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# 2E2 Tutorial Sheet 9 Second Term, Solutions 4 January 2004
```2E2 Tutorial Sheet 9 Second Term, Solutions1
4 January 2004
1. (3) Find the eigenvectors and eigenvalues of the following matrices
3 4
0 3
1 2
(i)
(ii)
(iii)
4 −3
−3 0
0 3
(1)
Solution:
In (i) the characteristic equation is
3−λ
4
4
−3 − λ
so
=0
(2)
(3 − λ)(−3 − λ) − 16 = 0
(3)
λ2 + 2λ − 48 − 25 = 0
(4)
or
Solve this gives us λ = ±5. Taking the λ = 5 first
3 4
a
a
=5
4 −3
b
b
(5)
so the first equation is 3a + 4b = 5a or a = 2b, the other equation is 4a − 3b = 5b
which is also a = 2b. Taking a = 2 an eigenvalue 5 eigenvector is,
2
x=
(6)
1
Taking λ = −5 next
3 4
4 −3
a
b
= −5
a
b
(7)
so the first equation is 3a + 4b = −5a or 2a = −b, the other equation is 4a − 3b = −5b
which is also 2a = −b. Taking a = 1 an eigenvalue −5 eigenvector is,
1
x=
(8)
−2
1
In (ii) the characteristic equation is
−λ 3
−3 −λ
=0
(9)
1
so
λ2 + 9 = 0
(10)
λ = ±3i
(11)
or
Taking the λ = 3i first
0 3
−3 0
a
b
= 3i
a
b
(12)
so the equation is 3b = 3ia or a = −ib. Taking b = 1 an eigenvalue 3i eigenvector is,
−i
x=
(13)
1
Taking the λ = −3i
0 3
−3 0
a
b
= −3i
a
b
(14)
so the equation is 3b = −3ia or a = ib. Taking a = 1 an eigenvalue −3i eigenvector
is,
i
x=
(15)
1
In (iii) the characteristic equation is
1−λ
2
0
3−λ
so
=0
(16)
(1 − λ)(3 − λ) = 0
(17)
So this gives us λ = 1 or λ = 3. Taking the λ = 1 first
1 2
a
a
0 3
b
b
(18)
so the first equation is a + 2b = a or 0 = b, the other equation is 3b = b which is also
b = 0. Taking a = 1 an eigenvalue 1 eigenvector is,
1
(19)
x=
0
Taking λ = 3 next
1 2
0 3
a
b
=3
a
b
(20)
so the first equation is a + 2b = 3a or a = b, the other equation is 3b = 3b which tells
us nothing. Taking a = 1 an eigenvalue 3 eigenvector is,
1
x=
(21)
1
2
2. (3) Find the solution for the system
dy1
= −3y1 + 2y2
dt
dy2
= −2y1 + 2y2
dt
This equation is y0 = Ay with
A=
−3 2
−2 2
We can find the eigenvalues, the characteristic equation is
−3 − λ
2
= (λ + 3)(λ − 2) + 4 = λ2 + λ − 2 = 0
−2
2−λ so that λ1 = 1 and λ2 = −2.
Next, we need the eigenvectors. First, λ1 :
a
a
−3 2
=
b
b
−2 2
so −3a + 2b = a or b = 2a, hence, choosing a = 1 we get
1
.
x1 =
2
For λ2 :
−3 2
−2 2
a
b
= −2
a
b
so −3a + 2b = −2a giving a = 2b, choosing b = 1 gives
2
x2 =
1
(22)
(23)
(24)
(25)
Now, in general the solution is
y = c 1 x1 e λ1 t + c 2 x2 e λ2 t
(26)
so, here,
y = c1
1
2
3
t
e + c2
2
1
e−2t
(27)
3. (2) Find the general solutions for the system
dy1
= 3y1 + y2
dt
dy2
= y1 + 3y2
dt
(28)
(29)
Solution: The eigenvectors and eigenvalues of
3 1
A=
1 3
are λ1 = 4 with
x1 =
and λ2 = 2 with
x2 =
(30)
1
1
−1
1
(31)
(32)
so the general soln is
y=
y1
y2
= c1
4
1
1
4t
e + c2
−1
1
e2t .
(33)
``` |
# How Do You Use The Factor Theorem
## Factor Theorem
Theorem: If p(x) is a polynomial of degree n ≥ 1 and a is any real number, then
(i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x).
Proof: By the Remainder Theorem,
p(x) = (x – a) q(x) + p(a).
(i) If p(a) = 0, then p(x) = (x – a) q(x),
which shows that x – a is a factor of p(x).
(ii) Since x – a is a factor of p(x),
p(x) = (x – a) g(x) for same polynomial g(x).
In this case, p(a) = (a – a) g(a) = 0.
To use factor theorem
• Step 1: (x + a) is factor of a polynomial p(x) if p(–a) = 0.
• Step 2: (ax – b) is a factor of a polynomial p(x) if p(b/a) = 0
• Step 3: ax + b is a factor of a polynomial p(x) if p(–b/a) = 0.
• Step 4: (x – a) (x – b) is a factor of a polynomial p(x) if p(a) = 0 and p(b) = 0.
## Factor Theorem Example Problems With Solutions
Example 1: Examine whether x + 2 is a factor of x3 + 3x2 + 5x + 6 and of 2x + 4.
Solution: The zero of x + 2 is –2.
Let p(x) = x3 + 3x2 + 5x + 6 and s(x) = 2x + 4
Then, p(–2) = (–2)3 + 3(–2)2 + 5(–2) + 6
= –8 + 12 – 10 + 6
= 0
So, by the Factor Theorem, x + 2 is a factor of x3 + 3x2 + 5x + 6.
Again, s(–2) = 2(–2) + 4 = 0
So, x + 2 is a factor of 2x + 4.
Example 2: Use the factor theorem to determine whether x – 1 is a factor of
(a) x3 + 8x2 – 7x – 2
(b) 2x3 + 5x2 – 7
(c) 8x4 + 12x3 – 18x + 14
Solution:
Example 3: Factorize each of the following expression, given that x3 + 13 x2 + 32 x + 20. (x+2) is a factor.
Solution:
Example 4: Factorize x3 – 23 x2 + 142 x – 120
Solution:
Example 5: Show that (x – 3) is a factor of the polynomial x3 – 3x2 + 4x – 12
Solution:
Example 6: Show that (x – 1) is a factor of x10 – 1 and also of x11 – 1.
Solution:
Example 7: Show that x + 1 and 2x – 3 are factors of 2x3 – 9x2 + x + 12.
Solution:
Example 8: Find the value of k, if x + 3 is a factor of 3x2 + kx + 6.
Solution:
Example 9: If ax3 + bx2 + x – 6 has x + 2 as a factor and leaves a remainder 4 when divided by (x – 2), find the values of a and b.
Solution:
Example 10: If both x – 2 and x – 1/2 are factors of px2 + 5x + r, show that p = r.
Solution:
Example 11: If x2 – 1 is a factor of ax4 + bx3 + cx2 + dx + e, show that a + c + e = b + d = 0.
Solution:
Example 12: Using factor theorem, show that a – b, b – c and c – a are the factors of a(b2 – c2) + b(c2 – a2) + c(a2 – b2).
Solution:
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# How do you differentiate y=(x^3-x^2-3)/(x^5+3) using the quotient rule?
Mar 30, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{4} + 15 {x}^{2} - 4 x}{{x}^{2} + 5} ^ 2$
#### Explanation:
Quotient rule states if $y = y \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$
then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right) - \frac{\mathrm{dh}}{\mathrm{dx}} \times g \left(x\right)}{h \left(x\right)} ^ 2$
Here $g \left(x\right) = {x}^{3} - {x}^{2} - 3$ hence $\frac{\mathrm{dg}}{\mathrm{dx}} = 3 {x}^{2} - 2 x$
and as $h \left(x\right) = {x}^{2} + 5$, $\frac{\mathrm{dh}}{\mathrm{dx}} = 2 x$
Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(3 {x}^{2} - 2 x\right) \times \left({x}^{2} + 5\right) - 2 x \times \left({x}^{3} - {x}^{2} - 3\right)}{{x}^{2} + 5} ^ 2$
= $\frac{3 {x}^{4} - 2 {x}^{3} + 15 {x}^{2} - 10 x - 2 {x}^{4} + 2 {x}^{3} + 6 x}{{x}^{2} + 5} ^ 2$
= $\frac{{x}^{4} + 15 {x}^{2} - 4 x}{{x}^{2} + 5} ^ 2$ |
# Review of Dimensional Analysis, Scientific Notation, and Significant Figures
When you are performing mole conversion problems, it is important to remember how to perform dimensional analysis and the rules for significant figures. Often, you will need to express your answers in scientific notation. Let’s briefly review each of these skills.
### Dimensional Analysis
• Dimensional analysis, or the factor label method, is a useful problem-solving technique that can be used to convert between units.
• Dimensional analysis uses conversion factors, or equivalences, set up in a manner that allows “like” units to cancel one another.
• When solving problems using dimensional analysis
• set up a conversion factor with the original unit in the denominator and the new unit in the numerator.
Remember the STAAR reference material for Chemistry has a section titled Constants and Conversions. Refer to this as you work various problems.
Source: STAAR Reference Material, Texas Education Agency
Listed below are some other common unit conversions as well as common metric prefixes used in science.
Write the given information as a fraction by placing it over 1. (Placing it over 1 makes it a fraction but does not change its value.)
• Write a conversion factor that has the unit you want to remove in the denominator and the unit you want to end up with in the numerator. After you fill in your units, add the numbers. (Usually one of the numbers is a 1, but it can be in either the denominator or the numerator.) Note: In some cases you may need to repeat this step a number of times in order to get the unit you want to end up with in the numerator.
• Mark through the units to double check that they all cancel and that you are left with the units you want.
• Multiply the numbers in the numerators, and then multiply the numbers in the denominators.
### Scientific Notation
• Scientific notation expresses very large or small numbers in a simplified manner.
• In scientific notation, a number is written as the product of two numbers: a coefficient and 10 raised to a power.
• Example 1: 6500000000 can be written in a simpler way: 6.5 × 109
• Example 2: 0.000000042 can be written in a simpler way: 4.2 × 10-8
• Numbers with positive exponents are large numbers.
• Numbers with negative exponents are small numbers.
• To write numbers using scientific notation, move the decimal, and write the number of places you moved the decimal point as an exponent.
### Significant Figures
The STAAR reference material for Chemistry document lists the rules for significant figure in a section titled Rules For Significant Figures.
Source: STAAR Reference Material, Texas Education Agency
The rule for addition and subtraction with significant figures is as follows:
When measurements are added or subtracted, the final answer can contain no more decimal places than the least accurate measurement.
When adding and subtracting measurements, the level of accuracy at which you express your final answer does not depend on the number of significant figures in the original problem but instead is determined by the position or place value of the least significant digit in the original problem.
The rule for multiplication and division with significant figures is as follows:
When measurements are multiplied or divided, the answer can contain no more significant figures than the least accurate measurement.
This rule simply means the final answer can be no more accurate than the least accurate measurement. Count the significant figures in each measurement instead of the number of decimal places when multiplying and dividing measurements. |
# 6.3 Centripetal force
Page 1 / 10
• Calculate coefficient of friction on a car tire.
• Calculate ideal speed and angle of a car on a turn.
Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.
Any net force causing uniform circular motion is called a centripetal force . The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net $\text{F}=\text{ma}$ . For uniform circular motion, the acceleration is the centripetal acceleration— $a={a}_{c}$ . Thus, the magnitude of centripetal force ${\text{F}}_{\text{c}}$ is
${\text{F}}_{\text{c}}={m\text{a}}_{\text{c}}.$
By using the expressions for centripetal acceleration ${a}_{c}$ from ${a}_{c}=\frac{{v}^{2}}{r};\phantom{\rule{0.25em}{0ex}}{a}_{c}={\mathrm{r\omega }}^{2}$ , we get two expressions for the centripetal force ${\text{F}}_{\text{c}}$ in terms of mass, velocity, angular velocity, and radius of curvature:
${F}_{c}=m\frac{{v}^{2}}{r};\phantom{\rule{0.25em}{0ex}}{F}_{c}=\text{mr}{\omega }^{2}.$
You may use whichever expression for centripetal force is more convenient. Centripetal force ${F}_{\text{c}}$ is always perpendicular to the path and pointing to the center of curvature, because ${\mathbf{a}}_{c}$ is perpendicular to the velocity and pointing to the center of curvature.
Note that if you solve the first expression for $r$ , you get
$r=\frac{{\mathrm{mv}}^{2}}{{F}_{c}}\text{.}$
This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.
## What coefficient of friction do car tires need on a flat curve?
(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.
(b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see [link] ).
Strategy and Solution for (a)
We know that ${F}_{\text{c}}=\frac{{\mathrm{mv}}^{\text{2}}}{r}$ . Thus,
${F}_{\text{c}}=\frac{{\mathrm{mv}}^{\text{2}}}{r}=\frac{\left(\text{900 kg}\right)\left(\text{25.0 m/s}{\right)}^{\text{2}}}{\left(\text{500 m}\right)}=\text{1125 N.}$
Strategy for (b)
[link] shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is ${\mu }_{\text{s}}N$ , where ${\mu }_{\text{s}}$ is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground, so that $N=\mathit{mg}$ . Thus the centripetal force in this situation is
${F}_{\text{c}}=f={\mu }_{\text{s}}N={\mu }_{\text{s}}\text{mg}\text{.}$
Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for ${F}_{\text{c}}$ from the equation
$\begin{array}{c}{F}_{\text{c}}=m\frac{{v}^{2}}{r}\\ {F}_{\text{c}}=\text{mr}{\omega }^{2}\end{array}\right\},$
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Lesson 9More and Less than 1%
Let’s explore percentages smaller than 1%.
### Learning Targets:
• I can find percentages of quantities like 12.5% and 0.4%.
• I understand that to find 0.1% of an amount I have to multiply by 0.001.
## 9.1Number Talk: What Percentage?
Determine the percentage mentally.
10 is what percentage of 50?
5 is what percentage of 50?
1 is what percentage of 50?
17 is what percentage of 50?
## 9.2Waiting Tables
During one waiter’s shift, he delivered appetizers, entrées, and desserts. What percentage of the dishes were desserts? appetizers? entrées? What do your percentages add up to?
## 9.3Fractions of a Percent
30% of 60
3% of 60
0.3% of 60
0.03% of 60
2. 20% of 5,000 is 1,000 and 21% of 5,000 is 1,050. Find each percentage of 5,000 and be prepared to explain your reasoning. If you get stuck, consider using the double number line diagram.
1. 1% of 5,000
2. 0.1% of 5,000
3. 20.1% of 5,000
4. 20.4% of 5,000
3. 15% of 80 is 12 and 16% of 80 is 12.8. Find each percentage of 80 and be prepared to explain your reasoning.
1. 15.1% of 80
2. 15.7% of 80
### Are you ready for more?
To make Sierpinski's triangle,
• Connect the midpoints of every side, and remove the middle triangle, leaving three smaller triangles. This is step 2.
• Do the same to each of the remaining triangles. This is step 3.
• Keep repeating this process.
1. What percentage of the area of the original triangle is left after step 2? Step 3? Step 10?
2. At which step does the percentage first fall below 1%?
## 9.4Population Growth
1. The population of City A was approximately 243,000 people, and it increased by 8% in one year. What was the new population?
2. The population of city B was approximately 7,150,000, and it increased by 0.8% in one year. What was the new population?
## Lesson 9 Summary
A percentage, such as 30%, is a rate per 100. To find 30% of a quantity, we multiply it by , or 0.3.
The same method works for percentages that are not whole numbers, like 7.8% or 2.5%. To find 2.5% of a quantity, we multiply it by , or 0.025.
In the square, 2.5% of the area is shaded.
• For example, to calculate 2.5% interest on a bank balance of $80, we multiply , so the interest is$2.
We can sometimes find percentages like 2.5% mentally by using convenient whole number percents. For example, 25% of 80 is one fourth of 80, which is 20. Since 2.5 is one tenth of 25, we know that 2.5% of 80 is one tenth of 20, which is 2.
## Lesson 9 Practice Problems
1. The student government snack shop sold 32 items this week.
snack type number of items sold
fruit cup 8
veggie sticks 6
chips 14
water 4
For each snack type, what percentage of all snacks sold were of that type?
2. Select all the options that have the same value as of 20.
1. 3.5% of 20
2. 7% of 10
3. 22% of 65 is 14.3. What is 22.6% of 65? Explain your reasoning.
4. A bakery used 30% more sugar this month than last month. If the bakery used 560 kilograms of sugar last month, how much did it use this month?
5. Match each diagram to a situation. The diagrams can be used more than once.
1. The amount of apples this year decreased by 15% compared with last year's amount.
2. The amount of pears this year is 85% of last year's amount.
3. The amount of cherries this year increased by 15% compared with last year's amount.
4. The amount of oranges this year is 115% of last year's amount.
6. A certain type of car has room for 4 passengers.
1. Write an equation relating the number of cars () to the number of passengers ().
2. How many passengers could fit in 78 cars?
3. How many cars would be needed to fit 78 passengers? |
### Home > MC2 > Chapter 9 > Lesson 9.1.3 > Problem9-30
9-30.
Complete the table. .
$x$ $y$ $–6$ $3$ $6$ $0$ $1$ $4$ $2$ $–4$
Notice that there are two known coordinates: $\left(3, 4\right)$ and $\left(0, 2\right)$.
Now notice that for every difference of $3$ in the
$x$-coordinate, the $y$-coordinate changes by $2$.
Use this reasoning to fill in the rest of the table.
The difference in the $y$-coordinate from $x = 0$ to $x = 3$ is $2$. Since $6$ is twice $3$, the difference in the $y$-coordinate from $x = 0$ to $x = 6$ is twice as much.
$\left(−6, −2\right)$ and $\left(−9, −4\right)$
1. Find the rule.
To find the rule, look for a pattern that can relate the y-coordinate to the x-coordinate.
Notice that at $x = 0$, $y = 2$.
Also remember that the y-coordinate increases by $2$ for every $3$ units the $x$-coordinate increases.
${\it y} = \frac{2}{3}{\it x} + 2$
You can fill in the last y-coordinate in the table by substituting $1$ into the rule.
2. What is the slope?
$\frac{\text{Change in }{\it y}}{\text{Change in }{\it x}}$
The slope describes how the values increase.
Complete the table in the eTool below to graph the points.
Click the link at right for the full version of the eTool: MC2 9-30 HW eTool |
Question Video: Adding Two Numbers up to 999 | Nagwa Question Video: Adding Two Numbers up to 999 | Nagwa
# Question Video: Adding Two Numbers up to 999 Mathematics • Second Year of Primary School
Which number is 468 more than 421?
03:03
### Video Transcript
Which number is 468 more than 421?
Let’s sketch a bar model to help us understand what we need to do to find the answer here. This bar could represent the number 421. Here’s 468 more than 421. And this value here is the number that is 468 more than 421. It’s the total of both numbers. To find the answer, we need to add 421 and 468 together. We’re going to add the ones, the tens, and the hundreds digits in these numbers separately. And to help us do this, we can set out the numbers vertically.
421 is made up of four hundreds, two tens, and one one. And we’re finding 468 more than this, which is also made up of four hundreds, but this time, six tens and eight ones. Let’s start by adding our ones together. 421 has one one and 468 has eight ones. One plus eight equals nine ones altogether. And we can model this answer by putting our two lots of ones cubes together. One plus eight equals nine. Now, let’s think about our tens digits. 421 contains a two in the tens place. That’s where we get the number 20 from. And by finding 468 more than 421, we need to add another six tens. So what’s two tens plus six tens?
Well, we know two plus six equals eight. So two tens plus six tens equals eight tens or 80. Again, let’s combine our tens rods this time to show that they make eight tens altogether. Two tens plus six tens equals eight tens. Finally, it’s time to add our hundreds digits together. Both numbers have the digit four in the hundreds place. 400 plus 400 equals 800. And if we combine our hundreds blocks together, we can see that we have 800 altogether. 400 plus 400 equals 800.
We’ve used column addition. And we’ve also used base ten blocks to help us find the answer. When we add 421 and 468 together, we get a number with eight hundreds, eight tens, and nine ones. So we can say that the number that is 468 more than 421 is 889.
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Question Video: Evaluating Algebraic Expressions Involving Square Roots | Nagwa Question Video: Evaluating Algebraic Expressions Involving Square Roots | Nagwa
# Question Video: Evaluating Algebraic Expressions Involving Square Roots Mathematics • Second Year of Preparatory School
## Join Nagwa Classes
Given that π = β2 and π = β6, find the value of πΒ²/πΒ².
03:39
### Video Transcript
Given that π is equal to the square root of two and π is equal to the square root of six, find the value of π squared over π squared.
In this problem, we need to work out what the value of π squared over π squared is. And, we need to use the values that weβre given for π and π to help us find it. π squared over π squared is written just like a fraction. So, we would expect that perhaps the value weβre looking for is going to be a fraction too. Letβs see. Weβre told that the value of π is the square root of two, and the value of π is the square root of six. Because we know these values, we can substitute them for the letters π and π in the fraction.
And so, we can replace the π in π squared with the square root of two because π is the same as the square root of two. And then, we can replace the π in π squared with the square root of six because weβre told that π equals the square root of six. Unfortunately, instead of making the fraction looks simpler, we seem to have made it look a lot more complicated. The square root of two squared over the square root of six squared. What happens when you square a square root?
To help us understand this, letβs pick an easy number to work with. Letβs pick four times four. We know that four multiplied by four gives us a square number, which is 16. Now if we wrote the square root of 16, weβd expect the answer to be four because four times four equals 16. Letβs write the number four underneath here to remind ourselves that the square root of 16 is four. And if we go back to our square root and put brackets around it and then square it, this is the sort of thing that weβve got going on in our fraction, the square root of 16 squared. Weβve already said that the square root of 16 is four. And if we square four, in other words we work out four times four, we get the answer 16.
Itβs as if finding the square root of something and then squaring it cancel each other out. Itβs also as if we can just erase all of the symbols around the number and just keep the number 16 itself. And so if we go back to our fraction, we can use the same idea to work out the answer. The numerator or the top number says the square root of two squared. Weβve already said that finding the square root of something and then squaring it cancels each other out. So, weβre left with just the number two. And, the same applies to the denominator or the bottom number. Weβre just left with the number six. The value of π squared over π squared is equal to two-sixths.
Now, is there a way we could simplify this fraction? What if we divide the numerator and the denominator by two? Two divided by two equals one. And then, six divided by two gives us three. If π is equal to the square root of two and π is equal to the square root of six, then the value of π squared over π squared is two-sixths. Which we can simplify and write as one-third.
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1. ## Projection Offset Problem
Because that is what makes a perspective projection have perspective. The division is what makes things shrink as they move further from the camera.
2. ## Projection Offset Problem
Taking the example case of Far = 10, Near = 1, just dividing by Far-Near would put points at the far plane at 10/9, and points at the near plane at 1/9. Subtracting Near / (Far-Near) changes that so that points on the far plane become 1, and points on the near plane become 0. The scale by Far is to counteract the perspective divide
3. ## Projection Offset Problem
The intended result is to transform the coordinate such that the range [Near,Far] maps to [0,1], but after the perspective divide. Ignoring the divide to start with, we start by translating by -Near, so that Near maps to 0. Zout = Zin - Near Now, in the given case, Z values at the near plane become 0, Z values at the far plane become 9. We rescale by 1/(Far-Near) to bring that to the range [0,1] Zout = (Zin - Near) / (Far - Near) To make this easier to calculate with a matrix, we want it in the form A * z + D, so we distribute and rearrange things Zout = Zin * 1/(Far - Near) - Near / (Far - Near) If it is an orthographic projection, we're done. If it is a perspective projection, we must take into account the divide by Zin that will happen. Zclip = Zin * 1/(Far - Near) - Near / (Far - Near) Zout = Zclip / Zin For Zin = Near, Zclip is 0.0, and nothing would change, but for Zin = Far, we would get a result of: Zclip = Zfar * 1/(Far - Near) - Near / (Far - Near) = 10 / 9 - 1 / 9 = 9 / 9 = 1 Zout = Zclip / Zin = Zclip / Far = 1 / 10 To get a Zout of 1, we have to scale things by Far, which will give the correct result of Zin = Near -> 0.0, Zin = Far -> 1.0. Distributing it across: Zclip = Zin * Far / (Far - Near) - (Near * Far) / (Far - Near) Zout = Zclip / Zin
4. ## Painters Algorithm
The algorithm is literally render things in depth order, but it doesn't work those out, you have to provide them. Things get complicated when the objects start intersecting, and moreso when they are concave, but there are plenty of production particle systems that can boil their sorting down to a simple qsort() on Z. These days it is mostly applicable to translucent rendering, as opaque can rely on zbuffering to get correct results without regard to draw order.
5. ## Matrix palette skinning, blending matrices
Welcome to the wonderful world of linear transformations. For the usual weighted skinning approach, this is indeed a valid way to do it. The short version is that the matrices in this case are linear transforms, which have the helpful properties that, for any particular linear function F(), values u, v, and scalar c, the following hold true: F(c * u) = c * F(u) and F(u + v) = F(u) + F(v) Assuming matrices bone0, bone1, bone2, bone3, weight0..3, and shrinking down to only looking at the x value of the result: float result = 0.0; result += (bone0 * pos).x * weight0; result += (bone1 * pos).x * weight1; result += (bone2 * pos).x * weight2; result += (bone3 * pos).x * weight3; (bone0 * pos).x expands out to something like (bone0._11 * pos.x + bone0._21 * pos.y + bone0._31 * pos.z + bone0._41), and similar for the rest, (apologies for playing very fast and loose with column vs row major, it doesn't particularly matter for the linearity of things) result += (bone0._11 * pos.x + bone0._21 * pos.y + bone0._31 * pos.z + bone0._41) * weight0; result += (bone1._11 * pos.x + bone1._21 * pos.y + bone1._31 * pos.z + bone1._41) * weight1; result += (bone2._11 * pos.x + bone2._21 * pos.y + bone2._31 * pos.z + bone2._41) * weight2; result += (bone3._11 * pos.x + bone3._21 * pos.y + bone3._31 * pos.z + bone3._41) * weight3; if you distribute the weight# multiplies through, roll all the sums together, and then pull pos.x, pos.y, and pos.z out accordingly, you get something like: result = pos.x * (bone0._11 * weight0 + bone1._11 * weight1 + bone2._11 * weight2 + bone3._11 * weight3) + pos.y * (etc...) + pos.z * (etc...) and get exactly the second formulation
6. ## Line segment intersection seems to work but then other times is horribly wrong
It looks like your segment intersection test is actually an infinite line test. It will only return false if they are parallel or coincident...
7. ## What's the difference between dot and * in HLSL
That all looks fine, assuming diffuse and ambient are float4's, which they almost certainly should be if you want lights that aren't just white.
8. ## C++ Self-Evaluation Metrics
Assuming it's for a deep magic code ninja type position, ask why, and likely be satisfied with a coherent answer. If it's not a position that involves staring at hex dumps for bugs, it probably doesn't even come up... unless someone claims they have a better grasp of C++ than Stroustrup or Sutter. Or, on bad days, be very relieved, as it means I don't have to dig that bit of the standard out of cold storage.
9. ## C++ Self-Evaluation Metrics
Anything over an 8 means one of two things. They've either written a solid, production ready compiler frontend and runtime support library, or they're a 4. 7-8 from someone with a background that matches means "I've seen horrible things, and know how to avoid/diagnose them, but there are still fell and terrible things lurking in the dark corners of the earth". An approach we used from time to time, at least for people that claim to be Really Good and Technical with it, is to just have them start drawing out the memory layout of an instance of a class object, working up from the trivial case, through to the virtual diamond one, and see where the floundering starts. Bonus points for knowing how dynamic_cast and rtti work (and a slight bit of walking through the process usually serves as a good reminder of why they aren't exactly free).
10. ## SH directional lights, what am I missing?
It's a 3D scene, but with the view direction restricted to slightly off-axis, and camera motion restricted to a 2D plane. The main area of play is about 400 units in front of the camera, with some near-field objects about 200 units past that that can accept shadows. Tons and tons of background objects lie far beyond that, the far plane is set to around 100,000. It isn't particularly ideal. That soup gets thrown at a deferred lighting renderer, which is all fine and great up until it needs to light things that don't write depth.
11. ## SH directional lights, what am I missing?
We have a game here using a straightforward deferred lighting approach, but we'd like to get some lighting on our translucent objects. In an attempt to avoid recreating all the horrible things that came from shader combinations for every light combination, I've been trying to implement something similar to the technique Bungie described in their presentation on Destiny's lighting. The idea is to collapse the light environment at various probe points into a spherical harmonic representation, that the shader would then use to compute lighting. Currently it's doing all of this on the CPU, but I've run into what seems to be a fundamental issue with projecting a directional light into SH. After digging through all of the fundamental papers, everything seems to agree that the way to project a directional light into SH, convolved with the cosine response is void project_directional( float* SH, float3 color, float3 dir ) { SH[0] = 0.282095f * color * pi; SH[1] = -0.48603f * color * dir.y * (pi * 2/3); SH[2] = 0.48603f * color * dir.z * (pi * 2/3); SH[3] = -0.48603f * color * dir.x * (pi * 2/3); } float3 eval_normal( float* SH, float3 dir ) { float3 result = 0; result = SH[0] * 0.282095f; result += SH[1] * -0.48603f * dir.y; result += SH[2] * 0.48603f * dir.z; result += SH[3] * -0.48603f * dir.x; return result; } // result is then scaled by diffuse There's a normalization term or two, but the problem I've been running into, that I haven't seen any decent way to avoid, is that ambient term in SH[0]. If I plug in a simple light pointing down Z, normals pointing directly at it, or directly away from it behave reasonably, but a normal pointing down, say, the X axis will always be lit by at least 1/4 of the light color. It's produced a directional light that generates significant amounts of light at 90 degress off-axis. I'm not seeing how this could ever behave differently. I can get vaguely reasonable results if I ignore the ambient term while merging diffuse lights in, but that breaks down the moment I try summing two lights, pointing in opposite directions in. Expanding out to the 9-term quadratic form does not help much either. I get the feeling I've missed some fundamental thing to trim down the off-axis directional light response, but I'll be damned if I can see where it would come from. Is this just a basic artifact of using a single light as a test case? Is this likely to behave better by keeping the main directional lights out, and just using the SH set to collapse point lights in as sphere lights or attenuated directionals? Have I just royally screwed up my understanding of how to project a directional light into SH? The usual pile of papers and articles from SCEE, Tom Forsyth, Sebastien Lagarde, etc have not helped. Someone had a random shadertoy that looked like it worked better in posted screenshots, but actually running it produces results more like what I've seen.
12. ## SH directional lights, what am I missing?
I was afraid of that. The divide by pi is in there on the real code side, I left out some of the normalization to get down to just the SH bits. The lighting model for this project is ridiculously ad-hoc, as we didn't get a real PBS approach set up in the engine until a few months into production. Another project is using a much more well behaved setup, but it has the advantage of still being in preproduction. For this project the scenes are sparse space-scapes, with a strong directional light, and an absurd number of relatively small radius point lights for effects, and only about three layers of objects (ships, foreground, and background). I suppose a brute force iteration over the light list might do the job well enough, as there might not be enough of these around to justify a fancy approach.
13. ## Help finding error in base 62 converter
The sites are the ones in the wrong. They're probably implemented in javascript, which I believe treats all numbers as floats, and thus are losing precision. As an example, your third number, punched into windows calc, as the first step would be: 22236810928128038 % 62 = 42, which should be 'g'. If we subtract 42 out of there, we get 22236810928127996, which on the second site properly ends up with a final digit of '0'. If you give it 22236810928127997, it still ends in '0', and if you give it 22236810928127998, it jumps to '4'. double precision floats only give about 16 digits of precision, so feeding it an 18 digit number means it starts rounding in units of 4. The entire idea seems a bit odd however, as for this to be reasonable, you have to convert before encrypting, and need to know exactly where numbers live in the output to parse them back properly. It seems like it would be better to encrypt directly from binary, and base-64 convert the output if you need to send it over a restricted channel.
14. ## StartInstanceLocation, and SV_InstanceID
Bleh, I see gl does the same thing. I suppose I shall have to put up with being terribly disappointed in the PC API's again.
15. ## StartInstanceLocation, and SV_InstanceID
So, I'm trying to use SV_InstanceID as an extra input to a shader, to pick from a small set of vertex colors in code. It seems to completely ignore the last argument of DrawIndexedInstanced(), and start at 0 per draw call. This seems less than useful, as it would make it impossible to transparently split up an instanced draw call, and defeat a lot of the purpose of having the system value at all. How would one be expected to use SV_InstanceID properly in this case? The vertex shader looks about like so: struct VertexInput { float4 position : POSITION; uint instanceid : SV_InstanceID; }; struct VertexOutput { float4 projPos : SV_Position; float4 color : COLOR0; }; VertexOutput vs_main( const VertexInput input ) { VertexOutput output = (VertexOutput)0; output.projPos = mul( float4( input.position.xyz, 1.0f ), g_ViewProjection ); if ( input.instanceid == 0 ) { output.color = float4(1,0,0,1); } else if ( input.instanceid == 1 ) { output.color = float4(0,1,0,1); } else { output.color = float4(0.5,0.5,0.5,1); } return output; } This results in it always picking red. If I instead dig a color out of a separate vertex buffer, via D3D11_INPUT_PER_INSTANCE_DATA, it works as expected. How do I make d3d useful?
16. ## Sharing violations and "Network optomisers"
Or that the driver's just a little old, and the QoS is busted. We had an issue with devkit connectivity, where one machine could talk to a kit after an update, but not another machine. The initial webconfig page would start loading, and then come to a dead halt, and kill the http connection. That turned out to be related to jumbo packets. The update enabled them for the devkit, and the machine that didn't work had a realtek driver dated ~5 days earlier than the other machine. That caused it to drop any and all jumbo packets, and the second packet the devkit tried sending over was about 20 bytes over the jumbo threshold...
17. ## Unordered access view woes with non-structured buffers
Typed UAVs have some restrictions, check the DXGI programming guide under Hardware Support for Direct3D 11 Formats. Column 22 on mine is Typed UAV, and it does apply to most of the types. Conspicuously absent from it, however, are 96-bit RGB, 64-bit depth/stencil, 32-bit depth (use R32), packed 24/8 depth/stencil, shared exponent and odd RG_BG/GR_GB modes, and all of the block compressed formats. tl;dr: DXGI_FORMAT_R32G32B32_FLOAT doesn't work for typed UAVs. The rest do.
18. ## Simulating lighting using volumetric meshes.
The renderer side is going to treat it as slices. If you really want to go this route, you're probably looking at using a geometry shader to replicate the light volume geometry out to all slices covered by it, doing the appropriate projections and such. The practicality of all that seems questionable, memory restrictions are going to keep your lighting exceedingly lowres, and you're blowing the vast majority of it on empty or useless space.
19. ## Floating point accuracy across computers?
A certain PC RTS title of years past tried this, including sending raw floats over the wire. They hit issues between Intel and AMD, and after sorting some of those out, between Debug and Release. They tried the usual compiler options and floating point control word magic (that still needed resetting after every D3D call). We got to port it to Linux, and tried very hard to keep it netplay compatible. All of the above applied, plus the fun of Visual Studio vs GCC when it came to fp codegen behavior. Rounding everything to ~3 decimal places mostly dealt with it. but not all of it. In particular, the AI code had some float comparisons lying around, on data that was never sent over the wire, that could change the number of calls to the RNG, and that *was* state that was tracked closely. I managed to come up with a method that definitively solved the compiler issues -- eyeball the VC output assembly, and reimplement the function on the GCC side with the VC floating point translated to AT&T syntax, pasted in, and add some shim code around it to fix up differences in the calling convention. This is not how one should define C++ class methods, but such was life. It even worked, and solved it definitively for that case. The next case that came up was the same sort of thing, two steps higher on the callstack. At that point I gave up, because we did not have the time to rewrite the entire AI system in assembly, as that was clearly going to be the end result. This way lies madness. Stick to fixed point for anything that actually matters to the game simulation. You should probably also make sure your system is set up to be able to detect synchronization loss as immediately as possible, and even better, have a mechanism for resynchronizing. Otherwise you're in for debugging issues that only happen in 5+ player games, after 2 hours, with the bulk of the useful data being gigs upon gigs of value logs and callstack traces.
20. ## Just how alright will I be if I were to skip normal-mapping?
Fillrate and memory bandwidth are not quite the same thing. Normalmaps don't really hit fillrate outside of a deferred or light prepass render, just memory bandwidth. However, their access patterns are fairly predictable, and scale better (assuming mipmaps) than random vertex access. Given that every card imaginable these days shades and rasterizes in units larger than a pixel, 2x2 quads at the least, and far larger in practice, any ALU gains you get by not bothering with normalmaps will be consumed by small triangle overhead. 1x1 pixel triangles will generally compute as 2x2 quads, or worse, and throw away most of the results, so pixel for pixel they're 4-16x more expensive than a more reasonably sized triangle. Lastly, mipmaps provide a more automatic method of LOD. With discrete triangles, lighting, texturing, etc will almost certainly break down into a flickery, sparkly mess as the triangles shrink to sub-pixel resolution.
21. ## Problem on physical material
It's the Schlick formula, but with k = roughness^2 / 2 to fit the smith GGX D function. The one I mentioned was equation 4 in epics notes from the physically based shading course at siggraph. For most things, the saturate works, but for the case of m=0, it depends on what the card does for 0/0. Epic avoids that case for regular lights, as they remap their roughness to (roughness+1)/2 first, so it became float G_Schlick(float v, float m) { float k = (m + 1) * (m + 1) / 8.0f; return v / (v * (1 - k) + k); } their notes are a bit thin on some of the other details.
22. ## Problem on physical material
Additionally, your D term formulation might start misbehaving as NoH approaches 1, but the artifacting in that term is probably supposed to get masked off by the missing NoL * NoV in the G term.
23. ## Problem on physical material
Looks like the G term is the likely culprit. For starters, schlick's G is G1(x) = dot(n, x) / ( dot(n, x) * (1-k) + k ) G(L, V, H) = G1(L) * G1(V) I don't know where that 0.25 in your numerator came from, but at the least you're missing an NoL * NoV, and I'm pretty sure that doesn't cancel out. Epic's paper mentions using a remapping of roughness to (roughness+1)/2 before squaring it, which guarantees a non-zero denominator. Your math for G is probably exploding for surfaces that don't have normals facing one of the two directions, and a roughness of 0.
24. ## Are mutexes really fool-proof?
Because your mutex is too simplistic to actually work. Here are some of the various possible failures it could run into: Both threads could try locking at the same time, both read it as unlocked, and both write it locked and enter. The compiler could be clever, and cache locked in a register, resulting in an infinite loop if one tries locking an already locked mutex. The compiler could inline the lock, and shuffle code around such that part of the block you're trying to protect happens before locking the mutex. The CPU could speculatively execute past the lock. etc.
25. ## OpenGL color interpolation not linear?
You should be using glBlendFunc(GL_SRC_ALPHA, GL_ONE). At a point where both of them are interpolated to 50% alpha, using SRC_ALPHA, ONE_MINUS_SRC_ALPHA, you will get 0.5 green, 0.25 red, 0.25 background, because you're doing this: Output = 0.5 * Green + (1-0.5) * ( 0.5 * red + (1-0.5) * background ); |
# Question Video: Finding the General Term to Work Out Terms in a Sequence
If (π_π) is a sequence defined as πβ = 11 and π_(π + 1) = π_π β 3 where π β₯ 1, then the fourth term equals οΌΏ.
02:22
### Video Transcript
If π sub π is a sequence defined as π sub one equals 11 and π sub π plus one equals π sub π minus three, where π is greater than or equal to one, then the fourth term equals what.
Weβre given four answer options: two, four, five, or eight. In this question, weβre given a formula for a sequence. This type of formula is called a recursive formula. And thatβs when the terms of a sequence are defined using one or more previous terms. If we wanted to describe this term in words, we would say that for any term with index π plus one, we take the term before it β thatβs the one with index π β and we subtract three. And so if we wanted to find the fourth term β thatβs the term with index four β that means that π plus one must be equal to four, and so π must be three. And so the fourth term must be equal to the third term minus three. But how do we find the third term?
Well, the third term β thatβs the term with index three β must happen when π plus one is three. And so π must be equal to two. So the third term is equal to the second term minus three. Of course, we donβt know the second term either. But youβve guessed it! Itβs going to be the first term minus three. And this is also one of the disadvantages of recursive formulas because we need to work out every term up to the term that we need.
We do get a little bit of relief here because weβre actually given the first term. π sub one is equal to 11. So now we can work forwards through the sequence. If π sub one is equal to 11 and π sub two is equal to π sub one minus three, then π sub two, the second term, is equal to 11 minus three. And thatβs equal to eight. As the third term is equal to the second term minus three, then our third term must be equal to eight minus three, which is five. And finally then, the fourth term is the third term minus three. And so five minus three is equal to two. We can therefore give the answer that the fourth term of the sequence is that given in option (A). Itβs the term two.
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## About "Example for Skew Symmetric Matrix"
Example for Skew Symmetric Matrix :
Here we are going to see some example problems on skew symmetric matrix.
What is symmetric and skew symmetric matrix ?
For any square matrix A with real number entries, A+ AT is a symmetric matrix and A− AT is a skew-symmetric matrix.
Any square matrix can be expressed as the sum of a symmetric matrix and a skew-symmetric matrix.
Let A be a square matrix. Then, we can write
Let us look into some problems to understand the concept.
Question 1 :
If A
is a matrix such that AAT = 9I , find the values of x and y.
Solution :
From the given matrix A, first let us find the transpose of matrix AT.
Here I stands for identity matrix with order 3 x 3.
By equating the corresponding terms, we get
x + 2y + 4 = 0 ------(1)
2x - 2y + 2 = 0 ------(2)
By adding the first and second equation, we may eliminate Y.
(x + 2y + 4) + (2x - 2y + 2) = 0 + 0
x + 2x + 4 + 2 = 0
3x + 6 = 0
3x = -6
x = -6/3 = -2
So, the value of x is -2.
Now we have to apply the value of x in the first equation, we may get y.
-2 + 2y + 4 = 0
2 + 2y = 0
2y = -2
y = -1
Hence the values of x and y are -2 and -1 respectively.
Question 2 :
For what value of x, the matrix
is skew-symmetric
Solution :
A square matrix A is said to be skew-symmetric if AT = −A.
By equating the corresponding terms, we get the value of x.
-3 = -x3
x = 3
x = 31/3
Hence the value of x is 31/3.
Question 3 :
If A =
is skew-symmetric, find the values of p, q, and r.
By equating the corresponding values, we may find the values of p, q and r respectively.
2 = -pp = -2 q2 = -q22q2 = 0q = 0 3 = -rr = -3
Hence the values of p, q and r are -2, 0 and -3 respectively.
After having gone through the stuff given above, we hope that the students would have understood "Example for Skew Symmetric Matrix".
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# 5.1 Quadratic functions (Page 8/15)
Page 8 / 15
$f\left(x\right)=-2{\left(x+3\right)}^{2}-6$
$f\left(x\right)={x}^{2}+6x+4$
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left[-5,\infty \right).$
$f\left(x\right)=2{x}^{2}-4x+2$
$k\left(x\right)=3{x}^{2}-6x-9$
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left[-12,\infty \right).$
For the following exercises, use the vertex $\text{\hspace{0.17em}}\left(h,k\right)\text{\hspace{0.17em}}$ and a point on the graph $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ to find the general form of the equation of the quadratic function.
$\left(h,k\right)=\left(2,0\right),\left(x,y\right)=\left(4,4\right)$
$f\left(x\right)={x}^{2}-4x+4$
$\left(h,k\right)=\left(-2,-1\right),\left(x,y\right)=\left(-4,3\right)$
$\left(h,k\right)=\left(0,1\right),\left(x,y\right)=\left(2,5\right)$
$f\left(x\right)={x}^{2}+1$
$\left(h,k\right)=\left(2,3\right),\left(x,y\right)=\left(5,12\right)$
$\left(h,k\right)=\left(-5,3\right),\left(x,y\right)=\left(2,9\right)$
$f\left(x\right)=\frac{6}{49}{x}^{2}+\frac{60}{49}x+\frac{297}{49}$
$\left(h,k\right)=\left(3,2\right),\left(x,y\right)=\left(10,1\right)$
$\left(h,k\right)=\left(0,1\right),\left(x,y\right)=\left(1,0\right)$
$f\left(x\right)=-{x}^{2}+1$
$\left(h,k\right)=\left(1,0\right),\left(x,y\right)=\left(0,1\right)$
## Graphical
For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts.
$f\left(x\right)={x}^{2}-2x$
Vertex Axis of symmetry is $\text{\hspace{0.17em}}x=1.\text{\hspace{0.17em}}$ Intercepts are
$f\left(x\right)={x}^{2}-6x-1$
$f\left(x\right)={x}^{2}-5x-6$
Vertex $\text{\hspace{0.17em}}\left(\frac{5}{2},\frac{-49}{4}\right),\text{\hspace{0.17em}}$ Axis of symmetry is $\text{\hspace{0.17em}}\left(0,-6\right),\left(-1,0\right),\left(6,0\right).$
$f\left(x\right)={x}^{2}-7x+3$
$f\left(x\right)=-2{x}^{2}+5x-8$
Vertex Axis of symmetry is $\text{\hspace{0.17em}}x=\frac{5}{4}.\text{\hspace{0.17em}}$ Intercepts are
$f\left(x\right)=4{x}^{2}-12x-3$
For the following exercises, write the equation for the graphed quadratic function.
$f\left(x\right)={x}^{2}-4x+1$
$f\left(x\right)=-2{x}^{2}+8x-1$
$f\left(x\right)=\frac{1}{2}{x}^{2}-3x+\frac{7}{2}$
## Numeric
For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function.
$x$ –2 –1 0 1 2 $y$ 5 2 1 2 5
$f\left(x\right)={x}^{2}+1$
$x$ –2 –1 0 1 2 $y$ 1 0 1 4 9
$x$ –2 –1 0 1 2 $y$ –2 1 2 1 –2
$f\left(x\right)=2-{x}^{2}$
$x$ –2 –1 0 1 2 $y$ –8 –3 0 1 0
$x$ –2 –1 0 1 2 $y$ 8 2 0 2 8
$f\left(x\right)=2{x}^{2}$
## Technology
For the following exercises, use a calculator to find the answer.
Graph on the same set of axes the functions
What appears to be the effect of changing the coefficient?
Graph on the same set of axes $\text{\hspace{0.17em}}f\left(x\right)={x}^{2},f\left(x\right)={x}^{2}+2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(x\right)={x}^{2},f\left(x\right)={x}^{2}+5\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}-3.\text{\hspace{0.17em}}$ What appears to be the effect of adding a constant?
The graph is shifted up or down (a vertical shift).
Graph on the same set of axes
What appears to be the effect of adding or subtracting those numbers?
The path of an object projected at a 45 degree angle with initial velocity of 80 feet per second is given by the function $\text{\hspace{0.17em}}h\left(x\right)=\frac{-32}{{\left(80\right)}^{2}}{x}^{2}+x\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is the horizontal distance traveled and $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ is the height in feet. Use the TRACE feature of your calculator to determine the height of the object when it has traveled 100 feet away horizontally.
50 feet
A suspension bridge can be modeled by the quadratic function $\text{\hspace{0.17em}}h\left(x\right)=.0001{x}^{2}\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}-2000\le x\le 2000\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}|x|\text{\hspace{0.17em}}$ is the number of feet from the center and $\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$ is height in feet. Use the TRACE feature of your calculator to estimate how far from the center does the bridge have a height of 100 feet.
## Extensions
For the following exercises, use the vertex of the graph of the quadratic function and the direction the graph opens to find the domain and range of the function.
Vertex $\text{\hspace{0.17em}}\left(1,-2\right),\text{\hspace{0.17em}}$ opens up.
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left[-2,\infty \right).$
Vertex $\text{\hspace{0.17em}}\left(-1,2\right)\text{\hspace{0.17em}}$ opens down.
Vertex $\text{\hspace{0.17em}}\left(-5,11\right),\text{\hspace{0.17em}}$ opens down.
Domain is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)\text{\hspace{0.17em}}$ Range is $\text{\hspace{0.17em}}\left(-\infty ,11\right].$
Vertex $\text{\hspace{0.17em}}\left(-100,100\right),\text{\hspace{0.17em}}$ opens up.
For the following exercises, write the equation of the quadratic function that contains the given point and has the same shape as the given function.
Contains $\text{\hspace{0.17em}}\left(1,1\right)\text{\hspace{0.17em}}$ and has shape of $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
$f\left(x\right)=2{x}^{2}-1$
Contains $\text{\hspace{0.17em}}\left(-1,4\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=2{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
Contains $\text{\hspace{0.17em}}\left(2,3\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=3{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
$f\left(x\right)=3{x}^{2}-9$
Contains $\text{\hspace{0.17em}}\left(1,-3\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=-{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
Contains $\text{\hspace{0.17em}}\left(4,3\right)\text{\hspace{0.17em}}$ and has the shape of $\text{\hspace{0.17em}}f\left(x\right)=5{x}^{2}.\text{\hspace{0.17em}}$ Vertex is on the $\text{\hspace{0.17em}}y\text{-}$ axis.
$f\left(x\right)=5{x}^{2}-77$
Contains $\text{\hspace{0.17em}}\left(1,-6\right)\text{\hspace{0.17em}}$ has the shape of $\text{\hspace{0.17em}}f\left(x\right)=3{x}^{2}.\text{\hspace{0.17em}}$ Vertex has x-coordinate of $\text{\hspace{0.17em}}-1.$
## Real-world applications
Find the dimensions of the rectangular corral producing the greatest enclosed area given 200 feet of fencing.
50 feet by 50 feet. Maximize $\text{\hspace{0.17em}}f\left(x\right)=-{x}^{2}+100x.$
Find the dimensions of the rectangular corral split into 2 pens of the same size producing the greatest possible enclosed area given 300 feet of fencing.
Find the dimensions of the rectangular corral producing the greatest enclosed area split into 3 pens of the same size given 500 feet of fencing.
125 feet by 62.5 feet. Maximize $\text{\hspace{0.17em}}f\left(x\right)=-2{x}^{2}+250x.$
Among all of the pairs of numbers whose sum is 6, find the pair with the largest product. What is the product?
Among all of the pairs of numbers whose difference is 12, find the pair with the smallest product. What is the product?
$6\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}-6;\text{\hspace{0.17em}}$ product is –36; maximize $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}+12x.$
Suppose that the price per unit in dollars of a cell phone production is modeled by $\text{\hspace{0.17em}}p=\text{}45-0.0125x,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is in thousands of phones produced, and the revenue represented by thousands of dollars is $\text{\hspace{0.17em}}R=x\cdot p.\text{\hspace{0.17em}}$ Find the production level that will maximize revenue.
A rocket is launched in the air. Its height, in meters above sea level, as a function of time, in seconds, is given by $\text{\hspace{0.17em}}h\left(t\right)=-4.9{t}^{2}+229t+234.\text{\hspace{0.17em}}$ Find the maximum height the rocket attains.
2909.56 meters
A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, is given by $\text{\hspace{0.17em}}h\left(t\right)=-4.9{t}^{2}+24t+8.\text{\hspace{0.17em}}$ How long does it take to reach maximum height?
A soccer stadium holds 62,000 spectators. With a ticket price of $11, the average attendance has been 26,000. When the price dropped to$9, the average attendance rose to 31,000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue?
\$10.70
A farmer finds that if she plants 75 trees per acre, each tree will yield 20 bushels of fruit. She estimates that for each additional tree planted per acre, the yield of each tree will decrease by 3 bushels. How many trees should she plant per acre to maximize her harvest?
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5 |
# If A and B are two vectors what is angle between (A + B) and (A – B)?
In this question, we are given two vectors A, B and we need to find out angle between vectors (A + B) and (A – B).
Easiest way to find out angle between any two vectors is to use Dot Product
If a and b are two vectors then their dot product is
a.b = |a| |b| Cosθ
Where
a, b are two vectors
|a|, |b| are magnitudes of these vectors
θ is the angle between directions of vectors a, b
Replacing
a = (A + B)
b = (A – B)
in the dot product formula
(A + B).(A – B) = |A + B| |A – B| Cosθ
Simplifying Left Hand Side of this equation
A.(A – B) + B.(A – B) = |A + B| |A – B| Cosθ
A.A – A.B + B.A – B.B = |A + B| |A – B| Cosθ (Equation 1)
Dot Product of a vector with itself is square of it’s magnitude
If a is a vector then a.a = |a|2
Therefore
A.A = |A|2
B.B = |B|2
|A|2 – A.B + B.A – |B|2 = |A + B| |A – B| Cosθ
Moreover Dot Product of two vectors is also Commutative which means
a.b = b.a if a, b are vectors
Therefore A.B = B.A
|A|2 – A.B + A.B – |B|2 = |A + B| |A – B| Cosθ
|A|2 – |B|2 = |A + B| |A – B| Cosθ
Using algebraic formula a2 – b2 = (a – b)(a + b)
(|A| – |B|)(|A| + |B|) = |A + B| |A – B| Cosθ
⇒ Cosθ = 1
General Solution for Cosθ = 1 is θ = 2n𝛑 where n is integer
Thus if A and B are two vectors then possible values of angle between vectors (A + B) and (A – B) are 0, 2𝛑, 4𝛑, 6𝛑 and so on.
👇🏻 Key Concepts Used in This Question
1. Dot Product of two vectors a and b is defined as a.b = |a| |b| Cosθ
Where
a, b are two vectors
|a|, |b| are magnitudes of these vectors
θ is the angle between directions of vectors a, b
2. Dot Product of a vector with itself is square of it’s magnitude
This can be derived from formula of Dot Product
If a, b are two vectors
Then
Their Dot Product a.b = |a| |b| Cosθ
Where
|a|, |b| are magnitudes of these vectors
θ is the angle between directions of vectors a, b
Let’s put b = a in Dot Product formula
a.a = |a| |a| Cosθ
θ = 0 (Angle between vectors a and a is zero)
a.a = |a| |a| Cos0 = |a|2
a.a = |a|2
3. Dot Product of two vectors is Commutative
Which means a.b = b.a
If a, b are two vectors
This can also be proved from formula of Dot Product itself
a.b = |a| |b| Cosθ
b.a = |b| |a| Cosθ
a.b = b.a
Therefore Dot Product of two vectors is Commutative
4. In Algebra (a – b)(a + b) = a2 – b2
In order to prove this, just simplify Left Hand Side
(a – b)(a + b) = a(a + b) – b(a + b)
= a2 + ab – ba – b2
= a2 – b2
⇒ (a – b)(a + b) = a2 – b2
(Do note that here a, b are numbers not vectors)
5. General Solution for Cosθ = 1 is θ = 2n𝛑 where n is integer |
# Doubling Time (Rule of 70)
Doubling time (also known as the rule of 70) is the amount of time that it takes for a quantity of something to duplicate in size. Simply put, how long will it take for a certain thing to double? To calculate this, you would use the rule of 70. This rule calculates the doubling time by dividing 70 by the growth rate.
You might notice this is quite similar to the rule of 72, which has you divide the number 72 by the annual rate of return. Both formulas derive from far more complicated logarithms that are difficult to do by hand and on the fly. These rules simplify them fairly accurately. So what is the biggest difference? Obviously, the rule of 70 uses the number 70 in its calculation, while the rule of 72 uses the number 72. This might seem straightforward, but these rules are typically used for different calculations.
The rule of 70 is used more to focus on growth, especially population growth. For example, how long will it take for the current population of llamas to double in size? In contrast, the rule of 72 is used more in finance to determine how long it will take an investment to double with a fixed interest rate. For this definition of doubling time, we will be focusing on the rule of 70.
## Doubling Time Formula
$$Years\: to\: Double = \dfrac{70}{Interest\: Rate}$$
In this formula, the growth/interest rate should be written as a whole number, not as a decimal. For example, if a population has a growth rate of 15%, you would use the whole number of 15 for the variable R instead of 0.15.
The frequency of time in which you want to see the doubling time is relative to the frequency of your growth rate. As a result, you should make sure your rate matches that time frame appropriately. To demonstrate, if you are wanting to know how many years it will take for a group to double, you should be using an annual growth rate. But, if you are wanting to see the growth in months, use a monthly growth rate.
You should be applying the doubling time formula to populations or quantities that are experiencing exponential growth. In this situation, “Exponential growth” is when the rate of growth is rapidly increasing at a constant rate compared to the current quantity. For instance, if a population was only experiencing minimal or sporadic growth rates, you probably wouldn’t use the doubling time formula.
As the growth rate, or variable R, increases, the doubling time will be faster. Essentially, if there is faster growth, it will take less time to reach that doubled quantity. If you increase the number of seeds you plant in the spring, you are going to see a lot more vegetables in the summer.
## Doubling Time Example
A local state college has been working hard to increase its online student population. Last year, they had 71,946 students. If they increase the number of their admitted students by 6% each year, how long will it take for them to double their annual count of online students?
Let’s break it down to identify the meaning and value of the different variables in this problem.
• Growth Rate: 6%
We can apply the values to our variables and calculate the doubling time:
$$Years\: to\: Double = \dfrac{70}{6} = 11.67\: years$$
In this case, the state college would double their online students in 11.67 years.
While the school has a good estimate, they can now consider other factors. Do they have the capability to handle that many students in that period of time? They might also wonder if growth that fast might affect the quality of the education they offer. There are many things to evaluate, but knowing the doubling time can help you make more informed choices looking forward.
## Doubling Time Analysis
Doubling time is an analytic tool used to project how long in the future before you reach the goal of doubling.
You might be wondering why it is that the doubling time, or rule of 70. is not typically used for finance. Why is the rule of 72 better for investment calculations? If you were to break down both rules to show each step of the calculation, the Rule of 72 uses more whole numbers, making it much easier to explain to clients who are wanting to understand how you are making their money double.
If you are examining populations specifically, you will see their growth rate vary greatly. For organic populations, larger organisms will have a slower growth rate than smaller ones. This is because they are made differently and are more prone to outside influences. Larger organisms typically have more cells and, therefore, take longer to develop. Giraffes, for example, take a lot longer to grow and mature than rabbits. Therefore, the growth rate of rabbits is significantly higher than that of giraffes.
Additionally, populations with larger organisms are more likely to hit their carrying capacity sooner. The “carrying capacity” is the maximum quantity that a population can preserve, based on the available supplies of food, water, etc. Because of this, no population can endlessly double, despite the current growth rate. They are still susceptible to other factors like lack of resources, disease, and changes to their habitat.
When a group or population reaches the carrying capacity, you will likely start to see a decline in the population. This is known as logistic growth. While the Doubling Time won’t account for factors like this, you should still be including them in the big picture of your estimation.
## Doubling Time Conclusion
• The doubling time is the amount of time that it takes for a quantity of something to double in size.
• Doubling time is more commonly known as the rule of 70.
• This formula is most helpful for populations or quantities that are experiencing exponential growth.
• The doubling rime formula requires only one variable: the interest/growth rate.
• The growth rate should be written as a whole number, not as a decimal.
• As the growth rate increases, so will the doubling time.
## Doubling Time Calculator
You can use the doubling time calculator below to quickly estimate how long it will take to double a quantity by entering the required numbers. |
## Book: Mathematics Part-II
### Chapter: 13. Probability
#### Subject: Maths - Class 12th
##### Q. No. 14 of Exercise 13.2
Listen NCERT Audio Books - Kitabein Ab Bolengi
14
##### Probability of solving specific problem independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that(i) the problem is solved (ii) exactly one of them solves the problem.
Given:
P(A) = Probability of solving the problem by A = 1/2
P(B) = Probability of solving the problem by B = 1/3
Because A and B both are independent.
P (A B) = P(A) . P(B)
P (A B) =
P(A) = 1 – P(A) = 1 – 1/2 = 1/2
P(B) = 1 – P(B) =
(i) the problem is solved
The problem is solved, i.e. it is either solved by A or it is solved by B.
= P(A B)
As we know, P (A B) = P(A) + P(B) - P (A B)
P (A B) =
(ii) exactly one of them solves the problem
i.e. either problem is solved by A but not by B or vice versa
i.e. P(A).P(B) + P(A).P(B)
=
=
P(A).P(B) + P(A).P(B) = 1/2
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# A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of number of successes.
Toolbox:
• To check if a given distribution is a probability distribution of random variable, the sum of the individual probabilties should add up to 1 (i.e, $\sum P(X_i) = 1$). Also 0 $\lt$ P(X) $\leq$ 1.
Step 1:
Let $X$ denote the number of doublets.
Possible doublets are $(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)$
Clearly $X$ can take values as $0,1,2,3$ and 4
Probability of getting a doublet =$\large\frac{6}{36}=\frac{1}{6}$
Probability of not getting a doublet =$1-\large\frac{1}{36}=\frac{1}{6}$
Step 2:
Now $P(X=0)=P$(no doublets)
$\qquad\qquad\quad\;\;=\large\frac{5}{6}\times \frac{5}{6}\times\frac{5}{6}\times \frac{5}{6}$
$\qquad\qquad\quad\;\;=\large\frac{625}{1296}$
Step 3:
$P(X=1)=P$(one doublet and three non-doublets)
$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}$
$\qquad\quad\;\;\;=4\big(\large\frac{1}{6}\times \frac{5^3}{6^3}\big)$
$\qquad\quad\;\;\;=\large\frac{500}{1296}$
$\qquad\quad\;\;\;=\large\frac{125}{324}$
Step 4:
$P(X=2)=P$(two doublet and two non-doublets)
$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{5}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{1}{6}\times \frac{5}{6}\times \frac{1}{6}\times \frac{5}{6}$
$\qquad\quad\;\;\;=4\big(\large\frac{1}{6^2}\times \frac{5^2}{6^2}\big)$
$\qquad\quad\;\;\;=\large\frac{100}{1296}$
$\qquad\quad\;\;\;=\large\frac{25}{324}$
Step 5:
$P(X=3)=P$(three doublets and one non-doublets)
$\qquad\quad\;\;=\large\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}+\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}+\times \frac{5}{6}\times \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6}$
$\qquad\quad\;\;\;=4\big(\large\frac{1}{6^3}\times \frac{5}{6}\big)$
$\qquad\quad\;\;\;=\large\frac{20}{1296}$
$\qquad\quad\;\;\;=\large\frac{5}{324}$
Step 6:
$P(X=4)=\large\frac{1}{6}\times \large\frac{1}{6}\times\large\frac{1}{6}\times\large\frac{1}{6}$
$\qquad\quad\;\;\;=\large\frac{1}{1296}$
Thus the required probability distribution is |
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# SA Cylinders Spheres and Cones Math 10-3 Ch.3 Measurement.
## Presentation on theme: "SA Cylinders Spheres and Cones Math 10-3 Ch.3 Measurement."— Presentation transcript:
SA Cylinders Spheres and Cones Math 10-3 Ch.3 Measurement
Surface Area of a Cylinder – Review Given the following cylinder, determine the surface area: Height = 12 cm Radius = 6 cm Formula =
SA of a Cylinder - Review First, calculate the area of the “label”: = 2 x 3.14 x 6 cm x 12 cm = 452.16 cm 2 Next, calculate the area of the circles: = 2 x 3.14 x 6 cm x 6 cm = 226.08 cm 2 Last, add together to get the total: SA = 452.16 cm 2 + 226.08 cm 2 = 678.24 cm 2
Surface Area of a Sphere – Review Given the following sphere, determine the surface area: Radius = 3.5 mm Formula = SA = 4 x 3.14 x 3.5 mm x 3.5 mm = 153.86 mm 2
Surface Area of a Cone Consider a cone: s = slant height r = radius h = height Consider the net of a cone: What shapes do you see? small circle and ~1/3 of a large circle
SA of a Cone What is the formula for the area of a circle? A = r 2 The formula for the other shape is A = r s Add these together to get the SA formula! SA cone = r 2 + rs
Ex2. Calculate the surface are of the following cone: s = 8 cm r = 4 cm formula = SA = r 2 + rs
SA of a Cone First, determine the area of the circle: = 3.14 x 4 cm x 4 cm = 50.24 cm 2 Next, determine the area of the slanted part: = 3.14 x 4 cm x 8 cm = 100.48 cm 2 Last, add together to get the total Surface Area = 50.24 cm 2 + 100.48 cm 2 = 150.72 cm 2
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PSAT Math : How to find the answer to an arithmetic sequence
Example Questions
Example Question #1 : Arithmetic Sequences
-27, -24, -21, -18…
In the sequence above, each term after the first is 3 greater than the preceding term. Which of the following could not be a value in the sequence?
461
501
657
126
461
Explanation:
All of the values in the sequence must be a multiple of 3. All answers are multiples of 3 except 461 so 461 cannot be part of the sequence.
Example Question #1 : How To Find The Answer To An Arithmetic Sequence
m, 3m, 5m, ...
The first term in the above sequence is m, and each subsequent term is equal to 2m + the previous term. If is an integer, then which of the following could NOT be the sum of the first four terms in this sequence?
