)
O
x
(Ϫ1, 3); consistent and independent 75. 5
76. 8
77. Ϫ2
78. Ϫ1
Lesson 6-2
Solving Quadratic Equations by Graphing
Pages 297–299
2. Sample answer: f (x) ϭ 3x 2 ϩ
2x Ϫ 1; 3x 2 ϩ 2x Ϫ 1 ϭ 0
1a. The solution is the value that satisfies an equation.
1b. A root is a solution of an equation. 1c. A zero is the x value of a function that makes the function equal to 0.
1d. An x-intercept is the point at which a graph crosses the x-axis. The solutions, or roots, of a quadratic equation are the zeros of the related quadratic function. You can find the zeros of a quadratic function by finding the x-intercepts of its graph.
143
Algebra 2
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3. The x-intercepts of the related function are the solutions to the equation. You can estimate the solutions by stating the consecutive integers between which the x-intercepts are located.
4. Ϫ4, 1
5. Ϫ2, 1
6. Ϫ4
7. Ϫ7, 0
8. Ϫ4, 6
9. Ϫ7, 4
10. Ϫ5
11. between Ϫ2 and Ϫ1, 3
12. between Ϫ1 and 0; between
1 and 2
13. Ϫ2, 7
14. 0, 6
15. 3
16. Ϫ2, 1
17. 0
18. Ϫ , 3
19. no real solutions
20. 0, 3
21. 0, 4
22. between Ϫ5 and Ϫ4; between 0 and 1
23. between Ϫ1 and 0; between
2 and 3
24. Ϫ4, 5
25. 3, 6
26. Ϫ7
27. 6
28. Ϫ1 , 3
1
2
1
2
1
2
1
2
29. Ϫ , 2
1
2
1
2
30. Ϫ4, 1
˛
1
2
31. Ϫ2 , 3
32. between Ϫ4 and Ϫ3; between 0 and 1
33. between 0 and 1; between 3 and 4
34. between Ϫ1 and 0, between
2 and 3
35. between Ϫ3 and Ϫ2; between 2 and 3
36. no real solutions
37. no real solutions
38. Ϫ8, Ϫ9
˛
144
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40. Let x be the first number.
Then, Ϫ9 Ϫ x is the other number. x(Ϫ9 Ϫ x) ϭ 24
2
Ϫx Ϫ 9x Ϫ 24 ϭ 0
39. Let x be the first number.
Then, 7 Ϫ x is the other number. x(7 Ϫ x) ϭ 14
2
Ϫx ϩ 7x Ϫ 14 ϭ 0
y
y
2
y ϭ Ϫx ϩ 7x Ϫ 14
O
O
x
x
2
y ϭ Ϫx Ϫ 9x Ϫ 24
Since the graph of the related function does not intersect the x-axis, this equation has no real solutions. Therefore, no such numbers exist.
Since the graph of the related function does not intersect the x-axis, this equation has no real solutions. Therefore, no such numbers exist.
41. Ϫ2, 14
42. 4 s
43. 3 s
44. about 12 s
45. about 35 mph
46. about 8 s
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48. Answers should include the following. • h (t )
2
47. Ϫ4 and Ϫ2; The value of the function changes from negative to positive, therefore the value of the function is zero between these two numbers. h (t ) ϭ Ϫ16t ϩ 185
180
160
140
120
100
80
60
40
20
0
1
2 3 4 5 t
• Locate the positive x-intercept at about 3.4.
This represents the time when the height of the ride is 0. Thus, if the ride were allowed to fall to the ground, it would take about
3.4 seconds.
49. A
50. B
51. Ϫ1
52. Ϯ3
53. 3, 5
54. Ϫ9, 1
55. Ϯ1.33
56. no real solutions
57. 4, x ϭ 3; 3
58. Ϫ1; x ϭ 1; 1 f (x)
f(x)
(1, 3)
f (x) ϭ Ϫ4x 2 ϩ 8x Ϫ 1
x
O
O
x
f (x) ϭ x 2 Ϫ 6x ϩ 4
(3, Ϫ5)
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60.
1
5
62.
59. 4; x ϭ Ϫ6; Ϫ6
1
13
f(x)
3
5
ϩ i
8 f (x) ϭ 1 x 2 ϩ 3x ϩ 4
4
4
O
Ϫ12 Ϫ8
x
Ϫ4
Ϫ4
(Ϫ6, Ϫ5)
61.
10
13
ϩ
2 i 13
ϩ
5 i 13
63. 24
64. Ϫ8
65. Ϫ60
66. \$500
67. x(x ϩ 5)
68. (x Ϫ 10)(x ϩ 10)
69. (x Ϫ 7)(x Ϫ 4)
70. (x Ϫ 9)2
71. (3x ϩ 2)(x ϩ 2)
72. 2(3x ϩ 2)(x Ϫ 3)
Lesson 6-3
Solving Quadratic Equations by Factoring
Pages 303–305
1. Sample answer: If the product of two factors is zero, then at least one of the factors must be zero.
2. Sample answer: roots
6 and Ϫ5; x 2 Ϫ x Ϫ 30 ϭ 0
3. Kristin; the Zero Product
Property applies only when one side of the equation is 0.
4. {0, 11}
5. {Ϫ8, 2}
6. {Ϫ7, 7}
7. {3}
8. eϪ , 4 f
3
4
10. x 2 ϩ 3x Ϫ 28 ϭ 0
9. {Ϫ3, 4}
11. 6x 2 Ϫ 11x ϩ 4 ϭ 0
12. 15x 2 ϩ 14x ϩ 3 ϭ 0
13. D
14. {Ϫ8, 3}
15. {Ϫ4, 7}
16. {Ϫ5, 5}
17. {Ϫ9, 9}
18. {Ϫ6, 3}
147
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20. e 0, f
5
3
19. {Ϫ3, 7}
21. e 0, Ϫ f
3
4
22. {6}
24.
23. {8}
25.
e
1
,
4
eϪ2,
1 f 4
26. eϪ , Ϫ f
4f
1
2
3
2
27. eϪ , Ϫ f
28. eϪ , Ϫ f
29. e ,
30. {2, 4}
31. {Ϫ3, 1}
32. 0, Ϫ6, 5
33. 0, Ϫ3, 3
34. x 2 Ϫ 9x ϩ 20 ϭ 0
35. x 2 Ϫ 5x Ϫ 14 ϭ 0
36. x 2 ϩ x Ϫ 20 ϭ 0
37. x 2 ϩ 14x ϩ 48 ϭ 0
38. 2x 2 Ϫ 7x ϩ 3 ϭ 0
39. 3x 2 Ϫ 16x ϩ 5 ϭ 0
40. 12x 2 Ϫ x Ϫ 6 ϭ 0
41. 10x 2 ϩ 23x ϩ 12 ϭ 0
42.
43. Ϫ14, Ϫ16
44. 12 cm by 16 cm
45. B ϭ D 2 Ϫ 8D ϩ 16
46. 4; The logs must have a diameter greater than 4 in. for the rule to produce positive board feet values.
47. y ϭ (x Ϫ p)(x Ϫ q) y ϭ x 2 Ϫ px Ϫ qx Ϫ pq y ϭ x 2 Ϫ (p ϩ q)x Ϫ pq a ϭ 1, b ϭ Ϫ(p ϩ q), c ϭ Ϫpq axis of symmetry:
48. Ϫ1
2
3
8
3
3
2
3 9 f 4 4
1
4
2
3
s
b
2a
Ϫ(p ϩ q)
Ϫ
2(1) pϩq 2
xϭϪ xϭ xϭ
148
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The axis of symmetry is the average of the x-intercepts.
Therefore the axis of symmetry is located halfway between the x-intercepts.
49. Ϫ6
50. Answers should include the following. • Subtract 24 from each side of x 2 ϩ 5x ϭ 24 so that the equation becomes x 2 ϩ 5x Ϫ 24 ϭ 0. Factor the left side as (x Ϫ 3)
(x ϩ 8). Set each factor equal to zero. Solving each equation for x. The solutions to the equation are 3 and Ϫ8. Since length cannot be negative, the width of the rectangle is
3 inches, and the length is
3 ϩ 5 or 8 inches.
• To use the Zero Product
Property, one side of the equation must equal zero.
51. D
52. B
53. Ϫ5, 1
54. Ϫ
55. between Ϫ1 and 0; between
3 and 4
56. min.; Ϫ19
57. 322 Ϫ 223
58. 523
1
2
59. 33 ϩ 20 22
60. (Ϫ4, Ϫ4)
66. 5i 22
61. (3, Ϫ5)
62. a , 2b
63. 222
64. 225
1
3
67. 2i 23
65. 323
˛
68. 4i23
˛
149
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Chapter 6
Practice Quiz 1
Page 305
2. max.;
1. 4; x ϭ 2; 2 f(x) 4
O
37
4
or 9
1
4
f (x) ϭ 3x 2 Ϫ 12x ϩ 4
4
8
12
x
Ϫ4
Ϫ8
(2, Ϫ8)
4. e Ϫ5, f
1
2
1
2
3. 1 , 4
5. 3x 2 ϩ 11x Ϫ 4 ϭ 0
Lesson 6-4 Completing the Square
Pages 310–312
1. Completing the square allows you to rewrite one side of a quadratic equation in the form of a perfect square. Once in this form, the equation is solved by using the Square Root
Property.
2. Never; the value of c that makes ax 2 ϩ bx ϩ c a perfect square trinomial is
3. Tia; before completing the square, you must first check to see that the coefficient of the quadratic term is 1. If it is not, you must first divide the equation by that coefficient.
4. {Ϫ10, Ϫ4}
5. e
b
the square of and the
2
square of a number can never be negative.
4 Ϯ 22 f 3
9. 54 Ϯ 256
7.
9
, ax
4
10. 5Ϫ1 Ϯ i 256
6. 36; (x Ϫ 6)2
Ϫ b
3 2
2
8. {Ϫ6, 3}
150
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11. e
3 Ϯ 233 f 4
16. 5Ϫ4 Ϯ 27, Ϫ4 Ϫ 276
12. Jupiter
13. Earth: 4.5 s, Jupiter: 2.9 s
15. {Ϫ2, 12}
Ϫ5 Ϯ 211 f 3
17. 53 Ϯ 2226
19. e
7 Ϯ 25 f 2
14. {3, Ϫ7}
18. e
20. e Ϫ ,
5
4
1 f 4
21. {Ϫ1.6, 0.2}
22. 25 ft
23. about 8.56 s
24. 64; (x ϩ 8)2
25. 81; (x Ϫ 9)2
26.
27.
49
; ax
4
Ϫ b
7 2
2
15 2 b 2
34. 5Ϫ1 Ϯ 276
30.
35. 52 Ϯ 236
25
, ax
16
Ϫ
28. 0.09; (x ϩ 0.3)2
29. 1.44; (x Ϫ 1.2)2
31.
225
; ax
4
ϩ b
5 2
4
16
; ax
9
Ϫ b
4 2
3
32. {3, 5}
33. {Ϫ12, 10}
36. 52 Ϯ i 6
5 Ϯ 213 f 6
37. {Ϫ3 Ϯ 2i}
38. e Ϫ , 1 f
39. e , 1 f
40. e
41. e
42. e
5
2
2 Ϯ 210 f 3
1
2
43. e
Ϫ5 Ϯ i 123 f 6
46. e Ϯ 23 f
47. e Ϯ 22 f
44. {Ϫ2, 0.6}
1
3
45. {0.7, 4}
3
4
49.
1 ϩ 25
2
48. 5
x
1
,
1 xϪ1
7 Ϯ i 247 f 4
50.
151
1
2
in. by 5
1
2
in.
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51. Sample answers: The golden rectangle is found in much of ancient Greek architecture, such as the Parthenon, as well as in modern architecture, such as in the windows of the United
Nations building. Many songs have their climax at a point occurring 61.8% of the way through the piece, with 0.618 being about the reciprocal of the golden ratio. The reciprocal of the golden ratio is also used in the design of some violins.
52a. n ϭ 0
52b. n Ͼ 0
52c. n Ͻ 0
53. 18 ft by 32 ft or 64 ft by 9 ft
54. To find the distance traveled by the accelerating race car in the given situation, you must solve the equation t 2 ϩ 22t ϩ 121 ϭ 246 or t 2 ϩ 22t Ϫ 125 ϭ 0.
Answers should include the following. • Since the expression t 2 ϩ 22t Ϫ 125 is prime, the solutions of t 2 ϩ 22t ϩ
121 ϭ 246 cannot be obtained by factoring.
• Rewrite t 2 ϩ 22t ϩ 121 as
(t ϩ 11)2. Solve (t ϩ 11)2 ϭ
246 by applying the
Square Root Property.
Then, subtract 11 from each side. Using a calculator, the two solutions are about 4.7 and
Ϫ26.7. Since time cannot be negative, the driver takes about 4.7 seconds to reach the finish line.
55. D
56. D
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57. x 2 Ϫ 3x ϩ 2 ϭ 0
58. x 2 Ϫ 6x Ϫ 27 ϭ 0
59. 3x 2 Ϫ 19x ϩ 6 ϭ 0
60. 12x 2 ϩ 13x ϩ 3 ϭ 0
61. between Ϫ4 and Ϫ3; between 0 and 1
62. 6, 8
1
2
3
63. Ϫ4, Ϫ1
64. 57
65. (2, Ϫ5)
66. a , b
43 6
21 7
67. 0 x Ϫ (Ϫ257)0 ϭ 2
68. greatest: Ϫ255ЊC; least:
Ϫ259ЊC
69. 37
70. Ϫ16
71. 121
72. 0
Lesson 6-5
The Quadratic Formula and the Discriminant
Pages 317–319
2. The square root of a negative number is a complex number.
y
x
O
1b. Sample answer: y O
x
153
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1c. Sample answer: y O
x
3. b 2 Ϫ 4ac must equal 0.
4a. 484
4b. 2 rational
4c.
2 Ϯ 22
2
1
,
4
5
2
Ϫ
5a. 8
5b. 2 irrational
6a. 0
6b. one rational
5c.
6c. Ϫ
1
2
Ϫ3 Ϯ i 23
2
10. 1 Ϯ 23
8. 0, Ϫ8
7a. Ϫ3
7b. two complex
7c.
Ϫ5 Ϯ i 22
2
9. Ϫ3, Ϫ2
11.
12. at about 0.7 s and again at about 4.6 s
Ϫ3 Ϯ 221
2
13. No; the discriminant of
Ϫ16t 2 ϩ 85t ϭ 120 is Ϫ455, indicating that the equation has no real solutions.
14a. 21
14b. 2 irrational
15a. 240
15b. 2 irrational
15c. 8 Ϯ 2215
16a. Ϫ16
16b. 2 complex
16c. 1 Ϯ 2i
17a. Ϫ23
17b. 2 complex
18a. 121
18b. 2 rational
17c.
18c. Ϫ ,
14c.
1 Ϯ i 223
2
1 2
4 3
19a. 49
19b. 2 rational
20a. 20
20b. 2 irrational
154
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20c. Ϫ2 Ϯ 25
21c. Ϫ1 Ϯ 26
19c. Ϫ2,
1
3
22a. 0
22b. one rational
21a. 24
21b. 2 irrational
22c.
23a. 0
23b. one rational
1
3
9 Ϯ i 231
8
28
9
24a. Ϫ31
24b. 2 complex
5
2
24c.
23c. Ϫ
Ϫ1 Ϯ i 215
4
2 Ϯ 4 27
9
25a. Ϫ135
26a.
25b. 2 complex
26b. 2 irrational
25c.
26c.
Ϫ1 Ϯ 2 20.37
0.8
27a. 1.48
27b. 2 irrational
27c.
33.
35.
30. 2 Ϯ i 23
221
7
Ϫ3 Ϯ 215
2
29. Ϯi
31.
28. Ϫ2, 32
34. Ϫ3 Ϯ i 27
36. 4 Ϯ 27
32. Ϯ22
5 Ϯ 246
3
9
2
3
10
37. 0, Ϫ
38. 3 Ϯ 222
39. Ϫ2, 6
40.
41. This means that the cables do not touch the floor of the bridge, since the graph does not intersect the x-axis and the roots are imaginary.
42. domain: 0 Յ t Յ 25, range:
73.7 Յ A(t ) Յ 1201.2
43. 1998
44. about 40.2 mph
155
Ϸ Ϫ0.00288
Algebra 2
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46. The person’s age can be substituted for A in the appropriate formula, depending upon their gender, and their average blood pressure calculated. See student’s work.
• If a woman’s blood pressure is given to be 118, then solve the equation 118 ϭ
0.01A2 ϩ 0.05A ϩ 107 to find the value of A. Use the
Quadratic Formula, substituting 0.01 for a, 0.05 for b, and Ϫ11 for c. This gives solutions of about
Ϫ35.8 or 30.8. Since age cannot be negative, the only valid solution for A is 30.8.
45a. k ϭ Ϯ6
45b. k Ͻ Ϫ6 or k Ͼ 6
45c. Ϫ6 Ͻ k Ͻ 6
49. Ϫ14, Ϫ4
50. 4 Ϯ 27
51.
52. Ϫ2, 0
1 Ϯ 2 22
2
47. D
48. C
2
,
3
53. Ϫ2, 7
54.
55. a 4b10
56. 10p6 0 q 0
57. 4b 2c 2
5
58. 7.98 ϫ 106
59.
60.
y
y xϭ1 xϩyϭ9 8 yϪxϭ4 6
4
2
Ϫ6 Ϫ4
O
Ϫ4
Ϫ6
yϭx
2 4 6 8 xϪy ϭ 3
x
x
O
y ϭ Ϫ1
61. no
62. yes; (x Ϫ 7)2
63. yes; (2x ϩ 3)2
64. yes; (5x ϩ 2)2
65. no
66. yes; (6x Ϫ 5)2
156
Algebra 2
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Lesson 6-6
Analyzing Graphs of Quadratic Functions
Pages 325–328
1a.
1b.
1c.
1d.
1e.
y ϭ 2(x ϩ 1)2 ϩ 5 y ϭ 2(x ϩ 1)2 y ϭ 2(x ϩ 3)2 ϩ 3 y ϭ 2(x Ϫ 2)2 ϩ 3
Sample answer: y ϭ 4(x ϩ 1)2 ϩ 3
1f. Sample answer: y ϭ (x ϩ 1)2 ϩ 3
1g. y ϭ Ϫ2(x ϩ 1)2 ϩ 3
2. Substitute the x-coordinate of the vertex for h and the y -coordinate of the vertex for k in the equation y ϭ a(x Ϫ h)2 ϩ
k. Then substitute the x-coordinate of the other point for x and the y-coordinate for y into this equation and solve for a. Replace a with this value in the equation you wrote with h and k.
3. Sample answer: y ϭ 2(x Ϫ 2)2 Ϫ1
4. Jenny; when completing the square is used to write a quadratic function in vertex form, the quantity added is then subtracted from the same side of the equation to maintain equality.
5. (Ϫ3, Ϫ1); x ϭ Ϫ3; up
6. y ϭ (x ϩ 4)2 Ϫ 19,
(Ϫ4, Ϫ19); x ϭ Ϫ4; up
7. y ϭ Ϫ3(x Ϫ 3)2 ϩ 38;
(Ϫ3, 38); x ϭ Ϫ3; down
8.
y
y ϭ 3(x ϩ 3)2
O x
9.
10.
y
y
O
y ϭ Ϫ 2x 2 ϩ 16x Ϫ 31 x 1 y ϭ 3 (x Ϫ 1)2 ϩ 3
O
x
12. y ϭ Ϫ(x ϩ 3)2 ϩ 6
11. y ϭ 4(x Ϫ 2)2
157
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1
2
13. y ϭ Ϫ (x ϩ 2)2 Ϫ 3
14. h(d ) ϭ Ϫ2d 2 ϩ 4d ϩ 6
15. (Ϫ3, 0); x ϭ Ϫ3 down
16. (1, 2); x ϭ 1; up
17. (0,Ϫ6); x ϭ 0 up
18. (0, 3); x ϭ 0; down
19. y ϭ Ϫ(x ϩ 2)2 ϩ 12;
(Ϫ2, 12); x ϭ Ϫ2; down
20. y ϭ (x Ϫ 3)2 Ϫ 8;
(3, Ϫ8); x ϭ 3; up
21. y ϭ Ϫ3(x Ϫ 2)2 ϩ 12;
(2, 12); x ϭ 2; down
22. y ϭ 4(x ϩ 3)2 Ϫ 36;
(Ϫ3, Ϫ36); x ϭ Ϫ3; up
23. y ϭ 4(x ϩ 1)2 Ϫ 7; (Ϫ1, Ϫ7); x ϭ Ϫ1; up
24. y ϭ Ϫ2(x Ϫ 5)2 ϩ 15;
(5, 15); x ϭ 5; down
25. y ϭ 3 ax ϩ b Ϫ ;
26. y ϭ 4 ax Ϫ b Ϫ 20;
1 2
2
1 aϪ ,
2
3 2
2
7
4
Ϫ b; x ϭ Ϫ ; up
7
4
3 a ,
2
1
2
27.
Ϫ20b; x ϭ , up
3
2
28.
y
y
O
y ϭ 4(x ϩ 3)2 ϩ 1
29.
y ϭ Ϫ(x Ϫ 5)2 Ϫ 3
x
x
O
30.
y
y
O
x
1 y ϭ 4 (x Ϫ 2)2 ϩ 4
31.
1 y ϭ 2 (x Ϫ 3)2 Ϫ 5
x
O
32.
y
y
O x
y ϭ x 2 Ϫ 8x ϩ 18
O
x
y ϭ x 2 ϩ 6x ϩ 2
158
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33.
y
34.
y ϭ Ϫ4x 2 ϩ 16x Ϫ 11
y ϭ Ϫ5x 2 Ϫ 40x Ϫ 80
y
O
O
35.
y
x
x
36.
y ϭ Ϫ 1 x 2 ϩ 5x Ϫ 27
2
2
y
x
O
O
x
y ϭ 1 x 2 Ϫ 4x ϩ 15
3
37. Sample answer: the graph of y ϭ 0.4(x ϩ 3)2 ϩ 1 is narrower than the graph of y ϭ 0.2(x ϩ 3)2 ϩ 1.
38. Sample answer: the graphs have the same shape, but the graph of y ϭ 2(x Ϫ 4)2 ϩ 1 is
1 unit to the left and 5 units below the graph of y ϭ
2(x Ϫ 5)2 Ϫ 4
39. y ϭ 9(x Ϫ 6)2 ϩ 1
40. y ϭ 3(x ϩ 4)2 ϩ 3
2
3
41. y ϭ Ϫ (x Ϫ 3)2
42. y ϭ Ϫ3(x Ϫ 5)2 ϩ 4
5
2
1
3
43. y ϭ x 2 ϩ 5
44. y ϭ (x ϩ 3)2 Ϫ 2
45. y ϭ Ϫ2x 2
46. y ϭ (x ϩ 3)2 Ϫ 4
47. 34,000 feet; 32.5 s after the aircraft begins its parabolic flight 48. about 1.6 s
49. d (t ) ϭ Ϫ16t 2 ϩ 8t ϩ 50
50. about 2.0 s
51. Angle A; the graph of the equation for angle A is higher than the other two since 3.27 is greater than
2.39 or 1.53.
52. Angle B; the vertex of the equation for angle B is farther to the right than the other two since 3.57 is greater than 3.09 or 3.22.
4
3
159
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53.
54. All quadratic equations are a transformation of the parent graph y ϭ x 2. By identifying these transformations when a quadratic function is written in vertex form, you can redraw the graph of y ϭ x 2.
Answers should include the following. • In the equation y ϭ a(x Ϫ h)2 ϩ k, h translated the graph of y ϭ x 2 h units to the right when h is positive and h units to the left when h is negative. The graph of y ϭ x 2 is translated k units up when k is positive and k units down when k is negative. When a is positive, the graph opens upward and when a is negative, the graph opens downward. If the absolute value of a is less than 1, the graph will be narrower than the graph of y ϭ x 2, and if the absolute value of a is greater than 1, the graph will be wider than the graph of y ϭ x 2.
• Sample answer: y ϭ 2(x ϩ 2)2 Ϫ 3 is the graph of y ϭ x 2 translated
2 units left and 3 units down. The graph opens upward, but is narrower than the graph of y ϭ x 2.
y ϭ ax 2 ϩ bx ϩ c y ϭ a ax 2 ϩ xb ϩ c b a
y ϭ a cx 2 ϩ x ϩ a b d ϩ b a
b 2
2a
c Ϫ aa b
b 2
2a
y ϭ a ax ϩ
b 2 b 2a
ϩcϪ
b2
4a
The axis of symmetry is x ϭ h or Ϫ
b
.
2a
55. D
56. B
160
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60. 5Ϫ5 Ϯ 2226
59. Ϫ23; 2 complex
61. 53 Ϯ 3i 6
63. 2t 2 ϩ 2t Ϫ
Ϫ2 Ϯ 213 f 2
58. 225; 2 rational
57. 12; 2 irrational
62. e
3
tϪ1
64. t 2 Ϫ 2t ϩ 1
65. n 3 Ϫ 3n 2 Ϫ 15n Ϫ 21
66. y 3 ϩ 1 Ϫ
67a. Sample answer using (1994,
76,302) and (1997, 99,448): y ϭ 7715x Ϫ 15,307,408
67b. 161,167
68. yes
69. no
4 yϩ3 70. yes
71. no
Chapter 6
Practice Quiz 2
Page 328
1. 5Ϫ7 Ϯ 2236
2. e
3. Ϫ11; 2 complex
4. 100; 2 rational
5. e
6. e
Ϫ9 Ϯ 5 25 f 2
7. y ϭ 1x Ϫ 22 2 Ϫ 5
2
3
8. y ϭ (x ϩ 4)2 ϩ 2; (Ϫ4, 2), x ϭ Ϫ4; up
9. y ϭ Ϫ1x Ϫ 62 2; 16, 02, x ϭ 6; down ©Glencoe/McGraw-Hill
2 Ϯ 2i 22 f 3
1 Ϯ 3i f 2
10. y ϭ 2(x ϩ 3)2 Ϫ 5; (Ϫ3, Ϫ5), x ϭ Ϫ3; up
161
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Lesson 6-7
Graphing and Solving Quadratic Inequalities
Pages 332–335
1. y Ն 1x Ϫ 32 2 Ϫ 1
2. Sample answer: one number less than Ϫ3, one number between Ϫ3 and 5, and one number greater than 5
4.
3a. x ϭ Ϫ1, 5
3b. x Յ Ϫ1 or x Ն 5
3c. Ϫ1 Յ x Յ 5
y
O
5.
12
8
4
Ϫ4
Ϫ2
Ϫ4
Ϫ8
Ϫ12
y
12
y ϭ Ϫ2x 2 Ϫ 4x ϩ 3
x
y
O
Ϫ20
7.
6.
y
y ϭ x 2 Ϫ 10x ϩ 25
2
4x
O
x
y ϭ x 2 Ϫ 16
8. x Ͻ 1 or x Ͼ 5
y ϭ Ϫx 2 ϩ 5x ϩ 6
8
4
Ϫ2
O
2
4
12. 5x 0 Ϫ 23 Յ x Յ 236
6x
9. 5x 0 Ϫ1 Ͻ x Ͻ 76
10. 5x 0 x Ͻ Ϫ3 or x Ͼ 46
11. л
162
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13. about 6.1 s
14.
15
y
5
Ϫ8
Ϫ4
Ϫ5
O
8x
4
Ϫ15
y ϭ x 2 ϩ 3x Ϫ 18
Ϫ25
15.
y
16.
y ϭ Ϫx 2 ϩ 7x ϩ 8
y
12
8
4
Ϫ4
O
17.
4
x
8
y ϭ x 2 ϩ 4x ϩ 4
18.
y
5
Ϫ8
O
x
O
y
8x
4
Ϫ4 O
Ϫ10
x
Ϫ20
Ϫ30
y ϭ x 2 ϩ 4x
y ϭ x 2 Ϫ 36
19.
20.
y
y ϭ Ϫx 2 Ϫ 3x ϩ 10
14
y
10
O
x
6
2
Ϫ6
y ϭ x 2 ϩ 6x ϩ 5
163
Ϫ4
Ϫ2 O
2x
Algebra 2
Chapter 6
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21.
y ϭ Ϫx 2 Ϫ 7x ϩ 10
22.
y
y y ϭ Ϫx 2 ϩ 10x Ϫ 23
20
12
x
O
4
Ϫ12 Ϫ8
23.
y
O
Ϫ4
Ϫ4
4x
24.
y ϭ Ϫx 2 ϩ 13x Ϫ 36
y
4
6
2
O
Ϫ2
2
6
10
x
2
O
4
x
Ϫ4
Ϫ4
Ϫ8
y ϭ 2x 2 ϩ 3x Ϫ 5
Ϫ8
25.
26. 5
y
x
O
y ϭ 2x 2 ϩ x Ϫ 3
27. Ϫ2 Յ x Յ 6
28. x Ͻ Ϫ3 or x Ͼ 3
29. x Ͻ Ϫ7 or x Ͼ Ϫ3
30. 5x 0 x Ͻ Ϫ3 or x Ͼ 66
33. 5x 0 x Յ Ϫ6 or x Ն 46
34. 5x 0 Ϫ4 Յ x Յ 36
31. 5x 0 Ϫ7 Ͻ x Ͻ 46
32. 5x 0 Ϫ1 Յ x Յ 56
35. 5x 0 x Յ Ϫ7 or x Ն 16
36. e x ` x ϭ
37. all reals
38. л
39. 5x 0 x ϭ 76
40. all reals
42. 5x 0 Ϫ4 Ͻ x Ͻ 1 or x Ͼ 36
41. л
43. 0 to 10 ft or 24 to 34 ft
1 f 3
44a. 0.98, 4.81; The owner will break even if he charges
\$0.98 or \$4.81 per square foot. 164
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44b. 0.98 Ͻ r Ͻ 4.81; The owner will make a profit if the rent is between \$0.98 and \$4.81.
44c. 1.34 Ͻ r Ͻ 4.45; If rent is set between \$1.34 and \$4.45 per sq ft, the profit will be greater than \$10,000
44d. r Ͻ 1.34 or r Ͼ 4.45; If rent is set between \$0 and \$1.34 or above \$4.45 per sq ft, the profit will be less than \$10,000.
45. The width should be greater than 12 cm and the length should be greater than 18 cm
46. P(n) ϭ n[15 ϩ 1.5(60 Ϫ n)] Ϫ
525 or Ϫ1.5n 2 ϩ 105n Ϫ 525
47. 6
48. \$1312.50; 35 passengers
49.
50. Answers should include the following. • Ϫ16t 2 ϩ 42t ϩ 3.75 Ͼ 10
• One method of solving this inequality is to graph the related quadratic function h(t ) ϭ Ϫ16t 2 ϩ 42t ϩ
3.75 Ϫ 10. The interval(s) at which the graph is above the x-axis represents the times when the trampolinist is above 10 feet. A second method of solving this inequality would be to find the roots of the related quadratic equation Ϫ16t 2 ϩ
42t ϩ 3.75 Ϫ 10 ϭ 0 and then test points in the three intervals determined by these roots to see if they satisfy the inequality. The interval(s) at which the inequality is satisfied represent the times when the trampolinist is above
10 feet.
y y ϭ Ϫx 2 ϩ 4
O
x y ϭ x2 Ϫ 4
165
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52. A
51. C
53. 5x 0 all reals, x
54. 5x 0 Ϫ7 Ͻ x Ͻ 76
26
55. 5x 0 x Ͻ Ϫ9 or x Ͼ 36
56. 5x 0 x Յ Ϫ3.5 or x Ն Ϫ2.56
57. 5x 0 Ϫ1.2 Յ x Յ Ϫ0.46
58. no real solutions
60. y ϭ Ϫ21x Ϫ 42 2; 14, 02, x ϭ 4; down 59. y ϭ (x Ϫ 1)2 ϩ 8; (1, 8), x ϭ 1; up
Ϫ5 Ϯ i 23
2
1
2
61. y ϭ (x ϩ 6)2; (Ϫ6, 0), x ϭ Ϫ6; up
63.
64.
66. Ϫ6x 3 Ϫ 4x 2y ϩ 13xy 2
65. 4a 2b 2 ϩ 2a 2b ϩ 4ab 2 ϩ
12a Ϫ7b
67. xy 3 ϩ y ϩ
69. B
1 x 68. Ϫ15a 2 ϩ 14a Ϫ 3
Ϫ21 48
R
Ϫ13 22
70. 3Ϫ54 64
71. 0x Ϫ 0.008 0 Յ 0.002;
0.078 Յ x Յ 0.082
Ϫ3 Ϯ 2 26
3
62. Ϫ4, Ϫ8
166
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Chapter 7 Polynomial Functions
Lesson 7-1 Polynomial Functions
Pages 350–352
1. 4 ϭ 4x 0; x ϭ x 1
2. Sample answer: Evendegree polynomial functions with positive leading coefficients have graphs in which f (x ) S ϩϱ as x S ϩϱ and as x SϪϱ. Odd-degree polynomial functions with positive leading coefficients have graphs in which f (x ) S ϩϱ as x S ϩϱ and f (x ) S Ϫϱ as x S Ϫϱ.
3. Sample answer given.
4. Sometimes; a polynomial function with 4 real roots may be a sixth-degree polynomial function with
2 imaginary roots. A polynomial function that has
4 real roots is at least a fourth-degree polynomial.
f (x)
O
x
5. 6; 5
6. 5; Ϫ3
7. Ϫ21; 3
8. 4; 12
9. 2a 9 ϩ 6a 3Ϫ12
10. 100a 2 ϩ 20
11. 6a 3 Ϫ 5a 2 ϩ 8a Ϫ 45
12a. f (x ) S Ϫϱ as x S ϩϱ, f (x ) S ϩϱ as x S Ϫϱ
12b. odd
12c. 3
13a. f(x ) S ϩϱ as x S ϩϱ, f(x ) S ϩϱ as x S Ϫϱ
13b. even
13c. 0
14a. f (x ) S ϩϱ as x S ϩϱ, f (x ) S Ϫϱ as x S Ϫϱ
14b. odd
14c. 1
15. 109 lumens
16. 1; Ϫ1
17. 3; 1
18. No, the polynomial contains two variables, a and b.
19. 4; 6
20. 3; Ϫ5
167
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21. No, this is not a polynomial
22. Ϫ2; 4
1
because the term cannot c be written in the form x n, where n is a nonnegative integer. 23. 12; 18
24. 125; Ϫ37
25. 1008; Ϫ36
26. Ϫ166; 50
27. 86; 56
28. 100; 4
29. 7; 4
30. 27a3 ϩ 3a ϩ 1
31. 12a 2 Ϫ 8a ϩ 20
32. 3a4 Ϫ 2a 2 ϩ 5
33. 12a6 Ϫ 4a3 ϩ 5
34. x 3 ϩ 3x 2 ϩ 4x ϩ 3
35. 3x 4 ϩ 16x 2 ϩ 26
36. 6x 2 ϩ 44x ϩ 90
37. Ϫx 6 ϩ x 3 ϩ 2x 2 ϩ 4x ϩ 2
38. 9x 4 Ϫ 12x 2 Ϫ 8x ϩ 50
39a. f (x ) S ϩϱ as x S ϩϱ, f (x ) S Ϫϱ as x S Ϫϱ
39b. odd
39c. 3
40a. f (x) S ϩϱ as x S ϩϱ, f (x) S ϩϱ as x S Ϫϱ
40b. even
40c. 4
41a. f (x) S ϩϱ as x S ϩϱ, f (x) S Ϫϱ as x S Ϫϱ
41b. even
41c. 0
42a. f (x ) S ϩϱ as x S ϩϱ, f (x ) S ϩϱ as x S Ϫϱ
42b. odd
42c. 5
43a. f (x) S ϩϱ as x S ϩϱ, f (x) S Ϫϱ as x S Ϫϱ
43b. odd
43c. 1
44a. f (x ) S Ϫϱ as x S ϩϱ, f (x ) S Ϫϱ as x S Ϫϱ
44b. even
44c. 2
45. 5.832 units
46. even
47. f (x ) S Ϫϱ as x S ϩϱ; f(x ) S Ϫϱ as x S Ϫϱ
48. Sample answer: Decrease; the graph appears to be turning at x ϭ 30 indicating a relative maximum at that point. So attendance will decrease after 2000.
49.
1
2
50. Ϫ1, 0, 4
168
Algebra 2
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2
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3
2
51. f(x) ϭ x 3 Ϫ x 2 Ϫ 2x
52.
8
f (x )
f (x) ϭ 1 x 3 Ϫ 3 x 2Ϫ 2x
4 2
2
Ϫ4 Ϫ2
O
2
x
Ϫ4
Ϫ8
53. 4
54. 16 regions
55. 8 points
56. Many relationships in nature can be modeled by polynomial functions; for example, the pattern in a honeycomb or the rings in a tree trunk. Answers should include the following.
• You can use the equation to find the number of hexagons in a honeycomb with 10 rings and the number of hexagons in a honeycomb with 9 rings.
The difference is the number of hexagons in the tenth ring.
• Other examples of patterns found in nature include pinecones, pineapples, and flower petals.
57. C
58. C
60. 5x 0 x Յ Ϫ9 or x Ն 76
59. 5x 0 2 Ͻ x Ͻ 66
169
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61. e x ` Ϫ1 Յ x Յ f
4
5
62.
y y ϭ Ϫ2(x Ϫ 2)2 ϩ 3
x
O
63.
64.
y
y
2
O
Ϫ12 Ϫ8
x
y ϭ 1 x2 ϩ x ϩ 3
Ϫ2
y ϭ 1 (x ϩ 5)2 Ϫ 1
3
x
O
2
2
Ϫ4
65. 54 Ϯ 3226
66. e Ϫ ,
7
6
67. 23,450(1 ϩ p);
23,450(1 ϩ p)3
68.
5 f 6 y y ϭ x2 ϩ 4
x
O
69.
70.
y
8
y
4
O
x
Ϫ8
Ϫ4 O
4
8x
Ϫ4
1
y ϭ 2 x 2 ϩ 2x Ϫ 6
y ϭ Ϫx 2 ϩ 6x Ϫ 5
170
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Lesson 7-2
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Graphing Polynomial Functions
Pages 356–358
1. There must be at least one real zero between two points on a graph when one of the points lies below the x-axis and the other point lies above the x-axis.
2. 4
3.
4.
f (x)
x
p
f(x )
8
Ϫ3 Ϫ20
O
Ϫ2
x
Ϫ1
0
f (x ) ϭ x 3 Ϫ x 2 Ϫ 4x ϩ 4
1
2
3
5. x p
0
f (x )
4
6
4
Ϫ4
Ϫ2 O
0
2
4x
Ϫ4
0
f(x) ϭ x 3 Ϫ x 2 Ϫ 4x ϩ 4
10
6. between Ϫ1 and 0 f(x )
f (x)
8
Ϫ3 20
Ϫ2 Ϫ9
Ϫ1 Ϫ2
0
5
1
0
2 Ϫ5
3 26
f (x)
4
Ϫ4
Ϫ2 O
2
4x
x
O
Ϫ4
f (x ) ϭ x 3 Ϫ x 2 ϩ 1
Ϫ8
f (x ) ϭ x 4 Ϫ 7x 2 ϩ x ϩ 5
8.
7. between Ϫ2 and Ϫ1, between
Ϫ1 and 0, between 0 and 1, and between 1 and 2
8
f (x)
4
f (x)
Ϫ4
Ϫ2
O
2
4x
Ϫ4
O
Ϫ8
x
f (x ) ϭ x 3 ϩ 2x 2 Ϫ 3x Ϫ 5
Sample answer: rel. max. at x ϭ Ϫ2, rel. min.: at x ϭ 0.5
f (x ) ϭ x 4 Ϫ 4x 2 ϩ 2
171
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10.
f (x)
C (t )
12000
8
4
O
Ϫ2
2
Cable TV Systems
Ϫ4
10000
4x
Ϫ4
f (x ) ϭ x 4 Ϫ 8x 2 ϩ 10
Sample answer: rel. max. at x ϭ 0, rel. min. at x ϭ Ϫ2 and at x ϭ 2
8000
6000
C(t) ϭ Ϫ43.2t 2 ϩ 1343t ϩ 790
4000
2000
O
x
p
14a. f (x )
Ϫ5 25
Ϫ4
0
Ϫ3 Ϫ9
Ϫ2 Ϫ8
Ϫ1 Ϫ3
0
0
1 Ϫ5
2 Ϫ24
x f (x )
2
Ϫ2
4x
Ϫ4
Ϫ8
f (x) ϭ Ϫx 3 Ϫ 4x 2
8
4
Ϫ4
Ϫ2 O
2
4x
Ϫ4
f (x) ϭ x 3 Ϫ 2x 2ϩ 6
14b. between Ϫ2 and Ϫ1
14c. Sample answer: rel. max. at x ϭ 0, rel. max. at x ϭ 1
13b. at x ϭ Ϫ4 and x ϭ 0
13c. Sample answer: rel. max. at x ϭ 0, rel. min. at x ϭ Ϫ3
p
f (x )
f (x )
Ϫ2 Ϫ10
Ϫ1
3
0
6
1
5
2
6
3 15
4 38
4
O
t
8
12
16
Years Since 1985
12. The number of cable TV systems rose steadily from
1985 to 2000. Then the number began to decline.
11. rel. max. between x ϭ 15 and x ϭ 16, and no rel. min.; f(x ) S Ϫϱ as x S Ϫϱ, f(x ) S Ϫϱ as x S ϩϱ.
13a.
4
172
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f (x )
f (x )
Ϫ2 Ϫ18
Ϫ1 Ϫ2
0
2
1
0
2 Ϫ2
3
2
4 18
x
x
O
Ϫ2
3
Ϫ1 Ϫ5
0 Ϫ9
1 Ϫ3
2 19
f (x) ϭ x 3 Ϫ 3x 2ϩ 2
x
p
Ϫ1 75
0 16
1 Ϫ3
2
0
3
7
4
0
5 Ϫ39
x
f (x )
Ϫ2
Ϫ1
0
1
2
3
4
5
4
Ϫ4
Ϫ2
O
2
4x
Ϫ4
Ϫ8
f (x) ϭ Ϫ3x 3 ϩ 20x 2 Ϫ 36x ϩ 16
17b. between 0 and 1, at x ϭ 2, and at x ϭ 4
17c. Sample answer: rel. max. at x ϭ 3, rel. min. at x ϭ 1
4
Ϫ8
Ϫ4
O
4
8x
Ϫ4
Ϫ8
f (x ) ϭ x 3 ϩ 5x 2 Ϫ 9
16b. between Ϫ5 and Ϫ4, between Ϫ2 and Ϫ1, and between 1 and 2
16c. Sample answer: rel. max. at x ϭ Ϫ3, rel. min. at x ϭ 0
18a.
f (x )
f (x )
f (x )
Ϫ5 Ϫ9
Ϫ4
7
Ϫ3
9
15b. at x ϭ 1, between Ϫ1 and 0, and between 2 and 3
15c. Sample answer: rel. max. at x ϭ 0, rel. min. at x ϭ 2
17a.
p
16a.
p
f (x )
Ϫ29
Ϫ8
Ϫ1
Ϫ2
Ϫ5
Ϫ4
7
34
f (x ) x O
f (x ) ϭ x 3 Ϫ 4x 2ϩ 2x Ϫ 1
18b. between 3 and 4
18c. Sample answer: rel. max. at x ϭ 0.5, rel. min. at x ϭ 2.5
173
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19a.
p
x
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f (x )
Ϫ3 73
Ϫ2
8
Ϫ1 Ϫ7
0 Ϫ8
1 Ϫ7
2
8
3 73
x
f (x )
Ϫ4
O
Ϫ2
2
x
x
Ϫ4
f (x) ϭ x 4 Ϫ 8 Ϫ8
p
Ϫ4 Ϫ169
Ϫ3 Ϫ31
Ϫ2
7
Ϫ1
5
0 Ϫ1
1
1
2 Ϫ1
3 Ϫ43
8
x
f (x )
Ϫ2
O
2
4x
Ϫ4
Ϫ8
f (x ) ϭ Ϫx 4 ϩ 5x 2Ϫ 2x Ϫ 1
21b. between Ϫ3 and Ϫ2, between
Ϫ1 and 0, between 0 and 1, and between 1 and 2
21c. Sample answer: rel. max. at x ϭ Ϫ2 and at x ϭ 1.5, rel. min. at x ϭ 0
f (x )
8
Ϫ4
Ϫ2
O
x
2
Ϫ8
Ϫ16
f (x ) ϭ x 4 Ϫ 10x 2ϩ 9
p
f (x )
Ϫ3 Ϫ39
Ϫ2
5
Ϫ1
3
0 Ϫ3
1
5
2 21
3 15
4 Ϫ67
4
Ϫ4
16
20b. at x ϭ Ϫ3, x ϭ Ϫ1, x ϭ 1, and x ϭ 3
20c. Sample answer: rel. max. at x ϭ 0, rel. min. at x ϭ Ϫ2 and x ϭ 2
22a.
f (x )
f (x )
Ϫ3
0
Ϫ2 Ϫ15
Ϫ1
0
0
9
1
0
2 Ϫ15
3
0
4 105
4
19b. between Ϫ2 and Ϫ1 and between 1 and 2
19c. Sample answer: no rel. max., rel. min. at x ϭ 0
21a.
p
20a.
24
f (x )
16
8
Ϫ4
Ϫ2
O
2
4x
Ϫ8
f (x) ϭ Ϫx 4 ϩ x 3ϩ 8x 2Ϫ 3
22b. between Ϫ3 and Ϫ2, between
Ϫ1 and 0, between 0 and 1, and between 3 and 4
22c. Sample answer: rel. max. at x ϭ Ϫ1.5 and at x ϭ 2.5, rel. min. at x ϭ 0.
174
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24a.
p
x f (x )
Ϫ1 65
0
6
1 Ϫ1
2
2
3 Ϫ3
4 Ϫ10
5 11
x
f (x )
4
Ϫ2
O
2
4
Ϫ2
Ϫ1
0
1
2
3
x
Ϫ4
Ϫ8
f (x) ϭ x 4 Ϫ 9x 3 ϩ 25x 2 Ϫ 24x ϩ 6
p
f (x )
2
O
4
6x
Ϫ4
Ϫ8
f (x) ϭ 2x 4 Ϫ 4x 3 Ϫ 2x 2 ϩ 3x Ϫ 5
p
x f (x )
24
Ϫ2
Ϫ1
16
0
8
Ϫ4
Ϫ2
O
1
2
2
4x
3
f (x) ϭ x 5 ϩ 4x 4 Ϫ x 3 Ϫ 9x 2 ϩ 3
4
25b. between Ϫ4 and Ϫ3, between
Ϫ2 and Ϫ1, between Ϫ1 and
0, between 0 and 1, and between 1 and 2
25c. Sample answer: rel. max. at x ϭ Ϫ3 and at x ϭ 0, rel. min. at x ϭ Ϫ1 and at x ϭ 1
Ϫ2
f (x )
26a.
x f (x )
Ϫ4 Ϫ77
Ϫ3 30
Ϫ2
7
Ϫ1 Ϫ2
0
3
1 Ϫ2
2 55
45
Ϫ4
Ϫ5
Ϫ6
Ϫ7
40
4
24b. between Ϫ2 and Ϫ1, and between 2 and 3
24c. Sample answer: rel. max. at x ϭ 0.5; rel. min. at x ϭ
Ϫ0.5 and at x ϭ 1.5
23b. between 0 and 1, between 1 and 2, between 2 and 3, and between 4 and 5
23c. Sample answer: rel. max. at x ϭ 2, rel. min. at x ϭ 0.5 and at x ϭ 4
25a.
p
f (x )
5
40
Ϫ88
5
f (x )
20
Ϫ6
5
20
Ϫ4
Ϫ2
O
2
4x
Ϫ20
Ϫ3
Ϫ10
269
Ϫ40
f (x) ϭ x 5 Ϫ 6x 4 ϩ 4x 3 ϩ 17x 2 Ϫ 5x Ϫ 6
26b. between Ϫ2 and Ϫ1, between
Ϫ1 and 0, between 0 and 1, between 2 and 3, and between 4 and 5
26c. Sample answer: rel. max. at x ϭ Ϫ1 and at x ϭ 2, rel. min. at x ϭ 0 and at x ϭ 3.5
175
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27. highest: 1982; lowest: 2000
28. Rel. max. between 1980 and
1985 and between 1990 and
1995, rel. min. between 1975 and 1980 and between 1985 and 1990; as the number of years increases, the percent of the labor force that is unemployed decreases.
29. 5
30. Sample answer: increase, based on the past fluctuations of the graph
31.
x
0 2 4 6 8 10
B(x) 25 34 40 45 50 54
G(x) 26 33 39 44 49 53
32. The growth rate for both boys and girls increases steadily until age 18 and then begins to level off, with boys averaging a height of
71 in. and girls a height of
60 in.
x 12 14 16 18 20
B(x) 59 64 68 71 71
G(x) 56 59 61 61 60
Average Height (in.)
y
B (x )
70
65
60
55
G (x )
50
45
40
35
30
25
20
0
2
4
6
8 10 12 14 16 18 x
Age (yrs)
33. 0 and between 5 and 6
34. 5 s
35. 3 s
36.
y
O
176
x
Algebra 2
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y
37.
38.
x
O
O
x
O
y
39.
y
40. The turning points of a polynomial function that models a set of data can indicate fluctuations that may repeat. Answers should include the following.
• To determine when the percentage of foreign-born citizens was at its highest, look for the rel. max. of the graph, which is at t ϭ 5.
The lowest percentage is found at t ϭ 75, the rel. min. of the graph.
• Polynomial equations best model data that contain turning points, rather than a constant increase or decrease like linear equations. x
41. D
42. B
43. Ϫ1.90; 1.23
44. 3.41; 0.59
45. 0; Ϫ1.22, 1.22
46. 0.52; Ϫ0.39, 1.62
47. 24a3 Ϫ 4a 2 Ϫ 2
48. 10c 2 Ϫ 25c ϩ 20
49. 8a4 Ϫ 10a 2 ϩ 4
50. 3x 3 Ϫ 10x 2 ϩ 11x Ϫ 6
51. 2x 4 ϩ 11x 2 ϩ 16
52. 4x 4 Ϫ 9x 3 ϩ 28x 2 Ϫ 33x ϩ 20
177
Algebra 2
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53.
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54.
y
y ϭ x 2 Ϫ 4x ϩ 6
x
O
x
O
y
y ϭ Ϫx 2 ϩ 6x Ϫ 3
55.
56. (7, Ϫ4)
y
x
O
2
y ϭ x Ϫ 2x
57. (Ϫ3, Ϫ2)
58. (4, Ϫ2)
59. (1, 3)
60. 1 ft
61. (x ϩ 5)(x Ϫ 6)
62. (2b Ϫ 1)(b Ϫ 4)
63. (3a ϩ 1)(2a ϩ 5)
64. (2m ϩ 3)(2m Ϫ 3)
65. (t Ϫ 3)(t 2 ϩ 3t ϩ 9)
66. (r 2 ϩ 1)(r ϩ 1)(r Ϫ 1)
Lesson 7-3 Solving Equations Using
Pages 362–364
16x 4 Ϫ 12x 2 ϭ 0;
4[4(x 2)2 Ϫ 3x 2] ϭ 0
2. The solutions of a polynomial equation are the points at which the graph intersects the x-axis.
3. Factor out an x and write the equation in quadratic form so you have x[(x 2)2 Ϫ 2(x 2) ϩ
1] ϭ 0. Factor the trinomial and solve for x using the
Zero Product Property. The solutions are Ϫ1, 0, and 1.
4. not possible
178
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5. 84(n 2 )2 Ϫ 62(n 2)
6. 0, Ϫ5, Ϫ4
7. Ϫ4, Ϫ1, 4, 1
8. 6, Ϫ3 ϩ 3i13, Ϫ3 Ϫ 3i13
9. 64
10. 8 feet
11. 2(x 2)2 ϩ 6(x 2) Ϫ 10
12. not possible
13. 11(n 3 )2 ϩ 44(n 3)
14. b[7(b 2)2 Ϫ 4(b 2 ) ϩ 2)]
15. not possible
16. 6 (x 5 )2 Ϫ 4 (x 5 ) Ϫ 16 ϭ 0
17. 0, Ϫ4, Ϫ3
18. 0, Ϫ1, Ϫ5
19. Ϫ13, 13, Ϫi 13, i 13
21. 2, Ϫ2, 212, Ϫ212
22. 12, Ϫ12, 3, Ϫ3
23. Ϫ9,
24. 8, Ϫ4 ϩ 4i13, Ϫ4 Ϫ 4i13
1
9 ϩ 9i 13 9 Ϫ 9i 13
,
2
2
1
20. 0, Ϫ4, 4, Ϫ4i, 4i
30. 8, i 23, Ϫi 23
25. 81, 625
26. Ϫ343, Ϫ64
27. 225, 16
28. 400
29. 1, Ϫ1, 4
31. w ϭ 4 cm, / ϭ 8 cm, h ϭ 2 cm
32. x 4 Ϫ 7x 2 ϩ 9 ϭ 27
33. 3 ϫ 3 in.
34. 6 ϫ 6 in.
35. h 2 ϩ 4, 3h ϩ 2, h ϩ 3
36. The height increased by 3, the width increased by 2, and the length increased by 4.
37. Write the equation in quadratic form, u 2 Ϫ 9x ϩ
8 ϭ 0, where u ϭ |a Ϫ 3|.
Then factor and use the Zero
Product Property to solve for a; 11, 4, 2, and Ϫ5.
38. Answers should include the following. • Solve the cubic equation
4x 3 ϩ (Ϫ164x 2) ϩ
1600x ϭ 3600 in order to determine the dimensions of the cut square if the desired volume is 3600 in3.
Solutions are 10 in. and
31Ϫ 2601
2
in.
• There can be more than one square cut to produce the same volume because the height of the box is not specified and 3600 has a variety of different factors.
179
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39. D
40.
p
1
18
42.
41. x f (x )
f (x )
Ϫ2 Ϫ21
Ϫ1 Ϫ1
0
5
1
3
2 Ϫ1
3 Ϫ1
4
9
5 35
O
x
Ϫ1 15
0 Ϫ3
1
1
2
3
3
3
4 25
x
f (x) ϭ x 3 Ϫ 4x 2 ϩ x ϩ 5
1715
;
3
x
O
f (x) ϭ x 4 Ϫ 6x 3 ϩ 10x 2 Ϫ x Ϫ 3
44. 262; 2
Ϫ2 Ϫ3
3
46. B
R
1 Ϫ3 Ϫ1 y 48.
43. 17; 27
45.
p
f (x )
f (x )
135
47. A¿(Ϫ1, Ϫ2), B¿(3, Ϫ3),
C¿(1, 3)
C'
A
O
A'
B
C x
B'
64 xϩ2 49. x 2 ϩ 5x Ϫ 4
50. 4x 2 Ϫ 16x ϩ 27 Ϫ
51. x 3 ϩ 3x 2 Ϫ 2
52. x 3 ϩ 2x 2 Ϫ 10x ϩ15 Ϫ
180
Algebra 2
21 xϩ1 Chapter 7
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Chapter 7
Practice Quiz 1
Page 364
1. 2a3 Ϫ 6a 2 ϩ 5a Ϫ 1
2. f(x ) S Ϫϱ as x S ϩϱ, f(x ) S ϩϱ as x S Ϫϱ; odd; 3
3. Sample answer: maximum at x ϭ Ϫ2, minimum at x ϭ 0.5
4. (6x 3)2 ϩ 3(6x 3) ϩ 5 or
3
3
36(2x)2 ϩ 18(2x) ϩ 5
8
1
1
f (x )
4
Ϫ4
O
Ϫ2
4x
2
Ϫ4
Ϫ8
f (x) ϭ x 3 ϩ 2x 2 Ϫ 4x Ϫ 6
5. Ϫ3, 3, Ϫi13, i13
Lesson 7-4
The Remainder and Factor Theorems
Pages 368–370
1. Sample answer: f(x ) ϭ x 2 Ϫ 2x Ϫ 3
2. 4
3. dividend: x 3 ϩ 6x ϩ 32; divisor: x ϩ 2; quotient: x 2 Ϫ 2x ϩ 10; remainder: 12
4. 7, Ϫ91
5. 353, 1186
6. x ϩ 1, x Ϫ 3
7. x Ϫ 1, x ϩ 2
8. 2x ϩ 1, x Ϫ 4
9. x Ϫ 2, x 2 ϩ 2x ϩ 4
10. \$2.894 billion
11. \$2.894 billion
12. Sample answer: Direct substitution, because it can be done quickly with a calculator. 13. Ϫ9, 54
14. 37, Ϫ19
15. 14, Ϫ42
16. 55, 272
181
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17. Ϫ19, Ϫ243
18. 267, 680
19. 450, Ϫ1559
20. 422, 3110
21. x ϩ 1, x ϩ 2
22. x Ϫ 4, x ϩ 2
23. x Ϫ 4, x ϩ 1
24. x Ϫ 3, x Ϫ 1
25. x ϩ 3, x Ϫ
1
2
26. x Ϫ 1, x ϩ
or 2x Ϫ 1
4
3
or 3x ϩ 4
27. x ϩ 7, x Ϫ 4
28. x Ϫ 1, x ϩ 6
29. x Ϫ 1, x 2 ϩ 2x ϩ 3
30. 2x Ϫ 3, 2x ϩ 3, 4x 2 ϩ 9
31. x Ϫ 2, x ϩ 2, x 2 ϩ 1
32.
33. 3
34. 8
35. 1, 4
36. Ϫ3
37.
38. Yes; 2 ft lengths. The binomial x Ϫ 2 is a factor of the polynomial since f (2) ϭ 0.
5 1 Ϫ14 69 Ϫ140 100
5 Ϫ45 120 Ϫ100
1 Ϫ9 24 Ϫ20
0
8
1 Ϫ4 Ϫ29 Ϫ24
8
32
24;
1 4
3
0
(x ϩ 3)(x ϩ 1)
39. 7.5 ft/s, 8 ft/s, 7.5 ft/s
40. 0; The elevator is stopped.
41. By the Remainder Theorem, the remainder when f(x ) is divided by x Ϫ 1 is equivalent to f(1), or a ϩ b ϩ c ϩ d ϩ e. Since a ϩ b ϩ c ϩ d ϩ e ϭ 0, the remainder when f (x ) is divided by x Ϫ 1 is 0. Therefore, x Ϫ 1 is a factor of f(x ).
42. \$31.36
43. \$16.70
44. B(x) ϭ 2000x 5 Ϫ 340(x 4 ϩ x 3 ϩ x 2 ϩ x ϩ 1)
182
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45. No, he will still owe \$4.40.
46. Using the Remainder
Theorem you can evaluate a polynomial for a value a by dividing the polynomial by x Ϫ a using synthetic division. Answers should include the following.
• It is easier to use the
Remainder Theorem when you have polynomials of degree 2 and lower or when you have access to a calculator.
• The estimated number of international travelers to the
U.S. in 2006 is 65.9 million.
47. D
48. x Ϫ 2, x ϩ 2, x ϩ 1, x 2 ϩ 1
49. (x 2)2 Ϫ 8(x 2) ϩ 4
50. 9(d 3)2 ϩ 5(d 3) Ϫ 2
51. not possible
52. Sample answer: rel. max, at x ϭ 0.5, rel. min. at x ϭ 3.5 f (x )
16
f (x) ϭ x 3 Ϫ 6x 2 ϩ 4x ϩ 3
8
Ϫ2
O
2
2 2mrFc
Fc
4
x
Ϫ8
Ϫ16
54. T ϭ
53. Sample answer: maximum at x ϭ Ϫ1, rel. max. at x ϭ 1.5, rel. min. at x ϭ 1 f (x )
8
4
Ϫ4
Ϫ2
O
2
4x
Ϫ4
f (x) ϭ Ϫx 4 ϩ 2x 3 ϩ 3x 2 Ϫ 7x ϩ 4
183
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55. (4, Ϫ2)
Ϫ7 Ϯ 117
2
56. (Ϫ3, Ϫ1)
9 Ϯ 157
6
Ϫ3 Ϯ i 17
4
57. A
58. C
59. S
60.
61.
62.
Lesson 7-5 Roots and Zeros
Pages 375–377
1. Sample answer: p(x ) ϭ x 3 Ϫ
6x 2 ϩ x ϩ 1; p(x ) has either
2 or 0 positive real zeros, 1 negative real zero, and 2 or
0 imaginary zeros.
2. An odd-degree function approaches positive infinity in one direction and negative infinity in the other direction, so the graph must cross the x-axis at least once, giving it at least one real root.
3. 6
4. 2i, Ϫ2i; 2 imaginary
5. Ϫ7, 0, and 3; 3 real
6. 2 or 0; 1; 2 or 0
7. 2 or 0; 1; 2 or 4
8. Ϫ4, 1 ϩ 2i, 1 Ϫ 2i
5 Ϯ i 271
;
4
10. 2i, Ϫ2i, 3
9. 2, 1 ϩ i, 1 Ϫ i
12. f (x ) ϭ x 3 Ϫ 2x 2 ϩ 16x Ϫ 32
11. 2 ϩ 3i, 2 Ϫ 3i, Ϫ1
8
3
13. Ϫ ; 1 real
14.
15. 0, 3i, Ϫ3i; 1 real, 2 imaginary
16. 3i, 3i, Ϫ3i, and Ϫ3i; 4 imaginary 17. 2, Ϫ2, 2i, and Ϫ2i; 2 real,
2 imaginary
18. Ϫ2, Ϫ2, 0, 2, and 2, 5 real
19. 2 or 0; 1; 2 or 0
20. 2 or 0; 1; 2 or 0
21. 3 or 1; 0; 2 or 0
22. 1; 3 or 1; 2 or 0
23. 4, 2, or 0; 1; 4, 2, or 0
24. 5, 3, or 1; 5, 3, or 1; 0, 2, 4,
6, or 8
25. Ϫ2, Ϫ2 ϩ 3i, Ϫ2 Ϫ 3i
26. 4, 1 ϩ i, 1 Ϫ i
184
2 imaginary
Algebra 2
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i
,
2
27. 2i, Ϫ2i,
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i
2
28. 5i, Ϫ5i, 7
Ϫ
3
2
1
,
2
29. Ϫ , 1 ϩ 4i, 1 Ϫ 4i
30.
31. 4 Ϫ i, 4 ϩ i, Ϫ3
32. 3 Ϫ i, 3 ϩ i, 4, Ϫ1
33. 3 Ϫ 2i, 3 ϩ 2i, Ϫ1, 1
34. 5 Ϫ i, 5 ϩ i, Ϫ1, 6
35. f(x ) ϭ x 3 Ϫ 2x 2 Ϫ 19x ϩ 20
36. f(x ) ϭ x 4 Ϫ 10x 3 ϩ 20x 2 ϩ
40x Ϫ 96
37. f(x ) ϭ x 4 ϩ 7x 2 Ϫ 144
38. f(x ) ϭ x 5 Ϫ x 4 ϩ 13x 3 Ϫ
13x 2 ϩ 36x Ϫ 36
39. f(x ) ϭ x 3 Ϫ 11x 2 ϩ 23x Ϫ 45
40. f(x) ϭ x 3 Ϫ 10x 2 ϩ 32x Ϫ 48
41a.
42. (3 Ϫ x)(4 Ϫ x)(5 Ϫ x) ϭ 24
f (x )
x
O
41b.
4 ϩ 5i, 4 Ϫ 5i
f (x )
x
O
41c.
f (x )
O
x
185
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43. 1 ft
44. V(r ) ϭ r 3 ϩ 17r 2
45. radius ϭ 4 m, height ϭ 21 m
46. 1; 2 or 0; 2 or 0
47. Ϫ24.1, Ϫ4.0, 0, and 3.1
48. Nonnegative roots represent times when there is no concentration of dye registering on the monitor.
[Ϫ30, 10] scl: 5 by [Ϫ20, 20] scl: 5
49. Sample answer: f(x) ϭ x 3 Ϫ
6x 2 ϩ 5x ϩ 12 and g(x) ϭ
2x 3 Ϫ 12x 2 ϩ 10x ϩ 24 each have zeros at x ϭ 4, x ϭ Ϫ2, and x ϭ 3.
50. One root is a double root.
Sample graph: f (x )
O
186
x
Algebra 2
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51. If the equation models the level of a medication in a patient’s bloodstream, a doctor can use the roots of the equation to determine how often the patient should take the medication to maintain the necessary concentration in the body.
Answers should include the following. • A graph of this equation reveals that only the first positive real root of the equation, 5, has meaning for this situation, since the next positive real root occurs after the medication level in the bloodstream has dropped below 0 mg. Thus according to this model, after 5 hours there is no significant amount of medicine left in the bloodstream. • The patient should not go more than 5 hours before taking their next dose of medication. 52. A
53. C
54. Ϫ127, 41
55. Ϫ254, 915
56. 36 in.
57. min.; Ϫ13
58. max.; 32
59. min.; Ϫ7
60. 5ab 2(3a Ϫ c 2)
61. (6p Ϫ 5)(2p Ϫ 9)
Ϫ3
2
63. C 3 Ϫ4 S
Ϫ2
9
62. 4y (y ϩ 3)2
11
5
64. C 7
0S
4 Ϫ5
187
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29 Ϫ8
65. £ 8
16 Ϫ16
1
2
2
3
66. y Ն Ϫ x Ϫ 1
5
2
68. Ϯ
67. Ϯ , Ϯ1, Ϯ , Ϯ5
1
9
1
,
14
1
16
1
3
1
7
2
7
1
2
1
8
1
4
1
2
Ϯ , Ϯ , Ϯ , Ϯ1, Ϯ2
70. Ϯ , Ϯ , Ϯ , Ϯ , Ϯ1, Ϯ2, Ϯ4
69. Ϯ , Ϯ , Ϯ1, Ϯ3
Lesson 7-6 Rational Zero Theorem
Pages 380–382
1. Sample answer: You limit the number of possible solutions.
2. Sample answer: 2x 2 ϩ x ϩ 3
3. Luis; Lauren found numbers q p in the form , not as p q
Luis did according to the
Rational Zero Theorem.
4. Ϯ1, Ϯ2, Ϯ5, Ϯ10
1
2
1
3
1
6
2
3
2 Ϫ3 Ϯ 217
,
3
4
5. Ϯ1, Ϯ2, Ϯ , Ϯ , Ϯ , Ϯ
6. Ϫ4, 2, 7
7. Ϫ2, Ϫ4, 7
8. 2, Ϫ2, 3, Ϫ3
9. Ϫ2, 2,
7
2
10.
11. 10 cm ϫ 11 cm ϫ 13 cm
12. Ϯ1, Ϯ2
13. Ϯ1, Ϯ2, Ϯ3, Ϯ6
14. Ϯ1, Ϯ3, Ϯ5, Ϯ15, Ϯ , Ϯ
15. Ϯ1, Ϯ2, Ϯ3, Ϯ6, Ϯ9, Ϯ18
16. Ϯ1, Ϯ , Ϯ3
1
3
1
3
5
3
1
3
1
9
17. Ϯ1, Ϯ , Ϯ , Ϯ3, Ϯ9, Ϯ27
18. Ϫ6, Ϫ5, 10
19. Ϫ1, Ϫ1, 2
20. 1, Ϫ1
21. 0, 9
22.
23. 0, 2, Ϫ2
24. 0, 3
25. Ϫ2, Ϫ4
26. Ϫ7, 1, 3
188
1
,
2
Ϫ1, 1
Algebra 2
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27.
1
,
2
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1
3
1 1
2 5
Ϫ , Ϫ2
28. Ϫ , , 2
1 1 1 3
2 3 2 4
30. Ϫ2, ,
5 Ϯ i 13
2
35. 2, Ϫ2 Ϯ i 13; 2
31.
4
,
5
0,
2 Ϫ3 Ϯ 213
3
2
4 Ϫ3 Ϯ i
3
2
29. Ϫ , , ,
2
3
32. 3, , Ϫ ,
1
3
4
3
34. V ϭ r 3 ϩ r 2
33. Ϫ1, Ϫ2, 5, i, Ϫi
˛
˛
36. r ϭ 2 in., h ϭ 6 in.
37. V ϭ 2h3 Ϫ 8h 2 Ϫ 64h
38. / ϭ 36 in., w ϭ 48 in., h ϭ 32 in.
1
3
1
3
39. V ϭ / 3 Ϫ 3/ 2
40. 6300 ϭ /3 Ϫ 3/2
41. / ϭ 30 in., w ϭ 30 in., h ϭ 21 in.
42. k ϭ Ϫ3; Ϫ3, Ϫ6, 5
43. The Rational Zero Theorem helps factor large numbers by eliminating some possible zeros because it is not practical to test all of them using synthetic substitution.
Answers should include the following. • The polynomial equation that represents the volume of the compartment is
V ϭ w 3 ϩ 3w 2 Ϫ 40w.
• Reasonable measures of the width of the compartment are, in inches, 1, 2, 3, 4, 6, 7, 9,
12, 14, 18, 21, 22, 28, 33,
36, 42, 44, 63, 66, 77, and
84. The solution shows that w ϭ 14 in., / ϭ 22 in., and d ϭ 9 in.
44. D
45. Sample answer: x 5 Ϫ x 4 Ϫ 27x 3 ϩ 41x 2 ϩ
106x Ϫ 120
46. Ϫ6, Ϫ3, 5
189
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53. Ϯ3xy 22x
47. Ϫ4, 2 ϩ i, 2 Ϫ i
48. Ϫ5, 3i, Ϫ3i
49. Ϫ7, 5 ϩ 2i, 5 Ϫ 2i
50. 4x ϩ 3, 5x Ϫ 1
51. x Ϫ 4, 3x 2 ϩ 2
52. 725
55. 6 cm, 8 cm, 10 cm
54. 0 4x Ϫ 5 0
56. x 3 ϩ 4x 2 Ϫ 6
57. 4x 2 Ϫ 8x ϩ 3
58. x 3 ϩ 5x 2 ϩ x Ϫ 10
59. x 5 Ϫ 7x 4 Ϫ 8x 3 ϩ 106x 2 Ϫ
85x ϩ 25
60. x Ϫ 9 ϩ
61. x 2 ϩ x Ϫ 4 ϩ
33 xϩ7 5 xϩ1 Chapter 7
Practice Quiz 2
Page 382
1. Ϫ930, Ϫ145
2. 0, Ϫ180
3. x 4 Ϫ 4x 3 Ϫ 7x 2 ϩ
22x ϩ 24 ϭ 0
4.
4
5
3
2
5. Ϫ
Lesson 7-7
Operations on Functions
Pages 386–389
1. Sometimes; sample answer:
If f(x ) ϭ x Ϫ 2, g(x ) ϭ x ϩ 8, then f ؠg ϭ x ϩ 6 and g ؠf ϭ x ϩ 6.
2. Sample answer: g(x ) ϭ
{(Ϫ2, 1), (Ϫ1, 2), (4, 3)}, f (x ) ϭ {(1, 7),(2, 9), (3, 3)}
3. Danette; [g ؠf ](x ) ϭ g [f (x )] means to evaluate the f function first and then the g function. Marquan evaluated the functions in the wrong order.
4. 4x ϩ 9; 2x Ϫ 1;
3x 2 ϩ 19x ϩ 20; 3x ϩ 4,
xϩ5
x
190
Ϫ5
Algebra 2
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5. x 2 ϩ x Ϫ 1; x 2 Ϫ x ϩ 7; x 3 Ϫ 4x 2 ϩ 3x Ϫ 12; x2 ϩ 3
,
xϪ4
x
6. {(Ϫ5, 7), (4, 9)}; {(4, 12)}
4
7. {(2, Ϫ7)}; {(1, 0), (2, 10)}
8. 6x Ϫ 8; 6x Ϫ 4
9. x 2 ϩ 11; x 2 ϩ 10x ϩ 31
10. 30
11. 11
12. 1
3
4
13. p(x ) ϭ x; c (x ) ϭ x Ϫ 5
14. \$32.49; price of CD when
25% discount is taken and then the coupon is subtracted 15. \$33.74; price of CD when coupon is subtracted and then 25% discount is taken
16. Discount first, then coupon; sample answer: 25% of
49.99 is greater than 25% of
44.99.
18. 6x ϩ 6; Ϫ2x Ϫ 12;
8x 2 ϩ 6x Ϫ 27;
17. 2x ; 18; x 2 Ϫ 81;
xϩ9
,
xϪ9
x
9
2x Ϫ 3
,
4x ϩ 9
19. 2x 2 Ϫ x ϩ 8; 2x 2 ϩ x Ϫ 8;
Ϫ2x 3 ϩ 16x 2;
21.
2x 2
,
8Ϫx
x
8
xϩ3
,
2
22.
Ϫ1; x 2 Ϫ x ; x
x3 ϩ x2 Ϫ x ϩ 1
,
x
x
9
4
Ϫ
20. x 2 ϩ 8x ϩ 15; x 2 ϩ 4x ϩ 3;
2x 3 ϩ 18x 2 ϩ 54x ϩ 54;
x3 ϩ x2 Ϫ 1
, x Ϫ1 xϩ1 x 3 ϩ x 2 Ϫ 2x Ϫ 1
,
xϩ1
x
x
x
Ϫ3
x 3 ϩ x 2 Ϫ 7x Ϫ 15
,x
xϩ2 x 3 ϩ x 2 Ϫ 9x Ϫ 9
,x
xϩ2
x 2 Ϫ 6x ϩ 9, x x 2 ϩ 4x ϩ 4, x
Ϫ1
0
Ϫ2;
Ϫ2;
2;
Ϫ2, 3
23. {(1, Ϫ3), (Ϫ3, 1), (2, 1)};
{(1, 0), (0, 1)}
24. {(2, 4), (4, 4)}; {(1, 5), (3, 3),
(5, 3)}
25. {(0, 0), (8, 3), (3, 3)};
{(3, 6), (4, 4), (6, 6), (7, 8)}
26. {(4, 5), (2, 5), (6, 12), (8, 12)}; does not exist
27. {(5, 1), (8, 9)}; {(2, Ϫ4)}
28. {(2, 3), (2, 2)}; {(Ϫ5, 6), (8, 6),
(Ϫ9, Ϫ5)}
191
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29. 8x Ϫ 4; 8x Ϫ 1
30. 15x Ϫ 5; 15x ϩ 1
31. x 2 ϩ 2; x 2 ϩ 4x ϩ 4
32. 3x 2 Ϫ 4; 3x 2 Ϫ 24x ϩ 48
33. 2x 3 ϩ 2x 2 ϩ 2x ϩ 2;
8x 3 ϩ 4x 2 ϩ 2x ϩ 1
34. 2x 2 Ϫ 5x ϩ 9; 2x 2 Ϫ x ϩ 5
35. Ϫ12
36. 50
37. 39
38. 68
39. 25
40. Ϫ48
41. 2
42. 1
43. 79
44. 104
45. 226
46. 36
47. P(x ) ϭ Ϫ50x ϩ 1939
48. 939,000
49. p(x ) ϭ 0.70x; s(x ) ϭ 1.0575x
50. s[p(x )]; The 30% would be taken off first, and then the sales tax would be calculated on this price.
51. \$110.30
52. [K ؠC](F ) ϭ
53. 373 K; 273 K
54. 309.67 K
55. \$700, \$661.20, \$621.78,
\$581.73, \$541.04
56. 244
57. Answers should include the following. • Using the revenue and cost functions, a new function that represents the profit is p(x ) ϭ r(c(x )).
• The benefit of combining two functions into one function is that there are fewer steps to compute and it is less confusing to the general population of people reading the formulas. 58. A
59. C
60. Ϯ1, Ϯ2, Ϯ4, Ϯ8
1
2
192
5
9
(F Ϫ 32) ϩ 273
Algebra 2
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2
12:02 PM
1
4
Page 193 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
3
2
1
3
61. Ϯ1, Ϯ , Ϯ , Ϯ2, Ϯ3, Ϯ ,
1
9
62. Ϯ1, Ϯ , Ϯ
3
Ϯ , Ϯ6
4
63. x 3 Ϫ 4x 2 Ϫ 17x ϩ 60
64. x 3 Ϫ 3x 2 Ϫ 34x Ϫ 48
65. 6x 3 Ϫ 13x 2 ϩ 9x Ϫ 2
66. x 3 Ϫ 6x 2 ϩ 4x Ϫ 24
67. x 3 Ϫ 9x 2 ϩ 31x Ϫ 39
68. x 4 ϩ x 3 Ϫ 14x 2 ϩ 26x Ϫ 20
5 Ϫ6
1
70. Ϫ B
R
2 Ϫ7
8
72. does not exist
69. 10 ϩ 2j
3 Ϫ2
71. c d Ϫ1
1
73. Ϫ B
Ϫ1 Ϫ2
R
Ϫ3 Ϫ4
1
2
1
16
75. Ϫ
74. does not exist
B
Ϫ5 Ϫ2
R
Ϫ3
2
76. x ϭ
6 ϩ 3y
2
Ϫ2
3 ϩ 7y
77. y ϭ
1 Ϫ 4x 2
Ϫ5x
78. x ϭ
79. t ϭ
I pr 80. F ϭ C ϩ 32
81. m ϭ
9
5
Fr 2
GM
Lesson 7-8
Inverse Functions and Relations
Pages 393–394
1. no
2. Switch x and y in the equation and solve for y.
3. Sample answer: f (x) ϭ 2x, f Ϫ1(x) ϭ 0.5x; f [f Ϫ1(x )] ϭ f Ϫ1[f (x )] ϭ x
4. n is an odd whole number.
5. {(4, 2), (1, Ϫ3), (8, 2)}
6. {(3, 1), (Ϫ1, 1), (Ϫ3, 1), (1, 1)}
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1
3
8. g Ϫ1(x ) ϭ x Ϫ
7. f Ϫ1(x ) ϭ Ϫx
4
f (x )
Ϫ4
O
Ϫ2
g (x )
f (x) ϭ Ϫx f Ϫ1(x ) ϭ Ϫx
2
g (x ) ϭ 3x ϩ 1
2
4x
2
Ϫ4
O
Ϫ2
Ϫ2
2
4x
Ϫ2
gϪ1(x) ϭ 1 x Ϫ 1
3
3
Ϫ4
10. yes
9. y ϭ 2x Ϫ 10
12
1
3
y y ϭ 1x ϩ 5
2
8
4
O
4
Ϫ4
8
y
Ϫ1
12
x
ϭ 2x Ϫ10
11. no
12. 32.2 ft/s2
13. 15.24 m/s2
14. {(6, 2), (5, 4), (Ϫ1, Ϫ3)}
15. {(8, 3), (Ϫ2, 4), (Ϫ3, 5)}
16. {(Ϫ4, 7), (5, 3), (4, Ϫ1), (5, 7)}
17. {(Ϫ2, Ϫ1), (Ϫ2, Ϫ3),
(Ϫ4, Ϫ1), (6, 0)}
18. {(11, 6), (7, Ϫ2), (3, 0),
(3, Ϫ5)}
19. {(8, 2), (5, Ϫ6), (2, 8), (Ϫ6, 5)}
20. x ϭ Ϫ3 y 4 x ϭ Ϫ3
2
Ϫ4
Ϫ2 O
2
4x
Ϫ2
Ϫ4
194
y ϭ Ϫ3
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1
2
22. f Ϫ1(x ) ϭ x ϩ 5
21. gϪ1(x ) ϭ Ϫ x g (x )
4
g (x ) ϭ Ϫ 1 x
2
2
Ϫ1
2
O
Ϫ2
2
Ϫ4
4
Ϫ2 O
g (x ) ϭ Ϫ2x
24. f Ϫ1(x ) ϭ x Ϫ 1
g (x )
4
f (x )
4
2 g (x ) ϭ x ϩ 4
O
2
2 f (x) ϭ 3x ϩ 3
4x
Ϫ4
gϪ1(x) ϭ xϪ2 4
Ϫ
Ϫ2 O
Ϫ4
1
2
1
2
25. y ϭ Ϫ x Ϫ
Ϫ1
Ϫ2
y Ϫ1 ϭ 3x
ϭ Ϫ 1x Ϫ 1
2
2
O
f Ϫ1(x) ϭ 1 x Ϫ1
3
2
4
y
2
Ϫ4
4x
Ϫ2 y ϭ Ϫ2x Ϫ 1
Ϫ4
4x
26. y ϭ 3x
y
4
Ϫ4
2
Ϫ2
Ϫ4
y
4
1
3
23. gϪ1(x ) ϭ x Ϫ 4
Ϫ2
2
Ϫ2 f (x) ϭ x Ϫ 5
Ϫ4
Ϫ2
Ϫ4
f Ϫ1(x) ϭ x ϩ 5 x x
Ϫ4
f (x )
4
y ϭ 1x
3
195
O
Ϫ2
2
4x
Ϫ2
Ϫ4
Algebra 2
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8
5
27. f Ϫ1(x ) ϭ x
28. f Ϫ1(x ) ϭ 3x Ϫ 12 f (x )
8
7 f (x) ϭ 1 x ϩ 4
3
6
5
f (x )
4
2
Ϫ4
O
Ϫ2
4x
2
3
2
1
f (x) ϭ 5 x
8
Ϫ2
O
Ϫ4
1 2 3 4 5 6 7 8x
f Ϫ1(x ) ϭ 3x Ϫ 12
f Ϫ1(x) ϭ 8 x
5
5
4
29. f Ϫ1(x ) ϭ x ϩ
35
4
30. gϪ1(x ) ϭ 3x Ϫ
f (x)Ϫ1 ϭ 5 x ϩ 35
4
4
Ϫ40
4
f (x )
O x
Ϫ30 Ϫ20 Ϫ10
Ϫ4
Ϫ2
O
Ϫ20
Ϫ30
4
2
4x
Ϫ2
Ϫ4
f (x) ϭ 4 x Ϫ 7
5
Ϫ40
8
7
g (x )
ϩ2 g (x) ϭ 2x 6 3
Ϫ10
31. f Ϫ1(x ) ϭ x ϩ
3
2
gϪ1(x) ϭ 3x Ϫ 3
2
4
7
32. yes
f (x )
f Ϫ1(x) ϭ 8 x ϩ2 4
7
7
Ϫ4
O
Ϫ2
Ϫ2
2
4x
Ϫ f (x) ϭ 7x 8 4
Ϫ4
33. no
34. no
35. yes
36. yes
37. yes
38. y ϭ
1
2
39. y ϭ x Ϫ
11
2
4(x ϩ 7) Ϫ 6
2
40. 12
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9
5
41. I(m) ϭ 320 ϩ 0.04m; \$4500
42. C Ϫ1(x ) ϭ x ϩ 32;
C [CϪ1(x )] ϭ C Ϫ1[C (x )] ϭ x
43. It can be used to convert
Celsius to Fahrenheit.
44. Sample answer: f(x ) ϭ x and f Ϫ1(x ) ϭ x or f(x ) ϭ Ϫx and f Ϫ1(x ) ϭ Ϫx
45. Inverses are used to convert between two units of measurement. Answers should include the following.
• Even if it is not necessary, it is helpful to know the imperial units when given the metric units because most measurements in the
U.S. are given in imperial units so it is easier to understand the quantities using our system.
• To convert the speed of light from meters per second to miles per hour,
3.0 ϫ 108 meters
ؒ
1 second
3600 seconds
1mile
ؒ
1 hour
1600 meters
46. A
f (x) Ϸ
Ϸ
675,000,000 mi/hr
47. B
48. g [h(x)] ϭ 4x ϩ 20; h[g(x)] ϭ 4x ϩ 5
49. g[h(x )] ϭ 6x Ϫ 10; h[g(x )] ϭ 6x
50. g [h(x)] ϭ x 2 Ϫ 3x Ϫ 24; h[g(x)] ϭ x 2 ϩ 5x Ϫ 24
51. Ϫ7, Ϫ2, 3
52. Ϫ , ,
53. 64
54. 32
55. 3
56. 4
57. 117
58. 196
1 4 5
4 3 2
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60. ٠
59. Ϫ7
61.
25
4
Lesson 7-9
2. Both have the shape of the graph of y ϭ 2x, but y ϭ 2x Ϫ 4 is shifted down
4 units, and y ϭ 2x Ϫ 4 is shifted to the right 4 units.
Square Root Functions and Inequalities
Pages 397–399
1. In order for it to be a square root function, only the nonnegative range can be considered. 3. Sample answer: y ϭ 22x Ϫ 4
4.
8
7
6
5
4
3
2
1
y
y ϭ ͙x ϩ 2
1 2 3 4 5 6 7 8x
O
D: x Ն 0, R: y Ն 2
5.
8
7
6
5
4
3
2
1
6.
y
4
x
O
y ϭ ͙4x
O
4
8
12
Ϫ2
1 2 3 4 5 6 7 8x
D: x Ն 0; R: y Ն 0
8
7
6
5
4
3
2
1
y ϭ 3 Ϫ ͙x
2
O
7.
y
D: x Ն 0; R: y Յ 3
8.
y
y ϭ ͙x Ϫ 1 ϩ 3
1 2 3 4 5 6 7 8x
8
7
6
5
4
3
2
1
O
y
y ϭ͙x Ϫ 4 ϩ 1
1 2 3 4 5 6 7 8x
D: x Ն 1; R: y Ն 3
198
Algebra 2
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10.
y
6
y ϭ ͙2x ϩ 4
4
O
11.
4
3
2
1
O
Ϫ2
2
y
4
y y ϭ 3 Ϫ ͙5x ϩ 1
12. v ϭ 22gh
Ϫ1 O
2
Ϫ2
4
3
2
1
1 2 3 4 5 6 7x
Ϫ2
Ϫ3
Ϫ4
6x
y ϭ ͙x ϩ 2 Ϫ 1
1 2 3 4 5 6x
Ϫ2
Ϫ3
Ϫ4
14.
13. Yes; sample answer: The advertised pump will reach a maximum height of 87.9 ft.
8
7
6
5
4
3
2
1
y
y ϭ ͙3x
1 2 3 4 5 6 7 8x
O
D: x Ն 0, R: y Ն 0
15.
16.
y
O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8
1 2 3 4 5 6 7 8x
y ϭ Ϫ͙5x
1 2 3 4 5 6 7 8x
y ϭ Ϫ4͙x
D: x Ն 0, R: y Յ 0
D: x Ն 0, R: y Յ 0
y
O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8
199
Algebra 2
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18.
y
8
7
6
5
4
3
2
1
2
2
1 2 3 4 5 6 7 8x
20.
1 2 3 4 5 6 7 8x
y ϭ Ϫ͙2x ϩ 1
22.
4
y
2
x
O
Ϫ4
Ϫ6
y ϭ ͙5x Ϫ 3
y ϭ ͙x ϩ 6 Ϫ 3 Ϫ4
1 2 3 4 5 6 7 8x
23.
8
D: x Ն Ϫ6, R: y Ն Ϫ3
24.
y
6
4
y ϭ 5 Ϫ͙x ϩ 4
2
O
Ϫ2
2
4x
8
7
6
5
4
3
2
1
O
D: x Ն Ϫ4, R: y Յ 5
2
Ϫ2
Ϫ2
D: x Ն 0.6, R: y Ն 0
Ϫ4
1 2 3 4 5 6 7 8x
D: x Ն Ϫ0.5, R: y Յ 0
y
O
6x
4
y
O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8
D: x Ն 7, R: y Ն 0
8
7
6
5
4
3
2
1
2
D: x Ն Ϫ2, R: y Ն 0
y ϭ ͙x Ϫ 7
O
O
Ϫ2
y
8
7
6
5
4
3
2
1
y ϭ ͙x ϩ 2
4
y ϭ 1 ͙x
D: x Ն 0, R: y Ն 0
21.
y
6
O
19.
8
y
y ϭ ͙3x Ϫ 6 ϩ 4
1 2 3 4 5 6 7 8x
D: x Ն 2, R: y Ն 4
200
Algebra 2
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25.
8
Page 201 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-07:
26.
y
y
O
y ϭ 2͙3 Ϫ 4x ϩ 3 6
4
2
Ϫ3
Ϫ2
x
O
Ϫ1
1 2 3 4 5 6 7 8x
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12 y ϭ Ϫ6͙x
Ϫ14
Ϫ16
D: x Յ 0.75, R: y Ն 3
27.
8
28.
y
8
6
4
4
2
y ϭ ͙x ϩ 5
6
2
O
Ϫ4
29.
8
7
6
5
4
3
2
1
O
31.
8
7
6
5
4
3
2
1
O
Ϫ2
y
2
y ϭ ͙2x ϩ 8
O
x
Ϫ4
30.
y
y ϭ ͙5x Ϫ 8
1 2 3 4 5 6 7 8x
8
7
6
5
4
3
2
1
O
Ϫ2
2
4x
y
y ϭ ͙x Ϫ 3 ϩ 4
1 2 3 4 5 6 7 8x
32. 125 ft
y
y ϭ ͙6x Ϫ 2 ϩ 1
1 2 3 4 5 6 7 8x
33. 317.29 mi
34. 119 lb
35. See students’ work.
36. If a is negative, the graph is reflected over the x-axis. The larger the value of a, the less steep the graph. If h is positive, the origin is
201
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translated to the right, and if h is negative, the origin is translated to the left. When k is positive, the origin is translated up, and when k is negative, the origin is translated down.
37. Square root functions are used in bridge design because the engineers must determine what diameter of steel cable needs to be used to support a bridge based on its weight. Answers should include the following.
• Sample answer: When the weight to be supported is less than 8 tons.
• 13,608 tons
38. C
39. D
40. yes
41. no
42. yes
43. 2x ϩ 2; 8; x 2 ϩ 2x Ϫ 15;
44. 11x Ϫ 22; 9xϪ18;
10x 2 Ϫ 40x ϩ 40; 10; x
45.
xϩ5
,x 3 xϪ3 8x 3 ϩ 12x 2 Ϫ 18x Ϫ 26
,
2x ϩ 3
3
x Ϫ ;
2
3 ϩ 12x 2 Ϫ 18x Ϫ 28
8x
,
2x ϩ 3
3
3 x Ϫ ; 2x Ϫ 3, x Ϫ ;
2
2
2
46. 4; If x is your number, you can write the expression
3x ϩ x ϩ 8
,
xϩ2
which equals 4
after dividing the numerator and denominator by the GCF, x ϩ 2.
