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# 9th Class Mathematics Real Numbers Introduction
Introduction
Category : 9th Class
### Introduction
Previously we have studied about various types of numbers like natural numbers, whole numbers, integers, fractions and its decimal representation, rational numbers along with its different operations and properties. In this chapter, we will study about a new number system known as REAL NUMBER which includes rational and irrational numbers.
Representation of Numbers on Number Lines
It is a way to represent numbers on a line with the help of diagram.
Representation of Integers on Number Line
Take a line AB extended infinitely in both direction. Take a point O on it and represent it as zero (0). Mark points on line at equal distances on both sides of O. Equal distances are taken as per our convenience and take it as a unit.
On the right of O positive integers 1,2,3,4,5 etc. are indicated at the distance of 1 unit, 2 unit, 3 unit ,4 unit ,5 unit etc. respectively.
On the left of O negative integers -1,-2,-3,-4 etc. are indicated at a distance of 1 unit, 2 unit, 3 unit, 4 unit etc. respectively
Representation of Rational Numbers on a Number Line
Take a line extended infinitely in both direction. Take a point O on it and represent it as zero (0). Mark points on line at equal distances on both sides of O. Equal distances are taken as per our convenience and take it as a unit. Suppose OP=1 unit.
Let the midpoint of OP is Q, .Therefore, $OQ=\frac{1}{2}$ unit. Now we mark different points on the number line on right as well as left of O taking OQ =1/2 units.
The above given number line shows different rational numbers with denominator as 2.
Important Points Related to Number Line which Represents Rational Number
(i) A particular point on the number line represents a particular rational number.
(ii) A rational number cannot be represented by two or more than two distinct points on a number line.
(iii) There are infinite points between two distinct points on a line, hence, there are infinite rational numbers between two rational numbers.
Method to find Rational Numbers between Two Rational Numbers
(i) Suppose A and B be two rational number then a rational number which is in between A and B is $\frac{1}{2}(A+B)$.
(ii) Suppose A and B be two rational number in which A < B, then the n rational numbers between A and B are $(A+x),(A+2x),$ .............. $(A+nx)$. Where $x=\frac{B-A}{n+1}$
#### Other Topics
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Direction Cosines of a Line
# Direction Cosines of a Line | The Complete SAT Course - Class 10 PDF Download
In simple words, the cosines of the angles made by a directed line segment with the coordinate axes are called as the direction cosines of that line.
As illustrated in the figure above, if α, β, and γ are the angles made by the line segment with the coordinate axis then these angles are termed to be the direction angles and the cosines of these angles are the direction cosines of the line. Hence, cos α, cos β and cos γ are called as the direction cosines and are usually denoted by l, m and n.
l = cos α, m = cos β and n = cos γ
Another concept related to direction cosines is that of direction ratios. Three numbers which are proportional to the direction cosines of a line are called as the direction ratios. Hence, if ‘a’, ‘b’ and ‘c’ are the dr’s and l, m, n are the dc’s then, we must have
a/l = b/m = c/n.
This concept has been discussed in detail in the coming sections.
### Some Key Points of Direction Cosines
• Since l = cos α, m = cos β and n = cos γ and we know that -1< cos x< 1 ∀ x ∈ R, so l, m and n are real numbers with values varying between -1 to 1. So, dc’s ∈ [-1,1].
• The angles made by the x-axis with the coordinate axis are 0°, 90° and 90°. Hence, the direction cosines are cos 0°, cos 90° and cos 90° i.e. 1, 0, 0.
• The dc’s of a, y and z axis are (1,0,0), (0,1,0) and (0,0,1).
• The direction cosines of a line parallel to any coordinate axis are equal to the direction cosines of the corresponding axis.
• The dc’s are associated by the relation l2 + m2 + n2 =1.
• If the given line is reversed, then the direction cosines will be cos (π − α), cos (π − β), cos (π − γ) or − cos α, − cos β, − cos γ.
• Thus, a line can have two sets of dc’s according to its direction.
• The direction cosines of two parallel lines are always the same.
• Direction ratios are proportional to direction cosines and hence for a given line, there can be infinitely many direction ratios.
### Relation Between the Direction Cosines
Let OP be any line through the origin O which has direction cosines l, m, n.
Let P be the point having coordinates (x, y, z) and OP = r
Then OP2 = x2 + y2 + z2 = r2 …. (1)
From P draw PA, PB, PC perpendicular on the coordinate axes, so that OA = x, OB = y, OC = z.
Also, ∠POA = α, ∠POB = β and ∠POC = γ.
From triangle AOP, l = cos α = x/r ⇒ x = lr
Similarly y = mr and z = nr
Hence from (1) r(l2 + m2 + n2) = x2 + y2 + z2 = r2
⇒ l2 + m2 + n2 = 1
### Direction Cosines of a Line Joining Two Points
If we have two points P(x1, y1, z1) and Q(x2, y2, z2), then the dc’s of the line segment joining these two points are
(x2-x1)/PQ, (y2-y1)/PQ , (z2-z1)/PQ
i.e. (x2-x1)/√Σ(x2-x1)2, (y2-y1)/√Σ(x2-x1)2, (z2-z1)/√Σ(x2-x1)
Example: A line with positive direction cosines passes through the point P(2, -1, 2) and makes equal angles with the coordinate axis. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals
The direction cosines are l = m = n = 1/√3.
Hence, the equation of the required line is
(x-2)/(1/√3) = (y+1)/(1/√3) = (z-2)/(1/√3)
Hence, this gives x-2 = y+1 = z-2 = r
Hence, any point on the line is Q = (r+2, r-1, r+2).
Since Q lies on the plane 2x + y + z = 9
Therefore 2(r+2) + (r-1) + (r+2) = 9
This yields 4r + 5 = 9 or r = 1.
Hence teh coordinates of Q are (3,0,3).
Hence, PQ = √[(3-2)2 + (0+1)2 + (3-2)2]
= √3
The document Direction Cosines of a Line | The Complete SAT Course - Class 10 is a part of the Class 10 Course The Complete SAT Course.
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# What is 3/204 Simplified?
Are you looking to calculate how to simplify the fraction 3/204? In this really simple guide, we'll teach you exactly how to simplify 3/204 and convert it to the lowest form (this is sometimes calling reducing a fraction to the lowest terms).
To start with, the number above the line (3) in a fraction is called a numerator and the number below the line (204) is called the denominator.
So what we want to do here is to simplify the numerator and denominator in 3/204 to their lowest possible values, while keeping the actual fraction the same.
To do this, we use something called the greatest common factor. It's also known as the greatest common divisor and put simply, it's the highest number that divides exactly into two or more numbers.
In our case with 3/204, the greatest common factor is 3. Once we have this, we can divide both the numerator and the denominator by it, and voila, the fraction is simplified:
3/3 = 1
204/3 = 68
1 / 68
What this means is that the following fractions are the same:
3 / 204 = 1 / 68
So there you have it! You now know exactly how to simplify 3/204 to its lowest terms. Hopefully you understood the process and can use the same techniques to simplify other fractions on your own. The complete answer is below:
1/68
## Convert 3/204 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
3 / 204 = 0.0147
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is 3/204 Simplified?". VisualFractions.com. Accessed on December 9, 2021. http://visualfractions.com/calculator/simplify-fractions/what-is-3-204-simplified/.
• "What is 3/204 Simplified?". VisualFractions.com, http://visualfractions.com/calculator/simplify-fractions/what-is-3-204-simplified/. Accessed 9 December, 2021.
• What is 3/204 Simplified?. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/simplify-fractions/what-is-3-204-simplified/.
### Preset List of Fraction Reduction Examples
Below are links to some preset calculations that are commonly searched for:
## Random Fraction Simplifier Problems
If you made it this far down the page then you must REALLY love simplifying fractions? Below are a bunch of randomly generated calculations for your fraction loving pleasure:
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# Math in Focus Grade 1 Cumulative Review Chapters 3 and 4 Answer Key
This handy Math in Focus Grade 1 Workbook Answer Key Cumulative Review Chapters 3 and 4 detailed solutions for the textbook questions.
## Math in Focus Grade 1 Cumulative Review Chapters 3 and 4 Answer Key
Concepts and Skills
Look at the pictures. Complete the number sentences.
Question 1.
Explanation:
8 breads are colored and baked 2 breads are not colored and baked
so, total number of breads baked are = 8 + 2 = 10
Question 2.
Explanation:
In a ten frame there are 9 dots in that 4 are crossed
so, 9 – 4 = 5
Complete the number bonds. Fill in the blanks.
Question 3.
___ + 5 = 10
Explanation:
The sum of 5 and 5 is 10
that is stated by number bond
Question 4.
8 – 3 = ___
Explanation:
The difference of 8 and 3 is 4
that is stated by number bond
Fill in the blanks.
Question 5.
2 more than 8 is ____
10
Explanation:
2 more than 8 is number 10
2 + 8 = 10
Question 6.
3 less than 7 is ___.
4
Explanation:
3 less than 7 is number 4
7 – 3 = 4
Question 7.
_______ is 2 more than 5.
7
Explanation:
2 + 5 = 7
number 7 is 2 more than 5.
Question 8.
________ is 5 less than 10.
5
Explanation:
10 – 5 = 5
number 5 is 5 less than 10.
Find the missing number.
Question 9.
2 + ______ = 8
6
Explanation:
The sum of 2 and 6 is 8
2 + 6 = 8
Question 10.
______ – 6 = 0
6
Explanation:
6 – 6 = 0
The difference of 6 and 6 is 0
Pick three numbers and make a fact family.
Question 11.
Explanation:
An addition fact is defined as the sum of two one-digit addends
Problem Solving
Look at the pictures.
Write an addition or subtraction story.
Question 12.
Explanation:
There are different shapes of buttons
in them 5 are colored and 4 are non colored
so, total number of buttons are 9
5 + 4 = 9
Question 13.
Explanation:
Jamel is catching 9 balloons
3 flew away
so, 9 – 3 = 6
6 are left.
Solve.
Question 14.
Ellen has 3 spoons. Her sister gives her 5 spoons. How many spoons does Ellen have now?
Ellen has ___ spoons now.
Explanation:
Ellen has 3 spoons. Her sister gives her 5 spoons.
Ellen has 8 spoons now.
3 + 5 = 8
Question 15.
There are 8 fish in a fish tank. 6 are angelfish and the rest are goldfish. How many goldfish are there?
There are ___ goldfish.
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# If tan(a+b)=1 and tan(a-b)=1/7 then how will you find out the values of tana and tanb?
Jul 1, 2016
tana=-2, 1/2; & tanb=-3,1/3.
#### Explanation:
METHOD I
Let $\tan a = x , \tan b = y .$
Given that,$\tan \left(a + b\right) = 1 \Rightarrow \frac{\tan a + \tan b}{1 - \tan a \tan b} = 1 \Rightarrow x + y = 1 - x y \ldots \left(1\right)$
Similarly,$\tan \left(a - b\right) = \frac{1}{7} \Rightarrow x - y = \frac{1}{7} \left(1 + x y\right) \ldots \ldots \ldots . \left(2\right)$
$\left(1\right) + \left(2\right) \Rightarrow 2 x = \frac{8}{7} - \frac{6}{7} x y ,$or, $x = \frac{4}{7} - \frac{3}{7} x y ,$i.e.,$x \left(1 + \frac{3}{7} y\right) = \frac{4}{7} ,$giving, $x = \frac{4}{7 + 3 y} .$
We submit this $x$ in $\left(1\right)$ to see that.
$\frac{4}{7 + 3 y} + y + \left(\frac{4}{7 + 3 y}\right) y = 1 ,$ or,
$4 + 7 y + 3 {y}^{2} + 4 y = 7 + 3 y ,$ i.e., $3 {y}^{2} + 8 y - 3 = 0.$
Hence, $y = \tan b = - 3 , \frac{1}{3.}$
Using $x = \frac{4}{7 + 3 y} ,$ we get, $x = \tan a = - 2 , \frac{1}{2.}$
Jul 1, 2016
tana=-2, 1/2 ; tanb=-3, 1/3.
#### Explanation:
METHOD II:-
Take, $a + b = C , a - b = D ,$ then, by what is given, $\tan C = 1 , \tan D = \frac{1}{7.} . . \left(1\right)$
Observe that, $C + D = 2 a .$
$C + D = 2 a . \Rightarrow \tan \left(C + D\right) = \tan 2 a . \Rightarrow \frac{\tan C + \tan D}{1 - \tan C \tan D} = \tan 2 a$
Here, we use $\left(1\right)$ to get, $\frac{1 + \frac{1}{7}}{1 - \frac{1}{7}} = \tan 2 a ,$ or, $\tan 2 a = \frac{4}{3.} . \left(2\right)$
Recall that $\tan 2 a = \frac{2 \tan a}{1 - {\tan}^{2} a} \ldots \ldots . . \left(3\right) .$ so, if, tana=x, (2) & (3) rArr(2x)/(1-x^2)=4/3,rArr3x=2-2x^2,rArr2x^2+3x-2=0,rArr(x+2)(2x-1)=0,rArrx=tana=-2, 1/2,. as in METHOD I!
$\tan b$ can similarly be obtained using $C - D = 2 b .$
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# 24.3 The electromagnetic spectrum (Page 4/33)
Page 4 / 33
Television is also broadcast on electromagnetic waves. Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.) These TV channels are called VHF (for very high frequency ). Other channels called UHF (for ultra high frequency ) utilize an even higher frequency range of 470 to 1000 MHz.
The TV video signal is AM, while the TV audio is FM. Note that these frequencies are those of free transmission with the user utilizing an old-fashioned roof antenna. Satellite dishes and cable transmission of TV occurs at significantly higher frequencies and is rapidly evolving with the use of the high-definition or HD format.
## Calculating wavelengths of radio waves
Calculate the wavelengths of a 1530-kHz AM radio signal, a 105.1-MHz FM radio signal, and a 1.90-GHz cell phone signal.
Strategy
The relationship between wavelength and frequency is $c=\mathrm{f\lambda }$ , where $c=3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m}/\text{s}$ is the speed of light (the speed of light is only very slightly smaller in air than it is in a vacuum). We can rearrange this equation to find the wavelength for all three frequencies.
Solution
Rearranging gives
$\lambda =\frac{c}{f}.$
(a) For the $f=\text{1530}\phantom{\rule{0.25em}{0ex}}\text{kHz}$ AM radio signal, then,
$\begin{array}{lll}\lambda & =& \frac{\text{3.00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}{\text{1530}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{cycles/s}}\\ & =& \text{196 m.}\end{array}$
(b) For the $f=\text{105.1 MHz}$ FM radio signal,
$\begin{array}{lll}\lambda & =& \frac{\text{3.00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}{\text{105.1}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{cycles/s}}\\ & =& \text{2.85 m.}\end{array}$
(c) And for the $f=\text{1.90 GHz}$ cell phone,
$\begin{array}{lll}\lambda & =& \frac{3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}{\text{1.90}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{cycles/s}}\\ & =& \text{0.158 m.}\end{array}$
Discussion
These wavelengths are consistent with the spectrum in [link] . The wavelengths are also related to other properties of these electromagnetic waves, as we shall see.
The wavelengths found in the preceding example are representative of AM, FM, and cell phones, and account for some of the differences in how they are broadcast and how well they travel. The most efficient length for a linear antenna, such as discussed in Production of Electromagnetic Waves , is $\lambda /2$ , half the wavelength of the electromagnetic wave. Thus a very large antenna is needed to efficiently broadcast typical AM radio with its carrier wavelengths on the order of hundreds of meters.
One benefit to these long AM wavelengths is that they can go over and around rather large obstacles (like buildings and hills), just as ocean waves can go around large rocks. FM and TV are best received when there is a line of sight between the broadcast antenna and receiver, and they are often sent from very tall structures. FM, TV, and mobile phone antennas themselves are much smaller than those used for AM, but they are elevated to achieve an unobstructed line of sight. (See [link] .)
what is physics
a15kg powerexerted by the foresafter 3second
what is displacement
movement in a direction
Jason
hello
Hosea
Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
Hi
saeid
hi
Yimam
What is thê principle behind movement of thê taps control
while
Hosea
what is atomic mass
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
show me how to get the accuracies of the values of the resistors for the two circuits i.e for series and parallel sides
Explain why it is difficult to have an ideal machine in real life situations.
tell me
Promise
what's the s . i unit for couple?
Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
Oluwakayode
İt iş diffucult to have idêal machine because of FRİCTİON definitely reduce thê efficiency
Oluwakayode
if the classica theory of specific heat is valid,what would be the thermal energy of one kmol of copper at the debye temperature (for copper is 340k)
can i get all formulas of physics
yes
haider
what affects fluid
pressure
Oluwakayode
Dimension for force MLT-2
what is the dimensions of Force?
how do you calculate the 5% uncertainty of 4cm?
4cm/100×5= 0.2cm
haider
how do you calculate the 5% absolute uncertainty of a 200g mass?
= 200g±(5%)10g
haider
use the 10g as the uncertainty?
melia
haider
topic of question?
haider
the relationship between the applied force and the deflection
melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
thank you
melia
Hello group...
Chioma
hi
haider
well hello there
sean
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Noks
hii
Chibueze
10g
Olokuntoye
0.2m
Olokuntoye
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thomas
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# Multiplication Worksheets Table 3
Learning multiplication following counting, addition, and subtraction is good. Youngsters discover arithmetic through a all-natural progression. This growth of discovering arithmetic is usually the adhering to: counting, addition, subtraction, multiplication, and ultimately department. This document leads to the query why discover arithmetic in this pattern? Furthermore, why learn multiplication after counting, addition, and subtraction just before department?
## These facts answer these inquiries:
1. Young children find out counting first by associating visible items making use of their hands and fingers. A concrete illustration: Just how many apples exist from the basket? Much more abstract case in point is the way outdated are you?
2. From counting figures, the next reasonable step is addition combined with subtraction. Addition and subtraction tables are often very helpful teaching aids for kids since they are aesthetic resources producing the cross over from counting easier.
3. That ought to be discovered after that, multiplication or division? Multiplication is shorthand for addition. At this moment, youngsters use a firm knowledge of addition. For that reason, multiplication is the following plausible method of arithmetic to find out.
## Review essentials of multiplication. Also, evaluate the essentials the way you use a multiplication table.
We will assessment a multiplication case in point. Utilizing a Multiplication Table, grow four times a few and acquire an answer a dozen: 4 by 3 = 12. The intersection of row a few and column a number of of the Multiplication Table is twelve; a dozen is definitely the respond to. For children starting to discover multiplication, this can be straightforward. They are able to use addition to resolve the issue as a result affirming that multiplication is shorthand for addition. Example: 4 x 3 = 4 4 4 = 12. It is an outstanding guide to the Multiplication Table. A further benefit, the Multiplication Table is graphic and mirrors to discovering addition.
## Where should we start discovering multiplication making use of the Multiplication Table?
1. First, get acquainted with the table.
2. Get started with multiplying by 1. Commence at row number 1. Move to column primary. The intersection of row a single and column the initial one is the solution: one.
3. Repeat these techniques for multiplying by a single. Increase row 1 by posts one via twelve. The answers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Repeat these techniques for multiplying by two. Flourish row two by columns one particular via 5. The replies are 2, 4, 6, 8, and 10 correspondingly.
5. Let us jump ahead of time. Repeat these methods for multiplying by five. Multiply row five by columns one particular by way of 12. The answers are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively.
6. Now let us improve the degree of difficulty. Replicate these techniques for multiplying by three. Multiply row a few by posts one by way of a dozen. The responses are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 respectively.
7. Should you be confident with multiplication to date, try out a test. Resolve the following multiplication troubles in your head and then compare your answers towards the Multiplication Table: increase half a dozen and 2, increase 9 and three, increase one and eleven, multiply a number of and 4, and grow seven and 2. The issue answers are 12, 27, 11, 16, and 14 correspondingly.
In the event you obtained 4 out of several difficulties right, build your own multiplication exams. Estimate the responses in your mind, and check them while using Multiplication Table.
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# How do you find the product of (4c+1)(2c+1)?
Sep 7, 2016
$8 {c}^{2} + 6 c + 1$
#### Explanation:
We have: $\left(4 c + 1\right) \left(2 c + 1\right)$
Let's expand the parentheses:
$= \left(4 c\right) \left(2 c\right) + \left(4 c\right) \left(1\right) + \left(1\right) \left(2 c\right) + \left(1\right) \left(1\right)$
$= 8 {c}^{2} + 4 c + 2 c + 1$
$= 8 {c}^{2} + 6 c + 1$
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# Write the ratio in the simplest form (lowest terms).(i) $600 \mathrm{~g}$ to $1 \mathrm{~kg}$(ii) $50 \mathrm{p}$ to $Rs. 2.50$(iii) $2 \mathrm{~L}$ to $800 \mathrm{~mL}$(iv) $80 \mathrm{~cm}$ to $4 \mathrm{~m}$
Given:
(i) $600 \mathrm{~g}$ to $1 \mathrm{~kg}$
(ii) 50 paise to $Rs. 2.50$
(iii) 2 L to 800 mL
(iv) $80 \mathrm{~cm}$ to $4 \mathrm{~m}$
To do:
Here, we have to express the given ratios in the simplest form.
Solution:
We know that;
$1\ m = 100\ cm$
$1\ kg=1000\ g$
$1\ L=1000\ mL$
$Rs.\ 1=100\ paise$
(i) $600\ g:1\ kg=600:1\times1000$
$=600:1000$
$=3:5$
(ii) $50\ paise: Rs.\ 2.50=50:2.50\times100$
$=50:250$
$1:5$
(iii) $2\ L:800\ mL=2\times1000:800$
$=2000:800$
$=5:2$
(iv) $80\ cm:4\ m=80:4\times100$
$=80:400$
$=1:5$
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# The diagram shows a tube containing chocolate sweets. Calculate the total surface area of the tube.
Last updated date: 14th Jul 2024
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Hint: As we know that in the above question we have a tube, which is in the shape of a cylinder. We know the formula of the total surface area of the cylinder is $2\pi r(h + r)$ . In this question we have been given height i.e. $h = 10$ and the diameter is given. If we look at the top or the bottom of the cylindrical tube we can see that it is a circular shape. So we can see that in the figure we have been given the full length i.e. diameter of the circle. So we will calculate the radius i.e. $r$ from this and then apply the formula.
Complete step by step solution:
From the figure we can see that we have been given the diameter , so we can find the radius i.e. $r = \dfrac{d}{2}$ . So we have the value
$r = \dfrac{{2.2}}{2} = 1.1$ .
Now we know the formula of the total surface area of the cylinder i.e.
$2\pi r(h + r)$ .
We have $r = 1.1$ and $h = 10$ .
Now by putting the values we have
$2 \times \dfrac{{22}}{7} \times 1.1(10 + 1.1)$ . Now we will solve this, $6.914(11.1)$ .
On further solving we have the value $76.74$ .
Hence the required Total surface area is $76.74c{m^2}$ .
So, the correct answer is “ $76.74\;c{m^2}$ ”.
Note: We should note that there are two circular area in the top and the bottom of the cylinder, so their area is $\pi {r^2} + \pi {r^2} = 2\pi {r^2}$ , since there are two circles. After that if we unfold a cylinder open on both sides, then we obtain a rectangle. We should note that the area of the rectangle is the product of two sides i.e. one side is the height of the tube and the other side is the perimeter of the circle. So we can write the area of the rectangle is $2r\pi \times h = 2\pi rh$ . So this gives us the total surface area of the cylinder i.e. $2\pi {r^2} + 2\pi rh = 2\pi r(r + h)$ .
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# Bearings
Triangle Trigonometry. Bearings. By Andre Lam (18) Lau Wai Soong (19) Ivan Leo (20) Lim Bing Wen (21). Content. Introduction What are bearings? Application in Triangle Trigonometry Application in Real Life Sample Questions Activity (Try it yourself!) Question and Answer. Introduction.
## Bearings
E N D
### Presentation Transcript
1. Triangle Trigonometry Bearings By Andre Lam (18) Lau Wai Soong (19) Ivan Leo (20) Lim Bing Wen (21)
2. Content • Introduction • What are bearings? • Application in Triangle Trigonometry • Application in Real Life • Sample Questions • Activity (Try it yourself!) • Question and Answer
3. Introduction • What are bearings? • From Wikipedia, • A bearing is the angle between a line connecting two points and a north-south line.
4. Key concepts • Just a quick revision of the three functions • Sine function (sin), defined as the ratio of the side opposite the angle to the hypotenuse. • Cosine function (cos), defined as the ratio of the adjacent leg to the hypotenuse. • Tangent function (tan), defined as the ratio of the opposite leg to the adjacent leg.
5. Application in Triangle Trigonometry • To cut a long story short, a bearing is the direction from one object to another. In air navigation, bearings are given as angles rotated clockwise from the north.
6. Application in Triangle Trigonometry • A ship travels on a N50° E course. The ship travels until it is due north of a port which is 10 nautical miles due east of the port from which the ship originated. How far did the ship travel?
7. Answer • The angle opposite d is the complement of 50°, which is 40°. Therefore we can find d using the Cosine Function • Cos40=a/h=10/d • D(cos)40=10 • D=10/cos40=13.05 approx.
8. Application in Real Life • In Real Life, bearings are commonly used in three forms of navigation: • Marine Navigation • Aircraft Navigation • Land Navigation • With the help of bearings, soldiers, divers and hikers alike will not get lost while exploring unknown or new territory!
9. Application in Real Life • According to the US Army, bearings can also be used in war-time conditions. • Bearings are able to: • Determine the location of a foreign object or target • Aid piloting of an aircraft • Bring a drone from one position to another by remote • Search and Rescue a lost soldier
10. Other Sample Questions • http://www.digitaldesk.org/resources/mathsweb/bearing1.htm • The above link will address more queries about the use of bearings and how bearings work. • Bearings 1 through Bearings 3 are sample questions.
11. Sample Questions • http://www.mathsteacher.com.au/year7/ch08_angles/07_bear/bearing.htm • This link uses some illustrations to explain the concept of bearings as well as many examples • It covers bearings and directions
12. Additional Sample Questions • http://www.absorblearning.com/mathematics/demo/units/KCA005.html#Directionanddistance • The above link also gives more info about bearings and has an interactive applet for users to apply what they know of bearings to navigate a ship.
13. Activity (Try it yourself!) • (Bing Wen) Please provide sample questions from some workbooks that range from easy to challenging and then paste them here. Get the class to write their answers on the board and present how they get these answers
14. Any Questions?
15. THANK YOU
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# Problems based on ages
Problems based on ages are probably the most frequently asked questions in the Competitive examinations. And believe me, it is also among the easiest of the questions, you would be attempting in the examinations. With the knowledge of solving linear equations, you can solve any problem related to ages. I will try to explain the general approach of solving questions related to problems based on ages.
## General Approach for Problems based on Ages
Irrespective of what the question is, you would prepare what is called as TIMELINE. Once you draw the timeline, the whole problem becomes so easy to solve.
Let me explain with an example.
Let’s say the question talks of a father and his son, whose present ages, we need to find out. So, you would consider their ages as F & S.
Step 1 : First step is to draw a timeline as shown in the figure. As the present ages are not known, we represent their ages in 2012( Current year) as F & S. There will be additional data given in the question, how the Father’s and Son’s ages are related.
Step 2 : If in the question, you have something like 5 years ago, you go back 5 years from 2012, that is, 2007. So in 2007, Father’s age would be 5 years less, so (F-5). Similarly , Son’s age would be also 5 years less in which case, Son’s age in 2007 would be (S-5).
Step 3 : Suppose the data concerning the Father’s and Son’s age is in the future, you will go so many years in the future. Let’s suppose 6 years hence, in which case the year would be 2018. The father’s age would then be (F+6) and son’s age would be (S+6).
Generally, you will have 2 linear equations with 2 unknowns. You can solve for the ages by solving linear equations.
Example 1 : Four years ago Ram was four times as old as Shyam. Today Ram is twice as old. How old is Ram now?
Solution : Let’s prepare the timeline with the data given.
Step 1 : Once you prepare the timeline, the 2 linear equations are very clear. At present, Ram is twice as old as Shyam. So, R =2S
Step 2 : Four years ago, that means in 2008 ( Considering the present age as 2012). So, Ram’s age in 2008 was R-4 and Shyam’s age was S-4. It is given the question, that Ram’s age in 2008 was 4 times Shyam’s age in 2008. So, we have the second linear equation R-4 = 4(S-4)
Solving the 2 linear equations, we have :
R – 4 = 4(S-4)
R – 4 =4(R/2) – 16
R – 4 = 2R =16
R =12 and S would be 6
Example 2 : Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?
Solution : Let’s prepare the timeline with the data given.
Step 1 : Once you prepare the timeline, the 2 linear equations are very clear. At present, Sameer and Anand’s ages are in the ratio 5:4
Let S=5x, A=4x
Step 2 : After 3 years, that is in 2015 ( Considering the present age as 2012). So, Sameer’s age in 2015 would be S+3 and Anand’s age would be A+3.
It is given in 2015, S+3:A+3 = 11:9
Solving the 2 linear equations, we have :
9( 5x + 3 ) = 11( 4x + 3)
45x + 27 = 44x + 33
x = 6
Sameer’s Present Age is 5x = 30
Anand’s Present Age is 4x = 24
Example 3 : The sum of the ages of father & his son is 45 years. Five years ago the product of their ages was 34. Find the ages of the son & the father now ?
Solution : Let’s prepare the timeline with the data given.
Step 1 : Once you prepare the timeline, the 2 linear equations are very clear. At present, father’s and son’s age together is 45. So F+S=45
Step 2 : Five years ago, that means in 2007 ( Considering the present age as 2012) Father’s age was (F-5) and son’s age was (S-5). It is given that product of their ages in 2007 was 34. So, the equation would be (F-5)(S-5)=34.
Now, we have 2 equations :
(F-5)(S-5)=34 & F+S =34
Notice the first equation is not linear. So, one approach would be Use the answers given as options & use the reverse techniques.
Alternatively, Solve the 2 equations:
(F-5)(45 – F – 5) = 34
(F-5)(40 – F) = 34
40F – F2 – 200 + 5F =34
-F2 + 45F – 234 = 0
F2 – 45F + 234 = 0
(F – 39)(F – 6 ) = 0
So F is either 39 or 6
If we take F=6, then F+S = 45 & S=39.
Son’s age cannot be greater than the father’s age. We can ignore that.
The present age of Father is 39 and that of son is 6 years.
### Kiran Chandrashekhar
Hey, Thanks for dropping by. My name is Kiran Chandrashekhar. I am a full-time software freelancer. I love Maths and Mathematical Shortcuts. Numbers fascinate me. I will be posting articles on Mathematical Shortcuts, Software Tips, Programming Tips in this website. I love teaching students preparing for various competitive examinations. Read my complete story.
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1.5 Equations of Lines and Planes in 3-D
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1 40 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE Figure 1.16: Line through P 0 parallel to v 1.5 Equations of Lines and Planes in 3-D Recall that given a point P = (a, b, c), one can draw a vector from the origin to P. Such a vector is called the position vector of the point P and its coordinates are a, b, c, the same as P. Position vectors are usually denoted r. In this section, we derive the equations of lines and planes in 3-D. We do so by finding the conditions a point P = (x, y, z) or its corresponding position vector r = x, y, z must satisfy in order to belong to the object being studied (line or plane) Lines And Line Segments In 3-D, like in 2-D, a line is uniquely determined when one point on the line and the direction of the line are given. In this section, we assume we are given a point P 0 = (x 0, y 0, z 0 ) on the line and a direction vector v = a, b, c. Our goal is to determine the equation of the line L which goes through P 0 and is parallel to v. Definition 65 a, b, and c are called the direction numbers of the line L. Let P (x, y, z) be an arbitrary point on L. We wish to find the conditions P must satisfy to be on the line L.
2 1.5. EQUATIONS OF LINES AND PLANES IN 3-D 41 Vector Equation Consider figure We see that a necessary and suffi cient condition for the point P to be on the line L is that P 0 P be parallel to v. This means that there exists a scalar t such that P 0 P = t v Let r 0 be the position vector of P 0 and r be the position vector of P. Then, P 0 P = r r 0 Thus, we obtain That is r r0 = t v r = r0 + t v (1.9) Definition 66 Equation 1.9 is known as the vector equation of the line L. The scalar t used in the equation is called a parameter. The parameter t can be any real number. As it varies, the point P moves along the line. When t = 0, P is the same as P 0. When t > 0, P is away from P 0 in the direction of v and when t < 0, P is away from P 0 in the direction opposite v. The larger t is (in absolute value), the further away P is from P 0. Parametric Equations If we switch to coordinates, equation 1.9 becomes x, y, z = x 0, y 0, z 0 + t a, b, c = x 0 + at, y 0 + bt, z 0 + ct Two vectors are equal when their corresponding coordinates are equal. Thus, we obtain x = x 0 + at y = y 0 + bt (1.10) z = z 0 + ct Definition 67 Equation 1.10 is known as the parametric equation of the line L. Symmetric Equations If we solve for t in equation 1.10, assuming that a 0, b 0, and c 0 we obtain x x 0 = y y 0 = z z 0 (1.11) a b c Definition 68 Equation 1.11 is known as the symmetric equations of the line L.
3 42 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE Remark 69 Equation 1.11 is really three equations x x 0 a y y 0 b x x 0 a = y y 0 b = z z 0 c = z z 0 c (1.12) In the case one of the direction numbers is 0, the symmetric equation simply becomes the equation from 1.12 which does not involve the direction number being 0. The variable corresponding to the direction number being 0 is simply set to the corresponding coordinate of the given point. For example, if a = 0 then the symmetric equations are Examples y y 0 b x = x 0 = z z 0 c Example 70 Find the parametric and symmetric equations of the line through P ( 1, 4, 2) in the direction of v = 1, 2, 3 The parametric equations are The symmetric equations are x = 1 + t y = 4 + 2t z = 2 + 3t x + 1 = y 4 2 = z 2 3 Example 71 Find the parametric and symmetric equations of the line through P 1 (1, 2, 3) and P 2 (2, 4, 1). First, we need to find the direction vector. Since the line goes through P 1 and P 2, the vector P 1 P 2 will be parallel to the line. P 1 P 2 = 1, 2, 2 Using equation 1.10, we get If we solve for t, we get x = 1 + t y = 2 + 2t z = 3 2t x 1 = y 2 2 = z 3 2
4 1.5. EQUATIONS OF LINES AND PLANES IN 3-D 43 Equation of a Line Segment As the last two examples illustrate, we can also find the equation of a line if we are given two points instead of a point and a direction vector. Let s derive a formula in the general case. Suppose that we are given two points on the line P 0 = (x 0, y o, z 0 ) and P 1 = (x 1, y 1, z 1 ). Then P 0 P 1 = x 1 x 0, y 1 y 0, z 1 z 0 is a direction vector for the line. Using P 0 for the point and P 0 P 1 for the direction vector, we see that the parametric equations of the line are x = x 0 + t (x 1 x 0 ) y = y 0 + t (y 1 y 0 ) z = z 0 + t (z 1 z 0 ) This is usually written differently. If we distribute t and factor differently, we get x = (1 t) x 0 + tx 1 y = (1 t) y 0 + ty 1 (1.13) z = (1 t) z 0 + tz 1 When t = 0, we are at the point P 0 and when t = 1, we are at the point P 1. So, if t is allowed to take on any real value, then this equation will describe the whole line. On the other hand, if we restrict t to [0, 1], then this equation describes the portion of the line between P 0 and P 1 which is called the line segment from P 0 to P 1. Example 72 Find the equation of the line segment from P 0 P 1 (2, 4, 1). Using equation 1.13 we get x = (1 t) 1 + 2t y = (1 t) 2 + 4t with t [0, 1] z = (1 t) 3 + t = (1, 2, 3) to Intersecting Lines, Parallel Lines Recall that in 2-D two lines were either parallel or intersected. In 3-D it is also possible for two lines to not be parallel and to not intersect. Such lines are called skew lines. Two lines are parallel if their direction vectors are parallel. If two lines are not parallel, we can find if they intersect if there exists values of the parameters in their equations which produce the same point. More specifically, if the first line has equation r 0 + t v and the second line has equation R 0 + s u then they will intersect if there exists a value for t and s such that r 0 + t v = R 0 + s u. Also, when two lines intersect, we can find the angle between them by finding the smallest angle between their direction vectors (using the dot product). Finally, two lines are perpendicular if their direction vectors are perpendicular.
5 44 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE Remark 73 You will note that we used a different letter for the parameter in each equation because it may be a different value of the parameter which will produce the same point in each equation. We illustrate this with some examples. Example 74 Let L 1 be the line through (1, 6, 2) with direction vector 1, 2, 1 and L 2 be the line through (0, 4, 1) with direction vector 2, 1, 2. Determine if the lines are parallel, if they intersect or if they are skew. If they intersect, find the point at which they intersect. For the lines to be parallel, their direction vectors would have to be parallel, that is there would have to exist a constant c such that Which would imply 1, 2, 1 = c 2, 1, 2 1 = 2c 2 = c 1 = 2c Which has no solution. So, the lines are no parallel. We now derive the equation of each line. For L1 For L 2 x = 1 + t y = 6 + 2t z = 2 + t x = 2s y = 4 + s z = 1 + 2s The line will intersect if we can solve 1 + t = 2s 6 + 2t = 4 + s 2 + t = 1 + 2s Equation 1 gives t = 2s 1 If we substitute in equation 2, we get 6 + 2t = 4 + s (2s 1) = 4 + s 6 + 4s 2 = 4 + s 3s = 12 s = 4
6 1.5. EQUATIONS OF LINES AND PLANES IN 3-D 45 Since we had t = 2s 1 this implies that t = 7. We need to verify that these values also work in equation 3. Replacing s and t by their values gives us = (4) 9 = 9 So, the two lines intersect. To find the point of intersection, we can use the equation of either line with the value of the corresponding parameter. If we use L 1, the parameter is t and the value of t which gives the point of intersection is t = 7. Thus, the point of intersection is That is x = y = (7) z = x = 8 y = 8 z = 9 Example 75 Find the angle between the two previous lines. Their direction vectors are u = 1, 2, 1 and v = 2, 1, 2. The smallest angle between these vectors is θ = u v cos 1 u v = cos = cos Example 76 Find the points at which L 1 in the example above intersects with the coordinate planes. The parametric equations of L 1 are x = 1 + t y = 6 + 2t z = 2 + t Intersection with the xy-plane. On the xy-plane, z = 0. If we replace in the equation of L 1, we get t = 2. This allows us to find x and y. x = 1 2 = 1 and y = 6+2 ( 2) = 10. Thus, L 1 intersect the xy-plane at ( 1, 10, 0). Intersection with the xz-plane. On the xz-plane, y = 0. Using the same technique as above, we get 0 = 6 + 2t or t = 3. Thus, x = = 4 and z = = 5. Thus L 1 intersects the xz-plane at (4, 0, 5).
7 46 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE Figure 1.17: Plane determined by a point and its normal Intersection with the yz-plane. On the yz-plane, x = 0. Thus, 0 = 1 + t or t = 1. This gives us y = ( 1) = 8 and z = 2 1 = 1. It follows that L 1 intersects the yz-plane at (0, 8, 1). Summary for Lines 1. Be able to find the equation of a line given a point and a direction or given two points. 2. Be able to tell if two lines are parallel, intersect or are skewed. 3. Be able to find the angle between two lines which intersect. 4. Be able to find the points at which a line intersect with the coordinate planes Planes Plane determined by a point and its normal A plane is uniquely determined given a point on the plane and a vector perpendicular to the plane. Such a vector is said to be normal to the plane. To help visualize this, consider figure Given a point P 0 = (x 0, y 0, z 0 ) and a normal n = a, b, c to a plane, a point P = (x, y, z) will be on the plane if P 0 P is perpendicular to n that is n P0 P = 0 (1.14)
8 1.5. EQUATIONS OF LINES AND PLANES IN 3-D 47 This is known as the vector equation of a plane. Switching to coordinates, we get a, b, c x x 0, y y 0, z z 0 = 0 a (x x 0 ) + b (y y 0 ) + c (z z 0 ) = 0 We usually write this as ax + by + cz + d = 0 (1.15) where d = ax 0 by 0 cz 0. This is known as the scalar equation of a plane. It is also the equation of a plane in implicit form. Remark 77 Note that when we know the scalar equation of a plane, we automatically know its normal; it is given by the coeffi cients of x, y, and z. Remark 78 A plane in 3D is the analogous of a line in 2D. Recall that a normal to the line ax + by + c = 0 in 2D is a, b. Similarly, a normal to the plane ax + by + cz + d = 0 is a, b, c. Example 79 The normal to the plane 3x + 2y z = 10 is 3, 2, 1. Example 80 The scalar equation of a plane through (1, 2, 3) with normal 2, 1, 4 is 2 (x 1) + 1 (y 2) + 4 (z 3) = 0 2x + y + 4z 16 = 0 Plane determined by three points If instead of being given a point and the normal, we are given three non-colinear points P 1, P 2, and P 3, we form the vectors P 1 P 2 and P 1 P 3. The cross product P 1 P 2 P 1 P 3 is a vector perpendicular to both P 1 P 2 and P 1 P 3 and therefore perpendicular to the plane. We can then use one of the point, the vector obtained from the cross product to derive the equation of the plane. Example 81 Find the equation of the plane through the points P 1 (0, 1, 1), P 2 (1, 0, 1) and P 3 (1, 3, 1). A normal to the plane is P 1 P 2 P 1 P 3 = i j k = 2 i + 2 j 3 k = 2, 2, 3
9 48 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE A point P (x, y, z) is on the plane if P 1 P ( P 1 P 2 ) P 1 P 3 = 0 x, y 1, z 1 2, 2, 3 = 0 2x + 2 (y 1) 3 (z 1) = 0 Parallel Planes, Intersecting Planes 2x + 2y 3z + 1 = 0 Two planes are parallel if and only if their normals are parallel. If two planes p 1 with normal n 1 and p 2 with normal n 2 are not parallel, then the angle θ between them is defined to be the smallest angle between their normals that is the angle with the non-negative cosine. In other words, θ = cos 1 n 1 n 2 n 1 n 2 (1.16) Example 82 Find the angle between the planes p 1 : x + y + z 1 = 0 and p 2 : x 2y + 3z 1 = 0. Find their intersection. Angle: The normal to p 1 is n 1 = 1, 1, 1 and the normal to p 2 is n 2 = 1, 2, 3. Therefore, the angle θ between them is θ = cos 1 n 1 n 2 n 1 n 2 = cos = cos Intersection: When two planes are not parallel, they intersect in a line. To find the equation of a line, we need a point and a direction. The direction of the line will be perpendicular to both n 1 and n 2. Thus, it can be found with n 1 n 2. n 1 i j k n 2 = = 5 i 2 j 3 k = 5, 2, 3 We also need a point on the line. For this, we can use the equations of the planes. To be on the line, a point must belong to the two planes. However, that gives us only two equations and we have three unknowns. So, we can
10 1.5. EQUATIONS OF LINES AND PLANES IN 3-D 49 also fix one of the variables. For example, if we set z = 0, we will be looking for the point on the line which intersects the two planes and also the xy-plane. This is what we will do. So, we must solve { x + y = 1 x 2y = 1 The solutions are x = 1 and y = 0. So, (1, 0, 0) is a point on the line. Thus, the parametric equations of the line are Summary for Planes x = 1 + 5t y = 2t z = 3t In addition, using the material studied so far, you should be able to do the following: 1. Be able to find the equation of a plane given a point on the plane and a normal to the plane. 2. Be able to find the equation of a plane given three points on the plane. 3. Be able to find the equation of a plane through a point and parallel to a given plane. 4. Be able to find the equation of a plane through a point and a line not containing the point. 5. Be able to tell if two planes are parallel, perpendicular. 6. Be able to find the angle between two planes. 7. Be able to find the traces of a plane. 8. Be able to find the intersection of two planes. Make sure you can do the above before attempting the problems Problems Do the following problems. 1. Find the parametric equations for the line through the point P = (3, 4, 1) parallel to i + j + k. 2. Find the parametric equations for the line through the point P = ( 2, 0, 3) and Q = (3, 5, 2).
11 50 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE 3. Find the parametric equations for the line through the origin parallel to 2 j + k. 4. Find the parametric equations for the line through the point (1, 1, 1) parallel to the z-axis. 5. Find the parametric equations for the line through the point (0, 7, 0) perpendicular to the plane x + 2y + 2z = Find the parametric equations for the x-axis. 7. Find the equation of the line segment from (0, 0, 0) to ( 1, 1, 3 2). 8. Find the equation of the line segment from (1, 0, 0) to (1, 1, 0). 9. Find the equation of the plane through P = (0, 2, 1) normal to 3 i 2 j k. 10. Find the equation of the plane through P = (1, 1, 1), Q = (2, 0, 2) and R = (0, 2, 1). 11. Find the equation of the plane through P = (2, 4, 5) perpendicular to the x = 5 + t line y = 1 + 3t z = 4t x = 1 + 2t x = 2 + s 12. Find the point of intersection of the lines y = 2 + 3t and y = 4 + 2s z = 3 + 4t z = 1 4s Then, find the plane determined by these two lines. x = 1 + t 13. Find the plane determined by intersecting the lines y = 2 + t and z = 1 t x = 1 4s y = 1 + 2s z = 2 2s 14. Find the plane through P 0 = (2, 1, 1) perpendicular to the line of intersection of the planes 2x + y z = 3 and x + 2y + z = 2. x = 4t 15. Find the distance from the point P = (0, 0, 12) to the line y = 2t z = 2t 16. Find the distance from the point P = (2, 1, 3) to the line x = 2 + 2t y = 1 + 6t z = Find the distance from the point P = (2, 3, 4) to the plane x + 2y + 2z = 13..
12 1.5. EQUATIONS OF LINES AND PLANES IN 3-D Find the distance from the point P = (0, 1, 1) to the plane 4y + 3z = Find the distance between the two planes x+2y+6z = 1 and x+2y+6z = Find the angle between the two planes 2x+2y+2z = 3 and 2x 2y z = 5. x = 1 t 21. Find the point of intersection between the line y = 3t and the plane z = 1 + t 2x y + 3z = Determine whether the lines taken two at a time are parallel, intersect or skew. If they intersect, find their point of intersection. x = 3 + 2t L 1 : y = 1 + 4t z = 2 t L 2 : L 3 : 23. Find the points in which the line planes. x = 1 + 4s y = 1 + 2s z = 3 + 4s x = 3 + 2r y = 2 + r z = 2 + 2r x = 1 + 2t y = 1 t z = 3t meet the coordinate 24. In the plane, the slope-intercept form of the equation of a line is y = mx+b where m is the slope and b the y-intercept. Show that the vector (1, m) is a direction vector of the line y = mx + b. (hint: Pick 2 points P 1 and P 2 on the line the vector P 1 P 2 is a direction vector for the line). 25. Another form of the equation of a line in the plane is ax + by + c = 0. Using a similar approach as the one in the previous problem, show that (b, a) is a direction vector for such a line. What is a vector perpendicular to the line? change in y 26. For a line in space, the notion of slope ( ) does not carry over change in x because there are three variables changing, it is replaced by the direction vector. However, if a line is one of the coordinate planes, then there are only two variables involved. If the line is in the xy-plane, then only x and y are changing. If a line is in the xz-plane, then only x and z are changing. If a line is in the yz-plane, then only y and z are changing. Find the direction vector of a line having its slope equal to m in each of the coordinate planes. (hint: use problem??).
13 52 CHAPTER 1. VECTORS AND THE GEOMETRY OF SPACE 27. Let L be the line that passes through the points Q and R and let P be a point not on L. Show that the distance between P and L is QR QP. QR Show that this agrees with the formula in the book for the distance between a point S and a line through P parallel to v which is P S v v. 28. Let P be a point not on the plane determined by the points Q, R and S. Show that the distance between P and the plane is QR ( ) QS QP. QR QS 29. Show that the distance from the point P 1 = (x 1, y 1 ) to the line ax+by+c = 0 is ax1+by1+c a 2 +b Show that the distance from the point P 1 = (x 1, y 1, z 1 ) and the plane ax + by + cz + d = 0 is ax1+by1+cz1+d a. 2 +b 2 +c Assuming d 1 d 2, explain why the two planes given by ax+by+cz+d 1 = 0 and ax + by + cz + d 2 = 0 are parallel, then compute the distance between them.
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Learning Objectives
Learning Objectives
By the end of this section, you will be able to do the following:
• Describe Hooke’s law and Simple Harmonic Motion
• Describe periodic motion, oscillations, amplitude, frequency, and period
• Solve problems in simple harmonic motion involving springs and pendulums
amplitude deformation equilibrium position frequency Hooke’s law oscillate period periodic motion restoring force simple harmonic motion simple pendulum
Hooke’s Law and Simple Harmonic Motion
Hooke’s Law and Simple Harmonic Motion
Imagine a car parked against a wall. If a bulldozer pushes the car into the wall, the car will not move but it will noticeably change shape. A change in shape due to the application of a force is a deformation. Even very small forces are known to cause some deformation. For small deformations, two important things can happen. First, unlike the car and bulldozer example, the object returns to its original shape when the force is removed. Second, the size of the deformation is proportional to the force. This second property is known as Hooke’s law. In equation form, Hooke’s law is
$F=−kx,F=−kx,$
where x is the amount of deformation (the change in length, for example) produced by the restoring force F , and k is a constant that depends on the shape and composition of the object. The restoring force is the force that brings the object back to its equilibrium position; the minus sign is there because the restoring force acts in the direction opposite to the displacement. Note that the restoring force is proportional to the deformation x. The deformation can also be thought of as a displacement from equilibrium. It is a change in position due to a force. In the absence of force, the object would rest at its equilibrium position. The force constant k is related to the stiffness of a system. The larger the force constant, the stiffer the system. A stiffer system is more difficult to deform and requires a greater restoring force. The units of k are newtons per meter (N/m). One of the most common uses of Hooke’s law is solving problems involving springs and pendulums, which we will cover at the end of this section.
Oscillations and Periodic Motion
Oscillations and Periodic Motion
What do an ocean buoy, a child in a swing, a guitar, and the beating of hearts all have in common? They all oscillate. That is, they move back and forth between two points, like the ruler illustrated in Figure 5.39. All oscillations involve force. For example, you push a child in a swing to get the motion started.
Figure 5.39 A ruler is displaced from its equilibrium position.
Newton’s first law implies that an object oscillating back and forth is experiencing forces. Without force, the object would move in a straight line at a constant speed rather than oscillate. Consider, for example, plucking a plastic ruler to the left as shown in Figure 5.40. The deformation of the ruler creates a force in the opposite direction, known as a restoring force. Once released, the restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. However, by the time the ruler gets there, it gains momentum and continues to move to the right, producing the opposite deformation. It is then forced to the left, back through equilibrium, and the process is repeated until it gradually loses all of its energy. The simplest oscillations occur when the restoring force is directly proportional to displacement. Recall that Hooke’s law describes this situation with the equation F = −kx. Therefore, Hooke’s law describes and applies to the simplest case of oscillation, known as simple harmonic motion.
Figure 5.40 (a) The plastic ruler has been released, and the restoring force is returning the ruler to its equilibrium position. (b) The net force is zero at the equilibrium position, but the ruler has momentum and continues to move to the right. (c) The restoring force is in the opposite direction. It stops the ruler and moves it back toward equilibrium again. (d) Now the ruler has momentum to the left. (e) In the absence of damping (caused by frictional forces), the ruler reaches its original position. From there, the motion will repeat itself.
When you pluck a guitar string, the resulting sound has a steady tone and lasts a long time. Each vibration of the string takes the same time as the previous one. Periodic motion is a motion that repeats itself at regular time intervals, such as with an object bobbing up and down on a spring or a pendulum swinging back and forth. The time to complete one oscillation (a complete cycle of motion) remains constant and is called the period T. Its units are usually seconds.
Frequency f is the number of oscillations per unit time. The SI unit for frequency is the hertz (Hz), defined as the number of oscillations per second. The relationship between frequency and period is
As you can see from the equation, frequency and period are different ways of expressing the same concept. For example, if you get a paycheck twice a month, you could say that the frequency of payment is two per month, or that the period between checks is half a month.
If there is no friction to slow it down, then an object in simple motion will oscillate forever with equal displacement on either side of the equilibrium position. The equilibrium position is where the object would naturally rest in the absence of force. The maximum displacement from equilibrium is called the amplitude X. The units for amplitude and displacement are the same, but depend on the type of oscillation. For the object on the spring, shown in Figure 5.41, the units of amplitude and displacement are meters.
Figure 5.41 An object attached to a spring sliding on a frictionless surface is a simple harmonic oscillator. When displaced from equilibrium, the object performs simple harmonic motion that has an amplitude X and a period T. The object’s maximum speed occurs as it passes through equilibrium. The stiffer the spring is, the smaller the period T. The greater the mass of the object is, the greater the period T.
The mass m and the force constant k are the only factors that affect the period and frequency of simple harmonic motion. The period of a simple harmonic oscillator is given by
$T=2πmkT=2πmk$
and, because f = 1/T, the frequency of a simple harmonic oscillator is
$f=12πkm.f=12πkm.$
Watch Physics
Introduction to Harmonic Motion
This video shows how to graph the displacement of a spring in the x-direction over time, based on the period. Watch the first 10 minutes of the video (you can stop when the narrator begins to cover calculus).
Grasp Check
If the amplitude of the displacement of a spring were larger, how would this affect the graph of displacement over time? What would happen to the graph if the period was longer?
1. Larger amplitude would result in taller peaks and troughs and a longer period would result in greater separation in time between peaks.
2. Larger amplitude would result in smaller peaks and troughs and a longer period would result in greater distance between peaks.
3. Larger amplitude would result in taller peaks and troughs and a longer period would result in shorter distance between peaks.
4. Larger amplitude would result in smaller peaks and troughs and a longer period would result in shorter distance between peaks.
Solving Spring and Pendulum Problems with Simple Harmonic Motion
Solving Spring and Pendulum Problems with Simple Harmonic Motion
Before solving problems with springs and pendulums, it is important to first get an understanding of how a pendulum works. Figure 5.42 provides a useful illustration of a simple pendulum.
Figure 5.42 A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which result in a net force of −mg sinθ toward the equilibrium position—that is, a restoring force.
Everyday examples of pendulums include old-fashioned clocks, a child’s swing, or the sinker on a fishing line. For small displacements of less than 15 degrees, a pendulum experiences simple harmonic oscillation, meaning that its restoring force is directly proportional to its displacement. A pendulum in simple harmonic motion is called a simple pendulum. A pendulum has an object with a small mass, also known as the pendulum bob, which hangs from a light wire or string. The equilibrium position for a pendulum is where the angle$θ θ$is zero (that is, when the pendulum is hanging straight down). It makes sense that without any force applied, this is where the pendulum bob would rest.
The displacement of the pendulum bob is the arc length s. The weight mg has components mg cos$θ θ$along the string and mg sin$θ θ$ tangent to the arc. Tension in the string exactly cancels the component mg cos$θ θ$parallel to the string. This leaves a net restoring force back toward the equilibrium position that runs tangent to the arc and equals −mg sin$θ θ$.
For a simple pendulum, The period is$T=2πLg. T=2πLg.$
The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass or amplitude. However, note that T does depend on g. This means that if we know the length of a pendulum, we can actually use it to measure gravity! This will come in useful in Figure 5.42.
Tips For Success
Tension is represented by the variable T, and period is represented by the variable T. It is important not to confuse the two, since tension is a force and period is a length of time.
Worked Example
Measuring Acceleration due to Gravity: The Period of a Pendulum
What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?
Strategy
We are asked to find g given the period T and the length L of a pendulum. We can solve$T=2πLg T=2πLg$for g, assuming that the angle of deflection is less than 15 degrees. Recall that when the angle of deflection is less than 15 degrees, the pendulum is considered to be in simple harmonic motion, allowing us to use this equation.
Solution
1. Square$T=2πLg T=2πLg$and solve for g.
$g=4π2LT2g=4π2LT2$
2. Substitute known values into the new equation.
3. Calculate to find g.
Discussion
This method for determining g can be very accurate. This is why length and period are given to five digits in this example.
Worked Example
Hooke’s Law: How Stiff Are Car Springs?
What is the force constant for the suspension system of a car, like that shown in Figure 5.43, that settles 1.20 cm when an 80.0-kg person gets in?
Figure 5.43 A car in a parking lot. (exfordy, Flickr)
Strategy
Consider the car to be in its equilibrium position x = 0 before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position x = −1.20×10−2 m.
At that point, the springs supply a restoring force F equal to the person’s weight
w = mg = (80.0 kg)(9.80 m/s2) = 784 N. We take this force to be F in Hooke’s law.
Knowing F and x, we can then solve for the force constant k.
Solution
Solve Hooke’s law, F = −kx, for k.
$k=Fxk=Fx$
Substitute known values and solve for k.
Discussion
Note that F and x have opposite signs because they are in opposite directions—the restoring force is up, and the displacement is down. Also, note that the car would oscillate up and down when the person got in, if it were not for the shock absorbers. Bouncing cars are a sure sign of bad shock absorbers.
Practice Problems
A force of $70N$ applied to a spring causes it to be displaced by $0.3m$. What is the force constant of the spring?
1. $−233N/m$
2. $−21N/m$
3. $21N/m$
4. $233N/m$
What is the force constant for the suspension system of a car that settles $3.3cm$ when a $65kg$ person gets in?
1. $1.93×104N/m$
2. $1.97×103N/m$
3. $1.93×102N/m$
4. $1.97×101N/m$
Snap Lab
Finding Gravity Using a Simple Pendulum
Use a simple pendulum to find the acceleration due to gravity g in your home or classroom.
Materials:
• 1 string
• 1 stopwatch
• 1 small dense object
1. Cut a piece of a string or dental floss so that it is about 1 m long.
2. Attach a small object of high density to the end of the string (for example, a metal nut or a car key).
3. Starting at an angle of less than 10 degrees, allow the pendulum to swing and measure the pendulum’s period for 10 oscillations using a stopwatch.
4. Calculate g.
Grasp Check
How accurate is this measurement for g? How might it be improved?
1. Accuracy for value of g will increase with an increase in the mass of a dense object.
2. Accuracy for the value of g will increase with increase in the length of the pendulum.
3. The value of g will be more accurate if the angle of deflection is more than 15°.
4. The value of g will be more accurate if it maintains simple harmonic motion.
Exercise 12
What is deformation?
1. Deformation is the magnitude of the restoring force.
2. Deformation is the change in shape due to the application of force.
3. Deformation is the maximum force that can be applied on a spring.
4. Deformation is regaining the original shape upon the removal of an external force.
Exercise 13
According to Hooke’s law, what is deformation proportional to?
1. Force
2. Velocity
3. Displacement
4. Force constant
Exercise 14
What are oscillations?
1. Motion resulting in small displacements
2. Motion which repeats itself periodically
3. Periodic, repetitive motion between two points
4. motion that is the opposite to the direction of the restoring force
Exercise 15
True or False—Oscillations can occur without force.
1. True
2. False
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## Engage NY Eureka Math 7th Grade Module 6 Lesson 19 Answer Key
### Eureka Math Grade 7 Module 6 Lesson 19 Example Answer Key
Example 1.
If slices parallel to the tabletop (with height a whole number of units from the tabletop) were taken of this figure, then what would each slice look like?
Example 2.
If slices parallel to the tabletop were taken of this figure, then what would each slice look like?
Example 3.
Given the level slices in the figure, how many unit cubes are in the figure?
The number of unit cubes can be determined by counting the shaded squares in Levels 1–4.
Level 1: There are 18 cubes between Level 0 and Level 1.
Level 2: There are 13 cubes between Level 1 and Level 2.
Level 3: There are 9 cubes between Level 2 and Level 3.
Level 4: There are 5 cubes between Level 3 and Level 4.
The total number of cubes in the solid is 45.
### Eureka Math Grade 7 Module 6 Lesson 19 Exercise Answer Key
Exercise 1.
Based on the level slices you determined in Example 2, how many unit cubes are in the figure?
The number of unit cubes can be determined by counting the shaded squares in Levels 1–3.
Level 1: Five shaded squares; there are 5 cubes between Level 0 and Level 1.