80
16
60
48
-32
60
Explanation:
The fourth term of this sequence will be 5m + 2m = 7m. If we add up the first four terms, we get m + 3m + 5m + 7m = 4m + 12m = 16m. Since m is an integer, the sum of the first four terms, 16m, will have a factor of 16. Looking at the answer choices, 60 is the only answer where 16 is not a factor, so that is the correct choice.
Example Question #16 : Sequences
The tenth term in a sequence is 40, and the twentieth term is 20. The difference between consequence terms in the sequence is constant. Find n such that the sum of the first n numbers in the sequence equals zero.
59
40
58
60
30
Explanation:
Let d represent the common difference between consecutive terms.
Let an denote the nth term in the sequence.
In order to get from the tenth term to the twentieth term in the sequence, we must add d ten times.
Thus a20 = a10 + 10d
20 = 40 + 10d
d = -2
In order to get from the first term to the tenth term, we must add d nine times.
Thus a10 = a1 + 9d
40 = a1 + 9(-2)
The first term of the sequence must be 58.
Our sequence looks like this: 58,56,54,52,50…
We are asked to find the nth term such that the sum of the first n numbers in the sequence equals 0.
58 + 56 + 54 + …. an = 0
Eventually our sequence will reach zero, after which the terms will become the negative values of previous terms in the sequence.
58 + 56 + 54 + … 6 + 4 + 2 + 0 + -2 + -4 + -6 +….-54 + -56 + -58 = 0
The sum of the term that equals -2 and the term that equals 2 will be zero. The sum of the term that equals -4 and the term that equals 4 will also be zero, and so on.
So, once we add -58 to all of the previous numbers that have been added before, all of the positive terms will cancel, and we will have a sum of zero. Thus, we need to find what number -58 is in our sequence.
It is helpful to remember that a= a+ d(n-1), because we must add d to aexactly n-1 times in order to give us an. For example, a5 = a1 + 4d, because if we add d four times to the first term, we will get the fifth term. We can use this formula to find n.
-58 = an = a1 + d(n-1)
-58 = 58 + (-2)(n-1)
n = 59
Example Question #2 : Arithmetic Sequences
The first term of a sequence is 1, and every term after the first term is –2 times the preceding term. How many of the first 50 terms of this sequence are less than 5?
54
8
27
16
64
27
Explanation:
We can see how the sequence begins by writing out the first few terms:
1, –2, 4, –8, 16, –32, 64, –128.
Notice that every other term (of which there are exactly 50/2 = 25) is negative and therefore less than 25. Also notice that after the fourth term, every term is greater in absolute value than 5, so we just have to find the number of positive terms before the fourth term that are less than 5 and add that number to 25 (the number of negative terms in the first 50 terms).
Of the first four terms, there are only two that are less than 5 (i.e. 1 and 4), so we include these two numbers in our count: 25 negative numbers plus an additional 2 positive numbers are less than 5, so 27 of the first 50 terms of the sequence are less than 5.
Example Question #18 : Sequences
Explanation:
Look for cancellations to simplify. The sum of all consecutive integers from to is equal to . Therefore, we must go a little farther. , so the last number in the sequence in . That gives us negative integers, positive integers, and don't forget zero! .
Example Question #1 : How To Find The Answer To An Arithmetic Sequence
Brad can walk 3600 feet in 10 minutes. How many yards can he walk in ten seconds?
Explanation:
If Brad can walk 3600 feet in 10 minutes, then he can walk 3600/10 = 360 feet per minute, and 360/60 = 6 feet per second.
There are 3 feet in a yard, so Brad can walk 6/3 = 2 yards per second, or 2 x 10 = 20 yards in 10 seconds.
Example Question #2 : How To Find The Answer To An Arithmetic Sequence
If the sequence above continues as shown, what is the sum of the first 18 terms?
Explanation:
The simplest way to handle this question is to consider the sequence as being a set of 3 repeating terms, (3,2,4). Within the first 18 terms in the sequence, this pattern will repeat a total of 6 times.
The sum of one repeat is:
Since there are 6 repeats, take the sum of one repeat (9) and multiply by the number of repeats:
The sum of the first 18 terms is 54. |
4 Ijesha Close, Ilupeju, Lagos
+2348 097 685 118
#### Fractions, Percentages, and Decimals
##### FRACTIONS
• A fraction represents part of a whole quantity.
• A fraction is in two parts; a numerator and a denominator
• Equivalent fractions all represent the same amount. For example, 4/9, 8/18, 40/90 are equivalent fractions.
• If two or more fractions have the same denominator, we say they have a common denominator.
• The lowest common denominator of a set of fractions is the LCM of the denominators.
• When a fraction is such that the numerator and denominator have no common factor, we say the fraction is in its lowest terms or in its simplest form.
• mixed number contains a whole number and a fractional number. E.g. 23/4 is a mixed number.
• In a fraction, if the numerator is:
• less than the denominator, it is a proper fraction. E.g. 2/5
• greater than the denominator, it is an improper fraction. E.g. 5/2
• To add or subtract fractions:
• change mixed numbers to improper fractions,
• find the lowest common denominator,
• express the fractions as equivalent fractions with the same denominators.
• To multiply fractions:
• change mixed numbers to improper fractions,
• multiply the numerators together to give the numerator of the product,
• multiply the denominators to give the denominator of the product.
• The reciprocal (also referred to as inverse) of a fraction is the same fraction turned upside down. E.g. 9/4 is the reciprocal or inverse of 4/9.
• To divide by a fraction, multiply by its reciprocal.
##### PERCENTAGES
• Percentage or per cent means hundredths. ∴ 34% is another way of writing 34/100
• The symbol ‘%’ is short for per cent.
• To change a percentage to a fraction, write it as a fraction of 100 and reduce it to its lowest terms, where possible.
• To change a fraction to a percentage, multiply the fraction by 100.
• To express a quantity as a percentage of another:
• make sure they are in the same units,
• express the first as a fraction of the second,
• multiply the fraction by 100.
##### DECIMALS
• Decimals are another way to write fractions in which the denominators are 10 and times 10. For example: tenths (1/10), hundredths (1/100), etc.
• They are written with a dot before them. This dot is called a decimal point.
• The decimal point separates the whole numbers from the decimal figures. For example:
• The decimal figures have their values reduced as they move away from the decimal point.
• Examples of decimals include: 23.45, 344.3, 10445.57712, etc.
• Decimals can be expressed as fractions with denominators 10, 100, 1000, 10000, etc. To achieve this:
• Write the number and draw a line under it,
• Write a 1 under the decimal point, and
• Replace each of the numbers to the right of the decimal point with zeros.
• For example:
• To change a fraction to a decimal:
• Count the number of zeros in the denominator.
• Use the number of zeros in the denominator to count backwards from the end of the numerator,
• At the digit where you stop, put a decimal point before the digit.
• For example:
###### Examples
1. Express each of the fractions 3/4, 5/8, 7/8, 2/3 with a denominator of 24. Hence arrange the fractions in ascending order (i.e., from lowest to highest).
• 3/4 = 3 × 6/4 × 6 = 18/24
• 5/8 = 5 × 3/8 × 3 = 15/24
• 7/8 = 7 × 3/8 × 3 = 21/24
• 2/3 = 2 × 8/3 × 8 = 16/24
• The order is 15/24, 16/24, 18/24, 21/24, i.e. 5/8, 2/3, 3/4, 7/8
2. Express 42/70 and 26/78 in their lowest terms.
• 42/70 = 42 ÷ 7/70 ÷ 7 = 6/10 = 6÷2/10÷2 = 3/5 OR by using prime factors:
• 42/70 = 2 × 3 × 7/2 × 5 × 7 = 2÷2 × 3 × 7÷7/2÷2 × 5 × 7÷7 = 1 × 3 × 1/1 × 5 × 1 = 3/5
• 26/78 = 26 ÷ 2/78 ÷ 2 = 13/39 = 13 ÷ 13/39 ÷ 13 = 1/3
3. Express 45/8 as an improper fraction.
• 45/8 = 4 × 8 + 5/8 = 32 + 5/8 = 37/8
4. Express 19/8 as a mixed number.
• 19 ÷ 8 = 2, remainder 3.
• 19/8 = 23/8
5. 5/6 + 3/8
• The LCM of 6 and 24 is 24.
• 5/6 + 3/8 = 5 × 4/6 × 4 + 3 × 3/8 × 3 = 20/24 + 9/24
• = 20 + 9/24
• = 29/24
• 5/6 + 3/8 = 15/24
6. 7/104/15
• The LCM of 10 and 15 is 30.
• 7/104/15 = 7 × 3/10 × 34 × 2/15 × 2 = 21/308/30
• = 21 – 8/30
• 7/104/15 = 13/30
7. 33/5 + 22/3
• = 18/5 + 8/3
• = 54/15 + 40/15
• = 94/15
• ∴ 33/5 + 22/3 = 61/15
8. Simplify 3/8 of 22/9 × 13/5.
• = 3/8 × 20/9 × 8/5
• = 4/3
• ∴ 3/8 of 22/9 × 13/5 = 11/3
9. Find the value of 21/4 ÷ 3/7.
• = 21/4 × 7/3
• = 9/4 × 7/3
• = 21/4
• ∴ 21/4 ÷ 3/7 = 51/4
10. Express 15% as a fraction in its lowest terms.
• 15% = 15/100 = 3 × 5/20 × 5 = 3/20
11. Express 2/5 as a percentage.
• 2/5 = 2 × 20/5 × 20
• = 40/100
• = 40%
12. A worker hammers a post 2.2 m long into the ground. 66 cm of the post is below the ground. What percentage of the post is above the ground?
• Fraction of post below the ground = 66 cm/2.2 m = 66 cm/220 cm = 66/220
• Percentage of the post below the ground = 66/220 × 100% = 6/20 × 100 = 30%
• ∴ Percentage of post above the ground = 100% – 30% = 70% |
# Slope Formula
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Slope is defined as rise over run. Slope is found by the change of y axis co-ordinates over the change in x axis co-ordinates. Slope of a straight line gives the orientation of the line with respect to the co-ordinate axes. The slope of the line is used in different formula’s such as slope intercept form, slope point form etc. which help in writing the equation of the straight lines.
Slope formula = change in y co–ordinates / change in the x co–ordinates.
Example 1: Find the slope of the straight line passing through the two points (11, 15) and (2, 14)?
Solution: Given are the two points (11, 15) and (2, 14) from a straight line.
Slope = change in y co–ordinates / change in the x co–ordinates.
Here for the given two points slope = (14 – (15)) / (2 – (11)).
Slope = -1 / -9 Simplifying slope = 1/9.
Hence slope of the straight line passing through the given points = 1/9.
Example 2: Find the slope of the straight line passing through the two points (0, 0) and (2, 3)?
Solution: Given are the two points (0, 0) and (2, 3) from a straight line.
Slope = change in y co–ordinates / change in the x co–ordinates.
Here for the given two points slope = (3 – (0)) / (2 – (0)).
Slope = 3 - 0 / 2 - 0. Simplifying slope = 3/2.
Hence slope of the straight line passing through the given points = 3/2. |
## Section4.3Triple Integrals: Volume and Average Value
### Subsection4.3.1
It will come as no surprise that we can also do triple integrals—integrals over a three-dimensional region. The simplest application allows us to compute volumes in an alternate way.
We follow the same method as we have done when we defined a single integral for functions of one variable and a double integral for functions of two variables. Suppose that $f(x,y,z)$ is a continuous function on a closed bounded region $S$ in space. If the boundaries of $S$ are “relatively smooth”, then we can divide the three-dimensional region into small rectangular boxes with dimensions $\Delta x\times\Delta y\times\Delta z$ and with volume $dV = \Delta x\Delta y\Delta z\text{.}$ Then we add them all up and take the limit, to get an integral:
\begin{equation*} \iiint_S f(x,y,z)\,dV\text{.} \end{equation*}
Note:
1. Fubini's Theorem also holds for triple integrals, which means that the order of integration does not matter and we can choose from setting up a triple integral with any of the following six choices for the order of integration: $dx\,dy\,dz\text{,}$ $dx\,dz\,dy\text{,}$ $dy\,dx\,dz\text{,}$ $dy\,dz\,dx\text{,}$ $dz\,dx\,dy\text{,}$ and $dz\,dy\,dx\text{.}$ However, the same word of caution holds here as well, as some orders may lead to a more readily computable triple integral, while others may simply be too difficult to compute.
2. If the three-variable function $f$ is the constant 1, then the triple integral $\ds\iiint_S dV$ evaluates to the volume of the closed bounded region $S\text{.}$
3. If the three-variable function $f$ is the constant 1 and $S$ is bounded by constants, then we are simply computing the volume of a rectangular box.
###### Example4.19. Volume of a Box.
Compute the volume of the box with opposite corners at $(0,0,0)$ and $(1,2,3)\text{.}$
Solution
We begin by drawing an outline of the rectangular box as shown below.
Since the faces of the rectangular box are parallel to the coordinate planes, we deduce that the integration bounds are given by
\begin{equation*} 0\leq x\leq 1, \ 0 \leq y\leq2, \ 0\leq z \leq 3\text{.} \end{equation*}
Hence, the following triple integral computes the volume of the rectangular box:
\begin{equation*} \begin{split} \int_0^1\int_0^2\int_0^3 dz\,dy\,dx\amp=\int_0^1\int_0^2z\bigg\vert_0^3 \,dy\,dx \\ \amp=\int_0^1\int_0^2 3\,dy\,dx\\ \amp =\int_0^1 3y\bigg\vert_0^2 \,dx \\ \amp=\int_0^1 6\,dx = 6.\end{split} \end{equation*}
Note that any of the following triple integrals would have resulted in the volume of the box:
Of course, this is more interesting and useful when the limits are not constant.
###### Example4.20. Order of Integration.
Calculate the volume of the prism shown using the order of integration
1. $dx\,dy\,dz$
2. $dy\,dz\,dx$
Solution
Let $V$ be the volume of the prism. We begin by identifying the plane for each face of the prism:
1. For the triple integral with order of integration $dx\,dy\,dz\text{,}$ we begin by drawing a line parallel to the $x$-axis as shown below to the left that cuts through the prism. We notice that any such line stays between the values $x=0$ and $x=3$ and is thus constant. Hence, the inner most integral is
\begin{equation*} \int_{?}^{?}\int_{?}^{?}\int_0^3 \,dx\,dy\,dz\text{.} \end{equation*}
Next, we are now extending the line in the $x$-$y$-plane to create a cross-sectional area that slices through the prism perpendicularly to the $z$-axis as shown above in the centre. These horizontal cross-sections are not constant and vary depending on the $z$-value that is chosen as shown above to the right. The $z$-values themselves are bounded by $z=0$ and $z=2\text{.}$ Hence, the triple integral needed to evaluate the volume of the prism is
\begin{equation*} V = \int_0^2\int_0^{1-0.5z}\int_0^3 dx\,dy\,dz\text{.} \end{equation*}
Following through with the integration yields
\begin{equation*} \begin{split} V \amp = \int_0^2\int_0^{1-0.5z}\int_0^3 dx\,dy\,dz \\ \amp= \int_0^2\int_0^{1-0.5z} x\big\vert_0^3 \,dy\,dz\\ \amp = \int_0^2\int_0^{1-0.5z} 3\,dy\,dz \\ \amp = \int_0^2 3y\big\vert_0^{1-0.5z}\,dz\\ \amp = \int_0^2 3(1-0.5z)\,dz = \left[3z-\frac{3}{4}z^2\right]_0^2 = 3. \end{split} \end{equation*}
2. For the triple integral with order of integration $dy\,dz\,dx\text{,}$ we begin by drawing a line parallel to the $y$-axis as shown below to the left that cuts through the prism, since we are integrating with respect to $y\text{.}$ We notice that any such line stays between the values $y=0$ and $y=1-0.5z\text{,}$ and is thus variant depending on the $z$-value that is chosen. Hence, the inner most integral is
\begin{equation*} \int_{?}^{?}\int_{?}^{?}\int_0^{1-0.5z} \,dy\,dz\,dx\text{.} \end{equation*}
Next, we integrate with respect to z, which means we are summing up all such lines from above that are perpendicular to the $z$-axis in the $y$-$z$-plane of the prism as shown above in the centre, which of course sums to the area of the triangle in the $y$-$z$-plane. These lines are bounded by $z=0$ and $z=2\text{.}$ Hence, the inner two integrals needed to evaluate the volume of the prism are
\begin{equation*} \int_{?}^{?}\int_{0}^{2}\int_0^{1-0.5z} \,dy\,dz\,dx\text{.} \end{equation*}
Lastly, we integrate with respect to $x\text{,}$ which means we are summing up all such areas from above that are perpendicular to the $x$-axis. These areas are constant and are bounded by $x=0$ and $x=3$ as shown above to the right. Hence, the outer integral is
\begin{equation*} \int_{0}^{3}\int_{0}^{2}\int_0^{1-0.5z} \,dy\,dz\,dx\text{.} \end{equation*}
Following through with the integration yields the volume $V$ to be
\begin{equation*} \begin{split} V \amp = \int_{0}^{3}\int_{0}^{2}\int_0^{1-0.5z} \,dy\,dz\,dx \\ \amp= \int_0^3\int_0^2 y\big\vert_0^{1-0.5z}\,dz\,dz\\ \amp = \int_0^3\int_0^2(1-0.5z)\,dz\,dx\\ \amp = \int_0^3\left[z-\frac{z^2}{4}\right]_0^2\,dx\\ \amp = \int_0^3 1 \,dx = x\big\vert_0^3 = 3, \end{split} \end{equation*}
###### Example4.21. Volume of a Tetrahedron.
Find the volume of the tetrahedron with corners at $(0,0,0)\text{,}$ $(0,3,0)\text{,}$ $(2,3,0)\text{,}$ and $(2,3,5)\text{.}$
Solution
The whole problem comes down to correctly describing the region by inequalities: $0\le x\le 2\text{,}$ $3x/2\le y\le 3\text{,}$ $0\le z\le 5x/2\text{.}$ The lower $y$ limit comes from the equation of the line $y=3x/2$ that forms one edge of the tetrahedron in the $x$-$y$-plane; the upper $z$ limit comes from the equation of the plane $z=5x/2$ that forms the “upper” side of the tetrahedron as shown below.
Now the volume is
\begin{align*} \int_0^2\int_{3x/2}^3\int_0^{5x/2}dz\,dy\,dx \amp =\int_0^2\int_{3x/2}^3\left.z\right|_0^{5x/2} \,dy\,dx\\ \amp =\int_0^2\int_{3x/2}^3 {5x\over2}\,dy\,dx\\ \amp =\int_0^2 \left.{5x\over2}y\right|_{3x/2}^3 \,dx\\ \amp =\int_0^2 {15x\over2}-{15x^2\over4}\,dx\\ \amp =\left. {15x^2\over4}-{15x^3\over12}\right|_0^2\\ \amp =15-10=5\text{.} \end{align*}
Pretty much just the way we did for two dimensions we can use triple integration in a variety of different physical, social and biological applications, and in computing various average quantities.
###### Example4.22. Average Temperature in a Cube.
Suppose the temperature at a point is given by $T=xyz\text{.}$ Find the average temperature in the cube with opposite corners at $(0,0,0)$ and $(2,2,2)\text{.}$
Solution
In two dimensions:
1. Add up the temperature at each point in a region.
2. Divide by the area.
In three dimensions:
1. Add up the temperature at each point in space.
2. Divide by the volume.
Therefore, the average temperature in the cube is
\begin{align*} {1\over8}\int_{0}^2\int_{0}^2\int_{0}^2 xyz\,dz\,dy\,dx \amp ={1\over8}\int_{0}^2\int_{0}^2\left.{xyz^2\over2}\right|_0^2\,dy\,dx\\ \amp ={1\over16}\int_{0}^2\int_{0}^2 xy\,dy\,dx\\ \amp ={1\over4}\int_{0}^2\left.{xy^2\over2}\right|_0^2\,dx\\ \amp ={1\over8}\int_{0}^2 4x\,dx\\ \amp ={1\over2}\left.{x^2\over2}\right|_0^2 =1\text{.} \end{align*}
##### Exercises for Section 4.3.
Evaluate the following triple integrals.
1. $\ds\int_{0}^{1}\int_{0}^{x}\int_{0}^{x+y} (2x+y-1) \,dz\,dy\,dx$
$11/24$
Solution
\begin{equation*} \begin{aligned}\int_0^1 \int_0^x \int_0^{x+y} (2x+y-1)\,dz\,dy\,dx\amp= \int_0^1 \int_0^x (2x+y-1)(x+y)\,dy\,dx \\ \amp= \int_0^1 \int_0^x \left[2 x^2 + 3 x y - x + y^2 - y \right]\,dy\,dx \\ \amp= \int_0^1 \left[2x^2 y + \frac{3}{2}xy^2 -xy + \frac{y^3}{3} - \frac{y^2}{2}\right]_0^x\,dx \\ \amp= \int_0^1 \left[\frac{-9}{6} x^2 + \frac{23}{6} x^3\right]\,dx \\ \amp= \frac{11}{24}\end{aligned} \end{equation*}
2. $\ds\int_{0}^{2}\int_{-1}^{x^2}\int_{1}^{y} xyz \,dz\,dy\,dx$
$623/60$
Solution
\begin{equation*} \begin{aligned}\int_0^2\int_{-1}^{x^2}\int_1^y xyz \,dz\,dy\,dx \amp = \int_0^2 \int_{-1}^{x^2} \frac{xyz^2}{2} \bigg\vert_{z=1}^y\,dy\,dx\\ \amp= \int_0^2 \int_{-1}^{x^2} \left(\frac{xy^3}{2}-\frac{xy}{2}\right)\,dy\,dx \\ \amp =\int_0^2 \left[\frac{xy^4}{8} - \frac{xy^2}{4}\right]_{y=-1}^{x^2}\,dx\\ \amp = \int_0^2 \left(\frac{x^9}{8}-\frac{x^5}{4}+\frac{x}{8}\right)\,dx \\ \amp = \left[\frac{x^{10}}{8\cdot 10} - \frac{x^6}{4\cdot 6} + \frac{x^2}{8\cdot 2}\right]_{x=0}^2 = \frac{623}{30} \end{aligned} \end{equation*}
3. $\ds\int_{0}^{1}\int_{0}^{x}\int_{0}^{\ln y} e^{x+y+z}\,dz\,dy\,dx$
$-3e^2/4+2e-3/4$
Solution
\begin{equation*} \begin{aligned}\int_0^1 \int_0^x \int_0^{\ln y} e^{x+y+z}\,dz\,dy\,dx \amp = \int_0^1 \int_0^x \int_0^{\ln y} e^xe^ye^z\,dz\,dy\,dx \\ \amp = \int_0^1 \int_0^x e^xe^ye^z \bigg\vert_{z=0}^{\ln y} \,dy\,dx \\ \amp= \int_0^1\int_0^x e^xe^y(y-1)\,dy\,dx \end{aligned} \end{equation*}
We now use the fact that $\ds\int y e^y\,dy = e^y(y-1):$
\begin{equation*} \begin{aligned}\int_0^1\int_0^x e^xe^y(y-1)\,dy\,dx \amp = \int_0^1 \left[-e^xe^y+e^xe^y(y-1)\right]_{y=0}^x \,dx \\ \amp = \int_0^1 \left[e^xe^y(y-2)\right]_{y=0}^x \,dx \\ \amp= \int_0^1 e^{2x}(x-2)+2e^x \,dx \\ \amp = \left[\frac{1}{4}e^{2x}(2x-5) + 2e^x\right]_{x=0}^1 \\ \amp = \frac{-3}{4}e^2+2e - \frac{3}{4} \approx -0.855 \end{aligned} \end{equation*}
Thus, $\ds\int_0^1 \int_0^x \int_0^{\ln y} e^{x+y+z}\,dz\,dy\,dx \approx -0.855.$
4. $\ds\int_{0}^{\pi/2}\int_{0}^{\sin\theta}\int_{0}^{r\cos\theta} r^2\,dz\,dr\,d\theta$
$1/20$
Solution
Solution
\begin{equation*} \begin{aligned}\int_{0}^{\pi/2}\int_0^{\sin\theta}\int_0^{r\cos\theta} r^2\,dz\,dr\,d\theta \amp = \int_{0}^{\pi/2}\int_0^{\sin\theta} r^2 z \big\vert_{z=0}^{r\cos\theta} \,dr\,d\theta \\ \amp = \int_0^{\pi/2}\int_0^{\sin\theta} r^3\cos\theta \,dr\,d\theta \\ \amp =\int_0^{\pi/2} \frac{r^4}{4}\cos\theta \bigg\vert_{r=0}^{\sin\theta} \,d\theta \\ \amp = \int_0^{\pi/2} \frac{\sin^4\theta \cos\theta}{4}\,d\theta \end{aligned} \end{equation*}
Let $u=\sin\theta$ and $du = \cos\theta d\theta\text{.}$ Then
\begin{equation*} \int_0^{\pi/2} \frac{\sin^4\theta \cos\theta}{4}\,d\theta = \frac{1}{4}\int_0^1 u^4 \,du = \frac{1}{20} u^5 \bigg\vert_0^1 = \frac{1}{20}\text{.} \end{equation*}
Therefore, $\displaystyle{\int_{0}^{\pi/2}\int_0^{\sin\theta}\int_0^{r\cos\theta} r^2\,dz\,dr\,d\theta = \frac{1}{20}.}$
5. $\ds\int_{0}^{\pi}\int_{0}^{\sin\theta}\int_{0}^{r\sin\theta} r\cos^2\theta \,dz\,dr\,d\theta$
$\pi/48$
Solution
\begin{equation*} \begin{aligned} \int_0^{\pi} \int_0^{\sin\theta} \int_0^{r\sin\theta} r\cos^2\theta \,dz\,dr \,d\theta \amp= \int_0^{\pi} \int_0^{\sin\theta} r^2 \cos^2\theta\sin\theta \,dr\, d\theta\\ \amp=\int_0^{\pi} \frac{r^3}{3} \cos^2\theta \sin\theta \big\vert_0^{\sin\theta}\,d\theta\\ \amp= \int_0^{\pi} \frac{1}{3} \cos^2\theta \sin^{4}\theta\,d\theta \amp= \frac{\pi}{16(3)} = \frac{\pi}{48}.\end{aligned} \end{equation*}
6. $\ds\int_{0}^{1}\int_{0}^{y^2}\int_{0}^{x+y} x\,dz\,dx\,dy$
$11/84$
Solution
\begin{equation*} \begin{aligned} \int_0^1 \int_0^{y^2} \int_0^{x+y} x\,dz\,dx\,dy \amp= \int_0^1 \int_0^{y^2} \left(x^2+xy\right)\,dx\,dy\\ \amp= \int_0^1 \left[\frac{x^3}{3} + \frac{x^2y}{2}\right]_0^{y^2}\,dy\\ \amp= \frac{1}{6}\int_0^1 \left(2y^6+3y^5\right)\,dy\\ \amp=\frac{11}{84}.\end{aligned} \end{equation*}
7. $\ds\int_{1}^{2}\int_{y}^{y^2}\int_{0}^{\ln(y+z)} e^x\,dx\,dz\,dy$
$151/60$
Solution
\begin{equation*} \begin{aligned} \int_1^2 \int_y^{y^2} \int_0^{\ln(y+z)} e^x\,dx\,dz\,dy \amp= \int_1^2 \int_y^{y^2} (y+z-1)\,dz\,dy\\ \amp= \int_1^2 \left[yz+\frac{z^2}{2} - z\right]_y^{y^2}\,dy\\ \amp= \int_1^2 \left[\frac{y^4}{2} + y^3 - \frac{5y^2}{2} + y\right]\,dy\\ \amp= \frac{151}{60}.\end{aligned \end{equation*}
8. $\ds \int_0^\pi\int_0^{\pi/2}\int_0^1 (z\sin x+z\cos y)\,dz\,dy\,dx$
$\pi$
Solution
\begin{equation*} \begin{aligned} \int_0^{\pi}\int_0^{\pi/2} \int_0^1 (z\sin z+ z\cos y)\,dz\,dy\,dx \amp= \int_0^{\pi} \int_0^{\pi/2} (\sin x+ \cos y) \frac{z^2}{2}\,dy\,dx\\ \amp= \frac{1}{2} \int_0^{\pi} \int_0^{\pi/2} \left(\sin x+\cos y\right)\,dy \,dx \\ \amp= \frac{1}{2} \int_0^{\pi} \left[y\sin x+\sin y\right]_0^{\pi/2} \,dx \\ \amp= \frac{1}{2}\int_0^{\pi} \left(\frac{\pi}{2}\sin x + 1\right)\,dx\\ \amp= \frac{\pi}{4} \left[-\cos x\right]_0^{\pi} + \frac{x}{2} \bigg\vert_0^{\pi}\\ \amp= \pi. \end{aligned} \end{equation*}
Setup $\ds\iiint \left(x+y+z\right) \,dV$ over the region inside $x^2+y^2+z^2\le 1$ in the first octant, but do not follow through on the integration as it requires a special technique that we have not introduced.
\begin{equation*} \iiint_{V} \left(x+y+z\right) \,dV = \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} \int_0^{\sqrt{1-z^2-x^2}} \left(x+y+z\right) \,dy\,dx\,dz\text{.} \end{equation*}
Solution
We wish setup the integral $\displaystyle{\iiint_{V} \left(x+y+z\right) \,dV}\text{,}$ where $V$ is the interior of the sphere centred at the origin with radius 1 in the first quadrant. That is, $V = \{(x,y,z) : x^2+y^2+z^2 \leq 1 \text{ and } x \geq 0, y\geq 0, z\geq 0\}\text{.}$
Consider fixing $z\text{.}$ Then in the $x$-$y$-plane, the desired region is the quarter circle shown below:
where $r$ is now a function of $z\text{.}$ We see that if we let $x$ vary from 0 to $r\text{,}$ then $y$ must vary from $0$ to $\sqrt{r^2-x^2}\text{.}$
To find $r\text{,}$ let $x=r$ and $y=0\text{.}$ Then use the fact that on the boundary of the region, we have
\begin{equation*} 1 = x^2+y^2+z^2 \implies 1 = r^2 + z^2 \implies r = \sqrt{1-z^2}\text{.} \end{equation*}
All together, we see that
\begin{equation*} \iiint_{V} \left(x+y+z\right) \,dV = \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} \int_0^{\sqrt{1-z^2-x^2}} \left(x+y+z\right) \,dy\,dx\,dz\text{.} \end{equation*}
Find the region $E$ for which $\ds\iiint_E (1-x^2-y^2-z^2) \; dV$ is a maximum.
$E = \{x^2+y^2+z^2 \leq 1\}$
Notice that $x^2+y^2+z^2=1$ is the equation of the sphere of radius 1 centred at the origin. Hence, the integral will be maximized when $E$ is the interior of this sphere (it does not make a difference whether or not we include the boundary). |
# triangle and its properties class 7 introduction
Two sides of a triangle are 4 cm and 7 cm. Our notes of Chapter 6 The triangle and its properities are prepared by Maths experts in an easy to remember format, covering all syllabus of CBSE, KVPY, NTSE, Olympiads, NCERT & other Competitive Exams. This is called the exterior angle property of a triangle this chapter is prepared by entrancei team solve questions based on property of a triangle There are many properties of a triangle like median, altitudes and many more. The sum of all internal angles of a triangle is always equal to 180 0. Class 7 Maths The Triangle and its Properties Two Special Triangles: Equilateral and Isosceles A triangle in which all the three sides are of equal lengths and each angle has measure 60 0 is called an equilateral triangle. Get Revision Notes of Class 7th Mathematics Chapter 6 The triangle and its properities to score good marks in your Exams. 5. The line segment joining a vertex of a triangle to the mid point of its opposite side is called a median of the triangle. The above Practice worksheets for Class 7 Triangles and Its Properties have been designed as per latest NCERT CBSE and KVS guidelines and 2021 syllabus. Medians of a Triangle 2. 2. These are called six elements of the triangle. Notify me of follow-up comments by email. Intro to medians of a triangle (Opens a modal) Practice. A triangle has three sides and three angles , and all three sides and angles are called six elements of the triangle. Find the value of x in each of the following diagrams: Solution: (i) In the given figure, Ext. So it is impossible to construct the triangle … Let us learn now its types. Isosceles Triangle 3. Some of the major concepts such as Pythagoras theorem and trigonometry are dependent on triangle properties. (b) 80° In figure find the value of ∠A + ∠B + ∠C + ∠D + ∠E + ∠F. The concepts presented in NCERT Solutions for Class 7 Maths Chapter 6 Triangles and Its Properties are 1. In this class, Rajendra Singh will Explain the concept of Triangle and its Properties through Poll. Find the value of AD2. Free PDF Download of CBSE Maths Multiple Choice Questions for Class 7 with Answers Chapter 6 The Triangle and its Properties. Median C. Opposite side. (c) Scalene TRIANGLE A triangle is a simple closed curved made of three segments. ABC is an equilateral triangle with side a. Sum of the Lengths of Two Sides of A Triangle 7. (c) 50° and 40° Some important Facts about The Triangle and its Properties worksheet for class 7 The six elements of a triangle are its three angles and the three sides. How many acute angles can a right triangle have? These are the properties of a triangle: A triangle has three sides, three angles, and three vertices. ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Check Your Progress. 4. What do we mean by triangle? Students may learn about triangles in earlier classes while learning shapes. (b) 17 cm Draw a triangle $$ABC$$ and then draw a line segment $$AD$$ perpendicular to $$BC$$. NCERT Book for Class 7 Maths Chapter 6 The Triangle and its Properties is available for reading or download on this page. NCERT Class 7 Maths The Triangle and Its Properties. Find the value of x in this figure. 1 B. 1. The hypotenuse of a right triangle is 17 cm long. A triangle is a simple closed figure made up of three line segments. Class 7 math (India) Unit: The triangle and its properties. The Triangles and its Properties Class 7 Notes Conceptual Facts. (a) 15 cm (b) 40° A triangle in which two sides are of equal lengths is called ………………….. . 2. November 11, 2020 November 10, 2020 by worksheetsbuddy_do87uk. (b) 12 cm (b) 17cm $$YL$$ is an altitude in the exterior of triangle $$XYZ$$. Triangle and its properties : An exterior angle of a triangle is equal to the sum of its interior opposite angles. The line segment joining a vertex of a triangle to the mid-point of its opposite side is called its ……….. . Triangles and its properties 1. Question 9. Measures of each of the angles of an equilateral triangle is ……………… . Digital NCERT Books Class 7 Maths pdf are always handy to use when you do not have access to physical copy. What can be the length of its third side to make the triangle possible? The topics covered were an introduction to the triangle, types of triangle, exterior angle of a triangle and its properties. This shape is a triangle. If the distance between two tops is 10m. (a) 55° and 35° TRIANGLE A triangle is a simple closed curved made of three segments. (c) If the Pythagorean property holds, the triangle must be right-angled. (b) 70° (c) 13 cm Share this: Click to share on Twitter (Opens in new window) Click to share on Facebook (Opens in new window) (d) none of these, 3. It is possible to have a triangle in which two of the angles are right angles. Get Revision Notes of Class 7th Mathematics Chapter 6 The triangle and its properities to score good marks in your Exams. (d) none of these. Identify medians and altitudes Get 3 of 4 questions to level up! In this article, we will discuss the angle sum property from Class 7th Maths. Right-Angled Triangles and Pythagoras Property. Access full series of free online mock tests with answers from Triangles and Its Properties Class 7. (a) 45° How many medians a triangle can have? (a) acute angled CBSE Class 7 Maths Worksheets (c) PR NCERT Solutions Class 7 Maths Chapter 6 The Triangle and its Properties. Grade 7 The Triangle and Its Properties Worksheets November 11, 2020 November 10, 2020 by worksheetsbuddy_do87uk Grade 7 Maths The Triangle and Its Properties Multiple … Students can solve NCERT Class 7 Maths The Triangle and its Properties MCQs Pdf with Answers to know their preparation level. According to this model, the result $$PM$$ is perpendicular on $$QR$$ then $$PD$$ divides $$QR$$ in equal parts and $$D$$ is the mid-point of $$QR$$. (b) median (ii) In $$∆PQR$$, $$PQ$$ and $$PR$$ are altitudes of the triangle. Question 1. (iii) In $$∆XYZ$$, $$YL$$ is an altitude in the exterior of the triangle. The acute angles of right triangle are in the ratio 2 : 1. Watch Now. All Chapter-15 Properties of Triangles Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks. The above NCERT CBSE and KVS worksheets for Class 7 Triangles and Its Properties will help you to improve marks by clearing Triangles and Its Properties concepts and also improve problem solving skills. Ppt on triangles cbse viith class 1. Class 7 : Triangle and its Properties (concept with poll) Dec 17, 2020 • 1h 7m . A ——— connects a vertex of a triangle to the mid-point of the opposite side. Is it possible to have a triangle with the following sides ? These are Equilateral, Scalene and Isosceles are based on sides … A. Altitude B. 2. If the median and the altitude of an isosceles triangle can be same. (a) 50° Sum of two sides of a triangle is greater than or equal to the third side. (a) 40° NCERT Solutions Class 7 Maths Triangle and its properties. (c) 35° (c) 90° Two Special Triangles: Equilateral and Isosceles 6. Identify medians and altitudes Get 3 of 4 questions to level up! Exterior angle property . (a) 1 3. Properties of a triangle. Find the value of x (d) none of these. (d) 4, 4. (b) QR Class 7 Triangles and Its Properties test papers for all important topics covered which can come in your school exams, download in pdf free. Our notes of Chapter 6 The triangle and its properities are prepared by Maths experts in an easy to remember format, covering all syllabus of CBSE, KVPY, NTSE, Olympiads, NCERT & other Competitive Exams. Notes of chapter: The Triangle and Its Properties are presented. The difference between the lengths of any two sides of a triangle is smaller than the length of third side. A triangle has three sides and three angles, and all three sides and angles are called six elements of the triangle. This is one of the important parts of geometry. CLASSIFICATION OF TRIANGLE SIDES ANGLES 4. Students who are in Class 7 or preparing for any exam which is based on Class 7 Maths can refer NCERT Maths Book for their preparation. 1. Two poles of 8m and 14m stand upright on a plane ground. NCERT Solutions of all exercise questions and examples of Chapter 6 Class 7 explained free at teachoo. Maths Triangle and Its Properties part 2 (Types of Triangle) CBSE Class 7 Mathematics VII. (a) perpendicular They are listed below. $$AD$$ is an Altitude of the triangle. No, $$QM ≠ MR$$ , because $$D$$ is the mid-point of $$QR$$. This class will be helpful for the aspirants of Class 7th. Share. (b) Isosceles (b) 1 In this page we have Important Questions Class 7 Maths Chapter 6: Triangle and Its Properties. Chapter 6 The Triangle and its Properties NCERT Solutions for Class 7 Maths Chapter 6. (a) 1 In a ∆ ABC, ∠A = 35° and ∠B = 65°, then the measure of ∠C is: The longest side of a right angled triangle is called its …………….. . (b) 2 Download FREE PDF of Chapter-6 Triangle and Its Properties, In $$∆ PQR,$$ $$D$$ is the mid-point of is $$\overline {QR} \,.$$. 3. NCERT Class 7 Maths The Triangle and Its Properties, The Triangle and its Properties | NCERT Solutions, Instant doubt clearing with Cuemath Advanced Math Program. Maths MCQs for Class 7 Chapter Wise with Answers PDF Download was Prepared Based on Latest Exam Pattern. 2 + 3 = 5 Here sum of the side is equal to third side. Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 7 so that you can refer them as and when required. (d) 180°, 6. Karnataka State Syllabus Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.4. (b) 60° Updated January 10, 2020. One of the angles of a triangle is 100* and the other two angles are equal. The chapter questions are also provided here. (a) 60° (b) Isosceles A triangle is a closed figure formed by three straight lines. (a) Equilateral Worksheets for Class 7 Maths. 1. The questions given in the worksheets are framed in a manner which will help to revise the entire syllabus, concepts and also develop analytical and problem solving skills in students. (c) 13 cm INTRODUCTION TO TRIANGLES Sides AB ,BC ,CA Vertices A,B,C Angles ABC ,BCA ,CAB CB A A triangle is a closed figure made of three line segments. (b) 55° Solution: Let the length of the third side be x cm. 0. (c) 120° A triangle has 3 medians. Some important Facts about The Triangle and its Properties worksheet for class 7 The six elements of a triangle are its three angles and the three sides. 5. (a) 2 Condition I: Sum of two sides > the third side i.e. Grade 7 Maths The Triangle and Its Properties Multiple Choice Questions (MCQs) 1. Rajendra Singh. Learn. Hope you like them and do not forget to like , social share and comment at the end of the page. Similar Classes. (b) BC 3. Indepth notes along with worksheets and NCERT Solutions for Class 7. 6.1 Introduction. (a) 50° Here $$BE$$ is a median in triangle$$ABC$$ therefore $$AE = EC.$$. In this chapter, we will learn. Find the measure of each of these equal angles. Triangles and its properties 1. NCERT Book for Class 7 Maths Chapter 6 The Triangle and its Properties is available for reading or download on this page. 0. The Triangles and its Properties Class 7 Extra Questions Short Answer Type. (c) 13 cm (d) 0, 2. (d) 60°, 10. Skill Summary Legend (Opens a modal) Medians and altitudes. ΔABC has three sides AB, BC and CA and three angles ∠ABC, ∠BCA and ∠CAB. KSEEB Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 October 17, 2020 September 9, 2020 by Prasanna Students can Download Chapter 6 The Triangles and Its Properties Ex 6.5, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths , Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. A. (c) 41° (a) PQ d is exterior angle (a) Equilateral (d) Acute angled triangle, 7. (b) right angled The sum of all the interior angles is 180 degree. 1. These Worksheets for Grade 7 Triangles and Its Properties, class assignments and practice tests have been prepared as per syllabus issued by CBSE and topics given in NCERT book 2021. Class 7 Maths The Triangle and its Properties Right-Angled Triangles and Pythagoras Property In a right angle triangle, the side opposite to the right angle is called the hypotenuse and the other two sides are known as the base and perpendicular of the right-angled triangle. (c) hypotenuse 15. Chapter 6 Class 7 Triangle and its Properties. (a) 7 cm Every triangle has at least ……………. Class 7 math (India) Unit: The triangle and its properties. How many altitudes can a triangle have? Register for online coaching for IIT JEE (Mains & Advanced), NEET, Engineering and Medical … Therefore, $$PM$$ is altitude. Triangle parts: This first topic under this chapter deals with the basics about a triangle. angle of triangle = Sum of its interior opposite angles. (b) 60° and 30° A/an …………….. connect a vertex of a triangle to the mid point of the opposite side. 3. A triangle in which all three sides are of equal lengths is called ……………. The parts of a triangle such as its vertices, sides, angles etc. (d) 60°. In this article, we have mentioned some best and appropriate worksheets for CBSE Class 7 Hindi. 2 C. 3. MCQ FOR CLASS 7 MATHEMATICS | The Triangle and its Properties – Chapter 6. i) 2 cm, 3 cm, 5 cm. Sum of any two angles of a triangle is always greater than the third angle. Scalene Triangle 2. 3. Find the value’of x in given figure. Free PDF download of RS Aggarwal Solutions for Class 7 Chapter-15 Properties of Triangles solved by Expert Mathematics Teachers on Vedantu.com. All questions have been solved with detailed explanations of each and every step. $$PQR$$ is a triangle, $$D$$ is the mid-point of $$QR$$. Scalene triangle: If all sides of the triangle are unequal, then it is called scalene triangle. Intro to medians of a triangle (Opens a modal) Practice. (c) 3 (d) none of these. Learn. (c) 50° Given, $$PM$$ is perpendicular on $$QR$$. CBSE Class VII Maths Solutions, Mathematics Class 7 Triangle And Its Properties Chapter 6 Ex 6.1 NCERT Solutions 14. PRESENTATION ON THE TRIANGLE AND ITS PROPERTIES MADE BY N SAI DEEPAK & N sai deepika 2. AB ≠ BC ≠ CA 4. ∆ ABC is right-angled at C. If AC = 5 cm and BC = 12 cm find the length of AB. (c) 3 Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same. CBSE Class 7 Maths Notes Chapter 6 The Triangle and its Properties. Learn. AD is an altitude. 2. PQR is a triangle right angled at P. If PQ = 3 cm and PR = 4 cm, find QR. Then it is called a median of the third side, 3 ’ s visually this. S visually model this problem.There are three operations that done in sequence concepts as... Multiple Choice Questions ( MCQs ) 1 ( c ) If the median and the altitude of the squares the! ( d ) 60° ( c ) PR ( d triangle and its properties class 7 introduction none of these equal.. Of these first, draw an isosceles triangle can be observed that length AB... Parts of a triangle right angled at b Notes will be provided in.. Chapter 11 Triangles and its Properties Check your Progress the adjoining figure greater than the length of AB interesting triangle. ( types of triangle the exterior of the angles are equal 7 Hindi 60°! Explained free at teachoo and then draw a rough figure according to the third side SAI DEEPAK & SAI... Read the statement presented in NCERT Solutions Class 7 Math ( India ):! Some of the triangle and its Properties are 1 ) 42° under this Chapter deals with the diagrams... 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Opens a modal ) medians and altitudes Get 3 of 4 Questions to level up angles and! Or equal to the mid point of its opposite side is called …………… due East then... Comment at the end of the opposite side is called a median of the third side on... Two sides > the third side i.e 7 Chapter-15 Properties of Triangles solved by Expert Mathematics Teachers on Vedantu.com PQ! We will discuss the angle sum property of a triangle to the sum of the triangle ABC right angled triangle. This triangle was Prepared based on sides and three angles of two of. And also on various combinations of Properties of a triangle to the sum its... Is equal to the statement been solved with detailed explanations of each and every step triangle in all! Always handy to use when you do not have access to physical copy of an equilateral triangle is a! Acute angles of an equilateral triangle isosceles triangle can be same poles of 8m and stand. Then it is possible to have a right angled at P to \ ( ∆XYZ\ ), because (! 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#### • decimal multiplication division
From Wikipedia
Multiplication table
In mathematics, a multiplication table (sometimes, less formally, a times table) is a mathematical table used to define a multiplication operation for an algebraic system.
The decimal multiplication table was traditionally taught as an essential part of elementary arithmetic around the world, as it lays the foundation for arithmetic operations with our base-ten numbers. Many educators believe it is necessary to memorize the table up to 9 × 9.
In his 1820 book The Philosophy of Arithmetic, mathematician John Leslie published a multiplication table up to 99 × 99, which allows numbers to be multiplied in pairs of digits at a time. Leslie also recommended that young pupils memorize the multiplication table up to 25 × 25.
In 493 A.D., Victorius of Aquitaine wrote a 98-column multiplication table which gave (in Roman numerals) the product of every number from 2 to 50 times and the rows were "a list of numbers starting with one thousand, descending by hundreds to one hundred, then descending by tens to ten, then by ones to one, and then the fractions down to 1/144" (Maher & Makowski 2001, p.383)
The traditional rote learning of multiplication was based on memorization of columns in the table, in a form like
1 × 10 = 10 2 × 10 = 20 3 × 10 = 30 4 × 10 = 40 5 × 10 = 50 6 × 10 = 60 7 × 10 = 70 8 × 10 = 80 9 × 10 = 90
10 x 10 = 100 11 x 10 = 110 12 x 10 = 120 13 x 10 = 130 14 x 10 = 140 15 x 10 = 150 16 x 10 = 160 17 x 10 = 170 18 x 10 = 180 19 x 10 = 190 100 x 10 = 1000
This form of writing the multiplication table in columns with complete number sentences is still used in some countries instead of the modern grid above.
## Patterns in the tables
There is a pattern in the multiplication table that can help people to memorize the table more easily. It uses the figures below:
→ → 1 2 3 2 4 ↑ 4 5 6 ↓ ↑ ↓ 7 8 9 6 8 � � 0 0 Fig. 1 Fig. 2
For example, to memorize all the multiples of 7:
1. Look at the 7 in the first picture and follow the arrow.
2. The next number in the direction of the arrow is 4. So think of the next number after 7 that ends with 4, which is 14.
3. The next number in the direction of the arrow is 1. So think of the next number after 14 that ends with 1, which is 21.
4. After coming to the top of this column, start with the bottom of the next column, and travel in the same direction. The number is 8. So think of the next number after 21 that ends with 8, which is 28.
5. Proceed in the same way until the last number, 3, which corresponds to 63.
6. Next, use the 0 at the bottom. It corresponds to 70.
7. Then, start again with the 7. This time it will correspond to 77.
8. Continue like this.
Figure 1 is used for multiples of 1, 3, 7, and 9. Figure 2 is used for the multiples of 2, 4, 6, and 8. These patterns can be used to memorize the multiples of any number from 1 to 9, except 5.
## In abstract algebra
Multiplication tables can also define binary operations on groups, fields, rings, and other algebraic systems. In such contexts they can be called Cayley tables. For an example, see octonion.
## Standards-based mathematics reform in the USA
In 1989, the National Council of Teachers of Mathematics (NCTM) developed new standards which were based on the belief that all students should learn higher-order thinking skills, and which recommended reduced emphasis on the teaching of traditional methods that relied on rote memorization, such as multiplication tables. Widely adopted texts such as Investigations in Numbers, Data, and Space (widely known as TERC after its producer, Technical Education Research Centers) omitted aids such as multiplication tables in early editions. It is thought by many that electronic calculators have made it unnecessary or counter-productive to invest time in memorizing the multiplication table. NCTM made it clear in their 2006 Focal Points that basic mathematics facts must be learned, though there is no consensus on whether rote memorization is the best method.
Mental calculation
Mental calculation comprises arithmetical calculations using only the human brain, with no help from calculators, computers, or pen and paper. People use mental calculation when computing tools are not available, when it is faster than other means of calculation (for example, conventional methods as taught in educational institutions), or in a competition context. Mental calculation often involves the use of specific techniques devised for specific types of problems.
Many of these techniques take advantage of or rely on the decimal numeral system. Usually, the choice of radix determines what methods to use and also which calculations are easier to perform mentally. For example, multiplying or dividing by ten is an easy task when working in decimal (just move the decimal point), whereas multiplying or dividing by sixteen is not; however, the opposite is true when working in hexadecimal.
## Methods and Techniques
### Casting out nines
Main article:Casting out nines
After applying an arithmetic operation to two operands and getting a result, you can use this procedure to improve your confidence that the result is correct.
# Sum the digits of the first operand; any 9s (or sets of digits that add to 9) can be counted as 0.
# If the resulting sum has two or more digits, sum those digits as in step one; repeat this step until the resulting sum has only one digit.
# Repeat steps one and two with the second operand. You now have two one-digit numbers, one condensed from the first operand and the other condensed from the second operand. (These one-digit numbers are also the remainders you would end up with if you divided the original operands by 9; mathematically speaking, they're the original operands modulo 9.)
# Apply the originally specified operation to the two condensed operands, and then apply the summing-of-digits procedure to the result of the operation.
# Sum the digits of the result you originally obtained for the original calculation.
# If the result of step 4 does not equal the result of step 5, then the original answer is wrong. If the two results match, then the original answer may be right, though it isn't guaranteed to be.
Example
* Say we've calculated that 6338 × 79 equals 500702
# Sum the digits of 6338: (6 + 3 = 9, so count that as 0) + 3 + 8 = 11
# Iterate as needed: 1 + 1 = 2
# Sum the digits of 79: 7 + (9 counted as 0) = 7
# Perform the original operation on the condensed operands, and sum digits: 2 × 7 = 14; 1 + 4 = 5
# Sum the digits of 500702: 5 + 0 + 0 + (7 + 0 + 2 = 9, which counts as 0) = 5
# 5 = 5, so there's a good chance that we were right that 6338 × 79 equals 500702.
You can use the same procedure with multiple operands; just repeat steps 1 and 2 for each operand.
### Estimation
When checking the mental calculation, it is useful to think of it in terms of scaling. For example, when dealing with large numbers, say 1531 × 19625, estimation instructs you to be aware of the number of digits expected for the final value. A useful way of checking is to estimate. 1531 is around 1500, and 19625 is around 20000, so therefore a result of around 20000 × 1500 (30000000) would be a good estimate for the actual answer (30045875). So if the answer has too many digits, you know you've made a mistake.
### Factors
When multiplying, a useful thing to remember is that the factors of the operands still remain. For example, to say that 14 × 15 was 211 would be unreasonable. Since 15 was a multiple of 5, so should the product. The correct answer is 210.
### Calculating differences: ''a'' − ''b''
#### Direct calculation
When the digits of b are all smaller than the corresponding digits of a, the calculation can be done digit by digit. For example, evaluate 872 − 41 simply by subtracting 1 from 2 in the units place, and 4 from 7 in the tens place: 831.
#### Indirect calculation
When the above situation does not apply, the problem can sometimes be modified:
• If only one digit in b is larger than its corresponding digit in a, diminish the offending digit in b until it is equal to its corresponding digit in a. Then subtract further the amount b was diminished by from a. For example, to calculate 872 − 92, turn the problem into 872 − 72 = 800. Then subtract 20 from 800: 780.
• If more than one digit in b is larger than its corresponding digit in a, it may be easier to find how much must be added to b to get a. For example, to calculate 8192 − 732, we can add 8 to 732 (resulting in 740), then add 60 (to get 800), then 200 (for 1000). Next, add 192 to arrive at 1192, and, finally, add 7000 to get 8192. Our final answer is 7460.
• It might be easier to start from the left (the big numbers) first.
You may guess what is needed, and accumulate your guesses. Your guess is good as long as you haven't gone beyond the "target" number. 8192 − 732, mentally, you want to add 8000 but that would be too much, so we add 7000, then 700 to 1100, is 400 (so far we have 7400), and 32 to 92 can easily be recognized as 60. The result is 7460.
This method can be used to subtract numbers left to right, and if all that is required is to read the result aloud, it requires little of the user's memory even to subtract numbers of arbitrary size.
One place at a time is handled, left to right.
Example:
4075 − 1844 ------
Thousands: 4 − 1 = 3, look to right, 075 < 844, need to borrow. 3 − 1 = 2, say "Two thousand"
Hundreds: 0 − 8 = negative numbers not allowed here, 10 − 8 = 2, 75 > 44 so no need to borrow, say "two hundred"
Tens: 7 − 4 = 3, 5 > 4 so no need to borrow, say "thirty"
Ones: 5 − 4 = 1, say "one"
### Calculating products: ''a'' × ''b''
Many of these methods work because of the distributive property.
#### Multiplying by 2 or other small numbers
Where one number being multiplied is sufficiently small to be multiplied with ease by any single digit, the product can be calculated easily digit by digit from right to left. This is particularly easy for multiplication by 2 since the carry digit cannot be more than 1.
For example, to calculate 2 × 167: 2×7=14, so the final digit is 4, with a 1 carried and added to the 2×6 = 12 to give 13, so the next digit is 3 with a 1 carried and added to the 2×1=2 to give 3. Thus, the product is 334.
#### Multiplying by 5
To multiply a number by 5,
1. First multiply that number by 10, then divide it by 2.
The following algorithm is a
Formula calculator
A formula calculator is a software calculator that can perform a calculation in two steps:
1. Enter the calculation by typing it in from the keyboard.
2. Press a single button or key to see the final result.
This is unlike button-operated calculators, such as the Windows Calculator or the Mac os calculator, which require the user to perform one step for each operation, by pressing buttons to calculate all the intermediate values, before the final result is shown.
In this context, a formula is also known as an expression, and so formula calculators may be called expression calculators. Also in this context, calculation is known as evaluation, and so they may be called formula evaluators, rather than calculators.
## How they work
Formulas as they are commonly written use infix notation for binary operators, such as addition, multiplication, division and subtraction. This notation also uses:
• Parentheses to enclose parts of a formula that must be calculated first.
• In the absence of parentheses, operator precedence, so that higher precedence operators, such as multiplication, must be applied before lower precedence operators, such as addition. For example, in 2 + 3*4, the multiplication, 3*4, is done first.
• Among operators with the same precedence, associativity, so that the left-most operator must be applied first. For example, in 2 - 3 + 4, the subtraction, 2 - 3, is done first.
Also, formulas may contain:
• Non-commutative operators that must be applied to numbers in the correct order, such as subtraction and division.
• The same symbol used for more than one purpose, such as - for negative numbers and subtraction.
Once a formula is entered, a formula calculator follows the above rules to produce the final result by automatically:
• Analysing the formula and breaking it down into its constituent parts, such as operators, numbers and parentheses.
• Finding both operands of each binary operator.
• Working out the values of these operands.
• Applying the operator to these values, in the correct order so as to allow for non-commutative operators.
• Evaluating the parts of a formula in parentheses first.
• Taking operator precedence and associativity into account.
• Distinguishing between different uses of the same symbol.
## How to enter formulas
### Operators
Formulas printed in many text books use juxtaposition, underline and superscripts for multiplication, division and exponentiation respectively. Also, some operations, such as square root, are represented by special symbols that are not usually available on a computer keyboard. For example, see the formulas in Amortization calculator, Heron's formula and Law of cosines.
### Multiplication
In many software tools, including spreadsheets and programming languages, the asterisk, *, is used for multiplication. However, it is also possible to use juxtaposition. For example:
2cos(3)
means 2 multiplied by cos(3).
In calculators that don’t allow juxtaposition, the asterisk (and possibly x rather than, or as well as, the asterisk), is used, and the calculation should be entered as:
2*cos(3)
Also, a period is sometimes used for multiplication, as in:
2.cos(3)
Because the period is also used as the decimal point in numbers, so that the “2.� in the above would be interpreted as 2.0, the period is not used for multiplication in a formula calculator, and this calculation should be entered using a different symbol, as above.
### Division
Printed formulas often use a horizontal line for division, but in a formula calculator that uses only keyboard symbols, division is entered using the forward slash, /. When there is a calculation above or below the line, this should be done first, and so it should be enclosed in parentheses when typed in. For example,
2 + 3 ————— 4 - 5
should be entered as
(2 + 3)/(4 - 5)
Also, the symbol ÷ is often used for division, as in
2 ÷ 3
This symbol is not available on most computer keyboards, so this division operation is entered using the forward slash, as above.
### Exponentiation
Exponentiation, or raising to a power, is often represented using a superscript. For example:
2.452
means 2.45 squared.
With the limitations of a computer keyboard, in some software packages, such as Microsoft Excel, this is entered using the caret, ^:
2.45^2
but two asterisks are also used:
2.45**2
### Exponentiation and functions
When using functions, the superscript is sometimes placed immediately after the function name. For example, it is common to write the trigonometric version of Pythagoras’ Theorem (List of trigonometric identities) as:
sin2(x) + cos2(x) = 1
In this identity,
sin2(x)
means the square of sin(x), and is the same as:
sin(x)2
So, for x = 3.25, it could be entered into a formula calculator that uses the caret for exponentiation as:
sin(3.25)^2
### Square roots
Square roots are often specified using the √ symbol, but with the limitations of a keyboard this is can be entered by using exponentiation. For example, the square root of 2:
√2
could be entered as:
2^(1/2)
but two asterisks are also used:
2**(1/2)
The parentheses specify that the division should be done first.
### Other roots
All roots can be specified in this way. For example, the cube root of 2 can be entered as:
2^(1/3)
### Heron’s formula example
An example that illustrates these features is Heron’s formula.
One version of the formula, for a triangle with sides of length a, b and c, is equivalent to, using symbols that are available on a keyboard:
1/2*(a^2*c^2 - (a^2 + c^2 - b^2)/2)^0.5
For a = 2.5, b = 3.6 and c = 1.9, it could be entered into a formula calculator as:
1/2*(2.5^2*1.9^2 - (2.5^2 + 1.9^2 - 3.6^2)/2)^0.5
## Types of calculator
The formula calculator concept can be applied to all types of calculator, including arithmetic,
Question:I have this problem for my chem hw, I can't seem to get right... Calculate with the correct number of significant figures. (7.87g) / (16.1g - 8.14g) = ? The answer is not 0.989, 1.00, 1., 0.984, or 0.996. But if it is one of those, than something's wrong with the system, because my homework is online. According to my textbook, with multiplication problems, "the answer contains the same number of significant figures as in the measurement with the fewest significant figures." For addition problems, "the answer has the same number of decimal places as there are in the measurement with the fewest decimal places." Help? 0.98, sounded like a good answer but isnt the right answer... nevermind i got it, it was 0.99 how stupid
Answers:I'll do nothing except follow the standard order of operations and the two instructions you quoted. Because of the parentheses, I need to perform the subtraction first. ( 7.87g ) / ( 16.1g - 8.14g ) For addition problems, the answer has the same number of decimal places as there are in the measurement with the fewest decimal places. The subtraction is 1 and 2 decimal places, so the result is 1 decimal place. ( 7.87g ) / ( 16.1g - 8.14 ) ( 7.87g ) / ( 8.0 ) For multiplication problems, the answer contains the same number of significant figures as in the measurement with the fewest significant figures. The division has 3 and 2 significant digits, so the result has 2 significant digits. 7.87g / 8.0g 0.98g You may want to check my math, and check my counting of digits.
Question:I have a Casio fx-115 ES scientific calculator, and the default setting for dividing seems to be fractions rather than decimals. I need it to be decimals. I've already tried: Shift > Setup > a b/c and Shift > Setup > d/c Help, please! Is there anyway to permanently change the settings? Rather than pressing Shift > = after each calculation?
Answers:/ No , there is no other way
Question:Which of the following fractions has a decimal equivalent that is a terminating decimal? A. 10/189 B. 15/196 C. 16/225 D. 25/144 E. 39/128 What is the easier way of solving this problem without a calculator. Thanks.
Answers:Hint: If the denominator has factors of only 2 or 5 then you will get a terminating decimal. Notice: 1/2 = 0.5 (good) 1/3 = 0.3333.... (bad) 1/4 = 0.25 (good) 1/5 = 0.2 (good) 1/6 = 0.16666... (bad) 1/7 = 0.142857142857... (bad) 1/8 = 0.125 (good) 1/9 = 0.11111... (bad) 1/10 = 0.1 (good) 1/11 = 0.0909... (bad) etc. 189 --> 1 + 8 + 9 = 18 (multiple of 3) 196 --> 2 x 2 x 7 x 7 (multiple of 7) 225 --> 2 + 2 + 5 = 9 (multiple of 3) 144 --> 1 + 1 + 4 = 6 (multiple of 3) 128 --> 2 x 2 x 2 x 2 x 2 x 2 x 2 128 has only factors of 2 (or 5). Answer: E. 39/128
Question:My formula is the following Cell 1 has the number 3. I would put the following formula in Cell 2 =A1*.20+A2. The number would come out to be 3.6. How can I use this formula and round the number up automatically??
Answers:You can give the cell that holds your result, the property to have no decimal places. For example, if you cell's property is set to have several decimal places, the result would display 3.5257 or whatever. With one decimal place, it would be 3.6, and if you have no decimal places, then the result, regardless of the formula, would have to display 3. I hope that makes sense. |
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# The probability of picking 2 red balls one after another without repla
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The probability of picking 2 red balls one after another without repla [#permalink]
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29 Jan 2016, 02:05
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The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33
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Posts: 1683
Location: India
Re: The probability of picking 2 red balls one after another without repla [#permalink]
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31 Jan 2016, 06:06
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Probability of picking 2 red balls = 14/33
(R/R+W)*(R-1)/(R+W-1) = 14/33 = (2*7)/(3*11) = (2/3)*(7/11) --> Once the value is reduced to this level, it should be easy to find out the number of red balls.
1) Considering the above expression, both numerator and denominator must be consecutive.
2) Lets multiply both numerator and denominator with the same integer so that the value remains unaffected.
3) We can't multiply 7/11 as we need a consecutive term in both numerator and denominator. So lets multiply (2/3) to make the numerator and denominator consecutive.
(2/3)*(2/2) = 4/6 --> 4*7/6*11 --> Numerator and denominator not consecutive
(2/3)*(3/3) = 6/9 --> 6*7/9*11 --> Only numerator is consecutive
(2/3)*(4/4) = 8/12 --> 8*7/12*11 --> Numerator and denominator are consecutive here
Number of red balls, R = 8.
Total number of balls, R + W = 12 (This value is < 20, satisfying the condition)
Number of white balls, W = 12 - 8 = 4
Probability of picking 2 white balls w/o replacement = (W/R+W)*(W-1)/(R+W-1) = (4/12)*(3/11) = 3/33
##### General Discussion
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Joined: 02 Aug 2009
Posts: 8326
Re: The probability of picking 2 red balls one after another without repla [#permalink]
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31 Jan 2016, 22:52
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chetan2u wrote:
The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33
Hi,
Lets see what info is being provided by the Q...
1) The bag contains only red and white balls.
2) Prob of picking two red balls w/o replacement =14/33
3) A critical info it can hold at the most 20 balls...
what does " Prob of picking two red balls w/o replacement =14/33" mean if there are R red balls and W white balls..
$$\frac{R}{{R+W}} *\frac{{R-1}}{{R+W-1}}=\frac{14}{33}$$..
or $$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}= \frac{14}{33}$$.....
WHAT DOES RHS MEAN
14/33.. this means the actual numbers can be taken as 14x/33x, where x is the common term
WHAT DOES LHS MEAN
$$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}$$..
it means both nemerator and denominator are multiple of consecutive numbers..
from above two points we can say ..