8x 3 ϩ 12x 3 Ϫ 18x Ϫ 27, x 3
2
Ϫ
47. 2x 2 Ϫ 4x Ϫ 16
48. 6p 2 Ϫ 2p Ϫ 20
49. a3 Ϫ 1
202
Algebra 2
Chapter 7
Chapter 8 Conic Sections
Lesson 8-1 Midpoint and Distance Formulas
Pages 414–416
1. Since the sum of the x-coordinates of the given points is negative, the x-coordinate of the midpoint is negative. Since the sum of the y-coordinates of the given points is positive, the y-coordinate of the midpoint is positive. Therefore, the midpoint is in Quadrant II.
2. all of the points on the perpendicular bisector of the segment 3. Sample answer: (0, 0) and
(5, 2)
4. aϪ2,
5. (2.5, 2.25)
6. 10 units
8. 12.61 units
7. 1122 units
13 b 2
10. (12, 5)
9. D
12. (2, Ϫ6)
11. (Ϫ4, Ϫ2)
17 27 b 2 2
13. a ,
14. (0.075, 3.2)
15. (3.1, 2.7)
16. a , Ϫ b
1
24
17. a ,
5
12
5
24
1 13
5
b, a ,
2 2
2
5 b 8
18. a ,
1
2
2b, a5, b
19. (7, 11)
20. around 8th St. and 10th Ave.
21. Sample answer: Draw several line segments across the U.S. One should go from the NE corner to the SW corner; another should go from the SW corner to the
NW corner; another should go across the middle (east to west); and so on. Find the midpoints of these segments.
Locate a point to represent all of these midpoints.
22. near Lebanon, Kansas
203
Algebra 2
Chapter 8
25. 25 units
26. 12 units
27. 3117 units
28. 0.75 unit
30. 165 units
29. 170.25 units
24. 13 units
23. See students’ work.
1813
12
32. 1271 units
36. 165 ϩ 2 12 ϩ 1122 ϩ
1277 units
35. 712 ϩ 158 units, 10 units2
31. 1 unit
37. 1130 units
33.
34. 6110 units, 90 units2
units
38. about 85 mi
39. about 0.9 h
40. 14 in.
41. The slope of the line through
42. The formulas can be used to decide from which location an emergency squad should be dispatched. Answers should include the following.
• Most maps have a superimposed grid. Think of the grid as a coordinate system and assign approximate coordinates to the two cities. Then use the Distance Formula to find the distance between the points with those coordinates. • Suppose the bottom left of the grid is the origin. Then the coordinates of Lincoln are about (0.7, 0.3); the coordinates of Omaha are about (4.6, 3.3); and the coordinates of Fremont are about (1.5, 4.6). The distance from Omaha to
y Ϫy
1
(x1, y1) and (x2, y2) is 2 x2 Ϫ x1 and the point-slope form of the equation of the line is
y2 Ϫ y1
(x Ϫ x1). x2 Ϫ x1 x ϩ x2 y1 ϩ y2
Substitute ¢ 1
,
2
2
y Ϫ y1 ϭ
into this equation. The left
y2 Ϫ y1
.
2 y Ϫ y1
The right side is 2 x2 Ϫ x1 y2 Ϫ y1 x2 Ϫ x1 x1 ϩ x2
¢
Ϫ x1≤ ϭ
¢ x2 Ϫ x1
2
2 y Ϫ y1 or 2
. Therefore, the
2
side is
y1 ϩ y2
2
Ϫ y1 or
point with coordinates
¢
x1 ϩ x2 y1 ϩ y2
,
2
2
lies on the
line through (x1, y1) and
(x2, y2). The distance from
B
¢
x1 ϩ x2 y1 ϩ y2
,
2
2
y1 Ϫ y2 2 x1 Ϫ x2 2
≤ ϩ¢
≤ . The
B
2
2
¢x1 Ϫ
or
to (x1, y1) is
¢
y1 ϩ y2 2 x1 ϩ x2 2
≤ ϩ ¢y1 Ϫ
2
2
102(1.5 Ϫ 4.6)2 ϩ (4.6 Ϫ 3.3)2
or 34 miles. The distance from Lincoln to Fremont is
204
Algebra 2
Chapter 8
distance from ¢
B
to (x2, y2) is
x1 ϩ x2 y1 ϩ y2
,
2
2
y2 Ϫ y1 2 x2 Ϫ x1 2
≤ ϩ¢
B
2
2
¢x1 Ϫ
102(1.5 Ϫ 0.7)2 ϩ (4.6 Ϫ 0.3)2
or 44 miles. Since Omaha is closer than Lincoln, the helicopter should be dispatched from Omaha.
y1 ϩy2 x1 ϩ x2 b ϩ ¢y2 Ϫ
≤ ϭ
2
2
2
y1 Ϫ y2 2 x1 Ϫ x2 2
≤ ϩ¢
≤.
B
2
2
¢
¢
2
or
Therefore, the point with coordinates ¢
x1 ϩ x2 y1 ϩ y2
,
2
2
is equidistant from (x1, y1) and (x2, y2).
43. C
44. B
45. on the line with equation y ϭ x
ឈជ
46. Ϫ1; AA ¿ is perpendicular to the line with equation y ϭ x, which has slope 1.
48. D ϭ 5x 0 x Ն 06,
R ϭ 5y 0 Ն Ϫ16
47. D ϭ 5x 0 x Ն 26, R ϭ 5y 0 y Ն 06 y y
y ϭ ͙x Ϫ 1
y ϭ ͙x Ϫ 2
O
x
x
O
49. D ϭ 5x 0 x Ն 06, R ϭ 5y 0 y Ն 16
50. no
y
y ϭ 2͙x ϩ 1
O
x
51. Ϫ1 ϩ 13i
52. 6 Ϫ 2i
53. 4 Ϫ 3i
54. y ϭ (x ϩ 3)2
205
Algebra 2
Chapter 8
55. y ϭ (x Ϫ 2)2 Ϫ 3
56. y ϭ 2(x ϩ 5)2
57. y ϭ 3(x Ϫ 1)2 ϩ 2
58. y ϭ Ϫ(x ϩ 2)2 ϩ 10
59. y ϭ Ϫ3(x ϩ 3)2 ϩ 17
Lesson 8-2 Parabolas
Pages 423–425
15
16
1. (3, Ϫ7), a3, Ϫ6 b, x ϭ 3, y ϭϪ7
2. Sample answer: x ϭϪy 2
1
16
4. y ϭ 2(x Ϫ 3)2 Ϫ 12
3. When she added 9 to complete the square, she forgot to also subtract 9. The standard form is y ϭ (x ϩ 3)2 Ϫ
9 ϩ 4 or y ϭ (x ϩ 3)2 Ϫ 5.
3
4
5. (3, Ϫ4), a3, Ϫ3 b, x ϭ 3,
1
4
7
8
y ϭ Ϫ4 , upward, 1 unit
y ϭ 2 , upward,
y
16
14
12
10
8
6
4 y ϭ 2(x ϩ 7)2 ϩ 3 2
x
O
Ϫ Ϫ12Ϫ Ϫ8 Ϫ6 Ϫ4 Ϫ2
14
10
y ϭ (x Ϫ 3) 2 Ϫ 4
1
8
6. (Ϫ7, 3), aϪ7, 3 b, x ϭ Ϫ7,
206
1
2
unit
y
2x
Algebra 2
Chapter 8
4
3
2
3
7 ϭϪ ,
12
4
3
3
4
4
3
3 9
2 2
7. aϪ , Ϫ b, aϪ , Ϫ b, x ϭ Ϫ , y downward,
y ϭ Ϫ3x 2 Ϫ 8x Ϫ 6
1
3
9 9
8 2
9
2
8. aϪ , b, aϪ , b, y ϭ ,
15
8
x ϭ Ϫ , right,
unit
3
2
units
y
y x O
x ϭ 2 y 2 Ϫ 6y ϩ 12
3
x
O
1
8
1
8
9. y ϭ (x Ϫ 3)2 ϩ 6
10. x ϭ Ϫ (y ϩ 1)2 ϩ 5 y y
8
8
x
O
11. x ϭ
x
O x ϭ Ϫ 1 (y ϩ 1) 2 ϩ 5
y ϭ 1 (x Ϫ 3) 2 ϩ 6
1 2 y 24
12. y ϭ (x Ϫ 3)2 ϩ 2
Ϫ6
1
2
14. y ϭ (x ϩ 12)2 Ϫ 80
13. x ϭ (y ϩ 7)2 Ϫ 29
5 2
3
2
1
12
15. x ϭ 3 ay ϩ b Ϫ 11
6
3
2
16. (0, 0), a0, Ϫ b, x ϭ 0, y ϭ , downward, 6 units y Ϫ6y ϭ x 2
O
207
x
Algebra 2
Chapter 8
1
2
1
2
17. (0, 0), a , 0b, y ϭ 0, x ϭ Ϫ , right, 2 units
3
4
18. (Ϫ6, 3), aϪ6, 3 b, x ϭ Ϫ6,
1
4
y ϭ 2 , upward, 3 units
y
y y 2 ϭ 2x x O
3(y Ϫ 3) ϭ (x ϩ 6)2
O x
1
2
1
2
19. (1, 4), a1, 3 b, x ϭ 1, y ϭ 4 , downward, 2 units
20. (2, Ϫ3), (3, Ϫ3), y ϭ Ϫ3, x ϭ 1, right, 4 units y y
x
O
4(x Ϫ 2) ϭ (y ϩ 3)2
O
Ϫ2(y Ϫ 4) ϭ (x Ϫ 1)2
x
16
14
12
10
8
6
4
2
Ϫ4 Ϫ3Ϫ2Ϫ1
3
4
22. (6, Ϫ16), a6, Ϫ15 b, x ϭ 6,
21. (4, 8), (3, 8), y ϭ 8, x ϭ 5, left, 4 units
1
4
y ϭ Ϫ16 , upward, 1 unit
y
Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12
Ϫ14
Ϫ16
(y Ϫ 8)2 ϭ Ϫ4(x Ϫ 4)
O
1 2 3 4x
208
y
O2 4 6 8 10 12 14 x
y ϭ x 2 Ϫ 12x ϩ 20
Algebra 2
Chapter 8
5
115
144
, Ϫ b, a ,
4
2
5
287
1
, right,
10
5
3
4
24. a
23. (Ϫ24, 7), aϪ23 , 7b, y ϭ 7,
1
4
x ϭ Ϫ24 , right, 1 unit
24
y
y
x ϭ y 2 Ϫ 14y ϩ 25 16
Ϫ8
25. (4, 2), a4, 2
11
12
1 b, 12
unit
x ϭ 5y 2 ϩ 25y ϩ 60
2
1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
8x
O
Ϫ8
5
2
O 10 20 30 40 50 60 70 80 x
8
Ϫ24 Ϫ16
5
2
Ϫ b, y ϭ Ϫ ,
5
4
y ϭ 1 , upward,
1
3
55
5
5
8
4
4
27
1
Ϫ , downward, unit
4
2
26. a , Ϫ b, a , Ϫ7b, x ϭ ,
x ϭ 4,
unit
y
O
y
Ϫ2
2
x
4
Ϫ4 x ϭ Ϫ2x 2 ϩ 5x Ϫ 10
Ϫ8
Ϫ12
y ϭ 3x 2 Ϫ 24x ϩ 50
Ϫ16
x
O
17 3
67 3
4 4
16 4
69
1 x ϭ , left,
4
16
3
4
1
2
27. a , b, a , b, y ϭ ,
1
2
28. (3, 5), a3, 5 b, x ϭ 3, y ϭ 4 , upward, 2 units
unit
y
y x ϭ Ϫ4y 2 ϩ 6y ϩ 2
O
y ϭ 1 x 2 Ϫ 3x ϩ 19
2
x
O
209
2
x
Algebra 2
Chapter 8
1
4
3 x ϭ 123 , left,
4
29. (123, Ϫ18), a122 , Ϫ18b, y ϭ Ϫ18,
20
Ϫ120 Ϫ60
O
30.
y x ϭ 3y 2 ϩ 4y ϩ 1
3 units
y
O
x
120x
60
Ϫ20
Ϫ40
Ϫ60
x ϭ Ϫ 1 y 2 Ϫ 12y ϩ 15
3
1
3
31. 1
32. Ϫ1 and Ϫ
1
3
2
3
2
3
33. y ϭ Ϫ
34. aϪ , Ϫ b
35. 0.75 cm
36. y ϭ
1 2 x 16
ϩ1
y ϭ 1 x2 ϩ 1
16
y
x
O
1
(y
24
37. x ϭ Ϫ y O
1
8
38. x ϭ (y ϩ 2)2 Ϫ 6
Ϫ 6)2 ϩ 8
y
14
12
10
8
6 x ϭ Ϫ 1 (y Ϫ 6)2ϩ 8
24
4
2
x
O
x ϭ 1 (y ϩ 2)2Ϫ 6
8
1 2 3 4 5 6 7 8x
Ϫ2
210
Algebra 2
Chapter 8
39. y ϭ
1
(x
16
1
6
40. y ϭ Ϫ (x ϩ 7)2 ϩ 4
Ϫ 1)2 ϩ 7
8
6
4
2
Ϫ4 Ϫ3Ϫ2Ϫ1
Ϫ2
Ϫ4
Ϫ6
Ϫ8
y
8
y ϭ Ϫ 1 (x ϩ 7)2ϩ 4
6
y
4
y ϭ 1 (x Ϫ 1)2ϩ 7
16
O
O1 2 3 4 5 6 x
Ϫ12
Ϫ8
Ϫ4
x
Ϫ4
XBox.
Ϫ8
2
9
1
4
42. y ϭ x 2 Ϫ 2
41. x ϭ (y Ϫ 3)2 ϩ 4 y x ϭ 1 (y Ϫ 3)2ϩ 4
4
x
O
1
(x
100
43. about y ϭ Ϫ0.00046x 2 ϩ 325
1
x2
26,200
45. y ϭ Ϫ
44. y ϭ Ϫ
Ϫ 50)2 ϩ 25
46. x ϭ (y Ϫ 3)2 ϩ 4
ϩ 6550
47. A parabolic reflector can be used to make a car headlight more effective. Answers should include the following.
• Reflected rays are focused at that point.
• The light from an unreflected bulb would shine in all directions. With a parabolic reflector, most of the light can be directed forward toward the road.
48. B
49. A
50. 13 units
52. 234 units
51. 10 units
211
Algebra 2
Chapter 8
53.
54. 2.016 ϫ 105
y
O
y ϭ͙x ϩ 1
x
55. 4
56. 5
57. 9
58. 12
59. 223
60. 322
61. 423
62. 622
Lesson 8-3 Circles
Pages 428–431
1. Sample answer: (x Ϫ 6)2 ϩ
(y ϩ 2)2 ϭ 16
2. (x ϩ 3)2 ϩ (y Ϫ 1)2 ϭ 64; left
3 units, up 1 unit
3. Lucy; 36 is the square of the radius, so the radius is 6 units. 4. (x Ϫ 3)2 ϩ (y ϩ 1)2 ϭ 9
5. (x ϩ 1)2 ϩ (y ϩ 5)2 ϭ 4
6. x 2 ϩ (y ϩ 2)2 ϭ 25
7. (x Ϫ 3)2 ϩ (y ϩ 7)2 ϭ 9
8. (4, 1), 3 units y O
x
(x Ϫ 4)2 ϩ (y Ϫ 1)2 ϭ 9
212
Algebra 2
Chapter 8
9. (0, 14), 234 units
24
y
10. (4, 0),
(x Ϫ 4)2 ϩ y 2 ϭ 16
25
8
x
O
O
Ϫ8
unit
y
x 2 ϩ (y Ϫ 14)2 ϭ 34
16
Ϫ16
4
5
16x
8
Ϫ8
2 1
3 2
212
3
11. aϪ , b
12. (Ϫ4, 3), 5 units
unit
y
y
(x ϩ 4)2 ϩ (y Ϫ 3)2 ϭ 25
x
O
(x ϩ 2 ) ϩ (y Ϫ 1 )
3
2
2
2
ϭ 8
O
9
x
14. x 2 ϩ y 2 ϭ 42,2002
13. (Ϫ2, 0), 223 units y x
O
(x ϩ 2)2 ϩ y 2 ϭ 12
16. (x ϩ 1)2 ϩ (y Ϫ 1)2 ϭ 16
15.
Earth
y
Satellite
35,800 km x
42,200 km 213
Algebra 2
Chapter 8
17. (x Ϫ 2)2 ϩ (y ϩ 1)2 ϭ 4
18. x 2 ϩ (y Ϫ 3)2 ϭ 49
1
4
23. (x ϩ 213)2 ϩ (y Ϫ 42)2 ϭ
1777
20. (x ϩ 1)2 ϩ (y Ϫ 4)2 ϭ 20
1945
4
22. (x Ϫ 8)2 ϩ (y ϩ 9)2 ϭ 1130
25. (x Ϫ 4)2 ϩ (y Ϫ 2)2 ϭ 4
26. (x Ϫ 1)2 ϩ (y Ϫ 4)2 ϭ 16
27. (x ϩ 5)2 ϩ (y Ϫ 4)2 ϭ 25
28. x 2 ϩ y 2 ϭ 18
29. (x ϩ 2.5)2 ϩ (y ϩ 2.8)2 ϭ 1600
30. (0, Ϫ2), 2 units
19. (x ϩ 8)2 ϩ (y Ϫ 7)2 ϭ
1 2
2
21. (x ϩ 1)2 ϩ ay ϩ b ϭ
24. (x ϩ 8)2 ϩ (y ϩ 7)2 ϭ 64
y x 2 ϩ (y ϩ 2)2 ϭ 4
O
32. (3, 1), 5 units
31. (0, 0), 12 units
16
y
x
y
x 2 ϩ y 2 ϭ 144
8
O
Ϫ16 Ϫ8
8
16x
(x Ϫ 3)2 ϩ (y Ϫ 1)2 ϭ 25
Ϫ8
x
O
Ϫ16
214
Algebra 2
Chapter 8
34. (3, 0), 4 units
33. (Ϫ3, Ϫ7), 9 units
(x ϩ 3)2 ϩ (y ϩ 7)2 ϭ 81 4
2
y
y
Ϫ12Ϫ10Ϫ8Ϫ6Ϫ4Ϫ2 O2 4 6 8 x
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12
Ϫ14
Ϫ16
(x Ϫ 3)2 ϩ y 2 ϭ 16
36. (Ϫ25, 4), 5 units
35. (3, Ϫ7), 522 units
2
Ϫ6Ϫ4 Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12
Ϫ14
x
O
y
y (x Ϫ 3)2 ϩ (y ϩ 7)2 ϭ 50
O
2 4 6 8 10 x
37. (Ϫ2, 23), 229 units
O
x
38. (Ϫ7, Ϫ3), 222 units
y
y
O x
O
x
215
Algebra 2
Chapter 8
40. (Ϫ1, 0), 211 units
39. (0, 3), 5 units y y
O
41. (9, 9), 2109 units
O
18
16
14
12
10
8
6
4
2
Ϫ2O
Ϫ2
x
9
2
42. aϪ , 4b,
y
2129
2
x
units y 3 217
2
2 4 6 8 10 12 14 16 18 x
3
2
43. a , Ϫ4b,
4
2
Ϫ6 Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12
O x
44. (6, 8), 4 units
units
y
O
16
14
12
10
8
6
4
2
2 4 6 8 10 x
Ϫ2
Ϫ2
216
y
O
2 4 6 8 10 12 14 x
Algebra 2
Chapter 8
45. (Ϫ1, Ϫ2), 214 units
46. (Ϫ2, 1), 22 units
y
O
y
x
O
47. a0, Ϫ b, 219 units
9
2
x
48. about 109 mi
y
O
x
49. (x ϩ 1)2 ϩ (y ϩ 2)2 ϭ 5
50. A circle can be used to represent the limit at which planes can be detected by radar. Answers should include the following.
• x 2 ϩ y 2 ϭ 2500
• The region whose boundary is modeled by x 2 ϩ y 2 ϭ 4900 is larger, so there would be more planes to track.
51. A
52. D
54. y ϭ 216 Ϫ (x ϩ 3)2,
53. y ϭ Ϯ216 Ϫ (x ϩ 3)2
y ϭ Ϫ216 Ϫ (x ϩ 3)2
217
Algebra 2
Chapter 8
56. x ϭ Ϫ3 Ϯ 216 Ϫ y 2;
The equations with the ϩ symbol and Ϫ symbol represent the right and left halves of the circle, respectively. 55.
[Ϫ10, 10] scl:1 by [Ϫ10, 10] scl:1
1
12
11
12
left,
1
3
1
4
58. (3, Ϫ2), a3, Ϫ2 b, x ϭ 3,
57. (1, 0), a , 0b, y ϭ 0, x ϭ 1 ,
3
4
y ϭ Ϫ1 , downward, 1 unit
unit y y
O
x ϭ Ϫ3y 2 ϩ 1
y ϩ 2 ϭ Ϫ(x Ϫ 3)2
x
x
O
3
4
59. (Ϫ2, Ϫ4), aϪ2, Ϫ3 b, x ϭϪ2,
1
4
60. (4, Ϫ4)
y ϭ Ϫ4 , upward, 1 unit y 2
y ϭ x ϩ 4x
O
x
3
2
61. (Ϫ1, Ϫ2)
62. a , 6b
63. Ϫ4, Ϫ2, 1
64. Ϫ , 2, 3
65. 28 in. by 15 in.
66. 12
67. 6
68. 4
69. 25
70. 225
1
2
71. 222
218
Algebra 2
Chapter 8
Chapter 8
Practice Quiz 1
Page 431
2. 2226 units
1. 13 units
1
2
1
Ϫ1 ,
2
1
4
4. (Ϫ4, 4), aϪ4, 4 b, x ϭ Ϫ4,
3. (0, 0), a1 , 0b, y ϭ 0,
3
4
y ϭ 3 , upward, 1 unit
right, 6 units
y
y y 2 ϭ 6x x O
y ϭ x 2 ϩ 8x ϩ 20
O x
5. (0, 4), 7 units
12
10
8
6
4
2
Ϫ8Ϫ6Ϫ4Ϫ2O
Ϫ2
Ϫ4
y
2 4 6 8x
x 2 ϩ (y Ϫ 4)2 ϭ 49
219
Algebra 2
Chapter 8
Lesson 8-4 Ellipses
Pages 437–440
2. Let the equation of a circle be (x Ϫ h)2 ϩ (y Ϫ k)2 ϭ r 2.
Divide each side by r 2 to get
1. x ϭ Ϫ1, y ϭ 2
(y Ϫ k)2
(x Ϫ h)2 ϩ r2 r2 ϭ1. This
is the equation of an ellipse with a and b both equal to r. In other words, a circle is an ellipse whose major and minor axes are both diameters.
4.
(x Ϫ
4
5.
2)2
(y ϩ 4)2
36
5)2
ϩ
(y ϩ
1
ϩ
(x Ϫ 2)2
4
x2
36
6.
y2
100
ϭ1 ϭ1 7. (0, 0): (0, Ϯ3); 612; 6
ϩ
y2
20
ϩ
ϭ1
x2
36
ϭ1
8. (1, Ϫ2); (5, Ϫ2), (Ϫ3, Ϫ2);
415; 4
y
y
O
x
O
y2 x2 ϩ ϭ1 18
9
x
( y ϩ 2)2
(x Ϫ 1)2 ϩ ϭ1
4
20
9. (0, 0); (Ϯ2, 0); 412; 4
10. (4, Ϫ2); (4 Ϯ 216, Ϫ2); 10; 2 y y
x
O
x
O
2
2
4x ϩ 8y ϭ 32
220
Algebra 2
Chapter 8
x2
1.32 ϫ 1015
12.
y2
64
14.
(x Ϫ 5)2
64
ϩ
(y Ϫ 4)2
9
ϭ1
16.
(x ϩ 2)2
81
ϩ
(y Ϫ 5)2
16
ϭ1
ϭ1
18.
(y Ϫ 2)2
100
ϩ
(x Ϫ 42 2
9
ϭ1
ϭ1
20.
(x Ϫ 1)2
81
ϩ
(y Ϫ 2)2
56
ϭ1
22.
x2
324
24.
x2
193,600
ϩ
y2
279,312.25
26.
(x Ϫ 1)2
30
ϩ
(y ϩ 1)2
5
ϩ
y2
1.27 ϫ 1015
ϩ
x2
39
ϭ1
ϭ1
13.
x2
16
ϩ
y2
7
15.
y2
16
ϩ
(x ϩ 2)2
4
17.
(y Ϫ 4)2
64
ϩ
19.
(x Ϫ 5)2
64
ϩ
21.
x2
169
ϩ
ϭ1
y2
25
ϭ1
(x Ϫ 2)2
4
(y Ϫ 4)2
81
4
ϭ1
x2
2.02 ϫ 1016
ϩ
y2
2.00 ϫ 1016
ϩ
y2
196
ϭ1 ϭ1 ϭ1
25.
y2
20
ϩ
x2
4
ϭ1
27. (0, 0); (0, Ϯ15); 2110; 215
28. (0, 0); (Ϯ4, 0); 10; 6
y
O
y
x
x
O
y2 x2 ϩ ϭ1 10
5
ϭ1
x2 y2 ϩ ϭ1 25
9
221
Algebra 2
Chapter 8
29. (Ϫ8, 2); (Ϫ8 Ϯ 3 17, 2); 24; 18
16
30. (5, Ϫ11); (5, Ϫ11 Ϯ 123);
24; 22
y
4
8
Ϫ12Ϫ8Ϫ4 O4 8 12 16 20 x
Ϫ4
Ϫ8
Ϫ12
Ϫ16
Ϫ20
Ϫ24
Ϫ28
(y ϩ 11)2
(x Ϫ 5)2 ϩ ϭ1
O
Ϫ24 Ϫ16
8x
Ϫ8
y
32. (0, 0); (0, Ϯ 16); 6; 213
Ϫ8
(x ϩ 8) 2
(y Ϫ 2)2 ϩ ϭ1
Ϫ16
144
81
144
31. (0, 0); (Ϯ16, 0); 6; 213
121
y
y
x
O
3x 2 ϩ 9y 2 ϭ 27
27x 2 ϩ 9y 2 ϭ 81
34. (0, 0); (Ϯ315, 0); 18; 12
33. (0, 0); (0, Ϯ17); 8; 6 y 8
6
4
2
x
O
2
x
O
Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
2
16x ϩ 9y ϭ 144
y
O
2 4 6 8x
36x 2 ϩ 81y 2 ϭ 2916
36. (Ϫ2, 7); (Ϫ2 Ϯ 412, 7);
4110 ; 412
35. (Ϫ3, 1); (Ϫ3, 5), (Ϫ3, Ϫ3);
416; 412 y 12
y
8
4
O
O
x
Ϫ8
4
Ϫ4
x
Ϫ4
222
Algebra 2
Chapter 8
37. (2, 2); (2, 4), (2, 0); 217; 213
38. (Ϫ1, 3); (2, 3), (Ϫ4, 3); 10; 8
y
y
x
O
x
O
39.
x2
12
ϩ
y2
9
ϭ1
40. Knowledge of the orbit of
Earth can be used in predicting the seasons and in space exploration. Answers should include the following.
• Knowledge of the path of another planet would be needed if we wanted to send a spacecraft to that planet. • 1.55 million miles
42. B
41. C
x2
1.35ϫ1019
ϩ
44. (x Ϫ 3)2 ϩ (y ϩ 2)2 ϭ 25
y2
1.26 ϫ 1019
ϭ1
45. (x Ϫ 4)2 ϩ (y Ϫ 1)2 ϭ 101
46. (x ϩ 1)2 ϩ y 2 ϭ 45
47. (x Ϫ 4)2 ϩ (y ϩ 1)2 ϭ 16
48. y ϭ (x Ϫ 3)2 ϩ 1
1
2
y
O
223
1
2
y ϭ 2 (x Ϫ 3) ϩ 1 x
Algebra 2
Chapter 8
49.
50. Sample answer using
(0, 104.6) and (10, 112.6): y ϭ 0.8x ϩ 104.6
People (millions)
Married Americans
120
118
116
114
112
110
108
106
104
0
0 2 4 6 8 10 12 14 16 18 20
51. Sample answer: 128,600,000
52.
y
O
y ϭ 2x
53.
y y ϭ Ϫ 1x
2
x
O
x
O
55.
54.
y
x
y ϭ Ϫ2x
56.
y
y
y ϭ 1x
2
O
x
O
x y ϩ 2 ϭ 2(x Ϫ1)
224
Algebra 2
Chapter 8
57.
y
O
x y ϩ 2 ϭ Ϫ2(x Ϫ 1)
Lesson 8-5 Hyperbolas
Pages 445–448
1. sometimes
5.
x2
1
Ϫ
y2
15
2. As k increases, the branches of the hyperbola become wider. x2
4
Ϫ
y2
9
6. (0, Ϯ3 22); (0, Ϯ238);
4.
ϭ1
ϭ1
y2
4
Ϫ
x2
21
yϭϮ
ϭ1
3110 x 10
8
6
4
2
Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
225
y
y2
x2
Ϫ 20 ϭ 1
18
O2 4 6 8 x
Algebra 2
Chapter 8
7. (1, Ϫ6 Ϯ 2 25);
(1, Ϫ6 Ϯ 3 25); y ϩ 6 ϭϮ
2 25
(x
5
8. (Ϯ6, 0); (Ϯ237, 0);
1
6
yϭϮ x
Ϫ 1)
16
y
2
2
8 x Ϫ 36y ϭ 36
x
O
y
Ϫ16
O
Ϫ8
8
16x
Ϫ8
2
(y ϩ 6)
2
Ϫ
20
(x Ϫ 1)
25
ϭ1
Ϫ16
9. (4 Ϯ 2 25, Ϫ2);
(4 Ϯ 3 25, Ϫ2); yϩ2ϭϮ 25
(x
2
3
4
10. (0, Ϯ15); (0, Ϯ25); y ϭ Ϯ x
20
15
10
5
Ϫ 4)
y
16
12
8
4
13.
x2
4
Ϫ
y2
12
ay
11 ϩ 2b
12.
Ϫ
15.
x2
25
17.
(x Ϫ 2)2
49
Ϫ
19.
x2
16
y2
36
(x ϩ 2)2
4
ϭ1
(x Ϫ 3)2
4
Ϫ
(y ϩ 5)2
9
ϭ1
16.
y2
16
ϭ1
18.
(y Ϫ 5)2
16
Ϫ
20.
ϭ1
ϭ1
Ϫ
Ϫ
y2
36
ϭ1
2
25
4
Ϫ
(y Ϫ 3)2
1
14.
ϭ1 x2 6
y2
9
ϭ1
x2
Ϫ 400 ϭ 1
225
y2
Ϫ Ϫ Ϫ Ϫ5 O5 10 15 20x
20 15 10
Ϫ5
Ϫ10
Ϫ15
Ϫ20
Ϫ12Ϫ8Ϫ4 O4 8 12 16 20 x
Ϫ4
Ϫ8
Ϫ12
Ϫ16
11.
y
(y ϩ 3)2
4
ϭ1
226
Ϫ
Ϫ
x2
49
x2
4
(x ϩ 4)2
81
ϭ1
Algebra 2
Chapter 8
21. (Ϯ9, 0); (Ϯ 2130, 0);
22. (0, Ϯ6); (0, Ϯ2210); y ϭϮ3x
7
9
yϭϮ x
8
6
4
2
2
2
y x Ϫ y ϭ1
81
49
16
12
8
4
4
5
24. (Ϯ3, 0); ( Ϯ234, 0); y ϭ Ϯ x
y
y x2 y2
Ϫ 25 ϭ 1
9
Ϫ8 Ϫ6 Ϫ4Ϫ2 O2 4 6 8 x
Ϫ2
Ϫ4
Ϫ6
Ϫ8 y2 x2
Ϫ 25 ϭ 1
16
22
Ϯ x
2
26. (Ϯ2, 0); (Ϯ222, 0); y ϭ Ϯx y x 2 Ϫ y 2ϭ 4
y x 2 Ϫ 2y 2 ϭ 2
x
O
25. (Ϯ 22, 0); (Ϯ23, 0);
O
x2
Ϫ 4 ϭ1
5
3
23. (0, Ϯ4); (0, Ϯ241); y ϭ Ϯ x
y2
36
Ϫ4 Ϫ3 Ϫ2Ϫ1 O1 2 3 4 x
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ16Ϫ12Ϫ8Ϫ4 O4 8 12 16x
Ϫ4
Ϫ8
Ϫ12
Ϫ16
8
6
4
2
y
O
x
x
227
Algebra 2
Chapter 8
27. (0, Ϯ6); (0, Ϯ3 25); y ϭ Ϯ2x
16
y
8
Ϫ8
Ϫ16
Ϫ4
yϭϮ
y
y 2 ϭ 36 ϩ 4x 2
O
16x
8
6y 2 ϭ 2x 2 ϩ 12
30. (2, Ϫ2), (2, 8); (2, 3 Ϯ 241);
Ϫ12
5
4
4
3
y Ϫ 3 ϭ Ϯ (x Ϫ 2)
(Ϫ2, 9); y Ϫ 4 ϭ Ϯ (x ϩ 2)
12
y
y
10
8
(x Ϫ 2)2
)
(y Ϫ 362 ϭ1 4 Ϫ
16
25
2
O
2 4 6 8 10 x
Ϫ6 Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
8
(y Ϫ 4
16
(x ϩ 2)2
Ϫ 49
O
Ϫ4
ϭ1
(Ϫ1 Ϯ 213, Ϫ3);
Ϫ8
4
x
(Ϫ6 Ϯ 3 25, Ϫ3);
Ϫ4
31. (Ϫ3, Ϫ3), (1, Ϫ3); yϩ3ϭ x
O
29. (Ϫ2, 0), (Ϫ2, 8); (Ϫ2, Ϫ1),
)2
23 x 3
28. (0, Ϯ22); (0, Ϯ222);
3
Ϯ (x
2
32. (Ϫ12, Ϫ3), (0, Ϫ3);
1
2
ϩ 1)
y ϩ 3 ϭ Ϯ (x ϩ 6)
y
O
6
4
2
x
y
O
2x
Ϫ Ϫ Ϫ Ϫ8 Ϫ6 Ϫ4Ϫ2
14 12 10
Ϫ2
Ϫ4
Ϫ6
2
(y ϩ 3)Ϫ8
(x ϩ 6)2 ϭ Ϫ
9 Ϫ10 1
36
(y ϩ 3)2
(x ϩ 1)2 ϭ1 Ϫ
9
4
228
Algebra 2
Chapter 8
34. (Ϫ4, 0), (6, 0); (1 Ϯ 229, 0);
33. (1, Ϫ3 Ϯ 2 26);
(1, Ϫ3 Ϯ 4 22);
2
5
y ϭ Ϯ (x Ϫ 1)
y ϩ 3 ϭ Ϯ23(x Ϫ 1)
6
4
2
8
6
4
2
y
Ϫ8 Ϫ6 Ϫ4Ϫ2 O2 4 6 8 x
Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
y
Ϫ8 Ϫ6 Ϫ4Ϫ2 O2 4 6 8 x
Ϫ2
Ϫ4
4x 2 Ϫ 25y 2 Ϫ 8x Ϫ 96ϭ 0
Ϫ6
Ϫ8
y 2 Ϫ 3x 2 ϩ 6y ϩ 6x Ϫ 18 ϭ 0
35.
x2
1.1025
Ϫ
y2
7.8975
36.
ϭ1
y
Station
38.
37. 120 cm, 100 cm
229
(x Ϫ 2)2
4
O
Ϫ
Station x
(y Ϫ 3)2
4
ϭ1
Algebra 2
Chapter 8
39. about 47.32 ft
40. Hyperbolas and parabolas have different graphs and different reflective properties.
Answers should include the following. • Hyperbolas have two branches, two foci, and two vertices. Parabolas have only one branch, one focus, and one vertex.
Hyperbolas have asymptotes, but parabolas do not.
• Hyperbolas reflect rays directed at one focus toward the other focus.
Parabolas reflect parallel incoming rays toward the only focus.
41. C
42. B
43.
44. (22, 22), (Ϫ22, Ϫ22)
y xy ϭ 2
x
O
45.
46. The graph of xy ϭ Ϫ2 can be obtained by reflecting the graph of xy ϭ 2 over the x-axis or over the y-axis. The graph of xy ϭ Ϫ2 can also be obtained by rotating the graph of xy ϭ 2 by 90Њ.
y xy ϭ Ϫ2 x O
47.
(x Ϫ 5)2
16
ϩ
(y Ϫ 2)2
1
48.
ϭ1
230
(y Ϫ 1)2
16
ϩ
(x ϩ 3)2
9
ϭ1
Algebra 2
Chapter 8
49.
(x Ϫ 1)2
25
ϩ
(y Ϫ 4)2
9
ϭ1
50. (5, Ϫ1), 2 units y x
O
3
2
51. Ϫ4, Ϫ2
Ϫ7 0
S
53. C
5 20
55. about 5,330,000 subscribers per year
52. Ϫ7,
57. 2x ϩ 17y
58. 2, 3, Ϫ5
59. 1, Ϫ2, 9
60. Ϫ3, 1, 2
61. 5, 0, Ϫ2
62. 1, 0, 0
54. [13 Ϫ8 1]
56. Ϫ5, 4
63. 0, 1, 0
2. (4, Ϫ2); (4 Ϯ 2 22, Ϫ2); 6; 2
Chapter 8
Practice Quiz 2
Page 448
1.
(y Ϫ 1)2
81
ϩ
(x Ϫ 3)2
32
ϭ1
y
O
x
( y ϩ 2) 2
(x Ϫ 4)2 ϭ1 ϩ
1
9
231
Algebra 2
Chapter 8
3. (Ϫ1, 1); (Ϫ1, 1 Ϯ 211); 8;
4.
x2
9
Ϫ
y2
16
ϭ1
225 y x
O
5.
(x Ϫ 2)2
16
Ϫ
(y Ϫ 2)2
5
ϭ1
Lesson 8-6 Conic Sections
Pages 450–452
2x 2 ϩ 2y 2 Ϫ 1 ϭ 0
2. 2x 2 Ϫ 4x ϩ 7y ϩ 1 ϭ 0
3. The standard form of the equation is (x Ϫ 2)2 ϩ
(y ϩ 1)2 ϭ 0. This is an equation of a circle centered at (2, Ϫ1) with radius 0. In other words, (2, Ϫ1) is the only point that satisfies the equation. 4. y ϭ ax ϩ b Ϫ , parabola
5.
y2
16
Ϫ
x2
8
3 2
2
y
Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
x
O
1 2
2
9
4
6. ax Ϫ b ϩ y 2 ϭ , circle
ϭ 1, hyperbola
8
6
4
2
5
4
y
y
O
2 4 6 8x
O
232
x
Algebra 2
Chapter 8
7.
(x ϩ 1)2
4
ϩ
(y Ϫ 3)2
1
8. parabola
ϭ 1,
ellipse y x
O
9. ellipse
11.
10
8
6
4
2
Ϫ2
Ϫ4
Ϫ6
10. hyperbola
12. x 2 ϩ y 2 ϭ 27, circle
y
8
y
4
O2 4 6 8 10 12 14
x
Ϫ8
Ϫ4
O
8x
4
Ϫ4
Ϫ8
13.
y2
4
ϩ
x2
2
1
8
14. y ϭ x 2, parabola
ϭ 1, ellipse
y
y
O
O
x
233
x
Algebra 2
Chapter 8
15.
x2
4
Ϫ
y2
1
16.
ϭ 1, hyperbola
(x Ϫ 1)2
36
Ϫ
(y Ϫ 4)2
4
ϭ 1,
hyperbola
y
12
y
8
x
O
4
O
Ϫ8
4
Ϫ4
8
12x
Ϫ4
1
9
18. x ϭ (y Ϫ 4)2 ϩ 4, parabola
17. y ϭ (x Ϫ 2)2 Ϫ 4, parabola y y
x
O
x
O
20. x 2 ϩ (y ϩ 3)2 ϭ 36, circle
19. (x ϩ 2)2 ϩ (y Ϫ 3)2 ϭ 9, circle y
y
4
O
Ϫ8
4
Ϫ4
8x
Ϫ4
Ϫ8
O
x
234
Algebra 2
Chapter 8
21.
(x ϩ 4)2
32
Ϫ
y2
32
22.
ϭ 1, hyperbola
8
6
4
2
(x Ϫ 1)2
9
ϩ
y
24.
y
(y ϩ 1)2
25
Ϫ
8
6
4
2
x2
4
ϩ
Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
x
(y ϩ 1)2
3
x
O
23. x 2 ϩ (y Ϫ 4)2 ϭ 5, circle
25.
26.
ϭ 1, ellipse
(x ϩ 1)2
16
ϩ
x2
9
O
2 4 6 8x
(y Ϫ 1)2
4
ϭ 1, ellipse
y
x
O
ϭ 1, hyperbola
y
y
O
ϭ 1, ellipse
9
2
y
Ϫ12Ϫ10Ϫ8Ϫ6 Ϫ4Ϫ2 O 2 4 x
Ϫ2
Ϫ4
Ϫ6
Ϫ8
O
y2
235
x
Algebra 2
Chapter 8
27. y ϭ Ϫ(x ϩ 4)2 Ϫ 7, parabola
Ϫ12
Ϫ8
y
O
Ϫ4
28.
(x Ϫ 2)2
5
Ϫ
(y ϩ 1)2
6
ϭ 1,
hyperbola
4x
y
Ϫ4
Ϫ8
x
O
Ϫ12
Ϫ16
29.
(x Ϫ 3)2
25
ϩ
(y Ϫ 1)2
9
30. parabolas and hyperbolas
ϭ 1, ellipse
y
O
x
32.
31. hyperbola
y
x
O
33. circle
34. hyperbola
35. parabola
36. ellipse
37. ellipse
38. circle
39. parabola
40. hyperbola
41. b
42. a
43. c
44. 2 intersecting lines
236
Algebra 2
Chapter 8
45. The plane should be vertical and contain the axis of the double cone.
46. If you point a flashlight at a flat surface, you can make different conic sections by varying the angle at which you point the flashlight.
Answers should include the following. • Point the flashlight directly at a ceiling or wall. The light from the flashlight is in the shape of a cone and the ceiling or wall acts as a plane perpendicular to the axis of the cone.
• Hold the flashlight close to a wall and point it directly vertically toward the ceiling.
A branch of a hyperbola will appear on the wall. In this case, the wall acts as a plane parallel to the axis of the cone.
47. D
48. C
49. 0 Ͻ e Ͻ 1, e Ͼ 1
50.
51.
(x Ϫ 3)2
9
Ϫ
(y ϩ 6)2
4
52. (3, Ϫ4); (3 Ϯ 25, Ϫ4); 6; 4
ϭ1
(y Ϫ 4)2
36
Ϫ
(x Ϫ 5)2
16
ϭ1
y x O
54. m12n
53. x12
55.
x7 y4 56. 196 beats per min
57. (2, 6)
58. (3, 2)
59. (0, 2)
237
Algebra 2
Chapter 8
Lesson 8-7
Pages 458–460
2. The vertex of the parabola is on the ellipse. The parabola opens toward the interior of the ellipse and is narrow enough to intersect the ellipse in two other points.
Thus, there are exactly three points of intersection.
1a. (Ϫ3, Ϫ4), (3, 4) y 4x Ϫ 3y ϭ 0
O
x
x 2 ϩ y 2 ϭ 25
1b. (Ϯ1, 4) y y ϭ 2x 2 ϩ 2
y ϭ 5 Ϫ x2
O
x
3. Sample answer: x 2 ϩ y 2 ϭ 40, y ϭ x 2 ϩ x
4. (Ϯ4, 5)
5. (Ϫ4, Ϫ3), (3, 4)
6. no solution
7. (1, Ϯ5), (Ϫ1, Ϯ5)
8.
14
12
10
8
6
4
2
Ϫ10Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
238
y
O
2 4 6 8 10 x
Algebra 2
Chapter 8
9.
10. (40, 30)
y
O
x
13. (Ϫ1 ϩ 217, 1 ϩ 217),
(Ϫ1 Ϫ 217, 1 Ϫ 217)
3
2
12. a ,
15. ( 25, 25), (Ϫ25, Ϫ25)
11. (2, 4), (Ϫ1, 1)
9 b, 2
(Ϫ1, 2)
123
11
,Ϫ b
2
4
14. no solution
16. no solution
17. (5, 0), (Ϫ4, Ϯ6)
18. (0, 3), aϮ
19. (Ϯ8, 0)
20. (0, Ϯ5)
21. no solution
22. (4, Ϯ3), (Ϫ4, Ϯ3)
23. (Ϫ5, 5), (Ϫ5, 1), (3, 3)
24. (6, 3), (6, 1), (Ϫ4, 4), (Ϫ4, 0)
5
3
7
3
25. aϪ , Ϫ b, (1, 3)
26. (3, Ϯ4), (Ϫ3, Ϯ4)
27. 0.5 s
29. a
x2 y2 ϩ ϭ 1,
36
16 x2 (y Ϫ 2)2 ϩ ϭ
16
4 x2 y2 ϩ ϭ1
2
16
40 Ϫ 24 25 45 Ϫ 12 25 b ,
5
5
1,
30. (39.2, Ϯ4.4)
239
Algebra 2
Chapter 8
32.
31. No; the comet and Pluto may not be at either point of intersection at the same time. y
x
O
33.
34.
y
y
x
O
x
O
35.
36.
y
37.
O
Ϫ8Ϫ6Ϫ4Ϫ2
Ϫ2
Ϫ4
Ϫ6
Ϫ8
x
O
y
2 4 6 8x
38. k Ͻ Ϫ3, Ϫ2 Ͻ k Ͻ 2, or k Ͼ 3
y
O
8
6
4
2
x
39. none
40. k ϭ Ϯ2 or k ϭ Ϯ3
41. none
42. Ϫ3 Ͻ k Ͻ Ϫ2 or 2 Ͻ k Ͻ 3
240
Algebra 2
Chapter 8
43. Systems of equations can be used to represent the locations and/or paths of objects on the screen.
Answers should include the following. • y ϭ 3x, x 2 ϩ y 2 ϭ 2500
• The y-intercept of the graph of the equation y ϭ 3x is 0, so the path of the spaceship contains the origin. • (Ϫ5 210, Ϫ15 210) or about (Ϫ15.81, Ϫ47.43)
45. B
44. A
46. Sample answer: y ϭ x 2, x ϭ (y Ϫ 2)2
x 2 ϩ y 2 ϭ 36,
(x ϩ 2)2
16
Ϫ
y2
4
ϭ1
49. Sample answer: x 2 ϩ y 2 ϭ 81, x2 4
ϩ
y2
100
x2
16
x 2 ϩ y 2 ϭ 100,
ϭ1
y2 x2 Ϫ
64
16
ϩ
x2
64
y2
4
ϩ
ϭ1
y2
16
ϭ 1,
ϭ1
52. (x ϩ 2)2 ϩ (y ϩ 1)2 ϭ 11, circle 51. impossible
y
O
241
x
Algebra 2
Chapter 8
53.
(y Ϫ 3)2
9
ϩ
x2
4
54. (0, Ϯ2); (0, Ϯ4); y ϭ Ϯ
ϭ 1, ellipse
y
23 x 3
y
6y 2 Ϫ 2x 2 ϭ 24
x
O
O
x
55. Ϫ7, 0
56. 0, 3
57. Ϫ7, 3
58. 7, Ϫ5
4
3
59. Ϫ
3
4
60. Ϫ ,
210
5
1
2
2i 23
3
61a. 40
61b. two real, irrational
62a. Ϫ48
62b. two imaginary
61c. Ϯ
62c. 1 Ϯ
63. 2 ϩ 9i
65.
8
5
64. 29 Ϫ 28i
1
5
66. about 1830 times
Ϫ i
67. 6
68. Ϫ2
69. Ϫ51
70. (5, 3, 7)
71. y ϭ 3x Ϫ 2
72. y ϭ Ϫ x Ϫ
5
3
242
4
3
Algebra 2
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Chapter 9 Rational Expressions and Equations
Lesson 9-1 Multiplying and Dividing Rational Expressions
Pages 476–478
4
6
1. Sample answer: ,
4(x ϩ 2)
6(x ϩ 2)
2. To multiply rational numbers or rational expressions, you multiply the numerators and multiply the denominators. To divide rational numbers or rational expressions, you multiply by the reciprocal of the divisor. In either case, you can reduce your answer by dividing the numerator and the denominator of the results by any common factors.
4.
3. Never; solving the equation using cross products leads to
15 ϭ 10, which is never true.
9m
4n4
5.
1 aϪb 6.
3y 2 yϩ4 7.
3c
20b
8.
5
12x
9.
6
5
10.
pϩ5 pϩ1 11. cd 2x
12.
2y(y Ϫ 2)
3(y ϩ 2)
13. D
14.
5c
2b
n2
7m
16. Ϫ3x 4y
15. Ϫ
17.
s
3
18.
5 tϩ1 19.
1
2
20.
yϩ2
3y Ϫ 1
21.
aϩ1
2a ϩ 1
22.
3x 2
2y
Glencoe/McGraw-Hill
243
Algebra 2
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4bc
27a
23. Ϫ
24. Ϫf
25. Ϫ2p 2
26.
xz
8y
27.
b3 x 2y 2
28. 3
29.
4
3
30.
2
3
31. 1
32.
5(x Ϫ 3)
2(x ϩ 1)
3(r ϩ 4) rϩ3 33.
wϪ3 wϪ4 34.
35.
2(a ϩ 5)
(a Ϫ 2)(a ϩ 2)
36. Ϫ
3n m 37. Ϫ2p
38.
mϩn m2 ϩ n2
39.
2x ϩ y
2x Ϫ y
40. y ϩ 1
41.
4
3
42. d ϭ Ϫ2, Ϫ1 or 2
43. a ϭ Ϫb or b
45.
44.
6827 ϩ m
13,129 ϩ a
6827
13,129
46. 2x ϩ 1 units
1
aϪ2
47. (2x 2 ϩ x Ϫ 15)m 2
48.
49. A rational expression can be used to express the fraction of a nut mixture that is peanuts. Answers should include the following.
• The rational expression
8ϩx
is in simplest form
50. C
13 ϩ x
because the numerator and the denominator have no common factors.
244
Algebra 2
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8ϩx
13 ϩ x ϩ y could be used to represent the fraction that is peanuts if x pounds of peanuts and y pounds of cashews were added to the original mixture. 51. A
52. (Ϫ1, Ϯ4), (5, Ϯ2)
53. (Ϯ217, Ϯ222)
54. x ϭ
1
3
(y ϩ 3)2 ϩ 1; parabola
y
O
x x ϭ 1 (y ϩ 3)2 ϩ 1
3
55.
(x Ϫ 7)2 (y Ϫ 2)2
9
1
56. even; 2
ϭ 1;
hyperbola
8
y
4
x
O
4
Ϫ4
8
(x Ϫ 7)
9
2
12
2
Ϫ
(y Ϫ 2) ϭ1 1
Ϫ8
AA
57. odd; 3
C A BLACK
58. even; 0
1 1
6 3
59. Ϫ1, 4
60. Ϫ ,
61. 0, 5
62. 4.99 ϫ 102 s or about 8 min
19 s
63. л
64.
1
9
11
24
65. Ϫ1
3 19
,
2 16
66. Ϫ1
245
Algebra 2
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4
15
11
16
67. 1
68. Ϫ
11
18
70.
69. Ϫ
Lesson 9-2
1
6
Adding and Subtracting Rational Epressions
Pages 481–484
1. Catalina; you need a common denominator, not a common numerator, to subtract two rational expressions. 2. Sample answer: d 2 Ϫ d, d ϩ 1
3a. Always; since a, b, and c are factors of abc, abc is always a common denominator of
4. 12x 2y 2
1
1
1 ϩ ϩ . a c b 3b. Sometimes; if a, b, and c have no common factors, then abc is the LCD of
1
1
1
ϩ ϩ . a c b 3c. Sometimes; if a and b have no common factors and c is a factor of ab, then ab is the
1
1
1
LCD of ϩ ϩ . a b
c
b
c
3d. Sometimes; if a and c are factors of b, then b is the
1
1
1
LCD of ϩ ϩ . a 3e. Always; since
1
1
1 ϩ ϩ ϭ a c b ac ab bc ϩ ϩ
, the sum abc abc abc bc ϩ ac ϩ ab
.
always abc is
5. 80ab3c
7.
6. x(x Ϫ 2)(x ϩ 2)
2 Ϫ x3 x 2y
8.
246
42a 2 ϩ 5b 2
90ab 2
Algebra 2
Chapter 9
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37
42m
10.
5d ϩ 16
(d ϩ 2) 2
11.
3a Ϫ 10
(a Ϫ 5)(a ϩ 4)
12.
8
5
13.
13x 2 ϩ 4x Ϫ 9
2x (x Ϫ 1)(x ϩ 1)
9.
14. 70s 2t 2
units
15. 180x 2yz
16. 420a3b3c 3
17. 36p 3q 4
18. 4(w Ϫ 3)
19. x 2(x Ϫ y)(x ϩ y)
20. (2t ϩ 3)(t Ϫ 1)(t ϩ 1)
21. (n Ϫ 4)(n Ϫ 3)(n ϩ 2)
22.
6 ϩ 8b ab 23.
31
12v
24.
5 ϩ 7r r 25.
2x ϩ 15y
3y
26.
9x 2 Ϫ 2y 3
12x 2y
27.
25b Ϫ 7a3
5a 2b 2
28. Ϫ
29.
110w Ϫ 423
90w
30.
13 yϪ8 31.
aϩ3 aϪ4 32.
5m Ϫ 4
3(m ϩ 2)(m Ϫ 2)
33.
y (y Ϫ 9)
(y ϩ 3)(y Ϫ 3)
34.
7x ϩ 38
2(x Ϫ 7)(x ϩ 4)
35.
Ϫ8d ϩ 20
(d Ϫ 4)(d ϩ 4)(d Ϫ 2)
36.
Ϫ4h ϩ 15
(h Ϫ 4)(h Ϫ 5)2
37.
x2 Ϫ 6
(x ϩ 2)2(x ϩ 3)
38. 0
39.
2y 2 ϩ y Ϫ 4
(y Ϫ 1)(y Ϫ 2)
40.
1 bϩ1 42.
2s Ϫ 1
2s ϩ 1
aϩ7 aϩ2 44.
3x Ϫ 4
2x (x Ϫ 2)
45. 12 ohms
46.
24 h x
3
20q
41. Ϫ1
43.
247
Algebra 2
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47.
49.
24 h xϪ4
2md
(d Ϫ L)2(d ϩ L) 2
2md
2
(d Ϫ L2 ) 2
48.
or
1
1
, xϩ1 xϪ2
51. Subtraction of rational expressions can be used to determine the distance between the lens and the film if the focal length of the lens and the distance between the lens and the object are known. Answers should include the following.
• To subtract rational expressions, first find a common denominator.
Then, write each fraction as an equivalent fraction with the common denominator. Subtract the numerators and place the difference over the common denominator. If possible, reduce the answer. •
1
1
1 ϭ Ϫ q 10
60
48(x Ϫ 2) h x(x Ϫ 4)
52. B
could be used
to determine the distance between the lens and the film if the focal length of the lens is 10 cm and the distance between the lens and the object is 60 cm.
54.
53. C
248
4
15xyz 2
Algebra 2
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55.
a(a ϩ 2) aϩ1 56.
y
8
x 2 ϩ y 2 ϭ 16
8x
O
Ϫ8
Ϫ8
9x 2 ϩ y 2 ϭ 81
57.
58. 2.5 ft
y
2
(y Ϫ 3) ϭ x ϩ 2
x
O
x2ϭ yϩ 4
59.
60.
y
y
15
6
2
Ϫ8
10
O
5
8x
Ϫ2
Ϫ10 Ϫ5 O
2
2
x Ϫ6 y ϭ 1
Ϫ 20
5
10x
Ϫ5
16
Ϫ10
y2 x2
Ϫ15 ϭ 1
Ϫ
49
61.
10
8
6
4
2
25
y
O
Ϫ2
2 4 6x
Ϫ6
Ϫ4
2
2
(x ϩ 2)
(y Ϫ 5) ϭ1 Ϫ Ϫ8
16
25
249
Algebra 2
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Chapter 9
Practice Quiz 1
Page 484 tϩ2 tϪ3
2.
c
6b 2
3. Ϫ
y2
32
4.
7
2
5. (w ϩ 4)(3w ϩ 4)
6. x Ϫ 1
1.
7.
4a ϩ 1 aϩb 9.
n Ϫ 29
(n ϩ 6)(n Ϫ 1)
8.
Lesson 9-3
10.
1
4
Graphing Rational Functions
Pages 488–490
1. Sample answer: f(x) ϭ
6ax ϩ 20by
15a 2b3
2. Each of the graphs is a straight line passing through
(Ϫ5, 0) and (0, 5). However,
1
(x ϩ 5)(x Ϫ 2)
the graph of f (x) ϭ
(x Ϫ 1)(x ϩ 5) xϪ1 has a hole at (1, 6), and the graph of g(x ) ϭ x ϩ 5 does not have a hole.
3. x ϭ 2 and y ϭ 0 are asymptotes of the graph. The y-intercept is 0.5 and there is no x-intercept because y ϭ 0 is an asymptote.
4. asymptote: x ϭ 2
5. asymptote: x ϭ Ϫ5; hole: xϭ1 6.
f (x )
O
f (x ) ϭ
250
x x x ϩ1
Algebra 2
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7.
8.
f (x )
4
f (x )
10
2
6
O
f (x ) ϭ
8x
4
Ϫ2
Ϫ8
Ϫ4
2
6
(x Ϫ 2)(x ϩ 3)
Ϫ4 O
10.
f (x )
4
2
Ϫ8
O
Ϫ2
10
x
4
( x Ϫ 1)2
8x
4
f (x ) ϭ
x Ϫ5 x ϩ1
Ϫ4
O
C
11.
6
f (x ) f (x ) ϭ
Ϫ4
2
Ϫ4
Ϫ4
9.
x 2 Ϫ 25
f (x ) ϭ x Ϫ 5
C
12. 100 mg
f (x)
O
x
x f(x) ϭ
13.
x ϩ2 x2 Ϫ x Ϫ 6
14. y ϭ Ϫ12, C ϭ 1; 0; 0
C
10
6
y y ϩ 12
8
16 y
2
O
Ϫ16 Ϫ8
Ϫ4
15. y Ͼ 0 and 0 Ͻ C Ͻ 1
16. asymptotes: x ϭ 2, hole: xϭ3 17. asymptotes: x ϭ Ϫ4, x ϭ 2
18. asymptotes: x ϭ Ϫ4, hole: x ϭ Ϫ3
19. asymptotes: x ϭ Ϫ1, hole: xϭ5 20. hole: x ϭ 4
251
Algebra 2
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22.
21. hole: x ϭ 1
f (x )
1
f (x ) ϭ x
x
O
23.
24.
f (x )
f (x ) f (x ) ϭ
1 xϩ2 x
O
x
O
3
f (x ) ϭ x
25.
26.
f (x )
6
f (x ) ϭ
Ϫ5 x ϩ1
2
Ϫ8
x
O
Ϫ4
f (x )
4
x
O
8
Ϫ4
f (x ) ϭ
x x Ϫ3
Ϫ8
27.
8
C
28.
f (x ) f (x ) ϭ
f (x )
C
f (x ) ϭ
5x x ϩ1
Ϫ3
( x Ϫ 2)2
x
O
4
Ϫ8
Ϫ4
O
4
8x
Ϫ4
29.
30.
f (x ) f (x ) ϭ
1
( x ϩ 3)2
f (x ) f (x ) ϭ
x ϩ4 x Ϫ1
6
2
Ϫ8
Ϫ4 O
4
8x
Ϫ4
O
x
Ϫ8
252
Algebra 2
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31.
32.
f (x )
4
Ϫ8
x
O
f (x ) ϭ
f(x) ϭ
f(x)
x
4
Ϫ4 O
x 2 Ϫ 36 Ϫ4 x ϩ6
x Ϫ1 x Ϫ3
Ϫ8
Ϫ12
33.
34.
f (x) f(x) ϭ
f (x )
x2 Ϫ 1 x Ϫ1
O
x
O
f (x ) ϭ
35.
f (x ) f (x ) ϭ
O
x
Ϫ1
( x ϩ 2)( x Ϫ 3)
37.
38.
f (x ) f (x ) ϭ
x x2 Ϫ 1
x
O
f (x ) ϭ
3
( x Ϫ 1)( x ϩ 5)
36.
f (x )
x
x Ϫ1 x2 Ϫ 4
f (x ) f (x ) ϭ
O
x
O
6
(x Ϫ 6)2
253
x
Algebra 2
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39.
40.
f (x ) f (x ) ϭ
f (x )
1
(x ϩ 2)2
f (x ) ϭ
64 x 2 ϩ 16
x
O
x
O
41. The graph is bell-shaped with a horizontal asymptote at f(x) ϭ 0.
Ϫ64
64
ϭ Ϫa 2 b, x ϩ 16 x ϩ 16
Ϫ64
graph of f (x) ϭ 2 x ϩ 16
42. Since the 2
would be a reflection of the graph of f (x) ϭ the x-axis.
43.
20
4
Vf ϭ
m1 Ϫ 7 m1 ϩ 7
5
O
8 m1
Ϫ16 Ϫ8 Ϫ4
45. about Ϫ0.83 m/s
f(x) ϭ f (x) ϭ f (x) ϭ
47.
8
P (x ) ϭ
Ϫ12
Ϫ8
over
44. m1 ϭ Ϫ7; 7; Ϫ5
Vf
12
64 x ϩ 16
2
xϩ2
,
(x ϩ 2)(x Ϫ 3)
2(x ϩ 2)
,
(x ϩ 2)(x Ϫ 3)
5(x ϩ 2)
(x ϩ 2)(x Ϫ 3)
48. the part in the first quadrant
P (x )
6ϩx
10 ϩ x
4
Ϫ4 O
4x
Ϫ4
Ϫ8
Glencoe/McGraw-Hill
254
Algebra 2
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49. It represents her original freethrow percentage of 60%.
50. y ϭ 1; This represents 100%, which she cannot achieve because she has already missed 4 free throws.
51. A rational function can be used to determine how much each person owes if the cost of the gift is known and the number of people sharing the cost is s. Answers should include the following.
c
52. A
100
50
sϭ0
O
Ϫ100 Ϫ50
150 s 50 100 s
Ϫ50 c ϭ 0
Ϫ100
• Only the portion in the first quadrant is significant in the real world because there cannot be a negative number of people nor a negative amount of money owed for the gift.
58. (Ϫ2, 0); 113
54.
55.
3x Ϫ 16
(x ϩ 3)(x Ϫ 2)
57. (6, 2); 5 y 3m ϩ 4 mϩn 56.
53. B
5(w Ϫ 2)
(w ϩ 3) 2
y
(x Ϫ 6)2 ϩ ( y Ϫ 2)2ϭ 25
O
O
Glencoe/McGraw-Hill
x
x x 2 ϩ y 2 ϩ 4x ϭ 9
255
Algebra 2
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Ϫ7 Ϯ 3 213
2
59. \$65,892
60. Ϫ4 Ϯ 2i
61. Ϫ12, 10
62.
63. 4.5
64. 1.4
65. 20
66. 12
Lesson 9-4
Direct, Joint, and Inverse Variation
Pages 495–498
1a. inverse
1b. direct
2. Both are examples of direct variation. For y ϭ 5x, y increases as x increases. For y ϭ Ϫ5x, y decreases as x increases. 3. Sample answers: wages and hours worked, total cost and number of pounds of apples; distances traveled and amount of gas remaining in the tank, distance of an object and the size it appears 4. inverse; 20
5. direct; Ϫ0.5
6. joint;
7. 24
8. Ϫ45
9. Ϫ8
1
2
10. P ϭ 0.43d
11. 25.8 psi
12. about 150 ft
13.
14. direct; 1.5
p
Depth(ft) Pressure(psi)
0
0
1
0.43
2
0.86
3
1.29
4
1.72
Glencoe/McGraw-Hill
256
Algebra 2
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P
P ϭ 0.43d
O
d
15. joint; 5
16. inverse; Ϫ18
17. direct; 3
18. inverse; 12
19. direct; Ϫ7
20. joint;
21. inverse; 2.5
22. V ϭ
23. V ϭ kt
24. directly; 2
25. 118.5 km
26. 60
27. 20
28. 216
29. 64
30. 25
31. 4
32. 1.25
33. 9.6
34. Ϫ12.6
35. 0.83
36. 2
37.
1
3
k p 1
4
1
6
38. 30 mph
39. 100.8 cm3
40. See students’ work.
41. m ϭ 20sd
42. joint
43. 1860 lb
44. / ϭ15md
45. joint
46. See students’ work.
47. I ϭ
k d2 48.
I
I ϭ 16
2
d
O
Glencoe/McGraw-Hill
257
d
Algebra 2
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1
0.02P1P2
49. The sound will be heard as
4
intensely.
50. 0.02; C ϭ
51. about 127,572 calls
52. about 601 mi
53. no; d
54. Sample answer: If the average student spends \$2.50 for lunch in the school cafeteria, write an equation to represent the amount s students will spend for lunch in d days. How much will 30 students spend in a week? a ϭ 2.50sd; \$375
0
d2
55. A direct variation can be used to determine the total cost when the cost per unit is known. Answers should include the following.
• Since the total cost T is the cost per unit u times the number of units n or
C ϭ un, the relationship is a direct variation. In this equation u is the constant of variation.
• Sample answer: The school store sells pencils for 20¢ each. John wants to buy
5 pencils. What is the total cost of the pencils? (\$1.00)
56. D
57. C
58. asymptote: x ϭ 1; hole x ϭ Ϫ1
59. asymptotes: x ϭ Ϫ4, x ϭ 3
60. hole: x ϭ Ϫ3 t 2 Ϫ 2t Ϫ 2
(t ϩ 2)(t Ϫ 2)
61.
x yϪx 62.
63.
m (m ϩ 1) mϩ5 64. 9.3 ϫ 107
65. 0.4; 1.2
66. 3; 7
3
5
67. Ϫ ; 3
68. C
69. A
70. S
258
Algebra 2
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71. P
72. A
73. C
Chapter 9
Practice Quiz 2
Page 498
1.
2.
f (x )
f (x )
O
f (x ) ϭ x Ϫ 1 xϪ4 x
x
O
f (x ) ϭ
3. 49
Ϫ2 x Ϫ 6x ϩ 9
2
4. 4.4
5. 112
Lesson 9-5 Classes of Functions
Pages 501–504
2. constant (y ϭ 1), direct variation (y ϭ 2x), identity (y ϭ x)
P
O
d
This graph is a rational function. It has an asymptote at x ϭ Ϫ1.
3. The equation is a greatest integer function. The graph looks like a series of steps.
4. greatest integer
5. inverse variation or rational
6. constant
259
Algebra 2
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7. c
8. b
9. identity or direct variation
y
y
y ϭ Ϫx 2 ϩ 2
x
O
yϭx x O
12. A ϭ r 2; quadratic; the graph is a parabola
11. absolute value y yϭ xϩ2
O
x
13. absolute value
14. square root
15. rational
16. direct variation
18. constant
19. b
20. e
21. g
22. a
23. constant
24. direct variation y y
y ϭ 2.5x x O
O
x
y ϭ Ϫ1.5
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25. square root
26. inverse variation or rational
y
y yϭ4 x
y ϭ ͙9x
x
O
x
O
AA
27. rational
C A BLACK
28. greatest integer
y
y
2 yϭx Ϫ1
xϪ1
y ϭ 3[x ]
x
O
x
O
29. absolute value
30. quadratic y y
y ϭ 2x
O
y ϭ 2x 2 x O
x
31. C ϭ 4.5 m
32. direct variation
33. a line slanting to the right and passing through the origin
34. similar to a parabola
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35.
36. The graph is similar to the graph of the greatest integer function because both graphs look like a series of steps. In the graph of the postage rates, the solid dots are on the right and the circles are on the left. However, in the greatest integer function, the circles are on the right and the solid dots are on the left.
y
Cost (cents)
160
120
80
40
0
x
2
4
6
Ounces
8
10
37a. absolute value
37c. greatest integer
37d. square root
38. A graph of the function that relates a person’s weight on
Earth with his or her weight on a different planet can be used to determine a person’s weight on the other planet by finding the point on the graph that corresponds with the weight on Earth and determining the value on the other planet’s axis. Answers should include the following.
• The graph comparing weight on Earth and Mars represents a direct variation function because it is a straight line passing through the origin and is neither horizontal nor vertical.
• The equation V ϭ 0.9E compares a person’s weight on Earth with his or her weight on Venus.
V
Venus
80
60
40
20
0
262
E
20
40 60
Earth
80
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39. C
40. D
41. 22
42.
f (x )
3
f (x ) ϭ x ϩ 2
x
O
43.
44.
f (x )
f (x )
2
f (x ) ϭ x Ϫ 5x ϩ 4
xϪ4
x
O
f (x ) ϭ (
8 x Ϫ 1)(x ϩ 3)
O
46. aϪ3 , 1b ; aϪ2 , 1b;
45. (8, Ϫ1); a8, Ϫ b ; x ϭ 8; yϭ 14
12
10
8
6
4
2
Ϫ2
Ϫ2
7
8
1
1
Ϫ1 ; up;
8
2
x
1
4
y ϭ 1; x
unit
1
4
1 ϭ Ϫ4 ;
4
right; 4 units
y
y
1
1
x ϭ 4 y2 Ϫ 2 y Ϫ 3
1(
y ϩ 1) ϭ (x Ϫ 8)2
2
O
2 4 6
10 12
O
x
x
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47.
48. c
(5, Ϫ4); a5 , Ϫ4b ; y ϭ Ϫ4;
3
4
1
4
Ϫ25
23 Ϫ54 d 66 Ϫ26
57
x ϭ 4 ; right; 3 units y O
x
3x Ϫ y 2 ϭ 8y ϩ 31
49. impossible
50. (7, Ϫ5)
51. a , 2b
52. (2, Ϫ2)
53. 1
54. 12
1
3
17
6
55. Ϫ
56. 60a3b 2c 2
57. 45x 3y 3
58. 15(d Ϫ 2)
59. 3(x Ϫ y)(x ϩ y)
60. (a Ϫ 3)(a ϩ 1)(a ϩ 2)
61. (t Ϫ 5)(t ϩ 6)(2t ϩ 1)
Lesson 9-6
Solving Rational Equations and Inequalities
Pages 509–511
1
5
ϩ
2 aϩ2 2. 2(x ϩ 4); Ϫ4
ϭ1
3. Jeff; when Dustin multiplied by 3a, he forgot to multiply the 2 by 3a.
4. 3
5. 2, 6
6.
7. Ϫ6, Ϫ2
8. Ϫ2 Ͻ c Ͻ 2
264
2
3
Algebra 2
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2
9
1
6
10. 2 h
9. v Ͻ 0 or v Ͼ 1
4
3
11. 2
12. Ϫ
13. Ϫ6, 1
14. Ϫ3, 2
15. Ϫ1 Ͻ a Ͻ 0
16. Ϫ1 Ͻ m Ͻ 1
17. 11
18. 3
19. t Ͻ 0 or t Ͼ 3
20. 0 Ͻ b Ͻ 1
21. 0 Ͻ y Ͻ 2
22. p Ͻ 0 or p Ͼ 2
23. 14
24.
Ϫ3 Ϯ 3 22
2
1
2
3
2
1 Ϯ 2145
4
25. л
26. л
27. 7
28.
29.
30.
7
3
31. 32
32. 2 or 4
33. band, 80 members; chorale,
50 members
34. 4.8 cm/g
35. 24 cm
36. 15 km/h
37. 5 mL
38. 5
39. 6.15
40.
41. If something has a general fee and cost per unit, rational equations can be used to determine how many units a person must buy in order for the actual unit price to be a given number. Answers should include the following.
42. B
• To solve
500 ϩ 5x x b bc ϩ 1
ϭ 6,
multiply each side of the equation by x to eliminate the rational expression.
265
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Then subtract 5x from each side. Therefore, 500 ϭ x.
A person would need to make 500 minutes of long distance calls to make the actual unit price 6¢.
• Since the cost is 5¢ per minute plus \$5.00 per month, the actual cost per minute could never be 5¢ or less.
43. C
y
y ϭ 2x 2 ϩ 1 x O
46. direct variation
45. square root
y
y
O
y ϭ 0.8x
y ϭ 2͙x
51. 2137
O
x
x
47. 36
48. 33.75
49. 22130
50. 225
53. 5x 0 0 Յ x Յ 46
54. e b ` Ϫ1 Ͻ b Ͻ 2 f
52. 5x 0 x Ͻ Ϫ11 or x Ͼ 36
1
2
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Chapter 10 Esponential and Logarithmic Relations
Lesson 10-1 Exponential Functions
Pages 527–530
2a.
2b.
2c.
2d.
1. Sample answer: 0.8
quadratic exponential linear exponential 4. a
3. c
6. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}
5. b
y
y ϭ 3(4)x
O
7. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}
x
8. growth
y
( 1 )x
3
yϭ2
O
x
10. growth
9. decay
13. 2227 or 427
1 x
2
11. y ϭ 3a b
15. 3322 or 2722
16. Ϫ9
17. x Յ 0
18. 2
19. y ϭ 65,000(6.20)x
20. 22,890,495,000
12. y ϭ Ϫ18132 x
14. a4
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21. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}
22. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}
y
y
y ϭ 5(2)x y ϭ 2(3)x x O
x
O
23. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}
24. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͼ 0}
y
y
(1)
3
yϭ4
)x
y ϭ 0.5(4
x
O
x
O
25. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͻ 0}
x
26. D ϭ {x 0 x is all real numbers.},
R ϭ {y 0 y Ͻ 0}
y
y
x
O
O
x
y ϭ Ϫ2.5(5)x yϭϪ ( 1 )x
5
27. growth
28. growth
29. decay
30. growth
31. decay
32. decay
1 x
4
33. y ϭ Ϫ2 a b
34. y ϭ 3(5)x
35. y ϭ 7(3)x
36. y ϭ Ϫ5 a b
37. y ϭ 0.2(4)x
38. y ϭ Ϫ0.3(2)x
1 x
3
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40. x115
41. 7412
42. y 2 13
43. n 2ϩ
44. 25
45. n ϭ 5
46.
47. 1
48. n Յ Ϫ2
39. 54 or 625
8
3
2
3
49. Ϫ
50. 0
51. n Ͻ 3
52.
53. Ϫ3
54. p Ն Ϫ2
55. 10
56. Ϫ3, 5
57. y ϭ 100(6.32)x
59. y ϭ 3.93(1.35)x
60. 9.67 million; 17.62 million;
32.12 million; These answers are in close agreement with the actual populations in those years.
61. 2144.97 million; 281.42 million; No, the growth rate has slowed considerably.
The population in 2000 was much smaller than the equation predicts it would be.
62. Exponential; the base, 1 ϩ , n is fixed, but the exponent, nt, is variable since the time t can vary.
63. A(t ) ϭ 1000(1.01)4t
65. s . 4x
64. \$2216.72
67. Sometimes; true when b Ͼ 1, but false when b Ͻ 1.
68. The number of teams y that could compete in a tournament with x rounds can be expressed as y ϭ 2x.
The 2 teams that make it to the final round got there as a result of winning games played with 2 other teams, for a total of 2 и 2 ϭ 22 or 4 games played in the previous rounds. Answers should include the following.
5
3
r
66. 1.5 three-year periods or 4.5 yr
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• Rewrite 128 as a power of
2, 27. Substitute 27 for y in the equation y ϭ 2x. Then, using the Property of
Equality for Exponents, x must be 7. Therefore, 128 teams would need to play
7 rounds of tournament play. • Sample answer: 52 would be an inappropriate number of teams to play in this type of tournament because 52 is not a power of 2.
69. A
70. 780.25
71.
72.
[Ϫ5, 5] scl: 1 by [Ϫ1, 9] scl: 1
[Ϫ5, 5] scl: 1 by [Ϫ1, 9] scl: 1
The graphs have the same shape. The graph of y ϭ 3xϩ1 is the graph of y ϭ 3x translated one unit to the left. The asymptote for the graph of y ϭ 3x and for y ϭ 3xϩ1 is the line y ϭ 0.
The graphs have the same domain, all real numbers, and range, y Ͼ 0. The y-intercept of the graph of y ϭ 3x is 1 and for the graph of y ϭ 3xϩ1 is 3.
The graphs have the same shape. The graph of y ϭ 2x ϩ 3 is the graph of y ϭ 2x translated three units up. The asymptote for the graph of y ϭ 2x is the line y ϭ 0 and for y ϭ 2x ϩ 3 is the line y ϭ 3. The graphs have the same domain, all real numbers, but the range of y ϭ 2x is y Ͼ 0 and the range of y ϭ 2x ϩ 3 is y Ͼ 3.
The y-intercept of the graph of y ϭ 2x is 1 and for the graph of y ϭ 2x ϩ 3 is 4.
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73.
74.
[Ϫ5, 5] scl: 1 by [Ϫ1, 9] scl: 1
[Ϫ5, 5] scl: 1 by [Ϫ3, 7] scl: 1
The graphs have the same shape. The graph of
The graphs have the same shape. The graph of yϭ 1 xϪ2 a b
5
1
4
x
1
4
x
y ϭ a b Ϫ1 is the graph of
is the graph of
x
y ϭ a b translated two units
y ϭ a b translated one
to the right. The asymptote
unit down. The asymptote
1
5
1
5
1
5
xϪ2
for y ϭ a b
for the graph of y ϭ a b is the line y ϭ 0 and
is the line
for the graph of y ϭ
y ϭ 0. The graphs have the same domain, all real numbers, and range, y Ͼ 0.
The y-intercept of the graph of y ϭ
1 x a b
5
graph of y ϭ
1 x a b
4
Ϫ1
is the line y ϭ Ϫ1. The graphs have the same domain, all real numbers, but the range of y ϭ
is 1 and for the
1 xϪ2 a b
5
x
1
4
x
for the graph of y ϭ a b and
1 x a b
4
1
4
x
is y Ͼ 0 and of y ϭ a b Ϫ 1
is 25.
is y Ͼ Ϫ1. The y-intercept
1
4
x
of the graph of y ϭ a b is 1 and for the graph of yϭ 75. For h Ͼ 0, the graph of y ϭ 2x is translated 0 h 0 units to the right. For h Ͻ 0, the graph of y ϭ 2x is translated
|h| units to the left. For k Ͼ
0, the graph of y ϭ 2x is translated 0k 0 units up. For k Ͻ 0, the graph of y ϭ 2x is translated 0k 0 units down.
1 x a b
4
Ϫ 1 is 0.
76. 1, 15
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13
3
77. 1, 6
78. Ϫ , 3
79. 0 Ͻ x Ͻ 3 or x Ͼ 6
80. square root
81. greatest integer
82. constant y yϭ8
O
83. B
1 0
R
0 1
85.
3
1
B
51 11
x
84. does not exist
Ϫ6
R
Ϫ5
86. about 23.94 cm
88. g [h(x)] ϭ x 2 ϩ 6x ϩ 9; h [g(x)] ϭ x 2 ϩ 3
87. g [h(x)] ϭ 2x Ϫ 6; h [g(x)] ϭ 2x Ϫ 11
89. g [h(x)] ϭ Ϫ2x Ϫ 2; h [g(x)] ϭ Ϫ2x ϩ 11
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Logarithms and Logarithmic Functions
Pages 535–538
1. Sample answer: x ϭ 5y and y ϭ log5 x
2. They are inverses.
3. Scott; the value of a logarithmic equation, 9, is the exponent of the equivalent exponential equation, and the base of the logarithmic expression, 3, is the base of the exponential equation. Thus x ϭ 39 or
19,683.
4. log5 625 ϭ 4
5. log7
1
49
6. 34 ϭ 81
ϭ Ϫ2
1
8. 4
7. 362 ϭ 6
9. Ϫ3
10. 21
11. Ϫ1
12. 27
13. 1000
14.
15.
1
,
2
1
2
ՅxՅ5
16. x Ͼ 6
1
17. 3
18. 1013
19. 107.5
20. 105.5 or about 316,228 times
21. log8 512 ϭ 3
22. log3 27 ϭ 3
23. log5
1
125
1
24. log 3 9 ϭ Ϫ2
ϭ Ϫ3
25. log100 10 ϭ
1
2
26. log2401 7 ϭ
28. 132 ϭ 169
27. 53 ϭ 125
29. 4Ϫ1 ϭ
1
4
1
4
1
30. 100Ϫ2 ϭ
1 Ϫ2
5
31. 83 ϭ 4
32. a b
33. 4
1
10
34. 2
2
273
ϭ 25
Algebra 2
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1
2
36.
5
2
37. Ϫ5
38. Ϫ4
39. 7
40. 45
41. n Ϫ 5
42. 3x ϩ 2
43. Ϫ3
44. 2x
45. 1018.8
46. 1010.67
47. 81
48. c Ͼ 256
49. 0 Ͻ y Յ 8
50. 125
51. 7
52. 0 Ͻ p Ͻ 1
53. x Ն 24
54. Ϯ3
55. 4
56. 11
57. 2
58. 25
59. 5
60. y Ն 3
61. a Ͼ 3
62. Ϯ8
?
63. log5 25 ϭ 2 log5 5
2 ?
1
log5 5 ϭ 2 log5 5
?
2 ϭ 2(1)
2 ϭ 2
64.
Original equation ?
log16 2 ؒ log2 16 ϭ 1
2
25 ϭ 5 and 5 ϭ 51
1
?
log16 164 ؒ log2 24 ϭ 1
Inverse
Prop. of
Exp. and
Logarithms
Simplify.
1
(4)
4
Original equation 1
2 ϭ 164 and 16 ϭ 24
?
ϭ 1 Inverse Prop. of Exp. and
Logarithms
1 ϭ 1
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66a.
65.
?
log7 [log3 (log2 8)] ϭ 0
Original equation ?
log7 [log3 (log2 23)] ϭ 0
8 ϭ 23
?
log7 (log3 3) ϭ 0
log7 (log3 31) ϭ 0
?
x
y ϭ log 1 x
2
66b. The graphs are reflections of each other over the line y ϭ x.
3 ϭ 31
log7 1 ϭ 0
( 1)
2
x
O
Inverse
Prop. of
Exp. and
Logarithms
?
?
y
Inverse
Prop. of
Exp. and
Logarithms
log7 70 ϭ 0
0 ϭ 0
1 ϭ 70
Inverse
Prop. of
Exp. and
Logarithms
68. 103 or 1000 times as great
67a. y y ϭ log2(x ϩ 2)
y ϭ log2x ϩ 3 y ϭ log2(x Ϫ 1)
O
x
y ϭ log2x Ϫ 4
67b. The graph of y ϭ log2 x ϩ 3 is the graph of y ϭ log2 x translated 3 units up. The graph of y ϭ log2 x Ϫ 4 is the graph of y ϭ log2 x translated 4 units down. The graph of log2 (x Ϫ 1) is the graph of y ϭ log2 x translated 1 unit to the right.
The graph of log2 (x ϩ 2) is the graph of y ϭ log2 x translated 2 units to the left.
69. 101.4 or about 25 times as great
70. 101.7 or about 50 times
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71. 2 and 3; Sample answer: 5 is between 22 and 23.
72. All powers of 1 are 1, so the inverse of y ϭ 1x is not a function. 73. A logarithmic scale illustrates that values next to each other vary by a factor of 10. Answers should include the following.
• Pin drop: 1 ϫ 100;
Whisper: 1 ϫ 102; Normal conversation: 1 ϫ 106;
Kitchen noise: 1 ϫ 1010;
Jet engine: 1 ϫ 1012
74. B
Pin drop Whisper
(4 feet)
Normal conversation 2 ϫ 10 11
0
4 ϫ 10 11
Kitchen noise 6 ϫ 10 11
Jet engine 8 ϫ 10 11 1 ϫ 10 12
• On the scale shown above, the sound of a pin drop and the sound of normal conversation appear not to differ by much at all, when in fact they do differ in terms of the loudness we perceive.
The first scale shows this difference more clearly.
76. x 216
75. D
5 Ϯ 273
4
77. b12
79. Ϫ3,
81.
83.
78. л
7
3
14
5
80. Ϯ
82.
6x Ϫ 58
(x Ϫ 3)(x ϩ 3)(x ϩ 7)
43
30y
84. \$2400, CD; \$1600, savings
85. x10
86. y 24
87. 8a6b 3
88. an6
89.
x3 y 2z 3
90. 1
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Chapter 10
Practice Quiz 1
Page–538
1. growth
2. y ϭ 2(4)x
3. log4 4096 ϭ 6
4. 92 ϭ 27
3
5.
4
3
6. 15
7.
3
5
8. n Յ Ϫ1
9. x Ͼ 26
10. 3
Lesson 10-3
Properties of Logarithms
Pages 544–546
1. properties of exponents
2log3 x ϩ log3 5; log3 5x 2
3. Umeko; Clemente incorrectly applied the product and quotient properties of logarithms. log7 6 ϩ log7 3 ϭ log7 (6 и 3) or log7 18 Product Property of
4. 1.1402
Logarithms
log7 18 Ϫ log7 2 ϭ log7 (18 Ϭ 2) or log7 9
Quotient Prop. of
Logarithms
5. 2.6310
6. Ϫ0.3690
7. 6
8. 2
9. 3
10. 4
11. pH ϭ 6.1 ϩ log10
B
C
12. 20:1
13. 1.3652
14. 1.2921
15. Ϫ0.2519
16. 0.2519
17. 2.4307
18. 2.1133
19. Ϫ0.4307
20. 0.0655
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21. 2
22. 3
23. 4
24. Ϫ2
25. 14
26. 12
27. 2
28. Ϯ4
29. л
30. 6
31. 10
32. 12
33.
x3
4
34.
35. False; log2 (22 ϩ 23) ϭ log2 12, log2 22 ϩ log2 23 ϭ 2 ϩ 3 or 5, and log2 12 Z 5, since 25 Z 12.
1
1x
2
Ϫ 12
36.
?
n logb x ϩ m logb x ϭ (n ϩ m)logb x
?
logb x n ϩ logb x m ϭ (n ϩ m)logb x
Power Prop. of Logarithms
?
logb (x n ؒ x m) ϭ (n ϩ m)logb x
Product Prop. of Logarithms
?
logb (x nϩm) ϭ (n ϩ m)logb x
Product of Powers Prop.
(n ϩ m)logb x ϭ (n ϩ m)logb
Power Prop. of Logarithms
C2
C1
37. 2
38. E ϭ 1.4 log
39. about 0.4214 kilocalories per gram 40. about 0.8429 kilocalories per gram 41. 3
42. 3
43. About 95 decibels;
L ϭ 10 log10 R, where L is the loudness of the sound in decibels and R is the relative intensity of the sound. Since the crowd increased by a factor of 3, we assume that the intensity also increases by a factor of 3. Thus, we need to find the loudness of 3R.
44. 5
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L ϭ 10 log10 3R
L ϭ 10 (log10 3 ϩ log10 R )
L ϭ 10 log103 ϩ 10 log10 R
L ഠ 10(0.4771) ϩ 90
L ഠ 4.771 ϩ 90 or about 95
45. 7.5
47. Let b x ϭ m and b y ϭ n. Then logb m ϭ x and logb n ϭ y.
48. Since logarithms are exponents, the properties of logarithms are similar to the properties of exponents. The
Product Property states that to multiply two powers that have the same base, add the exponents. Similarly, the logarithm of a product is the sum of the logarithms of its factors. The Quotient Property states that to divide two powers that have the same base, subtract their exponents. Similarly, the logarithm of a quotient is the difference of the logarithms of the numerator and the denominator. The Power
Property states that to find the power of a power, multiply the exponents. Similarly, the logarithm of a power is the product of the logarithm and the exponent. Answers should include the following.
• Quotient Property:
bx by ϭ
b xϪy ϭ
m n m n Quotient Prop.
m
Prop. of n Equality for
Logarithmic
Equations m logb Inverse n Prop. of
logb b x Ϫy ϭ logb
xϪyϭ
Exp. and
Logarithms
logb m Ϫ logb n ϭ logb
m
Replace x n with log m b and y with logbn. log2
32 a b
8
ϭ log2
25 a 3b
2
ϭ log2 2(5Ϫ3) ϭ 5 Ϫ 3 or 2
279
Replace 32 with 25 and 8 with 23.
Quotient of
Powers
Inverse Prop. of Exp. and
Logarithms
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log2 32 Ϫ log2 8 ϭ log2 25 Ϫ log2 23
Replace 32 with
25 and 8 with 23.
ϭ 5 Ϫ 3 or 2
Inverse Prop. of
Exp. and Logarithms
So, log2 a
32 b 8
ϭ log2 32 Ϫ log2 8
Power Property: log3 94 ϭ log3 (32)4
Replace 9 with 32.
ϭ log3 3(2ؒ4) ϭ 2 ؒ 4 or 8
Power of a Power
Inverse Prop. of Exp. and
Logarithms
4 log3 9 ϭ (log3 9) ؒ 4
Comm (ϫ)
ϭ (log3 32) ؒ 4 ϭ 2 ؒ 4 or 8
Replace 9 with 32.
Inverse Prop. of Exp. and
Logarithms
So, log3 94 ϭ 4 log3 9.
• The Product of Powers
Property and Product
Property of Logarithms both involve the addition of exponents, since logarithms are exponents.
