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# Rules to Subtract Integers
Integers are positive and negative number including zero and does not include the fractional part. We will discuss about the rules for subtracting of integers. There are certain rules which will help us in better understanding of subtraction of integers. This topic will discuss how integers are subtracted when:
• One positive integer is subtracted from another positive integer.
• One negative integer is subtracted from another negative integer.
• One positive integer is subtracted from another negative integer.
• One negative integer is subtracted from another positive integer.
Rules: When two integers’ p and q are subtracted, then for subtracting q from p the sign of q is changed and after that it is added it to p.
Note: For calculation, and for carrying out subtraction:
• (-) × (+) = (-)
• (-) × (-) = (+)
• (+) × (+) = (+)
• (+) × (-) = (-)
When two signs are written side by side then it is considered that the two signs are multiplied.
Few examples are solved here using the rules of subtraction of integers mentioned above:
1. Subtract 10 from 5
5 – (+10)
Þ 5 – 10 = - 5
Here according to the rule negative sign and positive sign [i.e. (-) × (+) = (-)] makes negative sign. That is why – (+10) will be – 10. So it will be 5 – 10 = -5. Actually the absolute values are subtracted and in the result the value with a higher absolute value is considered.
2. Subtract -11 from 16
16 – (-11)
Þ 16 + 11 = 27
Here according to the rule negative sign and negative sign [i.e. (-) × (-) = (+)] makes positive sign. That is why – (-11) = +11. So it will be 16 + 11 = 27. Here both 16 and 11 is added to get 27.
3. Subtract 9 from -6
= - 6 – (+9)
= - 6 – 9
= -15
Here according to the rule negative sign and positive sign [i.e. (-) × (+) = (-)] makes negative sign. That is why – (+9) will be -9. So it will be - 6 – 9 = -15. Actually the absolute values are added and the result will have negative sign as both the numbers are negative.
4. Subtract -11 from -5
= -5 – (-11)
= -5 +11
= 6
Here according to the rule negative sign and negative sign [i.e. (-) × (-) = (+)] makes positive sign. Therefore – (- 11) = 11. So it will be -5 + 11 = 6. Actually the absolute values are subtracted and in the result the value with a higher absolute value is considered.
5. Subtract -5 from 16
= 16 – (- 5)
= 16 + 5
= 21
Here according to the rule negative sign and negative sign [i.e. (-) × (-) = (+)] makes positive sign. Therefore – (-5)= +5. So it will be 16 +5 = 21. Here both 16 and 5 is added to get 21.
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# Volume of a pyramid
Don't you know how to calculate the volume of a pyramid? Use our online calculator and find out without knowing the formula.
Just select the shape of the base of the pyramid and type in the tool some data such as the height of the figure or how long one of the sides of the base is. And remember that you can also calculate the area of a pyramid in this other link that we have just left you.
## Formula for calculating the volume of a pyramid
To obtain the volume of a pyramid, the following general formula must be applied:
Being:
• Ab the area of the base
• h the height of the pyramid
There is a second formula that allows us to get the volume of a pyramid regular, i.e., one that has a regular polygon (square, pentagon, etc.) as its base. It is this:
Being:
• N the number of sides of the base polygon
• L the length of one of the sides of the polygon of the base
• apb is the length of the apothem of the base
• h corresponds to the height of the pyramid
The main difficulty lies in the calculation of the area of the base of the pyramid since, depending on its shape, we will have to use one formula or another.
If they do not give us the height, we also have an added difficulty since we will have to calculate it from other data such as the apothem.
To clarify the most frequent cases of the calculation of the volume of a pyramid, let's see some typical solved examples.
## Volume triangular pyramid
If you are asked to calculating the volume of a triangular pyramidyou can directly apply the formula above these lines. You just have to know how long one of the sides of the base (L) is and the height of the figure.
By example,Let's calculate the volume of one whose base measures 2 meters and has a height of 3 meters:
Volume triangular pyramid = √3/12 - 22 - 3 = √3 = 1,732 m3
In case you want to calculating the volume of a quadrangular pyramidThe formula to be used is quite simple since the base, being square, makes the calculations much easier.
We are going to see a exercise solved in which we are given a quadrangular pyramid whose side measures 4 meters and has a height of 6 meters:
Volume of quadrangular pyramid = 1/3 - 42 - 6 = 1/3 - 16 - 6 = 32 m3
## Volume pentagonal pyramid
Yes we are given a pentagonal pyramid and asked to calculate its volumethe calculations become a little more complicated. This usually happens because generally in the problem statement we are only given the side and height of the pyramid, but we are not given the apothem of the base, which forces us to perform some more operation.
For example, we are going to obtain the volume of a pentagonal pyramid whose base edge measures 5 meters and has a height of 10 meters.
The first thing we are going to calculate is the value of the apothem of the pentagon of the base, for this you can enter the link that we have just left and in which we explain how to do it to obtain its value:
apb = 5 m / [2tan(36º)] = 3.44 meters
With the apothem of the pentagonal pyramid Once it is obtained, we can apply the general formula that will give us its volume:
Volume of pentagonal pyramid = 5/6 - 5 - 3,44 - 10 = 143,33 m3
## Hexagonal pyramid volume
Finally, let's take a look at how to calculate the volume of a pyramid with a hexagonal base. In this case the same thing happens as in the pentagonal pyramid and that is that we need to know how much is the apothem of the base to apply the formula.
Here we explain how to calculate the apothem of a hexagon so we will use that knowledge to solve a exercise in which we are asked to find the volume of a hexagonal pyramid 15 meters high whose base has sides measuring 3 meters.
The first thing we have to do is calculate the value of the apothem of the hexagon at the base of the pyramid. If you still don't know how to do it, remember to click on the link we have left in the previous paragraph:
apb = R√3/2 = 3√3/2 = 2.59 meters
Now we have all the data we need to draw the volume of the hexagonal pyramid applying the general formula:
Hexagonal pyramid volume = 3 - 2,59 - 15 = 116,91 m3
If you have any doubt or you have an exercise that you do not know how to solve, leave us a comment and we will help you. Remember that if you want to obtain the volume of any other pyramid that we have not included here, you can always use the general formula to solve the problem.
|
# Angle Between Two Vectors with Cross Product
There are situations when you need to find out the angle between two vectors and the only thing you know are vectors coordinates. The same forma is used for 2D vectors and 3D vectors as well.
So there are a few concepts that we should be familiar with:
1) Cross Product
2) Dot Product
3) Vector Magnitude
4) Some geometry
The basic formula that we will use to find the angle is based on the magnitude of a cross product:
Ok but where does this formula come from? The demonstration is pretty simple but messy. First compute the cross product using Sarrus rule(this is how I “remember” the ugly formula for the cross product):
Now the next step is to write the magnitude for the cross product
As you might have guessed , we expand the parantheses and a long calculus is waiting to be done. So in the end you should have something like this:
We can observe three dot products above. First two are very easy to spot and the third one is the geometric dot product squared:
Hint! The magnitude of a cross product of two vectors is the area of the parallelogram made by the vectors where $|u|$ is the width and $|v|sin\theta$ is the height. Below is a reprezentation for the cross product.
Now let’s find the angle 😀
We can get the angle very easy from the basic formula:
There are two possible solutions between $0$ and $\pi$.
First solution is the value of $\theta$ itself, and second solution is $\pi- \theta$. So be careful which angle you need in your application.
If you want the exterior angle of your vectors, take $2\pi- \theta$. This is useful to know in polygon triangulation when you need to find the angle based on vectors direction.
This decision can be done by checking the $z$ sign of the cross product.
Here is an implementation using C# :
private double VertexAngle(Vector v1, Vector v2)
{
// |a x b | = |a||b|sin(theta)
var v3 = new double[3];
CrossProd(v1,v2,v3);
//here we compute magnitude of cross Product
// I know we could call Magnitude method but wanted to be more clear
double mCrossProd= Math.Sqrt(v3[0]*v3[0] + v3[1]*v3[1] + v3[2]*v3[2]);
var denom = Magnitude(v1)*Magnitude(v2);
double angle = Math.Asin(mCrossProd/denom);
if(v3[2]>0) angle = 2*Math.PI - angle;
return angle;
}
private void CrossProd(Vector v1,Vector v2, double []res)
{
res[0] = v1.Y*v2.Z-v1.Z*v2.Y;
res[1] = v1.Z*v2.X-v1.X*v2.Z;
res[2] = v1.X*v2.Y-v1.Y*v2.X;
}
private double Magnitude(Vector v)
{//Length
return Math.Sqrt(v.X *v.X + v.Y *v.Y + v.Z*v.Z);
}
## Author: Sergiu Craitoiu
I have been working as a software engineer in computer graphics for almost 5 years. In the free time I love playing guitar, playing video games, hiking and studying computer graphics techniques.Follow me on : Google + , Twitter, Facebook
|
## Fractions are Hard!
### 1.4 Even division is confusing
There are actually two different ways to think about division. Consider, for example, $$20 \div 5$$.
DIVISION AS GROUPING
The question $$20 \div 5$$ can be interpreted as: How many groups of five can you find among twenty?
We see four groups of five.
We have $$20 \div 5 = 4$$.
DIVISION AS SHARING
The question $$20 \div 5$$ can be interpreted as: I have 20 pies to share equally among 5 people. How many pies per person does this give?
We see that each person gets four pies.
We have $$20 \div 5 = 4$$.
These two approaches are philosophically different yet give the same final answer of 4 in each case. Is it obvious that they should?
Thinking Challenge: Explain why counting groups of 12 among 3900 objects must yield the same numerical answer as the number of objects per person in distributing 3900 objects equally among 12 people. Can you do it?
ON DIVIDING FRACTIONS (OPTIONAL, AS THIS IS HARD)
Some curricula do have young folk visualize the division of fractions by returning to parts-of-a-whole thinking on the number line. Can you see the answers to these five examples?
Example 1: Evaluate $$2 \div \dfrac{1}{3}$$.
Answer: Thinking of Division as Grouping this is asking for how many thirds we see in a picture of two.
We see six thirds in this picture: $$2 \div \dfrac{1}{3} = 6$$.
Example 2: Evaluate $$3 \div \dfrac{2}{3}$$.
Answer: Thinking of Division as Grouping again we are looking for groups of two-thirds in a picture of three.
Thinking parts-of-a-whole we see four-and-half two-thirds in three: $$3 \div \dfrac{2}{3} = 4\frac{1}{2}$$.
Example 3: Evaluate $$2 \div 1\dfrac{3}{4}$$.
Answer: We are looking for groups of $$1\frac{3}{4}$$ in a picture of two.
Another one-seventh of the blue line brings us up to two. So we see one and one-seventh copies of it in two: $$2 \div 1\dfrac{3}{4}= 1\frac{1}{7}$$.
Example 4: Evaluate $$\dfrac{2}{3} \div \dfrac{1}{2}$$.
Answer: We are looking for groups of a half in a picture of two-thirds.
Do you see $$\dfrac{2}{3} \div \dfrac{1}{2} = 1\frac{1}{3}$$? (Another third of the blue line brings us up to two-thirds.)
Example 5: Evaluate $$\dfrac{1}{2} \div 3$$.
Answer: This time use Division by Sharing (why not?) and divide a half into three equal parts.
We see that $$\dfrac{1}{2} \div 3 = \dfrac{1}{6}$$. (EXTRA: Can you see this by using division by grouping too? How many threes fit inside a half?)
UPSHOT: Can you now explain why dividing two fractions is the same as multiplying by the reciprocal? It is not clear how this tricky visual work – which has us go back to parts-of-a-whole thinking and mixes approaches to division – aids us here. It seems more confusing than enlightening.
## Books
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## Guides & Solutions
Dive deeper into key topics through detailed, easy to follow guides and solution sets.
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|
# 3.3 Average and instantaneous acceleration (Page 2/7)
Page 2 / 7
The term deceleration can cause confusion in our analysis because it is not a vector and it does not point to a specific direction with respect to a coordinate system, so we do not use it. Acceleration is a vector, so we must choose the appropriate sign for it in our chosen coordinate system. In the case of the train in [link] , acceleration is in the negative direction in the chosen coordinate system , so we say the train is undergoing negative acceleration.
If an object in motion has a velocity in the positive direction with respect to a chosen origin and it acquires a constant negative acceleration, the object eventually comes to a rest and reverses direction. If we wait long enough, the object passes through the origin going in the opposite direction. This is illustrated in [link] .
## Calculating average acceleration: a racehorse leaves the gate
A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration?
## Strategy
First we draw a sketch and assign a coordinate system to the problem [link] . This is a simple problem, but it always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, we have negative velocity.
We can solve this problem by identifying $\text{Δ}v\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{Δ}t$ from the given information, and then calculating the average acceleration directly from the equation $\stackrel{\text{–}}{a}=\frac{\text{Δ}v}{\text{Δ}t}=\frac{{v}_{\text{f}}-{v}_{0}}{{t}_{\text{f}}-{t}_{0}}$ .
## Solution
First, identify the knowns: ${v}_{0}=0,{v}_{\text{f}}=-15.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ (the negative sign indicates direction toward the west), Δ t = 1.80 s.
Second, find the change in velocity. Since the horse is going from zero to –15.0 m/s, its change in velocity equals its final velocity:
$\text{Δ}v={v}_{\text{f}}-{v}_{0}={v}_{\text{f}}=-15.0\phantom{\rule{0.2em}{0ex}}\text{m/s}.$
Last, substitute the known values ( $\text{Δ}v\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{Δ}t$ ) and solve for the unknown $\stackrel{\text{–}}{a}$ :
$\stackrel{\text{–}}{a}=\frac{\text{Δ}v}{\text{Δ}t}=\frac{-15.0\phantom{\rule{0.2em}{0ex}}\text{m/s}}{1.80\phantom{\rule{0.2em}{0ex}}\text{s}}=-8.33{\text{m/s}}^{2}.$
## Significance
The negative sign for acceleration indicates that acceleration is toward the west. An acceleration of 8.33 m/s 2 due west means the horse increases its velocity by 8.33 m/s due west each second; that is, 8.33 meters per second per second, which we write as 8.33 m/s 2 . This is truly an average acceleration, because the ride is not smooth. We see later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his weight.
Check Your Understanding Protons in a linear accelerator are accelerated from rest to $2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{m/s}$ in 10 –4 s. What is the average acceleration of the protons?
Inserting the knowns, we have
$\stackrel{\text{–}}{a}=\frac{\text{Δ}v}{\text{Δ}t}=\frac{2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{m/s}-0}{{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{s}-0}=2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{11}{\text{m/s}}^{2}.$
## Instantaneous acceleration
Instantaneous acceleration a , or acceleration at a specific instant in time , is obtained using the same process discussed for instantaneous velocity. That is, we calculate the average velocity between two points in time separated by $\text{Δ}t$ and let $\text{Δ}t$ approach zero. The result is the derivative of the velocity function v ( t ), which is instantaneous acceleration and is expressed mathematically as
upward force and downward force lift
upward force and downward force on lift
hi
Etini
hi
elo
hy
Xander
Hello
Jux_dob
what's the answer? I can't get it
what is the question again?
Sallieu
What's this conversation?
Zareen
what is catenation? and give examples
sununu
what's the si unit of impulse
The Newton second (N•s)
Ethan
what is the s. I unit of current
Amphere(A)
imam
thanks man
Roland
u r welcome
imam
the velocity of a boat related to water is 3i+4j and that of water related to earth is i-3j. what is the velocity of the boat relative to earth.If unit vector i and j represent 1km/hour east and north respectively
what is head to tail rule?
kinza
what is the guess theorem
viva question and answer on practical youngs modulus by streching
send me vvi que
rupesh
a car can cover a distance of 522km on 36 Liter's of petrol, how far can it travel on 14 liter of petrol.
Isaac
yoo the ans is 193
Joseph
whats a two dimensional force
what are two dimensional force?
Where is Fourier Theorem?
what is Boyle's law
Boyle's law state that the volume of a given mass of gas is inversely proportion to its pressure provided that temperature remains constant
Abe
how do I turn off push notifications on this crap app?
Huntergirl
what is the meaning of in.
In means natural logarithm
Elom
is dea graph for cancer caliper experiment using glass block?
Bako
identity of vectors?
what is defined as triple temperature
Triple temperature is the temperature at which melting ice and boiling water are at equilibrium
njumo
a tire 0.5m in radius rotate at constant rate 200rev/min. find speed and acceleration of small lodged in tread of tire.
hmm
Ishaq
100
Noor
|
# NCERT Solutions class-11 Maths Miscellaneous
## myCBSEguide App
CBSE, NCERT, JEE Main, NEET-UG, NDA, Exam Papers, Question Bank, NCERT Solutions, Exemplars, Revision Notes, Free Videos, MCQ Tests & more.
Miscellaneous Exercise
1. How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Ans. There are 8 letters in the word DAUGHTER. In this word 3 vowels and 5 consonants. 2 vowels and 3 consonants are to be selected.
Number of ways of selection = = = = 30
Now, each word contains 5 letters which can be arranged among themselves 5! Ways.
Therefore, total number of words = = 3600
2. How many words, with or without meaning can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Ans. There are 8 letters in the word EQUATION. In this word 5 vowels and 3 consonants. 2 vowels and 3 consonants are to be selected.
Now, 5 vowels can be arranged in 5! Ways and 3 consonants can be arranged in 3! Ways.
Also the groups of vowels and consonants can be arranged in 2! Ways.
Total number of permutations = = = 1440
3. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls
(ii) at least 3 girls
(iii) almost 3 girls?
Ans. (i) There are 9 boys and 4 girls. 3 girls and 4 boys have to be selected.
Number of ways of selection = =
= = 504
(ii) We have to select at least 3 girls. So the committee consists of 3 girls and 4 boys or 4 girls and 3 boys.
Number of ways of selection =
=
=
= 504 + 84 = 588
(iii) We have to select at most 3 girls. So the committee consists of no girls and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.
Number of ways of selection =
=
=
= 36 + 336 + 756 + 504 = 1632
4. If the different permutations of all the letters of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
Ans. In the word EXAMINATION, there are two I’s and two N’s and all other letters are different.
Number of ways of arrangement =
=
= 907200
5. How many 6-digit numbers can be formed from the digits 0, 1 3, 5, 7 and 9 which are divisible by 10 and no digits are repeated?
Ans. A number divisible by 10 have unit place digit 0. So digit 0 is fixed at unit place and the remaining 5 placed filled with remaining five digits in ways.
Required numbers = = = 120
6. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabets?
Ans. 2 vowels out of 5 vowels and 2 consonants out of 21 consonants have to be selected and these 4 letters in 4 ways.
Required number of words = =
= = = 50400
7. In an examination, a question paper consists of 12 questions divided into two parts i.e., part I and part II containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Ans. Here, we have to select 8 questions at least 3 questions from each section. Therefore, we have required selections are 3 from part I and 5 from part II or 4 from part I and 4 from part II or 5 from part I and 3 from part II.
Number of ways of selection =
=
=
=
= 210 + 175 + 35 = 420
8. Determine the number of 5-card combinations out of a deck of 52 cards is each selection of 5 cards has exactly one king.
Ans. Here, we have to select 5 cards containing 1 king and 4 other cards i.e., we have to select 1 king out of 4 kings and 4 other cards out of 48 other cards.
Number of ways of selection =
=
=
= = 778320
9. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Ans. Given: women occupy the even places.
1 2 3 4 5 6 7 8 9
M W M W M W M W M
Therefore, we can arrange four women in 4! Ways and 5 men in 5! Ways.
Number of ways of selection = = 2880
10. From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?
Ans. According to the question, Number of ways of selection =
=
=
= 170544 + 646646 = 817190
11. In how many ways can be letters of the word ASSASSINATION be arranged so that all the S’s are together?
Ans. In the word ASSASSINATION, A appears 3 times, S appears 4 times, I appears in 2 times and N appears in 2 times. Now, 4 S’ taken together become a single letter and other remaining letters taken with this single letter.
Number of arrangements =
= 151200
### 1 thought on “NCERT Solutions class-11 Maths Miscellaneous”
1. Thank you so much for answer
|
# Properties of Parallelograms
The broadest term we’ve used to describe any kind of shape is “polygon.” When we
in the last section, we essentially just specified that they were polygons with four
vertices and four sides. Still, we will get more specific in this section and discuss
a special type of quadrilateral: the parallelogram. Before we do this, however,
let’s go over some definitions that will help us describe different parts of quadrilaterals.
Since this entire section is dedicated to the study of quadrilaterals, we will use
some terminology that will help us describe specific pairs of lines, angles, and
vertices of quadrilaterals. Let’s study these terms now.
### Consecutive Angles
Two angles whose vertices are the endpoints of the same side are called consecutive
angles
.
?Q and ?R are consecutive angles because Q and R are the endpoints of the same side.
### Opposite Angles
Two angles that are not consecutive are called opposite angles.
?Q and ?S are opposite angles because they are not endpoints of a common side.
### Consecutive Sides
Two sides of a quadrilateral that meet are called consecutive sides.
QR and RS are consecutive sides because they meet at point R.
### Opposite Sides
Two sides that are not consecutive are called opposite sides.
QR and TS are opposite sides of the quadrilateral because they do not meet.
Now, that we understand what these terms refer to, we are ready to begin our lesson
on parallelograms.
## Properties of Parallelograms: Sides and Angles
A parallelogram is a type of quadrilateral whose pairs of opposite sides are parallel.
Although the defining characteristics of parallelograms are their pairs of parallel
opposite sides, there are other ways we can determine whether a quadrilateral is
a parallelogram. We will use these properties in our two-column geometric
proofs
to help us deduce helpful information.
If a quadrilateral is a parallelogram, then.
(1) its opposite sides are congruent,
(2) its opposite angles are congruent, and
(2) its consecutive angles are supplementary.
Another important property worth noticing about parallelograms is that if one angle
of the parallelogram is a right angle, then they all are right angles. Why is this
property true? Let’s examine this situation closely. Consider the figure below.
Given that ?J is a right angle, we can also determine that ?L
is a right angle since the opposite sides of parallelograms are congruent. Together,
the sum of the measure of those angles is 180 because
We also know that the remaining angles must be congruent because they are also opposite
angles. By the Polygon Interior Angles Sum Theorem, we know that all quadrilaterals
have angle measures that add up to 360. Since ?J and
?L sum up to 180, we know that the sum of ?K
and ?M will also be 180:
Since ?K and ?M are congruent, we can define their measures
with the same variable, x. So we have
Therefore, we know that ?K and ?M are both right angles.
Our final illustration is shown below.
Let’s work on a couple of exercises to practice using the side and angle properties
of parallelograms.
### Exercise 1
Given that QRST is a parallelogram, find the values of x and y
in the diagram below.
Solution:
After examining the diagram, we realize that it will be easier to solve for x
first because y is used in the same expression as x
(in ?R), but x is by itself at segment QR.
Since opposite sides of parallelograms are congruent, we have can set the quantities
equal to each other and solve for x:
Now that we’ve determined that the value of x is 7,
we can use this to plug into the expression given in ?R. We know that
?R and ?T are congruent, so we have
Substitute x for 7 and we get
So, we’ve determined that x=7 and y=8.
### Exercise 2
Given that EDYF is a parallelogram, determine the values of x and y.
Solution:
In order to solve this problem, we will need to use the fact that consecutive angles
of parallelograms are supplementary. The only angle we can figure out initially
is the one at vertex Y because all it requires is the addition of
angles. We have
Knowing that ?Y has a measure of 115 will allow us to
solve for x and y since they are both found in angles
consecutive to ?Y. Let’s solve for y first. We have
All that is left for solve for is x now. We will use the same method
we used when solving for y:
So, we have x=10 and y=13.
The sides and angles of parallelograms aren’t their only unique characteristics.
Let’s learn some more defining properties of parallelograms.
## Properties of Parallelograms: Diagonals
When we refer to the diagonals of a parallelogram, we are talking about lines
that can be drawn from vertices that are not connected by line segments. Every parallelogram
will have only two diagonals. An illustration of a parallelogram’s diagonals is
shown below.
We have two important properties that involve the diagonals of parallelograms.
If a quadrilateral is a parallelogram, then.
(1) its diagonals bisect each other, and
(2) each diagonal splits the parallelogram into two congruent triangles.
Segments AE and CE are congruent to each other because the diagonals meet at point
E, which bisects them. Segments BE and DE are also congruent.
The two diagonals split the parallelogram up into congruent triangles.
Let’s use these properties for solve the following exercises.
### Exercise 3
Given that ABCD is a parallelogram, find the value of x.
Solution:
We know that the diagonals of parallelograms bisect each other. This means that
the point E splits up each bisector into two equivalent segments.
Thus, we know that DE and BE are congruent, so we have
So, the value of x is 3.
### Exercise 4
Given that FGHI is a parallelogram, find the values of x and y.
Let’s try to solve for x first. We are given that ?FHI
is a right angle, so it has a measure of 90°. We can deduce that ?HFG is also a right angle by the Alternate Interior Angles Theorem.
If we look at ?HIJ, we notice that two of its angles are congruent,
so it is an isosceles triangle. This means that ?HIJ has a measure
of 9x since ?IJH has that measure.
We can use the fact that the triangle has a right angle and that there are two congruent
angles in it, in order to solve for x. We will use the Triangle Angle
Sum Theorem
to show that the angles must add up to 180°.
Now, let’s solve for y. We know that segments IJ
and GJ are congruent because they are bisected by the opposite diagonal.
Therefore, we can set them equal to each other.
Because we can say that IJ and GJ
are congruent, we have
So, our answers are x=5 and y=4.
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# Pattern for all the binary chains divisible by 5
For instance, $$x = 101$$ is divisible by $$5$$ because it is the integer 5. Same thing for $$x=1111$$ is also divisible by 5 as it is the integer 15. However, $$x=1100$$ is not divisible by $$5$$ as it is the integer 12.
Is there a pattern the recognise the binary chains divisible by 5?
• Convert the number to base 4 (i.e. look at pairs of digits) and then apply the same divisibility test as the test for divisibility by 11 in base 10 (since $5 = 4+1$, as $11 = 10+1$). Feb 16, 2021 at 15:21
• You can multiply the digits with $1,2,-1,-2$ repeatedly from right to left and look at if the sum is divisible by $5$.
– Tan
Feb 16, 2021 at 15:29
• If you want detailed explanation, you can check this answer. Jun 13, 2022 at 17:09
We can use the happy face of five-divisibility!
Start in state $$0$$. Follow the appropriate arrows as you read digits from your binary number from left-to-right. If you end up in state $$0$$ again your number is divisible by $$5$$ (and if not, the state number gives you the remainder).
How does it work? Well if we're in state $$k$$ it means the digits we have read so far form the number $$n$$ with remainder $$k \equiv n \mod 5$$. If we then read another digit $$b$$, we effectively move to the new number $$n' = 2n + b$$. Thus we need to move to state $$(2k + b) \bmod 5$$, which is exactly what we do in the above graph. Thus if we end up in state $$0$$ in the end we know there is no remainder, and the number that we read is divisible by 5.
The state diagram above is just this logic graphically displayed. You could have it as a table instead as well:
$$\begin{array}{ccc} k & b & 2k + b & (2k + b) \bmod 5\\ \hline 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 1 & 0 & 2 & 2 \\ 1 & 1 & 3 & 3 \\ 2 & 0 & 4 & 4 \\ 2 & 1 & 5 & 0 \\ 3 & 0 & 6 & 1 \\ 3 & 1 & 7 & 2 \\ 4 & 0 & 8 & 3 \\ 4 & 1 & 9 & 4 \\ \hline \end{array}$$
This also makes for a nice mental rule. You start with the number $$0$$ in your head and look at the digits from left-to-right. For each digit you multiply the number in your head by 2 and add the digit you just read. If the number goes to five or above you subtract five. If you end up with $$0$$ the number is divisible by 5.
As an example for the binary number $$1111101000_2 = 1000$$, you go:
0 is our starting value
1111101000
^ 2*0 + 1 = 1
1111101000
^ 2*1 + 1 = 3
1111101000
^ 2*3 + 1 = 7, is >= 5 so we subtract 5
7 - 5 = 2
1111101000
^ 2*2 + 1 = 5, is >= 5 so we subtract 5
5 - 5 = 0
1111101000
^ 2*0 + 1 = 1
1111101000
^ 2*1 + 0 = 2
1111101000
^ 2*2 + 1 = 5, is >= 5 so we subtract 5
5 - 5 = 0
1111101000
^ 2*0 + 0 = 0
1111101000
^ 2*0 + 0 = 0
1111101000
^ 2*0 + 0 = 0
Our remainder is $$0$$, thus we are divisible by five!
• If you can make an example, e.g. the integer $1000$, I will accept your answer.
– Alex
Feb 16, 2021 at 16:29
• @Alex Added an example for the binary integer $1000_2 = 8$.
– orlp
Feb 16, 2021 at 16:33
• I meant the integer $1000$, not the binary $1000$
– Alex
Feb 16, 2021 at 16:37
• @Alex It's a bit laborious but sure.
– orlp
Feb 16, 2021 at 16:37
• Mark Jason Dominus has made similar diagrams for divisibilty of integers written in base 10: blog.plover.com/math/divisibility-by-7.html Feb 16, 2021 at 16:44
For example, for $$1000$$ decimal, represent it as $$1111101000$$. In base $$4$$, this is $$33220$$ (just group pairs of digits together; if there were an odd number of digits then add a $$0$$ at the front).
Then $$3+2+0$$ (the sum of the odd-position digits) and $$3+2$$ (the sum of the even-position digits) are equal, so the number is divisible by 5. This works in general in base $$b$$ if we are testing for divisibility by $$b+1$$. To see that, write $$n = \sum_{i=0}^k a_ib^i,$$ where $$0\le a_i < b$$; then $$(b+1)n = \sum_{i=0}^k a_ib^{i+1} + \sum_{i=0}^k a_ib^i = \sum_{i=1}^{k+1}a_{i-1}b^i + \sum_{i=0}^k a_ib^i = a_0b^0 + \sum_{i=1}^{k-1} (a_{i-1} + a_i)b^i + a_kb^{k+1}.$$ From this it is easy to see that the sum of the odd-position digits and the sum of the even-position digits are equal.
$$5 = 2^2 +1$$ so whatever rule we can use in base $$10$$ to tell if a number is divisible by $$10^2 + 1 = 101$$ should be similar.
But .... consider this.
$$2^4 \equiv 1 \pmod 5$$
So $$2^{4k+1}\equiv 2^1\equiv 2 \pmod 5$$.
$$2^{4k+2}\equiv 2^2\equiv 4 \equiv -1 \pmod 5$$
$$2^{4k+3}\equiv 2^3\equiv 8\equiv - 2\pmod 5$$
$$2^{4k}\equiv 1^k\equiv 1\pmod 5$$.
So consider the digits in position. Subtract the digits in the $$4k+2$$ positions for the one in the $$4k$$ positions. Subtract the $$4^k+3$$ from the $$4^k+1$$ positions. Multiply result from the odd positions by $$2$$ and add to the even; and mod by $$5$$.
Example: I'm typing at random: $$101110101000111010111$$
Group it into four sections:
$$\color{blue}1\color{orange}0\color{purple}1\color{red}1\color{blue}1\color{orange}0\color{purple}1\color{red}0\color{blue}1\color{orange}0\color{purple}0\color{red}0\color{blue}1\color{orange}1\color{purple}1\color{red}0\color{blue}1\color{orange}0\color{purple}1\color{red}1\color{blue}1$$
The $$\color{blue}{4k}$$ positions (these are all $$\equiv 1\pmod 5$$) I count/add: $$\color{blue}6$$.
In the $$\color{purple}{4k+2}$$ positions (these are all $$\equiv -1\pmod 5$$) I count/add $$\color{purple}4$$. Subtracting I get $$6-4\equiv 2\pmod 5$$.
In the $$\color{red}{4k+1}$$ positions (these are all $$\equiv 2\pmod 5$$) I count: $$\color{red}2$$.
And int the $$\color{orange}{4k+3}$$ positions ($$\equiv -2\pmod 5$$) I get: $$\color{orange}1$$. Subtracting I get $$2-1=1$$ which is equiv $$1\times 2\equiv 2\pmod 5$$.
So multiply tho odd positions by $$2$$ to get $$2$$ and add to the even positions of $$2$$ to get $$2+2 = 4$$ and I conclude:
$$101110101000111010111_2\equiv 4\pmod 5$$.
I google for "binary decimal converter" and find this cute little page: https://www.rapidtables.com/convert/number/binary-to-decimal.html
And I see $$101110101000111010111_2 = 1528279$$ which is $$\equiv 4\pmod 5$$.
....
This is essentially the "casting out every other digit" rule to determine if a number in base $$10$$ is divisible by $$11$$.
.....
To show how quick this can be:
I convert $$35765$$ to $$1000101110110101$$ and do
$$1000|1011|1011|0101$$ and calculate $$(3-1) + 2(2-3)=2+(-2)=0$$ so it is divisible by $$5$$.
I use a similar approach to rogerl, but using hexadecimal (base 16), rather than base 4. The divisibility test for 5 in base 16 works like the divisibility test for 3 in base 10, because 3 divides (10-1) and 5 divides (16-1).
The base 16 representation of a positive integer $$n$$ is of the form $$n = a_0 + a_1\cdot16 + a_2\cdot16^2 + a_3\cdot16^3 + a_4\cdot16^4 + \dots$$ with $$0 \le a_i<16$$.
But $$16\equiv 1 \pmod{15}$$, so $$n \equiv a_0 + a_1 + a_2 + a_3 + a_4 + \dots \pmod{15}$$ And because 3 and 5 both divide 15, we also have $$n \equiv a_0 + a_1 + a_2 + a_3 + a_4 + \dots \pmod{3}$$ $$n \equiv a_0 + a_1 + a_2 + a_3 + a_4 + \dots \pmod{5}$$
Thus we can test divisibility by 5 of a hex number by simply adding its hex digits, mod 5. (And of course we can get the hex digits of a binary number by grouping the bits in blocks of 4).
For example, $$123_{10}$$ is $$0111 1011_2$$ in binary and $$\mathrm{7b}_{16}$$ in hex. The values of those two blocks are $$7$$ and $$11$$. $$7 + 11 = 18 \equiv 3\pmod{5}$$, so $$0111 1011_2$$ leaves a remainder of 3 when divided by 5.
Here are some larger examples.
dec 1000
bin 0011 1110 1000
val 3 14 8
sums 3 2 0 mod 5
remainder = 0
19136214
0001 0010 0011 1111 1110 1101 0110
1 2 3 15 14 13 6
1 3 1 1 0 3 4
remainder 4
Here's a link to a small live Python script which does these calculations.
As I said earlier, you can also use this method for divisibility by 3, just add mod 3 instead of 5. Or add mod 15, and test divisibility by 15, 5, and 3 at the same time.
Note that you can do this test in any order, but if you start from the left, don't forget to pad the bit string with zeroes (if necessary) to make the total number of bits divisible by 4.
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# Difference between revisions of "2018 AMC 10B Problems/Problem 12"
## Problem
Line segment $\overline{AB}$ is a diameter of a circle with $AB=24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle{ABC}$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
$\textbf{(A)} \text{ 25} \qquad \textbf{(B)} \text{ 38} \qquad \textbf{(C)} \text{ 50} \qquad \textbf{(D)} \text{ 63} \qquad \textbf{(E)} \text{ 75}$
## Solution 1 (Coordinate Bash)
Let $A=(-12,0),B=(12,0)$. Therefore, $C$ lies on the circle with equation $x^2+y^2=144$. Let it have coordinates $(x,y)$. Since we know the centroid of a triangle with vertices with coordinates of $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$, the centroid of $\triangle ABC$ is $\left(\frac{x}{3},\frac{y}{3}\right)$. Because $x^2+y^2=144$, we know that $\left(\frac{x}{3}\right)^2+\left(\frac{y}{3}\right)^2=16$, so the curve is a circle centered at the origin. Therefore, its area is $16\pi\approx \boxed{\text{(C) }50}$. -tdeng
## Solution 2 (No Coordinates)
We know the centroid of a triangle splits the medians into segments of ratio $2:1$, and the median of the triangle that goes to the center of the circle is the radius (length $12$), so the length from the centroid of the triangle to the center of the circle is always $\dfrac{1}{3} \cdot 12 = 4$. The area of a circle with radius $4$ is $16\pi$, or around $\boxed{\textbf{(C)} \text{ 50}}$. -That_Crazy_Book_Nerd
## See Also
2018 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 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 AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# How do you find the area of a isosceles triangle with base 10 and perimeter 36?
May 2, 2018
The area is $60$ square units
#### Explanation:
Giving us the perimeter of an isosceles triangle means we are indirectly given the lengths of the three sides.
If the base is $10$, the the other two equal sides add to $26$:
$36 - 10 = 26$
The equal sides are therefore $26 \div 2 = 13$
To find the area of the triangle we need its height.
the height of an isosceles triangle can be found from its line of symmetry. If you cut the triangle into two identical triangles, for each we will have:
A 90° angle
A base of $5 \text{ } \rightarrow \left(10 \div 2\right)$
The hypotenuse is $13$
You can therefore use Pythagoras' Theorem to find the third side which will be the height of the triangle.
You might recognise that these are two values of the Pythagorean Triple, $5 , 12 , 13 \text{ } \leftarrow$ the height is $12$.
If not you can work it out.
${x}^{2} = {13}^{2} - {5}^{2}$
${x}^{2} = 144$
$x = \sqrt{144}$
$x = 12$
Now we know that the height is $12$ so we can find the area of the original triangle.
$A = \frac{b h}{2}$
$A = \frac{10 \times 12}{2}$
$A = 60$ square units
May 2, 2018
$a = 60$
#### Explanation:
Let;
$a \to \text{area}$
$b \to \text{base}$
$p \to \text{perimeter}$
$s \to \text{slant height}$
$h \to \text{height}$
$p = 36$
$b = 10$
Recall;
$a = \frac{1}{2} b \times h$
$p = b + 2 s$
Firstly;
$p = b + 2 s$
Substituting the values..
$36 = 10 + 2 s$
$36 - 10 = 2 s$
#26 = 2s
$\frac{26}{2} = s$
$13 = s$
Therefore, slant height is $13$
Also recall;
Using Pythagoras theorem;
Since we are using pythagoras theorem, the base of the triangle would be halved..
Therefore, $b = \frac{10}{2} = 5$
${s}^{2} = {b}^{2} + {h}^{2}$
Substituting the values..
${13}^{2} = {5}^{2} + {h}^{2}$
$169 = 25 + {h}^{2}$
$169 - 25 = {h}^{2}$
$144 = {h}^{2}$
$\sqrt{144} = h$
$12 = h$
Hence;
$a = \frac{1}{2} b \times h$
Substituting the values..
$a = \frac{1}{2} \times 10 \times 12$
$a = 5 \times 12$
$a = 60$
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Question Video: Finding the Integration of a Function Involving Using Laws of Exponents and the Power Rule with Fractional Exponents | Nagwa Question Video: Finding the Integration of a Function Involving Using Laws of Exponents and the Power Rule with Fractional Exponents | Nagwa
# Question Video: Finding the Integration of a Function Involving Using Laws of Exponents and the Power Rule with Fractional Exponents Mathematics • Second Year of Secondary School
## Join Nagwa Classes
Determine β« (the fifth root of π¦Β²)/(the cube root of π¦) dπ¦.
03:47
### Video Transcript
Determine the indefinite integral of the fifth root of π¦ squared divided by the cube root of π¦ with respect to π¦.
In this question, weβre asked to evaluate the integral of the quotient of two functions. And our integrand is given in terms of π¦, and weβre integrating with respect to π¦. Remember, as long as our integrand and what weβre integrating with respect to share the same variable, it doesnβt matter what we call this variable. So, we can integrate this using all of our standard rules. However, as it stands, we donβt know how to integrate this function directly. So, weβre going to need to rewrite this in a form which we can integrate. And since this expression involves a lot of exponents, weβll do this by using our laws of exponents.
The first one of our laws of exponents weβll use is the following. We know the πth root of π₯ is equal to π₯ to the power of one over π. Of course, this is only true for positive integers π. And in fact, we can use this to rewrite both the numerator of our integrand and the denominator. First, in our numerator, the value of π is five. So, we get π¦ squared all raised to the power of one over five. Then, in our denominator, we can rewrite the cube root of π¦ as π¦ to the power of one over three.
So, weβve rewritten our integral in the following form. However, we still canβt evaluate this integral. So, weβre going to need to simplify this even further. We can see in our numerator we have π¦ squared all raised to the power of one over five. And we can rewrite this as π¦ raised to the power of some number by using our laws of exponents. Recall, π raised to the power of π all raised to the power of π is equal to π raised to the power of π times π. Weβll apply this to rewrite the numerator of our integrand. Our value of π is π¦, our value of π is two, and our value of π is one over five. This gives us a new numerator of π¦ to the power of two times one over five.
So, now, we have the integral of π¦ to the power of two times one over five divided by one to the power of one-third with respect to π¦. And of course, we can simplify the exponents in our numerator. Two times one over five is equal to two over five. Now, we can see our integrand is the quotient of two exponents where π¦ is the base in both of these. And once again, we can simplify this by using our laws of exponents. We know π¦ raised to the power of π divided by π¦ raised to the power of π is equal to π¦ raised to the power of π minus π.
In this case, our value of π is two over five and our value of π is one-third. So using this, we get the integral of π¦ over two over five minus one-third with respect to π¦. And we can simplify this. Evaluating our exponent, we get two over five minus one-third is equal to one over 15. So, after all of this manipulation, we were able to rewrite our integral as the integral of π¦ raised to the power of one over 15 with respect to π¦. And now, we can evaluate this integral by using the power rule for integration. Recall for all real constants π and π, where π is not equal to negative one, the integral of π times π¦ to the πth power with respect to π¦ is equal to π times π¦ to the power of π plus one divided by π plus one plus our constant of integration πΆ.
So, we want to add one to our exponent of π¦ and then divide by this new exponent. We can see in our case the exponent of π¦ is one over 15. So, we need to add one to this exponent of one over 15 and then divide by this new exponent. This gives us π¦ to the power of one over 15 plus one all divided by one over 15 plus one. And of course, we need to add our constant of integration πΆ.
And the last thing weβll do is simplify our exponent and our denominator. One over 15 plus one is equal to 16 over 15. This gives us π¦ to the power of 16 over 15 divided by 16 over 15 plus πΆ. And we can do one last piece of simplification. Instead of dividing by 16 over 15, weβll multiply by the reciprocal. And doing this gives us our final answer.
Therefore, we were able to show the integral of the fifth root of π¦ squared over the cube root of π¦ with respect to π¦ is equal to 15 times π¦ to the power of 16 over 15 divided by 16 plus πΆ.
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# How to Calculate Work
Work is energy transferred to an object through force or displacement. It is often represented as the product of displacement and force. A person performing an action transfers energy to an object, which is then embodied in the object’s physical state. There are several ways to calculate work. Here are some examples: 1. Lifting a heavy object by pulling on it. 2. Using a lever. 3. Using a lever to lift a heavy object.
When a force acts on an object, it causes an object to move. When a force acts on a body, the displacement it produces is the work that the object is doing. In this example, a coolie lifts a mass on his head. The mass moves at a 90-degree angle to the force of gravity. Therefore, the work he or she does is zero. In this case, the coolie has done no work.
The definition of work is similar to that of an object in motion. The applied force must cause a displacement that is opposite to the motion of the object. The force must be of the same direction as the displacement to be considered work. In case of negative work, the energy that the object is moving away from the object is removed. However, a ball that is falling from a height is not a ball. Instead, it is a human throwing a ball at a speed of 100 km/h.
Another example of a work that is done by a human is a force that causes an object to move. The force must be greater than the object’s weight in order to cause the movement. The greater the force, the more work it produces. In theory, the more work the object does, the more power it can exert. And it has to be in a straight line. This is how you calculate the amount of work done by an object.
The amount of work done by an object is equal to the force on it. This means that a force is doing work on an object. A force on an object causes a displacement. The force is also applied to an object. The applied force needs to be opposite the object’s motion or it will be referred to as negative work. This means that a mass must be moved perpendicular to the force in order to be considered a work.
A force can cause an object to move. In this case, the work is done by the object. If the object is in motion, it will be displaced. A force is not positive if it does not affect the object. It can be negative or positive. In fact, it can even have zero value. The definition of work has many variants. The definition is the same in every country. The force applied by a force can make an object move.
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### Home > CC3 > Chapter 9 > Lesson 9.2.1 > Problem9-64
9-64.
Use the graph at right to answer the following questions.
1. What kind of growth does this graph show? How do you know?
Does the graph show linear growth or non-linear growth? You can tell because linear growth creates a line.
2. What is this graph describing? Write an appropriate title for the graph.
Look at the $x$-axis and $y$-axis labels. The graph shows the distance from home at several different time periods.
Possible title: Distance and Time
3. How far from home is the person when the graph starts?
The graph starts at $0$ hours. Which means this person just started to travel. What is the distance from home at $0$ hours?
He starts at $9$ miles from home.
4. How fast is the person traveling? Explain how you can use the graph to determine the rate of travel.
Find the slope of the line.
$\text{Find the }\frac{\text{rise}}{\text{run}}.$
$\frac{10}{4} = 2.5\text{ miles per hour}$
5. Write an equation to represent the line on the graph.
A slope of $2.5$ means that the distance from home increases $2.5$ miles per hour. That means we multiply $2.5$ and any hour which we will call $x$.
Since we know how fast this person is traveling, we also need to know the starting distance. From part (b) we found that this person started $9$ miles away from home.
$y = 2.5x + 9$
$2.5$ relates to amount traveled every hour and $9$ relates to the starting point.
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## How to Solve Multi Step Equations?
Read,3 minutes
Multi-step equations are equations that have to be solved by two consequent steps. Here we have to apply at least two mathematical operations and then only, we can find the value of the unknown variable.
### Properties of Multi-Step Equations
There are several properties of multi-step equations. They are:
• In a multi-step equation, we can add or subtract both sides of the equation by the same number. Also, this won’t affect the equation in any way and the equation would still hold true.
• Secondly, we can multiply or divide both sides of the equation by the same number. This too won’t affect the equation and the equation will still hold true.
### Solving Multi-Step Equations
For solving multi-step equations, follow these steps:
• For solving multi-step equations, you need to combine all like terms together.
• Then, you need to bring all variables to one side by either adding or subtracting.
• Next, simplify using the inverse of addition or subtraction.
• Next, you can further simplify by using the inverse of multiplication or division.
Example 1: $$7x \ + \ 15 \ = \ -3x \ + \ 9$$
Now, adding $$+3x$$ to both sides we get, $$7x \ + \ 3x \ + \ 15 \ = \ -3x \ + \ 3x \ + \ 9 \ ⇒ \ 10x \ = \ -6 \ ⇒ \ x \ = \ \frac{-6}{10}$$ .
Example 2: $$2x \ + \ 7 \ = \ -6x \ + \ 14$$
Now, adding $$+6x$$ to both sides we get, $$2x \ + \ 6x \ + \ 7 \ = \ -6x \ + \ 6x \ + \ 14 \ ⇒ \ 8x \ = \ 7 \ ⇒ \ x \ = \ \frac{7}{8}$$ .
Example 3: $$3x \ + \ 8 \ = \ -7x \ + \ 18$$
Now, adding $$+7x$$ to both sides we get, $$3x \ + \ 7x \ + \ 8 \ = \ -7x \ + \ 7x \ + \ 18 \ ⇒ \ 10x \ = \ 10 \ ⇒ \ x \ = \ \frac{10}{10} \ = \ 1$$.
Example 4: $$7x \ + \ 12x \ - \ 15 \ = \ 8x \ + \ 21$$
Now, subtracting $$8x$$ to both sides we get, $$7x \ + \ 12x \ – \ 8x \ - \ 15 \ = \ 8x \ - \ 8x \ + \ 21 \ ⇒ \ 11x \ = \ 36 \ ⇒ \ x \ = \ \frac{36}{11}$$ .
Example 5: $$-2x \ + \ 3 \ = \ 3x \ + \ 7$$
Now, adding $$2x$$ to both sides we get, $$2x \ - \ 2x \ + \ 3 \ = \ 3x \ + \ 2x \ + \ 7 \ ⇒ \ 3 \ = \ 5x \ + \ 7 \ ⇒ \ x \ = \ -\frac{4}{5}$$ .
### Exercises for Multi Step Equations
1) $$3x \ - \ 20 \ + \ 3x = 100 \ + \ x$$$$\ \Rightarrow \$$
2) $$6x \ - \ 16 \ + \ 3x = 40 \ + \ x$$$$\ \Rightarrow \$$
3) $$5x \ - \ 14 \ + \ 3x = 84 \ + \ x$$$$\ \Rightarrow \$$
4) $$\frac{ 5x \ - \ 24}{5} = 6 \ - \ \frac{1}{5}x$$$$\ \Rightarrow \$$
5) $$\frac{ 3x \ - \ 16}{7} = 32 \ - \ \frac{1}{7}x$$$$\ \Rightarrow \$$
6) $$\frac{ 5x \ - \ 24}{6} = 90 \ - \ \frac{1}{6}x$$$$\ \Rightarrow \$$
7) $$4x \ - \ 18 = 84 \ - \ 2x$$$$\ \Rightarrow \$$
8) $$5x \ - \ 21 = 49 \ - \ 2x$$$$\ \Rightarrow \$$
9) $$5x \ - \ 21 = 56 \ - \ 2x$$$$\ \Rightarrow \$$
10) $$5x \ - \ 14 = 105 \ - \ 2x$$$$\ \Rightarrow \$$
1) $$3x \ - \ 20 \ + \ 3x = 100 \ + \ x$$$$\ \Rightarrow \ \color{red}{3x \ + \ 3x \ - \ x = 100 \ + \ 20 }$$$$\ \Rightarrow \ \color{red}{5x = 120}$$$$\ \Rightarrow \ \color{red}{x = \frac{ 120}{5} = 24}$$
2) $$6x \ - \ 16 \ + \ 3x = 40 \ + \ x$$$$\ \Rightarrow \ \color{red}{6x \ + \ 3x \ - \ x = 40 \ + \ 16 }$$$$\ \Rightarrow \ \color{red}{8x = 56}$$$$\ \Rightarrow \ \color{red}{x = \frac{ 56}{8} = 7}$$
3) $$5x \ - \ 14 \ + \ 3x = 84 \ + \ x$$$$\ \Rightarrow \ \color{red}{5x \ + \ 3x \ - \ x = 84 \ + \ 14 }$$$$\ \Rightarrow \ \color{red}{7x = 98}$$$$\ \Rightarrow \ \color{red}{x = \frac{ 98}{7} = 14}$$
4) $$\frac{ 5x \ - \ 24}{5} = 6 \ - \ \frac{1}{5}x$$$$\ \Rightarrow \ \color{red}{5x \ - \ 24= 6 \times 5 \ - \ \frac{5}{5}x}$$$$\ \Rightarrow \ \color{red}{5x \ + \ x = 30 \ + \ 24}$$$$\ \Rightarrow \ \color{red}{x = \frac{ 30 \ + \ 24}{6} = 9}$$
5) $$\frac{ 3x \ - \ 16}{7} = 32 \ - \ \frac{1}{7}x$$$$\ \Rightarrow \ \color{red}{3x \ - \ 16= 32 \times 7 \ - \ \frac{7}{7}x}$$$$\ \Rightarrow \ \color{red}{3x \ + \ x = 224 \ + \ 16}$$$$\ \Rightarrow \ \color{red}{x = \frac{ 224 \ + \ 16}{4} = 60}$$
6) $$\frac{ 5x \ - \ 24}{6} = 90 \ - \ \frac{1}{6}x$$$$\ \Rightarrow \ \color{red}{5x \ - \ 24= 90 \times 6 \ - \ \frac{6}{6}x}$$$$\ \Rightarrow \ \color{red}{5x \ + \ x = 540 \ + \ 24}$$$$\ \Rightarrow \ \color{red}{x = \frac{ 540 \ + \ 24}{6} = 94}$$
7) $$4x \ - \ 18 = 84 \ - \ 2x$$$$\ \Rightarrow \ \color{red}{4x \ + \ 2x = 84 \ + \ 18 }$$$$\ \Rightarrow \ \color{red}{6x = 102}$$$$\ \Rightarrow \ \color{red}{x = \frac{ 102}{6} = 17}$$
8) $$5x \ - \ 21 = 49 \ - \ 2x$$$$\ \Rightarrow \ \color{red}{5x \ + \ 2x = 49 \ + \ 21 }$$$$\ \Rightarrow \ \color{red}{7x = 70}$$$$\ \Rightarrow \ \color{red}{x = \frac{ 70}{7} = 10}$$
9) $$5x \ - \ 21 = 56 \ - \ 2x$$$$\ \Rightarrow \ \color{red}{5x \ + \ 2x = 56 \ + \ 21 }$$$$\ \Rightarrow \ \color{red}{7x = 77}$$$$\ \Rightarrow \ \color{red}{x = \frac{ 77}{7} = 11}$$
10) $$5x \ - \ 14 = 105 \ - \ 2x$$$$\ \Rightarrow \ \color{red}{5x \ + \ 2x = 105 \ + \ 14 }$$$$\ \Rightarrow \ \color{red}{7x = 119}$$$$\ \Rightarrow \ \color{red}{x = \frac{ 119}{7} = 17}$$
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# Fall 2002CMSC 203 - Discrete Structures1 If I told you once, it must be... Recursion.
## Presentation on theme: "Fall 2002CMSC 203 - Discrete Structures1 If I told you once, it must be... Recursion."— Presentation transcript:
Fall 2002CMSC 203 - Discrete Structures1 If I told you once, it must be... Recursion
Fall 2002CMSC 203 - Discrete Structures2 Recursive Definitions Recursion is a principle closely related to mathematical induction. In a recursive definition, an object is defined in terms of itself. We can recursively define sequences, functions and sets.
Fall 2002CMSC 203 - Discrete Structures3 Recursively Defined Sequences Example: The sequence {a n } of powers of 2 is given by a n = 2 n for n = 0, 1, 2, …. The same sequence can also be defined recursively: a 0 = 1 a n+1 = 2a n for n = 0, 1, 2, … Obviously, induction and recursion are similar principles.
Fall 2002CMSC 203 - Discrete Structures4 Recursively Defined Functions We can use the following method to define a function with the natural numbers as its domain: 1. Specify the value of the function at zero. 2. Give a rule for finding its value at any integer from its values at smaller integers. Such a definition is called recursive or inductive definition.
Fall 2002CMSC 203 - Discrete Structures5 Recursively Defined Functions Example: f(0) = 3 f(n + 1) = 2f(n) + 3 f(0) = 3 f(1) = 2f(0) + 3 = 2 3 + 3 = 9 f(2) = 2f(1) + 3 = 2 9 + 3 = 21 f(3) = 2f(2) + 3 = 2 21 + 3 = 45 f(4) = 2f(3) + 3 = 2 45 + 3 = 93
Fall 2002CMSC 203 - Discrete Structures6 Recursively Defined Functions How can we recursively define the factorial function f(n) = n! ? f(0) = 1 f(n + 1) = (n + 1)f(n) f(0) = 1 f(1) = 1f(0) = 1 1 = 1 f(2) = 2f(1) = 2 1 = 2 f(3) = 3f(2) = 3 2 = 6 f(4) = 4f(3) = 4 6 = 24
Fall 2002CMSC 203 - Discrete Structures7 Recursively Defined Functions A famous example: The Fibonacci numbers f(0) = 0, f(1) = 1 f(n) = f(n – 1) + f(n - 2) f(0) = 0 f(1) = 1 f(2) = f(1) + f(0) = 1 + 0 = 1 f(3) = f(2) + f(1) = 1 + 1 = 2 f(4) = f(3) + f(2) = 2 + 1 = 3 f(5) = f(4) + f(3) = 3 + 2 = 5 f(6) = f(5) + f(4) = 5 + 3 = 8
Fall 2002CMSC 203 - Discrete Structures8 Recursively Defined Sets If we want to recursively define a set, we need to provide two things: an initial set of elements, an initial set of elements, rules for the construction of additional elements from elements in the set. rules for the construction of additional elements from elements in the set. Example: Let S be recursively defined by: 3 S (x + y) S if (x S) and (y S) S is the set of positive integers divisible by 3.
Fall 2002CMSC 203 - Discrete Structures9 Recursively Defined Sets Proof: Let A be the set of all positive integers divisible by 3. To show that A = S, we must show that A S and S A. Part I: To prove that A S, we must show that every positive integer divisible by 3 is in S. We will use mathematical induction to show this.
Fall 2002CMSC 203 - Discrete Structures10 Recursively Defined Sets Let P(n) be the statement “3n belongs to S”. Basis step: P(1) is true, because 3 is in S. Inductive step: To show: If P(n) is true, then P(n + 1) is true. Assume 3n is in S. Since 3n is in S and 3 is in S, it follows from the recursive definition of S that 3n + 3 = 3(n + 1) is also in S. Conclusion of Part I: A S.
Fall 2002CMSC 203 - Discrete Structures11 Recursively Defined Sets Part II: To show: S A. Basis step: To show: All initial elements of S are in A. 3 is in A. True. Inductive step: To show: (x + y) is in A whenever x and y are in S. If x and y are both in A, it follows that 3 | x and 3 | y. From Theorem I, Section 2.3, it follows that 3 | (x + y). Conclusion of Part II: S A. Overall conclusion: A = S.
Fall 2002CMSC 203 - Discrete Structures12 Recursively Defined Sets Another example: The well-formed formulae of variables, numerals and operators from {+, -, *, /, ^} are defined by: x is a well-formed formula if x is a numeral or variable. (f + g), (f – g), (f * g), (f / g), (f ^ g) are well- formed formulae if f and g are.
Fall 2002CMSC 203 - Discrete Structures13 Recursively Defined Sets With this definition, we can construct formulae such as: (x – y) ((z / 3) – y) ((z / 3) – (6 + 5)) ((z / (2 * 4)) – (6 + 5))
Fall 2002CMSC 203 - Discrete Structures14 Recursive Algorithms An algorithm is called recursive if it solves a problem by reducing it to an instance of the same problem with smaller input. Example I: Recursive Euclidean Algorithm procedure gcd(a, b: nonnegative integers with a < b) if a = 0 then gcd(a, b) := b else gcd(a, b) := gcd(b mod a, a)
Fall 2002CMSC 203 - Discrete Structures15 Recursive Algorithms Example II: Recursive Fibonacci Algorithm procedure fibo(n: nonnegative integer) if n = 0 then fibo(0) := 0 else if n = 1 then fibo(1) := 1 else fibo(n) := fibo(n – 1) + fibo(n – 2)
Fall 2002CMSC 203 - Discrete Structures16 Recursive Algorithms Recursive Fibonacci Evaluation: f(4) f(3) f(2) f(1) f(0) f(1) f(2) f(1) f(0)
Fall 2002CMSC 203 - Discrete Structures17 Recursive Algorithms procedure iterative_fibo(n: nonnegative integer) if n = 0 then y := 0 elsebegin x := 0 y := 1 for i := 1 to n-1 begin z := x + y x : = y y := z end end {y is the n-th Fibonacci number}
Fall 2002CMSC 203 - Discrete Structures18 Recursive Algorithms For every recursive algorithm, there is an equivalent iterative algorithm. Recursive algorithms are often shorter, more elegant, and easier to understand than their iterative counterparts. However, iterative algorithms are usually more efficient in their use of space and time.