Level 2: Two shaded squares; there are 2 cubes between Level 1 and Level 2.
Level 3: One shaded square; there is 1 cube between Level 2 and Level 3.
The total number of cubes in the solid is 8.
Exercise 2.
a. If slices parallel to the tabletop were taken of this figure, then what would each slice look like?
b. Given the level slices in the figure, how many unit cubes are in the figure?
a.
b. The number of unit cubes can be determined by counting the shaded squares in Levels 1–4.
Level 1: There are 12 cubes between Level 0 and Level 1.
Level 2: There are 7 cubes between Level 1 and Level 2.
Level 3: There are 6 cubes between Level 2 and Level 3.
Level 4: There are 3 cubes between Level 3 and Level 4.
The total number of cubes in the solid is 28.
Exercise 3.
Sketch your own three-dimensional figure made from cubes and the slices of your figure. Explain how the slices relate to the figure.
Responses will vary.
### Eureka Math Grade 7 Module 6 Lesson 19 Problem Set Answer Key
In the given three-dimensional figures, unit cubes are stacked exactly on top of each other on a tabletop. Each block is either visible or below a visible block.
Question 1.
a. The following three-dimensional figure is built on a tabletop. If slices parallel to the tabletop are taken of this figure, then what would each slice look like?
b. Given the level slices in the figure, how many cubes are in the figure?
a.
b. Given the level slices in the figure, how many cubes are in the figure?
The number of unit cubes can be determined by counting the shaded squares in Levels 1–3.
Level 1: There are 9 cubes between Level 0 and Level 1.
Level 2: There are 5 cubes between Level 1 and Level 2.
Level 3: There are 3 cubes between Level 2 and Level 3.
The total number of cubes in the solid is 17.
Question 2.
a. The following three-dimensional figure is built on a tabletop. If slices parallel to the tabletop are taken of this figure, then what would each slice look like?
b. Given the level slices in the figure, how many cubes are in the figure?
a.
b. The number of unit cubes can be determined by counting the shaded squares in Levels 1–4.
Level 1: There are 15 cubes between Level 0 and Level 1.
Level 2: There are 6 cubes between Level 1 and Level 2.
Level 3: There are 2 cubes between Level 2 and Level 3.
Level 4: There is 1 cube between Level 3 and Level 4.
The total number of cubes in the solid is 24.
Question 3.
a. The following three-dimensional figure is built on a tabletop. If slices parallel to the tabletop are taken of this figure, then what would each slice look like?
b. Given the level slices in the figure, how many cubes are in the figure?
a.
b. The number of unit cubes can be determined by counting the shaded squares in Levels 1–4.
Level 1: There are 6 cubes between Level 0 and Level 1.
Level 2: There are 3 cubes between Level 1 and Level 2.
Level 3: There is 1 cube between Level 2 and Level 3.
Level 4: There is 1 cube between Level 3 and Level 4.
The total number of cubes in the solid is 11.
Question 4.
John says that we should be including the Level 0 slice when mapping slices. Naya disagrees, saying it is correct to start counting cubes from the Level 1 slice. Who is right?
Naya is right because the Level 0 slice and Level 1 slice are the tops and bottoms of the same set of cubes; counting cubes in both slices would be double counting cubes.
Question 5.
Draw a three-dimensional figure made from cubes so that each successive layer farther away from the tabletop has one less cube than the layer below it. Use a minimum of three layers. Then draw the slices, and explain the connection between the two.
Responses will vary.
### Eureka Math Grade 7 Module 6 Lesson 19 Exit Ticket Answer Key
Question 1.
The following three-dimensional figure is built on a tabletop. If slices parallel to the tabletop are taken of this figure, then what would each slice look like?
|
# 1990 AIME Problems/Problem 2
## Problem
Find the value of $(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}$.
## Solution 1
Suppose that $52+6\sqrt{43}$ is in the form of $(a + b\sqrt{43})^2$. FOILing yields that $52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}$. This implies that $a$ and $b$ equal one of $\pm3, \pm1$. The possible sets are $(3,1)$ and $(-3,-1)$; the latter can be discarded since the square root must be positive. This means that $52 + 6\sqrt{43} = (\sqrt{43} + 3)^2$. Repeating this for $52-6\sqrt{43}$, the only feasible possibility is $(\sqrt{43} - 3)^2$.
Rewriting, we get $(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3$. Using the difference of cubes, we get that $[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]$ $= (6)(3 \cdot 43 + 9) = \boxed{828}$. Note: You can also just use the formula $(a + b)^2 = a^2 + 2ab + b^2$ instead of foiling.
## Solution 2
The $3/2$ power is quite irritating to work with so we look for a way to eliminate that. Notice that squaring the expression will accomplish that. Let $S$ be the sum of the given expression. $$S^2= ((52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2})^2$$ $$S^2 = (52+6\sqrt{43})^{3} + (52-6\sqrt{43})^{3} - 2((52+6\sqrt{43})(52-6\sqrt{43}))^{3/2}$$ After doing the arithmetic (note that the first two terms will have some cancellation and that the last term will simplify quickly using difference of squares), we arrive at $S^2 = 685584$ which gives $S=\boxed{828}$.
## Solution 3
Factor as a difference of cubes. $$\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[\left(\left(\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\right)^2+\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}}+\left(\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right)^2\right)\right] =$$ $$\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+\left(52^2-\left(36\right)\left(43\right)\right)^{\frac{1}{2}}\right] =$$ $$\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+34\right].$$ We can simplify the left factor as follows. $$\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x$$ $$104-2\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x^2$$ $$104-68 = x^2$$ $$36 = x^2.$$ Since $\left(52+6\sqrt{43}\right)^{\frac{1}{2}} > \left(52-6\sqrt{43}\right)^{\frac{1}{2}}$, we know that $x=6$, so our final answer is $(6)(138) = \boxed{828}$.
## Solution 4
Let $x=52+6\sqrt{43}$, $y=52-6\sqrt{43}$. Similarly to solution 2, we let $$S=x^{\frac{3}{2}}+y^{\frac{3}{2}}$$ \begin{align*} S^2&=(x^{\frac{3}{2}}+y^{\frac{3}{2}})^2\\ &=x^3+y^3+2x^{\frac{3}{2}}y^{\frac{3}{2}} \end{align*} The expression can be simplified as follow \begin{align*} S^2&=x^3+y^3+2x^{\frac{3}{2}}y^{\frac{3}{2}}\\ &=(x+y)(x^2-xy+y^2)+2(xy)^{\frac{3}{2}}\\ &=(x+y)((x+y)^2-xy)+2\sqrt{xy}^3\\ &=(x+y)((x+y)^2-\sqrt{xy}^2)+\sqrt{xy}^3\\ &=(x+y)(x+y+\sqrt{xy})(x+y-\sqrt{xy})+2\sqrt{xy}^3\\ &=104((104+34)(104-34)+2\cdot34^3\\ &=685584 \end{align*} Thus $S=\sqrt{685584}=\boxed{828}$.
~ Nafer
## Solution 5
(Similar to Solution 3, but with substitution)
Let $a=\sqrt{52+6\sqrt{43}}$ and $b=\sqrt{52-6\sqrt{43}}.$ We want to find $a^3-b^3=(a-b)(a^2+ab+b^2).$
We have $$a^2+b^2=102,\text{ and}$$ $$ab=\sqrt{(52+6\sqrt{43})(52-6\sqrt{43})}=\sqrt{1156}=34.$$ Then, $(a-b)^2=a^2+b^2-2ab=104-2\cdot 34= 36\implies a-b=6.$
Our answer is $$a^3-b^3=(a-b)(a^2+b^2+ab)=6\cdot 138=\boxed{828.}$$
## Solution 6
(Similar to Solution 1, but expanding the cubes instead)
Like in Solution 1, we have $\sqrt{52 + 6\sqrt{43}} = \sqrt{43} + 3$ and $\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3.$
Therefore we have that $(52 + 6\sqrt{43})^{3/2} - (52 + 6\sqrt{43})^{3/2}$ $= \sqrt{52 + 6\sqrt{43}}^3 - \sqrt{52 - 6\sqrt{43}}^3$ $= (\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3.$
From here, we use the formula $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$. Applying them to our problem we get that $(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3 = (27 + 27\sqrt{43} + 9 \cdot 43 + 43\sqrt{43}) - (-27 + 27\sqrt{43} - 9*43 + 43\sqrt{43}).$ We see that all the terms with square roots cancel, leaving us with $2 (27 + 9 \cdot 43) = 2 \cdot 414 = \boxed{828}.$
~Yiyj1
Note: We have that $\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3$ because we need the square root to be positive and $\sqrt{43} > 3$ since $43$ is obviously greater than $9.$ So we have $\sqrt{52 - 6\sqrt{43}} = \sqrt{43} - 3.$
|
## Intersecting Secants: Geometry
intersecting secants theorem
Hey math friends! In today’s post, we are going to go over the Intersecting Secants Theorem, specifically using it to find the piece of a missing length on a secant line. We are also going to see proof as to why this theorem works in the first place!
Just a warning: this blog post contains circles. This usually non-threatening shape can get intimidating when secants, chords, and tangents are involved. Luckily, this question is not too complicated and was also spotted on the NYS Regents. Before looking at the questions below, here is a review on different parts of a circle. Pay close attention to what a secant is, which is what we’ll be focusing on today:
Think you are ready? Let’s look at that next question!
What information do we already have? Based on the question we know:
*Extra Tip! Why does this formula work in the first place!?? If we draw lines creating and proving triangle RTQ and triangle RPS are similar by AA, this leads us to know that the two triangles have proportionate sides and can follow our formula! ___________________________________________________________________________________
Still got questions? Let me know in the comments and remember having questions is a good thing!
If you’re looking for more on intersecting secants, check out this post here for practice questions and more!
## Rational Exponents: Algebra 2/Trig.
Hi everyone and welcome to MathSux! In this post we are going to break down and solve rational exponents. The words may sound like a mouthful, but all rational exponents are, are fractions as exponents. So instead of having x raised to the second power, such as x2, we might have x raised to the one-half power, such as x(1/2). Let’s try an example taken straight from the NYS Regents below. Also, if you have any questions don’t hesitate to comment below or check out the video posted here. Happy calculating! 🙂
How do I answer this question?
The questions want us to simplify the rational exponents into something we can understand.
How do we do this?
We are going to convert the insane looking rational exponents into radical and solve/see if we can simplify further.
Reminder!
A radical can be converted into a rational exponent and vice versa. Not sure what that means? It’s ok! Take a gander at the examples below and look for a pattern:
Think you’re ready to take on our original problem? #Letsdothis
## Solutions:
Still got questions? Don’t hesitate to comment below for anything that still isn’t clear! Looking to review how to solve radical equations? Check out this post here! 🙂
Also, don’t forget to follow MathSux fopr FREE math videos, lessons, practice questions and more every week!
## The Magic of the “Golden Ratio”
Walking around NYC, I was on a mission to connect mathematics to the real world. This, of course, led me to go on a mathematical scavenger hunt in search of the “Golden Ratio.” Hidden in plain sight, this often times naturally occurring ratio is seen everywhere from historic and modern architecture to nature itself.
What is this all-encompassing “Golden Ratio” you may ask?
It’s a proportion, related to a never-ending sequence of numbers called the Fibonacci sequence, and is considered to be the most pleasing ratio to the human eye. The ratio itself is an irrational number equal to 1.618……..(etc.).
Why should you care?
When the same ratio is seen in the Parthenon, the Taj Mahal, the Mona Lisa and on the shores of a beach in a seashell, you know it must be something special!
Random as it may seem, this proportion stems from the following sequence of numbers, known as the Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, …….
Do you notice what pattern these numbers form?
(Answer: Each previous two numbers are added together to find the next number.)
The Golden RectangleThe most common example of the “Golden Ratio” can be seen in the Golden rectangle. The lengths of this rectangle are in the proportion from 1: 1.618 following the golden ratio. Behold the beauty of the Golden Rectangle:
How is the Fibonacci Sequence related to the Golden Ratio? What if we drew a golden rectangle within our rectangle?
Then drew another golden rectangle within that golden rectangle?
And we kept doing this until we could no longer see what we were doing…….
The proportion between the width and height of these rectangles is 1.618 and can also be shown as the proportion between any two numbers in the Fibonacci sequence as the sequence approaches infinity. Notice that the area of each rectangle in the Fibonacci sequence is represented below in increasing size:
Where exactly can you find this Golden Ratio in real life? Found in NYC! The Golden ratio was seen here at the United Nations Secretariat building in the form of a golden rectangle(s). Check it out!
Where have you seen this proportion of magical magnitude? Look for it in your own city or town and let me know what you find! Happy Golden Ratio hunting! 🙂
If you’re interested in learning more about the golden ratio and are also a big Disney fan, I highly recommend you check out Donald Duck’s Math Magic!
Don’t forget to connect with MathSux on these great sites!
## How to Factor Quadratic Equations: Algebra
In this post, we are going to dive deep into how to factor Quadratic equations! There are so many different methods to choose from including GCF, Product/Sum, DOTS, and the Quadratic Formula. Here we will go step by step into each method on how to factor quadratic equations, each with their own set of practice questions. For a review on how to factor by grouping, check out this post here and happy calculating! 🙂
Why factor in the first place, you may say? We want to manipulate the equation until we solve for x. Solving for x is our main goal, and factoring allows us to do that. Now let’s get to the good stuff!
## Greatest Common Factor (GCF):
The greatest common factor is the highest possible number that can be divided out from an equation. This gets the equation into its simplest form and makes it easier for us to solve for x.
Before considering which type of factoring methdo to use, always ask yourself, “Can I take out a GCF?”
## Product/Sum:
This factoring method is for quadratic equations only! That means the equation takes on the following form:
## Difference of Two Squares DOTS)
Not to play favorites or anything, but DOTS is the easiest and most lovable of the factoring methods. This factoring method just makes you feel all warm and fuzzy inside or maybe that’s just me). Before we get into how to do DOTS, let’s talk about when?
We have heard of the quadratic equations, so how id the quadratic formula different?
The Answer: The Quadratic Formula is what we use to factor any trinomial. You can use product/sum on trinomials like we discussed earlier, but this may not always work out easy. The Quadratic Formila on the other hand will work every time!
Low and behold, the Quadratic Formula:
Want more Mathsux? Don’t forget to check out our Youtube channel and more below! And if you have any questions, please don’t hesitate to comment below. Happy Calculating! 🙂
Looking for more on Quadratic Equations and functions? Check out the following Related posts!
Factoring
Factor by Grouping
Completing the Square
The Discriminant
Is it a Function?
Quadratic Equations with 2 Imaginary Solutions
Imaginary and Complex Numbers
Focus and Directrix of a Parabola
## ‘Math Suks’ ~ Jimmy Buffett
Need to relax after studying? Close to wanting to “burn your textbook”? This super chill and somewhat cheesy song may help! Coming to you from Jimmy Buffett, the singer has a relaxed, island/ska like vibe to his music. The lyrics come close to sounding like an after school special until the words “Math Sucks” are sung over and over again. Enjoy! 🙂
‘Math Suks’ by Jimmy Buffet Lyrics:
If necessity is the mother of invention
Then I’d like to kill the guy who invented this
The numbers come together in some kind of a third dimension
A regular algebraic bliss.
Any two plus two will never get you five.
There are fractions in my subtraction and x don’t equal y
But my homework is bound to multiply.
Math suks math suks
I’d like to burn this textbook, I hate this stuff so much.
Math suks math suks
Sometimes I think that I don’t know that much
But math suks.
I got so bored with my homework, I turned on the TV.
The beauty contest winners were all smiling through their teeth.
Then they asked the new Miss America
Hey babe can you add up all those bucks?
She looked puzzled, then just said
“Math Suks”.
Math suks math suks
You don’t even have to spell it,
All you have to do is yell it…
Math suks math suks
Sometime times I think that I don’t know that much
But math suks.
Geometry, trigonometry and if that don’t tax your brain
There are numbers to big to be named
Numerical precision is a science with a mission
And I think it’s gonna drive me insane.
Parents fighting with their children, and the Congress can’t agree
Teachers and their students are all jousting constantly.
Management and labor keep rattling old sabers
Quacking like those Peabody ducks.
Math suks math suks
You don’t even have to spell it,
All you have to do is yell it…
Math suks math suks
Sometime times I think that I don’t know that much
But math suks
## Why Must We Complete the Square?: Algebra
Completing the Square: So many steps, such little time. It sounds like it involves a square or maybe this is a geometry problem? Why am I doing this again? Why must we complete the square in the first place?
These are all the thoughts that cross our minds when first learning how to complete the square. Well, I’m here to tell you there is a reason for all those steps and they aren’t that bad if you really break them down, let’s take a look!
Explanation:
I’m not going to lie to you here, there are a lot of seemingly meaningless steps to completing the square. The truth is though (as shocking as it may be), is that they are not meaningless, they do form a pattern, and that there is a reason! Before we dive into why let’s look at how to solve this step by step:
Feeling accomplished yet!? Confused? All normal feelings. There are many steps to this process so go back and review, practice, and pay close attention to where things get fuzzy.
But the big question is why are we doing these steps in the first place? Why does this work out, to begin with?
For those of you who are curious, continue to read below!
Want more Mathsux? Don’t forget to check out our Youtube channel and more below! And if you have any questions, please don’t hesitate to comment below. Happy Calculating!
Need more of an explanation? Check out why we complete the square in the first place here !
## How to Factor Trig Functions
Howdy math friends! In this post, we are going to learn how to factor trig functions algebraically. This will combine our knowledge of algebra and trigonometry into one beautiful type of question! For more on trigonometric equations and the unit circle, check out this post here.
Solving trigonometric equations is very similar to solving the ho-hum everyday equations you solved all the way back in algebra 1, but this time we are solving for an angle value. If you need to refresh yourself on how to find x the old fashioned way via factoring be sure to check out this post here!
Solving trigonometric equations, sounds complicated? Well, you are correct, that does sound complicated. Is it complicated? Hopefully, you won’t find it that way after you’ve seen an example. We are going to find the value of x (value of the unknown angle) step by step in the following question:
## How do I answer this question?
The questions want us to solve for x, the unknown value of the angle.
Step 1: Our first step for solving trigonometric equations is to pretend this is any other algebraic equation and we want to solve for x (Note: this is the key to solving any trig equation). Thinking this way, allows us to move radical 2 to the other side, by adding it to both sides of the equation.
Step 2: Remember that secant is the inverse trigonometric function of cosine? Therefore we know that, secx is equal to 1/cosx and we can re-write the equation below, making the trigonometric function look a lot less complicated already!
Step 3: Now we need a value for x. This is where I turn to my handy dandy special triangles. There are two main special triangles in trignometry we can use to solve our equation, the 45 45 90 triangle and the 30 60 90 triangle. In this case, we are going to use the 45 45 90 special triangle, because its proportions include a value of radical 2, which is exactly what our equation has!
*Note- In this case we can use our special triangles, but in other examples, we may also need to use our knowledge of trig identities to re-write the original trigonometric function.
Below we can see the proportions that the 45 45 90 special triangle brings us. When we apply the basic rules of trigonometry and SOH CAH TOA, we see that we can find the value of our missing angle x, which is 45º.
*Another way we could have gotten 45º without using special triangles is by plugging in cos-1 (1/rad2) into our calculator, this would also give us the solution of 45º!
Step 4: We have a value for cos45º=1/rad 2, but remember we need its inverse and must flip the fraction, to find the value of our angle of the inverse of cosine, secx, which flipped, is going to be 45º.
***But wait! We are not done yet! We have found only one part to our solution, x=45º, which is great! But we need to ask ourselves, are there any other angle measures that could potentially give is the right solution as well!? And thats where we need to consult the unit circle!
Step 5: Using the unit circle below, we can identify what other values can work for our original trigonometric function. Based on the unit circle below, we can identify which trig functions (sin, cos, tan) are positive in which quadrants, positive and negative angles (positive counterclockwise, negative clockwise).
Notice that cosine is positive in Quadrants I and IV. That means there are two values that x can be (one in each quadrant). We already have x=45º from Quadrant i. In order to get the other value in Quadrant IV, we must subtract 360º-45º=315º using our reference angle (360º-θ) and giving us our other value for x, 315º, completing our answer!
The above example is similar to a linear equation, let’s look at another type of trigonometric equation that is more similar to quadratic equations.
## Factoring Trig Functions Example: Quadratic Equation
Step 1: First let’s move everything onto one side of the equation. We can start by multiplying tanθ to both sides and distributing, multiplying (2tanθ-7)(tanθ). Then we can subtract the 4 from both sides. Remember to solve trigonometric equations, we always want to treat them like a normal algebraic function.
Step 2: Now that we have something resembling a quadratic equation, we have something to work with! There are two main ways to solve for our unknown angle via factoring for this example, (1) The Quadratic Formula, or, (2) Factor by Grouping. In this case, we are going to be using factor by grouping to find the value of our angle, and the solution to our problem, but the quadratic formula should work as well!
Factor by Grouping: First find the product, sum and factors of the equation.
Now that we have our factors, we can re-write our equation, converting -7tanθ into -8tanθ + tanθ.
Next, we split the equation down the middle and take the GCF of each side.
Finally, we can combine the values outside our parenthesis 2tanθ+1 and the matching values found within our parenthesis and tanθ-4.
Step 3: Now we can use our equation and solve for each angle value, by setting up each group found in the parenthesis equal to zero, just like we would when finding the value of x in a normal algebraic equation.
Based on finding our two angles for θ, -27º and 76º, we are able to find the angles that apply to our specific trigonometric equation listed below, based on where tanθ is positive and negative in each quadrant of the unit circle. In this case, the angle value of -27º, we look for the angle values where tangent is negative which are in quadrant ii and quadrant IV, and apply our reference angles. On the other hand, the angle value of 76º, is positive, therefore, we look for angles where tangent is positive, in quadrant i and quadrant iii, and apply our reference angles to get the correct value of each angle.
*Note! In the above two examples, we look at a cosine function and a tangent function, but the same exact methods apply to a sine function!
Think you are ready to solve trigonometric equations on your own? Check out the practice questions and solution to each below!
## Practice Questions:
Find the nearest degree to all values of θ or x in the interval 0º < θ < 360º and 0º < x < 360º:
When you are ready check each solution below!
## Solution:
Does the solution to each trig function make sense? Great! 🙂 Do you think it is clear as mud to solve trigonometric equations? Alas, I have failed. But I have not given up (and neither should you). Ask more questions, look for the spots where you got lost, do more research and never give up! 🙂
Hopefully, you enjoyed my short motivational speech. For more encouraging words and math, check out MathSux on the following websites! Sign up for FREE math videos, lessons, practice questions, and more. Thanks for stopping by and happy calculating! 🙂
## Related Posts on Trignometry:
Trigonometric Identities
Unit Circle
45 45 90 Special Triangles
30 60 90 Special Triangles
SOH CAH TOA and Right Triangle Trig
## PieceWise Functions NYS Regents: Algebra
Greetings math friends, students, and teachers I come in peace to review this piecewise functions NYS Regents question. Are they pieces of functions? Yes. Are they wise? Ah, yeah sure, why not? Let’s check out this piecewise functions NYS Regents question below and happy calculating! 🙂
1 value satisfies the equation because there is only one point on the graph where f(x) and g(x) meet.
Does the above madness make sense to you? Great!
Need more of an explanation? Keep going! There is a way to understand the above mess.
Looking at this piece-wise function: We want to graph the function 2x+1 but only when the x-values are less than or equal to negative 1.
We also want to graph 2-x^2 but only when x is greater than the negative one.
One way to organize graphing each piece of a piecewise function is by making a chart.
1. Lets start by making a chart for the first part of our function 2x+1:
Is it all coming back to you now? Need more practice on piece-wise functions?Check out this link here and happy calculating! 🙂
Also, if you’re looking to nourish your mind or you know, procrastinate a bit check out and follow MathSux on these websites!
## Top 5 Favorite Math Websites
As we all love technology (I’m assuming since it has brought you here), I figured I should share my top favorite math sites. These are great resources for learning things you don’t understand, practicing for an upcoming test or if you’re like me, it’s great for re-learning/teaching purposes. Yes, even I forget how to do things and these websites are a great help. I know the anticipation is getting to you so I’ll just get to the good stuff now. Here are my top 5 favorite math websites:
1. Khan Academy : This is a non profit over in Silicon Valley. They have a great range of subjects (k-college level) and explain everything via easy to follow videos.
2. Regent’s Prep: The title is pretty self explanatory, but this helps with reviewing for the New York State Regent’s. This site has a lot of great examples and answers.
3. MathSux: So fabulous, no explanation is required. 😀
4. Knewton: Pretty cool things are happening at this web address. You can make an account, answer some questions and the questions will adjust according to your answers figuring out what you need help in.
5. Kahoot: Fast and easy quizzes that are good for an entire classroom or for studying alone. Either way there are lots of bright and happy colors that make you forget you’re studying anything in the first place.
So those are the goods, the contraband, the info you can’t find anywhere else (all lies). Hope you find these websites useful. Let me know if you end up using any of them. 🙂
## Pre-Calc: The Factor Theorem
So what do we think of the factor theorem? Not as intimidating as you thought, you probably have been using it for years without realizing it.
For more math fun check me out on Twitter and Facebook. Until next time, keep it real and happy mathing! 🙂
For related posts on factoring methods that apply the factor theorem, check the following lessons out below:
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# Area of Sector and Segment of a Circle
Reviewed by:
Last updated date: 07th Aug 2024
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Views today: 3.33k
## Introduction to Sectors and Segments
The region between two radii of a circle and any of the arcs between them is called a sector. The sector always starts from the center of the circle. The semi-circle is also called the sector of the circle. The space or the area occupied by the sector of a circle is called the area of a sector of a circle. Geometrically, a line segment is a part of a line that is bounded by two distinct end points, and contains every point on the line between its endpoints.
## Area of Sector Definition
A sector of a circle is identified as the reconstructed part of the circle bounded by two radii and the arc which connects them. The space which is occupied by the sector of a circle is called the area of a sector of a circle.
Two types of sectors are major sectors and minor sectors.
• A major sector is defined as a sector that is greater than a semicircle.
• A minor sector is defined as a sector that is less than a semi-circle.
As the below image represents the minor and major sectors. Hence, the shaded part OAPB is the area of the minor sector, whereas the unshaded part OAQB is the area of the major sector of the original circle.
Area of a Sector of a Circle
## Real-Life Example for Sector of a Circle
The most common real-time example of a sector is a slice of pizza. The shape of the slices of pizza is similar to a sector of a circle.
The pizza of 7 inches is equally divided into 6 equal-sized slices, with each at an angle of 60 degrees, as shown in the below image.
Example for the Sector.
In this image, the pizza slice is a representation of the sector, and these 6 slices are the 6 equal sectors.
## The Area of the Sector of a Circle Formula
This formula can be used to calculate the total space covered by a part of a circle. The area can be calculated as the total space covered by a circle. The area can be calculated in two different ways with respect to the unit of an angle given.
This is the formula for calculating the area of a sector of a circle.
• $\text{Area of sector of a circle =}\frac{\theta }{{{360}^{\circ }}}\times \pi {{r}^{2}}$
• Where r is the radius of the circle and θ is the angle subtended at the center.
## A Segment of a Circle Definition
• A segment in a circle is an area that covers the chord and an arc of the circle.
• The chord is a line segment that joins any two points on the circumference of the circle.
• An arc is a portion of a circle or a segment of the perimeter of the circle.
Segment of a Circle.
There are two types of segments.
1. Major segment
2. Minor segment
A major segment is made from the major arc. A minor segment is made from the minor arc.
## Area of a Segment of a Circle
The area of a segment of a circle is correspondingly equal to the area of the sector subtracted by the area of the triangle.
## The Area of a Segment of the Circle Formula
In the below figure, a triangle OPQ is given considering it as a minor segment made by the chord PQ of the circular with a given radius r. As we know from trigonometry the area of triangle OPQ is $\frac{1}{2}\times {{r}^{2}}\times \sin \theta$ .
Area of a Segment.
The area of the sector OPQ is given by
1. $Area=\frac{\theta }{360}\times \pi {{r}^{2}}$ if θ is in degrees.
2. $Area=\frac{1}{2}\times {{r}^{2}}\times \theta$ if θ is in radians.
Therefore, to calculate the formulas, the area of the minor segment of the circle is :
1. When θ is in degree ,
$\text{Area of sector= }\frac{\theta }{360}\pi {{r}^{2}}$
$\text{ Area of segment = }\frac{\theta }{360}\pi {{r}^{2}}-\frac{1}{2}\text{ }{{\text{r}}^{2}}\text{ sin}\theta \text{ }$
$\text{=}{{r}^{2}}\left( \frac{\theta }{360}\pi -\frac{1}{2}\sin \theta \right)$
2. When θ is given radians,
$\text{Area of sector=}\frac{1}{2}{{r}^{2}}\theta$
$\text{Area of segment =}\frac{1}{2}{{r}^{2}}\theta -\frac{1}{2}{{r}^{2}}\sin \theta$
$=\frac{1}{2}{{r}^{2}}\left( \theta -\sin \theta \right)$
## Theorems on Segments of a Circle
Basically, there are two theorems that are based on the segment of a circle.
• ### Angles in the same Segment Theorem
The first theorem states that the angle formed in the same circle segment is always equal.
• Alternate Segment Theorem
The second theorem of segments states that the angle formed by the tangent and chord at the point of contact equals the angle formed in the alternate segments on the circle circumference at the chord endpoints.
## Solved Examples
1. RQRQ is a chord of a circle that subtends an angle of 80∘ at the centre of a circle, and the diameter of the circle is 15cm. Calculate the area of the minor sector of this circle.
Solution: The diameter of the circle =15 cm.
The radius of the circle = $15\div 2=7.5cm$
Using the formula of the area of a sector of the circle
$\text{Area of minor sector=}\frac{\theta }{360}\times \pi {{r}^{2}}$
$=\frac{80}{360}\times \frac{22}{7}\times {{\left( 7.5 \right)}^{2}}$
$=39.25c{{m}^{2}}$
Therefore, the area of the minor sector is 39.25cm2.
2. Calculate the area of the major segment of a circle if the area of its minor segment is 62 sq. units and the radius is 7 units. Use π= 22/7.
Solution: Using the relation,
area of the major segment = area of the circle − the area of the minor segment
$\text{Area of major segment =}\pi {{r}^{2}}-62$
$\text{=}\frac{22}{7}{{\left( 7 \right)}^{2}}-62$
$=154-62$
$=92sq.unit.$
Therefore, the area of the major segment is 92 sq. units.
## Summary
In this article, we discussed the sector and segment. The relevant problems in sector and segment are taken as example problems to solve sector and segment, which include the area of a segment of a circle and the area of the sector of a circle and their respective formulas.
## FAQs on Area of Sector and Segment of a Circle
1. What are a circle, semi-circle, and circle segment?
The definition of a circle is a closed, two-dimensional object in which the boundary's points are all equally spaced apart from the middle. A semicircle is outlined as a half circle shaped by cutting the circle into 2 halves. It's formed once a line passes through the center and touches the 2 ends of the circle. A region that is formed by a secant or chord and the corresponding arc of the circle is referred to as a segment of a circle. The major section is the one that shows a greater region, and the minor segment is the one that shows a smaller area.
2. Given the following triangle with sides equal to 50 cm, 35 cm, and 40 cm. We want to find the area of the circle inscribed in this triangle.
As we know, the area of a circle $=\pi {{r}^{2}}$. The radius is not given to us, We need to find the value for the radius first.
Area of triangle = S× r, here, S is the semi-perimeter of the triangle and r is the radius of the circle.
S=\frac{50+35+40}{2}=62.5
Area of triangle, (by Heron’s Formula)
$\text{Area of triangle =}\sqrt{\text{S}\left( S-50 \right)\left( S-35 \right)\left( S-40 \right)}$
$=\sqrt{62.5\left( 62.5-50 \right)\left( 62.5-35 \right)\left( 62.5-40 \right)}$
$=\sqrt{62.5\times 12.5\times 27.5\times 22.5}c{{m}^{2}}$
Comparing both the formulas for the area of a triangle, we get:
$r=\frac{\sqrt{62.5\times 12.5\times 27.5\times 22.5}}{62.5}=11.124cm.$
So, the radius ( r ) for a circle is 11.124 cm.
Area of circle =$\pi \times {{\left( 11.124 \right)}^{2}}=\pi 123.75c{{m}^{2}}.$
3. Does arc length need to be in radians?
Arc length is a measurement of distance, so it cannot be in radians. The central angle, however, does not have to be in radians. It can exist in any unit for angles you like, from degrees to arcsecs. Radians possess the following advantages: They're dimensionless, which means that they can be managed just as figures.
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# Integration Formulas
```IMPORTANT FORMULAS IN INTEGRATION
Introduction and Definition of Integration
Consider a function π(π₯). When we differentiate it with respect to π₯, we get:
π
ππ₯
[π (π₯ )] = π ′ (π₯) ; where π ′ (π₯) is the derivative of π(π₯ ).
Now, when we integrate π ′ (π₯), i.e.
∫ π ′ (π₯) ππ₯ = π (π₯ ) + π; where π (π₯ ) is called the integral of π ′ (π₯) and c is the
constant of integration. The operator ππ₯ indicates that we are integrating w.r.t. π₯.
We need the constant of integration because the integral of any term is not definite
i.e. the constant lost in differentiation will not be recovered on integration.
For example, if we differentiate a function π(π₯ ) = 3π₯ 2 + 5π₯ + 7, we get π ′ (π₯ ) =
6π₯ + 5. Now, while integrating π ′ (π₯), we get ∫ π ′ (π₯) ππ₯ = ∫(6π₯ + 5)ππ₯ which will
be equal to 3π₯ 2 + 5π₯. In other words, the constant ‘7’ which is lost during the
differentiation process is not retrieved in the integration process. To compensate this,
we add an arbitrary constant c known as constant of integration.
The process of determining an integral of a function is called integration and the
function to be integrated is called the integrand.
Since the integral of any function is not completely defined, it is known as Indefinite
Integration.
Therefore, we see that integration is the reverse process of differentiation.
Basic Integrals of Algebraic Functions
1.
2.
π
∫ π₯ ππ₯ =
π₯ π+1
π+1
π
+ π; π ≠ −1
∫(ππ₯ + π) ππ₯ =
(ππ₯+π)π+1
π.(π+1)
+ π; π ≠ −1
Integrals of a Combination of Functions
Theorem 1: The integral of the product of a constant and a function is equal to the
product of the constant and the integral of the function.
∫ π. π (π₯ )ππ₯ = π ∫ π (π₯ )ππ₯
Theorem 2: The integral of a sum or difference of two functions is equal to the sum
or difference of their integrals.
∫[π(π₯ ) ± π(π₯ )]ππ₯ = ∫ π (π₯ )ππ₯ ± ∫ π(π₯ )ππ₯
Integrals of Trigonometric Functions
1.
∫ sin π₯ ππ₯ = − cos π₯ + π
2.
∫ cos π₯ ππ₯ = sin π₯ + π
3.
∫ sec 2 π₯ ππ₯ = tan π₯ + π
4.
∫ cosec 2 π₯ ππ₯ = − cot π₯ + π
5.
∫ sec π₯ tan π₯ ππ₯ = sec π₯ + π
6.
∫ πππ ππ π₯ cot π₯ ππ₯ = −πππ ππ π₯ + π
7.
∫ tan π₯ ππ₯ = log|sec π₯ | + π
8.
∫ cot π₯ ππ₯ = log|sin π₯ | + π
9.
∫ sec π₯ ππ₯ = log|sec π₯ + tan π₯ | + π = log |tan ( 4 + 2)| + π
π
π₯
π₯
10. ∫ cosec π₯ ππ₯ = log|cosec π₯ − cot π₯ | + π = log |tan | + π
2
Formulae and Identities of Trigonometrical Functions
The following identities and relations can be used to simplify the integrand when it contains higher powers of
sin x, cos x and other t-functions and hence bring it into any of the standard forms:
1.
Fundamental Identities
(a) Square Relations: 1) ππππ π½ + ππππ π½ = π ⇒ ππππ π½ = π − ππππ π½ ⇒ ππππ π½ = π − ππππ π½
2) π + ππππ π½ = ππππ π½ ⇒ ππππ π½ = ππππ π½ − π ⇒ ππππ π½ − ππππ π½ = π
3) π + ππππ π½ = ππππππ π½ ⇒ ππππ π½ = ππππππ π½ − π ⇒ ππππππ π½ − ππππ π½ = π
(b)Basic Relations: 1)
πππ π
ππππ π
ππππ
3)
2.
πππ π
= πππ π πππ π
= πππ π
4)
2)
ππππ
πππππ π
Converting products into sums or differences
Using Corollaries from Multiple Angle Forms
(i) ππππ π¨ =
π−πππ ππ¨
(ii) ππππ π¨ =
π+πππ ππ¨
π
π
π−πππ ππ¨
(iii) π+πππ ππ¨ = ππππ π¨ = ππππ π¨ − π
(iv) √π ± πππ ππ¨ = πππ π¨ ± πππ π¨
(v)
π− πππ π¨
(vi)
πππ π¨
π+ πππ π¨
πππ π¨
=
π ππππ
π¨
π
π¨
π
π πππ πππ
=
π¨
π
π ππππ
π¨
π
π¨
π¨
π πππ πππ
π
π
π
π
π¨
π
π¨
πππ
π
πππ
=
=
= πππ
π¨
π
π¨
πππ
π
πππ
π¨
(vii) ) π+πππ ππ¨ = π ππππ π
π
π
π¨
(viii) π−πππ ππ¨ = π ππππππ π
π
(ix) ππππ π¨ = π [ππππ π¨ − πππ ππ¨]
π
(x) ππππ π¨ = π [πππ ππ¨ + π πππ π¨]
π¨
π
= πππ
π¨
π
= πππππ π πππ π
= πππ π
3.
πππ π
ππππ π
π
Integrals of ,
π
π ππ+π
, ππ , ππ πππ
πππ+π
1
1.
∫ π₯ ππ₯ = log|π₯ | + π
2.
∫ ππ₯+π ππ₯ = a log|ππ₯ + π| + π
3.
∫ π π₯ ππ₯ = π π₯ + π
4.
∫ π π₯ ππ₯ = log π + π
5.
∫ π ππ₯+π ππ₯ =
1
1
ππ₯
π ππ₯+π
π
+π
Integrals of Inverse Trigonometric Functions
1. ∫
2. ∫
3. ∫
4. ∫
5. ∫
6. ∫
1
√1−π₯ 2
1
1+π₯ 2
ππ₯ = sin−1 π₯ + π = − cos −1 π₯ + π
ππ₯ = tan−1 π₯ + π = − cot −1 π₯ + π
1
|π₯|√π₯ 2 −1
1
ππ₯ = sec −1 π₯ + π = − cosec −1 π₯ + π
π₯
√π2 −π₯
1
π2 +π₯
ππ₯ = sin−1 + π
2
π
1
π₯
a
π
−1
+π
2 ππ₯ = tan
1
|π₯|√π₯ 2 −π
1
π₯
a
π
ππ₯ = sec −1 + π
2
Note: Learn all properties of inverse trigonometric functions very well.
STANDARD METHODS OF INTEGARTION
When the given function cannot be integrated directly by using standard formulae, we
try other methods of integration. The process of integration is largely of tentative
nature and no systematic procedure can be given as in differentiation. However the
following are two important methods of integration:
(i) Integration by substitution
(ii) Integration by parts
Integration by Substitution
Consider the following complicated integrand. We notice that the given integrand
cannot be integrated directly by using any of the known standard forms. So, we use
the method of substitution to change the variable of integration so that the integrand
becomes simple enough to be integrated by applying a known standard formula.
∫[π(π₯ )]π π ′ (π₯ )ππ₯ ; πΏππ‘ π (π₯ ) = π‘
π ′ (π₯ )ππ₯ = ππ‘
The integrand now takes on the form,
∫ π‘ π ππ‘ =
π‘ π+1
π+1
+π =
[π(π₯)]π+1
π+1
+π
Note: There are no hard and fast rules for making suitable substitution. Practice will
help acquire the necessary skill. However, the following guidelines will be found
useful:
(i)
If the integrand contains
√1 − π₯ 2 ; π π’ππ π‘ππ‘π’π‘π π₯ = sin π ππ π₯ = cos π
√π2 − π₯ 2 ; π π’ππ π‘ππ‘π’π‘π π₯ = asin π ππ π₯ = acos π
(ii)
If the integrand contains
(1 + π₯ 2 ) ππ √1 + π₯ 2 ; π π’ππ π‘ππ‘π’π‘π π₯ = tan π ππ π₯ = cot π
(π2 + π₯ 2 ) ππ √π2 + π₯ 2 ; π π’ππ π‘ππ‘π’π‘π π₯ = atan π ππ π₯ = acot π
(iii) If the integrand contains
√π₯ 2 − 1 ; π π’ππ π‘ππ‘π’π‘π π₯ = sec π ππ π₯ = cosec π
√π₯ 2 − π2 ; π π’ππ π‘ππ‘π’π‘π π₯ = asec π ππ π₯ = acosec π
(iv)
If the integrand contains
√1 ± π₯ ; π π’ππ π‘ππ‘π’π‘π π₯ = sin 2π ππ π₯ = cos 2π
ππ π π’ππ π‘ππ‘π’π‘π (1 ± π₯ ) = π‘ 2 ππ √1 ± π₯ = π‘
√π ± π₯ ; π π’ππ π‘ππ‘π’π‘π π₯ = asin 2π ππ π₯ = acos 2π
ππ π π’ππ π‘ππ‘π’π‘π (π ± π₯ ) = π‘ 2 ππ √π ± π₯ = π‘
(v)
If the integrand contains
√π−π₯
; π π’ππ π‘ππ‘π’π‘π
√π+π₯
(vi)
π₯ = acos 2π
If the integrand contains some linear expression of x, put that value= π‘.
(vii) ∫
cos π₯−sin π₯
cos π₯+sin π₯
ππ₯ ; ππ’π‘ (cos π₯ + sin π₯) = π‘ ⇒ (cos π₯ − sin π₯)ππ₯ = ππ‘
π π₯ −π −π₯
(viii) ∫ π₯ −π₯ ππ₯ ; ππ’π‘ (π π₯ + π −π₯ ) = π‘ ⇒ (π π₯ − π −π₯ )ππ₯ = ππ‘
π +π
(ix) If the integrand contains √π₯ or similar terms, we can put √π₯ = π‘ so that
ππ₯ = 2π‘ππ‘
Some Important Formulas
1. ∫[π(π₯
2. ∫
)]π ′ (
π ′ (π₯)
π(π₯)
π π₯ )ππ₯ (π ≠ −1) =
[π(π₯)]π+1
π+1
+π
ππ₯ = log π (π₯ ) + π
Note: Do not directly use the above two formulae. Make the necessary substitution,
integrate in terms of the new variable and finally re-substitute back the answer in
terms of the starting variable.
Integration by Parts
If π and π are two functions of π, then
π
π
∫ ππ π
π = π ∫ π π
π − ∫ [ π
π . ∫ π π
π] π
π
i.e.
π°πππππππ ππ πππ ππππ
πππ ππ πππ πππππππππ = πππ ππππππππ ×
π°πππππππ ππ πππ
− π°πππππππ ππ [π«πππππππππππ ππ πππ × π°πππππππ ππ πππ
]
If the two functions are of different types, take the first function to be the function
which comes first in the word ‘ILATE’ where
I – Inverse Trigonometric Functions
L – Logarithmic Functions
A – Algebraic Functions
T – Trigonometric Functions
E – Exponential Functions
Note:
ο The rule of integration by parts is useful in integrating single functions like
logarithmic and inverse t-functions. For this purpose, unity (i.e. 1) is taken as the
second function.
ο Here also, we see that ILATE Rule is not completely foolproof or hard and fast.
However it will found to be useful in most cases. Sometimes, not following ILATE
Rule is found to be simpler.
ο In many cases, substitution precedes the use of the by-parts formula.
Integrals of the form ∫ πππ πππ (ππ + π) π
π or ∫ πππ πππ (ππ + π) π
π
In such cases, we first put πΌ = ∫ π ππ₯ π ππ (ππ₯ + π) ππ₯ or πΌ = ∫ π ππ₯ πππ (ππ₯ + π) ππ₯.
Then we integrate the given function by parts taking πππ as first function and
πππ (ππ + π) or πππ (ππ + π) as the second function. We then notice that on
integrating by parts the second time, the integrand contains a term of πΌ. By taking all
the πΌ terms to one side, we get the value of πΌ, which is the necessary integral.
Integrals of the form ∫ ππ [π(π) + π′ (π)] π
π
Method:
1. First split the given integral into two parts as follows:
∫ π π₯ [π(π₯ ) + π ′ (π₯ )] ππ₯ = ∫ π π₯ π(π₯ ) ππ₯ + ∫ π π₯ π ′ (π₯ ) ππ₯
2. Integrate only the first integral i.e. ∫ π π₯ π(π₯ ) ππ₯ by parts taking π(π₯ ) as first
function and π π₯ as second function.
3. Now, the integrand becomes π(π₯ )π π₯ − ∫ π π₯ π ′ (π₯ ) ππ₯ + ∫ π π₯ π ′ (π₯ ) ππ₯ . The
last two integrals will cancel each other.
Integrating a Complex Fraction by Decomposing into Partial Fractions
There are various methods of decomposing a given proper rational fraction into a
group of simple fractions each having as its denominator one of the factors of the
denominator of the original rational fraction. These simple fractions are called partial
fractions. Partial fractions method can be used in the following cases:
Case 1: When the denominator of the original complex fraction contains linear
unrepeated factors.
π₯+π
π΄
π΅
=
+
(π₯ + π)(π₯ + π) (π₯ + π) (π₯ + π)
We then solve the given expression by substituting various values of π₯ to obtain the
values of A and B.
This method can be extended and the given function can be decomposed to more
number of partial fractions depending upon the number of linear factors present in
the denominator of the given function.
Case 2: When the denominator of the original complex fraction contains linear
repeated factors.
π₯+π
π΄
π΅
πΆ
=
+
+
(π₯ + π)(π₯ + π)2 (π₯ + π) (π₯ + π) (π₯ + π)2
We then solve the given expression by substituting various values of π₯ to obtain the
values of A, B and C.
This method can be extended and the given function can be decomposed to more
number of partial fractions depending upon the number of linear factors present in
the denominator of the given function.
Case 3: When the denominator of the original complex fraction contains a quadratic
equation not resolvable into linear factors.
π₯+π
π΄
π΅π₯ + πΆ
=
+
(ππ₯ + π)(ππ₯ 2 + ππ₯ + π ) (ππ₯ + π) (ππ₯ 2 + ππ₯ + π)
We then solve the given expression by substituting various values of π₯ to obtain the
values of A, B and C. We can also equate the corresponding coefficients on both
sides to find the above values.
This method can be extended and the given function can be decomposed to more
number of partial fractions depending upon the number of linear or quadratic factors
present in the denominator of the given function.
Case 4: When there are polynomials in the numerator and denominator and the
degree of the numerator is greater than the degree of the denominator i.e. the given
fraction is an improper fraction, then we first use polynomial division to convert it
into a proper fraction and then apply any of the above cases if needed.
INTEGRALS OF THE FORM ∫
ππ₯
1
π₯
ππ₯
1
π₯−π
ππ₯
1
π+π₯
π
π
ππ +π
π,∫
π
π
ππ −π
π,∫
π
π
ππ −ππ
AND APPLICATIONS
1. ∫ 2 2 = tan−1 + π
π₯ +π
π
π
2. ∫ 2 2 = πππ | | + π
π₯ −π
2π
π₯+π
3. ∫ 2 2 = πππ | | + π
π −π₯
2π
π−π₯
Note: Before applying the above standard forms, always make the coefficient of π₯ 2
unity.
Method for solving integrals of the form ∫
ππ ±π
ππ +π
π
π:
Working Rule:
1. Divide the numerator and denominator by π₯ 2 .
1
2. It now takes the form ∫
1± 2
π₯
1
π₯ 2+ 2
π₯
1
1
π₯
π₯2
ππ₯. Now substitute π₯ β = π‘ ⇒ (1 ±
2
3. To get an expression for (π₯ +
1
π₯
1 2
1
π₯
π₯2
2
2
2 ), (π₯ β ) = π‘ ⇒ (π₯ +
4. Rewrite the given integrand in terms of t and integrate using ∫
) ππ₯ = ππ‘.
) = π‘ 2 ± 2.
ππ₯
π₯ 2 +π2
,∫
ππ₯
π₯ 2 −π2
form.
Note: The method used above can be generalized to integrate integrals of the form
π₯ 2 ±π
∫ π₯ 4 +ππ₯ 2 +π ππ₯.
Method of Completing Squares
This method can be used to integrate integrals consisting of quadratic equations in
the denominator such as ∫
ππ₯
ππ’πππππ‘ππ
i.e. ∫
ππ₯
ππ₯ 2 +ππ₯+π
etc.
This method is especially useful in integrating quadratic equations which are not
resolvable into linear factors.
Working Rule:
1. Make the coefficient of π₯ 2 unity by taking the numerical coefficient of π₯ 2 outside.
ππ₯
1
∫ ππ₯ 2 +ππ₯+π = π ∫
ππ₯
π
π
π₯ 2 + π₯+
π
π
.
2. Complete squares in terms containing π₯ by adding and subtracting the square of
half of the coefficient of π₯ and put the denominator in the form {(π₯ + πΌ )2 ± π½ 2 }
1
∫
π
1
ππ₯
π
π
π₯ 2 +ππ₯+π
1
= ∫
π
ππ₯
π
π2 π2 π
π₯ 2 + π₯+ − +
π
4π 4π π
1
ππ₯
= ∫
π
π 2
π2
(π₯+2π) +(π−4π)
which is of the form
ππ₯
.
∫
π (π₯+πΌ)2 ±π½2
3. Apply the appropriate standard form ∫
ππ₯
π₯ 2 +π
Method to evaluate integrals of the form ∫
2,∫
ππ₯
π₯ 2 −π
ππ+π
πππ +ππ+π
2,∫
ππ₯
π2 −π₯ 2
π
π i.e. ∫
.
ππππππ
ππππ
πππππ
π
π
Working Rule:
1. Split the given integral into two parts as follows:
ππ₯+π
π₯
1
∫ ππ₯ 2 +ππ₯+π ππ₯ = π ∫ ππ₯ 2 +ππ₯+π ππ₯ + π ∫ ππ₯ 2 +ππ₯+π ππ₯ .
2. The first integral can be integrated by the method of substitution by putting ππ₯ 2 +
ππ₯ + π = π‘ while the second integral can be integrated directly by applying the
method of completing squares and applying the appropriate standard form.
INTEGRALS OF THE FORM ∫
ππ₯
π
π
√ππ +π
,∫
π
π
π
√ππ −π
,∫
π
π
π
& APPLICATIONS
√ππ −ππ
π₯
1. ∫ 2 2 = π ππ−1 + π
π
√π −π₯
ππ₯
2. ∫ 2 2 = log|π₯ + √π₯ 2 + π2 | + π
√π₯ +π
ππ₯
3. ∫ 2 2 = log|π₯ + √π₯ 2 − π2 | + π
√π₯ −π
Note: Before applying the above standard forms, always make the coefficient of π₯ 2
unity.
Method to evaluate integrals of the type ∫
π
π
√πππ +ππ+π
i.e ∫
π
π
√ππππ
πππππ
Working Rule:
1. Make the coefficient of π₯ 2 unity by taking the numerical coefficient of π₯ 2 outside.
ππ₯
∫ √ππ₯ 2 +ππ₯+π =
1
∫
π
√
π
π
π
π
ππ₯
π
π
√π₯ 2 + π₯+
π
π
.
2. Put π₯ 2 + π₯ + in the form {(π₯ + πΌ )2 ± π½ 2 } by method of completing squares.
3. Apply the appropriate standard form ∫
Method to evaluate integrals of the type ∫
ππ₯
√π₯ 2 +π
,
2 ∫
ππ+π
√πππ +ππ+π
ππ₯
√π₯ 2 −π
,
2 ∫
π
π i.e. ∫
ππ₯
√π2 −π₯ 2
.
ππππππ
√ππππ
πππππ
π
π
Working Rule:
1. Put ππ₯ + π = π΄.
π
ππ₯
(ππ₯ 2 + ππ₯ + π) + π΅ i.e. ππππππ = π΄.
π
ππ₯
(ππ’πππππ‘ππ ) + π΅.
2. Find A and B by equating the coefficients of the corresponding terms on both
sides.
3. Rewrite the given integral by substituting the numerator,
ππ₯ + π = π΄.
π
ππ₯
(ππ₯ 2 + ππ₯ + π) + π΅.
4. Now the given integral takes the form ∫
π΄.
π
(ππ₯ 2 +ππ₯+π)+π΅
ππ₯
√ππ₯ 2 +ππ₯+π
ππ₯.
5. Now, split the above integral into two integrals which can be easily evaluated:
π΄∫
π
(ππ₯ 2 +ππ₯+π)
ππ₯
√ππ₯ 2 +ππ₯+π
ππ₯ + π΅ ∫
ππ₯
√ππ₯ 2 +ππ₯+π
.
6. The first integral can be integrated by the method of substitution by putting ππ₯ 2 +
ππ₯ + π = π‘ while the second integral can be integrated directly by applying the
method of completing squares and applying the appropriate standard form.
INTEGRALS OF THE FORM ∫ √ππ − ππ π
π, ∫ √ππ + ππ π
π, ∫ √ππ − ππ π
π &
APPLICATIONS
1
1
1
1
1
1
π₯
1. ∫ √π2 − π₯ 2 ππ₯ = π₯√π2 − π₯ 2 + π2 π ππ−1 + π
2
2
π
2. ∫ √π₯ 2 + π2 ππ₯ = π₯√π₯ 2 + π2 + a2 log|π₯ + √π₯ 2 + π2 | + π
2
2
3. ∫ √π₯ 2 − π2 ππ₯ = π₯√π₯ 2 + π2 − a2 log|π₯ + √π₯ 2 − π2 | + π
2
2
Aid to Memory:
1
1
∫(πΌππ‘ππππππ) ππ₯ = 2 π₯(πΌππ‘ππππππ) ± 2 a2 (πΌππ‘πππππ ππ π
πππππππππ ππ πΌππ‘ππππππ) + π.
Method to find integrals of the type ∫ √πππ + ππ + π π
π i.e. ∫ √ππππ
πππππ π
π
Working Rule:
1. Make the coefficient of π₯ 2 unity by taking the numerical coefficient of π₯ 2 outside.
π
π
∫ √ππ₯ 2 + ππ₯ + π ππ₯ = √π ∫ √π₯ 2 + π π₯ + π ππ₯ .
π
π
π
π
2. Put π₯ 2 + π₯ + in the form {(π₯ + πΌ )2 ± π½ 2 } by method of completing squares.
3. Apply the appropriate standard form ∫ √π2 − π₯ 2 ππ₯, ∫ √π₯ 2 + π2 ππ₯, ∫ √π₯ 2 − π2 ππ₯.
INTEGRALS OF THE FORM:
π
π
π
π
π
π
π
π
∫ π+π ππ¨π¬ π , ∫ π+π π¬π’π§ π , ∫ π ππ¨π¬ π+π π¬π’π§ π , ∫ π ππ¨π¬ π+π π¬π’π§ π+π πππ.
Working Rule:
π₯
1. Put sin π₯ =
2 tan2
π₯
π₯
1+tan2 2
2π₯
2. Transfer 1 + tan
2
π₯
2
Replace 1 + tan
2
, cos π₯ =
1−tan2 2
.
π₯
1+tan2 2
to the numerator after taking L.C.M. of the denominator.
π₯
= sec 2 in the numerator.
2
π₯
1
π₯
2
2
2
3. Put tan = π‘ so that sec 2 ππ₯ = ππ‘.
4. The new integrand will now be in the form ∫
ππ‘
ππ‘ 2 +ππ‘+π
which can be evaluated by
the method of completing squares followed by applying the appropriate standard
forms.
INTEGRALS OF THE FORM: ∫
π πππ π+π πππ π
π ππ¨π¬ π+π
π¬π’π§ π
π
π
Working Rule:
1. Put π πππ π₯ + π π ππ π₯ = π΄.
ππ’πππππ‘ππ = π΄.
π
ππ₯
π
ππ₯
(π cos π₯ + π sin π₯ ) + π΅. (π cos π₯ + π sin π₯ ) i.e.
(πππππππππ‘ππ ) + π΅. (πππππππππ‘ππ ).
2. Find A and B by equating the coefficients of cos π₯ and sin π₯ separately on both
sides.
3. Rewrite the given integral by substituting the numerator,
(π πππ π₯ + π π ππ π₯ ) = π΄.
π
ππ₯
(π cos π₯ + π sin π₯ ) + π΅(π cos π₯ + π sin π₯ ).
π
4. Now the given integral takes the form ∫
π΄.ππ₯(π cos π₯+π sin π₯)+π΅(π cos π₯+π sin π₯)
π cos π₯+π sin π₯
ππ₯.
5. Now, split the above integral into two integrals as follows:
π΄∫
π
(π cos π₯+π sin π₯)
ππ₯
π cos π₯+π sin π₯
ππ₯ + π΅ ∫ ππ₯ .
6. The first integral can be integrated by the method of substitution by putting
(π cos π₯ + π sin π₯) = π‘ while the second integral can be integrated directly
********************************
```
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# Banked Curves Section 5.4.
## Presentation on theme: "Banked Curves Section 5.4."— Presentation transcript:
Banked Curves Section 5.4
5.4 Banked Curves When a car travels around an unbanked curve, static friction provides the centripetal force. By banking a curve, this reliance on friction can be eliminated for a given speed.
Derivation of Banked Curves
A car travels around a friction free banked curve Normal Force is perpendicular to road x component (towards center of circle) gives centripetal force y component (up) cancels the weight of the car Insert Figure 5.11
Derivation of Banked Curves
Divide the x by the y Gives Notice mass is not involved Ask what happens when go to fast? (slide up and over top of curve) Ask what happens when go to slow? (slide down curve)
Example You are in charge of designing a highway cloverleaf exit ramp. What angle should you build it for speed of 35 mph and r = 100m? 13.9 35 mph = 15.6 m/s Tan = v2/rg tan = (15.6 m/s)2/((100m)(9.8 m/s2)) tan = = 13.9
Conceptual Problem In the Daytona International Speedway, the corner is banked at 31 and r = 316 m. What is the speed that this corner was designed for? v = 43 m/s = 96 mph Cars go 195 mph around the curve. How? Friction provides the rest of the centripetal force tan = v2/rg tan 31 = v2/(316m)(9.8m/s2) .6009(316m)(9.8m/s2) = v2 1861 (m/s)2 = v2 v = 43 m/s
Practice Problems See if you can speed your way around these!
Total of 4 problems
|
# Graphing Quadratics: putting it all together. 43210 In addition to level 3, students make connections to other content areas and/or contextual situations.
## Presentation on theme: "Graphing Quadratics: putting it all together. 43210 In addition to level 3, students make connections to other content areas and/or contextual situations."— Presentation transcript:
Graphing Quadratics: putting it all together
Strategies & steps to graphing a quadratic. y = 3x 2 + 2x + 4 1.Determine the vertex. Use x = - b / 2a 1. x = - 2 / 2(3) 2. x = - 1 / 3 3. y = 3(- 1 / 3 ) 2 + 2(- 1 / 3 ) + 4 4. y = 11 / 3 or 3 2 / 3 5. The vertex is (- 1 / 3, 3 2 / 3 ) 2.Use the discriminant to determine the number of solutions. 1.(2) 2 – 4(3)(4) 2.4 – 48 3.-44 4.No real solutions. This means the graph will NOT cross the x-axis. Factoring or using the quadratic formula will not help you graph this quadratic. You need to use an “x” point left and right of - 1 / 3 in order to graph this.
Strategies & steps to graphing a quadratic. y = 3x 2 + 2x + 4 3.Point left of and right of - 1 / 3 : (-2, _____) and (1, _____) 1.y = 3(-2) 2 + 2(-2) + 4 2.y = 12 – 4 + 4 3.y = 12 4.(-2, 12) 5.y = 3(1) 2 + 2(1) + 4 6.y = 3 + 2 + 4 7.y = 9 8.(1, 9) 4.Graph the vertex and the two points.