50. Let b x ϭ m, then logb m ϭ x.
(b x)p ϭ m p b xp ϭ m p Product of Powers logb b xp ϭ logb m p Prop. of Equality
49. A
for Logarithmic
Equations
xp ϭ logb m p
Inverse Prop. of Exp. and
Logarithms
plogb m ϭ logb m p
Replace x with logb m.
51. 4
52. Ϫ3
53. 2x
54. 6
55. Ϫ8
56. d Ͻ 4
57. odd; 3
58. even; 4
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59.
3b a 60. 2
61.
5
3x
62. 3.06 s
64. 5
63. 1
65. x Ͼ
3
4
5
3
66. Ϫ Ͻ x Ͻ 2
Lesson 10-4 Common Logarithms
Pages 549–551
1. 10; common logarithms
5x ϭ 2; x ഠ 0.4307
3. A calculator is not programmed to find base
2 logarithms.
4. 0.6021
5. 1.3617
6. Ϫ0.3010
8. {n 0 n Ͼ 0.4907}
7. 1.7325
9. 4.9824
10. Ϯ1.1615
12. {p 0 p Յ 4.8188}
11. 11.5665
13.
log 5
;
log 7
0.8271
14.
15.
log 9
;
log 2
3.1699
16. at most 0.00003 mole per liter log 42
;
log 3
3.4022
17. 0.6990
18. 1.0792
19. 0.8573
20. 0.3617
21. Ϫ0.0969
22. Ϫ1.5229
23. 11
24. 2.2
25. 2.1
26. 3.5
27. {x 0 x Ն 2.0860}
28. 2.4550
29. {a 0 a Ͻ 1.1590}
30. 0.5537
31. 0.4341
32. 4.8362
33. 4.7820
34. 8.0086
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36. Ϯ2.6281
35. Ϯ1.1909
37. {n 0 n Ͼ Ϫ1.0178}
38. 1.0890
40. {p 0 p Յ 1.9803}
39. 3.7162
42. 4.7095
41. 0.5873
44. 2.7674
43. Ϫ7.6377
45.
log 13 log 2
47.
log 3 log 7
49.
2log 1.6 log 4
46.
Ϸ 0.6781
log 8 log 3
50.
Ϸ 0.5646
log 20 log 5
48.
Ϸ 3.7004
0.5 log 5 log 6
Ϸ 1.8614
Ϸ 1.8928
Ϸ 0.4491
51. between 0.000000001 and
0.000001 mole per liter
52. 8
53. Sirius
54. Sirius: 1.45, Vega: 0.58
55. Vega
56a. 3;
1
3
3 2
56b. ;
2 3
56c. conjecture: loga b ϭ proof: ?
1 log b a
?
1 logb a
1
logb a
loga b ϭ logb b logb a
1
logb a
ϭ
Original statement
Change of Base
Formula
Inverse Prop. of
Exponents and
Logarithms
58. about 11.64 yr or 11 yr, 8 mo
57. about 3.75 yr or 3 yr 9 mo
ϭ
1
;
logba
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59. Comparisons between substances of different acidities are more easily distinguished on a logarithmic scale. Answers should include the following.
Tomatoes: 6.3 ϫ 10Ϫ5 mole per liter
Milk: 3.98 ϫ 10Ϫ7 mole per liter Eggs: 1.58 ϫ 10Ϫ8 mole per liter
• Those measurements correspond to pH measurements of 5 and 4, indicating a weak acid and a stronger acid. On the logarithmic scale we can see the difference in these acids, whereas on a normal scale, these hydrogen ion concentrations would appear nearly the same. For someone who has to watch the acidity of the foods they eat, this could be the difference between an enjoyable meal and heartburn.
60. A
61. C
62. 1.4248
63. 1.6938
64. 1.8416
65. 64
66. z Յ
67. 62
68. Ϫ22
69. (d ϩ 2)(3d Ϫ 4)
70. (7p ϩ 3)(6q Ϫ 5)
71. prime
72. 2x ϭ 3
73. 32 ϭ x
74. 53 ϭ 125
75. log5 45 ϭ x
76. log7 x ϭ 3
1
64
77. logb x ϭ y
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Base e and Natural Logarithms
Pages 557–559
1. the number e
2. e x ϭ 8
3. Elsu; Colby tried to write each side as a power of 10.
Since the base of the natural logarithmic function is e, he should have written each side as a power of e;
10ln 4x Z 4x.
4. 403.4288
5. 0.0334
6. 0.1823
7. Ϫ2.3026
8. x ϭ ln 4
9. e0 ϭ 1
10. 3
11. 5x
12. x Ͼ 3.4012
13. 1.0986
14. Ϫ0.8047
15. 0 Ͻ x Ͻ 403.4288
16. 2.4630
17. Ϯ90.0171
18. h ϭ Ϫ26200 ln
19. about 15,066 ft
20. 54.5982
21. 148.4132
22. 0.3012
23. 1.6487
24. 1.0986
25. 2.3026
26. 1.6901
27. Ϫ3.5066
28. \$183.21
29. about 49.5 cm
30. Ϫx ϭ ln 5
31. 2 ϭ ln 6x
32. e1 ϭ e
33. e x ϭ 5.2
34. 0.2
35. y
36. Ϫ4x
37. 45
38. 0.2877
39. Ϫ0.6931
40. x Ͻ 1.5041
41. x Ͼ 0.4700
42. 0.2747
43. 0.5973
44. x Ն 0.6438
45. x Ն Ϫ0.9730
46. 27.2991
284
P
101.3
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47. 49.4711
48. 1.7183
49. 14.3891
50. 232.9197
51. 45.0086
52. 2, 6
53. 1
54. about 19.8 yr
55. t ϭ
100 ln 2 r 56. 100 ln 2 Ϸ 70
57. t ϭ
110 r 58. about 7.33 billion
59. about 55 yr
60. about 32 students
61. about 21 min
62. always; log x log y
?
ϭ
In x
In y
Original statement log x logx ? log e ϭ log y logy log e
log x log y
log x log y
63. The number e is used in the formula for continuously compounded interest,
A ϭ Pe rt. Although no banks actually pay interest compounded continually, the equation is so accurate in computing the amount of money for quarterly compounding or daily compounding, that it is often used for this purpose.
ϭ
?
log x log e
ϭ
?
log x log y
Change of
Base Formula
ؒ
log e log y
Multiply log x by the log e reciprocal of log y
.
log e
Simplify.
64. B
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• If you know the annual interest rate r and the principal P, the value of the account after t years is calculated by multiplying P times e raised to the r times t power. Use a calculator to find the value of e rt.
• If you know the value A you wish the account to achieve, the principal P, and the annual interest rate r, the time t needed to achieve this value is found by first taking the natural logarithm of A minus the natural logarithm of P.
Then, divide this quantity by r.
66.
67.
log 0.047 log 6
ϭ Ϫ1.7065
log 68 log 4
ϭ 3.0437
68.
65. 1946, 1981, 2015; It takes between 34 and 35 years for the population to double.
log 23 log 50
ϭ 0.8015
69. 5
70. 4
71. inverse; 4
72. joint; 1
73. direct; Ϫ7
74. x ϭ
75. 3.32
76. 1.54
77. 1.43
78. 323.49
79. 13.43
80. 9.32
286
1 2 y 20
Ϫ5
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Chapter 10
Practice Quiz 2
Page 559
1.
log 5
;
log 4
2. e 2 ϭ 3x
1.1610
3. 3
4. x Ͼ 5.3219
5. 1.3863
Lesson 10-6
Exponential Growth and Decay
Pages 563–565
1. y ϭ a(1 ϩ r)t, where r Ͼ 0 represents exponential growth, and r Ͻ 0 represents exponential decay.
2. Take the common logarithm of each side, use the Power
Property to write log (1 ϩ r)t as t log(1 ϩ r), and then divide each side by the quantity log(1 ϩ r).
3. Sample answer: money in a bank 4. Decay; the exponent is negative. 5. about 33.5 watts
6. about 402 days
7. y ϭ 212,000e0.025t
8. about 349,529 people
9. C
10. \$1600
11. at most \$108,484.93
12. about 8.1 days
13. No; the bone is only about
21,000 years old, and dinosaurs died out
63,000,000 years ago.
14. more than 44,000 years ago
16. y ϭ ae0.0347t
17. \$12,565 billion
19. after the year 2182
20. 4.7%
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21. Never; theoretically, the amount left will always be half of the previous amount.
22. Answers should include the following. • Find the absolute value of the difference between the price of the car for two consecutive years. Then divide this difference by the price of the car for the earlier year.
• Find 1 minus the rate of decrease in the value of the car as a decimal.
Raise this value to the number of years it has been since the car was purchased, and then multiply by the original value of the car.
23. about 19.5 yr
24. D
25. ln y ϭ 3
26. ln 29 ϭ 4n Ϫ 2
27. 4x 2 ϭ e8
28. 1.5323
29. p Ͼ 3.3219
30. 9
31.
0.5 (0.08 p)
6
33.
p
150
ϩ
0.5 (0.08 p)
4
32.
p
60
34. hyperbola
35. ellipse
36. parabola
37. circle
38. 2.06 ϫ 108
39. 8 ϫ 107
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Chapter 11 Sequences and Series
Lesson 11-1 Arithmetic Sequences
Pages 580–582
1. The differences between the terms are not constant.
2. 95
3. Sample answer: 1, Ϫ4, Ϫ9,
Ϫ14, . . .
4. 24, 28, 32, 36
5. Ϫ3, Ϫ5, Ϫ7, Ϫ9
6. 5, 8, 11, 14, 17
7. 14, 12, 10, 8, 6
8. 43
10. 79
9. Ϫ112
11. 15
12. an ϭ 11n Ϫ 37
13. 56, 68, 80
14. \$12,000
15. 30, 37, 44, 51
16. 10, 3, Ϫ4, Ϫ11
17. 6, 10, 14, 18
18. 1, 4, 7, 10
19.
7
,
3
3,
11 13
,
3 3
20.
12
,
5
8 6
5 5
2, ,
21. 5.5, 5.1, 4.7, 4.3
22. 8.8, 11.3, 13.8, 16.3
23. 2, 15, 28, 41, 54
24. 41, 46, 51, 56, 61
25. 6, 2, Ϫ2, Ϫ6, Ϫ10
26. 12, 9, 6, 3, 0
27.
4
,
3
2 1
3 3
28.
1, , , 0
5
,
8
1,
11 7 17
, ,
8 4 8
29. 28
30. Ϫ49
31. 94
32. Ϫ175
33. 335
34. 340
35.
26
3
25
2
36. Ϫ
37. 27
38. Ϫ47
39. 61
40. 173
41. 37.5 in.
42. 304 ft
43. 30th
44. 19th
45. 82nd
46. an ϭ 9n Ϫ 2
47. an ϭ Ϫ7n ϩ 25
48. an ϭ Ϫ2n Ϫ 1
Glencoe/McGraw-Hill
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49. 13, 17, 21
50. pn ϭ 4n Ϫ 3
51. Yes; it corresponds to n ϭ 100.
52. 70, 85, 100
53. 4, Ϫ2
54. Ϫ5, Ϫ2, 1, 4
55. 7, 11, 15, 19, 23
56. z ϭ 2y Ϫ x
57. Arithmetic sequences can be used to model the numbers of shingles in the rows on a section of roof. Answers should include the following.
• One additional shingle is needed in each successive row.
• One method is to successively add 1 to the terms of the sequence: a8 ϭ
9 ϩ 1 or 10, a9 ϭ 10 ϩ 1 or 11, a10 ϭ 11 ϩ 1 or 12, a11 ϭ 12 ϩ 1 or 13, a12 ϭ
13 ϩ 1 or 14, a13 ϭ
14 ϩ 1 or 15, a14 ϭ 15 ϩ 1 or 16, a15 ϭ 16 ϩ 1 or 17.
Another method is to use the formula for the nth term: a15 ϭ 3 ϩ (15 Ϫ 1)1 or 17.
58. B
59. B
61. Ϫ0.4055
62. 0.4621
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63. 146.4132
64. 15
65. 2, 5, 8, 11
66. 5, 4, 3, 2
67. 11, 15, 19, 23, 27
Lesson 11-2 Arithmetic Series
Pages 586–587
1. In a series, the terms are added. In a sequence, they are not.
2. Sample answer: 0 ϩ 1 ϩ 2 ϩ
3ϩ4
3. Sample answer: a (3n ϩ 4)
4. 1300
5. 230
6. 1932
7. 552
8. 800
4
nϭ1
9. 260
10. 63
11. 95
12. 11, 20, 29
13. Ϫ6, 0, 6
14. 28
15. 344
16. 663
17. 1501
18. 2646
19. Ϫ9
20. Ϫ88
21. 104
22. 182
23. Ϫ714
24. 225
25. 14
26. Ϫ
27. 10 rows
28. 8 days
29. 721
30. 735
31. 162
32. Ϫ204
33. 108
34. Ϫ35
35. Ϫ195
36. 510
37. 315,150
38. 24,300
39. 1,001,000
40. 166,833
41. 17, 26, 35
42. Ϫ13, Ϫ8, Ϫ3
245
6
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43. Ϫ12, Ϫ9, Ϫ6
44. 13, 18, 23
45. 265 ft
46. True; for any series, 2a1 ϩ
2a2 ϩ 2a3 ϩ p ϩ 2an ϭ
2(a1 ϩ a2 ϩ a3 ϩ p ϩ an).
47. False; for example, 7 ϩ 10 ϩ
13 ϩ 16 ϭ 46, but 7 ϩ 10 ϩ
13 ϩ 16 ϩ 19 ϩ 22 ϩ 25 ϩ
28 ϭ 140.
48. Arithmetic series can be used to find the seating capacity of an amphitheater.
Answers should include the following. • The sequence represents the numbers of seats in the rows. The sum of the first n terms of the series is the seating capacity of the first n rows.
• One method is to write out the terms and add them:
18 ϩ 22 ϩ 26 ϩ 30 ϩ 34 ϩ
38 ϩ 42 ϩ 46 ϩ 50 ϩ 54 ϭ
360. Another method is to use the formula n Sn ϭ [2a1 ϩ (n Ϫ 1)d ]:
S10
2
10
ϭ
2
[2(18) ϩ
(10 Ϫ 1)4] or 360.
49. C
50. C
51. 5555
52. 3649
53. 6683
54. 111
60. 23
3 Ϯ 289
2
56. about 3.82 days
55. Ϫ135
62. 2 22
61. 26 221
16
3
9
2
58. Ϫ
57. Ϫ
59.
64. Ϫ54
63. 16
65.
2
27
292
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Lesson 11-3 Geometric Sequences
Pages 590–592
1a. Geometric; the terms have a common ratio of Ϫ2.
1b. Arithmetic; the terms have a common difference of Ϫ3.
3. Marika; Lori divided in the wrong order when finding r.
4. 67.5, 101.25
5. 2, Ϫ4
6. Ϫ2, Ϫ6, Ϫ18, Ϫ54, Ϫ162
7.
2 4 8
,
3 9 27
1, , ,
15
64
p
8. 56
10. an ϭ 4 и 2nϪ1
9. Ϫ4
11. 3, 9
12. A
13. 15, 5
14. 192, 256
15. 54, 81
16. 48, 32
17.
20 40
,
27 81
18.
125 625
,
24 48
19. Ϫ2.16, 2.592
20. Ϫ21.875, 54.6875
21. 2, Ϫ6, 18, Ϫ54, 162
22. 1, 4, 16, 64, 256
23. 243, 81, 27, 9, 3
24. 576, Ϫ288, 144, Ϫ72, 36
25.
3
16
26. 2592
27. 729
28. 1024
29. 243
30.
31. 1
32. 192
33. 78,125
34. 2
35. Ϫ8748
36.
37. 655.36 lb
38. \$46,794.34
39. an ϭ
nϪ1
5
72
1
4
1
36 a b
3
nϪ1
40. an ϭ 64 a b
42. an ϭ 4(Ϫ3)n Ϫ1
41. an ϭ Ϫ2(Ϫ5)n Ϫ1
1
4
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43. Ϯ18, 36, Ϯ72
44. Ϯ12, 36, Ϯ108
45. 16, 8, 4, 2
46. 6, 12, 24, 48
47. 8 days
48. 5 mg
49. False; the sequence 1, 4, 9,
16, p, for example, is neither arithmetic nor geometric. 50. False, the sequence 1, 1, 1,
1, p, for example, is arithmetic (d ϭ 0) and geometric (r ϭ 1).
51. The heights of the bounces of a ball and the heights from which a bouncing ball falls each form geometric sequences. Answers should include the following.
• 3, 1.8, 1.08, 0.648, 0.3888
• The common ratios are the same, but the first terms are different. The sequence of heights from which the ball falls is the sequence of heights of the bounces with the term 3 inserted at the beginning.
52. A
53. C
54. 632.5
55. 203
56. 19, 23
57. Ϫ12, Ϫ16, Ϫ20
58. 5 22 ϩ 3 210 units
59. 127
60.
61.
63
32
61
81
294
Algebra 2
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Chapter 11
Practice Quiz 1
Page 592
11
2
1. 46
2.
3. 187
4. 816
5. 1
Lesson 11-4 Geometric Series
Pages 596–598
2. The polynomial is a geometric series with first term 1, common ratio x, and
4 terms. The sum is
4ϩ2ϩ1ϩ
1
2
1(1 Ϫ x 4) x4 Ϫ 1 ϭ .
1Ϫx
xϪ1
4. 732
3. Sample answer: The first term is a1 ϭ 2. Divide the second term by the first to find that the common ratio is r ϭ 6. Therefore, the nth term of the series is given by
2 ؒ 6nϪ1. There are five terms, so the series can be written as a 2 ؒ 6nϪ1.
5
nϭ1
5. 39,063
6. 81,915
7. 165
8.
9. 129
10.
11.
1093
9
1330
9
31
4
12. 3
13. 3
14. 93 in. or 7 ft 9 in.
15. 728
16. 765
17. 1111
18. 300
295
Algebra 2
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19. 244
20. 1,328,600
21. 2101
22. 1441
23.
728
3
24.
215
4
25. 1040.984
26. 7.96875
27. 6564
28. Ϫ118,096
29. 1,747,625
30. \$10,737,418.23
31. 3641
32. 206,668
33.
182
9
5461
16
34. Ϫ
35. 2555
37.
36. Ϫ364
387
4
38.
58,975
256
39. 3,145,725
40. 86,093,440
41. 243
42. 1024
43. 2
44. 6
45. 80
46. 8
47. about 7.13 in.
48. If the first term and common ratio of a geometric series are integers, then all the terms of the series are integers. Therefore, the sum of the series is an integer.
49. If the number of people that each person sends the joke to is constant, then the total number of people who have seen the joke is the sum of a geometric series. Answers should include the following.
• The common ratio would change from 3 to 4.
• Increase the number of days the joke circulates so that it is inconvenient to find and add all the terms of the series.
50. A
296
Algebra 2
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51. C
52. Ϫ1,048,575
53. 3.99987793
54. 6.24999936
9
1 3
4 2
27
81
55. Ϯ , , Ϯ9
56. Ϫ3, Ϫ , Ϫ , Ϫ
2
4
8
57. 232
58. 192
59. Drive-In Movie Screens
60. Sample answer using
(1, 826) and (3, 750): y ϭ Ϫ38x ϩ 864
Screens
1000
900
800
700
600
0
0
1 2 3 4 5
Years Since 1995
6
61. Sample answer: 294
62. 2
63. 2
64.
65.
2
3
1
4
66. Ϫ2
67. 0.6
Lesson 11-5
Infinite Geometric Series
Pages 602–604
1 a b a 2 nϭ1 ϱ
n
2. 0.999999 . . . can be written as the infinite geometric
9
9
9
ϩ ϩ ϩp. series 10
100
1000
The first term of this series is
9
and the common ratio is
10
9
1
10
, so the sum is
1 or 1.
10
1 Ϫ 10
297
Algebra 2
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3. Beth; the common ratio for the infinite geometric series
4
3
4. 108
4
3
is Ϫ . Since ` Ϫ ` Ն 1, the series does not have a sum a and the formula S ϭ 1
1Ϫr
does not apply.
5. does not exist
7.
6. does not exist
3
4
8.
10.
73
99
5
9
12.
9. 100
11.
30
7
175
999
13. 96 cm
14. 14
15. does not exist
16. 7.5
17. 45
18. 64
19. Ϫ16
20. does not exist
21.
54
5
22. 3
23. does not exist
24. 1
25. 1
26. 7.5
27.
2
3
28. 144
29.
3
2
30. 6
33. 40 ϩ 20 22 ϩ 20 ϩ p
32. 30 ft
31. 2
34. 80 ϩ 4022 or about
136.6 cm
35. 900 ft
36. 27, 18, 12
37. 75, 30, 12
38. 24, 16 , 11 , 7
1
5
1
2
7
25
64
125
40.
39. Ϫ8, Ϫ3 , Ϫ1 , Ϫ
298
11
32
409
512
7
9
Algebra 2
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41.
1
9
42.
4
11
43.
82
99
44.
82
333
45.
427
999
46.
5
11
47.
229
990
48.
S ϭ a1 ϩ a1r ϩ a1r 2 ϩ a1r 3 ϩ p
(Ϫ)rS ϭ a1r ϩ a1r 2 ϩ a1r 3 ϩ a1r 4 ϩ p
S Ϫ rS ϭ a1 ϩ 0 ϩ 0 ϩ0 ϩ0 ϩ p
S(1 Ϫ r )ϭ a1 a1 Sϭ
1Ϫr
49. The total distance that a ball bounces, both up and down, can be found by adding the sums of two infinite geometric series. Answers should include the following.
• an ϭ a1 ؒ r nϪ1, Sn ϭ a1(1 Ϫ r n)
,
1Ϫr
or S ϭ
50. D
a1
1Ϫr
• The total distance the ball falls is given by the infinite geometric series 3 ϩ
3(0.6) ϩ 3(0.6)2 ϩ p . The sum of this series is
3
or 7.5. The total
1 Ϫ 0.6
distance the ball bounces up is given by the infinite geometric series 3(0.6) ϩ
3(0.6)2 ϩ 3(0.6)3 ϩ p .
The sum of this series is
3(0.6)
1 Ϫ 0.6
or 4.5. Thus, the total distance the ball travels is 7.5 ϩ 4.5 or 12 feet.
52. Ϫ182
51. C
53.
8744
81
54. 32.768%
56. Ϫ
55. 3
299
3
2
Algebra 2
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58.
59.
Ϫx ϩ 7
(x Ϫ 3)(x ϩ 1)
61. (x Ϫ 2)2 ϩ (y Ϫ 4)2 ϭ 36
1
2
63. Ϫ ,
Ϫ2a ϩ 5b
2
a b
60.
57. x Ն 5
3x ϩ 7
(x ϩ 4)(x ϩ 2)
62. (x Ϫ 3)2 ϩ (y ϩ 1)2 ϭ 32
3 7
,
2 2
1
2
1
3
64. Ϫ , Ϫ , 0,
1
2
65. x 2 Ϫ 36 ϭ 0
66. x 2 ϩ 9x ϩ 14 ϭ 0
67. x 2 Ϫ 10x ϩ 24 ϭ 0
68. about Ϫ180,724 visitors per year 69. The number of visitors was decreasing. 70. 2
71. 3
72. 2
1
2
74. 4
73.
75. Ϫ4
Lesson 11-6
Recursion and Special Sequences
Pages 608–610
1. an ϭ anϪ1 ϩ d; an ϭ r ؒ anϪ1
2. Sample answer: an ϭ 2anϪ1 ϩ anϪ2
3. Sometimes; if f(x) ϭ x 2 and x1 ϭ 2, then x2 ϭ 22 or 4, so x2 x1. But, if x1 ϭ 1, then x2 ϭ 1, so x2 ϭ x1.
4. 12, 9, 6, 3, 0
5. Ϫ3, Ϫ2, 0, 3, 7
6. 0, Ϫ4, 4, Ϫ12, 20
7. 1, 2, 5, 14, 41
8. 5, 11, 29
9. 1, 3, Ϫ1
10. 3, 11, 123
11. bn ϭ 1.05bnϪ1 Ϫ 10
12. \$1172.41
13. Ϫ6, Ϫ3, 0, 3, 6
14. 13, 18, 23, 28, 33
15. 2, 1, Ϫ1, Ϫ4, Ϫ8
16. 6, 10, 15, 21, 28
17. 9, 14, 24, 44, 84
18. 4, 6, 12, 30, 84
19. Ϫ1, 5, 4, 9, 13
20. 4, Ϫ3, 5, Ϫ1, 9
300
Algebra 2
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21.
7 7 7 7 7
, , , ,
2 4 6 8 10
22.
3 3 15 25 425
, ,
,
,
4 2 4 2
8
23. 67
24. Ϫ2.1
25. 1, 1, 2, 3, 5, . . .
26. the Fibonacci sequence
27. \$99,921.21,
\$99,762.21,
\$99,601.29,
\$99,438.44,
28. 1, 3, 6, 10, 15
\$99,841.95,
\$99,681.99,
\$99,520.11,
\$99,356.28
29. tn ϭ tnϪ1 ϩ n
30. 20,100
31. 16, 142, 1276
32. 5, 17, 65
33. Ϫ7, Ϫ16, Ϫ43
34. Ϫ4, Ϫ19, Ϫ94
35. Ϫ3, 13, 333
36. Ϫ1, Ϫ1, Ϫ1
37.
5 37 1445
,
,
2 2
2
38.
4 10 76
,
,
3 3 3
39. \$75.77
40. No; according to the first two iterates, f(4) ϭ 4. According to the second and third iterates, f(4) ϭ 7. Since f(x) is a function, it cannot have two values when x ϭ 4.
41. Under certain conditions, the
Fibonacci sequence can be used to model the number of shoots on a plant. Answers should include the following.
• The 13th term of the sequence is 233, so there are 233 shoots on the plant during the 13th month.
• The Fibonacci sequence is not arithmetic because the differences (0, 1, 1, 2, . . .) of the terms are not constant. The Fibonacci sequence is not geometric because the ratios
3
Q1, 2, , . . .R of the terms
42. D
2
are not constant.
301
Algebra 2
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44. 27
43. C
1
6
46.
12
5
47. Ϫ5208
48.
1093
243
49. 3x ϩ 7 units
50. 120
51. 5040
52. 6
53. 20
54. 126
45.
55. 210
Lesson 11-7 The Binomial Theorem
Pages 615–617
1. 1, 8, 28, 56, 70, 56, 28, 8, 1
2. n
3. Sample answer: (5x ϩ y)4
4. 40,320
5. 17,160
6. 66
7. p5 ϩ 5p4q ϩ 10p3q 2 ϩ
10p 2q 3 ϩ 5pq 4 ϩ q 5
8. t 6 ϩ 12t 5 ϩ 60t 4 ϩ
160t 3 ϩ 240t 2 ϩ 192t ϩ 64
9. x 4 Ϫ 12x 3y ϩ 54x 2y 2 Ϫ
108xy 3 ϩ 81y 4
10. 56a5b3
11. 1,088,640a6b4
12. 10
13. 362,880
14. 6,227,020,800
15. 72
16. 210
17. 495
18. 2002
19. a 3 Ϫ 3a 2b ϩ 3ab 2 Ϫ b 3
20. m4 ϩ 4m 3n ϩ 6m 2n 2 ϩ
4mn3 ϩ n4
21. r 8 ϩ 8r 7s ϩ 28r 6s 2 ϩ
56r 5s 3 ϩ 70r 4s 4 ϩ
56r 3s 5 ϩ 28r 2s 6 ϩ 8rs 7 ϩ s 8
22. m 5 Ϫ 5m 4a ϩ 10m 3a 2 Ϫ
10m 2a 3 ϩ 5ma 4 Ϫ a 5
23. x 5 ϩ 15x 4 ϩ 90x 3 ϩ
270x 2 ϩ 405x ϩ 243
24. a4 Ϫ 8a3 ϩ 24a2 Ϫ 32a ϩ 16
302
Algebra 2
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25. 16b4 Ϫ 32b3x ϩ
24b 2x 2 Ϫ 8bx 3 ϩ x 4
26. 64a6 ϩ 192a5b ϩ 240a4b 2 ϩ
160a3b3 ϩ 60a2b4 ϩ
12ab5 ϩ b6
28. 81x 4 ϩ 216x 3y ϩ
216x 2y 2 ϩ 96xy 3 ϩ 16y 4
27. 243x 5 Ϫ 810x 4y ϩ 1080x 3y 2
Ϫ 720x 2y 3 ϩ 240xy 4 Ϫ 32y 5
29.
a5
32
ϩ
5a4
8
30. 243 ϩ 135m ϩ 30m 2 ϩ
ϩ 5a3 ϩ
10m 3
3
20a2 ϩ 40a ϩ 32
ϩ
5m 4
27
ϩ
31. 27x 3 ϩ 54x 2 ϩ 36x ϩ 8 cm3
32. 1, 4, 6, 4, 1
33. 45
34. Ϫ126x 4y 5
35. 924x 6y 6
36. 280x 4
37. 5670a4
38. 1,088,640a6b4
39. 145,152x 6y 3
40.
35 4 x 27
42.
12!
7!5!
m5
243
63
8
41. Ϫ x 5
and
12!
6!6!
represent the 6th
and 7th entries in the row for n ϭ 12 in Pascal’s triangle.
13!
represents the seventh
7!6!
entry in the row for n ϭ 13.
13!
12!
Since
is below and 7!6!
7!5!
12! in Pascal’s triangle,
6!6!
12!
12!
13! ϩ ϭ
.
7!5!
6!6!
7!6!
43. The coefficients in a binomial expansion give the numbers of sequences of births resulting in given numbers of boys and girls. Answers should include the following.
• (b ϩ g)5 ϭ b5 ϩ 5b4g ϩ
10b 3g 2 ϩ 10b 2g 3 ϩ
5bg 4 ϩ g 5;
44. D
303
Algebra 2
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There is one sequence of births with all 5 boys, five sequences with 4 boys and
1 girl, ten sequences with
3 boys and 2 girls, ten sequences with 2 boys and
3 girls, five sequences with
1 boy and 4 girls, and one sequence with all 5 girls.
• The number of sequences of births that have exactly k girls in a family of n children is the coefficient of bnϪkgk in the expansion of (b ϩ g)n. According to the Binomial Theorem, this n! . coefficient is
(n Ϫ k)!k!
45. C
46. 7, 5, 3, 1, Ϫ1
47. 3, 5, 9, 17, 33
48. 125 cm
49.
log 5
;
log 2
2.3219
50.
51.
log 8
;
log 5
1.2920
52. asymptotes: x ϭ Ϫ2, x ϭ Ϫ3
1
;
log 3
2.0959
53. asymptotes: x ϭ Ϫ4, x ϭ 1
54. hole: x ϭ Ϫ3
55. hyperbola
56. parabola
57. yes
58. no
59. True;
61. True;
1(1 ϩ 1)
2
ϭ
12(1 ϩ 1)2
4
1(2)
2
ϭ
60. False;
or 1.
1(4)
4
2(3)
2
(1 ϩ 1)(2 ؒ 1 ϩ 1)
2
ϭ
or 3.
62. True; 31 Ϫ 1 ϭ 2, which is even. or 1.
304
Algebra 2
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Chapter 11
Practice Quiz 2
Page 617
1. 1,328,600
2. Ϫ364
3. 24
4.
5. 1, 5, 13, 29, 61
6. 2, 4, 8, 14, 22
7. 5, Ϫ13, 41
8. 243x 5 ϩ 405x 4y ϩ 270x 3y 2 ϩ
90x 2y 3 ϩ 15xy 4 ϩ y 5
9. a6 ϩ 12a5 ϩ 60a4 ϩ
160a3 ϩ 240a 2 ϩ 192a ϩ 64
Lesson 11–8
25
4
10. 4032a5b4
Proof and Mathematical Induction
Pages 619–621
1. Sample answers: formulas for the sums of powers of the first n positive integers and statements that expressions involving exponents of n are divisible by certain numbers
2. Mathematical induction is used to show that a statement is true. A counter example is used to show that a statement is false.
3. Sample answer: 3n Ϫ 1
4. Step 1: When n ϭ 1, the left side of the given equation is
1(1 ϩ 1)
1. The right side is or 2
1, so the equation is true for n ϭ 1.
Step 2: Assume 1 ϩ 2 ϩ
3ϩ p ϩkϭ
k(k ϩ 1)
2
for
some positive integer k.
Step 3: 1 ϩ 2 ϩ 3 ϩ p ϩ k ϩ 1k ϩ 12 ϭ ϭ ϭ ©Glencoe/McGraw-Hill
305
k(k ϩ 1) ϩ (k ϩ 1)
2
k(k ϩ 1) ϩ 2(k ϩ 1)
2
(k ϩ 1) ϩ (k ϩ 2)
2
Algebra 2
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The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore, 1 ϩ 2 ϩ 3 ϩ p ϩ nϭ n(n ϩ 1)
2
for all
positive integers n.
5. Step 1: When n ϭ 1, the left side of the given equation is
1
.
2
1
,
2
The right side is 1 Ϫ
1
2
6. Step 1: 41 Ϫ 1 ϭ 3, which is divisible by 3. The statement is true for n ϭ 1.
Step 2: Assume that 4k Ϫ 1 is divisible by 3 for some positive integer k. This means that 4k Ϫ 1 ϭ 3r for some whole number r.
Step 3: 4k Ϫ 1 ϭ 3r
4k ϭ 3r ϩ 1
4kϩ1 ϭ 12r ϩ 4
4kϩ1 Ϫ 1 ϭ 12r ϩ 3
4kϩ1 Ϫ 1 ϭ 314r ϩ 12
Since r is a whole number,
4r ϩ 1 is a whole number.
Thus, 4kϩ1 Ϫ 1 is divisible by
3, so the statement is true for n ϭ k ϩ 1. Therefore,
4n Ϫ 1 is divisible by 3 for all positive integers n.
or
so the equation is true for
n ϭ 1.
1
1 ϩ 2ϩ
2
2
1
1
1
ϩ p ϩ k ϭ 1Ϫ k for some
3
2
2
2
Step 2: Assume
positive integer k.
Step 3:
1
1
ϩ 2ϩ
2
2
1
1
1
ϩ p ϩ k ϩ kϩ1
3
2
2
2
1
1 ϭ 1 Ϫ k ϩ kϩ1
2
2
2
1 ϭ 1 Ϫ kϩ1 ϩ kϩ1
2
2
1
ϭ 1 Ϫ kϩ1
2
The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore,
ϭ
1
2n
1
1
1
1
ϩ 2ϩ 3ϩ pϩ n
2
2
2
2
for all positive
integers n.
306
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7. Step 1: 51 ϩ 3 ϭ 8, which is divisible by 4. The statement is true for n ϭ 1.
Step 2: Assume that 5k ϩ 3 is divisible by 4 for some positive integer k. This means that 5k ϩ 3 ϭ 4r for some positive integer r.
Step 3: 5k ϩ 3 ϭ 4r
5k ϭ 4r Ϫ 3
5kϩ1 ϭ 20r Ϫ 15
5kϩ1 ϩ 3 ϭ 20r Ϫ 12
5kϩ1 ϩ 3 ϭ 415r Ϫ 32
Since r is a positive integer,
5r Ϫ 3 is a positive integer.
Thus, 5kϩ1 ϩ 3 is divisible by
4, so the statement is true for n ϭ k ϩ 1.
Therefore, 5n ϩ 3 is divisible by 4 for all positive integers n.
8. Sample answer: n ϭ 2
9. Sample answer: n ϭ 3
10. Step 1: After the first guest has arrived, no handshakes have taken place.
1(1 Ϫ 1)
2
ϭ 0,
so the formula is correct for n ϭ 1.
Step 2: Assume that after k guests have arrived, a total of k(k Ϫ 1)
2
handshakes have
take place, for some positive integer k.
Step 3: When the (k ϩ 1)st guest arrives, he or she shakes hands with the k guests already there, so the total number of handshakes that have then taken place is k(k Ϫ 1)
2
307
ϩ k.
Algebra 2
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k(k Ϫ 1) k(k Ϫ 1) ϩ 2k ϩkϭ 2
2
k [(k Ϫ 1) ϩ 2] ϭ 2 k (k ϩ 1)
(k ϩ 1)k or ϭ
2
2
The last expression is the formula to be proved, where n ϭ k ϩ 1. Thus, the formula is true for n ϭ k ϩ 1.
Therefore, the total number of handshakes is
for
all positive integers n.
12. Step 1: When n ϭ 1, the left side of the given equation is
11. Step 1: When n ϭ 1, the left side of the given equation is
1. The right side is 1[2(1) Ϫ 1] or 1, so the equation is true for n ϭ 1.
Step 2: Assume 1 ϩ 5 ϩ 9 ϩ p ϩ (4k Ϫ 3) ϭ k (2k Ϫ 1) for some positive integer k.
Step 3: 1 ϩ 5 ϩ 9 ϩ p ϩ
(4k Ϫ 3) ϩ [4(k ϩ 1) Ϫ 3] ϭ k (2k Ϫ 1) ϩ
[4(k ϩ 1) Ϫ 3] ϭ 2k 2 Ϫ k ϩ 4k ϩ 4 Ϫ 3 ϭ 2k 2 ϩ 3k ϩ 1 ϭ (k ϩ 1)(2k ϩ 1) ϭ (k ϩ 1)[2(k ϩ 1) Ϫ 1]
The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore, 1 ϩ 5 ϩ 9 ϩ p ϩ
(4n Ϫ 3) ϭ n(2n Ϫ 1) for all positive integers n.
n(n Ϫ 1)
2
1[3(1) ϩ 1]
2. The right side is
2
or 2, so the equation is true for n ϭ 1.
Step 2: Assume 2 ϩ 5 ϩ
8 ϩ p ϩ (3k Ϫ 1) ϭ
k(3k ϩ 1)
2
for some positive integer k.
Step 3: 2 ϩ 5 ϩ 8 ϩ p ϩ
(3k Ϫ 1) ϩ [3(k ϩ 1) Ϫ 1] k(3k ϩ 1) ϩ [3(k ϩ 1) Ϫ 1]
2
k(3k ϩ 1) ϩ 2[3(k ϩ 1) Ϫ 1] ϭ 2
2
3k ϩ k ϩ 6k ϩ 6 Ϫ 2 ϭ 2
ϭ
3k 2 ϩ 7k ϩ 4
2
(k ϩ 1)(3k ϩ 4) ϭ 2
(k ϩ 1)[3(k ϩ 1) ϩ 1] ϭ 2
ϭ
308
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The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore, 2 ϩ 5 ϩ 8 ϩ p ϩ n(3n ϩ 1)
(3n Ϫ 1) ϭ for all
2
positive integers n.
14. Step 1: When n ϭ 1, the left side of the given equation is
12 or 1. The right side is
13. Step 1: When n ϭ 1, the left side of the given equation is
13 or 1. The right side is
12(1 ϩ 1)2
4
1[2(1) Ϫ 1][2(1) ϩ 1]
3
or 1, so the equation is true for n ϭ 1.
Step 2: Assume 13 ϩ 23 ϩ
33 ϩ p ϩ k 3 ϭ
k 2 1k ϩ 12 2
4
the equation is true for n ϭ 1.
Step 2: Assume
12 ϩ 32 ϩ 52 ϩ p ϩ(2k Ϫ 1)2 ϭ
for
some positive integer k.
Step 3: 13 ϩ 2 3 ϩ 33 ϩ p ϩ k 3 ϩ (k ϩ 1)3 ϭ ϭ ϭ ϭ ϭ ϭ
k (2k Ϫ 1)(2k ϩ 1)
3
for some
positive integer k.
Step 3:
12 ϩ 32 ϩ 52 ϩ p ϩ
(2k Ϫ 1)2 ϩ [2(k ϩ 1) Ϫ 1]2
k 2(k ϩ 1)2 ϩ (k ϩ 1)3
4
k 2(k ϩ 1)2 ϩ 4(k ϩ 1)3
4
2 2
(k ϩ 1) [k ϩ 4(k ϩ 1)]
4
2 2
(k ϩ 1) (k ϩ 4k ϩ 4)
4
2
(k ϩ 1) (k ϩ 2)2
4
2
(k ϩ 1) [(k ϩ 1) ϩ 1]2
4
ϭ
k(2k Ϫ 1)(2k ϩ 1) ϩ 3
[2(k ϩ 1) Ϫ 1]2 ϭ ϭ ϭ The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
or 1, so
ϭ ϭ ϭ
309
k(2k Ϫ 1)(2k ϩ 1) ϩ 3(2k ϩ 1)2
3
(2k ϩ 1)[k (2k Ϫ 1) ϩ 3(2k ϩ 1)]
3
2
(2k ϩ 1)(2k Ϫ k ϩ 6k ϩ 3)
3
2
(2k ϩ 1)(2k ϩ 5k ϩ 3)
3
(2k ϩ 1)(k ϩ 1)(2k ϩ 3)
3
(k ϩ1)[2(k ϩ1) Ϫ 1][2(k ϩ 1) ϩ1]
3
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The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1. Therefore,
12 ϩ 32 ϩ 52 ϩ p ϩ
Therefore, 13 ϩ 23 ϩ 33 ϩ p ϩ
2
2
n (n ϩ 1)
4
n3 ϭ
for all
positive integers n.
(2n Ϫ 1)2 ϭ
n(2n Ϫ 1)(2n ϩ 1)
3
for all positive integers n.
16. Step 1: When n ϭ 1, the left side of the given equation is
15. Step 1: When n ϭ 1, the left side of the given equation is
1
1
. The right side is a1
2
3
1
or , so the equation is
3
p ϩ
ϭ
1 a1 2
3
3
1
Ϫ kb for
3
true
or , so the equation is true
1
3k
ϩ
1
3
ϩ
1
32
ϩ
1
33
The right side is a1 Ϫ b
1
4
for n ϭ 1.
1
1
Step 2: Assume ϩ 2 ϩ
3
1
43
some
ϩp ϩ
Step 3:
ϩ p ϩ
1
4k
1
3k ϩ 1
ϩ
1
4
ϭ ϭ ϭ ϭ ϩ
1
42
ϩ
1
43
ϩ p ϩ
1 k ϩ1
4
1
1
1 kb ϩ k ϩ 1
3
4
4
1
1
1
Ϫ
k ϩ k ϩ1
3
3ؒ4
4
4k ϩ 1 Ϫ 4 ϩ 3
3 ؒ 4k ϩ 1
4k ϩ 1 Ϫ 1
3 ؒ 4k ϩ 1
1 4k ϩ 1 Ϫ 1 a b
3
4k ϩ 1
1
1
1 a2 Ϫ kb ϩ k ϩ 1
2
3
3
1
1
1
Ϫ
k ϩ k ϩ1
2
2ؒ3
3
3k ϩ 1 Ϫ 3 ϩ 2
2 ؒ 3k ϩ 1
3k ϩ 1 Ϫ 1
2 ؒ 3k ϩ 1
1 3k ϩ 1 Ϫ 1 a b
2
3k ϩ 1
ϭ a1 Ϫ
1
2
ϭ
4
4
1
1
1 ϭ a1 Ϫ kb k 4
3
4
for some positive integer k.
positive integer k.
Step 3:
1
4
1
.
4
1
3
for n ϭ 1.
1
1
1
Step 2: Assume ϩ 2 ϩ 2 ϩ
1
3k
1
3
Ϫ b
ϭ a1 Ϫ
ϭ a1 Ϫ
ϭ ϭ ϭ ϭ 1
3
1
b
3k ϩ 1
310
1
k ϩ 1b
4
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The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
1
1
1
1
Therefore, ϩ 2 ϩ 3 ϩ p ϩ n
The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore,
ϭ
1 a1 2
Ϫ
1
1
1
1
ϩ 2ϩ 3ϩpϩ n
3
3
3
3
1
b for all positive
3n
4
1
3
ϭ a1 Ϫ
1 b 4n
4
4
4
for all positive
integers n.
integers n.
17. Step 1: 81 Ϫ 1 ϭ 7, which is divisible by 7. The statement is true for n ϭ 1.
Step 2: Assume that 8k Ϫ 1 is divisible by 7 for some positive integer k. This means that 8k Ϫ 1 ϭ 7r for some whole number r.
Step 3: 8k Ϫ 1 ϭ 7r
8k ϭ 7r ϩ 1
8k ϩ1 ϭ 56r ϩ 8
8kϩ1 Ϫ 1 ϭ 56r ϩ 7
8k ϩ1 Ϫ 1 ϭ 7(8r ϩ 1)
Since r is a whole number,
8r ϩ 1 is a whole number.
Thus, 8kϩ1 Ϫ 1 is divisible by
7, so the statement is true for n ϭ k ϩ 1.
Therefore, 8n Ϫ 1 is divisible by 7 for all positive integers n.
18. Step 1: 91 Ϫ 1 ϭ 8, which is divisible by 8. The statement is true for n ϭ 1.
Step 2: Assume that 9k Ϫ 1 is divisible by 8 for some positive integer k. This means that 9k Ϫ 1 ϭ 8r for some whole number r.
Step 3: 9k Ϫ 1 ϭ 8r
9k ϭ 8r ϩ 1
9kϩ1 ϭ 72r ϩ 9
9kϩ1 Ϫ 1 ϭ 72r ϩ 8
9kϩ1 Ϫ 1 ϭ 8(9r ϩ 1)
Since r is a whole number,
9r ϩ 1 is a whole number.
Thus, 9k ϩ1 Ϫ 1 is divisible by 8, so the statement is true for n ϭ k ϩ 1.
Therefore, 9n Ϫ 1 is divisible by 8 for all positive integers n.
19. Step 1: 121 ϩ 10 ϭ 22, which is divisible by 11. The statement is true for n ϭ 1.
Step 2: Assume that 12k ϩ 10 is divisible by 11 for some positive integer k. This means that 12k ϩ 10 ϭ 11r for some positive integer r.
20. Step 1: 131 ϩ 11 ϭ 24, which is divisible by 12. The statement is true for n ϭ 1.
Step 2: Assume that 13k ϩ11 is divisible by 12 for some positive integer k. This means that 13k ϩ 11 ϭ 12r for some positive integer r.
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Step 3:
12k ϩ 10 ϭ 11r
12k ϭ 11r Ϫ 10
12kϩ1 ϭ 132r Ϫ 120
12kϩ1 ϩ 10 ϭ 132r Ϫ 110
12kϩ1 ϩ 10 ϭ 11(12r Ϫ 10)
Since r is a positive integer,
12r Ϫ 10 is a positive integer. Thus, 12k ϩ1 ϩ 10 is divisible by 11, so the statement is true for n ϭ k ϩ 1.
Therefore, 12n ϩ 10 is divisible by 11 for all positive integers n.
Step 3:
13k ϩ 11 ϭ 12r
13k ϭ 12r Ϫ 11
13k ϩ1 ϭ 156r Ϫ 143
13k ϩ1 ϩ 11 ϭ 156r Ϫ 132
13k ϩ1 ϩ 11 ϭ 12(13r Ϫ 11)
Since r is a positive integer,
13r Ϫ 11 is a positive integer. Thus, 13kϩ1 ϩ 11 is divisible by 12, so the statement is true for n ϭ k ϩ 1.
Therefore, 13n ϩ 11 is divisible by 12 for all positive integers n.
21. Step 1: There are 6 bricks in the top row, and 12 ϩ 5(1) ϭ
6, so the formula is true for n ϭ 1.
Step 2: Assume that there are k 2 ϩ 5k bricks in the top k rows for some positive integer k.
Step 3: Since each row has
2 more bricks than the one above, the numbers of bricks in the rows form an arithmetic sequence. The number of bricks in the
(k ϩ 1)st row is
6 ϩ [(k ϩ 1) Ϫ 1](2) or
2k ϩ 6. Then the number of bricks in the top k ϩ 1 rows is k 2 ϩ 5k ϩ (2k ϩ 6) or k 2 ϩ 7k ϩ 6. k 2 ϩ 7k ϩ 6 ϭ (k ϩ 1)2 ϩ
5(k ϩ 1), which is the formula to be proved, where n ϭ k ϩ 1. Thus, the formula is true for n ϭ k ϩ 1.
22. Step 1: When n ϭ 1, the left side of the given equation is
1
a1. The right side is a1(1 Ϫ r )
1Ϫr
or a1, so the equation is true for n ϭ 1.
Step 2: Assume a1 ϩ a1r ϩ a1r 2 ϩ p ϩ a1r kϪ1 ϭ a1(1 Ϫ r k )
1Ϫr
for some positive
integer k.
Step 3: a1 ϩ a1r ϩ a1r 2 ϩ p ϩ a1r kϪ1 ϩ a1r k ϭ ϭ ϭ ϭ
a1(1 Ϫ r k) ϩ a1r k
1Ϫr
a1(1 Ϫ r k ) ϩ (1 Ϫ r )a1r k
1Ϫr
k ϩ a r k Ϫ a r kϩ1 a1 Ϫ a1r
1
1
1Ϫr
a1(1 Ϫ r k ϩ 1)
1Ϫr
The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
312
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Thus, the equation is true for n ϭ k ϩ 1.
Therefore, a1 ϩ a1r ϩ a1r 2 ϩ p ϩ a1r nϪ1 ϭ
Therefore, the number of bricks in the top n rows is n 2 ϩ 5n for all positive integers n.
a1(1 Ϫ r n )
1Ϫr
for all positive integers n.
24. Step 1: The figure shows how to cover a 21 by 21 board, so the statement is true for n ϭ 1.
23. Step 1: When n ϭ 1, the left side of the given equation is
1
2
a1. The right side is [2a1 ϩ
(1 Ϫ 1)d ] or a1, so the equation is true for n ϭ 1.
Step 2: Assume a1ϩ
(a1 ϩ d ) ϩ (a1 ϩ 2d 2 ϩ p ϩ
[a1 ϩ (k Ϫ 1)d ] ϭ k
[2a1
2
ϩ (k Ϫ 1)d ] for
k
[2a1
2
ϩ (k Ϫ 1)d ] ϩ
k
[2a1
2
ϩ (k Ϫ 1)d ]
Step 2: Assume that a 2k by
2k board can be covered for some positive integer k.
some positive integer k.
Step 3: a1 ϩ (a1 ϩ d ) ϩ
(a1 ϩ 2d ) ϩ p ϩ [a1 ϩ
(k Ϫ 1)d ] ϩ
[a1 ϩ (k ϩ 1 Ϫ 1)d ] ϭ [a1 ϩ (k ϩ 1 Ϫ 1)d ] ϭ ϩ a1 ϩ kd
k [2a1 ϩ (k Ϫ 1)d ] ϩ 2(a1 ϩ kd )
2
2 k ؒ 2a1 ϩ (k Ϫ k)d ϩ 2a1 ϩ 2kd ϭ 2
(k ϩ 1)2a1 ϩ (k 2 Ϫ k ϩ 2k)d ϭ 2
(k ϩ 1)2a1 ϩ k(k ϩ 1)d ϭ 2 kϩ1 ϭ
(2a1 ϩ kd )
2
kϩ1
[2a1 ϩ (k ϩ 1 Ϫ 1)d ] ϭ 2
ϭ
Step 3: Divide a 2kϩ1 by
2kϩ1 board into four quadrants. By the inductive hypothesis, the first quadrant can be covered. Rotate the design that covers Quadrant
I 90Њ clockwise and use it to cover Quadrant II. Use the design that covers Quadrant
I to cover Quadrant III.
313
Algebra 2
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The last expression is the right side of the formula to be proved, where n ϭ k ϩ 1.
Thus, the formula is true for n ϭ k ϩ 1.
Therefore, a1 ϩ (a1 ϩ d ) ϩ
(a1 ϩ 2d ) ϩ p ϩ [a1 ϩ
(n Ϫ 1)d ] ϭ
Rotate the design that covers
Quadrant I 90Њ counterclockwise and use it to cover Quadrant IV. This leaves three empty squares near the center of the board, as shown. Use one more
L-shaped tile to cover these
3 squares. Thus, a 2kϩ1 by
2kϩ1 board can be covered.
The statement is true for n ϭ k ϩ 1.
Therefore, a 2n by 2n checkerboard with the top right square missing can be covered for all positive integers n.
n
[2a1 ϩ(n Ϫ 1)d ]
2
for all positive integers n.
25. Sample answer: n ϭ 3
26. Sample answer: n ϭ 4
27. Sample answer: n ϭ 2
28. Sample answer: n ϭ 3
29. Sample answer: n ϭ 11
30. Sample answer: n ϭ 41
31. Write 7n as (6 ϩ 1)n. Then use the Binomial Theorem.
7n Ϫ 1 ϭ (6 ϩ 1)n Ϫ 1
32. An analogy can be made between mathematical induction and a ladder with the positive integers on the steps. Answers should include the following.
• Showing that the statement is true for n ϭ 1 (Step 1).
• Assuming that the statement is true for some positive integer k and showing that it is true for k ϩ 1 (Steps 2 and 3).
ϭ 6n ϩ n ؒ 6nϪ1 ϩ
n(n Ϫ 1)
ؒ
2
6nϪ2 ϩ p ϩ n ؒ 6 ϩ 1 Ϫ 1 ϭ 6n ϩ n ؒ 6nϪ1 ϩ
n(n Ϫ 1)
ؒ
2
6nϪ2 ϩ p ϩ n ؒ 6
Since each term in the last expression is divisible by 6, the whole expression is divisible by 6. Thus, 7n Ϫ 1 is divisible by 6.
33. C
34. A
35. x 6 ϩ 6x 5y ϩ 15x 4y 2 ϩ
20x 3y 3 ϩ 15x 2y 4 ϩ 6xy 5 ϩ y 6
36. a7 Ϫ 7a6b ϩ 21a5b 2 Ϫ
35a4b3 ϩ 35a3b4 Ϫ
21a 2b 5 ϩ 7ab6 Ϫ b7
314
Algebra 2
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37. 256x 8 ϩ 1024x 7y ϩ
1792x 6y 2 ϩ 1792x 5y 3 ϩ
1120x 4y 4 ϩ 448x 3y 5 ϩ
112x 2y 6 ϩ 16xy 7 ϩ y 8
38. 4, 10, 28
39. 2, 14, 782
40. 12 h
41. 0, 1
42. Ϫ14
315
Algebra 2
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Chapter 12 Probability and Statistics
Lesson 12-1 The Counting Principle
Pages 634–637
1. HHH, HHT, HTH, HTT, THH,
THT, TTH, TTT
2. Sample answer: buying a shirt that comes in 3 sizes and 6 colors
3. The available colors for the car could be different from those for the truck.
4. independent
5. dependent
6. 30
7. 256
8. 20
10. dependent
9. D
11. independent
12. independent
13. dependent
14. 6
15. 16
16. 6
17. 30
18. 48
19. 1024
20. 240
21. 10,080
22. 151,200
23. 362,880
24. 17
25. 27,216
26. 160
27. 800
28. See students’ work.
316
Algebra 2
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29. The maximum number of license plates is a product with factors of 26s and 10s, depending on how many letters are used and how many digits are used.
Answers should include the following. 30. A
• There are 26 choices for the first letter, 26 for the second, and 26 for the third. There are 10 choices for the first number, 10 for the second, and 10 for the third. By the Fundamental
Counting Principle, there are 263 ؒ 103 or 17,576,000 possible license plates.
• Replace positions containing numbers with letters.
31. C
32. 45
33. 20 mi
34. Step 1: When n ϭ 1, the left side of the given equation is
4. The right side is
1[3(1) ϩ 5]
2
or 4, so the equation is true for n ϭ 1.
Step 2: Assume 4 ϩ 7 ϩ 10 ϩ p ϩ (3k ϩ 1) ϭ
k(3k ϩ 5)
2
for some positive integer k.
Step 3: 4 ϩ 7 ϩ 10 ϩ p ϩ
(3k ϩ 1) ϩ [3(k ϩ 1) ϩ 1] ϭ ϭ ϭ ϭ
317
k(3k ϩ 5) ϩ [3(k ϩ 1) ϩ 1]
2
k(3k ϩ 5) ϩ 2[3(k ϩ 1) ϩ 1]
2
2
3k ϩ 5k ϩ 6k ϩ 6 ϩ 2
2
2
3k ϩ 11k ϩ 8
2
Algebra 2
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ϭ ϭ (k ϩ 1)(3k ϩ 8)
2
(k ϩ 1)[3(k ϩ 1) ϩ 5]
2
The last expression is the right side of the equation to be proved, where n ϭ k ϩ 1.
Thus, the equation is true for n ϭ k ϩ 1.
Therefore, 4 ϩ 7 ϩ 10 ϩ p n(3n ϩ 5)
ϩ (3n ϩ 1) ϭ
2
for all positive integers n.
35. 28x 6 y 2
36. 280a3b4
37. 7
38. 5
39.
1
2
40. Ϫ1
41. Ϫ
x x ϩ 5y
42. 36 mi
43. Ϯ1, Ϯ2
44. 0, Ϫ2
45. y ϭ (x Ϫ 3)2 ϩ 2
46. y ϭ Ϫ2(x ϩ 1)2 ϩ 4
1
2
47. y ϭ Ϫ x 2 ϩ 8
48. 4
49. 3
50. 4
5
R
4
54. y ϭ Ϫ2x Ϫ 2
Ϫ1
R
3
53. no inverse exists
51.
1 1
B
7 4
2
3
55. y ϭ x ϩ
52.
1
3
1 Ϫ1
B
6 Ϫ2
56. 60
57. 30
58. 840
59. 720
60. 6
61. 15
62. 56
63. 1
318
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Permutations and Combinations
Pages 641–643
1. Sample answer: There are six people in a contest. How many ways can the first, second, and third prizes be awarded? 2. C (n, n Ϫ r) ϭ ϭ ϭ n!
[n Ϫ (n Ϫ r )]!(n Ϫ r )! n! r !(n Ϫ r )! n! (n Ϫ r )!r !
ϭ C(n, r )
3. Sometimes; the statement is only true when r ϭ 1.
4. 60
5. 120
6. 6
7. 6
8. combination; 15
10. permutation; 90,720
9. permutation; 5040
11. 84
12. 56
13. 9
15. 665,280
14. 2520
16. 10
17. 70
18. 792
19. 210
20. 27,720
21. 1260
22. permutation; 5040
23. combination; 28
24. permutation; 2520
25. permutation; 120
26. combination; 220
27. permutation; 3360
28. combination; 45
29. combination; 455
30. 11,880
31. 60
32. 75,287,520
33. 111,540
34. 267,696
35. 80,089,128
36. 528
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38. Permutations and combinations can be used to find the number of different lineups. Answers should include the following.
• There are 9! different
9-person lineups available:
9 choices for the first player, 8 choices for the second player, 7 for the third player, and so on.
So, there are 362,880 different lineups.
• There are C(16, 9) ways to choose 9 players from
37. C (n Ϫ 1,r ) ϩC (n Ϫ 1, r Ϫ 1) ϭ ϭ
(n Ϫ 1)! ϩ (n Ϫ 1 Ϫ r )!r !
(n Ϫ 1)!
[n Ϫ 1 Ϫ (r Ϫ 1)] !(r Ϫ 1)!
(n Ϫ 1)! ϩ (n Ϫ r Ϫ 1 )!r !
(n
ϭ
ϭ ϭ ϭ ϭ (n
(n
(n
(n
(n Ϫ 1)!
Ϫ r )! (r Ϫ 1)!
(n Ϫ 1)! nϪr ؒ ϩ Ϫ r Ϫ 1)!r ! n Ϫ r
(n Ϫ 1)! r ؒ
Ϫ r )!(r Ϫ 1)! r
Ϫ 1)!(n Ϫ r )
(n Ϫ 1)!r ϩ (n Ϫ r )!r !
(n Ϫ r )!r !
Ϫ 1)!(n Ϫ r ϩ r )
(n Ϫ r )!r !
16: C (16, 9) ϭ
11,440.
16!
7!9!
or
(n Ϫ 1)!n
(n Ϫ r )!r ! n! (n Ϫ r )!r !
ϭ C(n, r )
39.
D
40. A
41. 24
42. 6
43. 120
44. 8
45. 80
46. Sample answer: n ϭ 3
47. Sample answer: n ϭ 2
48. Ϫ1.0986
49. x Ͼ 0.8047
50. 21.0855
51. 20 days
53.
(y Ϫ 4)2
9
ϩ
52.
(x Ϫ 4) 2
4
ϭ1
x2
16
ϩ
y2
9
ϭ1
7 53
2 2
54. Ϫ ;
60. 0 x 3 0 y 3 23
55. Ϫ4; 128
56. {Ϫ4, 4}
57. {Ϫ2, 5}
58. e Ϫ3, f
1
3
59. 822
61. 425
62. (Ϫ1, 3)
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4
3
63. (0, 2)
64. Ϫ
65. Ϫ
6
7
66. 0
67. {Ϫ7, 15}
68. л
69.
3
5
70.
1
2
71.
1
5
72.
1
3
Lesson 12-3 Probability
Pages 647–650
1. Sample answer: The event
July comes before June has a probability of 0. The event
June comes before July has a probability of 1.
2.
3
5
3. There are 6 ؒ 6 or 36 possible outcomes for the two dice. Only 1 outcome, 1 and 1, results in a sum of 2,
4.
1
7
6.
4
7
1
so P(2) ϭ . There are 2
36
outcomes, 1 and 2 as well as
2 and 1, that result in a sum
2
1 of 3, so P(3) ϭ or .
36
5.
18
2
7
7. 8:1
8. 1:5
10.
9. 2:7
6
11
11.
10
11
12.
2
7
13.
1
8
14.
3
8
321
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15.
1
10
16.
21
50
17.
2
25
18.
1
50
19.
6
55
20.
21
55
21.
28
55
22.
14
575
23.
11
115
24.
7
115
25.
6
115
26.
132
575
27.
24
115
28.
6
115
29. 0
30.
1
22,957,480
31. 0.007
32. 0.623
33. 0.109
34. 1:1
35. 3:5
36. 11:1
37. 5:3
38. 4:3
39. 1:4
40. 4:7
41. 3:1
42.
6
7
43.
3
10
44.
5
11
45.
4
9
46.
9
17
47.
1
9
48.
7
16
49.
3
5
50.
1
10
51. 2:23
52. 1:999
53. 1:4
54.
322
540
1771
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55.
1
20
56.
9
20
57.
9
20
58.
1
20
59.
9
20
60.
9
20
61.
1
120
62.
Ϫ1
64. C
63. Probability and odds are good tools for assessing risk.
Answers should include the following. • P(struck by lightning) ϭ
1
s
, so ϭ sϩf
750,000
odds ϭ 1:(750,000 Ϫ 1) or
1:749,999. P(surviving a lightning strike) ϭ s sϩf
3
4
ϭ , so odds ϭ
3:(4 Ϫ 3) or 3 :1.
• In this case, success is being struck by lightning or surviving the lightning strike. Failure is not being struck by lightning or not surviving the lightning strike. 1
36
65. D
66. theoretical;
67. experimental; about 0.307
68. experimental;
69. theoretical;
1
17
1
5
70. permutation; 120
71. permutation; 1260
72. combination; 35
73. 16
74. 24
75. direct variation
76. square root
77. (4, 4)
78. (1, 3)
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79.
6
35
80.
3
14
81.
1
4
82.
2
21
83.
9
20
Chapter 12
Practice Quiz 1
Page 650
1. 24
2. 756
3. 18,720
4. 1320
5. 56
6. permutation; 40,320
7. combination; 20,358,520
8.
1
221
10.
8
663
9.
13
102
Lesson 12-4 Multiplying Probabilities
Pages 654–657
1. Sample answer: putting on your socks, and then your shoes 2. P(A, B, C, and D) ϭ
P(A) ؒ P(B) ؒ P(C) ؒ P(D)
3. Mario; the probabilities of rolling a 4 and rolling a 2 are
4.
1
36
1
6
both .
5.
1
4
6.
1
17
7.
4
663
8.
7
30
9.
1
4
10.
1
16
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21
220
12. independent;
13.
1
12
14.
1
36
15.
25
36
16.
1
36
17.
1
6
18.
1
6
19.
5
6
20.
1
42
21.
1
49
22.
25
49
23.
10
21
1
36
24. 0
26.
25. 0
1
15
1
10
27.
2
15
28.
29.
2
15
30. dependent;
31. independent;
25
81
3
28
32. independent;
168
4913
1
32
33. dependent;
1
21
34. independent;
35. dependent;
81
2401
36.
1
9
First Spin
Second Spin
R
R
P(R,B) ϭ
B
1
3
ϫ
1
3
or
1
9
B
Y
R
B
Y
R
B
Y
Y
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38.
First Spin
Blue Yellow Red
1
1
1
3
3
3
Blue
1
3
Second
Spin
BB
1
9
BY
1
9
BR
1
9
Yellow YB
1
1
3
9
YY
1
9
YR
1
9
RY
1
9
1
3
RR
1
9
Red
1
3
RB
1
9
39.
1
9
40.
1
635,013,559,600
41.
19
1,160,054
42.
1
158,753,389,900
43.
6327
20,825
44. a
99 4 b 100
1
1320
46.
47. no
48. no
49. Sample answer: As the number of trials increases, the results become more reliable. However, you cannot be absolutely certain that there are no black marbles in the bag without looking at all of the marbles.
50. 21
51. Probability can be used to analyze the chances of a player making 0, 1, or 2 free throws when he or she goes to the foul line to shoot 2 free throws. Answers should include the following.
52. D
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• One of the decimals in the table could be used as the value of p, the probability that a player makes a given free throw. The probability that a player misses both free throws is
(1 Ϫ p)(1 Ϫ p) or (1 Ϫ p) 2.
The probability that a player makes both free throws is p ؒ p or p 2.
Since the sum of the probabilities of all the possible outcomes is 1, the probability that a player makes exactly 1 of the 2 free throws is
1 Ϫ (1 Ϫ p) 2 Ϫ p 2 or
2p (1 Ϫ p).
• The result of the first free throw could affect the player’s confidence on the second free throw. For example, if the player makes the first free throw, the probability of making the second free throw might increase. Or, if the player misses the first free throw, the probability of making the second free throw might decrease.
54.
55.
3
340
57. 1440 ways
1
119
58. 6
59. 36
1
204
56.
53. C
60. x 2 Ϫ 4x ϩ 2
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62.
61. x, x Ϫ 4
y
x
O
y ϭ x2 ϩ x Ϫ 2
63.
64.
y
O
y
x
O
x y ϭ x 2 Ϫ 3x
y ϭ x2 Ϫ 4
65. 153
66. Ϫ9
69. (1, 2)
70. (13, 9)
67. 0 b 0
68. 5a 4 0 b 3 0
72.
71. (Ϫ2, 4)
2
3
5
4
73.
5
6
74.
75.
11
12
76. 1
1
6
5
12
77. 1
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Lesson 12-5 Adding Probabilities
Pages 660–663
1. Sample answer: mutually exclusive events: tossing a coin and rolling a die; inclusive events: drawing a 7 and a diamond from a standard deck of cards
2.
3. The events are not mutually exclusive, so the chance of rain is less than 100%.
4.
1
3
French and Algebra
French
150
5.
1
3
6.
1
2
8.
5
6
9.
2
3
Algebra
300
1
3
7.
400
11. inclusive;
10. mutually exclusive;
12.
13. 1
13
16
14.
4
13
2
13
1
6
15.
25
42
16.
37
42
17.
35
143
18.
105
143
19.
3
143
20.
84
143
21.
38
143
22.
32
39
23. mutually exclusive;
25. inclusive;
7
9
24. inclusive;
21
34
1
2
26. mutually exclusive;
329
4
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27.
4
13
28.
2
3
29.
55
221
30.
11
221
31.
188
663
32.
63
221
33.
1
8
34.
1
8
35.
1
4
36.
3
4
37.
1
780
38.
145
156
39.
9
130
40.
1
26
41.
11
780
42.
1
78
43.
3
5
44.
53
108
45.
17
27
46.
17
162
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47. Subtracting P(A and B) from each side and adding P (A or
B) to each side results in the equation P (A or B) ϭ P(A) ϩ
P(B) Ϫ P(A and B). This is the equation for the probability of inclusive events. If A and B are mutually exclusive, then P (A and B) ϭ 0, so the equation simplifies to P(A or B) ϭ
P(A) ϩ P(B), which is the equation for the probability of mutually exclusive events.
Therefore, the equation is correct in either case.
48. Probability can be used to estimate the percents of people who do the same things before going to bed.
Answers should include the following. • The events are inclusive because some people brush their teeth and set their alarm. Also, you know that the events are inclusive because the sum of the percents is not
100%.
• According to the information in the text and the table, P (read book) ϭ
38
and P (brush teeth) ϭ
100
81
. Since the events
100
are inclusive, P (read book and brush teeth) ϭ P (read book) ϩ P (brush teeth) Ϫ
P (read book and brush
38
81 ϩ Ϫ teeth) ϭ
100
1200
59
ϭ
.
2000
100
49. C
100
50. A
51.
1
216
52.
125
216
53.
1
216
54.
1
8
55. 4:1
56. 1:8
57. 2:5
58. 5:3
59. 254
60. 24
61. (Ϯ8, Ϫ10)
62. (Ϯ12, Ϯ5)
63. (x ϩ 1)2(x Ϫ 1)(x 2 ϩ 1)
64. min: (0, Ϫ5); max: (Ϫ1.33, Ϫ3.81)
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65. min: (Ϫ0.42, 0.62); max: (Ϫ1.58, 1.38)
66.
y
x
O
(0, 2), (2, 0), (0, Ϫ2); max: f(2, 0) ϭ 6; min: f (0, Ϫ2) ϭ Ϫ2
67.
68. d ϭ 12.79t
y
x
O
(1, 3), (1, Ϫ1), (3, 3), (3, 5); max: f(3, 5) ϭ 23; min: f (1, Ϫ1) ϭ Ϫ3
69. direct variation
70. 323.4, 298, no mode, 143
71. 35.4, 34, no mode, 72
72. 3.6, 3.45, 2.1, 3.6
73. 63.75, 65, 50 and 65, 30
74. 79.83, 89, 89, 57
75. 12.98, 12.9, no mode, 4.7
Lesson 12-6 Statistical Measures
Pages 666–670
1
a (x
B n iϭ1 i
2. Sample answer: The variance of the set {0, 1} is 0.25 and the standard deviation is 0.5.
{10, 10, 10, 10, 10, 10}
3. ϭ
n
Ϫ x )2
4. 40, 6.3
5. 8.2, 2.9
6. 424.3, 20.6
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7. \$7300.50, \$5335.25
8. The mean is more representative for the southwest central states because the data for the
Pacific states contains the most extreme value,
\$10,650.
9. 2500, 50
10. 1.6, 1.3
11. 3.1, 1.7
12. 4.8, 2.2
13. 37,691.2, 194.1
14. 569.4, 23.9
15. 82.9, 9.1
16. 43.6, 6.6
17. 114.5, 105, 23
18. The mean and median both seem to represent the center of the data.
19. Mean; it is highest.
20. Mode; it is lower and is what most employees make. It reflects the most representative worker.
21. \$1047.88, \$1049.50, \$695
22. Mode; it is the least expensive price.
23. Mean or median; they are nearly equal and are more representative of the prices than the mode.
24. 2,290,403; 2,150,000;
2,000,000
25. Mode; it is lowest.
26. Mean; it is highest.
27. 19.3
28. 28.9
29. 19.5
30. Washington; see students’ work. 31. 59.8, 7.7
32. 64%
33. 100%
34. Different scales are used on the vertical axes.
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35. Sample answer: The first graph might be used by a sales manager to show a salesperson that he or she does not deserve a big raise.
It appears that sales are steady but not increasing fast enough to warrant a big raise. 36. Sample answer: The second graph might be shown by the company owner to a prospective buyer of the company. It looks like there is a dramatic rise in sales.
37. A: 2.5, 2.5, 0.7, 0.8; B: 2.5,
2.5, 1.1, 1.0
38. The first histogram is lower in the middle and higher on the ends, so it represents data that are more spread out. Since set B has the greater standard deviation, set B corresponds to the first histogram and set A corresponds to the second.
39. The statistic(s) that best represent a set of test scores depends on the distribution of the particular set of scores.
Answers should include the following. • mean, 73.9; median, 76.5; mode, 94
• The mode is not representative at all because it is the highest score. The median is more representative than the mean because it is influenced less than the mean by the two very low scores of 34 and 19.
40. A
41. D
42. 3
43. 1.9
44. The mean deviations would be greater for the greater standard deviation and lower for the groups of data that have the smaller standard deviation. ©Glencoe/McGraw-Hill
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4
13
46. mutually exclusive;
47.
1
169
48.
4
663
49.
13
204
50.
3
7
1
16
51. 10, Ϯ92; 10, Ϯ21062; Ϯ
52. Ϫ3
53. 17
54. Ϫ2
9
5
56. 1Ϫ4, 62
55. 12 cm3
58. (3, 5)
57. (1, 5)
59. 136
60. 340
61. 380
62. 475
63. 396
64. 495
Chapter 12
Practice Quiz 2
Page 670
1.
3
20
2.
1
6
3.
2
9
4.
1
4
5.
1
6
6.
2
3
7.
3
4
8. 6.6, 2.6
9. 23.6, 4.9
10. 134.0, 11.6
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Lesson 12-7 The Normal Distribution
Pages 673–675
2. The mean of the three graphs is the same, but the standard deviations are different. The first graph has the least standard deviation, the standard deviation of the middle graph is slightly greater, and the standard deviation of the last graph is greatest. the use of cassettes since
CDs were introduced
3. Since 99% of the data is within 3 standard deviations of the mean, 1% of the data is more than 3 standard deviations from the mean. By symmetry, half of this, or
0.5%, is more than 3 standard deviations above the mean.
4. normally distributed
5. 68%
6. 13.5%
7. 95%
8. 6800
9. 250
10. 1600
11. 81.5%
12. positively skewed
13. normally distributed
14. Negatively skewed; the histogram is high at the right and has a tail to the left.
15. 68%
16. 34%
17. 0.5%
18. 16%
19. 50%
20. 50%
21. 95%
22. 50%
23. 815
24. 25
25. 16%
26. 652
27. The mean would increase by
25; the standard deviation would not change; and the
28. If a large enough group of athletes is studied, some of the characteristics may be
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graph would be translated 25 units to the right.
normally distributed; others may have skewed distributions. Answers should include the following.
10
Frequency
8
6
4
2
0
67 68 69 70 71 72 73 74 75 76 77 78 79 80
Height (in.)
• Since the histogram has two peaks, the data may not be normally distributed. This may be due to players who play certain positions tending to be of similar large sizes while players who play the other positions tend to be of similar smaller sizes.
29. A
30. D
31. 17.5, 4.2
32. 42.5, 6.5
33.
2
13
34.
35.
4
13
36. Ϫ5, 0, 1
38. 1, Ϫ1
37. Ϫ3, 2, 4
39.
1
,
4
4
13
40.
1
y
50
Ϫ2
1
O
Ϫ1
2t
Ϫ50
Ϫ100
y ϭ 216t 2 Ϫ 53
41. 0.76 h
42. 21a 5b 2
43. 56c 5d 3
44. 126x 5y 4
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Lesson 12-8 Binomial Experiments
Pages 678–680
1. Sample answer: In a 5-card hand, what is the probability that at least 2 cards are hearts? 2. RRRWW, RRWRW,
RRWWR, RWRRW,
RWRWR, RWWRR,
WRRRW, WRRWR,
WRWRR, WWRRR
3a. Each trial has more than two possible outcomes.
3b. The number of trials is not fixed. 3c. The trials are not independent. 4.
3
8
5.
1
8
6.
7
8
7.
1
28,561
8.
48
28,561
9.
27,648
28,561
12.
1
16
13.
1
16
14.
3
8
15.
1
4
16.
5
16
17.
11
16
18.
3125
7776
19.
125
3888
20.
625
648
21.
23
648
22.
243
1024
23.
1
1024
24.
15
1024
25.
135
512
26.
459
512
338
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27.
53
512
28.
105
512
29.
105
512
30.
319
512
31.
319
512
34.
560
2187
36.
1 3 15 5 15 3 1
, , , , , ,
64 32 64 16 64 32 64
37.
1
4
38. C (n, m)p m(1 Ϫ p)nϪm
40. 2
39. Getting a right answer and a wrong answer are the outcomes of a binomial experiment. The probability is far greater that guessing will result in a low grade than in a high grade. Answers should include the following.
• Use (r ϩ w)5 ϭ r 5 ϩ 5r 4w ϩ 10r 3w 2 ϩ
10r 2w 3 ϩ 5rw 4 ϩ w 5 and the chart on page 676 to determine the probabilities of each combination of right and wrong.
1 5
4
• P(5 right): r 5 ϭ P a b ϭ
1
1024
or about 0.098%; P (4 right,
1 wrong):
15
1024
1.5%; P (3 right, 2 wrong):
1 3 3 2
4
4
10r 2w 3 ϭ 10 a b a b ϭ
45
512
or about 8.8%; P (3 wrong,
2 right): 10r 2w 3 ϭ
1 2 3 3
4
4
10 a b a b ϭ
135
512
26.4%; P (4 wrong, 1 right):
339
Algebra 2
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1
4
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3 4
4
405
1024
243
1024
5r w 4 ϭ 5 a b a b ϭ
or
about 39.6%; P (5 wrong):
3
4
5
w5 ϭ a b ϭ
23.7%.
41. B
42. See students’ work.
43. normal distribution
44. 68%
45. 10
46. 16%
47. Mean; it is highest.
48.
y x ϭ Ϫ3 x O
49.
y
50.
xϩyϭ4
y y ϭ |5x |
x
O
O
51. 0.1
52. 0.05
53. 0.039
54. 0.027
55. 0.041
x
56. 0.031
340
Algebra 2
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Lesson 12-9 Sampling and Error
Pages 683–685
1. Sample answer: If a sample is not random, the results of a survey may not be valid.
2. Sample answer for good sample: doing a random telephone poll to rate the mayor’s performance; sample answer for bad sample: conducting a survey on how much the average person reads at a bookstore
3. The margin of sampling error decreases when the size of the sample n increases. As n p (1 Ϫ p) increases, decreases.
4. Yes; last digits of social security numbers are random. 5. No; these students probably study more than average.
n
9. The probability is 0.95 that the percent of Americans ages 12 and older who listen to the radio every day is between 72% and 82%.
11. No; you would tend to point toward the middle of the page. 12. Yes; all seniors would have the same chance of being selected. 13. Yes; a wide variety of people would be called since almost everyone has a phone.
14. No; freshmen are more likely than older students to be still growing, so a sample of freshmen would not give representative heights for the whole school.
341
Algebra 2
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28. 36 or 64
29. A political candidate can use the statistics from an opinion poll to analyze his or her standing and to help plan the rest of the campaign.
Answers should include the following. • The candidate could decide to skip areas where he or she is way ahead or way behind, and concentrate on areas where the polls indicate the race is close.
• The margin of error indicates that with a probability of 0.95 the percent of the Florida population that favored
Bush was between 43.5% and 50.5%. The margin of error for Gore was also about 3.5%, so with probability 0.95 the percent that favored Gore was between 40.5% and 47.5%.
Therefore, it was possible that the percent of the
Florida population that favored Bush was less than the percent that favored
Gore.
30. A
31. C
32.
1
32
5
32
34.
1
2
33.
35. 95%
36. 210
37. 97.5%
38. x Ϫ 2, x Ϫ 3
342
Algebra 2
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Chapter 13 Trigonometry
Lesson 13-1 Right Triangle Trigonometry
Pages 706–708
2.
1. Trigonometry is the study of the relationships between the angles and sides of a right triangle.
hypotenuse
opposite
8
15
; cos ϭ ;
17
17
8
17 tan ϭ ; csc ϭ ;
15
8
17
15 sec ϭ ; cot ϭ
15
8
3. Given only the measures of the angles of a right triangle, you cannot find the measures of its sides.
5. sin ϭ tan ϭ sec ϭ
285
;
11
285
;
6
11
;
6
7. cos 23Њ ϭ
csc ϭ
32
;
x
11 285
;
85
6
;
11
5 211
;
11
5
6
6 285
85
cos ϭ
cot ϭ
4. sin ϭ
6 211
;
11
6. sin ϭ ; cos ϭ tan ϭ sec ϭ
x Ϸ 34.8
8. tan x Њ ϭ
9. B ϭ 45Њ, a ϭ 6, c ϭ 8.5
15
;
21
211
;
6
211
5
6
5
csc ϭ ; cot ϭ
x Ϸ 36
10. A ϭ 34Њ, a Ϸ 8.9, b Ϸ 13.3
2105
;
11
11. a Ϸ 16.6, A Ϸ 67Њ, B Ϸ 23Њ
12. c Ϸ 19.1, A Ϸ 47Њ, B Ϸ 43Њ
13. 1660 ft
14. B
15. sin ϭ tan ϭ sec ϭ
Glencoe/McGraw-Hill
4 2105
; csc
105
4
;
11
cos ϭ
112105
;
105
ϭ
3
5
3 tan ϭ ; csc
4
5 sec ϭ ; cot
4
2105
4
11
;
4
cot ϭ
4
5
5 ϭ ;
3
4 ϭ 3
16. sin ϭ ; cos ϭ ;
343
Algebra 2
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17. sin ϭ
27
;
4
27
;
3
tan ϭ
4
3
tan ϭ
3 27
7
csc ϭ
sec ϭ ; cot ϭ
19. sin ϭ
4 27
;
7
3
4
18. sin ϭ
cos ϭ ;
25
;
2
csc ϭ
x
,
10
23. sin 54Њ ϭ
17.8
,
x
25. cos x Њ ϭ
15
,
36
27a. sin 30Њ ϭ sin 30Њ ϭ
sec ϭ
20. sin ϭ tan ϭ
cos 30Њ
x Ϸ 22.0
24. tan 17.5Њ ϭ
26. sin x Њ ϭ
16
,
22
28a. sin 45Њ ϭ
sine ratio
Replace opp with x and hyp with
2x.
sin 45Њ ϭ
1
Simplify.
2 adj cosine ratio ϭ hyp
23x Replace adj with 13x and ϭ 2x hyp with 2x.
cos 30Њ ϭ
215
;
7
8 215
;
15
7
8
7 215
15
cos ϭ ; csc ϭ
sin 45Њ ϭ
28b. cos 45Њ
cos 45Њ ϭ
Simplify.
cos 45Њ ϭ
344
x
;
23.7
x Ϸ 7.5
x Ϸ 47
opp hyp x
12x
1
12
12
2
sin 45Њ ϭ
cos 45Њ ϭ
5
9
3 x x Ϸ 65
23
2
215
;
8
cot ϭ
22. cos 60Њ ϭ , x ϭ 6
sin 30Њ ϭ
27b. cos 30Њ
2106
;
5
sec ϭ ; cot ϭ
x Ϸ 5.8
opp hyp x
2x
2106
;
9
9
5
tan ϭ ;
8
7
cot ϭ 2
21. tan 30Њ ϭ
5 2106
;
106
cos ϭ
25
2 25
;
; cos ϭ
5
5
1
; csc ϭ 25;
2
sec ϭ
9 2106
;
106
x
12x
1
12
12
2
Replace opp with x and hyp with 12x.
sine ratio
Simplify.
Rationalize the denominator. Replace adj with x and hyp with 12x. cosine ratio
Simplify.
Rationalize the denominator. Algebra 2
Chapter 13
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27c. sin 60Њ ϭ sin 60Њ ϭ sin 60Њ ϭ
23x
2x
opp hyp 23
2
Replace opp with 13x and hyp with 2x.
28c. tan 45Њ ϭ
sine ratio
tan 45Њ ϭ
opp adj tangent ratio
x x Replace opp with x and adj with x.
tan 45Њ ϭ 1
Simplify.
Simplify.
29. B ϭ 74Њ, a Ϸ 3.9, b Ϸ 13.5
30. A ϭ 63Њ, a Ϸ 13.7, c Ϸ 15.4
31. B ϭ 56Њ, b Ϸ 14.8, c Ϸ 17.9
32. A ϭ 75Њ, a Ϸ 24.1, b Ϸ 6.5
33. A ϭ 60Њ, a Ϸ 19.1, c ϭ 22
34. B ϭ 45Њ, a ϭ 7, b ϭ 7
35. A ϭ 72Њ, b Ϸ 1.3, c Ϸ 4.1
36. B ϭ 80Њ, a Ϸ 2.6, c Ϸ 15.2
37. A Ϸ 63Њ, B Ϸ 27Њ, a Ϸ 11.5
38. A Ϸ 26Њ, B Ϸ 64Њ, b Ϸ 8.1
39. A Ϸ 49Њ, B Ϸ 41Њ, a ϭ 8, c Ϸ 10.6
40. A Ϸ 19Њ, B Ϸ 71Њ, b Ϸ 14.1, c ϭ 15
41. about 300 ft
42. about 142.8 ft
44. about 3.2 in.
45. 93.54 units2
46. about 1.72 km high
47. The sine and cosine ratios of acute angles of right triangles each have the longest measure of the triangle, the hypotenuse, as their denominator. A fraction whose denominator is greater than its numerator is less than 1. The tangent ratio of an acute angle of a right triangle does not involve the measure of the hypotenuse,
48. When construction involves right triangles, including building ramps, designing buildings, or surveying land before building, trigonometry is likely to be used. Answers should include the following.
• If you view the ramp from the side then the vertical rise is opposite the angle that the ramp makes with the horizontal. Similarly, the horizontal run is the adjacent side. So the tangent of the angle is the ratio of the rise to the run or the slope of the ramp.
• Given the ratio of the slope of ramp, you can find the angle of inclination by calculating tan–1 of this ratio. opp
.
If the measure of the opposite side is greater than the measure of the adjacent side, the tangent ratio is greater than 1. If the measure of the opposite side is less than the measure of the adjacent side, the tangent ratio is less than 1.
345
Algebra 2
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49. C
50. 7.7
51. No; band members may be more likely to like the same kinds of music.
3
53.
8
15
55.
16
57. {Ϫ2, Ϫ1, 0, 1, 2}
52. Yes; this sample is random since different kinds of people go to the post office.
1
54.
16
59. 20 qt
60. 35,904 ft
61. 12 m2
62. 48 L
Lesson 13-2
56. {Ϯ222, 2i 22}
58. {121}
Angles and Angle Measure
Pages 712–715
1. reals
2. In a circle of radius r units, one radian is the measure of an angle whose rays intercept an arc length of r units. 3.
4.
y
y
290˚
70˚ x O
x
O
Ϫ70˚
5.
6.
y
y
300˚
570˚
O
x
O
346
x
Algebra 2
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8.
y
O
Ϫ45˚
13
18
10.
7.
97
36
x
18
9. Ϫ
11. 135Њ
12. Ϫ30Њ
13. 1140Њ
14. 420Њ, Ϫ300Њ
15. 785Њ, Ϫ295Њ
16.
17. 21 h
18. 2 h
19.
7
,
3
5
3
Ϫ
20.
y
235˚
y
270˚
x
O
21.
22.
y
O
24.
y
x
O
y
O
Ϫ150˚
x
380˚
790˚
23.
y
x
O
x
O
347
Ϫ50˚
x
Algebra 2
Chapter 13
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25.
26.
y
y
O
x
x
O
2
Ϫ
3
27.
2
3
28.
3
5
4
12
29. Ϫ
30. Ϫ
31.
11
3
32.
19
6
33.
79
90
34.
13
9
35. 150Њ
36. 495Њ
37. Ϫ45Њ
38. Ϫ60Њ
39. 1305Њ
40. 510Њ
41.
1620
Ϸ 515.7Њ
42.
540
Ϸ 171.9Њ
43. Sample answer: 585Њ, Ϫ135Њ
44. Sample answer: 390Њ, Ϫ330Њ
45. Sample answer: 345Њ, Ϫ375Њ
46. Sample answer: 220Њ, Ϫ500Њ
47. Sample answer: 8Њ, Ϫ352Њ
48. Sample answer: 400Њ, Ϫ320Њ
11
,
4
3
,
4
13
,
2
5
4
19
,
6
13
4
4
,
3
3
2
25
,
4
Ϫ
Ϫ
Ϫ
5
6
Ϫ
8
3
Ϫ
7
4
Ϫ
55. 2689Њ per second; 47 radians per second
56. 209.4 in2
57. about 188.5 m2
58. number 17
59. about 640.88 in2
60a. a 2 ϩ (Ϫb)2 ϭ a 2 ϩ b 2 ϭ 1
60b. b 2 ϩ a 2 ϭ a 2 ϩ b 2 ϭ 1
60c. b 2 ϩ (Ϫa)2 ϭ a 2 ϩ b 2 ϭ 1
348
Algebra 2
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61. Student answers should include the following.
• An angle with a measure of more than 180Њ gives an indication of motion in a circular path that ended at a point more than halfway around the circle from where it started.
• Negative angles convey the same meaning as positive angles, but in an opposite direction. The standard convention is that negative angles represent rotations in a clockwise direction.
• Rates over 360Њ per minute indicate that an object is rotating or revolving more than one revolution per minute.
62. C
63. D
64. a Ϸ 3.4, c Ϸ 6.0, B ϭ 56Њ
65. A ϭ 22Њ, a Ϸ 5.9, c Ϸ 15.9
66. A ϭ 35Њ, a Ϸ 9.2, b Ϸ 13.1
67. c Ϸ 0.8, A ϭ 30Њ, B ϭ 60Њ
70. permutation, 17,100,720
71. combination, 35
72. [g ؠh](x) ϭ 6x Ϫ 8,
[h ؠg](x) ϭ 6x Ϫ 4
73. [g ؠh](x) ϭ 4x 2 Ϫ 6x ϩ 23,
[h ؠg](x) ϭ 8x 2 ϩ 34x ϩ 44
74. 1041.8
75. 1418.2 or about 1418; the number of sports radio stations in 2008
76.
77.
78.
79.
81.
3 25
5
210
2
210
4
80.
349
2 23
3
2 26
3
214
2
Algebra 2
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Chapter 13
Practice Quiz 1
Page 715
10 2149
;
149
7 2149
;
149
1. A ϭ 42Њ, a Ϸ 13.3, c Ϸ 17.9
2. A Ϸ 59Њ, B Ϸ 31Њ, b Ϸ 10.8
3.
