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# Question #a22d8
##### 1 Answer
Jan 30, 2017
$C : y - 4 = \frac{1}{6} \left(x + 5\right)$
#### Explanation:
The equation of a line in $\textcolor{b l u e}{\text{point-slope form}}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y - {y}_{1} = m \left(x - {x}_{1}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the slope and $\left({x}_{1} , {y}_{1}\right) \text{ a point on the line}$
$\text{here "m=1/6" and } \left({x}_{1} , {y}_{1}\right) = \left(- 5 , 4\right)$
Substitute these values into the equation.
$y - 4 = \frac{1}{6} \left(x - \left(- 5\right)\right)$
$\Rightarrow y - 4 = \frac{1}{6} \left(x + 5\right) \leftarrow \textcolor{red}{\text{ in point-slope form}}$
We can distribute the bracket and simplify to obtain another version of the equation.
$y - 4 = \frac{1}{6} x + \frac{5}{6}$
$y = \frac{1}{6} x + \frac{5}{6} + 4$
$\Rightarrow y = \frac{1}{6} x + \frac{29}{6} \leftarrow \textcolor{red}{\text{ in slope-intercept form}}$
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Solving 3rd Order ODE by Reduction of Order
I am trying to solve the following ODE: $$x^3y''' -3x^2y'' +(6-x^2)xy' -(6-x^2)y=0$$ if $$y_1=x$$. The professor said that the solution could be found by reduction of order. If it's second order the solution this problem is quite easy, but I don't really know how to solve it if it's 3rd order. Any help will be appreciated.
• Let $y(x)=x z(x)$ and replace. It will be very simple. Commented Dec 31, 2021 at 9:54
Searching for the other homogeneous solutions by assuming it is the product of some function times another known solution is called variation of parameters. Formally we are plugging in the guess $$y = v\cdot y_1 = x\cdot v$$ into the equation $$D[y] = 0$$
$$\begin{cases}y = xv \\ y' = v+xv' \\ y'' = 2v' + xv'' \\ y'' = 3v'' + xv'''\end{cases} \implies x\cdot D[v] + (3x^3v''-6x^2v'+(6-x^2)xv) = 0$$
$$\implies (x^4v'''-3x^3v''+(6-x^2)x^2v'-(6-x^2)xv)+(3x^3v''-6x^2v'+(6-x^2)xv)$$
$$= x^4(v'''-v') = 0$$
In other words $$v$$ is the three homogeneous solutions to the above differential equation
$$v = C_1 e^x + C_2 e^{-x} + C_3$$
Your professor was justified in calling this a reduction of order because $$v'''-v'=0$$ is secretly a second order differential equation, with the constant solution already represented by the first homogeneous solution $$y_1 = x$$. This gives us our final solutions
$$y = C_1 x + C_2 x e^{x} + C_3xe^{-x}$$
• Thanks. I see the point now. Really appreciate your help. Commented Dec 31, 2021 at 11:11
It will be easier if you use the formula for the second solution: $$y_2=v(x)x$$, where x is your first solution.
Then
$$$$\begin{array} fy_2=vx\\ y_2'=v'x+v\\ y_2''=v''x+2v'\\ y_2'''=v'''x+3v'' \end{array}$$$$
Insert for every respective term of y in the original ode and get:
$$x^3(v'''x+3v'')-3x^2(v''x+2v')+(6-x^2)x(v'x+v)-(6-x^2)vx=0$$
which gives
$$x^4v'''+3x^3v''-3x^3v''-6x^2v'+(6xv'-x^3v')x=0$$
$$x^4v'''+3x^3v''-3x^2v''-6x^2v'+6x^2v'-x^4v'=0$$
Sum up each related term and cancel out $$-6x^2v'+6x^2v'$$ and $$3x^3v''-3x^3v''$$ and obtain
$$x^4v'''-x^4v'=0$$
Divide by $$x^4$$ and get the simplified form:
$$v'''-v'=0$$
Use reduction of order by letting $$u''=v'''$$ and get:
$$u''-u=0$$
This has the simple solution $$\lambda=\pm\sqrt{-4\cdot-1} \longrightarrow u(x)=e^{2x}+e^{-2x}$$. Replace u(x) back to v by integrating it once and get:
$$v(x)=\int u(x)dx=\int e^{2x}+e^{-2x}dx=\frac{1}{2}e^{2x}-\frac{1}{2}e^{-2x}+C$$
At last, go back to the form of $$y_2=v(x)x\rightarrow y_2(x)=\frac{x}{2}e^{2x}-\frac{x}{2}e^{-2x}+Cx$$
Nicely put:
$$$$y_2(x)=\frac{x}{2}\bigg(e^{2x}-e^{-2x}\bigg)+Cx$$$$
• I really appreciate your help. I tried the method suggested by Ninad and got the same result as yours. Commented Dec 31, 2021 at 11:22
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# Which number is equal to $(\dfrac{{0.1}}{{0.01}} + \dfrac{{0.01}}{{0.1}})$?$A)10.1$$B)1.10$$C)1.01$$D)10.01$
Last updated date: 24th Jul 2024
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Hint: First we have to define what the terms we need to solve the problem are.