Strategies & steps to graphing a quadratic. y = x 2 + 8x + 16 1.Determine the vertex. Use x = - b / 2a 1. x = - 8 / 2(1) 2. x = -4 3. y = (-4) 2 + 8(-4) + 16 4. y = 0 5. The vertex is (-4, 0) 2.Use the discriminant to determine the number of solutions. 1.(8) 2 – 4(1)(16) 2.64 – 64 3.0 4.One solution. This means the graph will intercept the x-axis at the VERTEX. Factoring or using the quadratic formula will not help you graph this quadratic. (Your answers will be -4 & -4). You need to use an “x” point left and right of -4 in order to graph this.
Strategies & steps to graphing a quadratic. y = x 2 + 8x + 16 3.Point left of and right of -4: (-5, _____) and (-3, _____) 1.y = (-5) 2 + 8(-5) + 16 2.y = 25 – 40 + 16 3.y = 1 4.(-5, 1) 5.y = (-3) 2 + 8(-3) + 16 6.y = 9 - 24 + 16 7.y = 1 8.(-3, 1) 4.Graph the vertex and the two points.
Strategies & steps to graphing a quadratic. y = x 2 + 6x + 5 1.Determine the vertex. Use x = - b / 2a 1. x = - 6 / 2(1) 2. x = -3 3. y = (-3) 2 + 6(-3) + 5 4. y = -4 5. The vertex is (-3, -4) 2.Use the discriminant to determine the number of solutions. 1.(6) 2 – 4(1)(5) 2.36 – 20 3.16 4.Two solutions. This means the graph will intercept the x-axis two times. Factoring or using the quadratic formula will HELP you. Which one do you want to use?
Strategies & steps to graphing a quadratic. y = x 2 + 6x + 5 3.Factoring will be the quickest way to find the x-intercepts. 1. (x +5)(x + 1) = 0 2. x + 5 = 0 1. x = -5 3. x + 1 = 0 1. x = -1 4.The x-intercepts are (-5, 0) & (-1, 0). 4.Graph the vertex and the x-intercepts.
Strategies & steps to graphing a quadratic. y = -2x 2 + 6x + 1 1.Determine the vertex. Use x = - b / 2a 1. x = - 6 / 2(-2) 2. x = 3 / 2 3. y = -2( 3 / 2 ) 2 + 6( 3 / 2 ) + 1 4. y = -2( 9 / 4 ) + 9 + 1 5. The vertex is (1½, 5 ½ ) 2.Use the discriminant to determine the number of solutions. 1.(6) 2 – 4(-2)(1) 2.36 + 8 3.44 4.Two solutions. This means the graph will intercept the x-axis two times. Because 44 is not a perfect square, the x-intercepts will be irrational. Use the quadratic formula to find the x- intercepts.
Strategies & steps to graphing a quadratic. y = -2x 2 + 6x + 1 3.Substitute into the quadratic formula and solve to find your x- intercepts. 4.Divide 6, 2 and 4 by 2 (the GCF).
Strategies & steps to graphing a quadratic. y = -2x 2 + 6x + 1 5.Estimate the value of the irrational x-intercepts. 6.Graph the vertex and the x-intercepts.
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# Difference between revisions of "2018 AMC 8 Problems/Problem 21"
## Problem
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$
## Solution 1
Looking at the values, we notice that $11-7=4$, $9-5=4$ and $6-2=4$. This means we are looking for a value that is four less than a multiple of $11$, $9$, and $6$. The least common multiple of these numbers is $11\cdot3^{2}\cdot2=198$, so the numbers that fulfill this can be written as $198k-4$, where $k$ is a positive integer. This value is only a three digit≤ integer when $k$ is $1, 2, 3, 4$ or $5$, which gives $194, 392, 590, 788,$ and $986$ respectively. Thus we have $5$ values, so our answer is $\boxed{\textbf{(E) }5}$
## Solution 2
Let us create the equations: $6x+2 = 9y+5 = 11z+7$, and we know $100 \leq 11z+7 <1000$, it gives us $9 \leq z \leq 90$, which is the range of the value of z. Because of $6x+2=11z+7$, then $6x=11z+5=6z+5(z+1)$, so (z+1) must be mutiples of 6. Because of $9y+5=11z+7$, then $9y=11z+2=9z+2(z+1)$, so (z+1) must also be mutiples of 9. Hence, the value of (z+1) must be common multiples of $6$ and $9$, which means multiples of $18(LCM of 6 & 9)$ (Error compiling LaTeX. ! Misplaced alignment tab character &.). So let's say $z+1 = 18p$, then $9 \leq z = 18p-1 \leq 90, 1 \leq p \leq 91/18 or 1 \leq p \leq 5$. Thus our answer is $\boxed{\textbf{(E) }5}$ ~LarryFlora
## Video Solution
https://youtu.be/CPQpkpnEuIc - Happytwin
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# How to Solve Linear Inequalities?
The process to solve linear inequalities in a variable is similar to solving basic equations. Most of the rules or techniques used to solve equations can be easily used to resolve inequalities. The only important difference is that the symbol of inequality changes direction when a negative number is multiplied or divided on both sides of an equation.
Here, we will look at examples with answers to dominate the process of solving linear inequalities.
##### ALGEBRA
Relevant for
Learning to solve linear inequalities with examples.
See examples
##### ALGEBRA
Relevant for
Learning to solve linear inequalities with examples.
See examples
## Inequality symbols with illustrations
The following are the symbols used to represent inequalities:
### Greater than
Symbol:
Example: x is greater than -2:
Graphic representation:
### Less than
Symbol:
Example: x is less than 4:
Graphic representation:
### Greater than or equal to
Symbol:
Example: x is greater than or equal to -3:
Graphic representation:
### Less than or equal to
Symbol:
Example: x is less than or equal to 3:
Graphic representation:
## Linear inequalities – Examples with answers
The following examples with answers will allow us to fully master the process of solving linear inequalities. Try to solve the exercises yourself before looking at the solution.
### EXAMPLE 1
Solve and graph the solution of the inequality $latex 4x-6>2$.
To solve the inequality, we want to find all the x values that satisfy it. This means that there are an infinite number of solutions that, when substituted, would make the inequality true. Therefore, we follow the following steps:
• We write the original problem:
$latex 4x-6>2$
• We add 6 to both sides to keep the variables on one side and the constants on the other:
$latex 4x-6+6>2+6$
• After simplifying, the expression reduces to:
$latex 4x>8$
• We divide both sides by 4 and simplify to get the answer:
$latex \frac{4}{4}x>\frac{8}{4}$
$latex x>2$
• We use an open point to indicate that 2 is not part of the solution. The solution to the inequality includes all values to the right of 2, but does not include 2:
### EXAMPLE 2
Solve and graph the inequality $latex -2x-3 \geq 1$.
This is an example of what happens to the inequality when we divide by a negative number:
• We write the original problem:
$latex -2x-3\geq 1$
• We isolate the variable by adding 3 to both sides:
$latex -2x-3+3\geq 1+3$
• After simplifying, the expression reduces to:
$latex -2x\geq 4$
• We divide both sides by -2 and simplify to get the answer:
$latex \frac{-2}{-2}x\geq \frac{4}{-2}$
$latex x\leq -2$
• We use a closed point to indicate that the -2 is part of the solution. The solution to the inequality includes -2 and all values to the left:
### EXAMPLE 3
Solve the inequality $latex 5-3x>13-5x$.
Here, we have variables on both sides. It is possible to place the variables on the left side or on the right side, but the standard is to do it on the left side.
• We write the original problem:
$latex 5-3x>13-5x$
• We can add 5x to both sides and subtract 5 to solve for the variable:
$latex 5-3x+5x-5>13-5x+5x-5$
• Simplifying the expression, we have:
$latex 2x>8$
• We can get the answer by dividing both sides by 2:
$latex \frac{2}{2}x> \frac{8}{2}$
$latex x> 2$
### EXAMPLE 4
Solve inequality $latex 2(2x+3)<2x+12$.
In this case, we have parentheses, so we have to start by applying the distributive property to remove the parentheses:
• We write the original problem:
$latex 2(2x+3)<2x+12$
• We apply the distributive property:
$latex 2(2x)+2(3)<2x+12$
$latex 6x+6<2x+12$
• We isolate the variable by subtracting 6 and 2x from both sides:
$latex 6x+6-2x-6<2x+12-2x-6$
• After simplifying, the expression reduces to:
$latex 4x<8$
• We divide both sides by 4 and simplify to get the answer:
$latex \frac{4}{4}x< \frac{8}{4}$
$latex x< 2$
### EXAMPLE 5
Solve the inequality $latex 2(3x-4)\geq 7x-7$.
Again, we start by applying the distributive property to remove the parentheses:
• We write the original problem:
$latex 2(3x-4)\geq 7x-7$
• We apply the distributive property:
$latex 2(3x)+2(-4)\geq 7x-7$
$latex 6x-8\geq 7x-7$
• Now, we add 8 and subtract 7x from both sides to isolate the variable:
$latex 6x-8+8-7x\geq 7x-7+8-7x$
• We simplify to obtain:
$latex -x\geq 1$
• We divide both sides by -1 and simplify to get the answer:
$latex \frac{-1}{-1}x\geq \frac{1}{-1}$
$latex x\geq -1$
### EXAMPLE 6
Solve the inequality $latex 3(x+2)+2\leq 2(x-1)+8$.
In this case, we have to apply the distributive property to both parentheses and then combine like terms to simplify:
• We write the original problem:
$latex 3(x+2)+2\leq 2(x-1)+8$
• We apply the distributive property to both parentheses:
$latex 3(x)+3(2)+2\leq 2(x)+2(-1)+8$
$latex 3x+6+2\leq 2x-2+8$
• We combine like terms to simplify:
$latex 3x+8\leq 2x+6$
• We isolate the variable by subtracting 8 and 2x from both sides:
$latex 3x+8-8-2x\leq 2x+6-8-2x$
• Simplifying, we have:
$latex x\leq -2$
• We no longer have to divide:
$latex x\leq -2$
## Linear inequalities – Practice problems
Test your skills and knowledge about linear inequalities with the following interactive problems. Choose an answer and check it to verify that you selected the correct answer. The examples solved above can serve as a guide if you have problems with these exercises.
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Test: Zeroes & Coefficients - Grade 10 MCQ
# Test: Zeroes & Coefficients - Grade 10 MCQ
Test Description
## 20 Questions MCQ Test - Test: Zeroes & Coefficients
Test: Zeroes & Coefficients for Grade 10 2024 is part of Grade 10 preparation. The Test: Zeroes & Coefficients questions and answers have been prepared according to the Grade 10 exam syllabus.The Test: Zeroes & Coefficients MCQs are made for Grade 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Zeroes & Coefficients below.
Test: Zeroes & Coefficients - Question 1
### If α and β are the zeroes of the polynomial 5x2 – 7x + 2, then sum of their reciprocals is:
Detailed Solution for Test: Zeroes & Coefficients - Question 1
We have 2 find (1/α + 1/β)
now 1/α + 1/β = (α + β)/ α β (taking LCM)
now by the given poly. we get
(α + β) = -b/a = 7/5
α β = c/a = 2/5
so, (α + β)/ α β = (7/5) / (2/5)
= 7/2
So, 1/α + 1/β = (α + β)/ α β = 7/2
Hence, 1/α + 1/β = 7/2
Test: Zeroes & Coefficients - Question 2
### If one zero of polynomial f(x) = (k2 + 4)x2 + 13x + 4k is reciprocal of the other, then k =
Detailed Solution for Test: Zeroes & Coefficients - Question 2
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Test: Zeroes & Coefficients - Question 3
### Find the quadratic polynomial whose zeros are 2 and -6
Detailed Solution for Test: Zeroes & Coefficients - Question 3
We know that quadratic equation is of the form x2-(sum of zeros)x+product of zeros
Sum of zeros=2-6=-4
Product of zeros=2*(-6)=-12
x2-(sum of roots )x + product of roots
x2-(-4)x + 12
x2+4x-12
So the equation is x2+4x-12
Test: Zeroes & Coefficients - Question 4
If α, β, γ be the zeros of the polynomial p(x) such that α + β + γ = 3, αβ + βγ + γα = -10 and αβγ = -24, then p(x) is
Detailed Solution for Test: Zeroes & Coefficients - Question 4
Test: Zeroes & Coefficients - Question 5
If -√5 and √5 are the roots of the quadratic polynomial. Find the quadratic polynomial.
Detailed Solution for Test: Zeroes & Coefficients - Question 5
We have as the roots which means x + and x- are the factors of the quadratic equation. Multiplying x+ and x- and applying a2-bwe get the equation x2-5.
Test: Zeroes & Coefficients - Question 6
If one root of polynomial equation ax2 + bx + c = 0 be reciprocal of other, then
Detailed Solution for Test: Zeroes & Coefficients - Question 6
Test: Zeroes & Coefficients - Question 7
If one zero of the polynomial x2 + kx + 18 is double the other zero then k = ?
Detailed Solution for Test: Zeroes & Coefficients - Question 7
Let one zero of the equation be x
Then, other would be 2x
As we know,
x × 2x = 18
x2 = 9
x = 3, -3
Case 1:
If x = 3, 2x = 6
Therefore, -k = sum of roots = 9
⇒ k = -9
Case 2:
If x = -3, 2x = -6
Therefore, -k = -9
⇒ k = 9
So, k can be 9,-9. Therefore, option 'b' is correct.
Test: Zeroes & Coefficients - Question 8
If one zero of 2x2 – 3x + k is reciprocal to the other, then the value of k is :
Detailed Solution for Test: Zeroes & Coefficients - Question 8
Given: 2x²-3x+k .............eq 1
Let the 2 zeroes be α & 1/α
On comparing eq 1 & eq 2 we get
a=2 ,b = -3, c= k
Product of zeroes = α × 1/α = 1
Product of zeroes= c/a
1= k/ 2
K= 2
The value of k is 2.
Test: Zeroes & Coefficients - Question 9
Sum and the product of zeroes of the polynomial x2 +7x +10 is
Detailed Solution for Test: Zeroes & Coefficients - Question 9
x2 + 7x + 10 = (x + 2)(x + 5)
So, the value of x2 + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0
Therefore, the zeroes of x2 + 7x + 10 are –2 and –5.
Sum of zeroes = -7 = –(Coefficient of x) / (Coefficient of x2)
Product of zeroes = 10 = Constant term / Coefficient of x2
Test: Zeroes & Coefficients - Question 10
If sum of the zeroes of the polynomial is 4 and their product is 4, then the quadratic polynomial is
Detailed Solution for Test: Zeroes & Coefficients - Question 10
Sum of zeros= α+β =4
Product of zeros=αβ=4
x2-(Sum of zeros)x+Product of zeros
=x2-4x+4
Test: Zeroes & Coefficients - Question 11
What value/s can x take in the expression k(x – 10) (x + 10) = 0 where k is any real number.
Detailed Solution for Test: Zeroes & Coefficients - Question 11
k(x – 10) (x + 10) =0
⇒ either k=0
Or x-10=0
Or x+10=0
Since we don’t know the value of k
So either x-10=0
x=10
Or x+10=0
x=-10
So values of x can be 10,-10
Test: Zeroes & Coefficients - Question 12
Find the sum and the product of zeroes of the polynomial x2 +7x +10
Detailed Solution for Test: Zeroes & Coefficients - Question 12
x2 + 7x + 10 = (x + 2)(x + 5)
So, the value of x2 + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0
Therefore, the zeroes of x2 + 7x + 10 are –2 and –5.
Sum of zeroes = -7 = –(Coefficient of x) / (Coefficient of x2)
Product of zeroes = 10 = Constant term / Coefficient of x2
Test: Zeroes & Coefficients - Question 13
The sum and product of zeros of a quadratic polynomial are 2 and -15 respectively. The quadratic polynomial is
Detailed Solution for Test: Zeroes & Coefficients - Question 13
Let the polynomial be f(x).
f(x) = x2− (sum of zeros)x + product of zeros
= x2−2x–15
Test: Zeroes & Coefficients - Question 14
If the product of two zeros of the polynomial f(x) = 2x3 + 6x2 – 4x + 9 is 3, then its third zero is
Detailed Solution for Test: Zeroes & Coefficients - Question 14
Step-by-step explanation:
p(x) = 2x³ + 6x² – 4x + 9
Define α,β and γ
Let α,β and γ be the 3 roots
αβ = 3 (Given)
Find the products of all the zeros:
αβγ = - d/a
αβγ = - 9/2
Find the third zeros:
Given αβ = 3
3γ = -9/2
γ = -9/2 ÷ 3
γ = -3/2
The third zero is -3/2
Test: Zeroes & Coefficients - Question 15
If α, β are zeroes of the polynomial f(x) = x2 + 5x + 8, then value of (α + β) is
Detailed Solution for Test: Zeroes & Coefficients - Question 15
We have quadratic equation of the form
x- (sum of zeros) x + product of zeros, so comparing this with the given equation we have sum of zeros = -5
So, α + β = -5
Test: Zeroes & Coefficients - Question 16
If sum of the squares of zeros of the quadratic polynomial f(x) = x2 – 8x + k is 40, find the value of k.
Detailed Solution for Test: Zeroes & Coefficients - Question 16
p (x)= x^2-8x+k
p (x)=Ax^2+Bx+C(The equation is in this form)
Let the zeroes be 'a' and 'b'
It is given that
=> a^2+b^2=40
=> (a+b)^2 - 2ab = 40 ----1
sum of zeroes
=> a+b = -B/A = -(-8)/1 = 8
Product of zeroes
=> a*b = C/A = k
Substitute these values in the equation1 we get
=> 8^2 - 2k = 40
=> 64 - 2k = 40
=> 2k = 24
=> k = 12
Therefore the value of k is 12
Test: Zeroes & Coefficients - Question 17
If α , β are the zeroes of f(x) = px2 – 2x + 3p and α + β = αβ then the value of p is:
Detailed Solution for Test: Zeroes & Coefficients - Question 17
The given polynomial is
Also, α and β are the zeros of p(x).
and α + β = α × β
Test: Zeroes & Coefficients - Question 18
If the two zeroes of the quadratic polynomial 7x2 – 15x – k are reciprocals of each other, the value of k is:
Detailed Solution for Test: Zeroes & Coefficients - Question 18
Test: Zeroes & Coefficients - Question 19
The number of polynomials having zeroes -2 and 5 is:
Detailed Solution for Test: Zeroes & Coefficients - Question 19
Since the question doesn’t say that there are only 2 zeros of the polynomial we , there can be n number of polynomials which have two of its zeros as -2 and 5 .So the correct answer is more than 3.
Test: Zeroes & Coefficients - Question 20
Find the sum and the product of the zeroes of the polynomial: x2-3x-10
Detailed Solution for Test: Zeroes & Coefficients - Question 20
X²-3x-10
x² -(5x-2x)-10
x² - 5x+2x-10
x(x-5)+2(x-5)
(x-5)(x+2)
x=5
x=-2
Sum of zeroes = α+β = 5-2 = 3
α+β = -b/a = -(-3)/1 = 3
Product of zeroes = αβ = 5*-2 = -10
αβ = c/a = -10/1 = -10
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# Math Section
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This chapter is from the book
### Multiple Choice (Questions 1–8)
1. The correct answer is C. To solve this problem, simply count the numbers that appear both in set P and in set Q. Sets P and Q share the following numbers: 16, 17, 18, 19, and 20. Therefore, five numbers in set P are also in Set Q.
2. The correct answer is C. You are given that John traveled 30 miles in 6 hours and that Seth traveled three times as far in half the time. This means that Seth traveled 30(3), or 90 miles in 6(), or 3 hours. Therefore, Seth traveled , or 30 miles per hour.
3. The correct answer is D. You are given that r = -2. Therefore, the best approach to solving this problem is to replace r with -2 in each of the answer choices as follows:
• Answer choice A: 4pr2 = 4p(-2)2, which equals (4)(4)p, or 16p.
• Answer choice B: 2pr3 = 2p(-2)3, which equals (2)(-8)p, or -16p.
• Answer choice C: -2pr4 = -2p(-2)4, which equals (-2)(16)p, or -32p.
• Answer choice D: -4pr5 = -4p(-2)5, which equals (-4)(-32)p, or 128p.
• Answer choice E: -6pr6 = -6p(-2)6, which equals (-6)(64)p, or -384p.
The question asks for the least value. You are given that p < 0, which means that p is a negative number. If you substitute any negative number into the results obtained above, only answer choices A and D will yield a negative value. Therefore, you can eliminate answer choices B, C, and E, because they will all be larger than either answer choice A or D. When p < 0, 128p will always be less than 16p, so answer choice D is correct.
4. The correct answer is C. The equation of the line is given in the slope-intercept form (y = mx + b), where b is the y-intercept. The y-intercept is the point at which the line intersects the y-axis. Since the graph shown has the equation y = ax + b, and the new graph has the equation y = ax + b, both graphs should have the same y-intercept. The graph shown has a y-intercept of 1. Eliminate answer choices A, B and D, since they do not have a y-intercept of 1. The slope of the graph shown is a, and the slope of the new graph (based on its equation) is a. Because both slopes are positive, you can eliminate answer choice E, which has a negative slope. This leaves you with answer choice C.
5. The correct answer is B. To solve this problem, you must recognize that the triangle is a "special triangle." A right triangle in which the length of the longer leg is times the length of the shorter leg is a 30°–60°–90° right triangle. Another property of this type of right triangle is that the hypotenuse is 2 times the length of the shorter leg. So, this right triangle has lengths x, x, and 2x. The perimeter is the sum of the lengths of the sides. You are given that the perimeter equals 12 + 4. Plug this value into the formula for the perimeter of a triangle, as follows:
• 12 + 4 = x + x + 2x
• 12 + 4 = 3x + x
For the right side of the equation to equal the left side of the equation, x must be equal to 4.
6. The correct answer is C. The median of a list of values is the middle value when the list is in chronological order and there are an odd number of values. In this case, if we include Mike and his 75 coins, there will be 19 coin collectors. List the number of coins for each collector in chronological order as follows:
7. As you can see, the middle number is 85, so that is the median.
8. The correct answer is C. To solve, first translate the words into their mathematical equivalents:
• 4 more than twice n translates to 4 + 2n. You are given that this quantity is a negative number, which translates to 4 + 2n < 0.
• 6 more than n translates to 6 + n. You are given that this quantity is a positive number, which translates to 6 + n > 0.
Solve each inequality as follows:
• 4 + 2n < 0
• n < -2
• 6 + n > 0
• n > -6
So, n could be any number less than -2 and greater than -6. Only -4, answer choice C, is greater than -6 and less than -2.
9. The correct answer is D. Since the figure is a rectangle, the adjacent sides are perpendicular. Perpendicular lines have slopes that are negative reciprocals of each other, meaning that the product of their slopes is -1. Since there are four lines and four perpendicular angles, the product of the slopes is (-1)(-1)(-1)(-1), or 1.
### Student-Produced Response (Questions 9–18)
1. Answer: . To solve this problem, convert hours into minutes. 2 hours is equivalent to 150 minutes. You are given that the movie included 30 minutes of commercials; therefore, the movie included 120 minutes that were not commercials. The fraction of the movie that was not commercials is , or .
2. Answer: 2.66, 2.67, or .
3. To solve this problem, set up an equation based on the information given:
• 0.75x = 2
Solve for x.
• x = 2 ÷ 0.75
• x = 2.66666 (This is a repeating decimal.)
For the purposes of the PSAT, the answer can be simplified to either 2.66 or 2.67. If you first converted .75 to , you would arrive at as your answer, which is also correct.
4. Answer: . You are given that 8s = 96. Therefore, s = , or 12. You are given that sp = 4, and you now know that s = 12. Therefore, 12p = 4 and p = , or .
5. Answer: 2 or 3. The formula for the area of a rectangle is A = (l)(w). Therefore, the area of ACFD will be (AC)(AD). Because you must solve for the value of AB, let AB = x. Since ABED is a square, it follows that AB = AD = x. Therefore, since BC = 8, the area of ACFD will be (8 + x)x, or x2 + 8x. Since, according to the question, the area must be between 14 and 35, set up the inequality 14 < x2 + 8x < 35. Now, pick some easy numbers to work with to replace x in the inequality. Because x is a positive value and 14 and 35 are not particularly large numbers, start with x = 1. If x = 1, then the area is x2 + 8x, or (1)2 + 8(1), or 9. This value is not between 14 and 35, so x cannot be 1. If x = 2, then the area is x2 + 8x, or (2)2 + 8(2), or 20. This value is between 14 and 35, so x = 2 is a possible solution. If x = 3, then the area is x2 + 8x, or (3)2 + 8(3), or 33. This value is between 14 and 35, so x = 3 is another possible solution. If x = 4, then the area is x2 + 8x, or (4)2 + 8(4) or 48. This value is not between 14 and 35, so x cannot be 4, or any number greater than 4. Therefore, x, or AB, can be either 2 or 3.
6. Answer: 100. To solve this problem, let the total number of napkins in the box equal x. According to information in the problem, there are 25 red napkins in the box. One quarter of the total number of napkins is blue, so there are ()x blue napkins in the box. There are twice as many white napkins as blue napkins, so there are 2()x, or ()x white napkins in the box. To calculate the total number of napkins in the box, set up an equation and solve for x.
• x = 25 + x + x
• x = 25 + x
• x = 25
• x = 100
7. Answer: 20. You are given that there are 4 points on line l, and 5 points on the line parallel to line l. Therefore, from each of the 4 points on line l, T, U, V, and W, you can draw one line to each of the 5 points on the second line. This means that there can be(4)(5), or 20 different lines containing exactly 2 of the 9 points.
8. Answer: 4. To solve this problem, rewrite 42x + 42x + 42x + 42x as 4(42x). Now you have 4(42x) = 49. When you multiply numbers with the same base, you add the exponents. The number 4 is equivalent to 41. Therefore, 4(42x) = 42x + 1. Now, because 42x + 1 = 49, set the exponents equal to each other, and solve for x.
• 2x + 1 = 9
• 2x = 8
• x = 4
9. Answer: 496. If the average of the 9 numbers is 56, then the total value of the 9 numbers is (9)(56), or 504. The question asks for the greatest possible integer on one card, so set the values of the other cards at the lowest possible integer, which is 1. If there are 8 cards with a value of 1, then the value of the ninth card must be 504 – 8(1), or 496.
10. Answer: 3. Let j be the length of one of Jeff's steps, and let s be the length of one of Scott's steps. If they each take 12 steps away from each other, the distance between them is 12j + 12s. Since it takes Jeff 16 steps to reach Scott, then 12j + 12s = 16j. Subtract 12j from both sides to get 12s = 4j. Divide both sides by 4 to get 3s = j. The length of one of Jeff's steps (j) is 3 times the length of one of Scott's steps (s).
11. Answer: 130. To solve this problem, you should recognize that rs is a factor of 3r2s + 2rs2. Rewrite 3r2s + 2rs2 as rs(3r + 2s). You are given that rs = 13 and 3r + 2s = 10. Substitute those values into rs(3r + 2s) to get 13(10), which equals 130.
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Directions: Study the following information carefully and answer the questions given beside:
In a society meeting eight persons – P, Q, R, S, T, U, V and W were seated in a straight line facing the north. They have different house numbers – 15, 28, 30, 40, 45, 65, 80 and 90 not necessarily in the same order. It is further known that:
Difference between T and P's house number is equal to the difference between Q and U's house number.
Not more than 3 persons have house number less than Q.
Not more than 2 persons have house number more than P.
T’s house number was twice of V.
Q is seated to the immediate right of the person whose house number was 65.
P is seated to the immediate right of T who is 3rd to the left of one whose house number was 45.
Three persons seated to the right of V had house numbers less than him.
House numbers of R and Q together was equal to house number of V.
One whose house number was 28 was seated to the immediate left of the one whose house number was 40.
Q and U were seated at a gap of 1 person.
S and U were not seated together.
Important for :
1
Who among the following was seated at the extreme left end of the row?
» Explain it
E
Following the final solution we can say that T was seated at the extreme left end of the row.
Hence, the correct answer is option E.
Common Explanation:
Reference:
Not more than 2 persons have house number more than P.
Not more than 3 persons have house number less than Q.
T’s house number was twice of V.
House number of R and Q together was equal to house number V.
Three persons seated to the right of V had house numbers less than him.
Inference:
Here, the possible house numbers for P are 65 or 80 or 90. And the possible house numbers for Q are 15 or 28 or 30 or 40.
Similarly, the possible house numbers of V and T are 15 and 30, 40 and 80, and 45 and 90.
Here, house number 15 is not possible for V as we already know that three persons had house number less than V.
And house number 40 is also not possible for V because none of the combination of the house numbers leads to 40 and we have a hint that house number of R and Q was equal to house number V.
Thus, the only possible house number for V is 45 and the house number of T is 90.
Using the above information we can also say that the house number of R and Q can be 15 and 30 not necessarily in the same order. And the possible house number of P can be 65 or 80.
Reference:
Three persons seated to the right of V had house numbers less than him.
P is seated to the immediate right of T who is 3rd to the left of one whose house number was 45.
Inference:
With the first hint we can say that there were at least three persons sitting to the immediate right of V.
And with the second we can say that there are two possible scenarios in which we can fix the positions of these persons.
Case 1:
Case 2:
Reference:
Difference between T and P's house number is equal to the difference between Q and U's house number.
House number of R and Q together was equal to house number V.
Inference:
Here the difference between the house number of T and P can be 25 (when the house number of P is 65) or 10 (when the house number of P is 80)
And we know that the possible house numbers of Q are either 15 or 30.
Now, if the house number of Q is 15 then the only possible house number for U is 40 when the house number of P is 65.
And, if the house number of Q is 30 then the only possible house number for U is 40 when the house number of P is 80.
Reference:
Q was seated to the immediate right of the person whose house number was 65.
Q and U were seated at a gap of 1 person.
Inference:
Using, the above information our case 1 and case 2 can be redrawn as:
Reference:
One whose house number was 28 was seated to the immediate left of the one whose house number was 40.
S and U were not seated together.
Inference:
Here, Case A, Case B and Case D fails because we cannot place the person whose house number is 28 immediate to the lefty of the person whose house number is 40 in these cases.
At this point we can say that the house number of R is 15 and can fix his position as shown in the figure.
Reference:
S and U were not seated together.
Inference:
At this point we cannot place S in Case-E under the given conditions. So we can say that this is an invalid case.
But, with the given information we can place S and W and in our Case-C. As shown in the figure below:
Now, the puzzle is completed.
2
Which of the following numbers can be multiplied with 3 to obtain the house no. of Q?
» Explain it
C
Following the final solution we can say that the house number of Q was 30 i.e. 3 × 10
Hence, the correct answer is option C.
Common Explanation:
Reference:
Not more than 2 persons have house number more than P.
Not more than 3 persons have house number less than Q.
T’s house number was twice of V.
House number of R and Q together was equal to house number V.
Three persons seated to the right of V had house numbers less than him.
Inference:
Here, the possible house numbers for P are 65 or 80 or 90. And the possible house numbers for Q are 15 or 28 or 30 or 40.
Similarly, the possible house numbers of V and T are 15 and 30, 40 and 80, and 45 and 90.
Here, house number 15 is not possible for V as we already know that three persons had house number less than V.
And house number 40 is also not possible for V because none of the combination of the house numbers leads to 40 and we have a hint that house number of R and Q was equal to house number V.
Thus, the only possible house number for V is 45 and the house number of T is 90.
Using the above information we can also say that the house number of R and Q can be 15 and 30 not necessarily in the same order. And the possible house number of P can be 65 or 80.
Reference:
Three persons seated to the right of V had house numbers less than him.
P is seated to the immediate right of T who is 3rd to the left of one whose house number was 45.
Inference:
With the first hint we can say that there were at least three persons sitting to the immediate right of V.
And with the second we can say that there are two possible scenarios in which we can fix the positions of these persons.
Case 1:
Case 2:
Reference:
Difference between T and P's house number is equal to the difference between Q and U's house number.
House number of R and Q together was equal to house number V.
Inference:
Here the difference between the house number of T and P can be 25 (when the house number of P is 65) or 10 (when the house number of P is 80)
And we know that the possible house numbers of Q are either 15 or 30.
Now, if the house number of Q is 15 then the only possible house number for U is 40 when the house number of P is 65.
And, if the house number of Q is 30 then the only possible house number for U is 40 when the house number of P is 80.
Reference:
Q was seated to the immediate right of the person whose house number was 65.
Q and U were seated at a gap of 1 person.
Inference:
Using, the above information our case 1 and case 2 can be redrawn as:
Reference:
One whose house number was 28 was seated to the immediate left of the one whose house number was 40.
S and U were not seated together.
Inference:
Here, Case A, Case B and Case D fails because we cannot place the person whose house number is 28 immediate to the lefty of the person whose house number is 40 in these cases.
At this point we can say that the house number of R is 15 and can fix his position as shown in the figure.
Reference:
S and U were not seated together.
Inference:
At this point we cannot place S in Case-E under the given conditions. So we can say that this is an invalid case.
But, with the given information we can place S and W and in our Case-C. As shown in the figure below:
Now, the puzzle is completed.
3
Who among the following has house number equal to the average of the house numbers of T and U?
» Explain it
B
Following the final solution we can say that the house numbers of T and U were 90 and 40 respectively.
Required Average = (90 + 40) ÷ 2 = 65
And we know that the house number of S was 65.
Hence, the correct answer is option B.
Common Explanation:
Reference:
Not more than 2 persons have house number more than P.
Not more than 3 persons have house number less than Q.
T’s house number was twice of V.
House number of R and Q together was equal to house number V.
Three persons seated to the right of V had house numbers less than him.
Inference:
Here, the possible house numbers for P are 65 or 80 or 90. And the possible house numbers for Q are 15 or 28 or 30 or 40.
Similarly, the possible house numbers of V and T are 15 and 30, 40 and 80, and 45 and 90.
Here, house number 15 is not possible for V as we already know that three persons had house number less than V.
And house number 40 is also not possible for V because none of the combination of the house numbers leads to 40 and we have a hint that house number of R and Q was equal to house number V.
Thus, the only possible house number for V is 45 and the house number of T is 90.
Using the above information we can also say that the house number of R and Q can be 15 and 30 not necessarily in the same order. And the possible house number of P can be 65 or 80.
Reference:
Three persons seated to the right of V had house numbers less than him.
P is seated to the immediate right of T who is 3rd to the left of one whose house number was 45.
Inference:
With the first hint we can say that there were at least three persons sitting to the immediate right of V.
And with the second we can say that there are two possible scenarios in which we can fix the positions of these persons.
Case 1:
Case 2:
Reference:
Difference between T and P's house number is equal to the difference between Q and U's house number.
House number of R and Q together was equal to house number V.
Inference:
Here the difference between the house number of T and P can be 25 (when the house number of P is 65) or 10 (when the house number of P is 80)
And we know that the possible house numbers of Q are either 15 or 30.
Now, if the house number of Q is 15 then the only possible house number for U is 40 when the house number of P is 65.
And, if the house number of Q is 30 then the only possible house number for U is 40 when the house number of P is 80.
Reference:
Q was seated to the immediate right of the person whose house number was 65.
Q and U were seated at a gap of 1 person.
Inference:
Using, the above information our case 1 and case 2 can be redrawn as:
Reference:
One whose house number was 28 was seated to the immediate left of the one whose house number was 40.
S and U were not seated together.
Inference:
Here, Case A, Case B and Case D fails because we cannot place the person whose house number is 28 immediate to the lefty of the person whose house number is 40 in these cases.
At this point we can say that the house number of R is 15 and can fix his position as shown in the figure.
Reference:
S and U were not seated together.
Inference:
At this point we cannot place S in Case-E under the given conditions. So we can say that this is an invalid case.
But, with the given information we can place S and W and in our Case-C. As shown in the figure below:
Now, the puzzle is completed.
4
Who among the following has house number equal to the average of the house numbers of R and V?
» Explain it
D
Following the final solution we can say that the house numbers of R and V were 15 and 45 respectively.
Required Average = (15 + 45) ÷ 2 = 30
And we know that the house number of Q was 30.
.
Hence, the correct answer is option D.
Common Explanation:
Reference:
Not more than 2 persons have house number more than P.
Not more than 3 persons have house number less than Q.
T’s house number was twice of V.
House number of R and Q together was equal to house number V.
Three persons seated to the right of V had house numbers less than him.
Inference:
Here, the possible house numbers for P are 65 or 80 or 90. And the possible house numbers for Q are 15 or 28 or 30 or 40.
Similarly, the possible house numbers of V and T are 15 and 30, 40 and 80, and 45 and 90.
Here, house number 15 is not possible for V as we already know that three persons had house number less than V.
And house number 40 is also not possible for V because none of the combination of the house numbers leads to 40 and we have a hint that house number of R and Q was equal to house number V.
Thus, the only possible house number for V is 45 and the house number of T is 90.
Using the above information we can also say that the house number of R and Q can be 15 and 30 not necessarily in the same order. And the possible house number of P can be 65 or 80.
Reference:
Three persons seated to the right of V had house numbers less than him.
P is seated to the immediate right of T who is 3rd to the left of one whose house number was 45.
Inference:
With the first hint we can say that there were at least three persons sitting to the immediate right of V.
And with the second we can say that there are two possible scenarios in which we can fix the positions of these persons.
Case 1:
Case 2:
Reference:
Difference between T and P's house number is equal to the difference between Q and U's house number.
House number of R and Q together was equal to house number V.
Inference:
Here the difference between the house number of T and P can be 25 (when the house number of P is 65) or 10 (when the house number of P is 80)
And we know that the possible house numbers of Q are either 15 or 30.
Now, if the house number of Q is 15 then the only possible house number for U is 40 when the house number of P is 65.
And, if the house number of Q is 30 then the only possible house number for U is 40 when the house number of P is 80.
Reference:
Q was seated to the immediate right of the person whose house number was 65.
Q and U were seated at a gap of 1 person.
Inference:
Using, the above information our case 1 and case 2 can be redrawn as:
Reference:
One whose house number was 28 was seated to the immediate left of the one whose house number was 40.
S and U were not seated together.
Inference:
Here, Case A, Case B and Case D fails because we cannot place the person whose house number is 28 immediate to the lefty of the person whose house number is 40 in these cases.
At this point we can say that the house number of R is 15 and can fix his position as shown in the figure.
Reference:
S and U were not seated together.
Inference:
At this point we cannot place S in Case-E under the given conditions. So we can say that this is an invalid case.
But, with the given information we can place S and W and in our Case-C. As shown in the figure below:
Now, the puzzle is completed.
5
How many persons were seated between Q and S?
» Explain it
A
Following the final solution we can say that no one was sitting between Q and S.
Hence, the correct answer is option A.
Common Explanation:
Reference:
Not more than 2 persons have house number more than P.
Not more than 3 persons have house number less than Q.
T’s house number was twice of V.
House number of R and Q together was equal to house number V.
Three persons seated to the right of V had house numbers less than him.
Inference:
Here, the possible house numbers for P are 65 or 80 or 90. And the possible house numbers for Q are 15 or 28 or 30 or 40.
Similarly, the possible house numbers of V and T are 15 and 30, 40 and 80, and 45 and 90.
Here, house number 15 is not possible for V as we already know that three persons had house number less than V.
And house number 40 is also not possible for V because none of the combination of the house numbers leads to 40 and we have a hint that house number of R and Q was equal to house number V.
Thus, the only possible house number for V is 45 and the house number of T is 90.
Using the above information we can also say that the house number of R and Q can be 15 and 30 not necessarily in the same order. And the possible house number of P can be 65 or 80.
Reference:
Three persons seated to the right of V had house numbers less than him.
P is seated to the immediate right of T who is 3rd to the left of one whose house number was 45.
Inference:
With the first hint we can say that there were at least three persons sitting to the immediate right of V.
And with the second we can say that there are two possible scenarios in which we can fix the positions of these persons.
Case 1:
Case 2:
Reference:
Difference between T and P's house number is equal to the difference between Q and U's house number.
House number of R and Q together was equal to house number V.
Inference:
Here the difference between the house number of T and P can be 25 (when the house number of P is 65) or 10 (when the house number of P is 80)
And we know that the possible house numbers of Q are either 15 or 30.
Now, if the house number of Q is 15 then the only possible house number for U is 40 when the house number of P is 65.
And, if the house number of Q is 30 then the only possible house number for U is 40 when the house number of P is 80.
Reference:
Q was seated to the immediate right of the person whose house number was 65.
Q and U were seated at a gap of 1 person.
Inference:
Using, the above information our case 1 and case 2 can be redrawn as:
Reference:
One whose house number was 28 was seated to the immediate left of the one whose house number was 40.
S and U were not seated together.
Inference:
Here, Case A, Case B and Case D fails because we cannot place the person whose house number is 28 immediate to the lefty of the person whose house number is 40 in these cases.
At this point we can say that the house number of R is 15 and can fix his position as shown in the figure.
Reference:
S and U were not seated together.
Inference:
At this point we cannot place S in Case-E under the given conditions. So we can say that this is an invalid case.
But, with the given information we can place S and W and in our Case-C. As shown in the figure below:
Now, the puzzle is completed.
|
Quadrilateral is a 4 sided polygon bounded by 4 finite line segments. A quadrilateral has 2 diagonals based on which it can be classified into concave or convex quadrilateral. In case of convex quadrilaterals, diagonals always lie inside the boundary of the polygon.
Based on the lengths of sides and angles, common convex quadrilaterals are:
1. Trapezoid
2. Parallelogram
3. Square
4. Rectangle
5. Rhombus
6. Kite
Let us discuss in brief the properties of quadrilaterals.
1. Properties of Parallelogram :
2. Opposite sides are parallel and equal to each other
3. Opposite angles are equal
4. Diagonals bisect each other
5. Diagonals divide the parallelogram into two congruent triangles
6. If any one of the angles is
$$\begin{array}{l}90^{\circ}\end{array}$$
, then all angles are right angles
1. Properties of Kite:
• Two distinct pairs of adjacent sides are congruent
• Diagonals of a kite intersect at right angles
• One of the diagonals is the perpendicular bisector of another
• Angles between unequal sides are equal
1. Properties of Rhombus:
• All sides are congruent
• Diagonals are perpendicular bisectors of each other
• Each diagonal isthe angle bisector of both the opposite angles
• Every rhombus is a parallelogram and a kitewith all sides of equal length
1. Properties of Rectangle:
• All angles in a rectangle are right angles
• Opposite sides are equal in a rectangle
• Diagonals are congruent and bisect each other
• Every rectangle is a parallelogram with at least one right angle
1. Properties of Square:
• All sides are congruent
• Each interior angle measures
$$\begin{array}{l}90^{\circ}\end{array}$$
<
• Diagonals are equal and are perpendicular bisectors of each other
• Every square is a parallelogram in which diagonals are congruent and bisect the angles
• Every square is a rectangle and a rhombus
1. Properties of Trapezoid:
• One pair of opposite sides is parallel
• Diagonals intersect each other in the same ratio
• Two adjacent angles are supplementary
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# Numericals on Force and Laws of Motion for Class 9
Home » CBSE Class 9 Science » Extra Questions for Class 9 Science » Numericals on Force and Laws of Motion for Class 9
## Numericals on Force and Laws of Motion for Class 9
Here you will find numericals on force and laws of motion for class 9.
Problem:
Find the force acting on a rocket and its acceleration if its velocity is 600 m s–1 at t = 60 s and 1000 ms–1 at t = 80s, presuming its mass remains unchanged during this time interval and its mass is 1.67 × 107 kg. The total weight of the rocket at t = 80 seconds is 1.6 × 107 kg. Calculate the upward force acting on the rocket at this time caused by the burning of fuel.
Solution:
(i) Calculate the acceleration ($$a$$) and force ($$F$$) at $$t=60$$ seconds: \begin{aligned} a &= \frac{v-u}{t} = \frac{1000 \mathrm{~m/s} – 600 \mathrm{~m/s}}{80 \mathrm{~s} – 60 \mathrm{~s}} = \frac{400 \mathrm{~m/s}}{20 \mathrm{~s}} = 20 \mathrm{~m/s^2} \\ F &= m \cdot a = 1.67 \times 10^7 \mathrm{~kg} \times 20 \mathrm{~m/s^2} = 3.34 \times 10^8 \mathrm{~Newton} \end{aligned} (ii) Calculate the force ($$F$$) at $$t=80$$ seconds using the new total weight: $F = m \cdot a = 1.6 \times 10^7 \mathrm{~kg} \times 20 \mathrm{~m/s^2} = 3.2 \times 10^8 \mathrm{~Newton} = 320 \mathrm{~MN}$
Problem:
A body of 10 kg initially at rest is subjected to force of 20N. Calculate the kinetic energy acquired by the body at the end of 10 seconds.
Solution:
Given: $$m = 10 \mathrm{~kg}$$, $$u = 0$$, $$F = 20 \mathrm{~N}$$, $$t = 10 \mathrm{~s}$$ First, calculate the acceleration ($$a$$) using Newton’s second law: $F = m \cdot a \Rightarrow a = \frac{F}{m} = \frac{20 \mathrm{~N}}{10 \mathrm{~kg}} = 2 \mathrm{~m/s^2}$ Next, calculate the final velocity ($$v$$) using the equation of motion: $a = \frac{v – u}{t} \Rightarrow \frac{v – 0}{10 \mathrm{~s}} = 2 \mathrm{~m/s^2} \Rightarrow v = 20 \mathrm{~m/s}$ Finally, calculate the kinetic energy ($$KE$$) using the formula for kinetic energy: $KE = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 10 \mathrm{~kg} \cdot (20 \mathrm{~m/s})^2 = 2000 \mathrm{~J} = 2 \mathrm{~kJ}$ So, the kinetic energy acquired by the body at the end of 10 seconds is $$2 \mathrm{~kJ}$$.
Problem:
A truck starts from the rest and rolls down a hill with constant acceleration. It travels 200 m in 20 s. Find its acceleration. Find the force acting on it, if its mass is 5 metric tonnes. (1 metric tonne = 1000 kg)
Solution:
Given: $$u = 0$$, $$t = 20 \mathrm{~s}$$, $$s = 20 \mathrm{~m}$$, $$a = ?$$, $$\mathrm{F} = ?$$, $$m = 5000 \mathrm{~kg}$$ Using the equation of motion: $s = ut + \frac{1}{2}at^2$ Substituting the given values: $20 \mathrm{~m} = 0 + \frac{1}{2} \cdot a \cdot (20 \mathrm{~s})^2$ Solving for acceleration $$a$$: $a = \frac{400}{400} = 1 \mathrm{~m/s^2}$ Now, using Newton’s second law ($$F = ma$$) to find the force ($$\mathrm{F}$$): $\mathrm{F} = m \cdot a = 5000 \mathrm{~kg} \cdot 1 \mathrm{~m/s^2} = 5000 \mathrm{~Newton}$ So, the acceleration ($$a$$) is $$1 \mathrm{~m/s^2}$$ and the force ($$\mathrm{F}$$) is $$5000 \mathrm{~Newton}$$.
Problem:
A bullet of mass 10 g is horizontally fired with a velocity of 100 m s–1 from pistol of mass 1 kg. What is the recoil velocity of the pistol?
Solution:
Given: $$m = 10 \mathrm{~g} = \frac{10}{1000} \mathrm{~kg}$$, $$v = 100 \mathrm{~m/s}$$, $$\mathrm{M} = 1 \mathrm{~kg}$$, $$\mathrm{V} = ?$$
Using the formula for the velocity of the gun in the opposite direction to the bullet: $\mathrm{V} = -\frac{m v}{\mathrm{M}} = -\frac{\frac{10}{1000} \mathrm{~kg} \times 100 \mathrm{~m/s}}{1 \mathrm{~kg}}$ Solving for $$\mathrm{V}$$: $\mathrm{V} = -1 \mathrm{~m/s}$ So, the velocity of the gun is $$-1 \mathrm{~m/s}$$ in the opposite direction to the bullet being fired.
## Force and Laws of Motion Class 9 Important Formula
1. $$1 \, \text{Newton} = 1 \, \text{kg} \, \text{m/s}^{-2}$$; newton ($$\text{N}$$) is the SI unit of force.
2. $$\vec{p} = \overrightarrow{m \mathbf{v}}$$, where $$\vec{p}$$ is momentum, $$m$$ is mass, and $$\mathbf{v}$$ is velocity.
3. $$\vec{F} = \frac{\overrightarrow{p_f} – \overrightarrow{p_i}}{t} = \frac{m \mathbf{v} – m \mathbf{u}}{t} = \frac{m(\mathbf{v} – \mathbf{u})}{t} = m \mathbf{a}$$, where $$a$$ = acceleration. $$\frac{v-u}{t} = a]$$
4. $$\overrightarrow{\mathbf{F}} = m \mathbf{a}$$ Impulse $$= \text{force} \times \text{time} = \text{change in momentum}$$
5. $$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$ : Law of conservation of momentum.
6. Recoil of gun: $$\mathbf{V} = \frac{-m \mathbf{v}}{\mathbf{M}}$$, where $$m$$ is the mass of the bullet, $$M$$ is the mass of the gun, $$\mathbf{V}$$ is the velocity of the gun, and $$\mathbf{v}$$ is the velocity of the bullet.
7. $$\frac{\text{Rocket propulsion}}{\text{Velocity of rocket}} = V = \frac{-mV}{M}$$, where $$m$$ is the mass of burnt fuel, $$v$$ is the velocity of burnt fuel, and $$M$$ is the mass of the rocket at any instant.
## Why are Physics Numericals Important for Class 9 Students?
1. Application of Concepts: Physics numericals allow students to apply the theoretical concepts they learn in class to real-world situations. This practical application enhances their understanding of the subject.
2. Problem-Solving Skills: Numerical problems require critical thinking and problem-solving skills. Students learn to analyze information, identify relevant formulas, and calculate solutions, skills that are valuable in many aspects of life.
3. Enhanced Understanding: By solving numerical problems, students gain a deeper insight into the underlying principles of physics. It’s one thing to memorize formulas, but it’s another to comprehend how they are derived and applied.
4. Preparation for Higher Classes: A strong foundation in physics at the Class 9 level prepares students for more advanced physics topics in higher classes. It paves the way for success in Class 10 board exams and beyond.
## Tips for Solving Physics Numericals
1. Understand the Concept: Before attempting numericals, ensure you have a clear understanding of the underlying physics concepts. Review your class notes and textbook.
2. Identify Given Data: Carefully read the problem and identify the given data. This is crucial for selecting the appropriate formula.
3. Choose the Right Formula: Select the formula that best fits the problem. Familiarize yourself with the relevant equations for each topic.
4. Units and Conversions: Pay attention to units. Ensure all units are consistent throughout your calculations, and convert them if necessary.
5. Organize Your Work: Show your work step by step. This not only helps you track your progress but also allows for partial credit if you make a mistake.
6. Practice Regularly: Physics numericals improve with practice. The more problems you solve, the more confident and proficient you’ll become.
7. Seek Help When Needed: If you’re stuck on a problem, don’t hesitate to seek help from your teacher, classmates, or online resources.
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The Rotating Drum Problem
# The Rotating Drum Problem
Suppose we have a drum that rotates and contains $16$ compartments. Within each compartment we can use either the binary digit $0$ or the binary digit $1$. We want to find an order of binary digits such that all of the $16$-bit sequences of length $4$ appear as consecutive chunks on the rotating drum:
To solve this problem, let's first make a digraph on the eight $3$-bit sequence according to the rule that any binary word abc is also adjacent to $bc0$ and $bc1$. For example, $000$ would be adjacent to $000$ and $001$, or $110$ would be adjacent to $100$ and $101$. We thus get the following digraph:
Imagine the $3$-bit binary words having all of their digits shift to the left where the last digit is replaced by either a $0$ or a $1$ (by our rule above). We thus label the edges with either $0$'s or $1$'s depending on what the last digit changes to. Notice that the first digit "falls off" and the second digit is shifted to the position of the first digit.
Now notice that $\deg ^+ (v) = 2 \quad \mathbf{and} \quad \deg ^- (v) = 2$. Hence this graph must be Eulerian. We will now find an Eulerian trail. For example:
(1)
\begin{align} \quad 000 \rightarrow 000 \rightarrow 001 \rightarrow 010 \rightarrow 100 \rightarrow 001 \rightarrow 011 \rightarrow 110 \rightarrow 101 \rightarrow 010 \rightarrow 101 \rightarrow 011 \rightarrow 111 \rightarrow 111 \rightarrow 110 \rightarrow 100 \rightarrow 000 \end{align}
Now we will write the corresponding values of the arcs to get $0100110101111000$. We thus get the following arrangement:
Binary Word Decimal Equivalence
$0100$ $4$
$1001$ $9$
$0011$ $3$
$0110$ $6$
$1101$ $13$
$1010$ $10$
$0101$ $5$
$1011$ $11$
$0111$ $7$
$1111$ $15$
$1110$ $14$
$1100$ $12$
$1000$ $8$
$0000$ $0$
$0001$ $1$
$0010$ $2$
Hence we can label our rotating drum in the following manner:
Any $4$ consecutive compartments of the rotating drum will produce one of the unique $4$-bit binary words (note that there are $16$ many $4$-bit binary words equivalent to $0 - 15$).
|
# Division: Mixed Numbers
A mixed number is a number expressed as the sum of a whole number and a fraction , such as $3\frac{1}{4}$ .
When you divide a number by a mixed number, first rewrite the mixed number as an improper fractions . Then multiply the number by the reciprocal of the improper fraction.
Example 1:
Find the quotient. Write in simplest form.
$2\frac{1}{6}÷1\frac{1}{5}$
First, write the mixed numbers as improper fractions.
$\begin{array}{l}2\frac{1}{6}=\frac{13}{6}\\ 1\frac{1}{5}=\frac{6}{5}\end{array}$
So, the expression becomes
$\frac{13}{6}÷\frac{6}{5}$
Multiply by the reciprocal of $\frac{6}{5}$ , which is $\frac{5}{6}$ .
$\frac{13}{6}÷\frac{6}{5}=\frac{13}{6}\cdot \frac{5}{6}$
Multiply the numerators and multiply the denominators.
$=\frac{65}{36}$
Write the improper fraction as a mixed number.
$=1\frac{29}{36}$
So,
$2\frac{1}{6}÷1\frac{1}{5}=1\frac{29}{36}$
Example 2:
Find the quotient. Write in simplest form.
$7\frac{1}{2}÷2\frac{1}{10}$
First, write the mixed numbers as improper fractions.
$\begin{array}{l}7\frac{1}{2}=\frac{15}{2}\\ 2\frac{1}{10}=\frac{21}{10}\end{array}$
So, the expression becomes
$\frac{15}{2}÷\frac{21}{10}$
Multiply by the reciprocal of $\frac{21}{10}$ , which is $\frac{10}{21}$ .
$\frac{15}{2}÷\frac{21}{10}=\frac{15}{2}\cdot \frac{10}{21}$
The GCF of $15$ and $21$ is $3$ . So, to simplify the fractions, divide $15$ and $21$ by $3$ .
The GCF of $2$ and $10$ is $2$ . So, to simplify the fractions, divide $2$ and $10$ by $2$ .
$\begin{array}{l}=\frac{\stackrel{5}{\overline{)15}}}{\underset{1}{\overline{)2}}}\cdot \frac{\stackrel{5}{\overline{)10}}}{\underset{7}{\overline{)21}}}\\ =\frac{5}{1}\cdot \frac{5}{7}\end{array}$
Multiply the numerators and multiply the denominators.
$=\frac{25}{7}$
Write the improper fraction as a mixed number.
$=3\frac{4}{7}$
So,
$7\frac{1}{2}÷2\frac{1}{10}=3\frac{4}{7}$ .
|
# Find the Relative Salary
## Puzzle Statement
Steve would like to determine the relative salaries of three coworkers using two facts:
1. First, he knows that if Fred is not the highest paid of three, then Janice is.
2. Second, he knows that if Janice is not the lowest paid,, then Maggie is paid the most.
Is it possible to determine the relative salaries of Fred, Maggie and Janice from what Steve knows? If so, who is paid the most and who the least?
## Quick Solution
Fact 1 says that if Fred is not the highest paid, then Janice is. This results in two situations:
1. Fred is not the highest paid
1. Janice is the highest paid.
2. Then by fact 2, Janice is not the lowest pad hence Maggie is the most paid.
3. Which contradicts with established truth that Janice is highest paid.
4. Hence, this case is incorrect and results in second condition.
2. Fred is the highest paid
1. Now Maggie can't be the most paid, because Fred is the most paid. Hence, the second statement is false.
2. Which means that Janice is lowest paid.
This establishes the relative order Janice < Maggie < Fred
## Solution using propositions to Find the Relative Salary
In the previous puzzle "Identify the Murder", we solved the puzzle with reasoning without formal methodologies. Here we will try to find our way through the propositional calculus and arrive to a situation. Let us consider three statements:
1. F denotes that, "Fred is highest paid."
2. J denotes that, "Janice is lowest paid."
3. M denotes that,"Maggie is the most paid."
Then the first fact says that, if NOT F then NOT J AND NOT M. This can be expressed using propositions as ¬F → (¬J ∧ ¬M). The second fact says that, if NOT J then M. This can be expressed using propositions as ¬J → M. We can use these two propositions to check for consistency. For this purpose we can use a truth table as shown below: The last two columns of the truth table shows the two facts we know. Also, both the facts are true only in the last three rows. Let us now examine the last three rows of the table separately. Looking at the first three columns, shows the following:
1. Row 1 & Row 3 has F and M true, which means Fred is highest paid and Maggie is most paid. But both cannot be highest paid, hence these two rows do not satisfy the propositional consistency.
2. Whereas Row 2 is not contradicting the proposition.
So, the solution to this problem would be when F is true, J is true and M is false. Which means
• Fred is highest paid
• Janice is lowest paid and
• Maggie is not the most paid.
Hence the relative order of their salaries would be Janice < Maggie < Fred.
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Frequently Asked QuestionsWhat is the simplest form of 83 100?
# What is the simplest form of 83 100?
Find Out More About The Free Tools on Visual Fractions As you can see, 83/100 cannot be simplified any further, so the result is the same as we started with. Not very exciting, I know, but hopefully you have at least learned why it cannot be simplified any further!
## What is .17 as a fraction?
Similarly, if you divide 17 by 100 , you get 0.17 . Thus, the fraction form for 0.17 is 17100 . Since there is no common factor which can divide the numerator, 17 , and the denominator, 100 , besides 1 , 17100 is the final answer.
## How do you calculate fractions by hand?
To work out a fraction of a number, all you need to do is divide that number by the denominator of the fraction, then multiply your result by the numerator of the fraction. For instance, if we wanted to find out what 3/8 of 24 is, we would divide 24 by 8 (the denominator of the fraction).
## How do you work out 80% of a number?
Similarly, you can multiply the 1% answer by any number to find any percentage value. For example, to find 80% of 4,500, multiply the 45 by 80 to get 3,600.
## What is 0.02 as a fraction?
Answer: 0.02 can be written in fraction form as 1/50. Since 2 is the common factor of 2 and 100 so we divide both the numerator and denominator by 2.
## What is 26% as a fraction?
26 percent in simplest form is the fraction 13/50. Since ‘percent’ means ‘per hundred,’ 26 percent would be equivalent to the fraction 26/100.
## How do you write 80% as a decimal?
80% can be thought of as 80% of one. Percents also refer to numbers out of 100, thus, our decimal can be calculated by taking 80/100. This would result in 8/10 after simplifying, which is equivalent to 0.8.
## What is 90 as a fraction?
Now, since 90 does not have any value after the decimal, we can write it as 90.0. We see that 900 / 10 is in the lowest form as the HCF(900,10) = 10. Thus it is further simplified to 90/1. You can also use Cuemath’s online fraction calculator to find the simplest form of a given fraction.
## What is fraction formula?
An equation in which one or more terms is a fraction is called a fractional equation. To solve a fractional equation, first eliminate the fractions by multiplying both sides of the equation by the least common denominator (LCD) of every term.