4. sin ϭ
y
cos ϭ
O
Ϫ60˚
x
tan ϭ sec ϭ
5.
19
18
6.
2149
;
7
10
;
7
csc ϭ
2149
;
10
cot ϭ
7
10
5
2
8. Ϫ396Њ
7. 210Њ
10.
9. 305Њ; Ϫ415Њ
5
;
3
3
Ϫ
Lesson 13-3 Trigonometric Functions of General Angles
Pages 722–724
1. False; sec 0Њ ϭ tan 0Њ ϭ
0 r r r or 1 and
2. Sample answer: 190Њ
or 0.
8
15
, cos ϭ Ϫ ,
17
17
8
17 tan ϭ Ϫ , csc ϭ ,
15
8
17
15 sec ϭ Ϫ , cot ϭ Ϫ
15
8
3. To find the value of a trigonometric function of , where is greater than 90Њ, find the value of the trigonometric function for Ј, then use the quadrant in which the terminal side of lies to determine the sign of the trigonometric function value of .
4. sin ϭ
350
Algebra 2
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22
,
2
22
,
2
tan ϭ 1, csc ϭ 22, sec ϭ 22, cot ϭ 1
5. sin ϭ 0, cos ϭ Ϫ1, tan ϭ 0, csc ϭ undefined, sec ϭ Ϫ1, cot ϭ undefined
6. sin ϭ
7. 55Њ
8.
y
cos ϭ
4
y
7
4
235˚
O
x
'
9. 60Њ
10. Ϫ
y
O
23
2
'
x
'
O
2 23
3
x
Ϫ240˚
11. Ϫ1
12. Ϫ23
13. Ϫ
14. sin ϭ
26
,
3
sec ϭ 23
15. sin ϭ Ϫ
cos ϭ
23
,
3
tan ϭ Ϫ22, cos ϭ Ϫ
24
,
25
24
tan ϭ ,
7
25 sec ϭ ,
7
17. sin ϭ
csc ϭ
23
,
2
2 23
,
3
cot ϭ Ϫ
26
,
2
23
3
tan ϭ Ϫ23, sec ϭ Ϫ2,
16. about 12.4 ft
7
,
25
25
csc ϭ ,
24
7 cot ϭ
24
25
, cos
5
1 ϭ , csc
2
18. sin ϭ
cos ϭ
tan
sin ϭ
351
25
,
5
2 25
,
5
ϭ 25,
ϭ
cos ϭ 2,
Algebra 2
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19. cos ϭ cos ϭ
8 289
,
89
5 289
,
89
289
,
5
8
5
tan ϭ Ϫ , csc ϭ Ϫ sec ϭ
3
4
5
5
3
5
tan ϭ Ϫ , csc ϭ Ϫ ,
4
3
5
4 sec ϭ , cot ϭ Ϫ
4
3
289
,
8
20. sin ϭ Ϫ , cos ϭ ,
5
8
cot ϭ Ϫ
21. sin ϭ Ϫ1, cos ϭ 0, tan ϭ undefined, csc ϭ Ϫ1, sec ϭ undefined, cot ϭ 0
22. sin ϭ 0, cos ϭ Ϫ1, tan ϭ 0, csc ϭ undefined, sec ϭ Ϫ1, cot ϭ undefined
23. sin ϭ Ϫ
24. sin ϭ Ϫ
22
,
2
22
,
2
26
,
3
tan ϭ 22, csc ϭ Ϫ
sec ϭ 22, cot ϭ Ϫ1 cos ϭ
tan ϭ Ϫ1, csc ϭ Ϫ22,
sec ϭ Ϫ23, cot ϭ
25. 45Њ
26. 60Њ
y
23
,
3
26
,
2
22
2
cos ϭ Ϫ
y
240˚
315˚
O
27. 30Њ
'
x
x
O
'
28. 55Њ
y
y
'
O
x
'
Ϫ210˚
352
x
O
Ϫ125˚
Algebra 2
Chapter 13
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29.
4
y
30.
6
5
4
7
33. Ϫ
'
32.
13
7
O
'
x
O
y
23
2
5
6
x
O
'
31.
y
3
y
x
O
'
x
2
Ϫ
3
34. Ϫ2
39. 23
22
2
35. Ϫ23
36. Ϫ23
37. undefined
38.
23
2
40.
1
2
41. undefined
42. 2
43.
44. Ϫ1
46. 6092.5 ft
45. 0.2, 0, Ϫ0.2, 0, 0.2, 0, and
Ϫ0.2; or about 11.5Њ, 0Њ,
Ϫ11.5Њ, 0Њ, 11.5Њ, 0Њ, and
Ϫ11.5Њ
4
5
5 csc ϭ Ϫ ,
4
3 cot ϭ Ϫ
4
sec ϭ
226
,
26
5 226
,
26
csc ϭ 226, sec ϭ Ϫ
4
3
5
,
3
47. sin ϭ Ϫ , tan ϭ Ϫ ,
48. sin ϭ
cos ϭ Ϫ
226
,
5
cot ϭ Ϫ5
353
Algebra 2
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2 22
,
3
49. cos ϭ Ϫ
22
,
4
2 25
,
5
3 22
,
4
tan ϭ Ϫ
25
,
5
50. sin ϭ Ϫ
csc ϭ 3, sec ϭ Ϫ
cos ϭ Ϫ
cot ϭ Ϫ222
csc ϭ Ϫ
3 210
,
10
51. sin ϭ Ϫ
cos ϭ Ϫ csc ϭ Ϫ
210
,
10
210
,
3
25
,
2
26
,
12
tan ϭ 2,
1
5
52. sin ϭ Ϫ , cos ϭ tan ϭ 3, cot ϭ
tan ϭ Ϫ
1
3
2 26
,
5
sec ϭ Ϫ25
sec ϭ
5 26
,
12
cot ϭ Ϫ226
53. about 173.2 ft
54. 45Њ; 2 ϫ 45Њ or 90Њ yields the greatest value for sin 2.
55. 9 meters
56. I, II
57. II
58. III
59. Answers should include the following. • The cosine of any angle is x defined as , where x is r the x-coordinate of any point on the terminal ray of the angle and r is the distance from the origin to that point. This means that for angles with terminal sides to the left of the y-axis, the cosine is negative, and those with terminal sides to the right of the y-axis, the cosine is positive. Therefore, the cosine function can be used to model real-world data that oscillate between being positive and negative. 60. C
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• If we knew the length of the cable we could find the vertical distance from the top of the tower to the rider.
Then if we knew the height of the tower we could subtract from it the vertical distance calculated previously. This will leave the height of the rider from the ground.
5
2
5 23 b 2
61. a , Ϫ
62.
2
63. 300Њ
64.
900Њ
65. sin 28Њ ϭ
67. sin x Њ ϭ
x
,
12
5
,
13
Ϸ 286.5Њ
66. cos 43Њ ϭ
5.6
x
,
83
60.7
68. 635
23
69. (7, 2)
70. (Ϫ4, 3)
71. (5, Ϫ4)
72. 4.7
73. 15.1
74. 2.7
75. 32.9Њ
76. 20.6Њ
77. 39.6Њ
Lesson 13-4 Law of Sines
Pages 729–732
1. Sometimes; only when A is acute, a ϭ b sin A, or a Ͼ b and when A is obtuse, a Ͼ b.
2. Sample answer: A ϭ 42Њ, a ϭ 2.6 cm, b ϭ 3.2 cm
C
3.2 cm
A
2.6 cm
3.9 cm
B
C
3.2 cm
A
0.9 cm
355
2.6 cm
B
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3. Gabe; the information given is of two sides and an angle, but the angle is not between the two sides, therefore the area formula involving sine cannot be used.
4. 57.5 in2
5. 6.4 cm2
6. C ϭ 30Њ, a Ϸ 2.9, c Ϸ 1.5
7. B ϭ 80Њ, a Ϸ 32.0, b Ϸ 32.6
8. B Ϸ 20Њ, A Ϸ 20Њ, a Ϸ 20.2
10. two; B Ϸ 42Њ, C Ϸ 108Њ, c Ϸ 5.7; B Ϸ 138Њ, C Ϸ 12Њ, c Ϸ 1.2
9. no solution
11. one; B ϭ 24Њ, C Ϸ 101Њ, c Ϸ 12.0
12. one; B Ϸ 19Њ, C Ϸ 16Њ, c Ϸ 8.7
13. 5.5 m
14. 43.1 m2
15. 19.5 yd2
16. 572.8 ft2
17. 62.4 cm2
18. 4.2 m2
19. 14.6 mi2
20. B ϭ 101Њ, c Ϸ 3.0, b Ϸ 3.4
21. C ϭ 73Њ, a Ϸ 55.6, b Ϸ 48.2
22. B Ϸ 21Њ, C Ϸ 37Њ, b Ϸ 13.1
23. B Ϸ 47Њ, C Ϸ 68Њ, c Ϸ 5.1
24. C ϭ 97Њ, a Ϸ 5.5, b Ϸ 14.4
25. A Ϸ 40Њ, B Ϸ 65Њ, b Ϸ 2.8
26. C Ϸ 67Њ, B Ϸ 63Њ, b Ϸ 2.9
27. A ϭ 20Њ, a Ϸ 22.1, c Ϸ 39.8
28. no
29. one; B Ϸ 36Њ, C Ϸ 45Њ, c Ϸ 1.8
30. two; B Ϸ 72Њ, C Ϸ 75Њ, c Ϸ 3.5; B Ϸ 108Њ, C Ϸ 39Њ, c Ϸ 2.3
31. no
32. one; B ϭ 90Њ, C ϭ 60Њ, c Ϸ 24.2
33. one; B Ϸ 18Њ, C Ϸ 101Њ, c Ϸ 25.8
34. two; B Ϸ 56Њ, C Ϸ 72Њ, c Ϸ 229.3; B Ϸ 124Њ,
C Ϸ 4Њ, c Ϸ 16.8
35. two; B Ϸ 85Њ, C Ϸ 15Њ, c Ϸ 2.4; B Ϸ 95Њ, C Ϸ 5Њ, c Ϸ 0.8
36. one; B Ϸ 23Њ, C Ϸ 129Њ, c Ϸ 14.1
37. two; B Ϸ 65Њ, C Ϸ 68Њ, c Ϸ 84.9; B Ϸ 115Њ, C Ϸ 18Њ, c Ϸ 28.3
38. 4.6 and 8.5 mi
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39. 7.5 mi from Ranger B,
10.9 mi from Ranger A
40. 690 ft
41. 107 mph
42a. 14.63 Ͻ b Ͻ 20
42b. b ϭ 14.63 or b Ն 20
42c. b Ͻ 14.63
43. Answers should include the following. • If the height of the triangle is not given, but the measure of two sides and their included angle are given, then the formula for the area of a triangle using the sine function should be used.
• You might use this formula to find the area of a triangular piece of land, since it might be easier to measure two sides and use surveying equipment to measure the included angle than to measure the perpendicular distance from one vertex to its opposite side.
1
• The area of ᭝ABC is ah.
44. D
2
C b a h A
sin B ϭ
Area ϭ
Area ϭ
B
c h or h ϭ c sin c 1 ah or
2
1 a (c sin B)
2
B
357
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47.
46.
49. 660Њ, Ϫ60Њ
51.
17
,
6
53.
23
2
48. 22
23
3
45. B ϭ 78Њ, a Ϸ 50.1, c Ϸ 56.1
50. 407Њ, Ϫ313Њ
7
6
55
221
52.
Ϫ
3
68
54. 780 ft
55. 5.6
56. 7.8
57. 39.4Њ
58. 136.0Њ
Lesson 13-5 Law of Cosines
Pages 735–738
1. Mateo; the angle given is not between the two sides, therefore the Law of Sines should be used.
2a. Use the Law of Cosines to find the measure of one angle. Then use the Law of
Sines or the Law of Cosines to find the measure of a second angle. Finally, subtract the sum of these two angles from 180Њ to find the measure of the third angle. 2b. Use the Law of Cosines to find the measure of the third side. Then use the Law of
Sines or the Law of Cosines to find the measure of a second angle. Finally, subtract the sum of these two angles from 180Њ to find the measure of the third angle. 358
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4. cosines; A Ϸ 76Њ, B Ϸ 69Њ, c Ϸ 6.5
15
9
13
5. sines; B ϭ 70Њ, a Ϸ 9.6, b ϭ 14
6. sines; C Ϸ 101Њ, B Ϸ 37Њ, c Ϸ 92.5
7. cosines; A ϭ 23Њ, B Ϸ 67Њ,
C Ϸ 90Њ
8. 19.5 m
10. sines; A ϭ 60Њ, b Ϸ 14.3, c Ϸ 11.2
9. 94.3Њ
11. cosines; A Ϸ 48Њ, B Ϸ 62Њ,
C Ϸ 70Њ
12. cosines; A Ϸ 46Њ, B Ϸ 74Њ,
C Ϸ 59.6
13. sines; B Ϸ 102Њ, C Ϸ 44Њ, b Ϸ 21.0
14. cosines; A Ϸ 56.8Њ, B Ϸ 82Њ, c Ϸ 11.5
15. sines; A ϭ 80Њ, a Ϸ 10.9, c Ϸ 5.4
16. cosines; A Ϸ 55Њ, C Ϸ 78Њ, b Ϸ 17.9
17. cosines; A Ϸ 30Њ, B Ϸ 110Њ,
C Ϸ 40Њ
18. no
19. sines; C Ϸ 77Њ, b Ϸ 31.7, c Ϸ 31.6
20. cosines; A Ϸ 103Њ, B Ϸ 49Њ,
C Ϸ 28Њ
21. no
22. cosines; A Ϸ 15Њ, B Ϸ 131Њ,
C Ϸ 34Њ
23. cosines; A Ϸ 52Њ, C Ϸ 109Њ, b Ϸ 21.0
24. sines; C ϭ 102Њ, b Ϸ 5.5, c Ϸ 14.4
25. cosines; A Ϸ 24Њ, B Ϸ 125Њ,
C Ϸ 31Њ
26. cosines; A Ϸ 107Њ, B Ϸ 35Њ, c Ϸ 13.8
27. cosines; B Ϸ 82Њ, C Ϸ 58Њ, a Ϸ 4.5
30. Since the step angle for the carnivore is closer to 180Њ, it appears as though the carnivore made more forward progress with each step than the herbivore did.
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31. 4.4 cm, 9.0 cm
32. about 1362 ft; about 81,919 ft 2
33. 91.6Њ
34. Since cos 90Њ ϭ 0, a 2 ϭ b 2 ϩ c 2 Ϫ 2bc cos A becomes a 2 ϭ b 2 ϩ c 2.
35. Answers should include the following. • The Law of Cosines can be used when you know all three sides of a triangle or when you know two sides and the included angle. It can even be used with two sides and the nonincluded angle. This set of conditions leaves a quadratic equation to be solved. It may have one, two, or no solution just like the SSA case with the
Law of Sines.
• Given the latitude of a point on the surface of
Earth, you can use the radius of the Earth and the orbiting height of a satellite in geosynchronous orbit to create a triangle. This triangle will have two known sides and the measure of the included angle. Find the third side using the Law of Cosines and then use the Law of
Sines to determine the angles of the triangle.
Subtract 90 degrees from the angle with its vertex on
Earth’s surface to find the angle at which to aim the receiver dish.
36. B
37. A
38. 100.0Њ
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Algebra 2
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39. Sample answer: 100.2Њ
40. By finding the measure of angle C in one step using the Law of Cosines, only the given information was used.
By finding this angle measure using the Law of
Cosines and then the Law of
Sines, a calculated value that was not exact was introduced; 100.0Њ.
41. one; B ϭ 46Њ, C ϭ 79Њ, c ϭ 9.6
42. no solution
12
5
, cos ϭ ,
13
13
12
13 tan ϭ , csc ϭ
5
12
13
5 sec ϭ , cot ϭ
5
12
44. sin ϭ
43. sin ϭ
45. sin ϭ
26
,
4
tan ϭ sec ϭ
215
,
5
2 210
,
5
cos ϭ
210
,
4
csc ϭ
csc ϭ cot ϭ
2 26
,
3
cot ϭ
47. {x 0 x Ͼ Ϫ0.6931}
cos ϭ
215
3
7 265
,
65
4 265
,
65
265
,
7
4
7
265
,
4
7
4
tan ϭ ,
sec ϭ
46. 1.3863
48. 4.3891
49. 405, Ϫ315Њ
50. 390Њ, Ϫ330Њ
51. 540Њ, Ϫ180Њ
52.
5
,
2
54.
10
,
3
53.
19
,
6
5
6
Ϫ
361
3
2
Ϫ
2
3
Ϫ
Algebra 2
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1. sin ϭ
3 213
,
13
2 213
,
13
cos ϭ Ϫ
213
,
2
3
2
tan ϭ Ϫ , csc ϭ sec ϭ Ϫ
2 23
3
Chapter 13
Practice Quiz 2
Page 738
213
,
3
2. Ϫ
2
3
cot ϭ Ϫ
3. 27.7 m2
4. two; B Ϸ 27Њ; C Ϸ 131Њ; c Ϸ 30.2; B Ϸ 153Њ;
C Ϸ 5Њ; c Ϸ 3.5
5. cosines; c Ϸ 15.9, A Ϸ 59Њ,
B Ϸ 43Њ
Lesson 13-6 Circular Functions
Pages 742–745
1. The terminal side of the angle in standard position must intersect the unit circle at P (x, y).
2. Sample answer: the motion of the minute hand on a clock; 60 s
3. Sample answer: The graphs have the same shape, but cross the x-axis at different points. 4. sin ϭ Ϫ , cos ϭ
5. sin ϭ
6.
22
;
2
cos ϭ
22
2
1
2
7. Ϫ
23
2
12
13
5
13
8. 720Њ
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10.
9. 2 s
h
3
O
2
1
3
Ϫ3
4
5
5
13
3
5
15. sin ϭ
23
;
2
15
;
17
cos ϭ
8
17
1
2
12
13
23
2
16. sin ϭ 0.8; cos ϭ 0.6
cos ϭ Ϫ
23
2
1
2
18.
19. Ϫ1
20.
21. 1
22. Ϫ
25.
1
2
14. sin ϭ Ϫ ; cos ϭ
17. Ϫ
23.
13
2
t
12. sin ϭ Ϫ ; cos ϭ Ϫ
11. sin ϭ ; cos ϭ Ϫ
13. sin ϭ
4
22
2
26. 23
1 Ϫ 23
2
1
4
9
4
24.
27. Ϫ323
28. 1
29. 6
30. 9
31. 2
32. 8
33.
1
440
34.
s
y
1
O
Ϫ1
363
1
440
1
220
Algebra 2
x
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1 13
1 13 b, aϪ , b, 2 2
2 2
35. a ,
13 b 2
1
2
(Ϫ1, 0), aϪ , Ϫ
1
a ,
2
Ϫ
13 b, 2
36. The population is around 425 near the 60th day of the year.
It rises to around 625 in
May/June. It falls to around 425 again by August/September.
It continues to drop to around
225 in November/December.
y x 41. 23
38. tan
43. sine: D ϭ {all reals},
R ϭ {Ϫ1 Յ y Յ 1}; cosine: D ϭ {all reals},
R ϭ {Ϫ1 Յ y Յ 1}
44. Answers should include the following. • Over the course of one period both the sine and cosine function attain their maximum value once and their minimum value once.
From the maximum to the minimum the functions decrease slowly at first, then decrease more quickly and return to a slow rate of change as they come into the minimum. Similarly, the functions rise slowly from their minimum. They begin to increase more rapidly as they pass the halfway point, and then begin to rise more slowly as they increase into the maximum. Annual temperature fluctuations behave in exactly the same manner.
37.
23
3
40. Ϫcot
39. Ϫ
42. Ϫ
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• The maximum value of the sine function is 1 so the maximum temperature would be 50 ϩ 25(1) or
75Њ F. Similarly, the minimum value would be
50 ϩ 25(Ϫ1) or 25Њ F. The average temperature over this time period occurs when the sine function takes on a value of 0. In this case that would be
50Њ F.
23
3
45. A
46.
47. cosines; c Ϸ 12.4, B Ϸ 59Њ,
A Ϸ 76Њ
48. cosines; A Ϸ 34Њ, B Ϸ 62Њ,
C Ϸ 84Њ
49. 27.0 in2
50. 12.5 m2
51. 6800
52. 9500
53. 5000
54. 5000
55. 250
56. 50
57. does not exist
58.
59. 8
60. 4x Ϫ 5
61. 2x ϩ 9
62. 5y 2 Ϫ 4y ϩ 4 Ϫ
63. 2y ϩ 7 ϩ
5 yϪ3 64
3
11 yϩ1 64. 20Њ
65. 110Њ
66. 73Њ
67. 80Њ
68. 56Њ
69. 89Њ
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Lesson 13-7
Inverse Trigonometric Functions
Pages 749–751
1. Restricted domains are denoted with a capital letter.
22
2
22
;
2
Cos 45Њ ϭ
CosϪ1
ϭ 45Њ
3. They are inverses of each other. 4. ϭ Arctan x
5. ␣ ϭ Arccos 0.5
6. 45Њ
7. 0Њ
8. Ϫ Ϸ Ϫ0.52
6
9. Ϸ 3.14
10. 0.22
11. 0.75
12. 0.66
13. 0.58
14. 30Њ
15.  ϭ Arcsin ␣
16. a ϭ Arctan b
17. y ϭ Arccos x
18. 30Њ ϭ Arcsin
19. Arccos y ϭ 45Њ
Ϫ
20. Arctan a b ϭ x
21. 60Њ
22. 30Њ
23. 45Њ
24. 30Њ
25. 45Њ
26. 90Њ
27. 2.09
28. does not exist
29. 0.52
30. 0.52
31. 0.5
32. 0.66
33. 0.60
34. 0.5
35. 0.8
36. 0.81
37. 0.5
38. 3
39. Ϫ0.5
40. 1.57
41. 0.71
42. does not exist
43. 0.96
44. 0.87
1
2
4
3
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45. 60Њ south of west
46. 83Њ
47. No; with this point on the terminal side of the throwing angle , the measure of is found by solving the equation
48. 60Њ
tan ϭ
17
. Thus
18
ϭ tanϪ1
or about 43.4Њ, which is greater than the 40Њ requirement. 17
18
49. 31Њ
50. 102Њ
51. Suppose P (x 1, y 1) and
Q (x 2, y 2) lie on the line y ϭ mx ϩ b. Then m ϭ y2 Ϫ y1
. The tangent of
52. Trigonometry is used to determine proper banking angles. Answers should include the following.
• Knowing the velocity of the cars to be traveling on a road and the radius of the curve to be built, then the banking angle can be determined. First find the ratio of the square of the velocity to the product of the acceleration due to gravity and the radius of the curve. Then determine the angle that had this ratio as its tangent. This will be the banking angle for the turn.
• If the speed limit were increased and the banking angle remained the same, then in order to maintain a safe road the curvature would have to be decreased. That is, the radius of the curve would also have to increase, which would make the road less curved.
x2 Ϫ x1
the angle the line makes with the positive x-axis is opp equal to the ratio or adj
y2 Ϫ y1
.
x2 Ϫ x 1
Thus tan ϭ m.
y
Q (x 2, y 2)
P (x , y 1)
O
x2 Ϫ x1
y2 Ϫ y1 x y ϭ mx ϩ b m 53. 37Њ
54. D
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23
1 22 23
1 22
Ϫ
Ϫ1
1 Ϫ Ϫ
2 2
2
2
2
2
y
2 2 2
2 2 2
2
2
2
56. SinϪ1 x ϩ CosϪ1 x ϭ for
2
all values of x.
55.
x 0
23
2
57. From a right triangle perspective, if an acute angle
has a given sine x, then the complementary angle
Ϫ has that same value
2
as its cosine. This can be verified by looking at a right triangle. Therefore, the sum of the angle whose sine is x and the angle whose cosine
is x should be .
58.
59. Ϫ1
60. 1
61. sines; B Ϸ 69Њ, C Ϸ 81Њ, c Ϸ 6.1 or B Ϸ 111Њ, C Ϸ 39Њ,
C Ϸ 3.9
62. cosines; A ϭ 13Њ, B ϭ 77Њ,
C ϭ 90Њ
63. 46, 39
64. Ϫ22, Ϫ57
65. 11, 109
66. 2.5 s
2
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Chapter 14 Trigonetmetric Graphs and Identities
Lesson 14-1 Graphing Trigonometric Functions
Pages 766–768
1. Sample answer: Amplitude is half the difference between the maximum and minimum values of a graph; y ϭ tan has no maximum or minimum value.
2. Sample answer: The graph repeats itself every 180Њ.
3. Jamile; The amplitude is 3 and the period is 3.
4. amplitude: ; period 360Њ or 2
1
2
y
2.5
2
1.5
1
0.5
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ1
Ϫ1.5
Ϫ2
Ϫ2.5
y
2
1.5
1
0.5
y ϭ 2 sin
O
90˚ 180˚ 270˚
6. amplitude: ; period 360Њ or 2
y
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
O
2
3
5. amplitude: 2; period: 360Њ or 2
5
4
3
2
1
1
y ϭ 2 sin
90˚ 180˚ 270˚
Ϫ0.5
Ϫ1
Ϫ1.5
Ϫ2
369
O
2 cos
3
90˚ 180˚ 270˚ 360˚
Algebra 2
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8. amplitude: does not exist; period: 180Њ or
7. amplitude: does not exist; period: 180Њ or
y
y
2
1.5
1
0.5
2
1.5
1
0.5
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ1
1
Ϫ1.5
y ϭ 4 tan
Ϫ2
O
90˚ 180˚ 270˚
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ1
Ϫ1.5 y ϭ csc 2
Ϫ2
10. amplitude: 4; period: 480Њ or
9. amplitude: 4; period: 180Њ or y 5
4
3
2
1
5
4
3
2
1
90˚ 180˚ 270˚ 360˚
y
1.25
1
0.75
0.5
0.25
30˚
60˚
O
Ϫ90˚
Ϫ0.5
Ϫ0.75
Ϫ1
Ϫ1.25
90˚ 120˚ 150˚
1
y ϭ 2 sec 3
13. 12 months; Sample answer:
The pattern in the population will repeat itself every 12 months. ©Glencoe/McGraw-Hill
3
4
2
3
2
1.5
1
0.5
Ϫ60˚ Ϫ30˚
Ϫ1
Ϫ1.5
Ϫ2
90˚ 180˚ 270˚ 360˚ 450˚
12. amplitude: ; period: 720Њ or 4
y
O
3
4
y ϭ 4 cos
O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
11. amplitude: does not exist; period: 120Њ or
8
2
y
y ϭ 4 sin 2
O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
Ϫ5
90˚ 180˚ 270˚
3
1
y ϭ 4 cos 2
90˚ 180˚ 270˚ 360˚ 450˚
14. 4250; June 1
370
Algebra 2
Chapter 14
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15. amplitude: 3; period: 360Њ or 2
16. amplitude: 5; period: 360Њ or 2
y
5
4
3
2
1
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
y
5
4
3
2
1
y ϭ 3 sin
O
90˚ 180˚ 270˚
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
y
y
10
8
6
4
2
5
4
3
2
1
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ2
y ϭ 2 csc
Ϫ3
Ϫ4
Ϫ5
90˚ 180˚ 270˚
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ6
Ϫ8
y ϭ 2 tan
Ϫ10
1
5
y
1
0.8
0.6
0.4
0.2
90˚ 180˚ 270˚
y
10
8
6
4
2
1 y ϭ 5 sin
O
O
20. amplitude: does not exist; period: 360Њ or 2
19. amplitude: ; period: 360Њ or 2
90˚ 180˚ 270˚
18. amplitude: does not exist; period: 180Њ or
17. amplitude: does not exist; period: 360Њ or 2
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ0.4
Ϫ0.6
Ϫ0.8
Ϫ1
y ϭ 5 cos
90˚ 180˚ 270˚
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ4
Ϫ6
1 y ϭ 3 sec
Ϫ8
Ϫ10
371
90˚ 180˚ 270˚
Algebra 2
Chapter 14
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21. amplitude: 1; period 90Њ or
2
22. amplitude: 1; period: 180Њ or y y
5
4
3
2
1
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
5
4
3
2
1
y ϭ sin 4
O
90˚ 180˚ 270˚
period: 36Њ or
O
Ϫ60˚ Ϫ30˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
5
y
30˚
60˚
O
Ϫ72˚ Ϫ54˚ Ϫ36˚ Ϫ18˚
Ϫ2
Ϫ3
Ϫ4
y ϭ cot 5
Ϫ5
25. amplitude: does not exist; period: 540Њ or 3
18˚
36˚
54˚
72˚
26. amplitude: does not exist; period: 360Њ or 2 y y
10
8
6
4
2
10
8
6
4
2
5
4
3
2
1
5
4
3
2
1
Ϫ810˚ Ϫ540˚Ϫ270˚
Ϫ4
Ϫ6
1
y ϭ 4 tan 3
Ϫ8
Ϫ10
90˚ 180˚ 270˚
24. amplitude: does not exist;
2
3
y
y ϭ sec 3
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
23. amplitude: does not exist; period: 120Њ or
y ϭ sin 2
O
270˚ 540˚ 810˚
O
Ϫ540˚ Ϫ360˚ Ϫ180˚
Ϫ4
Ϫ6
1
y ϭ 2 cot 2
Ϫ8
Ϫ10
372
180˚ 360˚ 540˚
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
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27. amplitude: 6; period: 540Њ or 3
28. amplitude: 3; period: 720Њ or 4
y
10
8
6
4
2
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ6
Ϫ8
Ϫ10
y
2
y ϭ 6 sin 3
O
90˚ 180˚ 270˚
10
8
6
4
2
O
Ϫ540˚Ϫ360˚
Ϫ180˚
Ϫ4
Ϫ6
Ϫ8
Ϫ10
29. amplitude: does not exist; period: 720Њ or 4
2
y
10
8
6
4
2
O
Ϫ540˚Ϫ360˚Ϫ180˚
Ϫ4
1 y ϭ 3 csc 2
Ϫ6
Ϫ8
Ϫ10
180˚ 360˚ 540˚
10
8
6
4
2
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ4
Ϫ6
1
y ϭ 2 cot 2
Ϫ8
Ϫ10
31. amplitude: does not exist; period: 180Њ or
O
45˚
90˚ 135˚
8
9
32. amplitude: ; period: 600Њ or
10
3
y
10
8
6
4
2
180˚ 360˚ 540˚
30. amplitude: does not exist;
period: 90Њ or
y
O
Ϫ270˚Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ6
2y ϭ tan
Ϫ8
Ϫ10
1
y ϭ 3 cos 2
y
90˚ 180˚ 270˚
5
4
3
2
1
Ϫ540˚Ϫ360˚Ϫ180˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
373
3
2
3 y ϭ 3 sin 5
4
O
180˚ 360˚ 540˚
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
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33.
34. y y
5
4
3
2
1
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
35.
3
5
5
4
3
2
1
3
y ϭ 5 sin 4
O
45˚
90˚ 135˚
7
y ϭ 8 cos 5
O
45˚
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
sin 4
1
107
7
8
90˚ 135˚
cos 5
36. y ϭ 0.25 sin 128t, y ϭ 0.25 sin 512t, y ϭ 0.25 sin 1024t
37. Sample answer: The amplitudes are the same. As the frequency increases, the period decreases.
38. f (x ) ϭ cos x and f (x ) ϭ sec x f (x )
5
4
3
2
1
f (x ) ϭ cos x f (x ) ϭ cos (Ϫx )
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
x
90˚ 180˚ 270˚
f (x ) f (x ) ϭ sec x f (x ) ϭ sec (Ϫx )
5
4
3
2
1
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
374
90˚ 180˚ 270˚
Algebra 2
x
Chapter 14
PQ245-6457F-P14[369-410].qxd
7/24/02
39. y ϭ 2 sin
5
2:03 PM
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40.
t
y
2.5
2
1.5
1
0.5
Ϫ0.5
Ϫ1
Ϫ1.5
Ϫ2
Ϫ2.5
y ϭ 2 sin 5 t
O
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
41. about 1.9 ft
42. Sample answer: Tides display periodic behavior.
This means that their pattern repeats at regular intervals.
Answers should include the following information.
• Tides rise and fall in a periodic manner, similar to the sine function.
• In f (x ) ϭ a sin bx, the amplitude is the absolute value of a.
43. A
t
44. C
22
2
22
2
45. 90Њ
46. Ϫ90Њ
47. 45Њ
48.
49.
50.
51.
13
16
1
2
52. 3, 11, 27, 59, 123
375
Algebra 2
Chapter 14
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7/24/02
53.
2:03 PM
Page 376 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
54.
y
15
13
11
9
7
5
3
1
Ϫ8
y ϭx
15
13
11
9
7
5
3
1
2
y ϭ 3x 2
O
Ϫ4
y
4
8
x
Ϫ8
Ϫ4
56.
y
y ϭ 2(x ϩ 1)2
Ϫ8
Ϫ4
15
13
11
9
7
5
3
1
O
y ϭ 2x 2
4
4
8
x
Lesson 14-2
x
8
y ϭ 3x 2 Ϫ 4
y
y ϭ x2 ϩ 2
Ϫ8
Ϫ4
15
13
11
9
7
5
3
1
O
y ϭ (x Ϫ 3)2 ϩ 2
4
x
8
Ϫ3
Ϫ5
Ϫ3
Ϫ5
Translations of Trigonometric Graphs
Pages 774–776
1. vertical shift: 15; amplitude: 3; period: 180Њ; phase shift: 45Њ
O
Ϫ3
Ϫ5
Ϫ3
Ϫ5
55.
y ϭ 3x 2
2. The midline of a trigonometric function is the line about which the graph of the function oscillates after a vertical shift.
376
Algebra 2
Chapter 14
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3. Sample answer: y ϭ sin ( ϩ 45Њ)
4. 1; 2;
2
y
(
y ϭ sin Ϫ 2
Ϫ
5. no amplitude; 180Њ; Ϫ60Њ
3
2
)
Ϫ Ϫ
2
5
4
3
2
1
O
2
Ϫ2
Ϫ3
Ϫ4
Ϫ5
3
2
6. 1; 360Њ; 45Њ y y
5
4
3
2
1
0.75
y ϭ cos ( Ϫ 45˚)
0.5
O
Ϫ270˚Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3 y ϭ tan ( ϩ 60˚) Ϫ4
Ϫ5
90˚ 180˚ 270˚
0.25
O
Ϫ45˚
Ϫ0.25
45˚ 90˚ 135˚ 180˚ 225˚
Ϫ0.5
Ϫ0.75
3
8.
7. no amplitude; 2; Ϫ
(
y ϭ sec ϩ 3
Ϫ
3
2
)
1
4
y ϭ ; 1; 360Њ y y
4
3
2
1
O
Ϫ Ϫ Ϫ1
2
Ϫ2
Ϫ3
Ϫ4
1
;
4
2
3
2
5
4
3
2
1
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
377
1
y ϭ cos ϩ 4
90˚ 180˚ 270˚
Algebra 2
Chapter 14
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9. Ϫ5; y ϭ Ϫ5; no amplitude;
360Њ
10. 4; y ϭ 4; no amplitude; 180Њ y 7
y
6
10
8
6
4
2
5
4
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ4
Ϫ6
Ϫ8 y ϭ sec Ϫ 5
Ϫ10
1
Ϫ135˚ Ϫ90˚ Ϫ45˚
y ϭ sin ϩ 0.25
0.5
90˚
180˚
270˚
45˚
y ϭ 3 sin [2( Ϫ 30˚)] ϩ 10
14
12
10
8
6
4
2
1
Ϫ0.5
O
Ϫ1
90˚
135˚
12. 10; 3; 180Њ; 30Њ
y
O
2
90˚ 180˚ 270˚
11. 0.25; y ϭ 0.25; 1; 360Њ
1.5
3
y ϭ tan ϩ 4
360˚
y
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ1
90˚ 180˚ 270˚
Ϫ1.5
2 4
14. 1; no amplitude; ;
13. Ϫ6; no amplitude; 60Њ; Ϫ45Њ y y
1
Ϫ45˚
O
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8
Ϫ9
Ϫ10
Ϫ11
45˚
4
3
2
1
Ϫ
3
8
Ϫ
4
Ϫ
Ϫ1O
8
Ϫ2
Ϫ3
1
Ϫ4 y ϭ 2 sec 4 Ϫ 4 ϩ 1
y ϭ 2 cot (3 ϩ 135˚) Ϫ 6
[(
378
8
4
3
8
)]
Algebra 2
Chapter 14
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7/24/02
2
3
2:03 PM
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6
15. Ϫ2; ; 4; Ϫ
16. 4; 1; 4 s
y
1
O
Ϫ3 Ϫ2 Ϫ
Ϫ1
2
3
Ϫ2
Ϫ3
2
[ 1(
y ϭ 3 cos 2 ϩ 6
)] Ϫ 2
17. h ϭ 4 Ϫ cos t or
2
h ϭ 4 Ϫ cos 90Њt
18.
h
h ϭ 4 Ϫ cos 2 t
6
5
4
3
2
1
O
Ϫ1
Ϫ2
Ϫ3
Ϫ4
19. 1; 360Њ; Ϫ90Њ
1
2
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
4
t
20. no amplitude; 180Њ; 30Њ y y
5
4
3
2
1
3
5
4
3
2
1
y ϭ cos ( ϩ 90˚)
90˚ 180˚ 270˚
O
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
379
45˚
90˚ 135˚
y ϭ cot ( Ϫ 30˚)
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
21. 1; 2;
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3
4
22. 1; 2; Ϫ
y
y
5
4
3
2
1
Ϫ
3
2
(
y ϭ sin Ϫ 4
5
4
3
2
1
)
O
Ϫ Ϫ
2 Ϫ2
Ϫ3
Ϫ4
Ϫ5
2
3
2
Ϫ
3
2
O
Ϫ Ϫ
2 Ϫ2
Ϫ3
Ϫ4
Ϫ5
5
4
3
2
1
5
4
3
2
1
45˚
90˚ 135˚
3
2
y ϭ 3 sin ( Ϫ 75˚)
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
1 y ϭ 4 tan ( ϩ 22.5˚)
90˚ 180˚ 270˚
26. 2; y ϭ 2; no amplitude; 360Њ
25. Ϫ1; y ϭ Ϫ1; 1; 360Њ y y
5
4
3
2
1
2
)
y
y
O
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
24. 3; 360Њ; 75Њ
23. no amplitude; 180Њ; Ϫ22.5Њ
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3 y ϭ sin Ϫ 1
Ϫ4
Ϫ5
(
y ϭ cos ϩ 3
5
4
3
2
1
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ2
Ϫ3
y ϭ sec ϩ 2
Ϫ4
Ϫ5
90˚ 180˚ 270˚
380
90˚ 180˚ 270˚
Algebra 2
Chapter 14
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Page 381 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
3
4
y
360Њ
2
1
y
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8
29.
1
;
2
90˚ 180˚ 270˚
5
4
3
2
1
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ2
Ϫ3
3 y ϭ csc Ϫ 4
Ϫ4
Ϫ5
y ϭ cos Ϫ 5
1 1
2 2
y
y
5
4
3
2
1
1
10
8
6
4
2
1
y ϭ 2 sin ϩ 2
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
90˚ 180˚ 270˚
32.
3
Ϫ2
4
O
Ϫ4
(
4
2
3
4
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
)
2 y ϭ 3 cos ( Ϫ 50˚) ϩ 2
90˚ 180˚ 270˚
translation 50Њ right and
2 units up with an amplitude
2
of unit
translation units left and
4
5 units up
90˚ 180˚ 270˚
y
5
4
3
2
1
y
y ϭ 5 ϩ tan ϩ 4
y ϭ 6 cos ϩ 1.5
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ6
Ϫ8
Ϫ10
31.
Ϫ
90˚ 180˚ 270˚
30. 1.5; y ϭ 1.5; 6; 360Њ
y ϭ ; ; 360Њ
18
16
14
12
10
8
6
4
2
3
4
28. Ϫ ; y ϭ Ϫ ; no amplitude;
27. Ϫ5; y ϭ Ϫ5; 1; 360Њ
3
381
Algebra 2
Chapter 14
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33. 1; 2; 120Њ; 45Њ
34. Ϫ5; 4; 180Њ; Ϫ30Њ y 5
4
3
2
1
y
10
8
6
4
2
y ϭ 2 sin [3( Ϫ 45˚)] ϩ 1
O
90˚ 180˚ 270˚
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ4
Ϫ6
Ϫ8
Ϫ10
35. Ϫ3.5; does not exist; 720Њ;
Ϫ60Њ
90˚ 180˚ 270˚
y ϭ 4 cos [2( ϩ 30˚)] Ϫ 5
36. 0.75; does not exist; 270Њ; 90Њ y 20
16
12
8
4
y
8
6
4
2
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12
O
90˚ 180˚ 270˚
[1 (
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ8
Ϫ12
Ϫ16
Ϫ20
)]
y ϭ 3 csc 2 ϩ 60˚ Ϫ 3.5
O
90˚ 180˚ 270˚
[2(
)]
y ϭ 6 cot 3 Ϫ 90˚ ϩ 0.75
1
4
38. Ϫ4; does not exist; 30Њ;Ϫ22.5Њ
37. 1; ; 180Њ; 75Њ
y
y
5
4
3
2
1
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
2
1
1 y ϭ 4 cos (2 Ϫ 150˚) ϩ 1
Ϫ22.5˚
90˚ 180˚ 270˚
O
22.5˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
Ϫ6
Ϫ7
Ϫ8
2 y ϭ 5 tan (6 ϩ 135˚) Ϫ 4
382
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
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2
3
4
40. 4; does not exist; 6; Ϫ
39. 3; 2; ; Ϫ
y
y
8
7
6
5
4
3
2
1
Ϫ
3
2
[(
O
Ϫ Ϫ
3
2
2
2 Ϫ2
16
14
12
10
8
6
4
2
y ϭ 3 ϩ 2 sin 2 ϩ 4
)]
O
Ϫ4 Ϫ2
Ϫ4
2
[1(
4
2
y ϭ 4 ϩ sec 3 ϩ 3
41.
42.
1
5
4
3
2
1
Ϫ
3
2
Ϫ Ϫ
2
y y ϭ 3 Ϫ 2 cos
1
y ϭ 3 ϩ cos ( ϩ )
2
Ϫ2
Ϫ3
Ϫ4
Ϫ5
3
2
)]
y
5
4
3
2
1
2
O
O
Ϫ4 Ϫ2
Ϫ2
Ϫ3
Ϫ4
Ϫ5
The graphs are identical.
[ 1 ( )]
1
3 y ϭ cos [ 4 ( ϩ 2 )] y ϭ Ϫsin 4 Ϫ 2
2
4
The graphs are identical.
43. c
44. 180; 5 yr
45. 300; 14.5 yr
46. Sample answer: When the prey (mouse) population is at its greatest the predator will consume more and the predator population will grow while the prey population falls.
47. h ϭ 9 ϩ 6 sin c (t Ϫ 1.5)d
48. a ϭ Ϫ1, b ϭ 1, h ϭ
49. Sample answer: You can use changes in amplitude and period along with vertical and horizontal shifts to show an animal population’s starting point and display changes to that population over a period of
50. B
9
383
2
Algebra 2
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time. Answers should include the following information.
• The equation shows a rabbit population that begins at 1200, increases to a maximum of 1450, then decreases to a minimum of 950 over a period of 4 years.
• Relative to y ϭ a cos bx, y ϭ a cos bx ϩ k would have a vertical shift of k units, while y ϭ a cos
[b (x Ϫ h)] has a horizontal shift of h units.
52. amplitude: does not exist; period: 360Њ or 2
51. D
y
5
4
3
2
1
O
Ϫ270˚Ϫ180˚Ϫ90˚
Ϫ2
Ϫ3
y ϭ 3 csc
Ϫ4
Ϫ5
period: 270Њ or
y
3
2
y
5
4
3
2
1
10
8
6
4
2
y ϭ sin 2
O
90˚ 180˚ 270˚
Ϫ360˚ Ϫ180˚
Ϫ4
Ϫ6
2
y ϭ 3 tan
3
Ϫ8
Ϫ10
O
180˚ 360˚
56. 0.57
55. 0.75
54. amplitude: does not exist;
53. amplitude: 1; period: 720Њ or
4
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
90˚ 180˚ 270˚
384
Algebra 2
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57. 0.83
58. 0.8
59. 35
60. 2.29
61. 0.66
62. 0.66
5a Ϫ 13
(a Ϫ 2)(a Ϫ 3)
64. Ϫ
65.
3y 2 ϩ 10y ϩ 5
2(y Ϫ 5)(y ϩ 3)
66. Ϫ
23
3
67. Ϫ1
69.
71.
23
2
1
4
63.
22
2
68. Ϫ1
1
2
70. 0
72. Ϫ
73. 1
Lesson 14-3 Trigonometric Identities
Pages 779–781
1. Sample answer: The sine function is negative in the third and fourth quadrants.
Therefore, the terminal side of the angle must lie in one of those two quadrants.
2. Sample answer: Pythagorean identities are derived by applying the Pythagorean
Theorem to trigonometric concepts. 3. Sample answer: Simplifying a trigonometric expression means writing the expression as a numerical value or in terms of a single trigonometric function, if possible.
4. Ϫ
7. 22
5
4
6.
5. Ϫ
3
5
8. 1
9. tan2
10. sec
12. sin ϭ cos
11. csc
23
3
385
v2 gR Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
13.
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14.
23
2
15. Ϫ25
17.
19.
21.
5
4
3
4
25
3
16. 2 22
1
2
18.
3
5
3 25
5
20. Ϫ
4 27
7
22.
4
5
4 217
17
23. Ϫ
24. Ϫ
25. cot
26. 1
27. cos
28. sin
29. 2
30. Ϫ3
31. cot2
32. tan
33. 1
34. cot2
35. csc2
36. 1
38. about 4 m/s
40. E ϭ
I tan cos
E
I sin simplifies to E ϭ
.
R2
42. P ϭ I 2R sin2 2ft
41. No; R 2 ϭ
43. P ϭ I 2R Ϫ
I 2R
.
1 ϩ tan2 2ft
44.
45. Sample answer: You can use equations to find the height and the horizontal distance of a baseball after it has been hit. The equations involve using the initial angle the ball makes with the ground with the sine function.
Answers should include the following information.
I cos
R2
9
16
46. B
386
Algebra 2
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• Both equations are quadratic in nature with a leading negative coefficient.
Thus, both are inverted parabolas which model the path of a baseball.
• model rockets, hitting a golf ball, kicking a rock
47. A
48. Ϫ1; y ϭ Ϫ1; 1; 360Њ y 5
4
3
2
1
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
49. 12; y ϭ 12; no amplitude; 180Њ
20
y ϭ sin Ϫ 1
y
5
4
3
2
1
15
10
5
50. amplitude: does not exist; period: 180Њ or
y
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ5
y ϭ tan ϩ 12
90˚ 180˚ 270˚
90˚ 180˚ 270˚
O
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3 y ϭ csc 2
Ϫ4
Ϫ5
387
45˚
90˚ 135˚
Algebra 2
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51. amplitude: 1; period: 120Њ or
2
3
52. amplitude: does not exist; period: 36Њ or
y
5
4
3
2
1
5
y
5
4
3
2
1
y ϭ cos 3
O
Ϫ135˚ Ϫ90˚ Ϫ45˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
45˚
90˚ 135˚
1
y ϭ 3 cot 5
O
Ϫ22.5˚
22.5˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
1
6
53. 93
54. y ϭ Ϫ (x Ϫ 11)2 ϩ
55. Symmetric (ϭ)
56. Substitution (ϭ)
57. Multiplication (ϭ)
1
2
58. Substitution (ϭ)
Chapter 14
Practice Quiz 1
Page 781
1.
3
,
4
2. Ϫ5, 2, 8,
720Њ or 4 y 5
4
3
2
1
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
y
3
6
4
2
1
y ϭ 4 sin 2
O
90˚ 180˚ 270˚
O
Ϫ4 Ϫ2
Ϫ4
Ϫ6
Ϫ8
Ϫ10
Ϫ12
Ϫ14
25
2
3
5
3. Ϫ
5.
4
4. Ϫ
388
213
3
2
4
[1 (
y ϭ 2 cos 4 Ϫ 4
Algebra 2
)] Ϫ 5
Chapter 14
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Lesson 14-4
Page 389 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
Verifying Trigonometric Identities
Pages 784–785
?
1. sin tan ϭ sec Ϫ cos
?
sin tan ϭ
1 cos
2. Sample answer: Use various identities, multiply or divide terms to form an equivalent expression, factor, and simplify rational expressions.
Ϫ cos
sec ϭ
1 cos
1 cos2
Ϫ
, cos cos
Multiply by the LCD, cos .
? 1 Ϫ cos2 tan ϭ cos
Subtract.
1 Ϫ cos 2
? sin2 tan ϭ ϭ sin2 cos sin
?
tan ϭ sin ؒ cos
Factor.
?
sin tan ϭ sin sin sin
sin tan ϭ sin tan
sin ϭ tan cos
?
3. Sample answer: sin2 ϭ 1 ϩ cos2 ; it is not an identity because sin 2 ϭ 1 Ϫ cos2 .
5.
4. tan (cot ϩ tan ) ϭ sec 2
?
1 ϩ tan2 ϭ sec2 sec2 ϭ sec2
?
tan2 cos2 ϭ 1 Ϫ cos2 sin2 cos2
6.
?
ؒ cos2 ϭ sin2 sin2 ϭ sin2
cos2
1 Ϫ sin
1 Ϫ sin2
1 Ϫ sin
(1 Ϫ sin )(1 ϩ sin )
1 Ϫ sin
?
ϭ 1 ϩ sin
?
ϭ 1 ϩ sin
?
ϭ 1 ϩ sin
1 ϩ sin ϭ 1 ϩ sin
389
Algebra 2
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1 ϩ tan2 csc2
ϭ tan2
sec2 csc2
7.
Page 390 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
ϭ tan2
?
8.
?
cos
1 cos 2 ? ϭ 1
?
2
tan2
sin cos
sin sec sin sec sin sec
2
ؒ sin ϭ tan tan2 ϭ tan2
9.
?
ϭ
?
ϭ
?
ϭ ϭ ?
ϭ ϭ sin cos sin2 ϩ cos2 sin cos
1
sin sec
10. D
?
?
12. cot (cot ϩ tan ) ϭ csc2
?
cot2 ϩ cot tan ϭ csc2
cos2 ϩ tan2 cos2 ϭ 1 cos2 ϩ
sin2 cos2
2
?
ؒ cos2 ϭ 1
2
cot 2 ϩ
?
cos ϩ sin ϭ 1
1ϭ1
13.
?
ϭ
tan sec ϩ 1
ؒ
sec Ϫ 1 sec ϩ 1 tan ؒ (sec ϩ 1) sec2 Ϫ 1 tan ؒ (sec ϩ 1) tan2 sec ϩ 1 tan
ϭ
sec ϩ 1 tan sec ϩ 1 tan sec ϩ 1 tan sec ϩ 1 tan
11.
tan sec Ϫ 1
?
sec ϩ 1 tan
cos
sin ?
1
ϭ sin2 ϩ cos2 sec
sin 2
1 cos2
1 sin ? ϭ sec tan ϩ cot sin ?
1
ϭ sin2 sin sec ϩ 14.
?
ؒ sin2 ϭ sec2
?
ϭ csc 2
?
sin sec cot ϭ 1 sin ؒ
1 cos
ؒ
cos sin
?
ϭ1
1ϭ1
?
1 ϩ tan2 ϭ sec2 sec2 ϭ sec2
cos sin
?
?
1 cos2
ؒ
cot2 ϩ 1 ϭ csc2 csc2 ϭ csc2
1 ϩ sec2 sin2 ϭ sec2
sin cos
390
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1 Ϫ 2 cos2 ? ϭ tan Ϫ cot sin cos
(1 Ϫ cos2 ) Ϫ cos2 ? ϭ tan Ϫ cot sin cos sin2 Ϫ cos2 ? ϭ tan Ϫ cot sin cos sin2 cos2
?
Ϫ ϭ tan Ϫ cot sin cos sin cos sin cos ?
Ϫ
ϭ tan Ϫ cot cos sin tan Ϫ cot ϭ tan Ϫ cot
16.
15.
1 Ϫ cos ? ϭ (csc Ϫ cot )2
1 ϩ cos
1 Ϫ cos ? ϭ csc2 Ϫ 2 cot csc
1 ϩ cos ϩ cot2
1 Ϫ cos ? 1 cos ϭ ؒ
Ϫ2ؒ
2
1 ϩ cos sin sin
1
cos2 ϩ sin sin2 cos2
1 Ϫ cos ? 1
2cos ϭ Ϫ ϩ 1 ϩ cos sin2 sin2 sin2
1 Ϫ cos ? 1 Ϫ 2 cos ϩ cos2 ϭ 1 ϩ cos sin2
1 Ϫ cos ? (1 Ϫ cos )(1 Ϫ cos ) ϭ 1 ϩ cos
1 Ϫ cos2
1 Ϫ cos ? (1 Ϫ cos )(1 Ϫ cos ) ϭ 1 ϩ cos (1 Ϫ cos )(1 ϩ cos )
1 Ϫ cos
1 Ϫ cos ϭ 1 ϩ cos
1 ϩ cos
17.
?
cot csc ϭ
cos
? sin
cot csc ϭ
cot csc
cot csc
18.
cot ϩ csc sin ϩ tan
?
sin ϩ cos ϭ
1
ϩ sin
?
sin ϩ cos ϭ
sin sin ϩ cos cos Ϫ 1 sin
?
ϭ sin cos ϩ sin cos cos ϩ 1 sin
?
ϭ sin (cos ϩ 1) cos
?
cos sin
1
sin
ϩ cos
1 cos sin ϩ cos cos
?
ϭ
1
cos
?
sin ϩ cos ϭ
sin ϩ cos cos
и cos
sin ϩ cos ϭ sin ϩ cos
cos ϩ 1
ؒ
sin cos sin (cos ϩ 1)
?
sin ϩ cos
1 ϩ tan sec
cot csc ϭ
cot csc ϭ
ؒ
1 sin
cot csc ϭ cot csc
391
Algebra 2
Chapter 14
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19.
sec sin
1
cos sin
7/24/02
Ϫ
sin cos
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?
20.
ϭ cot
sin
1 Ϫ cos ? ϩ ϭ 2csc
1 Ϫ cos sin sin sin
1 Ϫ cos 1 Ϫ cos ?
ؒ
ϩ
ؒ
ϭ 2csc sin 1 Ϫ cos
1 Ϫ cos sin
sin ?
Ϫ cos ϭ cot
1 sin cos
Ϫ
sin2 sin cos
1 Ϫ sin2 sin cos cos2 sin cos cos sin
1 Ϫ 2cos ϩ cos2 ? sin2 ϩ ϭ 2csc sin (1 Ϫ cos ) sin (1 Ϫ cos )
?
ϭ cot
sin2 ϩ cos2 ϩ 1 Ϫ 2cos ? ϭ 2csc sin (1 Ϫ cos )
2 Ϫ 2cos
?
ϭ 2csc sin (1 Ϫ cos )
?
ϭ cot
2(1 Ϫ cos ) ? ϭ 2csc sin (1 Ϫ cos )
?
ϭ cot
2 ? ϭ 2csc sin
?
ϭ cot
?
2csc ϭ 2csc
cot ϭ cot
21.
1 ϩ sin sin
1 ϩ sin sin
1 ϩ sin sin
1 ϩ sin sin
1 ϩ sin sin
1 ϩ sin sin
1 ϩ sin sin
1
?
ϭ
?
ϭ
?
ϭ
?
ϭ
cot2 csc Ϫ 1 cot2 csc ϩ 1
ؒ
csc Ϫ 1 csc ϩ 1 cot2 (csc ϩ 1) csc2 Ϫ 1 cot2 (csc ϩ 1) cot2
1 ϩ tan
1 ϩ cot
22.
1
?
ϭ ϭ sin ϩ cos cos
1 sin ϩ sin sin
1 ϩ sin sin
1
sec2
ϩ
csc2
?
24. 1 ϩ
?
23.
ϭ1
cos2 ϩ sin2 ϭ 1
1ϭ1
sin cos
?
sin cos
sin
ϩ cos
ϭ
cos sin sin ϩ cos cos
? sin ϭ cos sin ϩ cos sin
1
ϭ csc ϩ 1
?
?
ϭ
392
1 cos
1
cos
1
cos
1
cos
1
cos
1 cos
?
ϭ
?
ϭ
?
ϭ
ؒ
ϩ
sin sin ϩ cos sin cos
?
ϭ ϭ sin cos sin cos
tan2 sec Ϫ 1 tan2 sec ϩ 1
ؒ
sec Ϫ 1 sec ϩ 1 tan2 (sec ϩ 1) sec 2 Ϫ 1
2 (sec ϩ 1) tan ?
ϭ
tan2
?
ϭ sec ϩ 1 ϭ1ϩ 1 cos
Algebra 2
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25. 1 Ϫ tan4
26.
?
cos4 Ϫ sin4 ϭ cos2 Ϫ sin2
(cos2 Ϫ sin2 )(cos2 ϩ sin2 )
?
ϭ cos2 Ϫ sin2
(cos2 Ϫ sin2 ) ؒ 1
?
ϭ cos2 Ϫ sin2 cos2 Ϫ sin2 ϭ cos2 Ϫ sin2
?
ϭ 2 sec2 Ϫ sec4
(1 Ϫ tan2 )(1 ϩ tan2 )
?
ϭ sec2 (2 Ϫ sec2 )
[1 Ϫ (sec2 Ϫ 1)](sec2 )
?
ϭ (2 Ϫ sec2 )(sec2 )
(2 Ϫ sec2 )(sec2 ) ϭ (2 Ϫ sec2 )(sec2 )
1 Ϫ cos sin
sin
1 ϩ cos
?
sin
1 ϩ cos
?
sin
1 ϩ cos
ϭ
ؒ
1 ϩ cos
1 ϩ cos
ϭ
1 Ϫ cos2 sin (1 ϩ cos ) sin2 sin (1 ϩ cos ) sin
1 ϩ cos
29.
?
1 Ϫ cos sin
27.
ϭ
?
ϭ ϭ cos ? cos ϩ ϭ 2sec
1 ϩ sin
1 Ϫ sin cos
1 Ϫ sin cos
1 ϩ sin ?
ؒ
ϩ
ؒ
ϭ 2sec
1 ϩ sin 1 Ϫ sin 1 Ϫ sin 1 ϩ sin
28.
cos 11 Ϫ sin 2 ϩ cos 11 ϩ sin 2
11 ϩ sin 211 Ϫ sin 2
cos Ϫ sin cos ϩ cos ϩ sin cos ? ϭ 2 sec
1 Ϫ sin 2
2cos ? ϭ 2sec cos2
2
? ϭ 2sec cos2
2sec ϭ 2sec
sin
1 ϩ cos sin
1 ϩ cos
?
tan sin cos csc2 ϭ 1 sin cos
ؒ sin ؒ cos ؒ
1
2
sin2
1 Ϫ cos
30.
sin2
1 ϩ cos
ؒ
1 Ϫ cos 1 ϩ cos sin2 (1 ϩ cos )
1 Ϫ cos2 sin2 (1 ϩ cos )
?
sin
?
ϭ 2sec
ϭ1
1ϭ1
sin2
?
ϭ 1 ϩ cos
?
ϭ 1 ϩ cos
?
ϭ 1 ϩ cos
?
ϭ 1 ϩ cos
1 ϩ cos ϭ 1 ϩ cos
31.
2 v0 tan2
2g sec2
ϭ
sin2 cos2
1
2g cos2
2
v0
2 v0 ؒ
sin2 cos2
2g v 2 sin2
0
32. 598.7 m
ؒ
cos 2
1
2g
393
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Trigonometric identities are verified in a similar manner to proving theorems in geometry before using them.
Answers should include the following. • The expressions have not yet been shown to be equal, so you could not use the properties of equality on them.
• To show two expressions you must transform one, or both independently.
• Graphing two expressions could result in identical graphs for a set interval, that are different elsewhere. 33. Sample answer: Consider a right triangle ABC with right angle at C. If an angle A has a sine of x, then angle B must have a cosine of x.
Since A and B are both in a right triangle and neither is the right angle, their sum
2
must be .
35. D
36. B
37.
38.
[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1
[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1
may be
is not
40.
39.
[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1
[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1
may be
may be
394
Algebra 2
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41.
42.
25
2
25
3
[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1
[Ϫ360, 360] scl: 90 by [Ϫ5, 5] scl: 1
may be
is not
2193
12
43.
45.
44. Ϫ
46. Ϫ
27
4
48. 1: 360Њ; 45Њ
47. 1: 360Њ; 30Њ y 5
4
3
2
1
y
5
4
3
2
1
y ϭ cos ( Ϫ 30˚)
O
90˚ 180˚ 270˚
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
O
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ2
Ϫ3
Ϫ4
Ϫ5
2
49. 3; 2; Ϫ
50.
y ϭ sin ( Ϫ 45˚)
90˚ 180˚ 270˚
5
6
y
5
4
3
2
1
Ϫ
51.
53.
3
2
26
4
26
4
O
Ϫ Ϫ
2 Ϫ2
Ϫ3
Ϫ4
Ϫ5
ϩ
(
y ϭ 3 cos ϩ 2
2
3
2
)
22
2
52.
54.
395
22
4
2 Ϫ 23
4
Algebra 2
Chapter 14
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Sum and Difference of Angles Formulas
Pages 788–790
2. Use the formula sin(␣ ϩ ) ϭ sin ␣ cos  ϩ cos ␣ sin .
Since sin 105Њ ϭ sin(60Њ ϩ 45Њ), replace ␣ with 60Њ and  with 45Њ to get sin 60Њ cos 45Њ ϩ cos 60Њ sin 45Њ. By finding the sum of the products of the values,
1. sin (␣ ϩ ) Ϫ sin ␣ ϩ sin  sin ␣ cos  ϩ cos ␣ sin  sin ␣ ϩ 
26 ϩ 22
4
the result is
26 ϩ 22
4
26 Ϫ 22
4
3. Sometimes; sample answer:
The cosine function can equal 1.
4.
5.
6.
7.
23
2
8.
1
2
22 Ϫ 26
4
23
2
10. cos (270Њ Ϫ )
9. Ϫ
?
ϭ cos 270Њ cos ϩ sin 270Њ sin
?
ϭ 0 ϩ (Ϫ1 sin ) ϭ Ϫsin sin a ϩ b ϭ cos
2
11. sin cos
2
ϩ cos sin
2
?
12.
?
5 Ϫ 23
1 ϩ 5 23
23 sin
2
ϭ sin cos 30Њ ϩ cos cos 30Њ ϩ
?
cos cos 60Њ Ϫ sin sin 60Њ
ϭ cos
?
ϭ
?
sin ؒ 0 ϩ cos ؒ 1 ϭ cos cos ϭ cos
13.
sin( ϩ 30Њ) ϩ cos( ϩ 60Њ)
1 cos 2
?
1
2
23 sin 2
22
2
Ϫ
1
2
ϩ cos ϩ
1
2
ϭ cos ϩ cos ϭ cos
14.
396
Algebra 2
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PQ245-6457F-P14[369-410].qxd
15.
17.
19.
25.
27.
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22 Ϫ 26
4
Ϫ 26 Ϫ 22
4
16.
Ϫ 26 Ϫ 22
4
18.
22
2
Ϫ 26 Ϫ 22
4
Ϫ 26 Ϫ 22
4
22.
22 Ϫ 26
4
Ϫ 26 Ϫ 22
4
23
2
22
2
24. Ϫ
26.
22
2
22 Ϫ 26
4
20. Ϫ
22
2
21. Ϫ
23.
7/24/02
28. sin (270Њ Ϫ )
?
ϭ sin 270Њ cos Ϫ cos 270Њ sin
?
ϭ Ϫ1 cos Ϫ 0 ϭ Ϫcos
30. cos (90Њ Ϫ )
29. cos (90Њ ϩ )
?
?
ϭ cos 90Њ cos Ϫ sin 90Њ sin
?
ϭ 0 ؒ cos ϩ 1 ؒ sin ϭ sin
ϭ cos 90Њ cos Ϫ sin 90Њ sin
?
ϭ 0 Ϫ 1 sin ϭ Ϫsin
31.
?
32.
sin(90Њ Ϫ ) ϭ cos sin 90Њ cos Ϫ cos 90Њ sin
sin 1 ϩ sin cos
?
ϭ cos
3
2
3
2
2
?
ϭ Ϫcos
ϩ cos sin
3
2
?
?
ϭ Ϫcos
?
sin ؒ 0 ϩ cos (Ϫ1) ϭ Ϫcos
1 ؒ cos Ϫ 0 ؒ sin ϭ cos
?
cos Ϫ 0 ϭ cos
?
0 ϩ (Ϫcos ) ϭ Ϫcos
cos ϭ cos
Ϫcos ϭ Ϫcos
397
Algebra 2
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33.
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?
34.
cos ( Ϫ ) ϭ Ϫcos cos cos ϩ sin sin
?
cos(2 ϩ ) ϭ cos cos 2 cos Ϫ [sin 2 sin ]
?
?
ϭ cos
ϭ Ϫcos
?
Ϫ1 ؒ cos ϩ 0 ؒ sin ϭ Ϫcos
Ϫcos ϭ Ϫcos
?
1 ؒ cos Ϫ [0 ؒ sin ] ϭ cos
?
1 ؒ cos Ϫ 0 ϭ cos cos ϭ cos
?
35.
?
ϭ sin 60Њ cos ϩ cos 60Њ sin ϩ sin 60Њ cos Ϫ cos 60Њ sin
?
sin cos Ϫ [cos sin ] ϭ sin
23
2
ϭ 23 cos
?
0 ؒ cos Ϫ [Ϫ1 ؒ sin ] ϭ sin
?
ϭ
?
0 Ϫ [Ϫsin ] ϭ sin sin ϭ sin
37. sin a ϩ b Ϫ cos a ϩ b
3
23 sin ϩ cos 2
1
cos ϩ sin
2
1
? 1 ϭ sin ϩ sin
2
2
?
3
cos 6
6
1 ϭ 2
?
ϩ sin
23
2
3
sin 6
Ϫ
cos ϩ
cos Ϫ
1
2
1
2
sin ϩ
sin
38. sin ( ␣ ϩ ) sin ( ␣ Ϫ ) ϭ sin2 ␣ Ϫ sin2 
ϭ sin cos ϩ cos sin Ϫ cos
23
2
36. sin (60Њ ϩ ) ϩ sin (60Њ Ϫ )
sin( Ϫ ) ϭ sin
?
ϭ (sin ␣ cos  ϩ cos ␣ sin )
(sin ␣ cos  Ϫ cos ␣ sin )
?
ϭ sin2 ␣ cos2  Ϫ cos2 ␣ sin2 
?
ϭ sin2 ␣(1 Ϫ sin2 ) Ϫ
(1 Ϫ sin2 ␣) sin2 
?
ϭ sin2 ␣ Ϫ sin2 ␣ sin2  Ϫ
ϭ sin
sin2  ϩ sin2 ␣ sin2  ϭ sin2 ␣ Ϫ sin2 
398
Algebra 2
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39.
40.
?
cos (␣ ϩ ) ϭ
?
cos (␣ ϩ ) ϭ
1 Ϫ tan ␣ tan  sec ␣ sec 
1
sin ␣
y ϭ 10 sin (2t Ϫ 30°) ϩ 10 cos (2t ϩ 60°)
sin 
Ϫ cos ␣ ؒ cos 
1
cos ␣
ؒ
Ϫ180° Ϫ90°
1 cos 
O
90°
180° t
Ϫ2
Ϫ4
?
cos (␣ ϩ ) ϭ sin ␣
1 Ϫ cos ␣ ؒ
1
cos ␣
y
4
ؒ
sin  cos 
1 cos 
и
cos ␣ cos  cos ␣ cos 
?
cos (␣ ϩ ) ϭ cos ␣ cos  Ϫ sin ␣ sin 
1
cos (␣ ϩ ) ϭ cos (␣ ϩ )
41. Destructive; the resulting graph has a smaller amplitude than the two initial graphs. 42. 0.3681 E
43. 0.4179 E
44. 0.6157 E
45. 0.5563 E
46. tan (␣ ϩ ) ϭ ϭ
sin(␣ ϩ ) cos(␣ ϩ ) sin ␣ cos  ϩ cos ␣ sin  cos ␣ cos  Ϫ sin ␣ sin
399
sin ␣ cos  cos ␣ cos  cos ␣ cos  cos ␣ cos 
cos ␣ sin 
tan ␣ ϩ tan 
1 Ϫ tan ␣ tan 
ϩ cos ␣ cos  sin ␣ sin 
Ϫ cos ␣ cos 
Algebra 2
Chapter 14
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tan (␣ Ϫ ) ϭ ϭ
sin(␣ Ϫ ) cos(␣ Ϫ ) sin ␣ cos  Ϫ cos ␣ sin  cos ␣ cos  ϩ sin ␣ sin 
ϭ
sin ␣ cos  cos ␣ cos  cos ␣ cos  cos ␣ cos 
cos ␣ sin 
ϭ
tan ␣ Ϫ tan 
1 ϩ tan ␣ tan 
Ϫ cos ␣ cos  sin ␣ sin 
ϩ cos ␣ cos 
47. Sample answer: To determine communication interference, you need to determine the sine or cosine of the sum or difference of two angles. Answers should include the following information. • Interference occurs when waves pass through the same space at the same time. When the combined waves have a greater amplitude, constructive interference results and when the combined waves have a smaller amplitude, destructive interference results. 48. ϭA
49. C
50. cot ϩ sec
?
ϭ
?
ϭ
?
ϭ
cos2 ϩ sin sin cos cos2 sin ϩ sin cos sin cos cos sin
ϩ
1 cos
ϭ cot ϩ sec
400
Algebra 2
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51. sin2 ϩ tan2
52.
?
ϭ (1 Ϫ cos2 ) ϩ
?
ϭ sin2 ϩ
?
ϭ sin2 ϩ
?
ϭ sin2 ϩ
sec2 csc2
1
cos2 sin2 cos2
sec2
csc
?
sin (sin ϩ csc ) ϭ 2 Ϫ cos2
2
?
sin2 ϩ 1 ϭ 2 Ϫ cos2
Ϭ
?
1 Ϫ cos2 ϩ 1 ϭ 2 Ϫ cos2
2 Ϫ cos2 ϭ 2 Ϫ cos2
1 sin2
ϭ sin2 ϩ tan2
53.
1
Ϭ
cos
1
ؒ cos
sec tan sin cos cos sin
1
sin
?
ϭ csc
54. 1
?
ϭ csc
?
ϭ csc
?
ϭ csc
csc ϭ csc
3 234
,
34
5 234
,
34
55. 4
56. sec
57. 2 sec
58. sin ϭ Ϫ cos ϭ
sec ϭ
59.
4
5
3
5
sin ϭ Ϫ , cos ϭ Ϫ ,
4
3
5
3
5
3
cot ϭ Ϫ
60. sin ϭ 1, cos ϭ 0, tan ϭ undefined, csc ϭ 1, sec ϭ undefined, cot ϭ 0
5
4
3
4
tan ϭ , csc ϭ Ϫ , sec ϭ Ϫ , cot ϭ
234
,
3
234
,
5
csc ϭ Ϫ
3
5
tan ϭ Ϫ ,
61. 360
62. 3,991,680
2 25
2
63. 56
64. 210
65. about 228 mi
66.
67. Ϯ
68. Ϯ
y2
34
Ϫ
x2
6
ϭ1
3
5
401
Algebra 2
Chapter 14
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25
5
3 25
5
26
2
69. Ϯ
71. Ϯ
73. Ϯ
3
4
70. Ϯ
2 16Ϫ 12
2
72. Ϯ
74. Ϯ
Lesson 14-6
22Ϫ2 12
2
Double-Angle and Half-Angle Formulas
Pages 794–797
12
2
or 22
1. Sample answer: If x is in the x third quadrant, then is
2
between 90Њ and 135Њ. Use the half-angle formula for cosine knowing that the value is negative.
2. Sample answer: 45Њ; cos 2(45Њ) ϭ cos 90Њ or 0,
3. Sample answer: The identity used for cos 2 depends on whether you know the value of sin , cos , or both values.
4.
24
,
25
5.
6.
22 ϩ 13
2
4 25
,
9
1 230
,
9
6
Ϫ ,
3 27
,
8
28 Ϫ 2 17
,
4
7. Ϫ
1
8
Ϫ ,
22 Ϫ 13
2
Ϫ
Ϫ
2 cos 45Њ ϭ 2 ؒ
26
6
28 ϩ 2 17
4
7 25 2 25
,
25 5
5
Ϫ ,
23 1 22 Ϫ 13
, ,
,
2 2
2
8. Ϫ
22 Ϫ 13
2
?
sin 2x
1 Ϫ cos 2x
?
2 sin x cos x
1 Ϫ (1 Ϫ 2 sin2 x)
?
2 sin x cos x
2 sin2 x
?
9.
cos x sin x
10. cot x ϭ ϭ ϭ ϭ ϭ cot x
402
Algebra 2
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?
11. cos2 2x ϩ 4 sin2 x cos2 x ϭ 1
120 119 5 226 226
,
,
,
169 169
26
26
4 22
,
9
13. Ϫ
15.
Ϫ
16.
17 215 221
,
18
6
6
1 26 230
,
9 6
6
22.
25. Ϫ
20.
Ϫ ,
Ϫ ,
215
,
8
210
10
Ϫ
215
5
7 210 26
,
8
4
4
Ϫ
Ϫ ,
120 119 5 226
,
,
,
169 169
26
215 7
, ,
8
8
226
26
28 ϩ 2 115
,
4
28 Ϫ 2 115
4
Ϫ
4 221 17
, ,
5
25
Ϫ
25 110 ϩ 1210
,
10
Ϫ
25 110 Ϫ 1210
10
24. Ϫ
Ϫ ,
22 ϩ 13
2
22 ϩ 12
2
26.
22 Ϫ 12
2
27. Ϫ
23 210
,
25
5
24 7 3 210
, ,
,
25 25
10
18. Ϫ
23 16 ϩ 4 13
6
4 25
,
9
Ϫ
31.
23
3
14. Ϫ
4 22 7 23 16 Ϫ 4 13
, ,
9
9
6
21. Ϫ
29.
7 26
,
9 3
Ϫ ,
28 ϩ 155
4
235
,
18
Ϫ
23.
4 26
,
25
?
cos2 2x ϩ sin2 2x ϭ 1
1ϭ1
3 255 23 28 Ϫ 155
, ,
,
32
32
4
17. Ϫ
19.
12. 1.64
22 Ϫ 12
2
28. Ϫ
30. Ϫ
?
sin 2x ϭ 2 cot x sin2 x cos x
?
2 sin x cos x ϭ 2
ؒ sin2 x sin x
2 sin x cos x ϭ 2 sin x cos x
22 Ϫ 13
2
22 Ϫ 13
2
32.
2aϮ
3
2 cos2
x
2
?
ϭ 1 ϩ cos x
2
1 ϩ cos x b 2
1 ϩ cos x b 2
2a
?
ϭ 1 ϩ cos x
?
ϭ 1 ϩ cos x
1 ϩ cos x ϭ 1 ϩ cos x
403
Algebra 2
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?
1
2
?
1
2
?
1
2
34. sin2 x ϭ (1 Ϫ cos2 x)
33.
?
sin4 x Ϫ cos4 x ϭ 2 sin2 x Ϫ 1
(sin2 x Ϫ cos2 x)(sin2 x ϩ cos2 x)
sin2 x ϭ [1 Ϫ (1 Ϫ 2 sin2 x)]
?
ϭ 2 sin2 x Ϫ 1
(sin2 x Ϫ cos2 x) ؒ 1
sin2 x ϭ (2 sin2 x) sin2 x ϭ sin2 x
?
ϭ 2 sin2 x Ϫ 1
[sin2 x Ϫ (1 Ϫ sin2 x)] ؒ 1
?
ϭ 2 sin2 x Ϫ 1 sin2 x Ϫ 1 ϩ sin2 x
?
ϭ 2 sin2 x Ϫ 1
2 sin2 x Ϫ 1 ϭ 2 sin2 x Ϫ 1
35.
3
3
tan2
x
2
?
1 Ϫ cos x
1 ϩ cos x
?
1 Ϫ cos x
1 ϩ cos x
ϭ
x
sin2 2 cos 2 x
ϭ
36.
2
1 Ϫ cos x 2 b 2
? 1 Ϫ cos x
1 ϩ cos x 2 b 2
1 Ϫ cos x
1 ϩ cos x
39. 2 ϩ 23
ϭ
ϭ
1 cos x
Ϫ
sin x cos x sin x
1 Ϫ cos2 x sin x cos x sin2 x sin x cos x sin x cos x
1 ϩ cos x
1 Ϫ cos x
1 ϩ cos x
37. 46.3Њ
38.
40.
2 g 3
3
?
ϭ tan x
?
ϭ tan x
?
ϭ tan x
?
ϭ tan x
tan x ϭ tan x
1 Ϫ cos L
1 ϩ cos L
1 Ϫ cos L
1 ϩ cos L
v 2 (tan Ϫ tan sin2 )
?
2 g v 2 tan (1 Ϫ sin2 )
?
2 g v 2 tan cos2
?
2 2 v sin g v 2 sin 2 g ϭ ϭ ϭ ϭ ©Glencoe/McGraw-Hill
404
cos
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
41.
1 tan 4
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y
42.
Ϫ
2.5
2
1.5
1
0.5
3 Ϫ
2
Ϫ2
y ϭ sin2 x x O
2
Ϫ1
1
Ϫ1.5
y ϭ Ϫ 2 cos 2x
Ϫ2
3
2
y ϭ Ϫcos2 x
Ϫ2.5
Sample answer: They all have the same shape and are vertical translations of each other.
2
43. The maxima occur at x ϭ Ϯ and 3
Ϯ . The
2
44.
y
2.5
2
1.5
1
0.5
minima occur
at x ϭ 0, Ϯ and Ϯ2.
Ϫ270˚ Ϫ180˚ Ϫ90˚
Ϫ1
Ϫ1.5
Ϫ2
Ϫ2.5
y ϭ sin 2x x O
90˚ 180˚ 270˚
45. The graph of f(x) crosses the x-axis at the points specified in Exercise 41.
46. c ϭ 1 and d ϭ 0.5
47. Sample answer: The sound waves associated with music can be modeled using trigonometric functions.
Answers should include the following information.
• In moving from one harmonic to the next, the number of vibrations that appear as sine waves increase by 1.
48. D
405
Algebra 2
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• The period of the function as you move from the nth harmonic to the (n ϩ 1)th harmonic decreases from
2
2
.
to
26 Ϫ 22
4
n
49. B
51.
53. Ϫ
55.
26 ϩ 22
4
nϩ1
50.
23
2
54.
1
2
22
2
26 ϩ 22
4
52. Ϫ
56.
?
cot2 Ϫ sin2 ϭ
cos2 csc2 Ϫ sin2 sin2 csc2
1
2
2
?
cot Ϫ sin ϭ
cos2 sin2 Ϫ sin2
1
sin2 sin2
cot2 Ϫ sin2
1
cot2 Ϫ sin2 ϭ cot2 Ϫ sin2
?
cot2 Ϫ sin2 ϭ
58. 101 or 10
57. cos (cos ϩ cot )
?
ϭ cot cos 1sin ϩ 12
?
ϭ
cos cos sin ϩ cot cos sin
?
ϭ cos2 ϩ cot cos ϭ cos (cos ϩ cot )
59. 102.5 or about 316 times greater 60. Ϫ6, 5
61. 1, Ϫ1
62. 0, Ϫ2
63.
5
,
2
1 1
2 2
64. Ϫ ,
Ϫ2
1
2
65. 0, Ϫ
406
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Chapter 14
Practice Quiz 2
Page 797
?
1. sin sec ϭ tan sin ؒ
1
cos sin cos
?
2.
sec Ϫ cos ϭ sin tan cos ?
1
Ϫ cos ؒ ϭ sin tan cos cos cos2 ?
1
Ϫ ϭ sin tan cos cos
1 Ϫ cos2 ? ϭ sin tan cos sin2 ? ϭ sin tan cos sin ? ϭ sin tan sin cos sin tan ϭ sin tan
4.
sin (90Њ ϩ ) ϭ cos
?
ϭ tan
?
ϭ tan
?
tan ϭ tan
sin (cos ϩ 1) cos sin cos ϩ sin
?
sin ϩ tan ϭ cos sin
? sin cos sin ϩ tan ϭ ϩ cos cos sin ϩ tan ϭ sin ϩ tan
?
3. sin ϩ tan ϭ
?
sin 90Њ cos ϩ cos 90Њ sin ϭ cos
?
cos ϩ 0 ϭ cos cos ϭ cos
3
?
Ϫ b ϭ Ϫsin
2
3
3
? cos cos ϩ sin sin ϭ Ϫsin
2
2 cos a
5.
6. sin ( ϩ 30Њ) ϩ cos ( ϩ 60Њ)
?
?
ϭa
?
Ϫsin ϭ Ϫsin
9.
sin ϩ
1 a cos
2
23
2
?
1
2
Ϫ
9 282
82
13
2
1
2
cos b ϩ sin b
1
2
ϭ cos ϩ cos
22 Ϫ 13
2
13
2
ϭ (sin cos 30Њ ϩ cos sin 30Њ) ϩ
(cos cos 60Њ Ϫ sin sin 60Њ)
0 ϩ (Ϫ1 ؒ sin ) ϭ Ϫsin
7.
?
22 Ϫ 12
2
ϭ cos
8. Ϫ
10.
407
Algebra 2
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Lesson 14-7
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Solving Trigonometric Equations
Pages 802–804
2. Sample answer: The function is periodic with two solutions in each of its infinite number of periods.
1. Sample answer: If sec ϭ 0
1
then ϭ 0. Since no value cos
of makes
1 cos
ϭ 0. there
are no solutions.
3. Sample answer: sin ϭ 2
4. 60Њ, 120Њ, 240Њ, 300Њ
5. 135Њ, 225Њ
6.
7.
6
5 3
, , ,
6 2 6 2
8. 0 ϩ
9. 0 ϩ k
2k
3
10. 90Њ ϩ k ؒ 360Њ, 180Њ ϩ k ؒ 360Њ
12.
11. 60Њ ϩ k ؒ 360Њ, 300Њ ϩ k ؒ 360Њ
7
6
11
6
ϩ 2k,
ϩ 2k or
210Њ ϩ k ؒ 360Њ, 330Њ ϩ k ؒ 360Њ
13.
6
ϩ 2k,
5
6
ϩ 2k,
2
ϩ 2k
14. 31.3Њ
or 30Њ ϩ k ؒ 360Њ, 150Њ ϩ
360Њ, 90Њ ϩ k ؒ 360Њ
15. 60Њ, 300Њ
16. 240Њ, 300Њ
17. 210Њ, 330Њ
18. 30Њ, 150Њ, 210Њ, 330Њ
19.
5 3
, ,
6 6 2
20.
2
21.
7 11
,
6
6
22.
3 2 4
, , ,
2 2 3 3
23.
3
ϩ 2k,
5
3
ϩ 2k
24. ϩ 2k,
5
3
25.
2
3
27.
3
ϩ 2k,
ϩ 2k,
4
3
5
3
ϩ 2k
3
ϩ 2k,
ϩ 2k
26. 0 ϩ 2k
ϩ 2k
28. 0 ϩ k,
408
6
ϩ 2k,
5
6
ϩ 2k
Algebra 2
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29. 45Њ ϩ k ؒ 180Њ
30. 0Њ ϩ k ؒ 180Њ
31. 270Њ ϩ k ؒ 360Њ
32. 30Њ ϩ k ؒ 360Њ,
150Њ ϩ k ؒ 360Њ
33. 0Њ ϩ k ؒ 180Њ, 60Њ ϩ k ؒ 180Њ
34. 120Њ ϩ k ؒ 360Њ,
240Њ ϩ k ؒ 360Њ
3
ϩ 2k, ϩ 2k
2
2
35. 0 ϩ 2k,
36.
7
6
ϩ 2k,
11
6
ϩ 2k or
210Њ ϩ k ؒ 360Њ,
330Њ ϩ k ؒ 360Њ
or 0Њ ϩ k ؒ 360Њ, 90Њ ϩ k ؒ 360Њ,
270Њ ϩ k ؒ 360Њ
38.
37. 0 ϩ k or 0Њ ϩ k ؒ 180Њ
2
ϩ k,
2
3
ϩ 2k,
4
3
ϩ 2k
or 90Њ ϩ k ؒ 180Њ, 120Њ ϩ k ؒ
360Њ, 240Њ ϩ k ؒ 360Њ
39. 0 ϩ 2k,
3
ϩ 2k,
5
3
40.
ϩ 2,
2
ϩ 4k or 90Њ ϩ k ؒ 720Њ
or 0Њ ϩ k ؒ 360Њ, 60Њ ϩ k ؒ
360Њ, 300Њ ϩ k ؒ 360Њ
41. S ϭ
352 tan
43. y ϭ
3
2
y
4
3.5
3
2.5
2
1.5
1
0.5
O
Ϫ1
Ϫ1
ϩ
3
or S ϭ 352 cot
3
2
44. 10
sin (t)
3
y ϭ 2 ϩ 2 sin ( t )
1 2 3 4 5 6 7 8 9
t
Temperatures are cyclic and can be modeled by trigonometric functions.
Answers should include the following information.
45. (4.964, Ϫ0.598)
409
Algebra 2
Chapter 14
PQ245-6457F-P14[369-410].qxd
7/24/02
2:04 PM
Page 410 TF06 Manish 06:BOOKS:PRD:PQ245_64575F FINAL DELIVERY:PQ245-14:
• A temperature could occur twice in a given period such as when the temperature rises in the spring and falls in autumn.
24 7 210 3 210
, ,
,
25 25 10
10
5 211 7 23 233
, ,
,
18
18 6
6
1 1 23
23
,Ϫ , ,
2
2 2 2
7 25 2 25
24
,Ϫ ,
,
25
25 5
5
47. D
48. B
49.
50.
51.
53. Ϫ
23
2
52.
54.
22
2
55. b ϭ 11.0, c ϭ 12.2, mЄC ϭ 78 | 190,211 | 359,418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-49 | latest | en | 0.459325 |
https://www.coursehero.com/file/6646565/Differential-Equations-Lecture-Work-Solutions-314/ | 1,529,587,803,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864148.93/warc/CC-MAIN-20180621114153-20180621134153-00017.warc.gz | 799,857,940 | 68,755 | Differential Equations Lecture Work Solutions 314
# Differential Equations Lecture Work Solutions 314 - 1 Apply...
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1. Apply the ADI scheme to the 2-D heat equation and find u n +1 at the internal grid points in the mesh shown in figure 60 for r x = r y = 2 . The initial conditions are u n = 1 x 3∆ x along y = 0 u n = 1 y 2∆ y along x = 0 u n = 0 everywhere else and the boundary conditions remain fixed at their initial values. Step 1: u n +1 / 2 i j u n i j = r x 2 =1 u n +1 / 2 i +1 j 2 u n +1 / 2 i j + u n +1 / 2 i 1 j + r y 2 =1 u n i j +1 2 u n i j + u n i j 1 Step 2: u n +1 i j u n +1 / 2 i j = r x 2 =1 u n +1 / 2 i +1 j 2 u n +1 / 2 i j + u n +1 / 2 i 1 j + r y 2 =1 u n +1 i j +1
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Jill Tulane University ‘16, Course Hero Intern | 518 | 1,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2018-26 | latest | en | 0.91245 |
https://mvtrinh.wordpress.com/2014/02/07/postal-worker/ | 1,532,173,052,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592523.89/warc/CC-MAIN-20180721110117-20180721130117-00435.warc.gz | 733,860,969 | 15,219 | ## Postal Worker
A postal worker has $600$ houses on his route. During one snowy day, he noticed that every other house’s mailbox was blocked by snow, and he could not deliver the mail. He also noticed that he had no mail to deliver to every third house on the route. By the end of his route, how many households had mail that was still in his truck?
Source: NCTM Mathematics Teacher 2008
SOLUTION
Suppose there are $18$ households on the route.
B = blocked by snow
N = no mail
Number of households that have no mail, $18\div 3=6$, namely $\{3, 6, 9, 12, 15, 18\}$.
Number of households that have mail, $18-6=12$, namely {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17}.
Half of the $12$ households having mail are blocked by snow; the mail of $6$ households is still in his truck.
Number of households that have no mail, $600\div 3=200$.
Number of households that have mail, $600-200=400$.
Half of the $400$ households having mail are blocked by snow; the mail of $200$ households is still in his truck.
Answer: $200$ | 306 | 1,016 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-30 | latest | en | 0.989239 |
https://www.physicsforums.com/threads/force-between-polar-molecule-and-an-ion.803112/ | 1,508,200,177,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820487.5/warc/CC-MAIN-20171016233304-20171017013304-00585.warc.gz | 1,171,097,476 | 19,404 | # Force between polar molecule and an ion
1. Mar 14, 2015
### Westin
1. The problem statement, all variables and given/known data
Under certain conditions the interaction between a "polar" molecule such as HCl located at the origin and an ion located along the x axis can be described by a potential energy U=−b[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngx^2, [Broken] where b is a constant.
What is Fx, the x component of the force on the ion?
Fx=
What is Fy, the y component of the force on the ion?
Fy= 0
2. Relevant equations
U=-b/x^2
3. The attempt at a solution
The y component of the force on the ion was fairly obvious but I am getting stuck on the x component.
I know that the force must me positive but it was incorrect when I tried +b[PLAIN]https://s3.lite.msu.edu/adm/jsMath/fonts/cmmi10/alpha/144/char3D.pngx^3[/B][/B] [Broken]
Last edited by a moderator: May 7, 2017
2. Mar 14, 2015
### TESL@
I think you need to know the partial charges and their separation/distance.