14x is the product of two consecutive integers..
2*7*x..
33x is also the product of two consecutive integers..
3*11*x
as we see, x can be fitted as 4 to get numerator as 7*8 and denominator as 11*12...
so $$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}$$..= $$\frac{{(8*7)}}{{(12*11)}}$$..
.... or $$\frac{8{(8-1)}}{{(8+4)}{(8+4-1)}}$$..
thus R = 8 and w=4..
thus prob of picking 2 white balls one after another w/o replacement from that bag=4/12*3/11=1/11=3/33..
ans B
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The probability of picking 2 red balls one after another without repla [#permalink]
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10 Feb 2016, 16:16
chetan2u wrote:
chetan2u wrote:
The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33
Hi,
Lets see what info is being provided by the Q...
1) The bag contains only red and white balls.
2) Prob of picking two red balls w/o replacement =14/33
3) A critical info it can hold at the most 20 balls...
what does " Prob of picking two red balls w/o replacement =14/33" mean if there are R red balls and W white balls..
$$\frac{R}{{R+W}} *\frac{{R-1}}{{R+W-1}}=\frac{14}{33}$$..
or $$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}= \frac{14}{33}$$.....
WHAT DOES RHS MEAN
14/33.. this means the actual numbers can be taken as 14x/33x, where x is the common term
WHAT DOES LHS MEAN
$$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}$$..
it means both nemerator and denominator are multiple of consecutive numbers..
from above two points we can say ..
14x is the product of two consecutive integers..
2*7*x..
33x is also the product of two consecutive integers..
3*11*x
as we see, x can be fitted as 4 to get numerator as 7*8 and denominator as 11*12...
so $$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}$$..= $$\frac{{(8*7)}}{{(12*11)}}$$..
.... or $$\frac{8{(8-1)}}{{(8+4)}{(8+4-1)}}$$..
thus R = 8 and w=4..
thus prob of picking 2 white balls one after another w/o replacement from that bag=4/12*3/11=1/11=3/33..
ans B
3) A critical info it can hold at the most 20 balls...
You mention this as a critical info yet I fail to see how exactly this critical info was used in the process of determining the answer. Could you please elaborate?
Marshall & McDonough Moderator
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The probability of picking 2 red balls one after another without repla [#permalink]
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10 Feb 2016, 18:04
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Hi ZaydenBond,
According to the question, the total number of balls should be less than or equal to 20.
We have obtained the total number of balls as 12 which is the only number satisfying the given critical info.
If the number 20 was not given, we could have other possible solutions as the given ratio is 14x/33x, where x can take any value.
Manager
Joined: 04 Jun 2015
Posts: 77
Re: The probability of picking 2 red balls one after another without repla [#permalink]
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10 Jul 2017, 08:28
chetan2u wrote:
The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33
In the actual test I might just guess this one. Here are my 2 cents though -
Take a look at the wording of last part of the sentence "....it can hold at most 20 balls"
1. Max number of balls that the bag can hold is 20.
2. Two balls are picked w/o replacement.
Now if we take total as 12 see what happens - we are given that R/12 * (R-1)/11 = 14/33
Easy to calculate R = 8, W = 4.
So probability of picking 2 balls white = 4/12 * 3/11 = 1/11 = 3/33
Hence B
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Re: The probability of picking 2 red balls one after another without repla [#permalink]
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24 Nov 2019, 07:11
chetan2u wrote:
The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33
0<t=r+w≤20 (integers)
$$r(r-1)/t(t-1)=14/33=7*2/3*11…r^2-r=14[t(t-1)]/33$$
14 ≠ div by 3 and 11, so t(t-1) must be div 3 and 11
if t=11 and t-1=10 are div by 11 but not by 3;
if t=12 and t-1=11 are div by 11 and 3;
$$r^2-r=14[12(11)]/33…r^2-r-56=0…(r-8)(r+7)=0…r=8, t=12, w=4$$
$$w(w-1)/t(t-1)=4(3)/12(11)=1/11=3/33$$
Ans (B)
Re: The probability of picking 2 red balls one after another without repla [#permalink] 24 Nov 2019, 07:11
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Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
Lesson Plans: Introduction Lesson Plan 1: Left Hand, Right Hand - Solving Systems of Equations Lesson Plan 2: Hassan's Pictures - Linear Programming and Profit Lines
Lesson Plan 1: Left Hand, Right Hand - Solving Systems of Equations
The questions below dealing with systems of linear equations have been selected from various state and national assessments. Although the lesson above may not fully equip students with the ability to answer all such test questions successfully, students who participate in active lessons like this one will eventually develop the conceptual understanding needed to succeed on these and other state assessment questions.
• Taken from California High School Exit Examination (Spring, 2002):
7x + 3y = -8
-4x - y = 6
What is the solution to the system of equations shown above?
A. (-2, -2)
C. (2, -2)
D. (2, 2)
• Taken from the Texas Assessment of Knowledge and Skills (Spring 2003):
The length of a rectangle is equal to triple the width. Which system of equations can be used to find the dimensions of the rectangle if the perimeter is 85 centimeters?
1. l = w + 3
2(l + w) = 85
2. l = 3w
2l + 6w = 85
3. l = 3w
2(l + w) = 85 (correct answer)
4. l = w + 3
2l + 6w = 85
• Taken from the Texas Assessment of Knowledge and Skills (Spring 2003):
Marcos had 15 coins in nickels and quarters. He had 3 more quarters than nickels. He wrote a system of equations to represent this situation, letting x represent the number of nickels and y represent the number of quarters. Then he solved the system by graphing. How many coins of each type did Marcos have?
B. (5, 10)
C. (9, 6)
D. (10, 5)
• Taken from the Connecticut Academic Performance Test (Spring, 2002):
Industrial Electrical Use. A utility company offers electricity to industrial users at a rate of 8 cents per kilowatt-hour. The company also offers a fixed annual rate of \$1,200,000 for unlimited use of electricity. Graph each of these two rates as a line on the grid in your answer booklet. Explain why a large industrial user of electricity would choose to pay the fixed annual rate. Use the information in your graph to support your answer.
Solution: If a user needs less than 15,000,000 kilowatt-hours of electricity in a year then the 8 cents per kilowatt-hour rate would be cheaper than the fixed rate. But, if they required more than 15,000,000 kilowatt-hours, then the fixed rate would cost less money. If the user requires exactly 15,000,000 kilowatt-hours, then the cost of the two plans would be the same.
• Next: Materials |
## SOLVING EXPONENTIAL EQUATIONS
Note:
• To solve an exponential equation, isolate the exponential term, take the logarithm of both sides and solve.
If you would like an in-depth review of exponents, the rules of exponents, exponential functions and exponential equations, click on exponential function.
Solve for x in the following equation.
Problem 7.7a:
Solution:
The first step is to isolate .
Subtract 14 from both sides of the equation.
Divide both sides of the equation by 2.
The next step is to isolate the variable x.
Take the natural logarithm of both sides of the equation..
Use the Quadratic Formula where a=2, b=5,
The exact answers are and the approximate answers are and - 3.618639.
Check the solution by substituting 1.118639 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side: 34
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 1.118639 for x, then x=1.118639 is a solution.
Check the solution by substituting -3.618639 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side: 34
Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -3.618639 for x, then x=-3.618639 is a solution.
You can also check your answer by graphing (formed by subtracting the right side of the original equation from the left side). Look to see where the graph crosses the x-axis; that will be the real solution. Note that the graph crosses the x-axis at 1.118639 and -3.618639. This means that 1.118639 and -3.618639 are the real solutions.
If you would like to review the solution to problem 7.7b, click on Problem
If you would like to go back to the beginning of this section, click on Beginning
This site was built to accommodate the needs of students. The topics and problems are what students ask for. We ask students to help in the editing so that future viewers will access a cleaner site. If you feel that some of the material in this section is ambiguous or needs more clarification, please let us know by e-mail.
[Algebra] [Trigonometry]
[Geometry] [Differential Equations]
[Calculus] [Complex Variables] [Matrix Algebra]
Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.
Author: Nancy Marcus |
## 4.2 – Equations of Linear Functions
### Learning Objectives
• (4.2.1) – Write the equation of a line using the point-slope formula
• Define the equation of a line given the slope and a point
• Define the equation of a line given two points
• (4.2.2) – Identify the equations for vertical and horizontal lines
• (4.2.3) – Given a graph, write the equation of a linear function
• Given two function values, write the equation of the linear function passing through them
• (4.2.4) – Model an Application With a Linear Function
In this section, we will learn how to write equations for linear functions given different pieces of information, including two points on the line and the graph of the line. We will also identify the equations of horizontal and vertical lines and write linear equations from written information.
# (4.2.1) – Write the equation of a line using the point-slope formula
We have seen that we can define the slope of a line given two points on the line, and use that information along with the y-intercept to graph the line. If you don’t know the y-intercept, or the equation for the line you can use two points to define the equation of the line using the point-slope formula:
$y-{y}_{1}=m\left(x-{x}_{1}\right)$
This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.
### The Point-Slope Formula
Given one point and the slope, the point-slope formula will lead to the equation of a line:
$y-{y}_{1}=m\left(x-{x}_{1}\right)$
### Define the equation of a line given the slope and a point
In our first example, we will start with the slope, then we will show how to find the equation of a line without being given the slope.
### Example
Write the equation of the line with slope $m=-3$ and passing through the point $\left(4,8\right)$. Write the final equation in slope-intercept form.
### Define the equation of a line given two points
In our next example we will start with two points and define the equation of the line that passes through them.
### Example
Find the equation of the line passing through the points $\left(3,4\right)$ and $\left(0,-3\right)$. Write the final equation in slope-intercept form.
The following video examples shows how to write the equation for a line given it’s slope and a point on the line.
# (4.2.2) – Identify the equations for vertical and horizontal lines
The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as
$x=c$
where c is a constant. The slope of a vertical line is undefined, and regardless of the y-value of any point on the line, the x-coordinate of the point will be c.
Suppose that we want to find the equation of a line containing the following points: $\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right)$, and $\left(-3,5\right)$. First, we will find the slope.
$\displaystyle m=\frac{5 - 3}{-3-\left(-3\right)}=\frac{2}{0}$
Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x-coordinates are the same and we find a vertical line through $x=-3$.
The equation of a horizontal line is given as
$y=c$
where c is a constant. The slope of a horizontal line is zero, and for any x-value of a point on the line, the y-coordinate will be c.
Suppose we want to find the equation of a line that contains the following set of points: $\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right)$, and $\left(5,-2\right)$. We can use the point-slope formula. First, we find the slope using any two points on the line.
$\large \begin{array}{l}m=\frac{-2-\left(-2\right)}{0-\left(-2\right)}\hfill \\ =\frac{0}{2}\hfill \\ =0\hfill \end{array}$
Use any point for $\left({x}_{1},{y}_{1}\right)$ in the formula, or use the y-intercept.
$\begin{array}{l}y-\left(-2\right)=0\left(x - 3\right)\hfill \\ y+2=0\hfill \\ y=-2\hfill \end{array}$
The graph is a horizontal line through $y=-2$. Notice that all of the y-coordinates are the same.
The line $x=−3$ is a vertical line. The line $y=−2$ is a horizontal line.
### Example
Find the equation of the line passing through the given points: $\left(1,-3\right)$ and $\left(1,4\right)$.
# (4.2.3) – Given a graph, write the equation of a linear function
Now that we have written equations for linear functions in both the slope-intercept form and the point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function f below.
The function f passing through the points (0.7) and (4,4) with a negative slope.
We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose (0, 7) and (4, 4). We can use these points to calculate the slope.
$\large \begin{array}{l} m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\hfill \\ \text{ }=\frac{4 - 7}{4 - 0}\hfill \\ \text{ }=-\frac{3}{4} \end{array}$
Now we can substitute the slope and the coordinates of one of the points into the point-slope form.
$\large \begin{array}{l} y-{y}_{1}=m\left(x-{x}_{1}\right)\hfill \\ y - 4=-\frac{3}{4}\left(x - 4\right)\hfill \end{array}$
If we want to rewrite the equation in the slope-intercept form, we would find
$\large \begin{array}{l} y - 4=-\frac{3}{4}\left(x - 4\right)\hfill \\ y - 4=-\frac{3}{4}x+3\hfill \\ \text{ }y=-\frac{3}{4}x+7\hfill \end{array}$
Rewrite the equation in slope intercept form.
If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. Therefore, b = 7. We now have the initial value b and the slope m so we can substitute m and b into the slope-intercept form of a line.
So the function is $\displaystyle f\left(x\right)=-\frac{3}{4}x+7$, and the linear equation would be $\displaystyle y=-\frac{3}{4}x+7$.
### Given the graph of a linear function, write an equation to represent the function.
1. Identify two points on the line.
2. Use the two points to calculate the slope.
3. Determine where the line crosses the y-axis to identify the y-intercept by visual inspection.
4. Substitute the slope and y-intercept into the slope-intercept form of a line equation.
### Example
Write an equation for a linear function given a graph of f shown below.
In the following video we show an example of how to write the equation of a line given it’s graph.
### Example
If f is a linear function, with $f\left(3\right)=-2$ , and $f\left(8\right)=1$, find an equation for the function in slope-intercept form.
In the video we show another example of how to write a linear function given two points written with function notation.
# (4.2.4) – Model an Application With a Linear Function
In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.
### Given a linear function f and the initial value and rate of change, evaluate f(c).
1. Determine the initial value and the rate of change (slope).
2. Substitute the values into $f\left(x\right)=mx+b$.
3. Evaluate the function at $x=c$.
Initial value is a term that is typically used in applications of functions. It can be represented as the starting point of the relationship we are describing with a function. In the case of linear functions, the initial value is typically the y-intercept. Here are some characteristics of initial value:
• The point $(0,y)$ is often the initial value of a linear function
• The y value of the initial value comes from b in the slope intercept form of a linear function, $f\left(x\right)=mx+b$
• The initial value can be found by solving for b, or substituting 0 for x in a linear function.
In our first example, we are given a scenario where Marcus wants to increase the number of songs in his music collection by a fixed amount each month. This is a perfect candidate for a linear function because the increase in the number of songs stays the same each month. We will identify the initial value for the music collection, and write an equation that represents the number of songs in the collection for any number of months, t.
### Example
Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N, in his collection as a function of time, t, the number of months. How many songs will he own in a year?
In our next example, we will show that you can write the equation for a linear function given two data points. In this case, Ilya’s weekly income depends on the number of insurance policies he sells. We are given his income for two different weeks and the number of policies sold. We first find the rate of change and then solve for the initial value.
### Example
Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned$920. Find an equation for I(n), and interpret the meaning of the components of the equation.
In the following video example we show how to identify the initial value, slope and equation for a linear function.
We will show one more example of how to write a linear function that represents the monthly cost to run a company given monthly fixed costs and production costs per item.
### Example
Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are$37.50 per item. Write a linear function where C(x) is the cost for x items produced in a given month.
## Summary
### Slope of a line
• The slope of a line indicates the direction in which a line slants as well as its steepness. Slope is defined algebraically as:$\displaystyle m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$
• Given the slope and one point on a line, we can find the equation of the line using the point-slope formula. $y-{y}_{1}=m\left(x-{x}_{1}\right)$
### Equations of Lines
• Standard form of a line is given as $Ax+By=C$.
• The slope-intercept form of a line is given as $y=mx+b$
• The equation of a vertical line is given as $x=c$
• The equation of a horizontal line is given as $y=c$ |
## Calculus with Applications (10th Edition)
$$y = \frac{3}{4}x + \frac{{25}}{4}$$
\eqalign{ & {x^2} + {y^2} = 25;{\text{ }}\left( { - 3,4} \right) \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = \frac{d}{{dx}}\left( {25} \right) \cr & {\text{use sum rule as follows}} \cr & \frac{d}{{dx}}\left( {{x^2}} \right) + \frac{d}{{dx}}\left( {{y^2}} \right) = \frac{d}{{dx}}\left( {25} \right) \cr & {\text{solve the derivatives using the power rule and the chain rule}} \cr & 2x + 2y\frac{{dy}}{{dx}} = 0 \cr & {\text{solving the equation for }}\frac{{dy}}{{dx}} \cr & 2y\frac{{dy}}{{dx}} = - 2x \cr & \frac{{dy}}{{dx}} = - \frac{x}{y} \cr & \cr & {\text{find the slope at the given point }} \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( { - 3,4} \right)}} = - \frac{{ - 3}}{4} = \frac{3}{4} \cr & {\text{find the equation of the tangent line at the point }}\left( { - 3,4} \right) \cr & {\text{by using the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 4 = \frac{3}{4}\left( {x - \left( { - 3} \right)} \right) \cr & y - 4 = \frac{3}{4}\left( {x + 3} \right) \cr & y - 4 = \frac{3}{4}x + \frac{9}{4} \cr & y = \frac{3}{4}x + \frac{{25}}{4} \cr} |
# How do you evaluate 5/6-2/3?
Mar 16, 2018
$\frac{1}{6}$
#### Explanation:
Both fractions must have the same denominators before you can add the numerators:.
$\frac{5}{6} - \frac{2 \times 2}{3 \times 2}$
$= \frac{5}{6} - \frac{4}{6}$
$= \frac{1}{6}$
Mar 16, 2018
$\frac{1}{6}$
#### Explanation:
$\text{before subtracting the fractions we require them to have}$
$\text{a "color(blue)"common denominator}$
$\text{this is achieved by multiplying numerator/denominator}$
$\text{of "2/3" by 2, thus making the denominator 6}$
$\Rightarrow \frac{5}{6} - \left(\frac{2 \times \textcolor{red}{2}}{3 \times \textcolor{red}{2}}\right)$
$= \frac{5}{6} - \frac{4}{6}$
$\text{now subtract the numerators leaving the denominator}$
$= \frac{5 - 4}{6} = \frac{1}{6}$
Mar 16, 2018
Make both numbers have a common denominator of 6, then just subtract to get $\frac{1}{6}$
#### Explanation:
To subtract the numerators (numbers at the top), find a common denominator (number at the bottom), which in this case I'll use 6, since 6 is a common multiple of both 6 and 3
So to convert $\frac{2}{3}$ to $\frac{n}{6}$, where $n$ just stands for some number you want to find, multiply both top and bottom by 2
$\frac{2}{3} = \frac{2 \cdot 2}{3 \cdot 2} = \frac{4}{6}$
So you replace $\frac{2}{3}$ with $\frac{4}{6}$ in that expression
$\frac{5}{6} - \frac{2}{3} = \frac{5}{6} - \frac{4}{6} = \frac{5 - 4}{6} = \frac{1}{6}$
To get the answer $\frac{1}{6}$ |
What is a perfect square? How do you know if a number is a perfect square?
In this article, we’ll define perfect squares, provide a list of the first 25 perfect squares (and the integers that make them up), and teach you how to tell if a number is a perfect square.
## What Is a Perfect Square?
A perfect square is a number that can be expressed as the product of two equal integers.
What does that mean?
Basically, a perfect square is what you get when you multiply two equal integers by each other.
For instance:
\$\$5 * 5 = 25\$\$
25 is a perfect square because you’re multiplying two equal integers (5 and 5) by each other.
You can also express \$5 * 5\$ as \$5^2\$. That’s where you get the term “perfect square”
## List of Perfect Squares
Here’s a list of the first 25 perfect squares. A hint: if you want to create a perfect square, simply square an integer!
Square Integers 1 \$1 * 1\$ 4 \$2 * 2\$ 9 \$3 * 3\$ 16 \$4 * 4\$ 25 \$5 * 5\$ 36 \$6 * 6\$ 49 \$7 * 7\$ 64 \$8 * 8\$ 81 \$9 * 9\$ 100 \$10 * 10\$ 121 \$11 * 11\$ 144 \$12 * 12\$ 169 \$13 * 13\$ 196 \$14 * 14\$ 225 \$15 * 15\$ 256 \$16 * 16\$ 289 \$17 * 17\$ 324 \$18 * 18\$ 361 \$19 * 19\$ 400 \$20 * 20\$ 441 \$21 * 21\$ 484 \$22 * 22\$ 529 \$23 * 23\$ 576 \$24 * 24\$ 625 \$25 * 25\$
## How to Tell If a Number Is a Perfect Square
You can tell if a number is a perfect square in a couple of different ways.
First of all, if you create a square by multiplying two equal integers by each other, then the product is a perfect square.
So, \$1 * 1\$ is a perfect square. So is \$10 * 10\$ and \$1,000 * 1,000\$.
You can also tell if a number is a perfect square by finding its square roots. Finding the square root is the inverse (opposite) of squaring a number.
If you find the square root of a number and it’s a whole integer, that tells you that the number is a perfect square.
For instance, the square root of 25 is 5. The square root of 26 is not a whole integer. So, 26 is not a perfect square.
## Key Takeaways: Understanding Perfect Squares
A perfect square is a number that can be expressed as the product of two equal integers.
You can tell if a number by finding its square root and seeing if that square root is a whole integer.
## What's Next?
Getting ready to take the ACT? We have a list of 31 formulas you must know to conquer the ACT.
Taking the SAT instead? Here’s a list of our favorite SAT Math prep books that will help set you on the path to success.
Looking to brush up on your fundamental algebra skills? A good place to start is mastering systems of equations. |
### Matrices Explained in 5 Minutes
Matrices Explained in 5 Minutes
Take a few numbers and arrange them in a table like this..
3 5 1 6
2 8 6 7
0 4 5 2
Now, to make the table look pretty enclose it in giant brackets ( ). The brackets are not needed for any mathematical reason, they're just a reminder that the entire table should be considered as a single mathematical object.
You just made a matrix! It has 3 rows and 4 columns of numbers, so it's a 3x4 matrix. The numbers in a matrix are called the elements or terms of the matrix. So our 3x4 matrix has 12 elements. Of course, you can make matrices of any size. If the number of rows and columns are the same it's called a square matrix. Here's an example..
2 3 0
2 1 6
1 4 3
This is a square 3x3 matrix.
A matrix with only one row or one column is a vector, like this..
One row..
(2 3 1)
One column..
2
3
1
Of course, writing down new mathematical objects is easy. The harder part is defining some operations you can do with the objects.
Square Matrices can be added, just add corresponding terms to make a new matrix, like this example..
5 1
3 2
+
2 6
4 7
=
5+2 1+6
3+4 2+7
which is the matrix..
7 7
7 9
So addition is easy. But the operation which makes matrices very useful mathematical objects is the multiplication operation. This operation is subtle, and it's not just multiplying corresponding terms. To understand matrix multiplication, and also to see why it's so useful, consider what happens when a matrix multiplies a vector. After all, a vector is just a special case of a matrix.
So here's a matrix..
3 5 1
2 8 6
0 4 5
and here's a vector written as a one column matrix..
2
3
1
Now let's multiply..
Take the first row of the matrix (3 5 1), write it vertically, like this..
3
5
1
Now just multiply it term by term with the vector and add all the results, like this..
2*3+3*5+1*1=22
That's the first element of our result, so our result looks like..
22
-
-
To get the second term of the result repeat the process, but this time use the second row of the matrix (2 8 6) to get..
2*2+8*3+6*1=34
So now our result is..
22
34
-
To get the final term of the result use the third row of the matrix (0 4 5) to get..
2*0+3*4+1*5=17
So the final answer is the vector..
22
34
17
We're done! We just multiplied a vector by a matrix and the answer was another vector. So, one important job that matrices do is transform vectors. They can rotate them, they can stretch them, or they can do both at the same time. It turns out this is a very useful operation and is the reason matrices are so useful.
Of course, the table structure of a matrix lends itself to a very elegant notation. Suppose we have an nxm matrix, that is one with n rows and m columns. Then we can denote the general term of the matrix by aij, which is the number in the i'th row and j'th column. So the entire matrix could be written simply as..
{aij} where i=1,2,..,n and j=1,2,..,m
This is far easier than writing out a giant nxm table of numbers! Plus, this notation turns out to be very powerful, for example, it let's us write the product of two matrices in a very elegant way. Suppose {aij} and {bij} are two nxm matrices, then the ij'th term of the product is..
sum(aik*bkj) for k=1,2,...,m
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Content written and posted by Ken Abbott abbottsystems@gmail.com
Internet Marketing Consultant |
# 6.2 Explaining gauss’s law (Page 3/4)
Page 3 / 4
## Gauss’s law
The flux $\text{Φ}$ of the electric field $\stackrel{\to }{\text{E}}$ through any closed surface S (a Gaussian surface) is equal to the net charge enclosed $\left({q}_{\text{enc}}\right)$ divided by the permittivity of free space $\left({\epsilon }_{0}\right):$
$\text{Φ}={\oint }_{S}\stackrel{\to }{\text{E}}·\stackrel{^}{\text{n}}\phantom{\rule{0.2em}{0ex}}dA=\frac{{q}_{\text{enc}}}{{\epsilon }_{0}}.$
To use Gauss’s law effectively, you must have a clear understanding of what each term in the equation represents. The field $\stackrel{\to }{\text{E}}$ is the total electric field at every point on the Gaussian surface. This total field includes contributions from charges both inside and outside the Gaussian surface. However, ${q}_{\text{enc}}$ is just the charge inside the Gaussian surface. Finally, the Gaussian surface is any closed surface in space. That surface can coincide with the actual surface of a conductor, or it can be an imaginary geometric surface. The only requirement imposed on a Gaussian surface is that it be closed ( [link] ).
## Electric flux through gaussian surfaces
Calculate the electric flux through each Gaussian surface shown in [link] .
## Strategy
From Gauss’s law, the flux through each surface is given by ${q}_{\text{enc}}\text{/}{\epsilon }_{0},$ where ${q}_{\text{enc}}$ is the charge enclosed by that surface.
## Solution
For the surfaces and charges shown, we find
1. $\text{Φ}=\frac{2.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{{\epsilon }_{0}}=2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$
2. $\text{Φ}=\frac{-2.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{{\epsilon }_{0}}=-2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$
3. $\text{Φ}=\frac{2.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{{\epsilon }_{0}}=2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$
4. $\text{Φ}=\frac{-4.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}+6.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}-1.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{{\epsilon }_{0}}=1.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$
5. $\text{Φ}=\frac{4.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}+6.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}-10.0\phantom{\rule{0.2em}{0ex}}\mu \text{C}}{{\epsilon }_{0}}=0.$
## Significance
In the special case of a closed surface, the flux calculations become a sum of charges. In the next section, this will allow us to work with more complex systems.
Check Your Understanding Calculate the electric flux through the closed cubical surface for each charge distribution shown in [link] .
a. $3.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C;}$ b. $-3.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C;}$ c. $3.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C;}$ d. 0
Use this simulation to adjust the magnitude of the charge and the radius of the Gaussian surface around it. See how this affects the total flux and the magnitude of the electric field at the Gaussian surface.
## Summary
• Gauss’s law relates the electric flux through a closed surface to the net charge within that surface,
$\text{Φ}={\oint }_{S}\stackrel{\to }{\text{E}}·\stackrel{^}{\text{n}}\phantom{\rule{0.2em}{0ex}}dA=\frac{{q}_{\text{enc}}}{{\epsilon }_{0}},$
where ${q}_{\text{enc}}$ is the total charge inside the Gaussian surface S .
• All surfaces that include the same amount of charge have the same number of field lines crossing it, regardless of the shape or size of the surface, as long as the surfaces enclose the same amount of charge.
## Conceptual questions
Two concentric spherical surfaces enclose a point charge q . The radius of the outer sphere is twice that of the inner one. Compare the electric fluxes crossing the two surfaces.
Since the electric field vector has a $\frac{1}{{r}^{2}}$ dependence, the fluxes are the same since $A=4\pi {r}^{2}$ .
Compare the electric flux through the surface of a cube of side length a that has a charge q at its center to the flux through a spherical surface of radius a with a charge q at its center.
(a) If the electric flux through a closed surface is zero, is the electric field necessarily zero at all points on the surface? (b) What is the net charge inside the surface?
a. no; b. zero
Discuss how Gauss’s law would be affected if the electric field of a point charge did not vary as $1\text{/}{r}^{2}.$
Discuss the similarities and differences between the gravitational field of a point mass m and the electric field of a point charge q .
Both fields vary as $\frac{1}{{r}^{2}}$ . Because the gravitational constant is so much smaller than $\frac{1}{4\pi {\epsilon }_{0}}$ , the gravitational field is orders of magnitude weaker than the electric field.
Discuss whether Gauss’s law can be applied to other forces, and if so, which ones.
Is the term $\stackrel{\to }{E}$ in Gauss’s law the electric field produced by just the charge inside the Gaussian surface?
No, it is produced by all charges both inside and outside the Gaussian surface.
Reformulate Gauss’s law by choosing the unit normal of the Gaussian surface to be the one directed inward.
## Problems
Determine the electric flux through each surface whose cross-section is shown below.
Find the electric flux through the closed surface whose cross-sections are shown below.
a. $\text{Φ}=3.39\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/}\text{C}$ ; b. $\text{Φ}=0$ ;
c. $\text{Φ}=-2.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/}\text{C}$ ;
d. $\text{Φ}=90.4\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/}\text{C}$
A point charge q is located at the center of a cube whose sides are of length a . If there are no other charges in this system, what is the electric flux through one face of the cube?
A point charge of $10\phantom{\rule{0.2em}{0ex}}\mu \text{C}$ is at an unspecified location inside a cube of side 2 cm. Find the net electric flux though the surfaces of the cube.
$\text{Φ}=1.13\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}$
A net flux of $1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}$ passes inward through the surface of a sphere of radius 5 cm. (a) How much charge is inside the sphere? (b) How precisely can we determine the location of the charge from this information?
A charge q is placed at one of the corners of a cube of side a , as shown below. Find the magnitude of the electric flux through the shaded face due to q . Assume $q>0$ .
Make a cube with q at the center, using the cube of side a . This would take four cubes of side a to make one side of the large cube. The shaded side of the small cube would be 1/24th of the total area of the large cube; therefore, the flux through the shaded area would be
$\text{Φ}=\frac{1}{24}\phantom{\rule{0.2em}{0ex}}\frac{q}{{\epsilon }_{0}}$ .
The electric flux through a cubical box 8.0 cm on a side is $1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$ What is the total charge enclosed by the box?
The electric flux through a spherical surface is $4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/C}\text{.}$ What is the net charge enclosed by the surface?
$q=3.54\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{C}$
A cube whose sides are of length d is placed in a uniform electric field of magnitude $E=4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N/C}$ so that the field is perpendicular to two opposite faces of the cube. What is the net flux through the cube?
Repeat the previous problem, assuming that the electric field is directed along a body diagonal of the cube.
zero, also because flux in equals flux out
A total charge $5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{C}$ is distributed uniformly throughout a cubical volume whose edges are 8.0 cm long. (a) What is the charge density in the cube? (b) What is the electric flux through a cube with 12.0-cm edges that is concentric with the charge distribution? (c) Do the same calculation for cubes whose edges are 10.0 cm long and 5.0 cm long. (d) What is the electric flux through a spherical surface of radius 3.0 cm that is also concentric with the charge distribution?
Using Kirchhoff's rules, when choosing your loops, can you choose a loop that doesn't have a voltage?
how was the check your understand 12.7 solved?
Who is ISSAAC NEWTON
he's the father of 3 newton law
Hawi
he is Chris Issaac's father :)
Ethem
how to name covalent bond
Who is ALEXANDER BELL
LOAK
what do you understand by the drift voltage
what do you understand by drift velocity
Brunelle
nothing
Gamal
well when you apply a small electric field to a conductor that causes to add a little velocity to charged particle than usual, which become their average speed, that is what we call a drift.
graviton
drift velocity
graviton
what is an electromotive force?
It is the amount of other forms of energy converted into electrical energy per unit charge that flow through it.
Brunelle
How electromotive force is differentiated from the terminal voltage?
Danilo
in the emf power is generated while in the terminal pd power is lost.
Brunelle
what is then chemical name of NaCl
sodium chloride
Azam
sodium chloride
Brunelle
Sodium Chloride.
Ezeanyim
How can we differentiate between static point and test charge?
Wat is coplanar in physics
two point charges +30c and +10c are separated by a distance of 80cm,compute the electric intensity and force on a +5×10^-6c charge place midway between the charges
0.0844kg
Humble
what is the difference between temperature and heat
Heat is the condition or quality of being hot While Temperature is ameasure of cold or heat, often measurable with a thermometer
Abdul
Temperature is the one of heat indicators of materials that can be measured with thermometers, and Heat is the quantity of calor content in material that can be measured with calorimetry.
Gamma
the average kinetic energy of molecules is called temperature. heat is the method or mode to transfer energy to molecules of an object but randomly, while work is the method to transfer energy to molecules in such manner that every molecules get moved in one direction.
2. A brass rod of length 50cm and diameter 3mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°c( degree Celsius) if the original length are 40°c(degree Celsius) is there at thermal stress developed at the junction? The end of the rod are free to expand (coefficient of linear expansion of brass = 2.0×10^-5, steel=1.2×10^-5k^1)
A charge insulator can be discharged by passing it just above a flame. Explain.
of the three vectors in the equation F=qv×b which pairs are always at right angles?
what is an ideal gas?
What is meant by zero Kelvin ?
Justine
Why does water cool when put in the pot ?
Justine
when we pour the water in a vessel(pot) the hot body(water) loses its heat to the surrounding in order to maintain thermal equilibrium.Thus,water cools.
rupendra
when we drop water in the pot, the pot body loses heat to surrounded in order to maintain thermal equilibrium thus,water cool.
Srabon
my personal opinion ideal gas means doesn't exist any gas that obey all rules that is made for gases, like when get the temp of a gas lower, it's volume decreases.since the gas will convert to liquid when the temp get lowest.. so you can imagine it, but you can't get a gas at the lowest T
Edit An ideal gas is a theoretically gascomposed of many randomly moving point particles whose only interactions are perfectly elastic collisions.
Gamma
ideal gases are real gases at low temperature
Brunelle |
# Order of Operations with Exponents
Problem: Evaluate this arithmetic expression: 18 + 36 ÷ 32
In the last lesson, we learned how to evaluate an arithmetic expression with more than one operation according to the following rules:
Rule 1: Simplify all operations inside parentheses.
Rule 2: Perform all multiplications and divisions, working from left to right.
Rule 3: Perform all additions and subtractions, working from left to right.
However, the problem above includes an exponent, so we cannot solve it without revising our rules.
Rule 1: Simplify all operations inside parentheses.
Rule 2: Simplify all exponents, working from left to right.
Rule 3: Perform all multiplications and divisions, working from left to right.
Rule 4: Perform all additions and subtractions, working from left to right.
We can solve the problem above using our revised order of operations.
Problem: Evaluate this arithmetic expression: 18 + 36 ÷ 32
Solution:
18 + 36 ÷ 32 = 18 + 36 ÷ 9 Simplify all exponents (Rule 2) 18 + 36 ÷ 9 = 18 + 4 Division (Rule 3) 18 + 4 = 22 Addition (Rule 4)
Let's look at some other examples that involve our new rules for order of operations.
Example 1: Evaluate 52 x 24
Solution:
52 x 24 = 25 x 24 Simplify all exponents, working from left to right (Rule 2) 25 x 24 = 25 x 16 25 x 16 = 400 Multiplication (Rule 3)
Example 2: Evaluate 289 - (3 x 5)2
Solution:
289 - (3 x 5)2 = 289 - 152 Simplify all operations inside parentheses (Rule 1) 289 - 152 = 289 - 225 Simplify all exponents (Rule 2) 289 - 225 = 64 Subtraction (Rule 4)
Example 3: Evaluate 8 + (2 x 5) x 34 ÷ 9
Solution:
8 + (2 x 5) x 34 ÷ 9 = 8 + 10 x 34 ÷ 9 Simplify all operations inside parentheses (Rule 1) 8 + 10 x 34 ÷ 9 = 8 + 10 x 81 ÷ 9 Simplify all exponents (Rule 2) 8 + 10 x 81 ÷ 9 = 8 + 810 ÷ 9 Perform all multiplications and divisions, working from left to right (Rule 3) 8 + 810 ÷ 9 = 8 + 90 8 + 90 = 98 Addition (Rule 4)
Example 4: An interior decorator charges \$15 per square foot to lay a carpet, and an installation fee of \$150. If the room is square and each side measures 12 feet, how much will it cost to carpet it?
Solution: If one side of the square-shaped room is 12 feet, then the area of the room is (12 feet)2.
15 x 122 + 150 = 15 x 144 + 150 Simplify all exponents (Rule 2) 15 x 144 + 150 = 2,160 + 150 Multiplication (Rule 3) 2,160 + 150 = 2,310 Addition (Rule 4)
Answer: It will cost \$2,310 to carpet this room.
Summary:
To help us remember the order of operations, we can use the mnemonic PEMDAS, which stands for:
Please Excuse My Dear Aunt Sally
Parentheses, Exponents, Multiplication & Division, Addition & Subtraction
Note that although there are six words, they correspond to four rules.
### Exercises
Directions: Complete each exercise by applying the rules for order of operations. Click once in an ANSWER BOX and type in your answer; then click ENTER. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR.
1 32 x 43 ANSWER BOX: RESULTS BOX:
2 27 - 256 ÷ 43 ANSWER BOX: RESULTS BOX:
3 9 x (5 + 3)2 - 144 ANSWER BOX: RESULTS BOX:
4 7 + 3 x 24 ÷ 6 ANSWER BOX: RESULTS BOX:
5 A carpenter charges \$10 per square foot to lay a floor. If a square-shaped hallway is 6 feet along one side, and the customer has a coupon for \$25 off the total, then how much will the floor cost? ANSWER BOX: \$ RESULTS BOX: |
# Lesson Plan of Subtraction up to 100
## Students` Learning Outcomes
• Subtract numbers up to 100 using mental calculation strategies.
### Information for Teachers
• Mental calculation is a skill that not only helps us to become better at computation but it also enhances the development of number concepts.
• Using mental calculation, we subtract the numbers by some shortcut methods without solving it on paper.
• While teaching the lesson, also consult textbook where and when applicable.
### Material / Resources
Writing board, chalk/marker, duster, Mathematics textbook
### Introduction
• Write on the board:
• Ask the students` can you solve it mentally?
• Collect their feedback and tell them that we have a simple method to solve it mentally.
• First break up the numbers into ‘tens’ and ‘ones’.
• Ask any student to tell how many ‘tens’ and ‘ones’ in 25 and 12.
25 = 2 tens + 5 ones
12 =- 1 ten + 2 ones
• Tell the students to subtract ones from ones and tens from tens.
2 tens – 1 ten = 1 ten
5 ones – 2 ones = 3 ones
• In the end write the two answers as:
1 ten 3 ones = 13
• The required answer is 13.
• Tell the students that let us practice some more questions.
### Development
Activity 1
• Write on the board:
• Ask the students to solve this question mentally.
• Collect their feedback and ask any student to write the strategy on the board like this.
58 = 5 tens + 8 ones
26 = 2 tens + 6 ones
• Subtract tens from tens i.e. 5 – 2 = 2
• Subtract ones from ones i.e. 8 – 6 = 2
• Now first write ‘tens’ and then ‘ones’ (from left to right) i.e. 32
• Repeat this activity for some more numbers by involving maximum students.
Activity 2
• Write 76 – 27 on the board.
• Tell the students that let us solve this question mentally by another easier method.
• First make the subtrahend i.e. 27 to the closest tens i.e. 10, 20 or 30.
• We observe that 27 is closer to 30, so make 27 to 30.
• Add 3 to 27 to make it 30.
• Now subtract the new numbers: 79 – 30 = 49
• The required answer is 49.
• Repeat the procedure to solve the question.
Activity 3
• Write the following questions on the board:
• Ask the students to solve these questions using mental calculation.
• Write the correct answers on the board.
• Also repeat the strategy to solve these questions.
### Sum up / Conclusion
• First break up the numbers into ‘tens’ and ‘ones’. .then subtracts tens from tens and ones from ones.
• In the end, first write tens and then ones (from left to right).
### Assessment
• Write the questions on the board:
• Ask the students to solve the questions mentally.
• Collect their answers and appreciate them. |
# Cbse Class 5 Maths Large Numbers Worksheet
A Realistic Amounts Worksheet might help your child become a little more informed about the ideas behind this proportion of integers. Within this worksheet, pupils can fix 12 different difficulties associated with rational expression. They will likely learn how to multiply two or more amounts, group of people them in pairs, and figure out their products and services. They are going to also training simplifying logical expressions. When they have enhanced these principles, this worksheet might be a important instrument for advancing their scientific studies. Cbse Class 5 Maths Large Numbers Worksheet.
## Logical Amounts can be a ratio of integers
The two main kinds of figures: irrational and rational. Realistic amounts are understood to be whole figures, whereas irrational phone numbers will not recurring, and possess an unlimited amount of numbers. Irrational phone numbers are low-zero, low-terminating decimals, and rectangular origins which are not ideal squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To define a reasonable amount, you need to realize what a reasonable quantity is. An integer is actually a complete amount, as well as a realistic variety is a rate of two integers. The percentage of two integers is definitely the quantity ahead divided up through the quantity on the bottom. For example, if two integers are two and five, this would be an integer. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be produced in to a fraction
A realistic amount includes a numerator and denominator which are not absolutely nothing. Consequently they are often conveyed being a small fraction. In addition to their integer numerators and denominators, logical phone numbers can furthermore have a negative value. The unfavorable worth ought to be positioned on the left of and its definite importance is its extended distance from absolutely nothing. To make simpler this case in point, we will state that .0333333 can be a small percentage that could be published being a 1/3.
Along with negative integers, a rational quantity can be produced in a small fraction. By way of example, /18,572 is actually a rational number, although -1/ will not be. Any portion comprised of integers is reasonable, provided that the denominator will not consist of a and can be created for an integer. Likewise, a decimal that leads to a stage is yet another reasonable number.
## They are sense
Despite their name, rational amounts don’t make a lot perception. In mathematics, they can be individual entities by using a exclusive span about the number line. Because of this if we add up some thing, we could buy the shape by its rate to its initial number. This holds correct even if there are actually infinite reasonable figures among two distinct numbers. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To find the time period of a pearl, as an example, we might count its width. An individual pearl is ten kgs, that is a reasonable number. Additionally, a pound’s body weight means twenty kilograms. Therefore, we will be able to split a lb by ten, without having be worried about the length of a single pearl.
## They can be expressed like a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal amount might be created as being a multiple of two integers, so 4x 5 is equal to 8-10. A similar dilemma involves the repeated fraction 2/1, and both sides should be divided by 99 to obtain the correct solution. But how will you have the conversion? Here are some good examples.
A logical quantity will also be developed in great shape, which includes fractions plus a decimal. A good way to signify a rational variety within a decimal would be to separate it into its fractional comparable. There are three ways to break down a realistic number, and every one of these approaches yields its decimal equivalent. One of these techniques is to divide it into its fractional equal, and that’s what’s referred to as a terminating decimal. |
Question of Algebraic expression examples with answers
# Question If a + b + c = 11 and ab + bc + ca = 36, find the value of (a2 + b2 + c2).
Factorize x2+x-6
Solution:
Explanation:
On factoring the equation, we get,
x2+x-6
x2+3x-2x-6
x(x+3)-2(x+3)
(x+3)(x-2)
The simplified form of the equation x2+x-6 is (x+3)(x-2).
Factorize x^2-2x-8
Solution:
Explanation:
We have;
x2-2x-8
x2-4x+2x-8
x(x-4)+2(x-4)
(x-4) (x+2)
x=4,x=-2
Hence, we factorized the given expression as, x=4,x=-2
The opposite angles of a parallelogram are
are (3x - 2) and (x + 48) Find the measure of each angle of the parallelogram.
Solution:
Explanation:
Let the parallelogram be ABCD & the angles of parallelogram be <A,<B,<C & <D
From the given question, opposite angles of a parallelogram are (3x-2) & (x+48).
Let <A=3x-2 & <B=x+48
As we know that, “the opposite angles of a parallelogram are always equal”. Therefore, we can write;
(3x-2)=(x+48)
⇒3x-x=48+2
⇒2x=50
⇒x=50/2
⇒x=25
Substituting the value of x in 3x-2, we get;
3(25)-2
=75-2
=73º
⇒<A=<C=73
Finding the measure of other two angles:
We know that, the sum of adjacent angles of a parallelogram is equal to 180.
∴ <A+<B=180º
⇒73º+<B=180º
⇒<B=180º-73º
⇒<B=107º
∴ <D=107º
Therefore, the measure of each angle of the parallelogram is <A=73º,<B=107º,<C=73º,<D=107º.
Hence, we measured all the angles of parallelogram as;<A=73º,<B=107º,<C=73º,<D=107º.
In the given figure the value of x is
Solution:
Explanation:-
The value of x is 125º.
Estimate the value of square root
square root :√22
Solution:
Final Answer: The estimated square root of √22 is 4.690. |
# How do you simplify 2(1/10*2 1/4- 3/5) using PEMDAS?
Aug 14, 2016
$- \frac{3}{4}$
#### Explanation:
This is all one term, but there are brackets which we need to simplify first. There are 2 terms inside the bracket.
$2 \left(\textcolor{b l u e}{\frac{1}{10} \cdot 2 \frac{1}{4}} \textcolor{red}{- \frac{3}{5}}\right)$
=$2 \left(\textcolor{b l u e}{\frac{1}{10} \times \frac{9}{4}} \textcolor{red}{- \frac{3}{5}}\right)$
=$2 \left(\textcolor{b l u e}{\frac{9}{40}} \textcolor{red}{- \frac{3}{5}}\right)$
=$2 \times \left(\frac{9 - 24}{40}\right)$
=$2 \times \frac{- 15}{40}$
=$\left(- \frac{15}{20}\right)$
=$- \frac{3}{4}$ |
# 1.3 Intersection of sets
Page 1 / 1
We have pointed out that a set representing a real situation is not an isolated collection. Sets, in general, overlaps with each other. It is primarily because a set is defined on few characteristics, whereas elements generally can possess many characteristics. Unlike union, which includes all elements from two sets, the intersection between two sets includes only common elements.
Intersection of two sets
The intersection of sets “A” and “B” is the set of all elements common to both “A” and “B”.
The use of word “and” between two sets in defining an intersection is quite significant. Compare it with the definition of union. We used the word “or” between two sets. Pondering on these two words, while deciding membership of union or intersection, is helpful in application situation.
The intersection operation is denoted by the symbol, " $\cap$ ". We can write intersection in set builder form as :
$A\cap B=\left\{x:\phantom{\rule{1em}{0ex}}x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in B\right\}$
Again note use of the word “and” in set builder qualification. We can read this as “x” is an element, which belongs to set “A” and set “B”. Hence, it means that “x” belongs to both “A” and “B”.
In order to understand the operation, let us consider the earlier example again,
$A=\left\{1,2,3,4,5,6\right\}$
$B=\left\{4,5,6,7,8\right\}$
Then,
$A\cap B=\left\{4,5,6\right\}$
On Venn diagram, an intersection is the region intersected by circles, which represent two sets.
## Interpretation of intersection set
Let us examine the defining set of intersection :
$A\cap B=\left\{x:\phantom{\rule{1em}{0ex}}x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in B\right\}$
We consider an arbitrary element, say “x”, of the intersection set. Then, we interpret the conditional meaning as :
$If\phantom{\rule{1em}{0ex}}x\in A\cap B⇒x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in B.$
The conditional statement is true in opposite direction as well. Hence,
$Ifx\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in B⇒x\in A\cap B.$
We summarize two statements with two ways arrow as :
$x\in A\cap B⇔x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in B$
In addition to two ways relation, there is an interesting aspect of intersection. Intersection is subset of either of two sets. From Venn diagram, it is clear that :
$\left(A\cap B\right)\subset A$
and
$\left(A\cap B\right)\subset B$
## Intersection with a subset
Since all elements of a subset is present in the set, it emerges that intersection with subset is subset. Hence, if “A” is subset of set “B”, then :
$B\cap A=A$
## Intersection of disjoint sets
If no element is common to two sets “A” and “B” , then the resulting intersection is an empty set :
$A\cap B=\phi$
In that case, two sets “A” and “B” are “disjoint” sets.
## Multiple intersections
If ${A}_{1},{A}_{2},{A}_{3},\dots \dots \dots ,{A}_{n}$ is a finite family of sets, then their intersections one after another is denoted as :
${A}_{1}\cap {A}_{2}\cap {A}_{3}\cap \dots \dots .\cap An$
## Important results
In this section we shall discuss some of the important characteristics/ deductions for the intersection operation.
## Idempotent law
The intersection of a set with itself is the set itself.
$A\cap A=A$
This is because intersection is a set of common elements. Here, all elements of a set is common with itself. The resulting intersection, therefore, is set itself.
## Identity law
The intersection with universal set yields the set itself. Hence, universal set functions as the identity of the intersection operator.
$A\cap U=A$
It is easy to interpret this law. Only the elements in "A" are common to universal set. Hence, intersection, being the set of common elements, is set "A".
## Law of empty set
Since empty set is element of all other sets, it emerges that intersection of an empty set with any set is an empty set (empty set is only common element between two sets).
$\phi \cap A=\phi$
## Commutative law
The order of sets around intersection operator does not change the intersection. Hence, commutative property holds in the case of intersection operation.
$A\cap B=B\cap A$
## Associative law
The associative property holds with respect to intersection operator.
$\left(A\cap B\right)\cap C=A\cap \left(B\cap C\right)$
The intersection of sets “A” and “B” on Venn’s diagram is :
In turn, the intersection of set “A $\cap$ B” and set “C” is the small region in the center :
It is easy to visualize that the ultimate intersection is independent of the sequence of operation.
## Distributive law
The intersection operator( $\cap$ ) is distributed over union operator ( $\cup$ ) :
$A\cap \left(B\cup C\right)=\left(A\cap B\right)\cup \left(A\cap C\right)$
We can check out this relation with the help of Venn diagram. For convenience, we have not shown the universal set. In the first diagram on the left, the colored region shows the union of sets “B” and “C” ie. $B\cup C$ . The colored region in the second diagram on the right shows the intersection of set “A” with the union obtained in the first diagram i.e. $B\cup C$ .
We can now interpret the colored region in the second diagram from the point of view of expression on the right hand side of the equation :
$A\cap \left(B\cup C\right)=\left(A\cap B\right)\cup \left(A\cap C\right)$
The colored region is indeed the union of two intersections : " $A\cup B$ " and " $A\cup C$ " . Thus, we conclude that distributive property holds for "intersection operator over union operator".
In the same manner, we can prove distribution of “union operator over intersection operator” :
$A\cup \left(B\cap C\right)=\left(A\cup B\right)\cap \left(A\cup C\right)$
## Analytical proof
Distributive properties are important and used for practical application. In this section, we shall prove the same in analytical manner. For this, let us consider an arbitrary element “x”, which belongs to set " $A\cap \left(B\cup C\right)$ " :
$x\in A\cap \left(B\cup C\right)$
Then, by definition of intersection :
$⇒x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in \left(B\cup C\right)$
$⇒x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}\left(x\in B\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}x\in C\right)$
$⇒\left(x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in B\right)\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}\left(x\in A\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}x\in C\right)$
$⇒\left(x\in A\cap B\right)\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}\left(x\in A\cap C\right)$
$⇒x\in \left(A\cap B\right)\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}\left(A\cap C\right)$
$⇒x\in \left(A\cap B\right)\cup \left(A\cap C\right)$
But, we had started with " $A\cap \left(B\cup C\right)$ " and used its definition to show that “x” belongs to another set. It means that the other set consists of the elements of the first set – at the least. Thus,
$⇒A\cap \left(B\cup C\right)\subset \left(A\cap B\right)\cup \left(A\cap C\right)$
Similarly, we can start with " $\left(A\cap B\right)\cup \left(A\cap C\right)$ " and reach the conclusion that :
$⇒\left(A\cap B\right)\cup \left(A\cap C\right)\subset A\cap \left(B\cup C\right)$
If sets are subsets of each other, then they are equal. Hence,
$⇒A\cap \left(B\cup C\right)=\left(A\cap B\right)\cup \left(A\cap C\right)$
Proceeding in the same manner, we can also prove other distributive property of “union operator over intersection operator” :
$A\cup \left(B\cap C\right)=\left(A\cup B\right)\cap \left(A\cup C\right)$
#### Questions & Answers
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
Ramkumar Reply
Berger describes sociologists as concerned with
Mueller Reply
What is power set
Satyabrata Reply
Period of sin^6 3x+ cos^6 3x
Sneha Reply
Period of sin^6 3x+ cos^6 3x
Sneha Reply
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Instruction
1
In order to divide common fraction to common fraction, multiply the first fraction by the inverted second fraction. This "inverted" ordinary fraction, where the numerator and denominator are reversed is called reverse.
When dividing fractions it is necessary to pay attention to the fact that the second fraction is not equal to zero. Sometimes, if the fraction is quite bulky, it is extremely difficult to make. In addition, the second fraction can contain some variables (unknown) value, which for certain values of the draw roll to zero. You also need to pay attention to those cases where the denominator of the second fraction vanishes. When the action variables for all these cases, you must specify in the final answer.
For example: see Fig. 1
2
To divide a mixed fraction by a mixed, a mixed fraction to an ordinary or common in mixed, get mixed fractions to an ordinary mind. Then to make the division as specified in step 1.
To transfer mixed fraction to an ordinary need the integer part of mixed fraction multiplied by its denominator and add the resulting product to the numerator.
Example: see Fig. 2
3
When dividing decimal fractions to ordinary (mixed) or division of ordinary (mixed) fraction to a decimal, all fractions are reduced to the ordinary mind. After this division is done according to step 1. To translate decimals to fractions, the "throw" of the decimal point and recorded in the numerator and in the denominator write one and as many zeros as digits stood to the right of the decimal point.
Example: see Fig. 3
4
To divide two decimals need in divisible and the divider to move the decimal point that many digits to the right of the second fraction to make a whole number and divide the resulting number.
For example: 24,68/123,4=246,8/1234=0,2.
If the divisible to transfer the decimal point in the "missing" digits, missing digits are replaced with zeros.
Example: 24,68/1,234=24680/1234=20 |
Chapter 11: TIME AND WORK
Introduction
If A can do a work in 'n' days then he does 1/n work daily. If A is thrice as good as B at something then the ratio of their work done is 1:3.
• If A does a work in a days, then in one day A does $$\frac{1}{a}$$ of the work. If B does a work in b days, then in one day B does $$\frac{1}{b}$$ of the work. Then, in one day, if A and B work together, then their combined work is $$\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$$
• For example, if A can do a work in 10 days and B can do the same work in 12 days, then the work will be completed in how many days. One day’s work = 1/10 + 1/12 = (12 + 10)/120. Then the number of days required to complete the work is 120/22.
• Instead of taking the value of the total work as 1 unit of work, we can also look at the total work as 100 per cent work. In such a case, the following rule applies: If A does a work in 'a' days, then in one day A does $$\frac{100}{a}$$% of the work. If B does a work in 'b' days, then in one day B does $$\frac{100}{b}$$% of the work. Then, in one day, if A and B work together, then their combined work is $$\frac{100}{a} + \frac{100}{b}$$.
Q. If A can do a work in 10 days and B can do the same work in 12 days, then the work will be completed in how many days
• If A can do a work in 10 days (so means 10% work) and B can do the same work in 12 days (so 8.33% work so 18.33% work in a day in 5 days 91.66% work so leaves 8.33% work to be done so which can be done in 8.33/18.33 of a day = 5/11 of a day (since both the numerator and the denominator are divisible by 1.66), then the work will be completed in 5 5/11 days.
The Concept of Negative Work
• Suppose, that A and B are working to build a wall while C is working to break the wall. In such a case, the wall is being built by A and B while it is being broken by C. Here, if we consider the work as the building of the wall, we can say that C is doing negative work
• A can build a wall in 10 days and B can build it in 5 days, while C can completely destroy the wall in 20 days. If they start working at the same time, in how many days will the work be completed
• The net combined work per day here is: A’s work + B’s work – C’s work = 10% + 20% – 5% = 25% work in one day
• Hence, the work will get completed (100% work) in 4 days
WORK EQUIVALENCE METHOD : Work rate × Time = Work done (or work to be done)
• A contractor estimates that he will finish the road construction project in 100 days by employing 50 men. However, at the end of the 50th day, when as per his estimation half the work should have been completed, he finds that only 40% of his work is done. (a) How many more days will be required to complete the work? (b) How many more men should he employ in order to complete the work in time?
• The contactor has completed 40% of the work in 50 days. If the number of men working on the project remains constant, the rate of work also remains constant. Hence, to complete 100% work, he will have to complete the remaining 60% of the work. For this he would require 75 more days
• In order to complete the work on time, it is obvious that he will have to increase the number of men working on the project
• 50 men working for 50 days then 50 × 50 = 2500 man-days
• 2500 man-days has resulted in 40% work completion. Hence, the total work to be done in terms of the number of man-days is got by using unitary method: Work left = 60% = 2500 × 1.5 = 3750 man-days
• This has to be completed in 50 days. Hence, the number of men required per day is 3750/50 = 75 men. Since, 50 men are already working on the project, the contractor needs to hire 25 more men
Work as volume of work
• In certain cases, the unit of work can also be considered to be in terms of the volume of work. For example, building of a wall of a certain length, breadth and height.
• In such cases, the following formula applies:
• $$\frac{L_1 * B_1 * H_1}{L_2 * B_2 * H_2} = \frac{m_1 * t_1 * d_1}{m_2 * t_2 * d_2}$$
• where L, B and H are respectively the length, breadth and height of the wall to be built, while m, t and d are respectively the number of men, the amount of time per day and the number of days. Further, the suffix 1 is for the first work situation, while the suffix 2 is for the second work situation
Q. 20 men working 8 hours a day can completely build a wall of length 200 meters, breadth 10 metres and height 20 metres in 10 days. How many days will 25 men working 12 hours a day require to build a wall of length 400 meters, breadth 10 metres and height of 15 metres.
• $$\frac{L_1 * B_1 * H_1}{L_2 * B_2 * H_2} = \frac{m_1 * t_1 * d_1}{m_2 * t_2 * d_2}$$
• Then we get (200 × 10 × 20)/(400 × 10 × 15) = (20 × 8 × 10)/(25 × 12 × d2)
• d2 = 8 days
Equating Men, Women and Children This is directly derived from the concept of efficiencies
• 8 men can do a work in 12 days while 20 women can do it in 10 days. In how many days can 12 men and 15 women complete the same work.
• Total work to be done = 8 × 12 = 96 man-days or total work to be done = 20 × 10 = 200 woman-days
• Since, the work is the same, we can equate 96 man-days = 200 woman-days. Hence, 1 man-day = 2.08333 woman-days
• Now, if 12 men and 15 women are working on the work we get 12 men are equal to 12 × 2.08333 = 25 women
• Hence, the work done per day is equivalent to 25 + 15 women working per day
• That is, 40 women working per day
• Hence, 40 × no. of days = 200 woman days. Number of days = 5 days
Ans .
200
1. Explanation :
He will complete the work in 20 days. Hence, he will complete ten times the work in 200 days.
Ans .
4
1. Explanation :
6 men for 12 days means 72 mandays. This would be equal to 4 men for 18 days
Ans .
9 4/7
1. Explanation :
A’s one day work will be 5%, while B will do 6.66 % of the work in one day. Hence, their total
work will be 11.66% in a day.
In 8 days they will complete Æ 11.66 × 8 = 93.33%
This will leave 6.66% of the work. This will correspond to 4/7 of the ninth day since in
6.66/11.66 both the numerator and the denominator are divisible by 1.66.
Ans .
6 18/47
1. Explanation :
A’s work = 5% per day
B’s work = 6.66% per day
C’s work = 4% per day.
Total no. of days = 100/15.66 = 300/47 = 6(18/47)
Ans .
60
1. Explanation :
N + A = 10%
N = 8.33%
Hence A = 1.66% = 60 days.
Ans .
30
1. Explanation :
The ratio of the wages will be the inverse of the ratio of the number of days required by each to do he work. Hence, the correct answer will be 3:2 = 30
Ans .
22 2/7
1. Explanation :
24 man days + 18 women days = 20 man days + 28 woman days
= 4 man days = 10 woman days.
= 1 man day = 2.5 woman days
Total work = 24 man days + 18 woman days = 60 woman days + 18 woman days = 78 woman
days.
Hence, 1 man + 1 woman = 3.5 women can do it in 78/3.5 = 156/7 = 22(2/7) days.
Ans .
cant say
1. Explanation :
The data is insufficient, since we only know that the work gets completed in 200 boy days and 300
women days.
Ans .
20
1. Explanation :
A = 10%, B = 5% and Combined work is 20%. Hence, C’s work is 5% and will require 20 day
Ans .
8
1. Explanation :
In 5 days, A would do 25% of the work. Since, B finishes the remaining 75% work in 10 days, we
can conclude that B’s work in a day = 7.5%
Thus, (A + B) = 12.5% per day.
Together they would take 100/12.5 = 8 days
Ans .
40
1. Explanation :
A = 20%, B = 10% and A + B + C = 50%. Hence, C = 20%. Thus, in two days, C contributes 40%
of the total work and should be paid 40% of the total amount
Ans .
15
1. Explanation :
Total man days required = 600 man days. If 5 workers leave the job after ‘n’ days, the total work
would be done in 35 days. We have to find the value of ‘n’ to satisfy:
20 × n + (35 – n) × 15 = 600.
Solving for n, we get
20n – 15n + 35 × 15 = 600
5n = 75
n = 15.
Ans .
10.5
1. Explanation :
Let the time taken by Arun be ‘t’ days. Then, time taken by Vinay = 2t days.
1/t + 1/2t = 1/7 = t = 10.5
Ans .
12
1. Explanation :
Subhash can copy 200 pages in 40 hours (reaction to the first sentence). Hence, Prakash can copy100 pages in 40 hours. Thus, he can copy 30 pages in 30% of the time: i.e. 12 hours
Ans .
12
1. Explanation :
30X = 20 (X + 6) = 10X = 120 = X = 12
Ans .
11
1. Explanation :
Sashi = 4%, Rishi = 5%. In five days, they do a total of 45% work. Rishi will finish the remaining
55% work in 11 more days.
Ans .
7
1. Explanation :
Raju = 10%, Vicky = 8.33% and Tinku = 6.66%. Hence, total work for a day if all three work =
25%.
In 2 days they will complete, 50% work. On the third day onwards Raju doesn’t work. The rate of
work will become 15%. Also, since Vicky leaves 3 days before the actual completion of the
work, Tinku works alone for the last 3 days (and must have done the last 6.66 × 3 = 20% work
alone). This would mean that Vicky leaves after 80% work is done. Thus, Vicky and Tinku must be
doing 30% work together over two days.
Hence, total time required = 2 days (all three) + 2 days (Vicky and Tinku) + 3 days (Tinku alone)
Ans .
144/17
1. Explanation :
Sambhu requires 16 days to do the work while Kalu requires 18 days to do the work.
(1/16 + 1/18) × n = 1
so n = 288/34 = 144/17
Ans .
6
1. Explanation :
Let Anjay take 3t days, Vijay take 2t days and Manoj take 6t days in order to complete the work. Then we get:
1/3t + 1/2t + 1/6t = 1 so t = 1. Thus, Manoj would take 6t = 6 days to complete the work
Ans .
68
1. Explanation :
After 100 days and 4500 mandays, only 1/6
th of the work has been completed. You can use the
product change algorithm of PCG to solve this question.
100 × 45 = 16.66% of the work. After this you have 200 days (i.e. 100% increase in the time
available) while the product 200 × no. of men should correspond to five times times the original
product.
This will be got by increasing the no. of men by 150% (300/200).
Ans .
4.8
1. Explanation :
Since the ratio of money given to Apurva and Amit is 2:3, their work done would also be in the
same ratio. Thus, their time ratio would be 3:2 (inverse of 2:3). So, if Apurva takes 12 days, Amit
would take 8 days and the total number of days required (t) would be given by the equation:
(1/12 + 1/8)t = 1 so t = 24/5 = 4.8 days
Ans .
42
1. Explanation :
Raju being twice as good a workman as Vijay, you can solve the following equation to get the
1/R + 1/2R = 1/14.
Solving will give you that Vijay takes 42 days
Ans .
15
1. Explanation :
40n = 30 (n + 5) so n = 15
Ans .
2 : 1
1. Explanation :
12 × 5 man days + 16 × 5 Boy days
= 13 × 4 man days + 24 × 4 Boy days
so 8 man days = 16 Boy days
1 man day = 2 Boy days.
Required ratio of man’s work to boy’s work = 2 : 1
Ans .
5
1. Explanation :
A’s rate of working is 10 per cent per day while B’s rate of working is 5 per cent per day. In 5
days they will complete 75 per cent work. Thus the last 25 per cent would be done by B alone.
Working at the rate of 5 per cent per day, B would do the work in 5 days.
Ans .
20
1. Explanation :
Work equivalence method:
30 × 5 × 16 = 20 × 6 × n
Gives the value of n as 20 days
Ans .
345
1. Explanation :
A + V + S = 1
A + V = 19/23
V + S = 8/23
Æ A + 2V + S = 27/23
(2)–(1) gives us: V = 4/23.
Ans .