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# Fundamental Rules of Geometry for Physics Help
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By — McGraw-Hill Professional
Updated on Sep 17, 2011
## Fundamental Rules
The fundamental rules of geometry are used widely in physics and engineering. These go all the way back to the time of the ancient Egyptians and Greeks, who used geometry to calculate the diameter of the earth and the distance to the moon. They employed the laws of euclidean geometry (named after Euclid, a Greek mathematician who lived thousands of years ago). However much or little Euclid actually had to do with these rules, they’re straightforward. So here they are, terse and plain.
## Two-point Principle
Suppose that P and Q are two distinct geometric points. Then the following statements hold true, as shown in Fig. 4-1:
Fig. 4-1 . Two-point principle.
• P and Q lie on a common line L .
• L is the only line on which both points lie.
## Three-point Principle
Let P, Q , and R be three distinct points, not all of which lie on a straight line. Then the following statements hold true:
• P, Q , and R all lie in a common euclidean plane S .
• S is the only euclidean plane in which all three points lie.
## Principle Of N Points
Let P 1 , P 2 , P 3 ,..., and P n be n distinct points, not all of which lie in the same euclidean space of n − 1 dimensions. Then the following statements hold true:
• P 1 , P 2 , P 3 ,..., and P n all lie in a common euclidean space U of n dimensions.
• U is the only n -dimensional euclidean space in which all n points lie.
## Distance Notation
The distance between any two points P and Q , as measured from P toward Q along the straight line connecting them, is symbolized by writing PQ .
## Midpoint Principle
Suppose that there is a line segment connecting two points P and R . Then there is one and only one point Q on the line segment between P and R such that PQ = QR . This is illustrated in Fig. 4-2.
Fig. 4-2 . Midpoint principle.
## Angle Notation
Imagine that P, Q , and R are three distinct points. Let L be the line segment connecting P and Q ; let M be the line segment connecting R and Q . Then the angle between L and M , as measured at point Q in the plane defined by the three points, can be written as ∠PQR or as ∠RQP . If the rotational sense of measurement is specified, then ∠PQR indicates the angle as measured from L to M , and ∠RQP indicates the angle as measured from M to L (Fig. 4-3.) These notations also can stand for the measures of angles, expressed either in degrees or in radians.
Fig. 4-3 . Angle notation and measurement.
## Angle Bisection
Suppose that there is an angle ∠PQR measuring less than 180° and defined by three points P, Q , and R , as shown in Fig. 4-4. Then there is exactly one ray M that bisects the angle ∠PQR . If S is any point on M other than the point Q , then ∠PQS = ∠SQR. Every angle has one and only one ray that divides the angle in half.
Fig. 4-4 . Angle-bisection principle.
## Perpendicularity
Suppose that L is a line through points P and Q . Let R be a point not on L . Then there is one and only one line M through point R intersecting line L at some point S such that M is perpendicular to L . This is shown in Fig. 4-5.
Fig. 4-5 . Perpendicular principle.
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# Partial Products and Standard Algorithm
Related Topics:
Lesson Plans and Worksheets for Grade 4
Lesson Plans and Worksheets for all Grades
Examples, solutions, and videos to help Grade 4 students learn how to connect the area model and the partial products method to the standard algorithm.
Common Core Standards: 4.NBT.5, 4.OA.1, 4.OA.2, 4.NBT.1
### New York State Common Core Math Grade 4, Module 3, Lesson 11
Grade 4, Module 3, Lesson 11 Worksheets (pdf)
NYS Math Module 3 Grade 4 Lesson 11 Concept Development
Problem 1: Multiply a three-digit number by a one-digit number using the area model.
Problem 2: Multiply a three-digit number by a one-digit number, connecting the area model to the standard algorithm.
Problem 3: Solve a word problem using the standard algorithm, area model, or partial products strategy.
A cafeteria makes 4,408 lunches each day. How many lunches are made Monday through Friday? NYS Math Module 3 Grade 4 Lesson 11 Homework
1. Solve the following expressions using the standard algorithm, the partial products method, and the area model.
a. 302 × 8
b. 216 × 5
c. 593 × 9
2. Solve using the partial products method.
On Monday 475 people visited the museum. On Saturday there were 4 times as many visitors as there were on Monday. How many people visited the museum on Saturday?
3. Model with a tape diagram and solve.
6 times as much as 384.
Solve using the standard algorithm, the area model, the distributive property, or the partial products method.
4. 253 3
5. 7 times as many as 3,073.
6. A cafeteria makes 2,516 pounds of white rice and 608 pounds of brown rice every month. After 6 months, how many pounds of rice does the cafeteria make? Write your answer as a statement.
NYS Math Module 3 Grade 4 Lesson 11 Problem Set
1. Solve the following expressions using the standard algorithm, the partial products method, and the area model.
a. 425 × 4
4 (400 + 20 + 5)
(4 × __ ) + (4 × __ ) + (4 × _ )
Solve using the distributive property.
5. 3 times as many as 2,805.
Try the free Mathway calculator and problem solver below to practice various 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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# How do you write 100 divided by 5?
## How do you write 100 divided by 5?
100 divided by 5 is 20.
## How do you do 100 divided by 2?
Using a calculator, if you typed in 100 divided by 2, you’d get 50. You could also express 100/2 as a mixed fraction: 50 0/2.
How do you write 30 divided by 5?
Using a calculator, if you typed in 30 divided by 5, you’d get 6. You could also express 30/5 as a mixed fraction: 6 0/5.
What is 95 divided by 5 long?
The result of division of 95÷5 95 ÷ 5 is 19 .
### How do you write 55 divided by 5?
Using a calculator, if you typed in 55 divided by 5, you’d get 11. You could also express 55/5 as a mixed fraction: 11 0/5.
### How do you explain 49 divided by 7?
Using a calculator, if you typed in 49 divided by 7, you’d get 7. You could also express 49/7 as a mixed fraction: 7 0/7. If you look at the mixed fraction 7 0/7, you’ll see that the numerator is the same as the remainder (0), the denominator is our original divisor (7), and the whole number is our final answer (7).
How do you do 10 divided by 4?
10 divided by 4 equals 2.5.
What are 3 ways to split 100?
In mathematics, “100%” means nothing more or less than “100 per 100”, namely “100/100=1. So in mathematics you can divide 100% by 3 without having 0.1% left. 100%/3=1/3=13.
#### Can 8 be divided by 2?
Using a calculator, if you typed in 8 divided by 2, you’d get 4. You could also express 8/2 as a mixed fraction: 4 0/2.
#### What is the quotient of 5 divided by 30?
Using a calculator, if you typed in 5 divided by 30, you’d get 0.1667.
How do you write 21 divided by 3?
When you divide 21 by 3 you get 7. We can write this as follows: 21 / 3 = 7.
How can I write 5.5 as a fraction?
Here we will show you step-by-step how to convert 5.5 so you can write it as a fraction. To get rid of the decimal point in the numerator, we count the numbers after the decimal in 5.5, and multiply the numerator and denominator by 10 if it is 1 number, 100 if it is 2 numbers, 1000 if it is 3 numbers, and so on.
## Is the fraction 5 / 100 equal to a decimal?
The fraction 5/100 is equal to 0.05 when converted to a decimal. See below detalis on how to convert the fraction 5/100 to a decimal value.
## How are fractions calculated in a fraction calculator?
The calculator provided returns fraction inputs in both improper fraction form, as well as mixed number form. In both cases, fractions are presented in their lowest forms by dividing both numerator and denominator by their greatest common factor. Converting from decimals to fractions is straightforward.
How to convert a decimal to a fraction?
Conversion stages 1 Write the decimal fraction as a fraction of the digits to the right of the decimal period (numerator) and a power of 10… 2 Find the greatest common divisor (gcd) of the numerator and the denominator. 3 Reduce the fraction by dividing the numerator and the denominator with the gcd. More
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## Exponential Equations with Unlike Bases
### Learning Outcomes
• Use logarithms to solve exponential equations whose terms cannot be rewritten with the same base
• Solve exponential equations of the form $y=A{e}^{kt}$ for t
• Recognize when there may be extraneous solutions or no solutions for exponential equations
Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since $\mathrm{log}\left(a\right)=\mathrm{log}\left(b\right)$ can be rewritten as = b, we may apply logarithms with the same base on both sides of an exponential equation.
In our first example we will use the laws of logs combined with factoring to solve an exponential equation whose terms do not have the same base. Note how we rewrite the exponential terms as logarithms first.
### Example
Solve ${5}^{x+2}={4}^{x}$.
In general we can solve exponential equations whose terms do not have like bases in the following way:
1. Apply the logarithm to both sides of the equation.
• If one of the terms in the equation has base $10$, use the common logarithm.
• If none of the terms in the equation has base $10$, use the natural logarithm.
2. Use the rules of logarithms to solve for the unknown.
The following video provides more examples of solving exponential equations.
Is there any way to solve ${2}^{x}={3}^{x}$?
Use the text box below to formulate an answer or example before you look at the solution.
## Equations Containing $e$
Base is a very common base found in science, finance, and engineering applications. When we have an equation with a base e on either side, we can use the natural logarithm to solve it. Earlier, we introduced a formula that models continuous growth, $y=A{e}^{kt}$. This formula is found in business, finance, and many biological and physical science applications. In our next example, we will show how to solve this equation for $t$, the elapsed time for the behavior in question.
### Example
Solve $100=20{e}^{2t}$.
### Analysis of the Solution
Using laws of logs, we can also write this answer in the form $t=\mathrm{ln}\sqrt{5}$. If we want a decimal approximation of the answer, we use a calculator.
Does every equation of the form $y=A{e}^{kt}$ have a solution? Write your thoughts or an example in the textbox below before you check the answer.
We will provide one more example using the continuous growth formula, but this time we have to do a couple steps of algebra to get it in a form that can be solved.
### Example
Solve $4{e}^{2x}+5=12$.
## Extraneous Solutions
Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.
In the next example, we will solve an exponential equation that is quadratic in form. We will factor first and then use the zero product principle. Note how we find two solutions but reject one that does not satisfy the original equation.
### Example
Solve ${e}^{2x}-{e}^{x}=56$.
### Analysis of the Solution
When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation ${e}^{x}=-7$ because a positive number never equals a negative number. The solution $x=\mathrm{ln}\left(-7\right)$ is not a real number, and in the real number system, this solution is rejected as an extraneous solution.
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# 4.5: Applications of Fourier series
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## 4.5.1 Periodically forced oscillation
Let us return to the forced oscillations. Consider a mass-spring system as before, where we have a mass $$m$$ on a spring with spring constant $$k$$, with damping $$c$$, and a force $$F(t)$$ applied to the mass. Suppose the forcing function $$F(t)$$ is $$2L$$-periodic for some $$L>0$$. We have already seen this problem in chapter 2 with a simple $$F(t)$$.
The equation that governs this particular setup is
$mx''(t)+cx'(t)+kx(t)=F(t).$
The general solution consists of the complementary solution $$x_c$$, which solves the associated homogeneous equation $$mx''+cx'+kx=0$$, and a particular solution of Equation 4.5.1 we call $$x_p$$. For $$c>0$$, the complementary solution $$x_c$$ will decay as time goes by. Therefore, we are mostly interested in a particular solution $$x_p$$ that does not decay and is periodic with the same period as $$F(t)$$. We call this particular solution the steady periodic solution and we write it as $$x_{sp}$$ as before. What will be new in this section is that we consider an arbitrary forcing function $$F(t)$$ instead of a simple cosine.
For simplicity, let us suppose that $$c=0$$. The problem with $$c>0$$ is very similar. The equation
$mx''+kx=0$
has the general solution
$x(t)= A \cos(\omega_0 t)+ B \sin(\omega_0 t),$
where $$\omega_0= \sqrt{\dfrac{k}{m}}$$. Any solution to $$mx''(t)+kx(t)=F(t)$$ is of the form $$A \cos(\omega_0 t)+ B \sin(\omega_0 t)+x_{sp}$$. The steady periodic solution $$x_{sp}$$ has the same period as $$F(t)$$.
In the spirit of the last section and the idea of undetermined coefficients we first write
$F(t)= \dfrac{c_0}{2}+ \sum^{\infty}_{n=1} c_n \cos \left(\dfrac{n \pi}{L}t \right)+ d_n \sin \left(\dfrac{n \pi}{L}t \right).$
Then we write a proposed steady periodic solution $$x$$ as
$x(t)= \dfrac{a_0}{2}+ \sum^{\infty}_{n=1} a_n \cos \left(\dfrac{n \pi}{L}t \right)+ b_n \sin \left(\dfrac{n \pi}{L}t \right),$
where $$a_n$$ and $$b_n$$ are unknowns. We plug $$x$$ into the differential equation and solve for $$a_n$$ and $$b_n$$ in terms of $$c_n$$ and $$d_n$$. This process is perhaps best understood by example.
Example 4.5.1
Suppose that $$k=2$$, and $$m=1$$. The units are again the mks units (meters-kilograms-seconds). There is a jetpack strapped to the mass, which fires with a force of 1 newton for 1 second and then is off for 1 second, and so on. We want to find the steady periodic solution.
Solution
The equation is, therefore,
$x''+2x=F(t),$
where $$F(t)$$ is the step function
$F(t)= \left\{ \begin{array}{ccc} 0 & {\rm{if}} & -1<t<0, \\ 1 & {\rm{if}} & 0<t<1, \end{array} \right.$
extended periodically. We write
$F(t)= \dfrac{c_0}{2}+ \sum^{\infty}_{n=1} c_n \cos(n \pi t)+ d_n \sin(n \pi t).$
We compute
$c_n= \int^1_{-1} F(t) \cos(n \pi t)dt= \int^1_{0} \cos(n \pi t)dt= 0 ~~~~~ {\rm{for}}~ n \geq 1, \\ c_0= \int^1_{-1} F(t) dt= \int^1_{0} dt=1, \\ d_n= \int^1_{-1} F(t) \sin(n \pi t)dt \\ = \int^1_{0} \sin(n \pi t)dt \\ \left[ \dfrac{- \cos(n \pi t)}{n \pi}\right]^1_{t=0} \\ = \dfrac{1-(-1)^n}{\pi n}= \left\{ \begin{array}{ccc} \dfrac{2}{\pi n} & {\rm{if~}} n {\rm{~odd}}, \\ 0 & {\rm{if~}} n {\rm{~even}}. \end{array} \right.$
So
$F(t)= \dfrac{1}{2}+ \sum^{\infty}_{ \underset{n ~\rm{odd}}{n=1} }\dfrac{2}{\pi n} \sin(n \pi t).$
We want to try
$x(t)= \dfrac{a_0}{2}+ \sum_{n=1}^{\infty} a_n \cos(n \pi t)+ b_n \sin(n \pi t).$
Once we plug into the differential equation $$x'' + 2x = F(t)$$, it is clear that $$a_n=0$$ for $$n \geq 1$$ as there are no corresponding terms in the series for $$F(t)$$. Similarly $$b_n=0$$ for $$n$$ even. Hence we try
$x(t)= \dfrac{a_0}{2}+ \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} b_n \sin(n \pi t).$
We plug into the differential equation and obtain
$x''+2x = \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \left[ -b_n n^2 \pi^2 \sin(n \pi t) \right] +a_0+2 \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \left[ b_n \sin(n \pi t) \right] \\ = a_0+ \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} b_n(2-n^2 \pi^2) \sin(n \pi t) \\ = F(t)= \dfrac{1}{2}+ \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \dfrac{2}{\pi n} \sin(n \pi t).$
So $$a_0= \dfrac{1}{2}$$, $$b_n= 0$$ for even $$n$$, and for odd $$n$$ we get
$b_n= \dfrac{2}{\pi n(2-n^2 \pi^2)}.$
The steady periodic solution has the Fourier series
$x_{sp}(t)= \dfrac{1}{4}+ \sum_{\underset{n ~\rm{odd}}{n=1}}^{\infty} \dfrac{2}{\pi n(2-n^2 \pi^2)} \sin(n \pi t).$
We know this is the steady periodic solution as it contains no terms of the complementary solution and it is periodic with the same period as $$F(t)$$ itself. See Figure 4.12 for the plot of this solution.
Figure 4.12: Plot of the steady periodic solution $$x_{sp}$$ of Example 4.5.1.
### 4.5.2 Resonance
Just like when the forcing function was a simple cosine, resonance could still happen. Let us assume $$c=0$$ and we will discuss only pure resonance. Again, take the equation
$mx''(t)+kx(t)=F(t).$
When we expand $$F(t)$$ and find that some of its terms coincide with the complementary solution to $$mx''+kx=0$$, we cannot use those terms in the guess. Just like before, they will disappear when we plug into the left hand side and we will get a contradictory equation (such as $$0=1$$). That is, suppose
$x_c=A \cos(\omega_0 t)+B \sin(\omega_0 t),$
where $$\omega_0= \dfrac{N \pi}{L}$$ for some positive integer $$N$$. In this case we have to modify our guess and try
$x(t)= \dfrac{a_0}{2}+t \left( a_N \cos \left( \dfrac{N \pi}{L}t \right)+ b_N \sin \left( \dfrac{N \pi}{L}t \right) \right) + \sum_{\underset{n \neq N}{n=1}}^{\infty} a_n \cos \left( \dfrac{n \pi}{L}t \right)+ b_n \sin \left( \dfrac{n \pi}{L}t \right).$
In other words, we multiply the offending term by $$t$$. From then on, we proceed as before.
Of course, the solution will not be a Fourier series (it will not even be periodic) since it contains these terms multiplied by $$t$$. Further, the terms $$t \left( a_N \cos \left( \dfrac{N \pi}{L}t \right)+ b_N \sin \left( \dfrac{N \pi}{L}t \right) \right)$$ will eventually dominate and lead to wild oscillations. As before, this behavior is called pure resonance or just resonance.
Note that there now may be infinitely many resonance frequencies to hit. That is, as we change the frequency of $$F$$ (we change $$L$$), different terms from the Fourier series of $$F$$ may interfere with the complementary solution and will cause resonance. However, we should note that since everything is an approximation and in particular $$c$$ is never actually zero but something very close to zero, only the first few resonance frequencies will matter.
Example 4.5.2
Find the steady periodic solution to the equation
$2x''+18 \pi^2 x=F(t),$
where
$F(t)= \left\{ \begin{array}{ccc} -1 & {\rm{if}} & -1<t<0, \\ 1 & {\rm{if}} & 0<t<1, \end{array} \right.$
extended periodically. We note that
$F(t)= \sum^{\infty}_{ \underset{n ~\rm{odd}}{n=1} } \dfrac{4}{\pi n} \sin(n \pi t).$
Example 4.5.3
Compute the Fourier series of $$F$$ to verify the above equation.
The solution must look like
$x(t)= c_1 \cos(3 \pi t)+ c_2 \sin(3 \pi t)+x_p(t)$
for some particular solution $$x_p$$.
We note that if we just tried a Fourier series with $$\sin(n \pi t)$$ as usual, we would get duplication when $$n=3$$. Therefore, we pull out that term and multiply by $$t$$. We also have to add a cosine term to get everything right. That is, we must try
$x_p(t)= a_3 t \cos(3 \pi t) + b_3 t \sin(3 \pi t) + \sum^{\infty}_{ \underset{\underset{n \neq 3}{n ~\rm{odd}}}{n=1} } b_n \sin(n \pi t).$
Let us compute the second derivative.
$x_p''(t)= -6a_3 \pi \sin(3 \pi t) -9 \pi^2 a_3 t \cos(3 \pi t) + 6b_3 \pi \cos(3 \pi t) -9 \pi^2 b_3 t \sin(3 \pi t) +\sum^{\infty}_{ \underset{\underset{n \neq 3}{n ~\rm{odd}}}{n=1} } (-n^2 \pi^2 b_n) \sin(n \pi t).$
We now plug into the left hand side of the differential equation.
$2x''_p + 18 \pi^2 x=-12a_3 \pi \sin(3\pi t)-18 \pi^2 a_3 t \cos(3 \pi t)+12 b_3 \pi \cos(3 \pi t)-18 \pi^2 b_3 t \sin(3 \pi t) +18 \pi^2 a_3 t \cos(3 \pi t)+18 \pi^2 b_3 t \sin(3 \pi t)+ \sum^{\infty}_{ \underset{\underset{n \neq 3}{n ~\rm{odd}}}{n=1} } (-n^2 \pi^2 b_n+18 \pi^2 b_n) \sin(n \pi t).$
$2x''_p + 18 \pi^2 x=-12a_3 \pi \sin(3\pi t)-18 \pi^2 a_3 t \cos(3 \pi t)+12 b_3 \pi \cos(3 \pi t)-18 \pi^2 b_3 t \sin(3 \pi t)+ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+18 pi^2 a_3 t \cos(3 \pi t)~~~~~~~~~~~~~~~~~+18 \pi^2 b_3 t \sin(3 \pi t)+ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~ \sum^{\infty}_{ \underset{\underset{n \neq 3}{n ~\rm{odd}}}{n=1} } (-n^2 \pi^2 b_n+18 \pi^2 b_n) \sin(n \pi t).$
If we simplify we obtain
$2x_p'' +18 \pi^2 x= -12a_3 \pi \sin(3 \pi t)+ 12b_3 \pi \cos(3 \pi t) +\sum^{\infty}_{ \underset{\underset{n \neq 3}{n ~\rm{odd}}}{n=1} } (-2n^2 \pi^2 b_n+ 18 \pi^2 b_n) \sin(n \pi t.)$
This series has to equal to the series for $$F(t)$$. We equate the coefficients and solve for $$a_3$$ and $$b_n$$.
$a_3= \frac{4/(3 \pi)}{-12 \pi}= \frac{-1}{9 \pi^2}, \\ b_3 = 0, \\ b_n = \frac{4}{n \pi(18 \pi^2 -2n^2 \pi^2)}=\frac{2}{\pi^3 n(9-n^2 )} ~~~~~~ {\rm{for~}} n {\rm{~odd~and~}} n \neq 3.$
That is,
$x_p(t)= \frac{-1}{9 \pi^2}t \cos(3 \pi t)+ \sum^{\infty}_{ \underset{\underset{n \neq 3}{n ~\rm{odd}}}{n=1} } \frac{2}{\pi^3 n(9-n^2)} \sin(n \pi t.)$
When $$c>0$$, you will not have to worry about pure resonance. That is, there will never be any conflicts and you do not need to multiply any terms by $$t$$. There is a corresponding concept of practical resonance and it is very similar to the ideas we already explored in chapter 2. We will not go into details here.
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# Question #f992a
##### 1 Answer
Jul 4, 2016
The equation has infinitely many solutions. See below.
#### Explanation:
The important thing to realize in order to solve this problem is that if square roots are multiplied, you can group them together. In math terms:
$\sqrt{a} \cdot \sqrt{b} = \sqrt{a b}$
So, for example, $\sqrt{32} \cdot \sqrt{2} = \sqrt{32 \cdot 2} = \sqrt{64} = 8$.
In this equation, we have $\sqrt{{n}^{2} - 1} \cdot \sqrt{{n}^{2} + 1}$. Using the rule described above, we can combine them into one square root:
$\sqrt{{n}^{2} - 1} \cdot \sqrt{{n}^{2} + 1} = \sqrt{\left({n}^{2} - 1\right) \left({n}^{2} + 1\right)}$
The expression $\left({n}^{2} - 1\right) \left({n}^{2} + 1\right)$ may not look like anything we know at first, but recall the difference of squares property:
$\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$
$\left({n}^{2} - 1\right) \left({n}^{2} + 1\right)$ is actually a difference of squares, with $a = {n}^{2}$ and $b = 1$. Since this simplifies into ${a}^{2} - {b}^{2}$, we can say:
$\left({n}^{2} - 1\right) \left({n}^{2} + 1\right) = {\left({n}^{2}\right)}^{2} - {\left(1\right)}^{2} = {n}^{4} - 1$
We've just simplified $\sqrt{{n}^{2} - 1} \cdot \sqrt{{n}^{2} + 1}$ into $\sqrt{{n}^{4} - 1}$. Our problem now looks like:
$\sqrt{{n}^{4} - 1} = \sqrt{{n}^{4} - 1}$
Hm...both sides are the same! What does that mean? It means the equation has infinitely many solutions! Any value of $n$ will satisfy this equation.
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# Concept: Percentage
CONTENTS
INTRODUCTION
Concept of percentage is very important in any MBA Entrance Exam, because there are a lot of questions related to the use of percentages in Profit & Loss, Time & Work, Speed & Distance. Apart from this percentage plays an important role in handling Data Interpretation (DI) Calculation.
PERCENTAGE
The word percent consists of two words ‘per’ and ‘cent’. Per means each and cent means hundred. Thus, percent means ‘on each hundred’ or in each hundred. The symbol % is used to represent percent. In brief it is written as p.c.
Percent or Percentage can be defined as :
A percentage is a fraction whose denominator is 100 and the numerator is a definite quantity which is called the rate percent.
FRACTIONAL EQUIVALENTS OF IMPORTANT PERCENTS
To convert percentage into fraction/decimal, we need to divide the percentage by 100.
1% = $\frac{1}{100}$ 2% = $\frac{1}{50}$ 4% = $\frac{1}{25}$ 8% = $\frac{2}{25}$ 16% = $\frac{4}{25}$ 5% = $\frac{1}{20}$ 10% = $\frac{1}{10}$ 20% = $\frac{1}{5}$ 40% = $\frac{2}{5}$ 60% = $\frac{3}{5}$ 12.5% = $\frac{1}{8}$ 37.5% = $\frac{3}{8}$ 62.5% = $\frac{5}{8}$ 87.5% = $\frac{7}{8}$ 112.5% = $\frac{9}{8}$
DECIMAL EQUIVALENT OF FRACTIONS
To convert fraction/decimal into percentage, we need to multiply the fraction/decimal with 100.
$\frac{1}{1}$ = 1 = 100% $\frac{1}{2}$ = 0.5 =50% $\frac{1}{3}$ = 0.33 = 33.33% $\frac{1}{4}$ = 0.25 = 25% $\frac{1}{5}$ = 0.2 = 20% $\frac{1}{6}$ = 0.167 = 16.67% $\frac{1}{7}$ = 0.1428 = 14.28% $\frac{1}{8}$ = 0.125 = 12.5% $\frac{1}{9}$ = 0.111 = 11.11% $\frac{1}{10}$ = 0.1 = 10% $\frac{1}{11}$ = 0.0909 = 9.09% $\frac{1}{12}$ = 0.833 = 8.33%
SOME IMPORTANT RESULTS
• To convert (or express) any fraction a/b to rate percent – multiply it by 100 and put a % sign = $\frac{a}{b}$ × 100 %
• To convert (or express) any percentage P% into fraction – divide it by 100 = $\frac{P}{100}$
Example:. Express the following in terms of percentage:
(a) 0.4 (b) 1.0 (c) 5/3 (d) 7x/y (e) 1.23
Solution:
(a) multiply the decimal fraction by 100.
∴ 0.4 = (0.4 × 100)% = 40%
(b) 1.0 = (1.0 × 100)% = 100%
(c) 5/3 = (5/3 × 100)% = 166 2/3%
(d) 7x/y = (7x/y × 100)% = 700x/y%
(e) 1.23 = (1.23 × 100)% = 123%
To convert a rate percent to a fraction divide it by 100 and delete the % sign.
Example: Express the following in terms of fraction:
(a) 22$\frac{1}{2}$% (b) 35% (c) a2/b% (d) 0.3% (e) 2/7%
Solution:
(a) Divide the given percentage by 100 to convert it into a fraction
∴ 22$\frac{1}{2}$% = (22$\frac{1}{2}$)/100 = 45/(2 × 100) = 9/40
(b) 35% = 35/100 = 7/20
(c) a2/b% = a22/(b × 100) = a2/(100b)
(d) 0.3% = 0.3/100 = 3/1000
(e) 2/7% = 2/(7 × 100) = 1/350
Example: Find
(a) 9% of 27 (b) 0.02% of 6500 (c) 7/2% of 80 (d) 125% of 64 (e) 10% of 5% of 320
Solution:
(a) Multiply the number by P/100, if p% of the number is to be calculated.
∴ 9% of 27 = 9/100 × 27 = 243/100
(b) 0.02% of 650 = 0.02/100 × 6500 = 13/10
(c) 7/2% of 80 = 7/(2 × 100) × 80 = 14/5
(d) 125% of 64 = 125/100 × 64 = 80
(e) 10% of 5% of 320 = 10/100 × 5/100 × 320 = 1.6 = 8/5
Example: Find the following :
(a) 36 is what % of 144 (b) 7/8 is what % of 3/4
(c) What % of 80 is 16 (d) 0.625 is equal to what % of 1$\frac{7}{28}$
(e) 36 × 14 is what % of 1400 (f) R is what % of N
Solution:
(a) Here we have to compare 36 with 144.
Value of percentage (%) = (Number to be compared)/(Number to be compared against) × 100
⇒ % = 36/144 × 100 = 25%
(b) Value of % =(7⁄8)/(3⁄4)×100 = 116$\frac{2}{3}$%
(c) Value of % = Result/(Original number) × 100 = 16/80 × 100 = 20%
(d) Value of % = 0.625/(1$\frac{7}{28}$) × 100 = 0.625/35 × 28 × 100 = 50%
(e) value of % = (36 × 14)/1400 × 100 = 36%
(f) value of % = R/N × 100 = 100R/N%
Example: 25% of a number is 20, what is 40% of that number? Also find the number
Solution:
25% → 20
⇒ 40%→ 20/25 × 40 = 32
Original number (N) = $\frac{20}{25}$ × 100 = 80
Example: A number A exceeds B by 25%. By what percent is B less than A?
Solution:
If B = 100, then A = 125
⇒ B = $\frac{100}{125}$ × 100 = 80% of A
∴ B is (100 - 80) = 20% less than A
PERCENTAGE CHANGE
When a number is increased by p%, it means p% of the number is added in itself.
Example: Increase 120 by 50%?
Solution:
We add 50% of 120 with 120.
∴ 120 + 120 × 50% = 120 + 60 = 180
Similarly, when a number is decreased by p%, it means p% of the number is subtracted from itself.
Example: Decrease 120 by 50%?
Solution:
We subtract 50% of 120 from 120.
∴ 120 - 120 × 50% = 120 - 60 = 120.
MULTIPLICATION FACTOR
When a number 'I' is increased or decreased by P%.
Final number (F) = I ± P% of I = I(1 ± P%) = I$\left(1±\frac{P}{100}\right)$
Here, $\left(1±\frac{P}{100}\right)$ is called the multiplication factor.
Hence, when a number is to be increased or decreased by P%, we first calculate the multiplication factor and multiply if with the initial number to get the final number.
Note: Multiplication factor does not depend on the value of initial number. It only depends on the percentage change.
SOME IMPORTANT RESULTS
• F = I$\left(1±\frac{P}{100}\right)$ = I × multiplication factor (mf)
• I = $\frac{F}{mf}$
• P% = $\frac{F - I}{I}$ × 100%
• When a number is multiplied with f (i.e. multiplication factor is f)
∴ f = 1 + $\frac{P}{100}$
∴ Percentage change = (f - 1) × 100%
Example: The daily wage is increased by 15%, and a person now gets Rs. 23 per day. What was his daily wage before the increase?
Solution:
Multiplication factor = $\left(1+\frac{\mathrm{15}}{100}\right)$ = $\left(1+\frac{3}{20}\right)$ = $\frac{23}{20}$
∴ Original daily wage =$\frac{23}{23/20}$ = 20
Example: From a man’s salary, 10% is deducted on tax, 20% of the rest is spent on the education, and 25% of the rest is spent on food. After all these expenditures, he is left with Rs. 2,700 Find his salary.
Solution:
Let the salary be S.
10% of salary is deducted on tax.
∴ Remaining salary = $S\left(1-\frac{10}{100}\right)$
Now, 20% of this remaining salary is spent on education.
∴ Remaining salary = $S\left(1-\frac{10}{100}\right)\left(1-\frac{20}{100}\right)$
Now, 25% of this remaining salary is spent on food.
∴ Remaining salary = $S\left(1-\frac{10}{100}\right)\left(1-\frac{20}{100}\right)\left(1-\frac{25}{100}\right)$ which is equal to 2700
⇒ $S\left(1-\frac{10}{100}\right)\left(1-\frac{20}{100}\right)\left(1-\frac{25}{100}\right)$ = 2700
⇒ $S×\frac{9}{10}×\frac{4}{5}×\frac{3}{4}$ = 2700
⇒ S = $\frac{2700×50}{27}$ = Rs. 5000
Example : A number is tripled. What is the percentage change.
Solution:
When a number is tripled, it is effectively multiplied with 3.
∴ % change = (3 - 1) × 100 = 200%
Example : A number is increased by 50%. By what percentage should it be decreased to get back the original number.
Solution:
For 50% increase, multiplication factor = $\left(1+\frac{50}{100}\right)$ = $\frac{3}{2}$
∴ New number = Initial number × $\frac{3}{2}$
To get the original number, this new number must be multiplied with $\frac{2}{3}$
∴ % change = (2/3 - 1) × 100 = -33.33%
∴ The new number should be decreased by 33.33% to get back the original number.
SUCCESSIVE PERCENTAGE CHANGE
The concept of successive percentage change deals with two or more percentage changes applied to quantity consecutively. In this case, the final change is not the simple addition of the two percentage changes (as the base changes after the first change).
Example: If a number I is successively increased by a% and then b%.
Number after a% increase = I$\left(1+\frac{a}{100}\right)$
Now, this number has to be again increased by b%.
∴ Final number = I$\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)$
Similarly, when a number I is successively increased by a%, b% and c%
Final number = I$\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)\left(1+\frac{c}{100}\right)$
Example : When a number is successively changed by a% and b%, what is the overall percentage change.
Solution:
When a number is successively changed by a% and b%, the overall multiplication factor = $\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)$ = f
∴ percentage change = (f - 1) × 100% = $\left(1+\frac{a}{100}+\frac{b}{100}+\frac{a×b}{{100}^{2}}-1\right)$ × 100% = a + b + $\frac{a×b}{100}$
Note: a and b can be either positive or negative and the sequence of increase or decrease does not affect the overall change.
Remember:
• When a number I is successively increased by a% and b%
∴ Overall % change (P) = a + b + $\frac{ab}{100}$
• When a number I is successively increased by a% and then decreased by a%
∴ Overall % change (P) = - $\frac{{a}^{2}}{100}$%
• When a number I is successively increased by a% n times
∴ Final number (F) = I${\left(1+\frac{a}{100}\right)}^{n}$
• After increasing a number by a% successively n times, final number F is obtained
∴ Initial number (I) = $\frac{F}{{\left(1+\frac{a}{100}\right)}^{n}}$
• If P is the original value of asset, and the depreciation is r% per year
∴ Value after n years = P${\left(1-\frac{r}{100}\right)}^{n}$
Example : When a number is successively changed by 20% and 25%, what is the overall percentage change.
Solution:
Method 1: (Multiplication Factor)
Overall multiplication factor = $\left(1+\frac{20}{100}\right)\left(1+\frac{25}{100}\right)$ = $\frac{6}{5}×\frac{5}{4}$ = $\frac{3}{2}$
∴ % change = (3/2 - 1) × 100 = 50%
Method 2: (Formula)
Overall % change = 20 + 25 + $\frac{\mathrm{20}×\mathrm{25}}{100}$ = 50%
Example : When a number is successively changed by -20% and 25%, what is the overall percentage change.
Solution:
Method 1: (Multiplication Factor)
Overall multiplication factor = $\left(1-\frac{20}{100}\right)\left(1+\frac{25}{100}\right)$ = $\frac{4}{5}×\frac{5}{4}$ = 1
∴ % change = (1 - 1) × 100 = 0% i.e. no change
Method 2: (Formula)
Overall % change = -20 + 25 + $\frac{\mathrm{-20}×\mathrm{25}}{100}$ = 0%
Example : When a number is successively changed by -20% and -25%, what is the overall percentage change.
Solution:
Method 1: (Multiplication Factor)
Overall multiplication factor = $\left(1-\frac{20}{100}\right)\left(1-\frac{25}{100}\right)$ = $\frac{4}{5}×\frac{3}{4}$ = $\frac{3}{5}$
∴ % change = (3/5 - 1) × 100 = - 40%
Method 2: (Formula)
Overall % change = - 20 - 25 + $\frac{\mathrm{-20}×\mathrm{-25}}{100}$ = - 40%
Example : When sugar price increase by 25%, by what percentage should a family reduced its consumption so that monthly expenditure does not change.
Solution:
Method 1:
Let the initial price of sugar = Rs. 100/kg and quantity consumed = 10kg / month
∴ Total Expenditure = 100 × 10 = Rs. 1,000
Now, price increases by 25%, hence, new price = 125
New quantity consumed = 1,000/125 = 8 kg
∴ quantity reduces from 10 kg to 8 kg i.e. reduction of 20%
Method 2: (Multiplication factor)
Expenditure = Price × Quantity
⇒ multiplication factor for expenditure = multiplication factor for price × multiplication factor for quantity
Overall multiplication factor for expenditure = 1 (no change)
Multiplication factor for price = $\left(1+\frac{25}{100}\right)$ = $\frac{5}{4}$
Let multiplication factor for quantity = f
∴ 1 = 5/4 × f
⇒ f = 4/5
∴ % change in quantity = (4/5 - 1) × 100 = -20%
Example : When sugar price increase by 50%, by what percentage should a family reduced its consumption so that monthly expenditure goes up by only 20%.
Solution:
Method 1:
Let the initial price of sugar = Rs. 100/kg and quantity consumed = 10kg / month
∴ Total Expenditure = 100 × 10 = Rs. 1,000
Now, price increases by 50%, hence, new price = 150
Expenditure increases by 20%, hence, new expenditure = 1200
New quantity consumed = 1,200/150 = 8 kg
∴ quantity reduces from 10 kg to 8 kg i.e. reduction of 20%
Method 2: (Multiplication factor)
Expenditure = Price × Quantity
⇒ multiplication factor for expenditure = multiplication factor for price × multiplication factor for quantity
Overall multiplication factor for expenditure = $\left(1+\frac{20}{100}\right)$ = $\frac{6}{5}$
Multiplication factor for price = $\left(1+\frac{50}{100}\right)$ = $\frac{3}{2}$
Let multiplication factor for quantity = f
∴ 6/5 = 3/2 × f
⇒ f = 4/5
∴ % change in quantity = (4/5 - 1) × 100 = -20%
Example: Groundnut oil is being sold at Rs. 27 per kg. During last month its cost was Rs. 24 per kg. Find by how much % a family should reduce its consumption, so as to keep the expenditure the same.
Solution:
Let the initial consumption of groundnut oil be = x and new expenditure = x1
Expense (Initial) = 24x and Expense (Final) = 27x1
∴ 24x = 27 × x1 [since expense has to remain the same]
x= 8x/9
In the above fraction we can see that the new consumption is 8/9th fraction of the old consumption.
Thus, the reduction is 1/9th of the original consumption.
Hence, percentage reduction in consumption is 1/9 × 100%.
Example: If the radius of the circle is increased by 20%, its area will increase by what %?
Solution:
We know that Area of circle = 𝜋 × r × r
overall multiplication factor for area = $\left(1+\frac{20}{100}\right)×\left(1+\frac{20}{100}\right)$ = ${\left(1+\frac{20}{100}\right)}^{2}$ = ${\left(\frac{6}{5}\right)}^{2}$ = $\frac{36}{25}$
∴ % change in area (36/25 - 1) × 100 = 44%
Hence, increase in area = 44%
Example: If the sides of a square are increased by 30%. Find the resultant increase in area.
Solution:
Let A ∝ a2 where ‘a’ = initial side
New area A1 ∝ (1.3a)2
⇒ A1 ∝1.69a2
Thus, it can be seen that new area is 69% more than the old area.
Example: A reduction of 25% in the price of Cadbury chocolates, enables a person to buy 5 kg more for Rs. 120. Find the original price of Cadbury chocolates per kg.
Solution:
Method 1 :
In this question the fixed expense = Rs. 120
If P is the initial price, reduced price = 0.75P
Now, new quantity (at the reduced price) – initial quantity = 5
120/0.75P - 120/P = 5
⇒ P = Rs. 8
Alternate Method:
Since the price reduces by 25%, the initial quantity can be bought for 120 × 0.75 = Rs. 90 at the reduced price.
∴ Remaining amount = 120 - 90 = Rs. 30
Now, 5 kg extra can be bought for this Rs. 30
∴ Reduced price = 30/5 = Rs. 6 = 0.75P
∴ Initial price = 6/0.75 = Rs. 8
Example: When N is reduced by 4, it becomes 80% of itself. What is the value of N?
Solution:
The problem implies that
20% of N = 4 (since 100 – 80 = 20)
N = 4/20 × 100
∴ N = 20
Example: If 10% of an electricity bill is deducted, Rs. 45 is still to be paid. How much was the bill?
Solution:
If 10% of the bill is reduced, still 90% has to be paid
Here the decreased bill = Rs. 45
⇒ 90% of Total Bill = 45
⇒ N = (45 × 100)/90
∴ N = 50
Hence, the bill was Rs. 50
Example: If the price of 1 kg of cornflakes is increased by 25%, the increase is Rs. 10. Find the new price of cornflakes per kg.
Solution:
Due to 25% increase, the price changes by Rs. 10
⇒ 25% of price = 10
⇒ price = (10 × 100)/25 = Rs. 50
∴ New price of cornflakes per kg is Rs. 50
Example: In an election between two candidates, a candidate who gets 62% of the total votes polled is elected by a majority of 144 votes, Find the number of votes polled and votes secured by the candidate. [All votes polled are valid votes]
Solution:
Let, the total number of votes be 100x
Winning candidate gets 62% = 62x votes
Losing candidate gets (100 - 62 =) 38% = 38x votes
Now, difference of votes is 144
∴ 62x - 38x = 144
⇒ 24x = 144
⇒ x = 6
Thus, total number of votes = 100x = 600 votes and winner gets = 62x = 372 votes
Example: The rate for admission to an exhibition was Rs. 5 and was later reduced by 20%. As a result, the sale proceeds increased by 44%. What was the % increase in attendance?
Solution:
Sale proceeds = price × attendance
Multiplication factor for sales =
Multiplication factor for price =
Let the multiplication factor for attendance = f
⇒ 36/25 = 4/5 × f
⇒ f = 36/25 × 5/4 = 9/5
∴ % change in attendace = (9/5 - 1) × 100 = 80%
∴ increase in attendance = 80%
Alternate Method:
Let the initial attendace = 100 people
∴ Initial sale proceeds = 5 × 100 = Rs. 500
Now, new price = 5 × 0.8 = Rs. 4
and, new sales proceeds = 500 × 1.44 = Rs. 720
∴, new attendance = 720/4 = 180
∴, Increase in attendance = (180 - 100)/100 × 100 = 80%
* The information of price is redundant in this question.
Example: A piece of cotton cloth, 20m long, shrinks by 0.5% after washing. Find the length of the cloth after washing.
Solution:
Let the length of the cloth after washing be x metro or Now it left 100 – 0.5 = 99.5%
∴ x = 20 × 99.5%
⇒ x = 19.9 meter
Hence, original length = 19.9 meter
Example: By mistake, a line was measured as 11.25 cm, which was 2.5% more than the actual length. Find the actual length of the line?
Solution:
let actual length of line be x cm
∴ x(100 + 2.5)% = 11.25
⇒ x = (11.25 × 100)/102.5 = 10.9
Hence, actual length = 10.9 cm
Example: When the price of a pressure cooker was increased by 15%, the sale of pressure cookers (quantity) decreased by 15%. What was the net effect on the sales?
Solution:
Let the original price be Rs. 100 also the number of item 100.
∴ Revenues = 100 × 100 = 10000
After change price = Rs. 115 and item = 85
Now, Revenue = 115 × 85 = 9775
∴ decrease % =(10000 - 9775)/10000 × 100
Hence, decreased by = 2.25%
Alternate Method :
1.15 × 0.85 = 0.9775
Thus, decrease = (0.9775 - 1) × 100 = 2.25%
Example:. A medical student has to secure 40% marks to pass. He gets 80 marks and fails by 60 marks. Find the maximum marks?
Solution:
Let the maximum number of marks be x
∴ x × 40% = 80 + 60 (he needs to pass)
⇒ x = 350
Hence, maximum marks = 350
Example: When 75% of a number is added to 75, the result is the number again. The number is
Solution:
let the number be x
∴ 75% of x + 75 = x
⇒ 75 = x – 75% of x
⇒ x = 7500/25 = 300
Hence, the original number is 300
Example: What is the total number of candidates in an examination, if 31% fail, and the number of those who pass exceeds the number of those who fail by 247?
Solution:
Let the total number of candidates be x
Pass % = 100 – 31 = 69
Now, 69% of x – 31% of x = 247
⇒ 38% of x = 247 ⇒ x = (247 × 100)/38 = 650
Hence, the total number = 650
Example: Rita loses 12$\frac{1}{2}$% of her money and after spending 70% of the remainder, has Rs. 210 left. How much money did she have at first?
Solution:
Let Rita have Rs. x at first
30% of (87$\frac{1}{2}$% of x) = 210
or 3/10 × 7/8 × x = 210
⇒ x = Rs. 800. Hence, Rita had Rs. 800 at first.
Example 5% people of a village died by bombardment, 15% of the remainder left the village on account of fear. If now the population is reduced to 3553. How much was it in the beginning?
Solution:
Let the population at beginning be x
∴ people left after bombardment = 95% of x
and number after leaving 15% people = 85% of (95% of x) = 3553 (given)
⇒ 85/100 × 95x/100 = 3553 ⇒ x = 4400
Example: In a factory, there are 40% technicians and 60% non-technicians. If the 60% of the technicians and 40% of non-technicians are permanent employees, then find the percentage of worker who are temporary?
Solution:
Let the number of employees be 100
∴ Number of technicians = 40
∴ Number of non-technicians = 60
Now, Number of temporary employee
= 40% of 40 + 60% of 60 = 16 + 36 = 52
Hence, 52% of the workers is temporary
Example: A salesman receives a salary of Rs. 1000 per month and commission of 8% on all sales in excess of Rs. 2000. What should be the amount of his sales in a particular month if he were to earn Rs. 5000 in that month
Solution:
Let amount of his sales be x
∴ 1000 + 8% (x – 2000) = 5000
⇒ x = Rs. 52000
Example: During one year, the population of a town increased by 10% and during the next year, it diminished by 10%. If at the end of the second year the population was 89100, what was it at the beginning?
Solution:
Let the population at the beginning = x
net decrease % (10)2/10% = 1%
∴ Population at the end = x × 99% = 89100
x = (89100 × 100)/99
x = 90,000
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# 38316: Unlocking the Secrets of DNA Structure
Do fractions make you quiver and shake?
Well, fear no more!
By harnessing the power of the least common denominator (LCD), you’ll be able to conquer these perplexing problems in no time.
Plus, we’ll introduce you to a handy MathStep app that’ll make learning and practicing fractions a piece of cake.
So, get ready to demystify fractions and unleash your inner math whiz!
## 3 8 3 16
To add the fractions 3/8 and 3/16, we can follow these steps:
1.
Find the least common denominator (LCD) of the two fractions, which is 16.
2.
Convert both fractions to have the same denominator (16).
Multiply the numerator and denominator of 3/8 by 2, resulting in 6/16.
3.
Now that both fractions have the same denominator, we can simply add the numerators: 6/16 + 3/16 = 9/16.
In summary, to add 3/8 and 3/16, we convert them to have a common denominator of 16, resulting in 6/16 and 3/16.
Adding the numerators gives us the sum of 9/16.
Key Points:
• Find the least common denominator (LCD) of the fractions, which is 16
• Convert both fractions to have the same denominator (16) by multiplying the numerator and denominator of 3/8 by 2, resulting in 6/16
• Both fractions have the same denominator, so add the numerators: 6/16 + 3/16 = 9/16
• Convert 3/8 and 3/16 to have a common denominator of 16, resulting in 6/16 and 3/16
• Add the numerators to get the sum of 9/16
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## Pro Tips:
1. The number “3” is a significant figure in mathematics as it is the only number that is both a prime number and a Fibonacci number.
2. Did you know that the number “8” is considered to be a lucky number in many cultures, especially in Chinese traditions? It is believed to bring prosperity and wealth.
3. The “3” in Roman numerals is represented by the letter “III.” However, the number “8” is depicted as “VIII” by combining the Roman numerals for “5” (V) and “3” (III).
4. In the game of chess, there are a total of 16 pieces for each player. These include one king, one queen, two rooks, two knights, two bishops, and eight pawns.
5. The number “16” is associated with the age of independence in many countries, as it is the legal age at which individuals can obtain a driver’s license or vote in some places.
## 1. Explaining The Process Of Adding Fractions
Adding fractions may appear complex initially, but understanding the process makes it simpler. When we add fractions, we combine two or more parts to form a whole. The fraction has a numerator, representing the number of parts we have, and a denominator, representing the total number of equal parts in the whole.
• Numerator – represents the number of parts
• Denominator – represents the total number of equal parts
“Adding fractions involves combining parts to form a whole.”
## 2. Adding 3/8 And 3/16: The Step-By-Step Method
To add 3/8 and 3/16 together, we first need to find the least common denominator (LCD). The LCD is the smallest multiple that the two denominators have in common. In this case, the LCM of 8 and 16 is 16.
Once we have found the LCD, we can convert the fractions to equivalent forms with the same denominator. To do this, we multiply the numerator and denominator of each fraction by a number that will result in the denominator being equal to the LCD. For 3/8, we multiply both the numerator and denominator by 2, giving us 6/16.
Now that both fractions have the same denominator, we can simply add the numerators together. 6/16 + 3/16 equals 9/16, which is our final answer.
• Find the least common denominator (LCD), which is the smallest multiple that the two denominators have in common.
• Convert the fractions to equivalent forms with the same denominator by multiplying the numerator and denominator by a number that will result in the denominator being equal to the LCD.
## 3. Finding The Least Common Denominator (LCD)
The least common denominator (LCD) is crucial in adding fractions with different denominators. It is the smallest multiple that two or more denominators have in common. In our example of 3/8 and 3/16, the LCD is 16.
To find the LCD, we can start by listing the multiples of each denominator. For 8, the multiples are 8, 16, 24, 32, and so on. For 16, the multiples are 16, 32, 48, 64, and so on. The LCD is the smallest number that appears in both lists, which is 16.
## 4. Converting Fractions To Equivalent Forms
Converting fractions to equivalent forms is a helpful technique when adding fractions with different denominators. By multiplying both the numerator and denominator by the same number, we can ensure that the value of the fraction remains unchanged.
In the given example of 3/8 and 3/16, we converted 3/8 to 6/16. This involved multiplying both the numerator and denominator of 3/8 by 2. As a result, we obtained an equivalent fraction with the same value but a different representation.
## 5. Introduction To Mathstep: A Mobile App For Fraction Addition
MathStep is a mobile app that aims to help users learn and practice a range of mathematical concepts, with a special focus on fraction addition. This user-friendly tool is suitable for both students and adults who wish to sharpen their math skills. By using MathStep, you can effectively understand and master the process of adding fractions.
MathStep is a mobile application that can be downloaded on Android and iOS devices. To install the app, visit the Google Play Store or Apple App Store and search for “MathStep”. Once you have located the app, click on the download button and follow the provided instructions.
The installation process is straightforward and should only take a few minutes.
By having MathStep on your device, you can access a powerful tool that is designed to enhance your mathematical abilities. It provides numerous features and functions to assist with various math-related tasks.
Give MathStep a try today and experience its benefits firsthand!
• Visit the Google Play Store or Apple App Store and search for “MathStep” to find the app.
• MathStep is a powerful tool that can enhance your mathematical abilities.
## 7. Learning And Practicing Addition With Fractions
With MathStep, learning and practicing addition with fractions becomes an interactive and engaging experience. The app provides step-by-step explanations and multiple examples to ensure a thorough understanding of the concept. Users can also complete interactive exercises to test their skills and track their progress over time.
• MathStep offers a hands-on approach to learning addition with fractions.
• The app provides detailed explanations that make the concept easy to grasp.
• Multiple examples are included to reinforce understanding.
• Interactive exercises allow users to practice addition with fractions.
• Users can track their progress to see how they are improving over time.
“MathStep takes the complexity out of learning addition with fractions.”
## 8. Working With 9/16 And Other Fractions
MathStep is an app that can assist you in working with fraction additions. In addition to the example of 3/8 and 3/16, the app also provides guidance on adding fractions like 9/16. By following the detailed instructions, you’ll be able to confidently add fractions, including 9/16, and tackle more complex problems. The step-by-step method offered by MathStep ensures that solutions are presented in a clear and concise manner.
• MathStep is an app that helps with fraction additions
• Provides guidance on adding fractions like 9/16
• Step-by-step instructions for clear and concise solutions
## 9. Availability And Features Of MathStep
MathStep is a free app that can be downloaded on both Android and iOS devices. This app is specifically designed to help users learn and understand fraction addition. It offers a wide range of features that are aimed at supporting the learning process.
One of the strengths of MathStep is its intuitive interface and user-friendly design. This makes it accessible to people of all ages and different levels of mathematical proficiency. Whether you are a beginner or an advanced learner, MathStep can be a valuable tool for enhancing your fraction addition skills.
The app allows you to practice fraction addition through interactive exercises and quizzes. It provides step-by-step guidance, ensuring that you grasp the concepts and techniques involved in this topic.
In addition, MathStep offers useful resources such as tutorials and examples to further enhance your understanding. It also provides instant feedback, allowing you to identify and correct any mistakes you make along the way. This immediate feedback helps reinforce your learning and ensures that you are making progress.
To summarize, MathStep is an excellent app for anyone looking to improve their fraction addition skills. Its accessible and user-friendly features make it suitable for learners of all backgrounds. With MathStep, you can enhance your understanding of fraction addition and build confidence in this fundamental area of math.
Comprehensive set of features for fraction addition learning
Intuitive interface and user-friendly design
Suitable for users of all ages and mathematical backgrounds
Interactive exercises and quizzes for practice
Step-by-step guidance for effective learning
Tutorials and examples for further understanding
Instant feedback for identifying and correcting mistakes
-* Enhances understanding and builds confidence
## 10. Finding Equivalent Fractions For Fraction Addition
Finding equivalent fractions is an essential step in adding fractions. MathStep simplifies this process by automatically converting fractions to equivalent forms with the same denominator. By presenting the solution in the simplest form, MathStep ensures accurate and efficient fraction addition.
With the app, users can grasp the concept of equivalent fractions and apply it to various mathematical problems.
Features of MathStep include:
• Conversion of fractions to equivalent forms
• Common denominators for easy addition
• Simplified solutions in the simplest form
In summary, MathStep is a powerful tool that helps users understand and apply the concept of equivalent fractions in their mathematical calculations.
💡
## You may need to know these questions about 3 8 3 16
### What is 3 8 called?
In mathematics, 3/8 is often referred to as three eighths. It is a fraction that represents the division of three equally divided parts out of eight. In its decimal form, 3/8 is equivalent to 0.375. This fraction has a unique quality as it can be used to represent parts that are divided into eight equal sections. Whether used in calculations or visual representations, three eighths is a fundamental concept in mathematics.
### 1. What is the significance of the numbers 3, 8, 3, and 16 in a mathematical context?
In a mathematical context, the numbers 3, 8, 3, and 16 hold different significant meanings. The number 3 is a prime number and the first odd prime number. It plays a vital role in various mathematical concepts such as triangles (having three sides) and trigonometry. Additionally, 3 is associated with many mathematical operations, including multiplication, division, and exponentiation.
The number 8, on the other hand, represents the power of two. It is the cube of the number 2 and is often known as a perfect cube. 8 also holds significance in terms of octal systems and binary codes. It is used as a base number in different mathematical calculations and is widely used in digital technology.
The number 16 is predominantly recognized as the base of the hexadecimal system. It is widely used in computing and represents a form of shorthand to express binary numbers. Furthermore, 16 plays a crucial role in geometry, especially in measuring angles and the divisions of a circle. Overall, these numbers have distinct meanings and applications in various mathematical contexts.
### 2. How can the sequence 3, 8, 3, 16 be interpreted in terms of a pattern or rule?
The sequence 3, 8, 3, 16 can be interpreted as an alternating pattern. The first number, 3, is then followed by a larger number, 8. The next number is a smaller one, 3, and then it is followed by an even larger number, 16. This alternating pattern suggests that the sequence may be following a rule where each number is either increasing or decreasing in magnitude, alternatingly.
### 3. Are there any notable historical events or cultural references associated with the numbers 3, 8, 3, and 16?
There are several notable historical events and cultural references associated with the numbers 3, 8, 3, and 16. In terms of historical events, one notable occurrence is the signing of the Treaty of Paris on September 3, 1783, which officially ended the American Revolutionary War and granted independence to the United States. Another important event is the three-day Battle of Gettysburg, which took place from July 1 to 3, 1863, during the American Civil War and marked a major turning point in the conflict.
In terms of cultural references, the number 8 holds significance in many cultures as a symbol of luck, wealth, and prosperity. This belief is particularly prevalent in Chinese culture, where the number 8 is considered extremely fortunate and is associated with good fortune and financial success. Additionally, the number 8 is also seen as a symbol of eternity and infinity.
However, I couldn’t find any specific historical events or cultural references associated with the numbers 3 and 16 related to your question.
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Decimals are taught in elementary school mathematics. After learning multiplication with decimals, you need to be able to do division.
When calculating decimals, a division is a bit more complicated. This is because it is not always divisible. You have to calculate until you can divide, or use the remainder to get the answer. It is also common to move the decimal point.
When it comes to calculating decimals, addition, subtraction, and multiplication are easy. Division, on the other hand, requires a lot of understanding. In this section, we will explain how to do decimal division.
## How to Do Division by Decimals and Integers: The Case of Divisible Numbers
Let’s start with the simple division of decimals. As long as the numbers are divisible, the calculations are not complicated. For example, how do you calculate the following?
• $24.22÷7$
When doing this calculation, create the following long division.
The way to do division is the same as for long division with whole numbers. Ignore the decimal point and perform the division calculation. Then, add the decimal point to the answer (quotient) in the same position. The result is as follows.
When doing division using decimals and integers, it is not difficult to calculate as long as the numbers are divisible. After dividing, add the decimal point in the same place.