3. Mar 14, 2015
### Westin
these are the choices i have
A). +2b/x^3
B). 0
C). +b/x^3
D). -2b/x^3
E). -b/(2x)
F). -b/x^3
G). -b/x
4. Mar 14, 2015
### Staff: Mentor
What is the relation between force and potential?
5. Mar 14, 2015
### Staff: Mentor
Why?
6. Mar 14, 2015
### Staff: Mentor
Why do you say that?
7. Mar 14, 2015
### TESL@
The molecule will rotate in such way that H will be at the farthest and Cl at the closest point to the ion. Then the force will depend on partial charges and the distance between both H and Cl, and ion and molecule.
8. Mar 14, 2015
### Staff: Mentor
On can infer from the equation U=−b/x2 that the molecule is taken as a point dipole with its orientation assumed to be always favorable.
9. Mar 14, 2015
### Westin
Wait, wouldn't the answer be negative because the force on an object is the negative of the derivative of the potential function? The potential energy U is equal to the work you must do against that force to move an object from the U=0 reference point to the position r. The force you must exert to move it must be equal but oppositely directed, and that is the source of the negative sign.
10. Mar 14, 2015
### Westin
I'm almost positive the answer is -b/(2x) after what I just said. Any feedback?
11. Mar 14, 2015
### Staff: Mentor
Again, what is the equation that relates force and potential energy?
12. Mar 14, 2015
### Westin
F = -dU/dx
13. Mar 14, 2015
### Staff: Mentor
Great, now apply that equation to the problem given.
14. Mar 14, 2015
### Westin
Im confused with on how to.. F = −b/x^2 take derive then get -b/2x
15. Mar 14, 2015
### Staff: Mentor
I guess you mean U. What is the derivative of $x^{-2}$?
16. Mar 14, 2015
### Westin
F = -b/x^2 ---> -bx^-2 ---> 2b/x^3 this is it
17. Mar 14, 2015
### Westin
It was incorrect... I have one try left , I don't get what else it could be
18. Mar 14, 2015
### Staff: Mentor
Doesn't this contradict what you wrote earlier?
Before submitting answers, lets check if they make sense...
19. Mar 14, 2015
### Westin
Ahh yes it does,-2b/x^3 should be correct? Im sorry for my personal confusion. I'm in a poor experimental physics class at Michigan state where you have to teach yourself online everything..
20. Mar 14, 2015
### Staff: Mentor
It's ok to be confused. I'm just hoping that you can gain some good habits. From the formula for energy, you can see that the energy increases (decreases in magnitude but with a negative sign) when the ion moves away from the origin, so the force must be directed towards the molecule. That takes care of the sign. Then it is just a question of going back to fundamentals, ie, F = -dU/dx, to get the force. Then you can check that the answer makes sense, considering the physics (force must be directed towards the origin). It's easy to lose track of minus signs, especially in a case like this where you get 3 in a row. | 1,147 | 3,942 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-43 | longest | en | 0.914719 |
https://brainly.ph/question/95455 | 1,484,643,379,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279650.31/warc/CC-MAIN-20170116095119-00005-ip-10-171-10-70.ec2.internal.warc.gz | 808,551,665 | 10,377 | # What are changes in the form of mechanical energy
1
by charlie15
2015-01-05T17:16:21+08:00
n the physical sciences, mechanical energy is the sum of potential energy and kinetic energy. It is the energy associated with the motion and position of an object. The principle of conservation of mechanical energy states that in an isolated system that is only subject to conservative forces the mechanical energy is constant. If an object is moved in the opposite direction of a conservative net force, the potential energy will increase and if the speed (not the velocity) of the object is changed, the kinetic energy of the object is changed as well. In all real systems, however, non-conservative forces, like frictional forces, will be present, but often they are of negligible values and the mechanical energy's being constant can therefore be a useful approximation. In elastic collisions, the mechanical energy is conserved but in inelastic collisions, some mechanical energy is converted into heat. The equivalence between lost mechanical energy (dissipation) and an increase in temperature was discovered by James Prescott Joule.
Many modern devices, such as the electric motor or the steam engine, are used today to convert mechanical energy into other forms of energy, e.g. electrical energy, or to convert other forms of energy, like heat, into mechanical energy.
Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. Mechanical energy can be either kinetic energy (energy of motion) or potential energy (stored energy of position). Objects have mechanical energy if they are in motion and/or if they are at some position relative to a zero potential energy position (for example, a brick held at a vertical position above the ground or zero height position). | 389 | 1,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-04 | latest | en | 0.906007 |
http://www.mathworks.com/matlabcentral/answers/59034-subtract-from-each-element-its-mean-find-absolute-deviation-across-rows | 1,444,148,812,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736678861.8/warc/CC-MAIN-20151001215758-00078-ip-10-137-6-227.ec2.internal.warc.gz | 748,117,063 | 10,302 | ## subtract from each element its mean / find absolute deviation across rows
on 17 Jan 2013
### Walter Roberson (view profile)
```A = [1 2 NaN; 4 NaN 6; 7 8 9]
```
```A =
```
``` 1 2 NaN
4 NaN 6
7 8 9```
```mn = nanmean(A,2)
```
```mn =
```
``` 1.5000
5.0000
8.0000```
I would like to subtract each element by its corresponding row average as shown below(i.e find absolute deviations from mean),
```1-1.5 2-1.5 NaN
4-5 NaN 6-5
7-8 8-8 9-8
```
Note the below array has no negative values as I am trying to find modulus of deviations.
```0.5 0.5 NaN
1 NaN 1
1 0 1
```
Then sum up each row and divide by no. of elements.
```(-0.5+0.5)/2
(-1 + 1)/2
(-1 + 0 + 1)/3
```
How can I achieve this result vector using arrayfun or by any other method ?
I am trying to find teh absolute deviation of each element in the array. i.e std.dev without the sq
## Products
No products are associated with this question.
### Walter Roberson (view profile)
on 17 Jan 2013
```nanmean(abs(bsxfun(@minus A, nm)),2)
```
Ms. Mat
### Ms. Mat (view profile)
on 17 Jan 2013
Thank You !!!
nanmean(abs(bsxfun(@minus, A, nm)),2) | 399 | 1,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2015-40 | latest | en | 0.806865 |
https://course.dgist.ac.kr/index.php/Poincare_Sphere | 1,611,186,041,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522133.33/warc/CC-MAIN-20210120213234-20210121003234-00382.warc.gz | 280,727,418 | 11,247 | # Poincare Sphere
Poincare Sphere
관련코스 현대광학
소분류 수학, 물리
선행 키워드 Jones Matrix
연관 키워드
We can find $\psi$ and $\chi$ that satisfies
$S_0 = I$
$S_1 = Ip \cos2\psi\cos2\chi$
$S_2 = Ip\sin2\psi\cos2\chi$
$S_3 = Ip\sin2\chi$
Then $(S_1,S_2,S_3)$ can be regarded as a point on a sphere with radius $Ip$.
A good animated visualization of a Poincare Sphere can be found in [1] | 153 | 369 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2021-04 | latest | en | 0.399056 |
http://razhayesheitanparastan.com/uc95an/jbwckao2bb43zagr/ | 1,575,832,307,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540514475.44/warc/CC-MAIN-20191208174645-20191208202645-00558.warc.gz | 122,640,239 | 9,831 | # Ethos Pathos And Logos Persuasive Advertising Techniques 2019 Worksheet
Friday, August 30th 2019. | Multiplication Worksheets
## Continents And Oceans Worksheet Cut And Paste
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There's of multiplication games available . Demonstrate that multiplication is an instant means of adding groups. We've got a lot of unique to try. Additional Multiplication Tasks Choose another educational and enjoyable Activity to continue to your children busy. A Orientation may be only an problem of . The ordered drill System is in a manner where the educational child sees the Structure of multiplication table ahead of the student without any any Provided responses. It is to keep in mind identity house Of multiplication if demonstrated with a variety of examples.
## Cells Alive Cell Cycle Worksheet Answer Key
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###### Graphing Acceleration Worksheet
If students finish early, need to to Yet another problem. Before they could learn about multiplication sentences, they must the idea of a variety. Researching the subsequent few days students are vulnerable to a collection of different types of narrative troubles. Once they've the notion of multiplication through arrays, a very simple word problem can be used to anchor the concept. They require time and assistance to for themselves the relationship of equal numbers of units in each row a specific quantity of rows. The students create similar kinds of with images and also the problems to one another. They what they know about , the things they will need to know, then what they . | 479 | 2,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-51 | latest | en | 0.875143 |
https://communities.sas.com/t5/SAS-Statistical-Procedures/X-axis-tick-marks-in-lifetest/td-p/235162?nobounce | 1,516,566,460,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890874.84/warc/CC-MAIN-20180121195145-20180121215145-00417.warc.gz | 617,551,340 | 27,157 | ## X axis tick marks in lifetest.
Occasional Contributor
Posts: 6
# X axis tick marks in lifetest.
I'm capturing the output of proc lifetest through ods. I need to customize the tick interval on a survival plot for the x axis such that the ticks have values of 0 to 18 by intervals of 1.5.
The lifetest procedure code is:
/*** Compute Kaplan Meier Statistics ***/
ods rtf file = "(directory path)\figures\f4_1.rtf"
style=rtf;
ods graphics on;
proc lifetest data = km1
timelist=0 1.5 3 4.5 6 7.5 9 10.5 12 13.5 15 16.5 18
plots = survival(atrisk=0 1.5 3 4.5 6 7.5 9 10.5 12 13.5 15 16.5 18)
;
time months * event(0);
id usubjid;
strata TRT01PN;
title "Kaplan-Meier plot of PFS based on central review - Figure 4.1";
run;
ods graphics off;
ods listing;
run;
The at-risk table reports at the proper intervals but the ticks on the plot have intervals of 5 instead of 1.5. How do I make the ticks occur at 1.5 intervals?
Posts: 1,137
## Re: X axis tick marks in lifetest.
Try the intervals option in proc lifetest
ods rtf file = "(directory path)\figures\f4_1.rtf"
style=rtf;
ods graphics on;
proc lifetest data = km1 intervals=(0 to 18 by 1.5)
timelist=0 1.5 3 4.5 6 7.5 9 10.5 12 13.5 15 16.5 18
plots = survival(atrisk=0 1.5 3 4.5 6 7.5 9 10.5 12 13.5 15 16.5 18)
;
time months * event(0);
id usubjid;
strata TRT01PN;
title "Kaplan-Meier plot of PFS based on central review - Figure 4.1";
run;
ods graphics off;
ods listing;
run;
Thanks,
Jag
Occasional Contributor
Posts: 6
## Re: X axis tick marks in lifetest.
Thank you for responding. I tried your suggestion but the x axis still shows intervals of 5 instead of 1.5.
Posts: 1,137
## Re: X axis tick marks in lifetest.
it is strange i thought it will work. Could you please comment the timelist and try once.
Thanks,
Jag
Occasional Contributor
Posts: 6 | 622 | 1,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-05 | latest | en | 0.723813 |
https://astarmathsandphysics.com/university-maths-notes/number-theory/5215-numbers-which-are-both-squares-and-cubes-must-be-of-form-7k-7k-1.html | 1,718,700,214,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00082.warc.gz | 88,444,503 | 11,364 | ## Numbers Which Are Both Squares and Cubes Must be of Form 7k, 7k+1
Suppose a number
$x$
is both a square and a cube. Then we can write
$x=m^2=n^3$
. Only numbers of certain forms can be both a square and a cube. The table shows such numbers modulo 7.
$m, \; n$ $m^2$ $n^3$ 0 $0 \equiv 0 \; (mod \; 7)$ $0 \equiv 0 \; (mod \; 7)$ 1 $1 \equiv 1 \; (mod \; 7)$ $1 \equiv 1 \; (mod \; 7)$ 2 $4 \equiv 4 \; (mod \; 7)$ $8 \equiv 1 \; (mod \; 7)$ 3 $9 \equiv 2 \; (mod \; 7)$ $27 \equiv 6 \; (mod \; 7)$ 4 $16 \equiv 2 \; (mod \; 7)$ $64 \equiv 1 \; (mod \; 7)$ 5 $25 \equiv 4 \; (mod \; 7)$ $125 \equiv 6 \; (mod \; 7)$ 6 $36 \equiv 1 \; (mod \; 7)$ $216 \equiv 6 \; (mod \; 7)$ 7 $49 \equiv 0 \; (mod \; 7)$ $343 \equiv 0 \; (mod \; 7)$ 8 $64 \equiv 1 \; (mod \; 7)$ $512 \equiv 1 \; (mod \; 7)$ 9 $81 \equiv 4 \; (mod \; 7)$ $729 \equiv 1 \; (mod \; 7)$
From the table, only numbers which have remainder 0 or 1 on division by 7 can be both a square and a cube, so such numbers must be of the form
$7k$
or
$7k_1$
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# Earliest Known Uses of Some of the Words of Mathematics
BANACH SPACE. According to an Internet post, Banach wrote (in French) "spaces of type (B)". It has been suggested that he may have thus invited the subsequent term Banach space.
The term Banach space was coined by Maurice Fréchet (1878-1973), according to the University of St. Andrews website.
Banach space is found in National Mathematics Magazine in November 1945 in the title "Integrations of Functions in a Banach Space" by M. S. Macphail.
BANACH-TARSKI is found in Waclaw Sierpinski, "Sur le paradoxe de MM. Banach et Tarski," Fundamenta Mathematicae 33, pp 229-234 (1945) [James A. Landau].
BAR CHART occurs in Nov. 1914 in W. C. Brinton, "Graphic Methods for Presenting Data. IV. Time Charts," Engineering Magazine, 48, 229-241 (David, 1998).
The form of diagram, however, is much older; there is an example from William Playfair's Commercial and Political Atlas of 1786.
BAR GRAPH is dated 1924 in MWCD10.
Bar graph is found in 1925 in Statistics by B. F. Young: "Bar-graphs in the form of progress charts are used to represent a changing condition such as the output of a factory" (OED2).
The term BARYCENTRIC CALCULUS appears in 1827 in the title Der barycentrische calkul by August Ferdinand Möbius (1790-1868).
BASE (of a geometric figure) appears in English in 1570 in Sir Henry Billingsley's translation of Euclid's Elements (OED2).
BASE (in an isosceles triangle) is found in English in 1571 in Digges, Pantom.: "Isoscheles is such a Triangle as hath onely two sides like, the thirde being vnequall, and that is the Base" (OED2).
BASE (in logarithms) appears in Traité élémentaire de calcul différentiel et de calcul intégral (1797-1800) by Lacroix: "Et si a désigne la base du système, il en résulte l'équation y = ax, dans laquelle les logarithmes sont les abscisses."
Base is found in the 1828 Webster dictionary, in the definition of radix: "2. In logarithms, the base of any system of logarithms, or that number whose logarithm is unity."
BASE (of a number system). Radix was used in the sense of a base of a number system in 1811 in An Elementary Investigation of the Theory of Numbers by Peter Barlow [James A. Landau].
Base is found in the Century Dictionary (1889-1897): "The base of a system of arithmetical notation is a number the multiples of whose powers are added together to express any number; thus, 10 is the base of the decimal system of arithmetic."
BASE ANGLE is found in 1848 in "On the Formation of the Central Spot of Newton's Rings Beyond the Critical Angle" by Sir George Gabriel Stokes in the Transactions of the Cambridge Philosophical Society [University of Michigan Historic Math Collection].
BASIS (of a vector space). The term basis-system was used by Frobenius and Stickelberger in 1878 in Crelle, according to Moore (1896) [James A. Landau].
BAYES ESTIMATE, BAYES SOLUTION in statistical decision theory. Wald ("Contributions to the Theory of Statistical Estimation and Testing Hypotheses," Annals of Mathematical Statistics, 10, (1939), 299-326) originally used the term "minimum risk estimate" for what Hodges & Lehmann called a Bayes estimate ("Some Problems in Minimax Point Estimation," Annals of Mathematical Statistics, 21, (1950), 182-197.) Wald had used the term "Bayes solution" (in a more general setting) in his "An Essentially Complete Class of Admissible Decision Functions" Annals of Mathematical Statistics, 18, (1947), 549-555. Hodges & Lehmann (Annals of Mathematical Statistics, 19, 396-407) used the term BAYES RISK for a concept Wald had treated in 1939 without naming it [John Aldrich, based on David (2001)].
BAYES'S RULE is found in 1863 in An outline of the necessary laws of thought: a treatise on pure and applied logic by William Thomson: "The probability that there exist a cause of the reproduction of any event observed several times in succession is expressed by a fraction which has for its denominator the number 2 multiplied by itself as many times as the event has been observed, and for its numerator the same product minus one. This has been called Bayes's rule, and its validity is not so generally admitted as that of the preceding ones" [University of Michigan Historic Math Collection].
BAYES'S THEOREM. Règle de Bayes appears in 1843 in Exposition de la Théorie des Chances et des Probabilités by A. A. Cournot (David, 1998).
Bayes's Theorem appears in English in 1865 in A History of the Mathematical Theory of Probability by Isaac Todhunter (David, 1995).
BAYESIAN is found in 1950 in Contributions to Mathematical Statistics by R. A. Fisher. Fisher, a critic of the Bayesian approach, was distinguishing the probabilities used in the Bayesian argument from those generated by his own fiducial argument. The old names for the Bayesian argument, the "method of inverse probability" or "inverse method," have now disappeared, a change from the time De Morgan (An Essay on Probabilities (1838), p. vii) wrote "This [inverse] method was first used by the Rev. T. Bayes ... [who], though almost forgotten, deserves the most honourable remembrance from all who treat the history of this science" [John Aldrich, using David, 1998].
BELL-SHAPED. Bell-shaped parabola appears in 1857 in Mathematical Dictionary and Cyclopedia of Mathematical Science. The equation is ay2 - x2 + bx2 = 0.
Bell-shaped parabola appears in an 1860 translation of a Latin work of Isaac Newton, Sir Isaac Newton's Enumeration of lines of the third order, generation of curves by shadows, organic description of curves, and construction of equations by curves [University of Michigan Historic Math Collection].
Bell-shaped curve is found in 1876 in Catalogue of the Special Loan Collection of Scientific Apparatus at the South Kensington Museum by Francis Galton (David, 1998).
J. V. Uspensky, in Introduction to Mathematical Probability (1937), writes that "the probability curve has a bell-shaped form" [James A. Landau].
BELL CURVE is dated ca. 1941 in MWCD10.
BERNOULLI NUMBERS. According to Cajori (vol. 2, page 42), Leonhard Euler introduced the name "Bernoullian numbers."
The term appears in 1769 in the title "De summis serierum numeros Bernoullianos involventium" by Leonhard Euler.
According to the University of St. Andrews website, in its article on Johann Faulhaber, the Bernoulli numbers were "so named by [Abraham] de Moivre" (1667-1754).
BERNOULLI TRIAL is dated 1951 in MWCD10, although James A. Landau has found the phrases "Bernoullian trials" and "Bernoullian series of trials" in 1937 in Introduction to Mathematical Probability by J. V. Upensky.
BESSEL FUNCTION. Franceschetti (p. 56) implies that this term was introduced by Oskar Xavier Schlömilch in 1854.
Bessel'schen Functionen appears in 1868 in the title Studien über die Bessel'schen Functionen by Eugen Lommel.
Philosophical Magazine in 1872 has "The value of Bessel's functions is becoming generally recognized" (OED2).
Bessel function appears in 1894 in Ann. Math. IX. 27 in the heading "Roots of the Second Bessel Function" (OED2).
BETA DISTRIBUTION. Distribuzione [beta] is found in 1911 in C. Gini, "Considerazioni Sulle Probabilità Posteriori e Applicazioni al Rapporto dei Sessi Nelle Nascite Umane," Studi Economico-Giuridici della Università de Cagliari, Anno III, 5-41 (David, 1998).
The term BETTI NUMBER was coined by Henri Poincaré (1854-1912) and named for Enrico Betti (1823-1892), according to a history note by Victor Katz in A First Course in Abstract Algebra by John B. Fraleigh.
BETWEENNESS. The earliest citation in the OED2 for this word is in 1892 Monist II. 243: "In reality there are not two things and, in addition to them a betweenness of the two things."
Betweenness appears in G. B. Halsted, "The betweenness assumptions," Amer. Math. Monthly 9, 98-101.
The OED2 has a 1904 citation which makes reference to "Hilbert's betweenness assumptions."
BEZOUTIANT was "used by Sylvester and later writers" (Cajori 1919, page 249).
BIASED and UNBIASED. Biased errors and unbiased errors (meaning "errors with zero expectation") are found in 1897 in A. L. Bowley, "Relations Between the Accuracy of an Average and That of Its Constituent Parts," Journal of the Royal Statistical Society, 60, 855-866 (David, 1995).
Biased sample is found in 1911 An Introduction to the theory of Statistics by G. U. Yule: "Any sample, taken in the way supposed, is likely to be definitely biassed, in the sense that it will not tend to include, even in the long run, equal proportions of the A’s and [alpha]'s in the original material" (OED2).
Biased sampling is found in F. Yates, "Some examples of biassed sampling," Ann. Eugen. 6 (1935) [James A. Landau].
The term BICURSAL was introduced by Cayley (Kline, page 938).
In 1873 Cayley wrote, "A curve of deficiency 1 may be termed bicursal."
BIJECTION and BIJECTIVE are dated 1966 in MWCD10.
BILLION. See million.
BIMODAL appears in 1903 in S. R. Williams, "Variation in Lithobius Forficatus," American Naturalist, 37, 299-312 (David, 1998).
BINARY ARITHMETIC appears in English in 1796 A Mathematical and Philosophical Dictionary (OED2).
BINOMIAL. According to the OED2, the Latin word binomius was in use in algebra in the 16th century.
Binomial first appears as a noun in English in its modern mathematical sense in 1557 in The Whetstone of Witte by Robert Recorde: "The nombers that be compound with + be called Bimedialles... If their partes be of 2 denominations, then thei named Binomialles properly. Howbeit many vse to call Binomialles all compounde nombers that have +" (OED2).
BINOMIAL COEFFICIENT. According to Kline (page 272), this term was introduced by Michael Stifel (1487-1567) about 1544. However, Julio González Cabillón believes this information is incorrect. He says Stifel could not have used the word coefficient, which is due to Vieta (1540-1603).
Binomial coefficient is found in Rottock, "Ueber Reihen mit Binomialcoefficienten und Potenzen," Pr. d. G. Rendsburg (1868).
Binomial coefficient is found in English in an 1868 paper by Arthur Cayley [University of Michigan Historical Math Collection].
BINOMIAL DISTRIBUTION is found in 1911 in An Introduction to the Theory of Statistics by G. U. Yule: "The binomial distribution,..only becomes approximately normal when n is large, and this limitation must be remembered in applying the table..to cases in which the distribution is strictly binomial" (OED2).
BINOMIAL THEOREM appears in 1742 in Treatise of Fluxions by Colin Maclaurin (Struik, page 339).
In Gilbert and Sullivan's The Pirates of Penzance (1879), the song "I Am The Very Model of a Modern Major-General" includes the lines:
I'm very well acquainted, too, with matters mathematical,
I understand equations, both the simple and quadratical,
About binomial theorem I'm teeming with a lot o' news,
With many cheerful facts about the square of the hypotenuse. [...]
I'm very good at integral and differential calculus;
I know the scientific names of beings animalculous:
BINORMAL. According to Howard Eves in A Survey of Geometry, vol II (1965), "The name binormal was introduced by B. de Saint-Venant in 1845" [James A. Landau].
The term BIOMATHEMATICS was coined by William Moses Feldman (1880-1939), according to Garry J. Tee in "William Moses Feldman: Historian of Rabbinical Mathematics and Astronomy." The term appears in Feldman's textbook Biomathematics published in 1923.
BIOSTATISTICS appears in Webster's New International Dictionary (1909).
BIPARTITE. In 1858, Cayley referred to "bipartite binary quantics."
BIPARTITE CURVE appears in 1879 in George Salmon (1819-1904), Higher Plane Curves (ed. 3): "We shall then call the curve we have been considering a bipartite curve, as consisting of two distinct continuous series of points" (OED2).
BIQUATERNION. Hamilton used the term biquaternion in the sense of a quaternion with complex coefficients.
In the more recent sense, William Kingdon Clifford (1845-1879) coined the term. It appears in 1873 in Proc. London Math. Soc. IV. 386.
BISECT. According to the OED2, bisect is apparently of English formation. The word is dated ca. 1645 in MWCD10.
Bisection appears in 1656 in a translation of Hobbes's Elem. Philos. (1839) 307: "By perpetual bisection of an angle" (OED2).
In 1660, Barrow's translation of Euclid's Elements has "To bisect a right line."
Bisector appears in English in 1864 in The Reader 5 Oct. 483/2: "The internal and external bisectors of the angle" (OED2).
BIT was coined by John W. Tukey (1915-2000).
According to Niels Ole Finnemann in Thought, Sign and Machine, Chapter 6, "After some more informal contacts during the first war years, on the initiative of mathematician Norbert Wiener, a number of scientists gathered in the winter of 1943-44 at a seminar, where Wiener himself tried out his ideas for describing intentional systems as based on feedback mechanisms. On the same occasion J. W. Tukey introduced the term a 'bit' (binary digit) for the smallest informational unit, corresponding to the idea of a quantity of information as a quantity of yes-or-no answers."
Several Internet web pages say Tukey coined the term in 1946. Another web page says, "Tukey records that it evolved over a lunch table as a handier alternative to 'bigit' or 'binit.'"
Bit first appeared in print in July 1948 in "The Mathematical Theory of Communication" by Claude Elwood Shannon (1916-2001) in the Bell Systems Technical Journal. In the article, Shannon credited Tukey with the coinage [West Addison assisted with this entry.]
BIVARIATE is found in 1920 in Biometrika XIII. 37: "Thus in 1885 Galton had completed the theory of bi-variate normal correlation" (OED2).
BOOLEAN is found in 1851 in the Cambridge and Dublin Mathematical Journal vi. 192: "...the Hessian, or as it ought to be termed, the first Boolian Determinant" (OED2).
BOOLEAN ALGEBRA. Boolian algebra appears in the Century Dictionary (1889-1897):
Boolian algebra, a logical algebra, invented by the English mathematician George Boole (1815-64), for the solution of problems in ordinary logic. It has also a connection with the theory of probabilities.
According to E. V. Hutington in "New Sets of Independent Postulates for the Algebra of Logic with Special Reference to Whitehead and Russell's Principia Mathematica," Trans. Amer. Math. Soc. (1933), the term Boolean algebra was introduced by H. M. Sheffer in the paper "A Set of Five Independent Postulates for Boolean Algebras with Application to Logical Constants", Trans. Amer. Math. Soc., 14 (1913).
In an illuminating passage of "Algebraic Logic", Halmos writes (p. 11):
Terminological purists sometimes object to the Boolean use of the word "algebra". The objection is not really cogent. In the first place, the theory of Boolean algebras has not yet collided, and it is not likely to collide, with the theory of linear algebras. In the second place, a collision would not be catastrophic; a Boolean algebra is, after all, a linear algebra over the field of integers modulo 2. (...) While, to be sure, a shorter and more suggestive term than "Boolean algebra" might be desirable, the nomenclature is so thoroughly established that to change now would do more harm than good.
[Carlos César de Araújo]
BORROW is found in English in 1594 in Blundevil, Exerc.: "Take 6 out of nothing, which will not bee, wherefore you must borrow 60" (OED2).
In October 1947, "Provision for Individual Differences in High School Mathematics Courses" by William Lee in The Mathematics Teacher has: "The Social Mathematics course stresses understanding of arithmetic: 'carrying' in addition, 'regrouping' (not 'borrowing') in subtraction, 'indenting' in multiplication are analyzed and understood rather than remaining mere rote operations to be performed blindly."
BOYER'S LAW appears in H. C. Kennedy, "Boyer's Law: Mathematical formulas and theorems are usually not named after their original discoverers," Amer. Math. Monthly, 79:1 (1972), 66-67.
Boyer's theorem is found in 1968 in History of Mathematics by Barnabas Hughes.
The term BRACHISTOCHRONE was introduced by Johann Bernoulli (1667-1748). Smith (vol. 2, page 326) says the term is "due to the Bernoullis."
The terms BRA VECTOR and KET VECTOR were introduced by Paul Adrien Maurice Dirac (1902-1984). The terms appear in 1947 in Princ. Quantum Mech. by Dirac: "It is desirable to have a special name for describing the vectors which are connected with the states of a system in quantum mechanics, whether they are in a space of a finite or an infinite number of dimensions. We shall call them ket vectors, or simply kets, and denote a general one of them by a special symbol >|. ... We shall call the new vectors bra vectors, or simply bras, and denote a general one of them by the symbol <|, the mirror image of the symbol for a ket vector" (OED2).
BRIGGSIAN LOGARITHM. The phrase Briggs logarithm is found in the 1771 edition of the Encyclopaedia Britannica [James A. Landau].
BROKEN LINE is found in 1852 in Elements of the differential and integral calculus by Charles Davies: "But the arc POM can never be less than the chord PM, nor greater than the broken line PNM which contains it; hence, the limit of the ratio POM/PM = 1; and consequently, the differential of the arc is equal to the differential of the chord."
Broken line is found in 1852 in Legendre, A. M. (Adrien Marie): Elements of geometry and trigonometry, from the works of A. M. Legendre. Revised and adapted to the course of mathematical instruction in the United States, by Charles Davies" "5. A Straight Line is one which lies in the same direction between any two of its points. 6. A Broken Line is one made up of straight lines, not lying in the same direction."
Broken line is found in 1852 in Elements of plane trigonometry, with its application to mensuration of heights and distances, surveying and navigation by William Smyth: "Instead of a broken line, a field is sometimes bounded by a line irregularly curves, as by the margin of a brook, river, or lake. In this case (fig. 60) we run, as before, a chain line as near the boundary as possible, and by means of offsets determine a sufficient number of points in the curve to draw it." [These three citations were found using the University of Michigan Historic Math Collection.]
According to Schwartzman (page 38), the "broken line," meaning a curve composed of connected straight line segments, was adopted "around 1898" by David Hilbert (1862-1943).
BROWNIAN MOTION. In the course of the 20th century the physical phenomenon described by Brown in 1827 was described in mathematical terms and gradually "Brownian motion" came to refer as much to the mathematical formalism as to the phenomenon. Mathematical theories were developed by, inter alia, A. Einstein ("Zur Theorie der Brownschen Bewegung" (1905)). The "Brownian motion process" of J. L. Doob's Stochastic Processes (1954) is a type of stochastic process divested of physical application. Doob states that the process "was first discussed by Bachelier and later, more rigorously by Wiener. It is sometimes called the Wiener process." An earlier term in physics (and mathematics) was "Brownian movement." This slowly gave way to "Brownian motion," although David (2001) reports an early appearance of "Brownian motion" in 1892 in W. Ramsay's Report of a paper read to the Chemical Society, London. Nature, 45, 429/2. (See Wiener process.) [John Aldrich]
BRUN'S CONSTANT was coined by R. P. Brent in "Irregularities in the distribution of primes and twin primes," Math. Comp. 29 (1975), according to Algorithmic Number Theory by Bach and Shallit [Paul Pollack].
The term BYTE was coined in 1956 by Dr. Werner Buchholz of IBM. A question-and-answer session at an ACM conference on the history of programming languages included this exchange:
JOHN GOODENOUGH: You mentioned that the term "byte" is used in JOVIAL. Where did the term come from?
JULES SCHWARTZ (inventor of JOVIAL): As I recall, the AN/FSQ-31, a totally different computer than the 709, was byte oriented. I don't recall for sure, but I'm reasonably certain the description of that computer included the word "byte," and we used it.
FRED BROOKS: May I speak to that? Werner Buchholz coined the word as part of the definition of STRETCH, and the AN/FSQ-31 picked it up from STRETCH, but Werner is very definitely the author of that word.
SCHWARTZ: That's right. Thank you.
CALCULUS. In Latin calculus means "pebble." It is the diminutive of calx, meaning a piece of limestone.
In Latin, persons who did counting were called calculi. Teachers of calculation were known as calculones if slaves, but calculatores or numerarii if of good family (Smith vol. 2, page 166).
The Romans used calculos subducere for "to calculate."
In Late Latin calculare means "to calculate." This word is found in the works of the poet Aurelius Clemens Prudentius, who lived in Spain c. 400 (Smith vol. 2, page 166).
Calculus in English, defined as a system or method of calculating, is dated 1666 in MWCD10.
The earliest citation in the OED2 for calculus in the sense of a method of calculating, is in 1672 in Phil. Trans. VII. 4017: "I cannot yet reduce my Observations to a calculus."
The restricted meaning of calculus, meaning differential and integral calculus, is due to Leibniz.
A use by Leibniz of the term appears in the title of a manuscript Elementa Calculi Novi pro differentiis et summis, tangentibus et quadraturis, maximis et minimis, dimensionibus linearum, superficierum, solidorum, allisque communem calculum transcendentibus [The Elements of a New Calculus for Differences and Sums, Tangents and Quadratures, maxima and minima, the measurement of lines, surfaces and solids, and other things which transcend the usual sort of calculus]. The manuscript is undated, but appears to have been compiled sometime prior to 1680 (Scott, page 157).
Newton did not originally use the term, preferring method of fluxions (Maor, p. 75). He used the term Calculus differentialis in a memorandum written in 1691 which can be found in The Collected Correspondence of Isaac Newton III page 191 [James A. Landau].
Webster's dictionary of 1828 has the following definitions for calculus, suggesting the older meaning of simply "a method of calculating" was already obsolete:
1. Stony; gritty; hard like stone; as a calculous concretion.
2. In mathematics; Differential calculus, is the arithmetic of the infinitely small differences of variable quantities; the method of differencing quantities, or of finding an infinitely small quantity, which, being taken infinite times, shall be equal to a given quantity. This coincides with the doctrine of fluxions.
3. Exponential calculus, is a method of differencing exponential quantities; or of finding and summing up the differentials or moments of exponential quantities; or at least of bringing them to geometrical constructions.
4. Integral calculus, is a method of integrating or summing u moments or differential quantities; the inverse of the differential calculus.
5. Literal calculus, is specious arithmetic or algebra.
The 1890 Funk & Wagnalls Standard Dictionary has: "While calculus is sometimes used in this wide sense, it is commonly used, when without a qualifying word, for the infinitesimal calculus, and includes differential calculus and integral calculus."
The use of calculus without the definite article has become common only in the twentieth century. Some early titles in which "the" appears not to occur are Robinson's Differential and Integral Calculus for High Schools and Colleges (1868), Treatise on Infinitesimal Calculus by Price (1869), Differential Calculus with Numerous Examples by B. Williamson (1872), Calculus of Finite Differences by G. Boole (1872), Integral Calculus by W. E. Byerly (1898), The discovery of Calculus by A. C. Hathaway (1919).
The term CALCULUS OF DERIVATIONS was coined by Arbogast, according to the Mathematical Dictionary and Cyclopedia of Mathematical Science.
CALCULUS OF FINITE DIFFERENCES. An earlier term, method of increments, appears in 1715 in the title Methodus Incrementorum by Brook Taylor.
Method of increments appears in English in 1763 in the title The Method of Increments by W. Emerson.
The phrase finite difference appears in 1807 in the title An Investigation of the General Term of an important Series in the Inverse Method of Finite Differences by J. Brinkley.
Finite difference also appears in Sir John Frederick William Herschel, "On the development of exponential functions, together with several new theorems relating to finite differences," Trans. Phil. Soc., (1814), 440-468; (1816), 25-45.
Calculus of finite differences is found in 1820 in the title A Collection of Examples of the Applications of the Calculus of Finite Differences by Sir John Frederick William Herschel (1792-1871).
The term CALCULUS OF VARIATIONS was introduced by Leonhard Euler in a paper, "Elementa Calculi Variationum," presented to the Berlin Academy in 1756 and published in 1766 (Kline, page 583; DSB; Cajori 1919, page 251). Lagrange used the term method of variations in a letter to Euler in August 1755 (Kline).
Calculus of variations is found in English in 1810 in A Treatise on Isoperimetrical Problems, and the Calculus of Variations by Robert Woodhouse [James A. Landau].
The term CANONICAL FORM is due to Hermite (Smith, 1906).
Canonical form is found in 1851 in the title "Sketch of a Memoir on Elimination, Transformation, and Canonical Forms," by James Joseph Sylvester (1814-1897), Cambridge and Dublin Mathematical Journal 6 (1851).
CARDINAL. Glareanus recognized the metaphor between cardinal numbers and Cardinal, a prince of the church, writing in Latin in 1538.
The earliest citation in the OED2 is by Richard Percival in 1591 in Bibliotheca Hispanica: "The numerals are either Cardinall, that is, principall, vpon which the rest depend, etc."
CARDIOID was first used by Johann Castillon (Giovanni Francesco Melchior Salvemini) (1708-1791) in "De curva cardiode" in the Philosophical Transactions of the Royal Society (1741) [Julio González Cabillón and DSB].
CARMICHAEL NUMBER appears in H. J. A. Duparc, "On Carmichael numbers," Simon Stevin 29, 21-24 (1952).
CARRY (process used in addition). According to Smith (vol. 2, page 93), the "popularity of the word 'carry' in English is largely due to Hodder (3d ed., 1664)."
CARTESIAN COORDINATES. Hamilton used Cartesian method of coordinates in a paper of 1844 [James A. Landau].
Cartesian co-ordinates appears in 1885 in S. Newcomb, Elem. Analytic Geom. in the heading "Cartesian or bilinear co-ordinates" (OED2).
CARTESIAN GEOMETRY was used by Jean Bernoulli "as early as 1692," according to Boyer (page 484).
CARTESIAN PLANE appears in April 1956 in "Graphing in Elementary Algebra" by Max Beberman and Bruce E. Meserve in The Mathematics Teacher: "These axes are usually taken with a common origin, with the first co-ordinate referring to a horizontal axis having its unit point on the right of the origin, and with the second co-ordinate referring to a vertical axis having its unit point above the origin. We call such a co-ordinate plane a Cartesian plane."
CARTESIAN PRODUCT is found in Albert A. Bennett, "Concerning the function concept," The Mathematics Teacher, May 1956: "If A, B are sets, by "A X B" (called the "Cartesian product of A by B") is meant "the set of all ordered pairs (a, b), where a is an element of A, and b of B."
CASTING OUT NINES. Fibonacci called the excess of nines the pensa or portio of the number (Smith vol. 1, page 153).
Pacioli (1494) spoke of it as "corrente mercatoria e presta" (Smith vol. 1, page 153).
"Casting out the nines" is found in the first edition of the Encyclopaedia Britannica (1768-1771) in the article, "Arithmetick."
CATALECTICANT. Catalectic is found in English as early as 1589, describing verse, and meaning "lacking a syllable at the end or ending in an incomplete foot."
The OED2 shows a use of catalectic by James Joseph Sylvester in 1851 in the "The theory of the catalectic forms of functions of the higher degrees of two variables."
Catalecticant was coined by James Joseph Sylvester, who wrote:
Meicatalecticizant would more completely express the meaning of that which, for the sake of brevity, I denominate the catalecticant.
The quotation appears in "On the principles of the calculus of forms," Cambridge and Dublin Mathematical Journal 7 (1852), pp. 52-97, reprinted in Vol. 1 of Sylvester's Collected Papers as Paper 42, pp. 284-327. The quotation appears as a footnote to p. 293.
Bruce Reznick, who provided this quotation, writes, "Sylvester may appear a little pompous to us, but there is a reason for his language: a 'catalectic' verse is one in which the last line is missing a foot. A general homogeneous polynomial p(x,y) of degree 2k can be written as a sum of k+1 linear polynomials raised to the 2k-th power . . . unless its catalecticant vanishes, in which case it needs k linear polynomials, or fewer."
Meicatalecticizant probably did not appear anywhere in print again until Reznick used it in his monograph "Sums of even powers of real linear forms," which appeared as a Memoir of the American Mathematical Society, No. 463 in 1992.
In a letter to Thomas Archer Hirst dated Dec. 19, 1862, Sylvester wrote, "On further relfexion I retract my opinion expressed yesterday evening and reocmmend the continuance [illegible] of the word 'Catalecticant.' This sort of invariant is so important and stands in such close relation to the Canonizant that we cannot afford to let it go unnamed and as this name has been used by Cayley as well as myself it may as well remain. ... I took the Idea of the name from the Iambicus Trimeter Catalecticus."
CATASTROPHE THEORY is found in Thomas F. Banchoff, "Polyhedral catastrophe theory. I: Maps of the line to the line," Dynamical Syst., Proc. Sympos. Univ. Bahia, Salvador 1971, 7-21 (1973).
CATEGORICAL (AXIOM SYSTEM). This term was suggested by John Dewey (1859-1952) to Oswald Veblen (1880-1960) and introduced by the latter in his A system of axioms for geometry, Trans. Amer. Math. Soc. 5 (1904), 343-384, p. 346. Since then, the term as well as the notion itself has been attributed to Veblen. Nonetheless, the first proof of categoricity is due to Dedekind: in his Was sind und Was sollen die Zahlen? (1887) it was in fact proved that the now universally called "Peano axioms" are categorical - any two models (or "realizations") of them are isomorphic. In Dedekind's words:
132. Theorem. All simply infinite systems are similar to the number-series N and consequently (...) to one another.
(Strictly speaking, the categoricity in itself is not seem in this statement but in its proof.)
Instead of "categorical", the term "complete" is sometimes used, chiefly in older texts. The influence, in this case, comes from Hilbert's Vollständigkeitsaxiom ("completeness axiom") in his Über den Zahlbegriff (1900). Other names that were proposed for this concept are "monomorphic" (for categorical and consistent in Carnap's Introduction to symbolic logic, 1954) and "univalent" (Bourbaki), but these did not attain popularity. (It goes without saying that there is no connection with "Baire category", "category theory" etc.) The concept was somewhat shaken when Thoralf Skolem discovered (1922) that first-order set theory is not categorical. Facts like this have caused some confusion among mathematicians. Thus in his The Loss of Certainty (1980, p. 271) Morris Kline wrote:
Older texts did "prove" that the basic systems were categorical; (...) But the "proofs" were loose (...) No set of axioms is categorical, despite "proofs" by Hilbert and others.
This remark was corrected by C. Smorynski in an acrimonious review:
The fact is, there are two distinct notions of axiomatics and, with respect to one, the older texts did prove categoricity and not merely "prove".
[This entry was contributed by Carlos César de Araújo.]
CATENARY. According to E. H. Lockwood (1961) and the University of St. Andrews website, this term was first used (in Latin as catenaria) by Christiaan Huygens (1629-1695) in a letter to Leibniz in 1690.
According to Schwartzman (page 41) and Smith (vol. 2, page 327), the term was coined by Leibniz.
Maor (p. 142) shows a drawing by Leibniz dated 1690 which Leibniz labeled "G. G. L. de Linea Catenaria."
Huygens wrote "Solutio problematis de linea catenaria" in the Acta Eruditorum in 1691.
In 1727-41, Ephraim Chambers' Cyclopedia or Universal Dictionary of Arts and Sciences uses the Latin form catenaria in the article on the tractrix (OED2).
The OED shows a use of catenarian curve in English in 1751.
The 1771 edition of the Encyclopaedia Britannica uses the Latin form catenaria:
CATENARIA, in the higher geometry, the name of a curve line formed by a rope hanging freely from two points of suspension, whether the points be horizontal or not. See FLUXIONS.
In a letter to Thomas Jefferson dated Sept. 15, 1788, Thomas Paine, discussing the design of a bridge, used the term catenarian arch:
Whether I shall set off a catenarian Arch or an Arch of a Circle I have not yet determined, but I mean to set off both and take my choice. There is one objection against a Catenarian Arch, which is, that the Iron tubes being all cast in one form will not exactly fit every part of it. An Arch of a Circle may be sett off to any extent by calculating the Ordinates, at equal distances on the diameter. In this case, the Radius will always be the Hypothenuse, the portion of the diameter be the Base, and the Ordinate the perpendicular or the Ordinate may be found by Trigonometry in which the Base, the Hypothenuse and right angle will be always given.
In a reply to Paine dated Dec. 23, 1788, Thomas Jefferson used the word catenary:
You hesitate between the catenary, and portion of a circle. I have lately received from Italy a treatise on the equilibrium of arches by the Abbé Mascheroni. It appears to be a very scientifical work. I have not yet had time to engage in it, but I find that the conclusions of his demonstrations are that 'every part of the Catenary is in perfect equilibrium.'
The earliest citation for catenary in the OED2 is from the above letter.
CATHETUS. Nicolas Chuquet (d. around 1500), writing in French, used the word cathète (DSB).
Cathetus occurs in English in 1571 in A Geometricall Practise named Pantometria by Thomas Digges (1546?-1595) (although it is spelled Kathetus).
Cathetus is found in English in the Appendix to the 1618 edition of Edward Wright's translation of Napier's Descriptio. The writer of the Appendix is anonymous, but may have been Oughtred.
CAUCHY-SCHWARTZ INEQUALITY. Caucy-Schwarz inequality, Schwarz's inequality, and Schwarz's inequality for integrals appear in 1937 in Differential and Integral Calculus, 2nd. ed. by R. Courant [James A. Landau].
CAUCHY CONVERGENCE TEST. Cauchy's integral test is found in 1893 in A Treatise on the Theory of Functions by James Harkness and Frank Morley: "Cauchy's integral test for the convergence of simple series can be extended to double series."
Cauchy's convergence test and Cauchy test appear in 1937 in Differential and Integral Calculus, 2nd. ed. by R. Courant. Courant writes that the test is also called the general principle of convergence [James A. Landau].
The term CAUCHY SEQUENCE was defined by Maurice Fréchet (1878-1973) (Katz). The term is dated ca. 1949 in MWCD10.
CAUCHY'S THEOREM appears in 1868 in Genocchi, "Intorno ad un teorema di Cauchy," Brioschi Ann.
The term also appears in the title "Sur un théorème de Cauchy présenté par M. Hermite" (1868).
Cauchy's theorem appears in the third edition of An Elementary Treatise on the Theory of Equations (1875) by Isaac Todhunter.
CAYLEY'S SEXTIC was named by R. C. Archibald, "who attempted to classify curves in a paper published in Strasbourg in 1900," according to the St. Andrews University website.
CAYLEY'S THEOREM is found in J. W. L. Glaisher, "Note on Cayley's theorem," Messenger of Mathematics (1878).
Cayley's theorem, referring to a theorem given by Cayley in 1843, appears in 1897 in Abel's Theorem and the Allied Theory Including the Theory of the Theta Functions by H. F. Baker (1897).
The term Cayley's theorem (every group is isomorphic to some permutation group) was apparently introduced in 1916 by G. A. Miller. He wrote Part I of the book Theory and Applications of Finite Groups by Miller, Blichfeldt and Dickson. He liked the idea of listing the most important theorems, with names, so when this theorem had no name he introduced one. His footnote on p. 64 says:
This theorem is fundamental, as it reduces the study of abstract groups uniquely to that of regular substitution groups. The rectangular array by means of which it was proved is often called Cayley's Table, and it was used by Cayley in his first article on group theory, Philosophical Magazine, vol. 7 (1854), p. 49. The theorem may be called Cayley's Theorem, and it might reasonably be regarded as third in order of importance, being preceded only by the theorems of Lagrange and Sylow.
[Contributed by Ken Pledger]
The terms CEILING FUNCTION and FLOOR FUNCTION were coined by Kenneth E. Iverson, according to Integer Functions by Graham, Knuth, and Patashnik.
CENTRAL ANGLE is found in 1851 in The field practice of laying out circular curves for rail-roads by John Cresson Trautwine: "The deflexion angle of any curve is equal to the angle t c u, or t c s, ^c., at the centre of the circle, subtended by one of the equal chords t u, or t s. This angle at the centre, so subtended, is called the central angle. The tangential angle being always half the deflexion angle, is, of course, always half the central angle" [University of Michigan Digital Library].
CENTRAL LIMIT THEOREM. In 1919 R. von Mises called the limit theorems Fundamentalsätze der Wahrscheinlichkeitsrechnung in a paper of the same name in Math Z. 4, 1-97.
Central limit theorem appears in the title "Ueber den zentralen Grenzwertsatz der Wahrscheinlichkeitsrechnung," Math. Z., 15 (1920) by George Polya (1887-1985) [James A. Landau]. Polya apparently coined the term in this paper.
Central limit theorem appears in English in 1937 in Random Variables and Probability Distributions by H. Cramér (David, 1995).
CENTRAL TENDENCY is dated ca. 1928 in MWCD10.
Central tendency is found in 1929 in Kelley & Shen in C. Murchison, Found. Exper. Psychol. 838: "Some investigators have often preferred the median to the mean as a measure of central tendency" (OED2).
CENTROID is found in 1882 in Minchin, Unipl. Kinemat.: "To find..the position of the Centroid ('centre of gravity') of any plane area" (OED2).
The term CEPSTRUM was introduced by Bogert, Healey, and Tukey in a 1963 paper, "The Quefrency Analysis of Time Series for Echoes: Cepstrum, Pseudoautocovariance, Cross-Cepstrum, and Saphe Cracking." The word was created by interchanging the letters in the word "spectrum."
CEVIAN was proposed in French as cévienne in 1888 by Professor A. Poulain (Faculté catholique d'Angers, France). The word honors the Italian mathematician Giovanni Ceva (1647?-1734) [Julio González Cabillón].
An early use of the word in English is by Nathan Altshiller Court in the title "On the Cevians of a Triangle" in Mathematics Magazine 18 (1943) 3-6.
CHAIN. In his ahead-of-time Was sind und Was sollen die Zahlen? (1887), Richard Dedekind introduced the term chain (kette) with two related senses. Improving on his notation and style somewhat, let us take a function f : S ® S. According to him (§37), a "system" (his name for "set") K Ì S is a chain (under f) when f (K ) Ì K. (Incidentally, from such a "chain" one really gets a descending chain -in one of the more modern uses of this word -, namely, ...Ì f 3(K) Ì f 2(K) Ì f 1(K) Ì K.) Soon after (§44), he fixes A Ì S and defines the "chain of the system A" (under f ) as the intersection of all chains (under f ) K Ì S such that A Ì K. This formulation sounds familiar today, but in Dedekind's time it was a breakthrough! Now, it is easy to see (and he did it in §131) that the "chain of A" (under f ) is simply the union of iterated images A È f 1(A) È f 2(A) È f 3(A) È ..., a result which would yield a simpler definition. But what are the numbers 1, 2, 3, ...? This was precisely the question he intended to answer once and for all through his concept of chain! Gottlob Frege (in his Begriffsschrift, 1879) had similar ideas but his notation was strange and his terminology repulsively philosophic.
Dedekind's "theory of chains" would come to be quoted or used in many places: in proofs of the "Cantor-Bernstein" theorem (Dedekind-Peano-Zermelo-Whittaker), in Keyser's "axiom of infinity" (Bull. A. M. S., 1903, p. 424-433), in Zermelo's second proof of the well-ordering theorem (through his "q -chains", 1908) and in Skolem's first proof of Löwenheim theorem (1920) - to name only a few. All that said, it is simply wrong to say that "Dedekind's approach was so complicated that it was not accorded much attention." (Kline, Mathematical Thought from Ancient to Modern Times, p. 988.) Quite the contrary: the term "chain" in that sense did not survive, but the concept paved the way for the more general notion of closure (hull, span) of a set under an entire structure. [This article contributed by Carlos César de Araújo.]
CHAIN RULE. This term originally referred to a rule for calculating an equivalence in different units of measure when an intermediate unit of measure was involved.
In early Dutch books, it is called the chain rule, Den Kettingh-Regel and Den Ketting Reegel (Smith vol. 2, page 573).
Other names in various Dutch and Dutch-French books of the 17th and 18th century are Regula conjuncta, Regel conjoinct, Te Zamengevoegden Regel, Regel van Vergelykinge, and De Gemenghde Regel (Smith vol. 2, page 573).
Chain rule is found in English in 1842 in the title The Chain Rule; a Commercial Arithmetic by Charles L. Schönberg. [This title is listed in the 1850 Catalogue of the Mercantile library in New York, which was viewed at the University of Michigan Digital Library].
Kettenregel is found in 1877 in W. Simerka, "Die Kettenregel bei Congruenzen," Casopis.
In German, R. Just in Kaufmännisches Rechnen, I (1901) has "Gleichsam wie die Glieder einer 'Kette'" (Smith vol. 2, page 573).
In Differential and Integral Calculus (1902) by Virgil Snyder and J. I. Hutchinson, the calculus rule is shown but is not named.
In 1909, a Webster dictionary says the rule (in arithmetic) is also called Rees's rule, "for K. F. de Rees, its inventor."
In 1912 in Advanced Calculus by Edwin Bidwell Wilson, the calculus rule is referred to as "the rule for differentiating a function of a function."
Peter Flor has found Kettenregel in Höhere Mathematik (1921) by Hermann Rothe, where it is used in the calculus sense slightly differing from the present use, viz. only for composites of three or more functions. Flor writes, "Here the word 'chain' ('Kette', in German) is suggestive. I tried, rather perfunctorily, to pursue the term further back in time, without success. It seems that around 1910, most authors of textbooks as yet saw no problem in computing dz/dx = (dz/dy)*(dy/dx). On the other hand, when I was a student in Vienna and Hamburg (1953 and later), the word Kettenregel was a well-established part of elementary mathematical terminology, in German, for the rule on differentiating a composite of two functions. I guess that its use must have become general around 1930."
In 1922 in Introduction to the Calculus by William F. Osgood, the rule in calculus is not named.
Chain rule occurs in English in the calculus sense in 1937 in the Second English Edition of R. Courant, Differential and Integral Calculus, translated by E. J. McShane. Presumably the term appears in the German original, as well as in the 1st English edition of 1934.
Kettenregel appears in Differential und Integralrechnung by v. Mangoldt and Knopp in 1938 but is used only for composites of three or more functions.
Also in 1938, another classic appeared, the textbook of analysis by Haupt and Aumann, in which Kettenregel is used for the rule for the derivative of any composite function, exactly as we do now [Peter Flor].
Charles Hyman, ed., German-English Mathematical Dictionary, New York: Interlanguage Dictionaries Publishing Corp, 1960, has on page 59 the entry
kettenregel (f), kettensatz (m) [= English] chain rule
James A. Landau, who provided the last citation, suggests that "chain rule" is a German term which was at some point translated into English, possibly by Courant and McShane.
Chain rule appears with a different meaning in N. Chater and W. H. Chater, "A chain rule for use with determinants and permutations," Math. Gaz. 31, 279-287 (1947).
CHAOS appears in 1938 in Norbert Wiener, "The homogeneous chaos," Am. J. Math. 60, 897-936.
Chaos was coined as a mathematical term by James A. Yorke and Tien Yien Li in their classic paper "Period Three Implies Chaos" [American Mathematical Monthly, vol. 82, no. 10, pp. 985-992, 1975], in which they describe the behavior of some particular flows as chaotic [Julio González Cabillón].
It should be stressed that some mathematicians do not feel comfortable with the term "chaos". As an example we quote Paul Halmos in his Has Progress in Mathematics Slowed Down? (Am. Math. Monthly, 1990, p. 563):
Why the word "chaos" is used? The reason seems to be (...) a subjective (not really a mathematical) reaction to an unexpected appearance of discontinuity. A possible source of confusion is that the startling discontinuity can occur at two different parts of the theory. Frequently a dynamical system depends on some parameters (...), and, of course, (...) on the initial point. The startling change of the Hénon family (from periodic to strange attractor) is regarded as chaos - unpredictability - and the very existence of the Hénon strange attractor, not obviously visible in the definition of the dynamical system, is regarded as chaos - unpredictability. I would like to register a protest vote against the attitude that the terminology implies. The results of nontrivial mathematics are often startling, and when infinity is involved they are even more likely to be so. It's not easy to tell by looking at a transformation what its infinite iterates will do - but just because different inputs sometimes produce discontinuously outputs doesn't justify describing them as chaotic.
Probably having in mind such reservations, many prefer to use the term "deterministic chaos". That is to say, one is dealing with deterministic systems (such as a non-linear differential equation) which appear to behave in the long run in an unpredictable fashion. [Carlos César de Araújo]
CHARACTER (group character) appears in title of the paper "Uber die Gruppencharactere" by Ferdinand Georg Frobenius (1849-1917), which was presented to the Berlin Academy on July 16, 1896.
According to Shapir
However, in Dedekind's edition of Dirichlet's Vorlesungen ueber Zahlentheorie in 1894, Dedekind included a footnote in which he singled out the notion of "character," defined it explicitly, and denoted it by chi(n). [6.55]. However, he did not give the function a name. Weber's Lehrbuch der Algebra, II, 1899, defined the function chi(A) as a "Gruppencharakter," and developed some of its elementary properties. . . E. Landau's use of the symbol chi(n) in his texts, together with the terms "charakter and Hauptcharakter" most probably led to the subsequent widespread acceptance of the notation and terminology. Landau credited G. Torelli, 1901, with playing a major role in applying the theory of functions to the study of prime numbers [6.56]. Landau's treatment of characters [6.5.7] suggests that it was Torelli's use of notation that led to Landau's. This is further supported by a 1918 paper of Landau [6.58], where chi(n) is introduced in connection with a discussion of Torelli's results.
[Paul Pollack]
The term CHARACTERISTIC (as used in logarithms) was introduced by Henry Briggs (1561-1631), who used the term in 1624 in Arithmetica logarithmica (Cajori 1919, page 152; Boyer, page 345).
According to Smith (vol. 2, page 514), the term characteristic "was suggested by Briggs (1624) and is used in the 1628 edition of Vlacq." In a footnote, he provides the citation from Vlacq: "...prima nota versus sinistram, quam Characteristicam appellare poterimus..."
Scott (page 136) provides the following citation from Vlacq's Tabulae Sinuum, Tangentium et Secantium: "Here you will note that the first figure of the logarithm, which is called the characteristic is always less by unity than the nuber of figures in the number whose logarithm is taken" (p. xvii).
Scott (page 137) also provides this citation from Adriani Vlacq, Tabulae Sinuum, Tangentium et Secantium, et Logarithmorum. Sinuum, Tangentium et Numerorum ab Unitate ad 100000: "Si datur numerus 3.567894 = 3 567894/1000000 vel 35 67894/100000 vel 356 7894/10000 Logarithmi eorum iidem sunt, qui numeri integri 3567894, escepta tantum Characteristica aut prima figura, et modus eos inveniendi prorsus est idem." [Scott shows the decimal points as raised dots.]
The term index was another early term for the characteristic of a logarithm.
CHARACTERISTIC DETERMINANT, EQUATION, POLYNOMIAL, ROOT, VALUE, VECTOR. See Eigenvalue.
CHARACTERISTIC FUNCTION (1) of a random variable. The first person to apply characteristic functions was Laplace in 1810. Cauchy was probably the first to apply a name to the functions, using the term fonction auxiliaire. In 1919 V. Mises used the term komplexe Adjunkte.
The term characteristic function was first used by Jules Henri Poincaré (1854-1912) in Calcul des Probabilites in 1912. He wrote "fonction caracteristique." Poincare's usage corresponds with what is today called the moment generating function. This information is taken from H. A. David, "First (?) Occurrence of Common Terms in Mathematical Statistics," The American Statistician, May 1995, vol 49, no 2 121-133.
In 1922 P. Levy used the term characteristic function in the title Sur la determination des lois de probabilite par leurs fonctions characteristiques.
Characteristic function appears in English in 1934 in S. Kullback, "An Application of Characteristic Functions to the Distribution Problem of Statistics," Annals of Mathematical Statistics, 5, 263-307 (David, 1995).
CHARACTERISTIC FUNCTION (2) of a set A with respect to a "superset" U is widely used to designate the function from U to {0, 1} that is 1 on A and 0 on its complement. The name explains the common choice of the Greek letter [chi] (chi, which represents kh or ch) for this function. With this meaning, the term seems to have been introduced for the first time by C. de la Vallé Poussin (1866-1962) in Intégrales de Lebesgue, Fonctions d'ensemble, Classes de Baire (Paris, 1916), p. 7. This information is supported by references in Hausdorff's Set Theory (2d ed., Chelsea, 1962, pp. 22, 341, 342), where this function is denoted "simply by [A], omitting the argument x and thus emphasizing only its dependence on A."
Probably to avoid confusion with the other meaning (especially in probability theory, where both notions are useful), some prefer to use the term "indicator function". Besides, it is interesting no note that many logicians turn the usual order of things upside-down: for them, "characteristic function" of a set A (of natural numbers, 0 included) refers to the characteristic function of the complement! In his Foundations of mathematics (1968), W. S. Hatcher explains (p. 215):
In analysis, the characteristic function is usually 1 on the set and 0 off the set, but we generally reverse the procedure in number theory [[more precisely, in recursion theory]]. The reason stems from the minimalization rule and the fact that, when we treat characteristic functions in this way, a given problem often reduces to finding the zeros of some function. In analysis, we want the characteristic functions to be 1 on the set so that the measure of a set will be the integral of its characteristic function.
What is worse, the "characteristic function" of A in this sense is also called the "representing function" by many other logicians. The first logician to use this term seems to be Gödel in his Princeton lectures of 1934 (On undecidable propositions of formal mathematical systems, notes by S. C. Kleene and Barkeley Rosser). Having defined his (primitive) "recursive functions", he goes on to say that an n-place relation (essentially, a set of n-tuples of natural numbers) is "recursive" if its corresponding "representing function" is "recursive".
See also indicator function. [Hans Fischer, Brian Dawkins, Ken Pledger, Carlos César de Araújo]
The term CHARACTERISTIC TRIANGLE was used by Leibniz and apparently coined by him, as triangulum characteristicum.
The term CHINESE REMAINDER THEOREM is found in 1929 in Introduction to the theory of numbers by Leonard Eugene Dickson [James A. Landau].
CHI SQUARE. Karl Pearson introduced the chi-squared test and the name for it in an article in 1900 in The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science. Pearson had been in the habit of writing the exponent in the multivariate normal density as -1/2 chi-squared [James A. Landau, John Aldrich].
CHORD is found in English in 1551 in The Pathwaie to Knowledge by Robert Recorde:
Defin., If the line goe crosse the circle, and passe beside the centre, then is it called a corde, or a stryngline.
CHURCH'S THESIS. Martin Davis believes the term thesis first occurs in this connection in 1943 in Stephen Cole Kleene, "Recursive Predicates and Quantifiers," Transactions of the American Mathematical Society 53: "... led Church to state the following thesis ... Thesis I. Every effectively calculable function ... is general recursive."
Wilfried Sieg believes the first use of Church's thesis occurs in 1952 in Introduction to Metamathematics by Stephen Cole Kleene (1909-1994).
CIRCLE. According to Todhunter's translation of Euclid, Book 1 Def. 15 says "a circle is a plane figure bounded by one line, which is called the circumference ..." However Proposition 1 assumes circles consist of their circumferences: "From the point C, at which the circles cut one another, draw the straight lines ..." Heath's translation has the same problems: Def 15 "A circle is a plane figure contained by one line such that...", Prop 1 "... and from the point C, in which the circles cut one another, to the points A, B let the straight lines..." [John Harper].
A Mathematical and Philosophical Dictionary (1796) has, "The circumference or periphery itself is called the circle, though improperly, as that name denotes the space contained within the circumference."
Modern geometry texts define a circle as the set of points in a plane equidistant from a given point; the term disk is used for the circle and its interior.
CIRCLE GRAPH is dated 1928 in MWCD10.
CIRCLE OF CONVERGENCE appears in the Century Dictionary (1889-1897).
Circle of convergence appears in 1893 in A Treatise on the Theory of Functions by James Harkness and Frank Morley in the heading "The circle of convergence."
Circle of convergence also appears in 1898 in Introduction to the theory of analytic functions by Harkness and Morley: "Hence there is a frontier value R such that when |x| > R there is divergence. That is, with the circle (R) the series is absolutely convergent and without the circle it is divergent. The circle (R) is called the circle of convergence.
The term CIRCULAR COORDINATES was used by Cayley. Later writers used the term "minimal coordinates" (DSB).
CIRCULAR FUNCTION. Lacroix used fonctions circulaires in Traité élémentaire de calcul différentiel et de calcul intégral (1797-1800).
Circular function appears in 1831 in the second edition of Elements of the Differential Calculus (1836) by John Radford Young: "Thus, ax, a log x, sin x, &c., are transcendental functions: the first is an exponential function, the second a logarithmic function, and the third a circular function" [James A. Landau]
CIRCUMCENTER appears in the Century Dictionary (1889-1897).
CIRCUMCIRCLE was used in 1883 by W. H. H. Hudson in Nature XXVIII. 7: "I beg leave to suggest the following names: circumcircle, incircle, excircle, and midcircle" (OED2).
CIRCUMFERENCE. Periphereia was used by Heraclitus: "The beginning and end join on the circumference of the circle (kuklou periphereias)" (D. V. 12 B 103) (Michael Fried).
Periphereia was also used by Euclid.
Circumferentia is a Latin translation of the earlier Greek term periphereia.
Circumference is found in modern translations of the Bible, in 2 Chronicles 4:2, Jeremiah 52:21, and Ezekiel 48:35. However, the word does not appear in the King James version.
CIRCUMSCRIBE is found in English in 1570 in Billingsley's translation of Euclid: "How a triangle ... may be circumscribed about a circle" (OED2).
CIRCUMSCRIBED is found in its modern sense in 1571 in Digges Pantom.: "Circumscribed and inscribed bodies" (OED2).
CIRCUMSCRIBED POLYGON is found in 1850 in Lectures on the philosophy of arithmetic and the adaptation of that science to the business purposes of life: with numerous problems, curious and useful, solved by various modes; with explanations designed to make the study and application of arithmetic pleasant and profitable to such as have not the aid of a teacher; as well as to exercise advanced classes in schools by Uriah Parke: "The operator can know when he is correct to any given number of places by the length of the inscribed and circumscribed polygon coinciding to such extent; for as the circumference of the circle is between them it must be thus far the same. In VAN CEULEN's proportion given above, the first number gives the length of the inscribed, and the second the circumscribed polygon; the circle being between them" [University of Michigan Digital Library].
CISSOID. This term is mentioned by Geminus (c. 130 BC - c. 70 BC), according to Proclus, although the original work of Geminus does not survive.
Cissoid appears in Proclus (in Euclid, p.111, 152, 177...). It is not completely clear what curve Proclus was calling the cissoid (see W. Knorr, The Ancient Tradition in Geometric Problems, New York: Dover Publications, Inc., pp.246ff for a detailed discussion).
In the 17th century, cissoid became associated with a curve described by Diocles in his work, On Burning Mirrors.
Mathematics Dictionary (1949) by James says "the cissoid was first studied by Diocles about 200 B. C., who gave it the name 'Cissoid' (meaning ivy)"; however, according to Michael Fried, Diocles himself does not call his curve a cissoid.
The term CLASS (of a curve) is due to Joseph-Diez Gergonne (1771-1859). He used "curve of class m" for the polar reciprocal of a curve of order m in Annales 18 (1827-30) (Smith vol. I and DSB).
CLASS FIELD. The modern concept of a class field is due to Teji Takagi (1875-1960).
Leopold Kronecker (1823-1891) used the terminology "species associated with a field k."
Class field was introduced by Heinrich Weber (1842-1913) in Elliptische Funktionen und algebraische Zahlen in 1891. He originally only used the term for the Kronecker class field, but in 1896 enlarged the concept of a class field to fields K associated with a congruence class group in k, but only in the second edition of his Lehrbuch der Algebra was the term class field used to designate a general class field. (Günther Frei in "Heinrich Weber and the Emergence of Class Field Theory")
The terms CLASSICAL GROUP and CLASSICAL INVARIANT THEORY were coined by Hermann Weyl (1885-1955) and appear in The classical groups, their invariants and representations (1939).
CLASSICAL PROBABILITY. This term for probability as defined by Laplace and earlier writers came into use in the 1930s when alternative definitions were widely canvassed. J. V. Uspensky (Introduction to Mathematical Probability, 1937, p. 8) gave the "classical definition," which he favored, and criticized the "new definitions" (von Mises) and "the attempt to build up the theory of probability as an axiomatic science" (Kolmogorov) [John Aldrich].
CLASSICAL statistical inference. The polar pair "classical" and "Bayesian" have figured in discussions of the foundations of statistical inference since the 1960s. The body of work to which "classical" was attached went back only to the 1920s and -30s but, as Schlaifer wrote in 1959 (Probability and Statistics for Business Decisions, p. 607), "it is expounded in virtually every course on statistics [in the United States] and is adhered to by the great majority of practicing statisticians." Schlaifer and a few others were sponsoring a rejuvenated Bayesian alternative. The "classical" tag may have derived some authority from Neyman's "Outline of a Theory of Statistical Estimation based on the Classical Theory of Probability" (Philosophical Transactions of the Royal Society, 236, (1937), 333-380), one of the classics of classical statistics. The non-classical possibility Neyman had in mind and rejected was the Bayesian theory of Jeffreys. Confusingly Neyman's "classical theory of probability" has more to do with Kolmogorov and von Mises than with Laplace [John Aldrich].
CLELIA was coined by Guido Grandi (1671-1742). He named the curve after Countess Clelia Borromeo (DSB).
CLOSED (elements produced by an operation are in the set). Closed cycle appears in Eliakim Hastings Moore, "A Definition of Abstract Groups," Transactions of the American Mathematical Society, Vol. 3, No. 4. (Oct., 1902): "For in any finite set of elements with multiplication-table satisfying (1, 2) there exists a closed cycle of (one or more) elements, each of which is the square of the preceding element in the cycle...."
The phrase "closed under multiplication" appears in Saul Epsteen, J. H. Maclagan-Wedderburn, "On the Structure of Hypercomplex Number Systems," Transactions of the American Mathematical Society, Vol. 6, No. 2. (Apr., 1905).
CLOSED CURVE. In 1551 in Pathway to Knowledge Robert Recorde wrote, "Defin., Lynes make diuerse figures also, though properly thei maie not be called figures, as I said before (vnles the lines do close)" (OED2).
Closed curve is found in 1855 in An elementary treatise on mechanics, embracing the theory of statics and dynamics, and its application to solids and fluids by Augustus W. Smith: "Since the above principle is true, whatever be the number of sides of the polygon, it is true when the number becomes indefinitely great, or when the base becomes a continued closed curve, as a circle, an ellipse, &c.; or, the center of gravity of a cone, right or oblique, and on any base, is one fourth the distance from the center of gravity of the base to the vertex" [University of Michigan Digital Library].
CLOSED SET. Georg Cantor (1845-1918) in "De la puissance des ensembles parfaits de points," Acta Mathematica IV, March 4, 1884, introduced (in French) the concept and the term "ensemble fermé [Udai Venedem].
Closed is found in English in 1902 in Proc. Lond. Math. Soc. XXXI "Every example of such a set [of points] is theoretically obtainable in this way. For..it cannot be closed, as it would then be perfect and nowhere dense" (OED2).
CLUSTER ANALYSIS is found in 1939 in Cluster Analysis by R. C. Tryon [James A. Landau].
COCHLEOID (or COCHLIOID). In 1685 John Wallis referred to this curve as the cochlea:
... the Cochlea, or Spiral about a Cylinder, arising from a Circular motion about an Ax, together with a Rectilinear (in the Surface of the Cylinder) Perpendicular to the Plain of such Circle, (or, if the Cylinder be Scalene at such Angles with the Plain of the Circle, as is the Axis of that Cylinder) both motions being uniform, but not in the same Plain.
Some sources incorrectly attribute the term to Benthan and Falkenburg in 1884. While studying the processes of a mechanism of construction for steam engines, C. Falkenburg, Mechanical Engineer of the Actiengesellschaft Atlas in Amsterdam, rediscovered this curve. On March 25, 1883, he submitted an article titled "Die Cochleoïde", which was published in Archiv der Mathematik und Physik.
Er hat sie daher die Cochleoïde genannt, von *cochlea* = Schneckenhaus. [Therefore, it was christened the Cochleoid, from *cochlea* = snail's house.]
The reference for this citation is Nieuw Archief voor Wiskunde [Amsterdam: Weytingh & Brave], vol. 10, pp. 76-80, 1884. This entry was contributed by Julio González Cabillón.
COEFFICIENT. Cajori (1919, page 139) writes, "Vieta used the term 'coefficient' but it was little used before the close of the seventeenth century." Cajori provides a footnote reference: Encyclopédie des sciences mathématiques, Tome I, Vol. 2, 1907, p. 2. According to Smith (vol. 2, page 393), Vieta coined the term.
The term COEFFICIENT OF VARIATION appears in 1896 in Karl Pearson, "Regression, Heredity, and Panmixia," Philosophical Transactions of the Royal Society of London, Ser. A. 187, 253-318 (David, 1995). The term is due to Pearson (Cajori 1919, page 382). According to the DSB, he introduced the term in this paper.
CO-FACTOR is found in 1849 in Trigonometry and Double Algebra by Augustus De Morgan: "When an expression consists of terms, let them be called co-terms; when of factors, co-factors [University of Michigan Historic Math Collection].
The word COMBINANT was coined by James Joseph Sylvester (DSB).
The word appears in a paper by Sylvester in 1853 in Camb. & Dublin Math. Jrnl. VIII. 257: "What I term a combinant" (OED2).
COMBINATION was used in its present sense by both Pascal and Wallis, according to Smith (vol. 2, page 528).
In a letter to Fermat dated July 29, 1654, Pascal wrote a sentence which is translated from French:
If from any number of letters, as 8 for example, A, B, C, D, E, F, G, H, you take all the possible combinations of 4 letters and then all possible combinations of 5 letters, and then of 6, and then of 7, of 8, etc., and thus you would take all possible combinations, I say that if you add together half the combinations of 4 with each of the higher combinations, the sum will be the number equal to the number of the quaternary progression beginning with 2 which is half of the entire number.
This translation was taken from A Source Book in Mathematics by David Eugene Smith.
Combinations is found in English in 1673 in the title Treatise of Algebra...of the Cono-Cuneus, Angular Sections, Angles of Contact, Combinations, Alternations, etc. by John Wallis (OED2).
Leibniz used complexiones for the general term, reserving combinationes for groups of three.
Eberhard Knobloch writes in "The Mathematical Studies of G. W. Leibniz on Combinatorics," Historia Mathematica 1 (1974):
Leibniz's terminology for partitions, just as for symmetric functions, is not consistent. In his Ars Combinatoria he speaks of "discerptiones, Zerfällungen" as mentioned above, and defines them as special cases of "complexiones" (combinations). The Latin term "discerptio" he uses most, and it appears in numerous manuscripts up to his death. When he wants to refer to specific partitions into 1, 2, 3, 4 ... summands, he writes "uniscerptiones, biscerptiones, triscerptiones, quadriscerptiones..." and sometimes also "1scerptiones, 2scerptiones..." evidently following his former usage for combinations of certain sizes in the Ars Combinatoria. I have found only two places where Leibniz applies the general term "discerptio" to the special partition into two summands.
COMBINATORICS. Combinatorial was first used in the modern mathematical sense by Gottfried Wilhelm Leibniz (1646-1716) in his Dissertatio de Arte Combinatoria (Dissertation Concerning the Combinational Arts) (Encyclopaedia Britannica, article: "Combinatorics and Combinatorial Geometry").
Combinatorial analysis is found in English in 1818 in the title Essays on the Combinatorial Analysis by P. Nicholson (OED2).
An early use of the term combinatorics is by F. W. Levi in an essay entitled "On a method of finite combinatorics which applies to the theory of infinite groups," published in the Bulletin of the Calcutta Mathematical Society, vol. 32, pp. 65-68, 1940 [Julio González Cabillón].
COMMENSURABLE is found in English in 1557 in The Whetstone of Witte by Robert Recorde (OED2).
COMMON DIFFERENCE and COMMON RATIO are found in the 1771 edition of the Encyclopaedia Britannica in the article "Algebra" [James A. Landau].
COMMON FRACTION. Thomas Digges (1572) spoke of "the vulgare or common Fractions" (Smith vol. 2, page 219).
COMMON LOGARITHM appears in 1798 in Hutton, Course Math.: "When the radix r is = 10, then the index n becomes the common or Briggs's log. of the number N" (OED2).
Common system of logarithms appears in the 1828 Webster dictionary, in the definition of radix: "Thus in Briggs', or the common system of logarithms, the radix is 10; in Napier's, it is 2.7182818284."
Common logarithm appears in 1849 in An Introduction to the Differential and Integral Calculus, 2nd ed., by James Thomson: "Thus, in the common logarithms, in which 10, the radix of the decimal system of notation, is the base, we have...."
Common logarithm appears in a footnote in Elementary Illustrations of the Differential and Integral Calculus (1899) by Augustus de Morgan. This book is largely a reprint of a series of articles which appeared in the Library of Useful Knowledge in 1832, and thus the term may appear there.
COMMUTATIVE and DISTRIBUTIVE were used (in French) by François Joseph Servois (1768-1847) in a memoir published in Annales de Gergonne (volume V, no. IV, October 1, 1814). He introduced the terms as follows (pp. 98-99):
3. Soit
f(x + y + ...) = fx + fy + ...
Les fonctions qui, comme f, sont telles que la fonction de la somme (algébrique) d'un nombre quelconque de quantites est égale a la somme des fonctions pareilles de chacune de ces quantités, seront appelées distributives.
Ainsi, parce que
a(x + y + ...) = ax + ay + ...; E(x + y + ...) = Ex + Ey + ...; ...
le facteur 'a', l'état varié E, ... sont des fonctions distributives; mais, comme on n'a pas
Sin.(x + y + ...) = Sin.x + Sin.y + ...; L(x + y + ...) = Lx + Ly + ...;
...les sinus, les logarithmes naturels, ... ne sont point des fonctions distributives.
4. Soit
fgz = gfz.
Les fonctions qui, comme f et g, sont telles qu'elles donnent des résultats identiques, quel que soit l'ordre dans lequel on les applique au sujet, seront appelées commutatives entre elles.
Ainsi, parce que qu'on a
abz = baz ; aEz = Eaz ; ...
les facteurs constans 'a', 'b', le facteur constant 'a' et l'état varié E, sont des fonctions commutatives entre elles; mais comme, 'a' etant toujours constant et 'x' variable, on n'a pas
Sin.az = a Sin.z ; Exz = xEz ; Dxz = xDz [D = delta]; ...
il s'ensuit que le sinus avec le facteur constant, l'état varié ou la difference avec le facteur variable, ... n'appartiennent point a la classe des fonctions commutatives entre elles.
(These citations were provided by Julio González Cabillón).
COMPACT was introduced by Maurice René Fréchet (1878-1973) in 1906, in Rendiconti del Circolo Matematico di Palermo vol. 22 p. 6. He wrote:
Nous dirons qu'un ensemble est compact lorsqu'il ne comprend qu'un nombre fini d'éléments ou lorsque toute infinité de ses éléments donne lieu à au moins un élément limite.
This citation was provided by Mark Dunn.
In his 1906 thesis, Fréchet wrote:
A set E is called compact if, when {En} is a sequence of nomempty, closed subsets of E such that En+1 is a subset of En for each n, there is at least one element that belongs to all of the En's.
At the end of his life, Fréchet did not remember why he chose the term:
... jai voulu sans doute éviter qu'on puisse appeler compact un noyau solide dense qui n'est agrémenté que d'un fil allant jusqu'à l'infini. C'est une supposition car j'ai complétement oubliè les raisons de mon choix!" [Doubtless I wanted to avoid a solid dense core with a single thread going off to infinity being called compact. This is a hypothesis because I have completely forgotten the reasons for my choice!] (Pier, p. 440)
Some mathematicians did not like the term "compact." Schönflies suggested that what Fréchet called compact be called something like "lückenlos" (without gaps) or "abschliessbar" (closable) (Taylor, p. 266).
Fréchet's "compact" is the modern "relatively sequentially compact," and his "extremal" is today's "sequentially compact" (Kline, page 1078).
Compact is found in Paul Alexandroff and Paul Urysohn, "Mémoire sur les espaces topologiques compacts," Koninklijke Nederlandse Akademie van Vetenschappen te Amsterdam, Proceedings of the section of mathematical sciences) 14 (1929).
COMPLEMENT. "Complement of a parallelogram" appears in English in 1570 in Sir Henry Billingsley's translation of Euclid's Elements.
COMPLEMENTARY FUNCTION is found in 1841 in D. F. Gregory, Examples of Processes of Differential and Integral Calculus: "As operating factors of the form (d/dx)2 + n2 very frequently occur in differential equations, it is convenient to keep in mind that the complementary function due to it is of the form C cos nx + C' sin nx (OED2).
COMPLETE INDUCTION (vollständige Induktion) was the term employed by Dedekind in his Was sind und Was sollen die Zahlen? (1887) for what is nowadays called "mathematical induction", and whose "scientific basis" ("wissenschaftliche grundlage") he claimed to have established with his "Theorem of complete induction" (§59). Dedekind also used occasionally the phrase "inference from n to n + 1", but nowhere in his booklet did he try to justify the adjective "complete".
In Concerning the axiom of infinity and mathematical induction (Bull. Amer. Math. Soc. 1903, pp. 424-434) C. J. Keyser referred to "complete induction" as
a form of procedure unknown to the Aristotelian system, for this latter allows apodictic certainty in case of deduction only, while it is just characteristic of complete induction that it yields such certainty by the reverse process, a movement from the particular to the general, from the finite to the infinite.
Florian Cajori ("Origin of the name "mathematical induction," Amer. Math. Monthly, 1918, pp. 197-201) noted an earlier use of the term "vollständige Induktion" in the article "Induction" in Ersch and Gruber’s Encyklopädie (1840), but in an uninteresting and totally different "Aristotelian sense". According to Abraham Fraenkel (1891-1965) (Abstract Set Theory, 1953, p. 253),
[the] term "complete induction" used in most continental languages (...) [stress] the contrast with induction in natural science which is incomplete by its very nature, being based on a finite and even relatively small number of experiments.
This entry was contributed by Carlos César de Araújo. See also mathematical induction.
COMPLETE SOLUTION. The term complete solution or complete integral is due to Lagrange (Kline, page 532).
The term COMPLETENESS was used by Dedekind in 1872, both to describe the closure of a number field under arithmetical operations and as a synonym for "continuity" (Burn 1992).
COMPLETING THE SQUARE is found in 1806 in Hutton, Course Math.: "The general method of solving quadratic equations, is by what is called completing the square" (OED2).
COMPLEX FRACTION is found in English in 1827 in Hutton, Course Math.: "A Complex Fraction, is one that has a fraction or a mixed number for its numerator, or its denominator, or both" (OED2).
COMPLEX NUMBER. Most of the 17th and 18th century writers spoke of a + bi as an imaginary quantity. Carl Friedrich Gauss (1777-1855) saw the desirability of having different names for ai and a + bi, so he gave to the latter the Latin expression numeros integros complexos. Gauss wrote:
...quando campus arithmeticae ad quantitates imaginarias extenditur, ita ut absque restrictione ipsius obiectum constituant numeri formae a + bi, denotantibus i pro more quantitatem imaginariam , atque a, b indefinite omnes numeros reales integros inter - et + . Tales numeros vocabimus numeros integros complexos, ita quidem, ut reales complexis non opponantur, sed tamquam species sub his contineri censeatur.
The citation above is from Gauss's paper "Theoria Residuorum Biquadraticorum, Commentatio secunda," Societati Regiae Tradita, Apr. 15, 1831, published for the first time in Commentationes societatis regiae scientiarum Gottingensis recentiones, vol. VII, Gottingae, MDCCCXXXII (1832)]. [Julio González Cabillón]
The term complex number was used in English in 1856 by William Rowan Hamilton. The OED2 provides this citation: Notebook in Halberstam & Ingram Math. Papers Sir W. R. Hamilton (1967) III. 657: "a + ib is said to be a complex number, when a and b are integers, and i = [sqrt] -1; its norm is a2 + b2; and therefore the norm of a product is equal to the product of the norms of its factors."
COMPOSITE NUMBER (early meaning). According to Smith (vol. 2, page 14), "The term 'composite,' originally referring to a number like 17, 56, or 237, ceased to be recognized by arithmeticians in this sense because Euclid had used it to mean a nonprime number. This double meaning of the word led to the use of such terms as 'mixed' and 'compound' to signify numbers like 16 and 345." Smith differentiates between "composites" and "articles," which are multiples of 10.
COMPOSITE NUMBER (nonprime number). The OED2 shows numerus compositus Isidore III. v. 7.
Napier used the term numeri compositi.
Composite number appears in English in a dictionary of 1730-6 (OED2).
CONCAVE appears in English in 1571 in A Geometricall Practise named Pantometria by Thomas Digges (1546?-1595) (OED2).
CONCAVE POLYGON. Fibonacci referred to such a polygon as a figura barbata in Practica geomitrae.
Re-entering polygon is found in 1851 in Problems in illustration of the principles of plane coordinate geometry by William Walton [University of Michigan Historic Math Collection]. Another term is re-entrant polygon.
Concave polygon appears in 1899 in New Plane Geometry by Wooster Woodruff Beman and David Eugene Smith: "The word polygon is understood, in elementary geometry, to refer to a convex or concave polygon unless the contrary is stated" [University of Michigan Historic Math Collection].
CONCHOID (also known as CONCHLOID). Nicomedes (fl. ca. 250 BC) called various curves the first, second, third, and fourth conchoids (DSB). Pappus says that the conchoids were explored by Nicomedes in his work On Conchoid Lines [Michael Fried].
CONDITIONALLY CONVERGENT SERIES. Semi-convergent series appears in 1872 in J. W. L. Glaisher, "On semi-convergent series," Quart. J.
Conditionally convergent series and semi-convergent series appear in 1893 in A Treatise on the Theory of Functions by James Harkness and Frank Morley: "A series which converges, but does not converge absolutely, is called semi-convergent. ... A convergent series which is subject to the commutative law is said to be unconditionally convergent; otherwise it is said to be conditionally convergent. ... Semi-convergence implies conditional convergence."
CONDITIONAL PROBABILITY is found in J. V. Uspensky, Introduction to Mathematical Probability, New York: McGraw-Hill, 1937, page 31:
Let A and B be two events whose probabilities are (A) and (B). It is understood that the probability (A) is determined without any regard to B when nothing is known about the occurrence or nonoccurrence of B. When it is known that B occurred, A may have a different probability, which we shall denote by the symbol (A, B) and call 'conditional probability of A, given that B has actually happened.'
[James A. Landau]
CONE is defined in Euclid's Elements, XI, def.18, and it appears in a mathematical context in the presocratic atomist Democritus of Abdera, who wrote:
If a cut were made through a cone parallel to its base, how should we conceive of the two opposing surfaces which the cut has produced -- as equal or as unequal? If they are unequal, that would imply that a cone is composed of many breaks and protrusions like steps. On the other hand if they are equal, that would imply that two adjacent intersection planes are equal, which would mean that the cone, being made up of equal rather than unequal circles, must have the same appearance as a cylinder; which is utterly absurd (D. V. 55 B 155, translation by Philip Wheelwright in The Presocratics, Indianapolis: The Bobbs-Merrill Company, Inc., 1960, p.183).
(This entry was contributed by Michael Fried.)
CONFIDENCE INTERVAL was coined by Jerzy Neyman (1894-1981) in 1934 in "On the Two Different Aspects of the Representative Method," Journal of the Royal Statistical Society, 97, 558-625:
The form of this solution consists in determining certain intervals, which I propose to call the confidence intervals..., in which we may assume are contained the values of the estimated characters of the population, the probability of an error is a statement of this sort being equal to or less than 1 - (epsilon), where (epsilon) is any number 0 < (epsilon) < 1, chosen in advance.
CONFORMAL MAPPING. The term projectio conformis was introduced by F. T. Schubert in 1789 (DSB, article: "Euler").
Gauss used the term conforme Abbildung.
Cayley used the term orthomorphosis.
In 1956, Albert A. Bennett wrote: "Thus a function of one argument, or a mapping, is simply a one-valued, two-term relation. The term 'mapping' thus includes 'functional,' 'projectivity,' and so forth. Although the phrase 'conformal mapping' is old, the general use here mentioned is very recent and may be due to van der Waerden, 1937." (This quotation was taken from "Concerning the function concept," The Mathematics Teacher, May 1956.)
CONGRUENT (geometric figures). Congruere (Latin, "to coincide") was used by geometers of the sixteenth century in their editions of Euclid in quoting Common Notion 4: "Things which coincide with one another are equal to one another." ["Ea ... aequalia sunt, quae sibi mutuo congruunt."]
For instance, in 1539, Christoph Clavius (1537?-1612) writes:
...Hinc enim fit, ut aequalitas angulorum ejusdem generis requirat eandem inclinationem linearum, ita ut lineae unius conveniant omnino lineis alterius, si unus alteri superponatur. Ea enim aequalia sunt, quae sibi mutuo congruunt.
[Cf. page 363 of Clavius's "Euclidis", vol. I, Romae: Apvd Barthdomaevm Grassium, 1589]
As a more technical term for a relation between figures, congruent seems to have originated with Gottfried Wilhelm Leibniz (1646-1716), writing in Latin and French. His manuscript "Characteristica Geometrica" of August 10, 1679, is in his Gesammelte Werke, dritte Folge: mathematische Schriften, Band 5. On p. 150 he says that if a figure can be applied exactly to another without distortion, they are said to be congruent:
Quodsi duo non quidem coincidant, id est non quidem simul eundem locum occupent, possint tamen sibi applicari, et sine ulla in ipsis per se spectatis mutatione facta alterum in alterius locum substitui queat, tunc duo illa dicentur esse congrua, ut A.B et C.D in fig.39 ...
His Figure 39 shows two radii of a circle, with the center labelled both A and C. Later (p.154) he points out that "congruent" is the same as "similar and equal." He used "congruent" in the modern (Hilbert) sense, applied to line segments and various other things as well as triangles.