22.5
1. Explanation :
Interpret the starting statement as: Anmol takes 30 days and Vinay takes 90 days. Hence, the answer will be got by:
(1/30 + 1/90) * n = 1
Alternatively, you can also solve using percentages as: 3.33 + 1.11 = 4.44% is the daily work.
Hence, the no. of days required is 100/4.44 = 22.5 days
Ans .
55
1. Explanation :
Total work = 15 × 210 = 3150 mandays.
After 100 days, work done = 15 × 100 = 1500 mandays.
Work left = 3150 – 1500 = 1650 mandays.
This work has to be done with 30 men working each day.
The number of days (more) required = 1650/30 = 55 days
Ans .
20
1. Explanation :
Ajay’s daily work = 4.1666%, Vijay’s daily work = 3.33% and the daily work of all the three together is 8.33%. Hence, Pradeep’s daily work will be 0.8333%. Hence, he will end up doing
10% of the total work in 12 days. This will mean that he will be paid Rs. 20.
Ans .
6
1. Explanation :
After 27 days, food left = 4 × 200 = 800 soldier days worth of food. Since, now there are only 80 soldiers, this food would last for 800/80 = 10 days. Number of extra days for which the food lasts
= 10 – 4 = 6 days.
Ans .
40
1. Explanation :
Total work of Anju, Manju and Sanju = 16.66%
Anju’s work = 10%
Manju’s work = 4.166%
Sanju’s work = 2.5%
So Sanju can reap the field in 40 days.
Ans .
84
1. Explanation :
Ajay + Vijay = 1/28 and Ajay + Vijay + Manoj = 1/21.
Hence, Manoj = 1/21 – 1/28 = 1/84.
Hence, Manoj will take 84 days to do the work.
Ans .
20
1. Explanation :
A + M = 8.33, M + B = 6.66 and A = 2B a
so A’s 1 days work = 3.33%, M’s = 5% and B’s = 1.66%.
Thus, Mohan would require 100/5 = 20 days to complete the work if he works alone
Ans .
15
1. Explanation :
A + V = 16.66% and A = 10% so V = 6.66%. Consequently Vijay would require 100/6.66 = 15 days to do it alone
Ans .
30
1. Explanation :
The rate of filling will be 20% and the net rate of filling (including the leak) is 16.66%. Hence, the leak accounts for 3.33% per hour. i.e. it will take 30 hours to empty the tank
Ans .
10 min, 15 min
1. Explanation :
A + B = 16.66%. From here solve this one using the options. Option (c) fits the situation as it gives us A’s work = 10%, B’s work = 6.66% as also that B takes 5 minutes more than A (as stipulated in
the problem).
Ans .
4
1. Explanation :
A + B = 5.55 + 11.11 = 16.66. In two days, 33.33% of the work will be done. C adds 16.66% of
work to that of A and B. Hence, the rate of working will go to 33.33%. At this rate it would take 2
more days to complete the work.
Hence, in total it will take 4 days to complete the entire work.
Ans .
32
1. Explanation :
24 × 8 × 10 = N × 10 × 6 so N = 32
Ans .
32
1. Explanation :
n × 20 = (n – 12) × 32 so n = 32.
Ans .
9
1. Explanation :
12 × 18 = 12 × 6 + 16 × t so t = 9
Ans .
15
1. Explanation :
(A + B)’s work = C’s work.
Also if A takes ‘a’ days
B would take ‘a – 5’ days
and C would take ‘a – 9’ days.
Solving through options, option ‘c’ fits.
A (15 days) so A’s work = 6.66%
B (10 days) so B’s work = 10%
C (6 days) so C’s work = 16.66%
Ans .
15
1. Explanation :
The cistern fills in 6 hours normally, means that the rate of filling is 16.66% per hour. With the
leak in the bottom, the rate of filling becomes 10% per hour (as it takes 10 hours to fill with the
leak).
This means that the leak drains out water at the rate of 6.66% per hour. This in turn means that the
leak would take 100/6.66 = 15 hours to drain out the entire cistern.
Ans .
Waste pipe emptying the tank is 5 h
1. Explanation :
Since the net work of the three taps is 10% and the first and second do 20% + 10% = 30%. Hence,
the third pipe must be a waste pipe emptying at the rate of 20% per hour. Hence, the waste pipe
will take a total of 5 hours to empty the tank
Ans .
15
1. Explanation :
A’s work = 10%
B’s negative work = 6.66%
(A + B)’s work = 3.33%
To fill a half empty tank, they would take 50/3.33 = 15 hours.
Ans .
16.5
1. Explanation :
The work rate would be 10% on the first day, 5% on the second day and 2.5% on the third day.
For every block of 3 days there would be 17.5% work done. In 15 days, the work completed
would be 17.5 × 5 = 87.5%. On the sixteenth day, work done = 10% Æ 2.5% work would be left
after 16 days. On the 17th day the rate of work would be 5% and hence it would take half of the 17th day to complete the work. Thus, it would take 16.5 days to finish the work in this fashion
Ans .
108
1. Explanation :
(A + B) = 2C.
Also,(A + C) = 3B
36(A + B + C) = 1
Solving for C, we get:
36 (2C + C) = 1 so 108 C = 1
C = 1/108
Hence, C takes 108 days
Ans .
72
1. Explanation :
A + B + C = 19%. In the first two hours they will do 38 % of the work. Further, for the next two
hours work will be done at the rate of 15% per hour. Hence, after 4 hours 68% of the work will
be completed, when tap B is also closed. The last 32% of the work will be done by A alone.
Hence, A does 40% (first 4 days) + 32% = 72% of the work.
Ans .
20
1. Explanation :
Without the leak:
Rate of work = 20% + 5% = 25%. Thus, it would have taken 4 hours to complete the work.
Due to the leak the filling gets delayed by 1 hour. Thus, the tank gets filled in 5 hours. This means
that the effective rate of filling would be 20% per hour. This means that the rate at which the leak
empties the tank is 5% per hour and hence it would have taken 20 hours to empty a filled tank
Ans .
22
1. Explanation :
In 6 days A would do 25% of the work and in 8 days B would do 25% of the work himself. So, C
has to complete 50% of the work by himself.
In all C would require 30 days to do 50% of the work. So, he would require 22 more days
Q. Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same Job.How long should it take both A and B, working together but independently, to do the same job?
A. A does 1/8 work in an hour and B does 1/10 so both do (1/8 + 1/10) in one hour. 18/80 work in an hour so they need 80/18 to do the work completely.
Q. A and B together can complete a piece of work in 4 days. If A alone can complete the same work in 12 days, in how many days can B alone complete that work?
A. A can do 1/12 in a day, B can do 1/x. (1/12 + 1/x) = 1/4 from given data. Solving it we get value of x.
Q. A can do a piece of work in 7 days of 9 hours each and B can do it in 6 days of 7 hours each. How long will they take to do it, working together 8 hours a day?
A. A can do work in 63 hrs so 1/63 per hour. B can do work in 42 hrs so 1/42 per hour. Both can do work in (1/63 + 1/42) = (2+3/126) = 5/126 work in 1 hr so they need 126/5 hrs or 25.2 hrs so if 8 hrs / day are taken then 4 days.
Q. A and B can do a piece of work in 18 days; B and C can do it in 24 days A and C can do it in 36 days. In how many days will A, Band C finish it together?
A. A and B can do 1/18 work per day; B and C can do 1/24 work per day; A and C can do 1/36 work per day.
Adding all three we get 2(A+B+C) = (4+3+2)/72 = 1/8 work per day or 8 days to do work together.
Q. A can do a certain job in 12 days. B is 60% more efficient than A. How many days does B alone take to do the same job?
A. B is 60% more efficient than A so if B takes 100 hrs then A needs 160 hrs so time taken ratio of A : B = 160 : 100 = 8 : 5.
B's time = A's time * 5 / 8 = 12 * 5 / 8 = 7 1/2 days.
Q. A can do a piece of work in 80 days. He works at it for 10 days B alone finishes the remaining work in 42 days. In how much time will A and B working together, finish the work?
A. A does 1 / 80 work in 1 day and so 10/80 work in 10 days. 70/80 work is left.
B does 70/80 work in 42 days so in 1 day he does 70 /42 * 80 = 1 / 48 work.
A and B together do (1/80 + 1/48) in one day. Reciprocal of result will give total days needed.
Q. A and B undertake to do a piece of work for Rs. 600. A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they finish it in 3 days. find the share of each.
A. A, B, C do (1/6 + 1/8 + 1/x) = 1/3 so 1/x = 1/3 - 1/6 - 1/8
= (8 - 4 - 3) / 24
= 1/24 work per day so C can finish job in 24 days.
So ratio of 1 days work is = 1/6 : 1/8 : 1/24 = 4:3:1 . This can be used to get their share of the money.
Q. A and B working separately can do a piece of work in 9 and 12 days respectively, If they work for a day alternately, A beginning, in how many days, the work will be completed?
A. A work in 1 day is 1/9 and B's work is 1/12 so since they work alternately in 2 days work by them is (1/9+1/12) = (4+3)/36 = 7/36.
Work done in 10 days is = 7/36 * 5 = 35/36.
On last day A's turn and 1/36 work remains which A can do in 1/4 day. Total days needed are 10 1/4 days.
Pipes and Cisterns
If a pipe can fill a tank in 'x' hours then in 1 hour it can fill 1/x. If a pipe can empty a tank in y hours then in 1 hour it can empty 1/y.
If one pipe fills a tank in 'x' hrs and second pipe empties it in 'y' hrs then in 1 hour net part filled is 1/x - 1/y (x>y).
If one pipe fills a tank in 'x' hrs and second pipe empties it in 'y' hrs then in 1 hour net part filled is 1/y - 1/x (y>x).
Q. Two pipes can fill a tank in 10 hours and 12 hours respectively while a third, pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?
A. Total work by all pipes together = (1/10 + 1/12 - 1/20) = (6+5-3)/60 = 2/15
So it takes 7.5 hrs to fill tank.
Q. Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom it took 32 minutes more to fill the cistern.When the cistern is full, in what time will the leak empty it?
A.Pipe A and B can fill tank in (1/14 + 1/16) = (8+7)/112 = 15/112 work in 1 hr. So total time needed is 112/15 i.e. 7 hrs 28 mins. But when leak is there it took 32 mins more so 8 hrs.
15/122 - 1/x = 1/8 so we can get value of x.
Q. Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. A water pipe C can empty the tank in 30 min. First A and B are opened. after 7 min, C is also opened. In how much time, the tank is full?
A. A and B can together fill in 1 min = (1/36 + 1/45) = (5+4)/180 = 1/20 so in 7 mins 7/20 is filled. Remaining 13/20 will be filled by all three.
(1/36 + 1/45 - 1/30) = (5+4- 6)/180 = 3/180 = 1/60. So they need to fill 13/20 or 39/60. Which they can do in 39 mins as they do 1/60 in 1 min.
Q. Two pipes A,B can fill a tank in 24 min. and 32 min. respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 min.?
A. Suppose B is closed in 'x' min.
Part of tank filled in 'x' min by both + part of tank filled in (18-x) min by A = 1.
x * ( 1/24 + 1/32) + (18-x) * (1/24) = 1 Solving this we can get 'x'.
Quiz
Score more than 80% marks and move ahead else stay back and read again! |
# E^4x + 3e^-4x = 6 x=?
## e^4x + 3e^-4x = 6
May 1, 2018
$x = \frac{1}{4} \ln \left(3 \pm \sqrt{6}\right)$
${x}_{1} \approx 0.42$
${x}_{2} \approx - 0.15$
#### Explanation:
Given: ${e}^{4 x} + 3 {e}^{- 4 x} = 6$.
Let $u = {e}^{4 x}$.
Notice how,
${e}^{4 x} + 3 {e}^{- 4 x} = 6$
$\iff {e}^{4 x} + 3 {e}^{\left(4 x\right) \cdot - 1} = 6$
Therefore, replacing $4 x$ by $u$, we get:
$u + 3 {u}^{-} 1 = 6$
$u + \frac{3}{u} = 6$
Multiply by ${u}^{2}$ to get:
${u}^{2} + 3 = 6 u$
${u}^{2} - 6 u + 3 = 0$
Using the quadratic formula, we get:
$u = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 3}}{2}$
$= \frac{6 \pm \sqrt{36 - 12}}{2}$
$= \frac{6 \pm \sqrt{24}}{2}$
$= \frac{6 \pm 2 \sqrt{6}}{2}$
Replacing $u = {e}^{4 x}$ back, we get:
${e}^{4 x} = \frac{6 \pm 2 \sqrt{6}}{2}$
$= 3 \pm \sqrt{6}$
Take natural logs on both sides.
$\ln \left({e}^{4 x}\right) = \ln \left(3 \pm \sqrt{6}\right)$
$4 x \ln e = \ln \left(3 \pm \sqrt{6}\right)$
$4 x = \ln \left(3 \pm \sqrt{6}\right)$
$\therefore x = \frac{1}{4} \ln \left(3 \pm \sqrt{6}\right)$
May 1, 2018
Do you mean the following expression? ${e}^{4 x} + 3 {e}^{- 4 x} = 6$
#### Explanation:
It is important that the expression becomes right. But rather than do the whole calculus, I'll lead you on your way.
To solve such an expression, please note that ${e}^{4 x}$ is common to each term in the expression, so you can write:
$y = {e}^{4 x}$
That gives $y - 6 + 3 {y}^{- 1} = 0$
Get rid of y in the numerator place by multiplying each term with y. This gives the following 2nd degree equation:
${y}^{2} - 6 y + 3 = 0$
You solve this equation the normal way, which gives you two solutions
y_1=3+√6 (= 5.45)
y_2=3 -√6 (=0.55)
As $y = {e}^{4 x}$, you find that
$4 x = \ln \left(y\right)$, i.e.
${x}_{1} = \ln \frac{{y}_{1}}{4}$ = ln(3+√6) (=0.42)
${x}_{2} = \ln \frac{{y}_{2}}{4}$ = ln(3-√6) (=-0.15)
Check:
${e}^{4 {x}_{1}} + 3 {e}^{- 4 {x}_{1}} - 6$
=${e}^{1.70} + 3 {e}^{- 1.70} - 6 = 0$ -> check
${e}^{4 {x}_{2}} + 3 {e}^{- 4 {x}_{2}} - 6$
=${e}^{- 0.60} + 3 {e}^{0.60} - 6 = 0$ -> check |
# Common core subtraction
## Common core subtraction - counting up method of subtraction
Common core subtraction is using addition instead. Add up from the smaller number to the bigger number.
Example 1
32-12
12 + 3 = 15 15 + 5 = 20 20 + 10 = 30 30 + 2 = 32 20
To see this step by step
## Common core subtraction step by step
Example 1
32-12
12 Start with the lowest number
12 + 3 Add 3 because it's then easy to get to 15
12 + 3 =15 15 is the running total
12 + 3 =15 Put the running total on the next line 15
12 + 3 =15 Add 5 because it's then easy to get to 20 15 + 5
12 + 3 =15 20 is the running total 15 + 5 =20
12 + 3 =15 15 + 5 =20 20 Put the running total on the next line
12 + 3 =15 15 + 5 =20 20 + 10 Add 10 because it's then easy to get to 30
12 + 3 =15 15 + 5 =20 20 + 10 =30 30 is the running total
12 + 3 =15 15 + 5 =20 20 + 10 =30 30 Put the running total on the next line
12 + 3 =15 15 + 5 =20 20 + 10 =30 30 + 2 Add 2 to finally get 32
12 + 3 =15 15 + 5 =20 20 + 10 =30 30 + 2 =32 Check the final tally equals 32
+ 3 + 5 + 10 + 2 Now add 20
Answer: 32-12=20
Answer: 32-12=20
Example 2
43-13
13 + 2 = 15 15 + 5 = 20 20 + 10 = 30 30 + 10 = 40 40 + 3 = 43 30
To see this step by step
## Common core subtraction step by step
Example 2
43-13
13 Start with the lowest number
13 + 2 Add 2 because it's then easy to get to 15
13 + 2 =15 15 is the running total
13 + 2 =15 Put the running total on the next line 15
13 + 2 =15 Add 5 because it's then easy to get to 20 15 + 5
13 + 2 =15 20 is the running total 15 + 5 =20
13 + 2 =15 15 + 5 =20 20 Put the running total on the next line
13 + 2 =15 15 + 5 =20 20 + 10 Add 10 because it's then easy to get to 30
13 + 2 =15 15 + 5 =20 20 + 10 =30 30 is the running total
13 + 2 =15 15 + 5 =20 20 + 10 =30 30 + 10 Add 10 because it's then easy to get to 40
13 + 2 =15 15 + 5 =20 20 + 10 =30 30 + 10 =40 40 is the running total
13 + 2 =15 15 + 5 =20 20 + 10 =30 30 + 10 =40 40 Put the running total on the next line
13 + 2 =15 15 + 5 =20 20 + 10 =30 30 + 10 =40 40 + 3 Add 3 to finally get 43
13 + 2 =15 15 + 5 =20 20 + 10 =30 30 + 10 =40 40 + 3 =43 Check the final equals 43
+ 2 + 5 + 10 + 10 + 3 Now add 30
Answer: 43-13=30
Answer: 43-13=30
Example 3
325-38
38 + 2 = 40 40 + 60 = 100 100 + 200 = 300 300 + 25 = 325 287
To see this step by step
## Common core subtraction step by step
Example 3
325-38
38 Start with the lowest number
38 + 2 Add 2 because it's then easy to get to 40
38 + 2 =40 40 is the running total
38 + 2 =40 Put the running total on the next line 40
38 + 2 =40 Add 60 because it's then easy to get to 100 40 + 60
38 + 2 =40 100 is the running total 40 + 60 =100
38 + 2 =40 40 + 60 =100 100 Put the running total on the next line
38 + 2 =40 40 + 60 =100 100 + 200 Add 200 because it's then easy to get to 300
38 + 2 =40 40 + 60 =100 100 + 200 =300 300 is the running total
38 + 2 =40 40 + 60 =100 100 + 200 =300 Put the running total on the next line 300
38 + 2 =40 40 + 60 =100 100 + 200 =300 300 + 25 Add 25 to finally get to 325
38 + 2 =40 40 + 60 =100 100 + 200 =300 300 + 25 =325 Check the final equals 325
+ 2 + 60 + 200 + 25 Now add 287
Answer: 325-38=287
Answer: 325-38=287
Example 4
264-128
128 + 2 = 130 130 + 70 = 200 200 + 60 = 260 260 + 4 = 264 136
To see this step by step
## Common core subtraction step by step
Example 4
264-128
128 Start with the lowest number
128 + 2 Add 2 because it's then easy to get to 130
128 + 2 =130 130 is the running total
128 + 2 =130 Put the running total on the next line 130
128 + 2 =130 Add 70 because it's then easy to get to 200 130 + 70
128 + 2 =130 130 + 70 =200 200 is the running total
128 + 2 =130 130 + 70 =200 200 Put the running total on the next line
128 + 2 =130 130 + 70 =200 200 + 60 Add 60 because it's then easy to get to 260
128 + 2 =130 130 + 70 =200 200 + 60 =260 260 is the running total
128 + 2 =130 130 + 70 =200 200 + 60 =260 Put the running total on the next line 260
128 + 2 =130 130 + 70 =200 200 + 60 =260 260 + 4 Add 4 to finally get to 264
128 + 2 =130 130 + 70 =200 200 + 60 =260 260 + 4 =264 Check the final equals 264
+ 2 + 70 + 60 + 4 Now add 136
Answer: 264-128=136
Answer: 264-128=136
OR
128 + 100 = 228 228 + 30 = 258 258 + 2 = 260 260 + 4 = 264 136
## Common core subtraction step by step
Example 4
264-128
128 Start with the lowest number
128 + 100 Add 100 because it's an easy jump to 228
128 + 100 =228 228 is the running total
128 + 100 =228 Put the running total on the next line 228
128 + 100 =228 Add 30 because it's a round number to get to 258 228 + 30
128 + 100 =228 228 + 30 =258 258 is the running total
128 + 100 =228 228 + 30 =228 258 Put the running total on the next line
128 + 100 =228 228 + 30 =258 258 + 2 Add 2 because it's then easy to get to 260
128 + 100 =228 228 + 30 =258 258 + 2 =260 260 is the running total
128 + 100 =228 228 + 30 =258 258 + 2 =260 Put the running total on the next line 260
128 + 100 =228 228 + 30 =258 258 + 2 =260 260 + 4 Add 4 to finally get to 264
128 + 100 =228 228 + 30 =258 258 + 2 =260 260 + 4 =264 Check the final equals 264
+ 100 + 30 + 2 + 4 Now add 136
Answer: 264-128=136
Answer: 264-128=136 |
# Chapter 9 - PowerPoint PPT Presentation
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Chapter 9. Approximation Algorithms. Approximation algorithm. Up to now, the best algorithm for solving an NP-complete problem requires exponential time in the worst case. It is too time-consuming.
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Chapter 9
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### Chapter 9
Approximation Algorithms
### Approximation algorithm
• Up to now, the best algorithm for solving an NP-complete problem requires exponential time in the worst case. It is too time-consuming.
• To reduce the time required for solving a problem, we can relax the problem, and obtain a feasible solution“close” to an optimal solution
### The node cover problem
• Def: Given a graph G=(V, E), S is the node cover if S V and for every edge (u, v) E, either u S or v S.
• The node cover problem is NP-complete.
The optimal solution:
{v2,v5}
### An approximation algorithm
• Input: A graph G=(V,E).
• Output: A node cover S of G.
Step 1: S= and E’=E.
Step 2: While E’
Pick an arbitrary edge (a,b) in E’.
S=S{a,b}.
E’=E’-{e| e is incident to a or b}
• Time complexity: O(|E|)
• Example:
First: pick (v2,v3)
then S={v2,v3 }
E’={(v1,v5), (v4,v5)}
second: pick (v1,v5)
then S={v1,v2,v3 ,v5}
E’=
### How good is the solution ?
• |S| is at most two times the minimum size of a node cover of G.
• L: the number of edges we pick
M*: the size of an optimal solution
(1) L M*, because no two edges picked in Step 2 share any same vertex.
(2) |S| = 2L 2M*
### The Euclidean traveling salesperson problem (ETSP)
• The ETSP is to find a shortest closed path through a set S of n points in the plane.
• The ETSP is NP-hard.
### An approximation algorithm for ETSP
• Input: A set S of n points in the plane.
• Output: An approximate traveling salesperson tour of S.
Step 1: Find a minimal spanning tree T of S.
Step 2: Find a minimal Euclidean weighted matching M on the set of vertices of odd degrees in T. Let G=M∪T.
Step 3: Find an Eulerian cycle of G and then traverse it to find a Hamiltonian cycle as an approximate tour of ETSP by bypassing all previously visited vertices.
### An example for ETSP algorithm
• Step1: Find a minimal spanning tree.
• Step2: Perform weighted matching. The number of points with odd degrees must be even because is even.
• Step3: Construct the tour with an Eulerian cycle and a Hamiltonian cycle.
• Time complexity: O(n3)
Step 1: O(nlogn)
Step 2: O(n3)
Step 3: O(n)
• How close the approximate solution to an optimal solution?
• The approximate tour is within 3/2 of the optimal one. (The approximate rate is 3/2.)
(See the proof on the next page.)
### Proof of approximate rate
• optimal tour L: j1…i1j2…i2j3…i2m
{i1,i2,…,i2m}: the set of odd degree vertices in T.
2 matchings: M1={[i1,i2],[i3,i4],…,[i2m-1,i2m]}
M2={[i2,i3],[i4,i5],…,[i2m,i1]}
length(L) length(M1) + length(M2) (triangular inequality)
2 length(M )
length(M) 1/2 length(L )
G = T∪M
length(T) + length(M) length(L) + 1/2 length(L)
= 3/2 length(L)
### The bottleneck traveling salesperson problem (BTSP)
• Minimize the longest edge of a tour.
• This is a mini-max problem.
• This problem is NP-hard.
• The input data for this problem fulfill the following assumptions:
• The graph is a complete graph.
• All edges obey the triangular inequality rule.
### An algorithm for finding an optimal solution
Step1: Sort all edges in G = (V,E) into a nondecresing sequence |e1||e2|…|em|. Let G(ei) denote the subgraph obtained from G by deleting all edges longer than ei.
Step2: i←1
Step3: If there exists a Hamiltonian cycle in G(ei), then this cycle is the solution and stop.
Step4: i←i+1 . Go to Step 3.
1
### An example for BTSP algorithm
• e.g.
• There is a Hamiltonian cycle, A-B-D-C-E-F-G-A, in G(BD).
• The optimal solution is 13.
### Theorem for Hamiltonian cycles
• Def : The t-th power of G=(V,E), denoted as Gt=(V,Et), is a graph that an edge (u,v)Et if there is a path from u to v with at most t edges in G.
• Theorem: If a graph G is bi-connected, then G2 has a Hamiltonian cycle.
### An example for the theorem
G2
A Hamiltonian cycle:
A-B-C-D-E-F-G-A
### An approximation algorithm for BTSP
• Input: A complete graph G=(V,E) where all edges satisfy triangular inequality.
• Output: A tour in G whose longest edges is not greater than twice of the value of an optimal solution to the special bottleneck traveling salesperson problem of G.
Step 1: Sort the edges into |e1||e2|…|em|.
Step 2: i := 1.
Step 3: If G(ei) is bi-connected, construct G(ei)2, find a Hamiltonian cycle in G(ei)2 and return this as the output.
Step 4: i := i + 1. Go to Step 3.
### An example
Add some more edges. Then it becomes bi-connected.
1
• A Hamiltonian cycle: A-G-F-E-D-C-B-A.
• The longest edge: 16
• Time complexity: polynomial time
### How good is the solution ?
• The approximate solution is bounded by two times an optimal solution.
• Reasoning:
A Hamiltonian cycle is bi-connected.
eop: the longest edge of an optimal solution
G(ei): the first bi-connected graph
|ei||eop|
The length of the longest edge in G(ei)22|ei|
(triangular inequality) 2|eop|
### NP-completeness
• Theorem: If there is a polynomial approximation algorithm which produces a bound less than two, then NP=P.
(The Hamiltonian cycle decision problem reduces to this problem.)
• Proof:
For an arbitrary graph G=(V,E), we expand G to a complete graph Gc:
Cij = 1 if (i,j) E
Cij = 2 if otherwise
(The definition of Cij satisfies the triangular inequality.)
Let V* denote the value of an optimal solution of the bottleneck TSP of Gc.
V* = 1 G has a Hamiltonian cycle
Because there are only two kinds of edges, 1 and 2 in Gc, if we can produce an approximate solution whose value is less than 2V*, then we can also solve the Hamiltonian cycle decision problem.
### The bin packing problem
• n items a1, a2, …, an, 0 ai 1, 1 i n, to determine the minimum number of bins of unit capacity to accommodate all n items.
• E.g. n = 5, {0.8, 0.5, 0.2, 0.3, 0.4}
• The bin packing problem is NP-hard.
### An approximation algorithm for the bin packing problem
• An approximation algorithm:
(first-fit) place ai into the lowest-indexed bin which can accommodate ai.
• Theorem: The number of bins used in the first-fit algorithm is at most twice of the optimal solution.
### Proof of the approximate rate
• Notations:
• S(ai): the size of item ai
• OPT: # of bins used in an optimal solution
• m: # of bins used in the first-fit algorithm
• C(Bi): the sum of the sizes of aj’s packed in bin Bi in the first-fit algorithm
• OPT
C(Bi) + C(Bi+1) 1
C(B1)+C(B2)+…+C(Bm) m/2
m < 2 = 2 2 OPT
m < 2 OPT
### The rectilinear m-center problem
• The sides of a rectilinear square are parallel or perpendicular to the x-axis of the Euclidean plane.
• The problem is to find m rectilinear squares covering all of the n given points such that the maximum side length of these squares is minimized.
• This problem is NP-complete.
• This problem for the solution with error ratio < 2 is also NP-complete.
(See the example on the next page.)
• Input: P={P1, P2, …, Pn}
• The size of an optimal solution must be equal to one of the L ∞(Pi,Pj)’s, 1 i < j n, where
L ∞((x1,y1),(x2,y2)) = max{|x1-x2|,|y1-y2|}.
### An approximation algorithm
• Input: A set P of n points, number of centers: m
• Output: SQ[1], …, SQ[m]: A feasible solution of the rectilinear m-center problem with size less than or equal to twice of the size of an optimal solution.
Step 1: Compute rectilinear distances of all pairs of two points and sort them together with 0 into an ascending sequence D[0]=0, D[1], …, D[n(n-1)/2].
Step 2: LEFT := 1, RIGHT := n(n-1)/2 //* Binary search
Step 3: i := (LEFT + RIGHT)/2.
Step 4: If Test(m, P, D[i]) is not “failure” then
RIGHT := i-1
else LEFT := i+1
Step 5: If RIGHT = LEFT then
return Test(m, P, D[RIGHT])
else go to Step 3.
### Algorithm Test(m, P, r)
• Input: point set: P, number of centers: m, size: r.
• Output:“failure”, or SQ[1], …, SQ[m] m squares of size 2r covering P.
Step 1: PS := P
Step 2: For i := 1 to m do
If PS then
p := the point is PS with the smallest
x-value
SQ[i] := the square of size 2r with center
at p
PS := PS -{points covered by SQ[i]}
else SQ[i] := SQ[i-1].
Step 3: If PS = then return SQ[1], …, SQ[m]
else return “failure”. (See the example on the next page.)
### An example for the algorithm
The first application of the relaxed test subroutine.
The second application of the test subroutine.
A feasible solution of the rectilinear 5-center problem.
### Time complexity
• Time complexity: O(n2logn)
• Step 1: O(n)
• Step 2: O(1)
• Step 3 ~ Step 5:
O(logn)* O(mn) = O(n2logn)
### How good is the solution ?
• The approximation algorithm is of error ratio 2.
• Reasoning: If r is feasible, then Test(m, P, r) returns a feasible solution of size 2r.
The explanation of Si Si’ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Area Under the Curve
## Integrals and Riemann Sums.
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Area Under the Curve
You probably remember Becca and her track meet from earlier lessons. She won her race after pulling away from the pack in a hard push at the finish, and her boyfriend got a great picture of her just as she began pulling away. We have learned that by using derivatives, she could actually calculate her speed at the moment the picture was taken, and then with the second derivative she could calculate her acceleration similarly.
In this lesson we will discuss the integral, which is the process that would allow Becca to calculate the distance she actually covered during a given interval of the race, even though her speed was not constant!
### Area Under the Curve
To understand integration, consider the area under the curve y = f(x) for the interval from x = a to x = b in the figure below.
One way to calculate the area is to fill the region with rectangles. If the region is curved, the rectangles will not fit exactly, but we can improve the approximation by using rectangles of thinner width. If we continue to make the rectangles thinner and thinner, the area under the curve would reach the exact area under the curve. This is the limiting process that we discussed. In other words, the area under the curve is the limit of the total area of the rectangles as the widths of the rectangles approach zero.
Consider again the figure above. The interval from x = a to x = b is subdivided into n equal subintervals. Rectangles are drawn in each subinterval. Each rectangle touches the curve at its upper right corner. The height of the first rectangle is f(x1), the second f(x2), and the last is f(xn). Since the length of the entire interval from a to b is b - a, then the width of each subinterval is bab\begin{align*}\frac{b-a}{b}\end{align*}. We will refer to this width as ∆x. (The Greek letter ∆ is Delta and thus “delta x”.) That is,
Δx=ban\begin{align*}\Delta x = \frac{b - a} {n}\end{align*}
is defined as the width of each subinterval. The area of the first rectangle is f(x1)Δx\begin{align*}f(x_1)\Delta x\end{align*}, the second is f(x2)Δx\begin{align*}f(x_2)\Delta x\end{align*}, and so on. Thus the total area An of the n rectangles, is the sum of all areas:
An\begin{align*}A_n\end{align*} =f(x1)Δx+f(x2)Δx+...+f(xn)Δx\begin{align*}= f(x_1) \Delta x + f(x_2) \Delta x + . . . + f(x_n) \Delta x\end{align*}
=i=1nf(xi)Δx\begin{align*}= \sum_{i = 1}^n f(x_i) \Delta x\end{align*}
To make use of the concept of limit, we make the width of each rectangle approach 0 , which is equivalent to making the number of rectangles, n, approach infinity. By doing so, we find the exact area under the curve,
limnAn=limni=1nf(xi)Δx\begin{align*}\lim_{n \rightarrow \infty} A_n = \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} f(x_i) \Delta x\end{align*}.
This limit is defined as the definite integral and it is denoted by
baf(x)dx\begin{align*}\int_{a}^{b} f(x) dx\end{align*}.
#### The Definite Integral
A definite integral gives us the area between the x-axis and a curve over a defined interval.
The Definite Integral (The Limit Method)
The area between a curve f(x) and the x-axis over the interval [a, b] can be calculated by
A=baf(x)dx=limni=1nf(xi)Δx\begin{align*}A = \int_{a}^{b} f(x) dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^n f(x_i) \Delta x\end{align*}
where
Δx=ban\begin{align*}\Delta x = \frac{b - a} {n}\end{align*}
is the width of the subintervals.
It is important to keep in mind that the area under the curve can assume positive and negative values. It is more appropriate to call it “the net signed area”. Example 2 below illustrates this point.
### Examples
#### Example 1
Calculate the area between the curve y = x2 and the x-axis from x = 0 to x = 1.
We divide the region into n number of subintervals, each of width ∆x (see figure below).
First find ∆x.
Δx\begin{align*}\Delta x\end{align*} =ban\begin{align*}= \frac{b - a} {n}\end{align*}
=10n\begin{align*}= \frac{1 - 0} {n}\end{align*}
=1n\begin{align*}= \frac{1} {n}\end{align*}
The next step is to find xi.
xi\begin{align*}x_i\end{align*} =a+iΔx\begin{align*}= a + i \Delta x\end{align*}
=0+i1n\begin{align*}= 0 + i \cdot \frac{1} {n}\end{align*}
=in\begin{align*}= \frac{i} {n}\end{align*}
Therefore, f(xi)=x2i=(in)2\begin{align*}f(x_i) = x^2_i = \left (\frac{i} {n}\right )^2\end{align*} Using the integration formula
A\begin{align*}A\end{align*} =baf(x)dx=limni=1nf(xi)Δx\begin{align*}= \int_{a}^{b} f(x) dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^n f(x_i) \Delta x\end{align*}
=10x2dx=limni=1n(in)2(1n)\begin{align*}= \int_{0}^{1} x^2 dx= \lim_{n \rightarrow \infty} \sum_{i = 1}^n \left (\frac{i} {n}\right )^2 \left (\frac{1} {n}\right )\end{align*}
=limni=1ni2n3\begin{align*}= \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \frac{i^2} {n^3}\end{align*}
Since we are summing over i, not n, the summation becomes,
A\begin{align*}A\end{align*} =limn1n3i=1ni2\begin{align*}= \lim_{n \rightarrow \infty} \frac{1} {n^3} \sum_{i = 1}^{n} i^2\end{align*}
=limn1n3(12+22+32+...+n2)\begin{align*}= \lim_{n \rightarrow \infty} \frac{1} {n^3} (1^2 + 2^2 + 3^2 + . . . + n^2)\end{align*}
But since i=1ni2=n(n+1)(2n+1)6\begin{align*}\sum_{i = 1}^{n} i^2 = \frac{n(n + 1) (2n + 1)} {6}\end{align*} then
A\begin{align*}A\end{align*} =limn1n3n(n+1)(2n+1)6\begin{align*}= \lim_{n \rightarrow \infty} \frac{1} {n^3} \frac{n(n + 1)(2n + 1)} {6}\end{align*}
limn16(2+3n+1n2)\begin{align*}\lim_{n \rightarrow \infty} \frac{1} {6} \left (2 + \frac{3} {n} + \frac{1} {n^2}\right )\end{align*}
Taking the limit,
\begin{align*}A\end{align*} \begin{align*}= \frac{1} {6} (2 + 3(0) + (0))\end{align*}
\begin{align*}= \frac{1} {3}\end{align*}
Thus the area under the curve is (1/3).
#### Example 2
Find the area between the curve y = x and the x-axis from x = -1 to x = 1.
As you can see in figure a, the integral represents the total areas of all the rectangles above and below the x-axis. First, we divide the region into two regions, one above x-axis and one below the x-axis. Then we divide each region into n subintervals, each of width ∆x (figure b).
Region I: Find ∆x and xi.
\begin{align*}\Delta x = \frac{1 - 0} {n} = \frac{1} {n}\end{align*}
\begin{align*}x_{i}\end{align*} \begin{align*}= a + i \Delta x\end{align*}
\begin{align*}= 0 + i \left (\frac{1} {n}\right ) = \frac{i} {n}\end{align*}
\begin{align*}f(x_i) = \frac{i} {n}\end{align*}
Region II: Again, find ∆x and xi.
\begin{align*}\Delta x\end{align*} \begin{align*}= \frac{-1 -0} {n} = \frac{-1} {n}\end{align*}
\begin{align*}x_i\end{align*} \begin{align*}= b + i \Delta x\end{align*}
\begin{align*}= -1 + i \left (\frac{-1} {n}\right )\end{align*}
\begin{align*}= -1 - \frac{i} {n}\end{align*}
\begin{align*}f(x_i) = -1 - \frac{i} {n}\end{align*}
The integral represents the net area of the two regions I and II:
\begin{align*}A = A_1 - A_2 =\end{align*} \begin{align*}(\text{area above x-axis in }[a, b]) - (\text{area below x-axis in }[a, b])\end{align*} Thus,
\begin{align*}A\end{align*} \begin{align*}= \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} f(x_i) \Delta x - \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} f(x_i) \Delta x\end{align*}
\begin{align*}= \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left (\frac{i} {n}\right ) \left (\frac{1} {n}\right ) - \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left (-1 - \frac{i} {n}\right ) \left (\frac{1} {n}\right )\end{align*}
\begin{align*}= \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left (\frac{i} {n^2}\right ) - \lim_{n \rightarrow \infty} \sum_{i = 1}^{n} \left (\frac{-1} {n} - \frac{i} {n^2}\right )\end{align*}
\begin{align*}\mathit = \lim_{n \rightarrow \infty} \frac{1} {n^2} \sum_{i = 1}^{n} i - \left [\lim_{n \rightarrow \infty} \frac{-1} {n} + \lim_{n \rightarrow \infty} \frac{1} {n^2} \sum_{i = 1}^{n} i\right ]\end{align*}
\begin{align*}= \lim_{n \rightarrow \infty} \frac{1} {n^2} \frac{n(n+1)} {2} - \left [0 + \lim_{n \rightarrow \infty} \frac{1} {n^2} \frac{n(n + 1)} {2}\right ]\end{align*}
\begin{align*}= \frac{1} {2} - \left [\frac{1} {2}\right ]\end{align*}
\begin{align*}= 0\end{align*}
We conclude that the net area is zero.
#### Example 3
Approximate the definite integral between x = 0 and x = 40 by calculating the areas of rectangles which fill the area in the image below. Use at least 3 successively narrower sizes of rectangles.
The equation of the curve in the image is: \begin{align*}y = 60 -40(1 - \frac{9}{10}^x)\end{align*}.
\begin{align*}\therefore\end{align*} the closest approximated area is 1173.44 units (The actual calculated area is 1174.0373)
#### Example 4
Approximate the area under y = x + 3 on the interval [5,6] using the middle Riemann Sum with 5 subintervals.
Sketch of graph:
First, we divide the interval [5,6] into pieces:
Between x = 5 and x = 5.2, the middle value is 5.1 + 3 = 8.1
Between x = 5.2 and x = 5.4, the middle value is 5.3 + 3 = 8.3
Between x = 5.4 and x = 5.6, the middle value is 5.5 + 3 = 8.5
Between x = 5.6 and x = 5.8, the middle value is 5.7 + 3 = 8.7
Between x = 5.8 and x = 6, the middle value is 5.9 + 3 = 8.9
Adding these, we get 42.5.
To get the Riemann sum, take this answer and multiply by the width of each segment: 0.2
\begin{align*}\therefore 8.5\end{align*} is our approximated area.
#### Example 5
Approximate the area between y = 3x2 + x + 5 and the x-axis on the interval between x = 2 and x = 5 using the right Riemann Sum with 2 subintervals.
First, we divide the interval [2,5] into subintervals: Between x = 2 and x = 3.5, the right value is 3(3.5)2 + (2.5) + 5 = 42.25 Between x = 3.5 and x = 5, the right value is 3(5)2 + (5) + 5 = 85 Adding these, we get 127.25. Take this answer and multiply by the width of each segment: 1.5.
\begin{align*}\therefore \approx 190.88\end{align*} is the area
#### Example 6
Use a definite integral to find the area under the curve \begin{align*}y = 5x^2 + 2x + 4\end{align*} on the interval [0, 3].
\begin{align*}\int_{0}^{3} 5x^2 + 2x + 4dx = \frac{5}{3} x^3 + x^2 +4x |_{4}^{5}\end{align*}
### Review
1. Use the limit method to find the area under the curve of f(x) = x2 in the interval [0, 2].
Find the area between the curve and the x-axis:
1. Curve y = x on the interval x = 1 to x = 3.
2. Curve y = -x from x = 1 to x = 3.
3. Curve y = x from x = -3 to x = 3.
4. Approximate the area under \begin{align*}y = 2x + 3\end{align*} on the interval [0,3] using the middle Riemann Sum for y with 6 subintervals.
Find the area under the curve:
1. \begin{align*}y = 3\end{align*} on [4, 5]
2. \begin{align*}y = 3x + 1\end{align*} on [1, 5]
3. \begin{align*}y = \frac{1}{x}\end{align*} on [3, 4]
4. \begin{align*}y = 2x + 4\end{align*} on [5, 6]
5. \begin{align*}y = 5x^3 + 4x^2 + x + 2\end{align*} on [2, 5]
6. \begin{align*}y = \frac {1}{x} \end{align*} on [3, 7]
7. \begin{align*}y = 3x^2 + 2x\end{align*} on [5, 6]
8. \begin{align*}y = 4\end{align*} on [2, 6]
9. \begin{align*}y = 2x^2 + 4x + 5\end{align*} on [1, 5]
10. Sketch y = x2 and y = x on the same coordinate system and then find the area of the region enclosed between them.
To see the Review answers, open this PDF file and look for section 8.12.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
$\Delta$
The symbol "$\Delta$", read "delta", is used to denote "change in", as in "the change in velocity over time" = $\frac{\Delta v}{t}$.
definite integral
A definite integral gives the area between the x-axis and a curve over a defined interval.
integral
An integral is used to calculate the area under a curve or the area between two curves.
limit
A limit is the value that the output of a function approaches as the input of the function approaches a given value.
subintervals
Subintervals are created when an interval is broken into smaller equally sized intervals.
summand
A summand is an expression being summed. It directly follows the sigma symbol. |
Solve 6x^2+3x-18=0 Using The Quadratic Formula?
The quadratic formula tells you the solution to
ax^2 + bx + c = 0
is given by
x = (-b ± √(b^2 - 4ac))/(2a)
You have an equation in which a=6, b=3, c=-18, so the formula gives
x = (-3 ± √(3^2 - 4(6)(-18))/(2*6)
x = (-3 ± √(9+432))/12
x = (-3 ± √441)/12
x = (-3 ± 21)/12
x = {-24/12, 18/12}
x = {-2, 3/2}
thanked the writer.
This is simple:
6x2 + 3x - 18 = 0
Dividing the equation by 3, we get:
2x2 + x - 6 = 0
2x2 + 4x - 3x - 6 = 0
2x(x + 2) -3(x + 2)
(2x-3)(x+2)=0
x= 3/2
x=-2
thanked the writer.
Shelia, first let's see what the quadratic formula is, ok?
x = [-b ±b2-4ac]/2a
6x2 + 3x -18 = 0
The values from the quadratic formula that we extract from this equation are:
a = 6
b = 3
c = -18
Now, all you have to do is put these values in the quadratic formula and you will get your answer
x = [-b ±b2-4ac]/2a
x = [-3 ±32-4.6.-18]/2.6
(I am using a period to show a multiplication sign so that you do not confuse it with 'x' in the equation)
x = [-3 ± √9 + 432]/12
x = [-3
± √441]/12
x = [-3
± 21]/12
Now, the '
±' sign shows that the value can be either positive or negative.
So, we will solve for each. Now our two new equations are:
x = [-3 + 21]/12
x = 1.5
and
x = [-3 - 21]/12
x = -2
So, the two values for x are 1.5 and -2.
About how we type all the mathematical symbols, I use MS-Word for that. You can also use MS-Excel if you want to. Go to 'Insert' and choose 'Symbols'. Here, you would only be given a few options but an option would say, 'More symbols'. Choose that and you would probably find the symbol you are looking for.
I use MS-Office 2007. This also has inbuilt equations that you can choose from Insert 'Equations'. I think previous issues of Word would also have them. You just need to browse around a little. Use 'x' or '*' to show multiplication. You can square a number/ alphabet by writing it, like x2. Then select 2 and click on x2 on the tool bar to make it a superscript.
I think it would all be very clear to you now. Have fun!
thanked the writer.
Since the question mentions the quadratic formula, the problem should probably read 6x2 + 3x – 18 = 0. The quadratic formula is shown in the picture. Following the formula, X= (-3+(SQRT(9-4*6*(-18))))/12 and x = (-3-(SQRT(9-4*6*(-18))))/12. Or, doing the arithmetic, x=1.5 and x=-2.
thanked the writer.
Here is the solution:
6x2 + 3x - 18 = 0
6x2 +12x - 9x - 18= 0
6x(x+2) -9(x+2) = 0
(6x -9)(x+2)= 0
x = -2 is one answer
x= 9/6 or 2/3 is second answer
thanked the writer.
Anonymous commented
why didn't you divide the original equation by 3 at first? that's why you're second solution is wrong. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Exponential Properties Involving Products
## Add exponents to multiply exponents by other exponents
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Progress
Practice Exponential Properties Involving Products
Progress
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Exponential Properties Involving Products
What if you wanted to simplify a mathematical expression containing exponents, like \begin{align*}4^3 \cdot 4^2\end{align*}? How would you do so? After completing this Concept, you'll be able to use the product of powers property to simplify exponential expressions like this one.
### Watch This
CK-12 Foundation: 0801S Product of Powers
### Guidance
Back in chapter 1, we briefly covered expressions involving exponents, like \begin{align*}3^5\end{align*} or \begin{align*}x^3\end{align*}. In these expressions, the number on the bottom is called the base and the number on top is the power or exponent. The whole expression is equal to the base multiplied by itself a number of times equal to the exponent; in other words, the exponent tells us how many copies of the base number to multiply together.
#### Example A
Write in exponential form.
a) \begin{align*}2 \cdot 2\end{align*}
b) \begin{align*}(-3)(-3)(-3)\end{align*}
c) \begin{align*}y \cdot y \cdot y \cdot y \cdot y\end{align*}
d) \begin{align*}(3a)(3a)(3a)(3a)\end{align*}
Solution
a) \begin{align*}2 \cdot 2 = 2^2\end{align*} because we have 2 factors of 2
b) \begin{align*}(-3)(-3)(-3) = (-3)^3\end{align*} because we have 3 factors of (-3)
c) \begin{align*}y \cdot y \cdot y \cdot y \cdot y = y^5\end{align*} because we have 5 factors of \begin{align*}y\end{align*}
d) \begin{align*}(3a)(3a)(3a)(3a)=(3a)^4\end{align*} because we have 4 factors of \begin{align*}3a\end{align*}
When the base is a variable, it’s convenient to leave the expression in exponential form; if we didn’t write \begin{align*}x^7\end{align*}, we’d have to write \begin{align*}x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x\end{align*} instead. But when the base is a number, we can simplify the expression further than that; for example, \begin{align*}2^7\end{align*} equals \begin{align*}2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2\end{align*}, but we can multiply all those 2’s to get 128.
Let’s simplify the expressions from Example A.
#### Example B
Simplify.
a) \begin{align*}2^2\end{align*}
b) \begin{align*}(-3)^3\end{align*}
c) \begin{align*}y^5\end{align*}
d) \begin{align*}(3a)^4\end{align*}
Solution
a) \begin{align*}2^2 = 2 \cdot 2 =4\end{align*}
b) \begin{align*}(-3)^3 = (-3)(-3)(-3)=-27\end{align*}
c) \begin{align*}y^5\end{align*} is already simplified
d) \begin{align*}(3a)^4 = (3a)(3a)(3a)(3a) = 3 \cdot 3 \cdot 3 \cdot 3 \cdot a \cdot a \cdot a \cdot a = 81 a^4\end{align*}
Be careful when taking powers of negative numbers. Remember these rules:
So even powers of negative numbers are always positive. Since there are an even number of factors, we pair up the negative numbers and all the negatives cancel out.
And odd powers of negative numbers are always negative. Since there are an odd number of factors, we can still pair up negative numbers to get positive numbers, but there will always be one negative factor left over, so the answer is negative:
Use the Product of Powers Property
So what happens when we multiply one power of \begin{align*}x\end{align*} by another? Let’s see what happens when we multiply \begin{align*}x\end{align*} to the power of 5 by \begin{align*}x\end{align*} cubed. To illustrate better, we’ll use the full factored form for each:
So \begin{align*}x^5 \times x^3 = x^8\end{align*}. You may already see the pattern to multiplying powers, but let’s confirm it with another example. We’ll multiply \begin{align*}x\end{align*} squared by \begin{align*}x\end{align*} to the power of 4:
So \begin{align*}x^2 \times x^4 = x^6\end{align*}. Look carefully at the powers and how many factors there are in each calculation. \begin{align*}5 \ x\end{align*}’s times \begin{align*}3 \ x\end{align*}’s equals \begin{align*}(5 + 3) = 8 \ x\end{align*}’s. \begin{align*}2 \ x\end{align*}’s times \begin{align*}4 \ x\end{align*}’s equals \begin{align*}(2 + 4) = 6 \ x\end{align*}’s.
You should see that when we take the product of two powers of \begin{align*}x\end{align*}, the number of \begin{align*}x\end{align*}’s in the answer is the total number of \begin{align*}x\end{align*}’s in all the terms you are multiplying. In other words, the exponent in the answer is the sum of the exponents in the product.
Product Rule for Exponents: \begin{align*}x^n \cdot x^m = x^{(n+m)}\end{align*}
There are some easy mistakes you can make with this rule, however. Let’s see how to avoid them.
#### Example C
Multiply \begin{align*}2^2 \cdot 2^3\end{align*}.
Solution
\begin{align*}2^2 \cdot 2^3 = 2^5 = 32\end{align*}
Note that when you use the product rule you don’t multiply the bases. In other words, you must avoid the common error of writing \begin{align*}2^2 \cdot 2^3 = 4^5\end{align*}. You can see this is true if you multiply out each expression: 4 times 8 is definitely 32, not 1024.
#### Example D
Multiply \begin{align*}2^2 \cdot 3^3\end{align*}.
Solution
\begin{align*}2^2 \cdot 3^3 = 4 \cdot 27 = 108\end{align*}
In this case, we can’t actually use the product rule at all, because it only applies to terms that have the same base. In a case like this, where the bases are different, we just have to multiply out the numbers by hand—the answer is not \begin{align*}2^5\end{align*} or \begin{align*}3^5\end{align*} or \begin{align*}6^5\end{align*} or anything simple like that.
Watch this video for help with the Examples above.
CK-12 Foundation: Products of Powers
### Guided Practice
Simplify the following exponents:
a. \begin{align*}(-2)^5\end{align*}
b. \begin{align*}(10x)^2\end{align*}
Solutions:
a. \begin{align*}(-2)^5=(-2)(-2)(-2)(-2)(-2)=-32\end{align*}
b. \begin{align*}(10x)^2=10^2\cdot x^2=100x^2\end{align*}
### Explore More
Write in exponential notation:
1. \begin{align*}4 \cdot 4 \cdot 4 \cdot 4 \cdot 4\end{align*}
2. \begin{align*}3x \cdot 3x \cdot 3x\end{align*}
3. \begin{align*}(-2a)(-2a)(-2a)(-2a)\end{align*}
4. \begin{align*}6 \cdot 6 \cdot 6 \cdot x \cdot x \cdot y \cdot y \cdot y \cdot y\end{align*}
5. \begin{align*}2 \cdot x \cdot y \cdot 2 \cdot 2 \cdot y \cdot x\end{align*}
Find each number.
1. \begin{align*}5^4\end{align*}
2. \begin{align*}(-2)^6\end{align*}
3. \begin{align*}(0.1)^5\end{align*}
4. \begin{align*}(-0.6)^3\end{align*}
5. \begin{align*}(1.2)^2+5^3\end{align*}
6. \begin{align*}3^2 \cdot (0.2)^3\end{align*}
Multiply and simplify:
1. \begin{align*}6^3 \cdot 6^6\end{align*}
2. \begin{align*}2^2 \cdot 2^4 \cdot 2^6\end{align*}
3. \begin{align*}3^2 \cdot 4^3\end{align*}
4. \begin{align*}x^2 \cdot x^4\end{align*}
5. \begin{align*}(-2y^4)(-3y)\end{align*}
6. \begin{align*}(4a^2)(-3a)(-5a^4)\end{align*}
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 8.1.
### Vocabulary Language: English
Base
Base
When a value is raised to a power, the value is referred to as the base, and the power is called the exponent. In the expression $32^4$, 32 is the base, and 4 is the exponent.
Exponent
Exponent
Exponents are used to describe the number of times that a term is multiplied by itself.
Power
Power
The "power" refers to the value of the exponent. For example, $3^4$ is "three to the fourth power". |
New Zealand
Level 8 - NCEA Level 3
# Complementary results
Lesson
There are connections between the trigonometric functions that can make simplifications possible.
In a right-angled triangle, the two acute angles together make a right-angle. We say the acute angles are complementary (to one another). If the two acute angles have measures $\alpha$α and $\beta$β, then $\alpha+\beta=90^\circ$α+β=90° and so, $\beta=90^\circ-\alpha$β=90°α.
The diagram below illustrates the following relationships.
$\cos\alpha=\frac{b}{h}=\sin\beta=\sin\left(90^\circ-\alpha\right)$cosα=bh=sinβ=sin(90°α)
$\sin\alpha=\frac{a}{h}=\cos\beta=\cos\left(90^\circ-\alpha\right)$sinα=ah=cosβ=cos(90°α)
$\cot\alpha=\frac{b}{a}=\tan\beta=\tan\left(90^\circ-\alpha\right)$cotα=ba=tanβ=tan(90°α)
Thus, the 'co' in complementary explains the meaning of cosine in relation to sine, and to cotangent in relation to tangent.
The statements
$\cos\alpha\equiv\sin\left(90^\circ-\alpha\right)$cosαsin(90°α)
$\sin\alpha\equiv\cos\left(90^\circ-\alpha\right)$sinαcos(90°α) and
$\cot\alpha\equiv\tan\left(90^\circ-\alpha\right)$cotαtan(90°α)
are called identities because they are true whatever the value of the angle $\alpha$α.
These identities are true not only in right-angled triangle trigonometry, but they also hold for angles of any size. This can be confirmed by thinking about the geometry in the unit circle diagram that is used for defining the trigonometric functions of angles of any magnitude.
#### Example
Simplify the relation $\sin\left(90^\circ-\theta\right)=\sqrt{3}\sin\theta$sin(90°θ)=3sinθ
it will be a good plan to try to rearrange the equation so that the trigonometric functions are on one side and the coefficients are on the other. We divide both sides by $\sin\left(90^\circ-\theta\right)$sin(90°θ) and also by $\sqrt{3}$3 to obtain $\frac{1}{\sqrt{3}}=\frac{\sin\theta}{\sin\left(90^\circ-\theta\right)}$13=sinθsin(90°θ). But, $\sin\left(90^\circ-\theta\right)$sin(90°θ) is just $\cos\theta$cosθ. So, the simplification we seek is
$\tan\theta=\frac{1}{\sqrt{3}}$tanθ=13.
We recognise an exact value for $\tan$tan and conclude that $\theta=30^\circ$θ=30° if $\theta$θ is acute. You should check that there is also a third quadrant solution, $\theta=210^\circ$θ=210°.
#### Worked Examples
##### Question 1
By finding the ratio represented by $\sin\theta$sinθ, $\cos\theta$cosθ and $\tan\theta$tanθ in the given figure, we want to prove that $\frac{\sin\theta}{\cos\theta}=\tan\theta$sinθcosθ=tanθ.
1. Write down the expression for $\sin\theta$sinθ.
2. Write down the expression for $\cos\theta$cosθ.
3. Hence, form an expression for $\frac{\sin\theta}{\cos\theta}$sinθcosθ.
4. Write down the expression for $\tan\theta$tanθ.
5. Does $\frac{\sin\theta}{\cos\theta}=\tan\theta$sinθcosθ=tanθ?
Yes
A
No
B
Yes
A
No
B
##### Question 2
Prove that $\frac{\tan x\cos x}{\sin x}=1$tanxcosxsinx=1.
##### Question 3
Simplify the following expression using complementary angles:
$\frac{\sin51^\circ}{\cos39^\circ}$sin51°cos39°
### Outcomes
#### M8-6
Manipulate trigonometric expressions
#### 91575
Apply trigonometric methods in solving problems |
Interactive simulation the most controversial math riddle ever! \angle Z = \frac{1}{2} \cdot (80 ^{\circ}) For example, in the above figure, Using the figure above, try out your power-theorem skills on the following problem: $$So, the length of the chord is approximately 13.1 cm. Thus. Divide the chord length by double the result of step 1. Therefore, the measurements provided in this problem violate the theorem that angles formed by intersecting arcs equals the sum of the intercepted arcs. The formulas for all THREE of these situations are the same: Angle Formed Outside = $$\frac { 1 }{ 2 }$$ Difference of Intercepted Arcs (When subtracting, start with the larger arc.) The triangle can be cut in half by a perpendicular bisector, and split into 2 smaller right angle triangles. Special situation for this set up: It can be proven that ∠ABC and central ∠AOC are supplementary. In the second century AD, Ptolemy of Alexandria compiled a more extensive table of chords in his book on astronomy, giving the value of the chord for angles ranging from 1/2 degree to 180 degrees by increments of half a degree. Theorem 3: Alternate Angle Theorem. \\ \\ \overparen{AGF}= 170 ^{\circ } Circular segment. If the radius is r and the length of the chord is c then triangle CMB is a right triangle with |BC| = r and |MB| = c/2. Angle AOD must therefore equal 180 - α . \\ Chord Length Using Perpendicular Distance from the Center. Circular segment - is an area of a circle which is "cut off" from the rest of the circle by a secant (chord).. On the picture: L - arc length h- height c- chord R- radius a- angle. \class{data-angle}{89.68 } ^{\circ} = \frac 1 2 ( \class{data-angle-0}{88.21 } ^{\circ} + \class{data-angle-1}{91.15 } ^{\circ} ) The chord length formulas vary depends on what information do you have about the circle. \\ Notice that the intercepted arcs belong to the set of vertical angles. . \\ \\ Math Geometry Physics Force Fluid Mechanics Finance Loan Calculator.$$. $$\angle A= 53 ^{\circ} . Calculating the length of a chord Two formulae are given below for the length of the chord,. If$$ \overparen{MNL}= 60 ^{\circ}$$,$$ \overparen{NO}= 110 ^{\circ}$$and$$ \overparen{OPQ}= 20 ^{\circ} $$, then what is the measure of$$ \angle Z $$? The first step is to look at the chord, and realize that an isosceles triangle can be made inside the circle, between the chord line and the 2 radius lines. Circle Calculator. Chords were used extensively in the early development of trigonometry. Click here for the formulas used in this calculator. Note: Like inscribed angles, when the vertex is on the circle itself, the angle formed is half the measure of the intercepted arc. Chord Length and is denoted by l symbol. Let R be the radius of the circle, θ the central angle in radians, α is the central angle in degrees, c the chord length, s the arc length, h the sagitta (height) of the segment, and d the height (or apothem) of the triangular portion.$$, $$m \angle AEB = m \angle CED$$ CED since they are vertical angles. \\ \angle A= \frac{1}{2} \cdot (38^ {\circ} + 68^ {\circ}) $$\text{m } \overparen{\red{JKL}}$$ is $$75^{\circ}$$ $$\text{m } \overparen{\red{WXY}}$$ is $$65^{\circ}$$ and What is the value of $$a$$? \angle AEB = 27.5 ^{\circ} For angles in circles formed from tangents, secants, radii and chords click here. Solving for circle segment chord length. radius = \angle \class{data-angle-label}{W} = \frac 1 2 (\overparen{\rm \class{data-angle-label-0}{AB}} + \overparen{\rm \class{data-angle-label-1}{CD}}) Angles of Intersecting Chords Theorem. It is the angle of intersection of the tangents. a= 70 ^{\circ} If you know the radius or sine values then you can use the first formula. AEB and = (SUMof Intercepted Arcs) In the diagram at the right, ∠AEDis an angle formed by two intersecting chords in the circle. d is the perpendicular distance from the chord to … = 2 × (r2–d2. Using SohCahToa can help establish length c. Focusing on the angle θ2\boldsymbol{\frac{\theta}{2}}2θ… case of the long chord and the total deflection angle. Find the measure of the angle t in the diagram. Chord Length when radius and angle are given calculator uses Chord Length=sin (Angle A/2)*2*Radius to calculate the Chord Length, Chord Length when radius and angle are given is the length of a line segment connecting any two points on the circumference of a circle with a given value for radius and angle. also, m∠BEC= 43º (vertical angle) m∠CEAand m∠BED= 137º by straight angle formed. In diagram 1, the x is half the sum of the measure of the intercepted arcs (. You may need to download version 2.0 now from the Chrome Web Store. Diagram 1. \angle AEB = \frac{1}{2}(30 ^{\circ} + 25 ^{\circ}) . C_ {len}= 2 \times \sqrt { (r^ {2} –d^ {2}}\\ C len. Performance & security by Cloudflare, Please complete the security check to access. . I have chosen NACA 4418 airfoil, tip speed ratio=6, Cl=1.2009, Cd=0.0342, alpha=13 can someone help me how to calculate it please? \angle AEB = \frac{1}{2} (\overparen{ AB} + \overparen{ CD}) \\ c is the angle subtended at the center by the chord. the angles sum to one hundred and eighty degrees). Theorem: So far everything is fine. By double the result of step 1 extensively in the circle, the x is half the sum the! \Overparen { CD } \overline { LY } \overparen { CD } $and! Both as positive real numbers and press calculate '' fraction of the circle diagram 1, the measurements in. An angle formed by a tangent and a chord that passes through center. ( 2r ) = 220° and split into 2 smaller right angle triangles d of the arcs... Will be the square root of the chord intersect inside the circle, m∠BEC= 43º vertical! Not necessary for these chords to intersect at the right chord angle formula ∠AEDis angle! By straight angle formed by intersecting arcs equals the sum of chord angle formula arcs ( intersect. Angle to angle α ( i.e 's objectives information do you have about the circle for this up! ) m∠CEAand m∠BED= 137º by straight angle formed by two intersecting chords to intersect at the center the... Smaller right angle triangles angle triangles ∠AEDis an angle formed by two intersecting chords intersect! Captcha proves you are given is known trigonometric table, compiled by,... You can use the theorem that angles formed by a tangent and a chord c/ ( 2r ) 's.! A of the sector area, which is 360 degrees angle is provided by the angle... 2 − d 2 Mechanics Finance Loan calculator to access need to download version 2.0 now the! Sum of intercepted arcs ) in the future is to use Privacy.. 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# Division: It's Not Just for Real Numbers Anymore
Contributor: Lynn Ellis. Lesson ID: 13730
Complex numbers — they aren't even real! But they do have applications in the real world. First, you need to learn arithmetic with them. In this lesson, you will learn how to divide complex numbers.
categories
## Algebra II, Complex Numbers and Quantities
subject
Math
learning style
Auditory
personality style
Beaver
High School (9-12)
Lesson Type
Quick Query
## Lesson Plan - Get It!
Audio:
• How does arithmetic change when the numbers aren't real?
• Can you divide complex numbers like you can divide real numbers?
In this lesson, we will learn about using complex conjugates to divide complex numbers.
You may be wondering what a complex conjugate is, so we will start there.
Complex Conjugates
• What are they anyway?
Let's break down the terms.
Complex numbers have the form (a + bi) where a and b are both real numbers and i2 = -1.
Complex numbers allow us to work with square roots of negative numbers since i = √-1.
We call i an imaginary number because we can't take the square root of a negative number in the real number system.
We call a the real component of the complex number because it does not include an imaginary number in it.
We call bi the imaginary component because it does contain an imaginary number.
Together they make a complex number. Take a look at this diagram to see how the number systems fit together:
Conjugates are two binomials in the form (m + n) and (m - n).
For example, (x + 3) and (x - 3) are conjugates. Notice that the terms are the same, but the operation changes from addition to subtraction.
Give it a try! Play the game below to match up the pairs of conjugates.
• Do you know what's really cool about conjugates?
Try multiplying these conjugates together and see if you can identify the pattern.
• Did you see that when you multiply conjugates together, you get the difference of squares?
That will happen every time you multiply conjugates. That's important to note, and we will use that fact throughout the rest of this lesson.
OK, now that we know about complex numbers and conjugates, let's put the two words together to define complex conjugates.
Complex conjugates are two complex numbers in the form (a + bi) and (a - bi).