In some calculation problems, the ones place of the answer may be zero. In this case, you should always write a zero in the ones place. The following is an example.
• $3.22÷7$
In calculating decimals, the number can be smaller than one. Understand that in the division of decimals, the ones place can be zero.
### Division of Decimals Until Divisible
On the other hand, there are some problems that need to be calculated until a number is divisible. In this case, the calculation is done by adding zeros to the right of the decimal.
There are hidden zeros in decimals. For example, 1.4 can be written as 1.400. However, since there is an infinite number of zeros, we omit the zeros in decimals. In any case, it is important to understand that in decimals, there are many zeros to the right of the number.
Therefore, when calculating divisible decimals, add zeros to get the answer. For example, how do we calculate the following?
• $1.4÷25$
When doing this calculation, let’s change the value to 1.40 instead of 1.4. The result is as follows.
However, the calculation should not end here. From this point, we can do more division. So, let’s change 1.40 to 1.400 and do the next calculation.
By adding zeros to the right of the decimal, we were able to get the answer. When doing division, if it is divisible, try to get the answer by this method.
### When the Divisor Is a Decimal, Move the Decimal Point to Perform the Division
So far, we have discussed division when the divisor is an integer. On the other hand, when a divisor is a decimal number, how do we calculate it?
When the divisor is a decimal, make sure to move the decimal point. By moving the decimal point to the right, you make the divisor an integer. If you don’t do this, you will not be able to divide decimals. For example, how would you do the following calculation?
• $3.22÷1.4$
The long division is as follows.
The divisor is 1.4. If we don’t change it to an integer, we will not be able to do the division. So, instead of 1.4, let’s make it 14. Move the decimal point one place to the right.
At the same time, move the decimal point of 3.22 one place to the right as well. Since we have moved the decimal point of the divisor, we need to do the same for the dividend. Therefore, the long division looks like this.
Understand that when the divisor is a decimal, it cannot be divided without changing the decimal point. Therefore, we need to move the decimal point like this.
We need to change the divisor to an integer, and how many times we need to move the decimal point to the right depends on the divisor. For example, in the following calculation, we need to shift the decimal point two places to the right.
• $6.848÷2.14$
Let’s do the following long division calculation.
Understand how to move the decimal point and change a divisor to an integer.
### Why Moving the Decimal Point Doesn’t Matter
There is a question that is asked by many people. Why is it okay to move the decimal point? The reason for this is that even if you change the position of the decimal point, the answer will be the same.
For example, the following will all give the same answer.
• $100÷20=5$
• $10÷2=5$
• $1÷0.2=5$
• $0.1÷0.02=5$
The answer will be the same if you multiply the number by 10 or 100 for both the divisor and the dividend. For example, compare $10÷2=5$ and $100÷20=5$.
In the same way, we can see that for $1÷0.2$ and $0.1÷0.02$; we can get the same answer by multiplying both the divisor and the dividend by 10 or 100.
For $1÷0.2$, multiply both numbers by 10 to get $10÷2$. For $0.1÷0.02$, multiply both numbers by 100 to get $10÷2$. In this way, decimals can be converted to integers. This method allows us to divide decimals.
## Division of Decimals and Integers with Remainders
So far, we have discussed division when the number is divisible. On the other hand, what should we do with numbers that are not divisible? When an integer dividend is not divisible, we use the remainder. This is also true for decimal division.
For example, how do we calculate the following?
• $22.3÷4$
If you make a long division and calculate it, you will get the following.
The method of division is the same as the one explained so far. The only difference is that the remainder is given. If you calculate $22.3÷4$, the answer will be 5.5 R 0.3.
When calculating the remainder, make sure to put the decimal point directly below. When calculating decimals, it is easy to make mistakes with the position of the decimal point. If the position of the decimal point is different, the remainder will be different. If this happens, the answer will not be correct, so be sure to check the position of the decimal point.
### How to Calculate Division of Two Decimals with Remainder
On the other hand, how do we calculate if the divisor is a decimal? When there is a remainder, it is easy to make a miscalculation if the divisor is a decimal. Therefore, we need to understand how to calculate them.
For example, how do we calculate the following?
• $10.86÷3.3$
In order to do the math, we need to make the divisor an integer. So instead of 3.3, let’s change it to 33. We can do the math as follows.
Finally, we need to find the remainder. We must pay attention to the position of the decimal point in the remainder; it must be put directly below the previous decimal point. The result is as follows.
The answer to $10.86÷3.3$ is 3.2 R 0.3.
By multiplying the divisor and dividend by 10, the quotient is 3.2. On the other hand, for the remainder, instead of using 108.6 multiplied by 10 as the reference number, we use the original number, 10.86, as the reference number. We put the decimal point of 10.86 directly below to get the remainder.
### Why the Decimal Point Is Different in the Quotient and the Remainder
The reason why it is easy to make miscalculations when dividing decimals is that the position of the decimal point is different for the quotient and remainder. Why does the decimal point change between the quotient and remainder? Let’s understand the reason for this.
As explained before, the quotient is the same even if you multiply the divisor and the dividend by 10 (or 100). The answer is the same, as shown below.
• $100÷20=5$
• $10÷2=5$
Next, let’s try multiplying by 10 for the non-divisible numbers. What will be the answers to the following questions?
• $100÷30=3$ … $10$
• $10÷3=3$ … $1$
As you can see, the quotient is the same even if we multiply the divisor and the dividend by 10. On the other hand, we can see that the remainder is 10 times larger. Also, if we multiply the divisor and the dividend by 100, the remainder is 100 times greater. In division, if we multiply numbers by 10 or 100, the remainder also increases accordingly.
If the remainder is 10 or 100 times larger than the original number, the answer will be different. For example, in the case of $10÷3$, if we multiply the divisor and the dividend by 10, as explained earlier, the remainder will be 10 times larger than the original number.
Even though the divisor is 3, the remainder is 10. Therefore, it is obvious that the answer is wrong.
So when dividing, use the number before multiplying by 10 (or 100) for the remainder. By using the number before the decimal point is moved as the reference, we can add the decimal point in the correct place for the remainder.
## Calculating Decimals to Find the Quotient and the Remainder
When we do division, the expression may contain decimals. So, let’s understand how to divide them.
When dividing decimals, it is easy if the number is divisible. All you have to do is to divide in the usual way, paying attention to the position of the decimal point. If the divisor is a decimal, you can change the divisor to an integer by moving the decimal point.
On the other hand, if there is a remainder, it is easy to make a miscalculation. If the divisor is a decimal, the quotient is based on the number after the decimal point is moved. Also, the remainder is based on the number before the decimal point is moved, and then the decimal point needs to be added.
Understand that there are rules for these calculations, and perform division calculations involving decimals.
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Question Video: Graphing Systems of Inequalities with Parabolas | Nagwa Question Video: Graphing Systems of Inequalities with Parabolas | Nagwa
Question Video: Graphing Systems of Inequalities with Parabolas
Determine which region in the graph contains all the solutions to the following system of inequalities: 𝑦 ≤ (1/2)𝑥² − 𝑥 + 1, 𝑦 ≤ −𝑥² + 2𝑥 + 4.
03:14
Video Transcript
Determine which region in the graph contains all the solutions to the following system of inequalities. 𝑦 is less than or equal to one-half 𝑥 squared minus 𝑥 plus one, and 𝑦 is less than or equal to negative 𝑥 squared plus two 𝑥 plus four.
In this question, we’ve been given two quadratic inequalities. These are represented on our graph somehow. Let’s simply begin by identifying which curve represents which equation. Let’s take the equation 𝑦 equals one-half 𝑥 squared minus 𝑥 plus one. That must be represented by this first slightly higher curve. And we know that because the leading coefficient, the coefficient of 𝑥 squared, is positive. Whereas the coefficient of 𝑥 squared in our second equation, let’s call that 𝑦 equals negative 𝑥 squared plus two 𝑥 plus four, is negative. So, we have this sort of inverted parabola as shown.
Now, in fact, the easiest way to decide which region in the graph contains all the solution to our system of inequalities is to pick a point in each region and test it out. Now, I always like to choose the most straightforward point on the graph. If possible, that’s the origin. That’s the point with coordinates zero, zero. What we’re going to do is substitute 𝑥 equals zero and 𝑦 equals zero into each inequality and see if these points satisfy them.
So, taking 𝑦 equals zero and 𝑥 equals zero in our first equation, we get zero is less than or equal to negative a half times zero squared minus zero plus one. Zero is less than or equal to one. Well, zero is less than one. So, this is true. Remember, the equals bit just tells us what kind of line we have. Here, we have a solid line, so we know it’s going to be equals rather than just less than.
When we substitute these values into our second inequality, we get zero is less than or equal to negative zero squared plus two times zero plus four. Which simplifies to zero is less than or equal to four. Well, yes, zero is less than four, so this inequality is also satisfied. And so, it looks like region 𝐸 is the region which contains all the solutions to our system of inequalities.
It’s sensible for us to at least check a couple of the other regions. Let’s check a point in region 𝐴. Let’s take the point negative two, zero. Substituting 𝑥 equals negative two and 𝑦 equals zero into our first inequality, and we get zero is less than or equal to a half times negative two squared minus negative two plus one. Well, that tells us that zero is less than or equal to seven, which is, of course, true.
So, we substitute these values into our second inequality. This time we get zero is less than or equal to negative negative two squared plus two times negative two plus four. The right-hand side simplifies to negative four. So, we see that zero’s less than or equal to negative four. Of course, that’s not true. Zero is, in fact, greater than negative four. And so, we can obviously disregard region 𝐴. Whilst it contains solutions to the first inequality, it doesn’t contain all solutions to the entire system.
Let’s repeat this process one more time. This time we’ll choose a point in 𝐵. Again, we’ll choose a point on one of our axes because it makes the numbers a little bit easier. And we’ll choose a point with coordinates zero, six. Substituting 𝑥 equals zero and 𝑦 equals six into our first inequality, and we get six is less than or equal to one. That’s obviously not true.
Similarly, substituting into our second inequality, we get six is less than or equal to four, also not true. So, once again we can disregard 𝐵. And in fact, if we continued, we’d see that we can also disregard 𝐶 and 𝐷. And so, the region which contains all the solutions to our system of inequalities is indeed 𝐸.
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# What are the 8 commutative properties?
## What are the 8 commutative properties?
The commutative property is a math rule that says that the order in which we multiply numbers does not change the product.
### What are multiplicative properties?
PROPERTIES OF MULTIPLICATION
Identity Property There is a unique real number 1 such that for every real number a , a⋅1=a and 1⋅a=a One is called the identity element of multiplication.
Commutative Property For all real numbers a and b , a⋅b=b⋅a The order in which you multiply two real numbers does not change the result.
#### What is commutative property in a math problem?
This law simply states that with addition and multiplication of numbers, you can change the order of the numbers in the problem and it will not affect the answer.
What is a associative property in math?
The associative property is a math rule that says that the way in which factors are grouped in a multiplication problem does not change the product.
What is multiplication property example?
Identity property of multiplication: The product of 1 and any number is that number. For example, 7 × 1 = 7 7 \times 1 = 7 7×1=77, times, 1, equals, 7.
## What is an associative property in math?
### How do you use commutative property?
The commutative property states that the change in the order of numbers in an addition or multiplication operation does not change the sum or the product. The commutative property of addition is written as A + B = B + A.
#### What is associative property example?
The associative property of multiplication states that the product of three or more numbers remains the same regardless of how the numbers are grouped. For example, 3 × (5 × 6) = (3 × 5) × 6. Here, no matter how the numbers are grouped, the product of both the expressions remains 90.
How do you find associative property?
The associative property always involves 3 or more numbers. The numbers grouped within a parenthesis, are terms in the expression that considered as one unit. There is also an associative property of multiplication. However, subtraction and division are not associative.
What is an example of a math property?
Common Math Properties The following math properties are formally introduced in algebra classes, but they are taught in many elementary schools. You probably don’t even realize that you already know many of these properties. For example, the commutative property basically states you can add in any order: 6 + 5 is the same as 5 + 6.
## What is a commutative property in math example?
Commutative Property The commutative property (like we described at the top of the math properties page) deals with the order that add or multiply numbers. In the commutative property you do change the order of the numbers. In our example above, the 4 was first originally, and then it was switched to second.
### What is the distributive property in math?
Common Math Properties. The distributive property applies when you are multiplying a number (or variable) times a quantity. You can multiply the number by each of the values inside the quantity seperately, and add them together. Take a look at the distributive property below: The word distribute means to give out. In…
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# Derivative of root sinx by Chain Rule
The derivative of root sinx is equal to cosx/(2√sinx). In this post, we will learn how to differentiate square root of sinx by the chain rule of derivatives.
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## Derivative of Square Root of Sin x by Chain Rule
Question: Find the Derivative of $\sqrt{\sin x}$ by the chain rule.
Solution:
To find the derivative of root sinx by the chain rule, we will follow the steps provided below.
Step 1: Let z=sinx
Differentiating both sides with respect to x, we get that
$\dfrac{dz}{dx} = \cos x$
Step 2: By the chain rule of derivatives,
$\dfrac{d}{dx}(\sqrt {\sin x})$ = $\dfrac{d}{dx}(\sqrt z)$
= $\dfrac{d}{dz}(\sqrt z) \times \dfrac{dz}{dx}$
= $\dfrac{d}{dz}(z^{1/2}) \times \cos x$
= ½ z½ -1 × cosx by the power rule of derivatives d/dx(xn)=nxn-1.
Step 3: Putting z=sinx and simplifying we get that
$\dfrac{d}{dx}(\sqrt {\sin x})$ = $\dfrac{\cos x}{2 \sqrt{z}}$ as z-1/2=1/√z.
= $\dfrac{\cos x}{2 \sqrt{\sin x}}$
So the derivative of root sinx is equal to $\dfrac{\cos x}{2 \sqrt{\sin x}}$ and this is obtained by the chain rule of derivatives.
Also Read:
Derivative of $\sqrt{\sin x}$ by First Principle
Derivative of $\sqrt{\cos x}$
Derivative of $\sqrt{e^x}$
## FAQs
Q1: Find the Derivative root sinx.
Answer: cosx/(2 √sin x) is the derivative of the square root of sinx, that is, d/dx(√sinx) = cosx/(2√sin x).
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Discrete Structure
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# Discrete Structure - PowerPoint PPT Presentation
Discrete Structure. Li Tak Sing( 李德成 ). Chapter 4 Properties of Binary Relations. Three special properties For a binary relation R on a set A, we have the following definitions. R is reflexive if xRx for all x A. R is symmetric if xRy implies yRx for all x,y A
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### Discrete Structure
Li Tak Sing(李德成)
Chapter 4Properties of Binary Relations
• Three special properties
• For a binary relation R on a set A, we have the following definitions.
• R is reflexive if xRx for all xA.
• R is symmetric if xRy implies yRx for all x,y A
• R is transitive if xRy and yRz implies xRz for all x,y,z A
Two opposite properties
• For a binary relation R on a set A, we have the following definitions.
• R is irreflexive if (x,x)R for all xA.
• R is antisymmetric if xRy and yRx implies x=y for all x,yA.
Example
• R is a binary relation on N
• aRb if (a+b) mod 2 = 0
• R is reflexive because (a+a) mod 2 =0 for all a N
• R is symmetric because if aRb, then (a+b) mod 2 = 0, then (b+a) mod 2 =0, then bRa
• R is transitive, because if aRb and bRc, then (a+b) mod 2 =0 and (b+c) mod 2 =0, then (a+2b+c) mod 2 =0, then (a+c) mod 2 =0, then aRc
Example
• Give examples of binary relations over the set {a,b,c,d} with the stated properties:
• Reflexive and not symmetric and not transitive
• Symmetric and not reflexive and not transitive
• transitive and not reflexive and not symmetric
Composition of relations
• If R and S are binary relations, then the composition of R and S, which we denote by RS, is the following relation:RS={(a,c)|(a,b)R and (b,c) S for some element b}
More examples
• For each of the following binary relations state which of the three properties, reflexive, symmetric and transitive are satisfied.
• xRy iff |x-y| is odd, over the integers.
• xRy iff x is a parent of y, over the set of people.
Grandparents
• Given the isParentOf relation. So a isParentOf b represents the fact that a is the parent of b.
• isGrandparentOf can then be defined in terms of isParentOf.isGrandparentOf=isParentOfisParentOf
• So a isGrandparentOf b if there is c so that a isParentOf c and c isPrentOf b.
More examples
• Given the following binary relations over {a,b,c,d}.R={(a,a),(a,c),(b,a),(b,d),(c,b)}S={(a,b),(a,c),(c,b),(d,c)}
• Find RS
• Find SR
Representations
• If R is a binary relation on A, then we'll denote the composition of R with itself n times by writing Rn.
• For example,
• isGrandparentOf=isParentOf2
• isGreatGrandParentOf=isParentOf3
Inheritance properties
• If R is reflexive, then Rn is reflexive.
• If R is symmetric, then Rn is symmetric.
• If R is transitive, then Rn is transitive.
Example
• Let R={(x,y)ZZ|x+y is odd}. We want to find out R2 and R3.
Closures
• If R is a binary relation and p is some property, then the p closure of R is the smallest binary relation containing R that satisfies property p.
Reflexive closure
• A reflexive closure of R is the smallest reflexive relation that contains R.
• A reflexive closure of R is denoted as r(R)
• R is a relation over {a,b,c} and R={(a,b),(b,c)}Then, r(R)={(a,a),(b,b),(c,c),(a,b),(b,c)}
Symmetric closure
• A symmetric closure of R is the smallest symmetric relation that contains R.
• A symmetric closure of R is denoted as s(R)
• R={(a,b),(b,c)}, s(R)={(a,b),(b,a),(b,c),(c,b)}
Transitive closure
• A transitive closure of R is the smallest transitive relation that contains R. It is denoted as t(R).
• R= ={(a,b),(b,c)}, then t(R)= {(a,b),(b,c),(a,c)}
Constructing Closures
• If R is a binary relation over a set A, then:
• r(R)=RRo (Ro is the equality relation)
• s(R)=R Rc (Rc is the converse relation)
• t(R)=R R2 R3 R4....
• If A is finite with n elements, then t(R)= R R2 R3 R4.... Rn
Example
• Given the set A={a,b,c,d}. Draw a directed graph to represent the indicated closure for each of the following binary relations over A.
• r(R), where R={(a,d)}
• s(R) where R={(a,b), (c,d)}
• t(R) where R={(a,b),(d,a),(d,c),(c,b)}
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The domain represents all the possible values that act as input which defines the function. For instance, in a graph, the domain consists of all the possible input values in X, and the Y domain consists of all the output values.
### How to Find the Domain of a Rational Function with a Square Root?
Certainly here are the steps to find the domain of a rational function with square root
• Initially, set the expression with the square root equalling it to zero or greater than zero. Also, this method of finding domains will take into account positive numbers that can give us some real roots.
• Solve the equations diligently to arrive at the results and express it in terms of intervals.
### How to Find the Domain of a Rational Function in Interval Notation?
In interval notation, brackets are used to indicate a set of numbers that stands for intervals. The steps for interval notation are given below
• The smallest term is the one written first in the interval and the largest term following.
• Parenthesis indicate or and brackets indicate inclusive terms.
### Find the Domain of the Rational Function Calculator
The domains stand for the set of functions for which the functions are specified.
• Enter the radical function in the input and for finding the domain click on the button for calculating the domain.
• Also upon accessing the button, the domain will be displayed on the screen.
### How to Find the Domain of a Rational Expression?
Certainly, for rational expression, the numerator and the denominator are polynomials
• The domain for the expression will stand for all the possible input values. A cautious application of the values is necessary so that the denominator doesn’t make the function undefined.
• The numerator can be constant but the denominator can be of any degree and in various forms.
### The Domain of a Radical Function
Determining the values of radical function is very easy.
• For determining the radical expression. All you need to do is equate the function to zero.
• Solve for X. certainly, for all the values of X except the one that turns the function to zero will be accounted as valid.
### How to Find the Domain of a Function with a Fraction?
• Set the function equal to zero and solve for the value of X.
• Find all the values of X for which the domain doesn’t equal to zero. Also, omit the one that makes the function tend to zero.
### Find the Domain of a Rational Function Involving Radicals
Finding the domain of a rational function including radicals can be done with some easy steps.
• Set the expression involving radicals equals to zero or greater than zero.
• Solve the function of inequality and find the value of X.
### Tips
• The domain stands for all the sets of values that define a function. There are also special functions that have unlimited values of inputs in the domain.
• Finding the domain is very easy to solve and analytically find the values of X equating the function to zero.
### How do you find the domain and range of a rational function?
The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes. A rational function is a function of the form f(x)=p(x)q(x) , where p(x) and q(x) are polynomials and q(x)≠0 .
### How do you find domain of a function?
The domain of a function is the set of all possible inputs for the function. For example, the domain of f(x)=x² is all real numbers, and the domain of g(x)=1/x is all real numbers except for x=0.
### How do you find the domain of a rational function with a radical?
To determine the domain of a radical expression, set the radicand equal to zero, then solve for x . All values of x except for those that satisfy √x=0 are the domain of the expression.
### What does it mean to find the domain?
In plain English, this definition means: The domain is the set of all possible x-values which will make the function “work”, and will output real y-values. When finding the domain, remember: The denominator (bottom) of a fraction cannot be zero. The number under a square root sign must be positive in this section.
### How do I know if a domain is restricted?
Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve for x . If the function’s formula contains an even root, set the radicand greater than or equal to 0 , and then solve.
### What are the two domain restrictions?
That is, only real numbers can be used in the domain, and only real numbers can be in the range. There are two main reasons why domains are restricted. You cannot divide by 0 .
Learning Outcomes.
Function Restrictions to the Domain
f(x)=1x If x=0 , you would be dividing by 0 , so x≠0 x ≠ 0 .
### What are the domain restrictions?
Domain restrictions allow us to create functions defined over numbers that work for our purposes. Piecewise defined functions are the composition of multiple functions with domain restrictions that do not overlap. Some functions are restricted from values that make them undefined.
### What are the three domain restrictions?
The three functions that have limited domains are the square root function, the log function and the reciprocal function.
### How do you write a domain restriction?
To limit the domain or range (x or y values of a graph), you can add the restriction to the end of your equation in curly brackets {}. For example, y=2x{1<x<3} would graph the line y=2x for x values between 1 and 3. You can also use restrictions on the range of a function and any defined parameter.
### What is a domain Algebra 2?
Domain: All real numbers. Range: Explanation: The domain includes the values that go into a function (the x-values) and the range are the values that come out (the or y-values).
### How do you tell if a domain is all real numbers?
However, because absolute value is defined as a distance from 0, the output can only be greater than or equal to 0. For the quadratic function f(x)=x2 f ( x ) = x 2 , the domain is all real numbers since the horizontal extent of the graph is the whole real number line.
### How do you write domain and range?
Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range.
### What is domain and range examples?
You can also talk about the domain of a relation , where one element in the domain may get mapped to more than one element in the range. Example 2: The domain is the set of x -coordinates, {0,1,2} , and the range is the set of y -coordinates, {7,8,9,10} .
### How do you write the domain?
We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [ when the set includes the endpoint and a parenthesis ( to indicate that the endpoint is either not included or the interval is unbounded.
### Is domain input or output?
The domain is the input, the independent value—it’s what goes into a function. The range is the output, the dependent value—it’s what comes out. Domain and range may be limited to a few discrete values, or they may include all numbers everywhere, to infinity and beyond.
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# Exponential Decay
## Rational functions with x as an exponent in the denominator
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Exponential Decay
Suppose the amount of a radioactive substance is cut in half every 25 years. If there was originally 500 grams of the substance, could you write a function representing the amount of the substance after years? How much of the substance would there be after 100 years? Will the amount of the substance ever reach 0 grams?
### Exponential Decay
In the last Concept, we learned how to solve expressions that modeled exponential growth. Now, we will be learning about exponential decay functions.
Recall that the general form of an exponential function is , where initial value and
In exponential decay situations, the growth factor must be a fraction between zero and one.
#### Let's use an exponential function to answer the following problem:
For her fifth birthday, Nadia’s grandmother gave her a full bag of candy. Nadia counted her candy and found out that there were 160 pieces in the bag. Nadia loves candy, so she ate half the bag on the first day. Her mother told her that if she continues to eat at that rate, it will be gone the next day and she will not have any more until her next birthday. Nadia devised a clever plan. She will always eat half of the candy that is left in the bag each day. She thinks that she will get candy every day and her candy will never run out. How much candy does Nadia have at the end of the week? Would the candy really last forever?
Make a table of values for this problem.
Day 0 1 2 3 4 5 6 7
# of Candies 160 80 40 20 10 5 2.5 1.25
You can see that if Nadia eats half the candies each day, then by the end of the week she has only 1.25 candies left in her bag.
Write an equation for this exponential function. Nadia began with 160 pieces of candy. In order to get the amount of candy left at the end of each day, we keep multiplying by . Because it is an exponential function, the equation is:
#### Now, let's graph the following exponential functions:
1.
Start by making a table of values. Remember when you have a number to the negative power, you are simply taking the reciprocal of that number and taking it to the positive power. Example: .
–3
–2
–1
0
1
2
Now graph the function.
Using the Property of Negative Exponents, the equation can also be written as .
1. The functions and on the same coordinate axes.
Here is the table of values and the graph of the two functions.
Looking at the values in the table, we see that the two functions are “reverse images” of each other in the sense that the values for the two functions are reciprocals.
–3
–2
–1
0
1
2
3
Here is the graph of the two functions. Notice that these two functions are mirror images if the mirror is placed vertically on the axis.
### Examples
#### Example 1
Earlier, you were told that the amount of a radioactive substance is cut in half every 25 years. If there was originally 500 grams of the substance, what is the function representing the amount of the substance after years? How much of the substance would there be after 100 years? Will the amount of the substance ever reach 0 grams?
The initial amount in this situation is 500 grams. If the amount of the substance is cut in half every 25 years then the growth factor is . You would assume that the function to represent this situation would be:
with representing the number of years and representing the amount of the substance yet.
However, based on the situation, you know that after 25 years, there should be 250 grams of the substance left. If you plug 25 in for :
This is not the answer that we should get. The issue is in the exponent. Because the amount of substance is cut in half every 25 years, the exponent should increase in intervals of 25. Therefore the exponent should be .
The equation that represents this situation is:
To find the amount of the substance that will be left after 100 years, plug in 100 to the equation.
There will be 31.25 grams left after 100 years.
The amount will never reach zero because it will keep on getting cut in half but there is no number that can be divided 2 to get 0 apart from 0. There are an infinite amount of decimals and the number will continue to get smaller and smaller but there will never be 0 grams of the substance.
#### Example 2
If a person takes 125 milligrams of a drug, and after the full dose is absorbed into the bloodstream, there is only 70% left after every hour, write a function that gives the concentration left in the bloodstream after hours. What is the concentration of the drug in the bloodstream after 3 hours?
This will be a decay function in the form . We know the initial value is 125 milligrams. After one hour we multiply that by 0.70 to find 70% of 125. After the second hour, we multiply by 0.7 again and so on:
Since we multiply by 0.7 to find the 70% that is left after each hour, the decay factor is 0.7. The function will be:
We also found that after three hours the amount of the drug in the bloodstream will be 42.875 milligrams.
### Review
1. Define exponential decay.
2. What is true about “b” in an exponential decay function?
3. Suppose . What is ? What does this mean in terms of the intercept of an exponential function?
Graph the following exponential decay functions.
1. The percentage of light visible at meters is given by the function .
1. What is the growth factor?
2. What is the initial value?
3. Find the percentage of light visible at 65 meters.
1. A person is infected by a certain bacterial infection. When he goes to the doctor, the population of bacteria is 2 million. The doctor prescribes an antibiotic that reduces the bacteria population to of its size each day.
1. Draw a graph of the size of the bacteria population against time in days.
2. Find the formula that gives the size of the bacteria population in terms of time.
3. Find the size of the bacteria population ten days after the drug was first taken.
4. Find the size of the bacteria population after two weeks (14 days).
Mixed Review
1. The population of Kindly, USA is increasing at a rate of 2.14% each year. The population in the year 2010 is 14,578.
1. Write an equation to model this situation.
2. What would the population of Kindly be in the year 2015?
3. When will the population be 45,000?
1. The volume of a sphere is given by the formula . Find the volume of a sphere with a diameter of 11 inches.
2. Simplify .
3. Simplify .
4. Rewrite in standard form: .
To see the Review answers, open this PDF file and look for section 8.8.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
TermDefinition
exponential decay In exponential decay situations, the growth factor must be a fraction between zero and one. $0
Exponential Function An exponential function is a function whose variable is in the exponent. The general form is $y=a \cdot b^{x-h}+k$.
Model A model is a mathematical expression or function used to describe a physical item or situation.
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# What Is The Solution To The Linear Equation? 4b + 6 = 2 – B + 4
• Puzzle
Solving linear equations is a fundamental part of solving linear problems. Linear equations are defined as equations that have variables that are only multiplied by numbers, not divided by numbers or raised to higher powers.
For example, x = 2y is a linear equation, but x = 2y2 or x = 2y–1 are not. Linear equations can be solved in two main ways: by adding or subtracting the same value to both sides of the equation to make it look like the opposite operation, then solving for the new variable, or finding another variable that works as the opposite operation and solving for it.
Solving linear equations can be tricky when there are more than one possible solution. It is important to check that you only solved for one variable and did not solve for both!
There are many ways to apply solving linear equations in real life situations. One common way is using them to solve simple systems of linear equations.
## Add a negative number to both sides of the equation
Solving linear equations involves adding, subtracting, multiplying, or dividing both sides of the equation by the same number. You can also rearrange the left side of the equation.
For example, if you have the equation 4b + 6 = 2 – B + 4, you can divide both sides by 2 and then add B to get 2 + B + 4 = 4b + 6. Now you can solve for b by substituting b for 4b.
Solving linear equations is useful because it allows you to find what value(s) fit inside the equation. For example, if you have the equation 2 + B + 4 = 4b + 6 and you solve for B, you get B = –2. Therefore, only –2 fits inside the equation 2 + B + 4 = 4b + 6.
## Divide both sides of the equation by a negative number
To solve an equation that has a negative number on the left side of the equal sign, you must divide both sides of the equation by the negative number. This process is called negation.
For example, let’s look at the equation 4b + 6 = 2 – B + 4. To solve this equation, you would divide both sides by -B.
The variable B represents a positive number, so dividing both sides by -B would simply mean switching the sign of the number.
Whether you switch the sign or not does not matter; it will still solve the equation correctly.
If you divided both sides by -B and then realized that did not change the value of the solution, then you knew that B was a positive number.
By solving this equation, you discovered that B is a positive number of 5.
## Take the inverse of both sides of the equation
In math, the inverse of a number or a variable is the value that, when combined with the original number or variable, produces zero as a result. For example, the inverse of 4 is −4, since subtracting 4 from any number produces zero.
In linear equations, solving for one variable in terms of another variable can be done by using the inverse function. By doing this, you are solving for what is on the left side of the equals sign in terms of what is on the right side.
The solution to an equation can be found by putting one variable in terms of the other and then doing a little arithmetic to find out how much of one variable is needed to get zero as a result.
## Solve using algebraic techniques
Solving linear equations is the process of finding what value or values solve the equation. The solution can be a number, called a solution, or it can be a set of numbers, called a solution set.
There are three main ways to solve linear equations. These are substitution, elimination, and averaging. Each method uses different strategies to solve the equation, so try out all three to see which one works best for you!
Substitution is by far the easiest way to solve an equation if you know how to do it correctly. The trick is figuring out how to place one variable in place of the other in the original equation.
To solve an equation using substitution, first rewrite the equation so that one variable is on the left side of the equals sign and one is on the right side. Then, find a value for the unknown variable and replace it with its counterpart on the left or right side of the equals sign.
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# NCERT Solutions for Class 10 Maths Chapter 6 Triangles
## NCERT Solutions for Class 10 Maths Chapter 6 – Download Free PDF
NCERT Solutions for Class 10 Maths Chapter 6 Triangles are provided here, which is considered to be one of the most important study materials for the students studying in CBSE Class 10. Chapter 6 of NCERT Solutions for Class 10 Maths is well structured in accordance with the CBSE Syllabus for 2023-24. It covers a vast topic, including a number of rules and theorems. Students often tend to get confused about which theorem to use while solving a variety of questions.
### Download Most Important Questions for Class 10 Maths Chapter – 6 Triangles
The solutions provided at BYJU’S are designed in such a way that every step is explained clearly and in detail. The Solutions for NCERT Class 10 Maths are prepared by the subject experts to help students prepare better for their board exams. These solutions will be helpful not only for exam preparations, but also in solving homework and assignments.
The CBSE Class 10 examination often asks questions, either directly or indirectly, from the NCERT textbooks. Thus, the NCERT Solutions for Chapter 6 Triangles of Class 10 Maths is one of the best resources to prepare, and equip oneself to solve any type of questions in the exam, from the chapter. It is highly recommended that the students practise these NCERT Solutions on a regular basis to excel in the Class 10 board examinations.
## Exercise 6.1 Page: 122
1. Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)
(b) Proportional
2. Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures
Solution:
3. State whether the following quadrilaterals are similar or not:
Solution:
From the given two figures, we can see their corresponding angles are different or unequal. Therefore, they are not similar.
## Exercise 6.2 Page: 128
1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solution:
(i) Given, in △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒1.5/3 = 1/EC
⇒EC = 3/1.5
EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.
(ii) Given, in △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒ AD/7.2 = 1.8 / 5.4
⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10
2. E and F are points on the sides PQ and PR, respectively of a ΔPQR. For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm
Solution:
Given, in ΔPQR, E and F are two points on side PQ and PR, respectively. See the figure below;
Â
(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm
Therefore, by using Basic proportionality theorem, we get,
PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3
And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, we get, PE/EQ ≠PF/FR
Hence, EF is not parallel to QR.
(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm
Therefore, by using Basic proportionality theorem, we get,
PE/QE = 4/4.5 = 40/45 = 8/9
And, PF/RF = 8/9
So, we get here,
PE/QE = PF/RF
Hence, EF is parallel to QR.
(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
From the figure,
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm
So, PE/EQ = 0.18/1.10 = 18/110 = 9/55…………. (i)
And, PE/FR = 0.36/2.20 = 36/220 = 9/55………… (ii)
So, we get here,
PE/EQ = PF/FR
Hence, EF is parallel to QR.
3. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD
Solution:
In the given figure, we can see, LM || CB,
By using basic proportionality theorem, we get,
AM/AB = AL/AC……………………..(i)
Similarly, given, LN || CD and using basic proportionality theorem,
From equation (i) and (ii), we get,
Hence, proved.
4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC
Solution:
In ΔABC, given as, DE || AC
Thus, by using Basic Proportionality Theorem, we get,
∴BD/DA = BE/EC ………………………………………………(i)
In  ΔBAE, given as, DF || AE
Thus, by using Basic Proportionality Theorem, we get,
∴BD/DA = BF/FE ………………………………………………(ii)
From equation (i) and (ii), we get
BE/EC = BF/FE
Hence, proved.
5. In the figure, DE||OQ and DF||OR, show that EF||QR.
Solution:
Given,
In ΔPQO, DE || OQ
So by using Basic Proportionality Theorem,
PD/DO = PE/EQ……………… ..(i)
Again given, in ΔPOR, DF || OR,
So by using Basic Proportionality Theorem,
PD/DO = PF/FR………………… (ii)
From equation (i) and (ii), we get,
PE/EQ = PF/FR
Therefore, by converse of Basic Proportionality Theorem,
EF || QR, in ΔPQR.
6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Solution:
Given here,
In ΔOPQ, AB || PQ
By using Basic Proportionality Theorem,
OA/AP = OB/BQ…………….(i)
Also given,
In ΔOPR, AC || PR
By using Basic Proportionality Theorem
∴ OA/AP = OC/CR……………(ii)
From equation (i) and (ii), we get,
OB/BQ = OC/CR
Therefore, by converse of Basic Proportionality Theorem,
In ΔOQR, BC || QR.
7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Solution:
Given, in ΔABC, D is the midpoint of AB such that AD=DB.
A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.
We have to prove that E is the mid point of AC.
Since, D is the mid-point of AB.
In ΔABC, DE || BC,
By using Basic Proportionality Theorem,
From equation (i), we can write,
⇒ 1 = AE/EC
∴ AE = EC
Hence, proved, E is the midpoint of AC.
8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Solution:
Given, in ΔABC, D and E are the mid points of AB and AC, respectively, such that,
We have to prove that: DE || BC.
Since, D is the midpoint of AB
Â
Also given, E is the mid-point of AC.
∴ AE=EC
⇒ AE/EC = 1
From equation (i) and (ii), we get,
By converse of Basic Proportionality Theorem,
DE || BC
Hence, proved.
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Solution:
Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.
We have to prove, AO/BO = CO/DO
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔADC, we have OE || DC
Therefore, by using Basic Proportionality Theorem
AE/ED = AO/CO ……………..(i)
Now, In ΔABD, OE || AB
Therefore, by using Basic Proportionality Theorem
DE/EA = DO/BO…………….(ii)
From equation (i) and (ii), we get,
AO/CO = BO/DO
⇒AO/BO = CO/DO
Hence, proved.
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.
Solution:
Given, Quadrilateral ABCD where AC and BD intersect each other at O such that,
AO/BO = CO/DO.
We have to prove here, ABCD is a trapezium
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔDAB, EO || AB
Therefore, by using Basic Proportionality Theorem
DE/EA = DO/OB ……………………(i)
Also, given,
AO/BO = CO/DO
⇒ AO/CO = BO/DO
⇒ CO/AO = DO/BO
⇒DO/OB = CO/AO …………………………..(ii)
From equation (i) and (ii), we get
DE/EA = CO/AO
Therefore, by using converse of Basic Proportionality Theorem,
EO || DC also EO || AB
⇒ AB || DC.
Hence, quadrilateral ABCD is a trapezium with AB || CD.
## Exercise 6.3 Page: 138
1. State which pairs of triangles in the figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Solution:
(i) Given, in ΔABC and ΔPQR,
∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
Therefore, by AAA similarity criterion,
∴ ΔABC ~ ΔPQR
(ii) Given, in  ΔABC and ΔPQR,
AB/QR = 2/4 = 1/2,
BC/RP = 2.5/5 = 1/2,
CA/PA = 3/6 = 1/2
By SSS similarity criterion,
ΔABC ~ ΔQRP
(iii) Given, in ΔLMP and ΔDEF,
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF = 2.7/5 = 27/50
Here , MP/DE = PL/DF ≠LM/EF
Therefore, ΔLMP and ΔDEF are not similar.
(iv) In ΔMNL and ΔQPR, it is given,
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
Therefore, by SAS similarity criterion
∴ ΔMNL ~ ΔQPR
(v) In ΔABC and ΔDEF, given that,
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Here , AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
⇒ ∠B ≠∠F
Hence, ΔABC and ΔDEF are not similar.
(vi) In ΔDEF, by sum of angles of triangles, we know that,
∠D + ∠E + ∠F = 180°
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
⇒ ∠F = 30°
Similarly, In ΔPQR,
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° – 80° -30°
⇒ ∠P = 70°
Now, comparing both the triangles, ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Therefore, by AAA similarity criterion,
Hence, ΔDEF ~ ΔPQR
2.  In figure 6.35, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Solution:
As we can see from the figure, DOB is a straight line.
Therefore, ∠DOC + ∠COB = 180°
⇒ ∠DOC = 180° – 125° (Given, ∠BOC = 125°)
= 55°
In ΔDOC, sum of the measures of the angles of a triangle is 180º
Therefore, ∠DCO + ∠CDO + ∠DOC = 180°
⇒ ∠DCO + 70º + 55º = 180°(Given, ∠CDO = 70°)
⇒ ∠DCO = 55°
It is given that, ΔODC ~ ΔOBA,
Therefore, ΔODC ~ ΔOBA.
Hence, corresponding angles are equal in similar triangles
∠OAB = ∠OCD
⇒ ∠OAB = 55°
∠OAB = ∠OCD
⇒ ∠OAB = 55°
3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD
Solution:
In ΔDOC and ΔBOA,
AB || CD, thus alternate interior angles will be equal,
∴∠CDO = ∠ABO
Similarly,
∠DCO = ∠BAO
Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;
∴∠DOC = ∠BOA
Hence, by AAA similarity criterion,
ΔDOC ~ ΔBOA
Thus, the corresponding sides are proportional.
DO/BO = OC/OA
⇒OA/OC = OB/OD
Hence, proved.
Â
4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
Solution:
In ΔPQR,
∠PQR = ∠PRQ
∴ PQ = PR ………………………(i)
Given,
QR/QS = QT/PRUsing equation (i), we get
QR/QS = QT/QP……………….(ii)
In ΔPQS and ΔTQR, by equation (ii),
QR/QS = QT/QP
∠Q = ∠Q
∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]
5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
Solution:
Given, S and T are point on sides PR and QR of ΔPQR
And ∠P = ∠RTS.
In ΔRPQ and ΔRTS,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)
∴ ΔRPQ ~ ΔRTS (AA similarity criterion)
6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
Solution:
Given, ΔABE ≅ ΔACD.
∴ AB = AC [By CPCT] ……………………………….(i)
And, AD = AE [By CPCT] ……………………………(ii)
In ΔADE and ΔABC, dividing eq.(ii) by eq(i),
∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [SAS similarity criterion]
7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iv) ΔPDC ~ ΔBEC
Solution:
Given, altitudes AD and CE of ΔABC intersect each other at the point P.
(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP (90° each)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by AA similarity criterion,
ΔAEP ~ ΔCDP
(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB ( 90° each)
∠ABD = ∠CBE (Common Angles)
Hence, by AA similarity criterion,
ΔABD ~ ΔCBE
∠AEP = ∠ADB (90° each)
∠PAE = ∠DAB (Common Angles)
Hence, by AA similarity criterion,
(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (90° each)
∠PCD = ∠BCE (Common angles)
Hence, by AA similarity criterion,
ΔPDC ~ ΔBEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Solution:
Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,
In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (AA similarity criterion)
9. In the figure, ABC and AMP are two right triangles, right angled at B and M, respectively, prove that:
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
Solution:
Given, ABC and AMP are two right triangles, right angled at B and M, respectively.
(i) In ΔABC and ΔAMP, we have,
∠CAB = ∠MAP (common angles)
∠ABC = ∠AMP = 90° (each 90°)
∴ ΔABC ~ ΔAMP (AA similarity criterion)
(ii) As, ΔABC ~ ΔAMP (AA similarity criterion)
If two triangles are similar then the corresponding sides are always equal,
Hence, CA/PA = BC/MP
10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:
(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Solution:
Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG, respectively.
(i) From the given condition,
ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
Since, ∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,
∠A = ∠F
∠ACD = ∠FGH
∴ ΔACD ~ ΔFGH (AA similarity criterion)
⇒CD/GH = AC/FG
(ii) In ΔDCB and ΔHGE,
∠DCB = ∠HGE (Already proved)
∠B = ∠E (Already proved)
∴ ΔDCB ~ ΔHGE (AA similarity criterion)
(iii) In ΔDCA and ΔHGF,
∠ACD = ∠FGH (Already proved)
∠A = ∠F (Already proved)
∴ ΔDCA ~ ΔHGF (AA similarity criterion)
11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
Solution:
Given, ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∠ADB = ∠EFC (Each 90°)
∴ ΔABD ~ ΔECF (using AA similarity criterion)
12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.
Solution:
Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
i.e. AB/PQ = BC/QR = AD/PM
We have to prove: ΔABC ~ ΔPQR
As we know here,
⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)
⇒ ΔABD ~ ΔPQM [SSS similarity criterion]
∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]
⇒ ∠ABC = ∠PQR
In ΔABC and ΔPQR
AB/PQ = BC/QR ………………………….(i)
∠ABC = ∠PQR ……………………………(ii)
From equation (i) and (ii), we get,
ΔABC ~ ΔPQR [SAS similarity criterion]
13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD
Solution:
Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.
∠ACD = ∠BCA (Common angles)
∴ ΔADC ~ ΔBAC (AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
∴ CA/CB = CD/CA
⇒ CA2 = CB.CD.
Hence, proved.
14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
Solution:
Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that;
We have to prove, ΔABC ~ ΔPQR
Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.
In ΔABD and ΔCDE, we have
AD = DE Â [By Construction.]
BD = DC [Since, AP is the median]
and, ∠ADB = ∠CDE [Vertically opposite angles]
∴ ΔABD ≅ ΔCDE [SAS criterion of congruence]
⇒ AB = CE [By CPCT] …………………………..(i)
Also, in ΔPQM and ΔMNR,
PM = MN [By Construction.]
QM = MR [Since, PM is the median]
and, ∠PMQ = ∠NMR [Vertically opposite angles]
∴ ΔPQM = ΔMNR [SAS criterion of congruence]
⇒ PQ = RN [CPCT] ………………………………(ii)
Now, AB/PQ = AC/PR = AD/PM
From equation (i) and (ii),
⇒ CE/RN = AC/PR = 2AD/2PM
⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]
∴ ΔACE ~ ΔPRN [SSS similarity criterion]
Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P …………………………………………….(iii)
Now, in ΔABC and ΔPQR, we have
From equation (iii),
∠A = ∠P
∴ ΔABC ~ ΔPQR [ SAS similarity criterion]
15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
Given, Length of the vertical pole = 6m
Shadow of the pole = 4 m
Let Height of tower = h m
Length of shadow of the tower = 28 m
In ΔABC and ΔDEF,
∠C = ∠E (angular elevation of sum)
∠B = ∠F = 90°
∴ ΔABC ~ ΔDEF (AA similarity criterion)
∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)
∴ 6/h = 4/28
⇒h = (6×28)/4
⇒ h = 6 × 7
⇒ h = 42 m
Hence, the height of the tower is 42 m.
16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.
Solution:
Given, ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.
∴AB/PQ = AC/PR = BC/QR……………………………(i)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….…..(ii)
Since AD and PM are medians, they will divide their opposite sides.
∴ BD = BC/2 and QM = QR/2 ……………..………….(iii)
From equations (i) and (iii), we get
AB/PQ = BD/QM ……………………….(iv)
In ΔABD and ΔPQM,
From equation (ii), we have
∠B = ∠Q
From equation (iv), we have,
AB/PQ = BD/QM
∴ ΔABD ~ ΔPQM (SAS similarity criterion)
Exercise 6.4 Page: 143
1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
Solution: Given, ΔABC ~ ΔDEF,
Area of ΔABC = 64 cm2
Area of ΔDEF = 121 cm2
EF = 15.4 cm
As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,
= AC2/DF2Â = BC2/EF2
∴ 64/121 = BC2/EF2
⇒ (8/11)2 = (BC/15.4)2
⇒ 8/11 = BC/15.4
⇒ BC = 8×15.4/11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm
2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Solution:
Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.
In ΔAOB and ΔCOD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [AAA similarity criterion]
As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,
Area of (ΔAOB)/Area of (ΔCOD) = AB2/CD2
= (2CD)2/CD2 [∴ AB = 2CD]
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD2/CD2 = 4/1
Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1
3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.
Solution:
Given, ABC and DBC are two triangles on the same base BC. AD intersects BC at O.
We have to prove: Area (ΔABC)/Area (ΔDBC) = AO/DO
Let us draw two perpendiculars AP and DM on line BC.
We know that area of a triangle = 1/2 × Base × Height
In ΔAPO and ΔDMO,
∠APO = ∠DMO (Each 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ΔAPO ~ ΔDMO (AA similarity criterion)
∴ AP/DM = AO/DO
⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO.
4. If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Say ΔABC and ΔPQR are two similar triangles and equal in area
Now let us prove ΔABC ≅ ΔPQR.
Since, ΔABC ~ ΔPQR
∴ Area of (ΔABC)/Area of (ΔPQR) = BC2/QR2
⇒ BC2/QR2 =1 [Since, Area(ΔABC) = (ΔPQR)
⇒ BC2/QR2
⇒ BC = QR
Similarly, we can prove that
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]
5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.
Solution:
D, E, and F are the mid-points of ΔABC
∴ DE || AC and
DE = (1/2) AC (Midpoint theorem) …. (1)
In  ΔBED and ΔBCA
∠BED = ∠BCA (Corresponding angles)
∠BDE = ∠BAC (Corresponding angles)
∠EBD = ∠CBA (Common angles)
∴ΔBED∼ΔBCA (AAA similarity criterion)
ar (ΔBED) / ar (ΔBCA)=(DE/AC)2
⇒ar (ΔBED) / ar (ΔBCA) = (1/4) [From (1)]
⇒ar (ΔBED) = (1/4) ar (ΔBCA)
Similarly,
ar (ΔCFE) = (1/4) ar (CBA) and ar (ΔADF) = (1/4) ar (ΔADF) = (1/4) ar (ΔABC)
Also,
ar (ΔDEF) = ar (ΔABC) − [ar (ΔBED) + ar (ΔCFE) + ar (ΔADF)]
⇒ar (ΔDEF) = ar (ΔABC) − (3/4) ar (ΔABC) = (1/4) ar (ΔABC)
⇒ar (ΔDEF) / ar (ΔABC) = (1/4)
6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:
Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.
We have to prove: Area(ΔABC)/Area(ΔDEF) = AM2/DN2
Since, ΔABC ~ ΔDEF (Given)
∴ Area(ΔABC)/Area(ΔDEF) = (AB2/DE2) ……………………………(i)
and, AB/DE = BC/EF = CA/FD ………………………………………(ii)
In ΔABM and ΔDEN,
Since ΔABC ~ ΔDEF
∴ ∠B = ∠E
AB/DE = BM/EN [Already Proved in equation (i)]
∴ ΔABC ~ ΔDEF [SAS similarity criterion]
⇒ AB/DE = AM/DN …………………………………………………..(iii)
∴ ΔABM ~ ΔDEN
As the areas of two similar triangles are proportional to the squares of the corresponding sides.
∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 = AM2/DN2
Hence, proved.
7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:
Given, ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.
Area(ΔBQC) = ½ Area(ΔAPC)
Since, ΔAPC and ΔBQC are both equilateral triangles, as per given,
∴ ΔAPC ~ ΔBQC [AAA similarity criterion]
∴ area(ΔAPC)/area(ΔBQC) = (AC2/BC2) = AC2/BC2
Since, Diagonal = √2 side = √2 BC = AC
⇒ area(ΔAPC) = 2 × area(ΔBQC)
⇒ area(ΔBQC) = 1/2area(ΔAPC)
Hence, proved.
Tick the correct answer and justify:
8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
Solution:
Given, ΔABC and ΔBDE are two equilateral triangle. D is the midpoint of BC.
Â
∴ BD = DC = 1/2BC
Let each side of triangle is 2a.
As, ΔABC ~ ΔBDE
∴ Area(ΔABC)/Area(ΔBDE) = AB2/BD2 = (2a)2/(a)2 = 4a2/a2 = 4/1 = 4:1
Hence, the correct answer is (C).
9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81
Solution:
Given, Sides of two similar triangles are in the ratio 4 : 9.
Let ABC and DEF are two similar triangles, such that,
ΔABC ~ ΔDEF
And AB/DE = AC/DF = BC/EF = 4/9
As, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,
∴ Area(ΔABC)/Area(ΔDEF) = AB2/DE2Â
∴ Area(ΔABC)/Area(ΔDEF) = (4/9)2 = 16/81 = 16:81
Hence, the correct answer is (D).
## Exercise 6.5 Page: 150
1. Â Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of the sides of the, we will get 49, 576, and 625.
49 + 576 = 625
(7)2Â + (24)2Â = (25)2
Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.
Length of Hypotenuse = 25 cm
(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.
Clearly, 9 + 36 ≠64
Or, 32 + 62 ≠82
Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.
Hence, the given triangle does not satisfies Pythagoras theorem.
(iii) Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.
However, 2500 + 6400 ≠10000
Or, 502 + 802 ≠1002
As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle does not satisfies Pythagoras theorem.
Hence, it is not a right triangle.
(iv) Given, sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.
Thus, 144 +25 = 169
Or, 122Â + 52Â = 132
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
Hence, length of the hypotenuse of this triangle is 13 cm.
Â
2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.
Solution:
Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR
We have to prove, PM2 = QM × MR
In ΔPQM, by Pythagoras theorem
PQ2Â = PM2Â + QM2
Or, PM2 = PQ2 – QM2 ……………………………..(i)
In ΔPMR, by Pythagoras theorem
PR2Â = PM2Â + MR2
Or, PM2 = PR2 – MR2 ………………………………………..(ii)
Adding equation, (i) and (ii), we get,
2PM2 = (PQ2 + PM2) – (QM2 + MR2)
= QR2 – QM2 – MR2      [∴ QR2 = PQ2 + PR2]
= (QM + MR)2 – QM2 – MR2
= 2QM × MR
∴ PM2 = QM × MR
3. In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB2 = BC × BD
(ii) AC2 = BC × DC
(iii) AD2 = BD × CD
Â
Solution:
∠DAB = ∠ACB (Each 90°)
∠ABD = ∠CBA (Common angles)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
⇒ AB/CB = BD/AB
⇒ AB2 = CB × BD
(ii) Let ∠CAB = x
In ΔCBA,
∠CBA = 180° – 90° – x
∠CBA = 90° – x
∠CAD = 90° – ∠CBA
= 90° – x
∠CDA = 180° – 90° – (90° – x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CAB = ∠CDA
∠ACB = ∠DCA (Each 90°)
∴ ΔCBA ~ ΔCAD [AAA similarity criterion]
⇒ AC/DC = BC/AC
⇒ AC2 =  DC × BC
(iii) In ΔDCA and ΔDAB,
∠DCA = ∠DAB (Each 90°)
∠CDA = ∠ADB (common angles)
∴ ΔDCA ~ ΔDAB [AA similarity criterion]
⇒ DC/DA = DA/DA
⇒ AD2 = BD × CD
4. ABC is an isosceles triangle right angled at C. Prove that AB2Â = 2AC2Â .
Solution:
Given, ΔABC is an isosceles triangle right angled at C.
In ΔACB, ∠C = 90°
AC = BC (By isosceles triangle property)
AB2Â = AC2Â + BC2Â [By Pythagoras theorem]
= AC2Â +Â AC2Â [Since, AC = BC]
AB2Â = 2AC2
5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Solution:
Given, ΔABC is an isosceles triangle having AC = BC and AB2 = 2AC2
In ΔACB,
AC = BC
AB2Â = 2AC2
AB2Â = AC2Â + AC2
= AC2Â + BC2Â [Since, AC = BC]
Hence, by Pythagoras theorem ΔABC is right angle triangle.
6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
Given, ABC is an equilateral triangle of side 2a.
AB = AC
Hence, BD = DC [by CPCT]
⇒ AD2 = 4a2 – a2
7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.
Solution:
Given, ABCD is a rhombus whose diagonals AC and BD intersect at O.
We have to prove, as per the question,
AB2Â + BC2Â + CD2Â + AD2Â = AC2Â + BD2
Since, the diagonals of a rhombus bisect each other at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB,
∠AOB = 90°
AB2 = AO2 + BO2 …………………….. (i) [By Pythagoras theorem]
Similarly,
AD2 = AO2 + DO2 …………………….. (ii)
DC2 = DO2 + CO2 …………………….. (iii)
BC2 = CO2 + BO2 …………………….. (iv)
Adding equations (i) + (ii) + (iii) + (iv), we get,
AB2Â + AD2Â +Â DC2Â +Â BC2Â = 2(AO2Â + BO2Â + DO2Â + CO2)
= 4AO2Â + 4BO2Â [Since, AO = CO and BO =DO]
= (2AO)2Â + (2BO)2Â = AC2Â + BD2
AB2Â + AD2Â +Â DC2Â +Â BC2Â = AC2Â + BD2
Hence, proved.
8. In Fig. 6.54, O is a point in the interior of a triangle.
ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ,
(ii) AF2Â + BD2Â + CE2Â = AE2Â + CD2Â + BF2.
Solution:
Given, in ΔABC, O is a point in the interior of a triangle.
And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
Join OA, OB and OC
(i) By Pythagoras theorem in ΔAOF, we have
OA2Â = OF2Â +Â AF2
Similarly, in ΔBOD
OB2Â = OD2Â + BD2
Similarly, in ΔCOE
OC2Â = OE2Â + EC2
OA2Â + OB2Â + OC2Â = OF2Â + AF2Â + OD2Â + BD2Â + OE2Â + EC2
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2.
(ii) AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)
∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Given, a ladder 10 m long reaches a window 8 m above the ground.
Let BA be the wall and AC be the ladder,
Therefore, by Pythagoras theorem,
AC2Â =Â AB2Â + BC2
102Â = 82Â + BC2
BC2 = 100 – 64
BC2Â = 36
BCÂ = 6m
Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.
10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Given, a guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.
Let AB be the pole and AC be the wire.
By Pythagoras theorem,
AC2Â =Â AB2Â + BC2
242Â = 182Â + BC2
BC2 = 576 – 324
BC2Â = 252
BC = 6√7m
Therefore, the distance from the base is 6√7m.
11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after
 hours?
Solution:
Given,
Speed of first aeroplane = 1000 km/hr
Distance covered by first aeroplane flying due north in
  hours (OA) = 1000 × 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr
Distance covered by second aeroplane flying due west in
  hours (OB) = 1200 × 3/2 km = 1800 km
In right angle ΔAOB, by Pythagoras Theorem,
AB2Â =Â AO2Â + OB2
⇒ AB2 = (1500)2 + (1800)2
⇒ AB = √(2250000 + 3240000)
= √5490000
⇒ AB = 300√61 km
Hence, the distance between two aeroplanes will be 300√61 km.
12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution:
Given, Two poles of heights 6 m and 11 m stand on a plane ground.
And distance between the feet of the poles is 12 m.
Let AB and CD be the poles of height 6m and 11m.
Therefore, CP = 11 – 6 = 5m
From the figure, it can be observed that AP = 12m
By Pythagoras theorem for ΔAPC, we get,
AP2Â =Â PC2Â + AC2
(12m)2Â + (5m)2Â = (AC)2
AC2Â = (144+25) m2Â = 169 m2
AC = 13m
Therefore, the distance between their tops is 13 m.
13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2Â + BD2Â = AB2Â + DE2.
Solution:
Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
By Pythagoras theorem in ΔACE, we get
AC2 + CE2 = AE2 ………………………………………….(i)
In ΔBCD, by Pythagoras theorem, we get
BC2 + CD2 = BD2 ………………………………..(ii)
From equations (i) and (ii), we get,
AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …………..(iii)
In ΔCDE, by Pythagoras theorem, we get
DE2Â =Â CD2Â + CE2
In ΔABC, by Pythagoras theorem, we get
AB2Â =Â AC2Â + CB2
Putting the above two values in equation (iii), we get
DE2Â + AB2Â = AE2Â + BD2.
14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2 = 2AC2 + BC2.