Shortly afterwards, on September 8, 1679, he included a similar definition in a letter to Hugens (sic) van Zulichem. In his ges. Werke etc. as above, volume 2, p. 22, he illustrates congruence with a pair of triangles, and says that they "peuvent occuper exactement la meme place, et qu'on peut appliquer ou mettre l'un sur l'autre sans rien changer dans ces deux figures que la place." [Ken Pledger and Julio González Cabillón]
In English, writers commonly refer to geometric figures as equal as recently as the nineteenth century. In 1828, Elements of Geometry and Trigonometry (1832) by David Brewster (a translation of Legendre) has:
Two triangles are equal, when an angle and the two sides which contain it, in the one, are respectively equal to an angle and the two sides which contain it, in the other.
CONGRUENT (in modular arithmetic) was defined by Carl Friedrich Gauss (1777-1855) in 1801 in Disquisitiones arithmeticae: "Si numerus a numerorum b, c differentiam metitur, b et c secundum a congrui dicuntur."
CONIC SECTION is found in the title De sectionibus conicis by Claude Mydorge (1585-1647).
The term is also found in the title Essay on Conic Sections by Blaise Pascal (1623-1662) published in February 1640.
CONJECTURE. Isaac Newton used the term conjecture in 1672 in Phil. Trans. VII. 5084, although whether or not the term was used in a mathematical context is not clear: "I shall refer him to my former Letter, by which that conjecture will appear to be ungrounded."
Jacob Steiner (1796-1863) referred to a result of Poncelet as a conjecture. Poncelet showed in 1822 that in the presence of a given circle with given center, all the Euclidean constructions can be carried out with ruler alone (DSB, article: "Mascheroni").
In Récréations Mathématiques, tome II, Note II, Sur les nombres de Fermat et de Mersenne (1883), É. Lucas referred to "la conjecture de Fermat."
In his article "Conjecture" (Synthese 111, pp. 197-210, 1997), Barry Mazur writes (bottom of page 207):
Since I am not a historian of Mathematics I dare not make any serious pronouncements about the historical use of the term, but I have not come across any appearance of the word Conjecture or its equivalent in other languages with the above meaning [i.e., an opinion or supposition based on evidence which is admittedly insufficient] in mathematical literature except in the twentieth century. The earliest use of the noun conjecture in mathematical writing that I have encountered is in Hilbert's 1900 address, where it is used exactly once, in reference to Kronecker's Jugendtraum.
CONJUGATE. Augustin-Louis Cauchy (1789-1857) used conjuguées for for a + bi and a - bi in Cours d'Analyse algébrique (1821) (Smith vol. 2, page 267).
CONJUGATE ANGLE appears in the Century Dictionary (1889-1897).
The term also appears in Plane and Solid Geometry (1913) by Wentworth and Smith, and it may occur in the earlier 1888 edition, which has not been consulted.
CONSERVATIVE EXTENSION. Martin Davis believes the term was first used by Paul C. Rosenbloom. It appears in The Elements of Mathematical Logic, 1st ed., New York: Dover Publications, 1950.
CONSISTENCY. The term consistency applied to estimation was introduced by R. A. Fisher in "On the Mathematical Foundations of Theoretical Statistics" (Phil. Trans. R. Soc. 1922). Fisher wrote: "A statistic satisfies the criterion of consistency, if, when it is calculated from the whole population, it is equal to the required population."
In the modern literature this notion is usually called Fisher-consistency (a name suggested by Rao) to distinguish it from the more standard notion linked to the limiting behavior of a sequence of estimators. The latter is hinted at in Fisher's writings but was perhaps first set out rigorously by Hotelling in the "The Consistency and Ultimate Distribution of Optimum Statistics," Transactions of the American Mathematical Society (1930). [This entry was contributed by John Aldrich, based on David (1995).]
CONSTANT was introduced by Gottfried Wilhelm Leibniz (1646-1716) (Kline, page 340).
CONSTANT OF INTEGRATION. In 1807 Hutton Course Math. has: "To Correct the Fluent of any Given Fluxion .. The finding of the constant quantity c, to be added or subtracted with the fluent as found by the foregoing rules, is called correcting the fluent.
In 1831 Elements of the Integral Calculus (1839) by J. R. Young refers to "the arbitrary constant C."
Constant of integration is found in 1846 in "On the Rotation of a Solid Body Round a Fixed Point" by Arthur Cayley in the Cambridge and Dublin Mathematical Journal [University of Michigan Historical Math Collection].
In 1849 in An Introduction to the Differential and Integral Calculus, 2nd ed., by James Thomson, it is called the "constant quantity annexed."
CONTINGENCY TABLE was introduced by Karl Pearson in "On the Theory of Contingency and its Relation to Association and Normal Correlation," which appeared in Drapers' Company Research Memoirs (1904) Biometric Series I:
This result enables us to start from the mathematical theory of independent probability as developed in the elementary text books, and build up from it a generalised theory of association, or, as I term it, contingency. We reach the notion of a pure contingency table, in which the order of the sub-groups is of no importance whatever.
This citation was provided by James A. Landau.
The CONTINUED FRACTION was introduced by John Wallis (1616-1703) (DSB, article: "Cataldi").
Wallis used continue fracta in 1655 in Arithmetica Infinitorum Prop. CXCI.
The phrase "Esto igitur fractio eiusmode continue fracta quaelibet sic deignata..." is found in volume I of Opera Mathematica, a collection of Wallis' mathematical and scientific works published in 1693-1699.
The phrase "fractio, quae denominatorem habeat continue fractum" is found in Opera, I, 469 (Smith vol. 2, page 420).
In 1685 Wallis referred to Brouncker's continued fraction as "a fraction still fracted continually" in A Treatise of Algebra [Philip G. Drazin, David Fowler, James A. Landau, Siegmund Probst].
Continued fraction is found in English in 1811 An Elementary Investigation in the Theory of Numbers by Peter Barlow [James A. Landau].
CONTINUOUS. Euler defined a continuous curve in the second volume of his Introductio in analysin infinitorum (Katz, page 580).
CONTINUOUS CURVE is found in English in 1852 in Elements of the differential and integral calculus by Charles Davies [University of Michigan Digital Library].
CONTINUOUS FUNCTION is found in English in 1871 in A manual of spherical and practical astronomy, embracing the general problems of spherical astronomy, the special applications to nautical astronomy, and the theory and use of fixed and portable astronomical instruments, with an appendix on the method of least squares by William Chauvenet: "our geometrical representation should strictly consist of a number of isolated points; but, as these points will be more and more nearly represented by a continuous curve as we increase the accuracy of the observations, and thus diminish the intervals between the successive ordinates, we may, without hesitation, adopt such a continuous curve as expressing the law of error. We shall therefore regard [Greek letter capital delta] as a continuous variable, and [Greek letters small phi followed, no space, by capital delta] as a continuous function of it" [University of Michigan Digital Library].
CONTINUUM. According to the DSB, the term continuum appeared as early as the writings of the Scholastics, but the first satisfactory definition of the term was given by Cantor.
CONTINUUM HYPOTHESIS. In his 1900 Paris lecture, Hilbert titled his first problem "Cantor's Problem of the Power of the Continuum."
In his 1901 doctoral dissertation writen under Hilbert, Felix Bernstein used the term "Cantor's Continuum Problem" ("das Cantorsche Continuumproblem"). This is the first time the term "Cantor's Continuum Problem" appeared in print, according to Cantor's Continuum Problem by Gregory H. Moore, which also states that "presumbly Bernstein obtained the name 'Continuum Problem' by abbreviating Hilbert's title."
Continuum problem also appears in Felix Bernstein, "Zum Kontinuumproblem," Mathematische Annalen 60 (1905); Julius König, "Zum Kontinuum-Problem," Verhandlungen des dritten internat. Math.- Kongress (1905); and Julius König, "Zum Kontinuumproblem," Mathematische Annalen 60 (1905).
In his essay "On the Infinite" (1925) Hilbert referred to the question of whether continuum hypothesis is true as the "famous problem of the continuum"; the word "hypothesis" is not used. Two years later, in "The foundations of mathematics" (1927) he referred to "the proof or refutation of Cantor's continuum hypothesis."
Carlos César de Araújo believes that the use of "hypothesis" here became more popular and well-established only after the 1934 monograph of Sierpinski, "Hypothése du continu."
In the 1962 Chelsea translation of the 1937 3rd German edition of Hausdorff's Mengenlehre pp 45f is the following:
A conjecture that was made at the beginning of Cantor's investigations, and that remains unproved to this day, is that [alef] is the cardinal number next larger than [alef-null]; this conjecture is known as the continuum hypothesis, and the question as to whether it is true or not is known as the problem of the continuum
(Hausdorff used [alef] to mean the infinity of the continuum.)
Continuum hypothesis appears in Waclaw Sierpinski, "Sur deux propositions, dont l'ensemble équivaut à l'hypothése du conntinu," Fundamenta Mathematicae 29, pp 31-33 (1937).
Continuum hypothesis appears in the title "The consistency of the axiom of choice and of the generalized continuum-hypothesis" by Kurt Gödel, Proc. Nat. Acad. Sci., 24, 556-557 (1938) [James A. Landau, Carlos César de Araújo].
CONTRAPOSITIVE. Boethius wrote: " Est enim per contrapositionem conversio, ut si dicas omnis homo animal est, omne non animal non homo est."
Contraposition is found in English in 1551 in T. Wilson, Logike: "A conuersion by contraposition is when the former part of the sentence is turned into the last rehearsed part, and the last rehearsed part turned into the former part of the sentence, both the propositions being uniuersall, and affirmatiue, sauing that in the second proposition there be certaine negatiues enterlaced" (OED2).
Contrapose and contraposite other older terms in English.
De Morgan used the adverb contrapositively in 1858 in Trans. Camb. Philos. Soc. (OED2).
Contrapositive appears as an adjective in the preface to The Elements of Plane Geometry (1868) by R. P. Wright. The preface was written by T. A. Hirst (1830-1892): "The two theorems are, in fact, contrapositive forms, one of the other; the truth of each is implied, when that of the other is asserted, and to demonstrate both geometrically is more than superfluous; it is a mistake, since the true relation between the two is thereby masked."
Contrapositive was used as a noun in 1870 by William Stanley Jevons in Elementary Lessons in Logic (1880): "Convert and show that the result is the contrapositive of the original" (OED2).
CONVERGENCE (of a vector field) was coined by James Clerk Maxwell (Katz, page 752; Kline, page 785). It is the negative of the divergence, q.v.
See curl.
The terms CONVERGENT and DIVERGENT were used by James Gregory in 1667 in his Vera circuli et hyperbolae quadratura (Cajori 1919, page 228). Gregory wrote series convergens.
However, according to Smith (vol. 2, page 507), the term convergent series is due to Gregory (1668) and the term divergent series is due to Nicholas I Bernoulli (1713). In a footnote, he cites F. Cajori, Bulletin of the Amer. Math. Soc. XXIX, 55.
CONVERSE is first found in English in Sir Henry Billingsley's 1570 translation of Euclid's Elements (OED2).
CONVEX (curved outward) appears in English in 1571 in A Geometricall Practise named Pantometria by Thomas Digges (1546?-1595) (OED2).
CONVEX POLYGON. In 1828 in Elements of Geometry and Trigonometry (1832) by David Brewster (a translation of Legendre) is found the following:
We thought it better to restrict our reasoning to those lines which we have named convex, and which are such that a straight line cannot cut them in more than two points.
Convex polygon is found in English in 1852 in an adaptation by Charles Davies of Elements of geometry and trigonometry by Adrien Marie Legendre: "When this proposition is applied to polygons which have re-entrant angles, each re-entrant angle must be regarded as greater than two right angles. But to avoid all ambiguity, we shall henceforth limit our reasoning to polygons with salient angles, which are named convex polygons. Every convex polygon is such, that a straight line, drawn at pleasure, cannot meet the sides of the polygon in more than two points" [University of Michigan Digital Library].
In 1857 Mathematical Dictionary and Cyclopedia of Mathematical Sciences has convex polygon and the synonymous term salient polygon.
CONVOLUTION. In his 1933 book The Fourier Integral and Certain of its Applications Norbert Wiener uses convolutions (both the integral and sum types) but calls them by the German name Faltung, stating (p. 45) that there is no good English term [Yaakov Stein].
In 1934, Amer. Jrnl. Math. LVI. 662 has "Bernoulli convolutions" (OED2).
In 1935, Trans. Amer. Math. Soc. XXXVIII. 48 has "Distribution functions and their convolutions ('Faltungen')" (OED2).
The word COORDINATE was introduced by Gottfried Wilhelm Leibniz (1646-1716). He also used the term axes of co-ordinates. According to Cajori (1919, pages 175 and 211), he used the terms in 1692; according to Ball, he used the terms in a paper of 1694.
Leibniz used the term in "De linea ex lineis numero infinitis ordinatim ductis inter se concurrentibus formata, easque omnes tangente, ac de novo in ea re Analysis infinitorum usu," in Acta Eruditorum, vol. 11 (1692), pp. 168-171. On p. 170: "Verum tam ordinata quam abscissa, quas per x & y designari mos est (quas & coordinatas appellare soleo, cum una sit ordinata ad unum, altera ad alterum latus anguli, a duabus condirectricibus comprehensi) est gemina seu differentiabilis." The article is also printed in Leibniz, Mathematische Schriften (ed. Gerhardt), vol. 5, pp. 266-269 [Siegmund Probst].
Descartes did not use the term coordinate (Burton, page 350).
The term COORDINATE GEOMETRY is dated 1815-25 in RHUD2. An early use of the term is by Matthew O'Brien (1814-1855) in A treatise on plane co-ordinate geometry; or, The application of the method of co-ordinates to the solution of problems in plane geometry, Part 1, Cambridge: Deighton, 1844.
COPLANAR appears in Sir William Rowan Hamilton, Lectures on Quaternions (London: Whittaker & Co., 1853): "In that particular case, there was ready a known signification [36] for the product line, considered as the fourth proportional to the unit-line (assumed here on the last-mentioned axis), and to the two coplanar factor-lines" [James A. Landau].
COROLLARY. From the Latin corolla, a small garland. In an essay entitled "The Essence of Mathematics" (see James R. Newman’s anthology The world of mathematics), Charles Saunders Peirce (1839-1914) wrote:
(...) while all the "philosophers" follow Aristotle in holding no demonstration to be thoroughly satisfactory except what they call a "direct demonstration", or a "demonstration why" (...) the mathematicians, on the contrary, entertain a contempt for that style of reasoning, and glory in what the philosophers stigmatize as "mere indirect demonstrations", or "demonstrations that". Those propositions which can be deduced from others by reasoning of the kind that the philosophers extol are set down by mathematicians as "corollaries". That is to say, they are like those geometrical truths which Euclid did not deem worthy of particular mention, and which his editors inserted with a garland, or corolla, against each in the margin, implying perhaps that it was to them that such honor as might attach to these insignificants remarks was due. (...) we may say that corollarial, or "philosophical" reasoning is reasoning with words; while theorematic, or mathematical reasoning proper, is reasoning with specially constructed schemata.
[Carlos César de Araújo]
CORRELATION, CORRELATION COEFFICIENT and COEFFICIENT OF CORRELATION. Francis Galton introduced the measurement of correlation (Hald, p. 604). The index of co-relation appears in 1888 in his "Co-Relations and Their Measurement," Proc. R. Soc., 45, 135-145: "The statures of kinsmen are co-related variables; thus, the stature of the father is correlated to that of the adult son,..and so on; but the index of co-relation ... is different in the different cases" (OED2). "Co-relation" soon gave way to "correlation" as in W. F. R. Weldon's "The Variations Occurring in Certain Decapod Crustacea-I. Crangon vulgaris," Proc. R. Soc., 47. (1889 - 1890), pp. 445-453.
The term coefficient of correlation was apparently originated by Edgeworth in 1892, according to Karl Pearson's "Notes on the History of Correlation," (reprinted in Pearson & Kendall (1970). It appears in 1892 in F. Y. Edgeworth, "Correlated Averages," Philosophical Magazine, 5th Series, 34, 190-204.
Correlation coefficient appears in a paper published in 1895 [James A. Landau].
The OED2 shows a use of coefficient of correlation in 1896 by Pearson in Proc. R. Soc. LIX. 302: "Let r0 be the coefficient of correlation between parent and offspring." David (1995) gives the 1896 paper by Karl Pearson, "Regression, Heredity, and Panmixia," Phil. Trans. R. Soc., Ser. A. 187, 253-318. This paper introduced the product moment formula for estimating correlations--Galton and Edgeworth had used different methods.
Partial correlation. G. U. Yule introduced "net coefficients" for "coefficients of correlation between any two of the variables while eliminating the effects of variations in the third" in "On the Correlation of Total Pauperism with Proportion of Out-Relief" (in Notes and Memoranda) Economic Journal, Vol. 6, (1896), pp. 613-623. Pearson argued that partial and total are more appropriate than net and gross in Karl Pearson & Alice Lee "On the Distribution of Frequency (Variation and Correlation) of the Barometric Height at Divers Stations," Phil. Trans. R. Soc., Ser. A, 190 (1897), pp. 423-469. Yule went fully partial with his 1907 paper "On the Theory of Correlation for any Number of Variables, Treated by a New System of Notation," Proc. R. Soc. Series A, 79, pp. 182-193.
Multiple correlation. At first multiple correlation referred only to the general approach, e.g. by Yule in Economic Journal (1896). The coefficient arrives later. "On the Theory of Correlation" (J. Royal Statist. Soc., 1897, p. 833) refers to a coefficient of double correlation R1 (the correlation of the first variable with the other two). Yule (1907) discussed the coefficient of n-fold correlation R21(23...n). Pearson used the phrases "coefficient of multiple correlation" in his 1914 "On Certain Errors with Regard to Multiple Correlation Occasionally Made by Those Who Have not Adequately Studied this Subject," Biometrika, 10, pp. 181-187, and "multiple correlation coefficient" in his 1915 paper "On the Partial Correlation Ratio," Proc. R. Soc. Series A, 91, pp. 492-498.
[This entry was largely contributed by John Aldrich.]
The term CORRELOGRAM was introduced by H. Wold in 1938 (A Study in the Analysis of Stationary Time Series). There is a plot of empirical serial correlations, i.e. an empirical correlogram, in Yule's "Why Do We Sometimes Get Nonsense Correlations between Time-series ..." Journal of the Royal Statistical Society, 89, (1926), 1-69 (David 2001).
COSECANT. The Latin cosecans appears in Opus Palatinum de triangulis ("The Palatine Work on Triangles"), which was written by Georg Joachim von Lauchen Rheticus (1514-1574). This treatise was published after his death by his pupil Valentin Otto in 1596. According to Ball (page 243) and Smith (vol. 2, page 622), the term seems to have been first used by Rheticus.
The cosecant was called the secans secunda by Magini (1592) and Cavalieri (1643) (Smith vol. 2, page 622).
Some sources say the word cosecant was introduced by Edmund Gunter (1581-1626). This seems to be incorrect, as his use would likely have occurred after that of Rheticus.
COSET was used in 1910 by G. A. Miller in Quarterly Journal of Mathematics.
COSINE. Plato of Tivoli (c. 1120) used chorda residui for cosine.
Regiomontanus (c. 1463) used sinus rectus complementi.
Pitiscus wrote sinus complementi.
Rhaeticus (1551) used basis.
In 1558 Francisco Maurolyco used sinus rectus secundus for the cosine.
Vieta (1579) used sinus residuae.
Magini (1609) used sinus secundus (Smith vol. 2, page 619).
Cosine was coined in Latin by Edmund Gunter (1581-1626) in 1620 in Canon triangulorum, sive, Tabulae sinuum et tangentium artificialium ad radium 100000.0000. & ad scrupula prima quadrantis, Londini: Excudebat G. Iones, 1620. According to Smith (vol. 2, page 619), "Edmund Gunter (1620) suggested co.sinus, a term soon modified by John Newton (1658) into cosinus, a word which was thereafter received with general favor."
COTANGENT. Bradwardine used the term umbra recta.
Magini (1609) used tangens secunda.
Cotangent was coined in Latin by Edmund Gunter (1581-1626) in 1620 in Canon Triangulorum, or Table of Artificial Sines and Tangents. Gunter wrote cotangens.
The term COUNTABLE was introduced by Georg Cantor (1845-1918) (Kline, page 995). According to the University of St. Andrews website, he introduced the word in a paper of 1883.
COUNTING NUMBER is dated ca. 1965 in MWCD10.
COVARIANCE is found in 1930 in The Genetical Theory of Natural Selection by R. A. Fisher (David, 1998).
Earlier uses of the term covariance are found in mathematics, in a non-statistical sense.
COVARIANT was used in 1853 by James Joseph Sylvester (1814-1897) in Phil. Trans.: "Covariant, a function which stands in the same relation to the primitive function from which it is derived as any of its linear transforms do to a similarly derived transform of its primitive" (OED2).
According to Karen Hunger Parshall in James Joseph Sylvester: Life and Work in Letters, Sylvester coined this term.
Cayley at first used the term hyperdeterminant in this sense.
The term COVARIANT DIFFERENTIATION was introduced by Ricci and Levi-Civita (Kline, page 1127).
COVERING (Belegung, from the verb Belegen = cover) was used by Georg Cantor in his last works (1895-97) on set theory, as shown in the following passage from Philip Jourdain's translation (Contributions to the founding of the theory of transfinite numbers, Dover, 1915, p. 94):
By a "covering of the aggregate N with elements of the aggregate M," or, more simply, by a "covering of N with M," we understand a law by which with every element n of N a definite element of M is bound up, where one and the same element of M can come repeatedly into application. The element of M bound up with n is (...) called a "covering function of n". The corresponding covering of N will be called f (N).
Curiously, at the end of his Introduction Jourdain says that
The introduction of the concept of "covering" is the most striking advance in the principles of the theory of transfinite numbers from 1885 to 1895, (...)
Nevertheless, as everybody nowadays can see, a "covering of N with M" in Cantor's terminology is just a function f : N -> M; and his "covering of N" is nothing more than the direct image of N under f - a concept which was introduced for the first time (at least, in a mathematically recognizable form) in Dedekind's Was sind und Was sollen die Zahlen? (1887, §21) [Carlos César de Araújo].
CRAMER-RAO INEQUALITY in the theory of statistical estimation. The inequality was obtained independently by at least three authors in the 1940s. The name "Cramér-Rao inequality" appears in Neyman & Scott (Econometrica, 1948) and recognises the English language publications of Cramér (1946 Mathematical Methods of Statistics) and Rao (1945 Bull. Calcutta Math. Soc. 37, 81-91). L. J. Savage (Foundations of Statistics 1954) drew attention to the work of Fréchet (1943) and Darmois (1945) and "tentatively proposed" the impersonal information inequality. However the name "Cramér-Rao inequality" has remained popular, though the "Fréchet-Darmois-Cramer-Rao inequality" figures in some French literature. [This entry contributed by John Aldrich, with some information from David (1995).]
CRITERION OF INTEGRABILITY is found in 1816 in Edin. Rev. XXVII: "The theorem, which is called the Criterion of Integrability" (OED2).
The term CRITERION OF SUFFICIENCY was used by Sir Ronald Aylmer Fisher in his paper "On the Mathematical Foundations of Theoretical Statistics," in Philosophical Transactions of the Royal Society, April 19, 1922: "The complete criterion suggested by our work on the mean square error (7) is: -- That the statistic chosen should summarise the whole of the relevant information supplied by the sample. This may be called the Criterion of Sufficiency" [James A. Landau].
CRITICAL POINT is found in 1871 in A General Geometry and Calculus by Edward Olney [University of Michigan Historic Math Collection].
CRITICAL REGION is dated 1951 in MWCD10.
CROSS PRODUCT is found on p. 61 of Vector Analysis, founded upon the lectures of J. Willard Gibbs, second edition, by Edwin Bidwell Wilson (1879-1964), published by Charles Scribner's Sons in 1909:
The skew product is denoted by a cross as the direct product was by a dot. It is written
C = A X B
and read A cross b. For this reason it is often called the cross product.
(This citation contributed by Lee Rudolph.)
CROSS-RATIO. According to Taylor (p. 257), cross-ratio first appeared in Elements of Dynamic, Part 1, Kinematic (1878), p. 42, by William Kingdon Clifford (1845-1879). Clifford wrote "The ratio ab.cd : ac.bd is called a cross-ratio of the four points abcd ..."
CUBE. The word "cube" was used by Euclid. Heron used "hexahedron" for this purpose and used "cube" for any right parallelepiped (Smith vol. 2, page 292).
CUBE ROOT is found in English in 1679 in Moxon, Math. Dict. "Cube Root, the Root or Side of the third Power: So if 27 be the Cube, 3 is the Side or Root" (OED2).
The word CUBOCTAHEDRON was coined by Kepler, according to John Conway.
CUMULANT was used by James Joseph Sylvester in Phil. Trans. (1853) 1. 543: "The denominator of the simple algebraical fraction which expresses the value of an improper continued fraction" (OED2).
In statistics, cumulant is found in 1931 in R. A. Fisher and J. Wishart, "The Derivation of the Pattern Formulae of Two-Way Partitions from Those of Simpler Patterns," Proceedings of the London Mathematical Society, Ser. 2, 33, 195-208 (David, 1995).
Cumulant is a contraction of cumulative moment function, which Fisher used when he first discussed these quantities in his "Moments and Product Moments of Sampling Distributions," Proceedings of the London Mathematical Society, Series 2, 30, 199-238 (1929). The cumulative moment function of a particular order is a function of moments of the same and lower orders which motivates the name. The term "cumulant" was suggested by Hotelling (see J. Amer. Stat Assoc., 28, 1933, 374 and David 2001). Hald (pp. 344-9) describes how several earlier authors had used these quantities, most notably T. N. Thiele [John Aldrich].
CURL. James Clerk Maxwell wrote the following letter to Peter Guthrie Tait on Nov. 7, 1870:
Dear Tait,
What do you call this? Atled?
I want to get a name or names for the result of it on scalar or vector functions of the vector of a point.
Here are some rough-hewn names. Will you like a good Divinity shape their ends properly so as to make them stick?
(1) The result of applied to a scalar function might be called the slope of the function. Lamé would call it the differential parameter, but the thing itself is a vector, now slope is a vector word, wheras parameter has, to say the least, a scalar sound.
(2) If the original function is a vector then applied to it may give two parts. The scalar part I would call the Convergence of the vector function, and the vector part I would call the Twist of the vector function. Here the word twist has nothing to do with a screw or helix. If the word turn or version would do they would be better than twist, for twist suggests a screw. Twirl is free from the screw notion and is sufficiently racy. Perhaps it is too dynamical for pure mathematicians, so for Cayley's sake I might say Curl (after the fashion of Scroll). Hence the effect of on a scalar function is to give the slope of that scalar, and its effect on a vector function is to give the convergence and the twirl of that function. The result of 2 applied to any function may be called the concentration of that function because it indicates the mode in which the value of the function at a point exceeds (in the Hamiltonian sense) the average value of the function in a little spherical surface drawn round it.
Now if be a vector function of and F a scalar function of ,
F is the slope of F
V
. F is the twirl of the slope which is necessarily zero
S
. F = 2F is the convergence of the slope, which is the concentration of F.
Also S
is the convergence of
V
is the twirl of .
Now, the convergence being a scalar if we operate on it with , we find that it has a slope but no twirl.
The twirl of is a vector function which has no convergence but only a twirl.
Hence, 2 , the concentration of , is the slope of the convergence of together with the twirl of the twirl of , the sum of two vectors.
What I want to ascertain from you if there are any better names for these things, or if these names are inconsistent with anything in Quaternions, for I am unlearned in quaternion idioms and may make solecisms. I want phrases of this kind to make statements in electromagentism and I do not wish to expose either myself to the contempt of the initiated, or Quaternions to the scorn of the profane.
In 1873 by Maxwell wrote in A Treatise on Electricity and Magnetism "I propose (with great diffidence) to call the vector part...the curl."
CURRIED FUNCTION. According to an Internet web page, the term was proposed by Gottlob Frege (1848-1925) and first appears in "Uber die Bausteine der mathematischen Logik", M. Schoenfinkel, Mathematische Annalen. Vol 92 (1924). The term was named for the logician Haskell Curry.
CURVATURE. Nicole Oresme assumed the existence of a measure of twist called curvitas. Oresme wrote that the curvature of a circle is "uniformus" and that the curvature of a circle is proportional to the multiplicative inverse of its radius.
A translation of Isaac Newton in Problem 5 of his Methods of series and fluxions is:
A circle has a constant curvature which is inversely proportional to its radius. The largest circle that is tangent to a curve (on its concave side) at a point has the same curvature as the curve at that point. The center of this circle is the "centre of curvature" of the curve at that point.
Curvature appears in English in 1710 in Lexicon technicum, or an universal English dictionary of arts and sciences by John Harris, in which it is stated that "the Curvatures of different Circles are to one another Reciprocally as their Radii" (OED1).
CURVE FITTING appears in a 1905 paper by Karl Pearson. A footnote therein references a paper "Systematic Fittings of Curves" in Biometrika which may also contain the phrase [James A. Landau].
CURVE OF PURSUIT. The name ligne de poursuite "seems due to Pierre Bouguer (1732), although the curve had been noticed by Leonardo da Vinci" (Smith vol. 2, page 327).
CYCLE (in a modern sense) was coined by Edmond Nicolas Laguerre (1834-1886).
CYCLIC GROUP. The term cyclical group was used by Cayley in "On the substitution groups for two, three, four, five, six, seven, and eight letters," Quart. Math. J. 25 (1891).
The term also appears in 1898 in Introduction to the theory of analytic functions by J. Harkness and F. Morley: "Such a group is called a cyclic group and S is called the generating substitution of the group."
CYCLIC PERMUTATION. Permutation circulaire is found in Cauchy's 1815 memoir "Sur le nombre des valeurs q'une fonction peut acquérir lorsqu'on permute de toutes les manières possibles les quantités qu'elle renferme" (Journal de l'Ecole Polytechnique, Cahier XVII = Cauchy's Oeuvres, Second series, Vol. 13, pp. 64--96.) This usage was found by Roger Cooke, who believes this is the first use of the term.
CYCLIC QUADRILATERAL. Inscriptible polygon is found in about 1696 in Scarburgh, Euclid (1705): "Polygons do arise, that are mutually with a Circle, or with one another Inscriptible and Circumscriptible" (OED2).
Inscribable is found in the 1846 Worcester dictionary.
Inscriptible quadrilateral is found in 1857 in Mathematical Dictionary and Cyclopedia of Mathematical Science.
Cyclic quadrilateral is found in 1888 in Casey, Plane Trigonometry (OED2).
The CYCLOID was named by Galileo Galilei (1564-1642) (Encyclopaedia Britannica, article: "Geometry"). According to the website at the University of St. Andrews, he named it in 1599.
CYCLOTOMY and CYCLOTOMIC were used by James Joseph Sylvester in 1879 in the American Journal of Mathematics.
CYLINDER was used by Apollonius (262-190 BC) in Conic Sections. | 29,746 | 117,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2017-22 | longest | en | 0.838339 |
https://www.thecodingforums.com/threads/how-to-convert-infix-notation-to-postfix-notation.702951/ | 1,568,945,882,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573801.14/warc/CC-MAIN-20190920005656-20190920031656-00550.warc.gz | 1,049,018,817 | 24,403 | # How to convert Infix notation to postfix notation
Discussion in 'C Programming' started by Tameem, Oct 26, 2009.
1. ### TameemGuest
i have a string as (a+b)+8-(c/d) in Infix form.
how can i convert it to postfix form using C language,,,????
Tameem, Oct 26, 2009
2. ### spinoza1111Guest
I will assume that that string is not the ONLY string you have to
convert. If it is, then the answer is
int main()
{
printf("ab+8+cd/-\n");
}
Each one of the following productions should be written as a separate
C function.
expression -> additionFactor [ "+"|"-" expression ]
additionFactor -> multiplicationFactor [ "*"|"/" expression ]
multiplicationFactor -> term
term -> LETTER | NUMBER | "(" expression ")" // Rightmost must balance
Write code using the above "productions" that emits postfix code. For
each production write one C function. I won't write C for you because
I don't like C, I suck at it through disuse consequent on my dislike,
and it's your homework assignment, not mine.
"->" means "consists of the following material left to right"
"|" means "or"
Lower case and camelCase words are grammar categories that will
correspond to procedures in YOUR code
Upper case words are lexemes recognised by C code
Quoted material occurs as is.
Material in square brackets is optional. An expression could be just
multiplicationFactor and a multiplication factor a number. Therefore,
"3" is an expression.
When you parse a "term" as "(" expression ")" you must look ahead in
the overall algorithm I shall give you to find, not the first right
parenthesis, but the one that actually balances the left parenthesis.
To do this, maintain a counter. Increment it by one for each left
parenthesis the lookahead sees: decrement it by one for each right
parenthesis. When you see right parenthesis and the counter goes to
zero, you have found the right parenthesis.
OK, here's the untested and uncompiled C Sharp like pseudo code for
"expression": convert it to C:
string expressionMain(string strInstring)
{
int intIndex1 = 0;
string strPolish = "";
return expression(strInstring, ref intIndex1, ref strPolish);
}
string expression(string strInstring, ref int intIndex1, ref string
strPolish)
{
ref intIndex1,
ref strPolish)) return "Not valid";
if (strInstring[intIndex]=='+' || strInstring[intIndex]=='-')
{
int intIndexSave = intIndex1; intIndex1++;
if (expression(strInstring, ref intIndex1, ref strPolish))
strPolish += strInstring[intIndexSave];
else return("Not valid");
}
return strPolish;
}
C Sharp idioms: ref followed by type in a formal parameter list in a
function decl is like using asterisk in front of a parameter in C. It
means that the parameter is passed by reference. In C Sharp, a very
cool language, you must ALWAYS use the keyword ref in the "actual
parameters" passed when you call the function just so you know what
the hell you are doing, as opposed to C which likes to fool you.
Chapter 3 of my book Build Your Own .Net and Compiler (Edward G.
Nilges, Apress May 2004) provides a complete solution for this
homework but in Visual Basic .Net. Therefore this is not a commercial
promotion. If you find my book useful, buy it, check it out of the
library, or steal it.
spinoza1111, Oct 26, 2009
3. ### Michael SchumacherGuest
This has very little to do with the /language/ C, i.e., there's
no "built-in" way to accomplish this conversion. However, it's
of course possible to implement such a converter in C, e.g., by
writing a recursive-descent parser (for a good deal of details,
see <http://en.wikipedia.org/wiki/Operator-precedence_parser>).
You might also want to take a look at the GNU assembler (gas,
which is part of the GNU binutils package), which implements
such a recursive-descent parser to evaluate constant expressions.
mike
Michael Schumacher, Oct 26, 2009
4. ### GeneGuest
It this homework?
Gene, Oct 26, 2009
5. ### Ben BacarisseGuest
This is not the usual grammar for expressions and is poor advice for
someone learning this stuff.
To the OP: please don't; Spinoza1111 is leading you astray. Post in a
group about programming (comp.programming is a good one) for better
be open about that and explain how far you have been able to get on
<snip off-topic C#>
Ben Bacarisse, Oct 26, 2009
6. ### spinoza1111Guest
Ben, please indicate the errors and don't imitate Seebach by
discussing personalities instead of ideas. As it is, you sound like a
fundamentalist *imam*.
spinoza1111, Oct 27, 2009
7. ### spinoza1111Guest
I would have thought given your technical capabilities that you would
be able, instead of engaging in Fox news style politics of personal
destruction, to identify all flaws in my approach, whether in the
formal grammar of C# pseudo code.
If you cannot, I need an apology from you. NOW, punk.
C# is what C should be. As I indicated, C Sharp was used as pseudocode
because it is probably a homework assignment and the OP needs to do
this homework. The OP probably received poor algorithm guidance from
the teacher, especially if the teacher was like most people here.
spinoza1111, Oct 27, 2009
8. ### Ben BacarisseGuest
[The formatting errors in the above were added by you. My typing is
It's not topical here. There must be lots of groups where you can get
advice on writing a grammar for simple expressions, but I don't know,
off hand, where would be the best place. I suggested comp.programming
to the OP, but that is probably not the best place for your question.
That's OK then.
<snip>
Ben Bacarisse, Oct 27, 2009
9. ### spinoza1111Guest
You have not shown that it is INCORRECT. It's possible that the
"usual" grammar is incorrect (as in the case of left recursion) given
Gresham's Law that 99% of everything is crap. You need to show that
the grammar is either actually incorrect or not suitable for manual
interpretation, and I do not, personally think you qualified to do so.
This takes experience in writing an actual parser by hand.
In the first production it is obvious that on return from
greater than end. If the index is past the end, we're done. If it is
not, the parser can check either for a plus or a minus.
If either character occurs the parser then calls itself with a source
string that is necessarily smaller guaranteeing the safety of the
recursion. But, there must be an expression when there is an operator.
Disprove this informal logic and desist the Fox news crap, please.
His goal is to write a C program, but we don't want to give him the
homework solution. It appears that his instructor in fact knows about
as much as you about writing a recursive descent parser by hand.
Therefore he's in the right place.
spinoza1111, Oct 27, 2009
10. ### John BodeGuest
As others have said, there's no C-specific way of doing this.
However, here are the general steps you'll need to take:
First, you will need to separate the string into distinct *tokens*.
In your example, you have four kinds of tokens -- operators
('+','-','*','/'), separators ('(',')'), constants ('8'), and
identifiers ('a','b','c','d'). Generally, you'll have a set of rules
that determines what characters make up a particular kind of token
(for example, an identifier must start with a letter and consist of
only letters and digits: "a1", "foo", "x", etc.). As you scan each
character, you'll check against the currently active rule and if it
matches, you'll add it to the token.
There are two ways to go on the next step. You can either build a
full-blown recursive descent parser (a decent explanation appears on
Wikipedia), or you can use a simple stack-based converter. For the
stack-based converter, as you read the expression, you'll pass
constants and identifiers straight to output, and you'll push and pop
operators on the stack based on their precedence. If the operator on
the top of the stack has a lower precedence than the current operator,
then push the current operator. If the operator on the stack has an
equal or higher precedence than the current operator, then pop the
operator off the stack and write it to output, then push the current
operator on the stack. lparens are always pushed onto the stack;
rparens cause everything on the stack to be popped to the next
lparen. Neither lparens nor rparens are written to output.
Given your example, the operations would look something like this:
1. Read '(', push it onto the stack. Stack contents = {'('}
2. Read 'a', write it to output. Output string = "a"
3. Read '+', push it on the stack: {'(', '+'}
4. Read 'b', write it to output: "a b"
5. Read ')', pop everything off the stack to the next '(', but don't
write '(' to output. Output string is "a b +", stack contents = {}
6. Read '+', push it onto the stack: {'+'}
7. Read '8', write it to output: "a b + 8"
8. Read '-', has same precedence as '+', so pop '+' from stack, write
it to output, and push '-': "a b + 8 +", {'-'}
9. Read '(', push it onto the stack: {'-', '('}
10. Read 'c', write it to output: "a b + 8 + c"
11. Read '/', push onto stack: {'-', '(', '/'}
12. Read 'd', write to output: "a b + 8 + c d"
13. Read ')', pop everything to '(': "a b + 8 + c d /", {'-'}
14. End of string, pop remainder of stack: "a b + 8 + c d / -"
John Bode, Oct 27, 2009
11. ### Kenny McCormackGuest
Quibble. Sturgeon's law (although the story may be apocryphal and/or
tongue-in-cheek).
Gresham's law says that the bad drives out the good, which is, not
coincidentally, also very true and on display in comp.lang.c
Kenny McCormack, Oct 27, 2009
12. ### spinoza1111Guest
Might be overkill given that the problem statement implies that
symbols are a letter, and scanning non-integer tokens might be beyond
the requirements of the problem...since the OP's teacher hasn't told
him what a number is, and a fully general number recognizer that
recognizes signs, exponents, and decimal points is a project in
itself. Therefore a number is probably an integer, and this can be
recognized by adhoc code. A separate pass for lexical analysis is
probably overkill.
Very nice and correct, perhaps clearer than a grammar based solution,
with the caveats that when the top of the stack is an LP, anything
other than RP must be stacked and RP is an error. Also may not catch
some errors.
spinoza1111, Oct 28, 2009
13. ### spinoza1111Guest
Give me a break. What part of hyperbole don't you faggots understand?
Furthermore, any claims you might have had to being treated in a
politically correct fashion were made moot when you started to join
electronic mobs and to propagate lies about individuals here. Frankly,
I do not understand groups who form identities as "oppressed" and
their claims to be spoken about in a civil fashion when they do not
extend this civility towards others or their own members.
According to some of my sources, systematic "trashing" of targeted
individuals began in the women's movement as the complement to women's
demands to be spoken of in a new language cleansed of "honey" and
"babe". It was directed at women inside the movement. Likewise, Larry
Kramer, a gay man, has spoken of the way in which gay people demand
kid glove treatment and cleansed language from straights...while
trashing each other and infecting each other with AIDs.
Groups demand charity that is denied, that they deny, to individuals.
spinoza1111, Oct 28, 2009
14. ### spinoza1111Guest
Ben Bacarisse noticed that I used the wrong grammar and informed me by
email in a very professional way. The grammar should be the following
for direct conversion to code!!
additionFactor → multiplicationFactor [ "*"|"/" multiplicationFactor ]
*
multiplicationFactor → LETTER | NUMBER | "(" expression ")"
That is, an expression is a series of one or more additionFactors.
When there are at least two, they are separated by plus or minus. The
replacement of the right recursive call to expression by the iteration
over additionFactor (the asterisk expressing iteration) causes the
addition/subtraction to left associate naturally. Likewise for the
call to expression in the second production.
I also made this same error when developing the code for Build Your
Own .Net Language and Compiler but did not have Ben Bacarisse's expert
assistance. Instead, I discovered it in testing the code for the first
toy compiler in ch 3 and mention the problem, and its solution, in my
book. I was too lazy when posting to check my own goddamn book and
shall try to be more diligent in the future. However, such diligence
shall be pearls before swine in some cases and in others the exception
that proves the rule.
When I make an error I make each error twice, it seems to give the old
brain extra time to relearn things it learned in the past; in the past
week I have made the comma-as-operator-versus-separator as well as
this error twice. I do regret if the original post was useless to the
original poster, because as Ben pointed out, it right-associative.
But (big but): my post was of far greater quality, errors and all,
that the uncollegial crap which constitutes 99% of the postings here.
As was Ben's correction. This dialogue is what this facility should be
all the time, not denials that something is true unaccompanied by
information as to what is, and the politics of personal destruction.
Because of this, no response to this thread by Richard Heathfield will
be either read by me nor responded to by me.
Ben: thanks for your assistance and the very professional manner in
which you made it. If there are other errors please let me know.
Likewise for all other readers, except Richard Heathfield. I do not
think he's qualified to discuss this issue.
spinoza1111, Oct 28, 2009
15. ### spinoza1111Guest
Here is a C Sharp implementation of recursive descent to translate
infix to Polish notation. It runs as a command line program. If an
infix expression is entered in its command line, this expression is
converted to Polish notation and the result is displayed.
If the command line contains no operands, a test suite is run. It
should produce this output:
Expect "2 4 3 + /": "2 4 3 + /"
Expect "2 4 / 3 +": "2 4 / 3 +"
Expect "2 4 3 + /": "2 4 3 + /"
Expect "10 113 2 2 4 3 + / + 2 + 2 + - + a *": "10 113 2 2 4 3 + / + 2
+ 2 + - + a *"
Expect error: "Error detected at or near character 3 in "11+":
Expect error: "Error detected at or near character 2 in "11 +":
Expect "10 113 + a *": "10 113 + a *"
Expect "10 113 + a *": "10 113 + a *"
Expect "1": "1"
Expect "a b + c -": "a b + c -"
Expect "1 1 +": "1 1 +"
Here is the code. It should constitute the complete Program.CS file in
a .Net C Sharp command line mode application. It was tested under .Net
C Sharp Express, a free product available from Microsoft. Comments are
welcome.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace infix2Polish
{
class Program
{
static void Main(string[] strArgs)
{
if (strArgs.GetUpperBound(0) < 0)
Console.WriteLine(tester());
else
Console.WriteLine(infix2Polish(strArgs[0]));
return;
}
private static string tester()
{
return "Expect \"2 4 3 + /\": \"" +
infix2Polish("2/(4+3)") + "\"" +
Environment.NewLine +
"Expect \"2 4 / 3 +\": \"" +
infix2Polish("2/4+3") + "\"" +
Environment.NewLine +
"Expect \"2 4 3 + /\": \"" +
infix2Polish("((2/(4+3)))") + "\"" +
Environment.NewLine +
"Expect " +
"\"10 113 2 2 4 3 + / + 2 + 2 " +
"+ - + a *\": \"" +
infix2Polish
("((10+(113-(2+((((((2/(4+3)))))))+2+2))))*a") +
"\"" +
Environment.NewLine +
"Expect error: \"" +
infix2Polish("11+") + "\"" +
Environment.NewLine +
"Expect error: \"" +
infix2Polish("11 +") + "\"" +
Environment.NewLine +
"Expect \"10 113 + a *\": \"" +
infix2Polish("(10+113)*a") + "\"" +
Environment.NewLine +
"Expect \"10 113 + a *\": \"" +
infix2Polish("((10+113))*a") + "\"" +
Environment.NewLine +
"Expect \"1\": \"" +
infix2Polish("1") + "\"" +
Environment.NewLine +
"Expect \"a b + c -\": \"" +
infix2Polish("a+b-c") + "\"" +
Environment.NewLine +
"Expect \"1 1 +\": \"" +
infix2Polish("1+1") + "\"" +
Environment.NewLine;
}
private static string infix2Polish(string strInfix)
{
string strPolish = "";
int intIndex1 = 0;
string strErrorMessage = "";
if (!expression(strInfix,
ref strPolish,
ref intIndex1,
ref strErrorMessage))
return strErrorMessage;
return strPolish.Trim();
}
private static bool expression
(string strInfix,
ref string strPolish,
ref int intIndex,
ref string strErrorMessage)
{
return expression(strInfix,
ref strPolish,
ref intIndex,
ref strErrorMessage,
strInfix.Length - 1);
}
private static bool expression
(string strInfix,
ref string strPolish,
ref int intIndex,
ref string strErrorMessage,
int intEnd)
ref strPolish,
ref intIndex,
ref strErrorMessage,
intEnd))
return errorHandler
(strInfix,
intIndex,
ref strErrorMessage,
while (intIndex <= intEnd)
{
if ((chrAddOp = strInfix[intIndex]) != '+'
&&
return errorHandler
(strInfix,
intIndex,
ref strErrorMessage,
"Expected addition or subtraction " +
intIndex++;
if (intIndex > intEnd
||
ref strPolish,
ref intIndex,
ref strErrorMessage,
intEnd))
return errorHandler
(strInfix,
intIndex,
ref strErrorMessage,
"operator is not followed by " +
strPolish += chrAddOp.ToString() + ' ';
}
return true;
}
(string strInfix,
ref string strPolish,
ref int intIndex,
ref string strErrorMessage,
int intEnd)
{// addFactor -> mulFactor [ *|/ mulFactor ] *
if (!mulFactor(strInfix,
ref strPolish,
ref intIndex,
ref strErrorMessage,
intEnd))
return errorHandler
(strInfix,
intIndex,
ref strErrorMessage,
"Expected multiplication factor " +
char chrMulOp = ' ';
while (intIndex <= intEnd
&&
((chrMulOp = strInfix[intIndex]) == '*'
||
chrMulOp == '/'))
{
intIndex++;
if (intIndex > intEnd
||
!mulFactor(strInfix,
ref strPolish,
ref intIndex,
ref strErrorMessage,
intEnd))
return errorHandler
(strInfix,
intIndex,
ref strErrorMessage,
"Expected multiplication factor " +
strPolish += chrMulOp.ToString() + ' ';
}
return true;
}
private static bool mulFactor
(string strInfix,
ref string strPolish,
ref int intIndex,
ref string strErrorMessage,
int intEnd)
{// mulFactor -> LETTER|NUMBER|"(" expression ")"
if (strInfix[intIndex] >= 'a'
&&
strInfix[intIndex] <= 'z')
{
strPolish += strInfix[intIndex++].ToString() +
' ';
return true;
}
int intIndex1 = intIndex;
while(intIndex <= intEnd
&&
(strInfix[intIndex] >= '0'
&&
strInfix[intIndex] <= '9'))
strPolish += strInfix[intIndex++];
if (intIndex > intIndex1)
{
strPolish += ' ';
return true;
}
if (strInfix[intIndex] == '(')
{
intIndex++;
int intLevel = 1;
{
{
case '(':
{
intLevel++; break;
}
case ')':
{
intLevel--;
if (intLevel == 0) goto exit;
break;
}
default: break;
}
}
exit:
if (!expression(strInfix,
ref strPolish,
ref intIndex,
ref strErrorMessage,
return errorHandler
(strInfix,
intIndex,
ref strErrorMessage,
"Nested expression invalid");
intIndex++;
}
return true;
}
private static bool errorHandler
(string strInfix,
int intIndex,
ref string strBase,
string strMessage)
{
strBase +=
(strBase == "" ? "" : "; ") +
"Error detected at or near " +
"character " +
intIndex.ToString() + " " +
"in " + "\"" + strInfix + "\": " +
strMessage;
return false;
}
}
}
spinoza1111, Oct 31, 2009
16. ### spinoza1111Guest
I published it here to show the OP that the grammar based approach
works, and give him "pseudo code" for his C assignment, without doing
his homework for him.
Since by now he's probably handed in his assignment, I challenge you
to translate the C Sharp code to C. Should be easy since you're such a
Studly Dudley C programmer. >
spinoza1111, Oct 31, 2009
17. ### Ben BacarisseGuest
That would not be useful. Your code does not parse the language
defined by the grammar, and what it does do, it does in a rather
<snip>
Ben Bacarisse, Nov 1, 2009
18. ### spinoza1111Guest
If you think the grammar is correct, say so clearly. You imply it but
are too graceless to say it. Now the issue is whether the code
supports the correct grammar correctly.
You present no evidence for the first claim. As to the second, it
means you've never seen this simple algorithm.
spinoza1111, Nov 1, 2009
19. ### spinoza1111Guest
Did you write it or did you steal it?
I need to see you actually write new C code. You have a big mouth when
it comes to putting down people but the only time when you put your
big mouth on the line (the Spark Notes test) you failed. I am giving
you another chance.
I wrote and debugged a recursive descent parser for my book. I wrote
it again here after correcting the grammar because I think it's
incumbent on each one of us to be able to orally present an algorithm,
and writing it anew and posting it for comments simulates this oral
presentation.
It also demonstrates that the material is sufficiently complex for
smart people to make stupid mistakes (Knuth has himself said he
consistently gets binary search wrong the first time), and that the
stupid people aren't the ones making the mistakes.
It's the people too ready to trash and if possible render unemployable
their fellow human beings (by creating a record here) while explaining
away their own mistakes as you explained away your failure to get a
decent grade on the Spark Notes test.
You are as such contemptible.
spinoza1111, Nov 1, 2009
20. ### SeebsGuest
Presumably wrote it. I mean, come on. A parser is NOT that hard; I've
written one from scratch (maybe two) and done a few with yacc, and while
the code in question doubtless sucked (hint: everyone's first language
sucks), it worked fine.
Er, why?
.... Wait, SPARK NOTES?
You're telling me that you are comparing Richard Heathfield to SPARK NOTES
and you think that a discrepancy means he's wrong?
Please tell me that it's still Halloween where you are and you went as
a troll.
.... Wait, uhm.
Does that mean the one in the book is wrong?
If so, then how did this not show up in debugging?
Stupid people make mistakes too.
I am totally fascinated by the concept of such a test. Is it ...
Hmm.
Well, I went and looked, and found:
http://www.sparknotes.com/cs/pointers/review/quiz.html
This is a funny quiz. It is full of errors.
Is this the quiz you used? | 5,728 | 21,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-39 | longest | en | 0.868964 |
https://www.ibpsguide.com/sectional-ibps-clerk-online-test-series-day-44/ | 1,632,365,021,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057416.67/warc/CC-MAIN-20210923013955-20210923043955-00114.warc.gz | 819,788,696 | 73,489 | # Sectional IBPS Clerk Online Test Series – Day-44
## Sectional IBPS Clerk Online Test Series – Day-44:
Dear Readers, IBPS Clerk Examination for the year 2017 was approaching very shortly, for that we have given the Sectional IBPS Clerk Online Test Series which consist of questions from all the three sections such as, Quantitative Aptitude, Reasoning Ability, and English Language. This Sectional IBPS Clerk Test Series will be provided on daily basis kindly make use of it.
[WpProQuiz 741]
### Click “Start Quiz” to attend these Questions and view Solutions
1). The total number of men, women, and children working in a factory is 24. They earn Rs. 3330 in a day. If the sum of the wages of men to that of women to that of children is in the ratio of 21:10:6 and if the wages of man, a woman, and a child are in the ratio 7:5:2, then how much does a woman earns in a day?
1. Rs. 150
2. Rs. 142
3. Rs. 135
4. Rs. 169
5. Rs. 125
2). In a competitive examination, the average marks obtained was 45. It was later discovered that there was some error in computerization and the marks of 90 candidates had to be changed from 80 to 50, and the average came down to 40 marks. The total number of candidates appeared in the examination isÂ
1. 520
2. 550
3. 540
4. 525
5. None of these
3). A roller is 120 cm long and has diameter 84 cm. If it takes 500 complete revolutions to level a play-ground, then determine to cost of levelling at the rate of 30 paise per m2.
1. Rs. 475.40
2. Rs. 375.45
3. Rs. 375.20
4. Rs. 475.20
5. None of these
4). Out of 15 students studying in a class, 7 are from Maharashtra, 5 from Karnataka and 3 from Goa. Four students are to be selected at random. What are the chances that at least one is from Karnataka?
1. 12/13
2. 11/13
3. 100/15
4. 51/15
5. None of these
5). The compound interest earned by Suresh on a certain amount at the end of two years at the rate of 8 p.c.p.a. was 1414.4. What was the total amount that Suresh got back at the end of two years in the form of principal plus interest earned?
a) 8500
b) 9914.4
c) 9014.4
d) 8914.4
e) None of these
Directions (6 – 10): Each of the questions below consists of a question and two statements numbered I and II given below it: You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and Give answer—
a) if the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
b) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
c) if the data in Statement I alone or in Statement II alone are sufficient to answer the question.
d) if the data, in both the Statements I and II together are not sufficient to answer the question.
e) if the data in both the Statements I and II together are necessary to answer the question.
6). Who is the youngest among P, Q, R, S and T?
1. Q is younger than R and S but not as young as P.
2. T is not the youngest
7). How many children are there in the class?
1. If arranged in ascending order of height Suma is tenth from the top.
2. In order of height, Suma is five positions above Ranjit who is eighth from the bottom.
8).
 What is the code for ‘going’ in a code language?
1. In that code language ‘where are you going’ is written as ‘ma ka ta re’
2. In that code language ‘going to college’ is written as ‘ lope ta’
9). How is Shamim related to Mr Varghese?
1. Shamim’s son is the only grandson of Mr Varghese.
2. Mr. Varghese has only one son.
10). Village X is in which direction with respect to village Y?
1. X is to the East of Z which is to North of Y.
2. X is to the North of L which is to the East of Y.
Directions (11-15): Read the following passage carefully and answer the questions given after the passage. Certain words/phrases have been printed in bold to help you locate them while answering some of the questions.
According to an old saw, Britons are more likely to get divorced than ditch their bank. That may not be quite true (in 2010, 3.8% of customers changed their bank accounts, while 1.1% of married people divorced), but it’s certainly the case that the British are loath to swap their current accounts. While 10% of electricity users switch every year, one study suggests that by 2023 only around 5% of bank customers will move their money in any one year. Human inertia has a lot to do with this – there’s always something more exciting to do than read one’s statements, even if it’s only collecting lint. But it’s not the only factor: Sir Donald Cruickshank’s report into banking in 2000 rightly identified the numerous obstacles to switching. It was also a theme of the Vicker’s review – which has prompted this week’s overhaul of the current account switching service where before it could take anything up to 30 working days to move from one bank to another, the new standardized system will take a maximum of seven. It should also be better at handling the migration of direct debits. No doubt this will make the lives of customers easier. Even so, this particular overhaul deserves only one cheer. For one thing, the monitoring of its success is not quite as energetic as it might be. For another, it is hard to see what incentive there is to swap banks, when there are so few big banks to choose from. In that respect, it is of a piece with the other financial reforms announced since the crash: fine, as far as it goes, but not the answer. Let us deal with the monitoring point first: the payments council, which sets the strategy for how bank payments are made in the UK, is also largely funded by banks and building societies. Perhaps then it is unsurprising to see that its criteria for judging whether the new switching service is a success are rather modest. They comprise: whether customers know about the service; whether customers feel confident in the service; and how well it performs. Not included on that brief list is how many people actually use the service. This would appear to be a classic case of the financial industry setting its own homework and making sure the questions aren’t too hard. For a more searching investigation, we shall have to wait until the Office of Fair Trading look into things in 2015. Why the wait? Moreover, why not take up the challenge from the Treasury select committee, and allow customers to carry their account numbers with them wherever they go – which would have been a big step forward. And without the breaking up of the giant banks, all customers are really being offered is a slightly easier, shorter journey between a rock and a hard place.
11). What according to the author is true to make customers feel easy to handle their money?
1. Direct transfer of money in a day to bank accounts
2. New system that will take only a week to thirty days to transfer
3. Switching service that will help to deposit money in other banks
4. Financial industries setting up in their own cities
5. None of these
12). According to the passage, which of the following is not true?
1. It is difficult to choose banks as they are in small numbers.
2. Success of switching service depends on number of customers using it.
3. Only 3.8 percent of customers switch every year.
4. It is advisable to allow customers to carry their account numbers.
5. None of these
13). Which of the following would be a suitable title of the passage.?
1. New Financial industry- a boon for your money
2. Changing banking system
3. Current account money- safe or unsafe
4. Financial switching for customers
5. Current account switching-move your money
14). What is the central idea of the passage?
1. Current account switching should be encouraged
2. Banks are the sole entity to take up the decision of switching bank accounts
3. Switching bank account depends on how effective and easier the service is for customers.
4. Switching bank account service is easier and helpful for customers.
5. None of these
Choose the word which is most SIMILAR in meaning of the word printed in bold as used in the passage.
15). Overhaul
1. step
2. agenda
3. improve
4. break
5. report | 2,043 | 8,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-39 | latest | en | 0.945092 |
https://www.airmilescalculator.com/distance/dou-to-sdu/ | 1,726,736,330,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651995.50/warc/CC-MAIN-20240919061514-20240919091514-00055.warc.gz | 569,634,344 | 37,394 | # How far is Rio De Janeiro from Dourados?
The distance between Dourados (Dourados Airport) and Rio De Janeiro (Santos Dumont Airport) is 753 miles / 1212 kilometers / 655 nautical miles.
The driving distance from Dourados (DOU) to Rio De Janeiro (SDU) is 896 miles / 1442 kilometers, and travel time by car is about 17 hours 38 minutes.
753
Miles
1212
Kilometers
655
Nautical miles
1 h 55 min
130 kg
## Distance from Dourados to Rio De Janeiro
There are several ways to calculate the distance from Dourados to Rio De Janeiro. Here are two standard methods:
Vincenty's formula (applied above)
• 753.194 miles
• 1212.147 kilometers
• 654.507 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 751.999 miles
• 1210.226 kilometers
• 653.470 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Dourados to Rio De Janeiro?
The estimated flight time from Dourados Airport to Santos Dumont Airport is 1 hour and 55 minutes.
## Flight carbon footprint between Dourados Airport (DOU) and Santos Dumont Airport (SDU)
On average, flying from Dourados to Rio De Janeiro generates about 130 kg of CO2 per passenger, and 130 kilograms equals 287 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Dourados to Rio De Janeiro
See the map of the shortest flight path between Dourados Airport (DOU) and Santos Dumont Airport (SDU). | 432 | 1,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.809369 |
https://www.esaral.com/q/in-a-survey-of-600-students-in-a-school-98964 | 1,725,952,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00691.warc.gz | 722,258,471 | 11,699 | # In a survey of 600 students in a school,
Question:
In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Solution:
Let U be the set of all students who took part in the survey.
Let T be the set of students taking tea.
Let C be the set of students taking coffee.
Accordingly, $n(\mathrm{U})=600, n(\mathrm{~T})=150, n(\mathrm{C})=225, n(\mathrm{~T} \cap \mathrm{C})=100$
To find: Number of student taking neither tea nor coffee i.e., we have to find $n\left(T^{\prime} \cap C^{\prime}\right)$.
$n\left(\mathrm{~T}^{\prime} \cap \mathrm{C}^{\prime}\right)=n(\mathrm{~T} \cup \mathrm{C})^{\prime}$
$=n(\mathrm{U})-n(\mathrm{~T} \cup \mathrm{C})$
$=n(U)-n(T \cup C)$
$=n(\mathrm{U})-[n(\mathrm{~T})+n(\mathrm{C})-n(\mathrm{~T} \cap \mathrm{C})]$
$=600-[150+225-100]$
$=600-275$
$=325$
Hence, 325 students were taking neither tea nor coffee. | 339 | 1,001 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-38 | latest | en | 0.885927 |
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# As state governments become less and less able to support higher educa
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Updated on: 13 Feb 2018, 03:35
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As state governments become less and less able to support higher education in the coming years, universities have been becoming more and more dependent on alumni networks, corporate sponsorships, and philanthropists.
A. become less and less able to support higher education in the coming years, universities have been becoming
B. are becoming less and less able to support higher education in the coming years, universities have become
C. become less and less able to support higher education in the coming years, universities will become
D. become less and less able to support higher education in the coming years, universities have become
E. are becoming less and less able to support higher education in the coming years, universities will become
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Originally posted by carcass on 25 Feb 2016, 09:30.
Last edited by Mahmud6 on 13 Feb 2018, 03:35, edited 1 time in total.
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Re: As state governments become less and less able to support higher educa [#permalink]
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25 Feb 2016, 10:53
2
1
As state governments become less and less able to support higher education in the coming years, universities have been becoming more and more dependent on alumni networks, corporate sponsorships, and philanthropists.
become less and less able to support higher education in the coming years, universities have been becoming -- <The clue here is "in the coming years. Means "universities will become" is the right tense. For example - As I do become more and more weak in the coming days, I have been dying soon and resting in the grave. Doesn't make any sense. It should be "As I become weak, I will die soon" Simple future and present tense >
are becoming less and less able to support higher education in the coming years, universities have become The present continuous tense talks always about the now. The sentence states "In the coming years" So usage of present continuous is wrong here. "As I m becoming weak in the coming years" This doesn't convey any sense at all. Both the tenses are wrong. "Are becoming" and "Have "
become less and less able to support higher education in the coming years, universities will become <Conveys the intended meaning. Both tenses are correct which establish the correct sequence of the sentence. "Once the govt is less able to support, universities will become dependent on others" . In simple term "Once A happens, B will happen.">
become less and less able to support higher education in the coming years, universities have become Wrong usage of tense as stated above "have become"
are becoming less and less able to support higher education in the coming years, universities will become <Usage of wrong tense "Are becoming">
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Re: As state governments become less and less able to support higher educa [#permalink]
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25 Feb 2016, 11:32
1
IMHO (C) as well , with sowragu for the same reason posted above...
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Re: As state governments become less and less able to support higher educa [#permalink]
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03 Mar 2016, 08:10
E can be correct too
have done/past simple can go with simple present to make a sequence.
past perfect can go with past simple to make a sequence
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Re: As state governments become less and less able to support higher educa [#permalink]
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04 Mar 2016, 17:25
thangvietnam wrote:
E can be correct too
have done/past simple can go with simple present to make a sequence.
past perfect can go with past simple to make a sequence
Present continuous tense are becoming cannot be used to refer to a future period (in coming years) .
(In colloquial English, however, we often use that way - tomorrow I am going to school; nevertheless GMAT does not prefer this usage - we should say: tomorrow I shall go to school).
For this reason are becoming in coming years is incorrect. Another simpler example would probably make it clearer:
I am becoming older in coming years..definitely wrong.
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Re: As state governments become less and less able to support higher educa [#permalink]
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06 Dec 2017, 18:35
"in the coming years" & future tense "will" is a very common pattern in SC gmat questions.
C & E are both close, but E is incorrect b/c the tense is wrongly used. V-ing is normally avoided in gmat.
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Re: As state governments become less and less able to support higher educa [#permalink]
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06 Dec 2017, 22:07
As I am doing more hardworking in coming days , I will be able to crack GMAT is better
Or as I do more hardworking in coming days, I will be able to crack
Choice C is better
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Re: As state governments become less and less able to support higher educa [#permalink]
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03 Feb 2018, 10:41
Once A happens B will happens is the correct structure. C is correct.
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Re: As state governments become less and less able to support higher educa [#permalink] 03 Feb 2018, 10:41
Display posts from previous: Sort by | 1,720 | 7,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-43 | latest | en | 0.959782 |
http://digitalsolutions.org/api-storage-tanks.html | 1,716,286,686,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058442.20/warc/CC-MAIN-20240521091208-20240521121208-00339.warc.gz | 10,502,426 | 10,517 | Guidelines for Pressure Relief Design
7.0 API Storage Tanks
A. For API Standard 620 Tanks
API 620 regulates the manufacture of low-pressure oil storage tanks that are larger than 300 feet in diameter. By design, the configuration of API 620 dictates that you should have a flat or elevated bottom to store your tank. The tank should be made of a material with a minimum thickness of 3/16 inch and have a central, vertical axis of revolution.
B. For API 620 Tanks
(2.5-15 psig design pressure) the pressure relief devices must prevent the tank pressure from exceeding 105 % of the design pressure, unless the hazard is due to external heat, such as a fire. In that case additional pressure relief devices must prevent the pressure from exceeding 20% of the design pressure.
C. For API Standard 650 Tanks
API 650 applies to both aluminum and carbon stainless tanks that are usually used in pipelines and refineries. It sets standards for above-ground storage tanks both open and closed top and of different sizes. By design, these cylindrical tanks should not weigh more than the roof plates, meaning their pressure should not exceed the atmospheric pressure.
D. API 650 Tanks (atmospheric to 2.5 psig, design pressure)
The pressure relief device must prevent the tank pressure from exceeding the design pressure with sufficient pressure differential and device diameter to provide adequate emergency relief.
E. For API Standard 2000 Tanks (Full vacuum to 15 psig design pressure)
When determining the possible causes of overpressure or vacuum in a tank, sizing is based on either SCFH Air or NCMH Air, consider the following :
a. Inbreathing from liquid movement out of the tank
i. Vip = 8.02 Vpc , when
1. Vip = inbreathing rate (SCFH) and
2. Vpc = liquid discharge (gpm)
b. Out-breathing from movement of liquid into the tank
i. Vop = 8.02*Vpf, where
1. Vop out-breathing rate (SCFH Air),
2. Vfp max fill rate(gpm)
ii. Vop =16.04*Vpr, if fluid is volatile
c. Inbreathing from thermal effects
i. VIT = 3.08*C Vtk0.7*Ri where
1. VIT max thermal flowrate cooling (SCFH Air),
2. C = depends on storage factors.
3. Vtk tank volume(ft3),
4. Ri (insulation factor)
d. Out-breathing from thermal effects
i. VOT = 1.51* Y*Vtk0.9* Ri , whre
1. VoT is max thermal flowrate heating (SCFH Air),
2. Y (factor for latitude),
3. Ri (insulation factor)
e. Fire case
i. Q = 3.091 * (Q * F /L)*(T/M) where
1. q - required flow capacity for fire SCFH air
2. Q - heat input from fire (use table 4 API 2000 sec. 3.3.3.3 0)
3. F - Environmental factor (see API 2000 Sec 3.3.3.3. table 3)
4. L - Latent Heat of vaporization of stored liquid at relieving conditions
5. T - absolute temperature degree Rankine
6. M - Molecular weight of vapor
f. Other circumstance resulting from equipment failure or operator error not covered in standard.
F. High temperature failure – A problem can arise if the anticipated metal temperature during relief is greater than the vessel design temperature. Additional or better insulation or water sprays may reduce the potential for the design temperature being exceeded, or reducing the set pressure on the relief device(s) may be necessary for existing vessel. | 910 | 3,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-22 | latest | en | 0.70136 |
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Question Paper
Quantitative Methods-II (132): January 2005
\u2022\u2022 Marks are indicated against each question.
1.If a fair die is thrown twenty times, what is the probability of obtaining a minimum of 3 in a maximum of 17
throws?
(a) 0.01759
(b) 0.01428
(c) 0.98572
(d) 0.98241
(e) 0.9997.
(2 marks)
<
>
2.The probability that a continuous random variable will assume a particular value is
(a) Zero
(b) Between zero and 0.50
(c) Between 0.50 and 1.00
(d) Equal to 0.50
(e) Unknown.
(1 mark)
<
>
3.A random variable, Y, has the following probability distribution:
Y
20
40
60
80
100
Probability
0.15
0.20
0.30
0.25
0.10
15 values ofY have been
observed. What is the probability that less than 12 observations are such that40\u2264 Y\u2264 80?
(a) 0.8441
(b) 0.1559
(c) 0.0668
(d) 0.4613
(e) 0.5387.
(2 marks)
<
>
4.Z is a binomial random variable. For 25 trials the standard deviation of Zis
6
.
What will be the standard deviation ofZ for 75 trials?
(a) 6
(b)
6
(c)
3 2
(d) 18
(e)
9 2
.
(1 mark)
<
>
5.If two fair dice are thrown simultaneously ten times, what is the likelihood that a minimum of 3 can be
simultaneously observed in both the dice, on at least nine throws?
(a) 0.00376
(b) 0.00030
(c) 0.00406
(d) 0.99594
(e) 0.44444.
(2 marks)
<
>
6.Given a probability distribution for a random variable, the covariance of the variable with itself is equal to
(a) 0
(b) 1
(c) The variance of the variable
(d) The standard deviation of the variable
(e) The expected value of the variable.
(1 mark)
<
>
7.Which of the following is false with regard to a hypergeometric distribution?
(a) The trials are not independent
(b) The probability of success is variable
(c) The outcomes can be labeled as success or failure
<
>
(d) The composition of the population remains unchanged
(e) The population is finite.
(1 mark)
8.A continuous uniform probability distribution is described by
(a) The expected value of the distribution
(b) The variance of the distribution
(c) The upper and lower limits of range of values of the distribution
(d) The mid-point of the range of values of the distribution
(e) Only the lower limit of the range of values of the distribution.
(1 mark)
<
>
9. Xand Y are two dependent random variables. The following details are available:
Standard deviation ofX = 6
Standard deviation ofY = 8
Let Z = 5X \u2013 10Y
It is known that the standard deviation ofZ is 70.
What is the value of covariance between the variablesX andY?
(a) 36
(b) 24
(c) 64
(d) 240
(e) 2400.
(1 mark)
<
>
10.There are 25 students in a class, which consists of 14 boys and 11 girls. 5 students of the class were absent
on a particular day. What is the probability that maximum two girls were absent?
(a) 0.2072
(b) 0.3768
(c) 0.416
(d) 0.584
(e) 0.6232.
(2 marks)
<
>
11.A random variable, Z, has the following probability distribution:
Z
40
60
80
100
120
Probability
0.15
0.25
0.30
0.20
0.10If twenty values ofZ are observed, how many
observations are expected to be less than 80?
(a) 2
(b) 4
(c) 6
(d) 8
(e) 10.
(1 mark)
<
>
12.If each of the values of a random variable is transformed into a z-value, the distribution of the resulting
values will have a standard deviation equal to
(a) Zero
(b) One
(c) The mean of the original distribution
(d) The standard deviation of the original distribution
(e) A variable, depending upon the shape and spread of the original distribution.
(1 mark)
<
>
13.Which of the following is a required condition for the probability distribution of a discrete random variable?
(a) Sum of all the probabilities = 0
(b) Sum of all the probabilities = 1
(c) Sum of all the probabilities < 0
(d) Sum of all the probabilities < 1
(e) Sum of all the probabilities > 1.
(1 mark)
<
>
14.If a z-value is to the left of the mean in the standard normal distribution, then its value
<
>
(a) Is negative
(b) Is positive
(c) Lies between 0 and
\u221e
(d) Is zero
(e) Is 1.
(1 mark)
15.Which of the following statements is false?
(a) The expected value of the sum of two random variables is equal to the sum of the expected values of
the two random variables
(b) The variance of the sum of two random variables is equal to the sum of the variances of the two
random variables if the random variables are independent
(c) The expected value of the sum of two random variables is equal to zero if the expected value of each of
the random variables is zero
(d) The variance of the sum of two random variables is equal to zero if the random variables are
independent
(e) The covariance between two dependent random variables is not zero.
(1 mark)
<
>
16.The probability that a continuous random variable will assume a value within a specific interval is provided
by
(a) The height of the graph of the probability distribution at any point within the interval
(b) The height of the graph of the probability distribution at any point within the interval plus a constant
(c) The height of the graph of the probability distribution at any point within the interval minus a constant
(d) The area under the graph of the probability distribution corresponding to the specified interval
(e) The area under the graph of the probability distribution corresponding to the specified interval
subtracted from 1.
(1 mark)
<
>
17.Which criterion of decision-making is not applicable when the decision maker has insufficient information to
assign any probabilities of occurrence to the various states of nature?
(a) Maximax criterion
(b) Maximin criterion
(c) Hurwicz criterion
(d) Regret criterion
(e) Expected value criterion.
(1 mark)
<
>
18.A discrete uniform random variable N, assumes all the integers starting from 1 to 25. The characteristic of the
discrete uniform random variable is that all possible values of the variable have equal probability.
What is the variance of the random variableN?
(a) 12.5
(b) 25
(c) 52
(d) 624
(e) 625.
(1 mark)
<
>
19.A medical shop sells a variety of drugs and one of the drugs sold by the shop is Drug X. This drug has a short
shelf life and it is quite expensive. Each packet of Drug X costs Rs.300 and can be sold for Rs.400, The manager of the shop has observed that the demand for this drug before its expiry has the following probability distribution:
Demand level before expiry (number of packets)
300
400
500
600
700
Probability
0.15
0.25
0.40
0.15
0.05
Any packet of this drug not sold before expiry has no value and has to be rejected. It is assumed that the
demand for this drug will be one of the aforementioned values only.
How many packets of Drug X should be stocked by the shop?
(a) 300
(b) 400
(c) 500
(d) 600
(e) 700.
(1 mark)
<
>
20.A businessman has to decide on the next month\u2019s stock level of a particular product that he deals in. He has
observed that the monthly demand for the product has the following probability distribution:
< | 1,929 | 7,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2014-35 | latest | en | 0.794334 |
https://nl.mathworks.com/matlabcentral/answers/837143-how-to-find-points-between-two-intersecting-lines | 1,708,566,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473598.4/warc/CC-MAIN-20240221234056-20240222024056-00120.warc.gz | 450,159,017 | 27,300 | # How to find points between two intersecting lines?
3 views (last 30 days)
Niraj Bal Tamang on 22 May 2021
Commented: Niraj Bal Tamang on 26 May 2021
I have two straight lines defined by equations (say) x+y=3 and 2x-3y=4. I have to find out the points which lies between these two lines in the xy plot as in the attached figure. Can anyone suggest me how can i find the points between two lines in a plot?
Thank You
Image Analyst on 23 May 2021
Those lines are not on your graph.
Niraj Bal Tamang on 23 May 2021
Sorry. I just gave a random equations of straight line for assumption. I am attaching the whole code and variables if you want to give it a try. I am trying to get the points between each of those lines for classification.
scatter(usarea,slope);
set(gca,'xscale','log','yscale','log');
xlabel('Upstream Drainage Area (sq km)');
ylabel('Slope (m/m)');
hold on
xline(7*10^5,'Color','r','LineStyle','--');%vertical line
hold on
plot([1.1*10^6 10^10],[0.15 0.15],'--r');%horizontal line
hold on
plot([7*10^5 5*10^9],[0.2 10^-4],'--r');%diagonal line
hold off
grid on
G A on 23 May 2021
Edited: G A on 23 May 2021
myData = [X,Y];
y1 = 3 - X;
y2 = 2/3*X + 4/3;
Y1 = Y(Y < y2 & Y > y1);
X1 = X(Y < y2 & Y > y1); % or X1 = X(Y == Y1)
myData1 = [X1,Y1];
G A on 26 May 2021
scatter(usarea,slope);
set(gca,'xscale','log','yscale','log');
xlabel('Upstream Drainage Area (sq km)');
ylabel('Slope (m/m)');
hold on
xline(7*10^5,'Color','r','LineStyle','--');%vertical line
hold on
plot([1.1*10^6 10^10],[0.15 0.15],'--r');%horizontal line
hold on
plot([7*10^5 5*10^9],[0.2 10^-4],'--r');%diagonal line
grid on
X=usarea;
Y=slope;
Y1=ones(size(X))*0.15; % horizontal line
B=1e7; % vertical line, I shifted it to the right for demonstration purpose
% defining diagonal line as log(y)=a*log(x)+b
x1=log(7e5);
x2=log(5e9);
y1=log(0.2);
y2=log(1e-4);
a=(y2-y1)/(x2-x1);
b=y1-a*x1;
Y2=exp(a*log(X)+b); % diagonal line
Y3 = Y(Y < Y1 & Y > Y2 & X > B); % Y-data between lines
X3 = X(Y < Y1 & Y > Y2 & X> B); % X-array
plot(X3,Y3,'.r');
plot(X,Y1,'--b');
plot(X,Y2,'--g');
xline(B,'--m');
hold off
Niraj Bal Tamang on 26 May 2021
Thank you so much.It's working now. | 795 | 2,155 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-10 | latest | en | 0.745765 |
https://www.codespeedy.com/rearranging-spaces-between-words-in-python/ | 1,679,767,020,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945368.6/warc/CC-MAIN-20230325161021-20230325191021-00519.warc.gz | 793,240,494 | 16,310 | # Rearranging spaces between words in Python
In this tutorial, we will be solving a Leet code weekly contest problem in which we are required to rearrange the spaces between a given string text using Python programming.
Assume that we are given a string that has a bit of text and in between those words we find that there are a number of unhealthy spaces between them. This means that instead of conventionally having a single space character in between words, the spaces may be more than that.
Our goal can be thought of as something similar to text formatting. We will be required to parse through the text and split the spaces equally between the words such that the number of space characters between any two words in a given text will be equal.
### Procedure:
If we are mathematically unable to perform such an operation, we are allowed to put the extra spaces in the end. I have solved the problem in the manner given below and used Python for its implementation.
In a variable a, we extract the words alone and store it in a list which is achieved by the line:
`a=text.split()`
#### Counting spaces:
Our first goal would be to count the number of spaces in the given text. I handle texts with only one word in an easier fashion since all you’ve got to do it just add those extra spaces at the front to the back. This done using this following block of code:
```for i in list(text):
if(i==" "):
spaces+=1```
#### Splitting the spaces:
If the text contains more than one words, again the first thing I do is to count the number of total number of spaces. Next step is to count if the even split is possible. If there are enough words to split the spaces between the words, we can easily reconstruct the correct string with the number of spaces between each word calculated as spaces / (len(a)-1), where len(a) represents the number of words.
```if(spaces%(len(a)-1)==0):
for j in a:
mystr+=j + int((spaces/(len(a)-1)))*" "
return (mystr[0:len(mystr)-int((spaces/(len(a)-1)))])```
The format of every word should be
word + equal number of spaces.
The number of spaces after every word should be n = (spaces/(len(a)-1)). This space must not be added after the last word.
The above block of code does the job when spaces can be equally split between the words. We check for the condition if it’s possible to split spaces among words equally. This is carried out using the if statement. After that, we append to the empty string, each word and the number of spaces that need to follow it. The variable j represents the word and (spaces/(len(a)-1)) indicates the number of spaces. We use len(a)-1 because the spaces have to be split between the words. For eg., if there are 2 words, there is only 1 position in between 2 words for the spaces to fit in.
After that, we see that for the last word also we have added the spacing which needs to be removed before returning. Therefore in the return statement, we exclude the last n spaces using string slicing.
#### Extraspaces:
If this is not the case, we will have to add the extra spaces towards the end which can be calculated as the reminder of the above operation. The new string is reconstructed as per this format i.e. by splitting the spaces equally and finally adding the remaining spaces towards the end. This logic has been implemented in the Python code below:
```extraspaces=spaces%(len(a)-1)
spaces=spaces-extraspaces
print(extraspaces, spaces)
for j in a:
mystr+=j + int((spaces/(len(a)-1)))*" "
print(len(text), len(mystr[0:len(mystr)-int((spaces/(len(a)-1)))] + extraspaces*" "))
return (mystr[0:len(mystr)-int((spaces/(len(a)-1)))] + extraspaces*" ")
```
In the above block of code, we first calculate the number of spaces we need to add towards the end as the reminder of the total number of spaces when divided by the no. of positions between words. Now the total number of spaces that we need to split equally between the words comes down to spaces – extraspaces. As usual we split the new spaces equally and construct the string. We are not done yet. We need to add the extraspaces towards the end as extraspaces*” “. This is done in the return statement.
### Examples:
Input:
`welcome everyone`
Output:
`welcome everyone`
Input:
`" this is a sentence "`
Output:
`"this is a sentence"`
We see that we have split the spaces (9) equally between the three words.
Input:
`"Code Speedy Tech"`
Output:
`"Code Speedy Tech "`
Here, we have 5 spaces and 3 words. We can not equally split them. There is one extra space. So we places two spaces after every word and after the last word, place the extra space.
#### Here is the complete Python code to the above logic:
```class Solution:
def reorderSpaces(self, text: str) -> str:
spaces=0
print(list(text))
a=text.split()
print(a)
if(len(a)==1):
if(text[0]==" "):
for k in text:
if(k==" "):
spaces+=1
return str(a[0])+ spaces*" "
else:
for i in list(text):
if(i==" "):
spaces+=1
mystr=""
if(spaces%(len(a)-1)==0): #Condition when spaces can be equally split between the words
for j in a:
mystr+=j + int((spaces/(len(a)-1)))*" " #For every word, add the required number of spaces after it
return (mystr[0:len(mystr)-int((spaces/(len(a)-1)))])
else:
extraspaces=spaces%(len(a)-1)
spaces=spaces-extraspaces
print(extraspaces, spaces)
for j in a:
mystr+=j + int((spaces/(len(a)-1)))*" "
print(len(text), len(mystr[0:len(mystr)-int((spaces/(len(a)-1)))] + extraspaces*" "))
return (mystr[0:len(mystr)-int((spaces/(len(a)-1)))] + extraspaces*" ")
``` | 1,373 | 5,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-14 | latest | en | 0.91911 |
https://www.physicsforums.com/threads/proof-by-induction-help.258672/ | 1,571,396,868,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682037.37/warc/CC-MAIN-20191018104351-20191018131851-00255.warc.gz | 1,025,376,500 | 15,651 | # Proof by induction help.
#### Ed Aboud
1. Homework Statement
Prove by induction on k that for all integers
$$\frac{d}{dx} \prod_{i=1}^k f_i (x) = (\sum_{i=1}^k \frac{ \frac{d}{dx} f_i (x)}{f_i (x)} ) \prod_{i=1}^k f_i (x)$$
2. Homework Equations
Product rule
$$\frac{d}{dx} (uv) = u \frac{dv}{dx} + v \frac{du}{dx}$$
3. The Attempt at a Solution
I am honestly not sure how to start this. It is the first one like this that I have tried.
Related Calculus and Beyond Homework Help News on Phys.org
#### Dick
Homework Helper
This problem is harder to write than it is to solve. Let's write "'" for d/dx and call P(k) the product of the f_i and S(k) the sum of the f'_i/f_i. Then what you have above is (P(k))'=S(k)P(k). Assume that's true. Now you want to prove (P(k+1))'=S(k+1)P(k+1), correct? P(k+1)=P(k)*f_(k+1) and S(k+1)=S(k)+f'_(k+1)/f_(k+1), also ok? Apply the product rule to P(k)*f_(k+1) and see if you can match the two sides up. And don't forget to prove the n=1 case to start the induction.
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• Solo and co-op problem solving | 425 | 1,334 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-43 | latest | en | 0.889858 |
https://www.analystforum.com/t/duration-and-market-yield/71850 | 1,606,926,786,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00420.warc.gz | 583,342,546 | 3,832 | # Duration and market yield
Can someone explain why there is an inverse relationship between duration and market yield?
Good question… I think the misunderstanding stems from why interest rates have a inverse relationship with bond prices. The inverse relationship between bond prices and interest rates has to do with the fact that investors want to get the best return possible. If you have a zero coupon bond that trades at 960 with par value of 1000, it has a present rate of return of 1000/960 - 1 = 4.17%. If interest rates rise and you can get 10% return in the market, all things constant, why would you want to have a bond that returns 4.17% when you can get 10%? Thus, to compensate, the price of the bond will fall to 909 (to 10% yield) to attract more investors! (think about less liquidity and the fact that coupons are fixed and cannot be changed)
Now to your question, Duration is defined as a measure of price sensitivity to a 1% change in yield. Another way to look at it is that duration is the interest rate risk of a bond. the higher the yield, the less likely the price of the bond would have to change in order to attract investors, which in turn means that there is less interest rate risk (duration)! | 276 | 1,229 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-50 | latest | en | 0.9629 |
http://www.jiskha.com/display.cgi?id=1365627910 | 1,495,927,526,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609305.10/warc/CC-MAIN-20170527225634-20170528005634-00278.warc.gz | 669,555,055 | 3,916 | # Math
posted by on .
Darren is making a map of population density in Central Florida. The two isosceles triangles below show the most and least populated areas according to Darren's results. The base of the large triangle covers 96 miles and the base of the small triangle covers 24 miles.
If the two triangles are similar and the length of the leg of the small triangle covers 31 miles, what is the distance, in miles, covered by m on Darren's map?
• Math - ,
base of large triangle is to base of small triangle as 96 miles is to 31 miles.
base of large triangle is to base of small triangle as leg (m) of large triangle is to leg (31) of small triangle
Does this help?
• Math - ,
74
### Answer This Question
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Post a New Question | 188 | 825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-22 | latest | en | 0.941371 |
https://getacho.com/031-as-a-fraction/ | 1,685,695,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00050.warc.gz | 317,844,595 | 12,428 | # 031 as a Fraction
The number 031 can be written as a fraction in two different ways. It can be written as 31/100 or 31/1000. The former is an improper fraction, while the latter is a proper fraction. Both of these ways of expressing 031 as a fraction are useful in different situations, so it is important to know how to write 031 as a fraction.
## 31/100
Writing 031 as 31/100 is useful when dealing with percentages. For instance, the percent 031 would be equal to 31/100 or 0.31. This is because percentages are simply fractions with the denominator of 100, so 31/100 is equivalent to 031 as a percent. This fraction is also useful for expressing 031 as a decimal, since it is simply 0.31.
## 31/1000
Writing 031 as 31/1000 is useful when dealing with proportions or ratios. For instance, if you wanted to express 031 as a fraction of 1000, then 31/1000 would be the correct way to do so. This fraction is also useful if you wanted to express 031 as a decimal, since it is 0.031.
## Conclusion
In conclusion, the number 031 can be written as a fraction in two different ways. It can be written as 31/100 or 31/1000. The former is an improper fraction, while the latter is a proper fraction. Both of these ways of expressing 031 as a fraction are useful in different situations, so it is important to know how to write 031 as a fraction. | 344 | 1,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-23 | latest | en | 0.955992 |
http://dwd.wisconsin.gov/wc/medical/simple_dom_hand_calc.htm | 1,503,134,232,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105334.20/warc/CC-MAIN-20170819085604-20170819105604-00297.warc.gz | 120,359,497 | 4,016 | Department of Workforce Development - Worker's Compensation
# Example Of A Simple Dominant Hand Calculation
For this example, we assumed the injury took place in January 2010 and the employee is entitled to the maximum rate of \$282.00. The permanent partial disability rate is either the maximum rate for year of injury or employee’s temporary total disability rate, whichever is lower.
(view Maximum Weekly PPD Rates for other dates of injury)
Injury: Amputation of right middle finger at the distal joint, right hand dominant.
100% of 8 weeks (per schedule) = 8 weeks
plus 25% of 8 weeks (for dominant hand) = 2 weeks
8 wks (amputation) + 2 wks (dominant hand multiple) = 10 weeks X \$282.00 = \$2820.00
A proud partner of the network | 182 | 745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-34 | latest | en | 0.881052 |
http://mathhelpforum.com/calculus/134843-convergent-divergent.html | 1,480,811,073,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541142.66/warc/CC-MAIN-20161202170901-00039-ip-10-31-129-80.ec2.internal.warc.gz | 170,435,441 | 10,257 | 1. ## Convergent or divergent
Is this series convergent or diveregent...and from what method I can prove this.
$
\sum_{n=1}^{\infty} \frac{cos (n)}{n}
$
2. Originally Posted by racewithferrari
Is this series convergent or diveregent...and from what method I can prove this.
$
\sum_{n=1}^{\infty} \frac{cos (n)}{n}
$
To prove convergence, use Dirichlet's test.
If you want to know the sum of the series, notice that $\sum_{n=1}^\infty\frac{z^n}n = \log(1-z)$. If $z=e^{i\theta}$ then $\log(1-z) = \ln(2\sin\tfrac\theta2) + i(\theta+\pi)/2$. Take the real part to see that $\sum_{n=1}^\infty\frac{\cos n\theta}n = \ln(2\sin\tfrac\theta2)$. In particular, if $\theta=1$ then $\sum_{n=1}^\infty\frac{\cos n}n = \ln(2\sin\tfrac12)$. That argument is not rigorous, because the series for $\log(1-z)$ has radius of convergence 1, so you cannot assume that it behaves well when |z| = 1. However, once you have used the Dirichlet test to prove convergence, the rest of the argument works to give the correct result. | 340 | 1,013 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2016-50 | longest | en | 0.696457 |
http://www.conwaylife.com/forums/viewtopic.php?f=11&t=2238&start=200 | 1,569,051,059,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574286.12/warc/CC-MAIN-20190921063658-20190921085658-00193.warc.gz | 231,725,116 | 12,938 | Home • LifeWiki • Forums • Download Golly
## Thread for basic non-CGOL questions
For discussion of other cellular automata.