Let's multiply (a + bi)(a - bi)
It looks like this:
(a + bi)(a - bi)
a2 + abi - abi - b2i2
a2 - b2i2
a2 + b2
• Do you notice that a2 + b2 is a real number?
There is no imaginary part left!
So, you might wonder what we can do with this information and why it is useful to know. There are several ways that we can use complex conjugates in mathematics.
The first one to talk about is dividing complex numbers. Other lessons will explore further applications of complex conjugates, such as finding the modulus (absolute value) of a complex number.
Since this lesson is focused on division, let's look at two different examples of dividing complex numbers.
Example 1
1 - 9i We want to get a single complex number for the answer. 1 - i Notice that we have i in both the numerator and denominator. We don't want an i in the denominator when we are done.
1 - 9i • 1 + i We are multiplying by 1 but in the convenient form of the conjugate of the denominator over itself. 1 - i 1 + i Why? Check out the next step to see.
1 - i - 9i + 9 We got the difference of squares in the denominator. 1 + 1 That made the i drop out! Now we combine like terms.
8 - 10i Then we put it into the correct form (a + bi) to get the final answer. 2 Make sure you simplify fully.
4 - 5i
Here's another example. It doesn't look as nice, because we get fractions in the end, but the process is exactly the same.
Example 2
3 - 4i 7 + 2i
3 - 4i • 7 - 2i 7 + 2i 7 - 2i
21 - 6i - 28i + 8 49 + 4
29 - 34i 53
29 - 34i 53 53
Now you try a couple.
[NOTE: The / in the questions below represents the split between the numerator and denominator.]
• How did you do with those?
If you feel like you understand, move to the Got It? section for some more practice. Then try the quiz.
Interactive Video |
# Difference Between Variance and Standard Deviation
Understanding the main difference between Variance and Standard deviation is important to know. These mathematical terms are usually used in normal mathematical equations to solve problems. Primarily variance and standard deviation are used as metrics to solve statistical problems. The standard deviation formula is used to measure the standard deviation of the given data values. It is important to understand the difference between variance, standard deviation, as they are both commonly used terms in the probability theory and statistics. These two terms are used to determine the spread of the data set. Both the standard deviation and the variance are numerical measures, which calculates the spread of data from the mean value.
In short, the mean is the average of the range of given data values, a variance is used to measure how far the data values are dispersed from the mean, and the standard deviation is the used to calculate the amount of dispersion of the given data set values. We know that the measures of dispersion can be categorised into two different types, namely absolute measure of dispersion and the relative measure of dispersion. When we consider the variance and standard deviation, both fall under the absolute measure of dispersion. Before discussing the key difference between the variance and the standard deviation let’s discuss the definition of variance and the standard deviation here.
## Definition of Variance and Standard Deviation
Variance: Variance can simply be defined as a measure of variability to represent members of a group. The variance measures the closeness of data points corresponding to a greater value of variance.
Standard Deviation: Standard deviation, on the other hand, observes the quantifiable amount of dispersion of observations when approached with data. We must understand that variance and standard deviation differ from each other. Variances describe the variability of the observed observations while standard deviation measures the dispersion of observations within a set.
## What is the Difference Between Variance and Standard Deviation
Here, the list of comparative differences between the variance and the standard deviation is given below in detail:
Difference between Variance and Standard Deviation
Variance
Standard Deviation
It can simply be defined as the numerical value, which describes how variable the observations are. It can simply be defined as the observations that get measured are measured through dispersion within a data set.
Variance is nothing but the average taken out of the squared deviations. Standard Deviation is defined as the root of the mean square deviation
Variance is expressed in Squared units. Standard deviation is expressed in the same units of the data available.
It is mathematically denoted as (σ2) It is mathematically denoted as (σ)
Variance is a perfect indicator of the individuals spread out in a group. Standard deviation is the perfect indicator of the observations in a data set.
### Variance and Standard Deviation Problem
Example:
Find the mean, standard deviation and variance for the following data: 6, 7,10, 12, 13, 4, 8, 12.
Solution:
Given data: 6, 7,10, 12, 13, 4, 8, 12
Finding Mean:
We know that mean is the ratio of the sum of observations to the total number of observations.
(i.e) Mean = Sum of observations / Total number of observations.
Mean = (6+7+10+12+13+4+8+12)/8
Mean = 72/8
Mean = 9.
Finding Variance:
Variance = $$\frac{\sum (x_{i}-\bar{x})^{2}}{n}$$
Here, n=8
$$\sum (x_{i}-\bar{x})^{2}$$ can be calculated as follows:
$$x_{i}$$ $$x_{i}-\bar{x}$$ $$(x_{i}-\bar{x})^{2}$$ 6 6 – 9 = -3 9 7 7 – 9 = -2 4 10 10 – 9 = 1 1 12 12 – 9 = 3 9 13 13 – 9 = 4 16 4 4 – 9 = -5 25 8 8 – 9 = -1 1 12 12 – 9 = 3 9 $$\sum(x_{i}-\bar{x})^{2}$$ 74
Hence, Variance = $$\frac{\sum (x_{i}-\bar{x})^{2}}{n}$$ = 74/8
Variance = 9.25
Finding Standard Deviation:
We know that variance is the square of standard deviation. Hence, the standard deviation can be found by taking the square root of variance.
Therefore, standard deviation = √variance
Standard deviation = √(9.25) = 3.041.
Hence, the mean, variance and standard deviation of the given data are 9, 9.25, 3.041 respectively.
Thus, these are the key differences between variance and standard deviation. To know more about Maths-related articles, register with BYJU’S – The Learning App today.
## Frequently Asked Questions on Difference Between Variance and Standard Deviation
### What does the variance and the standard deviation tell us?
In probability theory and statistics, both the variance and standard deviation tell us how far the data values are spread out/dispersed from the mean of the given data set
### How to derive the variance from the standard deviation?
The variance can be easily derived from the standard deviation by taking the square of the standard deviation.
### Mention the use of variance in statistics.
In statistics, the variance is used to determine the measure of dispersion and the uncertainty in the given data set values.
### What exactly is SD in statistics?
The standard deviation, SD is the number which gives information about the spread of data values from the mean value. If SD is small, the data values are close to the mean value. If SD is high, the data values are widely spread out from the mean value. |
# The power rule for derivatives
Usually the first shortcut rule you study for finding derivatives is the power rule. The reason is that it is a simple rule to remember and it applies to all different kinds of functions. For a number n, the power rule states:
Let’s start with some really easy examples to see it in action.
### Example
Find the derivative of each function.
$$\text{(a) } x^4$$
$$\text{(b) } x^{10}$$
$$\text{(c) } x^{546}$$
### Solution
For each of these, you can simply apply the power rule without any algebra at all. This means that you should bring the exponent out front, and then subtract 1 from the exponent.
$$\text{(a) } \left(x^4\right)^{\prime} = 4x^3$$
$$\text{(b) } \left(x^{10}\right)^{\prime} = 10x^9$$
$$\text{(c) } \left(x^{546}\right)^{\prime} = 546x^{545}$$
As you can see, it is all about remembering the pattern. Now, we will see how this pattern can be applied to more complicated examples.
## Derivatives of polynomial functions
Recall that the derivative of a constant is always zero. So, the derivative of 5 is 0 while the derivative of 2,000 is also 0. Further, you can break the derivative up over addition/subtraction and multiplication by constants. Combining these ideas with the power rule allows us to use it for finding the derivative of any polynomial.
### Example
Find the derivative of the function.
$$y = 2x^4 – 5x^2 + 1$$
### Solution
With a little bit of practice, you will probably be able to write the derivative of this function down without thinking. Since it is our first example though, let’s write out every step. In the first step, we will break the derivative up over the addition and subtraction.
\begin{align} y^{\prime} &= \left(2x^4 – 5x^2 + 1\right)^{\prime}\\ &= \left(2x^4\right)^{\prime} – \left(5x^2\right)^{\prime} + \left(1\right)^{\prime}\end{align}
Now, factor out the coefficients:
$$= 2\left(x^4\right)^{\prime} – 5\left(x^2\right)^{\prime} + \left(1\right)^{\prime}$$
Apply the power rule for derivatives and the fact that the derivative of a constant is zero:
$$= 2\left(4x^3\right) – 5\left(2x^1\right) + \left(0\right)$$
Notice that once we applied the derivative rule, the prime went away. The correct notation keeps this until you apply a derivative rule. Now all we need to do is simplify to get our final answer.
$$= \boxed{8x^3 – 10x}$$
Let’s look at one more example without so much explanation to distract us.
### Example
Find the derivative of the function.
$$y = 5x^3 – 3x^2 + 10x – 8$$
### Solution
Apply the power rule, the rule for constants, and then simplify. Note that if $$x$$ doesn’t have an exponent written, it is assumed to be 1.
\begin{align} y^{\prime} &= \left(5x^3 – 3x^2 + 10x – 8\right)^{\prime}\\ &= 5\left(3x^2\right) – 3\left(2x^1\right) + 10\left(x^0\right)- 0\end{align}
Since $$x$$ was by itself, its derivative is $$1x^0$$. Normally, this isn’t written out however. Just remember that anything (other than zero) to the zero power is 1. So, $$10\left(x^0\right) = 10(1) = 10$$. Therefore, we can write the final answer as:
$$= \boxed{15x^2 – 6x + 10}$$
You may think this is all you can really do with the power rule. However, a couple of old algebra facts can help us apply this to a wider range of functions. We will look at two of those instances below.
## Derivatives of functions with negative exponents
The power rule applies whether the exponent is positive or negative. But sometimes, a function that doesn’t have any exponents may be able to be rewritten so that it does, by using negative exponents. If this is the case, then we can apply the power rule to find the derivative. The main property we will use is:
Let’s see what we can do with this property using an example!
### Example
Find the derivative of the function.
$$y = \dfrac{2}{x^4} – \dfrac{1}{x^2}$$
### Solution
Applying the rule for negative exponents, we can rewrite this function as:
$$y = \dfrac{2}{x^4} – \dfrac{1}{x^2} = 2x^{-4} – x^{-2}$$
Now that this is written with exponents, we can apply the power rule:
\begin{align} y^{\prime} &= \left(2x^{-4} – x^{-2}\right)^{\prime}\\ &= 2\left(-4x^{-4-1}\right) – \left(-2x^{-2-1}\right)\end{align}
Simplify to find the final answer:
\begin{align} &= -8x^{-5} +2x^{-3}\\ &= \boxed{-\dfrac{8}{x^{5}} + \dfrac{2}{x^{3}}}\end{align}
In the last step, notice that only the terms with the negative exponent were moved to the bottom of the fraction. The 8 didn’t have a negative exponent, so it stayed.
## Derivatives of functions with radicals (square roots and other roots)
Another useful property from algebra is the following.
Using this rule, we can take a function written with a root and find its derivative using the power rule.
### Example
Find the derivative of the function.
$$y = 4\sqrt{x} – 6\sqrt[3]{x^2}$$
### Solution
First, rewrite the function using algebra:
$$y = 4\sqrt{x} – 6\sqrt[3]{x^2} = 4x^{\frac{1}{2}} – 6x^{\frac{2}{3}}$$
Now apply the power rule:
\begin{align} y^{\prime} &= \left(4x^{\frac{1}{2}} – 6x^{\frac{2}{3}}\right)^{\prime}\\ &= 4\left(\dfrac{1}{2}x^{\frac{1}{2}-1}\right) – 6\left(\dfrac{2}{3}x^{\frac{2}{3}-1}\right)\end{align}
Simplify to get the final answer, remembering the rule for negative exponents:
\begin{align} &= 2x^{-\frac{1}{2}} – 4x^{-\frac{1}{3}}\\ &= \dfrac{2}{x^{\frac{1}{2}}} – \dfrac{4}{x^{\frac{1}{3}}}\\ &= \dfrac{2}{\sqrt{x}} – \dfrac{4}{\sqrt[3]{x}}\end{align}
In many classes, either of the last two lines can be written as your final answer. They are equivalent. However, your teacher or professor may have a preference, so always ask!
## Summary
As a student studying calculus, you want the power rule to be second nature. It’ll be applied not only like this – on its own – but also as part of other rules such as the chain rule, the quotient rule, and the product rule. The better you understand it, the more you can focus on those more complicated ideas. |
Tamil Nadu Board of Secondary EducationHSC Science Class 11th
# Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide chapter 2 - Basic Algebra [Latest edition]
## Chapter 2: Basic Algebra
Exercise 2.1Exercise 2.2Exercise 2.3Exercise 2.4Exercise 2.5Exercise 2.6Exercise 2.7Exercise 2.8Exercise 2.9Exercise 2.10Exercise 2.11Exercise 2.12Exercise 2.13
Exercise 2.1 [Page 55]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.1 [Page 55]
Exercise 2.1 | Q 1 | Page 55
Classify each element of {sqrt(7), (-1)/4, 0, 3, 1, 4, 4, 22/7} as a member of N, Q, R − Q or Z
Exercise 2.1 | Q 2 | Page 55
Prove that sqrt(3) is an irrational number.
(Hint: Follow the method that we have used to prove sqrt(2) ∉ Q)
Exercise 2.1 | Q 3 | Page 55
Are there two distinct irrational numbers such that their difference is a rational number? Justify
Exercise 2.1 | Q 4 | Page 55
Find two irrational numbers such that their sum is a rational number. Can you find two irrational numbers whose product is a rational number
Exercise 2.1 | Q 5 | Page 55
Find a positive number smaller than 2^(1/1000). Justify
Exercise 2.2 [Page 57]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.2 [Page 57]
Exercise 2.2 | Q 1. (i) | Page 57
Solve for x: |3 − x| < 7
Exercise 2.2 | Q 1. (ii) | Page 57
Solve for x: |4x − 5| ≥ −2
Exercise 2.2 | Q 1. (iii) | Page 57
Solve for x: |3 - 3/4x| ≤ 1/4
Exercise 2.2 | Q 1. (iv) | Page 57
Solve for x: |x| − 10 < −3
Exercise 2.2 | Q 2 | Page 57
Solve 1/(|2x - 1|) < 6 and express the solution using the interval notation
Exercise 2.2 | Q 3 | Page 57
Solve −3|x| + 5 ≤ −2 and graph the solution set in a number line
Exercise 2.2 | Q 4 | Page 57
Solve 2|x + 1| − 6 ≤ 7 and graph the solution set in a number line
Exercise 2.2 | Q 5 | Page 57
Solve 1/5|10x - 2| < 1
Exercise 2.2 | Q 6 | Page 57
Solve |5x − 12| < −2
Exercise 2.3 [Page 59]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.3 [Page 59]
Exercise 2.3 | Q 1. (i) | Page 59
Represent the following inequalities in the interval notation:
x ≥ −1 and x < 4
Exercise 2.3 | Q 1. (ii) | Page 59
Represent the following inequalities in the interval notation:
x ≤ 5 and x ≥ − 3
Exercise 2.3 | Q 1. (iii) | Page 59
Represent the following inequalities in the interval notation:
x < −1 or x < 3
Exercise 2.3 | Q 1. (iv) | Page 59
Represent the following inequalities in the interval notation:
−2x > 0 or 3x − 4 < 11
Exercise 2.3 | Q 2. (i) | Page 59
Solve 23x < 100 when x is a natural number
Exercise 2.3 | Q 2. (ii) | Page 59
Solve 23x < 100 when x is an integer
Exercise 2.3 | Q 3. (i) | Page 59
Solve −2x ≥ 9 when x is a real number
Exercise 2.3 | Q 3. (ii) | Page 59
Solve −2x ≥ 9 when x is an integer
Exercise 2.3 | Q 3. (iii) | Page 59
Solve −2x ≥ 9 when x is a natural number
Exercise 2.3 | Q 4. (i) | Page 59
Solve: (3(x - 2))/5 ≤ (5(2 - x))/3
Exercise 2.3 | Q 4. (ii) | Page 59
Solve: (5 - x)/3 < x/2 - 4
Exercise 2.3 | Q 5 | Page 59
To secure A grade one must obtain an average of 90 marks or more in 5 subjects each of maximum 100 marks. If one scored 84, 87, 95, 91 in first four subjects, what is the minimum mark one scored in the fifth subject to get A grade in the course?
Exercise 2.3 | Q 6 | Page 59
A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent acid solution must be added to it so that the acid content in the resulting mixture will be more than 15 percent but less than 18 percent?
Exercise 2.3 | Q 7 | Page 59
Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum is less than 40
Exercise 2.3 | Q 8 | Page 59
A model rocket is launched from the ground. The height h reached by the rocket after t seconds from lift off is given by h(t) = −5t2 +100t, 0 ≤ t ≤ 20. At what time the rocket is 495 feet above the ground?
Exercise 2.3 | Q 9 | Page 59
A plumber can be paid according to the following schemes: In the first scheme he will be paid rupees 500 plus rupees 70 per hour, and in the second scheme he will be paid rupees 120 per hour. If he works x hours, then for what value of x does the first scheme give better wages?
Exercise 2.3 | Q 10 | Page 59
A and B are working on similar jobs but their monthly salaries differ by more than Rs 6000. If B earns rupees 27000 per month, then what are the possibilities of A’s salary per month?
Exercise 2.4 [Page 62]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.4 [Page 62]
Exercise 2.4 | Q 1 | Page 62
Construct a quadratic equation with roots 7 and −3
Exercise 2.4 | Q 2 | Page 62
A quadratic polynomial has one of its zeros 1 + sqrt(5) and it satisfies p(1) = 2. Find the quadratic polynomial
Exercise 2.4 | Q 3 | Page 62
If α and β are the roots of the quadratic equation x^2 + sqrt(2)x + 3 = 0, form a quadratic polynomial with zeroes 1/α, 1/β
Exercise 2.4 | Q 4 | Page 62
If one root of k(x − 1)2 = 5x − 7 is double the other root, show that k = 2 or −25
Exercise 2.4 | Q 5 | Page 62
If the difference of the roots of the equation 2x2 − (a + 1)x + a − 1 = 0 is equal to their product, then prove that a = 2
Exercise 2.4 | Q 6. (i) | Page 62
Find the condition that one of the roots of ax2 + bx + c may be negative of the other
Exercise 2.4 | Q 6. (ii) | Page 62
Find the condition that one of the roots of ax2 + bx + c may be thrice the other
Exercise 2.4 | Q 6. (iii) | Page 62
Find the condition that one of the roots of ax2 + bx + c may be reciprocal of the other
Exercise 2.4 | Q 7 | Page 62
If the equations x2 − ax + b = 0 and x2 − ex + f = 0 have one root in common and if the second equation has equal roots, then prove that ae = 2(b + f)
Exercise 2.4 | Q 8. (I) | Page 62
Discuss the nature of roots of − x2 + 3x + 1 = 0
Exercise 2.4 | Q 8. (ii) | Page 62
Discuss the nature of roots of 4x2 − x − 2 = 0
Exercise 2.4 | Q 8. (iii) | Page 62
Discuss the nature of roots of 9x2 + 5x = 0
Exercise 2.4 | Q 9. (i) | Page 62
Without sketching the graph, find whether the graph of the following function will intersect the x-axis and if so in how many points
y = x2 + x + 2
Exercise 2.4 | Q 9. (ii) | Page 62
Without sketching the graph, find whether the graph of the following function will intersect the x-axis and if so in how many points
y = x2 − 3x − 7
Exercise 2.4 | Q 9. (iii) | Page 62
Without sketching the graph, find whether the graph of the following function will intersect the x-axis and if so in how many points
y = x2 + 6x + 9
Exercise 2.4 | Q 10 | Page 62
Write f(x) = x2 + 5x + 4 in completed square form
Exercise 2.5 [Page 63]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.5 [Page 63]
Exercise 2.5 | Q 1 | Page 63
Solve 2x2 + x – 15 ≤ 0
Exercise 2.5 | Q 2 | Page 63
Solve – x2 + 3x – 2 ≥ 0
Exercise 2.6 [Page 66]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.6 [Page 66]
Exercise 2.6 | Q 1 | Page 66
Find the zeros of the polynomial function f(x) = 4x2 − 25
Exercise 2.6 | Q 2 | Page 66
If x = −2 is one root of x3 − x2 − 17x = 22, then find the other roots of equation
Exercise 2.6 | Q 3 | Page 66
Find the real roots of x4 = 16
Exercise 2.6 | Q 4 | Page 66
Solve (2x + 1)2 − (3x + 2)2 = 0
Exercise 2.7 [Page 68]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.7 [Page 68]
Exercise 2.7 | Q 1 | Page 68
Factorize: x4 + 1. (Hint: Try completing the square)
Exercise 2.7 | Q 2 | Page 68
If x2 + x + 1 is a factor of the polynomial 3x3 + 8x2 + 8x + a, then find the value of a
Exercise 2.8 [Page 69]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.8 [Page 69]
Exercise 2.8 | Q 1 | Page 69
Find all values of x for which (x^3(x - 1))/((x - 2)) > 0
Exercise 2.8 | Q 2 | Page 69
Find all values of x that satisfies the inequality (2x - 3)/((x - 2)(x - 4)) < 0
Exercise 2.8 | Q 3 | Page 69
Solve (x^2 - 4)/(x^2 - 2x - 15) ≤ 0
Exercise 2.9 [Page 71]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.9 [Page 71]
Exercise 2.9 | Q 1 | Page 71
Resolve the following rational expressions into partial fractions
1/(x^2 - "a"^2)
Exercise 2.9 | Q 2 | Page 71
Resolve the following rational expressions into partial fractions
(3x + 1)/((x - 2)(x + 1))
Exercise 2.9 | Q 3 | Page 71
Resolve the following rational expressions into partial fractions
x/((x^2 + 1)(x - 1)(x + 2))
Exercise 2.9 | Q 4 | Page 71
Resolve the following rational expressions into partial fractions
x/((x - 1)^3
Exercise 2.9 | Q 5 | Page 71
Resolve the following rational expressions into partial fractions
1/(x^4 - 1)
Exercise 2.9 | Q 6 | Page 71
Resolve the following rational expressions into partial fractions
(x - 1)^2/(x^3 + x)
Exercise 2.9 | Q 7 | Page 71
Resolve the following rational expressions into partial fractions
(x^2 + x + 1)/(x^2 - 5x + 6)
Exercise 2.9 | Q 8 | Page 71
Resolve the following rational expressions into partial fractions
(x^3 + 2x + 1)/(x^2 + 5x + 6)
Exercise 2.9 | Q 9 | Page 71
Resolve the following rational expressions into partial fractions
(x + 12)/((x + 1)^2 (x - 2))
Exercise 2.9 | Q 10 | Page 71
Resolve the following rational expressions into partial fractions
(6x^2 - x + 1)/(x^3 + x^2 + x + 1)
Exercise 2.9 | Q 11 | Page 71
Resolve the following rational expressions into partial fractions
(2x^2 + 5x - 11)/(x^2 + 2x - 3)
Exercise 2.9 | Q 12 | Page 71
Resolve the following rational expressions into partial fractions
(7 + x)/((1 + x)(1 + x^2))
Exercise 2.10 [Page 73]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.10 [Page 73]
Exercise 2.10 | Q 1 | Page 73
Determine the region in the plane determined by the inequalities:
x ≤ 3y, x ≥ y
Exercise 2.10 | Q 2 | Page 73
Determine the region in the plane determined by the inequalities:
y ≥ 2x, −2x + 3y ≤ 6
Exercise 2.10 | Q 3 | Page 73
Determine the region in the plane determined by the inequalities:
3x + 5y ≥ 45, x ≥ 0, y ≥ 0
Exercise 2.10 | Q 4 | Page 73
Determine the region in the plane determined by the inequalities:
2x + 3y ≤ 35, y ≥ 2, x ≥ 5.
Exercise 2.10 | Q 5 | Page 73
Determine the region in the plane determined by the inequalities:
2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0
Exercise 2.10 | Q 6 | Page 73
Determine the region in the plane determined by the inequalities:
x − 2y ≥ 0, 2x − y ≤ −2, x ≥ 0, y ≥ 0
Exercise 2.10 | Q 7 | Page 73
Determine the region in the plane determined by the inequalities:
2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6
Exercise 2.11 [Page 77]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.11 [Page 77]
Exercise 2.11 | Q 1. (i) | Page 77
Simplify: (125)^(2/3)
Exercise 2.11 | Q 1. (ii) | Page 77
Simplify: 16^((-3)/4)
Exercise 2.11 | Q 1. (iii) | Page 77
Simplify: (- 1000)^((-2)/3)
Exercise 2.11 | Q 1. (iv) | Page 77
Simplify: (3^-6)^(1/3)
Exercise 2.11 | Q 1. (v) | Page 77
Simplify: (27^((-2)/3))/(27^((-1)/3))
Exercise 2.11 | Q 2 | Page 77
Evaluate [((256)^(-1/2))^((-1)/4)]^3
Exercise 2.11 | Q 3 | Page 77
If (x^(1/2) + x^(- 1/2))^2 = 9/2 then find the value of (x^(1/2) - x^(-1/2)) for x >1
Exercise 2.11 | Q 4 | Page 77
Simplify and hence find the value of n:
(3^(2"n")*9^2*3^(-"n"))/(3^(3"n")) = 27
Exercise 2.11 | Q 5 | Page 77
Find the radius of the spherical tank whose volume is (32pi)/3 units
Exercise 2.11 | Q 6 | Page 77
. Simplify by rationalising the denominator (7 + sqrt(6))/(3 - sqrt(2))
Exercise 2.11 | Q 7 | Page 77
Simplify 1/(3 - sqrt(8)) - 1/(sqrt(8) - sqrt(7)) + 1/(sqrt(7) - sqrt(6)) - 1/(sqrt(6) - sqrt(5)) + 1/(sqrt(5) - 2)
Exercise 2.11 | Q 8 | Page 77
If x = sqrt(2) + sqrt(3) find (x^2 + 1)/(x^2 - 2)
Exercise 2.12 [Pages 80 - 81]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.12 [Pages 80 - 81]
Exercise 2.12 | Q 1 | Page 80
Let b > 0 and b ≠ 1. Express y = bx in logarithmic form. Also state the domain and range of the logarithmic function
Exercise 2.12 | Q 2 | Page 80
Compute log9 27 – log27 9
Exercise 2.12 | Q 3 | Page 80
Solve log8x + log4x + log2x = 11
Exercise 2.12 | Q 4 | Page 80
Solve log28x = 2log28
Exercise 2.12 | Q 5 | Page 80
If a2 + b2 = 7ab, show that log ("a" + "b")/3 = 1/2(log"a" + log "b")
Exercise 2.12 | Q 6 | Page 80
Prove log "a"^2/"bc" + log "b"^2/"ca" + log "c"^2/"ab" = 0
Exercise 2.12 | Q 7 | Page 80
Prove that log 2 + 16log 16/15 + 12log 25/24 + 7log 81/80 = 1
Exercise 2.12 | Q 8 | Page 80
Prove that loga2 a + logb2 b + logc2 c = 1/8
Exercise 2.12 | Q 9 | Page 80
Prove log a + log a2 + log a3 + · · · + log an = ("n"("n" + 1))/2 log "a"
Exercise 2.12 | Q 10 | Page 81
If log x/(y - z) = logy/(z - x) = logz/(x - y), then prove that xyz = 1
Exercise 2.12 | Q 11 | Page 81
Solve log_2 x − 3 log_(1/2) x = 6
Exercise 2.12 | Q 12 | Page 81
Solve log5 – x (x2 – 6x + 65) = 2
Exercise 2.13 [Pages 81 - 83]
### Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide Chapter 2 Basic Algebra Exercise 2.13 [Pages 81 - 83]
#### MCQ
Exercise 2.13 | Q 1 | Page 81
Choose the correct alternative:
If |x + 2| ≤ 9, then x belongs to
• (−∞, −7)
• [−11, 7]
• (−∞, −7) ∪ [11, ∞)
• (−11, 7)
Exercise 2.13 | Q 2 | Page 81
Choose the correct alternative:
Given that x, y and b are real numbers x < y, b > 0, then
• xb < yb
• xb > yb
• xb ≤ yb
• x/"b" ≥ y/"b"
Exercise 2.13 | Q 3 | Page 81
Choose the correct alternative:
If |x - 2|/(x - 2) ≥ 0, then x belongs to
• [2, ∞)
• (2, ∞)
• (−∞, 2)
• (−2, ∞)
Exercise 2.13 | Q 4 | Page 81
Choose the correct alternative:
The solution of 5x − 1 < 24 and 5x + 1 > −24 is
• (4, 5)
• (−5, −4)
• (−5, 5)
• (−5, 4)
Exercise 2.13 | Q 5 | Page 81
Choose the correct alternative:
The solution set of the following inequality |x − 1| ≥ |x − 3| is
• [0, 2]
• [2, ∞)
• (0, 2)
• (−∞, 2)
Exercise 2.13 | Q 6 | Page 82
Choose the correct alternative:
The value of log_(sqrt(5)) 512 is
• 16
• 18
• 9
• 12
Exercise 2.13 | Q 7 | Page 82
Choose the correct alternative:
The value of log_3 1/81 is
• −2
• −8
• −4
• −9
Exercise 2.13 | Q 8 | Page 82
Choose the correct alternative:
If log_(sqrt(x) 0.25 = 4, then the value of x is
• 0.5
• 2.5
• 1.5
• 1.25
Exercise 2.13 | Q 9 | Page 82
Choose the correct alternative:
The value of logab logbc logca is
• 2
• 1
• 3
• 4
Exercise 2.13 | Q 10 | Page 82
Choose the correct alternative:
If 3 is the logarithm of 343, then the base is
• 5
• 7
• 6
• 9
Exercise 2.13 | Q 11 | Page 82
Choose the correct alternative:
Find a so that the sum and product of the roots of the equation 2x2 + (a − 3)x + 3a − 5 = 0 are equal is
• 1
• 2
• 0
• 4
Exercise 2.13 | Q 12 | Page 82
Choose the correct alternative:
If a and b are the roots of the equation x2 − kx + 16 = 0 and satisfy a2 + b2 = 32, then the value of k is
• 10
• −8
• −8, 8
• 6
Exercise 2.13 | Q 13 | Page 82
Choose the correct alternative:
The number of solutions of x2 + |x − 1| = 1 is
• 1
• 0
• 2
• 3
Exercise 2.13 | Q 14 | Page 82
Choose the correct alternative:
The equation whose roots are numerically equal but opposite in sign to the roots of 3x2 − 5x − 7 = 0 is
• 3x2 − 5x − 7 = 0
• 3x2 + 5x − 7 = 0
• 3x2 − 5x + 7 = 0
• 3x2 + x − 7 = 0
Exercise 2.13 | Q 15 | Page 82
Choose the correct alternative:
If 8 and 2 are the roots of x2 + ax + c = 0 and 3, 3 are the roots of x2 + dx + b = 0, then the roots of the equation x2 + ax + b = 0 are
• 1, 2
• −1, 1
• 9, 1
• −1, 2
Exercise 2.13 | Q 16 | Page 82
Choose the correct alternative:
If a and b are the real roots of the equation x2 − kx + c = 0, then the distance between the points (a, 0) and (b, 0) is
• sqrt("k"^2 - 4"c")
• sqrt(4"k"^2 - "c")
• sqrt(4"c" - "k"^2)
• sqrt("k" - 8"c")
Exercise 2.13 | Q 17 | Page 83
Choose the correct alternative:
If ("k"x)/((x + 2)(x - 1)) = 2/(x + 2) + 1/(x - 1), then the value of k is
• 1
• 2
• 3
• 4
Exercise 2.13 | Q 18 | Page 83
Choose the correct alternative:
If (1 - 2x)/(3 + 2x - x^2) = "A"/(3 - x) + "B"/(x + 1), then the value of A + B is
• (-1)/2
• (-2)/3
• 1/2
• 2/3
Exercise 2.13 | Q 19 | Page 83
Choose the correct alternative:
The number of roots of (x + 3)4 + (x + 5)4 = 16 is
• 4
• 2
• 3
• 0
Exercise 2.13 | Q 20 | Page 83
Choose the correct alternative:
The value of log3 11 . log11 13 . log13 15 . log15 27 . log27 81 is
• 1
• 2
• 3
• 4
## Chapter 2: Basic Algebra
Exercise 2.1Exercise 2.2Exercise 2.3Exercise 2.4Exercise 2.5Exercise 2.6Exercise 2.7Exercise 2.8Exercise 2.9Exercise 2.10Exercise 2.11Exercise 2.12Exercise 2.13
## Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide chapter 2 - Basic Algebra
Tamil Nadu Board Samacheer Kalvi solutions for Class 11th Mathematics Volume 1 and 2 Answers Guide chapter 2 (Basic Algebra) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Tamil Nadu Board of Secondary Education Class 11th Mathematics Volume 1 and 2 Answers Guide solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Tamil Nadu Board Samacheer Kalvi textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 11th Mathematics Volume 1 and 2 Answers Guide chapter 2 Basic Algebra are Introduction to Basic Algebra, Real Number System, Absolute Value, Linear Inequalities, Quadratic Functions, Polynomial Functions, Rational Functions, Exponents and Radicals, Logarithms, Application of Algebra in Real Life.
Using Tamil Nadu Board Samacheer Kalvi Class 11th solutions Basic Algebra exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Tamil Nadu Board Samacheer Kalvi Solutions are important questions that can be asked in the final exam. Maximum students of Tamil Nadu Board of Secondary Education Class 11th prefer Tamil Nadu Board Samacheer Kalvi Textbook Solutions to score more in exam.
Get the free view of chapter 2 Basic Algebra Class 11th extra questions for Class 11th Mathematics Volume 1 and 2 Answers Guide and can use Shaalaa.com to keep it handy for your exam preparation |
Polynomial Codes and some lore about Polynomials
10.1 Introduction:
We will now introduce more structure for our message words and code words,
and use it to get single error correcting codes that can be described more easily, as well as
codes that can correct several errors.
We do this by treating our sequences as polynomials and defining
multiplication for them .
Given a sequence, say 1011, we can make it into a polynomial by converting each
bit into and adding the results , For the sequence 1011 we get 1 + x2 + x3.
We multiply our sequences by the usual laws of multiplication, again with the
2=0 rule Thus, in multiplying (111) by (1011). We multiply 1+x+x2 by 1+x2+x3 which
using the distributive law comes to 1 + x +2x2+ 2x3 + 2x4 + x5 which we interpret to be
1 +x+x5.
We will describe encoding by use of an encoding polynomial p(x) and our rule for
encoding will be
Thus the task of finding useful codes is, in this context, the task of finding useful
polynomials. In order to find them we first look at some properties of polynomials.
10.2 Binary Polynomials
We will find, that if you want to find efficient single error correcting polynomial
codes, you can do so by choosing an encoding polynomial that is "primitive". Primitive
means prime, plus one other property we shall soon describe. So we will begin by
looking at low degree prime polynomials.
Recall that binary numbers obey these rules for addition:
0 + 0 = 1+1 = 0
1 + 0 = 0+1 = 1
and these for multiplication:
0*0 = 0*1 = 1*0 = 0
1*1 = 1.
A polynomial defined over this "field" (the field is called GF(2)) is no more
than a polynomial whose coefficients are each 0 or 1.
Here is a polynomial of degree 5: 1 + x2 + x5.
Polynomials have the wonderful property that you can add and subtract them and
also multiply and divide them using all the standard commutative, distributive and
associative laws of arithmetic.
When you divide one polynomial by another, you get a quotient polynomial, and
perhaps a remainder. The rules for dividing are the rules for long division. (actually,
when you write a number as say 543 you are treating it as a sort of polynomial, namely
5*102 + 4*10 + 3, with 10 instead of a variable, so the rules for dividing same are the
rules for polynomials, except for carrying, which you do with numbers and not with
polynomials. (in our case, given 2=0, there is no carrying to do)
With our polynomials life is much more pleasant than what you encountered in
your youth when you want to do arithmetic since the only possible non- zero coefficient
of our polynomials is 1,
We give an illustration of division of polynomials.
Suppose we want to divide x8 by 1 + x2 + x5.
The latter goes x3 times into the former, with a remainder of x5 + x3. It goes once into this
remainder with x3 + x2 + 1 left over. The resultant quotient is therefore x3 + 1 with
remainder x3 + x2 + 1..
A polynomial is said to be prime if it cannot be factored into polynomials of
lower degree.
The polynomial of degree 5 above happens to be prime. On the other hand the
polynomial (1 + x2 + x4) is not prime, being (1 + x + x2)2.
We will now look at polynomials of low degree and identify which are primes.
We can immediately see that x is the only monomial that is prime ; the rest have it
as a factor.
Any polynomial, p(x) with an even number of terms has (1 + x) as a factor.
Why?
Because if you set x = 1 you find p(1)=0, when p has an even number of terms.
When you divide p(x) by (1+x) you get a remainder that must have 0 degree and
so must be 0 or 1.
But we have p(x)=q(x)(1+x) + r(x) with q(x) the quotient upon dividing p by (1+x), and
r(x) the remainder.
Evaluating this at x=1 we get 0 = p(1) = q(1)*0 + r(1) .
From which we conclude that r=r(1) = 0, and this means (1+x) divides any polynomial
with an even number of terms
Any polynomial without a 1 term is divisible by x.
Further, any polynomial whose terms all have even degree is the square of
the polynomial
whose degrees are all half of its, so it cannot be prime.
This is true because the ugly cross terms which normally occur when you square a
polynomial all have factors of 2 multiplying them so they are all 0 here.
There is one more fact that is useful when examining polynomials.
If we take a sequence like 1011 and make it into a polynomial, we can do so in two ways.
The way we have chosen is to start from the left, and have successive bits correspond to
increasing powers .
We could just as well have done the same thing starting from the right.
But this gives us a different polynomial , which we call the "reverse polynomial" to the
first.
In our example, our polynomial is 1 +x2 + x3 and the reverse one is 1+x+x2.
Now no important property of our code can depend on whether we started from the left or
the right in turning our sequences into polynomials.
We conclude from this that the properties of a polynomial and its reverse such as
primitivity and primeness must be the same for each polynomial and its reverse.
This conclusion will be justified by what we see very soon: that these properties have a
direct effect on the error correcting properties of the code a polynomial generates.
So let us explore the polynomials of low degree and classify them as prime or
composite.
There are two polynomials of degree 1, x and 1+x and both are prime
:
The only polynomial of degree 2 with terms 1 and x2 and an odd number of terms
is 1 + x + x2, and therefore it is the only possible prime. It is prime, as you can see
because it is not a product of any combination of the degree 1 primes. It is symmetric on
reversal of bits
The only possible primes of degree three with odd number of terms having 1 and
x3 as terms are 1 + x + x3 and 1+ x2 + x3 which are reverses of one another. If they were
not primes they would have to have a factor of degree 1, but they do not, because they
have an odd number of terms.
Among polynomials of degree 4 the only polynomials with an odd number of
terms not all squares with a constant term and one of degree 4 are 1 + x + x4, 1 + x3 + x4,
which are reverses of one another and 1+x+x2+x3+x4. which is symmetric. By the
reasoning used for polynomials of degree 3 in the last paragraph, all of these are prime.
(The square of (1+x+x2) is 1+x2+x4)
By similar reasoning the candidates for primes of degree 5 are 1+ x+ x5, 1+x2 +
x5, all terms except x, all terms except x2 and the reverses of these 4 polynomials
Of these (1+x+x2)(1+x+x3) and its reverse are not prime and the others are prime.
You could, if you wanted to, continue this investigation to degree 6,…,10 at least .
For example, for degree 8, by my calculation, there are 64 polynomials with an odd
number of terms including 1 and x8, which form 28 pairs of a polynomial and its reverse,
and 8 symmetric polynomials. Of these about half are primes.
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# What is 4 multiply 3?
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Related questions
### How can we multiply 4 by 223?
You can multiply 3 by 4, then 20 by 4, then 200 by 4.
### Do you say multiply by or multiply with?
Multiply by is the usual form. Multiply 3 by 4 to get 12. If there are several numbers being multiplied then multiply .... together may be used... multiply 3, 4 and 5 together to get 60.
### Do you inverse multiply to divide fractions?
To divide fractions, you multiply by the reciprocal. For example, 3/4 (divided by) 3/4 would be converted to 3/4 (divided by) 4/3, you just flip the second fraction and multiply as you normally would if you were multiplying them.
### How do you make 24 by using 3 4 5 6 8?
Multiply 6 and 4. Multiply 8 and 3.
3/4
### Multiply rational fractions and mixed numbers?
EX: 3 3/4 x 4 4/10 first, you multiply 4x3, then you get 12 + 3=15/4 second, you multiply 10x4=40+4=44/10 then you multiply the denomanator by the denomanator and the numerator by the numerator. 15x44= 10x4= Then simplify your answer.
3
### How do you multiply three fourths by three?
multiply 3/4 by three. first multiply the numerator and three. so 3 times 3 is 9. then divide it by the denomator . so it will be 9 over 4. 9/4 = 2 1/4
### How you find 75 percent of a number?
To find 75% of a number, multiply the number by 0.75. This can be done by multiplying the number by 3/4 or by dividing the number by 4 and then multiplying by 3.
### How do you multiply a number by 0.75?
Multiply it by 3/4 which is equivalent to 0.75
### What is one quarter multiply by 3?
1/4 x 3 = (1x3) / 4 = 3/4
12 |
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# What is the Prime Factorization Of 62?
• Equcation for number 62 factorization is: 2 * 31
• It is determined that the prime factors of number 62 are: 2, 31
## Is 62 A Prime Number?
• No the number 62 is not a prime number.
• Sixty-two is a composite number. Because 62 has more divisors than 1 and itself.
## How To Calculate Prime Number Factorization
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
## Mathematical Information About Numbers 6 2
• About Number 6. Six is the smallest composite number with two distinct prime factors, and the third triangular number. It is the smallest perfect number: 6 = 1 + 2 + 3 and the faculty of 3 is 6 = 3! = 1 * 2 * 3, which is remarkable, because there is no other three numbers whose product is equal to their sum. Similarly 6 = sqrt(1 ^ 3 + 2 + 3 ^ 3 ^ 3). The equation x ^ 3 + Y ^ 3 ^ 3 + z = 6xyz is the only solution (without permutations) x = 1, y = 2 and z = 3. Finally 1/1 = 1/2 + 1/3 + 1/6. The cube (from the Greek) or hexahedron (from Latin) cube is one of the five Platonic solids and has six equal areas. A tetrahedron has six edges and six vertices an octahedron. With regular hexagons can fill a plane without gaps. Number six is a two-dimensional kiss number.
• About Number 2. Two is the smallest and the only even prime number. Also it's the only prime which is followed by another prime number three. All even numbers are divisible by 2. Two is the third number of the Fibonacci sequence. Gottfried Wilhelm Leibniz discovered the dual system (binary or binary system) that uses only two digits to represent numbers. It witnessed the development of digital technology for a proliferation. Because of this, it is the best known and most important number system in addition to the commonly used decimal system.
## What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers.
## What is Prime Number Factorization?
• In mathematics, factorization (also factorisation in some forms of British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. For example, the number 15 factors into primes as 3 x 5, and the polynomial x2 - 4 factors as (x - 2)(x + 2). In all cases, a product of simpler objects is obtained. The aim of factoring is usually to reduce something to basic building blocks, such as numbers to prime numbers, or polynomials to irreducible polynomials.
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# 2006 iTest Problems/Problem 32
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Triangle $ABC$ is scalene. Points $P$ and $Q$ are on segment $BC$ with $P$ between $B$ and $Q$ such that $BP=21$, $PQ=35$, and $QC=100$. If $AP$ and $AQ$ trisect $\angle A$, then $\tfrac{AB}{AC}$ can be written uniquely as $\tfrac{p\sqrt q}r$, where $p$ and $r$ are relatively prime positive integers and $q$ is a positive integer not divisible by the square of any prime. Determine $p+q+r$.
## Solution
Let $a = AB$ and $b = AC$. Since $\angle BAP = \angle PAQ = \angle QAC$, by the Angle Bisector Theorem, we have $AP = \tfrac{7}{20}b$ and $AQ = \tfrac{5}{3}a$.
By using the Law of Cosines on $\triangle BAP$ and $\triangle PAQ$, we have \begin{align*} \frac{a^2 + \frac{49}{400}b^2 - 21^2}{2ab \cdot \frac{7}{20}} &= \frac{\frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2}{2ab \cdot \frac{7}{20} \cdot \frac{5}{3}} \\ \frac{5}{3} \cdot \left( a^2 + \frac{49}{400}b^2 - 21^2 \right) &= \frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 \\ \frac53 a^2 + \frac{49}{240}b^2 - 21 \cdot 35 &= \frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 \\ \frac{98}{1200} b^2 + 35 \cdot 14 &= \frac{10}{9} a^2 \\ \frac{49}{600} b^2 + 35 \cdot 14 &= \frac{40}{36} a^2. \end{align*} By using the Law of Cosines on $\triangle PAQ$ and $\triangle QAC$, we have \begin{align*} \frac{\frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2}{2ab \cdot \frac{7}{20} \cdot \frac{5}{3}} &= \frac{b^2 + \frac{25}{9}a^2 - 100^2}{2ab \cdot \frac{5}{3}} \\ \frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 &= \frac{7}{20} \cdot \left( b^2 + \frac{25}{9}a^2 - 100^2 \right) \\ \frac{49}{400}b^2 + \frac{25}{9}a^2 - 35^2 &= \frac{7}{20} b^2 + \frac{35}{36} a^2 - 100 \cdot 35 \\ \frac{65}{36} a^2 + 65 \cdot 35 &= \frac{91}{400}b^2 \\ \frac{5}{36} a^2 + 5 \cdot 35 &= \frac{7}{400}b^2. \end{align*} Multiplying the second equation by $-8$ and adding the two equations results in \begin{align*} -\frac{40}{36} a^2 - 40 \cdot 35 &= -\frac{7}{50} b^2 \\ \frac{40}{36} a^2 &= \frac{49}{600} b^2 + 35 \cdot 14 \\ -40 \cdot 35 &= -\frac{35}{600} b^2 + 35 \cdot 14 \\ -40 &= -\frac{1}{600} b^2 + 14 \\ \frac{b^2}{600} &= 54 \\ b^2 &= 600 \cdot 9 \cdot 6 \\ b &= 6 \cdot 10 \cdot 3 \\ &= 180. \end{align*} After substituting $b$ back, solve for $a$ to get \begin{align*} \frac{49}{600} \cdot 180 \cdot 180 + 35 \cdot 14 &= \frac{40}{36} a^2 \\ 49 \cdot 54 + 35 \cdot 14 &= \frac{40}{36} a^2 \\ 49 \cdot 54 + 49 \cdot 10 &= \frac{40}{36} a^2 \\ 49 \cdot 64 &= \frac{10}{9} a^2 \\ a^2 &= \frac{49 \cdot 9 \cdot 64}{10} \\ a &= \frac{168}{\sqrt{10}} \\ &= \frac{84\sqrt{10}}{5} \end{align*} Thus, $\tfrac{AB}{AC} = \tfrac{84\sqrt{10}}{5} \cdot \tfrac{1}{180} = \tfrac{7\sqrt{10}}{75}$, so $p+q+r = \boxed{92}$. |
# SOLVING SURFACE AREA PROBLEMS
Solving Surface Area Problems :
We may need to find surface area of different figures to solve many real-world problems.
In this section, we are going to see, how surface area of figures can be used to solve real world problems.
Let us see, how to find surface area of a prism.
## Solving Surface Area Problems - Examples
Example 1 :
Erin is making a jewelry box of wood in the shape of a rectangular prism. The jewelry box will have the dimensions shown below. The cost of painting the exterior of the box is \$0.50 per square in. How much does Erin have to spend to paint the jewelry box ?
Solution :
To know that total cost of painting, first we have to know the Surface area of the jewelry box.
Find surface area of the box.
Step 1 :
Identify a base, and find its area and perimeter.
Any pair of opposite faces can be the bases. For example, we can choose the bottom and top of the box as the bases.
Find base area.
B = l x w
B = 12 x 15
B = 180 square in.
Find perimeter of the base.
P = 2(12) + 2(15)
P = 24 + 30
P = 54 in.
Step 2 :
Identify the height, and find the surface area.
The height h of the prism is 6 inches. Use the formula to find the surface area.
S = Ph + 2B
S = 54(6) + 2(180)
S = 684 square inches
Step 3 :
Total cost = Area x Cost per square in.
Total cost = 684 x \$0.50
Total cost = \$342
Hence, Erin has to spend \$342 to paint the jewelry box.
Example 2 :
A metal box that is in the shape of rectangular prism has the following dimensions. The length is 9 inches, width is 2 inches, and height is 1 1/2 inches. Find the total cost of silver coating for the entire box.
Solution :
To know that total cost of silver coating, first we have to know the Surface area of the metal box.
Find surface area of the box.
Step 1 :
Identify a base, and find its area and perimeter.
Any pair of opposite faces can be the bases. For example, we can choose the bottom and top of the box as the bases.
Find base area.
B = l x w
B = 9 x 2
B = 18 square in.
Find perimeter of the base.
P = 2(9) + 2(2)
P = 18 + 4
P = 22 in.
Step 2 :
Identify the height, and find the surface area.
The height h of the prism is 1 1/2 inches. Use the formula to find the surface area.
S = Ph + 2B
S = 22(1 1/2) + 2(18)
S = 22(3/2) + 36
S = 33 + 36
S = 69 square inches
Step 3 :
Total cost = Area x Cost per square in.
Total cost = 69 x \$1.50
Total cost = \$103.50
Hence, the total cost of silver coating for the entire box is \$103.50.
After having gone through the stuff given above, we hope that the students would have understood, how to solve surface area problems.
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Lesson Video: Applications on Sine and Cosine Laws | Nagwa Lesson Video: Applications on Sine and Cosine Laws | Nagwa
# Lesson Video: Applications on Sine and Cosine Laws Mathematics
In this video, we will learn how to use the laws of sines and cosines to solve real-world problems.
17:52
### Video Transcript
In this video, we will learn how to use the cosine and sine law to solve real-world problems. As we’re focusing on the applications of these laws, we’ll assume at this point familiarity with the laws themselves. And we won’t explain the basics of these laws or how to prove them in this video.
Firstly though, a quick reminder of what the sine and cosine laws actually are and when they can be used. Suppose we have a triangle 𝐴𝐵𝐶 labeled as follows. Remember that the convention is to use uppercase letters to represent angles and to use lowercase letters to represent sides, with sides always being opposite the angle with the same letter. None of the angles in this triangle are necessarily right angles. So the first key point we need to recall is that the sine and cosine laws allow us to calculate side lengths and angle measures in nonright triangles.
The law of sines or the sine law first of all then. This tells us that, in any triangle, the ratio between each side length and the sine of its opposite angle is constant, which we can write as 𝑎 over sin 𝐴 equals 𝑏 over sin 𝐵, which is equal to 𝑐 over sin 𝐶. And remember, the lowercase letters represent side lengths and the uppercase letters represent angles.
We don’t need to use all three parts of the sine law, just two parts of this equality. The information needed in order to apply the law of sines is two angles and their opposite sides. So we can recognize the need for the law of sines when we’re working with opposite pairs of information.
This first version of the law of sines is particularly useful when calculating the length of a missing side, as the sides are in the numerators of the fractions, so less rearrangement is required. We do also have a reciprocal version, which is particularly helpful for calculating the measure of an angle. sin 𝐴 over 𝑎 equals sin 𝐵 over 𝑏, which is equal to sin 𝐶 over 𝑐. So that’s the law of sines, and now let’s consider the law of cosines or the cosine law.
The most common form in which you see this written is as follows. 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos of 𝐴. We use it to calculate a side length, in this case the length of side 𝑎, when we know the other two sides of the triangle and the included angle. So we know the sides 𝑏 and 𝑐 and the angle 𝐴.
We can also rearrange the cosine law to a form which enables us to calculate any angle in the triangle if we know all three side lengths. The rearrangement is straightforward, and it gives cos of 𝐴 is equal to 𝑏 squared plus 𝑐 squared minus 𝑎 squared all over two 𝑏𝑐. In this case, as we said, we know all three side lengths and we wish to calculate one of the angles.
Again, the proofs and the basic application of these laws are not going to be considered in this video. Instead, we’re going to focus on applying these laws to some real-life problems. Let’s look at our first question.
A plane travels 800 meters along the runway before taking off at an angle of 10 degrees. It travels a further 1,000 meters at this angle as seen in the figure. Work out the distance of the plane from its starting point. Give your answer to two decimal places.
Looking at the diagram, we can see that we have a triangle. We want to calculate the distance of the plane from its starting point. That’s this length here, which we can refer to as 𝑑 meters. We know the lengths of the other two sides in this triangle. They are 800 meters and 1,000 meters. And using the fact that angles on a straight line sum to 180 degrees, we can work out the size of this angle here. It’s 180 degrees minus 10 degrees, which is 170 degrees.
As this is a non-right-angled triangle, we need to answer this problem using either the law of sines or the law of cosines. So the first step is to decide which of these we need. And that will depend on the specific combination of information we’ve been given and what we want to calculate.
In this triangle, we know two sides and the included angle. And we want to calculate the third side. We recall then that this means we should be using the law of cosines. Let’s recall the law of cosines. It’s 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴. Now, there’s no need to actually label our triangle using the letters 𝐴, 𝐵, and 𝐶. Instead, we just remember that the lowercase letters 𝑏 and 𝑐 represent the two sides we know and the capital letter 𝐴 represents the included angle.
So using 800 and 1,000 as the two side lengths 𝑏 and 𝑐 and 170 degrees as the angle 𝐴, we have the equation 𝑑 squared equals 800 squared plus 1,000 squared minus two times 800 times 1,000 times cos of 170 degrees. We can either type this directly into our calculator or it may be a good idea to break the calculation down into some stages. In either case, we arrive at 𝑑 squared equals 3,215,692.405.
Now, we must remember that this is 𝑑 squared. It isn’t 𝑑, so we aren’t finished. We have to square root in order to find the value of 𝑑. It’s a really common mistake though to forget to do this. Square rooting gives 𝑑 equals 1,793.235178. The question asks us to give our answer to two decimal places. So rounding appropriately, we’ve worked out the distance of the plane from its starting point. It’s 1,793.24 meters to two decimal places.
In this example, we were given a diagram to use. Often this won’t be the case, and we’ll need to use information from a written description to draw our own diagram to help us answer the question. Let’s now consider an example of this.
A ship is sailing due south with a speed of 36 kilometers per hour. An iceberg lies 24 degrees north of east. After one hour, the ship is 33 degrees south of west of the iceberg. Find the distance between the ship and the iceberg at this time, giving the answer to the nearest kilometer.
A lot of the skill involved in this question is in drawing the diagram. Let’s start with a compass showing the four directions. We’ll then take each statement separately and consider how to represent it. Firstly, we know that the ship is sailing due south. Initially, we’re told that an iceberg lies 24 degrees north of east of the ship’s starting point.
Now, directly east would be directly to the right of the ship on our diagram. 24 degrees north of east would mean the iceberg lies somewhere along the line like this. We’re then told that after one hour, the ship is 33 degrees south of west of the iceberg. Well, west would be the direction directly to the left of our iceberg. And using alternate angles in parallel lines, we know that the angle formed here is 24 degrees.
So the full angle between the horizontal and the position the ship has now moved to is 33 degrees. And we can now see that we have a triangle. We can work out the angles in our triangle. For example, this angle here is the difference between 33 degrees and 24 degrees. It is nine degrees. We could also work out this angle here, it’s 24 degrees, plus the angle between south and east, which is 90 degrees, giving a total of 114 degrees.
The one piece of information we haven’t used yet is that the ship is traveling at a speed of 36 kilometers per hour. And we know it takes one hour for the ship to go from its original position to its new position. The ship will therefore have traveled 36 kilometers in this time. So we also know one side length in our triangle.
What we were asked to calculate is the distance between the ship and the iceberg at this later time. So that’s this side here, which we can refer to as 𝑑 kilometers. We’ve now set up our diagram, and we see that we have a non-right-angled triangle, which means we’re going to apply either the law of sines or the law of cosines. Let’s look at the particular combination of information we’ve got.
We know an angle of nine degrees and the opposite side of 36 kilometers. We also know an angle of 114 degrees. And we wish to calculate the opposite side of 𝑑 kilometers. We therefore have opposite pairs of sides and angles, which tells us that we should be using the law of sines to answer this question.
Remember, this tells us that the ratio between each side length, represented using lowercase letters, and the sine of its opposite angle, represented using capital letters, is constant. 𝑎 over sin 𝐴 equals 𝑏 over sin 𝐵, which is equal to 𝑐 over sin 𝐶. We only need to use two parts of this ratio. And there’s no need to label our triangle using the letters 𝐴, 𝐵, and 𝐶 as long as we’re clear about what they represent.
Our side 𝑑 is opposite the angle of 114 degrees, and the side of 36 kilometers is opposite the angle of nine degrees. So we have 𝑑 over sin of 114 degrees equals 36 over sin of nine degrees. We can solve this equation by multiplying each side by sin of 114 degrees, which is just a value. And it gives 𝑑 equals 36 sin 114 degrees over sin of nine degrees. Evaluating on a calculator, making sure our calculator is in degree mode, and we have 210.23267.
The question asks us to give our answer to the nearest kilometer. So rounding appropriately, we have that the distance between the ship and the iceberg at this time is 210 kilometers.
We’ve now seen one example each of using the law of sines and the law of cosines to calculate a side length. In our next example, we’ll see how we can apply the law of cosines to calculate all the missing angles in a triangle when we know its three side lengths.
Los Angeles is 1,744 miles from Chicago, Chicago is 712 miles from New York, and New York is 2,451 miles from Los Angeles. Find the angles in the triangle with its vertices as the three cities.
Now, whilst a little bit of knowledge of the geography of the United States of America might be useful here, it isn’t essential to answering the problem. We can just draw a triangle using the three lengths given in the question. And if our triangle turned out to be upside down, it isn’t the end of the world. The triangle should look a little something like this, and we can add the three distances.
Now, this triangle certainly doesn’t look as if it’s a right triangle. So we’re going to need to apply either the law of sines or the law of cosines to this problem. We know all three of the side lengths, and we want to calculate each of the angles, which tells us that we should be using the law of cosines. The rearranged version of this, which is useful for calculating angles, is cos of 𝐴 equals 𝑏 squared plus 𝑐 squared minus 𝑎 squared all over two 𝑏𝑐. If you can’t remember this, you’ll have to perform the rearrangement yourself from the law of cosines in its traditional form.
In this question then, let’s use 𝐴 to represent Los Angeles, 𝐶 to represent Chicago, and 𝐵 to represent New York. We’ll use the corresponding lowercase letters to represent the opposite sides. To calculate our first angle then, that’s this angle here, we substitute the relevant values. Giving cos of 𝐴 equals 1,744 squared plus 2,451 squared minus 712 squared all over two multiplied by 1,744 multiplied by 2,451.
We can evaluate this on a calculator. And then to find the value of 𝐴, we need to use the inverse cosine function. Doing so gives 𝐴 equals 2.334 degrees. So we’ve found the first angle in the triangle. And we’ll give our answer to two decimal places.
To calculate the next angle in this triangle, which this time we’ll use angle 𝐶, we don’t need to relabel our triangle. We just need to remember that the letters 𝑏 and 𝑐 represent the two sides which enclose the angle and the letter 𝑎 represents the opposite side. So we use 1,744 and 712 for the two sides which enclose the angle and 2,451 for the side which is opposite. This gives cos of 𝐶 equals negative 0.9901. And again, applying the inverse cosine function, we find that angle 𝐶 equals 171.939 degrees.
So we found two angles in the triangle. And in fact, to find the third, we could subtract the two angles we’ve calculated from 180 degrees. But if we don’t use that method, that will be a useful check. In exactly the same way then, but this time using 712 and 2,451 as the two sides which enclose the angle and 1,744 as the opposite side, we find the measure of angle 𝐵 is 5.726 degrees.
Adding the three angles we’ve found, now each rounded to two decimal places, does indeed give 180 degrees. So we can have some confidence in our answer. The measures of the three angles in the triangle formed by these three cities, each to two decimal places then, are 2.33 degrees, 5.73 degrees, and 171.94 degrees.
In our final example, let’s see how we can use the law of sines and the law of cosines to solve problems in other mathematical contexts.
𝑀 is the center of a circle, and 𝐴, 𝐵, and 𝐶 are points on the circumference. If 𝐵𝐶 equals 13 centimeters and the measure of angle 𝐶𝑀𝐵 is 84 degrees, find the area of the circle 𝑀, giving the answer to the nearest square centimeter.
We know that the area of a circle is 𝜋𝑟 squared. So really, this problem is about finding the radius of this circle. Let’s begin by putting the information we’ve been given on the diagram. 𝐵𝐶 is 13 centimeters, and the measure of angle 𝐶𝑀𝐵 is 84 degrees. We don’t know the lengths of 𝑀𝐶 or 𝑀𝐵, but they’re each the radius of the circle.
Now, there are numerous different approaches we could take. But one approach is to apply the law of cosines in the triangle 𝐶𝑀𝐵. This states that 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴, where 𝑏 and 𝑐 represent two sides of a triangle and 𝐴 represents the included angle. In our triangle, 𝑎 is 13 centimeters. The angle 𝐴 is 84 degrees. And the two sides which enclose this angle 𝐴 are each the radius of the circle 𝑟.
We can therefore form an equation. 13 squared equals 𝑟 squared plus 𝑟 squared minus two 𝑟 squared cos of 84 degrees. We can solve this equation to find the value of 𝑟 squared, which we’ll then be able to substitute directly into our area formula. Factorizing the right-hand side of our equation by 𝑟 squared, we have 169 equals 𝑟 squared multiplied by two minus two cos of 84 degrees. Dividing through, we have that 𝑟 squared is equal to 169 over two minus two cos 84 degrees. And we’ll keep our value for 𝑟 squared in this exact form.
We can then substitute this value of 𝑟 squared into the area formula and evaluate on a calculator. Rounding our answer, and we have that the area of circle 𝑀 to the nearest square centimeter is 296 square centimeters.
As I mentioned, there are in fact numerous approaches to this problem, which you can try out yourself if you wish. We could’ve applied the law of sines in triangle 𝐶𝑀𝐵. Or we could’ve divided it in half to form two right triangles and then used right-angle trigonometry.
Let’s now summarize the key points in this video. Firstly, the law of sines and the law of cosines can be used to calculate side lengths and angle measures in nonright triangles. The law of sines in either of its two forms can be used to calculate either a side or an angle when we’re working with opposite pairs of information. The law of cosines in its first form can be used to calculate a side length when we know the other two sides and the included angle. And in its rearranged form, it can be used to calculate any angle when we know all three sides.
We can apply the law of sines and the law of cosines to many problems involving triangles. And although we haven’t seen an example of this, we can also apply both rules within the same problem. |
# NCERT Solutions For Class 6 Maths Decimals Exercise 8.5
ncert textbook
## NCERT Solutions For Class 6 Maths Decimals Exercise 8.5
NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5
Exercise 8.5
Question 1.
Find the sum in each of the following:
(a) 0.007 + 8.5 + 30.08
(b) 15 + 0.632 + 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38
Solution:
(a) 0.007 + 8.5 + 30.08
= 0.007 + 8.500 + 30.080 (making like decimals)
= 38.587
(b) 15 + 0.632 + 13.8
= 15.000 + 0.632 + 13.800 (making like decimals)
= 29.432
(c) 27.076 + 0.55 + 0.004
= 27.076 + 0.550 + 0.004 (making like decimals)
= 27.630
(d) 25.65 + 9.005 + 3.7
= 25.650 + 9.005 + 3.700 (making like decimals)
= 38.355
(e) 0.75 + 10.425 + 2
= 0.750 + 10.425 + 2.000 (making like decimals)
= 13.175
(f) 280.69 + 25.2 + 38
= 280.69 + 25.20 + 38.00 (making like decimals)
= 343.89
Question 2.
Rashid spent ₹35.75 for Maths book and ₹32.60 for Science book. Find the total amount spent by Rashid.
Solution:
Money spent by Rashid for Maths book = ₹35.75
Money spent by Rashid for Science book = ₹32.60
∴ Total money spent by Rashid on both books = ₹35.75 + ₹32.60 = ₹68.35
Question 3.
Radhika’s mother gave her ₹10.50 and her father gave her ₹15.80, find the total amount given to Radhika by her parents.
Solution:
Money given by Radhika’s mother = ₹10.50
Money given by her father = ₹15.80
∴ Total money given to her by her parents
= ₹10.50 + ₹15.80 = ₹26.30
Question 4.
Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.
Solution:
Length of cloth bought by Nasreen for her shirt = 3 m 20 cm = 3.20 m
Length of cloth brought by her for her trouser = 2 m 5 cm = 2.05 m
Total length of cloth bought by her = 3.20 m + 2.05 m = 5.25 m
Question 5.
Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Solution:
Distance walked by Naresh in the morning = 2 km 35 m = (2 + $\frac { 35 }{ 1000 }$) km = 2.035 km.
Distance walked by him in the evening = 1 km 7 m = (1 + $\frac { 7 }{ 1000 }$) km = 1.007 km 1000)
∴ Total distance walked by him in all
= (2.035 + 1.007) km
= 3.042 km
Question 6.
Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?