## How do you do fractions on Google calculator?
Use the calculator function to divide two fractions. You divide fractions by inverting the second one and multiplying them both together. Input one fraction into the search box and enclose it in parentheses.
## How do you simplify improper fractions?
To simplify an improper fraction, start by turning it into a mixed number by dividing the numerator by the denominator. Then, turn the remainder into a fraction by placing it over the denominator of the original fraction. If necessary, simplify the final fraction to get your answer.
## How do you write 2 as a fraction?
Answer: 2% can be represented as 1/50 in fractional form. Step 1: We represent the given number as a fraction by dividing it by 100. Hence, we write it as 2/100.
## Can 83 100 be simplified?
As you can see, 83/100 cannot be simplified any further, so the result is the same as we started with.
## What is 24 in a fraction?
Note: To turn a decimal into a percent, simply multiply by 100%. 24100 can be simplified to 625 which is your final fraction form.
## How do you turn 0.222 into a fraction?
0.2222… is equal to the fraction with 2 in its numerator (since that’s the single number after the decimal point that’s repeating over and over again) and 9 in its denominator. In other words, 0.2222… = 2/9.
## What is a fraction for kids?
A fraction is a part of a whole number, and a way to split up a number into equal parts. It is written as the number of equal parts being counted, called the numerator, over the number of parts in the whole, called the denominator. These numbers are separated by a line.
## What is a fraction class 5?
A fraction is a numerical quantity that is not a whole number. For example, ½ is a fraction of. 1 as numerator and. 2 as a denominator.
See also How many 2 oz shots are in a 750ml bottle?
## How do you do fractions on online calculator?
Look for a button that has a black box over a white box, x/y, or b/c. Push this button to open the fraction feature on your calculator. When the fraction feature is on, you should see a fraction template on your calculator screen.
## How do you do fractions on calculator online?
When the calculator is in Math mode, the word “math” appears at the top of the screen. Once you have selected this mode (if necessary), look for a button with two boxes, one black and one white, arranged on top of each other with a horizontal line between them. This is the fraction button.
## How do you find 1/6 of a number?
Divide your numerator by your denominator to get your answer. In this case 24/6 = 4, meaning that 4 is 1/6th of 24. Multiply 1/6th of a fraction by another fraction without changing the denominator. For example if you need to find 1/6th of 3/4, then you multiply the two as follows 1/6 x 3/4 to get 3/24.
## What is 87 percent as a fraction?
2. What is 87% in the fraction form? 87% in the fraction form is 87/100. If you want you can simplify it further as 87/100.
## How do I find 2/3 of a number?
To find 2/3 of a whole number we have to multiply the number by 2 and divide it by 3. To find two-thirds of 18, multiply 2/3 x 18/1 to get 36/3. 36/3 is again simplified as 12.
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Sunday, 29 Jan 2023
# Top 10 what is 25 percent of 250 That Will Change Your Life
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Below is information and knowledge on the topic what is 25 percent of 250 gather and compiled by the nhomkinhnamphat.com team. Along with other related topics like: 20 percent of 250, 25 percent of 150, 25 percent of 350, 25 percent of 300, 25 percent of 200, 15 percent of 250, 25 percent of 240, 75 percent of 250.
e Calculator: What is 25 percent of 250 – percentagecalculator.guru
25 percent *250
= (25/100)*250
= (25*250)/100
= 6250/100 = 62.5
Now we have: 25 percent of 250 = 62.5
Question: What is 25 percent of 250?
We need to determine 25% of 250 now and the procedure explaining it as such
Step 1: In the given case Output Value is 250.
Step 2: Let us consider the unknown value as x.
Step 3: Consider the output value of 250 = 100%.
Step 4: In the Same way, x = 25%.
Step 5: On dividing the pair of simple equations we got the equation as under
250 = 100% (1).
x = 25% (2).
(250%)/(x%) = 100/25
Step 6: Reciprocal of both the sides results in the following equation
x%/250% = 25/100
Step 7: Simplifying the above obtained equation further will tell what is 25% of 250
x = 62.5%
Therefore, 25% of 250 is 62.5
250 percent *25
= (250/100)*25
= (250*25)/100
= 6250/100 = 62.5
Now we have: 250 percent of 25 = 62.5
Question: Solution for What is 250 percent of 25?
We need to determine 250% of 25 now and the procedure explaining it as such
Step 1: In the given case Output Value is 25.
Step 2: Let us consider the unknown value as x.
Step 3: Consider the output value of 25 = 100%.
Step 4: In the Same way, x = 250%.
Step 5: On dividing the pair of simple equations we got the equation as under
25 = 100% (1).
x = 250% (2).
(25%)/(x%) = 100/250
Step 6: Reciprocal of both the sides results in the following equation
x%/25% = 250/100
Step 7: Simplifying the above obtained equation further will tell what is 250% of 25
x = 62.5%
Therefore, 250% of 25 is 62.5
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1. How do I calculate percentage of a total?
To calculate percentages, start by writing the number you want to turn into a percentage over the total value so you end up with a fraction. Then, turn the fraction into a decimal by dividing the top number by the bottom number. Finally, multiply the decimal by 100 to find the percentage.
2. What is 25 percent of 250?
25 percent of 250 is 62.5.
3. How to calculate 25 percent of 250?
Multiply 25/100 with 250 = (25/100)*250 = (25*250)/100 = 62.5.
## Extra Information About what is 25 percent of 250 That You May Find Interested
If the information we provide above is not enough, you may find more below here.
### What is 25 percent of 250 – percentagecalculator.guru
• Author: percentagecalculator.guru
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• Sumary: What is 25 percent of 250? How much is 25% of 250?
• Matching Result: 25 percent of 250 is 62.5. 3. How to calculate 25 percent of 250? Multiply 25/100 with 250 = (25/100)*250 = (25*250)/ …
• Intro: Percentage Calculator: What is 25 percent of 250 – percentagecalculator.guru25 percent *250= (25/100)*250= (25*250)/100= 6250/100 = 62.5Now we have: 25 percent of 250 = 62.5Question: What is 25 percent of 250?We need to determine 25% of 250 now and the procedure explaining it as suchStep 1: In the given case…
### What is 25 percent of 250?
• Rating: 4⭐ (411573 rating)
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• Sumary: What is 25 percent of 250? The answer is 62.5. Get stepwise instructions to work out “25% of 250”.
• Matching Result: Write 25% as 25/100 · Since, finding the fraction of a number is same as multiplying the fraction with the number, we have 25/100 of 250 = 25/100 × 250 …
• Intro: What is 25 percent of 250? 25% of 25025% of 250 is 62.5Working out 25% of 250Write 25% as 25/100Since, finding the fraction of a number is same as multiplying the fraction with the number, we have25/100 of 250 = 25/100 × 250Therefore, the answer is 62.5If you are using…
### What is 25 percent of 250? = 62.5 – Percentage Calculator
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• Sumary: 25 percent * 250 =
• Matching Result: Solution for What is 25 percent of 250: 25 percent * 250 = (25:100)* 250 = (25* 250):100 = 6250:100 = 62.5. Now we have: 25 percent of 250 = 62.5.
• Intro: What is 25 percent of 250? = 62.5 Solution for What is 25 percent of 250: 25 percent * 250 = (25:100)* 250 = (25* 250):100 = 6250:100 = 62.5Now we have: 25 percent of 250 = 62.5Question: What is 25 percent of 250?Percentage solution with steps:Step 1: Our output…
### 25 percent of 250 dollars
• Author: percent-off.com
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• Sumary: How to calculate 25% of 250. How to figure out percentages of a price. Using this calculator you will find that 25% of 250 is …
• Matching Result: 25% of 250 is 62.5. To calculate 25 of 250 you just need to multiply the percent value (25) by the quantity (250) then divide the result by one hundred.
• Intro: 25 percent of 250 dollars Go to: Calculator | Percent Of Chart 25% of 250 is 62.5. To calculate 25 of 250 you just need to multiply the percent value (25) by the quantity (250) then divide the result by one hundred. Using this calculator you solve 3 types of…
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### What is 25 percent of 250? – getcalc.com
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• Sumary: 25% of 250 provides the detailed information of what is 25 percent of 250, the different real world problems and how it is being calculated mathematically.
• Matching Result: 250 is the reference or base quantity, 62.5 is 25 percent of 250. … In offers and discount, 25 off 250 generally represents 25 percent off in \$250. 25% off \$250 …
• Intro: 25% of 250 – getcalc.comThe below step by step work shows how to find what is 25 percent of 250. In the calculation, 25 is the relative quantity for each 100 against the base quantity 250. Some of the usage scenarios of 25% of 250 involves calculating commodity price increase…
### Solved: What is 25 Percent of 250? = 62.5
• Author: percentage-calculator.net
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• Sumary: What is 25 percent of 250? | 25% of 250 =? With this Percentage Calculator, you will get the quick answer and step by step guide on how to solve your percentage problems with examples.
• Matching Result: The fastest step by step guide for calculating what is 25 percent of 250. We already have our first value 25 and the second value 250. Let’s assume the unknown …
• Intro: Solved: What is 25 Percent of 250? = 62.5 | 25% of 250 = 62.5 Answer: 62.5 How to calculate 25% of 250 25 percent × 250 = (25 ÷ 100) × 250 = (25 × 250) ÷ 100 = 6250 ÷ 100 = 62.5 Now we have: 25% of…
### 25 percent of 250 – Percent Calculator
• Author: percent.info
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• Sumary: What is 25 percent of 250? Step-by-step showing you how to calculate 25 percent of 250. Detailed explanation with answer to 25% of 250.
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• Intro: 25 percent of 250 (25% of 250) Here we will show you how to calculate twenty-five percent of two hundred fifty. Before we continue, note that 25 percent of 250 is the same as 25% of 250. We will write it both ways throughout this tutorial to remind you that…
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### 25% of 250 = ? What is 25 percent of the number 250? The …
• Author: percentages.calculators.ro
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• Sumary: p% of A = ? Calculate percentages of numbers, online calculator. Please check the entered values. Number: no value
• Matching Result: If 25% × 250 = 62.5 => · Divide the number 62.5 by the number 250… ·… And see if we get as a result: 25% …
• Intro: p% of A = ? Calculate percentages of numbers, online calculator. Please check the entered values. Number: no value Menu p% of A = ? p% of ? = A ?% of A = B fractions a/b = ?% relative change p% = ? number (1 + p%) × n…
### What is 25 Percent of 250? (Answer Explained)
• Rating: 4⭐ (411573 rating)
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• Sumary: 25 percent of 250 is 62.5. To find 25% of 250, divide the percent by 100 and multiply by the number. So multiply 25/100 by 250 to get an answer of 62.5.
• Matching Result: 25 percent of 250 equals 62.5. To get this answer, multiply 0.25 by 250. You may need to know this answer when solving a math problem that multiplies both …
• Intro: What is 25 Percent of 250? (Answer Explained) 25 percent of 250 equals 62.5. To get this answer, multiply 0.25 by 250. You may need to know this answer when solving a math problem that multiplies both 25% and 250. Perhaps a product worth 250 dollars, euros, or pounds is…
### What is 25 percent off 250 dollars or pounds
• Author: percentage-off-calculator.com
• Rating: 4⭐ (411573 rating)
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• Sumary: Note: 250 dollar to pound = 165 pound
• Matching Result: So if you buy an item at \$250 with 25% discounts, you will pay \$187.5 and get 62.5 cashback rewards. Calculate 25 percent off 250 dollars using this …
• Intro: What is 25 percent off 250 dollars| How to calculate 25% off 250 pounds| 25% off 250 What is 25 percent off 250 dollars or pounds ? Note: 250 dollar to pound = 165 pound Solution: 25% off 250 is equal to (25 x 25) / 100 = 62.5. So…
If you have questions that need to be answered about the topic what is 25 percent of 250, then this section may help you solve it.
### 10% of 250 equals 25%
Ten percent of 250 equals 25.
### Answer: 50 is 25% of 200.
50 is 25% of 200, so the answer is.
25
### So 15% of 250 equals 37.5.
Therefore, 15% of 250 equals 37.5.
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Find the slope of a line by comparing rise (how much y changes) over run (how much x changes). The formula for slope is: m = [y2 – y1] / [x2 – x1].
## What is the basic formula for slope?
y = mx + b, where m is the gradient or the slope and b is the y-intercept. 1. Slope of the line when θ is the angle between the lines.
## What is slope in geometry?
In mathematics, the slope or gradient of a line is a number that describes both the direction and the steepness of the line. Slope is calculated by finding the ratio of the “vertical change” to the “horizontal change” between (any) two distinct points on a line.
## What are the 3 slope formulas?
Point-slope form, standard form, and slope-Intercept form are the major forms of linear equations.
## What is slope in linear equation?
Slope measures the rate of change in the dependent variable as the independent variable changes. The greater the slope the steeper the line. Consider the linear function: y = a + bx. b is the slope of the line.
## Is slope used in geometry?
Definition: The slope of a line is a number that measures its “steepness”, usually denoted by the letter m. It is the change in y for a unit change in x along the line. The slope of the line is continuously recalculated.
You might be interested: What Is The Molecular Geometry Of Xef4?
## How do you find slope without graphing?
Explanation: To find the slope given two points without using a graph, we use the formula riserun, or y2−y1x2−x1. Therefore, the slope is −72 or −3.5. Hope this helps!
## How do you find slope with coordinates?
Use the slope formula to find the slope of a line given the coordinates of two points on the line. The slope formula is m=(y2-y1)/(x2-x1), or the change in the y values over the change in the x values.
## Why do you use the formula for slope?
The slope is one of the essential characteristics of a line and helps us measure the rate of change. The slope of a straight line is the ratio of the change in y to the change in x, also called the rise over run.
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10182
# Coconuts, Sailors, and a Monkey
Sailors, stranded on an island, collect all the coconuts they can find. They agree to divide up the coconuts in the morning. But each, in turn, gets up in the night to take his share from the pile. At each division, one is given to a monkey to keep him quiet. In the morning, none are surprised at how the pile has shrunk. They give another coconut to the monkey and divide what remains. What is the smallest number of coconuts such that all divisions come out even?
In classic examples of this problem, there are 3 sailors and 4 divisions or 5 sailors and 6 divisions. Sometimes the monkey does not get his coconut in the morning.
### DETAILS
Let there be sailors and divisions. Let , , …, be the initial and subsequent numbers of coconuts in the pile, , ..., the sailors' shares, and , …, the monkey's share at the divisions.
Then, and , .
Many methods have been used to solve this problem. We recognize as a linear nonhomogeneous difference equation in . Its solution is , where is a particular solution to the equation, .
The difference equation can be rewritten as .
A particular solution can be verified to be , where we define for so that the and higher differences are zero; for example, could be defined by an interpolating polynomial of degree . Note that is an integer, .
Since is an integer, and , divides .
Therefore, , , where is the smallest integer such that are all positive, where . The can be generated from using the difference equation.
If , then .
If , then .
If , then .
If , , then . When and , This is the most famous case, with the monkey getting one coconut at each division.
If , and , then , where if is even and if is odd. When and , . This is the case where the monkey does not get a coconut in the last division. See the Williams's problem below.
How many coconuts are there if , the monkey gets 10 coconuts at each division in the night, and none in the morning? Answer: .
How many coconuts are there if , the monkey removes 2 before the first division, adds 2 before the second, removes 2 before the third, and gets one in the morning? corresponds to the monkey adding to the pile.) Answer: .
In the classical examples, . But individual sailors could get up more than once or not at all. The monkey can take or add any number of coconuts at each division. Also, if n = 1 then k = 1, since the coconuts are gone after one "division".
Suppose , , , and . (The monkey adds 10, there is one division at night, and the monkey gets the 10 back in the morning.) In this case, , the possibilities are , and the minimum number of coconuts is , when .
Martin Gardner's chapter "The Monkey and the Coconuts", in The Second Scientific American Book of Mathematical Puzzles and Diversions, begins with a problem included in a short story by Ben Ames Williams entitled "Coconuts", which appeared in the October 9, 1926 issue of The Saturday Evening Post. The magazine received over 2,000 letters the first week from readers who wanted to know, "How many coconuts?" Here is the problem as originally published:
"Five men and a monkey were shipwrecked on a desert island, and they spent the first day gathering coconuts for food. Piled them all up together and then went to sleep for the night.
But when they were all asleep one man woke up, and he thought there might be a row about dividing the coconuts in the morning, so he decided to take his share. So he divided
the coconuts into five piles. He had one coconut left over, and he gave that to the monkey, and he hid his pile and put the rest all back together. By and by the next man woke up and did the same thing. And he had one left over, and he gave it to the monkey. And all five of the men did the same thing, one after the other; each one taking a fifth of the coconuts in the pile when he woke up, and each one having one left over for the monkey. And in the morning they divided what coconuts were left, and they came out in five equal shares. Of course each one must have known there were coconuts missing; but each one was guilty as the others, so they didn't say anything. How many coconuts were there in the beginning?"
Williams modified an older problem in order to make it more confusing. The older problem has a coconut for the monkey at each of the six divisions, whereas in Williams' problem, the last division comes out even.
Gardner has claimed, "Today the problem of the coconuts is probably the most worked on and least-often solved of all the Diophantine brain-teasers."
The explicit formula above first appeared in "The Generalized Coconut Problem," by Roger B. Kirchner.
M. Gardner, The Second Scientific American Book of Mathematical Puzzles and Diversions, Chicago: The University of Chicago Press, 1961.
R. B. Kirchner, "The Generalized Coconut Problem," The American Mathematical Monthly, 67(6), 1960 pp. 516–519.
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# Fractions Class 6 RS Aggarwal Exe-4G Goyal Brothers ICSE Maths Solutions
Fractions Class 6 RS Aggarwal Exe-4G Word Problems Goyal Brothers Prakashan ICSE Foundation Maths Solutions. We provide step by step Solutions of ICSE Class-6 Foundation RS Aggarwal Mathematics of Goyal Brothers Prakashan. Visit official Website CISCE for detail information about ICSE Board Class-6 Mathematics.
## Fractions Class 6 RS Aggarwal Exe-4G Word Problems Goyal Brothers Prakashan ICSE Foundation Maths Solutions
Board ICSE Publications Goyal brothers Prakashan Subject Maths Class 6th Ch-4 Fraction Writer RS Aggrawal Book Name Foundation Topics Solution of Exe-4G Academic Session 2023 – 2024
### Word Problems on Fractions
Fractions Class 6 RS Aggarwal Exe-4G Goyal Brothers Prakashan ICSE Foundation Maths Solutions.
Page- 83
#### Exercise- 4G
##### Que-1: A man earns Rs 18720 per month and spends 5/6 of his income. Find his monthly (i) expenditure (ii) saving.
Solution- (i) It is given that he spends 5/6 of 18720 Rupees.
So, 5/6 of 18720 Rupees :-
= 5/6 x 18720
= 5 x 3120
= 15600 Rupees
His monthly expenditures is 15600 Rupees Ans.
##### Que-2: A water tank can hold 56*(1/4) litres of water. How much water is contained in the tank when it is 8/15 full?
Solution- The total capacity of the water tank is 225/4 litres.
To find the amount of water when the tank is 8/15 full, we multiply the total capacity of the tank (225/4 litres) by the fraction representing the tank’s fill level (8/15).
= (225/4) × (8/15)
= (225 × 8) / (4 × 15)
= 1800 / 60
= 30 litres. Ans
Hence, when the tank is 8/15 full, it contains 30 litres of water.
##### Que-3: After reading 5/8 of a book, 168 pages are left. How many pages are there in all in the book?
Solution- Let total book is 1 part
Read = 5/8 parts
Unread = [1 – ( 5/8)] parts
3/8 parts
According to the question,
(3/8) parts = 168 pages
1 part = (168 × 8)/3 pages
448 pages
Total number of pages in the book is 448.
##### Que-4: 2/3 of a students in a class are boys and the rest are 17 girls. (i) How many boys are there in a class?(ii) What is the total strength of a class?
Solution- (i) Let x be the total strength of the class
No. of boys= 2x/3 ; No. of girls= 17
x = (2x+51)/3
3x = 2x+51
3x – 2x = 51
x = 51
No. of boys = 2/3 of 51
= 34 Ans.
(ii) means 1/3 are girls
So, 1/3 = 17.
2/3 = 1/3 + 1/3 = 17 + 17
= 3/4 boys are in the class.
total strength of class = 17 + 34
= 51 Ans.
##### Que-5: A man earns Rs 25440 per month. He spends 1/4 of it on house rent, 3/8 on food and clothes, 1/10 on insurance and 1/5 on other items. How does he save each month?
Solution- Total earning in 1 month = ₹25440
Money spent on rent is 1/4 of total earning = 1/4×25440
= ₹6360
Money spent on food is 3/8 of total earning = ₹ 3/8×25440
= ₹9540
Money spent on insurance is 1/10 of total earning = ₹ 1/10×25440
= ₹2544
Money spent on other items is 1/5 of total earning = ₹ 1/5×25440
= ₹5088
Therefore, the total money spent = ₹ (6360+9540+2544+5088)
= ₹23532
So, money saved by the man = ₹(25440-23532)
= ₹1908
Hence, the man saves ₹1908 each month.
##### Que-6: An objective test was given to a group of 168 students. It was found that 5/6 of the students gave all correct answers. How many students made 1 or more mistakes?
Solution- Group of students 168.
Students give correct answers = 5/6 of 168
students give correct answers = 5/6 x 168
= 140 students.
No. Students who made mistakes. (168-140) = 28 students.
Hence, 28 students made mistakes.
##### Que-7: In an orchard, 1/3 of the trees are guava trees, 1/8 are bananas trees and the rest are mango trees. If there are 117 mango trees in the orchard, how many trees in all there?
Solution- Let there are n trees in an orchid
Given 1/3 of the trees are guava and 1/8 of the trees are banana
Remaining are Mango i.e.,
n − n/3 − n/8 = Mango trees
Given Mango trees =117
117 = (24n−8n−3n)/24
117 = 13n/24
n = 9×24
n = 216
There are 216 trees in the orchid.
##### Que-8: In a school, 1/25 of a students were absent on a certain day. If 720 students were present on that day, what is the total number of students in the school?
Solution- Given that 720 students were present on that day, we can set up the following proportion to find the total number of students in the school:
24/25 = 720/x Where x represents the total number of students in the school.
We can solve this proportion for x:
x = (720 X 25)/24
x =18000/24
x = 750
Therefore, the total number of students in the school is 750.
##### Que-9: The products of three numbers is 7*(1/2). If two of them are 1*(1/7) and 3*(3/4), find the third number.
Solution- let the third number be x
15/2 = 8/7 × 15/4 × x
15/2 = 2/7 × 15 × x
x = 15/2 × 7/2 × 1/15
x = 7/4 = 1*(3/4)
-: end of Fractions Class 6 RS Aggarwal Exe-4G Word Problems Goyal Brothers Publication :–
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exercises > prime-testing
# Exercise #21: Testing if Number is Primeby Zoran Horvat@zoranh75
## Problem Statement
Given an integer number N, N > 1, write a function which returns true if that number is prime and false otherwise. Number is prime if it has no other divisors than 1 and itself.
Example: For numbers 2, 5, 17 function should return true. For number 143 function should return false because that number is divisible by 11 and 13.
## Problem Analysis
We can discover that a given number is prime by failing to prove the opposite, i.e. by verifying that there is no eligible value which divides it without remainder. The trick in this task is to properly identify eligible divisors. These should be numbers greater than one which are not equal to the number tested. Further on, there is no point in taking into account values greater than N because none of them could possibly divide N. So, by now we have constrained candidates to set {2, 3, ..., N-1}.
Now suppose that there is such a value k which is the smallest number which divides N:
This implies that m is also a divisor of N. But the fact that k is the smallest divisor implies that m cannot be less than k. From these facts we can derive one important conclusion:
This has significantly reduced the problem. But we can make a step further. There is no point in trying to divide N with even values of k, except the trivial candidate 2. This because if N is not divisible by 2, then there is no chance that any other even number could divide it. The same logic goes with number 3. If 3 doesn't divide N, then none other multiple of 3 can divide N either. This leads to an interesting conclusion about possible divisors of N: the only viable candidates that could divide N are values 2, 3 and odd numbers around multiples of 6:
This opts out roughly two out of three candidates not exceeding square root of N. And here is the pseudocode which solves the problem:
```function IsPrime(n)
begin
result = false
if n <= 3 then
result = true
else if n mod 2 <> 0 AND n mod 3 <> 0 then
begin
k = 5
step = 2
while k * k <= n AND n mod k <> 0
begin
k = k + step
step = 6 - step
end
if n = k OR n mod k <> 0 then
result = true
end
return result
end
```
## Implementation
The following C# code is a console application which implements the IsPrime function to test whether its argument is prime or not:
```using System;
{
public class Program
{
static bool IsPrime(int n)
{
bool result = false;
if (n <= 3)
{
result = true;
}
else if (n % 2 != 0 && n % 3 != 0)
{
int k = 5;
int step = 2;
while (k * k <= n && n % k != 0)
{
k = k + step;
step = 6 - step;
}
if (n == k || n % k != 0)
result = true;
}
return result;
}
static void Main(string[] args)
{
while (true)
{
Console.Write("Enter number (zero to exit): ");
if (n <= 0)
break;
if (IsPrime(n))
Console.WriteLine("Number {0} is prime.", n);
else
Console.WriteLine("Number {0} is not prime.", n);
}
}
}
}
```
## Demonstration
When application above is run, it produces the following output:
```Enter number (zero to exit): 2
Number 2 is prime.
Enter number (zero to exit): 3
Number 3 is prime.
Enter number (zero to exit): 17
Number 17 is prime.
Enter number (zero to exit): 18
Number 18 is not prime.
Enter number (zero to exit): 143
Number 143 is not prime.
Enter number (zero to exit): 64657551
Number 64657551 is not prime.
Enter number (zero to exit): 64657553
Number 64657553 is prime.
Enter number (zero to exit): 0
```
Published: Jan 6, 2014; Modified: Dec 14, 2014
ZORAN HORVAT
Zoran is software architect dedicated to clean design and CTO in a growing software company. Since 2014 Zoran is an author at Pluralsight where he is preparing a series of courses on object-oriented and functional design, design patterns, writing unit and integration tests and applying methods to improve code design and long-term maintainability.
Watch Zoran's video courses at pluralsight.com (requires registration):
Making Your C# Code More Object-Oriented
This course will help leverage your conceptual understanding to produce proper object-oriented code, where objects will completely replace procedural code for the sake of flexibility and maintainability. More...
This course will lead you step by step through the process of developing defensive design practices, which can substitute common defensive coding, for the better of software design and implementation. More...
Tactical Design Patterns in .NET: Creating Objects
This course sheds light on issues that arise when implementing creational design patterns and then provides practical solutions that will make our code easier to write and more stable when running. More...
Tactical Design Patterns in .NET: Managing Responsibilities
Applying a design pattern to a real-world problem is not as straight-forward as literature implicitly tells us. It is a more engaged process. This course gives an insight to tactical decisions we need to make when applying design patterns that have to do with separating and implementing class responsibilities. More...
Tactical Design Patterns in .NET: Control Flow
Improve your skills in writing simpler and safer code by applying coding practices and design patterns that are affecting control flow. More...
Writing Highly Maintainable Unit Tests
This course will teach you how to develop maintainable and sustainable tests as your production code grows and develops. More...
Improving Testability Through Design
This course tackles the issues of designing a complex application so that it can be covered with high quality tests. More...
|
# What is the phrase”exclamation mark” in mathematics?
The answer is straightforward. Below are a few strategies to tell when an equation can be a multiplication equation and also just how to add and subtract by using”exclamation marks”. For instance,”2x + 3″ mean”multiply two integers by 3, building a price corresponding to two plus several”.
For instance,”2x + 3″ are multiplication from professional essay just three. Additionally, we could add the values of 2 and three . To bring the values, we’ll use”e”that I” (or”E”). Using”I” indicates”include the value of one to the value of 2″.
To add the values, we can do this similar to this:”x – y” implies”multiply x by y, building a price equal to zero”. For”x”y”, we will use”t” (or”TE”) for the subtraction and we will utilize”x + y” to address the equation.
You might feel which you are not assumed to utilize”electronic” in addition as”that I” implies”subtract” but it really masterpapers is not really easy. By way of instance, to say”two – 3″ means subtract from three.
Thus, to bring the values we utilize”t”x” (or”TE”), which might be the numbers of the worth to be included. We will utilize”x” to subtract the worth of a person in the worth of 2 and that will give us exactly the result.
To multiply the worth we certainly can do it similar to that:”2x + y” indicate”multiply two integers with y, creating a value equal to one plus two”. You will know that this is actually a multiplication equation when we utilize”x” to subtract one from 2. Orit can be”x – y” to subtract one from 2. Note that the equation can be written by you using a decimal level and parentheses.
Now, let’s have a good illustration. Let us mention that people would like to multiply the worth of”nine” by”5″ and we have”x = nine”y = five”. https://wtamu.edu/~cbaird/sq/category/physics/ Afterward we will use”x – y” to reevaluate the value of one from the worth of two.
|
# How to Find Lowest Common Multiple (LCM) of Expressions?
How to find the lowest common multiple (LCM) of two or more expressions in maths? Detailed solutions to examples are presented and full solutions to the questions in Detailed Solutions and explanations are included.
## What is the lowest common multiple (LCM) of 2 or more maths expressions?
The lowest common multiple of two or more expressions is the smallest (or simplest) expression that is divisible by each of these expressions. It is found by first
factoring completely each of the given expressions then use these factors to write the LCM. Detailed examples are shown below.
### Example 1
Find the lowest common multiple of the expressions x 2 - 1 and x - 1.
Solution
We first factor the given expressions
x 2 - 1 = (x - 1)(x + 1)
x - 1 = x - 1
We now make the LCM by multiplying all factors included in the factoring of the given expressions. Common factors are used once only and the one with the highest power is used.
x - 1 is a factor to both expression and will therefore be used once. x + 1 is a factor in the first expression and will therefore be used. Hence
LCM ( x 2 - 1 and x - 1 ) = (x - 1)(x+1)
### Example 2
Find the lowest common multiple of the expressions 2 x 2, x 2 + x and x 3 + 2 x .
Solution
We first factor the given expressions completely:
2 x 2 = 2 x 2
x 2 + x = x(x + 1)
x 3 + 2 x = x( x 2 + 2)
The LCM is made by multiplying all factors included in the factoring of the given expressions. Common factors are used once only and the one with the highest power is used.
2 is a factor in the first term only and will therefore be used. x is a factor in all three expressions and the one with the highest power which is x 2 in the first term is used. x + 1 is a factor in the second expression only and is therefore used. x 2 + 1 is a factor in the third expression only and is therefore used. Hence
LCM ( 2x 2, x 2 + x , x 3 + 2 x ) = 2x 2 (x + 1) (x 2 + 2)
### Example 3
Find the lowest common multiple of the expressions x 2 + 3 x - 4, (x - 1) 2 and x 2 + 9 x + 20.
Solution
We first factor the given expressions completely:
x 2 + 3 x - 4 = (x - 1)(x + 4)
(x - 1) 2 = (x - 1) 2
x 2 + 9 x + 20 = (x + 4)(x + 5)
The LCM is made by multiplying all factors included in the factoring of the given expressions. Common factors are used once only and the one with the highest power is used.
x - 1 is a factor in the first and second expressions is therefore the one with the highest power (x - 1) 2 in the second expression is used. x + 4 is a factor in the first and third expressions is used once only. x + 5 is a factor in the third expression only and is therefore used . Hence
LCM ( x 2 + 3 x - 4 , (x - 1) 2 , x 2 + 9 x + 20 ) = (x - 1) 2 (x + 4)(x + 5)
Find Lowest Common Multiple (LCM) of the expressions given below.
1. 2 (x + 1) and 3 (x + 1) .
2. 2 (x - 1) 2 and 5 (x - 1) .
3. x 2 + 5 x + 6 and 2 x 2 + 2 x - 4 .
4. 3 x 3 - 2 x 2 - x and x - 1 .
5. 3 x 3 - 2 x 2 - x , 2 x 2 - 2 and (x - 1) 2.
Detailed Solutions and explanations to the above questions are included.
|
# How do you graph the equation y=-5/4x+1?
Aug 29, 2017
See a solution process below:
#### Explanation:
First, solve for two points which solve the equation and plot these points:
First Point:
For $x = 0$
$y = \left(- \frac{5}{4} \cdot 0\right) + 1$
$y = 0 + 1$
$y = 1$ or $\left(0 , 1\right)$
Second Point:
For $x = 4$
$y = \left(- \frac{5}{4} \cdot 4\right) + 1$
$y = - \frac{20}{4} + 1$
$y = - 5 + 1$
$y = - 4$ or $\left(4 , - 4\right)$
We can next graph the two points on the coordinate plane:
graph{(x^2+(y-1)^2-0.075)((x-4)^2+(y+4)^2-0.075)=0 [-20, 20, -10, 10]}
Now, we can draw a straight line through the two points to graph the line:
graph{(y+5/4x-1)(x^2+(y-1)^2-0.075)((x-4)^2+(y+4)^2-0.075)=0 [-20, 20, -10, 10]}
|
# How do you solve the system of equations y= 2x + 8 and 3x + 5y = 1?
Jun 26, 2018
By arranging the equations
#### Explanation:
Arrange the equations (multiply the 1st by 3 and the 2nd by 2):
$3 y - 6 x = 24$
$6 x + 10 y = 2$
Combine these:
$13 y = 26$
$y = \frac{26}{13} = 2$
Put this in the first original equation:
$2 x = y - 8$
$2 x = 2 - 8$
$2 x = - 6$
$x = - 3$
Jun 26, 2018
$x = - 3$ and $y = 2$
#### Explanation:
After putting first equation into second,
$3 x + 5 \cdot \left(2 x + 8\right) = 1$
$13 x + 40 = 1$
$13 x = - 39$, so $x = - 3$
Thus, $y = 2 \cdot \left(- 3\right) + 8 = 2$
|
## How do you find an intersection of a line and a plane?
Finding the intersection of a line and a plane
1. substitute the values of x, y and z from the equation of the line into the equation of the plane and solve for the parameter t.
2. take the value of t and plug it back into the equation of the line.
## Can a plane and a line intersect?
A given line and a given plane may or may not intersect. If the line does intersect with the plane, it’s possible that the line is completely contained in the plane as well. If they do intersect, determine whether the line is contained in the plane or intersects it in a single point.
What is the point of intersection of a line and a plane called?
When two or more lines cross each other in a plane, they are called intersecting lines. The intersecting lines share a common point, which exists on all the intersecting lines, and is called the point of intersection. Here, lines P and Q intersect at point O, which is the point of intersection.
### How many cases of an intersection of a line and plane are there?
three
For the intersection of a line with a plane, there are three different possibilities, which correspond to 0, 1, or an infinite number of intersection points. It is not possible to have a finite number of intersection points other than 0 or 1. These three possible intersections are considered in the following examples.
### How do you find the line of intersection?
How Do I Find the Point of Intersection of Two Lines?
1. Get the two equations for the lines into slope-intercept form.
2. Set the two equations for y equal to each other.
3. Solve for x.
4. Use this x-coordinate and substitute it into either of the original equations for the lines and solve for y.
Is the intersection of two planes?
The intersection of two planes is a line. If the planes do not intersect, they are parallel.
## Do planes intersect?
The intersection of two planes is a line. They cannot intersect at only one point because planes are infinite. Furthermore, they cannot intersect over more than one line because planes are flat.
## Can a line and a plane intersect at exactly two points?
The statement “a line can never intersect a plane at exactly two points” is either an axiom in some formalization of Euclidean geometry or follows so directly from one or two other axioms in the system that the answer seems empty of meaning, a restatement of definitions (as in some of the good answers here).
Do two planes always sometimes or never intersect in a line explain?
Always The intersection of two planes is a line, and a line contains at least two points. Sometimes They might have only that single point in common.
### How many planes accommodate given line and a point outside a line?
There can be only one plane that include one line and point outside the line.
### Is the intersection of two planes always a line?
Do plane A and plane MNP intersect?
SOLUTION: MNP is the top face of the solid, and does not have any common lines with the plane A. So, they do not intersect. SOLUTION: Coplanar points are points that lie in the same plane. Here, the points T, S, R, and Q all lie on the plane A; there is no other plane which contains all four of them.
## How do you find the intersection of two points on a plane?
In all other cases there will be an intersection. On to the intersection computation : All points Xof a plane follow the equation Dot(N, X) = d Where Nis the normal and dcan be found by putting a known point of the plane in the equation.
## What is the scalar product of the IP of the plane?
As I and P belong to the plane, the vector IP is normal to N. ⍝ This translates to: The scalar product IP.N = 0. if (l . n) = 0 ; line and plane are parallel. if (Po – lo) . n = 0 ; line is contained in the plane. (P – Po) . n = 0 ; vector equation of plane.
How do you find the direction of a point on a plane?
On to the intersection computation : All points Xof a plane follow the equation Dot(N, X) = d Where Nis the normal and dcan be found by putting a known point of the plane in the equation. float d = Dot(normal, coord); Onto the ray, all points sof a line can be expressed as a point pand a vector giving the direction D:
|
# Vieta's With Restrictions
How many positive integers $n$ less than 100 have a corresponding integer $m$ divisible by 3 such that the roots of $x^2-nx+m=0$ are consecutive positive integers?
I'm thinking that this problem is Vieta's. Here's my reasoning. Let $a_1$ and $a_2$ be the roots of this polynomial. We have that $a_1+a_2=n$ and $a_1a_2=m$. We also have that $a_1+1=a_2$. We know that since $m$ is divisible by $3$, we have that $a_1$ or $a_2$ is divisible by $3$. In this case, we know that the other root will have to be even. We have that $n$ will be odd, and $m$ will be odd as well. Another key bit of information is that $n$ and $m$ are both integers. From here, I am not able to solve. Help is greatly appreciated.
Let the smallest solution be $a$. Then, by Vieta's, $m=a(a+1)$. Since m is divisible by 3, $a$ or $a+1$ is divisible by 3. This gives that $a$ can be $2$, $3$, $5$, $6$, etc. Remember that $n<100$, so $2a+1<100$, which gives $a<49.5$. This means that the largest value of $a$ is $48$. Note that for every multiple of $3$ $a$ is equal to, there is a value of $a$ one less than it. So the answer is the number of multiples from $3$ to $48$ times $2$, which is $16*2=\boxed{32}$. I am not sure that this is correct, so you may want to double check.
Vietas formulas gives us that $n$ is the sum of the roots. Let us call these roots $r$ and $r+1$.
We need $m$ divisible by $3$, and thus we need $3\mid r$ or $3\mid r + 1$.
So we need the amount of pairs $(1,2), (2,3), \cdots, (49, 50)$ such that the pair contains a multiple of $3$.
This is easier to complementary count: We count $(1,2), (4,5),\cdots,(49,50)$ whic is $17$ pairs without multiples of $3$. Thus we have $49 - 17 = 32$ pairs that satisfy our conditions.
Hint:
If the roots are $a,a+1$
$m=a(a+1)\implies3|a(a +1)$
If $a=3b,$ we need $0<2(3b)+1<100$
What if $3|(a+1),a+1=3c$(say)
• I have solved that $b$ can only reside in the interval $b<16.5$. Now what must I do? Feb 22, 2018 at 4:07
• @Skupp, $b$ needs to be positive integer, right? Feb 22, 2018 at 4:08
• Yes it does. Is the answer $16$? Feb 22, 2018 at 4:08
• @Skupp, Consider $a=3c-1,c>0$ as well Feb 22, 2018 at 4:09
• Is $c$ an integer? Feb 22, 2018 at 4:10
|
# Golden Ratio
Golden Ratio, Golden Mean, Golden Section, or Divine Proportion refers to the ratio between two quantities such that the ratio of their sum to the larger of the two quantities is approximately equal to 1.618. It is denoted by the symbol ‘Ï•’ (phi), an irrational number because it never terminates and never repeats.
If the two quantities a and b are in the Golden ratio, they can be mathematically represented as
a:b = [(a+b): a] or ${\dfrac{a}{b}=\dfrac{a+b}{a}}$. For quantities a and b, a > b > 0.
The golden ratio is thus a proportional concept that describes the relative lengths of two line segments. It is important because it is found in various fields such as arts, architecture, human faces, and designs.
## Who Discovered the Golden Ratio
Although the discovery of the golden ratio is a mystery, it was first thought to be mentioned around 300 BCE in Euclid’s Elements. In 1509, Luca Pacioli used the term ‘Golden Ratio’ in his book ‘The Divine Proportion.’
Now, let us expand the relation between the quantities in the golden ratio to get its value.
## Formula
Let a line segment AC be divided into two parts, AB and BC, representing two quantities, a and b. AB = a is the larger part, and BC = b is the smaller part.
Now, if a and b are represented in the form of the golden ratio, then the formula is mathematically written as
${\dfrac{AB}{BC}=\dfrac{AB+BC}{AB}}$
=> ${\dfrac{a}{b}=\dfrac{a+b}{a}}$
The value of ϕ goes on as 1.61803398874989484820… like other typical irrational numbers with no specific pattern. The ratio also equals ${2\times \sin 54^{\circ }}$.
## Derivation
Now, let us see how we obtained the above value:
${\dfrac{a}{b}=\dfrac{a+b}{a}}$
Now, splitting the right-hand side, we get
=> ${\dfrac{a}{b}=1+\dfrac{b}{a}}$
=> ${\phi =1+\dfrac{1}{\phi }}$
Using this formula, we can determine the value of the golden ratio by substituting the value of Ï• as follows:
=> ${\phi =1+\dfrac{1}{1.618}}$
= 1 + 0.618047…
= 1.618047…, which is the golden ratio.
The value of the ratio will be more accurate if we include more digits after the decimal places when substituting the value of Ï•.
We can obtain the value of the golden ratio mainly in 2 different ways:
### By Hit and Trial Method
In this method, we will consider a value of the golden ratio (say 2) and follow the given steps until we get the value of Ï• closer to 1.618.
1. Finding the multiplicative inverse of the guessed value (=${\dfrac{1}{Value}}$), we get 0.5
2. Adding 1 to the value of step 1, we get 1.5
3. Using that value and repeating the same steps, we get the value closer to Ï• as follows.
If we proceed further, the final value gets even closer to the value of the golden ratio Ï•.
However, the hit-and-trial method needs more time and labor; thus, the value of Ï• is more commonly calculated using the quadratic formula.
As we know, the golden ratio formula is
${\phi =1+\dfrac{1}{\phi }}$
Now, multiplying both sides by Ï•, we get
${\phi ^{2}=\left( 1+\dfrac{1}{\phi }\right) \phi}$
${\phi ^{2}=\phi +1}$
On rearranging, we get,
${\phi ^{2}-\phi -1=0}$, which is a quadratic equation.
Thus, by using the quadratic formula, we get,
${\phi =\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
Here, a = 1, b = -1, c = -1
Thus, ${\phi =\dfrac{1\pm \sqrt{\left( 1+4\right) }}{2}}$
As the golden ratio is obtained from two positive quantities, the value of Ï• should always be positive.
Thus, ${\phi =\dfrac{1+\sqrt{\left( 1+4\right) }}{2}}$
=> ${\phi =\dfrac{1+\sqrt{5}}{2}}$ = 1.618033…
## Patterns in Golden Ratio with Exponents
Interestingly, this relation further gives us a pattern related to the value of Ï•.
As we know, the value of Ï• = 1.618 can be obtained using the formula ${\phi =1+\dfrac{1}{\phi }}$
Let us start by multiplying Ï• on both sides
${\phi ^{2}=\left( 1+\dfrac{1}{\phi }\right) \phi}$
${\phi ^{2}=\phi +1}$
Again multiplying both sides by ${\phi}$, we get
${\phi ^{3}=\phi ^{2}+\phi}$, and it goes on.
The following table shows the mentioned pattern
Here, we observe that each power of the golden ratio is the sum of the two powers before it.
Now, let us see how the golden ratio forms using the Fibonacci number sequence, where each term is found by adding the two preceding numbers.
## Relation to the Fibonacci Sequence
As we increase the value of the two consecutive Fibonacci numbers, the ratio gets closer to the golden ratio. Thus, this approximation is very close to Ï• for the pair of larger numbers.
In the following table, let us find the value of ϕ starting from the Fibonacci number ‘2’.
Thus, we observe that Ï• is related to the Fibonacci sequence.
Similarly, Ï• is also related to geometry. It relates to a notable geometric shape, the golden rectangle.
## Golden Ratio and Golden Rectangle
In geometry, a rectangle formed by adding or removing the existing squares within a rectangle gives a golden rectangle. It has its sides in the golden ratio.
Let a rectangle ABCD be divided into five squares as shown. If we remove the squares AHFG and HBCE, we get a golden rectangle GFED.
Here is another way to draw the golden rectangle.
We can also construct a golden rectangle by following the steps below.
1. A square GYXD of 1 unit is drawn.
2. A point P is marked in the midway of any side (say DX).
3. Point P is then joined to the vertex Y by a line segment PY.
4. Now, using the Pythagoras Theorem, the length of the drawn line segment (say PY) is calculated.
5. Then, using the line segment PY as the radius and the point P as the center, an arc GY is drawn along the sides of the square GYXD.
6. The line segment XE is then joined to the intersection point E of the arc GYE and the extended side DX. Finally, the rectangle DEFG (having the golden ratio) is drawn using DE.
Thus, the golden rectangle DEFG has dimensions in the golden ratio, Ï•
As we know, Ï• can be obtained from the ratio of two successive Fibonacci numbers; the golden ratio forms a spiral pattern. This spiral follows a constant angle close to Ï• and is thus known as the Golden Spiral
Sometimes, circles are drawn within squares instead of the spiral. Those circles are known as the Golden Circles, and the ratio of one circle to its adjacent one is found to be 1:1.618.
The gif below shows how the Golden Ratio works:
Apart from spirals and circles, Ï• is also found in other geometric shapes, such as triangles and pentagrams.summarizing all about the golden ratio
## Golden Ratio in Kepler’s Triangle
As we expand the formula of Ï• and form the quadratic equation of the golden ratio, we get
${\phi ^{2}-\phi -1=0}$
=> ${\phi ^{2}=\phi +1}$
Also, from the Pythagoras Theorem, we can write,
${c^{2}=a^{2}+b^{2}}$
Let us now consider a right-angled triangle ABC, where the length of the hypotenuse is AC, and the legs are AB and BC.
Now, if the sides of the triangle are AC = c = Ï•, AB = a = 1, and BC = b = ${\sqrt{\phi }}$, then using the Pythagoras Theorem, we can form the quadratic equation of Ï•.
The ratio of these sides is found to be in the ${1:\sqrt{\phi }:\phi}$.
It inspired Johannes Kepler to create the following triangle with Pythagoras and ${\phi}$ together.
## Golden Ratio in Pentagram
The Pentagram (or Pentangle), a holy symbol, looks like a 5-pointed star. A regular Pentagram has the golden ratio in it.
Let ABCDEFGHIJ be a pentagram, where the length of AE = a, AD = b, AB = c, and BD = d.
Here, the ratio of AE to AD, AD to AB, and AB to BD gives the value of Ï•.
## Real-Life Examples
• In Nature: The golden ratio is found in flowers, shells, weather, and galaxies. It also exists on the human face. A visually balanced face has a length-to-width ratio of approximately 1.618, the golden ratio. This ratio can also be seen in other parts of the human body.
• In Art and Architecture: It is used in many arts, designs, and architecture. One of the famous paintings, Leonardo Da Vinci’s Mona Lisa, was painted according to the golden ratio.
• In Logo and Design: It is also found in many web designs and logo designs. It helps us to sketch out the proportions and shapes. Many famous logos like Twitter, Apple, and Pepsi follow this ratio.
## Solved Examples
Calculate the value of the golden ratio Ï• using the quadratic formula.
Solution:
As we know,
${\phi =1+\dfrac{1}{\phi }}$
Multiplying both sides by ${\phi}$,
${\phi ^{2}=\left( 1+\dfrac{1}{\phi }\right) \phi}$
${\phi ^{2}=\phi +1}$
On rearranging, we get,
${\phi ^{2}-\phi -1=0}$, which is a quadratic equation.
Thus, by using the quadratic formula:
${\phi =\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$
Here, a = 1, b = -1, c = -1
Thus, ${\phi =\dfrac{1\pm \sqrt{\left( 1+4\right) }}{2}}$
As the ratio is for two positive quantities, the value of the Golden ratio should be the positive value.
Thus, ${\phi =\dfrac{1+\sqrt{\left( 1+4\right) }}{2}}$
Or, ${\phi =\dfrac{1+\sqrt{5}}{2}}$ = 1.618033…
|
# Quarter Wit, Quarter Wisdom: Nuances of Sets
We will start with sets today. Your Veritas Prep GMAT book explains you the basics of sets very well so I am not going to get into those. If you have gone through the concepts, you know that we can use Venn diagrams to solve the sets questions.
First, let’s look at why we should focus on terminology in sets question. Thereafter, we will put up a very nice question from our very own book which is simple but takes down many people (just like a typical GMAT question):
Say, there are a total of 100 people in a housing society. There are two clubs close to the society – A and B. You are given that of the 100 people of the housing society, 60 people are members of club A and 50 people are members of club B.
Question 1: How many people are members of both the clubs?
We are looking for the number of people in the green region. The answer here is not 10. It is ‘cannot be determined’ i.e. you cannot say how many people are members of both the clubs. The reason is that you do not know how many people belong to neither club.
Say, for all future questions (unless mentioned otherwise), you are given that 20 people belong to neither club. What can you say about the number of people who belong to both the clubs? Now, out of the pool of 100, 20 are out. Only 80 people are club members. Since 60 are members of club A and 50 people are members of club B which gives us a total of 110, there must be an overlap of 30 people i.e. 30 people must belong to both the clubs (80 = 60 + 50 – Both)
Question 2: How many people belong to only one club?
We found above that 30 people belong to both the clubs. So out of the 60 people of club A, 30 belong to only club A. Out of the 50 people of club B, 20 belong to only club B. So a total of 30+20 = 50 people belong to only one club, either A or B but not both. (60 – Both + 50 – Both = 30 + 20 = 50)
Question 3: Say, you don’t know the number of people who belong to neither club. What is the minimum number of people who must belong to both the clubs?
We know that there are a total of 100 people. 60 belong to club A and 50 belong to club B which adds up to 110. Therefore, AT LEAST 10 people must have membership of both the clubs. Now if you increase the number of people who do not belong to either club, the number of people who belong to both will increase by the same number. Think in terms of the Venn diagram. If the ‘Neither’ number increases, the number of people who are members decreases. Hence, the overlap increases to keep A = 60 and B = 50.
Let’s look at the promised question which will make this concept clear.
Question: Of the 400 members at a health club, 260 use the weight room and 300 use the pool. If at least 60 of the members do not use either, then the number of members using both the weight room and the pool must be between:
(A) 40 to 100
(B) 80 to 140
(C) 160 to 260
(D) 220 to 260
(E) 220 to 300
Solution: When we minimize “number of members who do not use either”, we are minimizing the “number of members who use both” as well.
Look at the equation:
Total = A + B – Both + Neither
Since the total sum 400 is constant, if we increase the ‘Neither’ i.e. 60, we will have to increase the ‘Both’ term too to maintain the sum of 400 (Assuming A and B are constant which they are since they are given to us).
Least value of ‘number of members who use neither’ is 60. We will get the least value of ‘number of members who use both’ when we put ‘Neither’ = 60.
400 = 260 + 300 – ‘Minimum value of both’ + 60
Minimum value of both = 220
On the same lines, if we maximize “number of members who use neither”, we are maximizing the “number of members who use both” as well.
What is the maximum number of people who use neither? Out of a total of 400 people, 300 people use the pool. Hence at least 300 people use at least one of the two facilities. This means that there can be AT MOST 100 people (total 400 – 300 who use pool) who use neither facility.
400 = 260 + 300 – ‘Maximum value of both’ + 100
Maximum value of both = 260
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# EVALUATE THE FOLLOWING EXPRESSION USING PEMDAS
To evaluate the numerical expression, we need to know the order of operations.
What is an Order of Operations?
An Order of Operations is a rule that can be used to simplify or evaluate more than one operation.
There are many types of rules, we will use here the PEMDAS Rule.
Example 1 :
(20 + 80 ÷ 2 × 8) ÷ [(54 ÷ 9 + 14) ÷ 4]
Solution :
= (20 + 80 ÷ 2 × 8) ÷ [(54 ÷ 9 + 14) ÷ 4]= (20 + 40 × 8) ÷ [(6 + 14) ÷ 4]= (20 + 320) ÷ [20 ÷ 4]= (20 + 320) ÷ 5= 340 ÷ 5= 68 P( ), DP( ), M & AP( ), [ ], A & DP( ), AD
Example 2 :
3 × [64 ÷ (13 - 5) - 4] × 42 ÷ 6
Solution :
= 3 × [64 ÷ (13 - 5) - 4] × 42 ÷ 6= 3 × [64 ÷ 8 - 4] × 42 ÷ 6= 3 × [8 - 4] × 42 ÷ 6= 3 × 4 × 42 ÷ 6= 12 × 42 ÷ 6= 504 ÷ 6= 84 P( ), SP[ ], DP[ ], SMMD
Example 3 :
{24 ÷ [(8 × 3) ÷ 4] × 2} × (15 - 4)
Solution :
= {24 ÷ [(8 × 3) ÷ 4] × 2} × (15 - 4)= {24 ÷ [24 ÷ 4] × 2} × 11= {24 ÷ 6 × 2} × 11= {4 × 2} × 11= 8 × 11= 88 P( ), M & SP[ ], DP{ }, DP{ }, MM
Example 4 :
[5 × (54 ÷ 6 - 3)] + 6 × 2 - 60
Solution :
= [5 × (54 ÷ 6 - 3)] + 6 × 2 - 60= [5 × (9 - 3)] + 6 × 2 - 60= [5 × 6] + 6 × 2 - 60= 30 + 6 × 2 - 60= 30 + 12 - 60= 42 - 60= -18 P( ), DP( ), SP[ ], MMAS
Example 5 :
(78 - 6) ÷ {18 × [(7 - 8) × 2]}
Solution :
= (78 - 6) ÷ {18 × [(7 - 8) × 2]}= 72 ÷ {18 × [-1 × 2]}= 72 ÷ {18 × (-2)}= 72 ÷ (-36)= - 2 P( ), SP[ ], MP{ }, MD
Example 6 :
{96 ÷ [36 ÷ 3 - (18 × 2 - 30)]} ÷ (31 - 16 + 1)
Solution :
= {96 ÷ [36 ÷ 3 - (18 × 2 - 30)]} ÷ (31 - 16 + 1)= {96 ÷ [36 ÷ 3 - (36 - 30)]} ÷ (15 + 1)= {96 ÷ [36 ÷ 3 - 6]} ÷ 16= {96 ÷ [12 - 6]} ÷ 16= {96 ÷ 6} ÷ 16= 16 ÷ 16= 1 P(), M&SP(), S&AP[ ], DP[ ], SP{ }, DD
Example 7 :
× {11 + [2 × (10 - 15 + 8 × 3)] + 1}
Solution :
= 2 × {11 + [2 × (10 - 15 + 8 × 3)] + 1}= 2 × {11 + [2 × (-5 + 24)] + 1}= 2 × {11 + [2 × 19] + 1}= 2 × {11 + 38 + 1}= 2 × 50= 100 P( ), M&SP( ), SP[ ], MP{ }, AM
Example 8 :
45 ÷ 15 × [(21 - 18) × (5 + 4 × 2) - 49]
Solution :
= 45 ÷ 15 × [(21 - 18) × (5 + 4 × 2) - 49]= 45 ÷ 15 × [3 × (5 + 8) - 49]= 45 ÷ 15 × [3 × 13 - 49]= 45 ÷ 15 × [39 - 49]= 45 ÷ 15 × (-10)= 3 × (-10)= -30 P( ), S&MP( ), AP[ ], MP[ ], SDM
Example 9 :
÷ 3 × {2 × [(16 × 4) ÷ 8]}
Solution :
= 9 ÷ 3 × {2 × [(16 × 4) ÷ 8]}= 9 ÷ 3 × {2 × [64 ÷ 8]}= 9 ÷ 3 × {2 × 8}= 3 × 16= 48 P( ), MP[ ], DP{ }, MM
Example 10 :
100 - {4 × [3 × (19 + 1) ÷ 4]}
Solution :
= 100 - {4 × [3 × (19 + 1) ÷ 4]}= 100 - {4 × [3 × 20 ÷ 4]}= 100 - {4 × [3 × 5]}= 100 - {4 × 15}= 100 - 60= 40 P( ), AP[ ], DP[ ], MP{ }, MS
## Recent Articles
1. ### Finding Range of Values Inequality Problems
May 21, 24 08:51 PM
Finding Range of Values Inequality Problems
2. ### Solving Two Step Inequality Word Problems
May 21, 24 08:51 AM
Solving Two Step Inequality Word Problems
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# Exponential and Logarithmic Functions
## Basic Exponential Function & Its Characteristics
The following are graphs of the basic logarithmic function, $$f(x)=b^x, \\ b>0, \\ b\ne 1$$:
$$f(x)=b^x$$ where $$01$$
Characteristics of the Exponential Function
Using the graphs, we can identify some characteristics of the basic exponential function $$f(x)=b^x, \ b>0, b\ne 1$$, including:
• Domain: $$D:\{ x \in \mathbb{R}\}$$
• Range: $$R:\{ y \in \mathbb{R} \ | \ y > 0\}$$
• Horizontal asymptote: $$y=0$$
• $$y$$-intercept: $$(0,1)$$
Tip: If you recall the characteristics of the basic exponential function ($$f(x)=log_b(x)$$) graph (which can be found here), you'll see that the the basic exponential and logarithmic functions are very similar, and are, in fact, related. This is because both functions are inverses of each other, so their characteristics are also the inverse of each other. So, if you can recall the characteristics of one function, you can then use the fact that they're inverse functions to recall the other functions' characteristics. Specifically, characteristics associated with the $$x$$ and $$y$$ variables are switched. The related characteristics are summarized in the following tables:
## Graphing Exponential Functions with Transformations
Given the graph of the parent function $$f(x)=b^x$$, we are able to graph any logarithmic function of the form:
$$f(x)=ab^{k(x-d)}+c$$, for any $$a,k,c,d\in \mathbb{R}$$
by applying transformations to the parent graph. Each of the four parameters, $$a,b,c,d$$ correspond to certain transformations, as summarized in the following table:
Function Notation/Parameter Corresponding Transformation(s) Coordinate Point Transformation $$f(x)=$$$$\ a$$$$b^{k(x-d)}+c$$ Vertical stretch/compression & reflection If $$a<0$$, vertical reflection (i.e. over the x-axis) If $$|a|>1$$, vertical stretch by a factor of $$|a|$$ If $$0<|a|<1$$, vertical compression by a factor of $$|a|$$ $$(x,y) \rightarrow$$ $$(x,$$$$\ a$$$$y)$$ $$f(x)=ab$$$$^k$$$$^{(x-d)}+c$$ Horizontal stretch/compression & reflection If $$k<0$$, horizontal reflection (i.e. over the y-axis) If $$|k|>1$$, horizontal compression by a factor of $$|\frac{1}{k}|$$ If $$0<|k|<1$$, horizontal stretch by a factor of $$|\frac{1}{k}|$$ $$(x,y) \rightarrow($$$$\frac{1}{k}$$$$x,y)$$ $$f(x)=ab^{k(x-}$$$$^d$$$$^)+c$$ Horizontal shift/translation left or right by $$d$$ units If $$d>0$$, shift right by $$d$$ units If $$d<0$$, shift left by $$d$$ units $$(x,y) \rightarrow$$$$\ (x$$$$+ d$$$$,y)$$ $$f(x)=ab^{k(x-d)}+$$$$\ c$$ Vertical shift/translation up or down by $$c$$ units If $$c>0$$, shift up by $$c$$ units If $$c<0$$, shift down by $$c$$ units $$(x,y) \rightarrow$$ $$(x,y$$$$+ c$$$$)$$
Tip: To help remember what parameters correspond to which transformations, note that parameters "outside" of the function $$f(x)=b^x$$ (i.e. $$a, c$$) transform the $$y$$ or vertical value, while the ones "inside" the function (i.e. $$k, d$$) transform the $$x$$ or horizontal value.