### Re: Thread for basic non-CGOL questions
What is the smallest (known) log-growth pattern in any isotropic non-totalistic Life-like cellular automaton (without B0)?
Bonus: And log(log)-growth?
2718281828
Posts: 738
Joined: August 8th, 2017, 5:38 pm
### Re: Thread for basic non-CGOL questions
2718281828 wrote:What is the smallest (known) log-growth pattern in any isotropic non-totalistic Life-like cellular automaton (without B0)?
4 cells with a 2-cell predecessor:
x = 41, y = 3, rule = B2ci3aer4eiqrz5i6cik8/S01c2cn3ek4nqtw5aekry6ekn78
39bo\$obo35bobo\$39bo!
source (obviously I remember everything I've created)
Moosey wrote:Where can I find the cblocks rule table?
I'm pretty sure cblocks allows unlimited instant pushing and so cannot be expressed by a rule table.
"Build a man a fire and he'll be warm for a day. Set a man on fire and he'll be warm for the rest of his life."
-Terry Pratchett
toroidalet
Posts: 1002
Joined: August 7th, 2016, 1:48 pm
Location: my computer
### Re: Thread for basic non-CGOL questions
toroidalet wrote:
2718281828 wrote:What is the smallest (known) log-growth pattern in any isotropic non-totalistic Life-like cellular automaton (without B0)?
4 cells with a 2-cell predecessor:
x = 41, y = 3, rule = B2ci3aer4eiqrz5i6cik8/S01c2cn3ek4nqtw5aekry6ekn78
39bo\$obo35bobo\$39bo!
source (obviously I remember everything I've created)
Long lifespan methuselah family:
x = 27, y = 28, rule = B2ci3aer4eiqrz5i6cik8/S01c2cn3ek4nqtw5aekry6ekn78
26bo27\$obo!
toroidalet wrote:
Moosey wrote:Where can I find the cblocks rule table?
I'm pretty sure cblocks allows unlimited instant pushing and so cannot be expressed by a rule table.
Oh. Where do I find it then, CA or no?
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Moosey
Posts: 2332
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board.
### Re: Thread for basic non-CGOL questions
What is the highest (known) period for an unloopable (loopability 1) RRO in isotropic non-totalistic Life-like cellular automaton (without B0)?
Current status: outside the continent of cellular automata. Specifically, not on the plain of life.
GUYTU6J
Posts: 670
Joined: August 5th, 2016, 10:27 am
Location: My Glimmering Garden
### Re: Thread for basic non-CGOL questions
GUYTU6J wrote:What is the highest (known) period for an unloopable (loopability 1) RRO in isotropic non-totalistic Life-like cellular automaton (without B0)?
I have a p1912 (or a predecessor of it) which is of loopability 1 in my signature.
That that is, is. That that is not, is not. Is that it? It is.
A predecessor to my favorite oscillator of all time:
x = 7, y = 5, rule = B3/S2-i3-y4i
4b3o\$6bo\$o3b3o\$2o\$bo!
Hdjensofjfnen
Posts: 1297
Joined: March 15th, 2016, 6:41 pm
Location: r cis θ
### Re: Thread for basic non-CGOL questions
Are there any guns for hybrid Gs in Quadlife or Immigration?
For instance, these:
x = 14, y = 3, rule = QuadLife
2B9.BC\$A.B8.A.C\$A10.A!
Or, perhaps my question would be better phrased:
Are there any synths for hybrid Gs in Quadlife? (since it can emulate immigration, I won't ask about immigration)
Here's a tricolor HF synth, to demonstrate that such things are possible:
x = 15, y = 18, rule = QuadLife
2.C\$3.C\$.3C10\$13.2B\$12.2B\$14.B\$.2D\$D.D\$2.D!
EDIT:
This R surprised me:
x = 3, y = 3, rule = QuadLife
EDIT:
More proof of concept-- a BH with a foreign cell:
x = 84, y = 42, rule = QuadLife
66.A\$65.A\$65.3A\$24.A.A\$25.2A\$25.A9\$54.2A\$46.B6.A2.A\$47.B5.A2.A\$3.A41.
3B6.2A26.B\$3.A.A75.B.B\$3.2A76.B.A\$82.B4\$49.3A\$2A49.A\$.2A47.A\$A12\$40.
2A\$41.2A\$40.A!
Similar:
x = 31, y = 24, rule = QuadLife
.A\$2.A\$3A9\$17.A\$18.A\$16.3A2\$24.A4.B\$25.A2.B\$23.3A2.3B4\$23.3C2.3D\$25.C
2.D\$24.C4.D!
x = 12, y = 10, rule = QuadLife
9.B\$9.B.B\$9.2B5\$.2A\$A.A\$2.A!
x = 8, y = 13, rule = QuadLife
.A\$2.A\$3A10\$5.3B!
Anyways, back to my question. Can anyone make a synth of this?
x = 3, y = 3, rule = QuadLife
A2B\$A\$.A!
EDIT:
If it helps, the distinct 2-color 2G collisions. Two asymmetric ones have multicolored output objects:
#C shamelessly copied from the golly pattern collection after edits
x = 379, y = 369, rule = QuadLife
154.A36.A18.A\$58.A18.A19.A16.A19.A18.A19.A16.A18.A36.A\$57.A18.A19.A
16.A19.A19.3A16.A17.3A16.3A15.A17.A\$57.3A16.3A17.3A14.3A17.3A36.3A51.
A18.3A\$226.3A4\$190.B\$58.3B17.3B89.B18.2B55.B\$58.B19.B20.B17.B20.B13.
2B15.2B18.B.B14.2B20.B16.2B\$59.B19.B18.2B16.2B19.2B13.B.B14.B.B34.B.B
18.2B16.B.B\$98.B.B15.B.B18.B.B12.B53.B20.B.B\$.A4.A3.A.A3.A.A.A2.A3.A
3.A3.A4.A4.A.A2\$.A.A2.A2.A3.A4.A4.A3.A3.A3.A.A2.A2.A2\$.A2.A.A2.A3.A4.
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22.3A18.A22.A17.A\$229.A15.A22.3A20.3A40.A22.A18.3A\$86.A16.A17.A16.A
16.A15.A17.A16.A22.3A13.3A86.3A20.3A\$66.A18.A16.A17.A16.A16.A15.A17.A
16.A\$65.A19.3A14.3A15.3A14.3A14.3A13.3A15.3A14.3A\$65.3A4\$224.3B15.3B
59.2B\$76.3B16.3B16.3B15.3B15.3B14.3B16.3B15.3B19.B17.B11.2B23.2B20.B.
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21.B21.2B12.2B26.2B\$56.B20.B18.B18.B17.B17.B16.B18.B17.B50.B24.B44.B
15.2B24.B\$55.B285.B23\$106.A157.A\$74.A30.A157.A38.A\$2.A.A4.A6.A3.A4.A
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2B119.A4.A.A.A44.B21.B\$108.B.B146.2B23.2B22.B\$66.2B21.2B17.B149.2B22.
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56.B21.2B15.2B23.B16.B.B14.2B102.2B16.2B17.B.B11.2B22.2B\$55.B23.2B15.
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A4.A.A.A32.3B165.A.A.A2.A.A3.A5.A\$59.B203.B\$60.B201.2B\$262.B.B17\$67.A
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A3.A3.A2.A2.A3.A3.A2.A2\$4.A.A.A2.A.A3.A3.A2.A203.A4.A.A.A2.A3.A2.A3.A
2.A.A4.A3.A3.A.A3.A\$58.2B239.B\$57.B.B238.2B\$59.B238.B.B16\$59.A\$58.A\$
58.3A264.A\$3.A.A.A4.A4.A.A.A2.A.A.A2.A.A193.A3.A4.A2.A.A.A2.A.A.A2.A.
A5.A.A3.A3.A4.A4.A4.A4.A.A3.A.A.A22.A\$35.A221.A66.3A\$3.A7.A.A5.A4.A6.
A195.A3.A.A2.A4.A4.A6.A6.A3.A2.A3.A3.A.A3.A.A2.A2.A7.A\$35.A221.A\$3.A.
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2.A2.A.A2.A2.A.A2.A.A.A\$57.B\$3.A6.A.A.A4.A4.A6.A2.A21.2B169.A3.A4.A4.
A4.A6.A2.A3.A3.A2.A3.A2.A.A.A2.A4.A2.A4.A2.A26.3B\$56.B.B265.B\$3.A.A.A
2.A3.A4.A4.A.A.A2.A3.A191.A3.A4.A4.A4.A.A.A2.A3.A3.A.A3.A3.A2.A3.A2.A
4.A4.A.A3.A.A.A23.B11\$305.2A\$306.A9\$55.A16.A237.A\$2.A.A5.A.A3.A4.A2.A
.A27.A16.A155.A.A.A2.A14.A.A3.A6.A3.A.A4.A.A.A2.A.A26.A\$6.A47.3A14.3A
210.A24.3A\$2.A6.A3.A2.A.A2.A2.A3.A200.A4.A7.A4.A7.A6.A3.A3.A2.A6.A\$6.
A277.A\$2.A.A4.A3.A2.A2.A.A2.A3.A200.A4.A5.A.A.A2.A2.A.A2.A6.A3.A3.A2.
A.A.A2.A.A2\$2.A6.A3.A2.A4.A2.A3.A200.A4.A7.A4.A4.A2.A6.A3.A3.A2.A6.A
2.A26.3B\$310.B\$2.A7.A.A3.A4.A2.A.A46.B155.A4.A.A.A10.A.A3.A.A.A2.A3.A
.A4.A.A.A2.A3.A26.B\$50.2B20.2B\$51.2B19.B.B\$50.B19\$2.A.A5.A7.A.A4.A7.A
.A4.A.A3.A3.A2\$2.A3.A3.A7.A3.A2.A6.A3.A2.A3.A2.A2.A28.A239.A\$77.A239.
A\$2.A.A5.A2.A.A2.A.A4.A6.A3.A2.A6.A.A28.3A237.3A\$229.A.A4.A.A3.A.A.A
3.A.A3.A5.A3.A3.A4.A3.A.A\$2.A3.A3.A7.A3.A2.A6.A3.A2.A3.A2.A2.A\$228.A
3.A2.A3.A4.A4.A3.A2.A.A.A.A3.A3.A.A2.A2.A3.A\$2.A.A5.A7.A.A4.A.A.A3.A.
A4.A.A3.A3.A\$228.A3.A2.A8.A4.A3.A2.A2.A2.A3.A3.A2.A.A2.A3.A\$77.3B\$77.
B150.A3.A2.A3.A4.A4.A3.A2.A5.A3.A3.A4.A2.A3.A\$78.B\$229.A.A4.A.A5.A5.A
.A3.A5.A3.A3.A4.A3.A.A26.2B\$307.B.B\$309.B7\$276.2A37.2A\$277.A38.A6\$
112.A13.A\$2.A6.A.A5.A4.A.A.A9.A.A4.A6.A3.A4.A2.A3.A2.A.A.A2.A.A32.A
13.A\$80.A30.3A11.3A\$2.A5.A3.A3.A.A3.A8.A4.A3.A2.A6.A3.A.A2.A2.A2.A3.A
6.A\$80.A\$2.A5.A3.A2.A3.A2.A.A.A2.A.A.A2.A.A4.A6.A3.A2.A.A2.A.A4.A.A.A
2.A.A2\$2.A5.A3.A2.A.A.A2.A8.A4.A3.A2.A6.A3.A4.A2.A2.A3.A6.A2.A2\$2.A.A
.A2.A.A3.A3.A2.A13.A.A4.A.A.A2.A3.A4.A2.A3.A2.A.A.A2.A3.A\$107.2B21.2B
195.A\$107.B.B20.B.B193.A\$107.B22.B98.A.A.A9.A.A3.A6.A3.A.A4.A.A.A2.A.
A7.A5.A2.A.A.A3.A.A3.A.A16.3A\$278.A\$233.A7.A7.A6.A3.A3.A2.A6.A9.A.A.A
.A2.A6.A5.A\$278.A\$229.A.A.A2.A.A2.A2.A.A2.A6.A3.A3.A2.A.A.A2.A.A7.A2.
A2.A2.A.A.A3.A.A3.A.A2\$229.A11.A4.A2.A6.A3.A3.A2.A6.A2.A6.A5.A2.A10.A
5.A\$324.B\$229.A.A.A9.A.A3.A.A.A2.A3.A.A4.A.A.A2.A3.A5.A5.A2.A.A.A3.A.
A3.A.A13.2B\$323.B.B7\$69.A\$68.A\$46.A21.3A34.A\$3.A5.A3.A3.A.A4.A.A18.A
58.A\$45.3A56.3A\$3.A.A.A.A3.A2.A6.A3.A2\$3.A2.A2.A3.A3.A.A3.A2\$3.A5.A3.
A6.A2.A3.A19.3B\$47.B24.2B17.3B177.2A32.2A\$3.A5.A3.A3.A.A4.A.A21.B23.B
.B18.B178.A33.A\$72.B19.B!
x = 13, y = 40, rule = QuadLife
11.A\$10.A\$10.3A9\$2.2B\$.B.B\$3.B16\$3.A\$2.A\$2.3A6\$.B\$2B\$B.B!
Can someone apgsearch quadlife?
EDIT:
Is there a thread for this rule? There was discussion of it while back in rules with interesting failed reps:
x = 3, y = 3, rule = B3-cky6cik/S23-ce4n7e
b2o\$obo\$b2o!
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Moosey
Posts: 2332
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board.
### Re: Thread for basic non-CGOL questions
Moosey wrote:Is there a thread for this rule? There was discussion of it while back in rules with interesting failed reps:
x = 3, y = 3, rule = B3-cky6cik/S23-ce4n7e
b2o\$obo\$b2o!
IT NEEDS TO BE
more normal rake
x = 80, y = 27, rule = B3-cky6cik/S23-ce4n7e
27bo\$26bobo\$2bo23bobo28bo5b2o3bo\$2obo12b3o5b2o3b2o25b4ob2o2bobobo2b3ob
4o\$o3bo11bob2o3bo7bo23b2obobo3bobo3bobobob2obo\$2obo12b3o5b2o3b2o25b4ob
2o2bobobo2b3ob4o\$2bo23bobo28bo5b2o3bo\$26bobo\$27bo3\$23b3o\$2b2o6bo12bobo
\$b4o4bob2o8b2o3b2o\$o3b2o2bo3bo8bo5bo\$b4o4bob2o8b2o3b2o\$2b2o6bo12bobo\$
23b3o5\$43bo5b2o3bo\$42b4ob2o2bobobo2b3ob4o\$41b2obobo3bobo3bobobob2obo\$
42b4ob2o2bobobo2b3ob4o\$43bo5b2o3bo!
x = 70, y = 29, rule = B3-cky6cik/S23-ce4n7e
28bo\$27bobo\$2b2o14bo8bobo\$b4o12bob2o4b2o3b2o15bo5b2o3bo\$o3b2o10bo3bo3b
o7bo13b4ob2o2bobobo2b3ob4o\$b4o12bob2o4b2o3b2o13b2obobo3bobo3bobobob2ob
o\$2b2o14bo8bobo16b4ob2o2bobobo2b3ob4o\$27bobo17bo5b2o3bo\$28bo4\$2bo\$2obo
4b3o6b2o\$o3bo3bob2o4bob2o\$2obo4b3o6b2o\$2bo4\$28bo\$27bobo17bo5b2o3bo\$2b
2o14bo8bobo16b4ob2o2bobobo2b3ob4o\$b4o12bob2o4b2o3b2o13b2obobo3bobo3bob
obob2obo\$o3b2o10bo3bo3bo7bo13b4ob2o2bobobo2b3ob4o\$b4o12bob2o4b2o3b2o
15bo5b2o3bo\$2b2o14bo8bobo\$27bobo\$28bo!
???
x = 151, y = 42, rule = B3-cky6cik/S23-ce4n7e
92bo\$63bo5b2o3bo19bo\$62b4ob2o2bobobo2b3ob4obobo4bo\$61b2obobo3bobo3bobo
bob2obo2bo3b4o11b2o\$62b4ob2o2bobobo2b3ob4obobo4bo10bo3bo4bo\$51b2o10bo
5b2o3bo19bo4b2o2bo11bo8b2o\$52bo39bo6bobo12b3o6bo2bo\$49bo49b2o2bo11bo8b
2o\$49b2o10b2o14b2o14b2o10bo3bo4bo\$38b3o20b2o14b2o14b2o12b2o\$28b2o8bobo
3b2o\$28bobo7bobo3bobo11b2o14b2o14b2o14b2o\$29bo9bo5bo11bo2bo12bo2bo12bo
2bo12bo2bo\$19bo17bo3bo16bobo13bobo13bobo13bobo\$18b3o16bobobo17bo15bo
15bo15bo\$b2o14b2o2bo\$o3b4o10bob3o\$o3bo2bo9b4obo\$o3b4o9b2o2b2o\$b2o17bob
obo12bo3bo11bo3bo11bo3bo11bo3bo11bo3bo11bo15bo15bo\$13b2o3b3ob2obo2b2o
6bobobobo9bobobobo9bobobobo9bobobobo9bobobobo9bobo13bobo13bobo\$13b2o3b
3ob2obo2b2o6bobobobo9bobobobo9bobobobo9bobobobo9bobobobo9bobo13bobo13b
obo\$b2o17bobobo12bo3bo11bo3bo11bo3bo11bo3bo11bo3bo11bo15bo15bo\$o3b4o9b
2o2b2o\$o3bo2bo9b4obo\$o3b4o10bob3o\$b2o14b2o2bo\$18b3o16bobobo17bo15bo15b
o15bo\$19bo17bo3bo16bobo13bobo13bobo13bobo\$29bo9bo5bo11bo2bo12bo2bo12bo
2bo12bo2bo\$28bobo7bobo3bobo11b2o14b2o14b2o14b2o\$28b2o8bobo3b2o60b2o\$
38b3o20b2o14b2o14b2o10b2o\$49b2o10b2o14b2o14b2o8bo3bo4bo\$49bo47b2o2bo
11bo8b2o\$52bo39bo4bobo12b3o6bo2bo\$51b2o10bo5b2o3bo19bo2b2o2bo11bo8b2o\$
62b4ob2o2bobobo2b3ob4obobo4bo8bo3bo4bo\$61b2obobo3bobo3bobobob2obo2bo3b
4o9b2o\$62b4ob2o2bobobo2b3ob4obobo4bo\$63bo5b2o3bo19bo\$92bo!
I like making color palettes for rules
Gustone
Posts: 421
Joined: March 6th, 2019, 2:26 am
### Re: Thread for basic non-CGOL questions
How many different conditions does the range-1 Moore 3D isotropic non-totalistic rulespace have?
Bored of using the Moore neighbourhood for everything? Introducing the Range-2 von Neumann isotropic non-totalistic rulespace!
muzik
Posts: 3466
Joined: January 28th, 2016, 2:47 pm
Location: Scotland
### Re: Thread for basic non-CGOL questions
Is PCA margolus
I like making color palettes for rules
Gustone
Posts: 421
Joined: March 6th, 2019, 2:26 am
### Re: Thread for basic non-CGOL questions
Gustone wrote:Is PCA margolus
I don't think so--the graphics don't help, but you may want to check golly's ruletable repository for Reversible world, which is a PCA. I don't believe it's margolus.
I am a prolific creator of many rather pathetic googological functions
My CA rules can be found here
Also, the tree game
Bill Watterson once wrote: "How do soldiers killing each other solve the world's problems?"
Moosey
Posts: 2332
Joined: January 27th, 2019, 5:54 pm
Location: A house, or perhaps the OCA board.
### Re: Thread for basic non-CGOL questions
Gustone wrote:Is PCA margolus
No. Partitioned CA do not use an alternating neighbourhood such as Margolus rules do. Instead, each cell is divided up (partitioned) into as many parts as the cell has neighbours. So for a 4-neighbour PCA, each cell has 4 parts. These parts are denoted as U(p), R(ight), D(own), and L(eft) in the paper referenced by the PCA thread. Each part has it's own state (0 or 1 for a 2 state PCA). A 2 state PCA with 4 neighbours is denoted as a 2PCA(4) cellular automata. The evolution of the CA is determined by a local function which maps from 1 part of each of the 4 neighbour cells to the 4 parts of the centre cell. Here's a little diagram which shows the four parts of a central cell and it's 4 neighbours. The state of the four parts in the central cell (u, r, d, and l) is entirely determined by the four parts labelled U, R, D, and L from the neighbouring cells (i.e. a cells current state does not influence it's resulting state)
. .
. . . .
D .
. . . . u .
. R . . L . --> . . l r . .
. . . . d .
U .
. . . .
. .
Because there are four parts which can take 2 states, every 2PCA(4) CA is equivalent to a regular CA with 2^4 states on a von Neumann neighbourhood.
The diagrams in the referenced papers make this easier to understand than I am able to explain here, so I suggest you have a look at them:
3-neighbour: http://bprentice.webenet.net/PCA/Two%20 ... e%20CA.pdf
4-neighbour: http://bprentice.webenet.net/PCA/16%20S ... tomata.pdf
The latest version of the 5S Project contains over 221,000 spaceships. Tabulated pages up to period 160 are available on the LifeWiki.
wildmyron
Posts: 1238
Joined: August 9th, 2013, 12:45 am
### Re: Thread for basic non-CGOL questions
muzik wrote:How many different conditions does the range-1 Moore 3D isotropic non-totalistic rulespace have?
I believe the answer is 8548, given by the sequence A054247: Number of n X n binary matrices under action of dihedral group of the square D_4.
This includes birth and survival configurations, so the number of neighbourhood configurations is 4274.
Edit: This is almost certainly wrong, see below
Last edited by wildmyron on September 18th, 2019, 9:24 am, edited 1 time in total.
The latest version of the 5S Project contains over 221,000 spaceships. Tabulated pages up to period 160 are available on the LifeWiki.
wildmyron
Posts: 1238
Joined: August 9th, 2013, 12:45 am
### Re: Thread for basic non-CGOL questions
Moosey wrote:Are there any guns for hybrid Gs in Quadlife or Immigration?
P22:
x = 45, y = 21, rule = QuadLife
18.2A\$19.A7.A\$19.A.A14.2A\$20.2A12.2A2.A\$24.3A7.2A.2A\$24.2A.2A7.3A\$24.
A2.2A12.2A\$25.2A14.A.A\$35.A7.A\$43.2A2\$2A\$.A\$.A.A13.3B\$2.2A3.B8.B3.B\$
6.B.2B6.B4.B\$5.B4.B6.2B.B\$6.B3.B8.B3.2A\$7.3B13.A.A\$25.A\$25.2A!
P30:
x = 36, y = 25, rule = QuadLife
26.2B\$9.2A14.B3.B\$9.A.A12.B5.B\$4.2A6.A11.B3.B.2B2.2A\$2A.A2.A2.A2.A11.
B5.B3.2A\$2A2.2A6.A12.B3.B\$9.A.A14.2B\$9.2A8\$16.5A\$15.A.3A.A\$16.A3.A\$
17.3A\$18.A4\$18.2A\$18.2A!
P36:
x = 75, y = 50, rule = QuadLife
25.A\$25.3A\$28.A\$27.2A7\$39.2A\$38.A2.A\$30.A6.A.A.A\$29.3A3.3A2.A\$28.2A2.
A\$30.3A4.2A\$38.A6\$27.A\$27.2A4.3B\$33.B2.2B\$25.A2.3A3.3B\$24.A.A.A6.B26.
2A\$24.A2.A33.A2.A\$25.2A35.A.A\$41.A16.4A.A\$41.A16.2A.2A\$42.A7.2A9.A\$
11.2A36.2A\$10.A2.A36.2A\$10.A.A38.A\$11.A3.2A42.A\$12.2A.A43.2A\$13.A6.3A
37.2A\$13.A5.2A.2A25.A9.2A\$19.2A2.A24.2A.2A18.2A\$21.2A24.A.4A18.A.A\$
16.2A28.A.A24.A\$15.A2.2A26.A2.A23.2A\$15.2A.2A5.A21.2A\$16.3A6.A\$2.2A
19.A.2A\$.A.A18.2A3.A\$.A24.A.A\$2A23.A2.A\$26.2A!
x = 5, y = 5, rule = B3-y/S234w
2b3o\$bo\$o3bo\$o2bo\$obo!
FWKnightship
Posts: 65
Joined: June 23rd, 2019, 3:10 am
Location: Behind you
### Re: Thread for basic non-CGOL questions
wildmyron wrote:
muzik wrote:How many different conditions does the range-1 Moore 3D isotropic non-totalistic rulespace have?
I believe the answer is 8548, given by the sequence A054247: Number of n X n binary matrices under action of dihedral group of the square D_4.
This includes birth and survival configurations, so the number of neighbourhood configurations is 4274.
Actually, this is clearly not the desired sequence, though I'm presuming that the term for n=2 matches the 1D case coincidentally. I'm fairly sure there'll be no such coincidence for n=4 - I see no reason why the number of 4x4 binary matrices under D_4 should be the same as the number of 3x3x3 binary matrices under the octahedral group O_h, which is what we actually want to know.
The latest version of the 5S Project contains over 221,000 spaceships. Tabulated pages up to period 160 are available on the LifeWiki.
wildmyron
Posts: 1238
Joined: August 9th, 2013, 12:45 am
Previous | 9,028 | 19,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-39 | latest | en | 0.869977 |
https://mail.haskell.org/pipermail/haskell-cafe/2006-September/017770.html | 1,508,398,417,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823255.12/warc/CC-MAIN-20171019065335-20171019085335-00613.warc.gz | 769,309,552 | 2,598 | [Haskell-cafe] Re: Exercise in point free-style
Thomas Davie tom.davie at gmail.com
Fri Sep 1 20:48:30 EDT 2006
```Shorter, although perhaps less insightful.
Bob
On 2 Sep 2006, at 01:36, Lennart Augustsson wrote:
> An easy way to solve this is to ask lambdabot. Log on to the
> lennart: @pl \ f l -> l ++ map f l
> lambdabot: ap (++) . map
>
> Notice how it's much shorter than the Hughes' solution. :)
>
> -- Lennart
>
> On Sep 1, 2006, at 13:11 , John Hughes wrote:
>
>>> From: Julien Oster <haskell at lists.julien-oster.de>
>>> Subject: [Haskell-cafe] Exercise in point free-style
>>>
>>> I was just doing Exercise 7.1 of Hal Daumé's very good "Yet Another
>>> Haskell Tutorial". It consists of 5 short functions which are to be
>>> converted into point-free style (if possible).
>>>
>>> It's insightful and after some thinking I've been able to come up
>>> with
>>> solutions that make me understand things better.
>>>
>>> But I'm having problems with one of the functions:
>>>
>>> func3 f l = l ++ map f l
>>>
>>> Looks pretty clear and simple. However, I can't come up with a
>>> solution.
>>> Is it even possible to remove one of the variables, f or l? If
>>> so, how?
>>>
>>> Thanks,
>>> Julien
>>
>> Oh, YES!!
>>
>> Two ways to remove l:
>>
>> func3a f = uncurry ((.map f).(++)) . pair
>> func3b f = uncurry (flip (++).map f) . pair
>>
>> And just to make sure they're right:
>>
>> propab new f l =
>> func3 f l == new f l
>> where types = f :: Int->Int
>>
>> quickCheck (propab func3a)
>> quickCheck (propab func3b)
>>
>> If you don't mind swapping the arguments, then
>>
>> propc f l =
>> func3 f l == func3c l f
>> where types = f :: Int->Int
>>
>> func3c l = (l++) . (`map` l)
>>
>> With the arguments swapped, you can even remove both!
>>
>> propd f l =
>> func3 f l == func3d l f
>> where types = f :: Int -> Int
>>
>> func3d = uncurry ((.(flip map)) . (.) . (++)) . pair
>>
>> MUCH clearer!
>>
>> The trick is to observe that l is duplicated, so you need to use a
>> combinator that duplicates something. The only one available here
>> is pair, which you then have to combine with uncurry.
>>
>> It would be nicer to have
>>
>> (f &&& g) x = (f x,g x)
>>
>> available. (&&& is one of the arrow combinators). Then you could
>> remove l by
>>
>> func3e f = uncurry (++) . (id &&& map f)
>>
>> which is sort of readable, and remove both by
>>
>> func3f = (uncurry (++).) . (id &&&) . map
>>
>> John
>>
>>
>> _______________________________________________ | 769 | 2,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-43 | latest | en | 0.85648 |
http://www.carrawaydavie.com/convert-oz-to-liters/ | 1,560,995,679,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999130.50/warc/CC-MAIN-20190620004625-20190620030625-00459.warc.gz | 211,601,261 | 8,728 | ## What is a gallon convert oz to liters?
Gallon is a royal as well as United States normal measurement system volume device. convert oz to liters? There is one kind of gallon in the imperial system and also 2 kinds (fluid and completely dry) in the US customary dimension system. 1 US liquid gallon is specified as 231 cubic inches, 1 US dry gallon is 268.8 cubic inches and 1 royal gallon is 277.4 cubic inches. The abbreviation is gall. convert oz to liters?
Gallons Conversion:
1 Gallon = 4 Quarts
1 Gallon = 8 Pints
1 Gallon = 16 Cups
1 Gallon = 256 Tablespoons
1 Gallon = 768 Teaspoons
convert oz to liters
## What is a liquid ounce (fl oz)convert oz to liters?
A liquid ounce is an imperial as well as US customary measurement system volume unit. 1 US fluid ounce equals to 29.5735 mL and 1 royal (UK) fluid ounce equals to 28.4131 mL. The acronym is fl oz. So, convert oz to liters?
There are 8 fluid ounces in a United States mug and also 10 imperial liquid ounces in an imperial cup.
1 United States Fluid Gallon = 128 US Fluid Ounces
1 US Dry Gallon = 148.946 United States Fluid Ounces
1 Imperial Gallon (UK) = 160 Imperial Fluid Ounces (UK).
convert oz to liters
## Liquid Ounce convert oz to liters
Liquid Ounce is utilized for volume, Ounce for mass, and also they are various.
As an example, 1 liquid ounce of honey has a mass of about 1,5 ounces!
But also for water, 1 liquid ounce has a mass of about 1 ounce.
convert oz to liters
If you imply an ounce of fluid says fluid ounce (fl oz).
In Summary:.
1 gallon = 4 quarts = 8 pints = 16 mugs = 128 fluid ounces.
convert oz to liters | 434 | 1,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-26 | longest | en | 0.862589 |
http://mathhelpforum.com/advanced-algebra/184607-how-do-you-prove-determinant-expand-leibniz-formula-laplace-formula.html | 1,498,284,904,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320226.61/warc/CC-MAIN-20170624050312-20170624070312-00188.warc.gz | 258,220,747 | 10,693 | # Thread: how do you prove determinant expand by the Leibniz formula or the Laplace formula ...
1. ## how do you prove determinant expand by the Leibniz formula or the Laplace formula ...
how do you prove determinant expand by the Leibniz formula or the Laplace formula is the same?
2. ## Re: how do you prove determinant expand by the Leibniz formula or the Laplace formula
Originally Posted by victorlui
how do you prove determinant expand by the Leibniz formula or the Laplace formula is the same?
It depends what your definition of the determinant is. Do you define it as the unique alternating multilinear form $K:\mathbb{R}^n\times\cdots\times\mathbb{R}^n\to \mathbb{R}$ for which $\det(I)=1$? If so, see my blog post here. If not, give us the definition you're using.
3. ## Re: how do you prove determinant expand by the Leibniz formula or the Laplace formula
Hello,
It sounds more difficult as it really is.
You know the determinant is a multilinear map and the definition of the adjugate matrix.
5. ## Re: how do you prove determinant expand by the Leibniz formula or the Laplace formula
Originally Posted by victorlui
Then I have no idea what your question means. What are you actually asking? | 306 | 1,213 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-26 | longest | en | 0.848685 |
http://www.jiskha.com/display.cgi?id=1289350553 | 1,498,319,635,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320264.42/warc/CC-MAIN-20170624152159-20170624172159-00431.warc.gz | 585,527,726 | 3,729 | # statistics
posted by .
Use Bayes' theorem to solve this problem. A storeowner purchases stereos from two companies. From Company A, 550 stereos are purchased and 1% are found to be defective. From Company B, 850 stereos are purchased and 6% are found to be defective. Given that a stereo is defective, find the probability that it came from Company A.
• statistics -
11/113 | 90 | 379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-26 | latest | en | 0.974312 |
https://de.scribd.com/presentation/124521671/Probabilty-LECTURE-1 | 1,563,232,457,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524254.28/warc/CC-MAIN-20190715215144-20190716000536-00066.warc.gz | 367,612,958 | 65,509 | You are on page 1of 83
# 5E Note 4
## Statistics with Economics and Business
Applications
Chapter 4 -5 Probability and Discrete
Probability Distributions
Experiment, Event, Sample space, Probability, Counting rules,
Conditional probability, Bayess rule, random variables, mean,
variance
5E Note 4
Probability
Will you pass this course .?
How much certain(sure) you are?
A quantitative measure?
Chance.
Probability (def)
is the chance of an event occurring
Note 5 of 5E
Basic Concepts
An experiment is the process by which
an observation (or outcome) is obtained.
An event is an outcome of an experiment,
usually denoted by a capital letter.
Note 5 of 5E
Experiments and Events
Experiment: flipping a coin
Out come
B: Tail
Experiment: rolling a die
out come {1,2,3,4,5,6}
A: observe an odd number
B: observe a number greater than 2
Note 5 of 5E
Basic Concepts
An event that cannot be decomposed is called
a simple event.
Denoted by E with a subscript.
Each simple event will be assigned a
probability, measuring how often it occurs.
The set of all simple events of an experiment is
called the sample space, S.
Note 5 of 5E
Example
The die roll:
Simple events: Sample space:
1
2
3
4
5
6
E
1
E
2
E
3
E
4
E
5
E
6
S ={E
1
, E
2
, E
3
, E
4
, E
5
, E
6
}
S
E
1
E
6
E
2
E
3
E
4
E
5
Note 5 of 5E
Drawing a card from a standard
deck
There will be 52 cards in the sample space:
king,
Clubs: 2,3,4,5,6,7,8,9,10, ace, jack, queen,
king,
Diamonds: 2,3,4,5,6,7,8,9,10, ace, jack,
queen, king,
Hearts: 2,3,4,5,6,7,8,9,10, ace, jack, queen,
king}
Note 5 of 5E
Example 3
The sample space of throwing a pair of dice is
Note 5 of 5E
Tree diagram
Is a device consisting of line segments
emanating(emerging) from a starting
point and also from the outcome point .
It is used to determine all possible
outcomes of a probability experiment
Total number of path in a tree is equal
to the total number of elements in the
sample space.
Note 5 of 5E
Tree diagram
O1
O2
O3
P1
P2
P1
P2
P1
P2
r1
r2
r3
r1
r2
r3
r1
r2
r3
r1
r2
r3
r1
r2
r3
r1
r2
r3
Note 5 of 5E
Tree diagram of the gender of three
children in a family
B
G
B
G
B
G
B
G
B
G
B
G
B
G
Note 5 of 5E
Basic Concepts
An event is a collection of one or more
simple events.
The die toss:
event A: an odd number
event B: a number > 2
S
A ={E
1
, E
3
, E
5
}
B ={E
3
, E
4
, E
5
,
E
6
}
B
A
E
1
E
6
E
2
E
3
E
4
E
5
Note 5 of 5E
Basic Concepts
Two events are mutually exclusive if,
when one event occurs, the other cannot,
and vice versa.
Experiment: Toss a die
A: observe an odd number
B: observe a number greater than 2
C: observe an odd no.
D: observe an even no
Not Mutually
Exclusive
Mutually
Exclusive
Note 5 of 5E
Equally likely events
Events that have the same probability
of occurring
Tossing a coin :H & T
Rolling a die
C: observe an odd no. 3
D: observe an even no. 3
Note 5 of 5E
The Probability
of an Event
The probability of an event A measures how
often A will occur. We write P(A).
Suppose that an experiment is performed n
times. The relative frequency for an event A is
n
f
n
=
occurs A times of Number
n
f
A P
n
lim
) (
=
If we let n get infinitely large,
Note 5 of 5E
The Probability
of an Event
P(A) must be between 0 and 1.
If event A can never occur, P(A) = 0.
If event A always occurs when the
experiment is performed, P(A) =1.
The sum of the probabilities for all
simple events in S equals 1.
The probability of an event A is found
by adding the probabilities of all the
simple events contained in A.
Note 5 of 5E
Suppose that 10% of the U.S. population has
red hair. Then for a person selected at random,
Finding Probabilities
Probabilities can be found using
Estimates from empirical studies
Common sense estimates based on
equally likely events.
P(Red hair) = .10
Examples:
Toss a fair coin.
Note 5 of 5E
Using Simple Events
The probability of an event A is equal to
the sum of the probabilities of the simple
events contained in A
If the simple events in an experiment are
equally likely, you can calculate
events simple of number total
A in events simple of number
) ( = =
N
n
A P
A
Note 5 of 5E
Example 1
Toss a fair coin twice. What is the probability
of observing at least one head?
H
1st Coin 2nd Coin E
i
P(E
i
)
H
T
T
H
T
HH
HT
TH
TT
1/4
1/4
1/4
1/4
= P(E
1
) + P(E
2
) + P(E
3
)
= 1/4 + 1/4 + 1/4 = 3/4
Note 5 of 5E
Example 2
A bowl contains three M&Ms
## , one red, one
blue and one green. A child selects two M&Ms
at random. What is the probability that at least
one is red?
1st M&M 2nd M&M E
i
P(E
i
)
RB
RG
BR
BG
1/6
1/6
1/6
1/6
1/6
1/6
P(at least 1 red)
= P(RB) + P(BR)+ P(RG)
+ P(GR)
= 4/6 = 2/3
m
m
m
m
m
m
m
m
m
GB
GR
Note 5 of 5E
Example 3
The sample space of throwing a pair of dice is
Note 5 of 5E
Example 3
Event Simple events Probability
Dice add to 3 (1,2),(2,1) 2/36
(4,2),(5,1)
5/36
Red die show 1 (1,1),(1,2),(1,3),
(1,4),(1,5),(1,6)
6/36
Green die show 1 (1,1),(2,1),(3,1),
(4,1),(5,1),(6,1)
6/36
Note 5 of 5E
Counting Rules
Sample space of throwing 3 dice has
216 entries, sample space of throwing
4 dice has 1296 entries,
At some point, we have to stop listing
and start thinking
We need some counting rules
Note 5 of 5E
The mn Rule
If an experiment is performed in two stages,
with m ways to accomplish the first stage and
n ways to accomplish the second stage, then
there are mn ways to accomplish the
experiment.
This rule is easily extended to k stages, with
the number of ways equal to
n
1
n
2
n
3
n
k
Example: Toss two coins. The total number of
simple events is:
2 2 = 4
Note 5 of 5E
Examples
Example: Toss three coins. The total number of
simple events is:
2 2 2 = 8
Example: Two M&Ms are drawn from a dish
containing two red and two blue candies. The total
number of simple events is:
6 6 = 36
Example: Toss two dice. The total number of
simple events is:
m
m
4 3 = 12
Example: Toss three dice. The total number of
simple events is:
6 6 6 = 216
Note 5 of 5E
Permutations
The number of ways you can arrange
n distinct objects, taking them r at a time
is
Example: How many 3-digit lock combinations
can we make from the numbers 1, 2, 3, and 4?
. 1 ! 0 and ) 1 )( 2 )...( 2 )( 1 ( ! where
)! (
!
=
=
n n n n
r n
n
P
n
r
24 ) 2 )( 3 ( 4
! 1
! 4
4
3
= = = P
The order of the choice is
important!
Note 5 of 5E
Examples
Example: A lock consists of five parts and
can be assembled in any order. A quality
control engineer wants to test each order for
efficiency of assembly. How many orders are
there?
120 ) 1 )( 2 )( 3 )( 4 ( 5
! 0
! 5
5
5
= = = P
The order of the choice is
important!
Note 5 of 5E
Combinations
The number of distinct combinations of n
distinct objects that can be formed,
taking them r at a time is
Example: Three members of a 5-person committee must
be chosen to form a subcommittee. How many different
subcommittees could be formed?
)! ( !
!
r n r
n
C
n
r
=
10
1 ) 2 (
) 4 ( 5
1 ) 2 )( 1 )( 2 ( 3
1 ) 2 )( 3 )( 4 ( 5
)! 3 5 ( ! 3
! 5
5
3
= = =
= C
The order of
the choice is
not important!
Note 5 of 5E
Example
A box contains six M&Ms
, four red
and two green. A child selects two M&Ms at
random. What is the probability that exactly
one is red?
The order of
the choice is
not important!
m
m
m m
m m
Ms. & M 2 choose to ways
15
) 1 ( 2
) 5 ( 6
! 4 ! 2
! 6
6
2
= = = C
M. & M green 1
choose to ways
2
! 1 ! 1
! 2
2
1
= = C
M. & M red 1
choose to ways
4
! 3 ! 1
! 4
4
1
= = C 4 2 =8 ways to
choose 1 red and 1
green M&M.
P(exactly one
red) = 8/15
Note 5 of 5E
Example
A deck of cards consists of 52 cards, 13 "kinds"
each of four suits (spades, hearts, diamonds, and
clubs). The 13 kinds are Ace (A), 2, 3, 4, 5, 6, 7,
8, 9, 10, Jack (J), Queen (Q), King (K). In many
poker games, each player is dealt five cards from
a well shuffled deck.
hands possible
960 , 598 , 2
1 ) 2 )( 3 )( 4 ( 5
48 ) 49 )( 50 )( 51 ( 52
)! 5 52 ( ! 5
! 52
are There
52
5
= =
= C
Note 5 of 5E
Example
Four of a kind: 4 of the 5 cards are the same
kind. What is the probability of getting
four of a kind in a five card hand?
and
There are 13 possible choices for the kind of
which to have four, and 52-4=48 choices for
the fifth card. Once the kind has been
specified, the four are completely determined:
you need all four cards of that kind. Thus there
are 1348=624 ways to get four of a kind.
The probability=624/2598960=.000240096
Note 5 of 5E
Example
One pair: two of the cards are of one kind,
the other three are of three different kinds.
What is the probability of getting one pair
in a five card hand?
kind that of cards four the of
two of choices possible 6 are there
choice, given the pair; a have which to of
kind for the choices possible 13 are There
4
2
= C
Note 5 of 5E
Example
There are 12 kinds remaining from
which to select the other three cards in
the hand. We must insist that the kinds
be different from each other and from
the kind of which we have a pair, or we
could end up with a second pair, three
or four of a kind, or a full house.
Note 5 of 5E
Example
422569 .
98960 1098240/25 y probabilit The
1,098,240. 64 220 6 13
is hands pair" one " of number the Therefore
three. all of suits for the choices 64 4 of
total a cards, three those of each of suit for the
choices 4 are There cards. three remaining the
of kinds pick the to ways 220 are There
3
12
3
=
= =
=
=
= C
Note 5 of 5E
S
Event Relations
The beauty of using events, rather than simple events, is
that we can combine events to make other events using
logical operations: and, or and not.
The union of two events, A and B, is the event that
either A or B or both occur when the experiment is
performed. We write
A B
A B
B A
Note 5 of 5E
S
A
B
Event Relations
The intersection of two events, A and B, is
the event that both A and B occur when the
experiment is performed. We write A B.
B A
If two events A and B are mutually
exclusive, then P(A B) = 0.
Note 5 of 5E
S
Event Relations
The complement of an event A consists of
all outcomes of the experiment that do not
result in event A. We write A
C
.
A
A
C
Note 5 of 5E
Example
Select a student from the classroom and
record his/her hair color and gender.
A: student has brown hair
B: student is female
C: student is male
What is the relationship between events B and C?
A
C
:
BC:
BC:
Mutually exclusive; B = C
C
Student does not have brown hair
Student is both male and female = C
Student is either male and female = all students = S
Note 5 of 5E
Calculating Probabilities for
Unions and Complements
There are special rules that will allow you to
calculate probabilities for composite events.
For any two events, A and B, the probability
of their union, P(A B), is
) ( ) ( ) ( ) ( B A P B P A P B A P + =
A
B
Note 5 of 5E
Example: Suppose that there were 120
students in the classroom, and that they
could be classified as follows:
Brown Not Brown
Male 20 40
Female 30 30
A: brown hair
P(A) = 50/120
B: female
P(B) = 60/120
P(AB) = P(A) + P(B) P(AB)
= 50/120 + 60/120 - 30/120
= 80/120 = 2/3
Check: P(AB)
= (20 + 30 + 30)/120
Note 5 of 5E
Example: Two Dice
A: red die show 1
B: green die show 1
P(AB) = P(A) + P(B) P(AB)
= 6/36 + 6/36 1/36
= 11/36
Note 5 of 5E
A Special Case
When two events A and B are
mutually exclusive, P(AB) = 0
and P(AB) = P(A) + P(B).
Brown Not Brown
Male 20 40
Female 30 30
A: male with brown hair
P(A) = 20/120
B: female with brown hair
P(B) = 30/120
P(AB) = P(A) + P(B)
= 20/120 + 30/120
= 50/120
A and B are mutually
exclusive, so that
Note 5 of 5E
Example: Two Dice
A and B are mutually
exclusive, so that
P(AB) = P(A) + P(B)
= 2/36 + 5/36
= 7/36
Note 5 of 5E
Calculating Probabilities
for Complements
We know that for any event A:
P(A A
C
) = 0
Since either A or A
C
must occur,
P(A A
C
) =1
so that P(A A
C
) = P(A)+ P(A
C
) = 1
P(A
C
) = 1 P(A)
A
A
C
Note 5 of 5E
Example
Brown Not Brown
Male 20 40
Female 30 30
A: male
P(A) = 60/120
B: female
P(B) = ?
P(B) = 1- P(A)
= 1- 60/120 = 60/120
A and B are
complementary, so that
Select a student at random from
the classroom. Define:
Note 5 of 5E
Calculating Probabilities for
Intersections
In the previous example, we found P(A B)
directly from the table. Sometimes this is
impractical or impossible. The rule for calculating
P(A B) depends on the idea of independent
and dependent events.
Two events, A and B, are said to be
independent if the occurrence or
nonoccurrence of one of the events does
not change the probability of the
occurrence of the other event.
Note 5 of 5E
Conditional Probabilities
The probability that A occurs, given
that event B has occurred is called
the conditional probability of A
given B and is defined as
0 ) ( if
) (
) (
) | ( =
= B P
B P
B A P
B A P
given
Note 5 of 5E
Example 1
Toss a fair coin twice. Define
HT
TH
TT
1/4
1/4
1/4
1/4
P(A|B) =
P(A|not B) =
HH
P(A) does not
change, whether
B happens or
not
A and B are
independent!
Note 5 of 5E
Example 2
A bowl contains five M&Ms
## , two red and
three blue. Randomly select two candies, and
define
A: second candy is red.
B: first candy is blue.
m
m
m
m
m
P(A|B) =P(2
nd
red|1
st
blue)= 2/4 = 1/2
P(A|not B) = P(2
nd
red|1
st
red) = 1/4
P(A) does change,
depending on
whether B happens
or not
A and B are
dependent!
Note 5 of 5E
Example 3: Two Dice
Toss a pair of fair dice. Define
A: red die show 1
B: green die show 1
P(A|B) = P(A and B)/P(B)
=1/36/1/6=1/6=P(A)
P(A) does not
change, whether
B happens or
not
A and B are
independent!
Note 5 of 5E
Example 3: Two Dice
Toss a pair of fair dice. Define
P(A|B) = P(A and B)/P(B)
=0/36/5/6=0
P(A) does change
when B happens
A and B are dependent!
In fact, when B happens,
A cant
Note 5 of 5E
Defining Independence
We can redefine independence in terms
of conditional probabilities:
Two events A and B are independent if and
only if
P(A,B) = P(A) or P(B|A) = P(B)
Otherwise, they are dependent.
Once youve decided whether or not two
events are independent, you can use the
following rule to calculate their
intersection.
Note 5 of 5E
The Multiplicative Rule for
Intersections
For any two events, A and B, the
probability that both A and B occur is
P(A B) = P(A) P(B given that A occurred)
= P(A)P(B|A)
If the events A and B are independent, then
the probability that both A and B occur is
P(A B) = P(A) P(B)
Note 5 of 5E
Example 1
In a certain population, 10% of the people can be
classified as being high risk for a heart attack. Three
people are randomly selected from this population.
What is the probability that exactly one of the three are
high risk?
Define H: high risk N: not high risk
P(exactly one high risk) = P(HNN) + P(NHN) + P(NNH)
= P(H)P(N)P(N) + P(N)P(H)P(N) + P(N)P(N)P(H)
= (.1)(.9)(.9) + (.9)(.1)(.9) + (.9)(.9)(.1)= 3(.1)(.9)
2
= .243
Note 5 of 5E
Example 2
Suppose we have additional information in the
previous example. We know that only 49% of the
population are female. Also, of the female patients, 8%
are high risk. A single person is selected at random. What
is the probability that it is a high risk female?
Define H: high risk F: female
From the example, P(F) = .49 and P(H|F) = .08.
Use the Multiplicative Rule:
P(high risk female) = P(HF)
= P(F)P(H|F) =.49(.08) = .0392
Note 5 of 5E
The Law of Total Probability
P(A) = P(A S
1
) + P(A S
2
) + + P(A S
k
)
= P(S
1
)P(A|S
1
) + P(S
2
)P(A|S
2
) + + P(S
k
)P(A|S
k
)
Let S
1
, S
2
, S
3
,..., S
k
be mutually exclusive
and exhaustive events (that is, one and only
one must happen). Then the probability of
any event A can be written as
Note 5 of 5E
The Law of Total Probability
A
A S
k
A S
1
S
2.
S
1
S
k
P(A) = P(A S
1
) + P(A S
2
) + + P(A S
k
)
= P(S
1
)P(A|S
1
) + P(S
2
)P(A|S
2
) + + P(S
k
)P(A|S
k
)
Note 5 of 5E
Bayes Rule
Let S
1
, S
2
, S
3
,..., S
k
be mutually exclusive and
exhaustive events with prior probabilities P(S
1
),
P(S
2
),,P(S
k
). If an event A occurs, the posterior
probability of S
i
, given that A occurred is
,...k , i
S A P S P
S A P S P
A S P
i i
i i
i
2 1 for
) | ( ) (
) | ( ) (
) | ( =
=
) | ( ) (
) | ( ) (
) (
) (
) | (
) | ( ) ( ) (
) (
) (
) | (
Proof
i i
i i i
i
i i i
i
i
i
S A P S P
S A P S P
A P
AS P
A S P
S A P S P AS P
S P
AS P
S A P
= =
= =
Note 5 of 5E
We know:
P(F) =
P(M) =
P(H|F) =
P(H|M) =
Example
From a previous example, we know that 49% of the
population are female. Of the female patients, 8% are
high risk for heart attack, while 12% of the male patients
are high risk. A single person is selected at random and
found to be high risk. What is the probability that it is a
male?
Define H: high risk F: female M: male
61 .
) 08 (. 49 . ) 12 (. 51 .
) 12 (. 51 .
) | ( ) ( ) | ( ) (
) | ( ) (
) | (
=
+
=
+
=
F H P F P M H P M P
M H P M P
H M P
.12
.08
.51
.49
Note 5 of 5E
Example
Suppose a rare disease infects one out of
every 1000 people in a population. And
suppose that there is a good, but not perfect,
test for this disease: if a person has the
disease, the test comes back positive 99% of
the time. On the other hand, the test also
produces some false positives: 2% of
uninfected people are also test positive. And
someone just tested positive. What are his
chances of having this disease?
Note 5 of 5E
We know:
P(A) = .001 P(A
c
) =.999
P(B|A) = .99 P(B|A
c
) =.02
Example
Define A: has the disease B: test positive
0472 .
02 . 999 . 99 . 001 .
99 . 001 .
) | ( ) ( ) | ( ) (
) | ( ) (
) | (
=
+
=
+
=
c
A B P
c
A P A B P A P
A B P A P
B A P
We want to know P(A|B)=?
Note 5 of 5E
Example
A survey of job satisfaction
2
of teachers was
taken, giving the following results
2
Psychology of the Scientist: Work Related Attitudes of U.S. Scientists
(Psychological Reports (1991): 443 450).
Satisfied Unsatisfied Total
College 74 43
117
High School 224 171
395
Elementary 126 140
266
Total 424 354 778
Job Satisfaction
L
E
V
E
L
Note 5 of 5E
Example
If all the cells are divided by the total number
surveyed, 778, the resulting table is a table of
empirically derived probabilities.
Satisfied Unsatisfied Total
College 0.095 0.055
0.150
High School 0.288 0.220
0.508
Elementary 0.162 0.180
0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
Note 5 of 5E
Example
For convenience, let C stand for the event that
the teacher teaches college, S stand for the
teacher being satisfied and so on. Lets look at
some probabilities and what they mean.
is the proportion of teachers who are
college teachers.
P(C) 0.150 =
is the proportion of teachers who are
satisfied with their job.
P(S) 0.545 =
is the proportion of teachers who are
college teachers and who are satisfied
with their job.
P(C S) 0.095 =
Satisfied Unsatisfied Total
College 0.095 0.055
0.150
High School 0.288 0.220
0.508
Elementary 0.162 0.180
0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
Note 5 of 5E
Example
is the proportion of teachers who
are college teachers given they are
satisfied. Restated: This is the
proportion of satisfied that are
college teachers.
P(C S)
P(C| S)
P(S)
0.095
0.175
0.545
=
= =
is the proportion of teachers who
are satisfied given they are
college teachers. Restated: This is
the proportion of college teachers
that are satisfied.
P(S C)
P(S| C)
P(C)
P(C S) 0.095
P(C) 0.150
0.632
=
= =
=
Satisfied Unsatisfied Total
College 0.095 0.055
0.150
High School 0.288 0.220
0.508
Elementary 0.162 0.180
0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
Note 5 of 5E
Example
P(C S) 0.095
P(C) 0.150 and P(C| S) 0.175
P(S) 0.545
= = = =
Satisfied Unsatisfied Total
College 0.095 0.055
0.150
High School 0.288 0.220
0.508
Elementary 0.162 0.180
0.342
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
P(C|S) = P(C) so C and S are dependent events.
Are C and S independent events?
Note 5 of 5E
Example
Satisfied Unsatisfied Total
College 0.095 0.055
0.150
High School 0.288 0.220
0.508
Elementary 0.162 0.180
0.658
Total 0.545 0.455 1.000
L
E
V
E
L
Job Satisfaction
P(C) = 0.150, P(S) = 0.545 and
P(CS) = 0.095, so
P(CS) = P(C)+P(S) - P(CS)
= 0.150 + 0.545 - 0.095
= 0.600
P(CS)?
Note 5 of 5E
Tom and Dick are going to take
a driver's test at the nearest DMV office. Tom
estimates that his chances to pass the test are
70% and Dick estimates his as 80%. Tom and
Dick take their tests independently.
Define D = {Dick passes the driving test}
T = {Tom passes the driving test}
T and D are independent.
P (T) = 0.7, P (D) = 0.8
Example
Note 5 of 5E
What is the probability that at most one of
the two friends will pass the test?
Example
P(At most one person pass)
= P(D
c
T
c
) + P(D
c
T) + P(D T
c
)
= (1 - 0.8) (1 0.7) + (0.7) (1 0.8) + (0.8) (1 0.7)
= .44
P(At most one person pass)
= 1-P(both pass) = 1- 0.8 x 0.7 = .44
Note 5 of 5E
What is the probability that at least one of
the two friends will pass the test?
Example
P(At least one person pass)
= P(D T)
= 0.8 + 0.7 - 0.8 x 0.7
= .94
P(At least one person pass)
= 1-P(neither passes) = 1- (1-0.8) x (1-0.7) = .94
Note 5 of 5E
Suppose we know that only one of the two
friends passed the test. What is the
probability that it was Dick?
Example
P(D | exactly one person passed)
= P(D exactly one person passed) / P(exactly one
person passed)
= P(D T
c
) / (P(D T
c
) + P(D
c
T) )
= 0.8 x (1-0.7)/(0.8 x (1-0.7)+(1-.8) x 0.7)
= .63
Note 5 of 5E
Random Variables
A quantitative variable x is a random variable if
the value that it assumes, corresponding to the
outcome of an experiment is a chance or random
event.
Random variables can be discrete or
continuous.
Examples:
x = SAT score for a randomly selected student
x = number of people in a room at a randomly
selected time of day
x = number on the upper face of a randomly
tossed die
Note 5 of 5E
Probability Distributions for
Discrete Random Variables
The probability distribution for a discrete
random variable x resembles the relative
frequency distributions we constructed in
Chapter 2. It is a graph, table or formula that
gives the possible values of x and the
probability p(x) associated with each value.
1 ) ( and 1 ) ( 0
have must We
= s s x p x p
Note 5 of 5E
Example
Toss a fair coin three times and
define x = number of heads.
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
P(x = 0) = 1/8
P(x = 1) = 3/8
P(x = 2) = 3/8
P(x = 3) = 1/8
HHH
HHT
HTH
THH
HTT
THT
TTH
TTT
x
3
2
2
2
1
1
1
0
x p(x)
0 1/8
1 3/8
2 3/8
3 1/8
Probability
Histogram for x
Note 5 of 5E
Example
Toss two dice and define
x = sum of two dice.
x p(x)
2 1/36
3 2/36
4 3/36
5 4/36
6 5/36
7 6/36
8 5/36
9 4/36
10 3/36
11 2/36
12 1/36
Note 5 of 5E
Probability Distributions
Probability distributions can be used to describe
the population, just as we described samples in
Chapter 2.
Shape: Symmetric, skewed, mound-shaped
Outliers: unusual or unlikely measurements
Center and spread: mean and standard
deviation. A population mean is called and a
population standard deviation is called o.
Note 5 of 5E
The Mean
and Standard Deviation
Let x be a discrete random variable with
probability distribution p(x). Then the
mean, variance and standard deviation of x
are given as
2
2 2
: deviation Standard
) ( ) ( : Variance
) ( : Mean
o o
o
=
=
=
x p x
x xp
Note 5 of 5E
Example
Toss a fair coin 3 times and
record x the number of heads.
x p(x) xp(x) (x-)
2
p(x)
0 1/8 0 (-1.5)
2
(1/8)
1 3/8 3/8 (-0.5)
2
(3/8)
2 3/8 6/8 (0.5)
2
(3/8)
3 1/8 3/8 (1.5)
2
(1/8)
5 . 1
8
12
) ( = = = x xp
) ( ) (
2 2
x p x o =
688 . 75 .
75 . 28125 . 09375 . 09375 . 28125 .
2
= =
= + + + =
o
o
Note 5 of 5E
Example
The probability distribution for x the
number of heads in tossing 3 fair
coins.
Shape?
Outliers?
Center?
Symmetric;
mound-shaped
None
= 1.5
o = .688
Note 5 of 5E
Key Concepts
I. Experiments and the Sample Space
1. Experiments, events, mutually exclusive events,
simple events
2. The sample space
II. Probabilities
1. Relative frequency definition of probability
2. Properties of probabilities
a. Each probability lies between 0 and 1.
b. Sum of all simple-event probabilities equals 1.
3. P(A), the sum of the probabilities for all simple events in A
Note 5 of 5E
Key Concepts
III. Counting Rules
1. mn Rule; extended mn Rule
2. Permutations:
3. Combinations:
IV. Event Relations
1. Unions and intersections
2. Events
a. Disjoint or mutually exclusive: P(A B) = 0
b. Complementary: P(A) = 1 P(A
C
)
)! ( !
!
)! (
!
r n r
n
C
r n
n
P
n
r
n
r
=
Note 5 of 5E
Key Concepts
3. Conditional probability:
4. Independent and dependent events
6. Multiplicative Rule of Probability:
7. Law of Total Probability
8. Bayes Rule
) (
) (
) | (
B P
B A P
B A P
=
) ( ) ( ) ( ) ( B A P B P A P B A P + =
) | ( ) ( ) ( A B P A P B A P =
Note 5 of 5E
Key Concepts
V. Discrete Random Variables and Probability
Distributions
1. Random variables, discrete and continuous
2. Properties of probability distributions
3. Mean or expected value of a discrete random
variable:
4. Variance and standard deviation of a discrete
random variable:
1 ) ( and 1 ) ( 0 = s s x p x p
2
2 2
: deviation Standard
) ( ) ( : Variance
o o
o
=
= x p x
) ( : Mean x xp = | 8,985 | 24,606 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-30 | latest | en | 0.877903 |
https://ontonixqcm.blog/2014/03/08/the-not-that-useful-definitions-of-complexity/ | 1,685,563,104,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647409.17/warc/CC-MAIN-20230531182033-20230531212033-00049.warc.gz | 487,824,394 | 32,707 | # The not-that-useful Definitions of Complexity
“Every few months seems to produce another paper proposing yet another measure of complexity, generally a quantity which can’t be computed for anything you’d actually care to know about, if at all. These quantities are almost never related to any other variable, so they form no part of any theory telling us when or how things get complex, and are usually just quantification for quantification’s own sweet sake”. Read more in: http://cscs.umich.edu/~crshalizi/notebooks/complexity-measures.html. The above mentioned abundance of candidate complexity measures – a clear reflection of the rampant fragmentation in the field – is summarized in: http://en.wikipedia.org/wiki/Complexity as follows: In several scientific fields, “complexity” has a specific meaning:
In computational complexity theory, the time complexity of a problem is the number of steps that it takes to solve an instance of the problem as a function of the size of the input (usually measured in bits), using the most efficient algorithm. This allows to classify problems by complexity class (such as P, NP) such analysis also exists for space, that is, the memory used by the algorithm.
In algorithmic information theory, the Kolmogorov complexity (also called descriptive complexity or algorithmic entropy) of a string is the length of the shortest binary program which outputs that string.
In information processing, complexity is a measure of the total number of properties transmitted by an object and detected by an observer. Such a collection of properties is often referred to as a state.
In physical systems, complexity is a measure of the probability of the state vector of the system. This is often confused with entropy, but is a distinct Mathematical analysis of the probability of the state of the system, where two distinct states are never conflated and considered equal as in statistical mechanics.
In mathematics, Krohn-Rhodes complexity is an important topic in the study of finite semigroups and automata.
In the sense of how complicated a problem is from the perspective of the person trying to solve it, limits of complexity are measured using a term from cognitive psychology, namely the hair limit.
Specified complexity is a term used in intelligent design theory, first coined by William Dembski.
Irreducible complexity is a term used in arguments against the generally accepted theory of biological evolution, being a concept popularized by the biochemist Michael Behe.
Unruly complexity denotes situations that do not have clearly defined boundaries, coherent internal dynamics, or simply mediated relations with their external context, as coined by Peter Taylor.
SDL complexity – the “simple” measure of complexity of Shiner, Davison and Landsberg (SDL). As an SDL classical complexity measure, the quantum complexity measure is defined by the weighted product of the quantum disorder by the quantum order.
And now, ask yourself this: can I use any of these measures to study the evolution of a corporation, of air-traffic, of a market? Can any of these ‘measures’ help identify a complex system and distinguish it from a “simple system”?
www.ontonix.com | 639 | 3,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-23 | latest | en | 0.925833 |
https://www.examrace.com/NTA-UGC-NET/NTA-UGC-NET-Previous-Years-Papers/Electronic-Science/NTA-UGC-NET-Electronic-Science-Paper-3-Solved-December-2013.html | 1,656,757,373,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104054564.59/warc/CC-MAIN-20220702101738-20220702131738-00064.warc.gz | 824,610,717 | 14,267 | 📣 Paper 3 has been removed from NET from 2018 (Notification)- now paper 2 and 3 syllabus is included in paper 2. Practice both paper 2 and 3 from past papers.
# NTA NET Electronic-Science December-2013 Solved Paper III
Online Paper 1 complete video lectures with Dr. Manishika Jain. Join now!
1. In an npn transistor, the expression for avalanche multiplication factor is given by
1. M = 1/ (1 + VCB/B VCBO) n
2. M = (1 + VCB/B VCBO) n
3. M = (1 – VCB/B VCBO) n
4. M = 1/ [1 – (VCB/B VCBO) n]
Where, VCB is collector to base voltage and BVCBO is maximum reverse biasing voltage which may be applied before breakdown between the collector and base terminals.
Answer: d
2. In a bipolar transistor, stability factor for a fixed bias circuit is given by
1. S = 1/ (1 + β)
2. S = 1/ (β – 1)
3. S = (β – 1) 2
4. S = β + 1
Answer: d
3. In a Wein bridge oscillator circuit, the value of frequency can be calculated by the following expression:
1. fo = π RC
2. fo = π LC
3. fo = π L1 (C1 + C2)
4. fo = 1 2π (R21 + R22) (C21 + C22)
Answer: a
4. A square coil has the dimensions 0.2 m × 0.2 m and carrying a current of 3.0 A in a field of 10 wb/m2. The value of Torque is given by
1. 10.2 N – m
2. 1.02 N – m
3. 0.12 N – m
4. 1.2 N – m
Answer: d
5. For a transmission line which isterminated in a normalised impedance Zn, VSWR = 2, the value of normalised impedance is given by
1. 2
2. 3
Answer: a
Answer: a
6. Which of the following circuit comes under the class of sequential logic circuits?
1. Multiplexer
2. ₹ Latch
3. Full Adder
4. ROM
Answer: b
7. The Hammng code for 0110 using even parity is
1. 0010110
2. 1010110
3. 1100110
4. 1110110
Answer: c
8. Phase Lock Loop (PLL) system is used for the detection of
1. PM
2. AM
3. FM
4. QAM
Answer: c
9. A transistor amplifier has a measured S/N of 10 at its input and 5 at its output. The transistor՚s Noise Figure (NF) in dB is
1. 2 dB
2. 3 dB
3. 6 dB
4. 10 dB
Answer: b
10. A signal varies from 20 Hz to 5 KHz is passed using pulse modulation scheme. Minimum sampling rate and number of channels that could be accommodated using TDM (assume each sample takes 10 μs) respectively will be
1. 5 KHz, 5
2. 10 KHz, 5
3. 5 KHz, 10
4. 10 KHz, 10
Answer: d
11. The term (1/jw) on the log-magnitude plot has a slope of
1. – 20 dB/decade
2. + 20 dB/decade
3. – 40 dB/decade
4. + 40 dB/decade
Answer: a
12. The Routh Array is as below: S6 1 8 20 6 S5 2 12 16 S4 2 12 16 S3 0 0 The row of zero of this ray will be replaced by coefficients of:
1. S4 + 12 S2 + 16
2. S3 + 3 S
3. S4 + 6 S2 + 8
4. S3 + 12 S
Answer: b
13. Zener breakdown mechanism occurs in reverse biased PN junction
1. When P and N regions are lightly doped.
2. When P and N regions are heavily doped.
3. Are of silicon material only.
4. When P and N regions are equally doped.
Answer: b
14. The operation of a Photo-diode involves
1. Photo-conductive effect
2. Photo-voltaic effect
3. Photo-emissive effect
4. Photo-multiplicative effect
Answer: c
15. A UJT has RBB = 10 K and RB2 = 4 K. Its intrinsic stand-off ratio is
1. 0.6
2. 0.4
3. 2.4
4. 3.5
Answer: a
16. The triggered voltage of a SCR is close to
1. 0.3 V
2. 0.7 V
3. 3 V
4. Breakdown voltage
Answer: b
17. A thyristor can be used as
1. An amplifier
2. A resistor
3. A switch
4. A power source
Answer: c
18. Which of the following is an advantage to use fiber optic data transmission?
1. Resistance to the data theft
2. Fast data transmission rate
3. Low noise level
4. All of the above
Answer: d
19. A breakdown which is caused by cumulative multiplication of carriers through field induced impact ionization occurs in
1. Avalanche diode
2. Tunnel diode
3. Varactor diode
4. Gunn diode
Answer: a
20. Transfer function of a system is necessary for the calculation of
1. The time constant
2. The output for a given input
3. The steady state gain
4. The order of the system
Answer: b
21. If one or more pairs of simple roots are located on imaginary axis of the s-plane but there are no roots in the right half of s-plane, the response due to initial condition will
1. Decrease to zero as time approaches infinity.
2. Increase as time approaches infinity.
3. Be undamped sinusoidal oscillations.
4. Be damped unsinusoidal oscillation and the damping factor will depend upon the relative location of the roots on the imaginary axis.
Answer: c
22. 8255A Programmable Peripheral Interface IC has got the 24 I/O lines in the following way (General form) :
1. Port A with 8 input lines, Port B with 8 output lines, Port C (upper nibble) with 4 input lines and Port C (lower nibble) with 4 output lines.
2. Port A with 8 input/out lines, Port B with 8 input lines and Port C with 8 output lines.
3. Port A with 8 input lines, Port B with 8 output lines and Port C with 8 bits as either input or output lines.
4. Port A with 8 lines as input/output lines, Port B with 8 lines as input/output lines, Port C (lower nibble) with 4 lines as input/output lines. Port C (upper nibble) with 4 lines as input/output lines.
Answer: d
23. C program is as follows: The result of the program is int i; / ⚹ … declare integer i … ⚹ /i = 10; / ⚹ … set i to 10 … ⚹ /i = i + ‘A’ / ⚹ … add character A … ⚹ // ⚹ … to integer i … ⚹ /print f ( “i =% d” i)
1. 55
2. 51
3. 71
4. 75
Answer: d
24. Consider the following conditional expression z = (x > y) ? x: y; if x = 2 and y = 8, the value of z is
1. 2
2. 8
3. 6
4. 10
Answer: b
25. int Net = 0; if x < 50 if (y > 5) Net = x + y; else Net = x – y; For this C program segment, x = 55 and y = 5, then the value Net is
1. 0
2. 60
3. 50
4. 55
Answer: a
26. In 8253 programmable interval timer, in which modes, the counting is neither enabled nor disabled?
1. 0 and 4
2. 1 and 5
3. 1 and 4
4. 3 and 5
Answer: b
27. In which mode of 8259A programmable interrupt controller, all the Interrupt Requests (IRs) are arranged from highest to lowest with IR0 as the highest and IR1 as the lowest?
1. Fully Nested Mode
2. Automatic Rotation Mode
3. Specific Rotation Mode
4. Simple Rotation Mode
Answer: a
28. In an OPAMP, following characteristics are given:
1. PSRR (Power Supply Rejection Ratio) is 0
2. Thermal Drift is defined in terms of μ Amperes/° C.
3. Thermal drift is defined in terms of μV/° C.
4. Slew rate is defined in terms of V/μs.
Which one of the following is correct?
1. 1 and 2 only
2. 1 and 3 only
3. 1,3 and 4
4. 1,2 and 4
Answer: c
29. Consider the following statement:
1. A flip-flop is used to store 1-bit of information.
2. Race-around condition occurs in J-K flip-flop when both the inputs are 1.
3. Master-slave configuration is used in flip-flop to store 2-bits of information.
4. A transparent latch consists of D type flip-flop.
Which of the following statements is/are true?
1. 1 only
2. 1,3 and 4
3. 1,2 and 4
4. 2 and 3 only
Answer: c
30. An AM demodulator can be implemented with
1. A linear multiplier followed by a low pass filter.
2. A linear multipler followed by a high-pass filter.
3. A diode followed by low pass filter.
4. A linear multiplier followed byband-stop filter.
The correct answer is:
1. 1 only
2. 3 only
3. 1 and 3
4. 4 only
Answer: c
31. In feedback control system, relative stability can be calculated using
1. Routh-Hurwitz Array
2. Nyquist plot
3. Polar plot
4. Root Locus Techniques
Correct answer is
1. All of the above
2. 1 and 2 only
3. 2 and 4 only
4. 2,3 and 4
Answer: c
32. In comparison to LED, LASER has
1. High emission frequency.
2. No tuning arrangement.
3. Wide spectral bandwidth.
4. Provision for confinement.
Of these statements:
1. 1,2 & 4 are correct.
2. 1,2, & 3 are correct.
3. 1 & 4 are correct.
4. 2 & 3 are correct.
Answer: c
33. The turn-off time of an SCR can be reduced by
1. Quick withdrawal of the gate voltage.
2. Reducing life-time by doping with gold.
3. Applying a negative voltage pulse to the gate.
Of these statements:
1. 1,2 and 3 are correct.
2. 1 and 2 are correct.
3. 1 and 3 are correct.
4. 2 and 3 are correct.
Answer: d
34. Consider the following statements regarding a semiconductor:
1. Acceptor level lies close to the valence band.
2. Donor level lies close to the valence band.
3. n-type semiconductor behavesas a conductor at zero Kelvin.
4. p-type semiconductor behaves as an insulator at zero Kelvin.
Of these statements:
1. 2 and 3 are correct.
2. 1 and 3 are correct.
3. 1 and 4 are correct.
4. 3 and 4 are correct
Answer: c
35. Transfer function for a control system is defined for where
1. Linear system
2. Nonlinear system
3. Time invariant system
4. Time variant system
1. 1 and 3
2. 2 and 4
3. 1 and 4
4. 2 and 3
Answer: a
36. Tri state Buffers provide
1. Reduction of current consumption in the circuit.
2. Isolation from input to output.
3. High impedance during OFF state.
4. Low impedance during ON state.
1. 1 and 4
2. 2 and 4
3. 1 and 2
4. 1 and 3
Answer: d
37. Consider the following:
1. CE stage
2. CC stage
3. OP amp
4. CB stage
The correct sequence of the input impedance in increasing order is:
1. 4,1, 2,3
2. 1,4, 2,3
3. 1,2, 4,3
4. 4,2, 1,3
Answer: a
38. The various components in super heterodyne receiver is arranged as
1. AM Detector
2. Mixer
3. RF Amplifier
4. AF Amplifier
The correct sequence is
1. 3,2, 1,4
2. 1,2, 4,3
3. 3,2, 4,1
4. 2,1, 3,4
Answer: a
39. Fora unity feedback control system having an open-loop transfer function G (S) = [K (S + 2) ] /S2 (S4 + 7 S + 12) The error constant Kp, Kv and Ka respectively are
1. ∞ ∞ K/6
2. 0,0, K/6
3. ∞ 0, K/6
4. 0, ∞ K/6
Answer: a
40. Arrange in ascending order the following logic families based on power delay products
1. ECL
2. TTL
3. CMOS
Codes:
1. 2,3, 1
2. 2,1, 3
3. 1,2, 3
4. 3,2, 1
Answer: a
41. Consider the following four common type of transistors:
1. Point Contact Transistor
2. Bipolar Junction Transistor
3. MOS Field Effect Transistor
4. Junction Field Effect Transistor
Correct arrangement of these transistors in the increasing order of input impedance is
1. 1,2, 4,3
2. 1,2, 3,4
3. 2,1, 3,4
4. 2,1, 4,3
Answer: d
42. If the various logic families are arranged in the ascending order of their fan-out capabilities, the sequence will be
1. TTL, DTL, ECL, MOS
2. DTL, TTL, MOS, ECL
3. MOS, DTL, TTL, ECL
4. ECL, TTL, DTL, MOS
Answer: c
43. Consider the following waves/rays:
1. UV Rays
2. X Rays
3. Visible light
4. UHF waves
The correct sequence of the descending order in terms of frequency is:
1. 3,1, 2,4
2. 4,3, 1,2
3. 2,1, 3,4
4. 2,4, 1,3
Answer: c
44. The structure of a decision table, divided into four parts like 1 4 2 3
1. Condition Stub
2. Condition Entries
3. Action Stub
4. Action Entries
1. 1,2, 4 and 3
2. 2,4, 3 and 1
3. 1,3, 4 and 2
4. 1,4, 2 and 3
Answer: c
45. The parameters associated with time response of control system are
1. Delay time (td)
2. Settling time (ts)
3. Rise time (tr)
4. Peak time (tp)
Arrange the above in the order, the parameter which is having minimum time to the parameter which has got the maximum time.
1. 1,4, 2 and 3
2. 1,3, 2 and 4
3. 1,4, 3 and 2
4. 1,3, 4 and 2
Answer: d
46. In the Assembly program, the following steps are to be followed. Find the sequence in which the program to be written for 8085 microprocessor.
1. Initialization of variables.
2. Initialization of stack.
3. Enable or disable interrupts.
4. Program should be completedwith last line as ‘END’ statement.
1. 1,3, 2 and 4
2. 1,2, 3 and 4
3. 2,1, 3 and 4
4. 2,3, 1 and 4
Answer: c
47. Match the following lists:
List-I List-II Alpha of TransistorBeta of TransistorCMRR (Common Mode Rejection Ratio)PSRR (Power Supply Rejection Ratio) Greater than 1∞Less than 1Equal to 0
• A
• B
• C
• D
• 3
• 1
• 2
• 4
• 4
• 1
• 2
• 3
• 1
• 2
• 3
• 4
• 2
• 1
• 3
• 4
Answer: a
48. Match the following lists:
List-I List-II MagnetronPin diodeKlystronGunn diode DetectorBunchingLow Power Oscillatorπ-mode
• A
• B
• C
• D
• 2
• 3
• 4
• 1
• 1
• 2
• 3
• 4
• 3
• 1
• 2
• 4
• 4
• 1
• 2
• 3
Answer: d
49. Match the following lists:
List-I List-II CompandingSquelchPreemphasisDouble conversion Improving image rejectionVariation of step rise in quantisationMuting the receiverBoosting of higher modulating frequencies at the transmitter
• A
• B
• C
• D
• 2
• 3
• 4
• 1
• 2
• 1
• 4
• 3
• 1
• 2
• 3
• 4
• 1
• 2
• 4
• 3
Answer: a
50. Match the following lists:
List-I List-II 74150741527415374157 Quad 2: 1 MultiplexerDual 4: 1 Multiplexer8: 1 Multiplexer16: 1 Multiplexer
• A
• B
• C
• D
• 1
• 2
• 3
• 4
• 4
• 3
• 2
• 1
• 4
• 2
• 3
• 1
• 2
• 1
• 4
• 3
Answer: b
51. Match the following lists:
List-I (Characteristic of the device) List-II (Device) Voltage controlled deviceCurrent controlled deviceConductivity modulated deviceNegative conductance device BJTUJTFETIMPATT
• A
• B
• C
• D
• 2
• 3
• 1
• 4
• 2
• 3
• 4
• 1
• 3
• 1
• 4
• 2
• 3
• 1
• 2
• 4
Answer: c
52. Match the following lists:
List-I List-II Optical fiber communicationMobile communicationDigital communicationAnalog communication FDMTDMCDMAWavelength-Division Multiplexing (WDM)
• A
• B
• C
• D
• 1
• 3
• 2
• 4
• 4
• 3
• 2
• 1
• 2
• 4
• 1
• 3
• 3
• 1
• 2
• 3
Answer: b
53. In the following question Match List-I with List-II
List-I List-II Bode՚s-Plot representationNyquist DagramNichols ChartsRoot-locus Method not a frequencydomaintechnique.low frequency and high frequency characteristics of the transfer function can be determineddecides the Stability criteriaalso known as Polar plots
• A
• B
• C
• D
• 4
• 3
• 1
• 2
• 4
• 3
• 2
• 1
• 3
• 4
• 1
• 2
• 2
• 3
• 4
• 1
Answer: d
54. The following are features of a micro controller 8051:
1. 4 k bytes of ROM or EPROM
2. 128 k bytes of data memory
3. Four programmable I/O ports
4. Three 16 bit timer/event counters.
1. 1,2 and 3
2. 1,2 and 4
3. 1,3 and 4
4. 2,3 and 4
Answer: a
55. Identify the peripheral devices for their applications.
List-I List-II 8155827982538251 Serial CommunicationTimers and CountersKeyboard and display interfaceAdditional input/output lines to processor
• A
• B
• C
• D
• 2
• 3
• 4
• 1
• 3
• 4
• 1
• 2
• 4
• 1
• 2
• 3
• 4
• 3
• 2
• 1
Answer: d
56. Match the following lists:
List-I List-II Stability of Control SystemUnstability of Control SystemFrequency responseDamped Oscillation Oscillation in which the amplitude decreases with timeAll roots of characteristic equation have negative real partSteady state responseAny root of characteristic equation has a positive real part
• A
• B
• C
• D
• 4
• 2
• 3
• 1
• 2
• 4
• 3
• 1
• 2
• 4
• 1
• 3
• 4
• 2
• 1
• 3
Answer: b
• Assertion (A) : In amplifiers, it is easy to compare two powers on a Logarithmic rather than on linear scale. It is called decibel.
• Reason (R) : Decibel is defined as N = 10 log P2/P1, where P2 is output power and P1 is input power.
1. Both A and R are true and R is the correct explanation of (A) .
2. Both A and R are true, but R is not the correct explanation of (A) .
3. A is true and R is false.
4. A is false and R is true.
Answer: a
• Assertion (A) : Magnetron is not cross field devices.
• Reason (R) : They make use of electric and magnetic fields simultaneously. The fields are perpendicular to each other.
1. Both A and R are true and R is the correct explanation of (A) .
2. Both A and R are true, but R is not the correct explanation of (A) .
3. A is true and R is false.
4. A is false and R is true.
Answer: d
• Assertion (A) : TDM can be employed to transmit channels having unequal bandwidths.
• Reason (R) : If sampling theorem is strictly followed, any analog signal can be reconstructed back from the samples.
1. Both A and R are true and R is the correct explanation of (A) .
2. Both A and R are true, but R is not the correct explanation of (A) .
3. A is true and R is false.
4. A is false and R is true.
Answer: b
• Assertion (A) : A NAND gate is called a Universal logic element.
• Reason (R) : Any logic function can be realized using NAND gates alone.
1. Both A and R are true and R is the correct explanation of (A) .
2. Both A and R are true, but R is not the correct explanation of (A) .
3. A is true and R is false.
4. A is false and R is true.
Answer: a
• Assertion (A) : The stability of a control system can be determined from the location of roots of characteristic equation.
• Reason (R) : For stability the roots should lie on the left half of s-plane.
1. Both A and R are true and R is the correct explanation of (A) .
2. Both A and R are true, but R is not the correct explanation of (A) .
3. A is true and R is false.
4. A is false and R is true.
Answer: a
• Assertion (A) : The intrinsic Fermi level of a semiconductor does not lie exactly at the middle of the energy band gap.
• Reason (R) : The densities of the available states in valence and conduction bands of an intrinsic semiconductor areequal.
1. Both A and R are true and R is the correct explanation of (A) .
2. Both A and R are true, but R is not the correct explanation of (A) .
3. A is true and R is false.
4. A is false and R is true.
Answer: d
• Assertion (A) : The operating principle of Laser is based on stimulated emission process.
• Reason (R) : In coherent radiation, the emitted photons have same phase, same polarization and same direction with the incident photon.
1. Both A and R are true and R is the correct explanation of (A) .
2. Both A and R are true, but R is not the correct explanation of (A) .
3. A is true and R is false.
4. A is false and R is true.
Answer: a
• Assertion (A) : A microprocessor interfaced with inputs, outputs and other peripheral devices is also called a micro controller system and is capable of controlling the process.
• Reason (R) : The microprocessor is a device which processes the instructions.
1. Both A and R are true and R is the correct explanation of (A) .
2. Both A and R are true, but R is not the correct explanation of (A) .
3. A is true and R is false.
4. A is false and R is true.
Answer: a
• Assertion (A) : In control system design, the transfer function consists of both poles and zeros. To make the system stable, the poles plotted in Bode plot should be near to origin.
• Reason (R) : The zeros must be also near to origin.
1. Both A and R are true and R is the correct explanation of (A) .
2. Both A and R are true, but R is not the correct explanation of (A) .
3. A is true and R is false.
4. A is false and R is true.
Answer: d
• Assertion (A) : A p-channel enhancement MOSFET based transistor can be turn on prematurely.
• Reason (R) : Most contaminants in MOS fabrication are mobile positively charged ions and they get trapped between the gate and the substrate in an n-channel enhancement MOSFET, whereas they are trapped on the other side of the substrate in the case of a p-channel enhancement MOSFET.
1. Both A and R are true and R is the correct explanation of (A) .
2. Both A and R are true, but R is not the correct explanation of (A) .
3. A is true and R is false.
4. A is false and R is true.
Answer: a
## Read the Paragraph and Answer the Question Nos. 71 to 75
The field effect transistor is a semiconductor device which depends for its operation on the control of current by an electric field. There are two types of field effect transistor, the Junction Field Effect Transistor (JFET) and Metal-Oxide-Semiconductor (MOSFET) FETs operation depends upon the flow of majority carriers only. It is therefore a unipolar device. BJT is a bipolar device. FET is relatively immune to radiation and it exhibits a high input resistance tipically many mega-ohms. It is less noisy than a tube or a bipolar transistor. It exhibits no offset voltage at zero drain current, and hence makes an excellent signal chopper. FETs are more temperature stable than BJTs. JFET is three terminal devices with gate applied potential control the flow of charges from source to drain. The n-channel MOSFET consists of a lightly p-type substrate into which two highly doped n + regions are diffused. These n + sections, which will act as the source and drain. A thin layer of insulating SiO2 is grown over the surface of the structure and holes are cut into the oxide layer, allowing contact with the source and drain. This layer results in an extremely high input resistance.
1. The point above the drain voltage, where there is no increase in drain current in a JFET is called as
1. Break down point
2. Pinch off point
3. Knee point
4. Critical point
Answer: b
2. FET is disadvantageous in comparison with BJT because of
1. High input impedance
2. Low noise
3. High gain bandwidth behaviour
4. Current controlled behaviour
Answer: d
3. For an n-channel silicon FET with α = 3 × 10 – 4 cms and ND = 1015 electron/cm3, find the pinch is of voltage. [∈ = 12 ∈ 0 π ∈ 0 = 9 × 109 (Newton – m2) /Coulomb]
1. 6.00 V
2. 5.4 V
3. 6.8 V
4. 4.5 V
Answer: c
4. An FET is a better chopper than BJT because it has
1. Higher series on resistance
2. Lower input current
3. Higher input impedance
4. Lower off-set voltage
Answer: d
5. For an n-channel enhancement mode MOSFET the drain current
1. Decreases with increase in drain current.
2. Decreases with decreases in drain voltage.
3. Increases with increase in drain voltage.
4. Increases with decrease in gate voltage.
Answer: c | 6,848 | 21,058 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2022-27 | latest | en | 0.870414 |
https://fxsolver.com/browse/?cat=2&p=54 | 1,620,296,758,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00104.warc.gz | 285,667,141 | 53,122 | '
# Search results
Found 589 matches
Triangular number
A triangular number or triangle number counts the objects that can form an equilateral triangle. The nth triangle number is the number of dots composing a ... more
Triangular Prism Volume
A triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides.
... more
Triangulation (surveying)
In surveying, triangulation is the process of determining the location of a point by measuring only angles to it from known points at either end of a fixed ... more
Triclinic crystal system (Unit cell's volume)
In crystallography, the triclinic crystal system is one of the 7 crystal systems. A crystal system is described by three basis vectors. In the triclinic ... more
Worksheet 542
Triple-angle's cosine (related to the cosine of the single angle)
rigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, ... more
Triple-angle's sine (related to the sine of the single angle)
Trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables. Geometrically, ... more
Uniform Circular Motion position (X - coordinate)
In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with ... more
Uniform Circular Motion position (Y - coordinate)
In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with ... more
Variance
The variance is a parameter that describes, in part, either the actual probability distribution of an observed population of numbers, or the theoretical ... more
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Create a new formula | 407 | 1,940 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-21 | latest | en | 0.877697 |
https://thesaurus.altervista.org/dict/en/parallelogram | 1,723,059,545,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640707024.37/warc/CC-MAIN-20240807174317-20240807204317-00560.warc.gz | 452,763,763 | 4,161 | English
parallelogram
Pronunciation
• (America) enPR: pâ'rə-lĕlʹə-grăm', IPA: /ˌpæ.ɹə.ˈlɛl.ə.ˌɡɹæm/
Noun
parallelogram (plural parallelograms)
1. (geometry) A convex quadrilateral in which each pair of opposite edges are parallel and of equal length.
2. (Gaelic games, dated) either of two rectangular areas (respectively the large parallelogram and the small parallelogram) abutting the goal line in front of the goal. (Since 1986 officially named the large rectangle and small rectangle, though the older names are still occasionally used.)
• 1907 GAA, Official Guide "Football Rules"; quoted in Joseph Lennon, The playing rules of football and hurling, 1884-1995 (Northern Recreation Consultants 1997) p.66 ISBN 978-1 902097 00 8:
2 Marking of ground — [...] A five yard square shall be marked in front of each goal, having the goal posts at adjacent angles. A five yards square shall be marked out in front of each point space, having a goal post and a point post at adjacent angles. There will be thus formed in front of scoring area a parallelogram fifteen yards by five yards.
• 1981 GAA, The Playing Rules of Football and Hurling; quoted in Joseph Lennon, The playing rules of football and hurling, 1884-1995 (Northern Recreation Consultants 1997) p.358 ISBN 978-1 902097 00 8:
Two parallelograms of the dimensions set out hereunder shall be formed in front of each scoring space. One parallelogram, 14 mts. by 4.5 mts. shall be formed by two lines 4.5 mts. long and at right angles to the end-line, being marked 3.80 mts. from each goal-post, and the ends of these lines being joined. A larger parallelogram 19 mts. by 13 mts. shall be formed by two lines 13 mts. long and at right angles to the end-line, being marked 6.4 mts. from each goal-post, and the ends of these lines being joined.
• 2009 June 13, "Classy Cork sink Kerry" ↗ GAA website:
Goulding was fouled in the small parallelogram and O’Connor nicked the penalty into the net off the post.
• 2011 September 4, Denis Walsh "Cats avoid making same mistakes but Tipp have the right balance" ↗ The Sunday Times
There are seven Kilkenny players inside the large parallelogram and only two from Tipperary. Yet Tipp have scored a goal.