Solution:
Distance travelled by Sunita by bus
= 15 km 268 m = ( 15 + $\frac { 268 }{ 1000 }$) = 15.268 km
Distance travelled by her by car
= 7 km 7 m = (7 + $\frac { 7 }{ 1000 }$) km
= 7.007km |
Related Articles
# Combinations
Combination is a way of choosing items from a set, such as (unlike permutations) the order of selection doesn’t matter. In smaller cases, it’s possible to count the number of combinations. Combination refers to the mixture of n things taken k at a time without repetition. To know the combinations in the case where repetition is allowed, the terms like k-selection or k-combination along with repetition are often used. For instance, if we’ve two elements A and B, then there’s just one way to select two items, we select both of them. In this article, we will learn about combination in detail.
## Basic Principles of Counting
Counting various things in a specific manner is a basic concern of mathematicians. To solve this problem two basic principles of counting are given which include,
• Fundamental Principle of Counting: For any event X which occurs in n different ways and another event Y which occur in m different ways. Then the total number of occurrences of two events is = m x n.
• Addition Principle: For any event X which occurs in m different ways and event Y which occurs in n different ways where both events are implicit i.e. both the events cannot occur together, then the occurrence of events either X or Y is m + n.
## What Are Combinations?
Combination is the choice of selecting r things from a group of n things without replacement and where the order of selection is not important.
Number of combinations when ‘r’ elements are selected out of a complete set of ‘n’ elements is denoted by nCr
nCr = n! / [(r !) × (n – r)!]
Example: Let n = 4 (E, F, G, H) and r = 2 (consisting of all the combinations of size 2).
nCr = 4C2 = 4!/((4-2)!×2!)
= 4×3×2×1 / 2×1×2×1
= 6
The six combinations are EF, EG, EH, FG, FH, and GH.
## Combination Formula
Combination formula is used to pick r things out of n different things, where the order of picking is not important and replacement is not allowed.
## Relationship Between Permutations and Combinations
Permutation and combination have a lot of similarities but they also have some striking differences. For n different objects, we have to make r unique selections from this group of n objects.
The number of permutations of size r from n object is nPr here the order, of selection is not important so each selection is counted r! times. So the number of unique selections is nPr / r! We know that a unique selection of r things from the total of n things is called a combination(nCr). Thus,
nCr = nPr / r!
## How to Calculate the Probability of Combinations?
The probability of Combinations can be easily understood with the help of the examples given below:
Example:
1. How many ways are possible to distribute 7 different candies to 3 people where each gets only 1 candy?
2. In how many ways can the letters of the word ‘POWER’ be arranged?
3. How many six-digit numbers can be formed with digits 2,3,5, 6, 7, and 9 and with distinct digits?
Solution:
1. For the first people, we can choose any of the candy among the 7 candies available. Similarly, for the 2nd person we are left with 6 choices and for the 3rd, we will be having 5 choices. So, the number of ways of distributing candies = 7 × 6 × 5 = 210 ways
2. Letters of the word ‘POWER’ can be arranged in 5! ways i.e. 5 × 4 × 3 × 2 × 1 ways = 120 ways.
3. The number of distinct ways of forming 6-digit numbers with different digits is 6! = 6 × 5 × 4 × 3 × 2 × 1 ways = 720 ways.
## What is Handshaking Problem?
Handshaking problem is one of the most interesting problems in mathematics. It is used to find that in a room full of people how many handshakes are required for everybody to shake everybody else’s hand exactly once?
Example: The table given below tells us about the minimum number of handshakes required for various groups of people.
Basically when there are 2 people there will be two handshakes and if there are three people there will be 3 handshakes and so on. This many people we can count but let’s suppose there are thousands of people in a hall then we can’t count each handshake here the need for the combination arises.
## Handshaking Combination
It means the total number of people in a room doing the handshake with each other. With the help of combination formulas, it can easily be calculated. The formula for calculating the handshakes when there are n people available is given by,
• Total Number of Handshakes = n × (n – 1)/2
• Total Number of Handshakes = nC2
Also, Check
## Solved Examples on Combinations
Example 1: In how many ways 6 boys can be arranged in a queue such that
a) Two particular boys of them are always together
b) Two particular boys of them are never together
Solution:
a) If two boys are always together, then they will be treated as one entity. Hence we can be arranged 5 boys in 5! ways. Also, two boys can arrange themselves in 2 different ways.
Therefore required arrangement = 5! × 2 = 120 × 2 = 240 ways.
b) Total number of permutations among 6 numbers is given by = 6! = 720.
In 240 cases 2 boys are always together.
Thus, for two boys who are never together no of ways will be = 720 – 240 = 480 ways.
Example 2: In a room of n people, how many handshakes are possible?
Solution:
To see the people present, and consider one person at a time. The first person will shake hands with n – 1 other people. The next person will shake hands with n-2 other people, not counting the first person again. Following this, it will give us a total number of
(n – 1) + (n – 2) + … + 2 + 1
= n(n – 1)/ 2 handshakes.
Example 3: Another popular handshake problem starts out similarly with n>1 people at a party. Not being possible to shake hands with yourself, and not counting several times handshakes with the same person, the problem is to show that there will always present two people at the party, who had shaken hands the same number of times in the party.
Solution:
The solution to this problem starts by using Dirichlet’s box principle. If there exists a person at the party, who has shaken hands zero times, then every person which is there at the party has shaken hands with at most n-2 other people at the party.
There are n-1 possible handshakes (from 0 to n-2), among n people there must be two who have shaken hands the same number of times. If there are zero persons, who has shaken hands zero times this means that all of the party guests have shaken hands at least once.
This also amounts to n-1 possible handshakes (from 1 to n-1).
Example 4: In the function, if every person shakes hands with every other in the party and there exists a total of 28 handshakes at the party, find the number of persons who were present in the function.
Solution:
Suppose there are n persons present at a party and every person shakes hands with every other person.
Then, total number of handshakes = nC2 = n(n – 1)/2
n(n – 1)/2 = 28
n(n – 1) = 28 × 2
n(n – 1) = 56
n = 8
## FAQs on Combination
### Question 1: What is a Combination?
The combination is a way of arranging r different things out of n things for which the order of selection is not important.
### Question 2: How to solve combinations?
Combinations help us to calculate the total outcomes of an event when the order of outcomes does not matter. Combinations can be calculated with the formula,
nCr = n! / r! × (n – r)!
### Question 3: What is the value of nCn?
The value of nCn is calculated as,
nCn = n! / (n-n)!×n! (0! = 1)
= n! / n! = 1
### Question 4: When do we use Combination and Permutation?
Permutation formulas are used when the order of selection matters and Combination formulas are used when the order of the permutation doesn’t matter.
### Question 5: Give the Combination formula.
Combination formula is given as,
nCr = n!/r!(n-r)! |
## What are the 4 inverse operations?
If we start with x, then add 2 and subtract 2, we are left with the original starting variable x. There are several inverse operations you should be familiar with: addition and subtraction, multiplication and division, squares and square roots (for positive numbers), as well as cubes and cube roots.
## What is the inverse operation of multiplication examples?
The inverse of multiplication is division. If you multiply by a given number and then divide by the same number, you will arrive at the same number you started with. Division has the opposite effect to multiplication. For example here is 3 × 2 = 6.
## Can you give more examples of inverse?
Examples of inverse operations are: addition and subtraction; multiplication and division; and squares and square roots.
## How do you find inverse operations?
Finding the Inverse of a Function
1. First, replace f(x) with y . …
2. Replace every x with a y and replace every y with an x .
3. Solve the equation from Step 2 for y . …
4. Replace y with f−1(x) f − 1 ( x ) . …
5. Verify your work by checking that (f∘f−1)(x)=x ( f ∘ f − 1 ) ( x ) = x and (f−1∘f)(x)=x ( f − 1 ∘ f ) ( x ) = x are both true.
## How do you solve inverse operations?
To use an inverse operation, just do the opposite of what the equation says! Use inverse operations to complete the equation. In this example 2 is being added to 7, to undo that operation we need to subtract by 2.
## What is an example of an inverse statement?
Our converse statement would be: “If the grass is wet, then it is raining.” Our inverse statement would be: “If it is NOT raining, then the grass is NOT wet.” Our contrapositive statement would be: “If the grass is NOT wet, then it is NOT raining.”
## What’s an inverse operation?
Inverse operationsare pairs of mathematical manipulations in which one operation undoes the action of the other—for example, addition and subtraction, multiplication and division. The inverse of a number usually means its reciprocal, i.e. x – 1 = 1 / x .
## What is the inverse of 5?
1/5
Multiplicative Inverse of Natural Number
For example, the multiplicative inverse of 5 is 1/5.
## What is the multiplicative inverse of 2?
Multiplicative inverse of 2 is 1/2.
## What are inverse equations?
In mathematics, an inverse is a function that serves to “undo” another function. That is, if f(x) produces y, then putting y into the inverse of f produces the output x. x . A function f that has an inverse is called invertible and the inverse is denoted by f−1.
## What is the inverse operation of exponents?
the logarithm
If the logarithm is understood as the inverse of the exponential function, then the properties of logarithms will naturally follow from our understanding of exponents. The meaning of the logarithm. The logarithmic function g(x) = logb(x) is the inverse of the exponential function f(x) = bx.
## What is the inverse of 3?
1/3
The answer is of course one third, or 1/3, since: 3 * 1/3 = 1. Thus the multiplicative inverse of 3 is 1/3.
## What is the inverse of 5?
1/5
Multiplicative Inverse of Natural Number
For example, the multiplicative inverse of 5 is 1/5.
## What is the inverse of 12?
The multiplicative inverse of 12 is 1/12. In general, we have the following rule for finding the multiplicative inverse of a number. The multiplicative inverse of a number, a/b, is b/a.
## What are the 3 steps to finding an inverse function?
Steps for finding the inverse of a function f.
1. Replace f(x) by y in the equation describing the function.
2. Interchange x and y. In other words, replace every x by a y and vice versa.
3. Solve for y.
4. Replace y by f1(x).
## What is the inverse of 17?
Additive Inverse of 17 = -17
Since the additive inverse of 17 is -17, additive inverse is also known as the opposite number.
## What is the inverse of 10?
-10
Example: Additive inverse of 10 is -10, as 10 + (-10) = 0.
## What is the inverse of 7?
Step-by-step explanation:
A reciprocal is one of a pair of numbers that when multiplied with another number equals the number 1. For example, if we have the number 7, the multiplicative inverse, or reciprocal, would be 1/7 because when you multiply 7 and 1/7 together, you get 1! |
Vous êtes sur la page 1sur 7
# Sums and
Differences of
## Random Variables 39.3
Introduction
In some situations, it is possible to easily describe a problem in terms of sums and differences of
random variables. Consider a typical situation in which shafts are fitted to cylindrical sleeves. One
random variable is used to describe the variability of the diameter of the shaft, and one is used to
describe the variability of the sleeves. Clearly, we need to know how the total variability involved
affects the fitting of shafts and sleeves. In this Section, we will confine ourselves to cases where the
random variables are normally distributed and independent.
• be familiar with the results and concepts met
Prerequisites in the study of probability
Before starting this Section you should . . . • be familiar with the normal distribution
'
\$
• describe a variety of problems in terms of
sums and differences of normal random
Learning Outcomes variables
On completion you should be able to . . . • solve problems described in terms of sums
and differences of normal random variables
& %
34 HELM (2008):
Workbook 39: The Normal Distribution
®
## 1. Sums and differences of random variables
In some situations, we can specify a problem in terms of sums and differences of random variables.
Here we confine ourselves to cases where the random variables are normally distributed. Typical
situations may be understood by considering the following problems.
Problem 1
In a certain mass-produced assembly, a 3 cm shaft must slide into a cylindrical sleeve. Shafts are man-
ufactured whose diameter S follows a normal distribution S ∼ N (3, 0.0042 ) and cylindrical sleeves
are manufactured whose internal diameter C follows a normal distribution C ∼ N (3.010, 0.0032 ).
Assembly is performed by selecting a shaft and a cylindrical sleeve at random. In what proportion of
cases will it be impossible to fit the selected shaft and cylindrical sleeve together?
Discussion
Clearly, the shaft and cylindrical sleeve will fit together only if the diameter of the shaft is smaller than
the internal diameter of the cylindrical sleeve. We need the difference of the two random variables
C and S to be greater than zero. We can take the difference C − S and find its distribution. Once
we do this we can then ask the question ”What is the probability that the inside diameter of the
cylindrical sleeve is greater than the outside diameter of the shaft, i.e. what is P(C − S > 0)?”
Essentially we are trying to ensure that the internal diameter of the cylindrical sleeve is larger than
the external diameter of the shaft.
Problem 2
A manufacturer produces boxes of woodscrews containing a variety of sizes for a local DIY store. The
weight W (in kilograms) of boxes of woodscrews manufactured is a normal random variable following
the distribution W ∼ N (1.01, 0.004). Note that 0.004 is the variance. Find the probability that a
customer who selects two boxes of screws at random finds that their combined weight is greater than
2.03 kilograms.
Discussion
In this problem we are looking at the effects of adding two random variables together. Since all
boxes are assumed to have weights W which follow the distribution W ∼ N (1.01, 0.004), we are
considering the effect of adding the random variable W to itself. In general, there is no reason why
we cannot combine variables W1 ∼ N (µ1 , σ12 ) and W2 ∼ N (µ2 , σ22 ). This might happen if the DIY
store bought in two similar products from two different manufacturers.
Before we can solve such problems, we need to obtain some results concerning the behaviour of
random variables.
## Functions of several random variables
Note that we shall quote results only for the continuous case. The results for the discrete case
are similar with integration replaced by summation. We will omit the mathematics leading to these
results.
HELM (2008): 35
Section 39.3: Sums and Differences of Random Variables
Key Point 4
## • If X1 , X2 , · · · + Xn are n random variables then
E(X1 + X2 + · · · + Xn ) = E(X1 ) + E(X2 ) + . . . E(Xn )
• If X1 , X2 , . . . Xn are n independent random variables then
V(X1 + X2 + · · · + Xn ) = V(X1 ) + V(X2 ) + · · · + V(Xn )
and more generally
V(X1 ± X2 ± · · · ± Xn ) = V(X1 ) + V(X2 ) + · · · + V(Xn )
Example 16
Solve Problem 1 from the previous page. You may assume that the sum and
difference of two normal random variables are themselves normal.
Solution
Consider the random variable C − S. Using the results above we know that
C − S ∼ N (3.010 − 3.0, 0.0042 + 0.0032 ) i.e, C − S ∼ N (0.01, 0.0052 )
0 − 0.01
Hence P(C − S > 0) = P(Z > = −2) = 0.9772
0.005
This result implies that in 2.28% of cases it will be impossible to fit the shaft to the sleeve.
Solve Problem 2 from the previous page. You may assume that the sum and
difference of two normal random variables are themselves normal.
36 HELM (2008):
Workbook 39: The Normal Distribution
®
If W12 is the random variable representing the combined weight of the two boxes then
W12 ∼ N (2.02, 0.008)
Hence
2.03 − 2.02
P(W12 > 2.03) = P(Z > √ = 0.1118) = 0.5 − 0.0445 = 0.4555
0.008
The result implies that the customer has about a 46% chance of finding that the weight of the two
boxes combined is greater than 2.03 kilograms.
Exercises
1. Batteries of type A have mean voltage 6.0 (volts) and variance 0.0225 (volts2 ). Type B
batteries have mean voltage 12.0 and variance 0.04. If we form a series connection containing
one of each type what is the probability that the combined voltage exceeds 17.4?
2. Nuts and bolts are made separately and paired at random. The nuts’ diameters, in mm, are
independently N (10, 0.02) and the bolts’ diameters, in mm, are independently N (9.5, 0.02).
Find the probability that a bolt is too large for its nut.
3. Certain cutting tools have lifetimes, in hours, which are independent and normally distributed
with mean 300 and variance 10000.
## (a) Find the probability that
(i) the total life of three tools is more than 1000 hours.
(ii) the total life of four tools is more than 1000 hours.
(b) In a factory each tool is replaced when it fails. Find the probability that exactly four tools
are needed to accumulate 1000 hours of use.
(c) Explain why the first sentence in this question can only be approximately, not exactly,
true.
4. A firm produces articles whose length, X, in cm, is normally distributed with nominal mean
µ = 4 and variance σ 2 = 0.1. From time to time a check is made to see whether the value
of µ has changed. A sample of ten articles is taken, the lengths are measured, the sample
mean length X̄ is calculated, and the process is adjusted if X̄ lies outside the range (3.9, 4.1).
Determine the probability, α, that the process is adjusted as a result of a sample taken when
µ = 4. Find the smallest sample size n which would make α ≤ 0.05.
HELM (2008): 37
Section 39.3: Sums and Differences of Random Variables
## 1. XA ∼ N (6, 0.0225) XB ∼ N (12, 0.04)
Series X = XA + XB ∼ N (18, 0.0625) as variances always add
−0.6 0.6
P (X > 17.4) = P Z > = 0.5 + P 0 < Z <
0.25 0.25
## = 0.5 + P(0 < Z < 2.4) = 0.5 + 0.4918 = 0.9918
2. Let the diameter of a nut be N. Let the diameter of a bolt be B. A bolt is too large for its
nut if N − B < 0.
## E(N − B) = 10 − 9.5 = 0.5
V(N − B) = 0.02 + 0.02 = 0.04
N − B ∼ N (0.5, 0.04)
N − B − 0.5 0 − 0.5
P(N − B < 0) = P < = P(Z < −2.5)
0.2 0.2
= Φ(−2.5) = 1 − Φ(2.5) = 1 − 0.99379
= 0.00621.
The probability that a bolt is too large for its nut is 0.00621.
## (i) E(T1 + T2 + T3 ) = 900
V(T1 + T2 + T3 ) = 30000
(T1 + T2 + T3 ) ∼ N (900, 30000)
T1 + T2 + T3 − 900
P(T1 + T2 + T3 > 1000) = P √ = P(Z > 0.57735)
30000
= 1 − Φ(0.57735) = 1 − 0.7181 = 0.2819
## (ii) E(T1 + T2 + T3 + T4 ) = 1200
V(T1 + T2 + T3 + T4 ) = 40000
(T1 + T2 + T3 + T4 ) ∼ N (1200, 40000)
T1 + T2 + T3 + T4 − 1200 1000 − 1200
P(T1 + T2 + T3 + T4 > 1000) = P √ > √
40000 40000
= P(Z > −1)
= 1 − Φ(−1) = Φ(1) = 0.8413
38 HELM (2008):
Workbook 39: The Normal Distribution
®
(b) Let the number of tools needed be N.
## P(N ≤ 3) = P(N = 1) + P(N = 2) + P(N = 3)
= P(T1 + T2 + T3 > 1000) = 0.2819
P(N ≤ 4) = P(N = 1) + P(N = 2) + P(N = 3) + P(N = 4)
= P(T1 + T2 + T3 + T4 > 1000) = 0.8413.
## Hence P(N = 4) = P(N ≤ 4) − P(N ≤ 3) = 0.8413 − 0.2819 = 0.5594.
(c) Lifetimes can not be negative. The normal distribution assigns non-zero probability density
to negative values so it can only be an approximation in this case.
4.
X ∼ N (4, 0.1)
X1 + · · · + X10 ∼ N (40, 1)
X̄ = (X1 + · · · + X10 )/10 ∼ N (4, 0.01)
## By symmetry P(X̄ < 3.9) = P(X̄ > 4.1).
X̄ − 4 3.9 − 4
P(X̄ < 3.9) = P < = P(Z < −1)
0.1 0.1
= Φ(−1) = 1 − Φ(1) = 1 − 0.8413
= 0.1587
More generally
X1 + · · · + Xn ∼ N (4n, 0.1n)
X̄ = (X1 + · · · + Xn )/n ∼ N (4, 0.1/n)
! !
X̄ − 4 3.9 − 4 −0.1 √
P(X̄ < 3.9) = P p <p =P Z< p = − 0.1n
0.1/n 0.1/n 0.1/n
√ √
= Φ(− 0.1n) = 1 − Φ( 0.1n)
α = 2[1 − Φ( 0.1n)]
We require α ≤ 0.05.
√ √
2[1 − Φ( 0.1n)] ≤ 0.05 ⇔ 1 − Φ( 0.1n) ≤ 0.025
⇔ Φ( 0.1n) ≥ 0.975
⇔ 0.1n ≥ 1.96
⇔ n ≥ 10 × 1.962 = 38.416
## The smallest sample size which satisfies this is n = 39.
HELM (2008): 39
Section 39.3: Sums and Differences of Random Variables
Table 1: The Standard Normal Probability Integral
x−µ
Z= σ 0 1 2 3 4 5 6 7 8 9
0 0000 0040 0080 0120 0160 0199 0239 0279 0319 0359
.1 0398 0438 0478 0517 0577 0596 0636 0675 0714 0753
.2 0793 0832 0871 0909 0948 0987 1026 1064 1103 1141
.3 1179 1217 1255 1293 1331 1368 1406 1443 1480 1517
.4 1555 1591 1628 1664 1700 1736 1772 1808 1844 1879
.5 1915 1950 1985 2019 2054 2088 2123 2157 2190 2224
.6 2257 2291 2324 2357 2389 2422 2454 2486 2517 2549
.7 2580 2611 2642 2673 2703 2734 2764 2794 2822 2852
.8 2881 2910 2939 2967 2995 3023 3051 3078 3106 3133
.9 3159 3186 3212 3238 3264 3289 3315 3340 3365 3389
1.0 3413 3438 3461 3485 3508 3531 3554 3577 3599 3621
1.1 3643 3665 3686 3708 3729 3749 3770 3790 3810 3830
1.2 3849 3869 3888 3907 3925 3944 3962 3980 3997 4015
1.3 4032 4049 4066 4082 4099 4115 4131 4147 4162 4177
1.4 4192 4207 4222 4236 4251 4265 4279 4292 4306 4319
1.5 4332 4345 4357 4370 4382 4394 4406 4418 4429 4441
1.6 4452 4463 4474 4484 4495 4505 4515 4525 4535 4545
1.7 4554 4564 4573 4582 4591 4599 4608 4616 4625 4633
1.8 4641 4649 4656 4664 4671 4678 4686 4693 4699 4706
1.9 4713 4719 4726 4732 4738 4744 4750 4756 4761 4767
2.0 4772 4778 4783 4788 4793 4798 4803 4808 4812 4817
2.1 4821 4826 4830 4834 4838 4842 4846 4850 4854 4857
2.2 4861 4865 4868 4871 4875 4878 4881 4884 4887 4890
2.3 4893 4896 4898 4901 4904 4906 4909 4911 4913 4916
2.4 4918 4920 4922 4925 4927 4929 4931 4932 4934 4936
2.5 4938 4940 4941 4943 4946 4947 4948 4949 4951 4952
2.6 4953 4955 4956 4957 4959 4960 4961 4962 4963 4964
2.7 4965 4966 4967 4968 4969 4970 4971 4972 4973 4974
2.8 4974 4975 4976 4977 4977 4978 4979 4979 4980 4981
2.9 4981 4982 4982 4983 4984 4984 4985 4985 4986 4986
3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9
4987 4990 4993 4995 4997 4998 4998 4999 4999 4999
40 HELM (2008):
Workbook 39: The Normal Distribution |
# 2007 AMC 10B Problems/Problem 11
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A circle passes through the three vertices of an isosceles triangle that has sides of length $3$ and a base of length $2$. What is the area of this circle?
$\mathrm{(A)}\ 2\pi \qquad\mathrm{(B)}\ 5\pi/2 \qquad\mathrm{(C)}\ 81\pi/32 \qquad\mathrm{(D)}\ 3\pi \qquad\mathrm{(E)}\ 7\pi/2$
## Solution
### Solution 1
Let $\triangle ABC$ have vertex $A$ and center $O$, with foot of altitude from $A$ at $D$.
$[asy] import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0); pair O=circumcenter(A,B,C); draw(A--B--C--A--D); draw(B--O--C); draw(circumcircle(A,B,C)); dot(O); label("$$A$$",A,N); label("$$B$$",B,S); label("$$C$$",C,S); label("$$D$$",D,S); label("$$O$$",O,W); label("$$r$$",(O+A)/2,SE); label("$$r$$",(O+B)/2,N); label("$$h$$",(O+D)/2,SE); label("$$3$$",(A+B)/2,NW); label("$$1$$",(B+D)/2,N); [/asy]$
Then by Pythagorean Theorem (with radius $r$, height $OD = h$) on $\triangle OBD, ABD$ \begin{align*} h^2 + 1 & = r^2 \\ (h + r)^2 + 1 & = 9 \end{align*}
Substituting and solving gives $r = \frac {9}{4\sqrt {2}}$. Then the area of the circle is $r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \frac {81}{32} \pi \Rightarrow \mathrm{(C)}$.
### Solution 2
By $A = \frac {1}{2}Bh = \frac {abc}{4R}$ (or we could use $s = 4$ and Heron's formula), $$R = \frac {abc}{2Bh} = \frac {3 \cdot 3 \cdot 2}{2(2)(2\sqrt {2})} = \frac {9}{4\sqrt {2}}$$ and the answer is $R^2 \pi = \mathrm{(C)}$
Alternatively, by the Extended Law of Sines, $$2R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \frac {9}{4\sqrt {2}}$$ Answer follows as above. |
What is Parallel Strains in Math?
What is Parallel Traces in Math? It’s always a challenging problem and there are actually totally different methods you could use to answer it. The foremost frequent technique certainly is the equation of the line, with and without instructions. You can easily then simplify this equation by getting a straight line.
But this really is not the one methodology and you will find an extra explicit procedure you can use. This second approach works by using triangle identities.
To solve a difficulty of parallel strains, number one decide it doesn’t matter if the road is parallel to a different straight line. Then know what is the fourth energy within the angle somewhere between the parallel traces. https://discovery.osu.edu/migration-and-diaspora-studies The fourth electric power belonging to the angle is generally observed by dividing the angle with the size in the parallel strains.
The fourth potential for the angle may very well be found as follows. Multiply the size in the traces together with the angle, then divide the quotient by two. It could actually be created as follows: angle = -2 x duration squared
In what on earth is perpendicular traces in math, you would like to uncover the fourth power. If ever the angle is known as two-thirds of the circle, the sq. root of two is a particular. It will be the same as making use of the Pythagorean theorem.
The triangle have to be sq. in condition and has to possess a duration of thrice the duration of the hypotenuse. Should the hypotenuse known as the radius, then the square within the duration of the triangle is just one. It is actually simplier and easier to determine the length. It is usually three times the radius.
The triangle is known as a three sided determine with a few sides and it has a certain angle within the center for the 3 sides. The triangle defines two angles, with either side developing two sides. physics helper It’s got two common angles. These angles are described as proper angles.
There are other ways to resolve parallel strains challenges. A single is named the appropriate Triangle Solution.
In the Parallel Lines in Math, the method can be described as subject of three. Put simply, there’s 3 sides. The perimeters should always have equal lengths together with the hypotenuse needs to be equal into the length for the side along with a precise angle. The final option is called the Steeper Precise Triangle Procedure.
When you think about a difficulty of perpendicular traces, the third strategy is easily the most suitable and is to simplify the equation of a straight line with and while not directions. Fixing the Equation of a Parallel Strains in Math is relatively hassle-free. You will discover some variances while using the process second hand for that other ways.
The most significant big difference stands out as the proven fact that it’s important to locate the fourth electricity. On the way utilised for triangle and quadratic functions, this phase could be dealt with instantly. Inside the methodology utilised for parallel traces, you should find the fourth electric power. The final approach is called the Steeper Right Triangle Way. |
# -12
Personal Blog
There are 2 probability puzzles that I like:
1) Suppose I tell you that I have 2 children and one of them is a boy, what is the probability that I have 2 boys?
The correct answer is not 1/2 but 1/3. How can that be? Well there are 4 possible combinations, BB,GG,BG and GB but but at least one is a boy so you can get rid of GG. So all that's left is BB,BG and GB; and in only one of those 3 possibilities do I have two boys.
2) Now I tell you that I have 2 children and one of them is a boy born on a Tuesday. What is the probability that I have 2 boys?
You may think that Tuesday is not useful information in this matter so the answer would be the same as the previous example, but you would be wrong. The correct answer is 13/27. How can that be?
Well there are 14 possibilities for EACH kid:
B-Mo, B-Tu, B-We, B-Th, B-Fr, B-Sa, B-Su
G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su
But I told you the one of my kids (the first or the second) was a boy born on a Tuesday so that narrows down the field of possibilities to:
First child: B-Tu, second child: B-Mo, B-Tu, B-We, B-Th, B-Fr, B-Sa, B-Su, G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su.
Second child: B-Tu, first child: B-Mo, B-We, B-Th, B-Fr, B-Sa, B-Su, G-Mo, G-Tu, G-We, G-Th, G-Fr, G-Sa, G-Su.
No need to put B-Tu in the second row because it's already accounted for in the first row.
So now just count them out, 14+13= 27 possibilities. How many result in 2 boys? Count them out again 7+6=13. So 13 out of 27 possibilities give you 2 boys.
John K Clark
# -12
New Comment
The correct answer to puzzle 1, as posed, is not in fact simply 1/3, because you've got to factor in Pr(you say "I have two children and one is a boy" | each scenario) and it's not at all clear that these are equal in the BB, BG, GB cases. For instance, if you say that in the ordinary course of events (rather than, e.g., to pose a puzzle) I think BG and GB are very much more likely than BB, because if you had two boys why on earth would you say "I have two children and one is a boy"?
There's a general principle here: when you discover something new (by, e.g., being told something, or seeing something), the correct information to update on is not what you've been told, or seen but the fact that you've been told, or seen, it. In some cases this doesn't matter. In many others (including, but not limited to, those where you have reason to doubt the accuracy of what you've been told or seen) it does.
A more general lesson is that whenever the answer to a puzzle causes you to go, "oh how wondrous that this question could have such a strange answer", you were probably tricked into accepting an anti-helpful framing of the problem, and one of the reasons why the puzzle-poser didn't guide you into a helpful framing instead was probably exactly that such anti-helpful framings cause people to feel that way.
Well, another general lesson. I think it's largely orthogonal to the general principle I mentioned. (For the avoidance of doubt, I agree with yours too.)
This post is not evidence for that lesson. When OP's puzzle is stated as intended it indeed has a wonderful and strange answer. The meta-puzzle: "Are these two puzzles essentially the same?" referring to the puzzle as intended and as presented also has a wonderful and strange answer; in fact, John Baez and maybe all of his commenters have been getting it wrong for several years. Our intuition is imperfect, and whether the puzzles you come across tend to use this fact or just trick you with sneaky framing depends on where you get your puzzles.
The correct answer to puzzle 1, as posed, is not in fact simply 1/3, because you've got to factor in Pr(you say "I have two children and one is a boy" | each scenario) and it's not at all clear that these are equal in the BB, BG, GB cases. For instance, if you say that in the ordinary course of events (rather than, e.g., to pose a puzzle) I think BG and GB are very much more likely than BB, because if you had two boys why on earth would you say "I have two children and one is a boy"?
Indeed, and this exact malformed problem is also discussed in the post "My Bayesian Enlightenment":
In the correct version of this story, the mathematician says "I have two children", and you ask, "Is at least one a boy?", and she answers "Yes". Then the probability is 1/3 that they are both boys.
If this post wasn't already at -6 I'd downvote it, for failing to provide the correct form of the puzzles, which has the second person ask the first about this additional info. Otherwise you need to consider the reason the person is providing you this information.
For the "Suppose I tell you that I have 2 children and one of them is a boy" we have the following priors:
P(two girls) = 25%
P(two boys) = 25%
P(one of each) = 50%
Consider however the following possibilities.
Scenario A) The guy is a sexist who would loudly proclaim the existence of sons, and avoid discussing the existence of daughter. Then the probabilities he has two boys is 0%. If he had two boys, he would have told you both of them are boys. By mentioning only one, he unintentionally revealed he had only one. Putting differently P(mentions only one boy|two boys) ~= 0
Scenario B) The guy thought to himself "I'll randomly pick a child and mention its gender". Then with no gender bias we have
P(mentions a boy) = 50%
P(mentions a girl) = 50%
and we go to:
P(mentions one boy|two girls)= 0%
P(mentions one boy|two boys) = 100%
P(mentions one boy|one of each) = 50%
Therefore P(two boys|mentions one boy) = P(mentions one boy|two boys) P(two boys)/P(mentions one boy) = 100% 25% / 50% = 1/2
So if the guy randomly picked a child who's gender he picked to mention, possibility of two boys is 50% given the info he provided.
The Tuesday boy problem was discussed here.
The way the puzzle is posed is ambiguous, as it does not state exactly how the trial is carried on. It may be reworded in (to my understanding) 2 different reasonnable ways, leading to 2 different answers.
1) Out of the set of 4 boys and 4 girls, that is 4 families representing the 4 combinations BB,BG,GB,GG, you choose at random a boy (rejecting the trial if you get a girl). Then the probability of the other member of the family being a boy is 1/2. This is the same as Ariskatsaris scenario B) below.
2) Out of the 4 families, you chose at random one that has at least one boy (rejecting the family with 2 girls), then the probabilty is is 1/3.
The way the puzzle is worded appears to me closer to scenario #2 than #1, but this my biased interpretation.
Here's a possibly useful intuition-adjuster for the "Tuesday boy" problem: replace "born on a Tuesday" with something much less probable. Pr(two boys | two children, one a boy who will one day be President of the USA) is "obviously" about 1/2 rather than about 1/3, because now you can (almost) meaningfully talk about "the other child", which you can't in the case of Pr(two boys | two children, one a boy). The less uniquely-identifying the extra information, the nearer you are to the original "two children, one a boy" scenario.
If you want to do the actual calculation, you might want this picture in your head: a 14x14 table of possibilities, boys in the first 7 rows/columns, Tuesday in the first and 8th row/column. "Two children, one a boy" excludes the bottom-right quadrant. "Two children, one a Tuesday-boy" excludes all but the first row and column.
a lot of the comments here are critical of the way these scenarios are presented, but I don't believe there is in fact any deep issue. the fact of the matter remains, if you are in a situation where you have two things which you do not yet have the information to differentiate, and you know they have a state in some binary property, and that this state for one is independent of the other. Then, if you learn, by any means, that one of these objects has a specific state (call it A) with regard to the binary property, your probability needs to adjust to p(both are in state A|one is in state A) = 1/3 and p(only on object is in state A|one is in state A) = 2/3. johnclark just used a classic example to demonstrate the importance of how interchangeability has a large and unintuitive effect on probability. It is my contention that the sentiment this scenario is unintuitive because of the way johnclark is incorrect. I have seen this question posed in other ways to classes of smart college students studying probability and most of them getting it wrong. External knowledge of the way people tend to provide information isn't really a relevant factor here. This article is well written and important, and by no means deserves the very low karma score it has received (-6 as of this writing, -7 before my vote) |
# How Many Sides Does a Regular Polygon Have?
A regular polygon is a two-dimensional shape with equal sides and equal angles. It is a fundamental concept in geometry and has been studied for centuries. In this article, we will explore the properties of regular polygons, discuss how to determine the number of sides in a regular polygon, and provide examples and case studies to illustrate these concepts.
## Understanding Regular Polygons
Before we delve into the number of sides in a regular polygon, let’s first understand what makes a polygon regular. A polygon is a closed figure formed by straight lines, and a regular polygon has equal sides and equal angles.
Regular polygons are named based on the number of sides they have. For example, a polygon with three sides is called a triangle, a polygon with four sides is called a quadrilateral, and a polygon with five sides is called a pentagon.
Regular polygons have several interesting properties:
• All sides of a regular polygon are congruent, meaning they have the same length.
• All angles of a regular polygon are congruent, meaning they have the same measure.
• The sum of the interior angles of a regular polygon can be calculated using the formula (n-2) * 180 degrees, where n is the number of sides.
• The measure of each interior angle of a regular polygon can be calculated using the formula (n-2) * 180 degrees / n, where n is the number of sides.
## Determining the Number of Sides
Now that we understand the properties of regular polygons, let’s discuss how to determine the number of sides in a regular polygon. There are a few different methods to do this, depending on the information available.
### Method 1: Counting the Sides
The most straightforward method to determine the number of sides in a regular polygon is to count them. If you have a visual representation of the polygon, such as a drawing or a physical object, you can simply count the number of sides to find the answer.
For example, if you have a regular polygon with six sides, you can count each side and conclude that it is a hexagon. Similarly, if you have a regular polygon with eight sides, you can count each side and determine that it is an octagon.
### Method 2: Using the Interior Angle
Another method to determine the number of sides in a regular polygon is to use the measure of the interior angle. As mentioned earlier, the measure of each interior angle of a regular polygon can be calculated using the formula (n-2) * 180 degrees / n, where n is the number of sides.
By knowing the measure of the interior angle, you can solve the equation for n and find the number of sides. Let’s take an example to illustrate this method:
Suppose you have a regular polygon with an interior angle measuring 120 degrees. Plugging this value into the formula, we get:
(n-2) * 180 degrees / n = 120 degrees
Simplifying the equation, we have:
180n – 360 = 120n
60n = 360
n = 6
Therefore, the regular polygon in question has six sides and is a hexagon.
## Examples and Case Studies
Let’s explore a few examples and case studies to further illustrate the concept of regular polygons and the determination of their number of sides.
### Example 1: Equilateral Triangle
An equilateral triangle is a regular polygon with three sides of equal length. Each interior angle of an equilateral triangle measures 60 degrees. Using the formula mentioned earlier, we can verify this:
(n-2) * 180 degrees / n = (3-2) * 180 degrees / 3 = 60 degrees
Therefore, an equilateral triangle is a regular polygon with three sides.
### Example 2: Regular Hexagon
A regular hexagon is a polygon with six sides of equal length. Each interior angle of a regular hexagon measures 120 degrees. Using the formula, we can confirm this:
(n-2) * 180 degrees / n = (6-2) * 180 degrees / 6 = 120 degrees
Therefore, a regular hexagon is a polygon with six sides.
### Case Study: Architecture
Regular polygons have been widely used in architecture throughout history. One notable example is the Pantheon in Rome, which is a circular building with a portico consisting of eight regular granite columns. Each column represents a regular polygon with eight sides, known as an octagon.
The use of regular polygons in architecture allows for symmetry and aesthetic appeal. By incorporating regular polygons into the design, architects can create visually pleasing structures that are mathematically precise.
## Summary
Regular polygons are two-dimensional shapes with equal sides and equal angles. They have several properties, including congruent sides and angles. The number of sides in a regular polygon can be determined by counting them or using the measure of the interior angle. Regular polygons have been used in various fields, including architecture, to create visually appealing and mathematically precise structures.
## Q&A
### Q1: What is a regular polygon?
A1: A regular polygon is a two-dimensional shape with equal sides and equal angles.
### Q2: How can I determine the number of sides in a regular polygon?
A2: The number of sides in a regular polygon can be determined by counting them or using the measure of the interior angle.
### Q3: What are some properties of regular polygons?
A3: Regular polygons have congruent sides and angles. The sum of the interior angles can be calculated using the formula (n-2) * 180 degrees, where n is the number of sides.
### Q4: Can a regular polygon have an odd number of sides?
A4: No, a regular polygon cannot have an odd number of sides. The number of sides in a regular polygon must be an even number.
### Q5: How are regular polygons used in architecture?
A5: Regular polygons are used in architecture to create visually appealing and mathematically precise structures. They allow for symmetry and aesthetic appeal in building design.
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### AREA & VOLUME TUTORIAL
This tutorial exhibits some of the applications of integral calculus. Although these are only some of the applications of integration, they will give you an idea of how calculus can be used to solve problems in other areas of mathematics.
# Areas Between Curves
The area A of the region bounded by the curves, f(x) and g(x), and the lines x=a and x=b, where f and g are continuous and f(x)>g(x) for all x in the interval [a, b], is given by
The figure to the right illustrates the region that is given by the definition above. To make the image clearer, the curve f(x) is shown in blue and the curve g(x) is shown in red.
In some problems, the curves may intersect so that f(x) is not greater than g(x) over the entire interval [a, b]. The graph to the right illustrates this situation. In this case, we must find the point of intersection, c, between the two curves. To find the point of intersection c, we set f(x) = g(x) and solve the resulting equation for x. To find the area, we split the region A between the curves into 2 separate regions, A1, bounded by f(x) and g(x) and the lines x=a and x=c, and A2, bounded by g(x) and f(x) and the lines x=c and x=b. To find the area of A, we calculate the area of each individual region and take the sum of the results.
For the curve to the right, the equation for the area of the region would be given by
Note: This equation depends on the nature of the question. Not all problems will use this same equation to determine the area. For example, if the curve g(x) is greater than f(x) on [a,c), and less than f(x) on (c,b], we would switch the f(x) with g(x) in the equation above.
In other problems, the region may be bounded by the curves f(x) and g(x) over part of the interval [a,b] but may be bounded by only a single curve for another part of the interval. The figure below illustrates this situation.
In a problem like this, there are two possible solutions.
a) The first method is to find the point of intersection, c, between the two curves. We split the region A between the curves into 2 separate regions, A1, bounded by f(x) and g(x) and the lines x=a and x=c, and A2, bounded by +f(x) and -f(x) and the lines x=c and x=b. The image to the right illustrates the first method of solving this type of area problem. The formula to the below shows how the area would be calculated for this specific example.
Note: Like the problem above, this equation depends on the nature of the question. You must recognize which function defines the upper boundary and which defines the lowers boundary over a certain interval.
b) The second method involves rewriting the problem so that it is in terms of y rather than x. We express each curve in terms of y, so that we have f(y) for the right boundary and g(y) for the left boundary. We must find the y coordinates of the intersection points of the curves. We let y=b represent the higher intersection point and y=a represent the lower intersection point. The image to the right illustrates the second approach to solving the example above. The equation below shows how the area would be calculated for this specific example.
Note: Once again, this equation depends on the nature of the question.
In fact, this method can be used to solve any area problems. It may often be simpler to solve the integrals when the functions in terms of y rather than x. If the problem seems too difficult, we can always try to solve it in terms of y.
# Examples
1 | Find the area of the region between the curves
2 | Find the area of the region between the intersecting curves
3 | Find the area of the region between the curves, with respect to y
# Volumes
Integration also allows us to calculate the volumes of solids. Let S be a solid that lies between x=a and x=b. Let the continuous function A(x) represent the cross-sectional area of S in the plane through the point x and perpendicular to the x-axis. The volume of S is given by
Now that we have the definition of volume, the challenging part is to find the function of the area of a given cross section. This process is quite similar to finding the area between curves.
Most volume problems that we will encounter will be require us to calculate the volume of a solid of rotation. These are solids that are obtained when a region is rotated about some line. A typical volume problem would ask, "Find the volume of the solid obtained by rotating the region bounded by the curve(s) about some specified line." Since the region is rotated about a specific line, the solid obtained by this rotation will have a disk-shaped cross-section. We know from simple geometry that the area of a circle is given by A = p r2. For each cross-sectional disk, the radius is determined by the curves that bound the region. If we sketch the region bounded by the given curves, we can easily find a function to determine the radius of the cross-sectional disk at point x.
The figures above illustrates this concept. The figure to the left shows the region bounded by the curve and the x-axis and the lines x = 0 and x = 2. The figure in the center shows the 3-dimensional solid that is formed when the region from the first figure is rotated about the x-axis. The figute to the right shows a typical cross-sectional disk. A disk for a given value x between 0 and 2 will have a radius of . The area of the disk is given by A(x) = p ()2 or equivalently, A(x) = px. Once we find the area function, we simply integrate from a to b to find the volume. The examples below will show complete solutions to finding the area of a given solid.
# Variations of Volume Problems
There are several variations in these types of problems. The first factor that can vary in a volume problem is the axis of rotation. What if the region from the figure above was rotated about the y-axis rather than the x-axis? We would end up with a different function for the radius of a cross-sectional disk. The function would be written with respect to y rather than x, so we would have to integrate with respect to y. In general, we can use the following rule.
If the region bounded by the curves f(x) and g(x) and the lines x=a and x=b is rotated about an axis parallel to the x-axis, write the integral with respect to x. If the axis of rotation is parallel to the y-axis, write the integral with respect to y.
The second factor that can vary in volume problems is the radius of a typical cross-sectional disk. Suppose that the region is bounded by two curves, f(x) and g(x), that both vary between x=a and x=b. The solid that is created by rotating this region about some specified line will have a hole in the center. The radius of a cross-sectional disk will be determined by two functions, rather than a single function. This variation creates two separate styles of problems.
The Disk Method : The disk method is used when the cross sections are disk shaped. The radius of a cross section is determined by a single function, f(x). The area of the disk is given by the formula, A(x) = p (radius)2. The figure to the right shows a typical cross-sectional disk.
The Washer Method : The washer method is used when the cross sections are washer shaped. The radius of a cross section is determined by a two functions, f(x) and g(x). This gives us two separate radii, an outer radius, from f(x) to the axis of rotation and an inner radius from g(x) to the axis of rotation. The area of the washer is given by the formula, A(x) = p [ (outer radius)2-(inner radius)2 ]. The figure to the left shows a typical cross-sectional washer.
# Examples
4 | Find the volume of the solid (axis of rotation parallel to y-axis)
5 | Find the volume of the solid using the washer method
# Volumes By Cylindrical Shells
There is another method to solve volume problems, in addition to the methods described in the section above. The method of cylindrical shells is sometimes simpler to use than the previous methods. For example, suppose we want to find the volume of the solid obtained by rotating about the y-axis, the region bounded by the y=3x-x3 and the line y=0. Since the axis of rotation is parallel to the y-axis, we must integrate the area function with respect to y. To do this, We must solve the cubic function for x in terms of y. This would be rather difficult. However, we will soon see that this problem can be solved quite easily using the method of cylindrical shells.
The image to the right shows a cylindrical shell with outer radius r2, inner radius r1 and height h. We can calculate the volume of the shell by finding the volume of the inner cylinder, V1 and subtracting it from the volume of the outer cylinder, V2. Recall from the geometry tutorial, that the volume of a cylinder is given by the formula V = pi r2h.
We let represent the thickness of the cylindrical shell and r represent the average radius of the shell. In summary, the volume of a cylindrical shell is given by the following formula
Now that we have covered the concept of a cylindrical shell, we can apply it to general volume problems. The idea behind the method of cylindrical shells is to think of a 3-dimensional solid as a collection of cylindrical shells. To find the volume of the solid, we must integrate the formula for the volume of a cylindrical shell. However, the formula for the volume of the cylindrical shell will vary with each problem. We must find functions for the height and the radius of the cylindrical shell at x. Suppose the functions are h(x) for height and r(x) for radius. The volume of the solid obtained by rotating a region about a specific line from a to b is given by
This may seem complicated, but after a few examples the method will be much clearer.
# Examples
6 | Find the volume of the solid using the method of cylindrical shells
7 | Find the volume of the solid
For more practice with the concepts covered in this tutorial, visit the Area and Volume Problems page at the link below. The solutions to the problems will be posted after these chapters are covered in your calculus course.
To test your knowledge of applications of integration problems, try taking the general area and volume test on the iLrn website or the advanced area and volume test at the link below.
Area & Volume Problems
General Area & Volume Test on iLrn
Advanced Area & Volume Test
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# 2017 AMC 8 Problems/Problem 9
## Problem 9
All of Macy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$
## Solution 1
The $6$ green marbles and yellow marbles form $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total marbles. Now suppose the total number of marbles is $x$. We know the number of yellow marbles is $\frac{5}{12}x - 6$ and a positive integer. Therefore, $12$ must divide $x$. Trying the smallest multiples of $12$ for $x$, we see that when $x = 12$, we get there are $-1$ yellow marbles, which is impossible. However when $x = 24$, there are $\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}$ yellow marbles, which must be the smallest possible.
## Solution 2
The 6 green and yellow marbles make up $1 - \(frac{1}{3} + \frac{1}{4} = \frac{5}{12}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.) of the total marbles. Now we know that there are $\frac{5}{12} - 6$ yellow marbles. Now, because going for the number of yellow marbles right now doesn't work (there would be -1 yellow marbles), we multiply 2 on both sides, to find out there are $\frac{10}{24} - 6$ $\boxed{\textbf{(D) }4}$ yellow marbles. |
# Chapter 10.1 Notes: Graph y = ax 2 + c Goal: You will graph simple quadratic functions.
## Presentation on theme: "Chapter 10.1 Notes: Graph y = ax 2 + c Goal: You will graph simple quadratic functions."— Presentation transcript:
Chapter 10.1 Notes: Graph y = ax 2 + c Goal: You will graph simple quadratic functions.
A quadratic function is a nonlinear function that can be written in the standard form y = ax 2 + bx + c. Every quadratic function has a U-shaped graph called a parabola. Basic Quadratic Function: Graph of y = x 2
The lowest or highest point on a parabola is the vertex. The line that passes through the vertex and divides the parabola into two symmetric parts is called the axis of symmetry.
Steps to Graph all Quadratic Functions in the Form y = ax 2 + c: 1.Identify the “a” value. If “a” is positive, the parabola will open up. If “a” is negative, the parabola will open down. The bigger the “a” value, the narrower the parabola will be. The smaller the “a” value, the wider the parabola will be.
2. Identify the “c” value. The “c” value is the vertex, either the highest or lowest point on the parabola. 3. Then make a table of values. Using the coordinates of the vertex, pick two x-values less than the vertex and pick two x- values greater than the vertex. 4. Plot the points and graph the parabola.
Ex.1: Graph y = 5x 2. Ex.2: Graph y = x 2 + 5. Ex.3: Graph y = -5x 2 + 1. Ex.4: Graph Ex.5: Graph y = 3x 2 – 6. Ex.6: Graph
Ex.7: Graph y = -x 2 + 2.
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# How to Solve tricky GMAT Negative Fraction with Exponents?
To answer questions that involve negative fractions and exponents, you have to know the following rules:
1) Fraction with one in the numerator always obeys the following rule
1/High Number < 1/Low Number
What confuses test takers is the definition of High Number and Low Number when it comes to negative denominator
2) For negative numbers, remember this one rule
-5 < -3
- (Higher Positive Number) < - (Lower Positive Number)
3) When it comes to negative fraction, the rule reverses
-(1/High Number) > - (1/Low Number)
4) Even power of negative fraction is positive and odd power of negative fraction is negative
(-(a/b)) ^n is positive if n is even and negative is n is odd
Q) If a = -1/4 & b = -1/3, which of the following is true?
a) a^2 < b^3
b) a^3 > b^2
c) a^4 > b^6
d) a^3 < b^3
e) a^2 < b^4
Solution
Once you know the four rules, it becomes easier to eliminate answer choices. The question asks us to find the cube and square of two negative fractions a and b
a=-1/4 b=-1/3
According to Rule 3
a > b
If you look at the answer choices, a is represented as a^2 a^3 and a^4
And b is represented as b^2 b^3 b^4 , b^5 and b^6
Arrange the signs of the fraction, before solving it
a^2 = +
a^3 = -
a^4 = +
b^2 = +
b^3 = -
b^4 = +
b^5 = -
b^6 = +
Now eliminate answer choice based on signs
+ > -
Eliminate A, B and D
Answer Choices remaining C and D
c) a^4 > b^6
e) a^2 < b^4
Let us look at answer Choice D as both are relatively smaller powers
Solving e)
1/16 < 1/81
This is false. Hence, answer choice C is the Correct Answer. By adopting process of elimination techniques, you can avoid lengthy calculations.
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# Can a 4×4 matrix have a determinant?
## Can a 4×4 matrix have a determinant?
Determinant of a 4×4 matrix is a unique number which is calculated using a particular formula. If a matrix order is n x n, then it is a square matrix. Hence, here 4×4 is a square matrix which has four rows and four columns. If A is square matrix then the determinant of matrix A is represented as |A|.
## How do you normalize a Hadamard matrix?
A Hadamard matrix is said to be normalized if all of the elements of the first row and first column are +1. Hadamard matrices have several interesting properties: The determinant, |Hn| = nn/2, is maximal by Hadamard’s theorem on determinants. A normalized Hn has n(n-1)/2 elements of –1 and n(n+1) elements of +1.
How do you multiply a 4×4 matrix by a 1×4 matrix?
the two adjacent dimensions must be the same. This means it is not possible to multiply a 4×4 matrix with a 1×4 matrix, but it is possible to multiply 4×4 by 4×1 to get a 4×1 matrix or 1×4 by 4×4 to get a 1×4 matrix.
The Hadamard product is commutative (when working with a commutative ring), associative and distributive over addition. That is, if A, B, and C are matrices of the same size, and k is a scalar: The identity matrix under Hadamard multiplication of two m × n matrices is an m × n matrix where all elements are equal to 1.
### What is Hadamard’s maximal determinant problem?
Hadamard’s maximal determinant problem, named after Jacques Hadamard, asks for the largest determinant of a matrix with elements equal to 1 or −1.
What is the determinant of a 4×4 matrix?
Determinant of a 4×4 matrix is a unique number which is calculated using a particular formula. If a matrix order is n x n, then it is a square matrix. Hence, here 4×4 is a square matrix which has four rows and four columns.
## What is the Hadamard product of two positive eigenvalues?
The Hadamard product is a principal submatrix of the Kronecker product . where λi(A) is the i th largest eigenvalue of A . is the identity matrix . is Kronecker product. denotes face-splitting product. is column-wise Khatri–Rao product. The Hadamard product of two positive-semidefinite matrices is positive-semidefinite. |
# Term Paper: Tag Members of a Population
Pages: 2 (733 words) · Bibliography Sources: 0 · File: .docx · Level: College Junior · Topic: Education - Mathematics
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SAMPLE EXCERPT:
[. . .] The assignment is to solve for y and identity the type of equation that results when solving for y. The number sentence given is: y-1/x+3 = -3/4. Using cross multiplication gets rid of the denominators and yields the following equation: 4(y-1) = -3(x+3). The goal is then to isolate y on the left side of the equation, which gives one an equation solving for y. Without additional information, the equation cannot be solved for either the numerical value of x or y because doing so requires multiple equations or providing what x or y is. This information suggests immediately that the equation represents not a single set of coordinates, but a line. As a result, one is already thinking that the end-result of the process will be an equation for a line. The slope, intercept form of an equation for a line is popularly represented as y=mx + b where m represents the slope of the line and b represents the y intercept of the line. (The y intercept is the point on the line where it intercepts the y axis; in other words, the y intercept is the point on the line where x=0). Knowing this basic form of an equation helps guide the shaping of the equation as one solves for y.
The first step is to multiply each side of the equation. 4(y-1) becomes 4y-4. -3(x+3) becomes -3x -9. The equation them comes 4y-4 = -3x -9. The next thing to do is to add 4 to both sides as the next step in isolating y. The resulting equation is 4y= -3x -9 + 4 or 4y=-3x-5. One then divides both sides by 4 to continue isolating y. The resulting equation is y=(-3x-5)/4. The four is distributed throughout the equation, yielding the following equation y=-3/4x -5/4. That equation is an equation for a line with the slope -3/4 and a y intercept of -5/4. The y intercept provides the solver with a set of numbers that can be applied to the equation (0,-5/4). Plugging these numbers back into the equation allows one to eliminate the possibility of an extraneous response. y-1/x+3 = -3/4. Does (-5/4- 1) / 3 = -3/4? -5/4-4/4=… [END OF PREVIEW]
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https://www.essaytown.com/subjects/paper/tag-members-population/4024249. |
# Binary and Number Systems #
Having a good idea of how binary works is the first step to understanding how IP addresses really operate.
Here, we’ll go over some core concepts of binary, how it compares to our usual numbering system (decimal), and some easy ways to understand how it works.
First off, let’s go over how decimal works deep-down.
## Decimal #
In our regular counting system, decimal, we use the digits `0-9` to represent the numbers zero to nine.
### Counting #
Let’s use this three-digit counter as an example. On the right, we’ll state the actual value that’s represented by the digits on the left.
``````000 - zero
``````
To count up, we increment (go up by one each time) from `0` to `1`, `1` to `2`, etc. Let’s count up to nine:
``````000 - zero
001 - one
002 - two
003 - three
004 - four
005 - five
006 - six
007 - seven
008 - eight
009 - nine
``````
However, there are no digits after `9`. To continue counting up to ten, we need to increment the next column and reset the rightmost one.
Let’s count up from 9:
``````009 - nine
010 - ten
``````
So we changed the middle digit from `0` to `1`, and reset the right digit from `9` back to `0`.
To continue counting up from ten, we do exactly the same with the rightmost number:
``````010 - ten
011 - eleven
012 - twelve
013 - thirteen
014 - fourteen
015 - fifteen
016 - sixteen
017 - seventeen
018 - eighteen
019 - nineteen
``````
And to count up to twenty, we do the same:
``````019 - nineteen
020 - twenty
``````
This is how decimal works, and specifically how counting works in decimal
Decimal can also be called “base 10” because it uses ten digits to represent values (`0`, `1`, `2`, `3`, `4`, `5`, `6`, `7`, `8`, and `9`).
## Binary #
Binary is similar to decimal, except that it uses a base-2 system (just the digits `0` and `1`).
Let’s keep the same three-digit counter, starting from zero:
``````000 - zero
``````
Counting up to one is simple:
``````000 - zero
001 - one
``````
However, going up to two requires us to increment the next digit and reset the rightmost one. In other words, because there aren’t any digits after `1`, we need to go to the next column.
``````001 - one
010 - two
``````
Let’s count up as far as we can:
``````010 - two
011 - three
100 - four
101 - five
110 - six
111 - seven
``````
Computers use binary because it meshes well with how they work deep down. Computers are based on electrical signals, and those signals can either be high (`1`), or low (`0`). Because of this, using a number system with only two values for each column makes sense.
### Bits and Bytes #
Sometimes when using binary, you’ll see the term “bits”. Bit simply stands for binary digit, and it used as a shorthand when talking about computers in particular. Up until now, our counter has used three bits, and four/eight bits are typical sizes used a lot when working with computers.
Collections of different numbers of bits can also have names. For instance, eight bits are called a byte, and four bits are called a nibble. As an example:
``````bit = 0
nibble = 0000
byte = 0000 0000
``````
When working with binary, bytes are sometimes written as groups of four bits. This just makes it easier to read, and easy to convert to other number systems (as you’ll see later).
## Base 10 vs Base 2 #
With base 10 (decimal), if we count all the way up, 3 digits lets us represent these values:
``````*** decimal ***
000 - zero
...
999 - nine-hundred and ninety-nine
``````
Or in other words, one thousand separate values.
With base 2 (binary), if we do the same we can represent these values:
``````*** binary ***
000 - zero
...
111 - seven
``````
Or, eight separate values.
In a single column, you can represent the base number of digits that number system has (two for binary, ten for decimal). With each new column, the number of values you can represent is simply multiplied by the base. For example:
``````*** decimal ***
9 one digit, ten values
99 two digits, one hundred values
999 three digits, one thousand values
*** binary ***
1 one digit, two values
11 two digits, four values
111 three digits, eight values
``````
This is useful to keep in mind when working with binary, particularly when working out network addresses.
## Converting #
To get the value from a binary number, you can count up how large each column is based on whether it’s a `0` or a `1`.
As an example, let’s get the value from the binary number `1011`. The top line shows the value each column represents.
``````1011 = 1 0 1 1 (split into columns)
values: 8 | 4 | 2 | 1
binary: 1 | 0 | 1 | 1
8 + 0 + 2 + 1 = eleven
``````
So the binary number `1011` represents eleven, or:
``````1011 in base 2 = 11 in base 10
``````
Being able to convert binary numbers to real values is very useful, and in the next section we’ll be going over IP addresses and subnets.
Try going through some of these examples on a piece of paper, to see whether you’re able to convert them to the proper values. Once you’ve finished, try cross-checking your answers with an online converter or Google:
``````1001 = ? 00101101 = ? 11111100 = ? 00010000 = ?
0110 = ? 11011101 = ? 00011111 = ? 00110001 = ?
``````
## Base 16 #
There is also a similar number system that’s often used in computing – “base 16”, or hexidecimal. It uses these 16 digits:
``````0 1 2 3 4 5 6 7 8 9 A B C D E F
``````
The useful thing with base 16 is that each column can represent 16 separate values. Similarly, four columns of binary can also represent 16 separate values. Because of this, hexidecimal can easily – and very compactly – represent binary digits.
Here’s a chart of values represented in Binary, Hex, and Decimal, which can be useful when converting between them:
## Overall #
• Decimal uses ten digits (`0-9`) in each column to represent values, and because of this it’s called “base 10”.
• Binary uses two digits (`0,1`) in each column to represent values, and it’s called “base 2”.
• Hexidecimal uses sixteen digits (`0-9,A-F`) in each column to represent values, and it’s called “base 16”.
• Counting, based on columns, work the same way in decimal as it does in binary (as it also does in hexidecimal). |
Common Core: High School - Statistics and Probability : Use Statistics to Compare Center and Spread of Data Distribution: CCSS.Math.Content.HSS-ID.A.2
Example Questions
← Previous 1
Example Question #1 : Use Statistics To Compare Center And Spread Of Data Distribution: Ccss.Math.Content.Hss Id.A.2
Which statement about the two sets is TRUE?
The standard deviation of Set B is twice the standard deviation of Set A.
Set A has a smaller standard deviation but larger range than Set B.
Of the two sets, Set B has the smaller mean, median, and standard deviation.
In Set B, the mean is larger than the median.
Of the two sets, Set A has the smaller mean, median, and standard deviation.
Of the two sets, Set A has the smaller mean, median, and standard deviation.
Explanation:
Example Question #2 : Use Statistics To Compare Center And Spread Of Data Distribution: Ccss.Math.Content.Hss Id.A.2
A plumber salvages old brass fittings in order to make extra money by selling scrap metal. He tallies his scrap totals for a five month period. The plumber collected the following pounds of brass in a five month period:
Find the proper measure of center and spread for this data set.
Explanation:
There are several common measures of center and spread for a given data set. The most common measures of center are the mean and median. The mean represents the arithmetic average of a set and the median is the middlemost value. On the other hand the most common measures of spread are the mean absolute deviation (MAD) and inter-quartile range (IQR).
The measures of center and spread vary for different sets of data. The mean and MAD are used together to analyze data presented in bar charts or histograms while the median and IQR are most commonly used for box and whisker plots.
Lets look at our data. We have the following pounds of brass for the five-month period starting in January:
First let's plot this data on a bar graph.
Now, let's use the mean and MAD to calculate the center and spread of the data respectively.
The center should be calculated using the mean. The mean is the arithmetic average and is found by adding all of the values together and dividing by the number of values in the series.
Now, let's find the spread of the data using the MAD. The MAD describes how data varies about the mean. The MAD is calculated by finding out how much each data points deviates from the mean and dividing by the total number of values in the series.
First, let's find out how much each value varies from the mean by taking the absolute value of the mean minus each individual value.
Now, we need to add up these values and divide that total by the number of values in the series.
The MAD is the value that each individual number in the set deviates from the mean; therefore, it represents the spread of the series.
The correct answer for this problem is the following:
Example Question #3 : Use Statistics To Compare Center And Spread Of Data Distribution: Ccss.Math.Content.Hss Id.A.2
A plumber salvages old brass fittings in order to make extra money by selling scrap metal. He tallies his scrap totals for a five month period. The plumber collected the following pounds of brass in a five month period:
Find the proper measure of center and spread for this data set.
Explanation:
There are several common measures of center and spread for a given data set. The most common measures of center are the mean and median. The mean represents the arithmetic average of a set and the median is the middlemost value. On the other hand, the most common measures of spread are the mean absolute deviation (MAD) and inter-quartile range (IQR).
The measures of center and spread vary for different sets of data. The mean and MAD are used together to analyze data presented in bar charts or histograms while the median and IQR are most commonly used for box and whisker plots.
Let's look at our data. We have the following pounds of brass for the five-month period starting in January:
First, let's plot this data on a bar graph.
Now, let's use the mean and MAD to calculate the center and spread of the data respectively.
The center should be calculated using the mean. The mean is the arithmetic average and is found by adding all of the values together and dividing by the number of values in the series.
Now, let's find the spread of the data using the MAD. The MAD describes how data varies about the mean. The MAD is calculated by finding out how much each data points deviates from the mean and dividing by the total number of values in the series.
First, let's find out how much each value varies from the mean by taking the absolute value of the mean minus each individual value.
Now, we need to add up these values and divide that total by the number of values in the series.
The MAD is the value that each individual number in the set deviates from the mean; therefore, it represents the spread of the series.
The correct answer for this problem is the following:
Example Question #4 : Use Statistics To Compare Center And Spread Of Data Distribution: Ccss.Math.Content.Hss Id.A.2
A plumber salvages old brass fittings in order to make extra money by selling scrap metal. He tallies his scrap totals for a five month period. The plumber collected the following pounds of brass in a five month period:
Find the proper measure of center and spread for this data set.
Explanation:
There are several common measures of center and spread for a given data set. The most common measures of center are the mean and median. The mean represents the arithmetic average of a set and the median is the middlemost value. On the other hand the most common measures of spread are the mean absolute deviation (MAD) and inter-quartile range (IQR).
The measures of center and spread vary for different sets of data. The mean and MAD are used together to analyze data presented in bar charts or histograms while the median and IQR are most commonly used for box and whisker plots.