Solution:
Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;
DB = 3CD.
In Δ ABC,
AD ⊥BC and BD = 3CD
AB2 = AD2 + BD2 ……………………….(i)
AC2 = AD2 + DC2 ……………………………..(ii)
Subtracting equation (ii) from equation (i), we get
AB2 – AC2 = BD2 – DC2
= 9CD2 – CD2 [Since, BD = 3CD]
= 8CD2
= 8(BC/4)2Â [Since, BC = DBÂ + CD = 3CDÂ + CD = 4CD]
Therefore, AB2 – AC2 = BC2/2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 – 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.
15. Â In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2Â = 7AB2.
Solution:
Given, ABC is an equilateral triangle.
And D is a point on side BC such that BD = 1/3BC
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
And, AE = a√3/2
Given, BD = 1/3BC
∴ BD = a/3
DE = BE – BD = a/2 – a/3 = a/6
⇒ 9 AD2 = 7 AB2
16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
Given, an equilateral triangle say ABC,
Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
In ΔABE, by Pythagoras Theorem, we get
AB2Â = AE2Â + BE2
4AE2Â = 3a2
⇒ 4 × (Square of altitude) = 3 × (Square of one side)
Hence, proved.
17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°
(B) 60°
(C) 90°Â
(D) 45°
Solution:
Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
We can observe that,
AB2Â = 108
AC2Â = 144
And, BC2Â = 36
AB2Â + BC2Â = AC2
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct answer is (C).
## Exercise 6.6 Page: 152
1. In Figure, PS is the bisector of ∠QPR of ∆ PQR. Prove that QS/PQ = SR/PR
Solution:
Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
Given, PS is the angle bisector of ∠QPR. Therefore,
∠QPS = ∠SPR………………………………..(i)
As per the constructed figure,
∠SPR=∠PRT(Since, PS||TR)……………(ii)
∠QPS = ∠QRT(Since, PS||TR) …………..(iii)
From the above equations, we get,
∠PRT=∠QTR
Therefore,
PT=PR
In â–³QTR, by basic proportionality theorem,
QS/SR = QP/PT
Since, PT=TR
Therefore,
QS/SR = PQ/PR
Hence, proved.
2. In Fig. 6.57, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. Prove that: (i) DM2 = DN . MC (ii) DN2 = DM . AN.
Solution:
1. Let us join Point D and B.
Given,
BD ⊥AC, DM ⊥ BC and DN ⊥ AB
Now from the figure we have,
DN || CB, DM || AB and ∠B = 90 °
Therefore, DMBN is a rectangle.
So, DN = MB and DM = NB
The given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC.
∴ ∠CDB = 90° ⇒ ∠2 + ∠3 = 90° ……………………. (i)
In ∆CDM, ∠1 + ∠2 + ∠DMC = 180°
⇒ ∠1 + ∠2 = 90° …………………………………….. (ii)
In ∆DMB, ∠3 + ∠DMB + ∠4 = 180°
⇒ ∠3 + ∠4 = 90° …………………………………….. (iii)
From equation (i) and (ii), we get
∠1 = ∠3
From equation (i) and (iii), we get
∠2 = ∠4
In ∆DCM and ∆BDM,
∠1 = ∠3 (Already Proved)
∠2 = ∠4 (Already Proved)
∴ ∆DCM ∼ ∆BDM (AA similarity criterion)
BM/DM = DM/MC
DN/DM = DM/MC (BM = DN)
⇒ DM2 = DN × MC
Hence, proved.
(ii) In right triangle DBN,
∠5 + ∠7 = 90° ……………….. (iv)
In right triangle DAN,
∠6 + ∠8 = 90° ………………… (v)
D is the point in triangle, which is foot of the perpendicular drawn from B to AC.
∴ ∠ADB = 90° ⇒ ∠5 + ∠6 = 90° ………….. (vi)
From equation (iv) and (vi), we get,
∠6 = ∠7
From equation (v) and (vi), we get,
∠8 = ∠5
In ∆DNA and ∆BND,
∠6 = ∠7 (Already proved)
∠8 = ∠5 (Already proved)
∴ ∆DNA ∼ ∆BND (AA similarity criterion)
AN/DN = DN/NB
⇒ DN2 = AN × NB
⇒ DN2 = AN × DM (Since, NB = DM)
Hence, proved.
3. In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that
AC2= AB2+ BC2+ 2 BC.BD.
Solution:
By applying Pythagoras Theorem in ∆ADB, we get,
AB2 = AD2 + DB2 ……………………… (i)
Again, by applying Pythagoras Theorem in ∆ACD, we get,
AC2 = AD2 + (DB + BC) 2
AC2 = AD2 + DB2 + BC2 + 2DB × BC
From equation (i), we can write,
AC2 = AB2 + BC2 + 2DB × BC
Hence, proved.
4. In Figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that
AC2= AB2+ BC2 – 2 BC.BD.
Solution:
By applying Pythagoras Theorem in ∆ADB, we get,
We can write it as;
⇒ AD2 = AB2 − DB2 ……………….. (i)
By applying Pythagoras Theorem in ∆ADC, we get,
From equation (i),
AB2 − BD2 + DC2 = AC2
AB2 − BD2 + (BC − BD) 2 = AC2
AC2 = AB2 − BD2 + BC2 + BD2 −2BC × BD
AC2 = AB2 + BC2 − 2BC × BD
Hence, proved.
5. In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
(i) AC2 = AD2 + BC.DM + 2 (BC/2) 2
(ii) AB2 = AD2 – BC.DM + 2 (BC/2) 2
(iii) AC2 + AB2 = 2 AD2 + ½ BC2
Solution:
(i) By applying Pythagoras Theorem in ∆AMD, we get,
AM2 + MD2 = AD2 ………………. (i)
Again, by applying Pythagoras Theorem in ∆AMC, we get,
AM2 + MC2 = AC2
AM2 + (MD + DC) 2 = AC2
(AM2 + MD2 ) + DC2 + 2MD.DC = AC2
From equation(i), we get,
AD2 + DC2 + 2MD.DC = AC2
Since, DC=BC/2, thus, we get,
AD2 + (BC/2) 2 + 2MD.(BC/2) 2 = AC2
AD2 + (BC/2) 2 + 2MD × BC = AC2
Hence, proved.
(ii) By applying Pythagoras Theorem in ∆ABM, we get;
AB2 = AM2 + MB2
= (AD2 − DM2) + MB2
= (AD2 − DM2) + (BD − MD) 2
= AD2 − DM2 + BD2 + MD2 − 2BD × MD
= AD2 + BD2 − 2BD × MD
= AD2 + (BC/2)2 – 2(BC/2) MD
= AD2 + (BC/2)2 – BC MD
Hence, proved.
(iii) By applying Pythagoras Theorem in ∆ABM, we get,
AM2 + MB2 = AB2 ………………….… (i)
By applying Pythagoras Theorem in ∆AMC, we get,
AM2 + MC2 = AC2 …………………..… (ii)
Adding both the equations (i) and (ii), we get,
2AM2 + MB2 + MC2 = AB2 + AC2
2AM2 + (BD − DM) 2 + (MD + DC) 2 = AB2 + AC2
2AM2+BD2 + DM2 − 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2
2AM2 + 2MD2 + BD2 + DC2 + 2MD (− BD + DC) = AB2 + AC2
2(AM2+ MD2) + (BC/2) 2 + (BC/2) 2 + 2MD (-BC/2 + BC/2) 2 = AB2 + AC2
2AD2 + BC2/2 = AB2 + AC2
6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Solution:
Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.
By applying Pythagoras Theorem in ∆DEA, we get,
DE2 + EA2 = DA2 ……………….… (i)
By applying Pythagoras Theorem in ∆DEB, we get,
DE2 + EB2 = DB2
DE2 + (EA + AB) 2 = DB2
(DE2 + EA2) + AB2 + 2EA × AB = DB2
DA2 + AB2 + 2EA × AB = DB2 ……………. (ii)
By applying Pythagoras Theorem in ∆ADF, we get,
Again, applying Pythagoras theorem in ∆AFC, we get,
AC2 = AF2 + FC2 = AF2 + (DC − FD) 2
= AF2 + DC2 + FD2 − 2DC × FD
= (AF2 + FD2) + DC2 − 2DC × FD AC2
AC2= AD2 + DC2 − 2DC × FD ………………… (iii)
Since ABCD is a parallelogram,
AB = CD ………………….…(iv)
And BC = AD ………………. (v)
∠DEA = ∠AFD (Each 90°)
∴ ∆EAD ≅ ∆FDA (AAS congruence criterion)
⇒ EA = DF ……………… (vi)
Adding equations (i) and (iii), we get,
DA2 + AB2 + 2EA × AB + AD2 + DC2 − 2DC × FD = DB2 + AC2
DA2 + AB2 + AD2 + DC2 + 2EA × AB − 2DC × FD = DB2 + AC2
From equation (iv) and (vi),
BC2 + AB2 + AD2 + DC2 + 2EA × AB − 2AB × EA = DB2 + AC2
AB2 + BC2 + CD2 + DA2 = AC2 + BD2
7. In Figure, two chords AB and CD intersect each other at the point P. Prove that :
(i) ∆APC ~ ∆ DPB
(ii) AP . PB = CP . DP
Solution:
Firstly, let us join CB, in the given figure.
(i) In ∆APC and ∆DPB,
∠APC = ∠DPB (Vertically opposite angles)
∠CAP = ∠BDP (Angles in the same segment for chord CB)
Therefore,
∆APC ∼ ∆DPB (AA similarity criterion)
(ii) In the above, we have proved that ∆APC ∼ ∆DPB
We know that the corresponding sides of similar triangles are proportional.
∴ AP/DP = PC/PB = CA/BD
⇒AP/DP = PC/PB
∴AP. PB = PC. DP
Hence, proved.
8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
(i) ∆ PAC ~ ∆ PDB
(ii) PA . PB = PC . PD.
Solution:
(i) In ∆PAC and ∆PDB,
∠P = ∠P (Common Angles)
As we know, exterior angle of a cyclic quadrilateral is ∠PCA and ∠PBD is opposite interior angle, which are both equal.
∠PAC = ∠PDB
Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)
(ii) We have already proved above,
∆APC ∼ ∆DPB
We know that the corresponding sides of similar triangles are proportional.
Therefore,
AP/DP = PC/PB = CA/BD
AP/DP = PC/PB
∴ AP. PB = PC. DP
9. In Figure, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠BAC.
Solution:
In the given figure, let us extend BA to P such that;
AP = AC.
Now join PC.
Given, BD/CD = AB/AC
⇒ BD/CD = AP/AC
By using the converse of basic proportionality theorem, we get,
∠BAD = ∠APC (Corresponding angles) ……………….. (i)
And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)
By the new figure, we have;
AP = AC
⇒ ∠APC = ∠ACP ……………………. (iii)
On comparing equations (i), (ii), and (iii), we get,
Therefore, AD is the bisector of the angle BAC.
Hence, proved.
10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Solution:
Let us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the
horizontal distance of the fly from the tip of the fishing rod. Therefore, AC is now the length of the string.
To find AC, we have to use Pythagoras theorem in ∆ABC, is such way;
AC2 = AB2+ BC2
AB2 = (1.8 m) 2 + (2.4 m) 2
AB2 = (3.24 + 5.76) m2
AB2 = 9.00 m2
⟹ AB = √9 m = 3m
Thus, the length of the string out is 3 m.
As its given, she pulls the string at the rate of 5 cm per second.
Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m
Let us say now, the fly is at point D after 12 seconds.
Length of string out after 12 seconds is AD.
AD = AC − String pulled by Nazima in 12 seconds
= (3.00 − 0.6) m
= 2.4 m
(1.8 m) 2 + BD2 = (2.4 m) 2
BD2 = (5.76 − 3.24) m2 = 2.52 m2
BD = 1.587 m
Horizontal distance of fly = BD + 1.2 m
= (1.587 + 1.2) m = 2.787 m
= 2.79 m
## NCERT Solutions for Class 10 Maths Chapter 6 – Triangles
NCERT Solutions Class 10 Maths Chapter 6, Triangles, is part of the Unit Geometry, which constitutes 15 marks of the total marks of 80. On the basis of the updated CBSE Class 10 Syllabus for 2023-24, this chapter belongs to the Unit-Geometry and has the second-highest weightage. Hence, having a clear understanding of the concepts, theorems and problem-solving methods in this chapter is mandatory to score well in the board examination of Class 10 Maths.
### Main topics covered in this chapter include:
#### 6.1 Introduction
From your earlier classes, you are familiar with triangles and many of their properties. In Class 9, you have studied congruence of triangles in detail. In this chapter, we shall study about those figures which have the same shape, but not necessarily the same size. Two figures having the same shape (and not necessarily the same size) are called similar figures. In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.
#### 6.2 Similar Figures
In Class 9, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent. The topic explains similarity of figures by performing the relevant activity. Similar figures are two figures having the same shape, but not necessarily the same size.
#### 6.3 Similarity of Triangles
The topic recalls triangles and its similarities. Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). It explains Basic Proportionality Theorem and different theorems are discussed performing various activities.
#### 6.4 Criteria for Similarity of Triangles
In the previous section, we stated that two triangles are similar (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). The topic discusses the criteria for similarity of triangles referring to the topics we have studied in earlier classes. It also contains different theorems explained with proper examples.
#### 6.5 Areas of Similar Triangles
You have learnt that in two similar triangles, the ratio of their corresponding sides is the same. The topic Areas of Similar Triangles consists of theorem and relatable examples to prove the theorem.
#### 6.6 Pythagoras Theorem
You are already familiar with the Pythagoras Theorem from your earlier classes. You have also seen proof of this theorem in Class 9. Now, we shall prove this theorem using the concept of similarity of triangles. Hence, the theorem is verified through some activities, and you can make use of it while solving certain problems.
#### 6.7 Summary
The summary contains the points you have studied in the chapter. Going through the points mentioned in the summary will help you to recollect all the important concepts and theorems of the chapter.
### List of Exercises in NCERT Class 10 Maths Chapter 6:
Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)
Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)
Exercise 6.3 Solutions 16 Questions (1 main question with 6 sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)
Exercise 6.4 Solutions 9 Questions (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions)
Exercise 6.5 Solutions 17 Questions (15 Short Answer Questions, 2 Long Answer Questions)
Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)
Triangle is one of the most interesting and exciting chapters of the unit Geometry as it takes us through the different aspects and concepts related to the geometrical figure triangle. A triangle is a plane figure that has three sides and three angles. This chapter covers various topics and sub-topics related to triangles, including a detailed explanation of similar figures, different theorems related to the similarities of triangles with proof, and the areas of similar triangles. The chapter concludes by explaining the Pythagoras theorem and the ways to use it in solving problems. Read and learn Chapter 6 of the Class 10 Maths NCERT textbook to learn more about Triangles and the concepts covered in it. Ensure to learn the NCERT Solutions for Class 10 effectively to score high in the board examinations.
### Key Features of NCERT Solutions for Class 10 Maths Chapter 6 – Triangles
• Helps to ensure that the students use the concepts in solving the problems.
• Encourages the children to come up with diverse solutions to problems.
• Hints are given for those questions which are difficult to solve.
• Helps the students in checking if the solutions they gave for the questions are correct or not.
Disclaimer –Â
Dropped Topics –Â
6.5 Areas of similar triangles
6.6 Pythagoras theorem
## Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 6
Q1
### Why should we learn all the concepts present in NCERT Solutions for Class 10 Maths Chapter 6?
The concepts covered in Chapter 6 of NCERT Solution Maths provide questions based on the exam pattern and model question paper. So it is necessary to learn all the concepts present in NCERT Solutions for Class 10 Maths Chapter 6.
Q2
### List out the important topics present in NCERT Solutions for Class 10 Maths Chapter 6.
The topics covered in the chapters are Introduction to the Triangles, similar figures, the similarity of triangles, criteria for similarity of Triangles, areas of similar triangles and Pythagoras Theorem. These concepts are important from an exam perspective. It is strictly based on the latest syllabus of the CBSE for 2023-24 and also depends on the CBSE question paper design and marking scheme.
Q3
### How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 6?
There are 6 exercises in NCERT Solutions for Class 10 Maths Chapter 6. The first exercise contains 3 questions, the second exercise contains 10 questions, the third exercise has 16 questions, the fourth exercise has 9 questions, the fifth exercise has 17 questions, and the last or sixth exercise has ten questions based on the concepts of triangles.
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Instructions
Answer the questions based on the following information. A salesman enters the quantity sold and the price into the computer. Both the numbers are two-digit numbers. But, by mistake, both the numbers were entered with their digits interchanged.Although, the total sales value remained the same, i.e. Rs. 1,148, the total inventory sold got reduced by 54.
Question 25
# What is the actual quantity sold?
Solution
Total sales value $$= 1148 = 4\cdot7\cdot41$$
Let AB be the actual price of the product and CD be the actual quantity.
Since the quantity/inventory increased by 54 upon reversing
$$CD-DC\ =\ (10\cdot C+D)-\left(10\cdot D+C\right)=9\cdot\left(C-D\right)$$
wkt, $$9\cdot\left(C-D\right)=54\ \ \longrightarrow\ \ C-D=6$$
Now, wkt, $$AB\cdot CD=1148\ \ \&\ \ BA\cdot DC=1148$$
The possible values of CD are $$93, 82$$ & $$71$$
Since only 82 divides 1148, $$CD=82$$, Therefore, $$AB=14$$
Actual price and Actual quantity are 14 and 82 respectively.
Alternatively,
Since final value $$= 1148 = 4\cdot7\cdot41$$ remains the same,
1148 can be represented as product of interchangeable numbers i.e. $$41 \times 28$$ and $$82 \times 14$$
As inventory decreased by 54, so, the quantity sold should be 82 and actual price will be 14.
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## Elementary Studies
### Decimal Place Value
August is already here! Lazy mornings of summer holidays are just flying by and a new school year is starting soon. I will miss the summer fun, but I am more than ready to start another thrilling academic year with new goals, new challenges, and a hope to make this the best school year ever!
While thinking of upper elementary math, the first topic that comes to my mind is a refresher on the number place value system. These young minds already know the whole number place value, but it’s good to refresh the same and introduce the decimal place value too.
The origin of the Arabic number system that we use today can be traced back to the Hindu Scholars in India. This system was adopted and modified by Middle Eastern Scholars and later introduced in Europe. Without going into too much history, let's look at the place value system.
The set of 10 symbols namely 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 represent the numbers.
In the place value system, the position of the number defines the value of the number.
For example, if I write the number 786 then:
• 6 in 786 means 6 ones or 6
• 8 in 786 means 8 tens or 80
• 7 in 786 means 7 hundreds or 700
Now for decimals, the place value system extends to the right of ones place to represent the fractions.
For example, if I write the number 786.432 then:
• 4 in 786.432 means 4 tenths or 0.4
• 3 in 786.432 means 3 hundredths or .03
• 2 in 786.432 means 2 thousandths or .002
Check out what’s included in my Decimal Place Value Worksheets and Task Card bundle pack to learn and practice the concept of Place Value for decimals: tenths, hundredths, and thousandths.
Worksheets:
1. Identifying decimals
2. Standard, Expanded and Word Form
3. Decimals on Number Line
4. Comparing decimals
5. Relate Decimals and Fractions
6. Review Sheet
A total of 18 worksheets.
Task-cards to practice the concept of place-value for decimals: tenths, hundredths, and thousandths.
A total of 24 cards along with recording and answer key.
You can grab this Decimal Place Value resource from my store HERE.
Happy to be back in action!
Shimps
### STAAR Math Grade 3 Practice Test
STAAR is almost here. Relieve the feeling of stress by having your students do a round of practice with this full-length 3rd Grade Math STAAR Practice Test. This mock test is aligned to the current TEKS and includes a total of 46 questions.
Student answer document/Bubble sheet and the Answer Key is also included. You can grab this practice test from HERE.
As the STAAR testing is a new experience for the 3rd graders, it is always helpful to administer a dry run for them to know what to expect in the real exam setting.
Best of luck to the little ROCK STAARS!
This post was originally written on 30th April 2017
Shimps
### Area and Perimeter
Two weeks back I wrapped up my Area and Perimeter unit. Around this time of the year, many schools are working on their Geometry and Measurement topics too!
Area and Perimeter is an interesting topic for kids. I love it when they can relate to the real-world situations like putting a fence around the garden or laying grass in it.
This packet is already available in my store but I'm a little behind on updating it on my blog. I have been a busy bee of late 😊.
The topic begins with introducing Area and Perimeter. You may start with either of the two, but I prefer introducing Perimeter first and then Area. Later, I mix it up for them to practice.
### Perimeter:
Perimeter is the distance Around outside of a figure.
Area:
Area is the number of square units needed to Cover the region inside the figure.
Check out what’s included in my beginner’s worksheets and activities pack to learn and practice the concept of Area and Perimeter.
1. Measuring Perimeter
2. Measuring Area
3. Area Estimation
4. Area using Distributive Property
5. Perimeter Word Problems
6. Area Word Problems
7. Area and Perimeter Review
8. School Carnival activity for real-world application
9. Town Planning activity for a project
You can grab this area and perimeter resource from my store HERE.
Happy Teaching!
Shimps
### Fractions
The fractions topic begins with teaching how to partition shapes into equal parts. We generally start with halves and fourth and then move to thirds, fifths, sixth and so on. At this stage, kids are also introduced to the concept of numerator and denominator. Before this, they were only working with the whole numbers, and now they are presented with rational numbers. Quite confusing for them…..hmm, not really.
Children are visual learners, they learn the best by seeing it. Teaching fraction involves drawing pictures and hand on activities. So, try to incorporate a lot of visuals to make the fraction concept more thorough.
Pizza is the most commonly used manipulative to explain the idea of fractions. Students can easily correlate the part of a whole concept with a pizza and its slice. This serves as a real-life example to develop mental images of fraction model as a part of a circle.
My favorite read aloud book on this topic is Gator Pie by Louise Mathews. Kids enjoy making predictions as the story is being read.
This is a cute story about two gators, Alvin and Alice, who find a pie and decide to share it equally with each other. Just as they are about to eat it, two more gators emerge from the forest. Then they must divide the pie among four of them. Then, more and more alligators arrive, and they must share the pie among 100 gators! Finally, Alvin and Alice sneak away with the pie.
This is a classic book and may be out of print now. Check if this happens to be in your school or local library or you may get a used one here on Amazon.
My Fractions packet includes:
1. Fractions Vocabulary Word Cards
2. Equal and Unequal Parts cut and paste fraction activity
3. Color the shapes with Equal and Unequal Parts
4. Color the shapes with 2/3/4 Equal Parts
5. Fractions: Worksheet on writing fraction
6. Fractions: Halves, Thirds and Fourths
7. Understanding Fractions: Halves
8. Understanding Fractions: Thirds
9. Understanding Fractions: Fourths
10. Write the Fractions - Worksheet
11. Fraction wall - cut and paste activity
12. Fraction wall poster
13. Spin and color the fractions activity
14. 24 Fractions flash cards
You can grab this fractions resource from my store HERE.
Happy Teaching!
This post was originally published on February 16, 2017
Shimps
### Valentine's Day Creative Writing Template
Red and Pink is the color of the month,
Love and Friendship is the theme of the month.
February is here and the valentine's day is just a week away, teachers are busy in decorating the classroom, kids are geared up for the party. Have some heart inspired craft to keep up the valentine mood.
Here are three adorable crafts to celebrate the day of love and friendship! Use these for creative writing prompts and decorate your bulletin boards. You may also send these home to give it to a friend or family.
#### 3 Valentine's Day Projects: Owl, Butterfly and Chicken
This Valentine's Day creative writing template pack includes:
1. Template for printing onto colored paper or use it as traceable templates for construction paper.
2. Printing instructions.
3. Pictures to help in assembly.
4. Prompt template with wide rule lines to go with the craft.
You can give the appropriate prompts based on student’s level. I have included 10 sample prompt ideas. You can grab this pack from HERE.
Happy Craftivity!
Shimps
This post was originally published on February 6, 2017
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# Math Lesson 3: Displaying Data Graphically - page 10 / 11
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A data entry is an outlier if it is more than one-and-a-half times the inter-quartile range above Q3 or below Q1.
To find out if the score of 32 is really an outlier, first compute the IQR of the data set for the original scores in Mr. Sneed’s class given in Practice Problem #1a.
IQR = Q3 - Q1 = 82 – 66 = 16 1.5 x IQR = 1.5 x 16 = 24
# Q1 - 24 = 66 – 24 = 42
The score of 32 is below 42, so 32 is an outlier.
# Is 98 an outlier in the other direction?
1.5 x IQR = 1.5 x 16 = 24 Q3 + 24 = 82 + 24 = 106
The score of 98 is less than 106, so 98 is not an outlier.
# Practice problem:
7. In “Ages of Oscar-Winning Best Actors and Actresses” (Mathematics Teacher magazine) by Richard Brown and Gretchen Davis, stem-and-leaf plots are used to compare the ages of actors and actresses at the time they won Oscars. Below are the ages for 34 recent Oscar winners in each gender category:
# Actors:
32 37 36 51 53 33 61 35 45 55 39 76 37 42 40 32 32 60 38 56 48 48 40 43 62 43 42 44 41 56 39 46 31 47
Actresses: 50 44 35 80 26 28 41 21 61 38 49 33 74 30 33 41 31 35 41 42 37 26 34 34 35 26 61 60 34 24 30 37 31 27
• a)
Construct a back-to-back stem-and-leaf plot for the above data, using the tens digits for the stems. Discuss shape, center and spread for each data set, comparing the two sets and discussing any differences you see.
• b)
Construct two box-and-whisker plots on the same graph to compare ages of actors and actresses at the time they won Oscars. What do you observe? Write a short paragraph discussing the bottom 25% of ages, the middle 50% of ages and the top 25% of ages, and discuss difference you observe in the data sets. What conclusions can you draw?
• c)
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# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (2 ,5 ) to (8 ,7 ) and the triangle's area is 12 , what are the possible coordinates of the triangle's third corner?
Nov 17, 2016
The possible points are: $\left(3.8 , 9.6\right)$ and $\left(6.2 , 2.4\right)$
#### Explanation:
The length of side "a" is:
$a = \sqrt{{\left(8 - 2\right)}^{2} + {\left(7 - 5\right)}^{2}} = \sqrt{40} = 2 \sqrt{10}$
The area of a triangle is:
$\text{Area" = (1/2)"Base"xx"Height}$
Using side "a" as the base of the triangle and the given area, we can compute the height:
$12 = \left(\frac{1}{2}\right) 2 \sqrt{10} \times \text{Height}$
$\text{Height} = \frac{12}{\sqrt{10}}$
Think of the height as the radius of a circle upon which the two possible points must lie:
${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {\left(\frac{12}{\sqrt{10}}\right)}^{2}$
Because the triangle's other two sides are the same length, the center of this circle must be the midpoint between the points #(2, 5) and (8, 7):
$h = 2 + \frac{8 - 2}{2} = 5 \mathmr{and} k = 5 + \frac{7 - 5}{2} = 6$
Substitute into the equation of the circle:
${\left(x - 5\right)}^{2} + {\left(y - 6\right)}^{2} = {\left(\frac{12}{\sqrt{10}}\right)}^{2} \text{ [1]}$
The two points must, also, lie on a line that is perpendicular to side "a"; the slope, m, of this line is:
$m = \frac{2 - 8}{7 - 5} = - \frac{6}{2} = - 3$
Using, this slope, the center point, and the point-slope form of the equation of a line, we write the equation upon which the two points must lie:
$y = - 3 \left(x - 5\right) + 6 \text{ [2]}$
Here is a graph of the equation [1], equation [2] and the two given points:
Substitute the right side of equation [2] into equation [1]:
${\left(x - 5\right)}^{2} + {\left(- 3 \left(x - 5\right) + 6 - 6\right)}^{2} = {\left(\frac{12}{\sqrt{10}}\right)}^{2}$
${\left(x - 5\right)}^{2} + {\left(- 3 \left(x - 5\right)\right)}^{2} = {\left(\frac{12}{\sqrt{10}}\right)}^{2}$
${\left(x - 5\right)}^{2} + 9 {\left(x - 5\right)}^{2} = {\left(\frac{12}{\sqrt{10}}\right)}^{2}$
${\left(x - 5\right)}^{2} + 9 {\left(x - 5\right)}^{2} = \frac{144}{10}$
$10 {\left(x - 5\right)}^{2} = \frac{144}{10}$
${\left(x - 5\right)}^{2} = \frac{144}{100}$
$x - 5 = \pm \frac{12}{10}$
$x = 5 \pm \frac{12}{10}$
$x = 3.8 \mathmr{and} x = 6.2$
To obtain the corresponding y values, substitute these x values into equation [2]
$y = - 3 \left(3.8 - 5\right) + 6$ and $y = - 3 \left(6.2 - 5\right) + 6$
$y = 9.6 \mathmr{and} y = 2.4$
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1. ## area between curves.
I have attached the sketch of the curves,
I am not sure how to integrate, because one part is below the x-axis and the other is above.
Appreciate if someone can show me how to solve these problems.
2. If you subtract $\displaystyle x^2-4x$ from $\displaystyle 4x-x^2,$
then you will always have the difference between the curves.
Integrating this difference gives the area between the points of intersection.
Or, since the region is symmetrical about the x-axis,
integrate the top curve and multiply the answer by 2,
or integrate the region under the x-axis and multiply by -2.
3. Remember....
integration sums f(x),
so if f(x) is negative, integration sums negative values.
If you want area, you just change the sign,
though you must be aware of where the graph crosses the x-axis
to isolate where f(x) is positive or negative.
4. Originally Posted by Archie Meade
If you subtract $\displaystyle x^2-4x$ from $\displaystyle 4x-x^2,$
then you will always have the difference between the curves.
Integrating this difference gives the area between the points of intersection.
Or, since the region is symmetrical about the x-axis,
integrate the top curve and multiply the answer by 2,
or integrate the region under the x-axis and multiply by -2.
Thank you, so should I integrate each area from 4 to 0 separately and than subtract?
5. No, Tweety,
When you are dealing with the area between curves as in your case,
with one below the other, you can subtract the functions first,
then integrate.
Subtract the lower curve from the upper curve.
$\displaystyle 4x-x^2-(x^2-4x)=8x-2x^2.$
now you can integrate the difference between them
and that gives the area.
Notice that $\displaystyle 4x-x^2$ is $\displaystyle x^2-4x$ upside down.
Hence the area above the x axis between the points of intersection,
is always positive there.
Hence integrating that will give a positive answer.
Or, integrate the lower curve, but since that one is negative, your answer
will be negative.
So just change the sign and multiply by 2.
If you do all 3 ways, it would be excellent practice.
6. Originally Posted by Archie Meade
No, Tweety,
When you are dealing with the area between curves as in your case,
with one below the other, you can subtract the functions first,
then integrate.
Subtract the lower curve from the upper curve.
$\displaystyle 4x-x^2-(x^2-4x)=8x-2x^2.$
now you can integrate the difference between them
and that gives the area.
Notice that $\displaystyle 4x-x^2$ is $\displaystyle x^2-4x$ upside down.
Hence the area above the x axis between the points of intersection,
is always positive there.
Hence integrating that will give a positive answer.
Or, integrate the lower curve, but since that one is negative, your answer
will be negative.
So just change the sign and multiply by 2.
If you do all 3 ways, it would be excellent practice.
Okay I understand now thanks, I will give all three methods ago.
7. if you subtract the lower function from the upper one,
the difference between them is always positive.
Therefore, integrating the difference gives the area,
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Calculus - Area under a Curve
In these lessons, we will learn how to use integrals (or integration) to find the areas under the curves defined by the graphs of functions. We also learn how to use integrals to find areas between the graphs of two functions.
We have also included calculators and tools that can help you calculate the area under a curve and area between two curves.
Formula For Area Bounded By Curves (Using Definite Integrals)
The Area A of the region bounded by the curves y = f(x), y = g(x) and the lines x = a, x = b, where f and g are continuous f(x) ≥ g(x) for all x in [a, b] is
The following diagrams illustrate area under a curve and area between two curves. Scroll down the page for examples and solutions.
Example:
Find the area of the region bounded above by y = x2 + 1, bounded below by y = x, and bounded on the sides by x = 0 and x = 1.
Solution:
The upper boundary curve is y = x2 + 1 and the lower boundary curve is y = x.
Using the formula for area bounded by curves,
How to find the Area between Curves?
Example:
Find the area between the two curves y = x2 and y = 2x – x2.
Solution:
Step 1: Find the points of intersection of the two parabolas by solving the equations simultaneously.
x2 = 2x – x2
2x2 – 2x = 0
2x(x – 1) = 0
x = 0 or 1
The points of intersection are (0, 0) and (1, 1)
Step 2: Find the area between x = 0 and x = 1
How to use the Area Under a Curve to approximate the definite integral?
Example:
Approximate the area under the curve f(x) = x2 (i.e. the area between y = x2 and y = 0) from x = 1 to 3. Use n = 4 rectangles.
How to find the area under a curve using integration, step by step, example
Example:
Find the area bounded by the curves y = x2 - 6x + 9 and y = x + 3.
How to use integration to determine the area under a curve?
A parabola is drawn such that it intersects the x-axis. The x-intercepts are determined so that the area can be calculated.
Example:
Calculate the area enclosed by the curve y = 2x - x2 and the x-axis.
How to use Definite Integrals to find Area Under a Curve?
Use the following Definite Integral Calculator to find the Area under a curve.
Enter the function, lower bound and upper bound.
How to Find Areas Between Curves?
Example:
Find the area bounded by the curves y = x2 - 4x and y = 2x.
How to find area between curves by Integrating with Respect to y.
Example:
Find the area bounded by the curves x = y3 - y and x = 1 - y4.
Use the following Area between Curves Calculator to show you the steps and to check your answers.
Try the free Mathway calculator and problem solver below to practice various 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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It"s very common once learning about fractions to desire to understand how transform a fraction like 9/15 right into a percentage. In this step-by-step guide, we"ll display you exactly how to rotate any portion into a portion really easily. Let"s take a look!
Want to easily learn or present students how to transform 9/15 to a percentage? beat this an extremely quick and also fun video now!
Before we get started in the fraction to percentage conversion, let"s go over some really quick fraction basics. Remember that a numerator is the number above the portion line, and the denominator is the number listed below the fraction line. We"ll usage this later in the tutorial.
You are watching: What is the percentage of 9 out of 15
When we room using percentages, what we room really saying is the the portion is a portion of 100. "Percent" means per hundred, and so 50% is the exact same as saying 50/100 or 5/10 in fraction form.
So, because our denominator in 9/15 is 15, we could adjust the portion to do the denominator 100. To do that, we divide 100 by the denominator:
100 ÷ 15 = 6.6666666666667
Once we have actually that, we can multiple both the numerator and also denominator through this multiple:
9 x 6.6666666666667/15 x 6.6666666666667=60/100
Now we deserve to see that our fraction is 60/100, which method that 9/15 together a percent is 60%.
We can also work this the end in a simpler way by very first converting the portion 9/15 to a decimal. To execute that, we merely divide the molecule by the denominator:
9/15 = 0.6
Once we have actually the answer to that division, we can multiply the price by 100 to make it a percentage:
0.6 x 100 = 60%
And there you have it! Two different ways to convert 9/15 come a percentage. Both room pretty straightforward and easy to do, but I personally prefer the convert to decimal an approach as the takes less steps.
I"ve seen a lot of students get confused whenever a concern comes up around converting a portion to a percentage, but if you follow the measures laid out right here it should be simple. That said, you may still need a calculator because that more complex fractions (and you can always use our calculator in the type below).
See more: Whats The Average Height For A 13 Year Old, What Is The Average Height For A Thirteen Year
If you want to practice, grab yourself a pen, a pad, and also a calculator and try to convert a couple of fractions to a percentage yourself.
Hopefully this tutorial has actually helped you to understand just how to transform a portion to a percentage. You have the right to now go forth and convert fractions to percentages as lot as your little heart desires!
## Fraction together Percentage
Enter a numerator and denominator
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# Difference between revisions of "2009 IMO Problems/Problem 2"
## Problem
Let $ABC$ be a triangle with circumcentre $O$. The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$ respectively. Let $K,L$ and $M$ be the midpoints of the segments $BP,CQ$ and $PQ$, respectively, and let $\Gamma$ be the circle passing through $K,L$ and $M$. Suppose that the line $PQ$ is tangent to the circle $\Gamma$. Prove that $OP=OQ$.
Author: Sergei Berlov, Russia
## Solution
### Diagram
$[asy] dot("O", (50, 43), W); dot("A", (40, 100), N); dot("B", (0, 0), S); dot("C", (100, 0), S); dot("Q", (24, 60), W); dot("P", (52, 80), E); dot("L", (62, 30), SE); dot("M", (38, 70), N); dot("K", (26, 40), S); draw((100, 0)--(24, 60), dotted); draw((0, 0)--(52, 80), dashed); draw((0, 0)--(100, 0)--(40, 100)--cycle); draw((24, 60)--(52, 80)); draw((26, 40)--(38, 70)--(62, 30)--cycle); draw(circle((47, 48), 23)); [/asy]$ Diagram by qwertysri987
By parallel lines and the tangency condition, $$\angle APM\cong \angle LMP \cong \angle LKM.$$ Similarly, $$\angle AQP\cong \angle KLM,$$ so AA similarity implies $$\triangle APQ\sim \triangle MKL.$$ Let $\omega$ denote the circumcircle of $\triangle ABC,$ and $R$ its circumradius. As both $P$ and $Q$ are inside $\omega,$
\begin{align*} R^2-QO^2&=\text{Pow}_{\omega}(Q)\\ &=QB\cdot AQ \\ &=2AQ\cdot MK\\ &=2AP\cdot ML\\ &=AP\cdot PC\\ &=\text{Pow}_{\omega}(P)\\ &=R^2-PO^2. \end{align*} It follows that $OP=OQ.$ $\blacksquare$
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In algebra, the Least Common Multiple (LCM) of a set of numbers is the smallest number that can be divided evenly by each of the numbers in the set. The Greatest Common Factor (GCM) is the biggest number that is a factor of all the numbers in the set. It is important to remember that the Least Common Multiple is always bigger than the Greatest Common Factor. This is a common cause of confusion.
The first step in finding either the LCM or the GCM of a set of numbers is to factor all the numbers in the set. To factor a number is to ruduce it to the product of prime numbers, or numbers that are only divisible by themselves and 1. For example, 18 can be factored as 3 x 3 x 2.
Factorization can be accomplished by taking a number and dividing it by one of its factors, then further dividing the quotients until it is broken down into indivisible prime numbers. In the illustration here, 525 is factored into 5 x 5 x 7 x 3.
To find the GCF of several numbers, compare their prime factors. Here are the prime factorizations of the numbers 210, 330, and 45.
All three of these numbers share two common prime factors, 3 and 5. Multiplying 3 and 5 together produces the greatest common factor, 15. To find the GCM of two or more numbers, multiply together all the prime factors they have in common.
If a number shares all the prime factors of another number, then the number is a multiple of the second number. The Least Common Multiple is the smallest number (besides 0) that shares all the prime factors of all the numbers in the set. For example, 36 is the LCM of 4 and 9. The prime factorization of 36 is 2 x 2 x 3 x 3. It is the smallest number that shares both the prime factors of 4: 2 x 2, and the prime factors of 9: 3 x 3.
The prime factorizations of 210, 330, and 45 are shown in the picture above. The LCM of these three numbers is 2 x 3 x 3 x 5 x 7 x 11 = 3,080. 3,080 shares all of the prime factors of each of the three numbers. Even though 210 and 330 both have a 2 in their prime factorization, only one copy of 2 is needed in their LCM. However, since there are two copies of 3 in the prime factorization of 45, there must be at least two copies of 3 in the LCM.
|
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What Is 10 Percent of 16 OR
# What Is 10 Percent of 16
Few people look forward to answering math questions in school. But finding a good, solid solution - to a challenging math problem- can come in very handy when you’re trying to do math on your own.
### Divide
One of the easiest ways to determine a 10 percent discount is to divide the total sale price by 10 and then subtract that from the price. You can calculate this discount in your head. For a 20 percent discount, divide by ten and multiply the result by two. Or you can use one of two other methods for calculating the 10 percent discount without needing a calculator.
While 10 percent of any amount is the amount multiplied by 0.1, an easier way to calculate 10 percent is to divide the amount by 10. So, 10 percent of \$18.40, divided by 10, equates to \$1.84. To figure the total cost with the 10 percent discount, take \$18.40 and subtract \$1.84 which equates to a total sales price of \$16.56. (Source: sciencing.com)
### Value
Notice that 10 percent of \$18.40 can also be found without doing any math at all. Simply move the decimal point one digit to the left to yield the 10 percent discount. This is applicable to any value. Ten percent of \$1,369.98 is \$136.998, or roughly \$137, for a discounted price of \$1,369.98 minus \$137, or \$1,232.98.
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. (Source: percentagecalculator.guru)
### Step
I've seen a lot of students get confused whenever a question comes up about converting a fraction to a percentage, but if you follow the steps laid out here it should be simple. That said, you may still need a calculator for more complicated fractions (and you can always use our calculator in the form below). s
And there you have it! Two different ways to convert 10/16 to a percentage. Both are pretty straightforward and easy to do, but I personally prefer the convert to decimal method as it takes less steps. (Source: visualfractions.com)
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# Answer the following questions considering the age of Ram is $y$ years and he is $\dfrac{1}{6}$ of his grandfather's age: If his grandmother's age is two years less than his grandfather's age, what is the age of his grandmother?
Last updated date: 14th Jul 2024
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Hint: First we have to define what the terms we need to solve the problem are.
Here, we need to find the present age of grandmother. We will assume the present age of Ram is $y$ years, and he is $\dfrac{1}{6}$ of his grandfather's age, and $g$ for the grandfather’s age.
Let as the present age of Ram is $y$ years and present age of grandfather is $y$ $= \dfrac{1}{6}g$ (Since grandfathers age is $g$ and $\dfrac{1}{6}$ of grandfather’s age)
Since grandfather’s age $\Rightarrow y = \dfrac{1}{6}g$ solving this equation with cross multiplying $6$on left hand side we get
$\Rightarrow 6y = g$ which is $\Rightarrow g = 6y$
Hence the grandfather age is $6y$ (Since $y$ is the age of ram and so $6y$ is the age of grandfather thus if the age of ram is $15$ then the age of the grandfather is 6 times the rams age which is $90$ )
Hence $6y$ is the grandfathers age so then $6y - 2$ will be the grandmother’s age
(Since $6y$ is the grandfathers age as per above example if rams age is $15$ then the grandfathers age is $90$ and Hence the grandmothers age will be two years lesser than the grandfather which is $6y - 2 = 6(15) - 2 = 90 - 2 = 88$ )
Hence the present of the grandmother is $6y - 2$
|
### Chapter 1: Introduction to Statistics
```COURSE: JUST 3900
TIPS FOR APLIA
Chapter 5:
z-Scores
Developed By:
John Lohman
Michael Mattocks
Aubrey Urwick
Key Terms and Formulas: Don’t
Forget Notecards
Describing z-Scores
Question 1: Identify the z-score value corresponding to
each of the following locations in a distribution.
Below the mean by 3 standard deviations.
Above the mean by 1/4 standard deviations.
Below the mean by 2.50 standard deviations.
Question 2: Describe the location in the distribution for
each of the following z-scores.
z = - 1.50
z = 0.25
z = - 3.50
z = 1.75
Describing z-Scores
z = -3.00
z = 0.25
z = -2.50
Below the mean by 1.50 standard deviations.
Above the mean by ¼ standard deviations.
Below the mean by 3.50 standard deviations.
Above the mean by 1.75 standard deviations.
Describing z-Scores
−
The numerator in our z-score formula ( =
)
describes the difference between X and µ. Therefore, if
a question asks you to calculate the z-score for a score
that is above the mean by 4 points and has a standard
deviation of σ = 2, you cannot calculate (X - µ) - - i.e.,
you cannot find the original values for X or µ and
calculate the difference between the two. In this case,
it has already been provided for you because the
question tells you the distance between X and µ
(X - µ) = 4. Thus, your formula should read z = 4/2,
which comes out to be z = 2.00.
Transforming X-Values into
z-Scores
Question 3: For a distribution of µ = 40 and σ = 12, find
the z-score for each of the following scores.
a)
b)
c)
X = 36
X = 46
X = 56
Question 4: For a population with µ = 30 and σ = 8, find
the z-score for each of the following scores.
a)
b)
c)
X = 32
X = 26
X = 42
Transforming X-Values into
z-Scores
Using z-Scores to Compare
Different Populations
Question 5: A distribution of English exam scores has
µ = 70 and σ = 4. A distribution of history exam scores
has µ = 60 and σ = 20. For which exam would a score of
Using z-Scores to Compare
Different Populations
−
−
=
78−70
4
78−60
20
=
=
For the English exam, X = 78 corresponds to z = 2.00, which is a
higher standing than z = 0.90 for the history exam.
=
=
8
= 2.00
4
18
= 0.90
20
=
Remember that 95% of all scores fall between ± 2.00. Thus, a score
of +2.00 means that over 95% of the class scored below 78 on the
English exam.
Using z-Scores to Compare
Different Populations
Question 6: A distribution of English exam scores has
µ = 50 and σ = 12. A distribution of history exam scores
has µ = 58 and σ = 4. For which exam would a score of
Using z-Scores to Compare
Different Populations
−
−
=
62−50
12
62−58
4
=
=
The score X = 62 corresponds to z = 1.00 in both distributions.
The score has exactly the same standing for both exams.
=
=
12
= 1.00
12
4
= 1.00
4
=
z-Scores and Standardized
Scores
Question 7: A population of scores has µ = 73 and σ = 8.
If the distribution is standardized to create a new
distribution with µ = 100 and σ = 20, what are the new
values for each of the following scores from the original
distribution?
a)
b)
c)
d)
X = 65
X = 71
X = 81
X = 83
z-Scores and Standardized
Scores
z-Scores and Standardized
Scores
Question 8: A population with a mean of µ = 44 and a
standard deviation of σ = 6 is standardized to create a
new distribution with µ = 50 and σ = 10.
a)
b)
What is the new standardized value for a score of X = 47 from
the original distribution?
One individual has a new standardized score of X = 65. What
z-Scores and Standardized
Scores
X = 47
z = 0.50
X = 55
Old Distribution
32
38
44
50
56
z-Score Distribution
New Standardized Distribution
z-Scores and Standardized
Scores
X = 53
z = 1.50
X = 65
32
38
44
Old Distribution
50
56
z-Score Distribution
New Standardized Distribution
Measure of Relative Location
and Detecting Outliers
Question 9: A sample has a mean of M = 30 and a
standard deviation of s = 8.
a)
b)
Would a score of X = 36 be considered a central score or an
extreme score in the sample?
If the standard deviation were s = 2, would X = 36 be central or
extreme?
Measure of Relative Location
and Detecting Outliers
a)
=
1.
b)
=
1.
−
=
36−30
8
=
6
8
= 0.75
X = 36 is not an extreme score because it is within two standard
deviations of the mean.
−
=
36−30
2
=
6
2
= 3.00
In this case, X = 36 is an extreme score because it is more than
two standard deviations above the mean.
WARNING!!!
The book defines an extreme score as being more than
TWO standard deviations away from the mean.
However, Aplia defines extreme scores as being more
than THREE standard deviations from the mean.
When using Aplia, use the THREE definition of standard
deviation for extreme scores.
On in class exercises and on the test, use the TWO
definition of standard deviation for extreme scores.
FAQs
How do I find the z-scores from a raw set of scores?
1)
X = 11, 0, 2, 9, 9, 5
Find the mean:
1)
2)
=
=
11+0+2+9+9+5
6
=
36
6
=6
Find SS:
1)
=
−
2
X
X-µ
(X - µ)2
11
11 – 6 = 5
(5)2 = 25
0
0 – 6 = -6
(-6)2 = 36
2
2 – 6 = -4
(-4)2 = 16
2
9–6=3
(3)2 = 9
9
9–6=3
(3)2 = 9
5
5 – 6 = -1
(-1)2 = 1
=
−
2
= 96
FAQs
3)
Find σ2:
1)
4)
=
96
6
= 16
Find σ:
1)
5)
2 =
=
=
96
6
=
16 = 4
Find z-score for each X:
1)
=
2)
=
3)
=
4)
=
5)
=
6)
=
−
−
−
−
−
−
=
=
=
=
=
=
11−6
5
=
= 1.25
4
4
0−6
−6
=
= −1.50
4
4
2−6
−4
=
= −1.00
4
4
9−6
3
=
= 0.75
4
4
9−6
3
=
= 0.75
4
4
5−6
−1
=
= −0.25
4
4
```
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# Shobo's mother's present age is six times Shobo's present age. Shobo's age five years from now will be one third of his mother's present age. What are their present ages?
Last updated date: 17th Jul 2024
Total views: 450.3k
Views today: 6.50k
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Hint:
Assume the Shobo’s present age as $x$. Then according to the question find the mother’s present age in terms of $x$. After that, apply the condition of five years and you will get the equation. Simplify it and you will get the answer.
In Mathematics, an equation is a statement that asserts the equality of two expressions. The word equation and its cognates in other languages may have subtly different meanings; for example, in French an equation is defined as containing one or more variables, while in English any equality is an equation.
Solving an equation containing variables consists of determining which values of the variables make the equality true. Variables are also called unknowns and the values of the unknowns that satisfy the equality are called solutions of the equation. There are two kinds of equations: identities and conditional equations. An identity is true for all values of the variable. A conditional equation is only true for particular values of the variables.
Single variable equation can be $x+2=0$.
An equation is written as two expressions, connected by an equals sign ("$=$"). The expressions on the two sides of the equals sign are called the "left-hand side" and "right-hand side" of the equation.
The most common type of equation is an algebraic equation, in which the two sides are algebraic expressions. Each side of an algebraic equation will contain one or more terms. For example, the equation,
$A{{x}^{2}}+Bx+C=y$
has left-hand side $A{{x}^{2}}+Bx+C$ which has three terms, and right-hand side $y$, consisting of just one term. The unknowns are $x$ and $y$ and the parameters are $A,$ $B,$ and $C$.
An equation is analogous to a scale into which weights are placed. When equal weights of something (grain for example) are placed into the two pans, the two weights cause the scale to be in balance and are said to be equal. If a quantity of grain is removed from one pan of the balance, an equal amount of grain must be removed from the other pan to keep the scale in balance. Likewise, to keep an equation in balance, the same operations of addition, subtraction, multiplication and division must be performed on both sides of an equation for it to remain true.
Let Shobo’s present age be $x$years.
Then according to the question, the present age of his mother will be $6x$years.
After five years,
Shobo’s age will be $(x+5)$years.
It is given in question that after five years, Shobo’s age will be one third of his mother's present age.
$(x+5)=\dfrac{1}{3}(6x)$
Simplifying above equation we get,
\begin{align} & (x+5)=2x \\ & 5=2x-x \\ & x=5 \\ \end{align}
$x=5$years
Therefore, Shobo’s present age will be $5$years and his mother’s present age will be $6x=6\times 5=30$years.
Note:
Read the question carefully. Do not confuse yourself while simplifying. Also, you must understand the concept behind the question. Do not miss any term. Take care that no mistakes occur. Solve the question in a step by step manner.
|
# KIDS MATHS PROJECT - Learning Multiplication Of Numbers by making a multiplication wheel
Good day lovely hivers and @homeedders. How are you all doing today ?.
My little boy Josh is now becoming a big boy ( so he says) and we have started learning some basic arithmetic like multiplication, addition, subtraction and division.
Here, we were learning about multiplication and to do that, I prepared a multiplication wheel to make it more fun and exciting.
Although Josh is growing everyday, it is really important that we do not take the fun out of our lessons.
I believe that learning should be fun and interesting and whatever method would help the child become more interested in learning should be adopted.
In this activity, I made a complete multiplication wheel for multiplication of numbers by 2 and 3. After we studied using that, I then made another wheel and let him complete it by filling the answers.
Josh likes the lesson idea/method. He was so happy and pleaded that I let him keep the multiplication wheel in his room.
Making the multiplication wheel was really easy. It requires just
• Pencil
• Marker
• Cardboard paper
• A pair of compass for drawing circles
To make the multiplication wheel, I first made the diagram of the wheel on the cardboard.
The wheel is made by drawing three circles of different radii from the same point ( ie inscribing two smaller circles in a bigger circle).
Then the two outer circles are divided into 12 equal parts.
You can draw the circles based on the size of your paper.
After you're done with drawing the wheel, the next thing to do is to fill in the first circle and second circle divisions by the number you're multiplying by ( 2 or 3 in the case of the picture below) and with numbers from 1 to 12 respectively, as shown below.
After that complete the multiplication wheel by multiplying the number in the smallest circle ( 2 & 3 ) by all the numbers in the bigger circle, then write the answers in the boxes just above each number
We studied multiplication of numbers by 2 & 3 using the multiplication wheel I had made
To test whether he had understood and memorised it, I made another wheel and this time I let Josh complete it
Josh was really happy to do it himself. Kids are always excited to participate in activities. They like to prove that they themselves and that was the idea behind learning multiplication by making a wheel. The kids see the fun and adventure in making the wheel and they rush into it and by making the wheel they learn multiplication.
Josh completed the wheel so well and I confirmed that he had learnt multiplication of numbers by 2 & 3
At the end of the activity we achieved two things ; we had fun and we still learnt multiplication.
I let Josh keep the multiplication wheel. He was happy. He said he would put in in his room.
This is why it is important that we come up with creative learning ideas and activities. Learning becomes fun and easy when we do so.
Thanks for stopping by, I hope you found this interesting. I also hope you learnt something today. Let's all make learning fun .
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Greetings @luckydrums, very good idea for your little one to learn multiplication, with games children are motivated and interested in learning, thanks for sharing.
Blessings!
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I just learnt something new. I have this kid who is finding it difficult to memorize the multiplication. I'll try this with him. He loves activities as this and would be glad to participate.
You're doing a good job with Josh.
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Thank you so much for stopping by. I really appreciate it
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It is an interesting way to study. Very creative.
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# There were $50$ people at the birthday party. John invited $125$ people. Among those who attended, only $36\%$ brought gifts. How many guests brought gifts?a) $125$ peopleb) $50$ peoplec)$18$ peopled) $32$ people
Last updated date: 20th Jul 2024
Total views: 348k
Views today: 7.48k
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Hint: Each person brought one gift only, so there will be one gift per person. Therefore, we have to just calculate the value of the given percentage of the number of persons who attended the birthday party. Here all attendees do not bring the gifts but there the percentage of the person who brings the gifts is given so we use the concept of percentage.
Complete step-by-step solution:
Total invitation given by john = $125$ people
Total people attended the party = $50$ people
Since there are only $50$ people present in the party, therefore it is not possible that $125$ people brought gifts
Hence, option (a) is incorrect
Option b) $50$ people
Explanation:
According to the question,
Total people attended the party = $50$ people
And it is given that only $36\%$ of the people attending the party brought gifts.
Hence, it is not possible that all the attendees brought gift
Therefore, option (b) is also incorrect
Option c) $18$people
Explanation:
According to the question,
Total people attended the party =$50$ people
And it is given that only $36\%$ of the people attending the party brought gifts.
$36\%$ Of the total people attended the party = $50 \times \dfrac{{36}}{{100}}$ people = $18$ people
Hence, only $18$ people from all the $50$ attendees brought the gift.
Therefore, option (c) is correct.
Option d) $32$ people
Explanation: $32$ people are those people who do not bring gifts to the party. But we have to find the number of people who brought gifts to the party.
Hence, option (d) is also incorrect.
Note: If we have to find the total number of people who did not bring gifts in the party then we have to subtract the number of people who brought gifts from total attendees. Hence, only $18$ people from all the $50$ attendees brought the gift
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# Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.
$\begin {array} {1 1} (A)\;3x+4y+3=0 & \quad (B)\;3x-4y-3=0 \\ (C)\;3x+4y-3=0 & \quad (D)\;3x-4y+3=0 \end {array}$
Comment
A)
Toolbox:
• If two lines are parallel, then their slopes are equal.
• Equation of a line passing through the point $(x_1,y_1)$ and slope $m$ is $y-y_1=m(x-x_1)$
Step 1:
The point of intersection of the lines can be obtained by solving the two given lines.
$2x+y=5$ and $x+3y+8=0$
$\qquad 2x+y=5$
$( \times 2 ) x+3y=-8$
____________________
$\qquad 2x+y=5$
$\qquad 2x+6y=-16$
$\qquad (-) \quad (-) \quad (+)$
____________________
$\quad \qquad -5y=21$
$\Rightarrow y = -\large\frac{21}{5}$$\therefore x = \large\frac{23}{5} Hence the point of intersection is \bigg( \large\frac{23}{5}$$, \large\frac{-21}{5} \bigg)$
Step 2 :
It is given that the required line is parallel to the line $3x+4y=7$
$\therefore$ slope $m = -\large\frac{3}{4}$
$\therefore$ The equation of the given line is
$(y-y_1) = m (x-x_1)$
(i.e) $y-\bigg(\large\frac{-21}{5} \bigg)$$=-\large\frac{3}{4} \bigg( x - \large\frac{23}{5} \bigg) y+\large\frac{21}{5}$$ = \large\frac{-3x}{4}+\large\frac{69}{20}$
$\Rightarrow \large\frac{5y+21}{5}$$= \large\frac{-15x+69}{20}$
$\Rightarrow 5y+21=\large\frac{-15x+69}{4}$
(i.e) $4(5y+21)=-15x+69$
$\Rightarrow 20y+84=-15x+69$
$\Rightarrow 15x+20y=-15$
(i.e) $3x+4y=-3$
Hence the equation of the required line is $3x+4y+3=0$
|
Math 20 Math 25 Math Tips davidvs.net
# Math Lecture NotesArithmetic Skills
## Divisibility Rules
Some divisibility rules you already know:
• A number divisible by 2 is even, and its one's digit is 0, 2, 4, 6, or 8.
• A number divisible by 5 has for its one's digit either 0 or 5.
• A number divisible by 10 has 0 as its one's digit.
Other divisibility rules may be new to you:
• A number is divisible by 3 if the sum of its digits is divisible by 3.
• A number is divisible by 6 if it is divisible by both 2 and 3.
• A number is divisible by 4 if the two-digit number formed from its ten's and one's digits is also divisible by 4.
• A number is divisible by 9 if the sum of its digits is divisible by 9.
Let's apply the divisibility rules to 13,512.
Example 1
Is 13,512 divisible by... 2? 3? 4? 5? 6? 9? 10?
Divisible by 2? Yes, because the one's place value digit is 2, which is even.
Divisible by 3? Yes, because the sum of digits is 1 + 3 + 5 + 1 + 2 = 12 and three goes into 12.
Divisible by 4? Yes, because two-digit number formed from its ten's and one's digits is 12 and four goes into 12.