Translations
This text is extracted from the Wiktionary and it is available under the CC BY-SA 3.0 license | Terms and conditions | Privacy policy 0.004 | 643 | 2,356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.918608 |
https://kids-math-games.com/subtraction/ | 1,679,651,584,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00715.warc.gz | 397,521,836 | 12,075 | # Subtraction
## Subtraction Fun Game For Kids
A great subtraction game for kids is Adder-Ladders, in which players race to the bottom of the game board by subtracting and adding. The game is fun and educational for kids and is a great way to practice math facts. It includes three levels of challenge, each requiring the players to learn their subtraction facts. The first two levels are for one-digit numbers, while the third level challenges them with two-digit numbers.
Another subtraction game that is sure to get kids involved is Around the World. Using a deck of cards and flashcards, this game requires students to figure out a subtraction equation on each card. The student who answers the question first moves on to the next student. This game is a great way to reinforce subtraction concepts and improve the students’ hand-eye coordination.
A number of subtraction math games are available online for children to practice their skills. Many games require little preparation and are ideal for first graders who are learning to subtract and second graders who need a review. Subtraction games are a great way to reinforce learning subtraction skills and are much more engaging than worksheets.
There are many types of subtraction games for kids, and they are all fun for all ages. Some of the more popular games are the following: a farm game, a game of subtraction, a game that involves matching objects, and a few others. A game that is challenging but fun for both kids and adults is a great way to teach kids how to think about math.
The first step in learning how to subtract is understanding the place value system. Students should be able to subtract a ten to a hundred, a ten to a ten, and a one from one. The students should be able to understand this relationship between addition and subtraction, and they should be able to perform multi-digit subtraction with regrouping.
Another fun subtraction game is one where students count the number of objects remaining after subtracting them. Using a picture of the objects is a fun way to teach kids how to perform subtraction with their fingers. In addition, a simple game called subtraction within ten requires students to write an equation for each number.
Author: Donald Young | 438 | 2,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2023-14 | latest | en | 0.959466 |
https://www.coursehero.com/file/21402464/Linear-Expansion-Coefficient-Lab/ | 1,610,812,781,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506697.14/warc/CC-MAIN-20210116135004-20210116165004-00303.warc.gz | 740,531,580 | 85,536 | Linear Expansion Coefficient Lab - Linear Expansion Coefficient Lab Physics 103A Mohammad Ihsan Julian Victoria Patrick Rojek Nirav Patel Professor
# Linear Expansion Coefficient Lab - Linear Expansion...
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Linear Expansion Coefficient Lab Physics 103A Mohammad Ihsan, Julian Victoria, Patrick Rojek, Nirav Patel Professor Huang 2/13/17
Objective The objective of this lab was to measure the coefficient of linear expansion for different metal rods. Introduction When the temperature of a certain substance is increased in a range that does not produce a change of phase, the linear dimension for that substance increases as well. The added heat increases the vibrational kinetic energy of the atoms in the material, which in turn increases the distance between the atoms. An equation is thus derived, where the increase in a linear dimension ∆ L (L-L 0 ) is proportional to the increase in temperature ∆T (T-T 0 ) and initial dimension L 0 . The equation is: ∆ L L 0 = α ∆T , where the proportionality α is called the coefficient of linear expansion. This coefficient depends on many factors, such as, but not limited to, the composition and purity of the material, and temperature. This was the main equation that we needed for this experiment.
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• Spring '16 | 302 | 1,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-04 | latest | en | 0.892471 |
https://fr.maplesoft.com/support/help/Maple/view.aspx?path=DEtools/convertsys | 1,718,588,292,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861674.39/warc/CC-MAIN-20240616233956-20240617023956-00098.warc.gz | 238,849,729 | 23,863 | convertsys - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
DEtools
convertsys
convert a system of differential equations to a first-order system
Calling Sequence convertsys(deqns, inits, vars, ivar, yvec, ypvec)
Parameters
deqns - ordinary differential equation, or set or list of equations; can be specified as expressions, which are assumed equal to zero inits - set or list of initial conditions vars - function or list or set of functions; dependent variables of the system ivar - independent variable yvec - (optional) name to be used for the solution vector in the first-order system ypvec - (optional) name to be used for the yvec' vector in the first-order system
Description
• DEtools[convertsys] converts a system of one or more ordinary differential equations to a system of first-order differential equations. Corresponding initial conditions (if specified) are also converted.
• The initial conditions inits must each be of the form function = expression. See examples below. Note: if no initial conditions are available, inits must still be specified as {} or [].
• The convertsys command returns a list $\left[\mathrm{eqnlist},\mathrm{Ydefs},\mathrm{x0},\mathrm{Y0}\right]$ where
* $\mathrm{eqnlist}$ is the list of equations representing the first-order system $Y'\left(x\right)=f\left(x,Y\left(x\right)\right)$ in which the $Y$ vector is specified by ${\mathrm{yvec}}_{1},...,{\mathrm{yvec}}_{\mathrm{neqns}}$ and the $Y'$ vector by ${\mathrm{ypvec}}_{1},...,{\mathrm{ypvec}}_{\mathrm{neqns}}$.
* $\mathrm{Ydefs}$ is the list of equations defining the ${\mathrm{yvec}}_{i}$ names in terms of the original functions.
* $\mathrm{x0}$ is the point at which the initial conditions are specified. It is returned as undefined if inits is the empty set.
* $\mathrm{Y0}$ is a list representing the vector of initial conditions (possibly empty). Only one of the initial conditions needs to be specified for this list to be nonempty. Note that the ordering of $\mathrm{Y0}$ matches that of $\mathrm{Ydefs}$, not that of inits.
• This function is part of the DEtools package, and so it can be used in the form convertsys(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[convertsys](..).
Examples
> $\mathrm{with}\left(\mathrm{DEtools}\right):$
> $\mathrm{deq1}≔\frac{{ⅆ}^{2}}{ⅆ{t}^{2}}y\left(t\right)=y\left(t\right)-x\left(t\right),\frac{ⅆ}{ⅆt}x\left(t\right)=x\left(t\right):$
> $\mathrm{init1}≔y\left(0\right)=1,\mathrm{D}\left(y\right)\left(0\right)=2,x\left(0\right)=3:$
> $\mathrm{convertsys}\left(\left\{\mathrm{deq1}\right\},\left\{\mathrm{init1}\right\},\left\{x\left(t\right),y\left(t\right)\right\},t,y,\mathrm{y_p}\right)$
$\left[\left[{{\mathrm{y_p}}}_{{1}}{=}{{y}}_{{1}}{,}{{\mathrm{y_p}}}_{{2}}{=}{{y}}_{{3}}{,}{{\mathrm{y_p}}}_{{3}}{=}{{y}}_{{2}}{-}{{y}}_{{1}}\right]{,}\left[{{y}}_{{1}}{=}{x}{}\left({t}\right){,}{{y}}_{{2}}{=}{y}{}\left({t}\right){,}{{y}}_{{3}}{=}\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\right]{,}{0}{,}\left[{3}{,}{1}{,}{2}\right]\right]$ (1)
> $\mathrm{deq2}≔{\mathrm{D}}^{\left(3\right)}\left(y\right)\left(x\right)=y\left(x\right)x:$
> $\mathrm{init2}≔y\left(0\right)=3,\mathrm{D}\left(y\right)\left(0\right)=2,{\mathrm{D}}^{\left(2\right)}\left(y\right)\left(0\right)=1:$
> $\mathrm{convertsys}\left(\mathrm{deq2},\left[\mathrm{init2}\right],y\left(x\right),x\right)$
$\left[\left[{{\mathrm{YP}}}_{{1}}{=}{{Y}}_{{2}}{,}{{\mathrm{YP}}}_{{2}}{=}{{Y}}_{{3}}{,}{{\mathrm{YP}}}_{{3}}{=}{{Y}}_{{1}}{}{x}\right]{,}\left[{{Y}}_{{1}}{=}{y}{}\left({x}\right){,}{{Y}}_{{2}}{=}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){,}{{Y}}_{{3}}{=}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right]{,}{0}{,}\left[{3}{,}{2}{,}{1}\right]\right]$ (2)
> $\mathrm{deq3}≔\frac{{ⅆ}^{2}}{ⅆ{t}^{2}}y\left(t\right)=100\left({ⅇ}^{-10t}+{ⅇ}^{10t}\right):$
> $\mathrm{convertsys}\left(\left\{\mathrm{deq3}\right\},\left[\right],y\left(t\right),t,V\right)$
$\left[\left[{{\mathrm{YP}}}_{{1}}{=}{{V}}_{{2}}{,}{{\mathrm{YP}}}_{{2}}{=}{100}{}{{ⅇ}}^{{-}{10}{}{t}}{+}{100}{}{{ⅇ}}^{{10}{}{t}}\right]{,}\left[{{V}}_{{1}}{=}{y}{}\left({t}\right){,}{{V}}_{{2}}{=}\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\right]{,}{\mathrm{undefined}}{,}\left[\right]\right]$ (3)
> $\mathrm{convertsys}\left(\left\{\mathrm{deq3}\right\},\mathrm{D}\left(y\right)\left(0\right)=1,y\left(t\right),t,V\right)$
$\left[\left[{{\mathrm{YP}}}_{{1}}{=}{{V}}_{{2}}{,}{{\mathrm{YP}}}_{{2}}{=}{100}{}{{ⅇ}}^{{-}{10}{}{t}}{+}{100}{}{{ⅇ}}^{{10}{}{t}}\right]{,}\left[{{V}}_{{1}}{=}{y}{}\left({t}\right){,}{{V}}_{{2}}{=}\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right)\right]{,}{0}{,}\left[{y}{}\left({0}\right){,}{1}\right]\right]$ (4) | 1,903 | 4,975 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 27, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-26 | latest | en | 0.651914 |
http://www.jiskha.com/display.cgi?id=1365635746 | 1,462,138,814,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860116929.30/warc/CC-MAIN-20160428161516-00176-ip-10-239-7-51.ec2.internal.warc.gz | 614,530,413 | 3,790 | Sunday
May 1, 2016
# Homework Help: Chemistry
Posted by ramire2 on Wednesday, April 10, 2013 at 7:15pm.
The reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15x10^2 at a given temperature. In a particular experiment, 3.0 mol of each component was added to a 1.5 L flask. What is the equilibrium concentration of hydrogen fluoride?
So I did 3/1.5L= 2M
H2 + F2 -------> 2HF
2 2 0
-x -x +2x
2x /(2-x)^2 = 1.15 x 10^2
From here, I don't really know how to solve for x to where the HF concentration would equal 5.056 M, which is the right answer. | 195 | 615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2016-18 | longest | en | 0.937937 |
https://hackaday.com/2024/05/29/the-genius-of-slide-rule-precision/ | 1,719,099,276,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00421.warc.gz | 252,828,484 | 31,543 | # The Genius Of Slide Rule Precision
Most people have heard of or seen slide rules, with older generations likely having used these devices in school and at their jobs. As purely analog computers these ingenious devices use precomputed scales on slides, which when positioned to a specific input can give the output to a wide range of calculations, ranging from simple divisions and multiplications to operations that we generally use a scientific calculator for these days. Even so, these simple devices are both very versatile and can be extremely precise, as [Bob, the Science Guy] demonstrates in a recent video.
Slide rules at their core are very simple: you got different scales (marked by a label) which can slide relative to each other. Simple slide rules will only have the A through D scales, with an input provided by moving one scale relative to the relevant other scale (e.g. C and D for multiplication/division) after which the result can be read out. Of course, it seems reasonable that the larger your slide rule is, the more precision you can get out of it. Except that if you have e.g. the W1 and W2 scales on a shorter (e.g. 10″) slide rule, you can use those to get the precision of a much larger (20″) slide rule, as [Bob] demonstrates.
Even though slide rules have a steeper learning curve than punching numbers into a scientific calculator, it is hard to argue the benefits of understanding such relationships between the different scales, and why they exist in the first place.
## 49 thoughts on “The Genius Of Slide Rule Precision”
1. AZdave says:
I got my EE degree with only a Post Versalog, and didn’t have access to a four function calculator until a year after I graduated and was working in industry. One of my professors had a Wang “portable” computer that took up half the space in the trunk of his car. It all seems really primitive in retrospect.
1. Johnu says:
You shouldn’t call another man’s Wang primitive, it’s not polite.
2. KDawg says:
The one time I used a slide rule it was no more accurate than using rounded numbers due to errors in printed numbers and parallax effect
These things are at best quick math for spot checks proper tools were used for precision
1. Rpol_404 says:
NASA used them extensively in the 60s – even for the Apollo missions. Seems quite capable to me…
1. Bill Hensley says:
This is correct. My dad used his Post Versalog slide rule when he was working at NASA on the Apollo program. Every engineer had one and used it extensively.
2. My Dad worked at NASA and when his old house was abandoned after my mother went to a nursng home, I went in there and got some of his things so they wouldn’t be stolen by crackheads. This included his 30-30 rifle, a surveyors’ compass, and two aluminum slide rules.
They were good enough for Lockheed to build an SR71
3. I Alone Possess the Truth says:
It’s a poor workman who forgets his lunch. Stupid grapes, probably sour anyway. “I tried once and underperformed so the tool is bad.” Words to something or other by. But we can all agree that improperly printed slide rules are at least a national problem and parallax poses a threat on the scale of Satanic abuse in our day care centers. Parallax lurks around every corner.
In my day we simply fed the homework to the dog. Team sliderule!
3. fabo says:
Not a big fan of slide rules, always messing up on the decimal point. Got a simple add-subtract-multiply-divide calculator (\$100+ at the time) to help with this. For intro lessons, some classrooms had a giant slide over the chalkboard. Graduated about the time scientific calculators came out, and never touched the slide rule again.
1. Charles Springer says:
That is one of the advantages of being proficient with a slide rule. Doing quick mental estimates and keeping track of orders of magnitude mentally is a great skill and prevents many errors. One might find they easily spot errors in the numbers in news reports and the flood of AI generated Youtube videos, etc. And they are fast, with no keypad.
2. Antron Argaiv says:
Dimensional analysis. If you don’t have a scientific calculator or an IBM 360, a slide rule is better and faster than doing it by hand. I was in the last freshman Chemistry class that had tomuse slide rules. A fellow student came to class in January with an HP-35. I had a Bowmar 4-banger, so I contijued to bring my slide rule for trig functions,roots and logs. Don’t sell the slide rule short, in skilled hands it can do an awful lot, but the scientific calculator is even faster and more accurate.
4. Tom G says:
20inch is NOT long!
I have slide rules with a 500inch (21ft, 12,7m) scale. I have too many of them; must sell some.
They were in production continuously for 94 years, until the HP35 stole their market.
1. a_do_z says:
I thought this was a slide rule/estimating orders of magnitude joke. But given the following, maybe it’s not:
“The Texas Magnum is 350 feet 6.6 inches and holds the world’s record for the longest linear slide rule.”
http://mit-a.com/TexasMagnum.shtml (note: there’s no s in http).a
5. cdilla says:
When I started my O-Levels we were recommended by the science teachers to buy a slide rule and a subscription to New Scientist magazine.
I still use one on a weekly basis. The other was retired a year later when my Grandad gifted me his Sinclair Scientific calculator after he had upgraded it for something less reverse polishly.
While it lasted I enjoyed using the slide rule. Better imo than the well thumbed log table booklets.
1. Julian Skidmore says:
Interesting. Aged about 7 I persuaded my parents to buy a Sinclair Scientific, because they’d come massively down in price, but neither I nor my parents understood RPN. After a while I gave up trying to understand how I was supposed to enter calculations and swapped it out for a Commodore 4-function calculator (with ‘%’ !!!).
It was only later during work experience in my fourth year of secondary school where I learned how to use a slide rule, because my line manager had a circular slide rule. Admittedly as I already understood logs etc it was pretty easy. I then bought a Thornton slide rule from a local Preedy’s stationary shop as they were being sold off cheaply (no-one needed a slide rule in 1983!). I still have that too, & it’s pretty complex, so I wonder if it’s got the Wx scales talked about here, or the equivalent?
2. . says:
Sinclair scientific was the fastest way to answer the question: “What is 22 222 ++ ?”
6. Joe says:
A slide rule got me through tech school in the seventies. Two place accuracy was acceptable for some courses, scientific notation took care of decimal places, often large ones.
Log tables were used for more precision. Beyond that it was hand calculations.
We had adding machine size calculators at work. When they became obsolete I diceted one out of curiosity. The most fascinating component was the buffer memory implement with a coil of piano wire.
7. thom says:
Still have one……….somewhere?!?
1. The Commenter Formerly Known As Ren says:
I’ve acquired around 30 over the years, they are in a tub in the garage.
8. The one absolutely overriding benefit of learning to use a slide rule is the discipline that how learning to use it imposes on one’s thinking and ability to calculate accurately and quickly–most specifically: keeping track of the decimal point.
Most students today cannot handle this simple task even with a calculator (scientific? four-function? Doesn’t matter.) This fact has forced me to impose a ‘no calculators or other electronic devices allowed for taking quizzes’ policy for all tests / quizzes I give. This policy also increase my work-load tremendously, as I must ensure that all problems require that students’ mental mathematical gymnastics are reduced to, at worst, dividing or multiplying by ten, or two. Even then, the amount of errors and wrong answers are–to put it mildly–stupefying.
(one doesn’t need any leaps of imagination to guess at the uproar and indignation that the appearance of this requirement, in the course’s syllabus, generates)
I will never forget one of the most prescient and accurate lectures I ever heard on engineering–in general–as a profession, and as an undertaking: “…never forget that in–in general–engineering, as in life: ‘close’ is good enough. The really good…the exceptional…engineer knows when this approach is NOT good enough“.
‘Smart’phones; ‘smart’ TVs; ‘smart’ electronic calculators…all have one thing in common: they are all oxymorons.
1. Miroslav says:
Exactly. If you can’t estimate properly without a calculator, you don’t know what you are doing.
9. prfesser says:
Used a pocket-sized circular slide rule in freshman chemistry. The C and D scales were accurate enough for three places. The inner scales weren’t much use; too small. For logarithms I used log tables instead. Learning where to place the decimal was a useful mental exercise.
This particular slide rule was all-purpose. The back had a periodic table, and there was an insert that could be removed, with conversion factors of all sorts. The chem prof said I could use the slide rule but not the insert… Everything was engraved so that it wouldn’t wear.
Calculators had barely come in. A guy in my dorm got a Sears calculator for high school graduation. Add, subtract, multiply, divide, and glorious! square roots! \$200 (1973). By the time I reached junior year a similar calculator was about \$20. A godsend during PChem.
1. Charles Springer says:
We must be the same age. I started physics with a Pickett and the CRC Handbook and the CRC Math Tables. When I was a junior I bought an HP45 by working afternoons in the Univ. print shop. It was fantastic and I have been firmly RPN ever since. Today I still have the Pickett and a Faber-Castell (which is fairly expensive today due to collectors) in my desk.
10. Keith says:
Still have my Lafayette 99-7031 from 1969 (high school physics). My dad, who was an electrical engineer, bought it for me. Usually use HP 35s now. Used to have the slide rule on my desk at work and younger engineers would sometimes look at it and ask, “What’s that?”
11. Oppy says:
I’ve still got a British Thornton AD 150 on my desk. I was the last year to be taught how to use one when at school. I’ve no practical justification for occasionally using it, I just enjoy it.
12. Fred says:
First off I admit I’m old. used a circular slide rule from grade 8-12. The first calculator’s came out when I was in grade 10. Very expensive, students where not allowed to use them . the thought being it would be unfair to the students that could not afford them .
I got very fast at using the device. We were grading tests of another class , basically lust taking the test score and coming up with a percentage. The teacher supervising the class had just bought a shiny , very expensive calculator. He spotted me using the slide rule and said “that must be pretty slow, and challenged me to a race . We started with 6 test papers each . He was just starting his second when I finished. For anyone that has used one, just had to move my curser to the total of the test and the percentage was on the next scale .
13. Paul says:
I had a professor who had a cylindrical (helical) slide rule in his office. It could muster four figures of precision… and he didn’t need it. He could compute anything (including natural logs and square roots) in his head faster than anybody could do on a slide rule, let alone punching it into a calculator. Impressive fellow.
1. The Commenter Formerly Known As Ren says:
1. The Commenter Formerly Known As Ren says:
Wow!
I had never heard of it.
Thanks!
14. craig says:
Like that this dude exists but unless you already super know what you are doing with a slide rule, this video is really, really tough to follow.
15. Chris Maple says:
Precision is repeatability. Accuracy is correctness. If you get the same wrong answer every time, you are precise but not accurate.
16. wetfewrtgfew says:
I see a half circle for sqrt(x) and many other functions
17. Dielectric says:
It must be slide rule week. I re-discovered this old Pickett 12-inch slide rule in my desk drawer a couple of days ago, and spend a fun half hour remembering how to use it. It really is a neat concept.
Then today, going through a drawer at work, I found an ancient Ohmite slide rule with parallel resistance (special marks for the E24 series!) and Ohm’s law scales. Also AB and CD, FWIW. Marked “Made in USA, 1970”, I saved it from a co-worker’s desk after he passed away.
1. The Commenter Formerly Known As Ren says:
That reminds me of a tip I read on USENET; to mark/highlight regular resistance values on a sliderule (1.0, 1.2, 4.7, 5.6, etc.) So if you are trying to find parallel resistance values you can see which values are closest to your objective.
18. a_do_z says:
Dang, this is an old bunch. I am not a graybeard myself. I look terrible with a beard.
19. LetItGo says:
Some of the comments make the logical fallacy – “argument from antiquity”.
just because slide rules are “good ‘ol” tech from days gone by doesn’t mean they are superior, or even necessarily suitable for use today.
Yes, they are simple, no batteries, accurate, etc. But today when scientific calculators are affordable and ubiquitous, the argument that a slide rule was good enough for [insert spacecraft, aircraft, bridge, skyscraper or other notable design here] doesn’t mean they still need to be used today.
Keep them around, keep the knowledge going, we need to preserve tech history. But no one needs to think they are relevant for day to day work or lament the fact that these young upstarts don’t know how to use them.
1. Rob says:
When you really appreciate and follow technology, you know when a tech’s time has passed. Doesn’t mean it was or is bad, it’s just the world has moved on.
20. There is an advantage that a sliderule has that a keyboard doesn’t. If you line up a ratio, then you have all the elements of the same ratio laid out before you. Choose those numerators or denominators that work for you.
1. Dielectric says:
Very easily overlooked until you’ve used one, for sure! I might commit to using my slide rule for a few weeks just for the mental exercise. Messing with it for a few minutes reinforced some numerical relationships I had sort of buried in my head.
2. spaceminions says:
Ratio’s aren’t bad, but the calculator gets there almost as fast with more precision. The thing I thought was cool was the special functions, starting with trigonometry. Because figuring a quick low precision answer to a large number of simple geometric problems is exactly what you want when you intend to immediately use that answer to tell you where to cut your lumber when building something. Especially if you plan on adjusting your values to what looks right or is convenient, as opposed to a one-and-done calculation. It does take a little focus to set up the function in your head if you haven’t done it in awhile though.
1. Suppose you have a resistance and you select a voltage, then you have a range of current values anywhere you slide the cursor. You don’t need to keep entering numbers. As for precision, in designing circuits with real valued voltages, currents and resistors, you shouldn’t expect any accuracy better than 3 significant figures, if that, in any real circuit. It’s like that in any common engineering problem.
1. spaceminions says:
I know that, although I also know that it’s easier to figure errors if they’re all from tolerances and not calculations. But it’s only a ratio; it doesn’t have to take long any way you do it.
And if you have an entire circuit full of unknowns to calculate, the modern calculator makes it easiest because you can just assign each a letter and type in a few current or voltage law equations, even using complex numbers for impedance, and then the calculator will solve either in numbers or each as a function of the remaining unknowns. And if you need more than that, you can use a simulator.
21. Robert says:
I wish i had kept my Fabar Castel rule. It served me well in the late 60’s during my engineering apprenticeship qualifications. With a good eye we could achieve an answer to 3 decimal places.
22. 12AU76L6GC says:
Using a sliderule is way different from a calculator because the brain is a lot more engaged.
The precision (or accuracy) of a sliderule is certainly sufficient to build all the electronic circuits I’m familiar with. 5% resistors, +100,,-20% capacitors, etc.
Really fast doing a microwave path loss.
23. n3hat says:
Engineering data is often not known to more than 3 digits of accuracy. A slide rule suffices.
I used a slide rule for the first year or 2 of engineering school. When the HP35 came out at \$400 the wealthy guys bought them. Not too long after, TI came out with their scientific calculator at half the price, and I got on board.
The discipline of using scientific notation and keeping track of the decimal place is useful, as others have stated.
Please be kind and respectful to help make the comments section excellent. (Comment Policy)
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 3,872 | 17,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | longest | en | 0.954976 |
https://www.championbets.com.au/horse-racing/cash-out/ | 1,519,082,630,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812855.47/warc/CC-MAIN-20180219231024-20180220011024-00540.warc.gz | 819,506,060 | 24,302 | The early ‘cash out’ feature has made its way across most of the major bookmakers and on most types of bets.
You can even cash out on many single bets now. I don’t know why you’d ever want to, but your friendly bookmaker will give you the option: with another clip of the ticket, of course!
The main thing to remember with most bookmaker gimmicks is that they’re generally aimed at the punters with whom they most want to engage: the mugs. The casual punter who doesn’t know (or probably care) about the concept of long-term value.
As a more serious punter, you should be looking at a consistent approach which, over time, will produce the most value. As you’re taking a long-term view, the wins and losses (variance) will even out and you’ll be better off.
Let’s look at an example:
Four leg multis on soccer are very popular at the moment, due to the number of “3 out of 4” offers around. We’ll assume we’ve already nailed our first three legs and have a cash-out offer prior to the last.
Bet: four-leg multi, \$100 stake, price \$7.30
Situation: 3/3 winners, alive into last leg (Stoke City to defeat Bournemouth)
Last leg: Stoke City \$1.60, Draw \$4.50, Bournemouth \$5.00
Cash-out offer: \$440
So your options are as follows:
Option 1, cash out: \$440 collect, \$340 profit
Option 2, let it ride: win (\$730 collect, \$630 profit) or lose (zero collect)
Value?
We know that the bookies rate Stoke a \$1.60 chance, which equates to a 62.5% chance of winning (1 / 1.6) but we will use 60% to allow for the bookies margin. So in mathematical terms, and at the bookie’s ratings, your real choice is between \$340 cash (option 1), or a 60% chance of \$630: \$378 (option 2).
There you can see the value the bookies generate when punters elect to cash out. A totally ‘fair’ cash out offer at their own odds would be \$378. They may have dozens of punters with the same offer: the more that take up the lower \$340 amount, the better off they’ll be in a mathematical sense.
The value bet is clearly to let it ride. As always, the key to value is in the long-term: in that if you were presented with this exact situation many times over, ‘letting it ride’ is clearly the most profitable option.
If you use this calculation each time you’re considering a ‘cash out’ offer from a bookie, you’ll see the value you’re getting (or more accurately, losing) each time.
Rated price
That’s fairly straightforward, however things change if your own ratings are more accurate than the bookmaker’s (which they need to be, if you are to be a successful punter). For example, you may assess Stoke’s true odds as \$1.45, with the associated adjustments to the odds for the draw or the Bournemouth win. That’ll obviously change the figures and allow you to assess you own true value, rather than the bookmaker’s.
Racing
Nick Aubrey is a former actuary who takes an extremely mathematical approach to his punting. We’ve spoken to him before on the Betting 360 podcast.
The basic concept of the ‘cash out’ is the same with a racing quaddie as a sports multi, except in a horse race, there’s usually far more outcomes (potential winners).
Nick explains:
“My overriding rule is that it’s all about the price.
“If you think the cash-out amount is better than your expected return (your estimated chance of collect, multiplied by your full payout) then cash out.
“If the last leg is a race field (eg quaddie), then cashing out is equivalent to having a field bet on the last leg.
“So you are betting on every runner, some of which you would never want to back.
“In this circumstance, a hedge (side) bet on those runners which you haven’t taken in your original bet BUT you think have got a chance would be a better strategy.
“The hedge bet(s) could just be to cover your total outlay so that you are profit neutral on these runners, but still can win big on your original bets.” | 960 | 3,894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-09 | longest | en | 0.909662 |
http://mathhelpforum.com/algebra/61025-factoring.html | 1,508,503,891,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824104.30/warc/CC-MAIN-20171020120608-20171020140608-00379.warc.gz | 228,367,452 | 10,687 | 1. ## factoring
a. f(x) = x2-5x-14 b. f(x) = 9x2 -4
c. f(x) = 6x2 -x -12
factor the following
2. Originally Posted by william
a. f(x) = x2-5x-14 b. f(x) = 9x2 -4
c. f(x) = 6x2 -x -12
factor the following
You should know how to factor.
a)
$f(x)= x^2 - 5x - 14$
$f(x)= (x+7)(x-2)$
Find the two numbers that multiply to make -14 and add to get -5.
3. Originally Posted by william
a. f(x) = x2-5x-14 b. f(x) = 9x2 -4
c. f(x) = 6x2 -x -12
factor the following
b)
$f(x)= 9x^2 - 4$
$f(x)= (3x+2)(3x-2)$
This is known as a difference of squares when factoring. When both numbers can be square root to get an even number, it is a difference of squares, notice how the even square root is in both brackets, with opposite signs distributing to;
$f(x)= 9x^2 - 4$
4. Originally Posted by william
a. f(x) = x2-5x-14 b. f(x) = 9x2 -4
c. f(x) = 6x2 -x -12
factor the following
c.
$f(x)= 6x^2 - x - 12$
$f(x)= (2x-3)(3x+4)$
Again, simply find the numbers that multiply to -12 and add to -x. Always remember the last sign in the equation, it is very important. If it is negative, the signs will be difference in the set of brackets. If it is positive the signs will be the same, they will both be what the first sign is.
5. Originally Posted by euclid2
You should know how to factor.
a)
$f(x)= x^2 - 5x - 14$
$f(x)= (x{\color{red}+}7)(x{\color{red}-}2)$
Find the two numbers that multiply to make -14 and add to get -5.
Small mistakes in red. It should be $f(x)= (x{\color{red}-}7)(x{\color{red}+}2)$.
Small mistakes in red. It should be $f(x)= (x{\color{red}-}7)(x{\color{red}+}2)$. | 592 | 1,578 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2017-43 | longest | en | 0.811023 |
https://studymaterialcenter.in/question/when-a-rubber-band-is-stretched-by-a-distance-x-it-exerts-restoring-force-of-magnitude-f-ax-bx-2-where-a-and-b-are-constants-the-work-done-in-stretching-the-unstretched-rubber-band-by-l-is-41656/ | 1,685,255,425,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00584.warc.gz | 582,285,338 | 24,604 | # When a rubber-band is stretched by a distance $x$, it exerts restoring force of magnitude $\mathrm{F}=\mathrm{ax}+\mathrm{bx}^{2}$ where $\mathrm{a}$ and $\mathrm{b}$ are constants. The work done in stretching the unstretched rubber-band by $\mathrm{L}$ is:
Question:
When a rubber-band is stretched by a distance $x$, it exerts restoring force of magnitude $\mathrm{F}=\mathrm{ax}+\mathrm{bx}^{2}$ where $\mathrm{a}$ and $\mathrm{b}$ are constants. The work done in stretching the unstretched rubber-band by $\mathrm{L}$ is:
1. $\mathrm{aL}^{2}+\mathrm{bL}^{3}$
2. $\frac{1}{2}\left(a L^{2}+b L^{3}\right)$
3. $\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{bL}^{3}}{3}$
4. $\frac{1}{2}\left(\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{bL}^{3}}{3}\right)$
JEE Main Previous Year Single Correct Question of JEE Main from Physics Work, Energy and Power chapter.
JEE Main Previous Year 2014
Correct Option: 3
Solution:
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https://math.answers.com/math-and-arithmetic/What_is_the_angle_of_each_exterior_angle_of_a_regular_polygon_with_15_sides | 1,725,763,017,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00316.warc.gz | 375,555,792 | 47,994 | 0
What is the angle of each exterior angle of a regular polygon with 15 sides?
Updated: 12/9/2022
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24 degrees
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Q: What is the angle of each exterior angle of a regular polygon with 15 sides?
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Related questions
How do you work out the number of sides in a regular polygon that has an exterior angle?
With a regular polygon: 360/exterior angle = number of sides
How do you work out the exterior and interior angles of a regular polygon with W sides?
Exterior angle regular polygon = 360° ÷ number of sides = 360° ÷ W Interior angle regular polygon = 180° - exterior angle regular polygon = 180° - (360° ÷ number of sides ) = 180° - (360° ÷ W)
how many sides does a regular polygon have if the exterior angle is 15?
by the number of sides, 'n' of that particular polygon. by the exterior angle. Therefore, the number of sides of the polygon is 24 sides.
What is the formula for the exterior angle of a regular polygon?
360/number of sides = exterior angle
What is the formula to find the sides of a regular polygon?
360/exterior angle = number of sides of a regular polygon
How many sides does a regular polygon have if the exterior angle measures 60?
A regular polygon that has an exterior angle of 60 degrees = 360/60 = 6 sides and it is an hexagon
How many sides has a rectangular polygon with an exterior angle of 40 degrees?
A regular polygon with an exterior angle of40O is a nonagon, and has 9 sides.
How many sides does a regular polygon have with an exterior angle measure of 1?
A regular polygon will have 360 sides with exterior angles of 181 degrees.
If each exterior angle of a regular polygon is 72 then how many sides does the regular polygon have?
5 sides, a regular pentagon.
How do you get the number of sides of a regular polygon with a given exterior angle?
By dividing the given exterior angle into 360 degrees tells you how many sides the polygon has.
What is the formula on how to get the measure of the exterior angle of the regular polygon?
360/number of sides = exterior angle
What is the exterior angle of a regular polygon with 24 sides?
Each exterior angle is 15 degrees | 536 | 2,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-38 | latest | en | 0.898235 |
http://rsmams.org/journals/articleinfo.php?articleid=472&tag=seajmams | 1,624,279,522,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488273983.63/warc/CC-MAIN-20210621120456-20210621150456-00082.warc.gz | 42,921,630 | 7,489 | Abstract
In this article we have discussed determination of distinct positive integers \$a,b,c,d\$ such that \$a+b, a+c, b+c, a+d, b+d, c+d\$ are cubes of positive integers with \\
(i) at least three numbers, say \$a,b,c\$ are positive.\\
(ii) all four numbers \$a,b,c,d\$ are positive.\\
We can obtain infinitely many four tuples from a single four-tuple.
Keywords and Phrases
Perfect squares, cubes, cubefree numbers, taxicab numbers, cubefree taxicab numbers, primes.
A.M.S. subject classification
11A67.
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http://www.urbandictionary.com/define.php?term=recursive&defid=48733 | 1,506,292,903,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690228.57/warc/CC-MAIN-20170924224054-20170925004054-00159.warc.gz | 584,304,845 | 25,003 | Top definition
recursive
by Anonymous May 11, 2003
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2
Please see recursive for a full explanation.
by bobbt.cum March 04, 2003
### The Urban Dictionary Mug
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3
"Wow, that's recursive."
by Ian Copp March 13, 2005
### The Urban Dictionary Mug
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1. Adj characterized by the recurrence or repetition of something.
Look at all those other definitions, they are recursive as fuck!
by Theron Valor October 19, 2011
### The Urban Dictionary Mug
One side has the word, one side has the definition. Microwave and dishwasher safe. Lotsa space for your liquids.
5
A procedure that is repeated over and over again.
This problem is a perfect example of a recursive formula.
by Rachel September 05, 2004
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Check out the spite of 'urban dictionary' entries on www.urbandictionary.com. The definitions are recursive, the act of viewing, reading and the emotions expressed are recursive.
by Jim Carhart January 26, 2007
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7
The reiteration of expletives to serve the interest of comprehension for the recipient and catharsis for the speaker.
She didn't just cuss me once, she got recursive on my ass.
by Harry Flashman June 28, 2003
### The Urban Dictionary Mug
One side has the word, one side has the definition. Microwave and dishwasher safe. Lotsa space for your liquids. | 443 | 1,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-39 | latest | en | 0.934547 |
http://www.codeproject.com/Articles/78486/Solar-Calculator-Calculate-Sunrise-Sunset-and-Maxi?msg=3588868 | 1,429,469,308,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246639325.91/warc/CC-MAIN-20150417045719-00155-ip-10-235-10-82.ec2.internal.warc.gz | 403,687,460 | 29,065 | # Solar Calculator - Calculate Sunrise, Sunset, and Maximum Solar Radiation
, 10 Jan 2011 CPOL
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A C# assembly for calculating Sunrise, Sunset, and Maximum Solar Radiation
## Introduction
I was looking for a .NET library to calculate the time the sun rises and sets. There were some libraries available, including here on The Code Project, but they had some problems. After trying some, I decided to create one myself. I needed the sunrise, sunset, and maximum solar radiation for a WordPress plug-in on my blog. The plug-in shows a map of the Netherlands together with the current weather conditions. The top of this post shows a screenshot of the plug-in. The map shows the real-time temperature including an icon. The icon is a translation of the measured solar radiation against the maximum solar radiation.
## Background
It is possible to calculate the sunrise, sunset, and maximum solar radiation using some known algorithms. For those of you who are interested in these algorithms, take a look at the following pages at Wikipedia: Declination of the Sun, Sunrise, Sunset. If you just want to calculate the sunrise, sunset, and maximum solar radiation, take a look below at how to use the code.
## Using the Code
The code is packed in a Visual Studio 2008 solution. It contains two assemblies: `Astronomy `and `AstronomyTest`. The assembly `Astronomy `contains the `SunCalculator` class which performs the actual calculation. The assembly `AstronomyTest `contains several unit-tests that validate the calculations against external sources.
`SunCalculator` is a single class that does not depend on external classes. Although this somewhat goes against the Single Responsibility Principle, it makes reuse of this class easier. The `SunCalculator` class needs the longitude, latitude, and time-zone of your location. You should also indicate whether to use daylight savings. An instance of the `SolarCalculator` can be created like this:
```const Double Longitute = 5.127869;
const Double Latitude = 52.108192;
const int LongituteTimeZone = 15;
const bool UseSummerTime = true;
SunCalculator sunCalculator = new SunCalculator(Longitute, Latitude,
LongituteTimeZone, UseSummerTime);```
You have to supply the longitude and the latitude from the location that you want to calculate the sunrise and sunset time. These are the two first arguments of the constructor. For locations that use daylight savings, you should set `UseSummerTime` to the actual daylight savings status. For locations that don't use daylight savings, set it to `false`.
The actual calculation of sunrise, sunset, and maximum solar radiation can be seen below:
```DateTime sunRise = sunCalculator.CalculateSunRise(new DateTime(2010, 4, 1));
DateTime sunSet = sunCalculator.CalculateSunSet(new DateTime(2010, 4, 1));
sunCalculator.CalculateMaximumSolarRadiation(new DateTime(2010, 1, 26, 16, 30, 0));```
The `DateTime` that is returned from `CalculateSunRise` and `CalculateSunSet` includes the sunrise and sunset time. For more information, take a look at the unit-tests in the assembly `AstronomyTest`.
## Points of Interest
The code first calculates the declination of the sun, cosine of the sun position, sinus of the sun position, and the difference between the solar and the actual time. All these parameters are used to calculate the sunrise and sunset times.
If you want to see the plug-in live, see my blog www.semanticarchitecture.net. The data that is retrieved and shown on the map comes from LetsGrow.com, the company that I work for.
## History
• v1.0 02/04/2010: Initial and first release
• v1.1 28/05/2010: Added a test case for Los Angeles, and a Console application that demonstrates the library in the source code
• v1.2 08/11/2011: Fixed failing tests
## Share
Architect http://www.hinttech.nl
Netherlands
Patrick Kalkman is a senior Software Architect with more than 20 years professional development experience. He works for Hinttech where he develops state of the art web applications.
Patrick enjoys writing his blog. It discusses software architectures using semantic web technologies. Patrick can be reached at pkalkie@gmail.com.
Published Windows 8 apps:
Published Windows Phone apps:
Awards:
Best Mobile article of March 2012
Best Mobile article of June 2012
First PrevNext
My vote of 5 ridoy 9-May-13 10:53
My vote of 5 Sperneder Patrick 30-Apr-13 5:02
Re: My vote of 5 Patrick Kalkman 30-Apr-13 5:22
How to calcualte LongituteTimeZone? fortuna_6 21-Jan-13 9:14
Re: How to calcualte LongituteTimeZone? Patrick Kalkman 30-Apr-13 5:24
My vote of 4 Christian Amado 23-Aug-12 12:54
Re: My vote of 4 Patrick Kalkman 30-Apr-13 5:18
Hi Patrick: squizfloats 17-Aug-12 13:53
Re: Hi Patrick: Patrick Kalkman 30-Apr-13 5:22
Solar radiation Ruben Tacq 3-Jul-12 22:11
Re: Solar radiation Patrick Kalkman 30-Apr-13 5:19
Re: Wrong result [modified] Patrick Kalkman 24-Mar-12 5:02
Hyperlink Failure Vasudevan Deepak Kumar 29-Feb-12 5:34
Re: Hyperlink Failure Patrick Kalkman 24-Mar-12 5:04
wrong calculation Member 8165547 16-Aug-11 12:08
Re: wrong calculation Patrick Kalkman 24-Mar-12 5:06
I see an "update" for Nov 8, 2011.... henslecd 3-Jun-11 3:08
Re: I see an "update" for Nov 8, 2011.... Patrick Kalkman 7-Jun-11 21:43
Not able to get corret sunrise and sunset time Paresh Rathod 11-May-11 10:56
Sky brightness Andrew Jones 19-Sep-10 5:55
Last Visit: 31-Dec-99 19:00 Last Update: 19-Apr-15 4:48 Refresh 12 Next » | 1,396 | 5,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2015-18 | latest | en | 0.877231 |
https://www.free-online-converters.com/short-tons-force/short-tons-force-into-onces-force.html | 1,660,029,969,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00431.warc.gz | 706,991,604 | 7,636 | Free Online Converters > Convert Short Tons-force Into Onces-force
Here you can Convert units of Short Tons-force to Onces-force units, find all information about Short Tons-force. So, enter your unit's value in Left Column like Short Tons-force(if you use standard resolution on most non-HD laptops. FULL HD resolution starts at 1920 x 1080). Otherwise, if you use a lower value, enter the value in the box above. The Result / another converted unit value shell appears in the Left or below Column.
# Convert Short Tons-force Into Onces-force
Short Tons-force
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##### conversion Table / conversion Chart
1 Short Tons-force = 32000 Onces-force
2 Short Tons-force = 64000 Onces-force
3 Short Tons-force = 96000 Onces-force
4 Short Tons-force = 128000 Onces-force
5 Short Tons-force = 160000 Onces-force
6 Short Tons-force = 192000 Onces-force
7 Short Tons-force = 224000 Onces-force
8 Short Tons-force = 256000 Onces-force
9 Short Tons-force = 288000 Onces-force
10 Short Tons-force = 320000 Onces-force
11 Short Tons-force = 352000 Onces-force
12 Short Tons-force = 384000 Onces-force
13 Short Tons-force = 416000 Onces-force
14 Short Tons-force = 448000 Onces-force
15 Short Tons-force = 480000 Onces-force
16 Short Tons-force = 512000 Onces-force
17 Short Tons-force = 544000 Onces-force
18 Short Tons-force = 576000 Onces-force
19 Short Tons-force = 608000 Onces-force
20 Short Tons-force = 640000 Onces-force
21 Short Tons-force = 672000 Onces-force
22 Short Tons-force = 704000 Onces-force
23 Short Tons-force = 736000 Onces-force
24 Short Tons-force = 768000 Onces-force
25 Short Tons-force = 800000 Onces-force
26 Short Tons-force = 832000 Onces-force
27 Short Tons-force = 864000 Onces-force
28 Short Tons-force = 896000 Onces-force
29 Short Tons-force = 928000 Onces-force
30 Short Tons-force = 960000 Onces-force
31 Short Tons-force = 992000 Onces-force
32 Short Tons-force = 1024000 Onces-force
33 Short Tons-force = 1056000 Onces-force
34 Short Tons-force = 1088000 Onces-force
35 Short Tons-force = 1120000 Onces-force
36 Short Tons-force = 1152000 Onces-force
37 Short Tons-force = 1184000 Onces-force
38 Short Tons-force = 1216000 Onces-force
39 Short Tons-force = 1248000 Onces-force
40 Short Tons-force = 1280000 Onces-force
41 Short Tons-force = 1312000 Onces-force
42 Short Tons-force = 1344000 Onces-force
43 Short Tons-force = 1376000 Onces-force
44 Short Tons-force = 1408000 Onces-force
45 Short Tons-force = 1440000 Onces-force
46 Short Tons-force = 1472000 Onces-force
47 Short Tons-force = 1504000 Onces-force
48 Short Tons-force = 1536000 Onces-force
49 Short Tons-force = 1568000 Onces-force
50 Short Tons-force = 1600000 Onces-force
50 Short Tons-force = 1600000 Onces-force
51 Short Tons-force = 1632000 Onces-force
52 Short Tons-force = 1664000 Onces-force
53 Short Tons-force = 1696000 Onces-force
54 Short Tons-force = 1728000 Onces-force
55 Short Tons-force = 1760000 Onces-force
56 Short Tons-force = 1792000 Onces-force
57 Short Tons-force = 1824000 Onces-force
58 Short Tons-force = 1856000 Onces-force
59 Short Tons-force = 1888000 Onces-force
60 Short Tons-force = 1920000 Onces-force
61 Short Tons-force = 1952000 Onces-force
62 Short Tons-force = 1984000 Onces-force
63 Short Tons-force = 2016000 Onces-force
64 Short Tons-force = 2048000 Onces-force
65 Short Tons-force = 2080000 Onces-force
66 Short Tons-force = 2112000 Onces-force
67 Short Tons-force = 2144000 Onces-force
68 Short Tons-force = 2176000 Onces-force
69 Short Tons-force = 2208000 Onces-force
70 Short Tons-force = 2240000 Onces-force
71 Short Tons-force = 2272000 Onces-force
72 Short Tons-force = 2304000 Onces-force
73 Short Tons-force = 2336000 Onces-force
74 Short Tons-force = 2368000 Onces-force
75 Short Tons-force = 2400000 Onces-force
76 Short Tons-force = 2432000 Onces-force
77 Short Tons-force = 2464000 Onces-force
78 Short Tons-force = 2496000 Onces-force
79 Short Tons-force = 2528000 Onces-force
80 Short Tons-force = 2560000 Onces-force
81 Short Tons-force = 2592000 Onces-force
82 Short Tons-force = 2624000 Onces-force
83 Short Tons-force = 2656000 Onces-force
84 Short Tons-force = 2688000 Onces-force
85 Short Tons-force = 2720000 Onces-force
86 Short Tons-force = 2752000 Onces-force
87 Short Tons-force = 2784000 Onces-force
88 Short Tons-force = 2816000 Onces-force
89 Short Tons-force = 2848000 Onces-force
90 Short Tons-force = 2880000 Onces-force
91 Short Tons-force = 2912000 Onces-force
92 Short Tons-force = 2944000 Onces-force
93 Short Tons-force = 2976000 Onces-force
94 Short Tons-force = 3008000 Onces-force
95 Short Tons-force = 3040000 Onces-force
96 Short Tons-force = 3072000 Onces-force
97 Short Tons-force = 3104000 Onces-force
98 Short Tons-force = 3136000 Onces-force
99 Short Tons-force = 3168000 Onces-force
100 Short Tons-force = 3200000 Onces-force
101 Short Tons-force = 3232000 Onces-force
102 Short Tons-force = 3264000 Onces-force
103 Short Tons-force = 3296000 Onces-force
104 Short Tons-force = 3328000 Onces-force
105 Short Tons-force = 3360000 Onces-force
106 Short Tons-force = 3392000 Onces-force
107 Short Tons-force = 3424000 Onces-force
108 Short Tons-force = 3456000 Onces-force
109 Short Tons-force = 3488000 Onces-force
110 Short Tons-force = 3520000 Onces-force
111 Short Tons-force = 3552000 Onces-force
112 Short Tons-force = 3584000 Onces-force
113 Short Tons-force = 3616000 Onces-force
114 Short Tons-force = 3648000 Onces-force
115 Short Tons-force = 3680000 Onces-force
116 Short Tons-force = 3712000 Onces-force
117 Short Tons-force = 3744000 Onces-force
118 Short Tons-force = 3776000 Onces-force
119 Short Tons-force = 3808000 Onces-force
120 Short Tons-force = 3840000 Onces-force
121 Short Tons-force = 3872000 Onces-force
122 Short Tons-force = 3904000 Onces-force
123 Short Tons-force = 3936000 Onces-force
124 Short Tons-force = 3968000 Onces-force
125 Short Tons-force = 4000000 Onces-force
126 Short Tons-force = 4032000 Onces-force
127 Short Tons-force = 4064000 Onces-force
128 Short Tons-force = 4096000 Onces-force
129 Short Tons-force = 4128000 Onces-force
130 Short Tons-force = 4160000 Onces-force
131 Short Tons-force = 4192000 Onces-force
132 Short Tons-force = 4224000 Onces-force
133 Short Tons-force = 4256000 Onces-force
134 Short Tons-force = 4288000 Onces-force
135 Short Tons-force = 4320000 Onces-force
136 Short Tons-force = 4352000 Onces-force
137 Short Tons-force = 4384000 Onces-force
138 Short Tons-force = 4416000 Onces-force
139 Short Tons-force = 4448000 Onces-force
140 Short Tons-force = 4480000 Onces-force
141 Short Tons-force = 4512000 Onces-force
142 Short Tons-force = 4544000 Onces-force
143 Short Tons-force = 4576000 Onces-force
144 Short Tons-force = 4608000 Onces-force
145 Short Tons-force = 4640000 Onces-force
146 Short Tons-force = 4672000 Onces-force
147 Short Tons-force = 4704000 Onces-force
148 Short Tons-force = 4736000 Onces-force
149 Short Tons-force = 4768000 Onces-force
150 Short Tons-force = 4800000 Onces-force
## how many Short Tons-force Into Onces-force
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How Many Zettanewtons in Short Tons-force | 2,803 | 9,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-33 | longest | en | 0.309414 |
http://en.wikipedia.org/wiki/Pl%c3%bccker_co-ordinates | 1,419,783,836,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447558091.146/warc/CC-MAIN-20141224185918-00054-ip-10-231-17-201.ec2.internal.warc.gz | 28,277,283 | 20,577 | # Plücker coordinates
(Redirected from Plücker co-ordinates)
In geometry, Plücker coordinates, introduced by Julius Plücker in the 19th century, are a way to assign six homogeneous coordinates to each line in projective 3-space, P3. Because they satisfy a quadratic constraint, they establish a one-to-one correspondence between the 4-dimensional space of lines in P3 and points on a quadric in P5 (projective 5-space). A predecessor and special case of Grassmann coordinates (which describe k-dimensional linear subspaces, or flats, in an n-dimensional Euclidean space), Plücker coordinates arise naturally in geometric algebra. They have proved useful for computer graphics, and also can be extended to coordinates for the screws and wrenches in the theory of kinematics used for robot control.
## Geometric intuition
Displacement and moment of two points on line
A line L in 3-dimensional Euclidean space is determined by two distinct points that it contains, or by two distinct planes that contain it. Consider the first case, with points x = (x1,x2,x3) and y = (y1,y2,y3). The vector displacement from x to y is nonzero because the points are distinct, and represents the direction of the line. That is, every displacement between points on L is a scalar multiple of d = yx. If a physical particle of unit mass were to move from x to y, it would have a moment about the origin. The geometric equivalent is a vector whose direction is perpendicular to the plane containing L and the origin, and whose length equals twice the area of the triangle formed by the displacement and the origin. Treating the points as displacements from the origin, the moment is m = x×y, where "×" denotes the vector cross product. For a fixed line, L, the area of the triangle is proportional to the length of the segment between x and y, considered as the base of the triangle; it is not changed by sliding the base along the line, parallel to itself. By definition the moment vector is perpendicular to every displacement along the line, so dm = 0, where "•" denotes the vector dot product.
Although neither d nor m alone is sufficient to determine L, together the pair does so uniquely, up to a common (nonzero) scalar multiple which depends on the distance between x and y. That is, the coordinates
(d:m) = (d1:d2:d3:m1:m2:m3)
may be considered homogeneous coordinates for L, in the sense that all pairs (λdm), for λ ≠ 0, can be produced by points on L and only L, and any such pair determines a unique line so long as d is not zero and dm = 0. Furthermore, this approach extends to include points, lines, and a plane "at infinity", in the sense of projective geometry.
Example. Let x = (2,3,7) and y = (2,1,0). Then (d:m) = (0:−2:−7:−7:14:−4).
Alternatively, let the equations for points x of two distinct planes containing L be
0 = a + ax
0 = b + bx .
Then their respective planes are perpendicular to vectors a and b, and the direction of L must be perpendicular to both. Hence we may set d = a×b, which is nonzero because a and b are neither zero nor parallel (the planes being distinct and intersecting). If point x satisfies both plane equations, then it also satisfies the linear combination
0 = a (b + b•x) − b (a + a•x) = (a b − b a)•x .
That is, m = a b − b a is a vector perpendicular to displacements to points on L from the origin; it is, in fact, a moment consistent with the d previously defined from a and b.
Example. Let a0 = 2, a = (−1,0,0) and b0 = −7, b = (0,7,−2). Then (d:m) = (0:−2:−7:−7:14:−4).
Although the usual algebraic definition tends to obscure the relationship, (d:m) are the Plücker coordinates of L.
## Algebraic definition
### Primal coordinates
In a 3-dimensional projective space P3, let L be a line through distinct points x and y with homogeneous coordinates (x0:x1:x2:x3) and (y0:y1:y2:y3). The Plücker coordinates pij are defined as follows:
$p_{ij} \,\!$ ${}= \begin{vmatrix} x_{i} & y_{i} \\ x_{j} & y_{j}\end{vmatrix} = x_{i}y_{j}-x_{j}y_{i} . \,\!$
This implies pii = 0 and pij = −pji, reducing the possibilities to only six (4 choose 2) independent quantities. The sixtuple
$(p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12}) \,\!$
is uniquely determined by L up to a common nonzero scale factor. Furthermore, not all six components can be zero. Thus the Plücker coordinates of L may be considered as homogeneous coordinates of a point in a 5-dimensional projective space, as suggested by the colon notation.
To see these facts, let M be the 4×2 matrix with the point coordinates as columns.
$M = \begin{bmatrix} x_0 & y_0 \\ x_1 & y_1 \\ x_2 & y_2 \\ x_3 & y_3 \end{bmatrix}$
The Plücker coordinate pij is the determinant of rows i and j of M. Because x and y are distinct points, the columns of M are linearly independent; M has rank 2. Let M′ be a second matrix, with columns x′ and y′ a different pair of distinct points on L. Then the columns of M′ are linear combinations of the columns of M; so for some 2×2 nonsingular matrix Λ,
$M' = M\Lambda . \,\!$
In particular, rows i and j of M′ and M are related by
$\begin{bmatrix} x'_{i} & y'_{i}\\x'_{j}& y'_{j} \end{bmatrix} = \begin{bmatrix} x_{i} & y_{i}\\x_{j}& y_{j} \end{bmatrix} \begin{bmatrix} \lambda_{00} & \lambda_{01} \\ \lambda_{10} & \lambda_{11} \end{bmatrix} .$
Therefore, the determinant of the left side 2×2 matrix equals the product of the determinants of the right side 2×2 matrices, the latter of which is a fixed scalar, det Λ. Furthermore, all six 2×2 subdeterminants in M cannot be zero because the rank of M is 2.
### Plücker map
Denote the set of all lines (linear images of P1) in P3 by G1,3. We thus have a map:
\begin{align} \alpha \colon \mathrm{G}_{1,3} & \rightarrow \mathbf{P}^5 \\ L & \mapsto L^{\alpha}, \end{align}
where
$L^{\alpha}=(p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12}) . \,\!$
### Dual coordinates
Alternatively, a line can be described as the intersection of two planes. Let L be a line contained in distinct planes a and b with homogeneous coefficients (a0:a1:a2:a3) and (b0:b1:b2:b3), respectively. (The first plane equation is ∑k akxk=0, for example.) The dual Plücker coordinate pij is
$p^{ij} \,\!$ ${}= \begin{vmatrix} a^{i} & a^{j} \\ b^{i} & b^{j}\end{vmatrix} = a^{i}b^{j}-a^{j}b^{i} . \,\!$
Dual coordinates are convenient in some computations, and they are equivalent to primary coordinates:
$(p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12})= (p^{23}:p^{31}:p^{12}:p^{01}:p^{02}:p^{03})$
Here, equality between the two vectors in homogeneous coordinates means that the numbers on the right side are equal to the numbers on the left side up to some common scaling factor $\lambda$. Specifically, let (i,j,k,l) be an even permutation of (0,1,2,3); then
$p_{ij} = \lambda p^{kl} .$
### Geometry
To relate back to the geometric intuition, take x0 = 0 as the plane at infinity; thus the coordinates of points not at infinity can be normalized so that x0 = 1. Then M becomes
$M = \begin{bmatrix} 1 & 1 \\ x_1 & y_1 \\ x_2& y_2 \\ x_3 & y_3 \end{bmatrix} ,$
and setting x = (x1,x2,x3) and y = (y1,y2,y3), we have d = (p01,p02,p03) and m = (p23,p31,p12).
Dually, we have d = (p23,p31,p12) and m = (p01,p02,p03).
## Bijection between lines and Klein quadric
### Plane equations
If the point z = (z0:z1:z2:z3) lies on L, then the columns of
$\begin{bmatrix} x_0 & y_0 & z_0 \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix}$
are linearly dependent, so that the rank of this larger matrix is still 2. This implies that all 3×3 submatrices have determinant zero, generating four (4 choose 3) plane equations, such as
$0 \,\!$ ${} = \begin{vmatrix} x_0 & y_0 & z_0 \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end{vmatrix}$ ${} = \begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix} z_0 - \begin{vmatrix} x_0 & y_0 \\ x_2 & y_2 \end{vmatrix} z_1 + \begin{vmatrix} x_0 & y_0 \\ x_1 & y_1 \end{vmatrix} z_2$ ${} = p_{12} z_0 - p_{02} z_1 + p_{01} z_2 . \,\!$ ${} = p^{03} z_0 + p^{13} z_1 + p^{23} z_2 . \,\!$
The four possible planes obtained are as follows.
$\begin{matrix} 0 & = & {}+ p_{12} z_0 & {}- p_{02} z_1 & {}+ p_{01} z_2 & \\ 0 & = & {}- p_{31} z_0 & {}- p_{03} z_1 & & {}+ p_{01} z_3 \\ 0 & = & {}+p_{23} z_0 & & {}- p_{03} z_2 & {}+ p_{02} z_3 \\ 0 & = & & {}+p_{23} z_1 & {}+ p_{31} z_2 & {}+ p_{12} z_3 \end{matrix}$
Using dual coordinates, and letting (a0:a1:a2:a3) be the line coefficients, each of these is simply ai = pij, or
$0 = \sum_{i=0}^3 p^{ij} z_i , \qquad j = 0,\ldots,3 . \,\!$
Each Plücker coordinate appears in two of the four equations, each time multiplying a different variable; and as at least one of the coordinates is nonzero, we are guaranteed non-vacuous equations for two distinct planes intersecting in L. Thus the Plücker coordinates of a line determine that line uniquely, and the map α is an injection.
The image of α is not the complete set of points in P5; the Plücker coordinates of a line L satisfy the quadratic Plücker relation
$0\,\!$ ${}= p_{01}p^{01}+p_{02}p^{02}+p_{03}p^{03} \,\!$ ${}= p_{01}p_{23}+p_{02}p_{31}+p_{03}p_{12} . \,\!$
For proof, write this homogeneous polynomial as determinants and use Laplace expansion (in reverse).
$0\,\!$ ${}= \begin{vmatrix}x_0&y_0\\x_1&y_1\end{vmatrix}\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}+ \begin{vmatrix}x_0&y_0\\x_2&y_2\end{vmatrix}\begin{vmatrix}x_3&y_3\\x_1&y_1\end{vmatrix}+ \begin{vmatrix}x_0&y_0\\x_3&y_3\end{vmatrix}\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}$ ${} = (x_0 y_1-y_0 x_1)\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}- (x_0 y_2-y_0 x_2)\begin{vmatrix}x_1&y_1\\x_3&y_3\end{vmatrix}+ (x_0 y_3-y_0 x_3)\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix} \,\!$ ${} = x_0 \left(y_1\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}- y_2\begin{vmatrix}x_1&y_1\\x_3&y_3\end{vmatrix}+ y_3\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}\right) -y_0 \left(x_1\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}- x_2\begin{vmatrix}x_1&y_1\\x_3&y_3\end{vmatrix}+ x_3\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}\right) \,\!$ ${} = x_0 \begin{vmatrix}x_1&y_1&y_1\\x_2&y_2&y_2\\x_3&y_3&y_3\end{vmatrix} -y_0 \begin{vmatrix}x_1&x_1&y_1\\x_2&x_2&y_2\\x_3&x_3&y_3\end{vmatrix} \,\!$
Since both 3×3 determinants have duplicate columns, the right hand side is identically zero.
Another proof may be done like this: Since vector
$d = \left( p_{01}, p_{02}, p_{03} \right)$
is perpendicular to vector
$m = \left( p_{23}, p_{31}, p_{12} \right)$
(see above), the scalar product of d and m must be zero! q.e.d.
### Point equations
Letting (x0:x1:x2:x3) be the point coordinates, four possible points on a line each have coordinates xi = pij, for j = 0…3. Some of these possible points may be inadmissible because all coordinates are zero, but since at least one Plücker coordinate is nonzero, at least two distinct points are guaranteed.
### Bijectivity
If (q01:q02:q03:q23:q31:q12) are the homogeneous coordinates of a point in P5, without loss of generality assume that q01 is nonzero. Then the matrix
$M = \begin{bmatrix} q_{01} & 0 \\ 0 & q_{01} \\ -q_{12} & q_{02} \\ q_{31} & q_{03} \end{bmatrix}$
has rank 2, and so its columns are distinct points defining a line L. When the P5 coordinates, qij, satisfy the quadratic Plücker relation, they are the Plücker coordinates of L. To see this, first normalize q01 to 1. Then we immediately have that for the Plücker coordinates computed from M, pij = qij, except for
$p_{23} = - q_{03} q_{12} - q_{02} q_{31} . \,\!$
But if the qij satisfy the Plücker relation q23+q02q31+q03q12 = 0, then p23 = q23, completing the set of identities.
Consequently, α is a surjection onto the algebraic variety consisting of the set of zeros of the quadratic polynomial
$p_{01}p_{23}+p_{02}p_{31}+p_{03}p_{12} . \,\!$
And since α is also an injection, the lines in P3 are thus in bijective correspondence with the points of this quadric in P5, called the Plücker quadric or Klein quadric.
## Uses
Plücker coordinates allow concise solutions to problems of line geometry in 3-dimensional space, especially those involving incidence.
### Line-line crossing
Two lines in P3 are either skew or coplanar, and in the latter case they are either coincident or intersect in a unique point. If pij and pij are the Plücker coordinates of two lines, then they are coplanar precisely when dm′+md′ = 0, as shown by
$0 \,\!$ ${} = p_{01}p'_{23} + p_{02}p'_{31} + p_{03}p'_{12} + p_{23}p'_{01} + p_{31}p'_{02} + p_{12}p'_{03} \,\!$ ${} = \begin{vmatrix}x_0&y_0&x'_0&y'_0\\x_1&y_1&x'_1&y'_1\\x_2&y_2&x'_2&y'_2\\x_3&y_3&x'_3&y'_3\end{vmatrix} .$
When the lines are skew, the sign of the result indicates the sense of crossing: positive if a right-handed screw takes L into L′, else negative.
The quadratic Plücker relation essentially states that a line is coplanar with itself.
### Line-line join
In the event that two lines are coplanar but not parallel, their common plane has equation
0 = (md′)x0 + (d×d′)•x ,
where x = (x1,x2,x3).
The slightest perturbation will destroy the existence of a common plane, and near-parallelism of the lines will cause numeric difficulties in finding such a plane even if it does exist.
### Line-line meet
Dually, two coplanar lines, neither of which contains the origin, have common point
(x0 : x) = (dm′:m×m′) .
To handle lines not meeting this restriction, see the references.
### Plane-line meet
Given a plane with equation
$0 = a^0x_0 + a^1x_1 + a^2x_2 + a^3x_3 , \,\!$
or more concisely 0 = a0x0+ax; and given a line not in it with Plücker coordinates (d:m), then their point of intersection is
(x0 : x) = (ad : a×ma0d) .
The point coordinates, (x0:x1:x2:x3), can also be expressed in terms of Plücker coordinates as
$x_i = \sum_{j \ne i} a^j p_{ij} , \qquad i = 0 \ldots 3 . \,\!$
### Point-line join
Dually, given a point (y0:y) and a line not containing it, their common plane has equation
0 = (ym) x0 + (y×dy0m)•x .
The plane coordinates, (a0:a1:a2:a3), can also be expressed in terms of dual Plücker coordinates as
$a^i = \sum_{j \ne i} y_j p^{ij} , \qquad i = 0 \ldots 3 . \,\!$
### Line families
Because the Klein quadric is in P5, it contains linear subspaces of dimensions one and two (but no higher). These correspond to one- and two-parameter families of lines in P3.
For example, suppose L and L′ are distinct lines in P3 determined by points x, y and x′, y′, respectively. Linear combinations of their determining points give linear combinations of their Plücker coordinates, generating a one-parameter family of lines containing L and L′. This corresponds to a one-dimensional linear subspace belonging to the Klein quadric.
#### Lines in plane
If three distinct and non-parallel lines are coplanar; their linear combinations generate a two-parameter family of lines, all the lines in the plane. This corresponds to a two-dimensional linear subspace belonging to the Klein quadric.
#### Lines through point
If three distinct and non-coplanar lines intersect in a point, their linear combinations generate a two-parameter family of lines, all the lines through the point. This also corresponds to a two-dimensional linear subspace belonging to the Klein quadric.
#### Ruled surface
A ruled surface is a family of lines that is not necessarily linear. It corresponds to a curve on the Klein quadric. For example, a hyperboloid of one sheet is a quadric surface in P3 ruled by two different families of lines, one line of each passing through each point of the surface; each family corresponds under the Plücker map to a conic section within the Klein quadric in P5.
### Line geometry
During the nineteenth century, line geometry was studied intensively. In terms of the bijection given above, this is a description of the intrinsic geometry of the Klein quadric.
#### Ray tracing
Line geometry is extensively used in ray tracing application where the geometry and intersections of rays need to be calculated in 3D. An implementation is described in Introduction to Pluecker Coordinates written for the Ray Tracing forum by Thouis Jones. | 5,158 | 16,039 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 41, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2014-52 | latest | en | 0.916175 |
http://blog.ihuxu.com/longest-substring-without-repeating-characters/ | 1,550,568,748,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489729.11/warc/CC-MAIN-20190219081639-20190219103639-00567.warc.gz | 32,678,789 | 17,299 | # 最长不重复子字符串问题（动态规划）
## 问题
Given a string, find the length of the longest substring without repeating characters.
Examples:
Given `"abcabcbb"`, the answer is `"abc"`, which the length is 3.
Given `"bbbbb"`, the answer is `"b"`, with the length of 1.
Given `"pwwkew"`, the answer is `"wke"`, with the length of 3. Note that the answer must be a substring`"pwke"` is a subsequenceand not a substring.
O(n)
## 代码
```/**
* https://leetcode.com/problems/longest-substring-without-repeating-characters/#/description
*
* dp
*/
#include <iostream>
#include <vector>
#include
<map>
#include <string>
using namespace std;
#define MAX 1000000001
class Solution {
public:
int lengthOfLongestSubstring(string s) {
map<char, int> dp;
int max_length = -1, cur_length = 0, cur_offset = 0, length = s.size(), last_offset = 0;
for (int i = 1; i < length + 1; ++i) {
map<char, int>::iterator it = dp.find(s[i - 1]);
if (it != dp.end() && it->second >= cur_offset) {
last_offset = it->second;
cur_offset = last_offset + 1;
cur_length = i - cur_offset + 1;
} else {
if (++cur_length > max_length) max_length = cur_length;
dp.insert(pair<char, int>(s[i - 1], i));
}
it->second = i;// re-write the current pos for the cur char
}
return max_length > 0 ? max_length : 0;
}
};
int main(int argc, char * argv[]) {
Solution * s = new Solution();
cout<<s->lengthOfLongestSubstring("qwnfenpglqdq");
}
```
## 1 评论
1. 匿名网友20170721
谢谢，用到了~~~~~~~~~~~~~~~~~~~~ | 460 | 1,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-09 | longest | en | 0.39117 |
https://www.numbersaplenty.com/56779 | 1,656,970,867,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104496688.78/warc/CC-MAIN-20220704202455-20220704232455-00388.warc.gz | 968,852,399 | 3,214 | Cookie Consent by FreePrivacyPolicy.com
Search a number
56779 is a prime number
BaseRepresentation
bin1101110111001011
32212212221
431313023
53304104
61114511
7324352
oct156713
985787
1056779
1139728
1228a37
131cac8
1416999
1511c54
hexddcb
56779 has 2 divisors, whose sum is σ = 56780. Its totient is φ = 56778.
The previous prime is 56773. The next prime is 56783. The reversal of 56779 is 97765.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 56779 - 25 = 56747 is a prime.
56779 is a lucky number.
It is a plaindrome in base 10 and base 14.
It is a nialpdrome in base 16.
It is not a weakly prime, because it can be changed into another prime (56773) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (11) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 28389 + 28390.
It is an arithmetic number, because the mean of its divisors is an integer number (28390).
256779 is an apocalyptic number.
56779 is a deficient number, since it is larger than the sum of its proper divisors (1).
56779 is an equidigital number, since it uses as much as digits as its factorization.
56779 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 13230, while the sum is 34.
The square root of 56779 is about 238.2834446620. The cubic root of 56779 is about 38.4352089542.
It can be divided in two parts, 567 and 79, that added together give a palindrome (646).
The spelling of 56779 in words is "fifty-six thousand, seven hundred seventy-nine". | 474 | 1,634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-27 | latest | en | 0.903294 |
https://www.tutorialspoint.com/logistic_regression_in_python/logistic_regression_in_python_quick_guide.htm | 1,696,238,466,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510983.45/warc/CC-MAIN-20231002064957-20231002094957-00119.warc.gz | 1,141,514,742 | 22,639 | # Logistic Regression in Python - Introduction
Logistic Regression is a statistical method of classification of objects. This chapter will give an introduction to logistic regression with the help of some examples.
## Classification
To understand logistic regression, you should know what classification means. Let us consider the following examples to understand this better −
• A doctor classifies the tumor as malignant or benign.
• A bank transaction may be fraudulent or genuine.
For many years, humans have been performing such tasks - albeit they are error-prone. The question is can we train machines to do these tasks for us with a better accuracy?
One such example of machine doing the classification is the email Client on your machine that classifies every incoming mail as “spam” or “not spam” and it does it with a fairly large accuracy. The statistical technique of logistic regression has been successfully applied in email client. In this case, we have trained our machine to solve a classification problem.
Logistic Regression is just one part of machine learning used for solving this kind of binary classification problem. There are several other machine learning techniques that are already developed and are in practice for solving other kinds of problems.
If you have noted, in all the above examples, the outcome of the predication has only two values - Yes or No. We call these as classes - so as to say we say that our classifier classifies the objects in two classes. In technical terms, we can say that the outcome or target variable is dichotomous in nature.
There are other classification problems in which the output may be classified into more than two classes. For example, given a basket full of fruits, you are asked to separate fruits of different kinds. Now, the basket may contain Oranges, Apples, Mangoes, and so on. So when you separate out the fruits, you separate them out in more than two classes. This is a multivariate classification problem.
# Logistic Regression in Python - Case Study
Consider that a bank approaches you to develop a machine learning application that will help them in identifying the potential clients who would open a Term Deposit (also called Fixed Deposit by some banks) with them. The bank regularly conducts a survey by means of telephonic calls or web forms to collect information about the potential clients. The survey is general in nature and is conducted over a very large audience out of which many may not be interested in dealing with this bank itself. Out of the rest, only a few may be interested in opening a Term Deposit. Others may be interested in other facilities offered by the bank. So the survey is not necessarily conducted for identifying the customers opening TDs. Your task is to identify all those customers with high probability of opening TD from the humongous survey data that the bank is going to share with you.
Fortunately, one such kind of data is publicly available for those aspiring to develop machine learning models. This data was prepared by some students at UC Irvine with external funding. The database is available as a part of UCI Machine Learning Repository and is widely used by students, educators, and researchers all over the world. The data can be downloaded from here.
In the next chapters, let us now perform the application development using the same data.
# Setting Up a Project
In this chapter, we will understand the process involved in setting up a project to perform logistic regression in Python, in detail.
## Installing Jupyter
We will be using Jupyter - one of the most widely used platforms for machine learning. If you do not have Jupyter installed on your machine, download it from here. For installation, you can follow the instructions on their site to install the platform. As the site suggests, you may prefer to use Anaconda Distribution which comes along with Python and many commonly used Python packages for scientific computing and data science. This will alleviate the need for installing these packages individually.
After the successful installation of Jupyter, start a new project, your screen at this stage would look like the following ready to accept your code.
Now, change the name of the project from Untitled1 to “Logistic Regression” by clicking the title name and editing it.
First, we will be importing several Python packages that we will need in our code.
## Importing Python Packages
For this purpose, type or cut-and-paste the following code in the code editor −
```In [1]: # import statements
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn import preprocessing
from sklearn.linear_model import LogisticRegression
from sklearn.model_selection import train_test_split
```
Your Notebook should look like the following at this stage −
Run the code by clicking on the Run button. If no errors are generated, you have successfully installed Jupyter and are now ready for the rest of the development.
The first three import statements import pandas, numpy and matplotlib.pyplot packages in our project. The next three statements import the specified modules from sklearn.
Our next task is to download the data required for our project. We will learn this in the next chapter.
# Logistic Regression in Python - Getting Data
The steps involved in getting data for performing logistic regression in Python are discussed in detail in this chapter.
If you have not already downloaded the UCI dataset mentioned earlier, download it now from here. Click on the Data Folder. You will see the following screen −
Download the bank.zip file by clicking on the given link. The zip file contains the following files −
We will use the bank.csv file for our model development. The bank-names.txt file contains the description of the database that you are going to need later. The bank-full.csv contains a much larger dataset that you may use for more advanced developments.
Here we have included the bank.csv file in the downloadable source zip. This file contains the comma-delimited fields. We have also made a few modifications in the file. It is recommended that you use the file included in the project source zip for your learning.
To load the data from the csv file that you copied just now, type the following statement and run the code.
```In [2]: df = pd.read_csv('bank.csv', header=0)
```
You will also be able to examine the loaded data by running the following code statement −
```IN [3]: df.head()
```
Once the command is run, you will see the following output −
Basically, it has printed the first five rows of the loaded data. Examine the 21 columns present. We will be using only few columns from these for our model development.
Next, we need to clean the data. The data may contain some rows with NaN. To eliminate such rows, use the following command −
```IN [4]: df = df.dropna()
```
Fortunately, the bank.csv does not contain any rows with NaN, so this step is not truly required in our case. However, in general it is difficult to discover such rows in a huge database. So it is always safer to run the above statement to clean the data.
Note − You can easily examine the data size at any point of time by using the following statement −
```IN [5]: print (df.shape)
(41188, 21)
```
The number of rows and columns would be printed in the output as shown in the second line above.
Next thing to do is to examine the suitability of each column for the model that we are trying to build.
# Logistic Regression in Python - Restructuring Data
Whenever any organization conducts a survey, they try to collect as much information as possible from the customer, with the idea that this information would be useful to the organization one way or the other, at a later point of time. To solve the current problem, we have to pick up the information that is directly relevant to our problem.
## Displaying All Fields
Now, let us see how to select the data fields useful to us. Run the following statement in the code editor.
```In [6]: print(list(df.columns))
```
You will see the following output −
```['age', 'job', 'marital', 'education', 'default', 'housing', 'loan',
'contact', 'month', 'day_of_week', 'duration', 'campaign', 'pdays',
'previous', 'poutcome', 'emp_var_rate', 'cons_price_idx', 'cons_conf_idx',
'euribor3m', 'nr_employed', 'y']
```
The output shows the names of all the columns in the database. The last column “y” is a Boolean value indicating whether this customer has a term deposit with the bank. The values of this field are either “y” or “n”. You can read the description and purpose of each column in the banks-name.txt file that was downloaded as part of the data.
## Eliminating Unwanted Fields
Examining the column names, you will know that some of the fields have no significance to the problem at hand. For example, fields such as month, day_of_week, campaign, etc. are of no use to us. We will eliminate these fields from our database. To drop a column, we use the drop command as shown below −
```In [8]: #drop columns which are not needed.
df.drop(df.columns[[0, 3, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19]],
axis = 1, inplace = True)
```
The command says that drop column number 0, 3, 7, 8, and so on. To ensure that the index is properly selected, use the following statement −
```In [7]: df.columns[9]
Out[7]: 'day_of_week'
```
This prints the column name for the given index.
After dropping the columns which are not required, examine the data with the head statement. The screen output is shown here −
```In [9]: df.head()
Out[9]:
job marital default housing loan poutcome y
0 blue-collar married unknown yes no nonexistent 0
1 technician married no no no nonexistent 0
2 management single no yes no success 1
3 services married no no no nonexistent 0
4 retired married no yes no success 1
```
Now, we have only the fields which we feel are important for our data analysis and prediction. The importance of Data Scientist comes into picture at this step. The data scientist has to select the appropriate columns for model building.
For example, the type of job though at the first glance may not convince everybody for inclusion in the database, it will be a very useful field. Not all types of customers will open the TD. The lower income people may not open the TDs, while the higher income people will usually park their excess money in TDs. So the type of job becomes significantly relevant in this scenario. Likewise, carefully select the columns which you feel will be relevant for your analysis.
In the next chapter, we will prepare our data for building the model.
# Logistic Regression in Python - Preparing Data
For creating the classifier, we must prepare the data in a format that is asked by the classifier building module. We prepare the data by doing One Hot Encoding.
## Encoding Data
We will discuss shortly what we mean by encoding data. First, let us run the code. Run the following command in the code window.
```In [10]: # creating one hot encoding of the categorical columns.
data = pd.get_dummies(df, columns =['job', 'marital', 'default', 'housing', 'loan', 'poutcome'])
```
As the comment says, the above statement will create the one hot encoding of the data. Let us see what has it created? Examine the created data called “data” by printing the head records in the database.
```In [11]: data.head()
```
You will see the following output −
To understand the above data, we will list out the column names by running the data.columns command as shown below −
```In [12]: data.columns
'job_housemaid', 'job_management', 'job_retired', 'job_self-employed',
'job_services', 'job_student', 'job_technician', 'job_unemployed',
'job_unknown', 'marital_divorced', 'marital_married', 'marital_single',
'marital_unknown', 'default_no', 'default_unknown', 'default_yes',
'housing_no', 'housing_unknown', 'housing_yes', 'loan_no',
'loan_unknown', 'loan_yes', 'poutcome_failure', 'poutcome_nonexistent',
'poutcome_success'], dtype='object')
```
Now, we will explain how the one hot encoding is done by the get_dummies command. The first column in the newly generated database is “y” field which indicates whether this client has subscribed to a TD or not. Now, let us look at the columns which are encoded. The first encoded column is “job”. In the database, you will find that the “job” column has many possible values such as “admin”, “blue-collar”, “entrepreneur”, and so on. For each possible value, we have a new column created in the database, with the column name appended as a prefix.
Thus, we have columns called “job_admin”, “job_blue-collar”, and so on. For each encoded field in our original database, you will find a list of columns added in the created database with all possible values that the column takes in the original database. Carefully examine the list of columns to understand how the data is mapped to a new database.
## Understanding Data Mapping
To understand the generated data, let us print out the entire data using the data command. The partial output after running the command is shown below.
```In [13]: data
```
The above screen shows the first twelve rows. If you scroll down further, you would see that the mapping is done for all the rows.
A partial screen output further down the database is shown here for your quick reference.
To understand the mapped data, let us examine the first row.
It says that this customer has not subscribed to TD as indicated by the value in the “y” field. It also indicates that this customer is a “blue-collar” customer. Scrolling down horizontally, it will tell you that he has a “housing” and has taken no “loan”.
After this one hot encoding, we need some more data processing before we can start building our model.
## Dropping the “unknown”
If we examine the columns in the mapped database, you will find the presence of few columns ending with “unknown”. For example, examine the column at index 12 with the following command shown in the screenshot −
```In [14]: data.columns[12]
Out[14]: 'job_unknown'
```
This indicates the job for the specified customer is unknown. Obviously, there is no point in including such columns in our analysis and model building. Thus, all columns with the “unknown” value should be dropped. This is done with the following command −
```In [15]: data.drop(data.columns[[12, 16, 18, 21, 24]], axis=1, inplace=True)
```
Ensure that you specify the correct column numbers. In case of a doubt, you can examine the column name anytime by specifying its index in the columns command as described earlier.
After dropping the undesired columns, you can examine the final list of columns as shown in the output below −
```In [16]: data.columns
'job_housemaid', 'job_management', 'job_retired', 'job_self-employed',
'job_services', 'job_student', 'job_technician', 'job_unemployed',
'marital_divorced', 'marital_married', 'marital_single', 'default_no',
'default_yes', 'housing_no', 'housing_yes', 'loan_no', 'loan_yes',
'poutcome_failure', 'poutcome_nonexistent', 'poutcome_success'],
dtype='object')
```
At this point, our data is ready for model building.
# Logistic Regression in Python - Splitting Data
We have about forty-one thousand and odd records. If we use the entire data for model building, we will not be left with any data for testing. So generally, we split the entire data set into two parts, say 70/30 percentage. We use 70% of the data for model building and the rest for testing the accuracy in prediction of our created model. You may use a different splitting ratio as per your requirement.
## Creating Features Array
Before we split the data, we separate out the data into two arrays X and Y. The X array contains all the features (data columns) that we want to analyze and Y array is a single dimensional array of boolean values that is the output of the prediction. To understand this, let us run some code.
Firstly, execute the following Python statement to create the X array −
```In [17]: X = data.iloc[:,1:]
```
To examine the contents of X use head to print a few initial records. The following screen shows the contents of the X array.
```In [18]: X.head ()
```
The array has several rows and 23 columns.
Next, we will create output array containing “y” values.
## Creating Output Array
To create an array for the predicted value column, use the following Python statement −
```In [19]: Y = data.iloc[:,0]
```
Examine its contents by calling head. The screen output below shows the result −
```In [20]: Y.head()
Out[20]: 0 0
1 0
2 1
3 0
4 1
Name: y, dtype: int64
```
Now, split the data using the following command −
```In [21]: X_train, X_test, Y_train, Y_test = train_test_split(X, Y, random_state=0)
```
This will create the four arrays called X_train, Y_train, X_test, and Y_test. As before, you may examine the contents of these arrays by using the head command. We will use X_train and Y_train arrays for training our model and X_test and Y_test arrays for testing and validating.
Now, we are ready to build our classifier. We will look into it in the next chapter.
# Logistic Regression in Python - Building Classifier
It is not required that you have to build the classifier from scratch. Building classifiers is complex and requires knowledge of several areas such as Statistics, probability theories, optimization techniques, and so on. There are several pre-built libraries available in the market which have a fully-tested and very efficient implementation of these classifiers. We will use one such pre-built model from the sklearn.
## The sklearn Classifier
Creating the Logistic Regression classifier from sklearn toolkit is trivial and is done in a single program statement as shown here −
```In [22]: classifier = LogisticRegression(solver='lbfgs',random_state=0)
```
Once the classifier is created, you will feed your training data into the classifier so that it can tune its internal parameters and be ready for the predictions on your future data. To tune the classifier, we run the following statement −
```In [23]: classifier.fit(X_train, Y_train)
```
The classifier is now ready for testing. The following code is the output of execution of the above two statements −
```Out[23]: LogisticRegression(C = 1.0, class_weight = None, dual = False,
fit_intercept=True, intercept_scaling=1, max_iter=100,
multi_class='warn', n_jobs=None, penalty='l2', random_state=0,
solver='lbfgs', tol=0.0001, verbose=0, warm_start=False))
```
Now, we are ready to test the created classifier. We will deal this in the next chapter.
# Logistic Regression in Python - Testing
We need to test the above created classifier before we put it into production use. If the testing reveals that the model does not meet the desired accuracy, we will have to go back in the above process, select another set of features (data fields), build the model again, and test it. This will be an iterative step until the classifier meets your requirement of desired accuracy. So let us test our classifier.
## Predicting Test Data
To test the classifier, we use the test data generated in the earlier stage. We call the predict method on the created object and pass the X array of the test data as shown in the following command −
```In [24]: predicted_y = classifier.predict(X_test)
```
This generates a single dimensional array for the entire training data set giving the prediction for each row in the X array. You can examine this array by using the following command −
```In [25]: predicted_y
```
The following is the output upon the execution the above two commands −
```Out[25]: array([0, 0, 0, ..., 0, 0, 0])
```
The output indicates that the first and last three customers are not the potential candidates for the Term Deposit. You can examine the entire array to sort out the potential customers. To do so, use the following Python code snippet −
```In [26]: for x in range(len(predicted_y)):
if (predicted_y[x] == 1):
print(x, end="\t")
```
The output of running the above code is shown below −
The output shows the indexes of all rows who are probable candidates for subscribing to TD. You can now give this output to the bank’s marketing team who would pick up the contact details for each customer in the selected row and proceed with their job.
Before we put this model into production, we need to verify the accuracy of prediction.
## Verifying Accuracy
To test the accuracy of the model, use the score method on the classifier as shown below −
```In [27]: print('Accuracy: {:.2f}'.format(classifier.score(X_test, Y_test)))
```
The screen output of running this command is shown below −
```Accuracy: 0.90
```
It shows that the accuracy of our model is 90% which is considered very good in most of the applications. Thus, no further tuning is required. Now, our customer is ready to run the next campaign, get the list of potential customers and chase them for opening the TD with a probable high rate of success.
# Logistic Regression in Python - Limitations
As you have seen from the above example, applying logistic regression for machine learning is not a difficult task. However, it comes with its own limitations. The logistic regression will not be able to handle a large number of categorical features. In the example we have discussed so far, we reduced the number of features to a very large extent.
However, if these features were important in our prediction, we would have been forced to include them, but then the logistic regression would fail to give us a good accuracy. Logistic regression is also vulnerable to overfitting. It cannot be applied to a non-linear problem. It will perform poorly with independent variables which are not correlated to the target and are correlated to each other. Thus, you will have to carefully evaluate the suitability of logistic regression to the problem that you are trying to solve.
There are many areas of machine learning where other techniques are specified devised. To name a few, we have algorithms such as k-nearest neighbours (kNN), Linear Regression, Support Vector Machines (SVM), Decision Trees, Naive Bayes, and so on. Before finalizing on a particular model, you will have to evaluate the applicability of these various techniques to the problem that we are trying to solve.
# Logistic Regression in Python - Summary
Logistic Regression is a statistical technique of binary classification. In this tutorial, you learned how to train the machine to use logistic regression. Creating machine learning models, the most important requirement is the availability of the data. Without adequate and relevant data, you cannot simply make the machine to learn.
Once you have data, your next major task is cleansing the data, eliminating the unwanted rows, fields, and select the appropriate fields for your model development. After this is done, you need to map the data into a format required by the classifier for its training. Thus, the data preparation is a major task in any machine learning application. Once you are ready with the data, you can select a particular type of classifier.
In this tutorial, you learned how to use a logistic regression classifier provided in the sklearn library. To train the classifier, we use about 70% of the data for training the model. We use the rest of the data for testing. We test the accuracy of the model. If this is not within acceptable limits, we go back to selecting the new set of features.
Once again, follow the entire process of preparing data, train the model, and test it, until you are satisfied with its accuracy. Before taking up any machine learning project, you must learn and have exposure to a wide variety of techniques which have been developed so far and which have been applied successfully in the industry. | 5,207 | 23,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-40 | latest | en | 0.935974 |
https://convertoctopus.com/476-months-to-years | 1,643,461,205,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00468.warc.gz | 235,773,233 | 8,214 | ## Conversion formula
The conversion factor from months to years is 0.083388698630137, which means that 1 month is equal to 0.083388698630137 years:
1 mo = 0.083388698630137 yr
To convert 476 months into years we have to multiply 476 by the conversion factor in order to get the time amount from months to years. We can also form a simple proportion to calculate the result:
1 mo → 0.083388698630137 yr
476 mo → T(yr)
Solve the above proportion to obtain the time T in years:
T(yr) = 476 mo × 0.083388698630137 yr
T(yr) = 39.693020547945 yr
The final result is:
476 mo → 39.693020547945 yr
We conclude that 476 months is equivalent to 39.693020547945 years:
476 months = 39.693020547945 years
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 year is equal to 0.025193345988676 × 476 months.
Another way is saying that 476 months is equal to 1 ÷ 0.025193345988676 years.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that four hundred seventy-six months is approximately thirty-nine point six nine three years:
476 mo ≅ 39.693 yr
An alternative is also that one year is approximately zero point zero two five times four hundred seventy-six months.
## Conversion table
### months to years chart
For quick reference purposes, below is the conversion table you can use to convert from months to years
months (mo) years (yr)
477 months 39.776 years
478 months 39.86 years
479 months 39.943 years
480 months 40.027 years
481 months 40.11 years
482 months 40.193 years
483 months 40.277 years
484 months 40.36 years
485 months 40.444 years
486 months 40.527 years | 460 | 1,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-05 | latest | en | 0.832661 |
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