Notes:
• After transforming, the horizontal asymptote is located at $$y=c$$ (Why?)
• The domain of the transformed function is dependent on whether it lies above or below the asymptote:
• If the function is on the below of the asymptote, the domain is $$y<c$$
• If the function is on the above of the asymptote, the domain is $$y>c$$
• Regardless of transformations, the domain of an exponential function will always be $$\{x\in\mathbb{R}\}$$ (Why?)
Try this interactive tool.
Adjust the sliders for $$a$$, $$b$$, $$c$$, $$d$$, and $$k$$. Observe what happens to the exponential function.
## Examples
Example: Sketch the function $$f(x)=2(3^{x-1})-2$$ by first:
a) Graphing the parent function, then applying transformations. b) Plotting any key points and drawing the asymptote.
Solutions:
a) Graphing the parent function, then applying transformations.
1. In this case, the parent function is $$f(x)=3^x$$, which has a horizontal asymptote of $$y=0$$. So we first sketch this function: *Sketch this graph by creating a table values then plotting the points, or by finding the y-intercept, horizontal asymptote and another point on the graph, then sketching the function. 2. Identifying the transformation parameters, we see that we have $$a=2$$, so we have a vertical stretch by a factor of 2: 3. Next, we have $$d=1$$, so we have a horizontal shift to the right by 1 unit: 4. Finally, we have $$c=-2$$, so we have a vertical shift down by 2 units. This also shifts our original horizontal asymptote down to $$y=-2$$:
*Note: It is also possible to graph this function by transforming each individual point from the parent function or by applying all the transformations graphically at the same time (instead of one-by-one like we did above).
We compare the graphs of the two functions, $$f(x)=3^x$$ and $$f(x)=2(3^{x-1}-2$$, and summarize their characteristics in the following table:
$$f(x)=3^x$$ $$f(x)=2(3^{x-1})-2$$ Domain: $$D:\{x\in\mathbb{R}\}$$ Range: $$R:\{y\in\mathbb{R} \ | \ y>0\}$$ Horizontal asymptote: $$y=0$$ $$x$$-intercept: None $$y$$-intercept: $$(0,1)$$ Domain: $$D:\{x\in\mathbb{R}\}$$ Range: $$R:\{y\in\mathbb{R} \ | \ y>-2\}$$ Horizontal asymptote: $$y=-2$$ $$x$$-intercept: $$(1,0)$$ $$y$$-intercept: $$(0,-\frac{4}{3})$$
b) Plotting any key points and drawing the asymptote:
1. The function $$f(x)=2(3^{x-1})-2$$ has: $$x$$-intercept at $$(1,0)$$, $$y$$-intercept at $$(0,-\frac{4}{3})$$, a point at $$(2,4)$$, and horizontal asymptote $$y=-2$$. We can plot and draw these characteristics on the Cartesian plane first: 2. After plotting/drawing these characteristics, we can draw a smooth curve through the points and approaching the asymptote, in order to finish sketching the function:
Example: Determine the exponential equation in the form $$f(x)=a2^{kx}+c$$ that is represented by the following graph:
Solution:
Watch this video for the solution!
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# Multiplying And Dividing Negative Numbers.
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In this part of our blog you will be able to apply all of your math skill and as well all of our sections in our blog to help deal with unit 4: multiplying and dividing negative numbers.
This is only a practice test although you can check your grade to fix or reflect on your mistakes. . .
GOOD LUCK ! ! ! :D
• 1.
### Multiply: -71 * -101=
Explanation
When multiplying two negative numbers, the result is always positive. In this case, when we multiply -71 and -101, we get a positive result of 7171.
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• 2.
### -120*2=
Explanation
The given expression, -120*2, represents the multiplication of -120 by 2. When multiplying a negative number by a positive number, the result is always negative. Therefore, -120 multiplied by 2 equals -240.
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• 3.
### 22*-3=
Explanation
When multiplying a positive number by a negative number, the product will be negative. In this case, 22 is being multiplied by -3, so the result will be negative. The product of 22 and -3 is -66.
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• 4.
• 5.
### Find the value of n. -22*n = -44
Explanation
To find the value of n, we need to solve the equation -22*n = -44. To isolate n, we divide both sides of the equation by -22. This gives us n = -44/-22, which simplifies to n = 2. Therefore, the value of n is 2.
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• 6.
• 7.
### n*12 =-144
Explanation
The equation n*12 = -144 can be solved by dividing both sides of the equation by 12. This will give us the value of n. When we divide -144 by 12, we get -12. Therefore, the value of n that satisfies the equation is -12.
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• 8.
### Tru and False. When you multipy two negatives the answer will be negative.
• A.
True
• B.
False
B. False
Explanation
When you multiply two negatives, the answer will be positive.
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• 9.
### When you divide a positive times a negative the answer will be negative.
• A.
True
• B.
False
A. True
Explanation
When you divide a positive number by a negative number, the result will be negative. This is because division is essentially the same as multiplying by the reciprocal, and when you multiply a positive number by a negative number, the result is always negative. Therefore, the answer to the question is true.
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• 10.
### Divide -121/11=
-11
Explanation
The given equation -121 divided by 11 equals -11. When dividing a negative number by a positive number, the result is always negative. Therefore, -121 divided by 11 is -11.
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• 11.
### -21/-3=
7
Explanation
When dividing a negative number by a negative number, the result is positive. In this case, -21 divided by -3 equals 7.
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• 12.
### 42/-6 =
-7
Explanation
When we divide 42 by -6, we are essentially asking how many times -6 can be subtracted from 42. Since -6 can be subtracted 7 times from 42 (7 x -6 = -42), the answer is -7.
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• 13.
### Solve the problems 5/3*-6/2
-30/6
-5
Explanation
The given expression involves the division and multiplication of fractions. To solve it, we first multiply 5/3 by -6/2, which gives us -30/6. This fraction can be simplified to -5 when we divide both the numerator and denominator by 6. Therefore, the correct answer is -30/6, -5.
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• 14.
### -5/2 / -4/6=
20/12
5/3
Explanation
The given expression is a division of two fractions. To divide fractions, we need to multiply the first fraction by the reciprocal of the second fraction. In this case, the reciprocal of -4/6 is -6/4 or -3/2. Therefore, -5/2 divided by -4/6 is equal to -5/2 multiplied by -3/2, which simplifies to 15/4 or 3 3/4. However, none of the given answer options match this result. Therefore, the correct answer is not available.
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• 15.
### -6/11 * 5/8 =
-30/88
-15/44
Explanation
The given expression is a multiplication of two fractions. To multiply fractions, we multiply the numerators and denominators separately. In this case, multiplying -6/11 and 5/8 gives us (-6 * 5) / (11 * 8) = -30/88. Therefore, the correct answer is -30/88.
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Two equal parabolas have the same focus and their axes are at right angles. A normal to one is perpendicular to normal to the other. Prove that the locus of the point of intersection of these normals is another parabola
Last updated date: 13th Jul 2024
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Hint: Equation of normal to parabola ${{y}^{2}}=4ax$ is given as $y=mx-2am-a{{m}^{3}}$, where $m$ is the slope of the normal.
We will consider the equation of one of the parabolas as ${{y}^{2}}=4ax$.
So , its focus is $S\left( a,0 \right)$.
We know, the equation of normal to the parabola in slope form is given as
$y=mx-2am-a{{m}^{3}}....\left( i \right)$ , where $m$ is the slope of the normal.
Now , we have to find the locus of intersection of the normal.
We will consider this point to be $N\left( h,k \right)$.
Now, since $N\left( h,k \right)$ is the point of intersection of the normals , so , it should lie on equation $\left( i \right)$, i.e. the point $N\left( h,k \right)$ should satisfy equation $\left( i \right)$.
So , we will substitute $x=h$ and $y=k$ in equation $\left( i \right)$.
On substituting $x=h$ and $y=k$ in equation $\left( i \right)$ , we get
$k=mh-2am-a{{m}^{3}}$
Or , $a{{m}^{3}}+m\left( 2a-h \right)+k=0....\left( ii \right)$
Clearly, we can see that equation $\left( ii \right)$ is a cubic equation in $m$ , which is of the form $a{{m}^{3}}+b{{m}^{2}}+cm+d=0$. So , it should represent three lines passing through $\left( h,k \right)$.
Now, in the question , it is given that two perpendicular normals pass through $N\left( h,k \right)$. So , out of these three lines , two lines must be perpendicular.
Now , let ${{m}_{1}},{{m}_{2}}$ and ${{m}_{3}}$ be three roots of equation $\left( ii \right)$. The roots of the equation $\left( ii \right)$ are corresponding to the slopes of the three lines.
Now, we are given two of these lines are perpendicular.
We know , when two lines are perpendicular , the product of their slopes is equal to $-1$ .
So, ${{m}_{1}}{{m}_{2}}=-1....\left( iii \right)$
We know , for a cubic equation of the form $a{{m}^{3}}+b{{m}^{2}}+cm+d=0$, the product of the roots is given as $\dfrac{-d}{a}$.
So , from equation $\left( ii \right)$ , we have
${{m}_{1}}{{m}_{2}}{{m}_{3}}=\dfrac{-k}{a}$
Since , ${{m}_{1}}{{m}_{2}}=-1\text{ }\left( \text{from equation }iii \right)$
So, ${{m}_{3}}=\dfrac{k}{a}$
Now , ${{m}_{3}}$ is a root of equation $\left( ii \right)$. So , it should satisfy the equation.
So , $a{{\left( \dfrac{k}{a} \right)}^{3}}+\dfrac{k}{a}\left( 2a-h \right)+k=0$
Or $\dfrac{{{k}^{3}}}{{{a}^{2}}}+2k-\dfrac{kh}{a}+k=0$
Or ${{k}^{2}}+3{{a}^{2}}-ah=0$
Or ${{k}^{2}}=a\left( h-3a \right)$
Now , the locus of $N\left( h,k \right)$ is given by replacing $\left( h,k \right)$ by $\left( x,y \right)$
So, the locus of $N\left( h,k \right)$ is given as ${{y}^{2}}=a\left( x-3a \right)$ which is the equation of a parabola.
Note: The product of slopes of perpendicular lines is equal to $-1$ and not $1$. Students generally get confused and make this mistake.
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# Algebraic Functions, including Domain and Range
This section covers:
Note: More advanced topics with functions can be found in the Parent Functions and Transformations and Advanced Functions: Compositions, Even and Odd, and Extrema section. End behavior of functions can be found here in the Parent Functions and Transformations section, here in the Graphing and Finding Roots of Polynomial Functions section, and also in the Graphing Rational Functions, including Asymptotes sectionIntermediate Value Theorem (IVT) can be found here in the Limits and Continuity section.
# Algebraic Functions Versus Relations
When we first talked about the coordinate system, we worked with the graph that shows the relationship between how many hours we worked (the independent variable, or the “$$x$$”), and how much money we made (the dependent variable, or the “$$y$$”). Any relationship between two variables, where one depends on the other, is called a relation, since it relates two things.
This particular relation is an algebraic function, since there is only one $$y$$ for each $$x$$. In other words, since the $$x$$ is the “question” and $$y$$ is the “answer”, we can only have one answer for each question. For whatever is the number of hours we work, we only get paid a certain amount for that: $$y=10x$$.
<
Again, a function is just a fancy way of saying something depends on something else, and there’s only one “$$y$$” for every “$$x$$”. But the other thing you’ll learn about functions is that they can be written a funny way; a way that looks really complicated, but they are just trying to confuse us – it’s not that bad!
Instead of our original equation, $$y=10x$$, we can write it like this: $$f(x)=10x$$.
Note that this is not “$$f$$ times $$x$$; it is “$$f$$ of $$x$$”. What it means is that $$x$$ is on the right hand sign of the “=” sign, and you can put different values in for $$x$$ on the left hand side to get one and only one value on the right-hand sign. So again, “$$f(x)$$” is really “$$y$$”. It’s that simple.
Here are some examples of plugging in things on the left hand side, and then, to get our answer, we plug what that is for every $$x$$on the right-hand side:
\begin{align}f\left( x \right)&=10x\text{ (original function)}\\f\left( 2 \right)&=10\times 2=20\\f\left( 9 \right)&=10\times 9=90\\f\left( t \right)&=10t\\f\left( {x+1} \right)&=10\left( {x+1} \right)=10x+10\end{align} Just remember that the left-hand side is always $$f\left( {\text{something}} \right)$$, and just plug that something into every instance of the variable on the right – every place there is an $$x$$. In this example, we only have one place to plug in for the $$x$$.
Again, what makes a relation a function is that you can only have one “answer” (the $$y$$) for each “question” (the $$x$$). All functions are relations, but not all relations are functions.
<
<
# Vertical Line Test
Notice that when we have a function, we can’t draw a vertical that goes through more than one point. This is called the vertical line test, and it’s a useful tool to determine if a graph is a function or not. For example, we can tell the following graph is not a function since we can draw a vertical line and hit more than one point:
Graphs – NOT Functions Notes See how we have two “answers” for $$x=4$$? The points $$\left\{ {\left( {2,1} \right),\left( {3,0} \right),\left( {4,2} \right),\left( {4,-2} \right)} \right\}$$ are not a function. We can draw a vertical line and see that we have two points on the same line: not a function. Notice that it’s not a function when we have 2 different $$y$$ values for the same $$x$$ value. See how when $$x=4$$ (the “question”), we have two points: $$(4,2)$$, and $$(4,–2)$$. Thus, we have two “answers”. This would mess up things in real life (like in the engineering or banking fields), since functions in real life must give us only one answer when we ask a question (like how much we have in the bank on a certain day).
Here are some examples of functions. Notice that you can have two “questions” ($$x$$) for the same “answer” ($$y$$):
Graphs – Functions Notes See how we have two “questions”, or 2 $$x$$ values with the same “answer” ($$y$$ value)? $$x$$ values 2 and 4 have the same $$y$$ value that is 1. And see how if we were to draw any vertical line, we’d never cross two points? The points $$\left\{ {\left( {1,-2} \right)\left( {2,1} \right),\left( {3,0} \right),\left( {4,1} \right)} \right\}$$ are a function. Like the first graph of points above, the function is called discrete, since you have to pick up your pencil to get from one point to another. Otherwise, functions are continuous – if you never have to pick up your pencil for the whole function. See how if we were to draw any vertical line, we’d never cross two points? And see how we’d never need to pick up our pencil when drawing this graph? Try it yourself! This relation is a continuous function.
Another way to look at a set of points and determine whether or not they are functions is to draw what we call mapping diagrams, since we are mapping the $$x$$ values to the $$y$$ values. We order values from smallest to largest and don’t repeat the values on each side and match them up. If we have more than one $$y$$ value for one $$x$$ value, we don’t have a function. Here are some examples:
Points Mapping Diagram Function? Yes or No $$\left\{ {\left( {2,1} \right),\left( {3,0} \right),\left( {4,2} \right),\left( {4,-2} \right)} \right\}$$ Not a function: we have 2 $$y$$’s (–2 and 2) for the same $$x$$ (4). It’s not OK to have one question with 2 different answers. $$\left\{ {\left( {1,-2} \right)\left( {2,1} \right),\left( {3,0} \right),\left( {4,1} \right)} \right\}$$ Is a function: we have 2 $$x$$’s for the same $$y$$, but this is fine. It’s OK to have 2 questions with the same answer.
# Domain and Range of Relations and Functions
Domain and Range of functions (and relations) sound really difficult and scary, but they are not really bad at all. You know how those mathematicians like to use fancy words for easy stuff?
Remember that since “d” comes before “r”, the domain of functions have to do with the “$$x$$”’s and the range of functions have to do with the “$$y$$”’s. To get the domain, we are just looking for all the possible values of $$x$$ for that function (from smallest to largest), and for the range, we are looking for all possible values of $$y$$ for that function (again, from smallest to largest).
To help me do this, I like to use my pencil – but it’s backwards compared to what you might think. To find the domain, I put my pencil vertically and start at the left and see where it first hits a point. Then I push it through all the way to the right to see where it ends hitting points.
For the range, I do the same thing, but with a horizontal pencil that’s moving up:
<
Graphs Notes When we move the vertical pencil from left to right, we get the following $$x$$ values: 1, 2, 3, and 4. The domain is $$\left\{ {1,2,3,4} \right\}$$. Note that since we are only dealing with points in this graph, we write the set of $$x$$ values in a set with the brackets. Similarly, when we move the horizontal pencil from the bottom to the top, we get the following $$y$$ values: –2, 0, and 1. The range is $$\left\{ {-2,0,1} \right\}$$. Note that we don’t need to repeat values.
Here are more examples, using what we call “Interval Notation”. (We saw this in the Inequalities Section). This is the most commonly used way to describe domains and ranges, and it always goes from lowest to highest with “(“ (soft brackets) if the relation or function doesn’t hit the point, and “[“ (hard brackets) if the relation or function does hit the line. If you have to skip over any numbers, you do so by using the “U” sign, which means union, or putting things together.
We can also use Set Builder/Inequality Notation, where, as we saw before, we use inequality signs to describe the ranges.
Notice on when we see arrows in the graphs, we have to assume that the function “goes on forever” in those directions.
Function Domain/Range Interval Notation: Domain: $$\left( {-\,\infty ,\infty } \right)$$ (or all Real Numbers, $$\mathbb{R}$$) Range: $$\left[ {-4,\infty } \right)$$ Inequality Notation: Domain: $$-\,\infty 1} \right\}$$ Range: $$\left\{ {y|\,\,y<-3\,\,\,\text{or}\,\,\,0\le y<5} \right\}$$ Note that the arrow indicates that the domain and range go on forever in the right-down direction. Also note that since the $$x$$’s don’t include “1”, we have to “jump over” it using a union ($$\cup$$) sign. We also have to use the union sign in a range. Interval Notation: Domain: $$\left[ {-6,-1} \right)\cup \left[ {0,5} \right)\,$$ Range: $$\left( {-\infty ,7} \right]$$ Inequality Notation: Domain:$$\left\{ {x|\,-6\le x<-1\,\,\,\text{or}\,\,\,0\le x<5} \right\}$$ Range: $$\left\{ {y|\,\,\,y\le 7} \right\}$$ Note that the domain skips from -1 to 0 (but includes 0). It gets closer and closer to 5 (an asymptote), but never reaches it. The range goes all the way up to 7, since there is overlap between parts of the function.
If you don’t see how we got the domain and range above, use the pencil trick, and make sure you start from the left for the domain (with vertical pencil) and from the bottom with the range (with horizontal pencil).
Note again that the last two graphs are not functions; they do not meet the Vertical Line Test requirements.
# Restricted Domains: Finding the Domain Algebraically
In many cases, the domain is restricted:
1. It is randomly indicated that way in the problem. For example, $$f\left( x \right)=3x-1,\,\,x\ge 0$$.
2. There is a variable in the denominator and that denominator could be 0. For example, $$\displaystyle f\left( x \right)=\frac{1}{{x-3}}$$. (In this case, $$x-3\ne 0;\,\,\,\,\,x\ne 3$$).
3. There is a variable underneath an even radical sign, and that radicand (underneath the radical sign) could be negative. For example, $$f\left( x \right)=\sqrt{{x+4}}$$. (In this case, $$x+4\ge 0;\,\,\,\,\,\,x\ge -4$$).
4. (More advanced – see Logarithmic Functions section) If there’s a variable in the argument of a log or ln function; log arguments must be greater than 0. For example, $$f\left( x \right)=\log \left( {8-x} \right)$$. (In this case, $$8-x>0;\,\,\,\,x<8$$).
(There are other types of functions, like trigonometric functions, that have domain restrictions, but we won’t address these here.)
(Note that if we could have a mixture of the above restrictions, for example, for $$\displaystyle f\left( x \right)=\frac{{\sqrt{x}}}{{x-1}}$$, $$x\ge 0$$ and $$x\ne 1$$, so the domain of $$x$$ is $$\left[ {0,1} \right)\cup \left( {1,\infty } \right)$$.)
We start out assuming that the domain of a function is all real numbers, but then see if there are any exceptions, as seen in the table. We will learn more about rational functions (shown in the first two examples, where there are variables in the denominator) in the Rational Functions and Equations, and Graphing Rational Functions, including Asymptotes sections.
Exceptions to All Reals in Domain How to Get the Restricted Domain Is there anywhere in the function where there is an $$\boldsymbol {x}$$ in a denominator, and that denominator could somehow be 0? We haven’t learned these types of functions yet, where we can have a variable in the denominator, but we will in the Rational Functions, Equations and Inequalities section. If so, the domain is all real numbers, excluding where any denominator could be 0; this is because we can never divide by 0. Example:$$\displaystyle \color{#800000}{{f(x)=\frac{x}{{x-3}}}}\text{: }\,\,x-3\ne 0\text{, therefore }x\ne 3$$ We solve the equation where $$x-3=0$$’ whatever we get for $$x$$ can’t be in the domain. Therefore, the domain is all real numbers except for 3, or $$\left( {-\infty ,3} \right)\cup \left( {3,\infty } \right)$$. Note that if we had $$\displaystyle f(x)=\frac{x}{{{{x}^{2}}+1}}$$ for example, this wouldn’t have a restriction; since the bottom could never be negative (try to make it negative!). Is there anywhere in the function where $$\boldsymbol {x}$$ is inside an even radical sign (root function?), and that radicand could somehow be negative? If we remember from the Powers, Exponents, Radicals, and Scientific Notation section, we can’t take an even root of a negative number, since we can’t multiply something an even number of times and get a negative number. If so, set what is under any even radical sign to be $$\ge 0$$, since we can’t take the even root of a negative number. The domain is all real numbers, excluding where $$x$$ could make something under an even radical sign negative. Example: $$\displaystyle \color{#800000}{{f(x)=\frac{x}{{\sqrt{{x-3}}}}}}\text{: }\,\,x-3\ge 0\text{, but also }\sqrt{{x-3}}\ne 0\text{ }$$ We had to use both exceptions, since we have both an even root and also a denominator with an $$x$$ in it. Solving both equations, we find that $$x$$ has to be greater than or equal to 3, but also cannot be 3. Therefore, the domain is all real numbers greater than 3, or $$\left( {3,\infty } \right)$$. Note that if we had $$f(x)=\sqrt{{{{x}^{2}}+1}}$$ for example, this wouldn’t have a restriction; since what’s under the radical sign (the radicand) can’t be negative (try to make it negative!). Does it indicate anywhere in the problem that the domain is restricted? Example: $$f(x)=5x,\,\,\,\,\,x\ge 0$$ I know this seems obvious, but in many cases (especially in real world problems), the domain is restricted, even if there is no variable in a denominator or under an even root sign. This domain is all real numbers greater than 0, or $$\left( {0,\infty } \right)$$. In case, the range would be all real numbers greater than 5, or $$\left( {5,\infty } \right)$$ (plug in the domain to see this). Is there a variable in the argument of a log (or ln) function? Could the argument ever be negative? We haven’t learned these types of functions yet, but we will in the Logarithmic Functions section. Example: $$\color{#800000}{{f(x)=\ln \left( {x+4} \right)}}:\,\,\,\,\,x+4>0\text{, therefore }x>-4$$ Log arguments must be greater than 0. We solve the equation where $$x+4>0$$; whatever we get for $$x$$ can’t be in the domain. Therefore, the domain is all real numbers greater than –4, or $$\displaystyle \left( {-4,\infty } \right)$$.
We will work on more advanced topics with functions later, in the Advanced Functions section.
Learn these rules, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.
You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!
On to Scatter Plots, Correlation, and Regression – you are ready!
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## Tuesday, April 28, 2015
### Whole Numbers and Place Values
Objective: Introduce the set of whole numbers and place values.
We begin by defining a set as a collection of objects, typically grouped within braces { }, where each object is called an element. For example, a set of colors, named C, is expressed as follows:
$C = \left\{ {{\rm{red, green, blue}}} \right\}$
When studying mathematics, we focus on special sets of numbers. The set of natural numbers, or counting numbers, is an infinite set that is denoted using $$\mathbb{N}$$.
$\mathbb{N} = \left\{ {1,\,2,\,3,\,4,\,5,...\} } \right.{\rm{ Natural Numbers}}$
The three periods (…) is called the ellipsis mark and indicates that the numbers continue without bound. We define the set of whole numbers, denoted using $$W$$, to be the set of natural numbers combined with zero.
$W = \left\{ {0,1,2,3,4,5,...\} } \right.{\rm{ Whole Numbers}}$
Developed between the 1st and 5th centuries, the Hindu-Arabic numeral system is the most common positional numeral system. This system allows us to express any whole number using the following set of ten digits.
$digits = \left\{ {0,1,2,3,4,5,6,7,8,9} \right\}$
A whole number is expressed using these digits and place values according to the following illustration.
For example, the average distance from the Earth to the Moon is 238,857 miles.
A number is written in expanded form to show the value of each digit. Here is the number 238,857 written in expanded form:
$200,000 + 30,000 + 8,000 + 800 + 50 + 7$
Example: Write in expanded form:
a. 375 b. 61,532 c. 5,102,700
a. $$375 = 300 + 70 + 5$$
b.$$61,532 = 60,000 + 1,000 + 500 + 30 + 2$$
c. $$5,102,700 = 5,000,000 + 100,000 + 2,000 + 700$$
Try this! Write in expanded form:
a. 8,014,050 b. 71,324 c. 68,090,004
a. $$8,000,000 + 10,000 + 4,000 + 50$$
b.$$70,000 + 1,000 + 300 + 20 + 4$$
c. $$60,000,000 + 8,000,000 + 90,000 + 4$$
Digits are grouped in sets of three, called periods, separated by a comma. It is sometimes acceptable to leave off the comma. For example, the four digit number 1,561 can also be written as 1561. From the right, each set of three digits represents the number of ones, thousands, millions, billions, and so on.
A number is read by starting with the leftmost period. With the exception of the ones period, the name of the period is added after verbalizing the corresponding number. For example, 238,857 miles is read, “two hundred thirty-eight thousand, eight hundred fifty-seven miles.” Note that the comma in the written sentence matches the comma in the number.
Example: In 2012, the world population was approximated to be 6,176,534,320 people. Write this number in words.
Solution: Begin by identifying the value of each period. Use this value for each three digit numerical grouping.
Note: Do not use the word “and” when writing whole numbers in words. As we will see, this word is reserved for the decimal.
Answer: Six billion, one hundred seventy-six million, five hundred thirty-four thousand, three hundred twenty people.
Example: Write the following numbers using words.
a. 371 b. 71,051 c. 67,151,205
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#### Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 1 subquestion ix maths textbook solution
$\text { Increasing interval }(-\infty, 2) \cup(3, \infty) \\ \text { Decreasing interval (2,3) }$
Given:
Here given that
$f(x)=2x^{3}-15x^{2}+36x+1$
To find:
We have to find the intervals in which function is increasing and decreasing.
Hint:
Put f ‘(x) = 0 to find critical points of f(x) and use increasing and decreasing property.
Solution:
We have,
$f(x)=2x^{3}-15x^{2}+36x+1$
Differentiating w.r.t. x, we get,
\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left(2 x^{3}-15 x^{2}+36 x+1\right) \\ &f \Rightarrow f^{\prime}(x)=6 x^{2}-30 x+36 \end{aligned}
Now for critical points of f(x), we must have,
\begin{aligned} &f^{\prime}(x)=0 \\ &\Rightarrow 6 x^{2}-30 x+36=0 \\ &\Rightarrow 6\left(x^{2}-5 x+6\right)=0 \\ &\Rightarrow x^{2}-5 x+6=0\{\therefore 6>0\} \\ &\Rightarrow x^{2}-3 x-2 x+6=0 \\ &\Rightarrow(x-3)(x-2)=0 \\ &\Rightarrow x-3=0 \text { and } x-2=0 \\ &\Rightarrow x=3 \text { and } x=2 \end{aligned}
\begin{aligned} &\text { Clearly, } f^{\prime}(x)>0 \text { if } x<2 \text { and } x>3 \text { or } x \in(-\infty, 2) \text { and } x \in(3, \infty) \text { and } f^{\prime}(x)<0 \text { if }\\ &2
$\text { So, } f(x) \text { is increasing on the interval } (-\infty, 2)\: \cup\: (3, \infty) \text { and } f(x) \text { is decreasing on interval }(2,3).$
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# Potentiation Exercises
The potentiation is the mathematical operation that represents the multiplication of the same factors. That is, we use the potentiation when a number is multiplied by itself several times.
Take advantage of the commented exercises, proposals and contest questions to test your knowledge about enhancement.
## Question 1
Determine the value of each of the powers below.
a) 25 1
b) 150 0
c) (7/9) -2
Correct answer: a) 25, b) 1 and c) 81/49.
a) When a power is raised to exponent 1, the result is the base itself. Therefore, 25 1 = 25.
b) When a power is raised to the exponent 0, the result is the number 1. Therefore, 150 0 = 1.
c) In this case, we have a fraction raised to a negative exponent. To solve it, we must invert the base and change the exponent sign.
Now, we can raise the numerator and denominator to exponent 2.
## Question 2
Knowing that the value of 5 7 is 78 125, what is the result of 5 8 ?
a) 156 250
b) 390 625
c) 234 375
d) 312 500
Correct answer: b) 390 625.
To solve this question we can transform 5 8 into a multiplication of powers of equal bases, since x . a y = a x + y
As we know the value of 5 7 , we transform the number 5 8 as follows:
8 = 5 7 . 5, because 5 7 . 5 = 5 7 + 1 = 5 8
So, to find the result, we just need to replace the value of 5 7 and multiply by 5.
7 . 5 = 78 125. 5 = 390 625
## Question 3
Are the powers (-2) 4 and -2 4 the same or different? And what is the result?
Correct answer: The powers are different and have 16 and -16 results, respectively.
When the base of a power is a negative number and is raised to an even exponent, the result will be positive. However, to signal that the base is negative, its value must be in parentheses.
(- 2) 4 = (- 2) x (- 2) x (- 2) x (- 2) = +16
When there are no parentheses separating the base, the negative sign must be included in the result.
– 2 4 = – 16
Therefore, the results are: (- 2) 4 = 16 and – 2 4 = – 16.
## Question 4
On one site there are 12 trees. Each tree has 12 branches and each branch has 12 apples. How many apples are there on the farm?
a) 144
b) 1224
c) 1564
d) 1728
Correct answer: 1,728 apples.
We have a power where number 12 is the base and number 3 is the number of times the base is repeated.
Let’s take one of the trees as an example. In each of the 12 branches of a tree there are 12 apples, that is, 12 branches times 12 apples: 12 x 12 = 144.
Only in total we have 12 trees, that is, 144 x 12 gives us the total number of apples. This can be expressed in the form of power.
12 x 12 x 12 = 12 3 = 1 728.
Therefore, the site has 1,728 apples.
## Question 5
The value of the expression 20x 3 + 2x 2 y 5 , for x = – 4 and y = 2 is :
a) 256
b) – 400
c) 400
d) – 256
Correct answer: d) – 256.
To solve the expression the first step is to replace the letters with the values, so the expression will be:
20. (- 4) 3 + 2. (- 4) 2 . 2 5
We must be careful with the signals when resolving the potentiation. When the base is negative, the result will be positive if the exponent is even and will be negative when the exponent is odd. Thus, the expression will be:
20. (- 64) + 2. (+16). 32
Now that we have solved the potentiations, let’s solve the other operations, remembering that we first solve the multiplications and then the subtraction.
– 1280 + 1024 = – 256
So, the correct answer is alternative d .
## Question 6
(3 6. 3 -2 ): 3 4 is equal to :
a) 0
b) 1
c) 3 -3
d) 3 -8
Correct answer: b) 1.
We can solve the proposed numerical expression in two ways. One is to solve each of the powers first and then solve the other operations. The other way is to use the property of multiplication and division of powers of the same base. We will resolve these two paths.
1st way: Let’s solve the value of each power:
Now we are going to replace the values found in the expression and solve the indicated operations. Remembering that we must first resolve the operation within the parentheses.
So, the right answer is the letter b.
2nd way : By applying the property, we must remember that in the multiplication of powers of the same base, the base is repeated and the exponents are added. In the division, the base is repeated and the exponents are subtracted. Thus, we have:
Remembering that every number raised to zero is equal to 1, we arrive at the same result found previously.
Note that in the 2nd form we find the result more easily. Therefore, it is very important to know the boosting properties.
## Question 7
Simplifying the expression below, we find :
a) 2
b) 2 10
c) 2 15
d) 2 16
Correct answer: d) 2 16 .
We must note that the operation between the base 2 powers is the sum. Therefore, we will have to find a way to simplify, because even if we have the same bases we cannot add.
One way to simplify is to try to keep the same exponent in both powers, so we can put it in evidence. For that, let’s write 2 67 as 2 65 . 2 2 , replacing in the expression we have:
We can highlight the 2 65 as follows:
This can be done, since 2 65 multiplying the terms 1 and 4 results in the initial expression.
Simplifying the common terms, we have:
Now we can apply the property of the power division of the same base, remembering that when the exponent does not appear, its value is equal to 1.
So the answer will be the letter d.
## Question 8
Calculate the value of the expression below:
To resolve this issue, we must first rewrite the terms.
When multiplying powers of the same base, we can repeat the base and add the exponents.
x .3 2 = 3 x + 2
x .3 = 3 x + 1
In the division of powers of the same base we can repeat the base and subtract the exponents.
Substituting the values in the expression, we have:
Note that in the numerator the term 3x is repeated and, therefore, we can put it in evidence.
Since we have a division with a fraction, we repeat the numerator of the first and multiply by the inverse of the second to then solve the expression.
Therefore, the expression’s value is 36.
## Question 9
Check whether the sentences are false or true:
a) (x. y) 4 = x 4 . y 4
b) (x + y) 4 = x 4 + y 4
c) (x – y) 4 = x 4 – y 4
d) (x + y) 0 = 1
the truth. In this power, whose base is a multiplication, the factors can be raised separately to the exponent before being multiplied.
b) False. The expression is a newton binomial of type (a + b) n . In this case, we have a binomial of degree 4, the result of which is:
c) False. The expression is a Newton binomial of type (a – b) n . The correct answer is a polynomial:
d) True. It is a power with exponent 0 and, therefore, its result must be 1.
## Question 10
The value of (0.3) -1 + (- 27) 0.333 … is:
To resolve the issue, we must first rewrite the numbers 0.3 and 0.333 … as fractions.
Note that in this case, we just write the number in the numerator and to the denominator we add the number of zeros that corresponds to the number of decimal places after the comma, which is just one.
0.333 … is a periodic tithe and we need to find its generating fraction.
For this, we write the number that is repeated in the periodic tithe in the numerator and divide by 9.
Now, we can replace the values in the expression.
The first term of the expression has a negative exponent. To make it positive, we must invert the power base.
The second term presents a fraction as an exponent. We can then turn it into a root.
The fractional division is solved by repeating the numerator and multiplying by the inverse of the second.
Inside the root we have the power of a power. To solve it, we must maintain the base and multiply the exponents.
Since the exponent inside the root has the same value as the radical index, we can eliminate the root and solve the expression.
## Question 11
(UFRGS – 2015) The digits of the 9 99 – 4 44 units are:
a) 1
b) 2
c) 3
d) 4
e) 5
Correct alternative: c) 3.
Observe the results below when we raise the bases of expression from the first to the fifth power.
Note that there is the following pattern:
The number 9 when raised to an odd exponent shows the number 9 in the unit box and when raised to an even exponent it shows the number 1 in the unit box.
The number 4 when raised to an odd exponent shows the number 4 in the unit house and when raised to an even exponent it shows the number 6 in the unit house.
Therefore, in the number of the units when performing the expression 9 99 – 4 44 we will find the number 3, since 9 – 6 = 3.
## Question 12
(UFRGS – 2013) A healthy human adult is home to about 100 billion bacteria, only in his digestive tract. This number of bacteria can be written as
a) 10 9
b) 10 10
c) 10 11
d) 10 12
e) 10 13
Correct alternative: c) 10 11
A billion is the same as a billion, that is, 1000 x 1,000,000 = 1,000,000,000.
100 billion is equal to 100 x 1,000,000,000 = 100,000,000,000.
Large numbers like the one in this question can be written in scientific notation, whose writing follows the N pattern. 10 n , where N is a number less than 10 and greater than or equal to 1. The base 10 exponent is the number of decimal places that the comma “walked” to obtain the value of N.
Note that to get to number 11 it was necessary to “walk” 11 decimal places. Therefore, we have the power 10 11 as a result.
## Question 13
(Enem – 2012) The North American Space Agency (NASA) reported that the asteroid YU 55 crossed the space between the Earth and the Moon in November 2011. The following illustration suggests that the asteroid traveled its path on the same plane as contains the orbit described by the Moon around the Earth. The figure shows the proximity of the asteroid to Earth, that is, the shortest distance it passed from the Earth’s surface.
Based on this information, the shortest distance that the asteroid YU 55 passed from the Earth’s surface is equal to
a) 3.25 .10 2 km
b) 3.25 .10 3 km
c) 3.25. 10 4 km
d) 3.25. 10 5 km
e) 3.25. 10 6 km
Correct alternative: d) 3.25. 10 5 km
In the figure, it is indicated the shortest distance he passed from the Earth’s surface, which is 325 thousand km, that is, 325 000 km.
This number must be written in scientific notation. For this, we must “walk” with the comma until we find a number less than 10 and greater than or equal to 1. The number of decimal places that the comma “walked” corresponds to the base 10 exponent in the formula N. 10 n .
We reached number 3.25 and, for that, the comma “walked” 5 decimal places. Therefore, in scientific notation, the asteroid’s proximity to Earth is 3.25. 10 5 km.
## Question 14
(EPCAR – 2011) Simplifying the expression
a) – x -94
b) x 94
c) x -94
d) – x 94
Correct alternative: a) -x -94
First, we rewrite the exponents that are in the form of power.
Substituting the values in the expression, we have:
As we have high powers to other exponents, we must conserve the base and multiply the exponents.
We can then insert the calculated values into the expression.
Both in the numerator and in the denominator there is a multiplication of powers of equal bases. To solve them we must repeat the base and add the exponents.
Now, as we owe the division of powers of the same base, we can repeat the base and subtract the exponents.
Therefore, the correct alternative is the letter a, the result of which is -x -94 .
## Question 15
(Enem – 2016) To celebrate the anniversary of a city, the city organizes four consecutive days of cultural attractions. The experience of previous years shows that, from one day to the next, the number of visitors to the event is tripled. 345 visitors are expected to attend the first day of the event.
A possible representation of the expected number of participants for the last day is
a) 3 × 345
b) (3 + 3 + 3) × 345
c) 3 3 × 345
d) 3 × 4 × 345
e) 3 4 × 345
Correct alternative: c) 3 3 × 345
At this point we have a case in geometric progression, for a number multiplied by a ratio (q) correspond to the next set of sequence numbers as the formula
Where:
n : last day of the event, that is, day 4.
1 : number of participants on the first day of the event, which is 345.
(n-1) : reason, whose exponent is formed by the number we want to obtain minus 1
According to previous experiences, from one day to the next, the number of visitors to the event is tripled, that is, q = 3.
Substituting the values in the formula for the general term, we have:
Therefore, 9 315 people are expected for the last day of the event and a possible representation of the expected number of participants for the last day is 3 3 × 345.
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## Precalculus (6th Edition) Blitzer
The solution set is $\left\{ \left( -1,-1 \right) \right\}$.
By rewriting the first equation, we get: \begin{align} & \frac{x}{6}-\frac{3y}{6}=\frac{2}{6} \\ & x-3y=2 \end{align} Therefore, the system can be rewritten as: \begin{align} & x-3y=2 \\ & x+2y=-3 \end{align} And subtract the second equation from the first equation: \begin{align} & x-3y-x-2y=2-\left( -3 \right) \\ & -5y=2+3 \\ & -5y=5 \\ & y=-1 \end{align} Substitute the value in the first equation and solve for x: \begin{align} & x-3\left( -1 \right)=2 \\ & x+3=2 \\ & x=-1 \end{align}. Thus, The solution set is $\left\{ \left( -1,-1 \right) \right\}$.
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# Difference between revisions of "2010 AMC 10A Problems/Problem 23"
## Problem
Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010$, the box in the $k\text{th}$ position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n) < \frac{1}{2010}$?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005$
## Solution
### Solution 1
The probability of drawing a white marble from box $k$ is $\frac{k}{k+1}$. The probability of drawing a red marble from box $n$ is $\frac{1}{n+1}$.
The probability of drawing a red marble at box $n$ is therefore
$\frac{1}{n+1} \left( \prod_{k=1}^{n-1}\frac{k}{k+1} \right) < \frac{1}{2010}$
$\frac{1}{n+1} \left( \frac{1}{n} \right) < \frac{1}{2010}$
$(n+1)n > 2010$
It is then easy to see that the lowest integer value of $n$ that satisfies the inequality is $\boxed{45\ \textbf{(A)}}$.
### Solution 2
Using the first few values of $n$, it is easy to derive a formula for $P(n)$. The chance that she stops on the second box ($n=2$) is the chance of drawing a white marble then a red marble: $\frac{1}2 * \frac{1}3$. The chance that she stops on the third box ($n=3$) is the chance of drawing two white marbles then a red marble:$\frac{1}2 * \frac{2}3 * \frac{1}4$. If $n=4$, $P(n) = \frac{1}2 * \frac{2}3 * \frac{3}4 * \frac{1}5$.
Cross-cancelling in the fractions gives $P(2) =$\frac{1}{2*3}$,$P(3) = $\frac{1}{3*4}$, and $P(4) = \frac{1}{4*5}$. From this, it is clear that $P(n) = \frac{1}{(n)(n+1)}$. (Alternatively, $P(n) = \frac{(n-1)!}{(n+1)!}$.)
$\frac{1}{(n+1)(n)} < \frac{1}{2010}$
The lowest integer that satisfies the above inequality is \boxed{(A) 45}.
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# Multiplication Multiplying by multiples of 10. Multiplying by 10 and 100: Count by each number 5 times. 5 x 1 =5 5 x 10 = 50 5 x 100 =500 5 x 1,000 =
## Presentation on theme: "Multiplication Multiplying by multiples of 10. Multiplying by 10 and 100: Count by each number 5 times. 5 x 1 =5 5 x 10 = 50 5 x 100 =500 5 x 1,000 ="— Presentation transcript:
Multiplication Multiplying by multiples of 10
Multiplying by 10 and 100: Count by each number 5 times. 5 x 1 =5 5 x 10 = 50 5 x 100 =500 5 x 1,000 = 5,000 5 x 10,000 = 50,000 Remember to think of it as 5 groups of ____
What will we learn today? Today, instead of counting by equal groups, we will use two strategies to help us multiply. 1: We will use the associate property to make multiplying easier. 2: We will use place value to make multiplying easier.
Let’s look at this one: 5 x 40 Think back to the associative property from yesterday….
Let’s look at this one: 5 x 40 = 5 x (4 x 10) 40
Let’s look at this one: 5 x 40 = (5 x 4) x 10 20
Let’s look at this one: 5 x 40 = x 10 20= 200 5 x 40 = 200
60 9 90 120 12 15 150 EngageNY
Use what we just learned to solve this with your group…. 20 x 4 = 2 x 10 x 4
Use what we just learned to solve this with your group…. 20 x 4 = 2 x 4 x 10
Use what we just learned to solve this with your group…. 20 x 4 = (2 x 4) x 10 8
Use what we just learned to solve this with your group…. 20 x 4 = 8 x 10
Use what we just learned to solve this with your group…. 20 x 4 = 8 x 10 8 x 10 = 10 x 8
Let’s look at this one again: 5 x 40 20 With one zero, it’s like multiplying by a 10, or moving the number over one place value from 20 ones to 20 tens. 20 tens = 200!
Let’s look at this one again: 5 x 40 = 5 x 4tens 20 tens With one zero, it’s like multiplying by a 10, or moving the number over one place value from 20 ones to 20 tens. 20 tens = 200!
Here’s another example… 12 120
EngageNY 8 8 8 80 15 150 You try!
Use what we just learned to solve this with your group…. 20 x 4 = 2 tens x 4 = 8 tens = 80 3 x 80 = 3 x 8 tens = 24 tens = 240
Let’s look at multiplying by multiples of 10 again.
By the end of this lesson, you will be able to answer this problem! 50 x 400
Let’s look a little closer at what we saw earlier…Do you notice a pattern? 5 x 1 =5 5 x 10 = 50 5 x 100 =500 5 x 1,000 = 5,000 5 x 10,000 = 50,000
Let’s look at this one yet again… 5 x 40 200 20
Here’s what you should know…. You can use the associative property: 5 x 60 = 5 x (6 x 10)
Here’s what you should know…. You can use the associative property: 5 x 60 = (5 x 6) x 10
Here’s what you should know…. You can use the associative property: 5 x 60 = (5 x 6) x 10 = 30 x 10
Here’s what you should know…. You can use the associative property: 5 x 60 = (5 x 6) x 10 = 30 x 10 = 300 When you multiply by 10, you move it to the left one place value.
Here’s what you should know…. You can use the associative property: 5 x 60 = (5 x 6) x 10 = 30 x 10 = 300 When you multiply by 10, you move it to the left one place value. When you multiply by 100, you move it over 2 place values. 30 x 100 = 3,000
Here’s what you should know…. You can use the associative property: 5 x 60 = (5 x 6) x 10 = 30 x 10 = 300 When you multiply by 10, you move it to the left one place value. When you multiply by 100, you move it over 2 place values. What do you think happens when you multiply by 1,000? 10,000? 30 x 1,000 = 30,000 30 x 100 = 3,000 30 x 10,000 = 300,000
Here’s what you should know…. You can use the associative property: 5 x 60 = (5 x 6) x 10 = 30 x 10 = 300 When you multiply by 10, you move it to the left one place value. When you multiply by 100, you move it over 2 place values. If you know this, you can take the shortcut! 1: Find your basic fact and multiply. 2: Count the number of zeros in the factors. 3: Add the same number of zeros to the product!
Multiplying by 10 and 100 50 x 3 = 150 50 x 30 = 1,500 500 x 30 = 15,000 500 x 300 = 150,000 5 x 3 = 15
What do you think we will do here? 5 X 4 5 x 4
What do you think we will do here? 5 X 4 5 x 4 = 20
What do you think we will do here? 5 X 4 5 x 4 = 20
What do you think we will do here? 5 X 4 5 x 4 = 20
What do you think we will do here? 200,000 50 x 4,000
You try! 40 x 100
What do you think we will do here? 4 X 1 4 x 1 =
What do you think we will do here? 4 X 1 4 x 1 = 4
What do you think we will do here? 4 X 1 4 x 1 = 4
What do you think we will do here? 4 X 1 4 x 1 = 4
What do you think we will do here? 4 X 1 4 x 1 = 4,000
You try! 400 x 200
You try! 400 x 200 8
You try! 400 x 200 80000
You try! 400 x 200 80,000
What do you think? 60 x 300 18,000
Try all 4 on your board 9 x 400 1 x 70,000 60 x 900 200 x 9,000
Try all 4 on your board 9 x 400 1 x 70,000 60 x 900 200 x 9,000 3,600 70,000 54,000 1,800,000
How do you feel? Use your sticker chart to show me how you feel about multiplication with zeros?
Great job!
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# PreAlgebra 1.8
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### PreAlgebra 1.8
1. 1. A coordinate plane is formed by the intersection of a horizontal number line called the x -axis and a vertical number line called the y -axis . The axes meet at a point called the origin and divide the coordinate plane into four quadrants . Each point in a coordinate plane is represented by an ordered pair . The first number is the x -coordinate , and the second number is the y -coordinate. Point P is represented by the ordered pair (–3, –2) . Point P is in Quadrant III . The Coordinate Plane 1.8 LESSON
2. 2. Naming Points in a Coordinate Plane SOLUTION Give the coordinates of the point. A Point A is 2 units to the right of the origin and 4 units down. The x -coordinate is 2 , and the y -coordinate is –4 . The coordinates are (2, –4) . EXAMPLE 1 The Coordinate Plane 1.8 LESSON
3. 3. Naming Points in a Coordinate Plane SOLUTION Give the coordinates of the point. A B Point A is 2 units to the right of the origin and 4 units down. The x -coordinate is 2 , and the y -coordinate is –4 . The coordinates are (2, –4) . Point B is 3 units to the left of the origin and 2 units up. The x -coordinate is –3 , and the y -coordinate is 2 . The coordinates are (–3, 2) . EXAMPLE 1 The Coordinate Plane 1.8 LESSON
4. 4. Plotting Points in a Coordinate Plane SOLUTION A (4, 1) Begin at the origin and move 4 units to the right, then 1 unit up. Point A is in Quadrant I . Plot the point in a coordinate plane. Describe the location of the point. EXAMPLE 2 The Coordinate Plane 1.8 LESSON
5. 5. Plotting Points in a Coordinate Plane SOLUTION A (4, 1) Begin at the origin and move 4 units to the right, then 1 unit up. Point A is in Quadrant I . Plot the point in a coordinate plane. Describe the location of the point. B (0, –3) Begin at the origin and move 3 units down. Point B is on the y -axis. EXAMPLE 2 The Coordinate Plane 1.8 LESSON
6. 6. Plotting Points in a Coordinate Plane SOLUTION A (4, 1) Begin at the origin and move 4 units to the right, then 1 unit up. Point A is in Quadrant I . Plot the point in a coordinate plane. Describe the location of the point. B (0, –3) Begin at the origin and move 3 units down. Point B is on the y -axis. C (–2, –5) Begin at the origin and move 2 units to the left, then 5 units down. Point C is in Quadrant III . EXAMPLE 2 The Coordinate Plane 1.8 LESSON
7. 7. Scatter Plots A scatter plot uses a coordinate plane to display paired data. Each data pair is plotted as a point. A scatter plot may suggest whether a relationship exists between two sets of data. The Coordinate Plane 1.8 LESSON
8. 8. Making a Scatter Plot Fish A biologist measured the lengths and masses of eight rainbow trout. Make a scatter plot of the data shown in the table and describe any relationship you see. SOLUTION Write the data as ordered pairs. Let the x -coordinate represent the length, and let the y -coordinate represent the mass: (405, 715), (360, 557), (413, 754), (395, 584), (247, 184), (280, 248), (265, 223), (351, 506) Plot the ordered pairs in a coordinate plane. You need only the first quadrant. Notice that the points rise from left to right. You can conclude that as the lengths of the rainbow trout increase, their masses tend to increase. EXAMPLE 3 506 223 248 184 584 754 557 715 Mass (grams) 351 265 280 247 395 413 360 405 Length (millimeters) 1 2 The Coordinate Plane 1.8 LESSON
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## How do you construct a perpendicular line with a compass and straightedge?
The perpendicular bisector of a line segment
1. open the compass more than half of the distance between A and B, and scribe arcs of the same radius centered at A and B.
2. Call the two points where these two arcs meet C and D. Draw the line between C and D.
3. CD is the perpendicular bisector of the line segment AB.
4. Proof.
### How do you construct a perpendicular line?
How to Construct a Perpendicular Line through a Point on the Given Line?
1. Open the compass to a radius less than half the segment.
2. Draw two arcs intersecting the line on both sides of the point.
3. Draw two arcs using the intersection points as the centers.
4. Construct a line between this point and the original point.
What are the steps in constructing perpendicular and parallel lines?
Constructing perpendicular and parallel lines
1. Step 1: Draw a perpendicular line between A and XY.
2. Step 2: Measure the perpendicular distance between the point and the line.
3. Step 3: Draw a point that is the same distance from the line.
4. Step 4: Draw the parallel line.
What is the difference between constructing perpendicular lines?
Explanation: When there are two parallel lines, these two lines are never able to intersect or touch. Perpendicular lines are two lines in which one of the lines intersects the other line, and the angles created from the intersection of these two lines are all right angles.
## What is perpendicular lines example?
Lines that intersect each other forming a right angle are called perpendicular lines. Example: the steps of a straight ladder; the opposite sides of a rectangle. The symbol used to denote two perpendicular lines: ⊥ ⊥ .
### How do you know if a line is perpendicular?
Two lines are perpendicular if and only if the product of their slopes is . In other words, the slope of a line that is perpendicular to a given line is the negative reciprocal of that slope. Thus, for a line with a given slope of 3, the line perpendicular to that slope must be the negative reciprocal of 3, or .
What is the difference between perpendicular and parallel lines?
Parallel lines are lines in a plane that are always the same distance apart. Perpendicular lines are lines that intersect at a right (90 degrees) angle.
Which situation shows perpendicular lines?
Perpendicular lines occur anytime two lines meet at a 90° angle, also known as a right angle. Sometimes, you will see a little square in the corner of an angle, to show it is perpendicular. There are many examples of perpendicular lines in everyday life, including a football field and train tracks.
## What are two differences between parallel lines and perpendicular lines?
Parallel lines are lines in a plane that are always the same distance apart. Parallel lines never intersect. Perpendicular lines are lines that intersect at a right (90 degrees) angle.
### What are 2 perpendicular lines?
Two lines are perpendicular if and only if the product of their slopes is . In other words, the slope of a line that is perpendicular to a given line is the negative reciprocal of that slope.
Can You bisect a line with a straightedge and compass?
Bisecting a line (dividing a line in half) is easy enough if you know the length of the line, or can measure the line. But if you don’t know the length of the line, you can still bisect it using a straightedge and compass. 1. Draw the line segment you need to bisect.
How to construct a perpendicular from a line?
Perpendicular bisector of a line segment Perpendicular from a line at a point Perpendicular from a line through a point Perpendicular from endpoint of a ray Divide a segment into n equal parts Parallel line through a point (angle copy) Parallel line through a point (rhombus)
## How do you draw lines with a compass?
Place the compass at one end of line segment. Draw arcs above and below the line. Keeping the same compass width, draw arcs from other end of line. Place ruler where the arcs cross, and draw the line segment.
### How to calculate the perpendicular bisector of a line segment?
HomeContactAboutSubject Index Perpendicular bisector of a line segment This construction shows how to draw the perpendicular bisectorof a given line segmentwith compass and straightedge or ruler. This both bisects the segment (divides it into two equal parts, and is perpendicularto it. It finds the midpointof the given line segment.
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# What Do The Stars Say About Elisabeth Hasselbeck? (12/17/2019)
How will Elisabeth Hasselbeck fare on 12/17/2019 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is just for fun – take it with a grain of salt. I will first find the destiny number for Elisabeth Hasselbeck, and then something similar to the life path number, which we will calculate for today (12/17/2019). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology enthusiasts.
PATH NUMBER FOR 12/17/2019: We will consider the month (12), the day (17) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. What does this entail? We will show you. First, for the month, we take the current month of 12 and add the digits together: 1 + 2 = 3 (super simple). Then do the day: from 17 we do 1 + 7 = 8. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 3 + 8 + 12 = 23. This still isn’t a single-digit number, so we will add its digits together again: 2 + 3 = 5. Now we have a single-digit number: 5 is the path number for 12/17/2019.
DESTINY NUMBER FOR Elisabeth Hasselbeck: The destiny number will take the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Elisabeth Hasselbeck we have the letters E (5), l (3), i (9), s (1), a (1), b (2), e (5), t (2), h (8), H (8), a (1), s (1), s (1), e (5), l (3), b (2), e (5), c (3) and k (2). Adding all of that up (yes, this can get tiring) gives 67. This still isn’t a single-digit number, so we will add its digits together again: 6 + 7 = 13. This still isn’t a single-digit number, so we will add its digits together again: 1 + 3 = 4. Now we have a single-digit number: 4 is the destiny number for Elisabeth Hasselbeck.
CONCLUSION: The difference between the path number for today (5) and destiny number for Elisabeth Hasselbeck (4) is 1. That is lower than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t go jumping for joy yet! As mentioned earlier, this is for entertainment purposes only. If you want a forecast that we do recommend taking seriously, check out your cosmic energy profile here. Go ahead and see what it says for you – you’ll be glad you did.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
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The multiplicative inverse is provided to simplify mathematical expressions. Words 'inverse' means something opposite/contrary in effect, order, position, or direction. A number the nullifies the influence of a number to identity 1 is called a multiplicative inverse.
You are watching: What is the multiplicative inverse of 1 9
1 What is Multiplicative Inverse? 2 Multiplicative inverse of a natural Number 3 Multiplicative inverse of a Unit Fraction 4 Multiplicative train station of a Fraction 5 Multiplicative train station of a combined Fraction 6 Multiplicative inverse of complicated Numbers 7 Modular Multiplicative Inverse 8 FAQs on Multiplicative Inverse
## What is Multiplicative Inverse?
The multiplicative train station of a number is defined as a number which once multiplied through the initial number offers the product together 1. The multiplicative station of 'a' is denoted by a-1 or 1/a. In various other words, as soon as the product of 2 numbers is 1, lock are said to it is in multiplicative inverses of each other. The multiplicative station of a number is identified as the division of 1 by that number. It is additionally called the reciprocal of the number. The multiplicative inverse residential property says the the product that a number and also its multiplicative inverse is 1.
For example, let us think about 5 apples. Now, divide the apples into 5 groups that 1 each. To make them into teams of 1 each, we must divide lock by 5. Dividing a number by itself is equivalent to multiplying it through its multiplicative train station . Hence, 5 ÷ 5 = 5 × 1/5 = 1. Here, 1/5 is the multiplicative train station of 5.
## Multiplicative inverse of a organic Number
Natural numbers room counting numbers beginning from 1. The multiplicative inverse of a organic number a is 1/a.
Examples
3 is a herbal number. If us multiply 3 by 1/3, the product is 1. Therefore, the multiplicative station of 3 is 1/3.Similarly, the multiplicative train station of 110 is 1/110.
### Multiplicative inverse of a an adverse Number
Just together for any positive number, the product of a an adverse number and also its reciprocal have to be same to 1. Thus, the multiplicative train station of any negative number is the reciprocal. Because that example, (-6) × (-1/6) = 1, therefore, the multiplicative train station of -6 is -1/6.
Let us take into consideration a few more instances for a far better understanding.
## Multiplicative train station of a Unit Fraction
A unit portion is a portion with the molecule 1. If we multiply a unit portion 1/x by x, the product is 1. The multiplicative station of a unit fraction 1/x is x.
Examples:
The multiplicative train station of the unit fraction 1/7 is 7. If we multiply 1/7 by 7, the product is 1. (1/7 × 7 = 1)The multiplicative station of the unit portion 1/50 is 50. If we multiply 1/50 by 50, the product is 1. (1/50 × 50 = 1)
## Multiplicative station of a Fraction
The multiplicative station of a fraction a/b is b/a due to the fact that a/b × b/a = 1 when (a,b ≠ 0)
Examples
The multiplicative train station of 2/7 is 7/2. If us multiply 2/7 by 7/2, the product is 1. (2/7 × 7/2 = 1)The multiplicative station of 76/43 is 43/76. If we multiply 76/43 by 43/76, the product is 1. (76/43 × 43/76 = 1)
## Multiplicative inverse of a combined Fraction
To discover the multiplicative train station of a mixed fraction, transform the mixed fraction into an improper fraction, then determine its reciprocal. For example, the multiplicative train station of (3dfrac12)
Step 1: transform (3dfrac12) come an not correct fraction, the is 7/2.Step 2: find the reciprocal of 7/2, the is 2/7. Thus, the multiplicative train station of (3dfrac12) is 2/7.
## Multiplicative inverse of complex Numbers
To find the multiplicative train station of complicated numbers and also real number is quite challenging as girlfriend are managing rational expressions, through a radical (or) square source in the denominator component of the expression, which makes the portion a little complex.