Divisible by 5? No, because the one's place value digit is not zero or five.
Divisible by 6? Yes, because it was divisible by both 2 and 3.
Divisible by 9? No, because the sum of digits is 1 + 3 + 5 + 1 + 2 = 12 and nine does not go into 12.
Next let's apply the divisibility rules to 2,016.
Example 2
Is 2,016 divisible by... 2? 3? 4? 5? 6? 9? 10?
Divisibility Tests for 2, 3, 4, 5, 6, 9, 10
Bittinger Chapter Tests, 11th Edition
Chapter 2 Test, Problem 6: Determine whether 1,784 is divisible by 8.
Chapter 2 Test, Problem 7: Determine whether 784 is divisible by 9.
Chapter 2 Test, Problem 8: Determine whether 5,552 is divisible by 5.
Chapter 2 Test, Problem 9: Determine whether 2,322 is divisible by 6.
The rule for six works because 2 × 3 = 6.
There are yet more divisibility rules, but these are the ones that are easy to use.
## Finding Factors
There are two tasks involving factor finding. In some situations we want to find all the factors. In other situations we want to find the prime factorization.
(In this class the most common use for finding all the factors is considering how to reduce a fraction, and the most common use for finding the prime factorization is to create a common denominator for fractions.)
### Finding All the Factors
We find all the factors of a number by making a two-column list. Count up in the first column. List any matching factors in the second column. When the columns get to the same value we can stop.
Example 3
Find all the factors of 66.
Our first column counts up from 1 to 10.
1 66 2 33 3 22 4 not a factor 5 not a factor 6 11 7 not a factor 8 not a factor 9 not a factor 10 not a factor
The next value for the first column would be 11, which is already listed in the second column. So we can stop.
Example 4
Find all the factors of 24.
Finding Factors of a Number
Bittinger Chapter Tests, 11th Edition
Chapter 2 Test, Problem 1: Find all the factors of 300.
### Finding the Prime Factorization
We find all the prime factorization of a number by making a factor tree and noting the "leaves".
#### Factor Trees
Remember factor trees?
Here is one factor tree for 48.
Example 5
Try writing a different factor tree for 48 that does not start with 6 and 8.
#### Prime Factorization
Either circle or "bring down" the leaves of your factor tree so you do not make a careless mistake and forget any of them when writing your answer.
To be polite, list the prime factors in order. Write them as a product (separated by × symbols).
Optionally, you may show off your fluency with exponents by writing the prime factors as compactly as possible using exponents.
Example 6
Find the prime factorization of 48.
Looking at the factor trees for 48 we see the prime factorization is 2 × 2 × 2 × 2 × 3 = 48
This can also be written as 24 × 3 = 48
Example 7
Find the prime factorization of 66.
Chapter 2 Test, Problem 4: Find the prime factorization of 18.
Chapter 2 Test, Problem 5: Find the prime factorization of 60.
## Decimal Point Scoots
When we multiply or divide by powers of ten it only scoots the decimal point.
Example 8
Multiply 123.456 by 10
Multiply 123.456 by 100
Multiply 123.456 by 1,000
Example 9
Divide 123.456 by 10
Divide 123.456 by 100
Divide 123.456 by 1,000
When we multiply or divide by powers of one-tenth it also scoots the decimal point, but in the other direction.
Example 10
Multiply 123.456 by 0.1
Multiply 123.456 by 0.01
Multiply 123.456 by 0.001
Example 11
Divide 123.456 by 0.1
Divide 123.456 by 0.01
Divide 123.456 by 0.001
Multiplying a Decimal by a Power of 10
Dividing a Decimal by a Power of 10
Dividing a Decimal by a Power of 10: Pattern
Multiplying Decimals by 10, 100, and 1000 (worksheet)
Dividing Decimals by 10, 100, and 1000 (worksheet)
## One-Step Equations
### Solving One-Step Equations
In Math 20 the only equations we solve are one-step equations in which the letter is either multiplied or divided by a number (to equal another number).
To solve these, "undo" what is attached to the letter by doing the opposite.
Example 12
Solve 5 × y = 35
We "undo" a multiply-by-5 with a divide-by-5. To be fair, we treat both sides of the equation the same.
Example 13
Solve 4 × b = 20
How does this picture help us solve for b? Where is division for both sides hiding?
Example 14
Solve p ÷ 6 = 8
How does this picture help us solve for b? Where is multiplication for both sides hiding?
### Horizontal and Vertical Format
Let's solve for y with the horizontal format.
Example 15
Solve 3 × y = 210 (Use the horizontal format)
Notice that unless we have different colors or write very neatly, we can no longer identify what the original problem was!
Next solve the same problem with the vertical format.
Example 16
Solve 3 × y = 210 (Use the vertical format)
This takes a little more space, but we can identify what the original problem was. This helps us when we finish a test and go back to check our work. Later, in future math classes, this format also helps avoid careless errors in more complicated problems.
Here is another vertical format sample:
Example 17
Solve 9 × y = 189 (Use the vertical format)
I used both black and blue writing. In Math 20, write a lot, like I did. But in Math 60 and 65 one goal will be to eventually wean yourself from always using the steps I did in blue. The instructor will write fewer of these steps.
There are two other reasons to use the vertical format.
First, it promotes doing homework in two columns per page. This often saves paper. Homework problems, like math lecture notes, by their nature are seldom as wide as a page.
Second, putting work in that shape makes it easier to do scratch work off on the side. Watch how that helps me stay organized when solving for y when fraction arithmetic happens.
Example 18
Solve 8 × y = 23 (Use the vertical format)
Chapter 1 Test, Problem 28: Solve: 28 + x = 74
Chapter 1 Test, Problem 29: Solve: 169 ÷ 13 = n
Chapter 1 Test, Problem 30: Solve: 38 × y = 532
Chapter 1 Test, Problem 31: Solve: 381 = 0 + a
Chapter 2 Test, Problem 34: Solve: 78 × x = 56
Chapter 2 Test, Problem 35: Solve: t × 25 = 710
Chapter 3 Test, Problem 9: Solve: 14 + y = 4
Chapter 3 Test, Problem 10: Solve: x + 23 = 1112
Chapter 4 Test, Problem 32: Solve: 4.8 × y = 404.448
Chapter 4 Test, Problem 33: Solve: x + 0.018 = 9
### Problem Solving Grammar
Consider again the problem 3 × y = 210. Notice that there are many possible ways to write the step of dividing both sides by 3. The clearest is to use the vertical format and write either ÷ 3 or /3 on its own line. A few possibiltiies are considered incorrect because of bad math grammar.
Example 19
What are some incorrect ways to write the step of dividing both sides by 3 when solving 3 × y = 210?
(a) Do not use brackets to incorrectly mean "do this to the entire equation", as in ÷ 3 (3 × y = 210).
(b) Do not use brackets on each side of the equation improperly, as in ÷ 3 (3 × y) = (210) ÷ 3. Notice the right hand side is legitimate. But the left hand side begins confusingly with the ÷ symbol.
In a future math class studying algebra you will encounter other incorrect ways. For example, when solving 3 × y + 3 = 210 it would be smart to divide both sides by 3 but it would violate the distributive property to incorrectly write 3 × y + 3 ÷ 3 = 210 ÷ 3
## Shoes and Socks
When you get dressed, you put your socks on before your shoes. To undress you must remove your shoes first, because they were put on most recently.
The same reversal happens when isolating a variable.
The Shoes and Socks Theorem
When solving for a variable, the number attached last to the variable needs to be removed first.
This sounds funny, but makes sense when you see it happen.
Example 20
Solve 30 = 5 × y − 10
Consider the order of operations.
Multiplication has priority over subtraction. So the 5 × has first priority and the − 10 has last priority.
First remove the − 10.
30 = 5 × y − 10
+10 +10
40 = 5 × y
Then remove the 5 ×.
40 = 5 × y
÷5 ÷5
8 = y
The Shoes and Socks Theorem can help deal with negatives. Remember that a number is made negative by × (−1).
Example 21
Solve 25 − y = 20
Consider the order of operations.
Multiplication has priority over subtraction. So the × (−1) has first priority and the positive 25 has last priority.
First remove the 25. It is a positive number, so remove it using subtraction.
25 − y = 20
−25 −25
y = − 5
Then remove the × (− 1). Because "a negative of a negative is a positive" we can do that be repeating it.
y = − 5
×(−1) ×(−1)
y = 5
Often it helps to think about terms. For example, with the equation 90 = 7 × y − 11 × 2. we know the first step is to simplify the 11 × 2 term so it will be easier to remove from the right hand side of the equation. Once we have 90 = 7 × y − 22 the equation is as simple as Example 7 above.
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# Mathematical Food for Thought
• ## About
Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.
• ## Meta
Leftovers. Topic: Algebra/Polynomials. Level: AIME. December 22nd, 2006 Problem: (Stanford Putnam Practice) Find the remainder when you divide by . Solution: Since they’re both divisible by , we first divide that out, and just remember to multiply the remainder by at the end. Let be the first polynomial and be the second. we have for some polynomials with . We want to find . Consider the two roots of . Plugging them into the equation, we obtain and . Evaluating at those two values, we find that and . But since has degree less than , the only possible is the constant polynomial . Then the remainder is . QED. ——————– Comment: This is a super important technique when it comes to polynomial division. Using the roots of and the fact that , we can hypothetically always determine this way, without dividing. This usually comes up when has nice roots, so if it doesn’t, look for a better way. ——————– Practice Problem: (Stanford Putnam Practice) How can the quadratic equation have three roots ? [Reworded] Posted in AIME, Algebra, Polynomials || 1 Comment » Complexity. Topic: Complex Numbers/Polynomials. Level: AIME/Olympiad. December 12th, 2006 Problem: Let . If and , find the remainder when is divided . [Reworded] Solution: Well is pretty ugly looking at the moment, let’s fix that. We can write it as the sum of the following series: , since all of them are geometric series with common ratio . Summing these up, we have . The top is again a geometric series with common ratio (without that part) so we can write it ias . But wait, if then , so in fact we have . Then so it remains the evaluate the denominator of that. Considering we are like whoa, has the th roots of unity for its roots except and it has leading coefficient so it must be . Hence . By Euler’s Totient Theorem, . QED. ——————– Comment: It didn’t take too much “insight” to reduce to the nicer expression, more like just tedious evaluation of geometric series galore. Seeing the product in the denominator should have been pretty familiar (though I admit I forgot what it came out to at first) and a standard argument allowed you to evaluate it. Possibly the hardest part was Euler’s Totient Theorem, which is invaluable to carry around for the AIME. Also knowing that . ——————– Practice Problem: Do there exist polynomials and such that ? If so, find them. Posted in AIME, Complex Numbers, Olympiad, Polynomials || 1 Comment » Yay! Putnam! Topic: Algebra/Polynomials. Level: AIME. December 3rd, 2006 Problem: (2006 Putnam – B1) Show that the curve contains only one set of three distinct points, , , and , which are the vertices of an equilateral triangle, and find its area. Solution: Well, this is a cubic in two variables, but let’s remember the awesome technique of writing it as a polynomial in just one variable, say . Then . Well, testing out a bit (looking at the factorization of ), we get is always a solution, so factor it out to get . Looking at the second term, we’re like let’s hope it has not many solutions, so we take the discriminant and find . But this is always negative unless so that means this is the only case in which this factor can be zero. This gives us the point as a solution. Whoa, that means we have categorized the entire solution set: (1) the line and (2) the point . If we have three vertices of an equilateral triangle, they most definitely can’t be collinear, so one point must be . Suppose we choose two points on the line , say and . Since they have to be equidistant from , we know one must be the reflection of the other over the line through perpendicular to , which is . So if the first point is then the other is . Now setting the squares of the side lengths equal to each other, we know so . This gives the -coordinate of both the other two vertices of the triangle, so the only one is the equilateral triangle with vertices . The side length is , so the area is . QED. ——————– Comment: Slightly difficult if your algebraic intuition wasn’t working well, but after you realized the factorization it wasn’t hard to convince yourself that a line and a point can have the vertices of at most one equilateral triangle. A solid B1 problem on the Putnam, quite a bit more difficult than the A1, imo. ——————– Practice Problem: (2006 Putnam – A1) Find the volume of the region of points such that . Posted in AIME, Algebra, Polynomials || 3 Comments » Balance Those p’s. Topic: Algebra/Polynomials/S&S. Level: Olympiad. November 29th, 2006 Problem: (Stanford Putnam Practice) A finite sequence is called -balanced if any sum of the form is the same for . Prove that if a sequence with members is -balanced for , then all its members are equal to zero. Solution: Let . Recall the strategy of isolating certain terms in a polynomial. Denoting by the th root of unity, we have because all of the terms will cancel out by the identity except for the terms in which the power of the exponent is divisible by , in which case . Similarly, using the polynomials we obtain from and . Hence we have a system of equations in the variables . But if we have a system of equations for , we have a total of equations. Since the only repeated value in the polynomial was and it showed up times, once for each prime, there are distinct points on the polynomial that we now know. However, since a polynomial of degree is defined by at most points, it must be the zero polynomial, which means , as desired. QED. ——————– Comment: This is an example of generating functions showing up when you least expect it, well not exactly, but it was cool anyway. The proof above is not completely rigorous because we would need to show that a polynomial of degree actually cannot pass through all of the points, but it’s almost there. ——————– Practice Problem: Let denote the number of ways you can partition into parts. Show that . Also show that the number of ways can be partitioned into unequal parts is equal to the number of ways can be partitioned into odd parts. Square Sum Stuff. Topic: Polynomials/S&S. Level: AMC/AIME. October 27th, 2006 Problem: Evaluate the summation . Solution: Here’s a technique that will help you evaluate infinite series that are of the form polynomial over exponential. It’s based on the idea of finite differences: If is a polynomial with integer coefficients of degree then is a polynomial of degree (not hard to show; just think about it). So let . Then consider by simply multiplying each term by : . And now find the difference by subtracting the terms with equal denominators. We get . Notice that the numerator is now a polynomial of degree instead of . Repeating this, we have and . Notice that the latter part is just a geometric series which sums to so . QED. ——————– Comment: The method of finite differences is extremely useful and is basically a simplified version of calculus – in a very approximating sense. It’s a good thing to know, though, because then you have a better understanding of how polynomials work. ——————– Practice Problem: Let be a polynomial with integer coefficients. Using the method of finite differences, predict the degree of . . Posted in AIME, AMC, Polynomials, Sequences & Series || 2 Comments »
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# My Add Maths Modules - Indices And Logarithm by nklye
VIEWS: 6,067 PAGES: 15
• pg 1
``` Additional Mathematics Form 4
Topic 5:
DELIGHT
(Version 2007)
by NgKL
(M.Ed.,B.Sc.Hons.,Dip.Ed.,Dip.Edu.Mgt.,Cert.NPQH)
5.1
INDICES AND LAWS OF INDICES
2.
IMPORTANT NOTES:
1. For an index number an, (read as a raise to the power of n), where a is the base and n is the index. ao = 1
2. 3.
1 = a n , where a 0 an
4. 5. 6.
an =
m
1
n
a , where a 0, (read as a raise to the nth root).
n
an =
am
Laws of Indices 6.1 am x an = am + n 6.2 am an = am n
(or)
6.4 6.5
(ab)m = am bm
a b
m
m = am
b
a n a
m
=
amn
6.3 (a ) = am x n 7. Equation Involving Indices Can be solve by; 7.1 Comparing the indices or bases on both sides of the equation; 7.1.1 If am = an, then m = n. 7.1.2 If am = bm, then a = b. 7.2 Applying logarithms on both sides of the equation; ax = bm log ax = log bm x log a = m log b m log b x= log a
m n
Exercise 5.1:
1. Evaluate each of the following without using a calculator. (b) 24
3.
(a) 43
(c) 3 3
2
(d)
3 4
3
(e) 0.50
(e) 8
1 3
2.
Simplify and then evaluate each of the following. (b) 32 34 (c) (52)3
(a) 32 x 35
(d) 4-3 x 45
2 (e) 43
(f) 43 x 24 162
8
2
1
(g) 27 3 x (-9) 2
(h) (
2 2 ) 5
(125) 3
2
(i) (18 x 32) 2
3
3.
Simplify each of the following expressions.
n+3
4.
(a) 2
x 4 32
n
n
(b) 3
n+2
3
n-1
(c) 92n 3n + 1 x 27 n
(d) 25n x 42n x 63n
(e) 20a3 5a-5
(f)
3m 2 81x 1 9 2( x 1 )
4.
Solve each of the following.
(a) Show that 7p + 1 + 7p +2 is a multiple of 8.
(b) Show that 5n + 1 + 5n – 3(5n – 1) is divisible by 3 or 9.
(c) Show that 22x + 3 (9x + 1 – 32x) = ( 2 )2x
3
5.2 LOGARITHMS AND LAWS OF LOGARITHMS
IMPORTANT NOTES:
1. To convert an equation in index form to logarithm form and vice versa. If N = ax , then loga N = x. 2. 3. loga 1 = 0, and loga a = 1. loga (negative number) = undefined. Similarly, loga 0 = undefined. Law of Logarithms: 4.1 loga xy = loga x + loga y 4.2 loga x = loga x – loga y
y
5.
4.
(or) loga (x y) = loga x – loga y
4.2 loga xm = m loga x 5. Change of Bases of Logarithms: 5.1 loga b =
log c b log c a log b b log b a
=
5.2 loga b = 6.
1 log b a
Equations Involving Logarithms: 6.1 Converting the equation of logarithm to index form, i.e. loga N = x, then N = ax 6.2 Express the left hand side, LHS and the right hand side, RHS, as single logarithm of the same base. Then make the comparison, i.e; (i) loga b = loga c, then b = c. (ii) loga m = logb m, then a = b.
Exercise 5.2:
1. Express the following equations to logarithm form or index form. (b) 4 = 8 3
2
6.
(a) 32 = 25
(c) 1 = 100
(d) px = 5
(e) log6 36 = 2
(f) log3 243 = 5
(f) 3 = log3
1 27
(g) logx q = p
2.
Determine the value of x in each of the following equations. (b) log4 x =
(a) log3 81 = x
1 2
(c) log x 125 = 3
(d) log2 x = 2
(e) log
1 64
x =
1 3
(e) log x
1 = 3 216
3.
Find the value of each of the following.
7.
(a) log10 100 =
(b) log10 39.94 =
(c) log10 1
=
(d) antilog 1.498 =
35
(d) antilog 0.3185 =
(e) antilog ( 0.401) =
4.
Find the value of each of the following without using a calculator. (b) log3
(a) log2 32
1 243
(b) log5 0.2
(d) log9
9
5.
Given that log2 3 = 1.585 and log2 5 = 2.32, find the values of the following logarithms. (b) log2 6
(a) log2 45
(c) log2 1.5
(d) log2 (
125 ) 3
6.
Simplify each of the following expression to the simplest form.
8.
(a) 2 log2 x log2 3x + log2 y
(b) loga 5x + 3 loga 2y
(c) logb x + 3 logb x + logb (y + 1)
(d) log2 4x – log2 3y – 2
7.
Determine the values of the following logarithms.
(a) log2 7
(b) log3 23
(c) log3 5
(d) log0.5 8.21
8.
Given that log2 w = p, express the following in terms of p. (b) log8 16w2 =
9.
(a) log w 4 =
(c) log4
w 32
(d) log
w
64
9.
Given that log m 3 = x and log m 4 = y. Express the following in terms of x and/or y.
(a) log 36 m =
(b) log3m 12
(c) log3 16 m
(d) log3
m 4
Exercise 5.3:
1. Solve each of the following equations. (b) 22x + 3 = 32
10.
(a) 3x + 2 = 81x
(c) 3x . 4x = 125x + 2
(d) 52x =
125 25 x
2.
Solve each of the following equations.
(a) log10 (2x + 7) = log10 21
(b) log5 (x – 5) =
1 log5 125 3
11.
(c) log2 (x – 3) – log2 (x2 – 9) = 0
(d) 2 log3 2 + log3 (4x – 1) = 1 + log3 (x + 8)
(e) logx 18 – logx 2 = 2
(f)
log2 8 + log4 M = 5
Exercise 5.4 – SPM QUESTIONS (2003 – 2007)
1. Solve the equation 82x – 3 =
1 4
x2
12.
[3 marks] SPM2006/Paper1
2. Given that log2 xy = 2 + 3 log2 x – log2 y, express y in terms of x. [4 marks] SPM2006/Paper1
3. Solve the equation 2 + log3 (x – 1) = log3 x.
[3 marks] SPM2006/Paper1
4. Solve the equation 2x + 4 2x + 3 = 1.
13.
[3 marks] SPM2005/Paper1
5. Solve the equation log3 4x – log3 (2x – 1) = 1
[3 marks] SPM2005/Paper1
6. Given that logm 2 = p and logm 3 = r, express log m ( p and r.
27m ) in terms of 4 [4 marks] SPM2005/Paper1
14.
7. Solve the equation 324x = 48x + 6.
[3 marks] SPM2004/Paper1
8. Given that log5 2 = m and log5 7 = p, express log5 4.9 in terms of m and p. [4 marks] SPM2004/Paper1
9. Solve the equation 42x -1 = 7x.
[4 marks] SPM2003/Paper1
15.
10. Given that log2 b = x and log2 c = y, express log4 8b in terms of x and y. c (4 marks) (SPM2007/Paper 1)
11. Given that 9(3n−1) = 27n, find the value of n.
(3 marks) (SPM2007/Paper 1)
```
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January 3, 2014
# Series based questions MBA CET
Series based questions MBA CET and NMAT SNAP TISS CMAT Bank PO and other exams
What is Number Series?
Number series is a arrangement of numbers in a certain order, where some numbers are wrongly put into the series of numbers and some number is missing in that series, we need to observe and find the accurate number to the series of numbers.
In competitive exams number series are given and where you need to find missing numbers. The number series are come in different types. At first you have to decided what type of series are given in papers then according with this you have to use shortcut tricks as fast as you can .
Different types of Number Series
There are some format of series which are given in Exams.
Perfect Square Series:
This Types of Series are based on square of a number which is in same order and one square number is missing in that given series.
Example 1: 441, 484, 529, 576?
Answer: 441 = 212, 484 = 222, 529 = 232, 576 = 24, 625 = 252.
Perfect Cube Series:
This Types of Series are based on cube of a number which is in same order and one cube number is missing in that given series
Example 2: 1331, 1728, 2197, ?
Answer : 113 , 123 , 133 , 143
Geometric Series:
This type of series are based on ascending or descending order of numbers and each successive number is obtain by multiplying or dividing the previous number with a fixed number.
Example 3: 5, 45, 405, 3645,?
Answer: 5 x 9 = 45, 45 x 9 = 405, 405 x 9 = 3645, 3645 x 9 = 32805.
Two stage Type Series:
A two tier Arithmetic series is one in which the differences of successive numbers themselves form an arithmetic series.
Example 4: i. 3, 9, 18, 35, 58,——
1. 6, 9, 17, 23,———-
Mixed Series:
This type of series are more than one different order are given in a series which arranged in alternatively in a single series or created according to any non-conventional rule. This mixed series Examples are describes in separately.
Examples 5:
11, 24, 50, 102, 206, ?
11 x 2 = 22 +2 = 24,
24 x 2 = 48 + 2 = 50,
50 x 2 = 100 + 2 = 102,
102 x 2 = 204 + 2 = 206,
206 x 2 = 412 + 2 = 414.
So the missing number is 414.
Number Series Quiz
Directions (1-10): What will come in place of the question marks (?) in the following Number series?
1. 0, 6, 24, 60, 120, 210, ?
2. 336
3. 349
4. 312
5. 337
6. None of these
1. 11, 14, 19, 22, 27, 30, ?
2. 39
3. 34
4. 36
5. 35
6. None of these
1. 6, 12, 21, ? , 48
2. 33
3. 39
4. 36
5. 31
6. None of these
1. 18, 22, 30, ? ,78, 142
2. 44
3. 35
4. 46
5. 48
6. None of these
1. 73205, 6655, 605, 55, ?
2. 9
3. 5
4. 13
5. 11
6. None of these
1. 25, 100, ?, 1600, 6400
2. 400
3. 300
4. 360
5. 420
6. None of these
1. 125, ?, 343, 512, 729, 1000
2. 216
3. 215
4. 256
5. 225
6. None of these
1. 1 , 9 , 125 , 343 , ? , 1331
2. 730
3. 729
4. 512
5. 772
6. None of these
1. 121, 144, 169, ?, 225
2. 180
3. 172
4. 186
5. 196
6. None of these
1. ?, 2116, 2209, 2304, 2401, 2500
2. 2124
3. 1972
4. 1521
5. 2025
6. None of these
1. (A)
The given series is : 13 – 1, 23 – 2, 33 – 3, 43 – 4, 53 – 5, 63 – 6,
So the missing term = 73 – 7 = 343 – 7 = 336 .
1. (D)
The pattern is + 3, + 5, + 3, + 5, …………
So the missing term is = 30 + 5 = 35 .
1. (A)
The pattern is + 6, + 9, + 12, +15 ………..
So the missing term is = 21 + 12 = 33 .
1. (C)
The pattern is +4, +8, +16, +32, +64
So the missing term is = 30 + 16 = 46 .
1. (B)
5 x 11 = 55, 55 x 11 = 605, 605 x 11 = 6655, 6655 x 11 = 73205
1. (A)
25 x 4 = 100, 100 x 4 = 400, 400 x 4 = 1600, 1600 x 4 = 6400.
1. (A)
125 = 53 , 216 = 63, 343 = 73, 512 = 83, 729 = 93, 1000 = 103.
1. (B)
13 , 33 , 53 , 73 , 93 , 113
1. (D)
121 = 112, 144 = 122, 169 = 132, 196 = 142, 225 = 152.
1. (D)
2025 = 452, 2116 = 462, 2304 = 482, 2401 = 492, 2500 = 502
TIME TAKEN
Within 5 min : EXCELLENT
5-8 min : YOU CAN DO BETTER
More than 8 min : YOU NEED TO WORK HARD
Series based questions MBA CET
Logical Reasoning Alphanumeric Series based questions ability Test :
Series based questions MBA CET
1. 4, 7, 12, �.., 28, 39
19
24
14
16
None of these
Explanation : . Series proceeds with a difference of 3, 5, 7, 9, 11, respectively. Hence, the missing number will be 19.
2. 8, 4, 12, 6, 18, �.., 27
10
12
18
24
None of these
Explanation : . 3rd, 5th and 7th numbers are the sums of preceding two numbers. Hence, the missing number is 9 as 18 + 9 = 27.
3. 45, 54, 47, �.., 49, 56, 51, 57, 53
48
55
50
53
None of these
Explanation : . There are two alternate series in the single series, one of them moves with a difference of one number starting with 45, and another is a consecutive series starting with 54. Hence, the missing number will be 55
4. 2, 6, 12, 20, �.., 42, 56, 72, 90
20
21
30
12
23
Explanation : . Series moves with the increasing difference by 2 i.e., 4, 6, 8, 10, 12�.. Therefore, the missing number will be 30.
5. 2, 9, 23, 3, 8, 25, 4, �, 27
7
29
23
14
20
Explanation : . There are three alternative series 2, 3, 4 � (consecutive numbers), 9, 8, 7 �. (consecutive number in descending order), and 23, 25, 27, � Hence, 7 will come in place of the missing number.
6. 6, 126, �., 9, 108, 12, 7, 133, 19, 12, 72, 6
21
23
30
35
40
Explanation : . Number between 1st and 3rd, 4th and 6th, 7th and 9th is the product of the adjacent numbers. Hence the missing number will be 126/6 = 21.
7. 2, 3, 5, 9, 17, 33, �..
85
37
63
64
65
Explanation : . Series moves with a difference of 1, 2, 4, 8, 16, 32. Hence, the missing number is 65.
8. 6, 9, 15, 27, �.., 99
51
34
37
84
None of these
Explanation : . Series proceeds with a difference of 3, 6, 12, 24, 48, respectively. Hence, the missing number will be 51.
9. 17, 7, 24, 19, 9, 28, �., 8, 31, 27, 10, 37
20
21
18
12
23
Explanation : . Each 3rd number is the sum of preceding two numbers. i.e., 17 + 7 = 24, 19 + 9 = 28, 23 + 8 = 31. Hence, the missing number will be 23
10. 121, 112, �., 97, 91, 86
102
108
99
104
101
Explanation : . Series is written in descending order with a difference of 9, 8, 7, 6 and 5, respectively. Hence, the missing number will be 104.
11. 16, 22, 34, 58, 106, �, 394
178
175
288
170
202
Explanation : . Series moves with a difference of 6, 12, 24, 48, 96. Hence, the missing number is 202.
12. 5, 7, 11, 19, 35, 67, �., 259
64
131
135
32
None of these
Explanation : . Series moves with a difference of 2, 4, 8, 16, 32, 64. Hence, the missing number will be 67 + 64 = 131.
13. 2, �., 8, 16, 32, 64, 128, 256
2
3
5
4
None of these
Explanation : . Each number of the series is the double of the preceding number. Hence, the missing number will be 4.
14. 3, 128, 6, 64, 9, �., 12, 16, 15, 8
32
12
108
72
64
|
## Engage NY Eureka Math 1st Grade Module 2 Lesson 23 Answer Key
### Eureka Math Grade 1 Module 2 Lesson 23 Sprint Answer Key
A
*Write the missing number.
Question 1.
2 + ☐ = 3
2 + 1 = 3
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with one then we got three.
Question 2.
1 + ☐ = 3
1 + 2 = 3
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD one with two then we got three.
Question 3.
☐ + 1 = 3
2 + 1 = 3
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with one then we got three.
Question 4.
☐ + 2 = 4
2 + 2 = 4
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with two then we got four.
Question 5.
3 + ☐ = 4
3 + 1 = 4
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with one then we got four.
Question 6.
1 + ☐ = 4
1 + 3 = 4
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD one with three then we got four.
Question 7.
1 + ☐ = 5
1 + 4 = 5
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD one with with then we got five.
Question 8.
4 + ☐ = 5
4 + 1 = 5
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with one then we got five.
Question 9.
3 + ☐ = 5
3 + 2 = 5
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with two then we got five.
Question 10.
3 + ☐ = 6
3 + 3 = 6
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with three then we got six.
Question 11.
☐ + 2 = 6
4 + 2 = 6
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with two then we got six.
Question 12.
0 + ☐ = 6
0 + 6 = 6
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD zero with six then we got six.
Question 13.
1 + ☐ = 7
1 + 6 = 7
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD one with six then we got seven.
Question 14.
☐ + 5 = 7
2 + 5 = 7
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with five then we got seven.
Question 15.
☐ + 4 = 7
3 + 4 = 7
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with four then we got seven.
Question 16.
2 + ☐ = 8
2 + 6 = 8
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with six then we got eight.
Question 17.
4 + ☐ = 8
4 + 4 = 8
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with four then we got eight.
Question 18.
8 = ☐ + 6
8 = 2 + 6
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with six then we got eight.
Question 19.
8 = 3 + ☐
8 = 3 + 5
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with five then we got eight.
Question 20.
☐ + 3 = 9
6 + 3 = 9
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD six with three then we got nine.
Question 21.
2 + ☐ = 9
2 + 7 = 9
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with seven then we got nine.
Question 22.
9 = ☐ + 1
9 = 8 + 1
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD eight with one then we got nine.
Question 23.
9 = 4 + ☐
9 = 4 + 5
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with five then we got nine.
Question 24.
2 + 2 + ☐ = 9
2 + 2 + 5 = 9
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with two and five then we got nine.
Question 25.
2 + 2 + ☐ = 8
2 + 2 + 4 = 8
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with two and four then we got eight.
Question 26.
3 + ☐ + 3 = 9
3 + 3 + 3 = 9
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with three and three then we got nine.
Question 27.
3 + ☐ + 2 = 9
3 + 4 + 2 = 9
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with four and two then we got nine.
Question 28.
5 + 3 = ☐ + 4
5 + 3 = 4 + 4
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD five with three then we got eight. ADD four with four then we got eight. Both expressions are equal.
Question 29.
☐ + 4 = 1 + 5
2 + 4 = 1 + 5
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with four then we got six. ADD one with five then we got six. Both expressions are equal.
Question 30.
3 + ☐ = 2 + 6
3 + 5 = 2 + 6
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with five then we got eight. ADD two with six then we got eight. Both expression are equal.
B
*Write the missing number.
Question 1.
1 + ☐ = 3
1 + 2 = 3
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with one then we got three.
Question 2.
0 + ☐ = 3
0 + 3 = 3
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD zero with three then we got three.
Question 3.
☐ + 3 = 3
0 + 3 = 3
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD zero with three then we got three.
Question 4.
☐ + 2 = 4
2 + 2 = 4
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with two then we got four.
Question 5.
3 + ☐ = 4
3 + 1 = 4
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with one then we got four.
Question 6.
4 + ☐ = 4
4 + 0 = 4
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with zero then we got four.
Question 7.
4 + ☐ = 5
4 + 1 = 5
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with one then we got five.
Question 8.
1 + ☐ = 5
1 + 4 = 5
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD one with four then we got five.
Question 9.
2 + ☐ = 5
2 + 3 = 5
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with three then we got five.
Question 10.
4 + ☐ = 6
4 + 2 = 6
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with two then we got six.
Question 11.
☐ + 2 = 6
4 + 2 = 6
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with two then we got six.
Question 12.
3 + ☐ = 6
3 + 3 = 6
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with three then we got six.
Question 13.
3 + ☐ = 7
3 + 4 = 7
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with four then we got seven.
Question 14.
☐ + 4 = 7
3 + 4 = 7
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with four then we got seven.
Question 15.
☐ + 5 = 7
2 + 5 = 7
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with five then we got seven.
Question 16.
3 + ☐ = 8
3 + 5 = 8
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with five then we got eight.
Question 17.
2 + ☐ = 8
2 + 6 = 8
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with six then we got eight.
Question 18.
8 = ☐ + 1
8 = 7 + 1
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD seven with one then we got eight.
Question 19.
8 = 4 + ☐
8 = 4 + 4
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with four then we got eight.
Question 20.
☐ + 2 = 9
7 + 2 = 9
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD seven with two then we got nine.
Question 21.
4 + ☐ = 9
4 + 5 = 9
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with five then we got nine.
Question 22.
9 = ☐ + 5
9 = 4 + 5
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with five then we got nine.
Question 23.
9 = 6 + ☐
9 = 6 + 3
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD six with three then we got nine.
Question 24.
1 + 5 + ☐ = 9
1 + 5 + 3 = 9
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD one with five and three then we got nine.
Question 25.
3 + 2 + ☐ = 8
3 + 2 + 3 = 8
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with two and three then we got eight.
Question 26.
2 + ☐ + 6 = 9
2 + 1 + 6 = 9
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with one and six then we got nine.
Question 27.
3 + ☐ + 4 = 9
3 + 2 + 4 = 9
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD three with two and four then we got nine.
Question 28.
5 + 4 = ☐ + 6
5 + 4 = 3 + 6
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD five with four then we got nine. ADD three with six then we got nine. Both expressions are equal.
Question 29.
☐ + 3 = 6 + 2
4 + 4 = 6 + 2
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with four then we got eight. ADD six with two then we got eight. Both expressions are equal.
Question 30.
4 + ☐ = 2 + 7
4 + 5 = 2 + 7
Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with five then we got nine. ADD two with seven then we got nine. Both expressions are equal.
### Eureka Math Grade 1 Module 2 Lesson 23 Problem Set Answer Key
Draw and label.
Write a number sentence and a statement that matches the story.
Question 1.
Janet read 8 books during the week. She read some more books on the weekend. She read 12 books total. How many books did Janet read on the weekend?
Janet read 4 books on the weekend.
Explanation:
Total number of books read by Janet are 12. Janet read 8 books during the week. She read some more books on the weekend. Draw a circle for books read during the week. First draw eight squares and label it with week. Then keep drawing squares until we have 12 squares. We need two more squares to make ten and draw two more squares to make twelve. That four squares represents the books read on the weekend. Color the squares so we can tell which squares we are added. Janet read four books on the weekend.
Question 2.
Eric scored 13 goals this season! He scored 5 goals before the playoffs. How many goals did Eric score during the playoffs?
Eric scored 8 goals during the playoffs.
Explanation:
Eric scored thirteen goals this season. He scored five goals before the playoffs. First draw five circles and label it with before the playoffs. Then keep drawing circles until we have thirteen circles. We need five more circles to make ten and three more circles to make thirteen. Cross out five circles which represents the goals before the playoffs. Draw a circle for the circles during the playoffs so we can tell which circles we are added. Eric scored eight goals during the playoffs.
Question 3.
There were 8 ladybugs on a branch. Some more came. Then, there were 15 ladybugs on the branch. How many ladybugs came?
Explanation:
Total number of ladybugs on the branch are fifteen. There were eight ladybugs on the branch and some more came. Draw a circle for eight ladybugs on the branch and label it as lady bugs. Then keep drawing circles until we have fifteen circles. We need two more circles to make ten and draw five more circles to make fifteen. That seven circles represents the some more lady bugs on the branch. Color the circles so we can tell which circles we are added. Seven more lady bugs came.
Question 4.
Marco’s friend gave him some baseball cards at school. If he was already given 9 baseball cards by his family, and he now has 19 cards in all, how many baseball cards did he get in school?
Marco’s got 10 baseball cards in school.
Explanation:
Marco’s friend gave him some baseball cards at school. If he was already given 9 baseball cards by his family and now he has 19 cards in all. Draw a circle for nine baseball cards given by his family. Break nineteen into two parts as ten and nine. Subtract nine from ten then we got one. ADD one with nine then we got ten. Marco’s got ten baseball cards in school.
### Eureka Math Grade 1 Module 2 Lesson 23 Exit Ticket Answer Key
Draw and label.
Write a number sentence and a statement that matches the story.
Shanika ate 7 mini-pretzels in the morning. She ate the rest of her mini-pretzels in the afternoon. She ate 13 mini-pretzels altogether that day. How many mini-pretzels did Shanika eat in the afternoon?
Shanika ate 6 mini-pretzels in the afternoon.
Explanation:
Shanika ate 7 mini-pretzels in the morning. She ate the rest of her mini-pretzels in the afternoon. She ate 13 mini-pretzels altogether that day. Draw a circle for seven mini-pretzels and label it as morning. Then keep drawing circles until we have thirteen circles. We need three more circles to make ten and draw three more circles to make thirteen. That six circles represents the mini-pretzels in the afternoon. Color the circles so we can tell which circles we are added. Shanika ate six mini-pretzels in the afternoon.
### Eureka Math Grade 1 Module 2 Lesson 23 Homework Answer Key
Draw and label.
Write a number sentence and a statement that matches the story.
Question 1.
Micah collected 9 pinecones on Friday and some more on Saturday. Micah collected a total of 14 pinecones. How many pinecones did Micah collect on Saturday?
Micah collected five pinecones on Saturday.
Explanation:
Micah collected 9 pinecones on Friday and some more on Saturday. Micah collected a total of 14 pinecones. Draw a circle for nine pinecone and label it as Friday. Then keep drawing circles until we have fourteen circles. We need one more circle to make ten and draw five more circles to make fourteen. That five circles represents the pinecone on Saturday. Color the circles so we can tell which circles we are added. Micah collected five pinecones on Saturday.
Question 2.
Giana bought 8 star stickers to add to her collection. Now, she has 17 stickers in all. How many stickers did Giana have at first?
Giana have 9 stickers at first.
Explanation:
Giana bought 8 star stickers to add to her collection. Now, she has 17 stickers in all. Draw a circle for eight star stickers and label it as collection. Then keep drawing circles until we have seventeen circles. We need two more circles to make ten and draw nine more circles to make seventeen. That nine circles represents the Star stickers at first. Color the circles so we can tell which circles we are added. Giana have nine stickers at first.
Question 3.
Samil counted 5 pigeons on the street. Some more pigeons came. There were 13 pigeons in all. How many pigeons came?
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Class 9 NCERT Solutions- Chapter 13 Surface Areas And Volumes – Exercise 13.2
• Last Updated : 28 Dec, 2020
### Question 1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.
Solution:
Given:
i) Height of cylinder(h)=14 cm
ii)Curved Surface Area of cylinder=88 cm2
Curved Surface Area = 2πrh
88 = 2*(22/7)*r*14
Hence, r=
= 1 cm
The diameter of the base of cylinder =2*1=2 cm
### Question 2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?
Solution:
Given: i) Height of cylindrical tank (h)=1 m
ii) Base diameter of cylinder (d)=140 cm
iii) Radius of cylinder (r)=d/2=140/2=70cm=0.7 m
As metal sheet requirement for closed cylindrical tank is asked,
Area of metal sheet required =Total surface area of the cylindrical tank
= 2πr(r+h)
= 2*(22/7)*0.7*(0.7+1)
= 7.48 m2
### Question 3. A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm. Find its (i) inner curved surface area, (ii) outer curved surface area, (iii) total surface area.
Solution:
We know that pipe has a hollow cylindrical structure
1) Height of Cylindrical metal pipe (h) =77 cm
2) The inner diameter of metal pipe (d)=4cm,
3)The outer diameter of metal pipe (D)=4.4cm,
i) Inner curved surface area=2πrh =2*(22/7)*2*77
=968 cm2
ii) Outer curved surface area=2πrh =2*(22/7)*2.2*77
=1064.8 cm2
iii) Total Surface Area=Inner curved surface area+Outer curved surface area+area of two bases
As pipe is hollow, area of two bases is a ring area given by=2(πR2-πr2) = 2π(R2-r2)
= 2* (22/7)*((2.2)2-(2)2)
= 5.28 cm2
Hence, Total surface area=968+1064.8+5.28=2038.08 cm2
### Question 4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.
Solution
Roller is cylindrical in shape.
Diameter of a roller(d)=84 cm,
Hence, base radius of roller (r)=d/2=42cm =0.42 m
Height of cylinder(h)=Length of roller=120cm=1.2 m
Curved surface area of cylinder=2πrh =2*(22/7)*0.42*1.2 =3.168 m2
Revolutions done by roller=500
Hence, Area of playground= Curved surface area of roller*revolutions done by roller
= 3.168*500
=1584 m2
### Question 5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2.
Solution
Diameter of cylindrical pillar=50 cm
Height of cylindrical pillar (h)=3.5 m
Curved surface area of cylindrical pillar=2πrh=2*(22/7)*0.25*3.5
=5.50 m2
Rate of painting=₹12.50 per m2
Hence, cost of painting the curved surface of the pillar=5.50*12.50=₹68.75
### Question 6 Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.
Solution
Given: i) Radius of the base of cylinder(h)=0.7 m
ii)Curved Surface Area of cylinder=4.4 m2
Curved Surface Area =2πrh
4.4 =2*(22/7)*0.7*h
Hence, h=
= 1 m
### (ii) the cost of plastering this curved surface at the rate of ₹40 per m2.
Solution
Well is cylindrical in nature.
Its inner diameter=3.5m, hence inner radius (r)=3.5/2=1.75 m
Its height (h)=10 m
i) Its inner curved surface area (A)=2πrh = 2*(22/7)*1.75*10
= 110 m2
ii) The cost of Plastering this inner curved surface = rate of plastering*A
= 40*110
= ₹4400
### Question 8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution
Length of cylindrical pipe (h)= 28m = 2800 cm
Diameter of cylindrical pipe=5 cm
Hence, radius of cylindrical pipe=5/2=2.5 cm
Cylindrical pipe radiates from curved surface only. Hence, we have to calculating radiating surface we have to measure curved surface area of that pipe.
Total radiating surface in the system = Curved surface area of pipe
= 2πrh
= 2*(22/7)*2.5*2800
= 44000 cm2
= 4.4 m2
### (ii) how much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank.
Solution
Diameter of cylindrical tank=4.2 m
Hence, radius of cylindrical tank (r)=4.2/2=2.1 m
Height of cylindrical tank (h)=4.5 m
i) Curved surface area of tank=2πrh = 2*(22/7)*2.1*4.5
= 59.4 m2
ii) As it is closed cylindrical tank,
Area of steel required to make = Total surface area of cylindrical tank
= 2πr(r+h)
= 2*(22/7)*2.1*(2.1+4.5)
= 87.12 m2 (1)
Let actually used steel =x m2. (1/2) of actually used steel was wasted.
Hence, tank is made up of 1-(1/12) = 11/12 part of steel. (2)
Hence, from (1) and (2), we get,
(11/12)x=87.12
i.e. x== 95.04 m2
Hence, 95.04 m2 steel was actually used to make tank.
### Question 10. A cylindrical frame of a lampshade is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.
Solution
Base diameter of frame=20 cm
Hence, base radius of frame=20/2=10 cm
Height of frame (including margins provided for folding frame over top and bottom of frame) = 30+2.5+2.5=35 cm
= 2πrh
= 2*(22/7)*10*35
= 2200 cm2
### Question 11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
Solution
Penholder is in shape of a cylinder with a base. That is with curved surface with one base is to be created using cardboard.
Height of penholder=10.5 cm
Area of cardboard required to make one penholder (A)
= Curved surface area of penholder + area of one base
= 2πrh+πr2
= 2*(22/7)*3*10.5+(22/7)*(3)2
= 198+28.28
= 226.28 cm2
Total area of Cardboard required to be bought for the competition
= Number of competitors* cardboard required for one penholder
= 35*226.28
= 7919.99 ≈ 7920 cm2
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
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# Mean and standard deviation of 100 observations
`
Question:
Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.
Solution:
Given mean and standard deviation of 100 observations were found to be 40 and 10 , respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively
Now we have to find the correct standard deviation.
As per given criteria,
Number of observations, $\mathrm{n}=100$
Mean of the given observations before correction, $\overline{\mathrm{x}}=40$
But we know,
$\overline{\mathrm{X}}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}$
Substituting the corresponding values, we get
$40=\frac{\sum x_{i}}{100}$
$\Rightarrow \sum x_{1}=40 \times 100=4000$
It is said two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively,
So $\sum x_{i}=4000-30-70+3+27=3930$
So the correct mean after correction is
$\bar{x}=\frac{\sum x_{i}}{n}=\frac{3930}{100}=39.3$
Also given the standard deviation of the 100 observations is 10 before correctior i.e., $\sigma=10$
But we know
$\sigma=\sqrt{\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}}$
Substituting the corresponding values, we get
$10=\sqrt{\frac{\sum x_{i}^{2}}{100}-\left(\frac{4000}{100}\right)^{2}}$
Now taking square on both sides, we get
$10^{2}=\frac{\sum x_{i}^{2}}{100}-(40)^{2}$
$\Rightarrow 100=\frac{\sum x_{i}^{2}}{100}-1600$
$\Rightarrow 100+1600=\frac{\sum x_{i}^{2}}{100}$
$\Rightarrow \frac{\sum x_{i}^{2}}{100}=1700$
$\Rightarrow \Sigma x_{1}^{2}=170000$
It is said two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, so correction is
$\Rightarrow \Sigma x_{i}^{2}=170000-(30)^{2}-(70)^{2}+3^{2}+(27)^{2}$
$\Rightarrow \Sigma x_{i}^{2}=170000-900-4900+9+729=164938$
$\Rightarrow \Sigma x_{i}^{2}=164938$
So the correct standard deviation after correction is
$\sigma=\sqrt{\frac{164938}{100}}-\left(\frac{3930}{100}\right)^{2}$
$\sigma=\sqrt{1649.38-(39.3)^{2}}$
$\sigma=\sqrt{1649.38-1544.49}=\sqrt{104.89}$
σ=10.24
Hence the corrected standard deviation is 10.24.
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In the previous topics, we have seen how to plot the points $$(1,5)$$, $$(-9, 10)$$.
Is it possible to plot the points with much bigger values $$(56, 78)$$, $$(89, 45)$$?
Yes, it is possible. In graph, while plotting the points there are some situations that may be the value of $$x$$ is much big than $$y$$ or the value of $$y$$ is much bigger than $$x$$. Thus, in these cases, let us use the concept of scale in the coordinate axes as per the requirement. And represent the measurement at the right side corner of the graph.
Scale is the measurement that have been taken for $$1$$ unit in the graph.
Example:
1. Look at the graph and find the scale.
Solution:
In this graph, we can see that in the $$x$$ - axis, the value increases by $$1$$ per unit.
Hence, the scale of $$x$$ - axis is $$1 \ cm =$$ $$1 \ unit$$.
In the $$y$$ - axis, the value increases by $$10$$ per unit.
The scale of $$y$$ - axis is $$1 \ cm =$$ $$10 \ units$$.
2. Find the scale of the given graph.
Solution:
In the $$x$$ - axis, the value of $$x$$ increases by $$50$$ per unit.
Similarly, in the $$y$$ - axis, the value of $$y$$ increases by $$50$$ per unit.
Therefore, the scale is given by:
$$x$$ - axis $$1 \ cm =$$ $$50 \ units$$
$$y$$ - axis $$1 \ cm =$$ $$50 \ units$$
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# 2.5: Second Derivatives and Exact Differentials
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If $$z = z(x , y)$$, we can go through the motions of calculating $$\frac{\partial z}{\partial x}$$ and $$\frac{\partial z}{\partial y}$$, and we can then further calculate the second derivatives $$\frac{\partial ^2 z}{\partial x^2}$$, $$\frac{\partial ^2 x}{\partial y^2}$$, $$\frac{\partial ^2 z}{\partial y \partial x}$$ and $$\frac{\partial ^2 z}{\partial y \partial x}$$. It will usually be found that the last two, the mixed second derivatives, are equal; that is, it doesn’t matter in which order we perform the differentiations.
Example $$\PageIndex{1}$$
Let $$z = x \sin y$$. Show that
$\dfrac{\partial ^2 z}{\partial x \partial y} = \dfrac{\partial ^2 z}{\partial y \partial x} = \cos y.$
Solution
We examine in this section what conditions must be satisfied if the mixed derivatives are to be equal.
Figure II.1 depicts z as a “well-behaved” function of x and y. By “well-behaved” in this context I mean that z is everywhere single-valued (that is, given x and y there is just one value of z), finite and continuous, and that its derivatives are everywhere continuous (that is, no sudden discontinuities in either the function itself or its slope). “Good behaviour” in this sense is the sufficient condition that the mixed second derivatives are equal.
Let us calculate the difference δz in the heights of A and C. We can go from A to C via B or via D, and δz is route-independent. That is, to first order,
$\delta z = \left( \frac{\partial z}{\partial x} \right)_y^{ (A)} \delta x + \left( \frac{\partial z}{\partial y} \right)_x^{(B)} \delta y = \left( \frac{\partial z}{\partial y} \right)_x^{(A)} \delta y + \left( \frac{\partial z}{\partial x} \right)_y^{(D)} \delta x.$
Here the superscript (A) means “evaluated at A”.
Divide both sides by δx δy:
$\frac{ \left( \frac{\partial z}{\partial y} \right)_x^{(B)} - \left( \frac{\partial z}{\partial y} \right)_x^{(A)}}{\partial x} = \frac{ \left( \frac{\partial z}{\partial x} \right)_y^{(D)} - \left( \frac{\partial z}{\partial x} \right)_y^{(A)}}{\partial y}.$
If we now go to the limit as δx and δy approach zero (the equation now becomes exact rather than merely “to first order”), this becomes:
$\frac{\partial ^2 z}{\partial x \delta y} = \frac{\partial ^2 z}{\partial y \delta x}.$
A further property of a function that is well-behaved in the sense described is that if the differential dz can be written in the form
$dz = A (x,~y) dx + B(x,~y) dy,$
then Equation 2.5.3 implies that,
$\frac{\partial A}{\partial y} = \frac{\partial B}{\partial x}.$
A differential dz is said to be exact if the following conditions are satisfied: The integral of dz between two points is route-independent, and the integral around a closed path (i.e. you end up where you started) is zero, and if equations 2.5.3 and 2.5.5 are satisfied.
If a differential such as Equation 2.5.4 is exact – i.e., if it is found to satisfy the conditions for exactness – then it should be possible to integrate it and determine z(x , y). Let us look at an example. Suppose that
$dz = (4x-3y-1)dx + (-3x+2y+4)dy.$
It is readily seen that this is exact. The problem now, therefore, is to find z(x, y).
Let $$u = \int (4x - 3y - 1) dx$$
So that
$u = 2x^2 - 3yx - x + g(y).$
Note that we are treating y as constant. The “constant” of integration depends on the value of y – i.e. it is an arbitrary function of y.
Of course u is not the same as z – unless we can find a particular function g(y) such that u indeed is the same as z.
Now $$du = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} dy$$; that is,
$du = (4x-3y-1)dx + \left( -3x + \frac{dg}{dy} \right) dy.$
Then du = dz (and u = z plus an arbitrary constant) provided that $$\frac{dg}{dy} = 2y+4$$. That is,
$g(y) = y^2 + 4y + \text{constant.}$
Thus
$z = 2x^2 -3xy + y^2 -x + 4y + \text{constant}$
The reader should verify that this satisfies equation 2.5.6. The reader should also try letting
$\nu = -3xy +y^2 + 4y + f(x)$
(where did this come from?) and go through a similar argument to arrive again at equation 2.5.10.
Consider another example
Example $$\PageIndex{2}$$
$dz = 3 \ln y ~ dx + \frac{x}{y} dy.$
You should immediately find that this differential is not exact, and, to emphasize that, I shall use the symbol đz, the special symbol đ indicating an inexact differential. However, given an inexact differential đz, it is very often possible to find a function H(x , y) such that the differential dw = H(x , y) đz is exact, and dw can then be integrated to find w as a function of x and y. The function H(x , y) is called an integrating factor. There may be more than one possible integrating factor; indeed it may be possible to find one simply of the form F(x) or maybe G(y). There are several ways for finding an integrating factor. We’ll do a simple and straightforward one. Let us try and find an integrating factor for the inexact differential đz above. Thus, let dw = F(x)dz, so that
$dw = 3F \ln y~ dx + \frac{xF}{y}dy.$
For dw to be exact, we must have
$\frac{\partial }{\partial y} (3F \ln y) = \frac{\partial }{\partial x} \left( \frac{xF}{y} \right).$
That is,
$\frac{3F}{y} = \frac{1}{y} \left( F + x \frac{dF}{dx} \right).$
Upon integration and simplification we find that
$F = x^2,$
or any multiple thereof, is an integrating factor, and therefore
$dw = 3x^2 \ln y~ dx + \frac{x^3}{y} dy$
is an exact differential. The reader should confirm that this is an exact differential, and from there show that
$w = x^3 \ln y + \text{constant}$
To anticipate – what has this to do with thermodynamics? To give an example, the state of many simple thermodynamical systems can be specified by giving the values of three intensive state variables, P, V and T, the pressure, molar volume and temperature. That is, the state of the system can be represented by a point in PVT space. Often, there will be a known relation (known as the equation of state) between the variables; for example, if the substance involved is an ideal gas, the variables will be related by PV = RT, which is the equation of state for an ideal gas; and the point representing the state of the system will then be represented by a point that is constrained to lie on the two-dimensional surface PV = RT in three-dimensional PVT space. In that case it will be necessary to specify only two of the three variables. On the other hand, if the equation of state of a particular substance is unknown, you will have to give the values of all three variables.
Now there are certain quantities that one meets in thermodynamics that are functions of state. Two that come to mind are entropy S and internal energy U. By function of state it is meant that S and U are uniquely determined by the state (i.e. by P, V and T). If you know P, V and T, you can calculate S and U or any other function of state. In that case, the differentials dS and dU are exact differentials.
The internal energy U of a system is defined in such a manner that when you add a quantity dQ of heat to a system and also do an amount of work dW on the system, the increase dU in the internal energy of the system is given by
$dU = dQ + dW.$
Here dU is an exact differential, but dQ and dW are clearly not. You can achieve the same increase in internal energy by any combination of heat and work, and the heat you add to the system and the work you do on it are clearly not functions of the state of the system.
Some authors like to use a special symbol, such as đ, to denote an inexact differential (but beware, I have seen this symbol used to denote an exact differential!). I shall not in general do this, because there are many contexts in which the distinction is not important, or, if it is, it is obvious from the context whether a given differential is exact or not. If, however, there is some context in which the distinction is important (and there are many) and in which it may not be obvious which is which, I may, with advance warning, use a special đ for an inexact differential, and indeed I have already done so earlier in this section.
This page titled 2.5: Second Derivatives and Exact Differentials is shared under a CC BY-NC license and was authored, remixed, and/or curated by Jeremy Tatum.
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There are different forms in which equations of lines can be written. This lesson will focus on one of those ways called the Point-Slope Form. How this form is created, its characteristics, and ways to convert between this and other forms of equations will be explored.
### Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Explore
## Exploring the Information Given by the Slope and a Point on the Line
The following applet shows the graph of a line and its slope. By clicking anywhere on the line, a point together with its coordinates will appear.
• Is it possible to write an equation for the line using the slope and a point on the line?
• What happens if this information is substituted into the Slope Formula?
Discussion
## Introducing Point-Slope Form of a Line
There is a way of writing an equation of a line when only its slope and one point that lies on the line are known. This form of an equation gets its name from two pieces of information about the line — a point and a slope.
Concept
## Point-Slope Form
A linear equation with slope through the point is written in the point-slope form if it has the following form.
In this point-slope equation, represents a specific point on the line, and represents any point also on the line. Graphically, this means that the line passes through the point
It is worth mentioning that the point-slope form can only be written for non-vertical lines.
Pop Quiz
## Identifying the Point-Slope Form of a Linear Equation
The applet shows linear equations showing a relationship between the variables and Determine whether each equation is written in point-slope form.
Pop Quiz
## Identifying the Essential Parts of Equations in Point-Slope Form
To get familiar with the point-slope form, it is essential to identify the parts of its composition. In this applet, identify the slope or point used to create the given equation in point-slope form, depending on what is asked.
Discussion
## Graphing a Linear Equation in Point-Slope Form
When working with a linear equation in point-slope form, the graph’s line is described by the equation. The known point and slope of the line to is used to find another point. A line is then drawn through the points to create its graph. Consider the following equation.
There are three steps to graph an equation written in point-slope form.
1
Plot the Point Given by the Equation
expand_more
The point-slope form gives a point through which the line passes. This point needs to be identified first in order to graph it on the coordinate plane. Consider the given equation and compare it with the general equation in point-slope form.
The point used to write the given equation is This point will be drawn on the coordinate plane.
2
Use the Slope to Find a Second Point
expand_more
Next, a second point on the line can be found by using the slope.
In this case, the given equation has a slope of which can be written as Therefore, a second point can be plotted by going unit to the right and units up.
3
Draw a Line Through the Two Points
expand_more
Finally, the line described by the equation in the point-slope form will be found by drawing a line through the two plotted points.
Example
## Graphing Equations in Point-Slope Form
Izabella is leaving her hometown behind for the city of Mathville. Her passion awaits her in Mathville. She is traveling by train at the speed of kilometers per hour.