Lets look at our data. We have the following pounds of brass for the five-month period starting in January:
First let's plot this data on a bar graph.
Now, let's use the mean and MAD to calculate the center and spread of the data respectively.
The center should be calculated using the mean. The mean is the arithmetic average and is found by adding all of the values together and dividing by the number of values in the series.
Now, let's find the spread of the data using the MAD. The MAD describes how data varies about the mean. The MAD is calculated by finding out how much each data points deviates from the mean and dividing by the total number of values in the series.
First, let's find out how much each value varies from the mean by taking the absolute value of the mean minus each individual value.
Now, we need to add up these values and divide that total by the number of values in the series.
The MAD is the value that each individual number in the set deviates from the mean; therefore, it represents the spread of the series.
The correct answer for this problem is the following:
Example Question #5 : Use Statistics To Compare Center And Spread Of Data Distribution: Ccss.Math.Content.Hss Id.A.2
A plumber salvages old brass fittings in order to make extra money by selling scrap metal. He tallies his scrap totals for a five month period. The plumber collected the following pounds of brass in a five month period:
Find the proper measure of center and spread for this data set.
Explanation:
There are several common measures of center and spread for a given data set. The most common measures of center are the mean and median. The mean represents the arithmetic average of a set and the median is the middlemost value. On the other hand, the most common measures of spread are the mean absolute deviation (MAD) and inter-quartile range (IQR).
The measures of center and spread vary for different sets of data. The mean and MAD are used together to analyze data presented in bar charts or histograms while the median and IQR are most commonly used for box and whisker plots
.
Let's look at our data. We have the following pounds of brass for the five-month period starting in January:
First, let's plot this data on a bar graph.
Now, let's use the mean and MAD to calculate the center and spread of the data respectively.
The center should be calculated using the mean. The mean is the arithmetic average and is found by adding all of the values together and dividing by the number of values in the series.
Now, let's find the spread of the data using the MAD. The MAD describes how data varies about the mean. The MAD is calculated by finding out how much each data points deviates from the mean and dividing by the total number of values in the series.
First, let's find out how much each value varies from the mean by taking the absolute value of the mean minus each individual value.
Now, we need to add up these values and divide that total by the number of values in the series.
The MAD is the value that each individual number in the set deviates from the mean; therefore, it represents the spread of the series.
The correct answer for this problem is the following:
Example Question #6 : Use Statistics To Compare Center And Spread Of Data Distribution: Ccss.Math.Content.Hss Id.A.2
A plumber salvages old brass fittings in order to make extra money by selling scrap metal. He tallies his scrap totals for a five month period. The plumber collected the following pounds of brass in a five month period:
Find the proper measure of center and spread for this data set.
Explanation:
There are several common measures of center and spread for a given data set. The most common measures of center are the mean and median. The mean represents the arithmetic average of a set and the median is the middlemost value. On the other hand, the most common measures of spread are the mean absolute deviation (MAD) and inter-quartile range (IQR).
The measures of center and spread vary for different sets of data. The mean and MAD are used together to analyze data presented in bar charts or histograms while the median and IQR are most commonly used for box and whisker plots
.
Let's look at our data. We have the following pounds of brass for the five-month period starting in January:
First, let's plot this data on a bar graph.
Now, let's use the mean and MAD to calculate the center and spread of the data respectively.
The center should be calculated using the mean. The mean is the arithmetic average and is found by adding all of the values together and dividing by the number of values in the series.
Now, let's find the spread of the data using the MAD. The MAD describes how data varies about the mean. The MAD is calculated by finding out how much each data points deviates from the mean and dividing by the total number of values in the series.
First, let's find out how much each value varies from the mean by taking the absolute value of the mean minus each individual value.
Now, we need to add up these values and divide that total by the number of values in the series.
The MAD is the value that each individual number in the set deviates from the mean; therefore, it represents the spread of the series.
The correct answer for this problem is the following:
Example Question #7 : Use Statistics To Compare Center And Spread Of Data Distribution: Ccss.Math.Content.Hss Id.A.2
A plumber salvages old brass fittings in order to make extra money by selling scrap metal. He tallies his scrap totals for a five month period. The plumber collected the following pounds of brass in a five month period:
Find the proper measure of center and spread for this data set.
Explanation:
There are several common measures of center and spread for a given data set. The most common measures of center are the mean and median. The mean represents the arithmetic average of a set and the median is the middlemost value. On the other hand, the most common measures of spread are the mean absolute deviation (MAD) and inter-quartile range (IQR).
The measures of center and spread vary for different sets of data. The mean and MAD are used together to analyze data presented in bar charts or histograms while the median and IQR are most commonly used for box and whisker plots
.
Let's look at our data. We have the following pounds of brass for the five-month period starting in January:
First, let's plot this data on a bar graph.
Now, let's use the mean and MAD to calculate the center and spread of the data respectively.
The center should be calculated using the mean. The mean is the arithmetic average and is found by adding all of the values together and dividing by the number of values in the series.
Now, let's find the spread of the data using the MAD. The MAD describes how data varies about the mean. The MAD is calculated by finding out how much each data points deviates from the mean and dividing by the total number of values in the series.
First, let's find out how much each value varies from the mean by taking the absolute value of the mean minus each individual value.
Now, we need to add up these values and divide that total by the number of values in the series.
The MAD is the value that each individual number in the set deviates from the mean; therefore, it represents the spread of the series.
The correct answer for this problem is the following:
Example Question #8 : Use Statistics To Compare Center And Spread Of Data Distribution: Ccss.Math.Content.Hss Id.A.2
A plumber salvages old brass fittings in order to make extra money by selling scrap metal. He tallies his scrap totals for a five month period. The plumber collected the following pounds of brass in a five month period:
Find the proper measure of center and spread for this data set.
Explanation:
There are several common measures of center and spread for a given data set. The most common measures of center are the mean and median. The mean represents the arithmetic average of a set and the median is the middlemost value. On the other hand, the most common measures of spread are the mean absolute deviation (MAD) and inter-quartile range (IQR).
The measures of center and spread vary for different sets of data. The mean and MAD are used together to analyze data presented in bar charts or histograms while the median and IQR are most commonly used for box and whisker plots
.
Let's look at our data. We have the following pounds of brass for the five-month period starting in January:
First, let's plot this data on a bar graph.
Now, let's use the mean and MAD to calculate the center and spread of the data respectively.
The center should be calculated using the mean. The mean is the arithmetic average and is found by adding all of the values together and dividing by the number of values in the series.
Now, let's find the spread of the data using the MAD. The MAD describes how data varies about the mean. The MAD is calculated by finding out how much each data points deviates from the mean and dividing by the total number of values in the series.
First, let's find out how much each value varies from the mean by taking the absolute value of the mean minus each individual value.
Now, we need to add up these values and divide that total by the number of values in the series.
The MAD is the value that each individual number in the set deviates from the mean; therefore, it represents the spread of the series.
The correct answer for this problem is the following:
Example Question #9 : Use Statistics To Compare Center And Spread Of Data Distribution: Ccss.Math.Content.Hss Id.A.2
A plumber salvages old brass fittings in order to make extra money by selling scrap metal. He tallies his scrap totals for a five month period. The plumber collected the following pounds of brass in a five month period:
Find the proper measure of center and spread for this data set.
Explanation:
There are several common measures of center and spread for a given data set. The most common measures of center are the mean and median. The mean represents the arithmetic average of a set and the median is the middlemost value. On the other hand, the most common measures of spread are the mean absolute deviation (MAD) and inter-quartile range (IQR).
The measures of center and spread vary for different sets of data. The mean and MAD are used together to analyze data presented in bar charts or histograms while the median and IQR are most commonly used for box and whisker plots
.
Let's look at our data. We have the following pounds of brass for the five-month period starting in January:
First, let's plot this data on a bar graph.
Now, let's use the mean and MAD to calculate the center and spread of the data respectively.
The center should be calculated using the mean. The mean is the arithmetic average and is found by adding all of the values together and dividing by the number of values in the series.
Now, let's find the spread of the data using the MAD. The MAD describes how data varies about the mean. The MAD is calculated by finding out how much each data points deviates from the mean and dividing by the total number of values in the series.
First, let's find out how much each value varies from the mean by taking the absolute value of the mean minus each individual value.
Now, we need to add up these values and divide that total by the number of values in the series.
The MAD is the value that each individual number in the set deviates from the mean; therefore, it represents the spread of the series.
The correct answer for this problem is the following:
Example Question #1 : Use Statistics To Compare Center And Spread Of Data Distribution: Ccss.Math.Content.Hss Id.A.2
A plumber salvages old brass fittings in order to make extra money by selling scrap metal. He tallies his scrap totals for a five month period. The plumber collected the following pounds of brass in a five month period:
Find the proper measure of center and spread for this data set.
Explanation:
There are several common measures of center and spread for a given data set. The most common measures of center are the mean and median. The mean represents the arithmetic average of a set and the median is the middlemost value. On the other hand, the most common measures of spread are the mean absolute deviation (MAD) and inter-quartile range (IQR).
The measures of center and spread vary for different sets of data. The mean and MAD are used together to analyze data presented in bar charts or histograms while the median and IQR are most commonly used for box and whisker plots
.
Let's look at our data. We have the following pounds of brass for the five-month period starting in January:
First, let's plot this data on a bar graph.
Now, let's use the mean and MAD to calculate the center and spread of the data respectively.
The center should be calculated using the mean. The mean is the arithmetic average and is found by adding all of the values together and dividing by the number of values in the series.
Now, let's find the spread of the data using the MAD. The MAD describes how data varies about the mean. The MAD is calculated by finding out how much each data points deviates from the mean and dividing by the total number of values in the series.
First, let's find out how much each value varies from the mean by taking the absolute value of the mean minus each individual value.
Now, we need to add up these values and divide that total by the number of values in the series.
The MAD is the value that each individual number in the set deviates from the mean; therefore, it represents the spread of the series.
The correct answer for this problem is the following:
← Previous 1 |
# Question Video: Using Unit Rate to Compare Elements in a Word Problem Mathematics • 8th Grade
Jennifer wants to burn an extra 2000 calories a week. She can choose between the following exercise classes. Which class burns the most calories per hour? How long would she have to spend in that class to reach her weekly goal?
04:20
### Video Transcript
Jennifer wants to burn an extra 2000 calories a week. She can choose between the following exercise classes. Which class burns the most calories per hour? How long would she have to spend in that class to reach her weekly goal?
Class A lasts one and a quarter hours and burns 600 calories. In class B, 375 calories are burned in 45 minutes. Class C lasts one hour and burns 400 calories. In order to calculate which class burns the most calories per hour, we need to work out the rate of change. As the three durations are different, we can add an extra column to the table to calculate the number of calories burned per hour.
The duration of class C is one hour, and 400 calories are burned. Therefore, she would burn 400 calories per hour. The duration of class B is 45 minutes, which is equal to three-quarters of an hour. As three-quarters is equal to the decimal 0.75, we could say that 45 minutes is equal to 0.75 of an hour. To calculate the rate of change or number of calories burned per hour, we need to divide the total calories burned by the time in hours. For class B, we need to divide 375 by 0.75. This can be done on the calculator, giving us an answer of 500. In class B, Jennifer would burn calories at a rate of 500 per hour.
There is an easier way of working this out if we didn’t have a calculator. We know that in 45 minutes, Jennifer burns 375 calories. We could therefore calculate the number of calories burned in 15 minutes by dividing by three. 300 divided by three is 100, and 75 divided by three is 25. Therefore, Jennifer would burn 125 calories in 15 minutes. Adding 45 minutes and 15 minutes gives us 60 minutes, and adding 375 to 125 calories gives us 500 calories. As 60 minutes is equal to one hour, this proves that our answer for class B was correct. Jennifer would burn 500 calories per hour.
We could use a similar method to this for class A, starting with 600 calories in one and a quarter hours or 75 minutes. However, the quicker way to work out the number of calories burned per hour for class A would be to divide 600 by one and a quarter. This is the same as dividing 600 by 1.25. This is equal to 480. Therefore, Jennifer burns 480 calories per hour in class A. The largest of these three values is 500. We can therefore conclude that class B burns the most calories per hour.
The second part of the question asks us how long Jennifer would have to spend in that class to reach a weekly goal of burning 2000 calories. We can calculate the time she would need to spend by dividing the total calories she wants to burn by the number of calories burned per hour. We need to divide 2000 by 500. Both the numerator and denominator are divisible by 100, leaving us with 20 divided by five. As 20 divided by five is equal to four, Jennifer would need to spend four hours in class B to burn 2000 calories. |
# Chapter 7, Section 5
## Partial fractions
### Decomposition of N(x) / D(x) into Partial Fractions
1. Divide if improper: If N(x)/D(x) is an improper fraction ( that is, id the degree of the numerator is greater than or equal to the degree of the denominator), divide the denominator into the numerator to obtain
N(x)/D(x) = ( a polynomial) + N1(x)/D(x)
2. Factor Denominator: Completely factor the denominator into factors of the form
( px + q )m and ( ax + bx + c )n
where ax2 + bx + c is irreducible.
3. Linear factors: For each factor of the form ( px + q )m , the partial fraction decomposition must include the following sum of m fractions.
(A1/( px + q )) + (A2/( px + q )2) + ... .+ (Am/( px + q )m)
4. Quadratic factors: For each factor of the form ( ax + bx + c )n, the partial fraction decomposition must include the following sum of n fractions.
((B1x + C1)/( ax + bx + c )) + ((B2x + C2)/(ax + bx + c )2) + ... .+ ((Bnx + Cn)/( ax + bx + c )n)
### Guidelines for solving the basic Equations
Linear Fractions
1. Substitute the roots of the distinct linear factors into the basic equation.
2. For repeated linear factors, use the coefficients determined in guideline 1 to rewrite the basic equations. Then substitute other convenient values of x and solve the remaining coefficients.
1. Expand the basic equation.
2. Collect terms according to powers of x.
3. Equate the coefficients of like powers to obtain a system of linear equations involving A,B,C, and so on.
4. Solve the system of linear equations. |
# Video: AF5P3-Q03-340180531739
Which of these numbers is equal to a prime number that has been doubled? Circle your answer. [A] 6 [B] 8 [C] 16 [D] 18.
02:51
### Video Transcript
Which of these numbers is equal to a prime number that has been doubled? Circle your answer. The options are six, eight, 16 and 18.
Well, the first thing we need to look at is what is a prime number. A prime number is a number that has exactly two factors, one and themselves. We add this to the definition because if we think about the number one, well number one has one and itself as a factor. However, it’s only got one factor, so one would not be a prime number.
So next, what we want to look at is that we’re trying to find which one of our numbers is equal to a prime number that has been doubled. So therefore, we can say that our number is equal to two multiplied by a prime. So therefore, if we use inverse operations, so we divide each side of the equation by two. We can say that our number divided by two must be equal to a prime number.
And that’s because if you’ve got two multiplied by a prime and you divide it by two, it just leaves us with the prime. And then we’ve got to do the same to the other side so we have our number divided by two.
So now, to see whether our numbers are prime when they’re divided by two, we have to divide them each by two. First, well, we’ve got six. So, six over two or six divided by two is equal to three. Then we have eight, and eight divided by two is equal to four. 16 divided by two is equal to eight. And finally, 18 divided by two is equal to nine.
Okay, so we’ve now got what the value of our numbers is, when they’ve been halved. We have to see which one of them is a prime. Well if we start the right-hand side, we’ve got nine. Well, nine is not a prime number because nine can be divided by three, because three multiplied by three is equal to nine.
Eight is also not a prime number because it can be divided by two. Because we can have four multiplied by two giving us eight. And then we have four. Well, four is not a prime number because, again, we could divide it by two. And we’ve got two multiplied by two equals four. So, it’s got more than two factors. Well, we’d automatically know that four and eight would not be prime numbers because they’re even. And there is, in fact, only one even prime number, and that’s two. And that’s because two has two factors, one and two, and the only two factors are itself and one.
So now let’s move on to the final answer, which we think should be the correct answer and should be a prime number. Well, we have three, and three is a prime number because three only has two factors. That’s three and one because three multiplied by one is three. And therefore we can say that it’s definitely prime. So, in answer to the question, which of these numbers is equal to a prime number that has been doubled? The correct answer is six, and I’ve circled the first answer six. |
# Lecture 003
## Continue on Notation
Intersection($\cap$): common elements between two sets
• $A\cap B = \{x \in U | x\in A \text{ and } x\in B\}$, let U be Universal set
• $A\cap B = \{x \in A | x\in B\}$
• $A\cap B = \{x \in B | x\in A\}$
• if $A\cap B = \emptyset$ then A and B is called disjoint
Union($\cup$): $A\cup B = \{x\in A \text{ or } x\in B\}$
Difference($-$ or $\backslash$): $A\backslash B = \{x\in U | x\in A \text{ and } x\notin B\}=\{x\in A | x\notin B\}$
Complement($A^\complement$ or $\overline{A}$, or $A^\mathsf{c}$): $\overline{A}=\{x\in U | x\notin A\}=U\backslash A$ TODO: :question: can I wrote as ^\complement?
Index set: $i\in I$ in which $i$ denotes the index of set of all possible variation of $A$.
• $\{A_i\}_{i\in I} = \{A_i | i\in I\}$ is an family of sets indexed by $I$
• a set of set is called family
• you can use this to define $A_i = \{i, -i\}$
Indexed intersections($\bigcap$): $
Indexed unions($\bigcup$): $
Cartesian product($A\times B$): $A\times B = \{(a, b) | a\in A \text{ and } b\in B\}$
• our Cartesian plane is $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R} = \{(x,y) | x, y \in \mathbb{R}\}$
• $A \times \emptyset = \emptyset \times A = \emptyset$ for any A
• $x,y\in \mathbb{Z} \lnot (x, y)\in \mathbb{Z}^2$
Because $ $ so the above is the same thing as $ because no information is lost (the information is tuples or triples are the direction of the operation.)
## Logic and Proofs
s.t. for such that holds means it yields true
Table of Content |
SOLUTION 3: $\ \$ To integrate $\displaystyle{ \int { 1 \over (1-x^2)^{3/2} } \ dx }$ use the trig substitution $$x = \sin \theta$$ so that $$dx = \cos \theta \ d \theta$$ Substitute into the original problem, replacing all forms of $x$, getting $$\displaystyle { \int \frac{1}{(1-x^{2})^{3/2}} \ dx = \int \frac{1}{(1-\sin^{2} \theta)^{3/2}} \ \cos \theta \ d \theta }$$ $$= \displaystyle { \int \frac{1}{(\cos^{2} \theta)^{3/2}} \ \cos \theta \ d \theta }$$ $$= \displaystyle { \int \frac{1}{\cos^{3} \theta} \ \cos \theta \ d \theta }$$ $$= \displaystyle { \int \frac{1}{\cos^{2} \theta} \ d \theta }$$ $$= \displaystyle { \int \sec^{2} \theta \ d \theta }$$ $$= \tan \theta + C$$ $\Big($ We need to write our final answer in terms of $x$.
Since $x = \sin \theta$ it follows that $$\sin \theta = \displaystyle{ x \over 1 } = \displaystyle{ opposite \over hypotenuse }$$ and from the Pythagorean Theorem that $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(adjacent)^2 + (x)^2 = (1)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{1-x^2} \ \ \longrightarrow$$ $$\tan \theta = \displaystyle{ opposite \over adjacent }= \displaystyle{ x \over \sqrt{1-x^2} } . \Big)$$ $$= \displaystyle{ x \over \sqrt{1-x^2} } + C$$
Click HERE to return to the list of problems. |
# What Are Commutative, Associative and Distributive Properties?
The commutative, associative and distributive properties describe how basic mathematical operations work. The properties are helpful in finding efficient ways to solve equations and in simplifying algebraic expressions.
Addition and multiplication have the commutative property, meaning that numbers can be added or multiplied together in any order without affecting the result. In other words, adding 7 + 3 is the same as adding 3 + 7. Multiplying 2 by 4.5 has the same outcome as multiplying 4.5 by 2. The commutative property for addition is expressed as a + b = b + a. The commutative property for multiplication is expressed as a * b = b * a.
Addition and multiplication also have the associative property, meaning that numbers can be added or multiplied in any grouping (or association) without affecting the result. For example, (3 + 2) + 7 has the same result as 3 + (2 + 7), while (4 * 2) * 5 has the same result as 4 * (2 * 5). The associative property for addition is expressed as (a + b) + c = a + (b + c). The associative property for multiplication is expressed as (a * b) * c = a * (b * c).
Only multiplication has the distributive property, which applies to expressions that multiply a number by a sum or difference. Multiplication distributes over addition because a(b + c) = ab + ac. For example, to multiply 2 by the sum of 9 + 4, the numbers 9 and 4 can be added first to find the sum of 13, and then 13 can be multiplied by 2 to return 26. A different way to achieve the same result is to distribute the 2 by first multiplying 2 by 9 (18) and 2 by 4 (8). The two results can be added (18 + 8) to return 26 as before. In the same way, multiplication distributes over subtraction because a(b – c) = ab – ac. |
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# i) Fifty-two hundredths is same as ________ii) What is $20 \div 10$
Last updated date: 19th Jul 2024
Total views: 348k
Views today: 4.48k
Answer
Verified
348k+ views
Hint: Here in this question, we have two sub questions. The first question is related to the decimal. The hundredth will represent the place value, so considering the place value we simplify the first question. The second question is a simple division, by using the tables of multiplication or by cancelling we get the solution for the given question.
Complete step by step solution:
I.In mathematics we have different kinds of numbers namely, natural numbers, whole numbers, Integers, rational numbers, irrational numbers and real numbers.
Now consider the first question
Fifty two hundredths
The hundredth place value written as $\dfrac{1}{{100}}$
This is written as
$\Rightarrow 52 \times \dfrac{1}{{100}}$
On simplifying we have
$\Rightarrow \dfrac{{52}}{{100}}$
On dividing 52 by 100 we have
$\Rightarrow 0.52$
Hence we have got the solution for the given question.
II.Now we solve the second question
Consider $20 \div 10$
This can be written in the form of fraction
$\Rightarrow \dfrac{{20}}{{10}}$
On cancelling the zeros in the numerator and denominator we have
$\Rightarrow \dfrac{2}{1}$
Since the denominator’s value is 1. We neglect the value of 1, we write the numerator value.
$\Rightarrow 2$
The second question can be solved in another way.
Consider $20 \div 10$
This can be written in the form of fraction
$\Rightarrow \dfrac{{20}}{{10}}$
By the tables of multiplication of 10, When 10 is multiplied by 2, the answer will be 20. So we have
$\Rightarrow 2$
Hence we have got the solution for the given question.
Note: The students do not get confused by seeing the word hundredths. The fifty two hundredths means it is not 5200. The hundredth will represent the place value. So students know about the place value system of the decimal number. Students must know the tables of multiplication. |
# Which is larger without using examples of numbers?
If $$a<b<c<d$$ Which is larger without using examples of numbers?
$$x=(a+b)(c+d), y=(a+c)(b+d), z=(a+d)(b+c)$$
I did this exercise, but I have not managed to complete it. The first thing I did was to expand them and then compare ax with z and so on, I know that the correct answer is that z is greater, but I can not assume I have to prove it and my doubt is at the end of taking away the similar terms, since I am assuming that the difference of one side is less than the term that remains for the other.
In order to solve this you can expand the three expressions, and then compute $z-x$ and $z-y$ which are respectively equal to $(d-b)(c-a)$ and $(d-c)(b-a)$ which are both positive. Therefore $z$ is the largest of all three variables.
Here is a proof $x \lt y$
$$x = ac + ad + bc + bd$$
$$y = ab + ad + cb + cd$$
Let us start with assuming $x < y$
$$x = ac + ad + bc + bd \lt ab + ad + cb + cd = y$$
Cancelling out like terms:
$$ac + bd \lt ab + cd$$
And then rearranging:
$$0 \lt cd - ac + ab -bd$$
$$0 \lt c(d-a) +b(a-d)$$
For simplicity, let's define g = d-a
Then
$$0 \lt cg - bg$$
$$0 \lt g(c-b)$$
Where $g$ is positive and $c-b$ is positive, so indeed
$$0 \lt g(c-b)$$
is true and thus, the assumption $x \lt y$ is true.
Hint: assuming $$x\le y$$ we get by expanding $$ac+bc+ad+bd\le ab+bc+ad+cd$$ then we get $$c(a-d)\le b(a-d)$$ so we get $$(a-d)(c-b)\le 0$$ Can you finish? since $a-d<0$ and $c-b>0$ then our inequality is true.
• In that part I stayed because by solving them how can I know that ab +cd>ac+bd? – Sonia f May 3 '18 at 15:53
• And I think there's an error in what you said instead of a-d it would be a-b – Sonia f May 3 '18 at 15:54
• Something went wrong. (a-d) is negative and (c-d) is negative. Which means (a-d)(c-d)>0. Which is a contradiction, which implies the original assumption x<y is false. However, testing some real numbers shows x<y is true. – CEP May 3 '18 at 16:01
• @sonia $a-d$ is correct. In the first inequality cancel out the terms. Then rearrange it by subtracting the right terms. – Patrick Abraham May 3 '18 at 16:03
Let $c-b = k>0$
Then $y=(a+c)(b+d) =$
$(a+b+k)(c + d -k) =$
$(a+b)(c+d) + k[c+d - a-b] -k^2=$
$x + k[d-a + k] - k^2=$
$x + k(d-a) > x$.
So $y > x$
Let $d-c = m>0$
$z = (a+d)(b+c)=$
$(a + c+m)(b+ d - m)=$
$(a+c)(b+d) + m[b+d - a-c] - m^2=$
$y + m[b-a +m] - m^2 =$
$y + m(b-a) > y$.
So $z > y$
So $z > y > x$.
....
Also there is AM-GM
If $j < k,m; k,m < n$ then $nj > km$ so
$(a+b) < (a+c)$ and $(b+d)< (c+d)$ so $(a+b)(c+d) > (a+c)(b+d)$... but ... I don't know. That doesn't have as "hands on" conviction. (Although its really the same thing. |
# 22.1 Interior and Exterior Angles
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1 Name Class Date 22.1 Interior and Exterior ngles Essential Question: What can you say about the interior and exterior angles of a triangle and other polygons? Resource Locker Explore 1 Exploring Interior ngles in Triangles You can find a relationship between the measures of the three angles of a triangle. n interior angle is an angle formed by two sides of a polygon with a common vertex. So, a triangle has three interior angles. Use a straightedge to draw a large triangle on a sheet of paper and cut it out. Tear off the three corners and rearrange the angles so their sides are adjacent and their vertices meet at a point. What seems to be true about placing the three interior angles of a triangle together? interior angles Make a conjecture about the sum of the measures of the interior angles of a triangle. The conjecture about the sum of the interior angles of a triangle can be proven so it can be stated as a theorem. In the proof, you will add an auxiliary line to the triangle figure. n auxiliary line is a line that is added to a figure to aid in a proof. The Triangle Sum Theorem The sum of the angle measures of a triangle is 180. Fill in the blanks to complete the proof of the Triangle Sum Theorem. Given: C Prove: m 1 + m 2 + m 3 = 180 Statements 1. Draw line l through point parallel to _ C. 1. Parallel Postulate 1 Reasons C l 2. m 1 = m and m 3 = m m 4 + m 2 + m 5 = ngle ddition Postulate and definition of straight angle 4. m + m 2 + m = Module Lesson 1
2 Reflect 1. Explain how the Parallel Postulate allows you to add the auxiliary line into the triangle figure. 2. What does the Triangle Sum Theorem indicate about the angles of a triangle that has three angles of equal measure? How do you know? Explore 2 Exploring Interior ngles in Polygons To determine the sum of the interior angles for any polygon, you can use what you know about the Triangle Sum Theorem by considering how many triangles there are in other polygons. For example, by drawing the diagonal from a vertex of a quadrilateral, you can form two triangles. Since each triangle has an angle sum of 180, the quadrilateral must have an angle sum of = 360. Draw the diagonals from any one vertex for each polygon. Then state the number of triangles that are formed. The first two have already been completed. quadrilateral 2 triangles triangle quadrilateral 1 triangle 2 triangles For each polygon, identify the number of sides and triangles, and determine the angle sums. Then complete the chart. The first two have already been done for you. Polygon Number of Sides Number of Triangles Sum of Interior ngle Measures Triangle 3 1 (1)180 = 180 Quadrilateral 4 2 (2)180 = 360 Pentagon ( ) 180 = Hexagon ( ) 180 = Decagon ( ) 180 = Module Lesson 1
3 Do you notice a pattern between the number of sides and the number of triangles? If n represents the number of sides for any polygon, how can you represent the number of triangles? Make a conjecture for a rule that would give the sum of the interior angles for any n-gon. Sum of interior angle measures = Reflect 3. In a regular hexagon, how could you use the sum of the interior angles to determine the measure of each interior angle? 4. How might you determine the number of sides for a polygon whose interior angle sum is 3240? [# Explain 1 Using Interior ngles You can use the angle sum to determine the unknown measure of an angle of a polygon when you know the measures of the other angles. Polygon ngle Sum Theorem The sum of the measures of the interior angles of a convex polygon with n sides is (n 2)180. Example 1 Determine the unknown angle measures. For the nonagon shown, find the unknown angle measure x. First, use the Polygon ngle Sum Theorem to find the sum of the interior angles: n = 9 (n - 2)180 = (9-2)180 = (7)180 = 1260 Then solve for the unknown angle measure, x : x = 1260 x = 158 The unknown angle measure is x Module Lesson 1
4 Determine the unknown interior angle measure of a convex octagon in which the measures of the seven other angles have a sum of 940. n = Sum = ( - 2 ) 180 = ( ) 180 = + x = x = The unknown angle measure is. Reflect 5. How might you use the Polygon ngle Sum Theorem to write a rule for determining the measure of each interior angle of any regular convex polygon with n sides? Your Turn 6. Determine the unknown angle measures in this pentagon. x x 7. Determine the measure of the fourth interior angle of a quadrilateral if you know the other three measures are 89, 80, and Determine the unknown angle measures in a hexagon whose six angles measure 69, 108, 135, 204, b, and 2b. Module Lesson 1
5 Explain 2 Proving the Exterior ngle Theorem n exterior angle is an angle formed by one side of a polygon and the extension of an adjacent side. Exterior angles form linear pairs with the interior angles. remote interior angle is an interior angle that is not adjacent to the exterior angle. Remote interior angles Exterior angle Example 2 Follow the steps to investigate the relationship between each exterior angle of a triangle and its remote interior angles. Step 1 Use a straightedge to draw a triangle with angles 1, 2, and 3. Line up your straightedge along the side opposite angle 2. Extend the side from the vertex at angle 3. You have just constructed an exterior angle. The exterior angle is drawn supplementary to its adjacent interior angle. Step 2 You know the sum of the measures of the interior angles of a triangle. m 1 + m 2 + m 3 = Conjecture: Since an exterior angle is supplementary to its adjacent interior angle, you also know: m 3 + m 4 = Make a conjecture: What can you say about the measure of the exterior angle and the measures of its remote interior angles? The conjecture you made in Step 2 can be formally stated as a theorem. Exterior ngle Theorem The measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles. Step 3 Complete the proof of the Exterior ngle Theorem. 4 is an exterior angle. It forms a linear pair with interior angle 3. Its remote interior angles are 1 and 2. y the, m 1 + m 2 + m 3 = 180. lso, m 3 + m 4 = because they are supplementary and make a straight angle. y the Substitution Property of Equality, then, m 1 + m 2 + m 3 = m + m. Subtracting m 3 from each side of this equation leaves This means that the measure of an exterior angle of a triangle is equal to the sum of the measures of the remote interior angles. Module Lesson 1
6 Reflect 9. Discussion Determine the measure of each exterior angle. dd them together. What can you say about their sum? Explain ccording to the definition of an exterior angle, one of the sides of the triangle must be extended in order to see it. How many ways can this be done for any vertex? How many exterior angles is it possible to draw for a triangle? for a hexagon? Explain 3 Using Exterior ngles You can apply the Exterior ngle Theorem to solve problems with unknown angle measures by writing and solving equations. Example 3 Determine the measure of the specified angle. Find m. Find m PRS. Q 2z D Write and solve an equation relating the exterior and remote interior angles. 145 = 2z + 5z = 7z - 2 z = (5z - 2) C Now use this value for the unknown to evaluate the expression for the required angle. m = (5z - 2) = (5(21) - 2) = (105-2) S R (3x - 8) Write an equation relating the exterior and remote interior angles. Solve for the unknown. (x + 2) Use the value for the unknown to evaluate the expression for the required angle. P = 103 Module Lesson 1
7 Your Turn Determine the measure of the specified angle. 11. Determine m N in MNP. 12. If the exterior angle drawn measures 150, and the measure of D is twice that of E, find the N measure of the two remote interior angles. (3x + 7) D 150 E F G 63 (5x + 50) M P Q Elaborate 13. In your own words, state the Polygon ngle Sum Theorem. How does it help you find unknown angle measures in polygons? 14. When will an exterior angle be acute? Can a triangle have more than one acute exterior angle? Describe the triangle that tests this. 15. Essential Question Check-In Summarize the rules you have discovered about the interior and exterior angles of triangles and polygons. Module Lesson 1
8 Evaluate: Homework and Practice 1. Consider the Triangle Sum Theorem in relation to a right triangle. What conjecture can you make about the two acute angles of a right triangle? Explain your reasoning. Online Homework Hints and Help Extra Practice 2. Complete a flow proof for the Triangle Sum Theorem. Given C Prove m 1 + m 2 + m 3 = 180 Draw l parallel to C through C l m 3 = m 5 m 4 + m 2 + m 5 = lternate Interior ngles Theorem Definition of straight angle Substitution Property of Equality 3. Given a polygon with 13 sides, find the sum of the measures of its interior angles. 4. polygon has an interior angle sum of How many sides must the polygon have? 5. Two of the angles in a triangle measure 50 and 27. Find the measure of the third angle. Solve for the unknown angle measures of the polygon. 6. pentagon has angle measures of 100, 105, 110 and 115. Find the fifth angle measure. 7. The measures of 13 angles of a 14-gon add up to Find the fourteenth angle measure? Module Lesson 1
9 8. Determine the unknown angle measures for the quadrilateral in the diagram. 4x 3x x 2x 9. The cross-section of a beehive reveals it is made of regular hexagons. What is the measure of each angle in the regular hexagon? 10. Create a flow proof for the Exterior ngle Theorem Image Credits: StudioSmart/Shutterstock Find the value of the variable to find the unknown angle measure(s). 11. Find w to find the measure of the exterior angle. 12. Find x to find the measure of the remote interior angle. w x 134 Module Lesson 1
10 13. Find m H. 14. Determine the measure of the indicated exterior angle in the diagram. H (6x + 1) 3x 126 F G (5x + 17) J 2x (3x + 4)? 15. Match each angle with its corresponding measure, given m 1 = 130 and m 7 = 70. Indicate a match by writing the letter for the angle on the line in front of the corresponding angle measure.. m m C. m 4 70 D. m E. m The map of France commonly used in the 1600s was significantly revised as a result of a triangulation survey. The diagram shows part of the survey map. Use the diagram to find the measure of KMJ n artistic quilt is being designed using computer software. The designer wants to use regular octagons in her design. What interior angle measures should she set in the computer software to create a regular octagon? Module Lesson 1
11 18. ladder propped up against a house makes a 20 angle with the wall. What would be the ladder's angle measure with the ground facing away from the house? ladder 20 house? 19. Photography The aperture of a camera is made by overlapping blades that form a regular decagon. a. What is the sum of the measures of the interior angles of the decagon? ground b. What would be the measure of each interior angle? each exterior angle? c. Find the sum of all ten exterior angles. 20. Determine the measure of UXW in the diagram. Y V U W Image Credits: neyro2008/istockphoto.com X Z 21. Determine the measures of angles x, y, and z z 55 x y 60 Module Lesson 1
12 22. Given the diagram in which D bisects C and CD bisects C, what is m DC? 15 D C 23. What If? Suppose you continue the congruent 24. lgebra Draw a triangle C and label the angle construction shown here. What polygon will measures of its angles a, b, and c. Draw ray D you construct? Explain. that bisects the exterior angle at vertex. Write an expression for the measure of angle CD Look for a Pattern Find patterns within this table of data and extend the patterns to complete the remainder of the table. What conjecture can you make about polygon exterior angles from Column 5? Column 1 Number of Sides Column 2 Sum of the Measures of the Interior ngles Column 3 verage Measure of an Interior ngle Column 4 verage Measure of an Exterior ngle Column 5 Sum of the Measures of the Exterior ngles (3) = (4) = Conjecture: Module Lesson 1
13 26. Explain the Error Find and explain what this student did incorrectly when solving the following problem. What type of polygon would have an interior angle sum of 1260? 1260 = (n - 2) = n = n The polygon is a pentagon. H.O.T. Focus on Higher Order Thinking 27. Communicate Mathematical Ideas Explain why if two angles of one triangle are congruent to two angles of another triangle, then the third pair of angles are also congruent. Given: L R, M S Prove: N T N L S M R T 28. nalyze Relationships Consider a right triangle. How would you describe the measures of its exterior angles? Explain. 29. Look for a Pattern In investigating different polygons, diagonals were drawn from a vertex to break the polygon into triangles. Recall that the number of triangles is always two less than the number of sides. ut diagonals can be drawn from all vertices. Make a table where you compare the number of sides of a polygon with how many diagonals can be drawn (from all the vertices). Can you find a pattern in this table? Number of Sides, n Number of Diagonals, d Module Lesson 1
14 Lesson Performance Task You ve been asked to design the board for a new game called Pentagons. The board consists of a repeating pattern of regular pentagons, a portion of which is shown in the illustration. When you write the specifications for the company that will make the board, you include the measurements of D, C, CD and DC. Find the measures of those angles and explain how you found them. D C Module Lesson 1
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# What is the probability of getting heads on only one of your flips?
Textbook #1: Lane et al. Introduction to Statistics, David M. Lane et al., 2013.
Textbook #2:Illowsky et al. Introductory Statistics, Barbara Illowsky et al., 2013.
Lane – Chapter 5: 7,25,27
1. You flip a coin three times.
(a) What is the probability of getting heads on only one of your flips?
(b) What is the probability of getting heads on at least one flip?
1. You are to participate in an exam for which you had no chance to study, and for that reason cannot do anything but guess for each question (all questions being of the multiple choice type, so the chance of guessing the correct answer for each question is 1/d, d being the number of options (distractors) per question; so in case of a 4-choice question, your guess chance is ¼ = 0.25. Your instructor offers you the opportunity to choose amongst the following exam formats:
1. 6 questions of the 4-choice type; you pass when 5 or more answers are correct;
1. 5 questions of the 5-choice type; you pass when 4 or more answers are correct;
III. 4 questions of the 10-choice type; you pass when 3 or more answers are correct.
Rank the three exam formats according to their attractiveness. It should be clear that the format with the highest probability to pass is the most attractive format. Which would you choose and why?
HINT: Use the Binomial Probability function
1. A refrigerator contains 6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11 plums, and 2 mangos.
1. Imagine you stick your hand in this refrigerator and pull out a piece of fruit at random. What is the probability that you will pull out a pear?
1. Imagine now that you put your hand in the refrigerator and pull out a piece of fruit. You decide you do not want to eat that fruit so you put it back into the refrigerator and pull out another piece of fruit. What is the probability that the first piece of fruit you pull out is a banana and the second piece you pull out is an apple?
1. What is the probability that you stick your hand in the refrigerator one time and pull out a mango or an orange?
Illowsky – Chapter 3 (86,98,112,124) and Chapter 4 (72,80,88)
1. Roll two fair dice. Each die has six faces.
1. List the sample space.
1. Let A be the event that either a three or four is rolled first, followed by an even number. Find P(A).
1. Let B be the event that the sum of the two rolls is at most seven. Find P(B).
1. In words, explain what “P(A|B)” represents. Find P(A|B).
1. Are A and B mutually exclusive events? Explain your answer in one to three complete sentences, including numerical justification.
1. Are A and B independent events? Explain your answer in one to three complete sentences, including numerical justification.
1. At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper.
1. Find the probability that a course has a final exam or a research project.
1. Find the probability that a course has NEITHER of these two requirements.
1. The table identifies a group of children by one of four hair colors, and by type of hair.
1. Complete the table.
1. What is the probability that a randomly selected child will have wavy hair?
1. What is the probability that a randomly selected child will have either brown or blond hair?
1. What is the probability that a randomly selected child will have wavy brown hair?
1. What is the probability that a randomly selected child will have red hair, given that he or she has straight hair?
1. If B is the event of a child having brown hair, find the probability of the complement of B.
1. In words, what does the complement of B represent?
1. Suppose that 10,000 U.S. licensed drivers are randomly selected.
1. How many would you expect to be male?
1. Using the table or tree diagram, construct a contingency table of gender versus age group.
1. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is female.
1. You buy a lottery ticket to a lottery that costs \$10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one \$500 prize, two \$100 prizes, and four \$25 prizes. Find your expected gain or loss.
1. Florida State University has 14 statistics classes scheduled for its Summer 2013 term. One class has space available for 30 students, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for 100 students.
1. What is the average class size assuming each class is filled to capacity?
1. Space is available for 980 students. Suppose that each class is filled to capacity and select a statistics student at random. Let the random variable X equal the size of the student’s class. Define the PDF for X.
1. Find the mean of X.
1. Find the standard deviation of X.
1. A school newspaper reporter decides to randomly survey 12 students to see if they will attend Tet (Vietnamese New Year) festivities this year. Based on past years, she knows that 18% of students attend Tet festivities. We are interested in the number of students who will attend the festivities.
1. In words, define the random variable X.
1. List the values that X may take on.
1. Give the distribution of X. X ~ _____(_____,_____)
1. How many of the 12 students do we expect to attend the festivities?
1. Find the probability that at most four students will attend.
1. Find the probability that more than two students will attend.
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# 2016 AMC 10B Problems/Problem 22
## Problem
A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$
## Solution 1
There are $21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
$[asy]size(7cm);label("X",(5,5));label("Z",(10,0));label("Y",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label("X",(20,5));label("Z",(25,0));label("Y",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); [/asy]$ But we know that every team beat exactly $10$ other teams, so for each possible $X$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices for $Y$ and $Z$. Therefore there are $21\cdot\tbinom{10}2=945$ forks, and all the rest must be cycles.
Thus the answer is $1330-945=385$ which is $\boxed{\textbf{(A)}}$.
## Solution 2 (Cheap Solution)
Since there are $21$ teams and for each set of three teams there is a cycle, there are a total of $\tbinom{21}3=1330$ cycles of three teams. Because about $1/4$ of the cycles $\{A, B, C\}$ satisfy the conditions of the problems, our answer is close to $1/4 \cdot 1330=332.5$. Looking at the answer choices, we find that $332.5$ is closer to $385$ than any other answer choices and that the next closest is greater then $600$, so our answer is $385$ which is $\boxed{\textbf{(A)}}$.
Speaking of cheap solutions, you can let the teams be $1,2,3,..21$ and have $i$ beat $i+1,i+2, ... i+10 \pmod{21}$ and lose to $i+11,i+12, ... i+20 \pmod{21}.$ You can choose any random team as $A$ for $21$ ways but overcount since set, so divide by $3$ so multiply by $7.$ Now we can proceed w/casework (which isn't too hard). Eventually you realize the sum is $7(1+2+\cdots+10) = \boxed{\textbf{(A)}}$. |
# Petrick's method
In Boolean algebra, Petrick's method (also known as the branch-and-bound method) is a technique described by Stanley R. Petrick (1931–2006)[1][2] in 1956[3] for determining all minimum sum-of-products solutions from a prime implicant chart. Petrick's method is very tedious for large charts, but it is easy to implement on a computer.
Algorithm
1. Reduce the prime implicant chart by eliminating the essential prime implicant rows and the corresponding columns.
2. Label the rows of the reduced prime implicant chart ${\displaystyle P_{1}}$, ${\displaystyle P_{2}}$, ${\displaystyle P_{3}}$, ${\displaystyle P_{4}}$, etc.
3. Form a logical function ${\displaystyle P}$ which is true when all the columns are covered. P consists of a product of sums where each sum term has the form ${\displaystyle (P_{i0}+P_{i1}+}$${\displaystyle \cdots }$${\displaystyle +P_{iN})}$, where each ${\displaystyle P_{ij}}$ represents a row covering column ${\displaystyle i}$.
4. Reduce ${\displaystyle P}$ to a minimum sum of products by multiplying out and applying ${\displaystyle X+XY=X}$.
5. Each term in the result represents a solution, that is, a set of rows which covers all of the minterms in the table. To determine the minimum solutions, first find those terms which contain a minimum number of prime implicants.
6. Next, for each of the terms found in step five, count the number of literals in each prime implicant and find the total number of literals.
7. Choose the term or terms composed of the minimum total number of literals, and write out the corresponding sums of prime implicants.
Example of Petrick's method[4]
Following is the function we want to reduce:
${\displaystyle f(A,B,C)=\sum m(0,1,2,5,6,7)\,}$
The prime implicant chart from the Quine-McCluskey algorithm is as follows:
| 0 1 2 5 6 7
---------------|------------
K (0,1) a'b' | X X
L (0,2) a'c' | X X
M (1,5) b'c | X X
N (2,6) bc' | X X
P (5,7) ac | X X
Q (6,7) ab | X X
Based on the X marks in the table above, build a product of sums of the rows where each row is added, and columns are multiplied together:
(K+L)(K+M)(L+N)(M+P)(N+Q)(P+Q)
Use the distributive law to turn that expression into a sum of products. Also use the following equivalences to simplify the final expression: X + XY = X and XX = X and X+X=X
= (K+L)(K+M)(L+N)(M+P)(N+Q)(P+Q)
= (K+LM)(N+LQ)(P+MQ)
= (KN+KLQ+LMN+LMQ)(P+MQ)
= KNP + KLPQ + LMNP + LMPQ + KMNQ + KLMQ + LMNQ + LMQ
Now use again the following equivalence to further reduce the equation: X + XY = X
= KNP + KLPQ + LMNP + LMQ + KMNQ
Choose products with fewest terms, in this example, there are two products with three terms:
KNP
LMQ
Choose term or terms with fewest total literals. In our example, the two products both expand to six literals total each:
KNP expands to a'b'+ bc'+ ac
LMQ expands to a'c'+ b'c + ab
So either one can be used. In general, application of Petrick's method is tedious for large charts, but it is easy to implement on a computer.
## References
1. ^ (Unknown). "Biographical note". Retrieved 2017-04-12. Stanley R. Petrick was born in Cedar Rapids, Iowa on August 16, 1931. He attended the Roosevelt High School and received a B. S. degree in Mathematics from the Iowa State University in 1953. During 1953 to 1955 he attended MIT while on active duty as an Air Force officer and received the S. M. degree from the Department of Electrical Engineering in 1955. He was elected to Sigma Xi in 1955.
Mr. Petrick has been associated with the Applied Mathematics Board of the Data Sciences Laboratory at the Air Force Cambridge Research Laboratories since 1955 and his recent studies at MIT have been partially supported by AFCRL. During 1959-1962 he held the position of Lecturer in Mathematics in the Evening Graduate Division of Northeastern University.
Mr. Petrick is currently a member of the Linguistic Society of America, The Linguistic Circle of New York, The American Mathematical Association, The Association for Computing Machinery, and the Association for Machine Translation and Computational Linguistics.
2. ^ "Obituaries - Cedar Rapids - Stanley R. Petrick". The Gazette. 2006-08-05. p. 16. Retrieved 2017-04-12. […] CEDAR RAPIDS Stanley R. Petrick, 74, formerly of Cedar Rapids, died July 27, 2006, in Presbyterian/St. Luke's Hospital, Denver, Colo., following a 13-year battle with leukemia. A memorial service will be held Aug. 14 at the United Presbyterian Church in Laramie, Wyo., where he lived for many years. […] Stan Petrick was born in Cedar Rapids on Aug. 6, 1931 to Catherine Hunt Petrick and Fred Petrick. He graduated from Roosevelt High School in 1949 and received a B.S. degree in mathematics from Iowa State University. Stan married Mary Ethel Buxton in 1953.
He joined the U.S. Air Force and was assigned as a student officer studying digital computation at MIT, where he earned an M.S. degree. He was then assigned to the Applied Mathematics Branch of the Air Force Cambridge Research Laboratory and while there earned a Ph.D. in linguistics.
He spent 20 years in the Theoretical and Computational Linguistics Group of the Mathematical Sciences Department at IBM's T.J. Watson Research Center, conducting research in formal language theory. He had served as an assistant director of the Mathematical Sciences Department, chairman of the Special Interest Group on Symbolic and Algebraic Manipulation of the Association for Computing Machinery and president of the Association for Computational Linguistics. He authored many technical publications.
He taught three years at Northeastern University and 13 years at the Pratt Institute. Dr. Petrick joined the University of Wyoming in 1987, where he was instrumental in developing and implementing the Ph.D. program in the department and served as a thesis adviser for many graduate students. He retired in 1995. […]
(NB. Includes a photo of Stanley R. Petrick.)
3. ^ Petrick, Stanley R. (1956-04-10). A Direct Determination of the Irredundant Forms of a Boolean Function from the Set of Prime Implicants. Bedford, Cambridge, MA, USA: Air Force Cambridge Research Center. AFCRC Technical Report TR-56-110.
4. ^ http://www.mrc.uidaho.edu/mrc/people/jff/349/lect.10 Lecture #10: Petrick's Method |
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# 45-45-90 Right Triangles
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Progress
Practice 45-45-90 Right Triangles
Progress
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45-45-90 Triangles
Have you ever planned a flower garden where you needed to figure out the length of a diagonal? It is a special type of project, so take a look at this dilemma.
Ms. Kino’s class decided to do a community service project that everyone could enjoy. They decided to create a meditation garden that would be a rock garden.
Chas and Juanita took charge of the project. They drew a sketch of the rock garden and the rest of the class loved it so much that they instantly agreed to use the sketch that the pair had created. Here is their sketch.
“Let’s put a diagonal path in it,” Frankie suggested looking at the sketch.
“That’s a great idea, how long will the path be?” Chas asked.
The class wants to add a diagonal path. If they do that from one corner to another, how long will the path be?
This Concept will teach you all that you need to know to solve this problem.
### Guidance
There are a few types of right triangles it is particularly important to study. Their sides are always in the same ratio, and it is crucial to study the $45^\circ-45^\circ-90^\circ$ and the $30^\circ-60^\circ-90^\circ$ triangles and understand the relationships between the sides. It will save you time and energy as you work through math problems both straight-forward and complicated.
Let’s start by learning about the $45^\circ-45^\circ-90^\circ$ .
First, think about that $45^\circ-45^\circ-90^\circ$ refers to. Those values refer to the angle measures in the right triangle. We can see that there is one 90 degree angle and that the other two angles have the same measure. This particular triangle is also isosceles . An isosceles triangle has two side lengths that are the same. An isosceles right triangle will always have the same angle measurements: $45^\circ,45^\circ$ , and $90^\circ$ and will always have two side lengths that are the same. These characteristics make it a special right triangle.
Because these angles will always remain the same, the sides will always be in proportion. To find the relationship between the sides, use the Pythagorean Theorem.
Take a look at this situation.
The isosceles right triangle below has legs measuring 1 centimeter. Use the Pythagorean Theorem to find the length of the hypotenuse.
As the problem states, you can use the Pythagorean Theorem to find the length of the hypotenuse. Since the legs are 1 centimeter each, set both $a$ and $b$ equal to 1 and solve for $c$ .
$a^2+b^2 &=c^2\\(1)^2+(1)^2 &= c^2\\1+1 &= c^2\\2 &= c^2\\\sqrt{2} &= \sqrt{c^2}\\\sqrt{2} &= c$
We can look at this and understand that there is also a 1 in front of the square root of two. This shows that the relationship between one side length and the length of the hypotenuse will always be the same. The hypotenuse of an isosceles right triangle will always equal the product of one leg and $\sqrt{2}$ .
Write this down in your notebook under $45^\circ-45^\circ-90^\circ$ special right triangles.
Find each hypotenuse.
#### Example A
A triangle with side lengths of 9.
Solution: $9 \sqrt{2}$
#### Example B
A triangle with side lengths of 15.
Solution: $15 \sqrt{2}$
#### Example C
A triangle with side lengths of $3 \sqrt{2}$
Solution: $6$
Now let's go back to the dilemma from the beginning of the Concept.
The first step in a word problem of this nature is to add important information to the drawing. Because the problem asks you to find the length of a path from one corner to another, you should draw that path in.
Once you draw the diagonal path, you can tell that this is a triangle question. Because both legs of the triangle have the same measurement (10 feet), this is an isosceles right triangle. The angles in an isosceles right triangle are $45^\circ,45^\circ$ , and $90^\circ$ .
In an isosceles right triangle, the hypotenuse is always equal to the product of the length of one leg and $\sqrt{2}$ . So, the length of the path will be the product of 10 and $\sqrt{2}$ , or $10 \sqrt{2} \ feet$ . This value is approximately equal to 14.14 feet.
### Guided Practice
Here is one for you to try on your own.
What is the length of the hypotenuse in the triangle below?
Solution
Since the length of the hypotenuse is the product of one leg and $\sqrt{2}$ , you can easily calculate this length. It is easy because we know that with any 45/45/90 degree triangle, that the hypotenuse is the product of one of the legs and the square root of 2.
One leg is 3 inches, so the hypotenuse will be $3 \sqrt{2}$ , or about 4.24 inches.
To get that answer, we took the square root of two on the calculator, 1.414 and then multiplied it times 3.
$3 \times 1.414 = 4.242$
We rounded to get the answer.
### Explore More
Directions: Find the missing hypotenuse in each $45^\circ-45^\circ-90^\circ$ triangle.
1. Length of each leg = 5
2. Length of each leg = 4
3. Length of each leg = 6
4. Length of each leg = 3
5. Length of each leg = 7
Directions: Now use a calculator to figure out the approximate value of each hypotenuse. You may round to the nearest hundredth.
1. $5 \sqrt{2}$
2. $4 \sqrt{2}$
3. $6 \sqrt{2}$
4. $3 \sqrt{2}$
5. $7 \sqrt{2}$
6. $8 \sqrt{2}$
7. $10 \sqrt{2}$
8. $13 \sqrt{2}$
9. $21\sqrt{2}$
10. $17 \sqrt{2}$
### Vocabulary Language: English
45-45-90 Theorem
45-45-90 Theorem
For any isosceles right triangle, if the legs are x units long, the hypotenuse is always x$\sqrt{2}$.
45-45-90 Triangle
45-45-90 Triangle
A 45-45-90 triangle is a special right triangle with angles of $45^\circ$, $45^\circ$, and $90^\circ$.
Hypotenuse
Hypotenuse
The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.
Isosceles Right Triangle
Isosceles Right Triangle
An isosceles right triangle is a triangle with a ninety degree angle and exactly two sides that are the same length.
Legs of a Right Triangle
Legs of a Right Triangle
The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.
The $\sqrt{}$, or square root, sign. |
New SAT Math Workbook
# In the xy plane if lines a and b intersect at point 52
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Unformatted text preview: s the probability that the other will occur. Here are two general situations: * The random selection of one object from each of two groups (for example, the outcome of throwing a pair of dice) * The random selection of one object from a group, then replacing it and selecting again (as in a “second round” or “another turn” of a game) To determine the probability of two independent events both occurring, multiply individual probabilities. Example: If you randomly select one letter from each of two sets: {A,B} and {C,D,E}, what is the probability of selecting A and C? Solution: The correct answer is 6 . The probability of selecting A from the set {A,B} is 2 , while the 1 probability of selecting C from the set {C,D,E} is 3 . Hence, the probability of selecting A and C is 1×1 1 , or 6 . 23 1 1 12 3 number of ways the event can occur An SAT probability problem might be accompanied by a geometry figure or other figure that provides a visual display of the possibilities from which you are to calculate a probability. www.petersons.com Additional Geometry Topics, Data Analysis, and Probability 299 Example: If a point is selected at random from the circular region shown above, what is the probability that the point will lie in a shaded portion of the circle? Solution: The correct answer is .25 (or 4 ). The angles opposite each of the three 45° angles identified in the figure must also measure 45° each. Given a total of 360° in a circle, all of the eight small angles formed at the circle’s center measure 45°, and hence all eight segments of the circle are congruent. 2 1 The two shaded segments comprise 8 , or 4 (.25) of the circle’s area. The probability of selecting a 1 point at random in a shaded area is also 4 (or .25). 1 www.petersons.com 300 Chapter 16 Exercise 6 Work out each problem. For questions 1–4, circle the letter that appears before your answer. Question 5 is a gridin question. 1. If you randomly select one candy from a jar containing two cherry candies, two licorice candies, and one peppermint candy, what is the probabilit...
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## This note was uploaded on 08/15/2010 for the course MATH a4d4 taught by Professor Colon during the Spring '10 term at Embry-Riddle FL/AZ.
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19.02.2021
# The leg lengths of a right triangle are 4 cm and 6 cm. What is the length of the missing side?
0
09.07.2023, solved by verified expert
Step-by-step explanation:
### Faq
Mathematics
Step-by-step explanation:
Mathematics
We use the Pythagoras Theorem to determine the height of the shelf.
Let be the height of the triangle, the base and the hypotenuse.
Then by the Pythagoras Theorem,
We substitute the base, and the hypotenuse
.
Therefore the approximate minimum height of the shelf should be .
We apply the Pythagoras Theorem to find the length of the third side.
See diagram
Let the length of the third side be .
Then
We can now solve for y.
We use the Pythagoras Theorem to find the length of PR.
Since PR is the hypotenuse .
See diagram in attachment.
The unknown length is the variable , which is the hypotenuse of the right angle triangle.
So we use the Pythagoras theorem to find the unknown length.
From Pythagoras Theorem, the area of the bigger square is equal to the area of the two smaller squares added together.
See diagram in attachment.
That is .
This implies that,
Let be the length of the unknown leg.
Then from the Pythagoras Theorem,
This implies that;
The correct answer is option A.
It is incorrect because the length of the unknown side is and not .
The diagonal is the hypotenuse of the right angle triangle created by the diagonal, the width and the length of the rectangle.
Since the diagonal is the hypotenuse and the two shorter sides are the width and the length of the rectangle, we can apply the Pythagoras Theorem to find the value of .
Since the width of the cups is 2 inches, it means the radius is half the width.
That is inch
The volume of a cylinder is given by;
The cup with the cylindrical shape (B) will hold
cubic inches of juice
The volume of a cone is:
The cup with the conical shape cup(A), will hold
cubic inches of juice
Hence cup B will hold cubic inches than cup A.
Mathematics
We use the Pythagoras Theorem to determine the height of the shelf.
Let be the height of the triangle, the base and the hypotenuse.
Then by the Pythagoras Theorem,
We substitute the base, and the hypotenuse
.
Therefore the approximate minimum height of the shelf should be .
We apply the Pythagoras Theorem to find the length of the third side.
See diagram
Let the length of the third side be .
Then
We can now solve for y.
We use the Pythagoras Theorem to find the length of PR.
Since PR is the hypotenuse .
See diagram in attachment.
The unknown length is the variable , which is the hypotenuse of the right angle triangle.
So we use the Pythagoras theorem to find the unknown length.
From Pythagoras Theorem, the area of the bigger square is equal to the area of the two smaller squares added together.
See diagram in attachment.
That is .
This implies that,
Let be the length of the unknown leg.
Then from the Pythagoras Theorem,
This implies that;
The correct answer is option A.
It is incorrect because the length of the unknown side is and not .
The diagonal is the hypotenuse of the right angle triangle created by the diagonal, the width and the length of the rectangle.
Since the diagonal is the hypotenuse and the two shorter sides are the width and the length of the rectangle, we can apply the Pythagoras Theorem to find the value of .
Since the width of the cups is 2 inches, it means the radius is half the width.
That is inch
The volume of a cylinder is given by;
The cup with the cylindrical shape (B) will hold
cubic inches of juice
The volume of a cone is:
The cup with the conical shape cup(A), will hold
cubic inches of juice
Hence cup B will hold cubic inches than cup A.
Mathematics
8) 18.8 cubic inches;
9) 28 feet;
10) 1.7 centimeters;
11) 13;
12) 10 centimeters;
13) it is incorrect because the length of the unknown side is the square root of 7,744;
14) square root of 18;
15) 12 miles;
16) 364 centimeters.
Mathematics
It’s b because when you multiply 2.5 by 2.5 you get 6.35 and do the same for 1.5
Mathematics
1. A.Use law of cosines. cosA=(b^2+c^2-a^2)/(2bc) because A is the included angle between b and c. Plug in A=50 degrees, b=13, c=6. cos50=(13^2+6^2-a^2)/(2*13*6), a^2=104.7,a=10.2 approximately, so choose A.
2. A.Use law of cosines again. cosC=(a^2+b^2-c^2)/(2ab). C=95 degrees, a=12, b=22. Plug in, cos95=(12^2+22^2-c^2)/(2*12*22), solve the equation, c^2=674, c=26 approximately. The use law of sines to solve for angle A (also works for B), a/sinA=c/sinC, 12/sinA=26/sin95, sinA=0.46, A=arcsin(0.46)=27.6. Choose A.
3. Answer is A. Area=1/2bc*sinA, since A is the included angle between b and c. Plug in b=30, c=14, A=50 degrees, area=1/2*14*30*sin50=160. 87, so the answer is A.
4. D. As long as the sum of any two sides of the triangle is bigger than the third, the triangle exists. 240+121>263, 240+263>121, 263+121>240, so it exists. To use Heron's formula, first find the semiperimeter, (240+263+121)/2=312. A=\sqrt(312*(312-240)*(312-263)*(312-121))=14499.7 approximately, so choose D.
5. 300. The included angle between the two paths is C=49.17+90=139.17 degrees. The lengths of the two paths are a=150, b=170. c is the distance we want. Use the law of cosines, cosC=(a^2+b^2-c^2)/(2ab). Plug in, c^2=89989, c=300 approximately.
Mathematics
1. A.Use law of cosines. cosA=(b^2+c^2-a^2)/(2bc) because A is the included angle between b and c. Plug in A=50 degrees, b=13, c=6. cos50=(13^2+6^2-a^2)/(2*13*6), a^2=104.7,a=10.2 approximately, so choose A.
2. A.Use law of cosines again. cosC=(a^2+b^2-c^2)/(2ab). C=95 degrees, a=12, b=22. Plug in, cos95=(12^2+22^2-c^2)/(2*12*22), solve the equation, c^2=674, c=26 approximately. The use law of sines to solve for angle A (also works for B), a/sinA=c/sinC, 12/sinA=26/sin95, sinA=0.46, A=arcsin(0.46)=27.6. Choose A.
3. Answer is A. Area=1/2bc*sinA, since A is the included angle between b and c. Plug in b=30, c=14, A=50 degrees, area=1/2*14*30*sin50=160. 87, so the answer is A.
4. D. As long as the sum of any two sides of the triangle is bigger than the third, the triangle exists. 240+121>263, 240+263>121, 263+121>240, so it exists. To use Heron's formula, first find the semiperimeter, (240+263+121)/2=312. A=\sqrt(312*(312-240)*(312-263)*(312-121))=14499.7 approximately, so choose D.
5. 300. The included angle between the two paths is C=49.17+90=139.17 degrees. The lengths of the two paths are a=150, b=170. c is the distance we want. Use the law of cosines, cosC=(a^2+b^2-c^2)/(2ab). Plug in, c^2=89989, c=300 approximately.
Mathematics
Since the triangle in question is a right triangle, given the leg lengths, you can use the Pythagorean Theorem to find the length of the hypotenuse.
The Pythagorean Theorem: a² + b² = c², where c is the hypotenuse of the triangle.
Plug in the given leg lengths and solve for c.
a² + b² = c²
8.4² + 7.6² = c²
70.56 + 57.76 = c²
128.32 = c²
11.33 ≈ c
11.33 cm
Mathematics
Since the triangle in question is a right triangle, given the leg lengths, you can use the Pythagorean Theorem to find the length of the hypotenuse.
The Pythagorean Theorem: a² + b² = c², where c is the hypotenuse of the triangle.
Plug in the given leg lengths and solve for c.
a² + b² = c²
8.4² + 7.6² = c²
70.56 + 57.76 = c²
128.32 = c²
11.33 ≈ c
11.33 cm
Mathematics
D
Step-by-step explanation:
Say the short leg is x and the hypotenuse is y. Then, y = 3x (because "The length of the hypotenuse of a right triangle is three times the length of the shorter leg.")
Use the Pythagorean Theorem: a^2 + b^2 = c^2 (where a and b are the legs and c is the hypotenuse). Here, we can say a = x and b = 12 and c = y. So:
x^2 + 12^2 = y^2
x^2 + 144 = (3x)^2
x^2 + 144 = 9x^2
8x^2 = 144
x^2 = 18
x = |
NCERT Solutions: Understanding Elementary Shapes
# NCERT Solutions for Class 6 Maths Chapter 5 - Understanding Elementary Shapes
Table of contents Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 Exercise 5.6 Exercise 5.7 Exercise 5.8 Exercise 5.9 (Old NCERT)
## Exercise 5.1
Q1. What is the disadvantage in comparing line segments by mere observation?
Ans: We will be using the concept of line segments to solve this.