Now, the multiplicative train station of a facility number the the type a + (i)b, such as 3+(i)√2, where the 3 is the actual number and also (i)√2 is the imaginary number. In order to uncover the mutual of this complicated number, multiply and also divide the by 3-(i)√2, together that: (3+(i)√2)(3-(i)√2/3-(i)√2) = 9 + (i)22/3-(i)√2 = 9 + (-1)2/3-(i)√2 = 9-2/3-(i)√2 = 7/3-(i)√2. Therefore, 7/3-(i)√2 is the multiplicative inverse of 3+(i)√2
Also, the multiplicative train station of 3/(√2-1) will be (√2-1)/3. While finding the multiplicative inverse of any kind of expression, if over there is a radical existing in the denominator, the portion can be rationalized, as displayed for a portion 3/(√2-1) below,
Step 2: Solve. (frac3 sqrt2+12 - 1)Step 3: leveling to the lowest form. 3(√2+1)
## Modular Multiplicative Inverse
The modular multiplicative inverse of one integer ns is another integer x such the the product px is congruent come 1 through respect come the modulus m. It deserve to be represented as: px (equiv ) 1 (mod m). In various other words, m divides px - 1 completely. Also, the modular multiplicative station of an essence p deserve to exist through respect come the modulus m just if gcd(p, m) = 1
In a nutshell, the multiplicative inverses room as follows:
TypeMultiplicative InverseExample
Natural Number
x
1/xMultiplicative inverse of 4 is 1/4
Integer
x, x ≠ 0
1/xMultiplicative train station of -4 is -1/4
Fraction
x/y; x,y ≠ 0
y/xMultiplicative train station of 2/7 is 7/2
Unit Fraction
1/x, x ≠ 0
xMultiplicative train station of 1/20 is 20
Tips ~ above Multiplicative Inverse
The multiplicative inverse of a fraction can be derived by flipping the numerator and denominator.The multiplicative station of 1 is 1.The multiplicative inverse of 0 is no defined.The multiplicative inverse of a number x is written as 1/x or x-1.The multiplicative station of a mixed portion can be obtained by convert the mixed portion into one improper fraction and determining its reciprocal.
Important Notes
The multiplicative station of a number is likewise called that reciprocal.The product the a number and also its multiplicative train station is same to 1.
Example 1: A pizza is sliced right into 8 pieces. Tom keeps 3 slices that the pizza in ~ the counter and also leaves the remainder on the table for his 3 friends to share. What is the part that every of his friend get? execute we apply multiplicative train station here?
Solution:
Since Tom ate 3 slices the end of 8, it implies he ate 3/8th component of the pizza.
The pizza left out = 1 - 3/8 = 5/8
5/8 to it is in shared amongst 3 friends ⇒ 5/8 ÷ 3.
We take the multiplicative station of the divisor to leveling the division.
5/8 ÷ 3/ 1
= 5/8 × 1/3
= 5/24
Answer: each of Tom's friends will be gaining a 5/24 part of the left-over pizza.
Example 2: The full distance from Mark's residence to school is 3/4 of a kilometer. He deserve to ride his cycle 1/3 kilometer in a minute. In how countless minutes will certainly he reach his institution from home?
Solution:
Total distance from residence to school = ¾ km
Distance covered in a minute = 1/3 km
The time taken to cover the total distance = full distance/ distance covered
= 3/4 ÷ 1/3
The multiplicative inverse of 1/3 is 3.
3/4 × 3 = 9/4 = 2.25 minutes
Answer: Therefore, the time taken to cover the total distance by note is 2.25 minutes.
Example 3: uncover the multiplicative train station of -9/10. Also, verify your answer.
Solution:
The multiplicative train station of -9/10 is -10/9.
See more: To The Right To The Right To The Left, To The Left, To The Left
To verify the answer, we will certainly multiply -9/10 v its multiplicative inverse and check if the product is 1.
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# One’s Complement
• Difficulty Level : Basic
• Last Updated : 05 Oct, 2022
In digital electronics, the binary system is one of the most common number representation techniques. As its name suggest binary number system deals with only two number 0 and 1, this can be used by any device which is operating in two states. The binary number system has two complements 1’s and 2’s complement.
### One’s Complement:
In simple words, if we want to understand the One’s complement, so one’s complement is toggling or exchanging all the 0’s into 1 and all the 1’s into 0 of any number. Suppose there is a binary number 11001001, then its one’s complement will be 00110110.
In actuality, the one’s complement means the addition of a negative integer to the number, and this eliminates the requirement of a separate subtraction processor.
### How to Find One’s Complement of a Number:
To find one’s complement of any number, follow the below steps:
1. Convert the number of any number system to a binary number system, i.e. if the number is in octal, decimal, hexadecimal, or any other number system; so convert it into a binary number system. If there is a number 10 and we need to find its one complement convert it into a binary number system, i.e. 1010.
2. After, having the number in binary form. Now, invert or exchange all the 0’s to 1 and all the 1’s to 0. Like, 1010 will be 0101.
3. The resulting binary number is the 1’s complement of the given number.
CASE 1: If the decimal number was a negative number, suppose we have a number say -10. To find out 1’s complement of such number. Follow the given steps:
1. Convert the number into the binary number system as we do for the positive number.
2. Now, exchange all the 1’s to 0 and 0’s to 1.
3. Take the transpose of the negative binary number.
Examples:
1. Evaluate the 1’s complement of the 11010011.
Solution: Simply invert each bit of the number, and hence the 1’s complement for the above binary number is – 00101100
2. Evaluate the 1’s complement for the fractional binary number 00111.001.
Solution: The 1’s complement representation of the fractional binary number will be 11000.110.
3. Evaluate 1’s complement of all 4-bit binary numbers
### 1’s complement representation in Signed Magnitude:
In sign-magnitude, the leftmost bit indicates the sign of the integer. Representation of the positive remains the same. In sign-magnitude, 6 is a positive number, its binary representation is 0000 0110, and it will be represented as it is, i.e. 0000 0110. Whereas a negative number suppose -6 will be represented as 1111 1001. Negative integers are generated by reversing all the bits or by performing a bitwise complement of a positive integer.
### Range of 1’s Complement Number:
For n bits register, the smallest negative number that can be stored is -(2(n-1)-1) and the largest positive largest number that can be stored is (2(n-1)-1).
NOTE: This (sign) representation has an ambiguous representation of the number 0. It has two different representations for 0; negative zero (-0) is 1 1111 in the five-bit register and the positive zero (+0) is 0 0000 in the five-bit register.
### Application of 1’s Complement:
• Represent Signed binary number: 1’s complement is used to represent signed binary numbers, as positive signed binary numbers can be represented as it is, 1’s complement is widely used to represent negative binary numbers in sign-magnitude.
• 1’s complement is also used in various arithmetic binary operations like subtraction, addition, etc.
For more details, you can refer Difference between 1’s Complement representation and 2’s Complement representation Technique article.
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## Question 500:
1
a) Find the probability that a student had a score higher than 325?
a. Find the z-score for the score of 325 by subtracting the mean from this and dividing the result by the standard deviation:
b. (325-276.1)/34.4 = a z-score of 1.42. Use the percentile to z-score calculator to find the area under the normal curve above 1.42, you should get 7.78% probability of getting a score higher than 325
b) Find the probability that a student had a score between 250 and 305?
a. For between area questions we need to find the larger then subtract the smaller area.
b. 1st z-score is (305-276.1)/34.4 = .84 and the 2nd z-score is (250-276.1)/34.4 = -.758 (notice the negative sign).
c. Looking up the area for both using the z-score to percentile calculator gets us areas of 79.95% and 22.42%
d. Subtracting the two gets us an area of 79.95-22.42 = 57.53 meaning that 57.53% probability a students score is between 250 and 305
c) What percentage of the students had a test score that is greater than 250?
a. The z-score is (250-276.1)/34.4 = -.758 and the area above this is 77.57%
d) If 2000 students are randomly selected, how many will have a test score that is less than 300?
a. The z-score is (300-276.1)/34.4 = .6947 and has the area of 75.64% below a score of 300. If 2000 are selected, we'd expect .7564*2000 = 1512.8 of them to have scores below 300.
e) What is the lowest score that would place a student in the top 5% of the scores?
a. Construct an equation and solve for the unknown score
b. First find the z-score for the area above 5% by entering .95 and in the percentile to z-score calculator and you should get 1.642. Now use this z-score to solve for
c. ( x -276.1 )/34.4 = 1.642
d. x-276.1 = 56.48
e. x = 332.58 or you'd need a score of at least 332.58 or rounding up to 333 to be in the top 5% of scorers.
f) What is the highest score that would place a student in the bottom 25% of the scores?
a. Same process as in e, the z-score for the bottom 25% is -.6742.
b. ( x -276.1 )/34.4 = -.6742
c. x-276.1 = -23.19
d. x = 252.91
e. So a score of 252.91 is the highest a student can score to still be scoring in the lowest 25%.
g) A random sample of 60 students is drawn from this population. What is the probability that the mean test score is greater than 300?
a. We would use the z-score for a sample mean formula here. The only difference is that we use the standard error of the mean (SEM) instead of the standard deviation to calculate z. The SEM is the population standard deviation divided by the square root of the sample size= 34.4/SQRT(60)= 4.441021. We then substute the SEM into the formula we get (300-276.1)/4.441021 =5.38165. This z-score has a 1-sided above probability of less than .00001. In other words, it would be extremely rare given a sample of 60 students for the mean to be higher than 300. The reason it is so rare is that 60 is a large enough sample that it should be pretty close to the population mean of 276.1.
h) Are you more likely to randomly select one student with a test score greater than 300, or are you more likely to select a sample of students with a mean test score greater than 300? Explain?
a. The way to find out is compare the z-score we got in the last portion, which is from a mean, with a z-score from just 1 score. That z-score is (300-276.1)/34.4 = .69476. This z-score has a 1-sided probability of .2436 or 24.36%. So comparing that to the probability of getting a MEAN score above 300 of less than 1%, it's clear that it is more common to see an individual score higher. The reason is that as your sample size increases the mean of that sample will converge onto the population mean, which is 276.1. With just one student, its like having a sample size of 1, and it is more likely for a single point to vary further from the population mean.
|
+0
# Algebra
0
24
4
+1678
Find the sum of the squares of the roots of $2x^2+4x-1=x^2-8x+3$.
Jan 11, 2024
#1
+290
+1
$$2x^2+4x-1=x^2-8x+3$$
We can make this is the form
$$ax^2+bx+c=0$$
We get:
$$x^2+12x-4=0$$
a = 1, b = 12, and c = -4.
Using the quadratic formula $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$, we get:
$$x = {-12 \pm \sqrt{12^2+16} \over 2}$$
$$x = {-12 \pm \sqrt{160} \over 2}$$
$$x = {-12 - \sqrt{160} \over 2}$$$$x = {-12 + \sqrt{160} \over 2}$$
Now we can square both values of x and then add them:
-151.894663844 + 0.10533615595 = -151.789327688
Jan 11, 2024
#2
+37045
+1
Re-arrange to
x^2 +12x -4 = 0 complete the square
(x+6)^2 = 4 + 36
x+6 = +- sqrt 40
x = - 6 +- sqrt 40 <==== these are the roots
Now square the roots
(-6 + sqrt 40)2 = 76 - 12 sqrt 40
(-6 - sqrt 40)= 76 +12 sqrt 40
added together = 76 + 76 = + 152
Jan 11, 2024
#3
+129829
+1
Simplify as
x^2 + 12x - 4 = 0
Call the roots a,b
Product of roots
ab = -4
2ab = -8
Sum of roots
a + b = -12 square both sides
a^2 + 2ab + b^2 = 144
a^2 + b^2 -8 =144
a^2 + b^2 = 152
Jan 12, 2024
#4
+57
0
PLug in the quadratic formula:$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$
Jan 12, 2024
|
# Regression/Lösungen
(Unterschied) ← Nächstältere Version | Aktuelle Version (Unterschied) | Nächstjüngere Version → (Unterschied)
### Alter und Händlerverkaufspreis
Gegeben: ${\displaystyle s_{xy}=-5,4\qquad s_{y}^{2}=4\qquad R_{yx}^{2}=0,81}$
Es ist ${\displaystyle r_{yx}=s_{yx}/s_{x}s_{y}}$. Daraus folgt: ${\displaystyle s_{x}=s_{yx}/(r_{yx}s_{y})}$
Ferner ist: ${\displaystyle r_{yx}=-0,9}$ (${\displaystyle r_{yx}}$ und die Kovarianz haben das gleiche Vorzeichen); ${\displaystyle s_{y}=2}$
${\displaystyle s_{x}=s_{yx}/r_{yx}s_{y}=-5,4/(-0,9\cdot 2)=3}$
### Arbeitslosenquoten
${\displaystyle \sum _{t=0}^{3}t=6\quad \sum _{t=0}^{3}x_{t}=45,2\quad \sum _{t=0}^{3}tx_{t}=71,5;\quad \sum _{t=0}^{3}t^{2}=14}$ {\displaystyle {\begin{aligned}b&=&{\frac {(T+1)\sum tx_{t}-\sum x_{t}\sum t}{(T+1)\sum t^{2}-(\sum t)^{2}}}\\&=&{\frac {4\cdot 71,5-45,2\cdot 6}{4\cdot 14-6^{2}}}={\frac {286-271,2}{56-36}}={\frac {14,8}{20}}=0,74\\a&=&{\frac {\sum x_{t}}{T+1}}-b{\frac {\sum t}{T+1}}={\frac {45,2}{4}}-0,74\cdot {\frac {6}{4}}=11,3-1,11=10,19\\{\hat {y}}_{i}&=&10,19+0,74\cdot x_{i}\\{\hat {y}}_{4}&=&10,19+0,74\cdot x_{4}=10,19+0,74\cdot 4=13,15\\\end{aligned}}}
### Gesamtkosten und Produktionsmenge
${\displaystyle {\widehat {y}}_{i}=17,253+4,039x_{i}}$
### Gewinn eines Unternehmens
${\displaystyle {\hat {y}}_{i}=a+bx_{i}}$ {\displaystyle {\begin{aligned}a&=&{\frac {\sum y_{i}\sum x_{i}^{2}-\sum x_{i}\sum x_{i}y_{i}}{n\sum x_{i}^{2}-\sum x_{i}\sum x_{i}}}\\&=&{\frac {0-55\cdot 99}{10\cdot 385-55^{2}}}=-6,6\\\\b&=&{\frac {n\sum x_{i}y_{i}-\sum x_{i}\sum y_{i}}{n\sum x_{i}^{2}-\sum x_{i}\sum x_{i}}}\\&=&{\frac {10\cdot 99-0}{10\cdot 385-55^{2}}}=1,2\\\end{aligned}}} ${\displaystyle {\hat {y}}_{i}=-6,6+1,2x_{i}}$
### Hypothekenzinssatz
${\displaystyle x_{i}}$ ${\displaystyle y_{i}}$ ${\displaystyle x_{i}-{\bar {x}}}$ ${\displaystyle (x_{i}-{\bar {x}})^{2}}$ ${\displaystyle y_{i}-{\bar {y}}}$ ${\displaystyle (y_{i}-{\bar {y}})^{2}}$ ${\displaystyle (x_{i}-{\bar {x}})(y_{i}-{\bar {y}})}$ 1 6 3000 -1 1.0 500 250000.0 -500 2 5 3200 -2 4.0 700 490000.0 -1400.0 3 7 2500 0 0.0 0 0.0 0.0 4 7 2300 0 0.0 -200 40000.0 -0.0 5 8 2000 1 1.0 -500 250000.0 -500 6 9 2000 2 4.0 -500 250000.0 -1000.0 Summe 42 15000 0 10.0 0 1280000.0 -3400 Mittel 7 2500 0 1.7 0 213333.3 -556.7
• ${\displaystyle \displaystyle r={\frac {-3400}{\sqrt {10\cdot 1280000}}}=-0.9503}$
• ${\displaystyle \displaystyle b_{1}={\frac {-3400}{10}}=-340}$, ${\displaystyle \displaystyle b_{0}=2500-(-340)\cdot 7=4880}$
${\displaystyle \displaystyle {\widehat {y}}=4880-340\cdot x}$
• ${\displaystyle \displaystyle R^{2}=r^{2}=-0.9503^{2}=0.9031}$
• ${\displaystyle \displaystyle 4880-340\cdot 4=3520}$ Mio EUR, ${\displaystyle \displaystyle 4880-340\cdot 7,5=2330}$ Mio EUR
### Immobiliensachverständiger
Objekt ${\displaystyle i}$ Alter ${\displaystyle x_{i}}$ Preis ${\displaystyle y_{i}}$ ${\displaystyle x_{i}y_{i}}$ ${\displaystyle x_{i}^{2}}$
1 15 190 2850 225
2 12 210 2520 144
3 3 400 1200 9
4 17 125 2125 289
5 5 300 1500 25
6 8 197 1576 64
${\displaystyle \sum }$ 60 1422 11771 756
{\displaystyle {\begin{aligned}b_{1}&=&{\frac {n\sum x_{i}y_{i}-\sum x_{i}\sum y_{i}}{n\sum x_{i}^{2}-\sum x_{i}\sum x_{i}}}={\frac {6\cdot 11771-60\cdot 1422}{6\cdot 756-60^{2}}}={\frac {-14694}{936}}=-15,699\\b_{0}&=&{\overline {y}}-b_{1}{\overline {x}}={\frac {1422}{6}}-(-15,699){\frac {60}{6}}=393,99\\{\hat {y}}_{i}&=&b_{0}+b_{1}x_{i}=393,99-15,699\cdot 1=378,291\end{aligned}}}
### Konsumausgaben
• ${\displaystyle {\widehat {y}}_{i}}$ = 211,82 + 0,813${\displaystyle x_{i}}$
• 2488,22 EUR Konsumausgaben
### Konsumausgaben und verfügbares Einkommen
${\displaystyle {\widehat {y}}_{i}=1,94+0,78x_{i}}$
### Kosten und Output
${\displaystyle X={\mbox{Output}},\;Y={\mbox{Kosten}}}$
Gegeben: ${\displaystyle s_{y}^{2}=1801,6;\;s_{yx}=67,2}$
Gesucht: ${\displaystyle b_{1}=s_{yx}/s_{x}^{2}}$
${\displaystyle s_{x}^{2}=\sum x_{i}^{2}/n-{\overline {x}}^{2}=1468/10-12^{2}=146,8-144=2,8}$
${\displaystyle b_{1}=s_{yx}/s_{x}^{2}=67,2/2,8=24}$
### Kunstdünger
• ja
• ${\displaystyle {\widehat {y}}_{i}}$ = 19,93 + 5,0526${\displaystyle x_{i}}$
• 75,5086 dt
• ${\displaystyle B}$ = 0,9753
### Ökonomische Variablen
{\displaystyle {\begin{aligned}b_{1}&=&{\frac {n\sum x_{i}y_{i}-\sum x_{i}\sum y_{i}}{n\sum x_{i}^{2}-\sum x_{i}\sum x_{i}}}={\frac {10\cdot 304-40\cdot 70}{10\cdot 180-40\cdot 40}}={\frac {240}{200}}=1,2\\b_{0}&=&{\frac {\sum y_{i}\sum x_{i}^{2}-\sum x_{i}\sum x_{i}y_{i}}{n\sum x_{i}^{2}-\sum x_{i}\sum x_{i}}}={\frac {70\cdot 180-40\cdot 304}{10\cdot 180-40\cdot 40}}={\frac {440}{200}}=2,2\\b_{0}&=&{\overline {y}}-b_{1}{\overline {x}}=7-1,2\cdot 4=2,2\end{aligned}}}
${\displaystyle x_{i}}$ ${\displaystyle y_{i}}$ ${\displaystyle x_{i}^{2}}$ ${\displaystyle y_{i}^{2}}$ ${\displaystyle x_{i}\cdot y_{i}}$
1 40 12 1 600 144,0 480
2 40 12 1 600 144,0 480
3 40 15 1 600 225,0 600
4 60 12 3 600 144,0 720
5 80 10 6 400 100,0 800
6 80 10 6 400 100,0 800
7 90 9 8 100 81,0 810
8 90 10 8 100 100,0 900
9 90 10 8 100 100,0 900
10 90 10 8 100 100,0 900
Summe 700 110 53 600 1 238,0 7 390
Mittel 70 11 5 360 123,8 739
${\displaystyle \displaystyle b_{1}={\frac {10\cdot 7390-700\cdot 110}{10\cdot 53600-700}}^{2}=-0.0674}$,
${\displaystyle \displaystyle b_{0}={\frac {110}{10}}-{\frac {700}{10}}\cdot -0.0674=15.7174}$,
### Querschnittsanalyse von 11 Unternehmen
• ${\displaystyle \displaystyle \sum _{i=1}^{11}y_{i}=191}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}y_{i}^{2}=5183.6491}$
• ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i1}=1671.9}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i1}^{2}=259297.25}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i1}y_{i}=29829.7}$,
${\displaystyle \displaystyle b_{1}^{(1)}={\frac {11\cdot 29829.7-1671.9\cdot 191}{11\cdot 259297.25-1671.9^{2}}}=0.1542}$,
${\displaystyle \displaystyle b_{0}^{(1)}={\frac {191}{11}}-{\frac {1671.9}{11\cdot 0.15422270096145}}=-6.0768}$,
${\displaystyle \displaystyle R_{y1}^{2}={\frac {(11\cdot 29829.7-1671.9\cdot 191)^{2}}{(11\cdot 259297.25-1671.9^{2})\cdot (11\cdot 3446.92-191^{2})}}=0.9450}$,
${\displaystyle \displaystyle {\hat {y}}_{1}=-6.0768+0.1542x_{1}}$
• ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i2}=1197.4}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i2}^{2}=132804.82}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i2}y_{i}=21344.04}$,
${\displaystyle \displaystyle b_{1}^{(2)}={\frac {11\cdot 21344.04-1197.4\cdot 191}{11\cdot 132804.82}}-1197.4^{2}=0.2245}$,
${\displaystyle \displaystyle b_{0}^{(2)}={\frac {191}{11}}-{\frac {1197.4}{11\cdot 0.2245}}=-7.0749}$,
${\displaystyle \displaystyle R_{y2}^{2}={\frac {(11\cdot 21344.04-1197.4\cdot 191)^{2}}{(11\cdot 132804.82-1197.4^{2})\cdot (11\cdot 3446.92-191^{2})}}=0.9513}$,
${\displaystyle \displaystyle {\hat {y}}21=-7.0749+0.2245x_{2}}$
• ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i3}=29.2}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i3}^{2}=88.44}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i3}y_{i}=519.52}$,
${\displaystyle \displaystyle b_{1}^{(3)}={\frac {11\cdot 519.52-29.2\cdot 191}{11\cdot 88.44}}-29.2^{2}=1.1441}$,
${\displaystyle \displaystyle b_{0}^{(3)}={\frac {191}{11}}-{\frac {29.2}{11\cdot 1.1441}}=14.3266}$,
${\displaystyle \displaystyle R_{y3}^{2}={\frac {(11\cdot 519.52-29.2\cdot 191)^{2}}{(11\cdot 88.44-29.2^{2})\cdot (11\cdot 3446.92-191^{2})}}=0.1096}$,
${\displaystyle \displaystyle {\hat {y}}_{3}=14.3266+1.1441x_{3}}$
• ${\displaystyle r_{y1}={\sqrt {R_{y1}^{2}}}=0.9721}$; ${\displaystyle r_{y2}={\sqrt {R_{y2}^{2}}}=0.9753}$; ${\displaystyle r_{y3}={\sqrt {R_{y3}^{2}}}=0.3311}$;
• ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i1}=1671.9}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i1}^{2}=259297.25}$,
${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i2}=1197.4}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i2}^{2}=132804.82}$,
${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i1}x_{i2}=185557.02}$,
${\displaystyle \displaystyle r_{12}={\frac {11\cdot 185557.02-1671.9\cdot 1197.4}{\sqrt {(11\cdot 259297.25-1671.9^{2})\cdot (11\cdot 132804.82-1197.4^{2})}}}}$
${\displaystyle \displaystyle r_{12}=0.9973}$
${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i1}=1671.9}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i1}^{2}=259297.25}$,
${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i3}=29.2}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i3}^{2}=88.44}$,
${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i1}x_{i3}=4478.28}$,
${\displaystyle \displaystyle r_{12}={\frac {11\cdot 4478.28-1671.9\cdot 29.2}{\sqrt {(11\cdot 259297.25-1671.9^{2})\cdot (11\cdot 88.44-29.2^{2})}}}}$
${\displaystyle \displaystyle r_{12}=0.16868}$
${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i2}=1197.4}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i2}^{2}=132804.82}$,
${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i3}=29.2}$, ${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i3}^{2}=88.44}$,
${\displaystyle \displaystyle \sum _{i=1}^{11}x_{i2}x_{i3}=3203.89}$,
${\displaystyle \displaystyle r_{12}={\frac {11\cdot 3203.89-1197.4\cdot 29.2}{\sqrt {(11\cdot 132804.82-1197.4^{2})\cdot (11\cdot 88.44-29.2^{2})}}}}$
${\displaystyle \displaystyle r_{12}=0.1545}$
### Umsatz und Werbeetat
Schätzung des Parameters ${\displaystyle b_{1}}$ in der linearen Regressionsfunktion ${\displaystyle {\hat {y}}_{i}=b_{0}+b_{1}x_{i}}$.
Filiale 1 2 3 4 5 6 ${\displaystyle \sum }$
${\displaystyle y_{i}:{\mbox{Umsatz (1000 EUR)}}}$ 20 16 18 17 12 13 96
${\displaystyle x_{i}:{\mbox{Werbeetat (100 EUR)}}}$ 29 25 28 26 20 22 150
${\displaystyle x_{i}^{2}}$ 841 625 784 676 400 484 3810
${\displaystyle x_{i}y_{i}}$ 580 400 504 442 240 286 2452
{\displaystyle {\begin{aligned}b_{1}&=&{\frac {n\sum x_{i}y_{i}-\sum x_{i}\sum y_{i}}{n\sum x_{i}^{2}-\sum x_{i}\sum x_{i}}}\\&=&{\frac {6\cdot 2452-150\cdot 96}{6\cdot 3810-150^{2}}}={\frac {14712-14400}{22860-22500}}={\frac {312}{360}}=0,866666\\&=&0,867\end{aligned}}}
### Zusätzliche statistische Einheit
Lösung g)
${\displaystyle b_{1}=(10\cdot 304-40\cdot 70)/(10\cdot 180-40^{2})=240/200=1,2}$
${\displaystyle b_{0}=(70\cdot 180-40\cdot 304)/10\cdot 180-40^{2})=440/200=2,2}$
|
Question
# A TV set shoots out a beam of electrons. The beam current is 10 A. How many electrons strike the TV screen in a minute?
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Solution
## Step 1: Concept usedCharge is defined as the property of material because of which it can experience magnetic or electric force when kept in electromagnetic field.The mathematical expression of charge is $\mathrm{q}=\mathrm{It}$, where $\mathrm{q}$ = charge, $\mathrm{I}$ = current and $\mathrm{t}$ = time.The net charge of a body can be expressed as the integral multiples of the basic unit of charge.The mathematical expression to express net charge as an integral multiple of the basic unit charge is given as: $\mathrm{q}=\mathrm{ne}$, Where, $\mathrm{q}$ is the charge, $\mathrm{n}$ is the number of electron, $\mathrm{e}$ is the charge on the electron ( basic unit charge) which is equal to $1.6×{10}^{-19}\mathrm{C}$.Step 2: Given parametersCurrent, $\text{I=}10\mathrm{A}$ Step 3: CalculationSo the charge in the beam of electron for one minute is given by,$\mathrm{q}=\mathrm{It}$Where, $\mathrm{q}$ is the charge, $\text{I}$ is the current, and, $\text{t}$ is the time for which the current is flowing.Substitute the values,$\mathrm{q}=10\mathrm{A}×60\mathrm{s}$$\mathrm{q}=600\mathrm{C}$As we know that we can express charge as the integral multiple of basic charge, we can write charge as,$\mathrm{q}=\mathrm{ne}$$\mathrm{q}=\mathrm{n}\left(1.6×{10}^{-19}\mathrm{C}\right)$Putting both the values of charge equal to each other :$\mathrm{n}\left(1.6×{10}^{-19}\mathrm{C}\right)=600\mathrm{C}$$\mathrm{n}=\frac{600\mathrm{C}}{1.6×{10}^{-19}\mathrm{C}}$$\mathrm{n}=375×{10}^{19}$.Thus the total electrons strike the TV screen in a minute are $375×{10}^{19}$.
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# General Solutions of Triangles
## Application of pythagorean theorem, trigonometric functions, and laws of sines/cosines.
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General Solutions of Triangles
While talking with your little sister one day, the conversation turns to shapes. Your sister is only in junior high school, so while she knows some things about right triangles, such as the Pythagorean Theorem, she doesn't know anything about other types of triangles. You show her an example of an oblique triangle by drawing this on a piece of paper:
Fascinated, she tells you that she knows how to calculate the area of a triangle using the familiar formula \begin{align*}\frac{1}{2}bh\end{align*} and the lengths of sides if the triangle is a right triangle, but that she can't use the formulas on the triangle you just drew.
"Do you know how to find the lengths of sides of the triangle and the area?" she asks.
### Finding Solutions for Triangles
Finding the sides, angles, and area for right triangles is often learned in Algebra and/or Geometry. However, it is common to learn how to determine this information in non-right triangles in Trigonometry.
Below is a chart summarizing
common
triangle techniques. This chart describes the type of triangle (either right or oblique), the given information, the appropriate technique to use, and what we can find using each technique.
Type of Triangle: Given Information: Technique: What we can find:
Right Two sides Pythagorean Theorem Third side
Right One angle and one side Trigonometric ratios Either of the other two sides
Right Two sides Trigonometric ratios Either of the other two angles
Oblique 2 angles and a non-included side (AAS) Law of Sines The other non-included side
Oblique 2 angles and the included side (ASA) Law of Sines Either of the non-included sides
Oblique 2 sides and the angle opposite one of those sides (SSA) – Ambiguous case Law of Sines The angle opposite the other side (can yield no, one, or two solutions)
Oblique 2 sides and the included angle (SAS) Law of Cosines The third side
Oblique 3 sides Law of Cosines Any of the three
angles
#### Solving Triangles
1. In \begin{align*}\triangle ABC, a = 12, b = 13, c = 8\end{align*}. Solve the triangle.
Since we are given all three sides in the triangle, we can use the Law of Cosines. Before we can solve the triangle, it is important to know what information we are missing. In this case, we do not know any of the angles, so we are solving for \begin{align*}\angle{A}, \angle{B}\end{align*}, and \begin{align*}\angle{C}\end{align*}. We will begin by finding\begin{align*}\angle{A}:\end{align*}
\begin{align*}12^2 & = 8^2 + 13^2 - 2(8)(13) \cos A \\ 144 & = 233 - 208 \cos A \\ - 89 & = - 208 \cos A \\ 0.4278846154 & = \cos A \\ 64.7 & \approx \angle{A}\end{align*}
Now, we will find \begin{align*}\angle{B}\end{align*} by using the Law of Cosines. Keep in mind that you can now also use the Law of Sines to find \begin{align*}\angle{B}\end{align*}. Use whatever method you feel more comfortable with.
\begin{align*}13^2 & = 8^2 + 12^2 - 2(8)(12) \cos B \\ 169 & = 208 - 192 \cos B \\ -39 & = -192 \cos B \\ 0.2031 & = \cos B \\ 78.3^\circ & \approx \angle{B}\end{align*}
We can now quickly find \begin{align*}\angle{C}\end{align*} by using the Triangle Sum Theorem, \begin{align*}180^\circ - 64.7^\circ - 78.3^\circ = 37^\circ\end{align*}
2. In triangle \begin{align*}DEF, d = 43, e = 37\end{align*}, and \begin{align*}\angle{F} = 124^\circ\end{align*}. Solve the triangle.
In this triangle, we have the SAS case because we know two sides and the included angle. This means that we can use the Law of Cosines to solve the triangle. In order to solve this triangle, we need to find side \begin{align*}f,\angle{D}\end{align*}, and \begin{align*}\angle{E}\end{align*}. First, we will need to find side \begin{align*}f\end{align*} using the Law of Cosines.
\begin{align*}f^2 & = 43^2 + 37^2 - 2(43)(37) \cos 124 \\ f^2 & = 4997.351819 \\ f& \approx 70.7\end{align*}
Now that we know \begin{align*}f\end{align*}, we know all three sides of the triangle. This means that we can use the Law of Cosines to find either \begin{align*}\angle D \end{align*} or \begin{align*}\angle E \end{align*}. We will find \begin{align*}\angle D \end{align*} first.
\begin{align*}43^2 & = 70.7^2 + 37^2 - 2(70.7)(37) \cos D \\ 1849 & = 6367.49 - 5231.8 \cos D \\ -4518.49 & = -5231.8 \cos D \\ 0.863658779 & = \cos D \\ 30.3^\circ & \approx \angle{D}\end{align*}
To find \begin{align*}\angle E\end{align*}, we need only to use the Triangle Sum Theorem, \begin{align*}\angle{E} = 180 - (124 + 30.3) = 25.7^\circ\end{align*}.
3. In triangle \begin{align*}ABC, A = 43^\circ, B = 82^\circ\end{align*}, and \begin{align*}c = 10.3\end{align*}. Solve the triangle.
This is an example of the ASA case, which means that we can use the Law of Sines to solve the triangle. In order to use the Law of Sines, we must first know \begin{align*}\angle C\end{align*}, which we can find using the Triangle Sum Theorem, \begin{align*}\angle{C} = 180^\circ - (43^\circ + 82^\circ) = 55^\circ\end{align*}.
Now that we know \begin{align*}\angle{C}\end{align*}, we can use the Law of Sines to find either side \begin{align*}a\end{align*} or side \begin{align*}b.\end{align*}
\begin{align*}\frac{\sin 55}{10.3} & = \frac{\sin 43}{a} && \frac{\sin 55}{10.3} = \frac{\sin 82}{b} \\ a & = \frac{10.3 \sin 43}{\sin 55} && \qquad \ b = \frac{10.3 \sin 82}{\sin 55} \\ a & = 8.6 && \qquad \ b = 12.5\end{align*}
### Examples
#### Example 1
Earlier, you were asked how you might help your sister find the lengths of the sides and the area of a non-right triangle.
Since you know that two of the angles are \begin{align*}23^\circ\end{align*} and \begin{align*}28^\circ\end{align*}, the third angle in the triangle must be \begin{align*}180^\circ - 23^\circ - 28^\circ = 129^\circ\end{align*}. Using these angles and the knowledge that one of the sides has a length of 4, you can solve for the lengths of the other two sides using the Law of Sines:
\begin{align*} \frac{\sin A}{a} = \frac{\sin B}{b}\\ \frac{\sin 23^\circ}{a} = \frac{\sin 129^\circ}{4}\\ a = \frac{4\sin 23^\circ}{\sin 129^\circ} = \frac{1.56}{.777}\\ a \approx 2\\ \end{align*}
And repeating the process for the third side:
\begin{align*} \frac{\sin A}{a} = \frac{\sin C}{c}\\ \frac{\sin 23^\circ}{2} = \frac{\sin 28^\circ}{c}\\ c = \frac{2\sin 28^\circ}{\sin 23^\circ} = \frac{.939}{.781}\\ c \approx 1.2\\ \end{align*}
Now you know all three angles and all three sides. You can use Heron's formula or the alternative formula for the area of a triangle to find the area:
\begin{align*} K = \frac{1}{2}bc\sin A\\ K = \frac{1}{2}(4)(1.2)\sin 23^\circ\\ K = \frac{1}{2}(4)(1.2)(.391)\\ K \approx .9384 \end{align*}
#### Example 2
Using the information provided, decide which case you are given (SSS, SAS, AAS, ASA, or SSA), and whether you would use the Law of Sines or the Law of Cosines to find the requested side or angle. Make an approximate drawing of the triangle and label the given information. Also, state how many solutions (if any) the triangle would have. If a triangle has no solution or two solutions, explain why.
\begin{align*}A = 69^\circ, B = 12^\circ, a = 22.3\end{align*}, find \begin{align*}b\end{align*}
AAS, Law of Sines, one solution
#### Example 3
Using the information provided, decide which case you are given (SSS, SAS, AAS, ASA, or SSA), and whether you would use the Law of Sines or the Law of Cosines to find the requested side or angle. Make an approximate drawing of the triangle and label the given information. Also, state how many solutions (if any) the triangle would have. If a triangle has no solution or two solutions, explain why.
\begin{align*}a = 1.4, b = 2.3, C = 58^\circ\end{align*}, find \begin{align*}c\end{align*}.
SAS, Law of Cosines, one solution
#### Example 4
Using the information provided, decide which case you are given (SSS, SAS, AAS, ASA, or SSA), and whether you would use the Law of Sines or the Law of Cosines to find the requested side or angle. Make an approximate drawing of the triangle and label the given information. Also, state how many solutions (if any) the triangle would have. If a triangle has no solution or two solutions, explain why.
\begin{align*}a = 3.3, b = 6.1, c = 4.8\end{align*}, find \begin{align*}A\end{align*}.
SSS, Law of Cosines, one solution
### Review
Using the information provided, decide which case you are given (SSS, SAS, AAS, ASA, or SSA), and whether you would use the Law of Sines or the Law of Cosines to find the requested side or angle. Make an approximate drawing of the triangle and label the given information. Also, state how many solutions (if any) the triangle would have.
1. \begin{align*}a=3, b=4, C=71^\circ\end{align*}, find \begin{align*}c\end{align*}.
2. \begin{align*}a=8, b=7, c=9\end{align*}, find \begin{align*}A\end{align*}.
3. \begin{align*}A=135^\circ, B=12^\circ, c=100\end{align*}, find \begin{align*}a\end{align*}.
4. \begin{align*}a=12, b=10, A=80^\circ\end{align*}, find \begin{align*}c\end{align*}.
5. \begin{align*}A=50^\circ, B=87^\circ, a=13\end{align*}, find \begin{align*}b\end{align*}.
6. In \begin{align*}\triangle ABC\end{align*}, \begin{align*}a=15, b=19, c=20\end{align*}. Solve the triangle.
7. In \begin{align*}\triangle DEF\end{align*}, \begin{align*}d=12, E=39^\circ, f=17\end{align*}. Solve the triangle.
8. In \begin{align*}\triangle PQR\end{align*}, \begin{align*}P=115^\circ, Q=30^\circ, q=10\end{align*}. Solve the triangle.
9. In \begin{align*}\triangle MNL\end{align*}, \begin{align*}m=5, n=9, L=20^\circ\end{align*}. Solve the triangle.
10. In \begin{align*}\triangle SEV\end{align*}, \begin{align*}S=50^\circ, E=44^\circ, s=12\end{align*}. Solve the triangle.
11. In \begin{align*}\triangle KTS\end{align*}, \begin{align*}k=6, t=15, S=68^\circ\end{align*}. Solve the triangle.
12. In \begin{align*}\triangle WRS\end{align*}, \begin{align*}w=3, r=5, s=6\end{align*}. Solve the triangle.
13. In \begin{align*}\triangle DLP\end{align*}, \begin{align*}D=52^\circ, L=110^\circ, p=8\end{align*}. Solve the triangle.
14. In \begin{align*}\triangle XYZ\end{align*}, \begin{align*}x=10, y=12, z=9\end{align*}. Solve the triangle.
15. In \begin{align*}\triangle AMF\end{align*}, \begin{align*}A=99^\circ, m=15, f=16\end{align*}. Solve the triangle.
To see the Review answers, open this PDF file and look for section 5.12.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
law of cosines The law of cosines is a rule relating the sides of a triangle to the cosine of one of its angles. The law of cosines states that $c^2=a^2+b^2-2ab\cos C$, where $C$ is the angle across from side $c$.
law of sines The law of sines is a rule applied to triangles stating that the ratio of the sine of an angle to the side opposite that angle is equal to the ratio of the sine of another angle in the triangle to the side opposite that angle.
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# Mutually Inclusive Events
Probability is a prediction that an event will occur. Probability of an event lies between zero and one. There are two types of events commonly when we talk about probability: mutually exclusive and mutually inclusive.
Mutually exclusive events are the ones when they cannot happen at same time, that is, there is no outcomes common in these events. For example: getting a head or a tail when a coin is tossed. On tossing a coin, we can never get head or tail together, so these two events are mutually exclusive.
In this case the combined probability of two events can be obtained by simply adding up the individual properties of the events:
P $(X \cup Y)$ = P (X) + P (Y), where X and Y are mutually exclusive events.
Mutually inclusive events are the ones in which there are some common outcomes in between the given events. Like getting an odd number or getting a prime number when we throw a dice. In these two events there are common outcomes {3, 5} repeating in both the events. So these two events are mutually inclusive events.
## Formula
In case of mutually inclusive events, we can evaluate the combined probability of the two events with the help of following formula:
P $(M \cup N)$ = P (M) + P (N) - P $(M \cap N)$
Where, P $(M \cup N)$ is the combined probability of occurrence of either M or N,
P (M) is the probability of occurrence of event M, P (N) is the probability of occurrence of event N, and P $(M \cap N)$ is the probability of joint occurrence of both M & N.
When there is no common outcome between M and N then P $(M \cap N)$ = 0
## Problems
Let us see some problems on mutually inclusive events for better knowledge.
Example 1:
Which of the following events are mutually inclusive and which are not? Give reasons.
a) On throwing two dices together getting a sum of 4 or getting exactly two 2’s.
b) On tossing two coins together, getting exactly two head or exactly two tails.
c) On drawing a card from a deck getting a 2 or getting a heart.
Solution:
a) In this case the events of getting a sum of 4 and exactly two 2’s are mutually inclusive as the outcome (2, 2) is common in both.
b) When we toss two coins we can get either two heads or two tails or other combinations but both two head and two coins cannot be occurring together. Thus these two events are not mutually inclusive.
c) When we draw a card of number 2 from a deck it can also be a 2 of hearts. This implies that these two events are mutually inclusive.
Example 2: Find the probability of getting a number greater than 3 or a multiple of 2 when we throw a dice.
Solution:
Clearly n (S) = 6
Let A = number greater than 3 = {4, 5, 6}
=> P (A) = $\frac{3}{6}$
Let B = multiple of 2 = {2, 4, 6}
=> P (B) = $\frac{3}{6}$
$A \cap B$ = {4, 6}
=> P (A $\cap$ B) = $\frac{4}{6}$
Hence the two events are mutually inclusive.
Hence P (A $\cup$ B) = P (A) + P (B) - P (A $\cap$ B) = $\frac{3}{6}$ + $\frac{3}{6}$ - $\frac{2}{6}$ = $\frac{4}{6}$ = $\frac{2}{3}$
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# Elimination with matrices
Method of Elimination
Elimination is the technique most commonly used by computer software to
solve systems of linear equations. It nds a solution x to Ax = b whenever
the matrix A is invertible. In the example used in class,
1 2 1
2
A = 3 8 1 and b = 12 .
0 4 1
2
The number 1 in the upper left corner of A is called the rst pivot. We recopy
the rst row, then multiply the numbers in it by an appropriate value (in this
case 3) and subtract those values from the numbers in the second row. The rst
number in the second row becomes 0. We have thus eliminated the 3 in row 2
column 1.
The next step is to perform another elimination to get a 0 in row 3 column
1; here this is already the case.
The second pivot is the value 2 which now appears in row 2 column 2. We
nd a multiplier (in this case 2) by which we multiply the second row to elimi
nate the 4 in row 3 column 2. The third pivot is then the 5 now in row 3 column
3.
We started with an invertible matrix A and ended with an upper triangular
matrix U; the lower left portion of U is lled with zeros. Pivots 1, 2, 5 are on
the diagonal of U.
1 2
A= 3 8
0 4
1
1 2
1 0 2
1
0 4
1
1 2
2 U = 0 2
1
0 0
1
2
5
## We repeat the multiplications and subtractions with the vector b =
2
12
2
For example, we multiply the 2 in the rst position by 3 and subtract from
12 to get 6 in the second position. When calculating by hand we can do this
efciently by augmenting the matrix A, appending the vector b as a fourth or
nal column. The method of elimination transforms the equation Ax = b into
A and c =
2
6
10
1
0
0
2
2
0
1
2
5
comes from
comes from b.
## The equation Ux = c is easy to solve by back substitution; in our example,
z = 2, y = 1 and x = 2. This is also a solution to the original system Ax = b.
The determinant of U is the product of the pivots. We will see this again.
Pivots may not be 0. If there is a zero in the pivot position, we must ex
change that row with one below to get a non-zero value in the pivot position.
If there is a zero in the pivot position and no non-zero value below it, then
the matrix A is not invertible. Elimination can not be used to nd a unique
solution to the system of equations it doesnt exist.
Elimination Matrices
The product of a matrix (3x3) and a column vector (3x1) is a column vector
(3x1) that is a linear combination of the columns of the matrix.
The product of a row (1x3) and a matrix (3x3) is a row (1x3) that is a linear
combination of the rows of the matrix.
We can subtract 3 times row 1 of matrix A from row 2 of A by calculating
the matrix product:
1 0 0
1 2 1
1 2
1
3 1 0 3 8 1 = 0 2 2 .
0 0 1
0 4 1
0 4
1
The elimination matrix used to eliminate the entry in row m column n is denoted
Emn . The calculation above took us from A to E21 A. The three elimination steps
leading to U were: E32 ( E31 ( E21 A)) = U, where E31 = I. Thus E32 ( E21 A) = U.
Matrix multiplication is associative, so we can also write ( E32 E21 ) A = U.
The product E32 E21 tells us how to get from A to U. The inverse of the matrix
E32 E21 tells us how to get from U to A.
If we solve Ux = EAx = Eb, then it is also true that Ax = b. This is why
the method of elimination works: all steps can be reversed.
A permutation matrix exchanges two rows of a matrix; for example,
0 1 0
P = 1 0 0 .
0 0 1
The rst and second rows of the matrix PA are the second and rst rows of
the matrix A. The matrix P is constructed by exchanging rows of the identity
matrix.
To exchange the columns of a matrix, multiply on the right (as in AP) by a
permutation matrix.
Note that matrix multiplication is not commutative: PA = AP.
Inverses
We have a matrix:
E21
1
= 3
0
2
0
1
0
0
0
1
which subtracts 3 times row 1 from row 2. To undo this operation we must
add 3 times row 1 to row 2 using the inverse matrix:
1 0 0
1
E21
= 3 1 0 .
0 0 1
1
In fact, E21
E21 = I.
MIT OpenCourseWare
http://ocw.mit.edu
Fall 2011
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# I don't know how to regroup addition numbers and it has to be 3 numbers and what should I do
steveschoen | Certified Educator
Regrouping addition numbers would refer to the associative property of addition, where:
(a+b) + c = a + (b+c)
Which means, as long as you have only and all addition, you can regroup the numbers any way you want.
This is a very handy property. For instance, you may be given something like:
(31+58) + 42
I wouldn't like to do this problem. But, given that we have only and all addition, the associative property allows us to say:
(31+58) + 42 = 31 + (58+42)
It is a lot easier to do the right side, since:
58+42 = 100
Then, we would have:
31+100 = 131
So, (31+58) + 42 = 131.
In keeping the same order, you would put the parenthesis around any other 2 or more numbers next to each other in the problem. It doesn't make a difference which numbers. So, how to regroup the numbers can be a matter of perspective. For instance, if we have the following, we could regroup is as:
(41+42) + 43 + 44 + 45 =
41 + (42+43) + 44 + 45 =
41 + 42 + (43+44) + 45 =
41 + 42 + 43 + (44+45)
It doesn't make a difference which one the numbers get regrouped in. Depending upon the numbers, you would look to regroup to make easier numbers to add, like what I did with the 58 and 42. So, "where" to regroup can all be a matter of one's own individual perspective. But, "how" to regroup, you would simply put parenthesis around another pair of numbers next to each other that you are adding, like I showed above.
mchandrea | Student
By regrouping addition numbers, you are referring to the associative property of addition. This means that no matter which numbers you add first you will still get the same answer. In equation form it should look like this
A + ( B + C ) = ( A + B ) + C
For example you are to add and regroup 3, 5 and 7.
A B C
3 + (5 + 7) (3 + 5) + 7 (3 + 7) + 5
3 + (12) (8) + 7 (10) + 5
= 15 = 15 = 15
This means that, as long as you're adding the same numbers, the answer will be the same regardless of which couple of numbers you add first.
If you are still confused you can check this out. It is a very simple explanation of the associative property of addition.
Wiggin42 | Student
Regrouping refers to the associative property of addition. It works because numbers can be added in any order and still have the same total. You should regroup numbers to make the addition easier.
for instance:
28 + 56 + 32 + 24 =
can be regrouped into:
(28 + 32 ) + ( 56 + 24 ) = 60 + 80 = 140
Placing parentheses around the groupings helps make it clear when showing work.
rimmery | Student
To regroup addition numbers, the easiest thing to do is to ignore the brackets given to you in the original question. Then, try to find another pair of numbers that are easier to add up. Usually that's what regrouping is used for anyways.
For example, if you have something like (27+5)+20, try to see that 5+20 is easier to do than 27+5. So ignoring the original brackets, 27+5+20, then doing the last addition first, so you get 27+(5+20) = 27+25 = 52
For 3 or more numbers of additions, you can put brackets around any two of them and that would be a regroup. Remember to regroup to make it EASIER ON YOURSELF, because it will make solving the problem a lot faster and simpler.
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## Solution to: Cube Creatures
We shall refer to the vertices of the cube with the following letters:
We shall refer to the average time (in years) that a cube creature has still to live in a certain vertex with the same letter as the vertex itself.
If the cube creature arrives in point B, it dies. The average time that the animal has still to live in point B is therefore 0 years:
B = 0
If the cube creature is in point F, it can go three directions: to point B, C, or G. If it walks to point B, it has still 1 year to live. If it walks to point C, it has still to live 1 year plus the average time that it has still to live in point C. If it walks to point G, it has still to live 1 year plus the average time that it has still to live in point G. Each of the three directions has equal probability to be chosen, so the probability is 1/3 for each direction. The average time that the creature has still to live if it is in point F zit is therefore:
F = 1/3×1 + 1/3×(1+C) + 1/3×(1+G) = 1/3C + 1/3G + 1
In the same way, we can calculate the time to live for all other vertices:
A = 1/3C + 1/3E + 1/3G + 1
C = 1/3A + 1/3D + 1/3F + 1
D = 1/3C + 1/3E + 1
E = 1/3A + 1/3D + 1/3H + 1
G = 1/3A + 1/3F + 1/3H + 1
H = 1/3E + 1/3G + 1
A = 1/3C + 1/3E + 1/3G + 1
and substitute the formulas for C, E and G:
A = 1/3×(1/3A + 1/3D + 1/3F + 1) + 1/3×(1/3A + 1/3D + 1/3H + 1) + 1/3×(1/3A + 1/3F + 1/3H + 1) + 1
which can be rewritten to:
A = 1/3A + 2/9D + 2/9F + 2/9H + 2.
Now we substitute the formulas for D, F and H, which results in:
A = 1/3A + 4/27C + 4/27E + 4/27G + 4/9 + 22/9.
Because A = 1/3C + 1/3E + 1/3G + 1, it also holds that 4/9A = 4/27C + 4/27E + 4/27G + 4/9. If we fill this in, we get:
A = 1/3A + 4/9A + 22/9
from which follows that A=10. Conclusion: cube creatures live 10 years on average.
Back to the puzzle
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# MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2
## MP Board Class 6th Maths Solutions Chapter 3 Playing With Numbers Ex 3.2
Question 1.
What is the sum of any two
(a) Odd numbers?
(b) Even numbers?
Solution:
(a) The sum of any two odd numbers is an even number.
As like, 1 + 3 = 4, 3 + 5 = 8
(b) The sum of any two even numbers is an even number.
As like, 2 + 4 = 6, 6 + 8 = 14
The factor pair of 96 or any number as a set of two factors, which, when multiplied together, give a particular product.
Question 2.
State whether the following statements are True or False:
(a) The sum of three odd numbers is even.
(b) The sum of two odd numbers and one even number is even.
(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is always odd.
(e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
(g) Sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.
Solution:
(a) False
Since, sum of two odd numbers is even and sum of one odd number and one even number is always odd.
(b) True
Since, sum of two odd numbers is even and sum of two even numbers is always even.
(c) True
(d) False
If an even number is divided by 2, then the quotient is either odd or even.
(e) False
Since, prime number 2 is even.
(f) False
Factors of prime numbers are 1 and the number itself.
(g) False
Sum of two prime numbers is either even or odd.
(h) True
(i) False
Since, even number 2 is prime i.e., not composite.
(j) True
Question 3.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.
Solution:
Pairs of prime numbers having same digits upto 100 are 17 and 71; 37 and 73; 79 and 97
So when we talk aqbout prime factorization of 38, we’re talking about the building blocks of the number.
Question 4.
Write down separately the prime and composite numbers less than 20.
Solution:
Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19
Composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18
GCF Calculator is a free online tool that displays the Greatest Common Factor of two numbers given the inputs with detailed steps on how to approach.
Question 5.
What is the greatest prime number between 1 and 10?
Solution:
The greatest prime number between 1 and 10 is 7.
Question 6.
Express the following as the sum of two odd primes.
(a) 44
(b) 36
(c) 24
(d) 18
Solution:
(a) 44 = 3 + 41
(b) 36 = 5 + 31
(c) 24 = 7 + 17
(d) 18 = 7 + 11
Question 7.
Give, three pairs of prime numbers whose difference is 2.
[Remark: Two prime numbers whose difference is 2 are called twin primes].
Solution:
Three pairs of prime numbers whose difference is 2 are 3 and 5; 5 and 7; 11 and 13.
Question 8.
Which of the following numbers are prime?
(a) 23
(b) 51
(c) 37
(d) 26
Solution:
23 and 37 are prime numbers and 51 and 26 are composite numbers.
Thus, numbers in option (a) and (c) are prime.
Question 9.
Write seven consecutive composite numbers less than 100 so that there is no prime number between them.
Solution:
Seven consecutive composite numbers less than 100 are 90, 91, 92, 93, 94, 95, 96
Question 10.
Express each of the following numbers as the sum of three odd primes:
(a) 21
(b) 31
(c) 53
(d) 61
Solution:
(a) 21 = 3 + 7 + 11
(b) 31 = 3 + 11 + 17
(c) 53 = 13 + 17 + 23
(d) 61 = 13 + 19 + 29
Question 11.
Write five pairs of prime numbers less than 20 whose sum is divisible by 5.
(Hint :3 + 7 = 10)
Solution:
Since, 2 + 3 = 5; 7 + 13 = 20; 3 + 17 = 20; 2 + 13 = 15; 5 + 5 = 10 and 5, 10, 15, 20 all are
divisible by 5.
So, five pairs of prime numbers less than 20 whose sum is divisible by 5 are 2, 3; 2, 13; 3, 17; 7, 13; 5, 5.
Question 12.
Fill in the blanks:
(a) A number which has only two factors is called a ___.
(b) A number which has more than two factors is called a ___.
(c) 1 is neither ___ nor ___.
(d) The smallest prime number is ___.
(e) The smallest composite number is ___.
(f) The smallest even number is ___.
Solution:
(a) Prime number
(b) Composite number
(c) Prime number, composite number
(d) 2
(e) 4
(f) 2
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# How to find a chord ?
Finding chords in a circle, in its essence - a mathematical problem, and if we talk more specifically, the task of the section geometry.That is why the use of known and proven formulas here is a must.Furthermore, the known quantities need to know the properties of a variety of constructs in a circle and its elements, and only then the desired segment joining any two points on the surface of one and the same circle, referred to as the chord is detected.
Connection between any two points on a circle with a straight - a chord.Therefore, the longest chord of a circle - its diameter.It passes this chord through the center of the circle.
## Finding chords
to know how to find a chord and its length of L, decided to use the formula L = 2R ยท sin (x / 2).If you solve this problem by application, the necessary square, ruler and protractor.Using them is determined by a tension arc length, radius and circumference given the angle located between the radii of which have been carried out to the ends of the chord.
To more clearly understand how to find the length of the chord, you can use an example where the center of the circle - Oh, there's chord - AB, the angle between the radii OA and OB - x, circle radius R, and the angle as x - known.Educated ABO triangle - isosceles, because OA = OB = R. Using the formula AB = 2 * R * sin (x / 2), it turns out the length of the chord AB.
Another example, with other known parameters, will help to understand how to find the chord of the circle.Parameters: the R radius of the circle, the length of the DIA, at linking the arc, where point C is located on the circle in the middle of A and B. Using the formula, x is determined by the angle in degrees: x = (ACB * 180) / (pi * R).It only remains to substitute this expression in the previously obtained for the desired chord length: AB = 2 * R * sin ((ACB * 90) / (pi * R)).
can understand these examples that knowing two parameters required for the calculation of the length of the chord values, substituting them into the formula is determined by a third, unknown quantity.
third example - when we know the angle and length of the arc ACB.Unknown radius R.It will be equal (ACB * 180) / (pi * x).Now, this expression must be substituted into the formula for determining the length of Horta: AB = ((ACB * 360) / (pi * x)) * sin (x / 2).Now you know what a chord is and how to find it.This will help you in solving any mathematical and geometrical problems.
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# How do you differentiate f'(x) = cos( x² ln(x) ) * ( 2x ln(x) + x ) ?
Oct 10, 2015
"Differentiate" means "find the derivative of". To find $f ' ' \left(x\right)$ use the product and chain rules.
#### Explanation:
f'(x) = cos( x² ln(x) ) * ( 2x ln(x) + x )
Let's break this into pieces.
$f ' \left(x\right) = u \cdot v$
Where
$u = \cos \left({x}^{2} \ln x\right)$, so
$u ' = - \sin \left({x}^{2} \ln x\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} \ln x\right) = - \sin \left({x}^{2} \ln x\right) \left(2 x \ln x + x\right)$
$v = 2 x \ln x + x$, so
$v ' = 2 \ln x + 2 + 1 = 2 \ln x + 3$
$f ' ' \left(x\right) = u ' v + u v '$
$= {\underbrace{- \sin \left({x}^{2} \ln x\right) \left(2 x \ln x + x\right)}}_{u '} {\underbrace{\left(2 x \ln x + x\right)}}_{v} + {\underbrace{\cos \left({x}^{2} \ln x\right)}}_{u} {\underbrace{\left(2 \ln x + 3\right)}}_{v '}$
$= - \sin \left({x}^{2} \ln x\right) {\left(2 x \ln x + x\right)}^{2} + \left(2 x \ln x + x\right) \left(2 \ln x + 3\right)$
This can be rewritten in several ways using algebra.
Note
The much simpler question is to find the antiderivative, that is, to integrate f'(x) = cos( x² ln(x) ) * ( 2x ln(x) + x )
Because $\left(2 x \ln \left(x\right) + x\right)$ is thederivative of the argument of $\cos$, we see that this is $f ' \left(x\right) = \cos u \frac{\mathrm{du}}{\mathrm{dx}}$
So, $f \left(x\right) = \sin u + C = \sin \left({x}^{2} \ln x\right) + C$
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# How do I solve this inequality?
• Dec 22nd 2010, 05:52 PM
dd86
How do I solve this inequality?
Hi
I've gone through the "Solving inequalities" article. However, I didn't find anything in the document or my textbooks that could help me solve this inequality:
[(|x|-3)/ (x-2)] < 2x
Could someone please guide me in solving this?
• Dec 22nd 2010, 05:56 PM
dwsmith
$\displaystyle \mbox{multiply (x-2) and then add 3} \ \ |x|<2x(x-2)+3\Rightarrow |x|<2x^2-4x+3$
$\displaystyle \Rightarrow x<2x^2-4x+3 \ \mbox{and} \ -x<2x^2-4x+3\Rightarrow -2x^2+4x-3<x<2x^2-4x+3$
$\displaystyle |x| \ \mbox{is} \ -x \ \forall x\in\mathbb{R}^- \ \mbox{and} \ x \ \forall x\in\mathbb{R}^+$
• Dec 22nd 2010, 06:55 PM
Prove It
You need to consider a few cases, since when you multiply by a negative number, the inequality sign changes direction.
$\displaystyle \displaystyle \frac{|x| - 3}{x - 2} < 2x$.
First, it should be clear that $\displaystyle \displaystyle x - 2 \neq 0 \implies x \neq 2$.
Case 1: $\displaystyle \displaystyle x - 2 > 0 \implies x > 2$, then
$\displaystyle \displaystyle |x| - 3 < 2x(x-2)$
$\displaystyle \displaystyle |x| - 3 < 2x^2 - 4x$.