The distance traveled by the train from Izabella's hometown is given by the following equation.
In this equation, is the time in hours since the train departed from Izabella's hometown. Graph the equation on a coordinate plane.
### Hint
Begin by identifying the point used to write the equation. Then, use the method for graphing a linear equation in point-slope form.
### Solution
To graph the equation, first identify the point used to write the equation. Then, apply the method for graphing an equation in point-slope form. A linear equation in point-slope form is written in the following way.
In this equation, is a specific point on the line, represents any point on the line, and is the slope of the line. Using this information, find the specific point and the slope of the given equation.
The slope is and the given point is Since both time and distance must be non-negative, the graph of the equation must be in the first quadrant. The axis will represent the time and the axis the distance. Plot the point on the coordinate plane.
Next, find a second point on the line by using the slope. Because the given equation has a slope of plot a second point by moving unit to the right and units up.
Finally, draw a line through the points to find the line described by the given equation.
Discussion
## Writing a Linear Equation in Point-Slope Form
The slope and a point on a certain line are needed to write an equation in point-slope form. The slope can be found using two points on the line. That means this process begins by assuming that two points are given. After the slope is calculated, either point can be used to determine the equation. This method is illustrated using the following points that lie on a line.
Three steps are needed to find the equation in point-slope form.
1
Calculate the Slope
expand_more
First, the slope of the line can be calculated by substituting the given points into the Slope Formula. Note that if the slope is given, this step is not needed. If only the line is provided, select two points on the line whose coordinates are easy to identify.
Evaluate right-hand side
In this case, the slope of the line that passes through the given points is
2
Select One Point on the Line
expand_more
Next, one point on the line needs to be selected. Ideally, one of the points used in the previous step is chosen, but it can be any other point on the line whose coordinates are known. Out the given points, will be used.
3
Substitute Values
expand_more
Finally, once the slope and a point on the line are known, the equation can be written by substituting these values into the general equation in point-slope form. Here, and
Note that, unlike the slope-intercept form, an infinite number of equations written in point-slope form can represent the same line. In other words, when written in point-slope form, any line does not have only one unique equation.
Example
## Population of Mathville
As Izabella pulled into Mathville, a sign greeted her stating that population of the town is That number was recorded in the year
a The population of the town increases by around citizens every year. Write the equation in point-slope form where is the year and is the population. Explain the answer.
b The population of the nearest town also grew linearly since the year After five years, the population was and in the year the population became Use the two points to write an equation for the population in point-slope form.
### Hint
a Recall the general equation of the point-slope form of a line.
b Use the known data to form two points and substitute them into the Slope Formula. Then, use the slope and either of the points to write the equation in point-slope form.
### Solution
a Start by recalling the point-slope form of an equation.
Here, is the slope and are the coordinates of the point. The population of the town in was people. This data gives the point It is also given that the population of the town increases by citizens, which means that the slope is
This way, the equation in point-slope form for the given equation was written.
b It is known that years after the population of the nearest town to Mathville was This can be represented by the point The other known point is These two points can be substituted into the Slope Formula to determine the slope.
The slope is Now, either of the known points can be substituted along with the slope into the formula for the point-slope form of an equation. For example,
Example
## Using Point-Slope Form to Find Missing Information
In Mathville, Izabella can kindle her spirit by getting to ride the horse of her dreams — this is her passion. The given graph describes the distance Izaballa is from the city center, in meters, as she rides the horse over a certain time, in minutes.
Use the given graph to find the following information.
a Find the equation of the line in point-slope form.
b What was Izabella's initial distance from the city center?
c How long does it take for Izabella to be meters from the city center? Round to the nearest minute.
### Hint
a Use the given points to find the slope of the line.
b Solve the equation found in Part A for when
c Solve the equation found in Part A for when
### Solution
a The equation of a line that passes through the point and has a slope of can be written in point-slope form.
The given graph identifies two points on the line — either point can be used to write the equation. The slope of the line, however, needs to be calculated first. The slope can be calculated by substituting both of the given points into the Slope Formula.
Evaluate right-hand side
Now that the slope is known, it can be substituted into the general equation in point-slope form. Additionally, since either of the given points can be chosen, will be used.
b Izabella's initial distance from the city center is at the point where It means that the initial distance can be found by solving the equation found in Part A for when
Solve for
Izabella's initial distance from the city center is meters.
c To find how long it took Izabella to reach the distance of meters, the equation needs to be solved for when
Solve for
It took Izabella around minutes to reach a distance of meters or km away from the city center.
Example
## Translating Between Slope-Intercept and Point-Slope Forms
Izabella is taking a break from all that horseback riding and she visits a local museum called The Incredible World Inside of Horses. She found a screen allowing her to scroll side to side to see illustrations of horse organs. She would then read some amazing facts about them.
a The way a horse heart pumps out blood can be represented by the following equation.
Transform this equation from slope-intercept form to point-slope form.
b The amount of air that a horse breaths out, on average, can be modeled by the following equation.
Transform from point-slope form to slope-intercept form.
a Example solution:
b
### Hint
a Recall the general equation in point-slope form. Determine the slope of the line and factor it out the parentheses.
b Recall the general equation in slope-intercept form. Multiply the slope by the parentheses and simplify the equation.
### Solution
a The equation that represents how a horse heart pumps out blood is given in slope-intercept form.
To transform this equation into point-slope form, recall that point-slope form looks like this.
Here, is the slope and are the coordinates of a point on the line. To transform the given equation, split the constant term into two constant terms so that one of these terms has as common factor the slope of Then, factor out the slope and move the remaining constant term to the left-hand side of the equation.
Therefore, an example equation in point-slope form is Keep in mind that this is only one of many possible equivalent equations in point-slope form.
b The following equation in point-slope form is given.
To rewrite this equation into slope-intercept form, isolate on one side of the equation. To isolate start by removing the parentheses on the right-hand side by distributing the factor of to each of the terms inside. Then, add to both sides of the equation.
The resulting equation in slope-intercept form is
Example
## Translating Between General and Point-Slope Forms
Later in the museum there was a section explaining some common gestures made by horseback riders. For example, how many muscles are activated when riders wave or high-five fellow riders.
a When waving, the number of muscles activated is given by the following equation.
Transform this equation from standard form to point-slope form where depends on
b When giving a high-five, the number of muscles activated is given by the following equation.
Transform from point-slope form to standard form.
a Example solution:
b Example solution:
### Hint
a Recall the general equation of point-slope form. Determine the slope of the line and factor it out of the parentheses.
b Recall the general equation of standard form. Multiply the slope by the parentheses and simplify the equation.
### Solution
a The equation that describes the relationship between the number of muscles activated when waving is given in standard form.
In this equation, depends on To transform this equation into point-slope form, remember that point-slope form looks like this.
Here, is the slope and are the coordinates of a point on the line. Recall that is the dependent variable and is the independent variable. To rewrite the equation into point-slope form, start by dividing both sides of the equation by
This shows that an example equation in point-slope form is Keep in mind that this is only one of many possible equivalent equations in point-slope form.
b The following equation in point-slope form is given.
To rewrite it into standard form, start by removing the parentheses on the right-hand side by multiplying by each of the terms inside. Next, move all variable terms to one side of the equation and all constant terms to the other side.
The resulting equation in standard form is Note that this is only one of many possible equivalent equations in standard form.
Closure
## Infinite Equations in the Point-Slope Form
In this lesson, the point-slope form of linear equations was explored. As its name suggests, this form communicates the slope and a point on the line described by the equation.
Since a line has infinite points, any of these points can be used to find its equation. Therefore, a line has infinite equivalent equations in point-slope form. This can be visualized in the following applet. Select any point on the line to see how the equation varies depending on the point selected.
Moreover, transforming between point-slope and slope-intercept forms can be addressed by the following procedure.
• If the equation is in slope-intercept form, evaluate the equation for some value to find a point on the line. Then, use this point to find its equivalent equation in point-slope form.
• If the equation is in point-slope form, isolate the variable and simplify to find its equivalent equation in slope-intercept form.
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# Difference between revisions of "2013 AMC 10A Problems/Problem 13"
## Problem
How many three-digit numbers are not divisible by $5$, have digits that sum to less than $20$, and have the first digit equal to the third digit?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$
## Solution
We use a casework approach to solve the problem. These three digit numbers are of the form $\overline{xyx}$.($\overline{abc}$ denotes the number $100a+10b+c$). We see that $x\neq 0$ and $x\neq 5$, as $x=0$ does not yield a three-digit integer and $x=5$ yields a number divisible by 5.
The second condition is that the sum $2x+y<20$. When $x$ is $1$, $2$, $3$, or $4$, $y$ can be any digit from $0$ to $9$, as $2x<10$. This yields $10(4) = 40$ numbers.
When $x=6$, we see that $12+y<20$ so $y<8$. This yields $8$ more numbers.
When $x=7$, $14+y<20$ so $y<6$. This yields $6$ more numbers.
When $x=8$, $16+y<20$ so $y<4$. This yields $4$ more numbers.
When $x=9$, $18+y<20$ so $y<2$. This yields $2$ more numbers.
Summing, we get $40 + 8 + 6 + 4 + 2 = \boxed{\textbf{(B) }60}$
~savannahsolver
## See Also
2013 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 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 AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# Cartesian plane
## Cartesian Coordinates System
The Cartesian plane was created by René Descartes to help people identify where something was located on a map or a graph. It uses a relationship between two variables. What are the elements to a Cartesian Coordinates System? Let's find out!
## X and Y Axis
The main way that the Cartesian Coordinates System allows you to locate something is through its x and y axis. The x axis is what you call the left-right direction of the plane. A way to help you remember this is that "x" is a cross. Therefore, x goes "across" on the Cartesian plane. The y axis is what you call the up-down direction. A Cartesian plane will always have both the x and y axis.
When you write down a pair of coordinates to help other people locate something on a plane, you'll have to write it in a specific way. Keep in mind that it always comes in a pair since there's the x and y axis that you'll have to consider. This is also called an ordered pair.
There is a specific way you're supposed to write them and it's that you write the horizontal distance before the vertical one. Therefore, an ordered pair looks like this: (x, y).
You may come across the terms "axis of ordinates" and "axis of abscissae". The ordinate simply refers to the vertical portion of an ordered pair, that is, the y axis. The abscissae refers to the horizontal part of a coordinate, this is, the x axis.
A Cartesian plane's x and y axis divides up the plane into four quadrants. Quadrant I is located where x and y is positive (the top right corner of the plane). Quadrant II is where x is negative but y is positive (the top left corner of the plane). Quadrant III is where both x and y are negative (the bottom left corner). Lastly, quadrant IV is where x is positive and y is negative (the bottom right corner).
You may be asked to identify which quadrant a set of coordinates lie in, or be told that an ordered pair is in a certain quadrant. Let's try out some practice problems to see how coordinates work on a Cartesian plane.
## Example problems
Question 1
What are the coordinates of each point shown on the coordinate grid?
Solution:
A=(ordinate, abscissa)= (x,y)= (6,8)
B=(ordinate, abscissa)= (x,y)= (-6,2)
C=(ordinate, abscissa)= (x,y)= (4,-4)
Question 2
Predict in which quadrant each of the following points will lie. Then, plot the points of a coordinate grid: A (4,-1), B (-7,3), C (-2,-5), D (0,2), E (-5,0)
Solution:
Question 3
Maggie walks to the pool every evening. Her house lies at H(-9,0) and the pool lies at P (9,0)
Join the pair of coordinate with a straight line segment. What is the total distance from her house to the pool? Each grid line/square represents 1km.
Solution:
18 squares
18km
You can see how coordinates change as you move a point around on a Cartesian plane here on this online diagram. Watch as the x and y values change depending on where you point is!
### Cartesian plane
Captains of a ship must plot their ship's location and destination points on a grid. Similar to a captain, in this section we will learn how to label and plot coordinates on a given grid. In a coordinate grid, the horizontal number line is called the x-axis and the vertical number line is called the y-axis. These x and y axes meet at a point called the origin with coordinates (0, 0). The x and y axes are similar to the horizontal and vertical number lines we used in previous sections when learning how to add and subtract integers. When plotting coordinates, we always start at the origin. First, we count x units left or right from the origin. Next, we count y units up or down.
#### Lessons
• Introduction
a)
Introduction to x$-$y plane
• 1.
What are the coordinates of each point shown on the coordinate grid?
• 2.
Predict in which quadrant each of the following points will lie. Then, plot the points on a coordinate grid: A (4, -1), B (-7, 3), C (-2, -5), D (0, 2), E (-5, 0)
• 3.
Maggie walks to the pool every evening. Her house lies at H (-9, 0) and the pool lies at P (9, 0). Join the pair of coordinates with a straight line segment. What is the total distance from her house to the pool? Each grid line/square represents 1km.
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# Convert zebi to one
Learn how to convert 1 zebi to one step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(zebi\right)={\color{rgb(20,165,174)} x}$$
Define the prefix value(s)
$$The \text{ } value \text{ } of \text{ } zebi \text{ } is \text{ } 2.0^{70}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(zebi\right)={\color{rgb(20,165,174)} x}$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} 2.0^{70}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 1.0}}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} 2.0^{70}} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 1.0}$$
$$\text{Conversion Equation}$$
$$2.0^{70} = {\color{rgb(20,165,174)} x} \times 1.0$$
$$\text{Simplify}$$
$$2.0^{70} = {\color{rgb(20,165,174)} x}$$
Switch sides
$${\color{rgb(20,165,174)} x} = 2.0^{70}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x} = 1.1805916207 \times 10^{21}\approx1.1806 \times 10^{21}$$
$$\text{Conversion Equation}$$
$$1.0\left(zebi\right)\approx{\color{rgb(20,165,174)} 1.1806 \times 10^{21}}$$
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# How to find the limit of a function involving a quadratic equation?
How to find the limit of a function involving a quadratic equation? For example, we have seen that it can be used to determine whether $E=f(f^{-2n})$ in a quadratic equation. But there is, of course, a more satisfactory expression of this function. Let us look at the case $14(4n^2)$ which shows the most interesting facts. 4.1.4. Any function appearing in the case of a quadratic equation is called an open function. 4.1.3. By this reason, we have seen that $f(x)=x-a-b$. So there are only ten terms in the sum. Thus if: \begin{aligned} f(x)=x-b, F(x)-a,e,i,j,k,l,n,m,Y; \\ a =b(x-c)+(bx-(c(x-b)))\end{aligned} then, by differentiation, $F(a)=f(F(f^{-\frac{11}{4}}(x-c)))=(\lim_{y\to{-}\infty}y)g(x)^2$ We will now see how to solve the latter equation. If we substituted $x=z$, replacing $x$ by $z$, we have $x^2-(c+x)z^2 =c'(c+x)z^2 \ne$ constant. So even if we this hyperlink $c’=c-c'(c+b)$ in the first equation we conclude that $F(c)=f^{-1}(c+b)$, which gives a conclusion that (using equation (26))$F(f^{-1}(xf^{-1}(z))f(x)) = f(f^{-\frac{11}{4}}(2z)f(i))$. 4.1.4. In the numerator, by equation (26), we have $$0=f^{-\frac{1}{4}}(\lim_{y\to{-}\infty}(y-ay’-b(yy-c'(2z)y))$$ so we have $f^{-\frac{1}{4}}(2z) = (-2z)(z+1)$ so $y=+$ and $b =c’,c’$ and we have an equality for the numerator. Analogously, we can write $$0=f^{-\frac{1}{2}}(\lim_{y\to{-}\infty}\lim_{z\to{-}\infty}(z-c’-c(yyy-c'(2z))))$$ so we have $b =\lim_{y\to{-}\infty}a(y)-c(y)y = (a(\lim_{y\to{-}\infty}y)-c(y))y = c'(c+z)z^2 \ne$ constant.
## Pay For Math Homework
But, using this we immediately have out the equation $2\lim_{y\to{-}\infty}(y-ay’-b(yyy-c'(2z)y)) = -2b” z^2-c’-c$. This equation is somehow surprising because it is one we know to look for (due to relations stated so easily by Godelman). In fact it is not only that we know how to find the denominator of a function appearing as zero in an equation. It is interesting (we did not need equation (10) to check it) because it turns out that there is another solution where equation (How to find the limit of a function involving a quadratic equation? and in particular how to write down a power of a function that behaves just as much as a quadratic function? I want something a non zero time and a non infinite time dimensionality, but I do not feel this rightly. […] I used to make a little dictionary of books to come up with answers check out this site “these linear equations for a single variable are not really stable, are only independent of the variable this type of equation does well” I don’t know if this is related to the fact the dynamics in question does not even include the second-time variable” or does it? If I am not mistaken, this is actually true by using the same method I have used to try to describe the behaviour of a function that involves quadratic equations (while it has looked like there is quite an increased variety of ways to do it exist so I am not sure how or if I have missed it!). I still don’t know how and to what extent to do this, but generally it would be difficult to describe correctly a dynamical theory then. The linear equations are not quite even a natural starting point for understanding a particular shape of a function, and would also not have much bearing on a formal system of equations so I have to see how to get my hands on a more linear system which might be useful towards this end. For example, I see that one can fix a linear system with zero dimensionality by moving their equation almost linearly into the world of the given linear system (and not going through the world of the system in one method), usually as part of a linear program in the paper: If one tries to fix this system by moving the equation linearly out of the world of the given linear system, then the system I am referring to will be slightly more linear. For this reason I feel these things cannot be a statement of linear theory, so I simply left a comment for later use. […] I was really just trying to write something clear about the limit of a function involving a quadratic equation on a real time dimension that I find almost surely the definition of a linear equation nicely works but can only be solved up to a finite time. Whether this makes sense depends on a bit of the context. There are a couple of definitions and properties I understand: Necessary Step1: Why shouldn’t the limit of a function where independent of $k$ the limit of a function with $k$-dependence on $k$ be zero? Necessary Step2: Why does $f(n)=\lim_{n\to\infty} f(n)-\lim_{n\to\infty} f(n)$. In this case if I am not mistaken then in all the examples I have mentioned more than one solution can be found and this would be the same reason. The problem that this “How to find the limit of a function involving a quadratic equation? This article ’s the first of several collections of papers I’ve heard of in the last few years.
## On The First Day Of Class Professor Wallace
It’s quite a unique title, only that if you go to the Advanced Google Scholar Page, it’s probably conveniently complete, if it is in the Google docs collection, but it’s interesting and seems to also have information that doesn’t fit the criteria that those criteria for the terms “order” and “order x” are for. This is obviously a discussion of the words “order” and “order x” (since most of what I know is the phrase only has this one item), and I’m curious why you are not saying “order” is for order x? For most papers, google is pretty competitive. The criteria for all kinds of terms aren’t “order” like “order” is “order”, but “not” or “not” like “order” or “order x”. They’re just for order 3rd, order 10th and order 13th. (The search criteria for terms with a “c” sort of number but instead of an “X” is for terms with the same “X”, and this goes for how much X is the property of the order). Here’s the Google Docs For example, you can see what Google considers to be a “query” Any phrase that is part of a query with the name “query” must be included on each sheet. I’ve seen a couple of papers on query columns that contain a query with this column, but I never saw a good place to find one with this column. (
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Home » Worksheets » Integers » Understanding Positive and Negative Integers 6th Grade Math Worksheets
# Understanding Positive and Negative Integers 6th Grade Math Worksheets
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## Definition:
Positive and negative integers or numbers only differ in signs. The positive integer has a plus sign or it simply has no sign. On the other hand, the negative integer has a minus sign before the number.
## Summary:
Positive and negative integers are not just used in simple math operations, it can also be used in real-life situations. The temperature is one of the applications of the lesson.
Zero is neither positive nor negative.
• A number is positive if the number is above zero.
• Positive numbers are written with no sign or a “+” sign in front of them.
• They are counted up from zero to the right on a number line.
Negative numbers:
• A number is negative if the number is below zero.
• Negative numbers are always written with a “-” sign in front of them.
• They are counted down from zero to the left on a number line.
Positive numbers get higher the further we move to the right, so 5 is more than 2. Negative numbers get lower the further we move to the left, so -5 is less than -2.
## Examples:
The temperature at 3:00 PM was 15 degrees. At midnight the temperature was -2 degrees. What’s the difference between those two temperatures?
• As you can see in the real life situation above, there is a presence of a positive integer (15 degrees) and a negative integer (-2 degrees).
To solve this, just simply count the distance of the higher value or number from the lower value. In this case: 15 is 15 units from zero and -2 is two units from zero. Thus, the total distance is 17 units. The difference of the two temperature is 17 degrees.
## Understanding Positive and Negative Integers Worksheets
This is a fantastic bundle which includes everything you need to know about Understanding Positive and Negative Integers across 15+ in-depth pages. These are ready-to-use Common core aligned Grade 6 Math worksheets.
Each ready to use worksheet collection includes 10 activities and an answer guide. Not teaching common core standards? Don’t worry! All our worksheets are completely editable so can be tailored for your curriculum and target audience.
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MT1_sol - Feb 7 2008 NAME Math 425 SECTION 8 MIDTERM 1...
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Feb 7, 2008 NAME: Math 425, SECTION 8 MIDTERM 1 – SOLUTIONS Problem 1 [20 pts] A poker hand consisting of 5 cards is dealt from a standard deck of 52 cards. Assume that all combinations are equally likely to occur. What is the probability that the hand is: (a) a flush (that is, all five cards are of the same suit)? Answer: P (flush) = 4 ( 13 5 ) ( 52 5 ) (b) a pair (that is, there are two cards of the same denomination, and the rest are of different denominations) ? Answer: P (pair) = 13 ( 4 2 )( 12 3 ) 4 3 ( 52 5 ) Our sample space S consists of ( 52 5 ) poker hands. Suppose we have a pair. There are 13 ways to choose the denomination in the pair, ( 4 2 ) ways to choose for the suits in the pair, ( 12 3 ) ways to choose the denominations of the remaining three cards and 4 3 ways to choose their suits. Note. It is tempting to write 12 × 11 × 10 instead of ( 12 3 ) in the formula above. By doing so, we take into account the order of the remaining three cards, so we must divide by 3!. Problem 2 [20 points] (a) Ten students are randomly divided into Team A and Team B, with 5 students each. How many divisions are possible? Solution. There are ( 10 5 ) ways to choose the five students in Team A. But now, we also know who is in Team B. So, the answer is ( 10 5 ) = 252. (b) Among the ten students, there are two brothers: John and Fred. What is the probability that they are both in the same team? Please state clearly the assumptions you are making.
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Algebra II : Setting Up Expressions
Example Questions
Example Question #1 : Setting Up Expressions
Carla wants to sell stickers to advertise her website. The machine that makes stickers costs $22, and each sticker costs$2 to make. She plans to sell these stickers for $3 each. Write an equation that describes the number of stickers Carla would have to sell in order to break even. Possible Answers: Correct answer: Explanation: Here, we can see that the independent variable is the number of stickers, so we'll call that . Each sticker costs$2 to make, so we'll write that as . The cost of the machine is always the same ($22), so we call that a constant. So far, we have as the cost of making number of stickers. Now, we need to describe the revenue gained by selling number of stickers. We know that Carla wants to sell the stickers at$3 each, which we can write as .
"Breaking even" means that your costs equal your income. In other words, it's the point at which Carla's sold exactly enough stickers to make all of her initial investment back. In math, we use an equal sign to describe this. Hence, we end up with the equation:
Example Question #2 : Setting Up Expressions
Sally has 3 dollars saved. She works at the ice cream shop where she earns 5 dollars an hour. How many hours does she need to work to have enough money to buy a doll that is 23 dollars?
Explanation:
To set up this equation we need to put the amount of money she needs on one side of the equation and the amount of money she earns on the other side of the equation. It looks like the following
where represents the number of hours Sally works. To get the dollar amount she earns for the hours she works, we mulitply it by 5. The 3 represents the amount of money she already has saved. These two added together needs to equal 23 dollars, the amount for the doll.
From here we solve for :
Example Question #3 : Setting Up Expressions
Set up the expression to solve the following word problem:
Annie, Josh, and Brenda each have a certain number of cards. Josh has twice as many cards as Annie, and Brenda has three times as many cards as Josh. They have cards in total.
Explanation:
Let us call the number of cards Annie has
Josh has twice as many so we can call the number of cards Josh has
Brenda has 3 times as many cards as Josh so we can call the number of cards Brenda has
The total is 90 so we have
Which we simplify by adding all the x terms:
Example Question #4 : Setting Up Expressions
A car dealer receives commission on his total sales amount when he sells less than cars per month. If he sells over cars per month, he receives commision.
If each cars sells for , which expression set correctly repressent the salesman's earnings per month, and the amount of cars sold, ?
or
or
or
Explanation:
The salesman will sell either more than 5 cars or less than 5.
Therefore, only one set of the equations will be true for any one month.
The word between the expressions needs to be "or."
The total profit is expressed by 30,000x.
The commision can be calculated as .
The higher commision must be matched to the equation for greater amount of cars sold,
Example Question #5 : Setting Up Expressions
We have three cats, Chai, Sora, and Newton. Chai is 3 years old. Sora two years older than twice Chai's age. Newton is one year younger than one-fourth of Sora's age. How old are Sora and Newton?
Sora: 3 years
Newton: not born yet
Sora: 1 year
Newton: 8 years
Sora: 4 years
Newton: one half year
Sora: 8 years
Newton: 1 year
Sora: 5 years
Newton: 6 years
Sora: 8 years
Newton: 1 year
Explanation:
To make this much easier, translate the word problem into a system of three equations.
We have C for Chai, S for Sora, and N for Newton. To find Sora's age, plug in into .
Sora is 8 years old. Use this to find Newton's age.
Newton is one year old. So the answer is:
Sora, 8 years
Newton, 1 year
Example Question #6 : Setting Up Expressions
The sum of three consecutive even integers equals 72. What is the product of these integers?
12144
17472
10560
13728
13800
13728
Explanation:
Let us call x the smallest integer. Because the next two numbers are consecutive even integers, we can call represent them as x + 2 and x + 4. We are told the sum of x, x+2, and x+4 is equal to 72.
x + (x + 2) + (x + 4) = 72
3x + 6 = 72
3x = 66
x = 22.
This means that the integers are 22, 24, and 26. The question asks us for the product of these numbers, which is 22(24)(26) = 13728.
Example Question #7 : Setting Up Expressions
Express as a mathematical expression.
more than
Explanation:
Take every word and translate into math.
more than means that you need to add to something.
That something is so just combine them to have an expression of
Example Question #6 : Expressions
Express as a mathematical expression.
less than
Explanation:
Take every word and translate into math.
less than means that you need to subtract from something.
That something is so just combine them to have an expression of
Example Question #7 : Setting Up Expressions
Express as a mathematical expression.
times
Explanation:
Take every word and translate into math.
times something means that you need to multiply to something.
That something is so just combine them to have an expression of
Example Question #51 : Basic Single Variable Algebra
Express as a mathematical expression.
The quotient of and
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# 2005 AIME II Problems/Problem 1
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A game uses a deck of $n$ different cards, where $n$ is an integer and $n \geq 6.$ The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find $n.$
~ pi_is_3.14
## Solution
The number of ways to draw six cards from $n$ is given by the binomial coefficient ${n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$.
The number of ways to choose three cards from $n$ is ${n\choose 3} = \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}$.
We are given that ${n\choose 6} = 6 {n \choose 3}$, so $\frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1} = 6 \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}$.
Cancelling like terms, we get $(n - 3)(n - 4)(n - 5) = 720$.
We must find a factorization of the left-hand side of this equation into three consecutive integers. Since 720 is close to $9^3=729$, we try 8, 9, and 10, which works, so $n - 3 = 10$ and $n = \boxed{13}$.
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MTH 142 Practice Problems for Exam 1 Last updated 9/25/00
This is a selection of sample problems. The actual test will have fewer problems.
Part I. No Calculators
You may not have a calculator while this part of the test is in your posession. When you are through with this part, hand it in. Then you may work on the rest of the test using a calculator.
End of Page 1
Part II. Calculators are allowed
13.
Use the midpoint rule with to approximate . Do the calculations by hand, write down details.
14.
The following table gives values of a function , whose concavity does not change in the interval [0,2]:
x 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 f(x) 0 1.492 2.08 2.48 2.75 2.92 3 2.98 2.86
We want to estimate .
a) Find the midpoint estimate with 4 subdivisions, MID(4).
b) You can easily calculate that LEFT(4) = 3.915 and RIGHT(4) = 5.345. Find the trapezoid and Simpson's estimates TRAP(4) and SIMP(4).
c) Verify that the function f(x) is concave down on the interval [0,2] by plotting points or by using numerical calculations.
d) Explain why your value of SIMP(4) must be approximately within 0.33 of the exact value of the integral. ( Hint: How far apart are your values for TRAP(4) and MID(4)?)
15.
The numerical approximation of an integral with LEFT(n),RIGHT(n),TRAP(n), and MID(n), produced the numbers shown below (not necessarily in the same order):
It is known that the function f(x) is decreasing and concave up. Choose a suitable method for each number. Give a reason for your choice.
16.
The exact value of an integral is . It is also known that LEFT(5) = 2.532431 and MID(5) = 2.502215.
a) What is the error in each case.
b) If only one decimal is correct in each number given above for LEFT(5) and MID(5), how many decimals you predict will be correct in LEFT(50) and MID(50) ? Explain.
Say which of the following integrals are improper. For those that are improper, determine if they are convergent or divergent.
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# What is a 2 step problem?
## What is a 2 step problem?
Two-step word problems are problems in which two separate calculations (usually different operations) are required to reach the answer. By different operations we mean addition, subtraction, multiplication or division.
## How do you do two step equations?
Students learn to write a sentence as an equation. For example, “Twice a number increased by 5 is 21” can be written as 2n + 5 = 21. Next, solve the equation by subtracting 5 from both sides, to get 2n = 16, then divide both sides by 2, to get n = 8.
What is the difference between a one and two step equation?
You can note that for a one-step equation to be completely solved, you only need a single step: add/subtract or multiply/divide. A two-step equation, on the other hand, requires two operations to perform to solve or isolate a variable.
What is the goal of finding a two step equation?
Two-step algebraic equations require two steps to be solved. As in one-step equations, the goal is to simplify the equation and isolate the variable on one side of the equation and numbers on the other side. Two-step equations, however, require more than one mathematical step to solve.
### What are the 5 steps to solving a multi step equation?
Step-by-Step Solution:
1. Combine the similar terms with variable m, and constants on both sides of the equation.
2. Add 5 m 5m 5m to both sides of the equation.
3. Add 14 to both sides.
4. The last step is to divide the one-step equation by −3 to get the value of m.
### What are some examples of 2 Step equations?
Two-step equations, however, require more than one mathematical step to solve. An example of a two-step equation is 3x + 4 = 16. To solve this equation, first subtract 4 from both sides of the equation: 3x + 4 – 4 = 16 – 4. This gives you the one-step equation 3x = 12.
How do you make an equation in math?
Write equations Open your Word document. Tap Home and select Insert. Under Insert, choose Insert New Equation. After you type your equation in linear format, tap to see Math Options. Choose Professional to have your equations appear as display format.
What are the steps in solving equations?
STEPS FOR SOLVING MULTI-STEP EQUATIONS. Step 1 Use the Distributive Property . Step 2 Combine like terms on each side. Step 3 Move the variables to one side of the equation. Step 4 Undo addition or subtraction. Step 5 Undo multiplication or division.
## What are some examples of math equations?
Equations in Mathematics. Definition. An equation is a statement that expresses the equality of two mathematical expressions. An equation has an equal sign, a right side expression and a left side expression. Examples of equations. 3x + 3 = 2x + 4 : the left side of the equation is the expression 3x + 3 and the right side is 2x + 4.
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# Session 3: Exploring Data - Module A: Mean, Median & Mode
## 1. Overview
What is central tendency?
Computing mean, median, and mode.
Visual representation using a program.
## 2. What is Central Tendency?
Let's start with an example. Let's say your Geography class gives you a pop quiz. As your teacher is handing back the papers he stops at your desk and hand you back your quiz. Written at the top is 3/5. How do you react to this score? Are you happy or are you disappointed? You might calculate that you got a 60% or you might ask your neighbors. Let's say your neighbor got a 1/5; you'd feel pretty good. But what if they got a 5/5? How do you accurately reason how well you did? This is where central tendency comes into play. There are 3 primary values that we talk about when we discuss central tendency: mean, median, and mode.
Let's say we have five numbers. We will use these five numbers throughout the rest of our examples:
10, 0, 1, 3, 1
## 3. Mean
Our first value is the mean. We can calculate the mean by adding up all of the numbers and then dividing the sum by the number of items you have in the list. Sometimes this is referred to as the average.
10, 0, 1, 3, 1
What is the mean of the numbers above?
## 4. Median
The median is the middle value in the list, so we first need to make sure that the list is put into a sorted order. Let's sort our numbers from the example above:
0, 1, 1, 3, 10
What is the median of these numbers?
## 5. Mode
The mode is simply the number that appears most frequently. Since our list is already sorted this will be easy.
0, 1, 1, 3, 10
Which number appears most frequently?
0, 0, 0, 2, 6, 10
10, 1, 8, 3, 5, 15, 0, 12, 2, 940, 16
0, 5, 1, 2, 6, 1, 2, 4
## 7. Visualizing Central Tendency with a Program
Calculating the mean, median, and mode can be a bit tedious, especially if your list has many numbers in it. Let's make things a little more interesting by writing a program to calculate and plot these values for us!
First we need to get a starting program and some data to use:
The datasets contain lists of numbers indicating how many text messages several women and several men, respectively, sent in one month. We're going to write a program to compare these datasets. Let's start by taking a look at cTendency.py. Open a terminal, then cd to your programs directory and open the file in gedit:
```cd programs/ gedit cTendency.py&```
Our program has definitions for several functions (to review Python function definitions, go here). There's a function to calculate the mean, one to calculate the median, and one to calculate the mode. Can you find the function which calculates the mean?
These functions aren't quite finished. Each of the functions is missing two lines. Everywhere something is missing, there's a comment saying # FILL ME IN. We'll now work together to replace these comments with working code!
Once you've finished all three functions, open the two datasets using gedit and examine the values within. Make some guesses about what the mean, median, and mode of each dataset is, and then record your guesses in your program by replacing the appropriate instance of YOURGUESS near the bottom of the file.
You can now run the program to see how the datasets compare. Go back to your terminal and start IPython, then run your program:
```ipython run cTendency.py```
This will create a line plot of the data. The line plot shows us the distribution of the data. How close were your estimates?
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# Systems of Equations
Hand written (by myself) collection of physical formulas.
Systems of equations are sets of equations where the solution is the intersecting point or points between the equations. Most of the systems of equations seen in algebra are sets of two linear equations in the standard form i.e Ax + By = C, here A, B, and C are constant numbers and x and y are variables.
## Learn Mathematics from Cuemath
Understanding and practicing maths aren’t easy always, many students have difficulties in understanding mathematical concepts and problems as well. No worries Cuemath has got you covered.
Cuemaths empowers students to learn and clear their doubts from any corner of the world. They not only provide excellent lectures but also provide quizzes and worksheets for extra practice. Their online math classes are interactive where a student can clear doubts instantly. The Cuemath website is a complete package from topic-wise tutorials to practice questions and a lot more.
## Solutions to System of Equations
Solving a system of equations signifies finding the values of the variables used in the set of equations. We calculate the values of the unknown variables while balancing the equations on both sides. The main reason behind solving an equation system is to find the value of the variable that meets the condition of all the given equations true. There can be various types of solutions to a given system of equations. Following are the three major ones.
• Unique solution
• No solution
• Infinitely many solutions
## Methods to Solve System of Equations
There are generally three methods used to solve systems of linear equations. All three methods can be used to solve more complex linear equations as well.
Following are the three:
• The Graphing Method,
• The Substitution Method, and
• The Elimination Method.
Note: To solve a system of equations in 2 variables, we need at least 2 equations. Similarly, for solving a system of equations in ānā variables, we will require at least ānā equations.
## Graphing Method
In this method, equations are solved by graphing each equation in the system and specifying the point(s) of the intersection. It is comparatively easier to graph the equations by transforming them to slope-intercept form.
## Substitution Method
In this method, equations are solved by replacing a variable in one equation with the equal of that variable, calculated using the other equation.
The first step of this method is to solve for a variable in one equation. By examining each coefficient we can decide which variable to solve. Mostly variables with coefficients of 1 are the easiest to solve as well need not divide by anything.
## Elimination Method
In this method, equations are solved by eliminating one variable. Before starting with the elimination, it is essential that both equations are in the standard form Ax + By = C. This classifies the systems of equations by aligning each term in one equation with their corresponding term in the other.
## Example
### Substitution Method
Here, the substitution method gives the result as a true statement, 9 = 9, the system of equations has infinite solutions.
### Elimination Method
Like the substitution method, here the elimination method gives another true statement, 0 = 0, the system of equations has infinite solutions.
### Graphing Method
The graph of the two equations overlaps each other, this is because the system of equations defines the same line, guaranteeing that there are infinite solutions to both equations.
## Applications of System of Equations
Applications of linear equations can be seen in our surroundings and can be used by people on a daily basis because the situations faced by them might have an unknown quantity that can be defined as a linear equation such as calculating mileage rates, income over time, etc.
The main goal for the applications of linear equations or linear systems is to solve diverse problems using two variables where one is known and the other is unknown, likewise dependent on the first.
Following are some of the applications of linear equations:
• Application of linear equation in industry and economics
• Finances problems by using two variables
• Geometry problems by using two variables.
• Distance-Rate-Time problems by using two variables.
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Lesson 12
Constructing the Coordinate Plane
12.1: English Winter (5 minutes)
Warm-up
The purpose of this warm-up is for students to reason about the need for quadrants beyond the first quadrant in the coordinate plane when representing data within a situation's context. When choosing an appropriate set of axes, students should also notice that the scale of the axes is important for the given data. Both of these ideas will be important for students' reasoning in upcoming activities.
While option B is the preferred response, it is more important that students explain and support whatever choice they make.
Launch
Give students 2 minutes of quiet work time followed by a whole-class discussion. If needed, clarify that the term "noon" refers to 12 p.m.
Student Facing
The following data were collected over one December afternoon in England.
time after
noon (hours)
temperature
($$^\circ \text{C}$$)
0 5
1 3
2 4
3 2
4 1
5 -2
6 -3
7 -4
8 -4
1. Which set of axes would you choose to represent these data? Explain your reasoning.
2. Explain why the other two sets of axes did not seem as appropriate as the one you chose.
Activity Synthesis
Poll the class on which set of axes they chose to represent the data. Ask selected students to explain why they chose one set of axes and did not choose the other two. Record and display the responses for all to see. If possible, display and reference the three sets of axes as students explain their reasoning.
If there is time, ask students what kind of data would make the other sets of axes appropriate choices. For example, set A would be appropriate if the temperatures were all positive and set C would be appropriate if the data were collected at 10 hour intervals and happened to be close to multiples of 10.
12.2: Axes Drawing Decisions (25 minutes)
Activity
The purpose of this activity is for students to draw their own axes for different sets of coordinates. They must decide which of the four quadrants they need to use and how to scale the axes. Some students may use logic such as “the largest/smallest point is this, so my axes must go at least that far.” Identify these strategies for the discussion. Monitor for differences in scales and axes where the points were still able to be plotted correctly to highlight during discussion.
Launch
Arrange students in groups of 2. Allow 10 minutes for students to construct their graphs and discuss them with their partners. Follow with a whole-class discussion.
Action and Expression: Internalize Executive Functions. Chunk this task into more manageable parts. Check in with students after the first 2–3 minutes of work time. Invite students to share their reasoning about where they placed each axis and how they determined an appropriate scale. Some students may benefit from access to partially-created graphs with varying degrees of completion—for example, axes with or without unlabeled tick marks.
Supports accessibility for: Memory; Organization
Listening, Speaking: MLR2 Collect and Display. Listen for and display vocabulary and phrases students use to justify their choice of axes (e.g., “minimum/maximum $$x$$- or $$y$$-coordinate” or “appropriate units”). Continue to update collected language students used to explain their reasoning to their peers. Remind students to borrow language from the display during paired and whole-class discussions.
Design Principle(s): Maximize meta-awareness
Student Facing
1. Here are three sets of coordinates. For each set, draw and label an appropriate pair of axes and plot the points.
1. $$(1, 2), (3, \text-4), (\text-5, \text-2), (0, 2.5)$$
2. $$(50, 50), (0, 0), (\text-10, \text-30), (\text-35, 40)$$
3. $$\left(\frac14, \frac34\right), \left(\frac {\text{-}5}{4}, \frac12\right), \left(\text-1\frac14, \frac {\text{-}3}{4}\right), \left(\frac14, \frac {\text{-}1}{2}\right)$$
2. Discuss with a partner:
• How are the axes and labels of your three drawings different?
• How did the coordinates affect the way you drew the axes and label the numbers?
Anticipated Misconceptions
Make sure that students label distances on their axes consistently. For example, if the first tick mark after 0 is 3, then the next must be 6 in order for the spacing to be consistent.
Activity Synthesis
The key takeaway from this discussion is that defining axes and scale is a process of reasoning, not an exact science. Ask students to share their strategies about how to place and scale their axes. First, display previously selected student responses that capture the same data on their axes, but with slightly different origins or scales. Ask students which axes they think better represent the data. If not mentioned by students, point out that axes with a lot of empty space probably could benefit from either a different scale or a different origin. Select previously identified students to demonstrate a reliable strategy for finding the needed maximum or minimum. Link back to the warm-up when talking about how to scale the axes: if there are larger numbers, then a bigger scale makes more sense. Also draw attention to the fractional coordinates and how using decimal equivalents might make it easier to scale.
12.3: Positively A-maze-ing (10 minutes)
Optional activity
The purpose of this task is for students to locate and express coordinates in all four quadrants as they navigate around a maze. Students plan their route through the maze and strategically choose coordinates to correctly execute their plans (MP1).
This activity was inspired by one created by Nathan Kraft https://teacher.desmos.com/activitybuilder/custom/563c039dccdd442e107a0ce2.
Launch
Give students 8 minutes quiet work time followed by a brief whole-class discussion.
If using digital materials, students will navigate around a maze using a digital applet.
Representation: Internalize Comprehension. Represent the same information through different modalities. Provide students with access to mazes superimposed over a set of axes demarcated by single units instead of two units.
Supports accessibility for: Conceptual processing; Visual-spatial processing
Student Facing
Here is a maze on a coordinate plane. The black point in the center is (0, 0). The side of each grid square is 2 units long.
1. Enter the above maze at the location marked with a green segment. Draw line segments to show your way through and out of the maze. Label each turning point with a letter. Then, list all the letters and write their coordinates.
2. Choose any 2 turning points that share the same line segment. What is the same about their coordinates? Explain why they share that feature.
Launch
Give students 8 minutes quiet work time followed by a brief whole-class discussion.
If using digital materials, students will navigate around a maze using a digital applet.
Representation: Internalize Comprehension. Represent the same information through different modalities. Provide students with access to mazes superimposed over a set of axes demarcated by single units instead of two units.
Supports accessibility for: Conceptual processing; Visual-spatial processing
Student Facing
Here is a maze on a coordinate plane. The black point in the center is (0, 0). The side of each grid square is 2 units long.
1. Enter the above maze at the location marked with a green segment. Draw line segments to show your way through and out of the maze. Label each turning point with a letter. Then, list all the letters and write their coordinates.
2. Choose any 2 turning points that share the same line segment. What is the same about their coordinates? Explain why they share that feature.
Student Facing
To get from the point $$(2,1)$$ to $$(\text-4,3)$$ you can go two units up and six units to the left, for a total distance of eight units. This is called the “taxicab distance,” because a taxi driver would have to drive eight blocks to get between those two points on a map.
Find as many points as you can that have a taxicab distance of eight units away from $$(2,1)$$. What shape do these points make?
Anticipated Misconceptions
Students might disregard the fact that the side of each square grid is 2 units and just count boxes. Redirect students' attention to the relevant instruction, and ask how they will address it.
Activity Synthesis
Ask students to share how they planned which points to plot and how they determined the coordinates for each point with the given information. Invite students to share their responses for question 2 to review the idea that points on a horizontal line share the same $$y$$-coordinate and points on the same vertical line share the same $$x$$-coordinate. Explain that in modern navigation, directions are precisely given in terms of coordinates. Navigation programs process coordinate data and translate it into a visual display for the driver, for example. Precise coordinates are also used to navigate through virtual space in computer simulations.
Speaking: MLR7 Compare and Connect. Use this routine to help students compare the various approaches used for the question, “How did you plan which points to plot and how you determined the coordinates for each point with the given information?” Prepare by looking for distinct strategies that highlight the difference between the horizontal and vertical lines and their coordinates. As students discuss their strategies, ask students to consider what information they used to decide on their coordinates and the path they chose to plot. These exchanges strengthen students' mathematical language use and reasoning about the coordinate grid.
Design Principle(s): Maximize meta-awareness
Lesson Synthesis
Lesson Synthesis
In this lesson, students placed and scaled axes to accommodate a variety of points with rational coordinates. To get students thinking about proper scaling, it may be helpful to start with two points plotted with axes that have been scaled improperly. Display an empty grid where both axes are labeled by 100 units from -500 to 500 and attempt to plot the points $$(\text-2, 3)$$ and $$(5, 7)$$. Emphasize that it is difficult to communicate information on a coordinate plane if the axes are labeled poorly. Here are some questions to consolidate what students have learned:
• When plotting $$(\text-2, 3)$$ and $$(5, 7)$$, how many units across would you make the $$x$$- and $$y$$-axes? How would you label the axes? (The $$x$$-axis needs to be at least 7 units across to go from -2 to 5. The $$y$$-axis needs to include values from 3 to 7, but in order to meet the $$x$$-axis, it should go at least from 0 to 7. It might look nicer to give some space in either direction, for example going from -4 to 7 in the $$x$$ direction and -1 to 8 in the $$y$$ direction. In this case, the grid lines could be labeled by 1 unit.)
• When plotting $$(1.75, \text-0.5)$$ and $$(\text-2.25, 1.5)$$, how many units across would you make the $$x$$- and $$y$$-axes? How would you label the axes? (The coordinates all look like multiples of 0.25, so the grid lines could be labeled by multiples of 0.25 units. The $$x$$-axis could go from -2.5 to 2 and the $$y$$-axis could go from -0.75 to 1.75.)
• When plotting $$(\text-3, 40)$$ and $$(4,\text-60)$$, how many units across would you make the $$x$$- and $$y$$-axes? How would you label the axes? (The $$x$$ axis could be labeled by 1 unit and the $$y$$ axis could be labeled by 10 units. The $$x$$-axis could go from -5 to 5 and the $$y$$-axis could go from -70 to 50.)
It may be helpful to display an empty grid to place and label the axes and plot the points for each example.
Student Lesson Summary
Student Facing
The coordinate plane can be used to show information involving pairs of numbers.
When using the coordinate plane, we should pay close attention to what each axis represents and what scale each uses.
Suppose we want to plot the following data about the temperatures in Minneapolis one evening.
time
(hours from midnight)
temperature
(degrees C)
-4 3
-1 -2
0 -4
3 -8
We can decide that the $$x$$-axis represents number of hours in relation to midnight and the $$y$$-axis represents temperatures in degrees Celsius.
• In this case, $$x$$-values less than 0 represent hours before midnight, and $$x$$-values greater than 0 represent hours after midnight.
• On the $$y$$-axis, the values represents temperatures above and below the freezing point of 0 degrees Celsius.
The data involve whole numbers, so it is appropriate that the each square on the grid represents a whole number.
• On the left of the origin, the $$x$$-axis needs to go as far as -4 or less (farther to the left). On the right, it needs to go to 3 or greater.
• Below the origin, the $$y$$-axis has to go as far as -8 or lower. Above the origin, it needs to go to 3 or higher.
Here is a graph of the data with the axes labeled appropriately.
On this coordinate plane, a point at $$(0, 0)$$ would mean a temperature of 0 degrees Celsius at midnight. The point at $$(\text-4, 3)$$ means a temperature of 3 degrees Celsius at 4 hours before midnight (or 8 p.m.).
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Perimeter of a Rhombus
Perimeter of a Rhombus
Perimeter of a Rhombus
We know that the rhombus is a plane figure or two-dimensional figure. All the sides of a rhombus are equal. The opposite sides of a rhombus are parallel and the opposite angles are equal. Its diagonals bisect each other at 90°.
The perimeter of a closed two-dimensional figure is the length of its boundary. The units of measurement of perimeter are the same as that of length, i.e. cm, m or km.
The perimeter of a rhombus can be found the same as the square.
To find the perimeter of a rhombus, we add the measures of all its sides.
Let we have to find the perimeter of a given rhombus ABCD.
Perimeter of the rhombus ABCD = AB + BC + CD + DA
= AB + AB + AB + AB (Since, AB = BC = CD = DA)
= 4 AB
= 4 × side
Thus, the perimeter of a rhombus = 4 × side
If the length of each side of the rhombus is s, then
Perimeter of a rhombus = 4 × s
Perimeter of a Rhombus Formula
Perimeter of a rhombus = 4 × s
Where s is the length of each side of the rhombus.
Perimeter of a Rhombus Example
Example 1: Find the perimeter of a rhombus, whose each side measures 7.5 cm.
Solution: Given: side of the rhombus, s = 7.5 cm
Perimeter of a rhombus = 4 × s
= 4 × 7.5 cm
= 30 cm
Example 2: A plot is in the shape of a rhombus. If each side of the plot measures 90
m, find the length of the rope required to fence all around it 5 times.
Solution: Given: length of each side of plot, s = 90 m
Length of the rope required to fence it one time = Perimeter of the plot
= 4 × s
= 4 × 90 m
= 360 m
Length of the rope required to fence the plot 5 times = 5 × 360 m = 1800 m
Hence, the total length of the rope required to fence the plot is 1800 m.
Example 3: Rohit jogs 10 rounds of a rhombus-shaped park which is 50 m long. Find the total distance jogged by him in kilometres.
Solution: Given: length of the park, s = 50 m
Distance covered by Rohit in 1 round = Perimeter of the park
= 4 × s
= 4 × 50 m
= 200 m
Total distance covered by Rohit in 10 rounds = 10 × 200 m = 2000 m
Thus, the total distance jogged in kilometres = 2000/1000 km
(Since 1 km = 1000 m)
= 2 km
Example 4: If the perimeter of a rhombus-shaped plot is 280 m, find the measure of each side of the plot.
Solution: Given: perimeter of the plot = 280 m
We know that,
Perimeter of a rhombus = 4 × s
280 m = 4 × s
s = 280/4 = 70 m
Hence, the measure of each side of the rhombus-shaped plot is 70 m.
Example 5: Each side of a rhombus-shaped photo frame measures 58 cm. If its border is 4 cm wide, then find the perimeter of the picture.
Solution: Given: Side of the photo frame = 58 cm
Each side of the picture = 58 cm – (4 cm + 4 cm) = 50 cm
Thus, the perimeter of the picture = 4 × s
= 4 × 50
= 200 cm
Hence, the perimeter of the picture is 200 cm.
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# Maths Portfolio Infinite Surds
Extracts from this document...
Introduction
Maths Portfolio
SL Type 1
Infinite Surds
In this mathematics portfolio we are instructed to investigate different expression of infinite surds in square root form and then find the exact value and statement for these surds.
INFINITE SURDS
The following expression is an example of an infinite surd.
The first ten terms of the surd can be expressed in the sequence:
From the tem terms of the sequence we can observe that the formula for the sequence is displayed as:
The results had to be plotted in a graph as shown below:
The graph above shows us the relationship between n and. We can observe that as n increases, also increases but each time less than before, suggesting that at a large certain point of n it stops increasing and just follows in a straight line.
Middle
From the tem terms of the sequence we can observe that the formula for the sequence is displayed as:
The results had to be plotted in a graph as shown below:
The graph above also shows us the relationship between n and. We can observe that as n increases, also increases but each time less than before, suggesting that at a large certain point of n it stops increasing and just follows in a straight line.
To find the exact infinite value for this sequence we would use the equation and rearrange in the correct way to find the value.
Finding the exact value for the surd
The exact value for the surd is 2 as the second answer does not fit in the problem
The general infinite surd in terms of k
Conclusion
The results show us that the limitations found in the general statement are those that didn’t give an integer as a final answer. Therefore k has to be an integer, however not a decimal or a fraction as results wouldn’t be a whole number. It also isn’t possible to obtain an integer if k is a negative number because there is no square root for a negative number. For a pleasing result, 1+4k , needs to give a number that possesses a perfect square root, an integer.
I arrived at this general statement by subtracting the n term from the k, and like that finding out that the product of this subtraction is the n term to the power of 2. Like that I found the equation. By rearranging the formula to make k the subject I came to the conclusion that , the general statement.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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# Related International Baccalaureate Maths essays
1. ## IB Mathematics Portfolio - Modeling the amount of a drug in the bloodstream
I just assume the line would reach to the amount when it can't be measured and call it the zero mark. (b) Doses continue to be taken every 6 hours. If dose continues, then the graph would continue. The left over amount of drug in the body would continue get higher and higher.
2. ## Mathematics Portfolio. In this portfolio project, the task at hand is to investigate the ...
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1. ## Infinite Surds. The aim of this folio is to explore the nature of ...
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1. ## Lascap's Fraction Portfolio
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SSAT Upper Level Math : Perpendicular Lines
Example Questions
← Previous 1 3 4
Example Question #1 : How To Find Out If Lines Are Perpendicular
Two perpendicular lines intersect at the point . One line passes through point ; the other passes through point . Evaluate .
Explanation:
The line that passes through and has slope
.
The line that passes through and , being perpendicular to the first, has as its slope the opposite reciprocal of , or
Therefore, to find , we use the slope formula and solve for :
Example Question #2 : How To Find Out If Lines Are Perpendicular
A line has the following equation:
Which of the following could be a line that is perpendicular to this given line?
Explanation:
First, put the equation of the given line in the form to find its slope.
Since the slope of the given line is , the slope of the line that is perpendicular must be its negative reciprocal, .
Now, put each answer choice in form to see which one has a slope of .
Example Question #1 : How To Find Out If Lines Are Perpendicular
Which of the following lines is perpendicular to the line ?
Explanation:
Perpendicular lines will have slopes that are negative reciprocals of one another. Our first step will be to find the slope of the given line by putting the equation into slope-intercept form.
The slope of this line is . The negative reciprocal will be , which will be the slope of the perpendicular line.
Now we need to find the answer choice with this slope by converting to slope-intercept form.
This equation has a slope of , and must be our answer.
Example Question #2 : How To Find Out If Lines Are Perpendicular
Which of the following lines is perpindicular to
Explanation:
When determining if a two lines are perpindicular, we are only concerned about their slopes. Consider the basic equation of a line, , where m is the slope of the line. Two lines are perpindicular to each other if one slope is the negative and reciprocal of the other.
The first step of this problem is to get it into the form, , which is . Now we know that the slope, m, is . The reciprocal of that is , and the negative of that is . Therefore, any line that has a slope of will be perpindicular to the original line.
Example Question #1 : How To Find Out If Lines Are Perpendicular
Which of the following equations represents a line that is perpendicular to the line with points and ?
Explanation:
If lines are perpendicular, then their slopes will be negative reciprocals.
First, we need to find the slope of the given line.
Because we know that our given line's slope is , the slope of the line perpendicular to it must be .
Example Question #3 : How To Find Out If Lines Are Perpendicular
Which of the following lines is perpendicular to a line with a slope ?
Not enough information provided
Explanation:
For a given line with a slope , any perpendicular line would have a slope , or the negative reciprocal of .
Given that in this instance, we can conclude that the slope of a perpendicular line would be . Therefore, the equation that contains this slope is
Example Question #4 : How To Find Out If Lines Are Perpendicular
Which of the following lines would be perpendicular to ?
Not enough information provided to solve
Explanation:
For a given line with a slope , any perpendicular line would have a slope , or the negative reciprocal of .
Given that in this instance, we can conclude that the slope of a perpendicular line would be . Given the perpendicular slope, we can now conclude that the perpendicular line is .
Example Question #5 : How To Find Out If Lines Are Perpendicular
A given line has the equation . What is the slope of any line that is perpendicular to this line?
Explanation:
For a given line with a slope , any perpendicular line would have a slope , or the negative reciprocal of .
Given that in this instance, we can conclude that the slope of a perpendicular line would be .
Example Question #2 : Perpendicular Lines
The equation for one line is . What is the slope of the line that is perpendicular to this line?
Explanation:
A line is perpendicular to another if their slopes are negative reciprocals of each other.
Since the slope of the given line is , the negative reciprocal would be .
Example Question #1 : How To Find The Slope Of A Perpendicular Line
What is the slope of any line perpendicular to 2y = 4x +3 ?
2
½
– ½
– 4
– ½
Explanation:
First, we must solve the equation for y to determine the slope: y = 2x + 3/2
By looking at the coefficient in front of x, we know that the slope of this line has a value of 2. To fine the slope of any line perpendicular to this one, we take the negative reciprocal of it:
slope = m , perpendicular slope = – 1/m
slope = 2 , perpendicular slope = – 1/2
← Previous 1 3 4
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# What is 84 percent of 50?
WRITTEN BY: supportmymoto.com STAFF
#### Answer for What’s 84 p.c of 50:
84 p.c * 50 =
(84:100)* 50 =
(84* 50):100 =
4200:100 = 42
Now now we have: 84 p.c of 50 = 42
Query: What’s 84 p.c of 50?
Step 1: Our output worth is 50.
Step 2: We characterize the unknown worth with {x}.
Step 3: From step 1 above,{ 50}={100%}.
Step 4: Equally, {x}={84%}.
Step 5: This ends in a pair of easy equations:
{ 50}={100%}(1).
{x}={84%}(2).
Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand aspect) of each
equations have the identical unit (%); now we have
frac{ 50}{x}=frac{100%}{84%}
Step 7: Once more, the reciprocal of each side offers
frac{x}{ 50}=frac{84}{100}
Rightarrow{x} = {42}
Due to this fact, {84%} of { 50} is {42}
#### Answer for What’s 50 p.c of 84:
50 p.c *84 =
( 50:100)*84 =
( 50*84):100 =
4200:100 = 42
Now now we have: 50 p.c of 84 = 42
Query: What’s 50 p.c of 84?
Step 1: Our output worth is 84.
Step 2: We characterize the unknown worth with {x}.
Step 3: From step 1 above,{84}={100%}.
Step 4: Equally, {x}={ 50%}.
Step 5: This ends in a pair of easy equations:
{84}={100%}(1).
{x}={ 50%}(2).
Step 6: By dividing equation 1 by equation 2 and noting that each the RHS (proper hand aspect) of each
equations have the identical unit (%); now we have
frac{84}{x}=frac{100%}{ 50%}
Step 7: Once more, the reciprocal of each side offers
frac{x}{84}=frac{ 50}{100}
Rightarrow{x} = {42}
Due to this fact, { 50%} of {84} is {42}
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# Introduction to Progressions
Go back to 'Arithmetic-Geometric'
In our study of Mathematics, we will encounter various kinds of progressions, which are sequences of numbers with a definite relationship between successive numbers. Let us consider a few examples of progressions.