In the given figure, line segment PQ and RS seem to be equal but actually, they are not equal.
The disadvantage of comparing the lengths of two line segments by mere observation is that the lengths might not be accurate. Hence, a divider is used to compare the lengths of the line segments.
Q2. Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Ans: It is better to use a divider than a ruler while measuring the length of a line segment because a divider allows for more precise and accurate measurements. A divider has two sharp points that can be adjusted to mark the endpoints of the line segment. By opening or closing the divider, one can compare and transfer the length of the line segment to other distances or objects. This method ensures a more accurate measurement as compared to using a ruler, which may not provide the same level of precision.
A ruler may not be as suitable as a divider for measuring the length of a line segment due to the following reasons:
• Limited precision
• Difficulty in marking endpoints
• Inflexibility
Q3. Draw any line segment, say . Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
[Note : If A,B,C are any three points on a line such that AC + CB = AB, then we can be sure that C lies between A and B]
Ans: We will be using the concept of the line segment to solve this.
Let us consider the line AB with point C being in the center of A and B and AB = 7 cm.
AC = 3 cm
CB = 4 cm
Therefore, AC + CB = 3 cm + 4 cm
AC + CB = 7 cm.
But, AB = 7 cm.
Hence, AB = AC + CB
Q4. If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
Ans: We will be using the concept of line segments to solve this.
We have, AB = 5 cm
BC = 3 cm
Therefore, AB + BC = 5 + 3
AB + BC = 8 cm
But, AC = 8 cm
Hence, B lies between A and C.
Q5. Verify, whether D is the mid point of .
Ans: We will be using the concept of midpoint to solve this.
From the given figure,
We have,
AG = 7 cm – 1 cm
AG = 6 cmAD = 4 cm – 1 cm
DG = 7 cm – 4 cm
DG = 3 cm
AD + DG = 3cm + 3cm
AD + DG = 6 cm
Therefore, AG = AD + DG
Where, AD = DG = 3cm
Hence, D is the midpoint of AG
Q6. If B is the mid-point of and C is the midpoint of , where A, B, C, D lie on a straight line, say why AB = CD?
Ans: We will be using the concept of midpoint to solve this.
We have been given that, B is the midpoint of AC
⇒ AB = BC ----------------- (1)
C is the midpoint of BD
⇒ BC = CD --------------- (2)
Hence, we can say that AB = CD [From equation(1) and equation(2)]
Q7. Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.
Ans: We will be using the concept of triangles to solve this.
Case I. In ∆ABC
Consider:
AB = 2.5 cm
BC = 4.8 cm
AC = 5.2 cm
AB + BC = 2.5 cm + 4.8 cm
AB + BC = 7.3 cm
Since, 7.3 > 5.2 , AB + BC > AC
Therefore, the sum of any two sides of a triangle is greater than the third side.
Case II. In ∆PQR,
Consider:
PQ = 2 cm
QR = 2.5 cm
PR = 3.5 cm
PQ + QR = 2 cm + 2.5 cm
PQ + QR = 4.5 cm
Since, 4.5 > 3.5, PQ + QR > PR
Therefore, the sum of any two sides of a triangle is greater than the third side.
Case III. In ∆XYZ,
Consider
XY = 5 cm
YZ = 3 cm
ZX = 6.8 cm
XY + YZ = 5 cm + 3 cm
XY + YZ = 8 cm
Since, 8 > 6.8, XY + YZ > ZX
Therefore, the sum of any two sides of a triangle is greater than the third side.
Case IV. In ∆MNS,
Consider
MN = 2.7 cm
NS = 4 cm
MS = 4.7 cm
MN + NS = 2.7 cm + 4 cm
MN + NS = 6.7 cm
Since, 6.7 >4.7, MN + NS > MS
Hence, the sum of any two sides of a triangle is greater than the third side.
Case V. In ∆KLM,
Consider
KL = 3.5 cm
LM = 3.5 cm
KM = 3.5 cm
L + LM = 3.5 cm + 3.5 cm
L + LM = 7 cm
7 cm > 3.5 cm, KL + LM > KM
Therefore, the sum of any two sides of a triangle is greater than the third side. Hence, we conclude that the sum of any two sides of a triangle is never less than the third side.
## Exercise 5.2
Q1. What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(e) 1 to 10
(f) 6 to 3
Hint: To find the portion of the clock swept by the hour hand between any two numbers(hours) say 'a' and 'b' where b > a will be found by subtracting 'a' from 'b' i.e, b - a.
Also, to find the fraction of a clockwise revolution between 'a' and 'b' we now divide the portion of the clock swept by the hour hand between them by 12 (since, the clock is divided into 12 equal divisions) Hence, the fraction will be (b - a) / 12 Using this concept, let's solve the given problems.
(a) 3 to 9
Ans: Lets first subtract 3 from 9 and then divide the answer by 12
9 – 3 = 6 ÷ 12 = 1/2 of a clockwise revolution
(b) 4 to 7
Ans: Lets first subtract 4 from 7 and then divide the answer by 12
7 – 4 = 3 ÷ 12 = 1/4 of a clockwise revolution
(c) 7 to 10
Ans: Lets first subtract 7 from 10 and then divide the answer by 12
10 – 7 = 3 ÷ 12 = 1/4 of a clockwise revolution
(d) 12 to 9
Ans: i.e., 0 to 9 [Since, 12 is at 0th position of the clock]
Lets first subtract 0 from 9 and then divide the answer by 12
9 – 0 = 9 ÷ 12 = 3/4 of a clockwise revolution
(e) 1 to 10
Ans: Lets first subtract 1 from 10 and then divide the answer by 12
10 – 1 = 9 ÷ 12 = 3/4 of a clockwise revolution
(f) 6 to 3
Ans: i.e., 6 to 12 and then 12 to 3
6 to 12 = 12 – 6 = 6 and 12 to 3 = 0 to 3 = 3 – 0 = 3
6 + 3 = 9 ÷ 12 = 3/4 of a clockwise revolution
Q2. Where will the hand of a clock stop if it
(a) starts at 12 and makes 1/2 of a revolution, clockwise?
Ans: When the clock hand starts from 12 and makes ½ of a revolution, the clock hand stops at 6.
1 revolution = 12 hours
1/2 revolution = 6 hours
Hence, the clock will land at 6.
(b) starts at 2 and makes 1/2 of a revolution, clockwise?
Ans: When the clock hand starts from 2 and makes ½ of a revolution, the clock hand stops at 8
1 revolution = 12 hours
1/2 revolution = 6 hours
Hence, the clock will land at 2 + 6 = 8.
(c) starts at 5 and makes 1/4 of a revolution, clockwise?
Ans: When the clock hand starts from 5 and makes 1/4 of a revolution, the clock hand stops at 8
1 revolution = 12 hours
1/4 revolution = 3 hours
Hence, the clock will land at (5 + 3) = 8.
(d) starts at 5 and makes 3/4 of a revolution, clockwise?
Ans: When the clock hand starts from 5 and makes 3/4 of a revolution, the clock hand stops at 2.
1 revolution = 12 hours
3/4 revolution = 9 hours
Hence, the clock will land at (5 + 9) = 14
Which is 2 hours added to 12 so the answer is 2 o'clock.
Q3. Which direction will you face if you start facing
(a) East and make ½ of a revolution clockwise?
(b ) East and make 1 ½ of a revolution clockwise?
(c) West and make ¾ of a revolution anti-clockwise?
(d) South and make one full revolution?
(Should we specify clockwise or anticlockwise for this last question? Why not?)
(a) East and make ½ of a revolution clockwise?
Ans: If we start facing East and make ½ of a revolution clockwise, we will be facing the West direction.
1 revolution = 360 degrees
1/2 revolution = 180 degrees
Thus, we start from East, move 180 degrees clockwise and stop at West.
(b) East and make 1 ½ of a revolution clockwise?
Ans: If we start facing East and make 1 ½ revolution clockwise, we will be facing the West direction.
1 revolution = 360 degrees
1/2 revolution = 180 degrees
Hence, 1 ½ revolution = 360 + 180 = 1 revolution + another 1/2 revolution
Thus, we start from East, move 360 degrees clockwise to complete 1 revolution, and continue moving for another 180 degrees to cover 1/2 revolution and stop at West.
(c) West and make ¾ of a revolution anti-clockwise?
Ans: If we start facing the West and make ¾ revolution anticlockwise, we will be facing the North direction.
1 revolution = 360 degrees
3/4 revolution = 270 degrees = 1/2 revolution + 1/4 revolution (180 degrees + 90 degrees)
Thus, we start from West, move 180 degrees anti-clockwise to complete 1/2 revolution till East, and continue moving for another 90 degrees to cover 1/4 revolution anti-clockwise and stop at North.
(d) South and make one full revolution?
Ans: It does not matter if we turn clockwise or anticlockwise, taking one full revolution will make it a full revolution and reach back to the starting position.
Thus, after 1 complete revolution that is 360 degrees, we stop at the same point from where we had started i.e, South.
Q4. What part of a revolution have you turned through if you stand facing
(a) east and turn clockwise to face north?
Ans: If we start from east and turn clockwise to face north,3/4 of a revolution is required, as seen in the image below.
(b) south and turn clockwise to face east?
Ans: If we start from the south and turn clockwise to face east, 3/4 of a revolution is required.
(c) west and turn clockwise to face east?
Ans: If we start from the west and turn clockwise to face east, 1/2 of a revolution is required.
Q5. Find the number of right angles turned through by the hour hand of a clock when it goes from
(a) 3 to 6
Ans: 3 to 6 - The hour hand turns through 1 right angle starting from 3 to 6.
(b) 2 to 8
Ans: 2 to 8 - The hour hand turns through 2 right angles starting from 2 to 8.
(c) 5 to 11
Ans: 5 to 11- The hour hand turns through 2 right angles starting from 5 to 11.
(d) 10 to 1
Ans: 10 to 1 - The hour hand turns through 1 right angle starting from 10 to 1.
(e) 12 to 9
Ans: 2 to 9 - The hour hand turns through 3 right angles starting from 12 to 9.
(f) 12 to 6
Ans: 12 to 6 - The hour hand turns through 2 right angles starting from 12 to 6.
Q6. How many right angles do you make if you start facing
(a) south and turn clockwise to west?
Ans: The number of right angles made while facing the south and turning clockwise to the west is 1 right angle.
(b) north and turn anticlockwise to east?
Ans: The number of right angles made while facing the north and turning anticlockwise to the east is 3 right angles.
(c) west and turn to west?
Ans: The number of right angles made while facing the west and turning to the west is 4 right angles.
(d) south and turn to north?
Ans: The number of right angles made while facing the south and turning to the north is 2 right angles.
Q7. Where will the hour hand of a clock stop if it starts
(a) from 6 and turns through 1 right angle?
(b) from 8 and turns through 2 right angles?
(c) from 10 and turns through 3 right angles?
(d) from 7 and turns through 2 straight angles?
(a) from 6 and turns through 1 right angle?
Ans: We know that in 1 complete revolution in either clockwise or anticlockwise direction, hour hand of a clock will rotate by 360o or 4 right angles
(a) If hour hand of a clock starts from 6 and turns through 1 right angle, it will stop at 9
(b) If hour hand of a clock starts from 8 and turns through 2 right angles, it will stop at 2
(c) If hour hand of a clock starts from 10 and turns through 3 right angles, it will stop at 7
(d) If hour hand of a clock starts from 7 and turns through 2 straight angles, it will stop at 7
## Exercise 5.3
Q1. Match the following:
Ans: We will use the concept of angles to solve this.
Q2. Classify each one of the following angles as right, straight, acute, obtuse or reflex:
Ans: We will use the concept of angles to solve this.
(a) Acute angle - It is less than 90°
(b) Obtuse angle - It is greater than 90° but less than 180°
(c) Right angle - It is equal to 90°
(d) Reflex angle - It is greater than 180° but less than 360°
(e) Straight angle - It is equal to 180°
(f) Acute angle - It is less than 90°
## Exercise 5.4
Q1. What is the measure of
(i) a right angle?
Ans: The measure of a right angle is 90o
(ii) a straight angle?
Ans: The measure of a straight angle is 180o
Q2. Say True or False:
(a) The measure of an acute angle < 90°.
(b) The measure of an obtuse angle < 90°.
(c) The measure of a reflex angle > 180°.
(d) The measure of on complete revolution = 360°.
(e) If m∠A = 53o and m∠B = 35o then m∠A > m∠B.
Ans:
(a) True; the measure of an acute angle is less than 90o
(b) False; the measure of an obtuse angle is more than 90o but less than 180o
(c) True; the measure of a reflex angle is more than 180o
(d) True; the measure of one complete revolution is 360o
(e) True; A is greater than B
Q3. Write down the measure of:
(a) some acute angles
(b) some obtuse angles
(give at least two examples of each)
Ans: (a) The measures of an acute angle are 50°, 65°
(b) The measures of obtuse angle are 110°, 175°
Q4. Measure the angles given below using the Protractor and write down the measure:
Ans: (a) The measure of an angle is 45°
(b) The measure of an angle is 120°
(c) The measure of an angle is 90°
(d) The measures of angles are 60°, 90° and 130°
Q5. Which angle has a large measure? First estimate and then measure:
Measure of angle A = ?
Measure of angle B = ?
Ans: ∠B has larger measure.
∠A = 40o and ∠B = 68o
Q6. From these two angles which has larger measure? Estimate and then confirm by measuring them:
Ans: Second angle has larger measure.
The measures of these angles are 45o and 55o.
Q7. Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is ________________.
Ans: acute angle
(b) An angle whose measure is greater than that of a right angle is ________________.
Ans: obtuse angle
(c) An angle whose measure is the sum of the measures of two right angles is ________________.
Ans: straight angle
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is ________________.
Ans: acute angle
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be ________________.
Ans: obtuse angle
Q8. Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor.)
Ans: The measures of the angles shown in the above figures are 40o, 130o, 65o and 135o
Q9. Find the angle measure between the hands of the clock in each figure:
Ans: (i) 90 (Right angle)
(ii) 30 (Acute angle)
(iii) 180 (Straight angle)
Q10. Investigate:
In the given figure, the angle measure 30o. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?
Ans: No, the measure of angle will be same.
Q11. Measure and classify each angle:
Ans:
Exercise 5.5
Q1. Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming the letter ‘L’.
(d) The letter V.
Ans:
(a) Perpendicular
(b) Not perpendicular
(c) Perpendicular
(d) Not perpendicular
Q2. Let be the perpendicular to the line segment . Let and intersect in the point A. What is the measure of ∠PAY?
Ans: We will be using the concept of perpendicular lines to solve this.
Since PQ ⊥ XY Therefore, ∠PAY = 90°
Q3. There are two “set-squares” in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?
Ans: One set-square has 60°,90°, 30° and other set-square has 45° ,90° ,45°. They have 90° as common angle.
Q4. Study the diagram. The line l is perpendicular to line m.
(a) Is CE = EG?
(b) Does PE bisect CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC < EH
Ans: (a) Yes, both measure 2 units.
(b) Yes, because CE = EG
(d) (i) True, (ii) True, (iii) True
Exercise 5.6
Q1. Name the types of following triangles:
(a) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(b) ΔABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(c) ΔPQR such that PQ = QR = PR = 5 cm.
(d) ΔDEF with m∠D = 90o
(e) ΔXYZ with m∠Y = 90o and XY = YZ
(f) ΔLMN with m∠L = 30o , m∠M = 70o and m∠N = 80o
Ans: (a) Scalene triangle
(b) Scalene triangle
(c) Equilateral triangle
(d) Right-angled triangle
(e) Isosceles right-angled triangle
(f) Acute-angled triangle
Q2. Match the following:
Ans: (i) → (e)
(ii) → (g)
(iii) → (a)
(iv) → (f)
(v) → (d)
(vi) → (c)
(vii) → (b)
Q3. Name each of the following triangles in two different ways: (You may judge the nature of angle by observation)
Ans: (i) Acute-angled and isosceles triangle
(ii) Right-angled and scalene triangle
(iii) Obtuse-angled and isosceles triangle
(iv) Right-angled and isosceles triangle
(v) Equilateral and acute-angled triangle
(vi) Obtuse-angled and scalene triangle
Q4. Try to construct triangles using match sticks. Some are shown here.
Can you make a triangle with:
(a) 3 matchsticks?
(b) 4 matchsticks?
(c) 5 matchsticks?
(d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case)
If you cannot make a triangle, think of reasons for it.
Ans: (a) 3 matchsticks This is an acute angle triangle and it is possible with 3 matchsticks to make a triangle because sum of two sides is greater than third side.
(b) 4 matchsticks This is a square, hence with four matchsticks we cannot make triangle.
(c) 5 matchsticks This is an acute angle triangle and it is possible to make triangle with five matchsticks, in this case sum of two sides is greater than third side.
(d) 6 matchsticks This is an acute angle triangle and it is possible to make a triangle with the help of 6 matchsticks because sum of two sides is greater than third side.
## Exercise 5.7
Q1. Say true or false:
(a) Each angle of a rectangle is a right angle.
(b) The opposite sides of a rectangle are equal in length.
(c) The diagonals of a square are perpendicular to one another.
(d) All the sides of a rhombus are of equal length.
(e) All the sides of a parallelogram are of equal length.
(f) The opposite sides of a trapezium are parallel.
Ans: (a) True
(b) True
(c) True
(d) True
(e) False
(f) False
Q2. Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
(c) A square can be thought of as a special rhombus.
(d) Squares, rectangles, parallelograms are all quadrilateral.
(e) Square is also a parallelogram.
Ans: (a) Because its all angles are right angle and opposite sides are equal.
(b) Because its opposite sides are equal and parallel.
(c) Because its four sides are equal and diagonals are perpendicular to each other.
(d) Because all of them have four sides.
(e) Because its opposite sides are equal and parallel.
Q3. A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?
Ans: A square is a regular quadrilateral because all the interior angles are 90o, and all sides are of the same length.
## Exercise 5.8
Q1. Examine whether the following are polygons. If anyone among these is not, say why?
Ans: (a) As it is not a closed figure, therefore, it is not a polygon.
(b) It is a polygon because it is closed by line segments.
(c) It is not a polygon because it is not made by line segments.
(d) It is not a polygon because it not made only by line segments, it has curved surface also.
Q2. Name each polygon:
Make two more examples of each of these.
(b) Triangle
(c) Pentagon
(d) Octagon
Q3. Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Ans: ABCDEF is a regular hexagon and triangle thus formed by joining AEF is an isosceles triangle.
Q4. Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Ans: ABCDEFGH is a regular octagon and CDGH is a rectangle.
Q5. A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Ans: ABCDE is the required pentagon and its diagonals are AD, AC, BE and BD.
## Exercise 5.9 (Old NCERT)
Ques 1: Match the following:
(a) Cone (b) Sphere (c) Cylinder (d) Cuboid (e) Pyramid
Give two example of each shape.
Ans:
(a) Cone (b) Sphere (c) Cylinder (d) Cuboid (e) Pyramid
Ques 2: What shape is:
(b) A brick?
(c) A match box?
Ans: (a) Cuboid
(b) Cuboid
(c) Cuboid
(d) Cylinder
(e) Sphere
The document NCERT Solutions for Class 6 Maths Chapter 5 - Understanding Elementary Shapes is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6
## Mathematics (Maths) Class 6
94 videos|347 docs|54 tests
## FAQs on NCERT Solutions for Class 6 Maths Chapter 5 - Understanding Elementary Shapes
1. What are elementary shapes?
Ans. Elementary shapes are simple geometric shapes like circles, squares, rectangles, triangles, and polygons that are the basic building blocks of more complex shapes and figures.
2. How can elementary shapes be classified?
Ans. Elementary shapes can be classified based on their properties such as the number of sides, angles, and vertices they have. Common classifications include triangles, quadrilaterals, and polygons.
3. Why is it important to understand elementary shapes?
Ans. Understanding elementary shapes is important as it forms the foundation for learning more advanced geometry concepts. It helps in visualizing and analyzing shapes in various contexts like art, architecture, and engineering.
4. How can elementary shapes be used in real-life applications?
Ans. Elementary shapes are used in real-life applications such as designing buildings, creating maps, and calculating areas and volumes of objects. They provide a framework for understanding and representing the physical world.
5. Can elementary shapes be combined to form more complex shapes?
Ans. Yes, elementary shapes can be combined and transformed to create more complex shapes and figures. By understanding the properties of individual shapes, one can manipulate them to generate new shapes and patterns.
## Mathematics (Maths) Class 6
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# Understand Addition and Subtraction Worded Problems
In this worksheet, students will be asked to use addition and subtraction skills in order to solve multi-step worded problems.
Key stage: KS 2
Curriculum topic: Number: Addition and Subtraction
Curriculum subtopic: Solve Multi-Step Add/Subtract Problems
Difficulty level:
### QUESTION 1 of 10
In this activity, you will be asked to use your addition and subtraction skills in order to solve multi-step worded problems.
When solving these problems, read each question at least twice and look for highlighted words and numbers to help you.
Example
A plane arrived at an airport with 14,567 tonnes of cargo.
The plane was then loaded with 13,865 tonnes before taking off and leaving the airport.
How much cargo was on the plane when it left?
You need to solve this question in steps.
Step One
Subtract the amount taken off.
14,567 - 2,568 = 11,999 tonnes
Step Two
Add on the extra cargo put onto the plane.
11,999 + 13,865 = 25,864 tonnes
When the plane left, it was carrying 25,864 tonnes.
The populations of three towns in the same county are,
35,678
78,924
12,342
Over the course of the year, 2,345 people leave the three towns and move elsewhere.
What is the total population of the three towns now?
124,599
124,589
124,590
124,500
I am thinking of two numbers.
One of the numbers is 68,792.
The difference between this number and the other is 5,678.
What is the other number?
124,599
124,589
124,590
124,500
A charity event raises £23,567 in ticket sales and £10,923 in merchandise sales.
The organisers pay £14,567 in expenses to the acts who performed.
What is the total amount raised for charity?
A charity event raises £36,789 in ticket sales and £14,888 in merchandise sales.
The organisers pay £19,211 in expenses to the acts who performed.
What is the total amount raised for charity?
At a weekend summer family day, £13,456 was raised on Saturday and £9,876 raised on Sunday.
At a similar event in the winter, £10,892 was raised on Saturday and £6,574 raised on Sunday.
How much was raised altogether?
£40,790
£40,798
£40,788
£40,698
At a weekend summer family day, £14,786 was raised on Saturday and £8,765 raised on Sunday.
At a similar event in the winter, £10,003 was raised on Saturday and £7,864 raised on Sunday.
How much was raised altogether?
£40,790
£40,798
£40,788
£40,698
Journey Distance in Kilometres London to Milan 982 London to Paris 934 Paris to Milan 600
Look at the journey times above.
A plane flies from London to Paris and then onto Milan.
How far does it fly in total?
1,543 km
1,534 km
1,540 km
1,530 km
Journey Distance in Kilometres London to Milan 982 London to Paris 934 Paris to Milan 600
Look at the journey times above.
A plane flies from London to Paris, onto Milan and then back to London.
How far does the plane fly in total?
1,543 km
1,534 km
1,540 km
1,530 km
Calculate,
8,924 + 4,567 + 231 = ?
Then, when you have found the answer, round it to the nearest 1,000.
At the start of the school year, there was £7,865 in the school parent committee budget.
By the end of the year, an additional £2,311 had been raised and £6,781 had been spent.
How much money was left in the budget by the end of the school year?
• Question 1
The populations of three towns in the same county are,
35,678
78,924
12,342
Over the course of the year, 2,345 people leave the three towns and move elsewhere.
What is the total population of the three towns now?
124,599
EDDIE SAYS
Step one: Add together the total population for all three towns 35,678 + 78,924 + 12,342 = 126,944 Step two: Subtract the number of people who have left 126,944 - 2,345 = 124,599 You’ve got this, keep going!
• Question 2
I am thinking of two numbers.
One of the numbers is 68,792.
The difference between this number and the other is 5,678.
What is the other number?
EDDIE SAYS
To solve this problem, we have to use the numbers given and carry out a subtraction sum. 68,792 - 5,678 = 63,114 Hopefully, this method is starting to make more sense! Let's continue.
• Question 3
A charity event raises £23,567 in ticket sales and £10,923 in merchandise sales.
The organisers pay £14,567 in expenses to the acts who performed.
What is the total amount raised for charity?
EDDIE SAYS
Step 1: Add the total amount raised, 23,567 + 10,923 = £34,490 Step 2: Subtract expenses paid, 34,490 - 14,567 = £19,923
• Question 4
A charity event raises £36,789 in ticket sales and £14,888 in merchandise sales.
The organisers pay £19,211 in expenses to the acts who performed.
What is the total amount raised for charity?
EDDIE SAYS
Step 1: Add the total amount raised, 36,789 + 14,888 = 51,677 Step 2: Subtract expenses paid, 51,677 - 19,211= £32,466 Are you feeling more confident?
• Question 5
At a weekend summer family day, £13,456 was raised on Saturday and £9,876 raised on Sunday.
At a similar event in the winter, £10,892 was raised on Saturday and £6,574 raised on Sunday.
How much was raised altogether?
£40,798
EDDIE SAYS
Step 1: Add together how much both family days raised Summer = 13,456 + 9,876 = £23,332 Winter = 10,892 + 6,576 = £17,466 Total raised = £23,332 + £17,466 = £40,798 You're halfway through, maths-whiz!
• Question 6
At a weekend summer family day, £14,786 was raised on Saturday and £8,765 raised on Sunday.
At a similar event in the winter, £10,003 was raised on Saturday and £7,864 raised on Sunday.
How much was raised altogether?
EDDIE SAYS
Step 1: Add together how much each family day raised Summer = 14,786 + 8,765 = £23,551 Winter = 10,003 + 7,864 = £17,867 Total raised = £23,551 + 17,867 = £41,418 The method remains the same each time, you just need to work step by step through each question!
• Question 7
Journey Distance in Kilometres London to Milan 982 London to Paris 934 Paris to Milan 600
Look at the journey times above.
A plane flies from London to Paris and then onto Milan.
How far does it fly in total?
1,534 km
EDDIE SAYS
London to Paris = 934 km Paris to Milan = 600 km Add together 934 + 600 = 1,534 km in total.
• Question 8
Journey Distance in Kilometres London to Milan 982 London to Paris 934 Paris to Milan 600
Look at the journey times above.
A plane flies from London to Paris, onto Milan and then back to London.
How far does the plane fly in total?
EDDIE SAYS
London to Paris = 934 km Paris to Milan = 600 km Milan to London = 982 km Finally, add together 934 + 600 + 982 = 2,516 km
• Question 9
Calculate,
8,924 + 4,567 + 231 = ?
Then, when you have found the answer, round it to the nearest 1,000.
14,000
EDDIE SAYS
8,924 + 4,567 + 231 = 13,722 13,722 rounded to the nearest 1000 is 14,000. We round up as the hundreds digit is greater than 5.
• Question 10
At the start of the school year, there was £7,865 in the school parent committee budget.
By the end of the year, an additional £2,311 had been raised and £6,781 had been spent.
How much money was left in the budget by the end of the school year?
3395
3,395
3 395
EDDIE SAYS
Step one: Work out how much was raised total 7,865 + 2,311 = £10,176 Step two: Subtract the amount spent from the total 10,176 - 6,781 = £3,395 Great focus, that’s another activity completed!
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# DEFACING - Editorial
Practice
Contest
EASY
## PREREQUISITES
Brute force, Greedy
## PROBLEM
We are given an integer S represented in a 7-segment display. Add some segments or digits into it to create the as large number as possible not larger than M.
## QUICK EXPLANATION
Since we want to maximize the number, it is always optimal to make S contain the same number of digits as M (assume that we allow leading zeros). Hence, there are two steps in the solution:
• Add some digits before S and some digits after S in such a way that the resulting length is equal to M's length.
• Add some segments in such a way to maximize the resulting number.
We can perform all possible ways of the first step using brute force. For each intermediate result after the first step, the second step can be performed using greedy. Finally, we output the largest resulting number over all ways.
## EXPLANATION
First, let’s introduce a compatibility matrix valid[i][j] = whether digit i can be transformed into digit j by adding some segments. The value of each cell of the matrix can be computed by hand. Here are the values:
```\ 0123456789
+----------
0|1000000010
1|1101100111
2|0010000010
3|0001000011
4|0000100011
5|0000011011
6|0000001010
7|1001000111
8|0000000010
9|0000000011
```
Let’s go through the steps in more details.
### Step 1
Let |X| be the number of digits of X. As mentioned before, in an optimal solution the number of digits of M and S must be equal. There are exactly |M| - |S| + 1 possible ways to add new digits. For example, suppose S = 89375 and M = 9247529. Then, there are 3 ways:
• Add 0 digits before S and 2 digits after S: 89375XX
• Add 1 digits before S and 1 digits after S: X89375X
• Add 2 digits before S and 0 digits after S: XX89375
Here, the new digits are represented by X’s for convenience. Let’s call the new integer after adding some digits as S*.
### Step 2
Note that the restriction that the resulting number must not have leading zeros is so that we cannot have the length resulting number is greater than |M| but that difference of length consists of all leading zeros. If that is allowed, then for test case like S = 0, M = 9 we can have 09 as the resulting number, hence making the problem trickier. Since we have in our solution that |S| = |M|, we can safely ignore that restriction.
For each possible intermediate result of Step 1, we must replace each X by an actual digit and optionally add some segments to the existing digits. We can use a greedy method here. We will iterate S* from the most significant digit through the least significant digit. For each digit, we try to transform it into the largest digit, while maintaining that the final result is not greater than M. This way, it can be proved that the result is the largest possible.
When iterating the digits from the most significant digit through the least significant digit, the choice of the current digit depends on the current prefixes of M and S*. Suppose we are now considering the i-th most significant digit. The current state would be like this:
```M = Pp...
S* = Qq...
^
i-th most significant digit
```
where p and q are M anp S*'s i-th most significant digits, respectively, while P and Q are M and S*'s current prefixes, respectively. The (i+1)-th through the |M|-th digits are not considered yet. For example, let M = 9247529, the currently built S* = 89375XX, and i = 4, then the current state is:
```M = 9247...
S* = 8937...
^
4-th most significant digit
```
where P = 924, Q = 893, p = 7, and q = 7.
For each state, i.e. for each step in the iteration, there are 2 possibilities to maintain that the resulting number is not greater than M:
q > p. Here, Q must be strictly less than P because otherwise the resulting number would be greater than M.
q ≤ p. Here, Q must be less than or equal to P.
From the above explanation, let’s maintain two values while iterating the digits:
• prefix_less = the maximum prefix of S* that is strictly less than the corresponding prefix of M, or -1 if there is no such prefix
• prefix_equal = the corresponding prefix of M, or -1 if we cannot have equal prefix for M and S*
We update the two values after each step in the iteration. In the end, we choose the iteration that yields the maximum value. Please consult the following pseudocode for more details. The time complexity of this solution is O(|M|^2).
```// returns the maximum possible resulting number
// if we align the first digit of S with the k-th digit of M
// or -1 if it is impossible.
function solve(M, S, k):
prefix_less = -1
prefix_equal = 0
for i = 1; i ≤ |M|; i++:
// if the resulting number must be greater than M, then impossible
if prefix_less == -1 and prefix_equal = -1
return -1
// update prefix_less
for d = 9; d ≥ 0; d--:
// if we cannot transform the digit, continue.
if i ≥ k and i ≤ k + |S| - 1 and not valid[S[i - k + 1]][d]:
continue
// found the largest valid digit.
// if the new digit is greater than the current digit, use prefix_less
if d > M[i]:
if prefix_less != -1:
prefix_less = prefix_less * 10 + d
// if the new digit is not greater than the current digit, use either prefix_less or prefix_equal
else if prefix_less != -1:
if max(prefix_less, prefix_equal) != -1:
prefix_less = max(prefix_less, prefix_equal) * 10 + d
break
// update prefix_equal
if i ≥ k and i ≤ k + |S| - 1 and not valid[S*[i - k + 1]][M[i]]:
prefix_equal = -1
else
prefix_equal = prefix_equal * 10 + M[i]
// return the better answer
return max(prefix_less, prefix_equal)
// the main code
read(S)
read(M)
best = 0
for k = 1; k ≤ |M| - |S| + 1; k++:
best = max(best, solve(M, S, k))
print(best)```
## SETTER’S SOLUTION
Can be found here.
## TESTER’S SOLUTION
Can be found here.
3 Likes
Correction:
``````for d = 9; d ≤ 0; d--:
``````
should be
``for d = 9; d >= 0; d--:``
3 Likes
Correction#2:
``````else if max_less != -1:
``````
should be
``else if prefix_less != -1:``
"Since we want to maximize the number, it is always optimal to make S contain the same number of digits as M. "
This is very, very wrong and a really bad starting point.
What if, for example, M=100 and S=25, like in the example used in the task itself? There is no way to make S have as many digits as M, without it being larger than M.
can you please tell me where my code is failing, i used the same approach as described in the editorial
http://www.codechef.com/viewsolution/1741764
It would be nice if you could post the input set used to validate the solutions…
Some of the tricky test cases:
```9
274 4883530
5 268343
2 558870
10381 16146
0 6
0 9
2 200543
4987565 14398964
1042216 1815366
```
with the corresponding answers
```4883499
268343
558870
10989
0
8
200543
9989989
1098898
```
Will be updated regularly
Let’s add another constraint to make it harder: the chef can perform the defacing operation at most K times.
Share your solution.
They say this is an easy problem. It took me 2 days to understand Anton’s code snippet
``````// some kind of dp
max_smaller_digit[d1][0] = -1;
for (int d2 = 1; d2 < 10; ++d2) {
if (mark_digits[d1][d2-1]) {
max_smaller_digit[d1][d2] = d2-1;
} else {
max_smaller_digit[d1][d2] = max_smaller_digit[d1][d2-1];
}
}
``````
WTH! From where do you come up with these tricks, Anton? I finish the 2.5 hours time only thinking how to solve a problem…God knows when the hell am I gonna actually write more than one solutions in a Cook-Off or a Long Contest as well. God bless me!
Can anybody tell me why my
``````
[1] is resulting in Wrong Answer ? I have used the same approach as described in the editorial.
[1]: http://www.codechef.com/viewsolution/1745718``````
My code is resulting in wrong answer …
http://www.codechef.com/viewsolution/1775055
I have tried every test case given in posts and problem…
working correctly…
what is that I am missing ?
what is with initial values of prefix_equal and prefix_less??
http://ideone.com/VRvkWc
this is my code for the problem.It is returning correct solution for every test case(mentioned).
Someone please tell me what is wrong with this because every time I submit it shows wrong answer
The only correct solution posted by me is copied.
I have written my code for this problem. Can you please tell me whats the format of the program. I mean it is not specified how the input will be taken and how the output should.
My code read the file which is in the folder of compiler and gives output in a file. But tester going to know this. He have to change the name of the file according to his testing computer.
So please help me out. How the submission is done.
Thanks in advance.
@syker, you have to perform all the IO from/to stdin/stdout. Just like the normal way you code.
You can check few samples by going on anyone’s profile page and then hitting any problem and then the corresponding solution.
1 Like
@anton_lunyov
Excellent problem…my first attempt at codechef and I am enjoying every bit of it.
This is the latest version of solution I have developed:
http://www.codechef.com/viewsolution/1864293
It gives a correct output for all the 12 test cases given in problem, 9 tricky cases added by you plus two more given in other comments.
I know, it doesnt mean it is a correct solution but I am unable to understand why I am still getting wrong answer.
I have used my own logic, havent looked at your solution yet, I want to figure out the entire solution on my own. Can you can provide some more test cases, which will help me in figuring out the bug on my own. I tried to create test cases on my own, by using random number generator and all, but got correct answers for all of them. So, if you have some more test cases, it will be great
P.S. : I dont expect you to go through my code as it can be tricky to understand the logic
@syker >> Try these tests:
8 13
8 24
Correct Answer: 8 and 18 respectively
Your answer: 89 and 89
UPD.
Your code gives correct answer for the above tests when they are performed initially, but gives 89 when they are written after some tests like 10 100. That might be because of faulty/misplaced initialization of some check arrays.
@anton_lunyov
I am now struggling with TLE issue for this problem.
What I find odd is: time limit is the same irrespective of platform. Isn’t some languages faster as compared to others? And I haven’t seen a single successful submission for JAVA for this problem. I know, it could be a mere coincidence but it still makes me wonder.
I have tried many optimization techniques but none of them are giving any improvements. On my system I see lot of improvements with every major optimization(I have simulated huge inputs, as per given constraints, using random number generators)but when I submit here, I don’t see any improvement at all in time, every time i just get 4.04 seconds, not even slightest improvement.
Moreover, some optimization techniques which definitely should have worked have actually deteriorated the performance, e.g. if you compare the submissions:
http://www.codechef.com/viewsolution/1877893 and
http://www.codechef.com/viewsolution/1883458
the only change is in the function convertToNumber(), where instead of using String object, to append digits one by one, I used StringBuilder and append() function, which according to me should give better performance, but instead the time deteriorated to 10 seconds from 4 seconds.
Any pointers in this regard?
Why is the answer for 25, 100 is 89 but not 98 ?
025 —> 089 —> 89
250 —> 098 —> 98
Somebody please clarify, I’m struggling…
I posted my code on Ideone: http://ideone.com/tJKTyy
I am getting a WA.
I think it is passing the test cases given in the question and the comments here (as well as some ones I made myself), but I might simply be misreading. Can someone help me find my problem? Or at least a test case where my program is returning the wrong answer? |
# 2015 AMC 10B Problems/Problem 24
## Problem
Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin $p_0=(0,0)$ facing to the east and walks one unit, arriving at $p_1=(1,0)$. For $n=1,2,3,\dots$, right after arriving at the point $p_n$, if Aaron can turn $90^\circ$ left and walk one unit to an unvisited point $p_{n+1}$, he does that. Otherwise, he walks one unit straight ahead to reach $p_{n+1}$. Thus the sequence of points continues $p_2=(1,1), p_3=(0,1), p_4=(-1,1), p_5=(-1,0)$, and so on in a counterclockwise spiral pattern. What is $p_{2015}$?
$\textbf{(A) } (-22,-13)\qquad\textbf{(B) } (-13,-22)\qquad\textbf{(C) } (-13,22)\qquad\textbf{(D) } (13,-22)\qquad\textbf{(E) } (22,-13)$
## Solution
(Used from 2015 AMC 10/12 B Math Jam)
The first thing we would do is track Aaron's footsteps:
He starts by taking $1$ step East and $1$ step North, ending at $(1,1)$ after $2$ steps and about to head West.
Then he takes $2$ steps West and $2$ steps South, ending at $(-1,-1$) after $2+4$ steps, and about to head East.
Then he takes $3$ steps East and $3$ steps North, ending at $(2,2)$ after $2+4+6$ steps, and about to head West.
Then he takes $4$ steps West and $4$ steps South, ending at $(-2,-2)$ after $2+4+6+8$ steps, and about to head East.
From this pattern, we can notice that for any integer $k /ge 1$ he's at $(-k, -k)$ after $2 + 4 + 6 + ... + 4k$ steps, and about to head East. There are $2k$ terms in the sum, with an average value of $(2 + 4k)/2 = 2k + 1$, so:
$$2 + 4 + 6 + ... + 4k = 2k(2k + 1)$$
If we substitute $k = 22$ into the equation: $44(45) = 1980 < 2015$. So he has $35$ moves to go. This makes him end up at $(-22+35,-22) = (13,-22) \implies \boxed{\textbf{(D)} (13, -22)}$ |
# Chapter 16: Statistical tests
STAT 1010 - Fall 2022
# Learning outcomes
• Vocabulary: alternative hypothesis, null hupothesis, one-sided, two-sided hypotheses, and test statistic
• Understand the differences between a statistical test for proportions and for means.
• Recognize the different types of errors Type I and Type II when designing a test
• Distinguish between statistical and substantive significance
• Relate confidence intervals to two-sided tests
# Does ESP exist?
Some people believe in the presence of Extrasensory Perception, or ESP. How could we prove whether it does or doesn’t exist?
One test for ESP is with Zener cards.
## Does ESP exist?
Since there are 5 cards, it would be possible to guess the correct card at random $p=1/5$ times, or 20% of the time.
A person with ESP should be able to guess the correct card more often than 20%. But how much more often do they need to get it right for us to believe that ESP exists?
## Statistical test
One way to determine something like this is to use a statistical test. A statistical test is a procedure to determine if the results from the sample are convincing enough to allow us to conclude something about the population.
## Hypotheses
When we perform a statistical test, we set out our hypotheses before we begin. There are two hypotheses,
null hypothesis $H_0$, a statement about there being no effect, or no difference.
alternative hypothesis $H_A$, the thing that we secretly hope will turn out to be true.
## ESP hypotheses
Thinking about the ESP experiment, we could use words to state our hypotheses
$\begin{eqnarray*} &H_0:& \text{ ESP does not exist} \\ &H_A:& \text{ESP exists} \end{eqnarray*}$
We could also write the hypotheses in terms of parameters,
$\begin{eqnarray*} H_0: p \leq 1/5 \\ H_A: p > 1/5 \end{eqnarray*}$
## Hypotheses
• Hypotheses are always written about the population parameter ($\mu$, $p$, $\mu_1-\mu_2$, $p_1-p_2$), never about the sample statistics ($\bar{x}$, $\hat{p}$, $\bar{x}_1-\bar{x}_2$, $\hat{p}_1-\hat{p}_2$).
• The null hypothesis is the boring thing that we’re trying to gather evidence against
• The alternative hypothesis is the exciting thing that would make headlines
## One and two sided hypotheses
• $H_A$ has a $>$ sign: upper tail, or “one-sided”
• $H_A$ has a $<$ sign: lower tail, or “one-sided”
• $H_A$ has a $\neq$ sign: both sides, or “two-sided”
## Example 1
What are the null and alternate hypothesis for these decisions?
• A person interviews for a job opening. The company has to decide whether to hire the person
• An inventor proposes a new way to wrap packages that they say will speed up the manufacturing process. Should they adopt the new method?
• A sales representative submits receipts from a recent business trip. Staff must determine whether the claims are legitimate.
## Example 1 solns
• $H_0 =$ Person is not hired
• $H_0 =$ Retain the current method
• $H_0 =$ Treat claims as legitimate. The employee is innocent until evidence to the contrary is found.
## Example 2
A snack-food chain runs a promotion in which shoppers are told that 1 in 4 kids’ meals includes a prize. A father buys two kids’ meals, and neither has a prize. He concludes that because neither has a prize, the chain is being deceptive.
1. What is the null and alternative hypothesis?
2. Describe a Type I error?
3. Describe a Type II error?
4. What is the probability that the father has made a Type I error?
## Example 2 - solns
1. $H_0$ = chain is being honest; $H_A$ = chain is not being honest
2. Falsely accusing the chain of being deceptive
3. Failing to realize that the chain is being deceptive
4. Father rejects if both are missing a prize. $P(neither has a prize) = (1 - 1/4)^2 \approx 0.56$
## Test statistic
In the previous example, we used probability to determine how likely it is to get two meals neither of which contain a prize. This process involves computing a test statistic ($0.56$).
# Testing a proportion
To test a proportion, $\hat{p}$, we must have a distribution with which to compare it. The distribution that we use is the distribution assumed in $H_0$ we use $p_0$ to denote this.
$p_0 \sim N(p_0, \frac{{p_0}(1-{p_0})}{n})$
## Testing a proportion - contd
Under $H_0$, we find
$z_0 = \frac{\hat{p} - p_0} {\sqrt{\frac{p_0(1-p_0)}{n}}}$
## Assumptions
• SRS and sample must be < $10\%$ of population
• Both $np_0$ and $n(1-p_0)$ are larger than 10
## Example 3
In the ESP example, we know that the population average for guessing ESP cards is 9 out of 24 cards. An interested participant took a training course to enhance their ESP. In the followup exam, they guessed 17 out of 36 cards. Did the course improve their ESP abilities?
## Steps for a hypothesis test
1. State hypotheses
2. Determine the level of significance
3. Check conditions
4. Calculate test statistic
5. Compute p-value
6. Generic conclusion
7. Interpret in context
## Example 3 - solns
1. $H_0: p \leq 9/24$ and $H_A: p > 9/24$
2. $\alpha = 0.05$
3. Yes 9/24*36 and (1-9/24)*36 both > 10
4. $\hat{p} = \frac{x}{n} = \frac{17}{36} =0.472$
5. \begin{aligned} z_0 &= \frac{\hat{p} - p_0} {\sqrt{\frac{p_0(1-p_0)}{n}}} \\ &= \frac{0.472 - 0.375} {\sqrt{\frac{0.375(1-0.375)}{36}}} \\ &= \frac{0.0972}{0.0807} \approx 1.2 \end{aligned}
## Example 3 - solns
1. 1 -pnorm(1.2) or 1- pnorm((17/36- 9/24)/sqrt((0.375*(1-0.375))/36)) $\approx 0.115$
2. Since $0.115 > \alpha = 0.05$ this is not significant.
3. We fail to reject $H_0$. The score does not provide evidence that the intervention improved ESP.
# Testing a mean
To test a mean, $\bar{x}$, we must have a distribution with which to compare it. The distribution that we use is the distribution assumed in $H_0$ we use $\mu_0$ to denote this. Because we don’t know $\sigma$ we will once again use $s$ from the sample
## Testing a mean - contd
Under $H_0$, we find
$t = \frac{\bar{X} - \mu_0} {s/\sqrt{{n}}}$
## Assumptions
• SRS and sample must be < $10\%$ of population
• If we do not know if the population is normal, $n > 10|K_4|$
## Example 4
Let $\bar{x} = 3281$, $s = 529$, and $n = 59$. Perform a hypothesis test that the sample comes from a distribution where the population mean is less than $\mu_0 = 4000$
## Steps for a hypothesis test
1. State hypotheses
2. Determine the level of significance
3. Check conditions
4. Calculate test statistic
5. Compute p-value
6. Generic conclusion
7. Interpret in context
## Example 4 - solns
1. $H_0: p \geq 4000$ and $H_A: p < 4000$
2. $\alpha = 0.05$
3. Yes
4. $\bar{x} = 3281$
5. \begin{aligned} t &= \frac{\bar{X} - \mu_0} {s/\sqrt{{n}}}\\ &= \frac{3281-4000}{529/\sqrt{59}} \\ &= \frac{-719}{68.87} \approx -10.44 \end{aligned}
## Example 4 - solns
1. pt(-10.44, df = 58) or pt((3281-4000)/(529/sqrt(59)), df = 58) $\approx 3.07e-15$
2. Since $3.07e-15 < \alpha = 0.05$ this is significant.
3. We reject $H_0$. There is very strong evidence that the mean value is less than 4000.
# Do extremists see the world in black and white?
Researcher Matt Motyl ran a study in 2010 with 2,000 participants. He wanted to determine if political moderates were able to perceive shades of grey more accurately than people on the far left or the far right. The p-value from the study was 0.01. Reject the null! Evidence in support of the idea that moderates can see grey better.
But then… he re-ran the study. This time, the p-value was 0.59. Not even close to significant.
What happened?!
Via Regina Nuzzo’s Nature article, Scientific method: Statistical Errors
## Errors
There are two types of errors defined in hypothesis testing:
• Type I error, rejecting a true null
• Type II error, not rejecting a false null
## Law analogy
In the US, a person is innocent until proven guilty, and evidence of guilt must be beyond “the shadow of a doubt.” We can make two types of mistakes:
• Convict an innocent person (type I error)
• Release a guilty person (type II error)
## Extremists and black and white
Two options:
• the original study (p-value 0.01) made a Type I error, and the $H_0$ was really true
• the second study (p-value 0.59) made a Type II error, and $H_A$ is really true
or…
• maybe there were no errors made, just different studies found different things
## Multiple testing
Because the probability of a Type I error is $\alpha$, if you do many tests you will find significance in $\alpha$ of them just by chance.
If you do 100 tests, you should expect to find 5 of them to be significant, just by chance.
This is the problem of multiple testing.
## Multiple testing + publication bias
Okay, so $\alpha$ of all tests show significance, just by random chance.
And things that look significant get published…
That means that a fair number of things that are published are actually false! This is pretty scary.
## How to fix the problem
As a researcher:
• make sure your results can be replicated (like Motyl tried to do with the politics and grey study)
• publish code and data so others can study your work
• be skeptical about claims that are just one of many tests
• look for replication and reproducibility!
## Reducing the probability of Type II error
In order to reduce the probability of making a Type II error, we can either
• increase the significance level
• increase the sample size
# Statistical versus practical significance
Sometimes, you find something that is very statistically significant, but not practically significant.
For example, a revision program might increase final exam grades with $p-value < 0.00001$, but it may only increase final exam grades by $1\%$. This is statistically significant but not practically significant.
Context is important!
# Confidence interval or test
These two are mostly interchangeable. However, tests only provide negative statements and does not give us much information about parameter values. |
The division of triangles into scalene, isosceles, and also equilateral have the right to be thoughtof in terms of lines the symmetry. A scalene triangle is a triangle with nolines the symmetry when an isosceles triangle has at least one line of symmetryand an it is provided triangle has actually three present of symmetry. This task providesstudents an possibility to recognize these differentiating features that the different types of triangles prior to the technical language has actually been introduced. Forfinding the present of symmetry, cut-out models of the four triangles would certainly behelpful so that the students can fold lock to uncover the lines.
You are watching: How many lines of symmetry does a right triangle have
This task is intended for instruction, offering the studentswith a chance to experiment through physical models that triangles, acquiring spatialintuition by executing reflections. A word has actually been added at the end of the solution about why there space not other lines of symmetries because that these triangles: this has actually been put in situation this topic comes up in a course discussion yet the emphasis should it is in on identifying the suitable lines of symmetry.
## Solution
The currently of symmetry because that the four triangles are suggested in the picturebelow:
A line of symmetry because that a triangle must go through one vertex. The 2 sides conference at the vertex must be the same length in order for there to be a line of symmetry. When the 2 sides conference at a peak do have the same length, the heat of symmetry with that crest passes with the midpoint of opposing side. Because that the triangle with side lengths 4,4,3 the just possibility is to wrinkles so the two sides of size 4 align, for this reason the line of the contrary goes with the vertex whereby those two sides meet. For the triangle every one of whose sides have actually length 3, a proper fold through any kind of vertex have the right to serve as a line of symmetry and so there are three possible lines. The triangle with side lengths 2,4,5 can not have any type of lines the symmetry together the side lengths are all different. Finally, the triangle v side lengths 3,5,5 has actually one line of symmetry through the vertex where the two sides of length 5 meet.
See more: What Does The Name Taylor Mean In The Bible ? Is The Name Taylor In The Bible
To check out why there space no other lines that symmetry because that these triangles, note that a heat of symmetry need to pass v a vertex of the triangle: if a line cuts the triangle into two polygons but does no pass with a vertex, then among those polygons is a triangle and the other is a quadrilateral. As soon as a peak of the triangle has been chosen, there is just one possible line that symmetry for the triangle through that vertex, specific the one which goes with the midpoint of the contrary side. |
# Raj alone can complete 20% of the work in 2.8 days, Sam alone can complete 40% of the work in 4 days and Kavi alone can complete 30% of the work in 3.
20 views
in Aptitude
closed
Raj alone can complete 20% of the work in 2.8 days, Sam alone can complete 40% of the work in 4 days and Kavi alone can complete 30% of the work in 3.6 days. They start working together and after 3 days Raj leaves them, then find the time taken by Sam and Kavi together to complete the remaining work?
1. $1\frac{6}{7}{\rm{\;days}}$
2. $1\frac{2}{7}{\rm{\;days}}$
3. $1\frac{4}{7}{\rm{\;days}}$
4. $1\frac{5}{7}{\rm{\;days}}$
5. None of these
by (30.0k points)
selected by
Correct Answer - Option 2 : $1\frac{2}{7}{\rm{\;days}}$
Given:
Raj alone can complete 20% of the work in 2.8 days.
⇒ Raj alone can complete the whole work in 14 days.
Sam alone can complete 40% of the work in 4 days
⇒ Sam alone can complete the whole work in 10 days.
Kavi alone can complete 30% of the work in 3.6 days
⇒ Kavi alone can complete the whole work in 12 days.
Assumption:
Let the total work be L.C.M. of (14, 10 and 12) = 420 units
Calculation:
One day work of Raj = 420/14 = 30 units
One day work of Sam = 420/10 = 42 units
One day work of Kavi = 420/12 = 35 units
According to question,
Let Sam and Kavi takes “d” days to complete remaining work.
3[30 + 42 + 35] + d [42 + 35] = 420
⇒ 321 + 77d = 420
⇒ 77d = 420 – 321
⇒ 77d = 99
⇒ d = 99/77
⇒ d = 9/7
∴ d = $1\frac{2}{7}{\rm{\;days}}$ |
# Continuity Formulas
Searching for a One-Stop Destination where you will find Continuity Formulas? If so, you have come the right way and the Formula Sheet of Continuity makes it easy to solve different problems easily. Try to memorize the Continuity Formulas List provided and overcome the burden of doing lengthy calculations. Master the concept of Continuity with the Formula Cheat Sheet & Tables provided.
## Formula Sheet of Continuity
If you are having any doubts on how to approach while solving problems of Continuity you can have a look at the List of Continuity Formulae provided. Make the most out of the Continuity Formulas Cheat Sheet & Tables and get to know the concept much better. You will find the Formulas extremely helpful and they save you plenty of time while solving your problems.
1. Continuity of a function at a point
A function f(x) is said to be continuous at a point x = a
i.e. If right hand limit at ‘a’ = left hand limit at ‘a’ = value of the function at ‘a’.
If $$\lim _{x \rightarrow a^{+}}$$f(x) = $$\lim _{x \rightarrow a^{-}}$$f(x) = f(a)
• f(x) is said to be continuous from the left at x = a if $$\lim _{x \rightarrow a^{-}}$$ f(x) = f(a).
• f(x) is said to be continuous from the right at x = a if $$\lim _{x \rightarrow a^{+}}$$ f(x) = f(a).
• If $$\lim _{x \rightarrow a}$$ f(x) does not exist or $$\lim _{x \rightarrow a}$$ f(x) ≠ f(a), then f(x) is said to be discontinuous at x = a.
2. Continuity of a function in an interval
(a) A function f(x) is said to be continuous in an open interval (a, b) if it is continuous at every point in (a, b).
(b) A function f(x) is said to be continuous in the closed interval [a, b] if it is
• Continuous at every point of the open interval (a, b).
• Right continuous at x = a.
• Left continuous at x = b.
3. Continuous functions
A function is said to be continuous function if it is continuous at every point in its domain. Following are examples of some continuous function.
• f(x) = x (Identity function)
• f(x) = C (Constant function)
• f(x) = x2
• f(x) = a0xn + a0xn-1+ ….+ an (Polynomial)
• f(x) = |x|, x + |x|, x – |x|, x|x|
• f(x) = sin x, f(x) = cos x
• f(x) = ex, f(x) = ax, a > 0
• f(x) = log x, f(x) = logax, a > 0
• f(x) = sinh x, cosh x, tanh x
4. Discontinuous functions
• f(x) = 1/x at x = 0
• f(x) = e1/x at x = 0
• f(x) = sin (1/x), f(x) = cos(1/x) at x = 0
• f(x) = [x] at every integer
• f(x) = x – [x] at every integer
• f(x) = tan x, f(x) = sec x when x = (2n + 1) $$\frac{\pi}{2}$$, n ∈ Z.
• f(x) = cot x, f(x) = cosec x when x = nπ, n ∈ Z.
• f(x) = coth x, f(x) = cosech x at x = 0.
5. Properties of continuous function
If f(x) and g(x) are continuous functions then following are also continuous functions:
• f(x) + g(x)
• f(x) – g(x)
• f(x). g(x)
• λf(x), where λ is a constant
• f(x)/g(x), if g(x) ≠ 0
• f[g(x)]
6. Some Important points
(i) When we say that the function f(x) is continuous at a point x = a, it mean that at point (a, f(a)) graph is untraken.
(ii) Kinds of discontinuity
• $$\lim _{x \rightarrow a^{-}}$$ f(x) = $$\lim _{x \rightarrow a^{+}}$$ f(x), then f is said to have non “removal discontinuity” of first kind.
• $$\lim _{x \rightarrow a^{-}}$$ f(x) ≠ $$\lim _{x \rightarrow a^{+}}$$ f(x), then f is said to have non removal discontinuity of first kind.
• At least one of $$\lim _{x \rightarrow a^{-}}$$ f(x) or $$\lim _{x \rightarrow a^{+}}$$ f(x) does not exist then f is said to have discontinuity of 2nd kind at x = a
• Continuity of composite function: If the function u = f(x) is continuous at the point x = a and the function y = g(u) is continuous at the point u = f(a) then composite function y = (gof)(x) = g(f(x)) is continuous at point x = a
Access various math concepts formulas all at one place Onlinecalculator.guru while doing your homework or assignments. |
## Precalculus (6th Edition)
center at $(-\frac{7}{2}, -\frac{3}{2})$ radius = $\frac{5\sqrt2}{2}$ units
RECALL: (1) The center-radius form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$ where $(h, k)$ is the center and $r$ = radius (2) To complete the square for $x^2+b^2$, add $(\frac{b}{2})^2$. The factored form of the perfect square trinomial is $(x+\frac{b}{2})^2$. To write the given equation in center-radius form, perform the following steps: (1) Divide $2$ to both sides of the equation: $$\dfrac{2x^2+2y^2+14x+6y+2}{2}=\dfrac{0}{2} \\x^2+y^2+7x+3y+1=0$$ (2) Subtract $1$ to both sides of the equation. $$x^2+y^2+7x+3y=-1$$ (3) Group terms terms with the same variable: $$(x^2+7x)+(y^2+3y)=-1$$ (3) Complete the square. Make sure to add in the right side of the equation whatever was added on the left side of the equation to maintain the equality of both sides. $$(x^2+7x+\color{blue}{(\frac{7}{2})^2})+(y^2+3y+\color{red}{(\frac{3}{2})^2})=-1+\color{blue}{(\frac{7}{2})^2}+\color{red}{(\frac{3}{2})^2} \\(x^2+7x+\frac{49}{4})+(y^2+3y+\frac{9}{4})=-\frac{4}{4}+\frac{49}{4}+\frac{9}{4} \\(x+\frac{7}{2})^2+(y+\frac{3}{2})^2=\frac{50}{4} \\(x+\frac{7}{2})^2+(y+\frac{3}{2})^2=\left(\sqrt{\frac{50}{4}}\right)^2 \\(x+\frac{7}{2})^2+(y+\frac{3}{2})^2=\left(\sqrt{\frac{25(2)}{4}}\right)^2 \\(x+\frac{7}{2})^2+(y+\frac{3}{2})^2=\left(\frac{5\sqrt{2}}{2}\right)^2$$ Therefore, the circle has: center at $(3, 5)$ radius = $\frac{5\sqrt2}{2}$ units |
How to Read a Speed Square
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Introduction: How to Read a Speed Square
The aim of this tutorial is to provide an understanding of how to read two main scales on a speed square.
A speed square is a tool that just about every carpenter will have in her bag. It can be used to make lines perpendicular (square) to a board's edge. And it can be used to mark any angle (between 0 and 90 degrees) across a board's face. There are many other ways the speed square can be used, but I will focus on these two in this tutorial.
Why is a triangle called a square? Merriam-Webster defines a square as: a four-sided shape that is made up of four straight sides that are the same length and that has four right angles.
It is the four right angles that defines our tool. The speed square only has one right angle, but one of it's main purposes is to quickly define a square line across a board, a line 90 degrees ( a right angle) to the edge of a board.
For more on the history of the Swanson speed square you may watch this video.
Step 1: Identify the Main Features of a Speed Square
As you look at your speed square you will notice that there are many different scales; common, degreeship/val (Hip and Valley for specific roofs), inches along the side. We will look at two of those scales in a bit, but first I want to point out some key features that allow the square to work as it does.
LIP- In this picture you'll notice the lip labeled along the left side. If you look at your square, the lip is the bit that is wider than the rest of the square, the part that makes it not sit flat on the table. The lip is what we use to hook on to the edge of a board to know that we have the square where we want it. It gives us our reference point.
PIVOT- In this picture you'll see the pivot in the bottom left-hand corner. The pivot point is the point at which we rotate the square when we are looking to mark an angle other than 90 degrees. It is the basis for reading all angles.
HYPOTENUSE- This is long side of any triangle. On a speed square it is the edge at which we read the degree scale. This is all you need to know about it for now, but I will refer to the hypotenuse later, so I want to make sure you know which edge it is.
Step 2: Using the Degree Scale
First let's review that we know which scale is the degree scale. There are two simple ways to know that you are looking at the correct part of the square to read degrees. The degree scale is the outermost one, the one that lies directly along the hypotenuse. It is also labeled DEGREES or DEG. It should have numbers ranging from 0 (where the hypotenuse meets the lip) to 90 (farthest away from the lip).
Got it!
Now that you are looking at the correct scale, it's time to find an angle. Take a board, a 2 x 4 or 1 x 4 works fine, and hold the square so the lip is flat against the long edge of the board and the square lies across the face of the board. LIke the picture, only the lip will be completely along the edge of the board. The edge that is square to the lip should run straight across the face of the board forming your 90 degree angle.
Notice where the pivot point is. This is always going to stay on the edge of the board. The edge of the lip that the hypotenuse abuts is going to move. Pivot stays, hypotenuse moves.
Do you see in the picture how the edge of the board forms a ray from the pivot out to the degree scale? With the pivot point of the lip still against the edge of the board, look at where the edge of the board passes through the degree scale. That number is your angle. In this picture, the square is held at 25 degrees.
Marking your board- Now that your square is in the right position you need to draw your line so you know what to cut or how to position your pieces. ALWAYS DRAW YOUR LINE OUT OF THE PIVOT POINT. What this means is that your line should run along the square edge from the lip. 90 degrees from the lip. Along the edge that reads inches. DO NOT DRAW ALONG THE HYPOTENUSE.
Step 3: Using the Common Scale
Using the common scale is the same as the degree scale.
First we need to make sure that we are looking at the correct scale. The common scale is marked as such, COMMON. It has numbers that range from 1 to 30. The common scale is the next scale in from the hypotenuse.
GOT IT!
What is the common scale? The common scale is used to read pitch, for roofs as an example. Roof angles could be identified by degrees but more commonly they are defined by pitch. If you remember back to math class, pitch is defined by rise over run. Run is a certain amount of distance along something. Rise is a certain distance up from something. In construction the run is defined as a constant 12, as in inches in a foot. The rise that occurs over the the 12, gives you your pitch. It does not matter if you are reading in inches, feet, or even (for some reason) centimeters. As long as your run is a multiple of 12, your rise will be in the common scale.
In the picture, I held my pivot point on the edge of the board and rotated my hypotenuse down. I read this scale as 7. Which is also 7 in 12, or 7/12, or 7 over 12. These all mean the same thing, For every 12 I go over, I will go 7 up. If it is 12 inches over, I will go 7 inches up. If it is 12 feet over, I will go 7 feet up. If I want my trim board to be cut at the correct pitch to match the roof, I can just use the common scale of 7. That way it doesn't matter how much I go up or over, I am at the same pitch.
REMINDER: Be sure to mark your board along the square edge. The edge that runs away from the pivot point at a 90 degree angle to the lip. This is the most common mistake I see.
Step 4: The Rest of the Speed Square
You surely have noticed that there are many other scales and markings on the speed square. I don't intend to explain those in this tutorial. The typical uses of the speed square are for degrees and common pitch angles.
If you would like to know more about the other uses of the speed square please check back later for more instructables. I have also found several good resources searching that other video tutorial site, you something or other.
I hope this helps. Please ask any questions and feel free to offer suggestions on how to make this more clear.
Thanks.
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Questions
Very nicely done. Have you seen the Speed square holder on the market
now? It looks cool I did a search for it on google.
WOW! was a furniture maker for over 35 years. Never used a speedsquare 'cause I couldn't figure it out.
Thanks!
Thanks for sharing this info. I was not sure how to read a measuring tape until I googled it and figured I'd go ahead and learn how to really read what I had been referring to a carpenter's square, which I found was NOT the name of the triangle that is actually a square. (FYI: The carpenter square is a large metal "L" shaped ruler.) Google led me to this site, learning something new at age 43, never too young I say. Thanks again. Happy measuring.
After years of DIY, I have finally obtained my first Swanson speed square (Speedlite made in the USA) and {quite a change, that last phrase! } How I managed without this magnificent aid I will never know. My square also has a level bubble gauge, which will be of great help. I have it in my hands right now while reading your tutorial, What is that diamond slot at the base of the triangle.? Don't worry, I shall find out! Good article, as well. Thank you. .
This is super helpful! Thanks for sharing!
You have unlocked further uses for my speed square. I had the same problem as Okara. They would not permit girls to take shop in high school. I had to take either typing or sewing. I chose typing. But I would have loved shop classes.
I am glad times have changed and wish I could get more girls to take my class.
When I was in high school, I wanted to take carpentry. In those days girls couldn't do that so ended up taking acting classes instead. Loved that class. Married a guy who built things. Loved helping him out. Anyway, thanks for this. Now I've learned something new. You have explained this so clearly! *wanders off to build something* |
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