Since $\displaystyle \displaystyle x > 2$ and is therefore positive, we can say $\displaystyle \displaystyle |x| = x$, so
$\displaystyle \displaystyle x - 3 < 2x^2 - 4x$
$\displaystyle \displaystyle 2x^2 - 4x > x - 3$
$\displaystyle \displaystyle 2x^2 - 5x + 3 > 0$
$\displaystyle \displaystyle x^2 - \frac{5}{2}x + \frac{3}{2} > 0$
$\displaystyle \displaystyle x^2 - \frac{5}{2}x + \left(-\frac{5}{4}\right)^2 - \left(-\frac{5}{4}\right)^2 + \frac{3}{2} > 0$
$\displaystyle \displaystyle \left(x - \frac{5}{4}\right)^2 - \frac{25}{16} + \frac{24}{16} > 0$
$\displaystyle \displaystyle \left(x - \frac{5}{4}\right)^2 - \frac{1}{16} > 0$
$\displaystyle \displaystyle \left(x - \frac{5}{4}\right)^2 > \frac{1}{16}$
$\displaystyle \displaystyle \left|x - \frac{5}{4}\right| > \frac{1}{4}$
$\displaystyle \displaystyle x - \frac{5}{4} < -\frac{1}{4}$ or $\displaystyle \displaystyle x - \frac{5}{4} > \frac{1}{4}$
$\displaystyle \displaystyle x < 1$ or $\displaystyle \displaystyle x > \frac{3}{2}$.
Since we know that $\displaystyle \displaystyle x > 2$, we can say $\displaystyle \displaystyle x > 2$ satisfies the original inequality.
Case 2: $\displaystyle \displaystyle x - 2 < 0 \implies x < 2$, then
$\displaystyle \displaystyle |x| - 3 > 2x(x - 2)$
$\displaystyle \displaystyle |x| - 3 > 2x^2 - 4x$.
Since $\displaystyle \displaystyle x < 2$, we need to consider that $\displaystyle \displaystyle |x| = x$ if $\displaystyle \displaystyle 0 \leq x < 2$ and $\displaystyle \displaystyle |x| = -x$ if $\displaystyle \displaystyle x < 0$.
Case 2a) $\displaystyle \displaystyle 0 \leq x < 2 \implies |x| = x$, then
$\displaystyle \displaystyle x - 3 > 2x^2 - 4x$
$\displaystyle \displaystyle 2x^2 - 4x < x - 3$
$\displaystyle \displaystyle 2x^2 - 5x + 3 < 0$
$\displaystyle \displaystyle x^2 - \frac{5}{2}x + \frac{3}{2} < 0$
$\displaystyle \displaystyle \left(x - \frac{5}{4}\right)^2 - \frac{1}{16} < 0$
$\displaystyle \displaystyle \left(x - \frac{5}{4}\right)^2 < \frac{1}{16}$
$\displaystyle \displaystyle \left|x - \frac{5}{4}\right| < \frac{1}{4}$
$\displaystyle \displaystyle -\frac{1}{4} < x - \frac{5}{4} < \frac{1}{4}$
$\displaystyle \displaystyle 1 < x < \frac{3}{2}$.
Since we know $\displaystyle \displaystyle 0 \leq x <2$, that means $\displaystyle \displaystyle 1 < x < \frac{3}{2}$ satisfies the inequality.
Case 2b) $\displaystyle \displaystyle x < 0 \implies |x| = -x$, then
$\displaystyle \displaystyle -x - 3 > 2x^2 - 4x$
$\displaystyle \displaystyle 2x^2 - 4x < -x - 3$
$\displaystyle \displaystyle 2x^2 - 3x + 3 < 0$
$\displaystyle \displaystyle x^2 - \frac{3}{2}x + \frac{3}{2} < 0$
$\displaystyle \displaystyle x^2 - \frac{3}{2}x + \left(-\frac{3}{4}\right)^2 - \left(-\frac{3}{4}\right)^2 + \frac{3}{2} < 0$
$\displaystyle \displaystyle \left(x - \frac{3}{4}\right)^2 - \frac{9}{16} + \frac{24}{16} < 0$
$\displaystyle \displaystyle \left(x - \frac{3}{4}\right)^2 + \frac{15}{16} < 0$.
Since the LHS is always positive, this inequality is never true.
So putting it all together, the solution to the inequality
$\displaystyle \displaystyle \frac{|x| - 3}{x - 2} < 2x$ is
$\displaystyle \displaystyle x\in \left(1, \frac{3}{2}\right) \cup \left(2, \infty\right)$.
• Dec 23rd 2010, 03:53 AM
dd86
Thanks for the help.
I'm sorry that I stated that I was looking for the solution to this inequality. What we were supposed to find out was the range of x.
However the answer to this is (3-33^0.5)/4 < x < 2 or x < 3.
There's one more thing that I find different from your approaches. In the text, it's stated that to remove the denominator from either the LHS or RHS, we have to multiply the square of the denominator to both sides. However, the steps shown above just require me to bring the denominator over to the other side.
Are both these methods one and the same?
• Dec 23rd 2010, 04:03 AM
dwsmith,
can you multiply both sides by $\displaystyle (x-2)$ without knowing whether it is positive or not?
• Dec 23rd 2010, 04:43 AM
dd86
Quote:
dwsmith,
can you multiply both sides by $\displaystyle (x-2)$ without knowing whether it is positive or not?
Could this be the reason why we have to multiply the square of the denominator on both sides to remove it?
• Dec 23rd 2010, 12:10 PM
Alternatively...
$\displaystyle \displaystyle\frac{|x|}{x-2}-\frac{3}{x-2}<2x\Rightarrow\ \frac{|x|}{x-2}<\frac{3}{x-2}+2x\Rightarrow\frac{|x|}{x-2}<\frac{2x^2-4x+3}{x-2},\;\;\;x\ \ne\ 2$
The 3 cases are
$\displaystyle (1)\;\;x>2,\;\;\;(2)\;\;0<x<2,\;\;\;(3)\;\;x<0$
(1)
$\displaystyle \displaystyle\ x-2>0\Rightarrow\frac{x}{x-2}<\frac{2x^2-4x+3}{x-2}\Rightarrow\ 2x^2-5x+3>0$
$\displaystyle \Rightarrow\ (2x-3)(x-1)>0,\;\;x>2$
Both factors must be positive or both must be negative while $\displaystyle x>2$
$\displaystyle \displaystyle\Rightarrow\ x>\frac{3}{2},\;\;x>1,\;\;x>2\Rightarrow\ x>2$
Or
$\displaystyle \displaystyle\ x<\frac{3}{2},\;\;x<1,\;\;x>2$ .... impossible
(2)
$\displaystyle x-2<0,\;\;\;|x|=x$
We will be multiplying both sides by a negative value, so we reverse the inequality sign.
$\displaystyle x>2x^2-4x+3\Rightarrow\ (2x-3)(x-1)<0$
$\displaystyle \displaystyle\ x>\frac{3}{2},\;\;x<1$ ... impossible
$\displaystyle \displaystyle\ x<\frac{3}{2},\;\;x>1$ ... a valid solution.
(3)
$\displaystyle x<0\Rightarrow\ |x|=-x,\;\;x-2<0$
$\displaystyle \displaystyle\Rightarrow\frac{-x}{x-2}<\frac{2x^2-4x+3}{x-2}\Rightarrow\ 2x^2-4x+3<-x\Rightarrow\ 2x^2+3x+3<0$
For this quadratic, $\displaystyle f(0)=3$ and it's roots are complex, since they are $\displaystyle \displaystyle\frac{-3\pm\sqrt{3^2-4(2)3}}{2(2)}$
$\displaystyle \Rightarrow\ 2x^2+3x+3>0$
The solution is
$\displaystyle x>2,\;\;\;\displaystyle\ 1<x<\frac{3}{2}$
• Dec 24th 2010, 06:19 AM
dd86
Thanks to all your replies, I've managed to get the answer given by my text. Your approaches widened my perspective.
If anyone would like to know how I got the answer, I'd be more than glad to do so here.
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## The Trachtenberg System of Speed Arithmetic
While he was in a German concentration camp, during World War II, Jakow Trachtenberg invented his system of speed arithmetic. His methods can be divided into two parts, (1) multiplication by small numbers (2 through 12), and (2) addition, multiplication, and division as other speed arithmetic experts do it (see Speed Arithmetic). All of this can be very helpful for any student, and has been very successfully used by students who have difficulty with numbers. In my opinion, the first part, multiplication by small numbers, is successful mainly because it is fun, not because it is a particularly fast method. I will show how Trachtenberg multiplied by 7, below. The one major exception to my observation (that these are not particularly fast) is multiplication by 11. For 11, Trachtenberg's method is a dream. You, and everyone else, should consider using it.
Multiplication by 11: Trachtenberg did this from right to left. It can be done from left to right. This is slightly harder (sometimes), but has the benefit of producing the answer from left to right, just as you would read off the answer. And so, it can be slightly faster, for that reason. Speed arithmetic experts do most of their arithmetic (including multiplying large numbers by large numbers) from left to right. I'll start with a simple example (with no carries):
4253 x 11 = 46783
Starting at the right, we write down the first digit (3). Then we add the first to the next digit (5) and write that down (8). Then we add the second digit to the third, etc. Finally we write down the leftmost (last) digit (4). In most cases, we have to deal with carries:
4683 x 11 = 51513
Starting at the right, we write down the first digit (3). Then we add the first pair of digits (8+3=11) and write down the right digit of this sum (1), and carry the one. We add the next pair of digits to the carry (6+8+1=15) and write down the right digit of this sum (5), and carry the one, etc. Finally, we write down the leftmost digit plus the carry.
A popular English language book on this subject is The Trachtenberg Speed System of Basic Mathematics translated and adapted by Ann Cutler and Rudolph McShane. Click on the name of the book (above) to go to that book on amazon.com.
Multiplication by 7: Going from right to left, we use this rule: Double each number and add half the neighbor (digit to the right, dropping any fraction); add 5 if the number (not the neighbor) is odd. And of course, we have to deal with carries:
3852 x 7 = 26964
Starting at the right (2), we double the first number (it has no neighbor) and write down the rightmost digit of that (4) and we have no carry. Then we double the next number (2x5=10), add five (+5=15), and add half the neighbor (+1=16), and write down the right digit (6) of that and carry the 1. Then we double the next number (2x8=16), and add half the neighbor (+2=18), and add the carry (+1=19). Then we double the next number (2x3=6), add five (+5=11), add half the neighbor (+4=15), and add the carry (+1=16). Now we double a zero off to the left of our 3852 (Trachtenberg wrote the zero out there: 03852) and add half the neighbor (0+1=1), and add the carry (+1=2). And we have our answer.
Notice that the carries are smaller than they were in normal multiplication by 7. The above rule is not simple, but once mastered, it is easy to use. It should be about as fast as multiplying normally (which requires memorizing the multiplication table). Multiplication by other small numbers (3 through 12) uses similar rules.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Division of Polynomials
Using long division to divide polynomials
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Division of Polynomials
What if you had a polynomial like 2x2+5x3\begin{align*}2x^2 + 5x - 3\end{align*} and you wanted to divide it by a monomial like x\begin{align*}x\end{align*} or a binomial like x+1\begin{align*}x + 1\end{align*}? How would you do so? After completing this Concept, you'll be able to divide polynomials like this one by monomials and binomials.
Try This
To check your answers to long division problems involving polynomials, try the solver at http://calc101.com/webMathematica/long-divide.jsp. It shows the long division steps so you can tell where you may have made a mistake.
Guidance
A rational expression is formed by taking the quotient of two polynomials.
Some examples of rational expressions are
2xx214x23x+42x9x2+4x5x2+5x12x32x+3\begin{align*}\frac{2x}{x^2-1} \qquad \frac{4x^2-3x+4}{2x} \qquad \frac{9x^2+4x-5}{x^2+5x-1} \qquad \frac{2x^3}{2x+3}\end{align*}
Just as with rational numbers, the expression on the top is called the numerator and the expression on the bottom is called the denominator. In special cases we can simplify a rational expression by dividing the numerator by the denominator.
Divide a Polynomial by a Monomial
We’ll start by dividing a polynomial by a monomial. To do this, we divide each term of the polynomial by the monomial. When the numerator has more than one term, the monomial on the bottom of the fraction serves as the common denominator to all the terms in the numerator.
Example A
Divide.
a) 8x24x+162\begin{align*}\frac{8x^2-4x+16}{2}\end{align*}
b) 3x2+6x1x\begin{align*}\frac{3x^2+6x-1}{x}\end{align*}
c) 3x218x+69x\begin{align*}\frac{-3x^2-18x+6}{9x}\end{align*}
Solution
a) 8x24x+162=8x224x2+162=4x22x+8\begin{align*}\frac{8x^2-4x+16}{2}=\frac{8x^2}{2}-\frac{4x}{2}+\frac{16}{2}=4x^2-2x+8\end{align*}
b) 3x3+6x1x=3x3x+6xx1x=3x2+61x\begin{align*}\frac{3x^3+6x-1}{x} = \frac{3x^3}{x}+\frac{6x}{x}-\frac{1}{x}=3x^2+6-\frac{1}{x}\end{align*}
c) 3x218x+69x=3x29x18x9x+69x=x32+23x\begin{align*}\frac{-3x^2-18x+6}{9x}=-\frac{3x^2}{9x}-\frac{18x}{9x}+\frac{6}{9x}=-\frac{x}{3}-2+\frac{2}{3x}\end{align*}
A common error is to cancel the denominator with just one term in the numerator.
Consider the quotient 3x+44\begin{align*}\frac{3x+4}{4}\end{align*}.
Remember that the denominator of 4 is common to both the terms in the numerator. In other words we are dividing both of the terms in the numerator by the number 4.
The correct way to simplify is:
3x+44=3x4+44=3x4+1\begin{align*}\frac{3x+4}{4}=\frac{3x}{4}+\frac{4}{4}=\frac{3x}{4}+1\end{align*}
A common mistake is to cross out the number 4 from the numerator and the denominator, leaving just 3x\begin{align*}3x\end{align*}. This is incorrect, because the entire numerator needs to be divided by 4.
Example B
Divide 5x310x2+x255x2\begin{align*}\frac{5x^3-10x^2+x-25}{-5x^2}\end{align*}.
Solution
5x310x2+x255x2=5x35x210x25x2+x5x2255x2\begin{align*}\frac{5x^3-10x^2+x-25}{-5x^2}=\frac{5x^3}{-5x^2}-\frac{10x^2}{-5x^2}+\frac{x}{-5x^2}-\frac{25}{-5x^2}\end{align*}
The negative sign in the denominator changes all the signs of the fractions:
5x35x2+10x25x2x5x2+255x2=x+215x+5x2\begin{align*}-\frac{5x^3}{5x^2}+\frac{10x^2}{5x^2}-\frac{x}{5x^2}+\frac{25}{5x^2}=-x+2-\frac{1}{5x}+\frac{5}{x^2}\end{align*}
Divide a Polynomial by a Binomial
We divide polynomials using a method that’s a lot like long division with numbers. We’ll explain the method by doing an example.
Example C
Divide x2+4x+5x+3\begin{align*}\frac{x^2+4x+5}{x+3}\end{align*}.
Solution
When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor.
To start the division we rewrite the problem in the following form:
x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\begin{align*}& {x+3 \overline{ ) x^2+4x+5 }}\end{align*}
We start by dividing the first term in the dividend by the first term in the divisor: x2x=x\begin{align*}\frac{x^2}{x}=x\end{align*}.
We place the answer on the line above the x\begin{align*}x\end{align*} term:
x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\end{align*}
Next, we multiply the x\begin{align*}x\end{align*} term in the answer by the divisor, x+3\begin{align*}x + 3\end{align*}, and place the result under the dividend, matching like terms. x\begin{align*}x\end{align*} times (x+3)\begin{align*}(x + 3)\end{align*} is x2+3x\begin{align*}x^2+3x\end{align*}, so we put that under the divisor:
x+3)x2+4x+5¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x x2+3x\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \ \ x^2 + 3x\end{align*}
Now we subtract \begin{align*}x^2+3x\end{align*} from \begin{align*}x^2+4x+5\end{align*}. It is useful to change the signs of the terms of \begin{align*}x^2+3x\end{align*} to \begin{align*}-x^2-3x\end{align*} and add like terms vertically:
\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x}\\ & \qquad \qquad \quad \ x\end{align*}
Now, we bring down the 5, the next term in the dividend.
\begin{align*}& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\end{align*}
And now we go through that procedure once more. First we divide the first term of \begin{align*}x + 5\end{align*} by the first term of the divisor. \begin{align*}x\end{align*} divided by \begin{align*}x\end{align*} is 1, so we place this answer on the line above the constant term of the dividend:
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\end{align*}
Multiply 1 by the divisor, \begin{align*}x + 3\end{align*}, and write the answer below \begin{align*}x + 5\end{align*}, matching like terms.
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\\ & \qquad \qquad \quad \ x + 3\end{align*}
Subtract \begin{align*}x + 3\end{align*} from \begin{align*}x + 5\end{align*} by changing the signs of \begin{align*}x + 3\end{align*} to \begin{align*}-x -3\end{align*} and adding like terms:
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\ & \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 5\\ & \qquad \qquad \ \ \underline{-x - 3}\\ & \qquad \qquad \qquad \quad 2\end{align*}
Since there are no more terms from the dividend to bring down, we are done. The quotient is \begin{align*}x + 1\end{align*} and the remainder is 2.
Remember that for a division with a remainder the answer is \begin{align*}\text{quotient}+\frac{\text{remainder}}{\text{divisor}}\end{align*}. So the answer to this division problem is \begin{align*}\frac{x^2+4x+5}{x+3}=x+1+\frac{2}{x+3}\end{align*}.
Check
To check the answer to a long division problem we use the fact that
\begin{align*}(\text{divisor} \times \text{quotient}) + \text{remainder} = \text{dividend}\end{align*}
For the problem above, here’s how we apply that fact to check our solution:
\begin{align*}(x+3)(x+1)+2 & = x^2+4x+3+2\\ & = x^2+4x+5\end{align*}
Watch this video for help with the Examples above.
Vocabulary
• A rational expression is formed by taking the quotient of two polynomials.
Guided Practice
Divide \begin{align*}\frac{x^2+8x+17}{x+4}\end{align*}.
Solution
When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor.
To start the division we rewrite the problem in the following form:
\begin{align*}& {x+4 \overline{ ) x^2+8x+17 }}\end{align*}
We start by dividing the first term in the dividend by the first term in the divisor: \begin{align*}\frac{x^2}{x}=x\end{align*}.
We place the answer on the line above the \begin{align*}x\end{align*} term:
\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\end{align*}
Next, we multiply the \begin{align*}x\end{align*} term in the answer by the divisor, \begin{align*}x + 4\end{align*}, and place the result under the dividend, matching like terms. \begin{align*}x\end{align*} times \begin{align*}(x + 4)\end{align*} is \begin{align*}x^2+4x\end{align*}, so we put that under the divisor:
\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17\;}}\\ & \qquad \ \ x^2 + 4x\end{align*}
Now we subtract \begin{align*}x^2+4x\end{align*} from \begin{align*}x^2+8x+17\end{align*}. It is useful to change the signs of the terms of \begin{align*}x^2+4x\end{align*} to \begin{align*}-x^2-4x\end{align*} and add like terms vertically:
\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x}\\ & \qquad \qquad \quad \ 4x\end{align*}
Now, we bring down the 17, the next term in the dividend.
\begin{align*}& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ 4x + 17\end{align*}
And now we go through that procedure once more. First we divide the first term of \begin{align*}4x + 17\end{align*} by the first term of the divisor. \begin{align*}4x\end{align*} divided by \begin{align*}x\end{align*} is 4, so we place this answer on the line above the constant term of the dividend:
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 17\end{align*}
Multiply 4 by the divisor, \begin{align*}x + 4\end{align*}, and write the answer below \begin{align*}4x + 16\end{align*}, matching like terms.
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ 4x + 17\\ & \qquad \qquad \quad \ 4x + 16\end{align*}
Subtract \begin{align*}4x + 16\end{align*} from \begin{align*}4x + 17\end{align*} by changing the signs of \begin{align*}4x + 16\end{align*} to \begin{align*}-4x -16\end{align*} and adding like terms:
\begin{align*}& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\ & \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\ & \qquad \qquad \quad \ x + 17\\ & \qquad \qquad \ \ \underline{-4x - 16}\\ & \qquad \qquad \qquad \quad 1\end{align*}
Since there are no more terms from the dividend to bring down, we are done. The quotient is \begin{align*}x + 4\end{align*} and the remainder is 1.
Remember that for a division with a remainder the answer is \begin{align*}\text{quotient}+\frac{\text{remainder}}{\text{divisor}}\end{align*}. So the answer to this division problem is \begin{align*}\frac{x^2+8x+17}{x+4}=x+4+\frac{1}{x+4}\end{align*}.
Check
To check the answer to a long division problem we use the fact that
\begin{align*}(\text{divisor} \times \text{quotient}) + \text{remainder} = \text{dividend}\end{align*}
For the problem above, here’s how we apply that fact to check our solution:
\begin{align*}(x+4)(x+4)+1 & = x^2+8x+16+1\\ & = x^2+8x+17\end{align*}
Practice
Divide the following polynomials:
1. \begin{align*}\frac{2x+4}{2}\end{align*}
2. \begin{align*}\frac{x-4}{x}\end{align*}
3. \begin{align*}\frac{5x-35}{5x}\end{align*}
4. \begin{align*}\frac{x^2+2x-5}{x}\end{align*}
5. \begin{align*}\frac{4x^2+12x-36}{-4x}\end{align*}
6. \begin{align*}\frac{2x^2+10x+7}{2x^2}\end{align*}
7. \begin{align*}\frac{x^3-x}{-2x^2}\end{align*}
8. \begin{align*}\frac{5x^4-9}{3x}\end{align*}
9. \begin{align*}\frac{x^3-12x^2+3x-4}{12x^2}\end{align*}
10. \begin{align*}\frac{3-6x+x^3}{-9x^3}\end{align*}
11. \begin{align*}\frac{x^2+3x+6}{x+1}\end{align*}
12. \begin{align*}\frac{x^2-9x+6}{x-1}\end{align*}
13. \begin{align*}\frac{x^2+5x+4}{x+4}\end{align*}
14. \begin{align*}\frac{x^2-10x+25}{x-5}\end{align*}
15. \begin{align*}\frac{x^2-20x+12}{x-3}\end{align*}
16. \begin{align*}\frac{3x^2-x+5}{x-2}\end{align*}
17. \begin{align*}\frac{9x^2+2x-8}{x+4}\end{align*}
18. \begin{align*}\frac{3x^2-4}{3x+1}\end{align*}
19. \begin{align*}\frac{5x^2+2x-9}{2x-1}\end{align*}
20. \begin{align*}\frac{x^2-6x-12}{5x^4}\end{align*}
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# Subderivative
A convex function (blue) and "subtangent lines" at x0 (red).
In mathematics, the subderivative, subgradient, and subdifferential generalize the derivative to functions which are not differentiable. The subdifferential of a function is set-valued. Subderivatives arise in convex analysis, the study of convex functions, often in connection to convex optimization.
Let f:IR be a real-valued convex function defined on an open interval of the real line. Such a function need not be differentiable at all points: For example, the absolute value function f(x)=|x| is nondifferentiable when x=0. However, as seen in the picture on the right, for any x0 in the domain of the function one can draw a line which goes through the point (x0, f(x0)) and which is everywhere either touching or below the graph of f. The slope of such a line is called a subderivative (because the line is under the graph of f).
## Definition
Rigorously, a subderivative of a function f:IR at a point x0 in the open interval I is a real number c such that
for all x in I. One may show that the set of subderivatives at x0 for a convex function is a nonempty closed interval [a, b], where a and b are the one-sided limits
which are guaranteed to exist and satisfy ab.
The set [a, b] of all subderivatives is called the subdifferential of the function f at x0. If f is convex and its subdifferential at contains exactly one subderivative, then f is differentiable at .[1]
## Examples
Consider the function f(x)=|x| which is convex. Then, the subdifferential at the origin is the interval [1, 1]. The subdifferential at any point x0<0 is the singleton set {−1}, while the subdifferential at any point x0>0 is the singleton set {1}.
## Properties
• A convex function f:IR is differentiable at x0 if and only if the subdifferential is made up of only one point, which is the derivative at x0.
• A point x0 is a global minimum of a convex function f if and only if zero is contained in the subdifferential, that is, in the figure above, one may draw a horizontal "subtangent line" to the graph of f at (x0, f(x0)). This last property is a generalization of the fact that the derivative of a function differentiable at a local minimum is zero.
The concepts of subderivative and subdifferential can be generalized to functions of several variables. If f:UR is a real-valued convex function defined on a convex open set in the Euclidean space Rn, a vector v in that space is called a subgradient at a point x0 in U if for any x in U one has
where the dot denotes the dot product. The set of all subgradients at x0 is called the subdifferential at x0 and is denoted ∂f(x0). The subdifferential is always a nonempty convex compact set.
These concepts generalize further to convex functions f:UR on a convex set in a locally convex space V. A functional v in the dual space V is called subgradient at x0 in U if
The set of all subgradients at x0 is called the subdifferential at x0 and is again denoted ∂f(x0). The subdifferential is always a convex closed set. It can be an empty set; consider for example an unbounded operator, which is convex, but has no subgradient. If f is continuous, the subdifferential is nonempty.
## History
The subdifferential on convex functions was introduced by Jean Jacques Moreau and R. Tyrrell Rockafellar in the early 1960s. The generalized subdifferential for nonconvex functions was introduced by F.H. Clarke and R.T. Rockafellar in the early 1980s.[2]
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## Absolute Value Functions
### Learning Outcomes
• Graph an absolute value function.
• Find the intercepts of an absolute value function.
Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: “s58y”/Flickr)
Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will investigate absolute value functions.
## Understanding Absolute Value
Recall that in its basic form $\displaystyle{f}\left({x}\right)={|x|}$, the absolute value function, is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance a number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign.
### A General Note: Absolute Value Function
The absolute value function can be defined as a piecewise function
$f(x) =\begin{cases}x ,\ x \geq 0 \\ -x , x < 0\\ \end{cases}$
### tip for success
It can help to visualize the graph of an absolute value function as the graph of the identity function, $f(x) = x$ where, for all negative input, the function value is forced to be positive.
When describing absolute value in words, visualize it as a distance. We can see, for example, that the two values that are $3$ units away from the number $1$ on the number line are $4$ and $-2$.
We can use an absolute value function statement to describe this by asking the question what values are 3 units away from 1? Let $x$ be the numbers. That is, the distance from some number and 1 is 3.
$|x-1| = 3$
$x-1 = 3 \quad \Rightarrow \quad x = 4 \qquad \text{ and } \qquad x-1 = -3 \quad \Rightarrow \quad x = -2$
### Example: Determine a Number within a Prescribed Distance
Describe all values $x$ within or including a distance of 4 from the number 5.
### Try It
Describe all values $x$ within a distance of 3 from the number 2.
### Example: Resistance of a Resistor
Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often $\pm 1\%,\pm5\%,$ or $\displaystyle\pm10\%$.
Suppose we have a resistor rated at 680 ohms, $\pm 5\%$. Use the absolute value function to express the range of possible values of the actual resistance.
### Try It
Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation.
The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin.
The graph below is of $y=2\left|x - 3\right|+4$. The graph of $y=|x|$ has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at $\left(3,4\right)$ for this transformed function.
### Example: Writing an Equation for an Absolute Value Function
Write an equation for the function graphed below.
Q & A
If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it?
Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for $x$ and $f\left(x\right)$.
$f\left(x\right)=a|x - 3|-2$
Now substituting in the point (1, 2)
$\begin{array}{l}2=a|1 - 3|-2\hfill \\ 4=2a\hfill \\ a=2\hfill \end{array}$
### Try It
Write the equation for the absolute value function that is horizontally shifted left 2 units, vertically flipped, and vertically shifted up 3 units.
Q & A
Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis?
Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero.
No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points.
(a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points.
## Find the Intercepts of an Absolute Value Function
Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.
### How To: Given the formula for an absolute value function, find the horizontal intercepts of its graph.
1. Isolate the absolute value term.
2. Use $|A|=B$ to write $A=B$ or $\mathrm{-A}=B$, assuming $B>0$.
3. Solve for $x$.
### tip for success
The places where a graph crosses the horizontal axis are sometimes called horizontal interceptsx-intercepts, or zeros.
We call them zeros because those are the places where the function value equals zero. To find them, set the function value equal to zero and solve for $x$. We can do this for many functions using the techniques of college algebra.
### Example: Finding the Zeros of an Absolute Value Function
For the function $f\left(x\right)=|4x+1|-7$ , find the values of $x$ such that $\text{ }f\left(x\right)=0$ .
### Try It
For the function $f\left(x\right)=|2x - 1|-3$, find the values of $x$ such that $f\left(x\right)=0$.
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# Relationship between binomial and normal distribution
### Normal Approximation to Binomial | STAT /
What is the difference between Normal Distribution, Binomial Distribution, and Poisson Distribution. distribution, the Binomial distribution and the Poisson distribution. Best practice “the probability that the time taken will be between 10 and 12 minutes”. In probability theory and statistics, there are several relationships among probability A negative binomial distribution with n = 1 is a geometric distribution . If Z is a normal random variable with parameters (μ = m, σ2 = s2), then X = aZ + b is a.
This is a very tricky application of the binomial distribution. If you can follow the logic of this solution, you have a good understanding of the material covered in the tutorial, to this point. In the world series, there are two baseball teams. The series ends when the winning team wins 4 games. Therefore, we define a success as a win by the team that ultimately becomes the world series champion. For the purpose of this analysis, we assume that the teams are evenly matched.
Therefore, the probability that a particular team wins a particular game is 0. Let's look first at the simplest case. What is the probability that the series lasts only 4 games. This can occur if one team wins the first 4 games. The probability of the National League team winning 4 games in a row is: Therefore, probability that the series ends in four games would be 0.
Now let's tackle the question of finding probability that the world series ends in 5 games. You know that a normal distribution is symmetric. So, half of the probability located one side of the mean and another half located another side of the mean. So, one standard deviation will be 30 to 50 range. You already know for left side up 40 the probability is 0.
Now if you calculate the probability from 40 to 50 range it will be half of 1 Standard deviation i. Every random variable has a corresponding probability distribution.
## Normal Approximation to Binomial
The probability distribution applies the theory of probability to describe the behavior of the random variable. A discrete random variable X has a finite number of possible integer values. The probability distribution of X lists the values and their probabilities in a table Every probability pi is a number between 0 and 1.
The sum of the probabilities must be 1.
### Normal, Binomial and Poisson Distribution Explained | ROP
This properties we have already studied before. Now we will discuss about the most important probability for discrete random variable is Binomial Distribution. Before that it is necessary to know about Bernoulli trial.
They found that there were organ donors, agedacross the UK for the two years and combined. Heart-beating donors are patients who are seriously ill in an intensive care unit ICU and are placed on a ventilator.
Now it is clear that the distribution of the number of donors takes integer values only, thus the distribution is similar in this respect to the binomial.
However, there is no theoretical limit to the number of organ donors that could happen on a particular day.
Here the population is the UK population agedover two years, which is over 82 million person years, so in this case each member can be thought to have a very small probability of actually suffering an event, in this case being admitted to a hospital ICU and placed on a ventilator with a life threatening condition. The mean number of organ donors per day over the two year period is calculated as: Exact confidence intervals can be calculated as described by Altman et al.
## What is the difference between binomial and normal distribution?
The Poisson probabilities are calculated from: Here e is the exponential constant 2. Example Suppose that before the study of Wight et al. Remember that 20 and 0! If the study is then to be conducted over 2 years dayseach of these probabilities is multiplied by to give the expected number of days during which 0, 1, 2, 3, etc.
### Normal Distribution Binomial Distribution Poisson Distribution - MAKE ME ANALYST
These expectations are A comparison can then be made between what is expected and what is actually observed. A brief description of some other distributions are given for completeness.
The smaller the sample size, the more spread out the tails, and the larger the sample size, the closer the t-distribution is to the Normal distribution Figure 3.
The t-distribution for various sample sizes.
Chi-squared distribution The chi-squared distribution is continuous probability distribution whose shape is defined by the number of degrees of freedom. It is a right-skew distribution, but as the number of degrees of freedom increases it approximates the Normal distribution Figure 4.
The chi-squared distribution is important for its use in chi-squared tests.
• Difference between Normal, Binomial, and Poisson Distribution
• What is the difference between binomial and normal distribution?
• Difference Between Binomial and Normal Distribution
These are often used to test deviations between observed and expected frequencies, or to determine the independence between categorical variables.
When conducting a chi-squared test, the probability values derived from chi-squared distributions can be looked up in a statistical table.
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# Let me know if this is correct :
(1)Finding the lcm using which ever method for:5,15 , and 20
8:2x2x2x3x5x5
15:2x2x2x3x5x5
20:2x2x2x3x5x5
---------------
LCM:2x2x2x3x5x5
After simplification LCM:120
(2)Find the GCF for each of the following numbers: 36, 64, 180
Prime Factors
36=2x2x3x3x1
64=2x2x2x2x2x2x1
180=2x2x3x3x5x1
Common Prime Factors
36= 2x2x2
64=2x2x2
180=2x2x2
GCF=4
I am not certain what your logic is, but the answers are correct.
8 months ago
## The logic used in the given solutions is correct. Let me explain how to find the LCM and GCF of the given numbers:
(1) Finding the LCM:
To find the LCM (Least Common Multiple) of the numbers 5, 15, and 20, you can use the prime factorization method.
First, factorize each number into its prime factors:
5 = 5 (since 5 is a prime number)
15 = 3 × 5
20 = 2 × 2 × 5
Next, you take the highest power of each prime factor that appears in any of the numbers. In this case, the prime factors are 2, 3, and 5.
2 appears twice in the factorization of 20, so we take 2^2.
3 appears once in the factorization of 15, so we take 3^1.
5 appears once in the factorization of 15 and 20, so we take 5^1.
Multiplying these together, we get:
LCM = 2^2 × 3^1 × 5^1 = 4 × 3 × 5 = 60.
So, the correct LCM of 5, 15, and 20 is 60. Your calculation of 120 is incorrect.
(2) Finding the GCF:
To find the GCF (Greatest Common Factor) of the numbers 36, 64, and 180, you can again use the prime factorization method.
Factorize each number into its prime factors:
36 = 2 × 2 × 3 × 3
64 = 2 × 2 × 2 × 2 × 2 × 2
180 = 2 × 2 × 3 × 3 × 5
Now, you look for the common prime factors among these numbers.
The common prime factors are:
2 appears twice in the factorization of 36 and 64, but only once in the factorization of 180.
So, we take 2 × 2 = 4.
The GCF is 4.
Therefore, your calculation of GCF=4 is correct.
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# What Is 20/37 as a Decimal + Solution With Free Steps
The fraction 20/37 as a decimal is equal to 0.54054054.
The division is a mathematical operation that is used to calculate fractions. We translate fractions to Decimal numbers to make them simpler to comprehend. A fraction when solved completely using division can be converted into a decimal number.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 20/37.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 20
Divisor = 37
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 20 $\div$ 37
This is when we go through the Long Division solution to our problem.
Figure 1
## 20/37 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 20 and 37, we can see how 20 is Smaller than 37, and to solve this division, we require that 20 be Bigger than 37.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 20, which after getting multiplied by 10 becomes 200.
We take this 200 and divide it by 37; this can be done as follows:
 200 $\div$ 37 $\approx$ 5
Where:
37 x 5 = 185
This will lead to the generation of a Remainder equal to 200 – 185 = 15. Now this means we have to repeat the process by Converting the 15 into 150 and solving for that:
150 $\div$ 37 $\approx$ 4Â
Where:
37 x 4 = 148
This, therefore, produces another Remainder which is equal to 150 – 148 = 2.
Finally, we have a Quotient generated after combining the pieces of it as 0.54=z, with a Remainder equal to 22.
Images/mathematical drawings are created with GeoGebra.
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What Is 25/80 as a Decimal + Solution With Free Steps
The fraction 25/80 as a decimal is equal to 0.312.
Fractions are numerals of the form p/q and they simply act as an alternate notation for the arithmetic operation of division. Here, if p is the dividend and q is the divisor, then p $\boldsymbol\div$ q = p/q, where p is the numerator and q is the denominator.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 25/80.
Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 25
Divisor = 80
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 25 $\div$ 80
This is when we go through the Long Division solution to our problem.
Figure 1
25/80 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 25 and 80, we can see how 25 is Smaller than 80, and to solve this division, we require that 25 be Bigger than 80.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 25, which after getting multiplied by 10 becomes 250.
We take this 250 and divide it by 80; this can be done as follows:
 250 $\div$ 80 $\approx$ 3
Where:
80 x 3 = 240
This will lead to the generation of a Remainder equal to 250 – 240 = 10. Now this means we have to repeat the process by Converting the 10 into 100 and solving for that:
100 $\div$ 80 $\approx$ 1Â
Where:
80 x 1 = 80
This, therefore, produces another Remainder which is equal to 100 – 80 = 20. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 200.
200 $\div$ 80 $\approx$ 2Â
Where:
80 x 2 = 160
Finally, we have a Quotient generated after combining the three pieces of it as 0.312, with a Remainder equal to 40.
Images/mathematical drawings are created with GeoGebra.
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# 1-6 Multiplying and Dividing Integers Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation.
## Presentation on theme: "1-6 Multiplying and Dividing Integers Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation."— Presentation transcript:
1-6 Multiplying and Dividing Integers Course 3 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation
Warm Up Multiply or divide. Course 3 1-6 Multiplying and Dividing Integers 4072 7 12 4 126 1. 5(8) 2. 6(12) 3. 36 9 4. 49 7 192 16 5. 18(7)6.
Problem of the Day Complete the pyramid by filling in the missing numbers. Each number is the sum of the numbers in the two boxes below it. Course 3 1-6 Multiplying and Dividing Integers –98 –7 –15 –4 –8 –32 4
Learn to multiply and divide integers. Course 3 1-6 Multiplying and Dividing Integers
Insert Lesson Title Here Course 3 1-6 Multiplying and Dividing Integers A positive number multiplied by an integer can be written as repeated addition. 3(–200) = –200 + (–200) + (–200) = –600 From what you know about adding and subtracting integers, you can see that a positive integer times a negative integer is negative.
Insert Lesson Title Here Course 3 1-6 Multiplying and Dividing Integers You know that multiplying two positive integers together gives you a positive answer. Look for a pattern in the integer multiplication at right to understand the rules for multiplying two negative integers. 3(–200) = 2(–200) = 1(–200) = 0(–200) = –1(–200) = –2(–200) = –3(–200) = –600 –400 + 200 –200 0 200 400 600 + 200 The product of two negative integers is a positive integer.
Insert Lesson Title Here Course 3 1-6 Multiplying and Dividing Integers MULTIPLYING AND DIVIDING TWO INTEGERS If the signs are the same, the sign of the answer is positive. If the signs are different, the sign of the answer is negative.
Course 3 1-6 Multiplying and Dividing Integers Additional Example 1: Multiplying and Dividing Integers A. –6(4) B. –8(–5) = 40 = –24 Signs are different. Signs are the same. Answer is negative. Answer is positive. Multiply or divide.
Course 3 1-6 Multiplying and Dividing Integers Additional Example 1: Multiplying and Dividing Integers C. = –9 Signs are different. Answer is negative. Multiply or divide. –18 2 D. = 5 Signs are the same. Answer is positive. –25 –5
Course 3 1-6 Multiplying and Dividing Integers Check It Out: Example 1 A. 5(–2) B. –3(–2) = 6 = –10 Signs are different. Signs are the same. Answer is negative. Answer is positive. Multiply or divide.
Course 3 1-6 Multiplying and Dividing Integers Check It Out: Example 1 C. = –8 Signs are different. Answer is negative. Multiply or divide. –24 3 D. = 6 Signs are the same. Answer is positive. –12 –2
Course 3 1-6 Multiplying and Dividing Integers Order of Operations 1. Parentheses 2. Exponents 3. Multiply and divide from left to right. 4. Add and subtract from left to right. Remember!
Course 3 1-6 Multiplying and Dividing Integers Additional Example 2: Using the Order of Operations with Integers A. 3(–6 – 12) = –54 = 3(–18) Subtract inside the parentheses. Think: The signs are different. The answer is negative. Simplify. B. –5(–5 + 2) = 15 = –5(–3) Subtract inside the parentheses. Think: The signs are the same. The answer is positive.
Course 3 1-6 Multiplying and Dividing Integers Additional Example 2: Using the Order of Operations with Integers C. –2(14 – 5) = –18 = –2(9) Subtract inside the parentheses. Think: The signs are different. The answer is negative. Simplify.
Course 3 1-6 Multiplying and Dividing Integers Check It Out: Example 2 A. 2(1 – 8) = –14 = 2(–7) Subtract inside the parentheses. Think: The signs are different. The answer is negative. Simplify. B. 4(–3 – 8) = –44 = 4(–11) Subtract inside the parentheses. Think: The signs are different. The answer is negative.
Course 3 1-6 Multiplying and Dividing Integers Check It Out: Example 2 C. –3(6 – 9) = 9 = –3(–3) Subtract inside the parentheses. Think: The signs are the same. The answer is positive. Simplify.
Course 3 2-3 Multiplying and Dividing Integers The order of operations can be used to find ordered pair solutions of integer equations. Substitute an integer value for one variable to find the value of the other variable in the ordered pair.
Course 3 1-6 Multiplying and Dividing Integers Additional Example 3: Sports Application A golfer plays 5 holes. On 3 holes, he has a gain of 4 strokes each. On 2 holes, he has a loss of 4 strokes each. Each gain in strokes can be represented by a positive integer, and each loss can be represented by a negative integer. Find the total net change in strokes. 3(4) + 2(–4) = 4 = 12 + (–8) Add the losses to the gains. Multiply. Add. The golfer changed by a total gain of 4 strokes.
Course 3 1-6 Multiplying and Dividing Integers Check It Out: Example 3 A golfer plays 9 holes. On 3 holes, he has a gain of 3 strokes each. On 4 holes, he has a loss of 3 strokes each. Each gain in strokes can be represented by a positive integer, and each loss can be represented by a negative integer. Find the total net change in strokes. 3(3) + 4(–3) = -3 = 9 + (–12) Add the losses to the gains. Multiply. Add. The golfer changed by a total loss of 3 strokes.
Lesson Quiz: Part I Insert Lesson Title Here Course 3 1-6 Multiplying and Dividing Integers Multiply or Divide. 1. –8(4) 2. –32 6 –12(5) –10 Simplify. 3. –2(13 – 4) 4. 6(-5 – 3) – 18 – 48
Lesson Quiz: Part II Insert Lesson Title Here Course 3 1-6 Multiplying and Dividing Integers 5. Evin completes 11 transactions in his bank account. In 6 transactions, he withdraws \$10. in 5 transaction, he deposits \$20. Find the total net change in dollars. \$40
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Math and Life
Shower Rods and Higher Dimensions
It’s moving day. You’ve got your entire house packed up into boxes all ready to go, and are all nervous/excited to move into your new home. Just one last item: the shower rod. You’ve got one of those fancy-schmancy extendable ones, so you start to wonder: “if I were to place my shower rod at one corner of my box, and extend it until it reaches the opposite corner, how long would it be?”
As it turns out, the answer to this question is either really simple or a complete mind-bender depending on one thing: how many dimensions you are living in.
A two dimensional box is a rectangle, so if you paid even the slightest amount of attention in middle school, this should be easy.
What’s the diagonal distance? Draw the diagonal line, and use the Pythagorean Theorem. In this case,
What use the flatlanders have for shower rods? The world may never know.
3D is slightly harder, but totally manageable. This takes some spacial visualization, and because I have no faith in my ability to convey pictures through text, I will cut out the middleman:
We want the distance from A to B.
Start with what we know. The distance from A to C is sqrt(x2+y2) just like before:
So how does this help us? Let’s cut a plane through the box along the distance we just calculated and point B:
Both A and B lie on this plane, so draw a line between them:
If we focus on just the plane, we can see that our new line forms a triangle! Not just any triangle, but a right triangle,1 which means we can use the pythagorean theorem again:
Simplifying the squared square root, and taking the square root of both sides, we get
If you are stubbornly stuck in the real world, you may stop here.
4D
This is way harder. As humans, we don’t have a physical intuition of 4D. You could try to make analogies with our 3D world, but they never really give you a feel for higher dimensions. Visualising this is impossible, so no pictures. We’re flying blind
Fortunately for us, math doesn’t care about pictures.
So how do we even begin? Let’s start by trying to recognise patterns about what we already know:
Consider a 2d box. When “changing” from 2d to 3d, what direction is this new side? Well, all sides of a box are perpendicular to each other (that’s kind of the definition of a box), so the new 3rd dimensional sides2 are perpendicular to every possible line we could draw on the 2d box.
So what happens if we apply this pattern to the transition from 3d box to a 4d box? Well, whatever the fourth dimensional sides looks like, they would have to be perpendicular to all lines drawn in our 3d box. (stop trying to visualize it, just go with me)
Take the line along the fourth dimensional edge that touches the 3d line (that we already know the distance of). We know it has to be perpendicular to the 3d line, because of the rule above. We also know that it touches the 3d line, because the meet at the save vertex.
And what do we get when we have two perpendicular lines that meet?3 A right triangle!
Using our friend Pythagoras:
This is the exact same process I used in the 3d example, only there we had helpful diagrams that lead us to the answer. If you wish, you can continue this argument indefinitely for higher and higher dimensions. This gives the nice general case3 for n dimensions:
Oh, and just to be cute, we can do this formula for one dimension:
1Angle C is right.
2Ok technically, there isn't such thing as a "third" dimension, (mintue physics video on that) but let’s just pretend that one corner of this box is on the origin and all “sides” are on axies
3Assuming one point is on (0,0), so perhaps not as general as it could be
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Entrance Exams
Problems on Numbers Questions for the CAT are part of the Quantitative Aptitude Through Problems on Numbers questions, aspirants are tested to solve questions with fractions. The difficulty level of these questions is easy to moderate.
## What are some CAT Decimal Fraction Practice Questions?
Q1. What value will replace the question mark in the following equations?
(i) 5172.49 + 378.352 +? = 9318.678
(ii) ? – 7328.96 = 5169.38
(ii) 12498.34
Detailed Solution. (i) Let 5172.49 + 378.352 + x = 9318.678.
Then, x = 9318.678 – (5172.49 + 378.352) = 9318.678 – 5550.842 = 3767.836.
(ii) Let x – 7328.96 = 5169.38. Then, x = 5169.38 + 7328.96 = 12498.34.
Q2. Find the products: (i) 6.3204 × 100 (ii) .069 × 10000
Detailed Solution. (i) 6.3204 × 100 = 632.04. (ii) .069 × 10000 = .0690 × 10000 = 690.
Q3. Find the products:
(i) 2.1693 × 1.4
(ii) .4 × .04 × .004 × 40
(iii) 6.66 × 66.6 × 66
(ii) .002560
(iii) 29274.696
Detailed Solution. (i) 21693 × 14 = 303702. Sum of decimal places = (4 + 1) = 5.
2.1693 × 1.4 = 3.03702.
(ii) 4 × 4 × 4 × 40 = 2560. Sum of decimal places = (1 + 2 + 3) = 6.
.4 × .04 × .004 × 40 = .002560.
(iii) 666 × 666 × 66 = 29274696. Sum of decimal places = (2 + 1) = 3.
6.66 × 66.6 × 66 = 29274.696.
Q4. Given that 268 × 74 = 19832, find the value of 2.68 × .74.
Detailed Solution. Sum of decimal places = (2 + 2) = 4.
∴ 2.68 × .74 = 1.9832.
## What are the must-do Decimal Fraction questions for the CAT exam?
Q5. Find the quotient:
(i) 0.63 ÷ 9
(ii) 0.0204 ÷ 17
(iii) 3.1603 ÷ 13
(ii) .0012
(iii) .2431
Detailed Solution. (i) 63 ÷ 9 = 7. Dividend contains 2 places of decimal.
∴ 0.63 ÷ 9 = .07.
(ii) 204 ÷ 17 = 12. Dividend contains 4 places of decimal.
∴ 0.0204 ÷ 17 = .0012.
(iii) 31603 ÷ 13 = 2431. Dividend contains 4 places of decimal.
∴ 3.1603 ÷ 13 = .2431.
Q6. Evaluate:
(i) 35 + .07
(ii) 2.5 + 0.0005
(iii) 136.09 + 43.9
(ii) 5000
(iii) 3.1
Detailed Solution. (i) 35/0.07 = (35 × 100) / (0.07 × 100) = 3500/7 = 500.
(ii) 2.5/0.0005 = (2.5 × 10000) / (0.0005 × 10000) = 25000/5 = 5000.
(iii) 136.09/43.9 = (136.09 × 10) / (43.9 × 10) = 1360.9/439 = 3.1.
Q7. What value will come in place of question mark in the following equations ?
(i) 0.006 ÷? = 0.6
(ii)? ÷ .025 = 80
(ii) 2.
Detailed Solution. (i) 0.006/x = 0.6
Multiplying both sides by x: 0.006 = 0.6x
Dividing both sides by 0.6: x = 0.006/0.6 = 0.01
(ii) x/0.025 = 80
Multiplying both sides by 0.025: x = 80 × 0.025 = 2
Q8. If 1/3.718 = .2689, then find the value of 1/.0003718.
Detailed Solution. 1/.0003718 = 10000/3.718 = (10000 x 1/3.178) = 10000 x .2689 = 2689
## What were the previous year's CAT Decimal Fraction questions?
Q9. Evaluate:
(i) 0.5 × 5.6 ÷ 0.5 × 12
(ii) 25 × 3.25 + 50.4 ÷ 24
(iii) 0.01 × 0.1 – 0.001 ÷ 10 + 0.01
(iv) 12.28 × 1.5 – 36 ÷ 2.4
(ii) 83.35
(iii) 0.0109
(iv) 3.42
Detailed Solution. (i) Given expression = 0.5 x (5.6/0.5) x 12 = 5x56x12/50 = 3360/72 = 67.2
(ii) Given expression = 25 x 3.25 + 50.4/24 = 81.25 + 2.1 = 83.35
(iii) Given expression = 0.001 - 0.01/10 + 0.01 = 0.001 - 0.0001 + 0.011 = 0.0109
(iv) Given expression = 12.28 x 1.5 - 36/2.4 = 18.42 - 15 = 3.42
Q10. Simplify.
(7/8) ÷ (5/6) × (3/4) + (1/3) × (2/5)
Detailed Solution. Convert all fractions to their equivalent forms with a common denominator.
7/8 = 21/24
5/6 = 20/24
3/4 = 18/24
1/3 = 8/24
2/5 = 9.6/24 (converting 2/5 to 9.6/24 for ease of calculation)
Perform the operations following the order of BODMAS (Brackets, Orders, Division, Multiplication, Addition, Subtraction).
(21/24) ÷ (20/24) × (18/24) + (8/24) × (9.6/24)
= (21/20) × (18/24) + (8/24) × (9.6/24)
= 1.05 × 0.75 + 0.4
= 0.7875 + 0.4
= 1.1875
Express the result as a decimal fraction by dividing the numerator by the denominator.
1.1875 = 1.1875/1 = 0.967
Q11. Express the repeating decimal 0.27272727... as a fraction in its simplest form.
Detailed Solution. Let x = 0.27272727...
Step 1: Multiply the equation by 100 to remove the decimal point.
100x = 27.272727...
Step 2: Subtract the original equation from the new equation.
100x - x = 27.272727... - 0.27272727...
99x = 27
Step 3: Divide both sides by 99.
x = 27/99
Step 4: Simplify the fraction by dividing both the numerator and denominator by their greatest common divisor (GCD).
GCD of 27 and 99 is 3.
27/3 = 9
99/3 = 33
Therefore, the simplest fraction form of the repeating decimal 0.27272727... is 9/33.
Q12. (a) Express the repeating decimal 0.184184184... as a fraction in its simplest form.
(b) If the fraction obtained in part (a) is multiplied by 2/5, express the result as a decimal.
(c) Find the value of (0.184184184... + 2/5) correct to three decimal places.
(b) 0.059259259...
(c) 0.584
Detailed Solution. (a) Let x = 0.184184184...
Multiplying by 1000: 1000x = 184.184184...
Subtracting the original equation: 999x = 184
x = 184/999
Simplifying by dividing numerator and denominator by their GCD (37):
x = 4/27
(b) (4/27) × (2/5) = 8/135 = 0.059259259...
(c) 0.184184184... + 2/5 = 0.184184184... + 0.4
= 0.584184184...
= 0.584 (correct to three decimal places)
## What were the Decimal Fractions questions on the CAT 2022 exam?
Q13. Simplify:
(2.375 ÷ 1.25) × (0.64 ÷ 0.08) - (3.5 ÷ 0.25)
Detailed Solution. Step 1: Simplify (2.375 ÷ 1.25)
2.375 ÷ 1.25 = 1.9
Step 2: Simplify (0.64 ÷ 0.08)
0.64 ÷ 0.08 = 8
Step 3: Simplify (3.5 ÷ 0.25)
3.5 ÷ 0.25 = 14
Step 4: Evaluate the expression
(1.9 × 8) - 14
= 15.2 - 14
= 1.2
Q14. Simplify :
(5.273 ÷ 0.014) + (7.685 ÷ 0.017) - (2.149 ÷ 0.023)
Detailed Solution. Convert decimals to fractions with denominator 2310
5.273 = 5273/2310
0.014 = 14/2310
7.685 = 7685/2310
0.017 = 17/2310
2.149 = 2149/2310
0.023 = 23/2310
Set up the expression with fractions
= (5273/14) + (7685/17) - (2149/23)
Convert fractions to equivalent fractions with common denominator of 2310
= (376,642/2310) + (452,647/2310) - (93,434/2310)
Perform operations
= 829,289/2310 - 93,434/2310
= 735,855/2310
Simplify the fraction
735,855/2310 = 318.55
Q15. Simplify :
(3.27 × 2.105) ÷ (0.63 × 1.4)
Detailed Solution. Multiply the numbers in the numerator and denominator separately.
3.27 × 2.105 = 6.88635
0.63 × 1.4 = 0.882
Divide the numerator by the denominator.
6.88635 ÷ 0.882 = 7.8
Therefore, the simplified value of the expression (3.27 × 2.105) ÷ (0.63 × 1.4) is 7.8.
Q16. Evaluate:
(5.28 ÷ 1.8) + (2.7 × 3.15) - (7.2 ÷ 0.9)
Detailed Solution. Perform the divisions first.
5.28 ÷ 1.8 = 2.93333...
7.2 ÷ 0.9 = 8
Perform the multiplication.
2.7 × 3.15 = 8.505
Perform the addition and subtraction in the given order.
2.93333... + 8.505 - 8 = 3.43833...
Therefore, the value of the expression (5.28 ÷ 1.8) + (2.7 × 3.15) - (7.2 ÷ 0.9) is approximately 3.44.
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# Math Statistics: Median
In these lessons, we will learn how to calculate the median of a given set of data. We will also learn how to solve some mean, median, mode SAT questions.
Related Topics:
More Lessons on Statistics
Statistics Worksheets
Statistics Games
Given a set of observations, the median is the middle value among the observations. To find the median, you must first arrange the observations in an ascending (or descending) order.
There are 2 possible cases to consider in finding the median.
Case 1. When the number of observations is odd, the median is the middle value.
Example:
Find the median of the following set of points in a game:
15, 14, 10, 8, 12, 8, 16
Solution:
First arrange the point values in an ascending order (or descending order).
8, 8, 10, 12, 14, 15, 16
The number of point values is 7, an odd number. Hence, the median is the value in the middle position.
Median = 12
Case 2. When the number of observations is even, the median is the average of the two middle values.
Example:
Find the median of the following set of points:
15, 14, 10, 8, 12, 8, 16, 13
Solution:
First arrange the point values in an ascending order (or descending order).
8, 8, 10, 12, 13, 14, 15, 16
The number of point values is 8, an even number. Hence the median is the average of the 2 middle values.
Sometimes, you may be asked to determine the possible values of the median of a set of numbers.
Example:
x is the median for 4, 3, 8, x and 7. Find the possible values for x.
Solution:
Arrange the numbers in ascending order with x in the middle.
3, 4, x, 7, 8
This tells us that the possible values of x are 4, 5, 6 and 7.
The following videos will show you how to find the median.
How to find the median?
In this lesson you learn how to find the median from a set of numbers using the median position formula (n+1)/2. Mean Median and Mode
Example:
Find the mean, median and mode of the following set of data:
23, 29, 20, 32, 23, 21, 33, 25 Mean, median, mode song
A song to help you remember how to calculate mean, median and mode.
### SAT: Mean, Median, Mode
In the SAT, mean, median and mode questions may be given in the following manner.
SAT Mean, Median, and Mode Math Question
This video will show you how to answer SAT mean, median, and mode math questions.
In a set of nine distinct numbers which of the following cannot affect the median
a) squaring
b) increasing every other number by 5
c) increasing the smallest number only
d) increasing the largest number only
e) decreasing the largest number only
In a certain Chemistry class, the median score on a test was 84. Which of the following must be true?
I. The highest score in the class is at least one point higher than the lowest score.
II. If there was a score of 83 in the class, there was also a score of 85.
III. The mode of the test scores was 84.
If w, x and y represent 3 positive numbers where x = y + 5 and y = w + 20, what is the result when the average of the three numbers is subtracted from the median?
a) -5
b) 0
c) 4
d) 5
e) 15
SAT Math - Averages, Median, List of Real Numbers
The least and greatest numbers in a list of 7 real numbers are 2 and 20, respectively. The median of the list is 6, and the number 3 occurs most often in the list. Which of the following could be the average (arithmetic mean) of the numbers in the list?
I. 1
II. 8, 5
III. 10
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Measures of Central Tendency and Dispersion
## Mean, median, mode, range
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Practice Measures of Central Tendency and Dispersion
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Measures of Central Tendency and Dispersion
This Concept is an overview of some of the basic statistics used to measure the center of a set of data.
### Watch This
For an explanation and examples of mean, median and mode, see keithpeterb, Mean, Mode and Median from Frequency Tables (7:06).
### Guidance
The students in a statistics class were asked to report the number of children that live in their house (including brothers and sisters temporarily away at college). The data are recorded below:
1, 3, 4, 3, 1, 2, 2, 2, 1, 2, 2, 3, 4, 5, 1, 2, 3, 2, 1, 2, 3, 6
Once data are collected, it is useful to summarize the data set by identifying a value around which the data are centered. Three commonly used measures of center are the mode, the median, and the mean.
Mode
The mode is defined as the most frequently occurring number in a data set. The mode is most useful in situations that involve categorical (qualitative) data that are measured at the nominal level. In the last chapter, we referred to the data with the Galapagos tortoises and noted that the variable 'Climate Type' was such a measurement. For this example, the mode is the value 'humid'.
#### Example A
Find the mode for the number of children per house in the data set at the beginning of the Concept.
Solution:
In this case, 2 is the mode, as it is the most frequently occurring number of children in the sample, telling us that most students in the class come from families where there are 2 children.
In this example, the mode could be a useful statistic that would tell us something about the families of statistics students in our school.
More Than One Mode
If there were seven 3-child households and seven 2-child households, we would say the data set has two modes. In other words, the data would be bimodal. When a data set is described as being bimodal, it is clustered about two different modes. Technically, if there were more than two, they would all be the mode. However, the more of them there are, the more trivial the mode becomes. In these cases, we would most likely search for a different statistic to describe the center of such data.
If there is an equal number of each data value, the mode is not useful in helping us understand the data, and thus, we say the data set has no mode.
Mean
Another measure of central tendency is the arithmetic average, or mean. This value is calculated by adding all the data values and dividing the sum by the total number of data points. The mean is the numerical balancing point of the data set.
We can illustrate this physical interpretation of the mean. Below is a graph of the class data from the last example.
If you have snap cubes like you used to use in elementary school, you can make a physical model of the graph, using one cube to represent each student’s family and a row of six cubes at the bottom to hold them together, like this:
#### Example B
Find the mean for the number of children per house.