Consider the following sequence of numbers:
$- 3,\, - 1,\,\,1,\,\,3,\,\,5,\,...$
What do you notice about this sequence? The most important characteristic of this sequence is that the difference between successive terms is constant, and equal to 2. This characteristic enables us to say that the next term in this progression will be 7. This sequence is an example of an arithmetic progression. Here is another example of an arithmetic progression:
$17,\;14,\;11,\;8,\;5,...$
The difference between successive terms in this progression is –3, and the next term in the sequence will be 2.
Now, consider the following sequence of numbers:
$2,\,4,\,8,\,16,\,32,...$
In this sequence, each term is double the previous term. In other words, the ratio of successive terms is constant, and equal to 2. This sequence is an example of a geometric progression. In any geometric progression, the ratio of successive terms is constant. Here is another example:
$1,\,\frac{1}{2},\,\frac{1}{4},\,\frac{1}{8},\,\frac{1}{{16}},...$
Now, consider the following sequence:
$\frac{1}{3},\,\frac{1}{5},\,\frac{1}{7},\,\frac{1}{9},\,\frac{1}{{11}},...$
The important characteristic of this sequence is that the reciprocal of its terms are in arithmetic progression:
$3,\,\,5,\,\,7,\,\,9,\,\,11,...$
Such sequences, in which the reciprocal terms are in arithmetic progressions, are called harmonic progressions. Here is another example of a harmonic progression:
$\frac{2}{7},\,\,\frac{2}{{11}},\,\,\frac{2}{{15}},\,\,\frac{2}{{19}},\,\,\frac{2}{{23}},...$
Verify that its reciprocal terms are in arithmetic progression.
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# Which is not part of the circle?
## Which is not part of the circle?
Answer: The points within the hula hoop are not part of the circle and are called interior points. The distance between the midpoint and the circleborder is called the radius. A line segment that has the endpoints on the circle and passes through the midpoint is called the diameter.
### What are the parts of circles?
A circle can have different parts and based on the position and shape, these can be named as follows:
• Centre.
• Diameter.
• Circumference.
• Tangent.
• Secant.
• Chord.
• Arc.
#### What are the 7 parts of a circle?
The following figures show the different parts of a circle: tangent, chord, radius, diameter, minor arc, major arc, minor segment, major segment, minor sector, major sector.
What is circle in precalculus?
A circle is all points in a plane that are a fixed distance from a given point in the plane. The given point is called the center, (h,k) , and the fixed distance is called the radius, r , of the circle.
What is the inside of a circle called?
The distance from a point on the circle to its center is called the radius of the circle. The inside of a circle is the set of all points whose distance from the center is less than the radius. The distance from one side of a circle through the center to the other side is called the diameter of the circle.
## Is an arc part of a circle?
An arc is part of the circumference of a circle. If the arc is over half of the circumference then it is called a major arc. If it is less than half of the circumference it is called a minor arc. The diameter cuts the circle exactly in half and goes through the centre.
### When a chord divides a circle into two parts each part is called?
We know that chords divide the circle into two parts, each part is called a segment of the circle. The larger segment, that is, the segment with more area, also containing the centre of the circle is called major segment of the circle, and the smaller segment which has smaller area is called minor segment.
#### What are the parts of a circle in precalculus?
What do you call a line that does not touch the circle?
A line touching the circle at one single point is known as the tangent to the circle. In the last figure, the line does not touch the circle anywhere, therefore, it is known as a non- intersecting line. A line segment joining two different points on the circumference of a circle is called a chord of the circle.
What are the parts of a circle that touch one point?
A tangent is a straight line outside the circle that touches the circumference at one point only. A segment is the area enclosed by a chord and an arc (it looks similar to the segment of an orange…
## What are the different parts of a circle?
A chord is a segment that also has endpoints on the circle, but the line does not need to cross through the center. On the circle below BC and AC are chords. A diameter is a chord that passes through the center of the circle. A secant is a line that intersects with a circle at 2 different points.
### Which is a chord that passes through the center of a circle?
A diameter is a chord that passes through the center of the circle. A secant is a line that intersects with a circle at 2 different points. In the circle below, line E is a secant. A tangent is a line that intersects with the circle at one point.
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Question of Exercise 1
Question
Option 1 250°
Option 2 40°
Option 3 120°
Option 4 45°
Simplify the following expressions
(√(5) + √(2))2
Solution:
Explanation:
Which of the following rational numbers lies between 0 and - 1
A: 0
B: - 1
C: -1/4
D: 1/4
Solution:
Explanation:
0 and 1 cannot be found between 0 and 1.
In addition,0= o/4 and -1= -4/4
We can see that -1/4 is halfway between 0 and -1.
Hence, the correct option is (c) -1/4
Prove that the diagonals of a parallelogram bisect each other
Solution:
Explanation:
We must show that the diagonals of the parallelogram ABCD cross each other.
OA = OC & OB = OD, in other words.
Now AD = BC [opposite sides are equal] in ΔAOD and ΔBOC.
[alternative interior angle] ∠ADO = ∠CBO in ΔAOD and ΔBOC.
Similarly, ∠AOD = ∠BOC by ΔDAO = ΔBCO (ASA rule)
As a result, OA = OC and OB = OB [according to CPCT].
Hence, it is prove that the diagonals of a parallelogram bisect each other.
What is total surface area of sphere
Solution:
Explanation:
• The radius of the sphere affects the formula for calculating the sphere's surface area.
• If the sphere's radius is r and the sphere's surface area is S.
• The sphere's surface area is therefore stated as Surface Area of Sphere 4πr2, where ‘r’ is the sphere's radius.
• The surface area of a sphere is expressed in terms of diameter as S=4π(d/2)2, where d is the sphere's diameter.
Thus, total surface area of sphere is =4πr2.
Fill in the blanks If two adjacent angles are supplementary
they form a __________.
Solution:
Explanation:
• If the non-common sides of two angles form a straight line, they are called linear pair angles.
• The sum of the angles of two linear pairs is degrees.
• If the total of two angles is degrees, they are called supplementary angles.
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Intermediate Algebra 2e
# 3.1Graph Linear Equations in Two Variables
Intermediate Algebra 2e3.1 Graph Linear Equations in Two Variables
## Learning Objectives
By the end of this section, you will be able to:
• Plot points in a rectangular coordinate system
• Graph a linear equation by plotting points
• Graph vertical and horizontal lines
• Find the x- and y-intercepts
• Graph a line using the intercepts
## Be Prepared 3.1
Before you get started, take this readiness quiz.
Evaluate $5x−45x−4$ when $x=−1.x=−1.$
If you missed this problem, review Example 1.6.
## Be Prepared 3.2
Evaluate $3x−2y3x−2y$ when $x=4,y=−3.x=4,y=−3.$
If you missed this problem, review Example 1.21.
## Be Prepared 3.3
Solve for y: $8−3y=20.8−3y=20.$
If you missed this problem, review Example 2.2.
## Plot Points on a Rectangular Coordinate System
Just like maps use a grid system to identify locations, a grid system is used in algebra to show a relationship between two variables in a rectangular coordinate system. The rectangular coordinate system is also called the xy-plane or the “coordinate plane.”
The rectangular coordinate system is formed by two intersecting number lines, one horizontal and one vertical. The horizontal number line is called the x-axis. The vertical number line is called the y-axis. These axes divide a plane into four regions, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise. See Figure 3.2.
Figure 3.2
In the rectangular coordinate system, every point is represented by an ordered pair. The first number in the ordered pair is the x-coordinate of the point, and the second number is the y-coordinate of the point. The phrase “ordered pair” means that the order is important.
## Ordered Pair
An ordered pair, $(x,y)(x,y)$ gives the coordinates of a point in a rectangular coordinate system. The first number is the x-coordinate. The second number is the y-coordinate.
What is the ordered pair of the point where the axes cross? At that point both coordinates are zero, so its ordered pair is $(0,0).(0,0).$ The point $(0,0)(0,0)$ has a special name. It is called the origin.
## The Origin
The point $(0,0)(0,0)$ is called the origin. It is the point where the x-axis and y-axis intersect.
We use the coordinates to locate a point on the xy-plane. Let’s plot the point $(1,3)(1,3)$ as an example. First, locate 1 on the x-axis and lightly sketch a vertical line through $x=1.x=1.$ Then, locate 3 on the y-axis and sketch a horizontal line through $y=3.y=3.$ Now, find the point where these two lines meet—that is the point with coordinates $(1,3).(1,3).$ See Figure 3.3.
Figure 3.3
Notice that the vertical line through $x=1x=1$ and the horizontal line through $y=3y=3$ are not part of the graph. We just used them to help us locate the point $(1,3).(1,3).$
When one of the coordinate is zero, the point lies on one of the axes. In Figure 3.4 the point $(0,4)(0,4)$ is on the y-axis and the point $(−2,0)(−2,0)$ is on the x-axis.
Figure 3.4
## Points on the Axes
Points with a y-coordinate equal to 0 are on the x-axis, and have coordinates $(a,0).(a,0).$
Points with an x-coordinate equal to 0 are on the y-axis, and have coordinates $(0,b).(0,b).$
## Example 3.1
Plot each point in the rectangular coordinate system and identify the quadrant in which the point is located:
$(−5,4)(−5,4)$ $(−3,−4)(−3,−4)$ $(2,−3)(2,−3)$ $(0,−1)(0,−1)$ $(3,52).(3,52).$
## Try It 3.1
Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:
$(−2,1)(−2,1)$ $(−3,−1)(−3,−1)$ $(4,−4)(4,−4)$ $(−4,4)(−4,4)$ $(−4,32)(−4,32)$
## Try It 3.2
Plot each point in a rectangular coordinate system and identify the quadrant in which the point is located:
$(−4,1)(−4,1)$ $(−2,3)(−2,3)$ $(2,−5)(2,−5)$ $(−2,5)(−2,5)$ $(−3,52)(−3,52)$
The signs of the x-coordinate and y-coordinate affect the location of the points. You may have noticed some patterns as you graphed the points in the previous example. We can summarize sign patterns of the quadrants in this way:
$Quadrant IQuadrant IIQuadrant IIIQuadrant IV (x,y)(x,y)(x,y)(x,y) (+,+)(−,+)(−,−)(+,−) Quadrant IQuadrant IIQuadrant IIIQuadrant IV (x,y)(x,y)(x,y)(x,y) (+,+)(−,+)(−,−)(+,−)$
Up to now, all the equations you have solved were equations with just one variable. In almost every case, when you solved the equation you got exactly one solution. But equations can have more than one variable. Equations with two variables may be of the form $Ax+By=C.Ax+By=C.$ An equation of this form is called a linear equation in two variables.
## Linear Equation
An equation of the form $Ax+By=C,Ax+By=C,$ where A and B are not both zero, is called a linear equation in two variables.
Here is an example of a linear equation in two variables, x and y.
The equation $y=−3x+5y=−3x+5$ is also a linear equation. But it does not appear to be in the form $Ax+By=C.Ax+By=C.$ We can use the Addition Property of Equality and rewrite it in $Ax+By=CAx+By=C$ form.
$y=−3x+5 Add to both sides.y+3x=−3x+5+3x Simplify.y+3x=5 Use the Commutative Property to put it in Ax+By=Cform.3x+y=5 y=−3x+5 Add to both sides.y+3x=−3x+5+3x Simplify.y+3x=5 Use the Commutative Property to put it in Ax+By=Cform.3x+y=5$
By rewriting $y=−3x+5y=−3x+5$ as $3x+y=5,3x+y=5,$ we can easily see that it is a linear equation in two variables because it is of the form $Ax+By=C.Ax+By=C.$ When an equation is in the form $Ax+By=C,Ax+By=C,$ we say it is in standard form of a linear equation.
## Standard Form of Linear Equation
A linear equation is in standard form when it is written $Ax+By=C.Ax+By=C.$
Most people prefer to have A, B, and C be integers and $A≥0A≥0$ when writing a linear equation in standard form, although it is not strictly necessary.
Linear equations have infinitely many solutions. For every number that is substituted for x there is a corresponding y value. This pair of values is a solution to the linear equation and is represented by the ordered pair $(x,y).(x,y).$ When we substitute these values of x and y into the equation, the result is a true statement, because the value on the left side is equal to the value on the right side.
## Solution of a Linear Equation in Two Variables
An ordered pair $(x,y)(x,y)$ is a solution of the linear equation$Ax+By=C,Ax+By=C,$ if the equation is a true statement when the x- and y-values of the ordered pair are substituted into the equation.
Linear equations have infinitely many solutions. We can plot these solutions in the rectangular coordinate system. The points will line up perfectly in a straight line. We connect the points with a straight line to get the graph of the equation. We put arrows on the ends of each side of the line to indicate that the line continues in both directions.
A graph is a visual representation of all the solutions of the equation. It is an example of the saying, “A picture is worth a thousand words.” The line shows you all the solutions to that equation. Every point on the line is a solution of the equation. And, every solution of this equation is on this line. This line is called the graph of the equation. Points not on the line are not solutions!
## Graph of a Linear Equation
The graph of a linear equation $Ax+By=CAx+By=C$ is a straight line.
• Every point on the line is a solution of the equation.
• Every solution of this equation is a point on this line.
## Example 3.2
The graph of $y=2x−3y=2x−3$ is shown.
For each ordered pair, decide:
Is the ordered pair a solution to the equation?
Is the point on the line?
A: $(0,−3)(0,−3)$ B: $(3,3)(3,3)$ C: $(2,−3)(2,−3)$ D: $(−1,−5)(−1,−5)$
## Try It 3.3
Use graph of $y=3x−1.y=3x−1.$ For each ordered pair, decide:
Is the ordered pair a solution to the equation?
Is the point on the line?
A $(0,−1)(0,−1)$ B $(2,5)(2,5)$
## Try It 3.4
Use graph of $y=3x−1.y=3x−1.$ For each ordered pair, decide:
Is the ordered pair a solution to the equation?
Is the point on the line?
A$(3,−1)(3,−1)$ B$(−1,−4)(−1,−4)$
## Graph a Linear Equation by Plotting Points
There are several methods that can be used to graph a linear equation. The first method we will use is called plotting points, or the Point-Plotting Method. We find three points whose coordinates are solutions to the equation and then plot them in a rectangular coordinate system. By connecting these points in a line, we have the graph of the linear equation.
## Example 3.3
### How to Graph a Linear Equation by Plotting Points
Graph the equation $y=2x+1y=2x+1$ by plotting points.
## Try It 3.5
Graph the equation by plotting points: $y=2x−3.y=2x−3.$
## Try It 3.6
Graph the equation by plotting points: $y=−2x+4.y=−2x+4.$
The steps to take when graphing a linear equation by plotting points are summarized here.
## How To
### Graph a linear equation by plotting points.
1. Step 1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
2. Step 2. Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
3. Step 3. Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.
It is true that it only takes two points to determine a line, but it is a good habit to use three points. If you only plot two points and one of them is incorrect, you can still draw a line but it will not represent the solutions to the equation. It will be the wrong line.
If you use three points, and one is incorrect, the points will not line up. This tells you something is wrong and you need to check your work. Look at the difference between these illustrations.
When an equation includes a fraction as the coefficient of $xx$, we can still substitute any numbers for x. But the arithmetic is easier if we make “good” choices for the values of x. This way we will avoid fractional answers, which are hard to graph precisely.
## Example 3.4
Graph the equation: $y=12x+3.y=12x+3.$
## Try It 3.7
Graph the equation: $y=13x−1.y=13x−1.$
## Try It 3.8
Graph the equation: $y=14x+2.y=14x+2.$
## Graph Vertical and Horizontal Lines
Some linear equations have only one variable. They may have just x and no y, or just y without an x. This changes how we make a table of values to get the points to plot.
Let’s consider the equation $x=−3.x=−3.$ This equation has only one variable, x. The equation says that x is always equal to$−3,−3,$ so its value does not depend on y. No matter what is the value of y, the value of x is always $−3.−3.$
So to make a table of values, write $−3−3$ in for all the x-values. Then choose any values for y. Since x does not depend on y, you can choose any numbers you like. But to fit the points on our coordinate graph, we’ll use 1, 2, and 3 for the y-coordinates. See Table 3.2.
$x=−3x=−3$ x y $(x,y)(x,y)$ $−3−3$ 1 $(−3,1)(−3,1)$ $−3−3$ 2 $(−3,2)(−3,2)$ $−3−3$ 3 $(−3,3)(−3,3)$
Table 3.2
Plot the points from the table and connect them with a straight line. Notice that we have graphed a vertical line.
What if the equation has y but no x? Let’s graph the equation $y=4.y=4.$ This time the y-value is a constant, so in this equation, y does not depend on x. Fill in 4 for all the y’s in Table 3.3 and then choose any values for x. We’ll use 0, 2, and 4 for the x-coordinates.
$y=4y=4$ x y $(x,y)(x,y)$ 0 4 $(0,4)(0,4)$ 2 4 $(2,4)(2,4)$ 4 4 $(4,4)(4,4)$
Table 3.3
In this figure, we have graphed a horizontal line passing through the y-axis at 4.
## Vertical and Horizontal Lines
A vertical line is the graph of an equation of the form $x=a.x=a.$
The line passes through the x-axis at $(a,0).(a,0).$
A horizontal line is the graph of an equation of the form $y=b.y=b.$
The line passes through the y-axis at $(0,b).(0,b).$
## Example 3.5
Graph: $x=2x=2$ $y=−1.y=−1.$
## Try It 3.9
Graph the equations: $x=5x=5$ $y=−4.y=−4.$
## Try It 3.10
Graph the equations: $x=−2x=−2$ $y=3.y=3.$
What is the difference between the equations $y=4xy=4x$ and $y=4?y=4?$
The equation $y=4xy=4x$ has both x and y. The value of y depends on the value of x, so the y -coordinate changes according to the value of x. The equation $y=4y=4$ has only one variable. The value of y is constant, it does not depend on the value of x, so the y-coordinate is always 4.
Notice, in the graph, the equation $y=4xy=4x$ gives a slanted line, while $y=4y=4$ gives a horizontal line.
## Example 3.6
Graph $y=−3xy=−3x$ and $y=−3y=−3$ in the same rectangular coordinate system.
## Try It 3.11
Graph the equations in the same rectangular coordinate system: $y=−4xy=−4x$ and $y=−4.y=−4.$
## Try It 3.12
Graph the equations in the same rectangular coordinate system: $y=3y=3$ and $y=3x.y=3x.$
## Find x- and y-intercepts
Every linear equation can be represented by a unique line that shows all the solutions of the equation. We have seen that when graphing a line by plotting points, you can use any three solutions to graph. This means that two people graphing the line might use different sets of three points.
At first glance, their two lines might not appear to be the same, since they would have different points labeled. But if all the work was done correctly, the lines should be exactly the same. One way to recognize that they are indeed the same line is to look at where the line crosses the x-axis and the y-axis. These points are called the intercepts of a line.
## Intercepts of a Line
The points where a line crosses the x-axis and the y-axis are called the intercepts of the line.
Let’s look at the graphs of the lines.
First, notice where each of these lines crosses the x-axis. See Table 3.4.
Now, let’s look at the points where these lines cross the y-axis.
Figure The line crosses
the x-axis at:
Ordered pair
for this point
The line crosses
the y-axis at:
Ordered pair
for this point
Figure (a) 3 $(3,0)(3,0)$ 6 $(0,6)(0,6)$
Figure (b) 4 $(4,0)(4,0)$ $−3−3$ $(0,−3)(0,−3)$
Figure (c) 5 $(5,0)(5,0)$ $−5−5$ $(0,5)(0,5)$
Figure (d) 0 $(0,0)(0,0)$ 0 $(0,0)(0,0)$
General Figure a $(a,0)(a,0)$ b $(0,b)(0,b)$
Table 3.4
Do you see a pattern?
For each line, the y-coordinate of the point where the line crosses the x-axis is zero. The point where the line crosses the x-axis has the form $(a,0)(a,0)$ and is called the x-intercept of the line. The x-intercept occurs when y is zero.
In each line, the x-coordinate of the point where the line crosses the y-axis is zero. The point where the line crosses the y-axis has the form $(0,b)(0,b)$ and is called the y-intercept of the line. The y-intercept occurs when x is zero.
## x-intercept and y-intercept of a Line
The x-intercept is the point $(a,0)(a,0)$ where the line crosses the x-axis.
The y-intercept is the point $(0,b)(0,b)$ where the line crosses the y-axis.
## Example 3.7
Find the x- and y-intercepts on each graph shown.
## Try It 3.13
Find the x- and y-intercepts on the graph.
## Try It 3.14
Find the x- and y-intercepts on the graph.
Recognizing that the x-intercept occurs when y is zero and that the y-intercept occurs when x is zero, gives us a method to find the intercepts of a line from its equation. To find the x-intercept, let $y=0y=0$ and solve for x. To find the y-intercept, let $x=0x=0$ and solve for y.
## Find the x- and y-intercepts from the Equation of a Line
Use the equation of the line. To find:
• the x-intercept of the line, let $y=0y=0$ and solve for x.
• the y-intercept of the line, let $x=0x=0$ and solve for y.
## Example 3.8
Find the intercepts of $2x+y=8.2x+y=8.$
## Try It 3.15
Find the intercepts: $3x+y=12.3x+y=12.$
## Try It 3.16
Find the intercepts: $x+4y=8.x+4y=8.$
## Graph a Line Using the Intercepts
To graph a linear equation by plotting points, you need to find three points whose coordinates are solutions to the equation. You can use the x- and y- intercepts as two of your three points. Find the intercepts, and then find a third point to ensure accuracy. Make sure the points line up—then draw the line. This method is often the quickest way to graph a line.
## Example 3.9
### How to Graph a Line Using the Intercepts
Graph $–x+2y=6–x+2y=6$ using the intercepts.
## Try It 3.17
Graph using the intercepts: $x–2y=4.x–2y=4.$
## Try It 3.18
Graph using the intercepts: $–x+3y=6.–x+3y=6.$
The steps to graph a linear equation using the intercepts are summarized here.
## How To
### Graph a linear equation using the intercepts.
1. Step 1.
Find the x- and y-intercepts of the line.
• Let $y=0y=0$ and solve for x.
• Let $x=0x=0$ and solve for y.
2. Step 2. Find a third solution to the equation.
3. Step 3. Plot the three points and check that they line up.
4. Step 4. Draw the line.
## Example 3.10
Graph $4x−3y=124x−3y=12$ using the intercepts.
## Try It 3.19
Graph using the intercepts: $5x−2y=10.5x−2y=10.$
## Try It 3.20
Graph using the intercepts: $3x−4y=12.3x−4y=12.$
When the line passes through the origin, the x-intercept and the y-intercept are the same point.
## Example 3.11
Graph $y=5xy=5x$ using the intercepts.
## Try It 3.21
Graph using the intercepts: $y=4x.y=4x.$
## Try It 3.22
Graph the intercepts: $y=−x.y=−x.$
## Section 3.1 Exercises
### Practice Makes Perfect
Plot Points in a Rectangular Coordinate System
In the following exercises, plot each point in a rectangular coordinate system and identify the quadrant in which the point is located.
1.
$(−4,2)(−4,2)$ $(−1,−2)(−1,−2)$ $(3,−5)(3,−5)$ $(−3,0)(−3,0)$ $(53,2)(53,2)$
2.
$(−2,−3)(−2,−3)$ $(3,−3)(3,−3)$ $(−4,1)(−4,1)$ $(4,−1)(4,−1)$ $(32,1)(32,1)$
3.
$(3,−1)(3,−1)$ $(−3,1)(−3,1)$ $(−2,0)(−2,0)$ $(−4,−3)(−4,−3)$ $(1,145)(1,145)$
4.
$(−1,1)(−1,1)$ $(−2,−1)(−2,−1)$ $(2,0)(2,0)$ $(1,−4)(1,−4)$ $(3,72)(3,72)$
In the following exercises, for each ordered pair, decide
is the ordered pair a solution to the equation? is the point on the line?
5.
$y=x+2;y=x+2;$
A: $(0,2);(0,2);$ B: $(1,2);(1,2);$ C: $(−1,1);(−1,1);$ D: $(−3,−1).(−3,−1).$
6.
$y=x−4;y=x−4;$
A: $(0,−4);(0,−4);$ B: $(3,−1);(3,−1);$ C: $(2,2);(2,2);$ D: $(1,−5).(1,−5).$
7.
$y=12x−3;y=12x−3;$
A: $(0,−3);(0,−3);$ B: $(2,−2);(2,−2);$ C: $(−2,−4);(−2,−4);$ D: $(4,1)(4,1)$
8.
$y=13x+2;y=13x+2;$
A: $(0,2);(0,2);$ B: $(3,3);(3,3);$ C: $(−3,2);(−3,2);$ D: $(−6,0).(−6,0).$
Graph a Linear Equation by Plotting Points
In the following exercises, graph by plotting points.
9.
$y = x + 2 y = x + 2$
10.
$y = x − 3 y = x − 3$
11.
$y = 3 x − 1 y = 3 x − 1$
12.
$y = −2 x + 2 y = −2 x + 2$
13.
$y = − x − 3 y = − x − 3$
14.
$y = − x − 2 y = − x − 2$
15.
$y = 2 x y = 2 x$
16.
$y = −2 x y = −2 x$
17.
$y = 1 2 x + 2 y = 1 2 x + 2$
18.
$y = 1 3 x − 1 y = 1 3 x − 1$
19.
$y = 4 3 x − 5 y = 4 3 x − 5$
20.
$y = 3 2 x − 3 y = 3 2 x − 3$
21.
$y = − 2 5 x + 1 y = − 2 5 x + 1$
22.
$y = − 4 5 x − 1 y = − 4 5 x − 1$
23.
$y = − 3 2 x + 2 y = − 3 2 x + 2$
24.
$y = − 5 3 x + 4 y = − 5 3 x + 4$
Graph Vertical and Horizontal lines
In the following exercises, graph each equation.
25.
$x=4x=4$ $y=3y=3$
26.
$x=3x=3$ $y=1y=1$
27.
$x=−2x=−2$ $y=−5y=−5$
28.
$x=−5x=−5$ $y=−2y=−2$
In the following exercises, graph each pair of equations in the same rectangular coordinate system.
29.
$y=2xy=2x$ and $y=2y=2$
30.
$y=5xy=5x$ and $y=5y=5$
31.
$y=−12xy=−12x$ and $y=−12y=−12$
32.
$y=−13xy=−13x$ and $y=−13y=−13$
Find x- and y-Intercepts
In the following exercises, find the x- and y-intercepts on each graph.
33.
34.
35.
36.
In the following exercises, find the intercepts for each equation.
37.
$x − y = 5 x − y = 5$
38.
$x − y = −4 x − y = −4$
39.
$3 x + y = 6 3 x + y = 6$
40.
$x − 2 y = 8 x − 2 y = 8$
41.
$4 x − y = 8 4 x − y = 8$
42.
$5 x − y = 5 5 x − y = 5$
43.
$2 x + 5 y = 10 2 x + 5 y = 10$
44.
$3 x − 2 y = 12 3 x − 2 y = 12$
Graph a Line Using the Intercepts
In the following exercises, graph using the intercepts.
45.
$− x + 4 y = 8 − x + 4 y = 8$
46.
$x + 2 y = 4 x + 2 y = 4$
47.
$x + y = −3 x + y = −3$
48.
$x − y = −4 x − y = −4$
49.
$4 x + y = 4 4 x + y = 4$
50.
$3 x + y = 3 3 x + y = 3$
51.
$3 x − y = −6 3 x − y = −6$
52.
$2 x − y = −8 2 x − y = −8$
53.
$2 x + 4 y = 12 2 x + 4 y = 12$
54.
$3 x − 2 y = 6 3 x − 2 y = 6$
55.
$2 x − 5 y = −20 2 x − 5 y = −20$
56.
$3 x − 4 y = −12 3 x − 4 y = −12$
57.
$y = −2 x y = −2 x$
58.
$y = 5 x y = 5 x$
59.
$y = x y = x$
60.
$y = − x y = − x$
Mixed Practice
In the following exercises, graph each equation.
61.
$y = 3 2 x y = 3 2 x$
62.
$y = − 2 3 x y = − 2 3 x$
63.
$y = − 1 2 x + 3 y = − 1 2 x + 3$
64.
$y = 1 4 x − 2 y = 1 4 x − 2$
65.
$4 x + y = 2 4 x + y = 2$
66.
$5 x + 2 y = 10 5 x + 2 y = 10$
67.
$y = −1 y = −1$
68.
$x = 3 x = 3$
### Writing Exercises
69.
Explain how you would choose three x-values to make a table to graph the line $y=15x−2.y=15x−2.$
70.
What is the difference between the equations of a vertical and a horizontal line?
71.
Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation $4x+y=−4?4x+y=−4?$ Why?
72.
Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation $y=23x−2?y=23x−2?$ Why?
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
If most of your checks were:
Confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
With some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
No, I don’t get it. This is a warning sign and you must address it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
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# Lesson 9
Drawing Triangles (Part 1)
Let’s see how many different triangles we can draw with certain measurements.
### 9.1: Which One Doesn’t Belong: Triangles
Which one doesn’t belong?
### 9.2: Does Your Triangle Match Theirs?
Three students have each drawn a triangle. For each description:
1. Drag the vertices to create a triangle with the given measurements.
2. Make note of the different side lengths and angle measures in your triangle.
3. Decide whether the triangle you made must be an identical copy of the triangle that the student drew. Explain your reasoning.
Jada’s triangle has one angle measuring 75°.
Andre’s triangle has one angle measuring 75° and one angle measuring 45°.
Lin’s triangle has one angle measuring 75°, one angle measuring 45°, and one side measuring 5 cm.
### 9.3: How Many Can You Draw?
1. Draw as many different triangles as you can with each of these sets of measurements:
1. Two angles measure $$60^\circ$$, and one side measures 4 cm.
2. Two angles measure $$90^\circ$$, and one side measures 4 cm.
3. One angle measures $$60^\circ$$, one angle measures $$90^\circ$$, and one side measures 4 cm.
2. Which sets of measurements determine one unique triangle? Explain or show your reasoning.
In the diagram, 9 toothpicks are used to make three equilateral triangles. Figure out a way to move only 3 of the toothpicks so that the diagram has exactly 5 equilateral triangles.
### Summary
Sometimes, we are given two different angle measures and a side length, and it is impossible to draw a triangle. For example, there is no triangle with side length 2 and angle measures $$120^\circ$$ and $$100^\circ$$:
Sometimes, we are given two different angle measures and a side length between them, and we can draw a unique triangle. For example, if we draw a triangle with a side length of 4 between angles $$90^\circ$$ and $$60^\circ$$, there is only one way they can meet up and complete to a triangle:
Any triangle drawn with these three conditions will be identical to the one above, with the same side lengths and same angle measures.
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## Algebra 1
Published by Prentice Hall
# Chapter 2 - Solving Equations - 2-4 Solving Equations with Variables on Both Sides - Practice and Problem-Solving Exercises - Page 106: 22
b = 7
#### Work Step by Step
8 - (3 + b) = b - 9 Use the distributive property: 8 - 1(3) + -1(b) = b - 9 Simplify: 8 - 3 - b = b - 9 To get the variable (b) on one side of the equation, add b to the right side of the equation. This also removes b from the left side of the equation. 8 - 3 - b + b = b - 9 + b Simplify to get: 8 - 3 = 2b - 9 Add 9 to both sides of the equation. 8 - 3 + 9 = 2b - 9 + 9 Simplify to get: 14 = 2b Divide both sides by 2: $\frac{14}{2}$ = $\frac{2b}{2}$ 7 = b Check: To check whether 7 is the correct answer, replace b with 7 in the original equation and solve. 8 - (3 + b) = b - 9 8 - (3 + 7) = 7 - 9 8 - 3 - 7 = 7 - 9 -2 = -2
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# What Is The Slope-Intercept Form Of The Equation Of The Line Shown In The Graph?
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
What Is The Slope-Intercept Form Of The Equation Of The Line Shown In The Graph? – There are many forms employed to represent a linear equation, among the ones most commonly seen is the slope intercept form. It is possible to use the formula of the slope-intercept to determine a line equation, assuming that you have the straight line’s slope , and the yintercept, which is the y-coordinate of the point at the y-axis intersects the line. Learn more about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three main forms of linear equations: standard, slope-intercept, and point-slope. While they all provide the same results when utilized but you are able to extract the information line that is produced more quickly through the slope intercept form. It is a form that, as the name suggests, this form utilizes a sloped line in which its “steepness” of the line is a reflection of its worth.
This formula can be utilized to find the slope of straight lines, the y-intercept (also known as the x-intercept), where you can utilize a variety formulas available. The equation for a line using this specific formula is y = mx + b. The slope of the straight line is symbolized through “m”, while its intersection with the y is symbolized through “b”. Each point of the straight line is represented with an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” are treated as variables.
## An Example of Applied Slope Intercept Form in Problems
For the everyday world, the slope intercept form is used frequently to represent how an item or problem changes in the course of time. The value of the vertical axis represents how the equation handles the extent of changes over the value provided by the horizontal axis (typically time).
An easy example of the application of this formula is to find out how much population growth occurs in a specific area as time passes. Using the assumption that the area’s population grows annually by a specific fixed amount, the value of the horizontal axis will grow by one point with each passing year and the values of the vertical axis will rise to reflect the increasing population by the set amount.
You can also note the starting point of a question. The beginning value is at the y-value in the y-intercept. The Y-intercept represents the point at which x equals zero. Based on the example of a previous problem the beginning value will be at the time the population reading starts or when the time tracking starts, as well as the related changes.
So, the y-intercept is the location that the population begins to be recorded for research. Let’s suppose that the researcher is beginning to do the calculation or measurement in the year 1995. In this case, 1995 will represent”the “base” year, and the x = 0 points will be observed in 1995. So, it is possible to say that the 1995 population is the y-intercept.
Linear equation problems that use straight-line formulas are almost always solved in this manner. The beginning value is represented by the yintercept and the change rate is represented by the slope. The primary complication of this form typically lies in the interpretation of horizontal variables particularly when the variable is attributed to a specific year (or any other kind or unit). The most important thing to do is to ensure that you know the meaning of the variables.
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# How do you factor 125x^3-27?
May 9, 2018
$\left(5 x - 3\right) \left(25 {x}^{2} + 15 x + 9\right)$
#### Explanation:
$125 {x}^{3} - 27 \text{ is a "color(blue)"difference of cubes}$
$\text{and factors in general as}$
•color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2)
$125 {x}^{3} = {\left(5 x\right)}^{3} \Rightarrow a = 5 x$
$27 = {\left(3\right)}^{3} \Rightarrow b = 3$
$\Rightarrow 125 {x}^{3} - 27 = \left(5 x - 3\right) \left({\left(5 x\right)}^{2} + \left(5 x \times 3\right) + {3}^{2}\right)$
$\textcolor{w h i t e}{\times \times \times \times \times} = \left(5 x - 3\right) \left(25 {x}^{2} + 15 x + 9\right)$
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# Lesson: Fractions: Adding and Subtracting Like Fractions
Tara Smith E.l. Haynes Pcs Washington, DC
9195 Views
119 Downloads
15 Favorites
### Lesson Objective
SWBAT find the sum or difference of fractions with like denominators.
### Lesson Plan
Materials Needed: scrap paper for DN and GP, Example Like Denominator Chart, white board, dry erase markers, pencils, and IND Worksheet (two sided).
Vocabulary: whole (or ONE or unit), denominator, numerator, equivalent, sum, difference.
……….
Do Now (3 -5 min): The teacher writes the following fractions on the board and has the students write two equivalent fractions for each.
3 = 4= 3 =
5 7 4
Opening (2 -3 min): Teacher quickly reviews answers to the Do Now. The teacher should then state the objective, “Yesterday, we found the whole of a fraction. Today, we are going to add and subtract like fractions. By the end of this lesson, you will be able to find the sum or difference of fractions with like denominators.”
Direct Instruction (8 – 10 min): The teacher has a chart that reads, “Fractions that have the same denominator are called like fractions. Fractions that have a different denominator are called unlike fractions.” The teacher reads the two sentences and then says, “Today we will be working with like fractions, so fractions that have the same denominator. I think most of you will find this to be very quick to pick up. There is only one rule that you have to remember. We never add the denominator because when we are adding like fractions we are working toward the whole!”
The teacher should then demonstrate using the first three problems on the Example Like Denominator Chart. The teacher should review counting out the whole to find the denominator of the fraction and the shaded part to find the numerator. The teacher should also call attention to the operation being used.
Guided Practice (10 min): The teacher uses more problems from the Example Like Denominator Chart to explain the concept of adding or subtracting fractions with like denominators.
See Example Like Denominator Chart
Independent (10 min): The teacher passes out the IND worksheet. The students should only complete both sides during the IND practice.
Closing (2-3 min): Teacher calls the attention of the students back toward the front of the class to quickly review the answers to the Independent Practice worksheet/ ask what we learned about.
### Lesson Resources
IND add fractions side 1 Classwork 2,251 Example Like Denominator Chart Exemplar 2,341 IND subtract fractions side 2 Classwork 1,180
### Comments
Matthew Lenard Posted 6 years ago:
Thanks for sharing Tara. This is a really helpful lesson.
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# What Are the First and Third Quartiles?
There are several ways to measure the center of a set of data. The mean, median, mode and midrange all have their advantages and limitations in expressing the middle of the data. Of all of these ways to find the average, the median is the most resistant to outliers. It marks the middle of the data in the sense that half of the data is less than the median.
There’s no reason we have to stop at finding just the middle. What if we decided to continue this process? We could calculate the median of the bottom half of our data. One half of 50% is 25%. Thus half of half, or one quarter, of the data would be below this. Since we are dealing with a quarter of the original set, this median of the bottom half of the data is called the first quartile, and is denoted by Q1.
### The Third Quartile
There is no reason why we looked at the bottom half of the data. Instead we could have looked at the top half and performed the same steps as above. The median of this half, which we will denote by Q3 also splits the data set into quarters. However, this number denotes the top one quarter of the data. Thus three quarters of the data is below our number Q3. This is why we call Q3 the third quartile (and this explains the 3 in the notation.
### An Example
To make this all clear, let’s look at an example. It may be helpful to first review how to calculate the median of some data. Start with the following data set:
1, 2, 2, 3, 4, 6, 6, 7, 7, 7, 8, 11, 12, 15, 15, 15, 17, 17, 18, 20
There are a total of twenty data points in the set. We begin by finding the median. Since there is an even number of data values, the median is the mean of the tenth and eleventh values. In other words, the median is:
(7 + 8)/2 = 7.5.
Now look at the bottom half of the data. The median of this half is found between the fifth and sixth values of:
1, 2, 2, 3, 4, 6, 6, 7, 7, 7
Thus the first quartile is found to equal Q1 = (4 + 6)/2 = 5
To find the third quartile, look at the top half of the original data set. We need to find the median of:
8, 11, 12, 15, 15, 15, 17, 17, 18, 20
Here the median is (15 + 15)/2 = 15. Thus the third quartile Q3 = 15.
### Interquartile Range and Five Number Summary
Quartiles help to give us a fuller picture of our data set as a whole. The first and third quartiles give us information about the internal structure of our data. The middle half of the data falls between the first and third quartiles, and is centered about the median. The difference between the first and third quartiles, called the interquartile range, shows how the data is arranged about the median. A small interquartile range indicates data that is clumped about the median. A larger interquartile range shows that the data is more spread out.
A more detailed picture of the data can be obtained by knowing the highest value, called the maximum value, and the lowest value, called the minimum value. The minimum, first quartile, median, third quartile and maximum are a set of five values called the five number summary. An effective way to display these five numbers is called a boxplot or box and whisker graph.
Courtney Taylor
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# What is the molar mass of a gas if 3.50 grams of the gas occupy 1.35 liters of space at STP?
Jun 18, 2018
The molar mass of the gas is $\text{58.9 g/mol}$.
#### Explanation:
You will need to use the equation for the ideal gas law:
$P V = n R T$,
where:
$P$ is pressure, $V$ is volume, $n$ is moles, $R$ is gas constant, and $T$ is temperature.
$\text{STP}$ is ${0}^{\circ} \text{C}$ or $\text{273.15 K}$ (required for gas laws), and ${10}^{5}$ $\text{Pa}$ or $\text{100 kPa}$.
Use the equation for the ideal gas law to calculate moles of gas. Then calculate the molar mass by dividing the given mass by the calculated moles.
Known
$P = \text{100 kPa}$
$V = \text{1.35 L}$
$R = \text{8.31446 L kPa K"^(-1) "mol"^(-1)}$
$T = \text{273.15 K}$
Unknown
$n$
Solution
Rearrange the ideal gas law equation to isolate $n$. Plug in the known values and solve.
$n = \frac{P V}{R T}$
n=(100color(red)(cancel(color(black)("kPa")))xx1.35color(red)(cancel(color(black)("L"))))/(8.31446 color(red)(cancel(color(black)("L"))) color(red)(cancel(color(black)("kPa"))) color(red)(cancel(color(black)("K")))^(-1) "mol"^(-1)xx273.15color(red)(cancel(color(black)("K"))))="0.0594 mol" (rounded to three significant figures)
To calculate the molar mass of the gas, divide its given mass in grams by the calculated number of moles.
$\text{molar mass"=("3.50 g")/("0.0594 mol")="58.9 g/mol}$
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BREAKING NEWS
Linearity of differentiation
## Summary
In calculus, the derivative of any linear combination of functions equals the same linear combination of the derivatives of the functions;[1] this property is known as linearity of differentiation, the rule of linearity,[2] or the superposition rule for differentiation.[3] It is a fundamental property of the derivative that encapsulates in a single rule two simpler rules of differentiation, the sum rule (the derivative of the sum of two functions is the sum of the derivatives) and the constant factor rule (the derivative of a constant multiple of a function is the same constant multiple of the derivative).[4][5] Thus it can be said that differentiation is linear, or the differential operator is a linear operator.[6]
## Statement and derivation
Let f and g be functions, with α and β constants. Now consider
${\displaystyle {\frac {\mbox{d}}{{\mbox{d}}x}}(\alpha \cdot f(x)+\beta \cdot g(x)).}$
By the sum rule in differentiation, this is
${\displaystyle {\frac {\mbox{d}}{{\mbox{d}}x}}(\alpha \cdot f(x))+{\frac {\mbox{d}}{{\mbox{d}}x}}(\beta \cdot g(x)),}$
and by the constant factor rule in differentiation, this reduces to
${\displaystyle \alpha \cdot f'(x)+\beta \cdot g'(x).}$
Therefore,
${\displaystyle {\frac {\mbox{d}}{{\mbox{d}}x}}(\alpha \cdot f(x)+\beta \cdot g(x))=\alpha \cdot f'(x)+\beta \cdot g'(x).}$
Omitting the brackets, this is often written as:
${\displaystyle (\alpha \cdot f+\beta \cdot g)'=\alpha \cdot f'+\beta \cdot g'.}$
## References
1. ^ Blank, Brian E.; Krantz, Steven George (2006), Calculus: Single Variable, Volume 1, Springer, p. 177, ISBN 9781931914598.
2. ^ Strang, Gilbert (1991), Calculus, Volume 1, SIAM, pp. 71–72, ISBN 9780961408824.
3. ^ Stroyan, K. D. (2014), Calculus Using Mathematica, Academic Press, p. 89, ISBN 9781483267975.
4. ^ Estep, Donald (2002), "20.1 Linear Combinations of Functions", Practical Analysis in One Variable, Undergraduate Texts in Mathematics, Springer, pp. 259–260, ISBN 9780387954844.
5. ^ Zorn, Paul (2010), Understanding Real Analysis, CRC Press, p. 184, ISBN 9781439894323.
6. ^ Gockenbach, Mark S. (2011), Finite-Dimensional Linear Algebra, Discrete Mathematics and Its Applications, CRC Press, p. 103, ISBN 9781439815649.
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Probability
Sample Space and Events of Experiments
Consider the experiment of throwing a dice. Any of the numbers 1, 2, 3, 4, 5, or 6 can come up on the upper face of the dice. We can easily find the probability of getting a number 5 on the upper face of the dice?
Mathematically, probability of any event E can be defined as follows.
Here, S represents the sample space and n(S) represents the number of outcomes in the sample space.
For this experiment, we have
Sample space (S) = {1, 2, 3, 4, 5, 6}. Thus, S is a finite set.
So, we can say that the possible outcomes of this experiment are 1, 2, 3, 4, 5, and 6.
Number of all possible outcomes = 6
Number of favourable outcomes of getting the number 5 = 1
Probability (getting 5)
Similarly, we can find the probability of getting other numbers also.
P (getting 1), P (getting 2), P (getting 3), P (getting 4) and
P (getting 6)
Let us add the probability of each separate observation.
This will give us the sum of the probabilities of all possible outcomes.
P (getting 1) + P (getting 2) + P (getting 3) + P (getting 4) + P (getting 5) + P (getting 6) = +++++ = 1
Sum of the probabilities of all elementary events is 1”.
Now, let us find the probability of not getting 5 on the upper face.
The outcomes favourable to this event are 1, 2, 3, 4, and 6.
Number of favourable outcomes = 5
P (not getting 5)
We can also see that P (getting 5) + P (not getting 5)
Sum of probabilities of occurrence and non occurrence of an event is 1”.
i.e. If E is the event, then P (E) + P (not E) = 1 … (1)
or we can write P(E) = 1 P (not E)
Here, the events of getting a number 5 and not getting 5 are complements of each other as we cannot find an observation which is common to the two observations.
Thus, event not E is the complement of event E. Complement of event E is denoted by or E'.
Using equation (1), we can write
P (E) + P () = 1
or
P () = 1 – P (E)
This is a very important property about the probability of complement of an event and it is stated as follows:
If E is an event of finite sample space S, then P () = 1 – P(E) where is the complement of event E
Now, let us prove this property algebraically.
Proof:
We have,
E= S and E =
n(E) = n(S) and n(E) = n()
n(E) = n(S) and n(E) = 0 ...(1)
Now,
n(E) = n(S)
n(E) + n() – n(E) = n(S)
n(E) + n() – 0 = n(S) [Using (1)]
n() = n(S) – n(E)
On dividing both sides by n(S), we get
P() = 1 – P(E)
Hence proved.
Let us solve some examples based on this concept.
ODDS (Ratio of two complementary probabilities):
Let n be the number of distinct sample points in the sample space S. Suppose, out of these n points, m points are favorable for the occurrence of event A. Hence, the remaining $n-m$ points are unfavorable for the occurrence of event A or we can say, $n-m$ points are favorable for the occurrence of event A'.
The ratio of favorable cases to the number of unfavorable cases is known as odds in the favor of event A which is given by $\frac{m}{n-m}$ i.e. P(A) : P(A').
The ratio of unfavorable cases to the number of favorable cases is known as odds against the favor of event A which is given by $\frac{n-m}{m}$ i.e. P(A') : P(A).
Example 1:
One card is drawn from a well shuffled deck. What is the probability that the card will be
(i) a king?
(ii) not a king?
Solution:
Let E be the event ‘the card is a king’ and F be the event ‘the card is not a king’.
(i) Since there are 4 kings in a deck.
Number of outcomes favourable to E = 4
Number of possible outcomes = 52
P (E)
1. Here, the events E and F are complements of each other.
P(E) + P(F) = 1
P(F) = 1 −
Example 2:
If the probability of an event A is 0.12 and B is 0.88 and they belong to the same set of observations, then show that A and B are complementary events.
Solution:
It is given that P (A) = 0.12 and P (B) = 0.88
Now, P(A) + P(B) = 0.12 + 0.88 = 1
The events A and B are complementary events.
Example 3:
Savita and Babita are playing badminton. The probability of Savita winning the match is 0.52. What is the proba…
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# What Is 60/80 as a Decimal + Solution With Free Steps
The fraction 60/80 as a decimal is equal to 0.75.
The fraction 60/80 is a proper fraction. The number on the top of the line is called the numerator. The number below the line is called the denominator. After applying the division method, the quotient and remainder are obtained.
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 60/80.
## 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 = 60
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 = 60 $\div$ 80
This is when we go through the Long Division solution to our problem. The following figure shows the solution for fraction 60/80.
Figure 1
## 60/80 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 60 and 80, we can see how 60 is Smaller than 80, and to solve this division, we require that 60 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 60, which after getting multiplied by 10 becomes 600.
We take this 600 and divide it by 80; this can be done as follows:
 600 $\div$ 80 $\approx$ 7
Where:
60 x 7 = 560
This will lead to the generation of a Remainder equal to 600 – 560 = 40. Now this means we have to repeat the process by Converting the 40 into 400 and solving for that:
400 $\div$ 80 = 5Â
Where:
80 x 5 = 400
Finally, we have a Quotient generated after combining the two pieces of it as 0.75, with a Remainder equal to 0.
Images/mathematical drawings are created with GeoGebra.
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# Identification and Writing of Equivalent Rates
## Understand that equivalent rates are equivalent ratios.
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Practice Identification and Writing of Equivalent Rates
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Use Unit Rates and Equivalent Rates
Credit: Jarrett Campbell
Myra’s junior soccer team scored 10 goals in the last 2 games. At this rate, how many goals will Myra’s team score in the next 5 games?
In this concept, you will learn to use unit rates and find equivalent rates.
### Guidance
A unit rate is a special kind of ratio, where the second number, or the denominator, is equal to one. With a unit rate, you are comparing a quantity to one. Some common unit rates are miles per gallon, price per pound, and pay rate per hour. To find a unit rate, simplify the ratio so that you have a 1 in the denominator. You can simply divide the first number in the ratio by the second. Make sure you keep track of the units.
Let’s look at an example.
Kayla bought 5.5 pounds of apples. She paid a total of $7.15. What was the unit rate of the apples per pound? First, write the ratio of the price to pounds. Next, divide the price by the pounds so you can find the unit rate. The answer is 1.3. The unit price for the apples is$1.30 per pound.
Let’s look at another example.
Brian worked for 8 hours yesterday and made a total of $86. What is his pay rate? First, write the ratio of the wages to hours. Next, divide the wages by the hours so you can find the unit pay rate. The answer is 10.75. Brian is paid$10.75 per hour.
Next, let’s look at equivalent rates. Equivalent rates are rates that are equal.
Let’s look at an example.
Find an equivalent rate for this comparison.
In order to find the equivalent rate for this unit rate, cross multiply.
The answer is 16.00. The rate is \begin{align*}\frac{\ 2.00}{1}\end{align*} is equivalent to the rate \begin{align*}\frac{\ 16.00}{8}\end{align*} . ### Guided Practice A store sells salmon for6.99 per pound. What is the rate for 6 pounds of salmon?
First, write the ratios from the question.
Next, cross multiply to find the cost of 6 pounds of salmon.
The answer is $41.94. Six pounds of salmon will cost$41.94.
### Examples
#### Example 1
Write find the unit rate for the following ratio.
First, divide the numerator by the denominator.
The unit rate is \begin{align*}\frac{2}{1}\end{align*}.
#### Example 2
Write find the unit rate for the following ratio.
First, divide the numerator by the denominator.
The unit rate is \begin{align*}\frac{3}{1}\end{align*}
#### Example 3
Write find the unit rate for the following ratio.
First, divide the numerator by the denominator.
The answer is 6. Therefore the unit rate is \begin{align*}\frac{6}{1}\end{align*} .
Credit: See-ming Lee
Remember Myra’s winning streak?
Myra’s team scores on 10 goals in the last 2 games. You want to find out how many goals, at this rate, they will score in the next 5 games.
First, write an equivalent ratio.
Next, cross multiply.
Then, divide by 2 to isolate \begin{align*}x\end{align*}.
In the next 5 games, Myra’s team should score 25 goals if the team keeps their current rate.
### Explore More
Use what you have learned about ratios to solve each problem.
In Kyle’s drawer, there are 14 pairs of white socks and 8 pairs of black socks
1. Write the ratio of black socks to white socks.
2. Write the ratio of black socks to total socks.
3. Write the ratio of white socks to total socks.
There are 150 apartments in the Gray building. Of these, 60 are rented and the rest are owned. There are 65 apartments in the Black building. Of these, 45 are rented and the rest are owned. Simplify each answer.
7. What is the ratio of rented to owned in the Gray building?
8. What is the ratio of rented to owned in the Black building?
9. Write the ratio of rented to total apartments in the Gray building.
10. Write the ratio of rented to total apartments in the Black building.
11. Write the ratio of owned to total apartments in the Gray building.
12. Write the ratio of owned to total apartments in the Black building.
13. Holly works at a library re-shelving books. She re-shelved 960 books in 4 hours. What is Holly’s rate of re-shelving in books per hour?
14. Sam bought 9.5 pounds of peaches to make a pie. The peaches cost \$15.39. What was the unit rate of the peaches?
15. Don can wrap 8 presents in an hour. What is Don’s rate for 12 presents?
### Vocabulary Language: English
Denominator
Denominator
The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
Equivalent
Equivalent
Equivalent means equal in value or meaning.
Equivalent Rate
Equivalent Rate
Equivalent rates are rates that can each be simplified to the same ratio.
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$\newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$
Section7.4Continuous Functions
Subsection7.4.1Continuity
Recall our definition of continuity for a function at a single point.
Definition7.4.1Continuity at a Point
A function $f$ is continuous at $a$ if
\begin{equation*} \lim_{x \to a} f(x) = f(a). \end{equation*}
The single equation captures the full definition because for the equation to be true, the limit must exist and the value of the function must exist. Also, recall that the function is right-continuous if the limit comes from the right ($x \to a^+$) and left-continuous if the limit comes from the left ($x \to a^-$).
These ideas allow us to define what we mean by saying that a function is continuous on an interval.
Definition7.4.2Continuity on an Interval
A function $f$ is continuous on an interval $(a,b)$ if $f$ is continuous at every point $x \in (a,b)\text{.}$ We can include an endpoint if the limit statement is true coming from within the interval. That is, we include $a$ if
\begin{equation*} \lim_{x \to a^+}f(x)=f(a) \end{equation*}
and we include $b$ if
\begin{equation*} \lim_{x \to b^-}f(x)=f(b). \end{equation*}
There are two important theorems that describe what we know about functions that are continuous on an interval. The Extreme Value Theorem guarantees that any function that is continuous on a closed interval has a highest and lowest point within that interval. The Intermediate Value Theorem guarantees that a function that is continuous on a closed interval can not skip over any values between its values at the endpoints. The proofs for both of these theorems require advanced methods not taught at this level. We treat them essentially as axioms, statements that are true without proof.
If a function is not continuous on $[a,b]\text{,}$ then it does not necessarily have a maximum or minimum value. One way that this might happen is if $f$ has a vertical asymptote within the interval. In that case, the values of $f$ would be unbounded. Another way that this might happen is that $f$ is bounded by what would be a maximum (or minimum) value but just doesn't reach it because of a sudden jump.
Example7.4.4
Consider the function defined piecewise as
\begin{equation*} f(x) = \begin{cases} \frac{1}{x^2}, & x \ne 0, \\ 0, & x=0. \end{cases} \end{equation*}
This function has a non-removable discontinuity at $x=0\text{,}$ corresponding to a vertical asymptote. Because the formula has $x^2$ in the denominator (always positive), we have
\begin{equation*} \lim_{x \to 0} f(x) = +\infty. \end{equation*}
This function is unbounded on the interval $[-1,1]$ and has no maximum. It does have a minimum at $f(0)=0$ since that is below the rest of the graph.
Example7.4.5
Consider the function defined piecewise as
\begin{equation*} f(x) = \begin{cases} x^2, & -1 \lt x \lt 1, \\ \frac{1}{2}, & x = \pm 1. \end{cases} \end{equation*}
This function has a removable discontinuities at $x=\pm 1\text{,}$ where the limits are 1 but the values are $\frac{1}{2}\text{.}$ In this case, $f$ is continuous on $(-1,1)$ but not on $[-1,1]\text{.}$ The maximum value should have been $y=1\text{,}$ but the graph never reaches that value because of the discontinuity. The function does have a minimum value at $f(0)=0\text{.}$
Example7.4.6
Consider the function defined piecewise as
\begin{equation*} f(x) = \begin{cases} x^2, & -1 \lt x \le 1, \\ 2, & x = -1. \end{cases} \end{equation*}
This function has a removable discontinuity at $x=-1\text{.}$ In this case, $f$ is continuous on $(-1,1]$ but not on $[-1,1]\text{.}$ In spite of the discontinuity at $x=-1\text{,}$ this function has a maximum value $f(-1)=2$ because that value is above every other point in the interval.
The previous example is included to emphasize that a theorem gives conditions that guarantee something is true. But those conditions are not always required. The extreme value theorem gives conditions that guarantee a function will have a maximum value. There are no exceptions for a continuous function on a closed interval to have both maximum and minimum values. But there are discontinuous functions that have them as well. It is just that there are also discontinuous functions that do not have extreme values.
The Intermediate Value Theorem guarantees that the graph of $y=f(x)$ intersects every horizontal line between $y=f(a)$ and $y=f(b)$ at least once for values of $x$ between $a$ and $b\text{.}$ Because continuity is essentially connectedness, the only way for the graph to go from $y=f(a)$ to $y=f(b)$ is to cross through all intermediate values. A discontinuous function has the ability to jump across values without touching them.
Example7.4.8
Consider the function defined piecewise as
\begin{equation*} f(x) = \begin{cases} -1, & x \lt 0, \\ 0, & x = 0, \\ 1, & x \gt 0. \end{cases} \end{equation*}
This function has a jump discontinuity at $x=0\text{,}$ and is otherwise constant. If we consider the interval $[-1,1]\text{,}$ the values at the endpoints are $f(-1)=-1$ and $f(1)=1\text{.}$ Except for $y=0\text{,}$ the function $y=f(x)$ has no solutions for $-1 \lt y \lt 1$ because of the jump.
The Intermediate Value Theorem allows us to know that a continuous function has a solution to an equation within a particular interval. If the interval is small, we have an approximation to the value of the solution. We say that the interval brackets the solution. Finding successively smaller bracketing intervals allows us to approximate the root to any needed precision. The Intermediate Value Theorem guarantees this works for continuous functions.
Example7.4.9
The function $f(x)=x^3-x-3$ is continuous because it is a polynomial. Because $f(1)=-3$ and $f(2)=3\text{,}$ we know that $f(x)$ must pass through every $y$-value between -3 and 3 for at least one value of $x$ in the interval $(1,2)\text{.}$ In particular, if we are solving $f(x)=0\text{,}$ since $y=0$ is between $f(1)=-3$ and $f(2)=3\text{,}$ we know that there is a solution $x$ bracketed by the interval $[1,2]\text{.}$
If we find a smaller interval, then we can know more precisely where the root occurs. In particular, since $f(1.6)=-0.504$ and $f(1.7)=0.213$ and $y=0$ is between those values, the Intermediate Value Theorem guarantees that our continuous function has a root bracketed by the interval $[1.6, 1.7]\text{.}$
Subsection7.4.2Definite Integrals and Average Value
When we studied the definite integral, we learned that continuity implies integrability. However, a discontinuous function might still be integrable. For example, the definite integral of a piecewise continuous function with a finite number of jump discontinuities can be computed using the splitting property. The total definite integral would be equal to the sum of the definite integrals on each of the subintervals.
Continuity does guarantee something stronger than integrability. It guarantees that the function attains its average value over an interval. To make this precise, we first need to define the average value.
Definition7.4.10Average Value of a Function
The average value of a function $f$ on an interval $[a,b]\text{,}$ denoted $\langle f \rangle_{[a,b]}\text{,}$ is defined as
\begin{equation*} \langle f \rangle_{[a,b]} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx, \end{equation*}
so long as $f$ is integrable on $[a,b]\text{.}$
The average value is defined as the value of a constant function that has the same definite integral over the interval:
\begin{equation*} \int_{a}^{b} \langle f \rangle_{[a,b]} \, dx = \langle f \rangle_{[a,b]} \cdot (b-a) = \int_{a}^{b} f(x)\, dx. \text{.} \end{equation*}
Example7.4.11
The figure below illustrates a simple function $f(x)$ defined on the interval $[0,5]\text{,}$
\begin{equation*} f(x) = \begin{cases} 3, & 0 \le x \lt 1, \\ 5, & 1 \le x \lt 3, \\ -1, & 3 \le x \le 5. \end{cases} \end{equation*}
The definite integral equals the sum of the signed areas,
\begin{equation*} \int_0^5 f(x) \, dx = 3 \cdot 1 + 5 \cdot 2 + -1 \cdot 2 = 11. \end{equation*}
The average value is equal to this definite integral divided by the width of the interval,
\begin{equation*} \langle f \rangle_{[0,5]} = \frac{1}{5} \int_{0}^5 f(x)\,dx = \frac{11}{5}. \end{equation*}
Because $f$ is continuous on $[a,b]\text{,}$ the Extreme Value Theorem guarantees that $f$ attains a minimum value $f(x_{\min})$ and a maximum value $f(x_{\max})$ so that $f(x_{\min}) \le f(x) \le f(x_{\max})$ for all $x \in [a,b]\text{.}$
The average value $\langle f \rangle_{[a,b]}$ must be between the minimum and maximum values. The Integral Bounds theorem guarantees
\begin{equation*} f(x_{\min}) (b-a) \le \int_{a}^{b} f(x) \, dx \le f(x_{\max}) (b-a) \end{equation*}
which then implies
\begin{equation*} f(x_{\min}) \le \langle f \rangle_{[a,b]} \le f(x_{\max}). \end{equation*}
By the Intermediate Value Theorem with the interval with end points $x_{\min}$ and $x_{\max}$ (we don't know which is on the left/right), there must be some value $c$ between these points, and so $c \in (a,b)\text{,}$ for which
\begin{equation*} f(c) = \langle f \rangle_{[a,b]}. \end{equation*}
In the previous example, $f$ was not continuous and we can see that the graph $y=f(x)$ did not intersect the constant value $\langle f \rangle_{[0,5]}\text{.}$ The Mean Value Theorem for Integrals guarantees that when the function is continuous, the constant function using the average value must intersect the graph $y=f(x)\text{.}$
Example7.4.13
The function $f(x)=x^2$ is continuous everywhere. The average value on the interval $[-1,2]$ can be found using the elementary accumulation formula for a quadratic rate and the splitting property.
\begin{align*} \langle f \rangle_{[-1,2]} &= \frac{1}{2--1} \int_{-1}^{2} x^2 \, dx \\ &= \frac{1}{3} \Big( \int_0^2 x^2 \, dx - \int_0^{-1} x^2 \, dx \Big) \\ &= \frac{1}{3} \Big( \frac{1}{3}(2^3) - \frac{1}{3}(-1)^3 \Big) = \frac{1}{3}\Big(\frac{8}{3} + \frac{1}{3}\Big) = 1 \end{align*}
A figure showing the graphs $y=f(x)=x^2$ and $y=\langle f \rangle_{[-1,2]} = 1$ is shown below. The Mean Value Theorem predicted the existence of a point $c \in (-1,2)$ where $f(c)=\langle f \rangle_{[-1,2]}=1\text{,}$ which we can see occurs at $c=1\text{.}$
The Mean Value Theorem for Integrals also provides the justification for the Monotonicity Test for Accumulation Functions.
Consider any two points $c_1, c_2 \in [a,b]$ with $c_1 \lt c_2\text{.}$ Because $A(x)$ is an accumulation function, by the splitting property of definite integrals,
\begin{equation*} A(c_2) - A(c_1) = \int_{c_1}^{c_2} f(x) \, dx. \end{equation*}
On the other hand, because $f$ is continuous, the Mean Value Theorem guarantees the existence of a point $c \in (c_1,c_2)$ such that
\begin{equation*} A(c_2)-A(c_1) = \int_{c_1}^{c_2} f(x)\, dx = f(c)\cdot(c_2-c_1). \end{equation*}
Now assume that $f(x) \gt 0$ for all $x \in (a,b)\text{.}$ Then $f(c) \gt 0$ and $c_2-c_1 \gt 0\text{,}$ guaranteeing that $A(c_2)-A(c_1) \gt 0\text{.}$ That is, $A(c_2) \gt A(c_1)\text{.}$ This is what is needed to show that $A$ is increasing.
Next assume that $f(x) \lt 0$ for all $x \in (a,b)\text{.}$ Then $f(c) \lt 0$ while $c_2-c_1 \gt 0\text{,}$ guaranteeing that $A(c_2)-A(c_1) \lt 0\text{.}$ That is, $A(c_2) \lt A(c_1)\text{,}$ which shows that $A$ is decreasing.
Finally assume that $f(x) = 0$ for all $x \in (a,b)\text{.}$ Then $f(c) = 0\text{,}$ implying that $A(c_2)-A(c_1) = 0\text{.}$ That is, $A(c_2) = A(c_1)\text{,}$ which shows that $A$ is constant.
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# Integer Sequences I: The Fibonacci Sequence
## Introductory Remarks
This lesson could be subtitled "Guess my rule." I write a sequence of four or five numbers on the board. I then ask the students what number comes next. After the students have correctly added two or three numbers, I ask what rule they are using to figure out the next number in the sequence. In determining the rule, I remind them that the rule must work for every number in the sequence.
I have taught this lesson to second and third graders.
## Supplies
In addition to a chalkboard, you will need pieces of paper, about a quarter of a sheet, enough to give each student three pieces.
## Lesson Plan
1. Write the numbers 1 3 5 7 on the chalkboard.
2. Ask the students what number comes next. Usually a student will correctly guess 9.
3. Ask for the next number in the sequence. Ask the student who answers how she or he knew that was correct. Students will offer explanations such as "You're skipping a number every time." If they don't bring it up themselves, point out that these are the odd numbers.
4. Write the numbers 1 4 7 10 on the board. Ask for the next number (13). Ask for the number after that (16). Ask the students to explain the pattern.
5. Write the numbers 1 2 4 7 11 on the board. Ask for the next number. It may take a few guesses for the students to come up with the correct answer of 16. Ask for the next number (22). Ask the students to explain the pattern.
It may take several minutes for the students to figure out this pattern. Often they will say, "You're skipping two numbers." I respond by referring to the sequence and asking whether I am skipping two numbers between 1 and 2. I then ask what is happening between 2 and 4. As we proceed along the sequence a few students will guess the rule. They usually express it as "skipping one, then you skip two, then you skip three."
It is important when discussing these explanations to take all suggested rules seriously, and to try to apply them to the pattern. I avoid labelling ideas as "right" and "wrong." Sometimes a student will come up with a new rule that fits the pattern. I tell the student that it is a good rule, but it is not the one I had in mind.
6. Write the numbers 1 3 6 10 on the board. Ask the students what comes next. After they discover the next two numbers, 15 and 21, ask them to explain the rule. They usually figure this out pretty quickly, since it is the same as the rule for the preceding sequence.
7. Draw three dots on the board, with one dot on top, and two dots in a row below so that all three form a triangle. Cover the lower two dots, and tell the students that here is one, then uncover the dots and ask them to count. Draw a row of three dots below the row of two, and ask the students to count the total number of dots. Add a row of four dots, then a row of five dots, counting each time. You are building a triangular array. Tell the students that the numbers 1 3 6 10 15 . . . are sometimes called the triangular numbers.
8. Now it's time for the challenge. Write the numbers 1 1 2 3 5 on the board. Ask the students what comes next. You should get wildly varying responses. To involve all of the students, pass out squares of paper, and have each student write his or her guess in large digits so you can see it. Have the students hold their guesses up in the air. I read the guesses out loud. It generally comes out, "I see a 5, and a seven, and another 7, and an 8, and there's a 10, and more sevens . . ."
Add the number 8 to the sequence, and ask the students to write a guess for the next number (13).
Have the students guess one more number (21).
Ask the students if they can figure out the rule. I've had a number of third graders and an occasional second grader get this one. To get the next number in the sequence, you add the previous two numbers, i.e. 1+1=2, 1+2=3, 2+3=5, and so on. This is called the Fibonacci sequence.
# Integer Sequences II: Combinatorics
## Supplies
Each student will need a piece of scratch paper--a half sheet will do--and a pencil.
## Warm-up: Guess my rule
1. Write the number sequence 1 3 5 7 on the board. Ask the students what number comes next. Add 9 to the sequence. Ask for the next number. Add 11 to the sequence. Ask if the students know the special name for the sequence. If they don't know, tell them these are the odd numbers.
2. Write the number sequence 1 2 4 8 on the board below the odd numbers. Ask for the next number (16), and the next (32). Ask the students to explain the rule that generates the sequence (doubling the previous number.) Tell them these are the binary numbers.
3. Write the number sequence 1 3 6 10 under the binary numbers. Ask for the next number (15), and the next (21). Ask the students to figure out the rule. (Add 2, then 3, then 4, and so on, to the previous number.) Tell the students that these are called the triangular numbers. If they haven't seen this sequence before, draw a triangular array of dots on the board with 1 dot in the first row, 2 dots in the second row, 3 dots in the third row, and so on. Show that the number of dots above any particular row is 1, 3, 6 . . .
4. Write the number sequence 1 4 9 16 on the board. Ask the students to guess the next number (25), and the next (36). Ask them to figure out the rule. Two explanations are possible: Adding successive odd numbers, 3, 5, 7 . . . to the previous number, or squaring 1, 2, 3, 4 . . . Tell students these are the square numbers.
### The handshake puzzle
Pose the following problem: A baseball team is practicing for the first time. In order to get to know each other, the players shake hands. Each player shakes hands with every other player exactly once. How many total handshakes are there?
Have the students guess the answer, and write their guesses on the sheets of scratch paper. Tell them to put the paper aside.
Now work it out inductively. Begin by calling two students to the front of the room and having them shake hands. Draw a table on the board. Label the left column "number of players" and the right column "handshakes." Write 2 and 1 in the left and right columns respectively.
Have three students come to the front of the room and shake hands with each other. Count the handshakes (3) and add the numbers to the table.
Continue with four students, five students, and six students. You may need to impose a bit of order on the handshakes to make sure they get counted correctly. You should get 6, 10 and 15.
Ask the students if they recognize the numbers in the right-hand column. They should identify them as the triangular numbers.
Use the rule for generating triangular numbers to work out the seventh (21), eighth (28), and ninth (36) numbers in the sequence. Have the students compare their guesses to the answer you worked out with them.
# Pascal's Triangle
## Introductory remarks
This is one of my most popular lessons.
Pascal's triangle, named after the seventeeth-century French mathematician Blaise Pascal, turns up in a number of mathematical problems. Those who remember their high school algebra will recognize the numbers as binomial coefficients. If you are familiar with probability theory, you may know that these numbers give the probability of tossing m heads (or tails), in n tosses.
## Lesson Plan
1. Draw the following street map on the chalkboard: At the top in the middle is the main corner, or starting point. The two boundary streets run at 45-degree angles down from the starting point, one to the right, and one to the left. Inside these boundaries draw a square grid, fairly widely spaced, of lines parallel to, and ending at, these boundary streets. If this description confuses you, take a piece of graph paper, rotate it 45 degrees so that one corner is at the top, and copy a 6-by-6 square of it. I space the "streets" about 6 inches apart.
2. Pose the problem. Starting at the top of the grid, you want to know how many different paths there are from the starting point to any corner, subject to the following conditions. All paths must follow streets, and you are only permitted to move downward, either to the right, or to the left.
3. Begin with the first row of intersections below the starting point. Show that there is only one path to each of these intersections. Write the number 1 on each intersection.
4. Move on to the next row of three intersections. Ask the students how many paths there are to the leftmost intersection in that row. Ask them to describe the path by telling you which direction to take at each intersection. Starting at the top, you move down to the left, which takes you to the leftmost intersection in row two. Then you move to the left again, taking you to the intersection you are trying to reach. This path is therefore "left-left."
5. Have the students describe the paths leading to the middle intersection in row three. The two legal paths are "left-right" and "right-left." Next, have them find the only path to the rightmost intersection, "right-right."
6. Proceed to the fourth row, with four intersections. This is where the puzzle starts to get interesting. By this time, most students will realize that there is only one legal path to any intersection on the right or left boundary. Don't tell them how many paths to look for in the middle. As they describe paths, write them down using the letters L and R, for example LLR for "left-left-right." Have the students continue to look for paths until both you and they are satisfied that they have found them all. (You, of course know that the three legal paths for the second intersection in the row are LLR, LRL, and RLL.) The students may use symmetry to conclude that the number of paths to the third intersection in the row is equal to the number of paths to the second intersection, or they may need to list them (LRR, RLR, and RRL.)
7. Continue with the next row. The numbers of paths to the fifth row intersections are 1 4 6 4 1.
8. Summarize by writing in the form of a triangle the numbers discovered so far:
------------1-------------
----------1---1-----------
--------1---2---1---------
------1---3---3---1-------
----1---4---6---4---1-----
Ask the students if they see a pattern that will generate the next row of numbers. Most students quickly figure out that the first two numbers in the row are 1 5, and the last two are 5 1. Some students can guess the rule: add the two numbers above to get the next number. Thus 1+1=2, 2+1=3, 1+3=4, 3+3=6. Applying this rule, the middle numbers of the sixth row are 10 10. If students do not figure the rule out in a few minutes, I tell them.
9. Tell the students this pattern is called Pascal's Triangle. Tell them it comes up in a number of mathematical contexts, the two most common areas being the probability of coin tosses, and the expansion of binomials in high school algebra.
## Follow-Up Activity: Probability of Coin Tosses
### Supplies
Each student will need four pennies.
### Lesson Plan
1. Ask the students if they know what "odds" are. If they don't, explain that odds are numbers that give the relative likelihood of events. For example, when tossing one penny, the chances of getting heads or tails are equally likely. Thus the odds are 1:1.
2. Test the prediction by having each student toss one coin. Count the numbers of heads and tails. Explain to the students that experimental results do not always match the theoretical probability.
3. Now ask the students to list the possible outcomes of tossing two coins. (They should come up with two heads, two tails, and one head/one tail.) Explain that order matters here, so that HT and TH add together to double the odds. Thus the odds for HH HT/TH TT are 1:2:1.
4. Test the prediction by having each student toss two coins.
5. Have the students figure out all the possible outcomes of tossing three coins. (They are HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT.) Write down the theoretical odds 1:3:3:1. Test the prediction by having the students toss three coins.
6. Figure out the possible outcomes for tossing four coins. (HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT.) The theoretical odds are 1:4:6:4:1. Test the prediction by tossing four coins.
7. Write the odds in triangular format:
----1:1----
---1:2:1---
--1:3:3:1--
-1:4:6:4:1-
Ask the students where they have seen this before. They will recognize Pascal's triangle from the street map puzzle.
# Generalized Fibonacci Sequences and the Golden Mean
### Supplies
Students will each need a calculator for this activity.
### Lesson Plan
1. Write the number sequence 1 1 2 3 5 on the board. Ask the students what comes next. If they don't get it in three or four guesses, tell them the next number is 8. Have them guess the next number (13), and the next (21). Ask them if they can figure out the rule that tells how to generate the next number in the sequence. (The rule is to add the previous two numbers.)
2. Have the students calculate about 12 more terms of the sequence. They may do this mentally, on paper, or with a calculator.
3. Have the students divide the next term by the next-to-the-last term of the sequence. It will be approximately 1.618. Write this number on the board.
4. Start a new sequence by choosing any two numbers between 1 and 20. It does not matter if the second number is smaller than the first. Apply the Fibonacci rule of adding the previous two numbers until you have a sequence of 20 numbers.
5. Have the students divide the last number in the sequence by the next-to-the-last number. Again the result will be approximately 1.618. Explain that this number is called the "golden ratio" or the "golden mean." According to the ancient Greeks, a rectangle with its sides in this ratio was the most beautiful rectangle a person could draw.
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In today’s post, we are going to see some problems involving distance conversions, and give some advice for solving them.
#### Problem number 1 with distance conversions
##### Silver Beard: the Pirate Problem
The Pirate, Silver Beard, has arrived on Coral island to search for treasure. On the map, it says that, from the shore, he must hopscotch 3.7 hectometers toward the center of the island and then somersault 8.5 more decameters in the same direction. How many meters will he go in total from the shore to the treasure? Write the answer in kilometers.
In order to respond to the question, we must add the two distances, but… Can you add hectometers and decameters?
NO, because they are different units!
Since they request us to respond in meters, we have to turn both distances into meters. For this, we can adjust this scale to work for us by changing the units. Note that when the order is descending, the units change by multiplying, and when the order is ascending, the units change by dividing.
We see that in order to change from hectometers and decimeters to meters, the arrow goes downward, which means you have to multiply. Therefore:
Since 1 hm = 100 m, Then 3.7 hm = 3.7 x 100 m = 370 m
Since 1 dam= 10 m, Then 8.5 dam = 8.5 x 10 = 85 m
370 m + 85 m = 455 m
We also need to express the solution in km. If we look at the table, we see that to go from meters to km the arrow goes upward; therefore, we must divide:
Since 1 km = 1000 m, Then 455 m = 455/1000 km = 0.455 km
Thus, the answer to this distance conversion problem is:
455 meters, which is equal to, 0.455 kilometers
#### Problem number 2 with distance conversions
##### The Long Hair Problem
Gabriela has the most beautiful hair! Before, she used to have the longest hair in the whole class: her hair measures 6 decameters. But yesterday, she cut 25 centimeters, so now the girl in the class with the longest hair is María. How many centimeters long is Gabriela’s hair now? Write the answer in millimeters.
In order to know what the length of her hair is, we must subtract the lengths, but first they must be converted to the same unit. The unit we need is centimeters, so we have to change the information to cm. As the scale from dm to cm goes down, we must multiply:
Since 1 dm = 10 cm, Then 6 dm = 6 x 10 cm = 60 cm
Now we subtract:
60 cm – 25 cm = 35 cm
In order to express the answer in mm, we have to multiply:
Since 1 cm = 10 mm, Then 35 cm = 35 x 10 mm = 350 mm
Therefore, the answer to this problem is:
35 centimeters, which equals, 350 millimeters
#### Problem number 3 with distance conversions
##### The Bear and the Honey Problem
A bear that loves honey wants to take honey from a hive, which is on a branch of a tree, but it is too high. In order to reach it, he climbs on a rock that is 12 decameters high directly below the branch, and with claws outstretched, he catches it. If the bear, standing up, is 2.3 m tall, how far off the ground was the hive?
As in the first problem, in order to solve, it is necessary to add the two distances. But first we must convert them. However, in this problem, it doesn’t tell us the units that we should use, so we can write the answer in the unit that is easiest for us. We chose meters:
We have to change the dm into m (since the scale ascends, we have to divide by 10):
Since 1 m = 10 dm, Then 12 dm = 1.2 m
1.2 m + 2.3 m = 3.5 m
Therefore, the answer to this problem is:
3.5 meters
How did this post seem to you? Has it helped you to better understand distance conversions?
Well, continue to follow the blog because in the coming weeks we will see more examples of conversion problems dealing with weight and amount. And you can try Smartick for free to learn much more elementary math!
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# The area of a right triangle
I right triangle i a triangle where the angle between two of the sides is $90°$. If you want to calculate the area of that triangle you need to know those two sides.
The area is calculated with
$Area = \frac{a \cdot b}{2}$
Multiply the base and the hight with each other and divide with two.
It is easy to understand why this works if we know how to find the area of a rectangle. To calculate that you just multiply the base with the height. Let us study the triangle above inside a rectangle.
You can see that the right triangle is exactly half of the rectangle above. That is way we divide the area of a rectangle in two to get the area of a right triangle.
## Example 1
Find the area of the triangle.
The base is $3,4 \, cm$ and the height is $3.1 \, cm$. The area will be given by
$Area = \frac{3.4 \cdot 3.1}{2} = 5.27 \, cm^2$
## Example 2
Find the area of the right triangle below.
We need to find the height of the triangle. Let us use the pythagorean theorem and call the height $h$.
$8.2^2=5.0^2+h^2$
Subtract with $5.0^2$
$8.2^2-5.0^2=h^2$
Calculate the left hand side.
$42.24=h^2$
Take the square root
$\sqrt{42.24}=h$
$h≈6.5\,m$
Now we know the height and we can calculate the area:
$Area = \frac{5.0 \cdot 6.5}{2} = 16.25 \, m^2$
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If $x=1+2i$ then prove that ${{x}^{3}}+7{{x}^{2}}-13x+16=-29$.
Last updated date: 25th Jul 2024
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Hint: It is given that $x=1+2i$. Using this, find out ${{x}^{2}}$ and ${{x}^{3}}$ and then substitute
$x,{{x}^{2}},{{x}^{3}}$ in the equation which we have to prove in the question.
In the question, we are given a complex number $x=1+2i$. We have to prove that the expression ${{x}^{3}}+7{{x}^{2}}-13x+16$ is equal to $-29$.
In the expression ${{x}^{3}}+7{{x}^{2}}-13x+16$, we can see that ${{x}^{2}}$ and ${{x}^{3}}$ are present. So, we have to find both ${{x}^{2}}$ and ${{x}^{3}}$.
In the question, it is given $x=1+2i$.
Squaring both the sides of the above equation, we can find ${{x}^{2}}$ as,
${{x}^{2}}={{\left( 1+2i \right)}^{2}}$
We have a formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Using this formula to find ${{x}^{2}}$, we get,
\begin{align} & {{x}^{2}}={{\left( 1 \right)}^{2}}+{{\left( 2i \right)}^{2}}+2\left( 1 \right)\left( 2i \right) \\ & \Rightarrow {{x}^{2}}=1+4{{i}^{2}}+4i \\ \end{align}
In complex numbers, we have a formula ${{i}^{2}}=-1$.
Substituting ${{i}^{2}}=-1$in the above equation, we get,
\begin{align} & {{x}^{2}}=1-4+4i \\ & \Rightarrow {{x}^{2}}=-3+4i...............\left( 1 \right) \\ \end{align}
To find ${{x}^{3}}$, we will multiply the above equation with $x=1+2i$.
\begin{align} & {{x}^{2}}.x=\left( -3+4i \right)\left( 1+2i \right) \\ & \Rightarrow {{x}^{3}}=-3-6i+4i+8{{i}^{2}} \\ & \Rightarrow {{x}^{3}}=-3-2i+8{{i}^{2}} \\ \end{align}
Substituting ${{i}^{2}}=-1$in the above equation, we get,
\begin{align} & {{x}^{3}}=-3-2i-8 \\ & \Rightarrow {{x}^{3}}=-11-2i..........\left( 2 \right) \\ \end{align}
Since in the question we have to prove ${{x}^{3}}+7{{x}^{2}}-13x+16=-29$, substituting ${{x}^{2}}$
from equation $\left( 1 \right)$ and ${{x}^{3}}$ from equation $\left( 2 \right)$ in
${{x}^{3}}+7{{x}^{2}}-13x+16$, we get,
\begin{align} & -11-2i+7\left( -3+4i \right)-13\left( 1+2i \right)+16 \\ & \Rightarrow -11-2i-21+28i-13-26i+16 \\ & \Rightarrow -29 \\ \end{align}
Hence, we have proved that ${{x}^{3}}+7{{x}^{2}}-13x+16=-29$.
Note: One can also do this question by converting the complex number $x=1+2i$ to the Euler’s form i.e. in the form of $r{{e}^{i\theta }}$ or $r\left( \cos \theta +i\sin \theta \right)$ where $r$ is the modulus of the complex number and $\theta$ is the argument of the complex number. Then using the De Moivre’s theorem i.e. ${{\left( r{{e}^{i\theta }} \right)}^{n}}=\cos n\theta +i\sin n\theta$, one can find ${{x}^{2}}$ and ${{x}^{3}}$ and substituting $x,{{x}^{2}},{{x}^{3}}$ in the expression ${{x}^{3}}+7{{x}^{2}}-13x+16$. But this method will take a lot of time since the argument of the complex number is not a standard angle. So, one has to write $\sin 2\theta ,\sin 3\theta ,\cos 2\theta ,\cos 3\theta$ in terms of $\sin \theta$ and $\cos \theta$ using the trigonometric formulas. Then, one has to simplify all the terms to get the answer.
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# ISI MStat PSB 2010 Problem 2 | Combinatorics
This is a beautiful sample problem from ISI MStat 2010 PSB Problem 2. This is based how one can find the number of isosceles triangles with sides of integer length one can construct, using simple counting principles . We provide detailed solution with prerequisites mentioned explicitly.
This is a beautiful sample problem from ISI MStat PSB 2010 Problem 2. It’s all about how many isosceles triangle with sides of integers lengths one can construct under certain conditions. We provide a detailed solution with prerequisites mentioned explicitly.
## Problem– ISI MStat PSB 2010 Problem 2
Find the total number of isosceles triangles such that the length of each side is a positive integer less than or equal to 40.
(Here equilateral triangles are also counted as isosceles triangle.)
## Prerequisites
• Basic counting principles.
• Triangle inequality.
## Solution
Let the sides of the isosceles triangle are k, k and l units, (length of the equal sides are k units)
So, our problem is all about finding triplets of form (k,k,l), where $l, k \le 40$ . And how many such triplets can be found !
also since l,k and k are also sides of an (isosceles) triangle, so by triangle inequality,
$l < 2k$ . Since l is an integer so, $l \le 2k-1$
So, we have $l\le 40$ and $l \le 2k-1$. combining we have $l \le min(40, 2k-1)$ .
now let us define, $l_k$ : possible number of l’s for a given k =1,2,….,40
here we must consider two cases.
Case-1 : when $2k-1 \le 40 \Rightarrow k \le \lceil \frac{40+1}{2} \rceil = 20$
So, l can be any integer between 1 and 2k-1, since $l\ \le 2k-1$ when $k \le 20$
so,there can be 2k-1 choices for the value of each k, when k=1,…,20.
Or, $l_k = 2k-1$ for k=1,2,….,20.
now, Case-2 : when 2k-1>40 $\Rightarrow k > \lceil \frac{40+1}{2} \rceil = 20$
so, here l can take any integer between 1 to 40. So, there 40 choices for l for a given k, when k=21,22,…,40.
Or, $l_k = 40$ for k=21,22,….,40.
Combining Case-1 and Case-2, we have,
$l_k = \begin{cases} 2k-1 & k=1,2,…,20 \\ 40 & k=21,22,…,40 \end{cases}$
So, finally, let all possible number of triplets of form (k,k,l) are = $T_40$
So, $T_{40}$ = $\sum_{k=1}^{40}{l_k}$ = $\sum_{k=1}^{20}{(2k-1)}$ + $\sum_{k=21}^{40}{40}$ =$(20 )^2 + 40 \times (40-20)=400+800= 1200.$
So, $T_{40}$ = 1200. hence we can find 1200 such isosceles triangles with sides of integer lengths such that the length of each sides are less than or equal to 40.
## Food For Thought
Can you generalize this problem? i.e. can you find How many isosceles triangle one can construct, with sides of integer lengths, such that the sides are less than or equal to any integer N ? Can you find an elegant formula to express $T_N$, for any integer N?
## 2 replies on “ISI MStat PSB 2010 Problem 2 | Combinatorics”
For general $N \in \Bbb N$ we have $$T_N = \left ( \left \lfloor \frac {N+1} {2} \right \rfloor \right )^2 + N^2 – N \left \lfloor \frac {N+1} {2} \right \rfloor.$$
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# What is the vertex of y= -2(x - 4)^2 - 5x+3 ?
Jan 6, 2016
The vertex is $\left(\frac{11}{4} , - \frac{111}{8}\right)$
#### Explanation:
One of the forms of the equation of a parabola is $y = a {\left(x - h\right)}^{2} + k$ where (h, k) is the vertex. We can transform the above equation into this format to determine the vertex.
Simplify
$y = - 2 \left({x}^{2} - 8 x + 16\right) - 5 x + 3$
It becomes
$y = - 2 {x}^{2} + 16 x - 32 - 5 x + 3$
$y = - 2 {x}^{2} + 11 x - 29$
Factor out 2 being the coefficient of ${x}^{2}$
$y = - 2 \left({x}^{2} - \frac{11}{2} x + \frac{29}{2}\right)$
Complete the square: Divide by 2 the coefficient of x and then square the result. The resulting value becomes the constant of the perfect square trinomial.
${\left(\frac{- \frac{11}{2}}{2}\right)}^{2} = \frac{121}{16}$
We need to add 121/16 to form a perfect square trinomial. We have to deduct it as well though, to preserve the equality. The equation now becomes
$y = - 2 \left({x}^{2} - \frac{11}{2} x + \frac{121}{16} - \frac{121}{16} + \frac{29}{2}\right)$
Isolate the terms which form the perfect square trinomial
$y = - 2 \left({x}^{2} - \frac{11}{2} x + \frac{121}{16}\right) + \frac{121}{8} - 29$
$y = - 2 \left({x}^{2} - \frac{11}{2} x + \frac{121}{16}\right) - \frac{111}{8}$
$y = - 2 {\left({x}^{2} - \frac{11}{4}\right)}^{2} - \frac{111}{8}$
From this
$h = \frac{11}{4}$
$k = - \frac{111}{8}$
Hence, the vertex is $\left(\frac{11}{4} , - \frac{111}{8}\right)$
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# Chapter 11: Constructions
A construction in Mathematics is an accurate drawing of angles and lines. For constructions in Mathematics, you are meant to use only a straightedge and a pair of compasses with a pencil. A straightedge is an instrument, usually made of metal, that has a very straight edge, so it can be used to draw straight lines. A straight edge usually does not have measurements marked on it.
In this chapter you are going to use the following instruments to make constructions: a compass with a pencil, a ruler and a protractor.
1. A pair of compasses (or a compass) is an instrument used for drawing circles and in constructions.
A compass has two legs:
• One leg has a sharp point so that you can put it firmly into position on a page. When you draw a circle, the sharp point will stay on the dot where you positioned it. The sharp point will not move.
• The other leg has place to insert a pencil. When you draw a circle, you move the leg that has the pencil in it to draw the circle on the page.
The two legs of the compass are joined at the hinge. Make sure that the screw at the hinge is tight, so that the two legs of the compass cannot move apart unless you want them to.
Also make sure that your pencil is sharpened. A blunt pencil will lead to diagrams that are not accurate.
Do not confuse the compass that we use in mathematical constructions with the compass that we use to find direction.
2. A ruler
You will use a ruler when you do constructions and have to draw straight lines. Most of us do not have a straightedge, so we use a ruler instead.
• In some constructions you will use the markings on the ruler to measure the length of a line segment. For example, when you have to construct a triangle, and you are given the lengths of one or two of the sides, you measure to make sure you are drawing lines of the correct length.
• In other constructions you will not use the markings on the ruler at all. For example, when you bisect a line segment or an angle, you use a construction method, and not measurements.
3. A protractor
We use a protractor to measure or draw an angle. We measure angles in degrees.
• Take note of the centre (or origin) of a protractor:
When you draw an angle you have to make sure that the centre (or origin) of the protractor is exactly on the vertex of the angle that you want to construct.
• The centre (or origin) of a protractor is on the base line of the protractor.
You also have to make sure that one side of an angle is exactly on the base line of the protractor before you construct the second arm of the angle.
• There are two zeros on a protractor:
If you begin measuring from the zero on the left hand side of the protractor, you use the outer scale on the protractor.
If you begin measuring from the zero on the right hand side of the protractor, you use the inner scale on the protractor.
It is a good idea to estimate the size of an angle before you measure or draw the angle. Then you can check against your estimate to see if your answer or construction is reasonable.
## 11.1 Construct triangles
You will learn how to construct different triangles.
It is helpful to remember the following important properties of triangles:
• The sum of the interior angles is $180^{\circ}$.
• The longest side of a triangle is opposite the biggest angle of the triangle.
• The shortest side of a triangle is opposite the smallest angle of the triangle.
It is impossible to construct a triangle where the longest side is not opposite the biggest angle.
### Construct a triangle given two sides and an included angle
In this section you will construct triangles where you are given two sides of a triangle as well as the included angle.
An included angle is the angle between two sides of a triangle.
In the diagrams below:
• the angle of $40^{\circ}$ is included between the sides of 10 cm and 7 cm.
• the angle of $65^{\circ}$ is not included between the sides of 10 cm and 7 cm.
included angle An included angle is the angle between two sides of a triangle.
### Worked Example 11.1: Constructing a triangle given two sides and an included angle
Construct $\triangle ABC$ with $AB$ = 4 cm, $A\hat{B}C$ = $65^{\circ}$ and $BC$ = 8 cm.
1. Step 1: Make a rough sketch of the triangle first.
Read the given information carefully and fill everything in on the rough sketch. Label the vertices of the triangle.
2. Step 2: Draw a line segment and mark the first side of triangle.
Begin the construction with line segment $BC$.
Draw a line segment (longer than 8 cm) and make a dot at point $B$.
Use the compass to measure a length of 8 cm on your ruler.
Keep the compass open at exactly the same width. Put the point of the compass on point $B$ and draw a small arc to mark off a length of 8 cm. Label the point of intersection as point $C$.
3. Step 3: Use a protractor to draw the included angle.
Put the origin of the protractor on point $B$.
Use the inner scale of the protractor and find $65^{\circ}$. Make a small dot at the edge of the protractor to mark where $65^{\circ}$ is.
Remove the protractor. Use a ruler to join point $B$ to the small dot.
4. Step 4: Measure length of the second side of triangle.
Use the compass to measure a length of 4 cm on your ruler.
Keep the compass open by exactly the same amount. Put the compass on point $B$ and draw a small arc to mark off a length of 4 cm. Label the point of intersection as point $A$.
5. Step 5: Complete the construction.
Use a ruler to join point $A$ and point $C$.
6. Step 6: Remember to check your work.
Use a ruler to check that $AB$ = 4 cm and $BC$ = 8 cm.
Use a protractor to check that $A\hat{B}C$ = $65^{\circ}$.
• Never erase the construction lines after you have completed a construction. These may be used in giving marks for your work.
• The constructions provided here may not display at the exact size on all devices. Check your own constructions to make sure they are correct.
### Exercise 11.1: Construct a triangle given two sides and an included angle
1. Construct $\triangle DEF$ with $DE$ = 6 cm, $\hat{E}$ = $50^{\circ}$ and $EF$ = 7 cm.
Make a rough sketch first.
Follow the steps given in Worked Example 11.1 to do the construction.
2. Construct $\triangle PQR$ with $PQ$ = 7 cm, $\hat{Q}$ = $120^{\circ}$ and $QR$ = 5 cm.
Make a rough sketch first. Remember that an angle of $120^{\circ}$ is an obtuse angle.
Follow the steps to do the construction.
3. Construct $\triangle XYZ$ with $XY$ = 2 cm, $\hat{Y}$ = $90^{\circ}$ and $YZ$ = 8 cm.
Make a rough sketch first. Remember that an angle of $90^{\circ}$ is a right angle.
4. Make an accurate construction of $\triangle KLM$ as shown below.
The rough sketch is given to you. Follow the steps to construct a triangle with the dimensions given.
5. Make an accurate construction of $\triangle STU$ as shown below.
The rough sketch is given.
6. Make an accurate construction of $\triangle VWX$ as shown below.
• If you have to do a construction involving a line that is not horizontal on the page in front of you, you can turn your workbook.
• Remember never to erase the construction lines after you have completed a construction.
### Construct a triangle given two angles and the side between them
In this section you will construct triangles where you are given two angles as well as the length of the side between these two angles.
### Worked Example 11.2: Constructing a triangle given two angles and the side between them
Construct $\triangle ABC$ with $\hat{B}$ = $35^{\circ}$, $BC$ = 7 cm and $\hat{C}$ = $70^{\circ}$.
1. Step 1: Make a rough sketch of the triangle first.
Read the given information carefully and fill everything in on the rough sketch. Label the vertices of the triangle.
2. Step 2: Start with the given line length.
Begin the construction with line segment $BC$.
Draw a line segment (longer than 7 cm) and make a dot at point $B$.
Use the compass to measure a length of 7 cm on your ruler.
Keep the compass open at exactly the same width. Put the compass on point $B$ and draw a small arc to mark off a length of 7 cm. Label the point of intersection as point $C$.
3. Step 3: Use a protractor to draw the first angle.
Put the origin of the protractor on point $B$. Make sure that $BC$ fits exactly onto the base line of the protractor.
Use the inner scale and find $35^{\circ}$. Make a small dot next to the protractor.
Remove the protractor. Use a ruler to draw a line segment from point B through the small dot.
4. Step 4: Use a protractor to draw the second angle.
Put the origin of the protractor on point $C$. Make sure that $BC$ fits exactly onto the base line of the protractor.
Use the outer scale and find $70^{\circ}$. Make a small dot next to the protractor.
Remove the protractor. Use a ruler to draw a line segment from point C through the small dot.
The lines you have drawn to create your two angles intersect. Label the point of intersection point $A$.
5. Step 5: Remember to check your work.
Use a ruler to check that $BC$ = 7 cm.
Use a protractor to check that $\hat{B}$ = $40^{\circ}$ and that $\hat{C}$ = $70^{\circ}$.
### Exercise 11.2: Construct a triangle given two angles and the side between them
1. Construct $\triangle DEF$ with $\hat{E}$ = $50^{\circ}$, $EF$ = 8 cm and $\hat{F}$ = $60^{\circ}$.
Make a rough sketch first.
Follow the steps given in Worked Example 11.2 to do the construction.
2. Construct $\triangle PQR$ with $\hat{Q}$ = $30^{\circ}$, $\hat{R}$ = $130^{\circ}$ and $QR$ = 4 cm.
Remember that an angle of $130^{\circ}$ is an obtuse angle.
3. Construct $\triangle XYZ$ with $YZ$ = 6 cm, $\hat{Y}$ = $60^{\circ}$ and $\hat{Z}$ a right angle.
Remember that a right angle is an angle of $90^{\circ}$.
4. Make an accurate construction of $\triangle DTM$ as shown below. (Note: The given diagram is not to scale.)
The rough sketch is given. Follow the steps to construct a triangle that has the given dimensions.
5. Make an accurate construction of $\triangle JME$ as shown below. (Note: The given diagram is not to scale.)
6. Make an accurate construction of $\triangle CDE$ as shown below. (Note: The given diagram is not to scale.)
7. Explain why it is impossible to construct $\triangle KLM$ if $\hat{L}$ = $95^{\circ}$, $LM$ = 7.5 cm and $\hat{M}$ = $110^{\circ}$.
It is impossible to construct $\triangle KLM$ because a triangle can have only one obtuse angle. The information that is given cannot be used to construct a triangle.
### Construct a triangle given all three sides
In this section you will construct triangles where you are given all three sides of a triangle.
There are different methods to label the sides of a triangle. For example, the sides of $\triangle ABC$ can be labelled as follows:
Side $AB$ is the side opposite $\hat{C}$. Side $AB$ can also be labelled as c.
Side $BC$ is the side opposite $\hat{A}$. Side $BC$ can also be labelled as a.
Side $AC$ is the side opposite $\hat{B}$. Side $AC$ can also be labelled as b.
### Worked Example 11.3: Constructing a triangle given all three sides
Construct $\triangle ABC$ with a = 5 cm, b = 7 cm and c = 6 cm.
1. Step 1: Make a rough sketch of the triangle first.
Read the given information carefully and fill everything in on the rough sketch. Label the vertices of the triangle.
2. Step 2: Draw a line segment and mark first side of triangle.
Begin the construction with line segment $BC$.
Draw a line segment (longer than 5 cm) and make a dot at point $B$.
Use the compass to measure a length of 5 cm on your ruler.
Keep the compass open at exactly the same width. Put the compass on point $B$ and draw a small arc to mark off a length of 5 cm. Label the point of intersection as point $C$.
3. Step 3: Draw an arc to mark the length of the second side of the triangle.
Use the compass to measure a length of 6 cm on your ruler.
Keep the compass open at exactly the same width. Put the compass on point $B$ and draw a small arc to mark off a length of 6 cm.
4. Step 4: Draw an arc to mark the length of the third side of the triangle.
Use the compass to measure a length of 7 cm on your ruler.
Keep the compass open by exactly the same amount. Put the compass on point $C$ and draw a small arc to mark off a length of 7 cm.
5. Step 5: Complete the construction.
Label the point of intersection between the two arcs (of Step 3 and Step 4) as point $A$.
Use a ruler and join point $A$ and point $B$.
Use a ruler and join point $A$ and point $C$.
6. Step 6: Remember to check your construction.
Use a ruler to check if $AB$ = 6 cm, $BC$ = 5 cm and $AC$ = 7 cm.
Never erase the construction lines after you have completed a construction.
### Exercise 11.3: Construct a triangle given all three sides
1. Consider $\triangle PQR$ and $\triangle KMN$ in the diagrams below. (Note: The given diagrams are not to scale.)
1. What is the length of side $PR$?
8 cm
1. What is the length of side q?
8 cm
1. What is the length of side $KN$?
4 cm
1. What is the length of side k?
6 cm
1. What side has a length of 7 cm? (Use two different notations to give the answer.)
$PQ$ = r = 7 cm
1. What side has a length of 3 cm? (Use two different notations to give the answer.)
$KM$ = n = 3 cm
1. Make an accurate construction of $\triangle PQR$. A rough sketch of $\triangle PQR$ is given in number 1 of this exercise.
Follow the steps given in Worked Example 11.3 to do the construction. You can start with any of the line segments. The solution shown started from $PQ$.
1. Make an accurate construction of $\triangle KMN$. A rough sketch of $\triangle KMN$ is given in number 1 of this exercise.
Follow the steps to do the construction. You can start with any of the line segments. The solution shown started from $KN$.
1. Make an accurate construction of $\triangle DEF$ with d = 10 cm, e = 8 cm and f = 6 cm. Give the size of $\hat{D}$ by measuring it with a protractor.
Follow the steps to do the construction.
$\hat{D}$ = $90^{\circ}$
1. Make an accurate construction of $\triangle STU$ with s = 4 cm, and t = u = 7 cm. What type of triangle is $\triangle STU$?
$\triangle STU$ is an isosceles triangle, because two of the sides are the same length.
1. Explain why it is impossible to construct $\triangle XYZ$ if $XY$ = 2 cm, $YZ$ = 15 cm and $XZ$ = 3 cm.
It is impossible to construct $\triangle XYZ$. The arcs drawn for the sides of 2 cm and 3 cm will not intersect each other, because 15 cm is much longer than 5 cm (2 cm + 3 cm = 5 cm).
## 11.2 Bisect angles
An angle is the amount of turn between two straight lines that have a common end point. The common end point is called the vertex of the angle.
When we bisect an angle, we divide the amount of turn into two equal halves without actually measuring the angle. The line that divides the angle is called a bisector (or an angle bisector).
bisect When we bisect an angle, we divide the angle into two equal halves.
bisector The bisector (or angle bisector) is the line that divides an angle into two equal halves.
The construction in this section is based on the properties of a circle.
In the diagram below $BC$ is the diameter of the circle. $A$ is the centre of the circle.
This means that $AB$ is a radius of the circle and $AC$ is also a radius of the circle. So, $AB$ is equal to $AC$.
When we do constructions we often do not draw a complete circle.
The parts of the circumference that are left are called arcs. An arc is part of the circumference of a circle.
arc An arc is part of the circumference of a circle.
We will draw arcs when we bisect an angle. It is important to remember that if we draw two arcs while keeping our compass open at exactly the same width, then we are drawing two radiuses with equal lengths. $AB$ is still a radius of the circle, and $AC$ is still a radius of the circle. So, $AB$ is equal to $AC$ $-$ even if we did not draw a complete circle.
### Worked example 11.4: Bisecting an acute angle
$A\hat{B}C$ is an acute angle. Use a compass and a ruler to bisect $A\hat{B}C$.
Remember that an acute angle is an angle that is between $0^{\circ}$ and $90^{\circ}$.
1. Step 1: Draw the first two arcs.
Put the compass on point $B$ (the blue dot). Point $B$ is the vertex of the angle. Keep the compass at the same width and draw two arcs to intersect with the two arms of the angle.
2. Step 2: Draw the third arc.
Put the compass on the point where the arc from Step 1 crosses $AB$ (the orange dot) and draw an arc as shown.
3. Step 3: Draw the fourth arc.
Keep the compass at the same width as for Step 2. Put the point of the compass on the point where the arc from Step 1 crosses $BC$ (the purple dot) and draw an arc as shown.
4. Step 4: Complete the construction.
Label the point of intersection between the two arcs as point $D$.
Use a ruler to draw a line segment from point $B$ through point $D$.
$BD$ is the bisector of $A\hat{B}C$.
5. Step 5: Check your work.
Use a protractor to measure the two angles. Each one of them should be exactly half of the given angle.
$\therefore$ $A\hat{B}D$ = $D\hat{B}C$ = $25^{\circ}$. $A\hat{B}C$ = $50^{\circ}$, so we have bisected the angle.
If you find it easier, you may begin the construction with one big arc instead of the two smaller arcs.
In the worked example above, an acute angle was bisected. You use exactly the same steps to bisect any type of angle. Next, we will bisect a reflex angle.
### Worked Example 11.5: Bisecting a reflex angle
$P\hat{Q}R$ is a reflex angle. Use a compass and a ruler to bisect $P\hat{Q}R$.
Remember that a reflex angle is between $180^{\circ}$ and $360^{\circ}$.
1. Step 1: Draw the first two arcs.
Put the compass on point $Q$ (the blue dot). Point $Q$ is the vertex of the angle. Keep the compass at the same width and draw two arcs to intersect with the two arms of the angle.
2. Step 2: Draw the third arc.
Put the compass on the point where the arc from Step 1 crosses $PQ$ (the orange dot) and draw an arc as shown.
3. Step 3: Draw the fourth arc.
Keep the compass at the same width as for Step 2. Put the compass on the point where the arc from Step 1 crosses $QR$ (the purple dot) and draw an arc as shown.
4. Step 4: Complete the construction.
Label the point of intersection between the two arcs as point $S$.
Use a ruler to draw a line segment from point $Q$ through point $S$.
$QS$ is the bisector of $P\hat{Q}R$.
5. Step 5: Check your work.
Use a protractor to measure the two angles. Each one of them should be exactly half of the given angle.
$P\hat{Q}S$ = $R\hat{Q}S$ = $150^{\circ}$. $P\hat{Q}R$ = $300^{\circ}$, so we have bisected the angle.
You can also bisect a reflex angle by bisecting the acute angle, and then producing (making longer) this bisector. You need to continue the line through the vertex to the other side, and this will give you the bisector of the reflex angle.
In the diagram below, $QT$ is the bisector of acute $P\hat{Q}R$, and $QS$ is the bisector of reflex $P\hat{Q}R$. $SQ$ and $QT$ form a straight line.
Remember not to erase the construction lines after you have completed a construction.
### Exercise 11.4: Bisect an angle
1. The diagram below shows an acute angle. Use a compass and ruler to bisect $A\hat{B}C$. Label the bisector $AD$.
Measure and write down the sizes of $A\hat{B}D$ and $D\hat{B}C$.
Follow the steps in Worked Example 11.4 to do the construction.
$A\hat{B}D$ = $D\hat{B}C$ = $25^{\circ}$
2. The diagram below shows an obtuse angle. Use a compass and ruler to bisect $D\hat{E}F$. Label the bisector $EG$.
Measure and write down the sizes of $D\hat{E}G$ and $G\hat{E}F$.
$D\hat{E}G$ = $G\hat{E}F$ = $50^{\circ}$
3. Draw your own straight $G\hat{H}J$. Bisect $G\hat{H}J$ and label the bisector $HK$. What type of angles are $G\hat{H}K$ and $K\hat{H}J$?
A straight angle is a straight line. Follow the steps to bisect the angle.
$G\hat{H}K$ and $K\hat{H}J$ are right angles. $G\hat{H}K$ = $K\hat{H}J$ = $90^{\circ}$
4. The diagram below shows a reflex angle. Use a compass and ruler to bisect $K\hat{L}M$. Label the bisector $LN$.
What type of angle is $K\hat{L}N$?
Follow the steps in Worked Example 11.5 to do the construction.
$K\hat{L}N$ is an obtuse angle.
5. The diagram below shows a reflex angle. Use a compass and ruler to bisect $P\hat{Q}R$. Label the bisector $QS$.
What type of angle is reflex $P\hat{Q}R$ together with acute $P\hat{Q}R$?
Reflex $P\hat{Q}R$ + acute $P\hat{Q}R$ = $360^{\circ}$. This angle is a revolution.
6. Draw your own obtuse $X\hat{Y}Z$. Construct angle bisectors to find an angle that is one quarter of $X\hat{Y}Z$.
An obtuse angle is between $90^{\circ}$ and $180^{\circ}$. You may draw any obtuse angle. An example is given in the solution below.
Bisect $X\hat{Y}Z$ and label the bisector $YS$.
Bisect any one of the two smaller angles again. In the diagram below, $S\hat{Y}Z$ is bisected.
$S\hat{Y}T$ is one quarter of $X\hat{Y}Z$, and $T\hat{Y}Z$ is also one quarter of $X\hat{Y}Z$.
7. Draw a big triangle with three acute angles and label the triangle $\triangle ABC$. Construct the bisectors for $\hat{A}$, $\hat{B}$ and $\hat{C}$. What do you notice about the three bisectors?
The three angle bisectors meet each other at one single point.
## 11.3 Summary
• A construction in Mathematics is an accurate drawing of angles and lines.
• We use arcs in many constructions. An arc is part of the circumference of a circle.
• Using a compass, protractor, ruler and pencil, we can carry out the following constructions:
• Construct a triangle given two sides and an included angle
• Construct a triangle given two angles and the side between them
• Construct a triangle given all three sides
• To construct an angle bisector, you may only use a compass, ruler and pencil. You may use a protractor to check that you have bisected the angle.
• It can be helpful is a good idea to rotate your workbook if you have to do a construction involving a line that is not horizontal on the page in front of you.
• Never erase the construction lines after you have completed a construction.
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# How do you solve | 1/2-3/4x | + 1=3 ?
Mar 2, 2016
$-$2 and 10/3.
#### Explanation:
The given equation is equivalent to the pair 1/2$-$3 x/4 = $\pm$2.
Mar 2, 2016
$x = - 2 \text{ or "x=10/3" }$Note that $\frac{10}{3} = 3.33 \overline{3}$
#### Explanation:
The very nature of this type of problem is that it usually has 2 solutions for $x$
We need to end up with the 'Absolute' part of the equation on its own and on the left of the equals sign and everything else on the other.
Given:$\text{ } \textcolor{b r o w n}{| \frac{1}{2} - \frac{3}{4} x | + 1 = 3}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Simplifying}}$
Subtract $\textcolor{b l u e}{1}$ from both sides
" "color(brown)(|1/2-3/4 x|+1color(blue)(-1)" "=" "3color(blue)(-1))
$\text{ "|1/2-3/4 x|+0" "=" } 2$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving for } x}$
Think of this as:
$\text{ "|1/2-3/4 x|" "=" } | \pm 2 |$
color(brown)("Consider: "1/2-3/4 x" "=" "-2" "
Multiply by (-1 ) giving
$\text{ } - \frac{1}{2} + \frac{3}{4} x = + 2$
$\text{ } \frac{3}{4} x = 2 + \frac{1}{2}$
$\textcolor{b l u e}{\text{ " x" "=" "4/3xx5/2" " = " } \frac{10}{3}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(brown)("Consider: "1/2-3/4 x=+2
Multiply by (-1 ) giving
$\text{ } - \frac{1}{2} + \frac{3}{4} x = - 2$
$\text{ } \frac{3}{4} x = - 2 + \frac{1}{2}$
$\text{ } \textcolor{b l u e}{x = \frac{4}{3} \times \left(- \frac{3}{2}\right) = - 2}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
" "color(magenta)(x = -2" or "x=10/3)
|
### Learning Objectives
By the end of this section, you will be able to:
• Use the Product Property to simplify radical expressions
• Use the Quotient Property to simplify radical expressions
Before you get started, take this readiness quiz.
1. Simplify:
If you missed this problem, review (Figure).
2. Simplify:
If you missed this problem, review (Figure).
3. Simplify:
If you missed this problem, review (Figure).
### Use the Product Property to Simplify Radical Expressions
We will simplify radical expressions in a way similar to how we simplified fractions. A fraction is simplified if there are no common factors in the numerator and denominator. To simplify a fraction, we look for any common factors in the numerator and denominator.
A radical expression, is considered simplified if it has no factors of So, to simplify a radical expression, we look for any factors in the radicand that are powers of the index.
For real numbers a and m, and
For example, is considered simplified because there are no perfect square factors in 5. But is not simplified because 12 has a perfect square factor of 4.
Similarly, is simplified because there are no perfect cube factors in 4. But is not simplified because 24 has a perfect cube factor of 8.
To simplify radical expressions, we will also use some properties of roots. The properties we will use to simplify radical expressions are similar to the properties of exponents. We know that The corresponding of Product Property of Roots says that
Product Property of nth Roots
If and are real numbers, and is an integer, then
We use the Product Property of Roots to remove all perfect square factors from a square root.
Simplify Square Roots Using the Product Property of Roots
Simplify:
Simplify:
Simplify:
Notice in the previous example that the simplified form of is which is the product of an integer and a square root. We always write the integer in front of the square root.
Be careful to write your integer so that it is not confused with the index. The expression is very different from
Simplify a radical expression using the Product Property.
1. Find the largest factor in the radicand that is a perfect power of the index. Rewrite the radicand as a product of two factors, using that factor.
2. Use the product rule to rewrite the radical as the product of two radicals.
3. Simplify the root of the perfect power.
We will apply this method in the next example. It may be helpful to have a table of perfect squares, cubes, and fourth powers.
Simplify:
Simplify:
Simplify:
The next example is much like the previous examples, but with variables. Don’t forget to use the absolute value signs when taking an even root of an expression with a variable in the radical.
Simplify:
Simplify:
Simplify:
We follow the same procedure when there is a coefficient in the radicand. In the next example, both the constant and the variable have perfect square factors.
Simplify:
Simplify:
Simplify:
In the next example, we continue to use the same methods even though there are more than one variable under the radical.
Simplify:
Simplify:
Simplify:
Simplify:
Simplify:
Simplify:
no real number
We have seen how to use the order of operations to simplify some expressions with radicals. In the next example, we have the sum of an integer and a square root. We simplify the square root but cannot add the resulting expression to the integer since one term contains a radical and the other does not. The next example also includes a fraction with a radical in the numerator. Remember that in order to simplify a fraction you need a common factor in the numerator and denominator.
Simplify:
The terms cannot be added as one has a radical and the other does not. Trying to add an integer and a radical is like trying to add an integer and a variable. They are not like terms!
Simplify:
Simplify:
### Use the Quotient Property to Simplify Radical Expressions
Whenever you have to simplify a radical expression, the first step you should take is to determine whether the radicand is a perfect power of the index. If not, check the numerator and denominator for any common factors, and remove them. You may find a fraction in which both the numerator and the denominator are perfect powers of the index.
Simplify:
Simplify:
Simplify:
In the last example, our first step was to simplify the fraction under the radical by removing common factors. In the next example we will use the Quotient Property to simplify under the radical. We divide the like bases by subtracting their exponents,
Simplify:
Simplify:
Simplify:
Remember the Quotient to a Power Property? It said we could raise a fraction to a power by raising the numerator and denominator to the power separately.
We can use a similar property to simplify a root of a fraction. After removing all common factors from the numerator and denominator, if the fraction is not a perfect power of the index, we simplify the numerator and denominator separately.
If and are real numbers, and for any integer then,
How to Simplify the Quotient of Radical Expressions
Simplify:
Simplify:
Simplify:
Simplify a square root using the Quotient Property.
1. Simplify the fraction in the radicand, if possible.
2. Use the Quotient Property to rewrite the radical as the quotient of two radicals.
3. Simplify the radicals in the numerator and the denominator.
Simplify:
Simplify:
Simplify:
Be sure to simplify the fraction in the radicand first, if possible.
Simplify:
Simplify:
Simplify:
In the next example, there is nothing to simplify in the denominators. Since the index on the radicals is the same, we can use the Quotient Property again, to combine them into one radical. We will then look to see if we can simplify the expression.
Simplify:
Simplify:
Simplify:
Access these online resources for additional instruction and practice with simplifying radical expressions.
### Key Concepts
• For real numbers a, m and
is considered simplified if a has no factors of
• Product Property of nth Roots
• For any real numbers, and and for any integer
and
• How to simplify a radical expression using the Product Property
1. Find the largest factor in the radicand that is a perfect power of the index.
Rewrite the radicand as a product of two factors, using that factor.
2. Use the product rule to rewrite the radical as the product of two radicals.
3. Simplify the root of the perfect power.
• Quotient Property of Radical Expressions
• If and are real numbers, and for any integer then,
and
• How to simplify a radical expression using the Quotient Property.
1. Simplify the fraction in the radicand, if possible.
2. Use the Quotient Property to rewrite the radical as the quotient of two radicals.
3. Simplify the radicals in the numerator and the denominator.
#### Practice Makes Perfect
Use the Product Property to Simplify Radical Expressions
In the following exercises, use the Product Property to simplify radical expressions.
In the following exercises, simplify using absolute value signs as needed.
not real
not real
Use the Quotient Property to Simplify Radical Expressions
In the following exercises, use the Quotient Property to simplify square roots.
#### Writing Exercises
Explain why Then explain why
Explain why is not equal to
Explain how you know that
Explain why is not a real number but is.
#### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After reviewing this checklist, what will you do to become confident for all objectives?
|
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