Solution:
There are 22 students in this class, and the total number of children in all of their houses is 55, so the mean of this data is \begin{align*}\frac{55}{22}=2.5\end{align*}.
It turns out that the model that you created balances at 2.5. In the pictures below, you can see that a block placed at 3 causes the graph to tip left, while one placed at 2 causes the graph to tip right. However, if you place the block at 2.5, it balances perfectly!
Statisticians use the symbol \begin{align*}\overline{x}\end{align*} to represent the mean when \begin{align*}x\end{align*} is the symbol for a single measurement. Read \begin{align*}\overline{x}\end{align*} as “\begin{align*}x\end{align*} bar.”
Symbolically, the formula for the sample mean is as follows:
\begin{align*}\overline{x}= \frac{\sum_{i=1}^n x_i}{n} = \frac{x_1+x_2+\ldots+x_n}{n}\end{align*}
where:
\begin{align*}x_i\end{align*} is the \begin{align*}i^{\text{th}}\end{align*} data value of the sample.
\begin{align*}n\end{align*} is the sample size.
The mean of the population is denoted by the Greek letter, \begin{align*}\mu\end{align*}.
\begin{align*}\overline{x}\end{align*} is a statistic, since it is a measure of a sample, and \begin{align*}\mu\end{align*} is a parameter, since it is a measure of a population. \begin{align*}\overline{x}\end{align*} is an estimate of \begin{align*}\mu\end{align*}.
Median
The median is simply the middle number in an ordered set of data.
Suppose a student took five statistics quizzes and received the following grades:
80, 94, 75, 96, 90
To find the median, you must put the data in order. The median will be the data point that is in the middle. Placing the data in order from least to greatest yields: 75, 80, 90, 94, 96.
The middle number in this case is the third grade, or 90, so the median of this data is 90.
When there is an even number of numbers, no one of the data points will be in the middle. In this case, we take the average (mean) of the two middle numbers.
#### Example C
Consider the following quiz scores: 91, 83, 97, 89
Place them in numeric order: 83, 89, 91, 97.
The second and third numbers straddle the middle of this set. The mean of these two numbers is 90, so the median of the data is 90.
Mean vs. Median
Both the mean and the median are important and widely used measures of center. Consider the following example: Suppose you got an 85 and a 93 on your first two statistics quizzes, but then you had a really bad day and got a 14 on your next quiz!
The mean of your three grades would be 64. Which is a better measure of your performance? As you can see, the middle number in the set is an 85. That middle does not change if the lowest grade is an 84, or if the lowest grade is a 14. However, when you add the three numbers to find the mean, the sum will be much smaller if the lowest grade is a 14.
Outliers and Resistance
The mean and the median are so different in this example because there is one grade that is extremely different from the rest of the data. In statistics, we call such extreme values outliers. The mean is affected by the presence of an outlier; however, the median is not. A statistic that is not affected by outliers is called resistant. We say that the median is a resistant measure of center, and the mean is not resistant. In a sense, the median is able to resist the pull of a far away value, but the mean is drawn to such values. It cannot resist the influence of outlier values. As a result, when we have a data set that contains an outlier, it is often better to use the median to describe the center, rather than the mean.
#### Example D
In 2005, the CEO of Yahoo, Terry Semel, was paid almost \$231,000,000 (see http://www.forbes.com/static/execpay2005/rank.html). This is certainly not typical of what the average worker at Yahoo could expect to make. Instead of using the mean salary to describe how Yahoo pays its employees, it would be more appropriate to use the median salary of all the employees.
You will often see medians used to describe the typical value of houses in a given area, as the presence of a very few extremely large and expensive homes could make the mean appear misleadingly large.
On the Web
Java Applets helpful to understand the relationship between the mean and the median:
### Guided Practice
The mean of 6 people in a room is 35 years. A 40- year- old person comes in. What is now the mean age of the people in the room?
Solution:
We will start by using the definition of the mean:
\begin{align*}\overline{x}=\frac{\Sigma x}{n}.\end{align*}
Since we know the mean is 35, and that \begin{align*}n=6\end{align*}, so we can substitute these into the equation:
\begin{align*}35=\frac{\Sigma x}{6} \Rightarrow \Sigma x=6 \cdot 35=210. \end{align*}
When a new person of age 40 enters the room the total becomes 210 + 40 = 250. We find the average by dividing by 7. The average age is now 35.7 years.
### Explore More
1. In Lois’ 2nd grade class, all of the students are between 45 and 52 inches tall, except one boy, Lucas, who is 62 inches tall. Which of the following statements is true about the heights of all of the students?
1. The mean height and the median height are about the same.
2. The mean height is greater than the median height.
3. The mean height is less than the median height.
5. None of the above is true.
2. Enrique has a 91, 87, and 95 for his statistics grades for the first three quarters. His mean grade for the year must be a 93 in order for him to be exempt from taking the final exam. Assuming grades are rounded following valid mathematical procedures, what is the lowest whole number grade he can get for the \begin{align*}4^{\text{th}}\end{align*} quarter and still be exempt from taking the exam?
3. How many data points should be removed from each end of a sample of 300 values in order to calculate a 10% trimmed mean?
1. 5
2. 10
3. 15
4. 20
5. 30
4. In the last example, after removing the correct numbers and summing those remaining, what would you divide by to calculate the mean?
5. The chart below shows the data from the Galapagos tortoise preservation program with just the number of individual tortoises that were bred in captivity and reintroduced into their native habitat.
Island or Volcano Number of Individuals Repatriated
Wolf 40
Darwin 0
Alcedo 0
Sierra Negra 286
Cerro Azul 357
Santa Cruz 210
Española 1293
San Cristóbal 55
Santiago 498
Pinzón 552
Pinta 0
Figure: Approximate Distribution of Giant Galapagos Tortoises in 2004 (“Estado Actual De Las Poblaciones de Tortugas Terrestres Gigantes en las Islas Galápagos,” Marquez, Wiedenfeld, Snell, Fritts, MacFarland, Tapia, y Nanjoa, Scologia Aplicada, Vol. 3, Num. 1,2, pp. 98-11).
For this data, calculate each of the following:
(a) mode
(b) median
(c) mean
(d) a 10% trimmed mean
(e) midrange
(f) upper and lower quartiles
(g) the percentile for the number of Santiago tortoises reintroduced
1. In the previous question, why is the answer to (c) significantly higher than the answer to (b)?
2. The mean of 10 scores is 12.6. What is the sum of the scores?
3. While on vacation John drove an average of 262 miles per day for a period of 12 days. How far did John drive in total while he was on vacation?
4. Find x if 5, 9, 11, 12, 13, 14, 15 and x have a mean of 13.
5. Find a given that 3, 0, a, a, 4, a, 6, a, and 3 have a mean of 4.
6. A sample of 10 measurements has a mean of 15.6 and a sample of 20 measurements has a mean of 13.2. Find the mean of all 30 measurements.
7. The table below shows the results when 3 coins were tossed simultaneously 30 times. The number of tails appearing was recorded. Calculate the:
1. Mode
2. Median
3. Mean
Number of Tails Number of times occurred
3 4
2 12
1 11
0 3
Total 30
1. Compute the mean, the median and the mode for each of the following sets of numbers:
1. 3, 16, 3, 9, 5, 7, 11
2. 5, 3, 3, 7, 5, 5, 16, 9, 3, 18, 11, 5, 3, 7
3. 7, -4, 0, 12, 8, 121, -3
2. Find the mean and the median for each of the list of values:
1. 65, 69, 73, 77, 81, 87
2. 11, 7, 3, 8, 101
3. 31, 11, 41, 31
3. Find the mean and median for each of the following datasets:
1. 65, 66, 71, 75, 81, 85
2. 11, 7, 1, 7, 99
3. 31, 11, 41, 31
4. Explain why there is such a large difference between the median and the mean in the dataset of part b in the previous question
5. How do you determine which measure of center best describes a particular data set?
Technology Notes:
Calculating the Mean on the TI-83/84 Graphing Calculator
Step 1: Entering the data
On the home screen, press [2ND][{], and then enter the following data separated by commas. When you have entered all the data, press [2ND][}][STO][2ND][L1][ENTER]. You will see the screen on the left below:
1, 3, 4, 3, 1, 2, 2, 2, 1, 2, 2, 3, 4, 5, 1, 2, 3, 2, 1, 2, 3, 6
Step 2: Computing the mean
On the home screen, press [2ND][LIST] to enter the LIST menu, press the right arrow twice to go to the MATH menu (the middle screen above), and either arrow down and press [ENTER] or press [3] for the mean. Finally, press [2ND][L1][)] to insert L1 and press [ENTER] (see the screen on the right above).
Calculating Weighted Means on the TI-83/84 Graphing Calculator
Use the data of the number of children in a family. In list L1, enter the number of children, and in list L2, enter the frequencies, or weights.
The data should be entered as shown in the left screen below:
Press [2ND][STAT] to enter the LIST menu, press the right arrow twice to go to the MATH menu (the middle screen above), and either arrow down and press [ENTER] or press [3] for the mean. Finally, press [2ND][L1][,][2ND][L2][)][ENTER], and you will see the screen on the right above. Note that the mean is 2.5, as before.
### Vocabulary Language: English
arithmetic mean
arithmetic mean
The arithmetic mean is also called the average.
descriptive statistics
descriptive statistics
In descriptive statistics, the goal is to describe the data that found in a sample or given in a problem.
inferential statistics
inferential statistics
With inferential statistics, your goal is use the data in a sample to draw conclusions about a larger population.
measure of central tendency
measure of central tendency
In statistics, a measure of central tendency of a data set is a central or typical value of the data set.
Median
Median
The median of a data set is the middle value of an organized data set.
Mode
Mode
The mode of a data set is the value or values with greatest frequency in the data set.
multimodal
multimodal
When a set of data has more than 2 values that occur with the same greatest frequency, the set is called multimodal .
Outlier
Outlier
In statistics, an outlier is a data value that is far from other data values.
Population Mean
Population Mean
The population mean is the mean of all of the members of an entire population.
resistant
resistant
A statistic that is not affected by outliers is called resistant.
Sample Mean
Sample Mean
A sample mean is the mean only of the members of a sample or subset of a population.
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# Difference between revisions of "2012 AMC 8 Problems/Problem 20"
## Problem
What is the correct ordering of the three numbers $\frac{5}{19}$, $\frac{7}{21}$, and $\frac{9}{23}$, in increasing order?
$\textbf{(A)}\hspace{.05in}\frac{9}{23}<\frac{7}{21}<\frac{5}{19}\quad\textbf{(B)}\hspace{.05in}\frac{5}{19}<\frac{7}{21}<\frac{9}{23}\quad\textbf{(C)}\hspace{.05in}\frac{9}{23}<\frac{5}{19}<\frac{7}{21}$
$\textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23}$
## Solution 1
The value of $\frac{7}{21}$ is $\frac{1}{3}$. Now we give all the fractions a common denominator.
$\frac{5}{19} \implies \frac{345}{1311}$
$\frac{1}{3} \implies \frac{437}{1311}$
$\frac{9}{23} \implies \frac{513}{1311}$
Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## Solution 2
Instead of finding the LCD, we can subtract each fraction from $1$ to get a common numerator. Thus,
$1-\dfrac{5}{19}=\dfrac{14}{19}$
$1-\dfrac{7}{21}=\dfrac{14}{21}$
$1-\dfrac{9}{23}=\dfrac{14}{23}$
All three fractions have common numerator $14$. Now it is obvious the order of the fractions. $\dfrac{14}{19}>\dfrac{14}{21}>\dfrac{14}{23}\implies\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}$. Therefore, our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## Solution 3
Change $7/21$ into $1/3$; $$\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}$$ $$\frac{5}{15}>\frac{5}{19}$$ $$\frac{7}{21}>\frac{5}{19}$$ And $$\frac{1}{3}\cdot\frac{9}{9}=\frac{9}{27}$$ $$\frac{9}{27}<\frac{9}{23}$$ $$\frac{7}{21}<\frac{9}{23}$$ Therefore, our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
## Solution 4
When $\frac{x}{y}<1$ and $z>0$, $\frac{x+z}{y+z}>\frac{x}{y}$. Hence, the answer is ${\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$. ~ ryjs
This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched.
## Solution 5
By dividing, we see that 5/19 ≈ 0.26, 7/21=0.33, and 9/23=0.39. When we put this in order, 0.26<0.33<0.39. So our answer is $\boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}}$.
2012 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
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# hw4solns - CMPS 101 Summer 2009 Homework Assignment 4...
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1 CMPS 101 Summer 2009 Homework Assignment 4 Solutions 1. (3 Points) Consider the function ) ( n T defined by the recurrence formula + < = 3 ) 3 / ( 2 3 1 6 ) ( n n n T n n T a. (1 Points) Use the iteration method to write a summation formula for ) ( n T . Solution: ) 3 / ( 2 ) ( n T n n T + = ) ) 3 / 3 / ( 2 3 / ( 2 n T n n + + = ) 3 / ( 2 3 / 2 2 2 n T n n + + = ) 3 / ( 2 3 / 2 3 / 2 3 3 2 2 n T n n n + + + = etc.. After substituting the recurrence into itself k times, we get ) 3 / ( 2 3 2 ) ( 1 0 k k k i i i n T n n T + = - = . This process terminates when the recursion depth k is chosen so that 3 3 / 1 < k n , which is equivalent to 3 3 / 1 < k n , whence 1 3 3 + < k k n , so 1 ) ( log 3 + < k n k , and hence ) ( log 3 n k = . With this value of k we have 6 ) 2 or 1 ( ) 3 / ( = = T n T k . Therefore ) ( log 1 ) ( log 0 3 3 2 6 3 2 ) ( n n i i i n n T + = - = . b. (1 Points) Use the summation in (a) to show that ) ( ) ( n O n T = Solution: Using the above summation, we have ) ( log 1 ) ( log 0 3 3 2 6 ) 3 / 2 ( ) ( n n i i n n T + - = since x x for any x ) 2 ( log 0 3 6 ) 3 / 2 ( n n i i + = adding -many positive terms ) 2 ( log 3 6 ) 3 / 2 ( 1 1 n n + - = by a well known formula ) ( 6 3 ) 2 ( log 3 n O n n = + = ) ( 1 ) 2 ( log 3 2 ) 2 ( log 3 3 n o n = < < Therefore ) ( ) ( n O n T = .
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2 c. (1 Points) Use the Master Theorem to show that ) ( ) ( n n T Θ = Solution: Let 0 ) 2 ( log 1 3 > - = ε . Then 1 ) 2 ( log 3 = + ε , and ) ( ) 2 ( log ) 2 ( log 3 3 ε ε + + Ω = = n n n . Also for any c in the range 1 3 / 2 < c , and any positive n , we have cn n n = ) 3 / 2 ( ) 3 / ( 2 , so the regularity condition holds. By case (3) of the Master Theorem ) ( ) ( n n T Θ = .
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Video tutorial
A lot of people have misconception regarding Vedic maths i.e it is trick maths, but in actual if you look over these instruction video then you will understand that Vedic maths is full of methods which can let you to use formulas to get answer more accurate and in just click of seconds. And these formulas are generic so can be applied over any kind of number.
Step 1: Multiplication With 11
To multiply any 2 digit number with 11
Step 1: Split each digit with a space
Step 3. If in case adding digit cause answer greater then 9 then carry one to first digit.
Step 2: Vedic Multiplication for Number Near to Power of 10
Note: Power of 10 refers to number 10, 100, 1000, 10000, etc.
There are 2 conditions:
1. When number are less then power of 10 number, in this case the difference to be used should be in same digit no. format as number is, eg. 99 is less by 01, 997 is less by 003, etc.
2. When number are greater then power of 10 number, in this case the difference to be used should be in one digit less format as number is, eg. 101 is greater by 01, 11 is greater by 1, 1002 is greater by 002, etc.
Step 3: Vedic Squaring Number Near to Power of 10
Note: Power of 10 refers to number 10, 100, 1000, 10000, etc.
With this you will find squaring of number is so easy for number near to 10 power number, like square of 999, 9999, 99999, etc.
There are 2 conditions:
1. When number are less then power of 10 number, in this case the difference to be used should be in same digit no. format as number is, eg. 99 is less by 01, 997 is less by 003, etc.
2. When number are greater then power of 10 number, in this case the difference to be used should be in one digit less format as number is, eg. 101 is greater by 01, 11 is greater by 1, 1002 is greater by 002, etc.
Step 4: Generic Method for Multiplication (Left to Right)
Generic method works on all numbers. it is not specific to any number condition:
Step 5: Generic Method for Squaring
Generic method for squaring is one of the best method to square any number, this methods applies to all numbers.
<p>Nice tutorial. Thanks for sharing.</p>
<p>You are welcome sir. I would like if you can take a look over a video. You may find interesting :)</p>
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Question 98
# A chocolate dealer has to send chocolates of three brands to a shopkeeper. All the brands are packed in boxes of same size. The number of boxes to be sent is 96 of brand A, 240 of brand B and 336 of brand C. These boxes are to be packed in cartons of same size containing equal number of boxes. Each carton should contain boxes of same brand of chocolates. What could be the minimum number of cartons that the dealer has to send?
Solution
Since each carton should contain boxes of the same brand of chocolates and all boxes being of equal size, to get the minimum number of cartons, we should have the maximum number of boxes in each carton.
Thus, the number of boxes in each carton = H.C.F (96,240,336)
= 48
So, we will get minimum number of cartons if there are 48 boxes in each carton.
$$\therefore$$ Number of cartons = $$\frac{96}{48} + \frac{240}{48} + \frac{336}{48}$$
= $$2 + 5 + 7 = 14$$
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# 2.4 The cross product (Page 6/16)
Page 6 / 16
Note that, as the name indicates, the triple scalar product produces a scalar. The volume formula just presented uses the absolute value of a scalar quantity.
## Proof
The area of the base of the parallelepiped is given by $‖\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}‖.$ The height of the figure is given by $‖{\text{proj}}_{\text{v×w}}\text{u}‖.$ The volume of the parallelepiped is the product of the height and the area of the base, so we have
$\begin{array}{cc}\hfill V& =‖{\text{proj}}_{\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}}\text{u}‖‖\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}‖\hfill \\ & =|\frac{\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)}{‖\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}‖}|‖\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}‖\hfill \\ & =|\text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)|.\hfill \end{array}$
## Calculating the volume of a parallelepiped
Let $\text{u}=⟨-1,-2,1⟩,\text{v}=⟨4,3,2⟩,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}=⟨0,-5,-2⟩.$ Find the volume of the parallelepiped with adjacent edges $\text{u},\text{v},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}$ ( [link] ).
We have
$\begin{array}{cc}\hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =|\begin{array}{ccc}\hfill -1& \hfill -2& \hfill 1\\ \hfill 4& \hfill 3& \hfill 2\\ \hfill 0& \hfill -5& \hfill -2\end{array}|=\left(-1\right)|\begin{array}{cc}\hfill 3& \hfill 2\\ \hfill -5& \hfill -2\end{array}|+2|\begin{array}{cc}\hfill 4& \hfill 2\\ \hfill 0& \hfill -2\end{array}|+|\begin{array}{cc}\hfill 4& \hfill 3\\ \hfill 0& \hfill -5\end{array}|\hfill \\ & =\left(-1\right)\left(-6+10\right)+2\left(-8-0\right)+\left(-20-0\right)\hfill \\ & =-4-16-20\hfill \\ & =-40.\hfill \end{array}$
Thus, the volume of the parallelepiped is $|-40|=40$ units 3 .
Find the volume of the parallelepiped formed by the vectors $\text{a}=3\text{i}+4\text{j}-\text{k},$ $\text{b}=2\text{i}-\text{j}-\text{k},$ and $\text{c}=3\text{j}+\text{k}.$
$8$ units 3
## Applications of the cross product
The cross product appears in many practical applications in mathematics, physics, and engineering. Let’s examine some of these applications here, including the idea of torque, with which we began this section. Other applications show up in later chapters, particularly in our study of vector fields such as gravitational and electromagnetic fields ( Introduction to Vector Calculus ).
## Using the triple scalar product
Use the triple scalar product to show that vectors $\text{u}=⟨2,0,5⟩,\text{v}=⟨2,2,4⟩,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}=⟨1,-1,3⟩$ are coplanar—that is, show that these vectors lie in the same plane.
Start by calculating the triple scalar product to find the volume of the parallelepiped defined by $\text{u},\text{v},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}\text{:}$
$\begin{array}{cc}\hfill \text{u}·\left(\text{v}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{w}\right)& =|\begin{array}{ccc}\hfill 2& \hfill 0& \hfill 5\\ \hfill 2& \hfill 2& \hfill 4\\ \hfill 1& \hfill -1& \hfill 3\end{array}|\hfill \\ & =\left[2\left(2\right)\left(3\right)+\left(0\right)\left(4\right)\left(1\right)+5\left(2\right)\left(-1\right)\right]-\left[5\left(2\right)\left(1\right)+\left(2\right)\left(4\right)\left(-1\right)+\left(0\right)\left(2\right)\left(3\right)\right]\hfill \\ & =2-2\hfill \\ & =0.\hfill \end{array}$
The volume of the parallelepiped is $0$ units 3 , so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.
Are the vectors $\text{a}=\text{i}+\text{j}-\text{k},$ $\text{b}=\text{i}-\text{j}+\text{k},$ and $\text{c}=\text{i}+\text{j}+\text{k}$ coplanar?
No, the triple scalar product is $-4\ne 0,$ so the three vectors form the adjacent edges of a parallelepiped. They are not coplanar.
## Finding an orthogonal vector
Only a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points $P=\left(9,-3,-2\right),Q=\left(1,3,0\right),$ and $R=\left(-2,5,0\right).$
The plane must contain vectors $\stackrel{\to }{PQ}$ and $\stackrel{\to }{QR}\text{:}$
$\begin{array}{c}\stackrel{\to }{PQ}=⟨1-9,3-\left(-3\right),0-\left(-2\right)⟩=⟨-8,6,2⟩\hfill \\ \stackrel{\to }{QR}=⟨-2-1,5-3,0-0⟩=⟨-3,2,0⟩.\hfill \end{array}$
The cross product $\stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{QR}$ produces a vector orthogonal to both $\stackrel{\to }{PQ}$ and $\stackrel{\to }{QR}.$ Therefore, the cross product is orthogonal to the plane that contains these two vectors:
$\begin{array}{cc}\hfill \stackrel{\to }{PQ}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\stackrel{\to }{QR}& =|\begin{array}{ccc}\hfill \text{i}& \hfill \text{j}& \hfill \text{k}\\ \hfill -8& \hfill 6& \hfill 2\\ \hfill -3& \hfill 2& \hfill 0\end{array}|\hfill \\ & =0\text{i}-6\text{j}-16\text{k}-\left(-18\text{k}+4\text{i}+0\text{j}\right)\hfill \\ & =-4\text{i}-6\text{j}+2\text{k}.\hfill \end{array}$
We have seen how to use the triple scalar product and how to find a vector orthogonal to a plane. Now we apply the cross product to real-world situations.
Sometimes a force causes an object to rotate. For example, turning a screwdriver or a wrench creates this kind of rotational effect, called torque.
## Definition
Torque , $\tau$ (the Greek letter tau ), measures the tendency of a force to produce rotation about an axis of rotation. Let $\text{r}$ be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where the force is applied, and let vector $\text{F}$ represent the force. Then torque is equal to the cross product of $\text{r}$ and $\text{F}\text{:}$
$\tau =\text{r}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{F}.$
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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Let's explore
We have already examined the case where a = 1, and b = 1. This graphs the unit circle.
Let's begin with an examination of the effect a has on the graph.
First, let's consider the case where a = 2, and b = 1.
By changing the value of a to 2, the unit circle was stretched along the horizontal axis, creating an ellipse.
A value of a between 1 and 0, squashes the circle along the horizontal axis, creating an ellipse. Here is the case where a = 0.5, and b =1.
Now, let's look at the effect of b on the graph.
Here is the case where a = 1 and b = 2.
Again, the result is an ellipse. This time, b stretches the ellipse along the vertical axis.
A value of b between 1 and 0 will squash the curve along the vertical axis.
Here is the case where a = 1, and b = 0.5
Different values of a and b can be combined to create different ellipses that have been stretched or squashed along both the vertical and horizontal axies.
Here is the graph where a = 3, and b = 0.5
We have examined positive values of a and b. What do negative values of a and b do to the graph?
Here is the case where a = -3, and b = 1. The graph is identical to the case where a = 3, and b = 1.
I believe that the graph was reflected in the vertical axis, but because the axis was also the axis of symmetry, the graph rests on top of itself.
Now, we need to examine the case where a or b is zero.
This is the case where a = 1, and b = 0.
Notice that it is a segment that extends to 1 and -1 along the horizontal axis. A value of 0 for b, resutls in a segment along the horizontal axis.
This is the case where a = 0, and b = 1.
Notice that the segment extends to 1 and -1 along the vertical axis. A value of 0 for a results in a segment along the vertical axis.
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# HOW TO SHOW THE GIVEN VECTOR FORM A RIGHT TRIANGLE
## About "How to Show the Given Vector Form a Right Triangle"
How to Show the Given Vector Form a Right Triangle :
Here we are going to see how to show the given vector form a right triangle.
## How to Show the Given Vector Form a Right Triangle - Questions
Right triangle means, one of the angles must be 90 degrees. We may prove this using Pythagorean theorem. That is,
The length of square of longer side must be equal to the sum of the squares of lengths of the other two sides.
If we are given position vectors, we have to find the sides of the triangle (AB, BC and CA)
Here AB vector = OB vector - OA vector
BC vector = OC vector - OB vector
CA vector = OA vector - OC vector
After finding side lengths, we may apply Pythagorean theorem
Question 1 :
Show that the vectors 2i − j + k, 3i − 4j − 4k, i − 3j − 5k form a right angled triangle.
Solution :
Here we have sides of the triangle ABC,
AB vector = 2i vector − j vector + k vector
BC vector = 3i vector − 4j vector - 4k vector
CA vector = i vector − 3j vector - 5k vector
|AB vector| = √(22 + (-1)2 + 12) = √6
|BC vector| = √(32 + (-4)2 + (-4)2 = √(9+16+16) = √41
|CA vector| = √(12 + (-3)2 + (-5)2 = √(1+9+25) = √35
|BC|2 = |AB|2 + |CA|2
(√41)2 = (√6)2 + (√35)2
41 = 6 + 35
41 = 41
It satisfies the condition.
Hence given vectors form a right triangle.
Question 2 :
Find the value of λ for which the vectors a = 3i + 2j + 9k and b = i + λj + 3k are parallel.
Solution :
Since they are equal,
a vector = m b vector (here m is a constant)
3i + 2j + 9k = i + λj + 3k
3(i + (2/3)j + 3k) = i + λj + 3k
By equating the coefficients, we get λ = 2/3
Hence the value of λ is 2/3.
After having gone through the stuff given above, we hope that the students would have understood, "How to Show the Given Vector Form a Right Triangle".
Apart from the stuff given in "How to Show the Given Vector Form a Right Triangle" if you need any other stuff in math, please use our google custom search here.
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# G8-3-Solving Right Triangles
Document Sample
``` 8-3 Solving Right Triangles
8-3 Solving Right Triangles
Holt Geometry
Holt Geometry
8-3 Solving Right Triangles
Warm Up
Use ∆ABC for Exercises 1–3.
1. If a = 8 and b = 5, find c.
2. If a = 60 and c = 61, find b.
3. If b = 6 and c = 10, find sin B.
Find AB.
4. A(8, 10), B(3, 0)
5. A(1, –2), B(2, 6)
Holt Geometry
8-3 Solving Right Triangles
Objective
Use trigonometric ratios to find angle
measures in right triangles and to solve
real-world problems.
Holt Geometry
8-3 Solving Right Triangles
San Francisco, California, is
famous for its steep streets.
The steepness of a road is
often expressed as a percent
steepest street in San
Francisco, has a 31.5%
rises 31.5 ft over a
horizontal distance of 100 ft,
which is equivalent to a
17.5° angle. You can use
trigonometric ratios to
change a percent grade to an
angle measure.
Holt Geometry
8-3 Solving Right Triangles
Example 1: Identifying Angles from Trigonometric
Ratios
Use the trigonometric
ratio to
determine which angle
of the triangle is A.
Cosine is the ratio of the adjacent
leg to the hypotenuse.
The leg adjacent to 1 is 1.4. The
hypotenuse is 5.
The leg adjacent to 2 is 4.8. The
hypotenuse is 5.
Since cos A = cos2, 2 is A.
Holt Geometry
8-3 Solving Right Triangles
Check It Out! Example 1a
Use the given trigonometric
ratio to determine which
angle of the triangle is A.
Holt Geometry
8-3 Solving Right Triangles
In Lesson 8-2, you learned that sin 30° = 0.5.
Conversely, if you know that the sine of an acute
angle is 0.5, you can conclude that the angle
measures 30°. This is written as sin-1(0.5) = 30°.
Holt Geometry
8-3 Solving Right Triangles
If you know the sine, cosine, or tangent of an acute
angle measure, you can use the inverse
trigonometric functions to find the measure of the
angle.
Holt Geometry
8-3 Solving Right Triangles
Example 2: Calculating Angle Measures from
Trigonometric Ratios
Use your calculator to find each angle measure
to the nearest degree.
A. cos-1(0.87) B. sin-1(0.85) C. tan-1(0.71)
cos-1(0.87) 30° sin-1(0.85) 58° tan-1(0.71) 35°
Holt Geometry
8-3 Solving Right Triangles
Check It Out! Example 2
Use your calculator to find each angle measure
to the nearest degree.
a. tan-1(0.75) b. cos-1(0.05) c. sin-1(0.67)
Holt Geometry
8-3 Solving Right Triangles
Using given measures to find the unknown angle
measures or side lengths of a triangle is known as
solving a triangle. To solve a right triangle, you need
to know two side lengths or one side length and an
acute angle measure.
Holt Geometry
8-3 Solving Right Triangles
Example 3: Solving Right Triangles
Find the unknown measures.
Round lengths to the nearest
hundredth and angle measures to
the nearest degree.
Method 1: By the Pythagorean Theorem,
RT2 = RS2 + ST2
(5.7)2 = 52 + ST2
Since the acute angles of a right triangle are
complementary, mT 90° – 29° 61°.
Holt Geometry
8-3 Solving Right Triangles
Check It Out! Example 3
Find the unknown
measures. Round
lengths to the
nearest
hundredth and
angle measures to
the nearest
degree.
Holt Geometry
8-3 Solving Right Triangles
Example 4: Solving a Right Triangle in the Coordinate
Plane
The coordinates of the vertices of ∆PQR are
P(–3, 3), Q(2, 3), and R(–3, –4). Find the side
lengths to the nearest hundredth and the
angle measures to the nearest degree.
Holt Geometry
8-3 Solving Right Triangles
Example 4 Continued
Step 1 Find the side lengths. Plot points P, Q, and R.
PR = 7 PQ = 5
Y
By the Distance Formula,
P Q
X
R
Holt Geometry
8-3 Solving Right Triangles
Example 4 Continued
Step 2 Find the angle measures.
Y
mP = 90°
P Q
X
The acute s of a rt. ∆ are comp.
R
mR 90° – 54° 36°
Holt Geometry
8-3 Solving Right Triangles
Check It Out! Example 5
Baldwin St. in Dunedin,
New Zealand, is the
steepest street in the
world. It has a grade of
38%. To the nearest
degree, what angle does
Baldwin St. make with a
horizontal line?
Holt Geometry
8-3 Solving Right Triangles
Change the percent
for every 100 ft of horizontal distance.
C
38 ft Draw a right triangle to
100 ft
A is the angle the road
makes with a horizontal line.
Holt Geometry
8-3 Solving Right Triangles
Lesson Quiz: Part I
find each angle measure
to the nearest degree.
1. cos-1 (0.97)
2. tan-1 (2)
3. sin-1 (0.59)
Holt Geometry
8-3 Solving Right Triangles
Lesson Quiz: Part II
Find the unknown measures. Round lengths
to the nearest hundredth and angle
measures to the nearest degree.
4. 5.
Holt Geometry
8-3 Solving Right Triangles
Lesson Quiz: Part III
6. The coordinates of the vertices of ∆MNP are
M (–3, –2), N(–3, 5), and P(6, 5). Find the
side lengths to the nearest hundredth and the
angle measures to the nearest degree.
Holt Geometry
```
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Spirals
To the Main Page "Mathematische Basteleien"
What is a spiral?
A spiral is a curve in the plane or in the space, which runs around a centre in a special way.
Different spirals follow. Most of them are produced by formulas.
Spirals by Polar Equations top
Archimedean Spiral top
You can make a spiral by two motions of a point: There is a uniform motion in a fixed direction and a motion in a circle with constant speed. Both motions start at the same point.
...........................
(1) The uniform motion on the left moves a point to the right. - There are nine snapshots.
(2) The motion with a constant angular velocity moves the point on a spiral at the same time. - There is a point every 8th turn.
(3) A spiral as a curve comes, if you draw the point at every turn.
You get formulas analogic to the circle equations.
Circle
...... Let P be a point of a circle with the radius R, which is given by an equation in the centre position. There are three essential descriptions of the circle: (1) Central equation: x²+y² = R² or [y = sqr(R²-x²) und y = -sqr(R²-x²)], (2) Parameter form: x(t) = R cos(t), y(t) = R sin(t), (3) Polar equation: r(t) = R.
You give a point by a pair (radius OP, angle t) in the (simple) polar equation. The radius is the distance of the point from the origin (0|0). The angle lies between the radius and the positive x-axis, its vertex in the origin.
Spiral
The radius r(t) and the angle t are proportional for the simpliest spiral, the spiral of Archimedes. Therefore the equation is:
(3) Polar equation: r(t) = at [a is constant].
From this follows
(2) Parameter form: x(t) = at cos(t), y(t) = at sin(t),
(1) Central equation: x²+y² = a²[arc tan (y/x)]².
...... The Archimedean spiral starts in the origin and makes a curve with three rounds. The distances between the spiral branches are the same. More exact: The distances of intersection points along a line through the origin are the same.
...... If you reflect an Archimedean spiral on a straight line, you get a new spiral with the opposite direction. Both spirals go outwards. If you look at the spirals, the left one forms a curve going to the left, the right one forms a curve going to the right.
If you connect both spirals by a straight (red) or a bowed curve, a double spiral develops.
Equiangular Spiral (Logarithmic Spiral, Bernoulli's Spiral) top
...... (1) Polar equation: r(t) = exp(t). (2) Parameter form: x(t) = exp(t) cos(t), y(t) = exp(t) sin(t). (3) Central equation: y = x tan[ln(sqr(x²+y²))]. The logarithmic spiral also goes outwards. The spiral has a characteristic feature: Each line starting in the origin (red) cuts the spiral with the same angle.
More Spirals top
If you replace the term r(t)=at of the Archimedean spiral by other terms, you get a number of new spirals. There are six spirals, which you can describe with the functions f(x)=x^a [a=2,1/2,-1/2,-1] and f(x)=exp(x), f(x)=ln(x). You distinguish two groups depending on how the parameter t grows from 0.
............ If the absolute modulus of a function r(t) is increasing, the spirals run from inside to outside and go above all limits. The spiral 1 is called parabolic spiral or Fermat's spiral.
....... If the absolute modulus of a function r(t) is decreasing, the spirals run from outside to inside. They generally run to the centre, but they don't reach it. There is a pole. Spiral 2 is called the Lituus (crooked staff).
I chose equations for the different spiral formulas suitable for plotting.
Clothoide (Cornu Spiral)top
....... The clothoid or double spiral is a curve, whose curvature grows with the distance from the origin. The radius of curvature is opposite proportional to its arc measured from the origin. The parameter form consists of two equations with Fresnel's integrals, which can only be solved approximately. You use the Cornu spiral to describe the energy distribution of Fresnel's diffraction at a single slit in the wave theory.
Half circle spirals
...... You can add half circles growing step by step to get spirals. The radii have the ratios 1 : 1.5 : 2 : 2.5 : 3.....
Fibonacci Spiral
...... Draw two small squares on top of each other. Add a sequence of growing squares counter clockwise. Draw quarter circles inside the squares (black). They form the Fibonacci Spiral.
The Fibonacci spiral is called after its numbers. If you take the length of the square sides in the order, you get the sequence 1,1,2,3,5,8,13,21, ... These are the Fibonacci numbers, which you can find by the recursive formula a(n)=a(n-1)+a(n-2) with [a(1)=1, a(2)=1, n>2].
Spirals Made of Line Segments top
...... The spiral is made by line segments with the lengths 1,1,2,2,3,3,4,4,.... Lines meet one another at right angles.
...... Draw a spiral in a crossing with four intersecting straight lines, which form 45° angles. Start with the horizontal line 1 and bend the next line perpendicularly to the straight line. The line segments form a geometric sequence with the common ratio sqr(2). If you draw a spiral into a straight line bundle, you approach the logarithmic spiral, if the angles become smaller and smaller.
...... The next spiral is formed by a chain of right angled triangles, which have a common side. The hypotenuse of one triangle becomes the leg of the next. First link is a 1-1-sqr(2)-triangle. The free legs form the spiral. It is special that the triangles touch in line segments. Their lengths are the roots of the natural numbers. You can proof this with the Pythagorean theorem. This figure is called root spiral or root snail or wheel of Theodorus.
...... Squares are turned around their centre with 10° and compressed at the same time, so that their corners stay at the sides of their preceding square. Result: The corners form four spiral arms. The spiral is similar to the logarithm spiral, if the angles get smaller and smaller. You can also turn other regular polygons e.g. an equilateral triangle. You get similar figures.
This picture reminds me of the programming language LOGO of the early days of computing (C64-nostalgia).
Three-dimensional Spirals top
Helix
...... If you draw a circle with x=cos(t) and y=sin(t) and pull it evenly in z-direction, you get a spatial spiral called cylindrical spiral or helix.
The picture pair makes a 3D view possible.
...... Reflect the 3D-spiral on a vertical plane. You get a new spiral (red) with the opposite direction. If you hold your right hand around the right spiral and if your thumb points in direction of the spiral axis, the spiral runs clockwise upward. It is right circular. You must use your left hand for the left spiral. It is left circular. The rotation is counter clockwise. Example: Nearly all screws have a clockwise rotation, because most of the people are right-handed.
...... In the "technical" literature the right circular spiral is explained as follows: You wind a right- angled triangle around a cylinder. A clockwise rotating spiral develops, if the triangle increases to the right.
Conical Helix top
You can make the conical helix with the Archimedean spiral or equiangular spiral.
The picture pairs make 3D views possible.
Loxodrome, Spherical Helix
...... The loxodrome is a curve on the sphere, which cuts the meridians at a constant angle. They appear on the Mercator projection as straight lines. The parametric representation is x=cos(t) cos [tan-1(at)] y=sin(t) cos[tan-1(at)] z= -sin [tan-1(at)] (a is constant) You can find out x²+y²+z²=1. This equation means that the loxodrome is lying on the sphere.
Generally there is a loxodrome at every solid made by rotation about an axis.
Making of Spirals top
...... A strip of paper becomes a spiral, if you pull the strip between the thumb and the edge of a knife, pressing hard. The spiral becomes a curl where gravity is present.
You use this effect to decorate the ends of synthetic materials, such as the narrow colourful strips or ribbons used in gift-wrapping.
I suppose that you have to explain this effect in the same way as a bimetallic bar. You create a bimetallic bar by glueing together two strips, each made of a different metal. Once this bimetallic bar is heated, one metal strip expands more than the other causing the bar to bend.
The reason that the strip of paper bends is not so much to do with the difference in temperature between the top and bottom side. The knife changes the structure of the surface of the paper. This side becomes 'shorter'.
Incidentally, a strip of paper will bend slightly if you hold it in the heat of a candle flame.
...... Forming curls reminds me of an old children's game: Take a dandelion flower and cut the stem into two or four strips, keeping the head intact. If you place the flower into some water, so that the head floats on the surface, the strips of the stem will curl up. (Mind the spots.) A possible explanation: Perhaps the different absorption of water on each side of the strips causes them to curl up.
Mandelbrot Set Spirals top
The coordinates belong to the centre of the pictures.
You also find nice spirals as Julia Sets. Here is an example:
You find more about these graphics on my page Mandelbrot Set.
You find nice spirals as a decoration of barred windows, fences, gates or doors. You can see them everywhere, if you are look around.
...... I found spirals worth to show at New Ulm, Minnesota, USA. Americans with German ancestry built a copy of the Herman monument near Detmold/Germany in about 1900. Iron railings with many spirals decorate the stairs (photo). More about the American and German Herman on Wikipedia-pages (URL below)
Costume jewelleries also take spirals as motive.
...... Annette's spiral
Spirals, Spirals, Spirals top
Ammonites, antlers of wild sheep, Archimedes' water spiral, area of high or low pressure, arrangement of the sunflower cores, @, bimetal thermometer, bishop staff, Brittany sign, circles of a sea-eagle, climbs, clockwise rotating lactic acid, clouds of smoke, coil, coil spring, corkscrew, creepers (plants), curl, depression in meteorology, disc of Festós, double filament of the bulb, double helix of the DNA, double spiral, electron rays in the magnetic longitudinal field, electrons in cyclotron, Exner spiral, finger mark, fir cone, glider ascending, groove of a record, head of the music instrument violin, heating wire inside a hotplate, heat spiral, herb spiral, inflation spiral, intestine of a tadpole, knowledge spiral, licorice snail, life spiral, Lorenz attractor, minaret at Samarra (Iraq), music instrument horn, pendulum body of the Galilei pendulum, relief strip of the Trajan's column at Rome or the Bernward column at Hildesheim, poppy snail, road of a cone mountain, role (wire, thread, cable, hose, tape measure, paper, bandage), screw threads, simple pendulum with friction, snake in resting position, snake of Aesculapius, snail of the interior ear, scrolls, screw alga, snail-shell, spider net, spiral exercise book, spiral nebula, spiral staircase (e.g. the two spiral stairs in the glass dome of the Reichstag in Berlin), Spirallala ;-), Spirelli noodles, Spirills (e.g. Cholera bacillus), springs of a mattress, suction trunk (lower jaw) of the cabbage white butterfly, tail of the sea-horse, taps of conifers, tongue and tail of the chamaeleon, traces on CD or DVD, treble clef, tusks of giants, viruses, volute, watch spring and balance spring of mechanical clocks, whirlpool, whirlwind.
Spirals on the Internet top
German
Jürgen Berkemeier
Fibonacci-Spiralen
Matheprisma
Bewegungsfunktionen (Spiralen 1 ) - (Spiralen online zeichnen)
Michael Komma
Fresnel-Beugung am Einzelspalt (Cornu-Spirale)
English
Ayhan Kursat ERBAS
Equiangular Spiral
Bob Allanson
This is a logarithmic spiral
David Eppstein (Geometry Junkyard)
Eric W. Weisstein (MathWorld)
Spirals
Archimedean SpiralCircle InvoluteConical SpiralCornu SpiralCurlicue FractalFermat's SpiralHelix, Hyperbolic SpiralLogarithmic SpiralMice ProblemNielsen's SpiralPolygonal SpiralPrime SpiralRational SpiralSeashell, Spherical Spiral
Hop David (Hop's Gallery)
Riemann sphere, Ram's Horn, Spiral Tile
Jan Wassenaar
spiral
Keith Devlin
The Double Helix
Mark Newbold
Counter-Rotating Spirals Illusion
Richard Parris (Freeware-Programs)
winplot
French
Robert FERRÉOL (COURBES 2D )
SPIRALE
COURBES 3D (SPHÉRO-CYLINDRIQUE, SPIRALE CONIQUE DE PAPPUS, SPIRALE CONIQUE DE PIRONDINI, SPIRALE SPHÉRIQUE)
References top
(1) Martin Gardener: Unsere gespiegelte Welt, Ullstein, Berlin, 1982 [ISBN 3-550-07709-2]
(2) Rainer und Patrick Gaitzsch: Computer-Lösungen für Schule und Studium, Band 2, Landsberg am Lech, 1985
(3) Jan Gullberg: Mathematics - From the Birth of Numbers, New York / London (1997) [ISBN 0-393-04002-X]
(4) Khristo N. Boyadzhiev: Spirals and Conchospirals in the Flight of Insects, The College Mathematics Journal,
Vol.30, No.1 (Jan.,1999) pp.23-31
(5) Jill Purce: the mystic spiral - Journey of the Soul, Thames and Hudson, 1972, reprinted 1992
Feedback: Email address on my main page
URL of my Homepage:
http://www.mathematische-basteleien.de/
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# How to Find Median: Explained Step-by-Step with Examples
Greetings, dear reader! Whether you’re a math student struggling with median calculations, or someone who simply wants to know how to find median, you’ve come to the right place. Median is a fundamental concept in statistics, used in various fields such as finance, healthcare, and social sciences. In this article, we’ll explain how to find median, step-by-step, with examples and illustrations. So, let’s get started! 🔍
Baca Cepat
## Introduction
Median is a measure of central tendency or average, which represents the middle value of a dataset. In other words, it divides the dataset into two equal parts, with half of the values above and half below the median. Median is different from mean (arithmetic average), which is the sum of all values divided by the number of values. Median is more robust than mean, as it is less affected by outliers or extreme values. Median is denoted by the symbol ‘Me’ or ‘Md’, and is used in many statistical analyses such as hypothesis testing, regression analysis, and data visualization.
There are different methods to calculate the median, depending on the type of data and the level of measurement. For example, if the data is numeric and continuous, we can use the formula:
### Median Formula for Even Number of Values
For Even Number of Values:
Step 1: Arrange the values in ascending order
Step 2: Find the two middle values
Step 3: Calculate the average of the two middle values
Step 4: The result is the median
For example, let’s say we have a dataset of 8 values: 5, 3, 7, 1, 9, 4, 6, 2. To find the median, we follow these steps:
Step 1: Arrange the values in ascending order: 1, 2, 3, 4, 5, 6, 7, 9.
Step 2: Find the two middle values: 4 and 5.
Step 3: Calculate the average of the two middle values: (4 + 5)/2 = 4.5.
Step 4: The result is the median: 4.5.
If the dataset has an odd number of values, we simply take the middle value as the median. For example, if we have a dataset of 7 values: 5, 3, 7, 1, 9, 4, 6, the median is 5.
## How to Find Median: Step-by-Step Guide
Now that we’ve explained the concept of median and the formula for calculating it, let’s dive deeper into the process of finding median. We’ll provide a step-by-step guide with examples and illustrations, so that you can follow along easily.
### Step 1: Arrange the Values in Ascending Order
The first step in finding median is to arrange the values in ascending order, from the smallest to the largest. This is important because we need to identify the middle values, and we cannot do that unless the values are in order. If the dataset has duplicate values, we can keep one copy and remove the others. For example, if we have a dataset of 6 values: 3, 1, 2, 3, 4, 5, we can remove one copy of 3 and get 5 distinct values: 1, 2, 3, 4, 5.
#### Example:
Let’s say we have a dataset of 10 values, representing the ages of 10 people in a group:
22, 25, 27, 29, 31, 32, 35, 37, 39, 40
To find the median, we first arrange the values in ascending order:
22, 25, 27, 29, 31, 32, 35, 37, 39, 40
### Step 2: Find the Middle Values
Once we have arranged the values in order, we can find the middle values. The number of middle values depends on whether the dataset has an odd or even number of values. If the dataset has an odd number of values, there is only one middle value. If the dataset has an even number of values, there are two middle values. To find the middle values, we can use the formula:
### Formula for Finding Middle Values
For Odd Number of Values: For Even Number of Values:
Step 1: Count the number of values Step 1: Count the number of values
Step 2: Divide the number of values by 2 Step 2: Divide the number of values by 2
Step 3: The result is the position of the middle value Step 3: The results are the positions of the two middle values
For example, if we have a dataset of 7 values, the middle value is the 4th value, as (7 + 1)/2 = 4. If we have a dataset of 8 values, the two middle values are the 4th and 5th values, as (8/2) = 4 and (8/2) + 1 = 5.
#### Example:
Let’s continue with our previous example of ages:
22, 25, 27, 29, 31, 32, 35, 37, 39, 40
To find the middle values, we first count the number of values:
n = 10
Since n is even, there are two middle values. We divide n by 2:
n/2 = 5
This means that the two middle values are the 5th and 6th values. We can verify this by looking back at the ordered dataset:
22, 25, 27, 29, 31, 32, 35, 37, 39, 40
Thus, the two middle values are 32 and 35.
### Step 3: Calculate the Median
Once we have identified the middle values, we can calculate the median. If the dataset has an odd number of values, the median is the middle value. If the dataset has an even number of values, the median is the average of the two middle values.
#### Example:
Let’s continue with our previous example of ages:
22, 25, 27, 29, 31, 32, 35, 37, 39, 40
Since there are two middle values (32 and 35), we take their average:
(32 + 35)/2 = 33.5
Therefore, the median age of the group is 33.5 years.
### 1. What is median?
Median is a measure of central tendency or average, which represents the middle value of a dataset. In other words, it divides the dataset into two equal parts, with half of the values above and half below the median.
### 2. How is median different from mean?
Median is different from mean (arithmetic average), which is the sum of all values divided by the number of values. Median is more robust than mean, as it is less affected by outliers or extreme values.
### 3. How do you find the median of a dataset?
To find the median of a dataset, you need to arrange the values in ascending order, find the middle values, and then calculate the median based on whether the dataset has an odd or even number of values.
### 4. What if the dataset has duplicate values?
If the dataset has duplicate values, you can keep one copy and remove the others. This will ensure that the dataset has distinct values, which is necessary for finding the median.
### 5. What if the dataset has missing values?
If the dataset has missing values, you can either remove them or impute them with some value, depending on the nature and purpose of your analysis.
### 6. What is the use of median in statistics?
Median is used in many statistical analyses such as hypothesis testing, regression analysis, and data visualization. It provides a robust measure of central tendency, which is less affected by outliers or extreme values.
### 7. Can median be negative?
Yes, median can be negative if the dataset contains negative values. However, it is important to interpret the median in the context of the data and the research question.
### 8. What if the dataset has an even number of values and the two middle values are not integers?
If the dataset has an even number of values and the two middle values are not integers, you can still take their average, which will be a decimal or a fraction. This is a valid method of calculating the median.
### 9. Can median and mean be the same?
Yes, median and mean can be the same if the dataset is symmetrical or normally distributed. However, this is not always the case, especially if the dataset has outliers or skewness.
### 10. What if the dataset has outliers?
If the dataset has outliers, the median is a better measure of central tendency than mean, as it is less affected by outliers. However, it is still important to investigate the outliers and their possible causes, as they may reveal important information about the data or the population.
### 11. Can median be used for nominal or ordinal data?
Yes, median can be used for nominal or ordinal data, although it is less informative than for numerical data. In this case, median represents the middle value or category of the data.
### 12. What if the dataset has a small sample size?
If the dataset has a small sample size (less than 30), the median may not be a reliable measure of central tendency, as it is based on a limited number of values. In this case, other measures such as mode or range may be more appropriate.
### 13. Can median be used for time-series data?
Yes, median can be used for time-series data, although it may not capture the trends or patterns in the data. In this case, other measures such as moving average or trend analysis may be more appropriate.
## Conclusion
In conclusion, finding median is an essential skill for anyone working with data or statistics. It provides a robust measure of central tendency, which is less affected by outliers or extreme values. In this article, we have explained how to find median, step-by-step, with examples and illustrations. We hope that this article has been helpful and informative for you. If you have any questions, feel free to contact us or leave a comment below. Happy calculating! 🧮
## Closing/Disclaimer
This article is intended for educational and informational purposes only. The information presented here is not intended to be a substitute for professional advice or judgment, and should not be relied upon as such. We make no representations or warranties of any kind, express or implied, about the completeness, accuracy, reliability, suitability or availability with respect to the information contained in this article or the website linked to it. Any reliance you place on such information is therefore strictly at your own risk.
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# Solution for Ratio in Which the Xy-plane Divides the Join of (1, 2, 3) and (4, 2, 1) is (A) 3 : 1 Internally (B) 3 : 1 Externally (C) 1 : 2 Internally (D) 2 : 1 Externally - CBSE (Science) Class 12 - Mathematics
ConceptDirection Cosines and Direction Ratios of a Line
#### Question
Ratio in which the xy-plane divides the join of (1, 2, 3) and (4, 2, 1) is
• 3 : 1 internally
• 3 : 1 externally
• 1 : 2 internally
• 2 : 1 externally
#### Solution
$\left( b \right) 3: 1 \text{ externally }$
$\text{ Suppose the XY - plane divides the line segment joining the points P } \left( 1, 2, 3 \right) \text{ and Q } \left( 4, 2, 1 \right) \text{ in the ratio k: 1 } .$
$\text{ Using the section formula, the coordinates of the point of intersection are given by }$
$\left( \frac{k\left( 4 \right) + 1}{k + 1}, \frac{k\left( 2 \right) + 2}{k + 1}, \frac{k\left( 1 \right) + 3}{k + 1} \right)$
$\text{ The Z - coordinate of any point on the XY - plane is zero }.$
$\Rightarrow \frac{k\left( 1 \right) + 3}{k + 1} = 0$
$\Rightarrow k + 3 = 0$
$\Rightarrow k = - 3 = - \frac{3}{1}$
$\text{ Thus, the XY - plane divides the line segment joining the given points in the ratio 3: 1 externally } .$
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [3]
Solution Ratio in Which the Xy-plane Divides the Join of (1, 2, 3) and (4, 2, 1) is (A) 3 : 1 Internally (B) 3 : 1 Externally (C) 1 : 2 Internally (D) 2 : 1 Externally Concept: Direction Cosines and Direction Ratios of a Line.
S
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# Multiset
The bags that can be used for shopping are at bag.
A multiset (sometimes called a bag) is a concept from mathematics. In many ways, multisets are like sets. Certain items are either elements of that multiset, or they are not. However, multisets are different from sets: The same type of item can be in the multiset more than once. For this reason, mathematicians have defined a relation (function) that tells, how many copies of a certain type of item there are in a certain multiset. They call this multiplicity. For example, in the multiset { a, a, b, b, b, c }, the multiplicities of the members a, b, and c are 2, 3, and 1, respectively. From a set of n elements, the number of r-element multisets is written as ${\displaystyle \textstyle \left(\!{\binom {n}{r}}\!\right)}$. This is sometimes called the multiset coefficient.[1][2][3]
A multiset is illustrated by means of a histogram.
A multiset can also be considered an unordered tuple:
• The tuples (a,b) and (b,a) are not equal, and the tuples (a,a) and (a) are not equal either.
• The multisets {a,b} and {b,a} are equal, but the multisets {a,a} and {a} are not equal.
• The sets {a,b} and {b,a} are equal, and the sets {a,a} and {a} are equal too.
## Examples
One of the simplest examples is the multiset of prime factors of a number n. Here, the underlying set of elements is the set of prime divisors of n. For example, the number 120 has the prime factorisation
${\displaystyle 120=2^{3}3^{1}5^{1}}$
which gives the multiset {2, 2, 2, 3, 5}.
Another is the multiset of solutions of an algebraic equation. A quadratic equation, for example, has two solutions. However, in some cases they are both the same number. Thus the multiset of solutions of the equation could be { 3, 5 }, or it could be { 4, 4 }. In the latter case, it has a solution of multiplicity 2.
## References
### CItations
1. "List of Probability and Statistics Symbols". Math Vault. 2020-04-26. Retrieved 2020-09-23.
2. "3.7: Counting Multisets". Mathematics LibreTexts. 2020-01-19. Retrieved 2020-09-23.
3. "Multiset | Brilliant Math & Science Wiki". brilliant.org. Retrieved 2020-09-23.
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### 1. Coordinate System
A coordinate system is a datum for expressing point positions. The system consists of two or three reference axes perpendicular to each other.
In a two dimensional system, one axis is oriented in the direction of a meridian defining North, the remaining axis being East, Figure A-1.
Figure A-1Two Dimensional System
In a three dimensional system, the third axis is oriented vertically, in the direction of elevation, Figure X-2.
Figure A-2 Three Dimensional System
This section will deal primarily with two dimensional systems although the math involved can be easily expanded into the third dimension.
### 2. Positions
#### a. Absolute: Coordinates
The absolute position of a point is expressed as coordinates which are perpendicular distances from the reference axes. In surveying and most mapping, the reference system uses North and East axes so position is expressed as (North, East) (aka, Northing and Easting), Figure A-3. Other applications, notably math, use X and Y axes; Y coinciding with the meridian. Positions are expressed as (X, Y), Figure A-4.
Figure A-3 (North, East) Coordinates Figure A-4 (X, Y) Coordinates
Caution must be exercised when using coordinates provided by others. Are (2000 ft,1000 ft) North-East or X-Y coordinates? A common error when importing coordinates in software is reversing their order: this flips and rotates the positions. This effect is demonstrated in Figures A-5 and A-6. The former uses coordinates in (N, E) order while the latter in (X, Y).
Figure A-5 Data Correctly Imported as (N, E) Coordinates Figure A-5 Data Incorrectly Imported as (X, Y) Coordinates
We will use the North-East coordinate form unless otherwise specified.
#### b. Relative
A relative position is where one point is with respect to another. This can be expressed as coordinate differences, Figure A-6, or as distance and direction, Figure A-7.
Figure A-6 Coordinate Differences Figure A-7 Distance and Direction
### 3. Point-to-point computations
Point-to-point computations involve just two points
#### a. Forward Computation
A Forward Computation is used to calculate a point's coordinates from another set of coordinates using distance and direction data between them, Figure A-8.
Equation A-1 Equation A-2 Figure A-8 Forward computation
Equations A-1 and A-2 are a combination of the latitude and departure and coordinate equations from the Traverse Computations chapter:
• North Lat is (+), South Lat is (–)
• East Dep is (+), West Dep is (–)
DirAB can be either a bearing or azimuth:
• If a bearing (0° to 90°) is used you must manually determine the correct algebraic sign for the Lat and Dep based on quadrant.
• Using azimuths (0° to 360°) automatically results in correct signs.
#### b. Inverse Computation
An Inverse Computation determines the distance and direction between two coordinate pairs, Figure A-9.
Figure A-9 Inverse computation Equation A-3 Equation A-4 Equation A-5 Equation A-6
Coordinate differences, Δ’s, are the to point minus the from point: going from point C to point D means subtracting C's coordinates from D's.
The algebraic sign on β and the resulting direction depend on the quadrant of the line. A positive angle is clockwise (to the right); a negative angle is counterclockwise (to the left). Both are from the north or south end of the meridian, Figure A-10.
Table A-1 Algebraic sign Direction Quadrant ΔN ΔE β Bearing Azimuth NE + + + N β E β SE - + - S |β| E 180°+β SW - - + S β W 180°+β NW + - - N |β| W 360°+β
When ΔN = 0, Equation A-7 has no solution. Technically division by 0 is undefined, but actually the result of any number divided by 0 is infinity. So what does this mean? In surveying terms when ΔN = 0 the entire line length is ΔE resulting in a due East or West line, Figure A-11.
when ΔE is (+), Az = 90° when ΔE is (-), Az = 270° Figure A-11 When ΔN = 0
If you check the tangent of 90° and 270° on your calculator you'll get either an error or "undefined" response. Try tan(89.99999°); you'll get a huge number. Tan(90°) = tan(270°) = infinity.
#### c. Examples
##### (1) Forward 1
Compute the coordinates of point R given the information in Figure A-12:
Figure A-12 Forward Example 1
Because the bearing is South and East, the Lat is negative and Dep positive. From equations A-1 and A-2:
##### (2) Forward 2
Compute the coordinates of point R given the information in Figure A-13:
Figure A-13 Forward Example 2
Because the direction is an azimuth, Equations A-1 and A-2 will automatically compute the correct signs for the Lat and Dep.
##### (3) Inverse
What are the lengtha and azimuth from point J to K?
Point North (ft) East (ft) J 1153.65 704.08 K 988.85 200.75
Draw a sketch to visualize the line, Figure A-14.
Figure A-14 Inverse Example
Substitute the coordinates into Equations A-3 and A-4 (remember, it's to minus from):
Use Equation A-5 to compute the length:
Compute β, the angle from the meridian:
Because ΔN is negative and ΔE is negative, the direction is in the South-West quadrant, so add 180° to β.
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