Since we need to know about the division and addition operator. A divisor is a number that will divide another number either completely (remainder zero) or with a remainder (real numbers)${\text{ }}\dfrac{{Dividend}}{{Divisor{\text{ }}}} = {\text{ }}Quotient$. And for the addition operator the things are very simple; Addition is the sum of two or more than two numbers, or values, or variables, and in addition if we sum the two or more numbers a new frame of the number will be found.
Complete step-by-step solution:
Since the given questions are in decimals and we will first convert into the whole numbers so that we can simply solve by using the division and addition operators;
Let us first convert; $\dfrac{{0.1}}{{0.01}} \times \dfrac{{100}}{{100}}$ (so the denominator will be a whole number and easy to divide)
Solving we get $\dfrac{{0.1}}{{0.01}} \times \dfrac{{100}}{{100}} = \dfrac{{10}}{1}$ and the answer is $10$ (by division we obtained this much simplification)
Now take that second term which is $\dfrac{{0.01}}{{0.1}} \times \dfrac{{10}}{{10}}$ and multiplied and divided by ten because the denominator needs to be one and hence solving, we get; $\dfrac{{0.01}}{{0.1}} \times \dfrac{{10}}{{10}} = \dfrac{{0.1}}{1} = 0.1$(by division algorithm)
Finally, we just need to add both the terms using the addition algorithm which is $(\dfrac{{0.1}}{{0.01}} + \dfrac{{0.01}}{{0.1}}) = 10 + 0.1 = 10.1$ and hence option $A)10.1$ is correct.
Since there isn’t any possibility of getting options like $B)1.10$, $C)1.01$, $D)10.01$
Because of the division algorithm and addition property; the first term gets ten and the second term gets zero point one; if we use it correctly, we only get option A
Note: We can also able to solve this problem (without converting) by just divide the terms inside like $\dfrac{{0.1}}{{0.01}} = 10$and $\dfrac{{0.01}}{{0.1}} = 0.1$ hence finally add them using the addition property and thus we get the same result.
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Apr 18
## From minimum to infimum: Math is just a logical game
The true Math is a continuous exercise in logic. A good teacher makes that logic visible and tangible. A genius teacher does the logic mentally and only displays the result. I am trying to be as good a teacher as humanly possible.
## Steps to develop a new definition
What is the minimum $m$ of the set $A=[1,\infty)$? In my class everybody says $m=1$.
Step 2. Give the most visual definition.
Here is a good candidate: the minimum $m$ of a set $A$ is its leftmost point.
Step 3. Formalize the definition you gave.
We said "its point". This means $m$ should be an element of $A$. "Leftmost" is formalized as $m\le a$ for any $a\in A$. Thus the formal definition should be: the number $m$ is called a minimum of a set $A$ if
1) $m\le a$ for any $a\in A$ and
2) $m\in A$.
Step 4. Look at bad cases when the definition fails. Try to come up with a generalized definition that would cover the bad cases.
## Here is a bad case: meet the infinity
Let $A=(1,\infty)$. In my class, some students suggested $m=0$ and $m=2$. Can you see why neither is good? See the explanation below if you can't. I would try $m=1$. It still satisfies part 1) of the above definition but does not belong to $A$. This is the place to stretch one's imagination.
Statement 1. No element of $A$ satisfies the formal definition above.
Proof. Take any element $a\in A$. Then it is larger than 1 and the number $a_1=\frac{a+1}{2}$ is halfway between $a$ and 1. We have shown that to the left of any $a\in A$ there is another element of $A$. So no element of $A$ is leftmost.
In the game of chess, the one who thinks several moves ahead wins. A stronger version of Statement 1 is the following.
Statement 2. To the left of any $a\in A$ there are infinitely many elements of $A$.
Proof. Indeed, set $a_n=1+1/n,\ n=1,2,...$ The numbers $a_n$ approach 1 from the right. As $n$ increases, they become closer and closer to 1. For any given $a\in A$, infinitely many of these numbers will satisfy $1.
Step 5. When thinking about a new definition, try to exclude undesirable outcomes.
Since for $m=1$ condition 2) is not satisfied, one might want to omit it altogether. However, leaving only part 1) would give a bad result. For example, $m=0$ would satisfy such a "definition", and it would be bad for two reasons. Firstly, there is no uniqueness. We could take any number $m<1$. Secondly, if you take any $m<1$, in the interval $(m,1)$ there are no elements of $A$, so there is no reason to call such a number a minimum of $A$.
Definition. A number $m$ is called an infimum of $A$ (denoted $m=\inf A$) if
I) $m\le a$ for any $a\in A$ and
II) in $A$, there is a sequence $\{a_n\}$ approaching $m$.
The above discussion shows that $\inf (1,\infty)=1$.
Exercise 1. Repeat all of the above for $A=(-\infty,-1)$. First define the maximum of a set. Then modify the definition to obtain what is called a supremum.
Exercise 2. Let $A=(e,\pi)$, where $e=2.71828...$ is the basis of the natural log and $\pi=3.1415...$ is the ratio Length of circumference/Diameter of that circumference. Find the infimum and supremum of $A$. Simply naming them is not enough; you have to prove that both parts of the definitions are satisfied.
Exercise 3. The infimum is also defined as the largest lower bound. Can you prove equivalence of the two definitions?
Exercise 4. The supremum is also defined as the least upper bound. Can you prove equivalence of the two definitions?
Exercise 5. Consider any two sets and their union. What is the relationship between the infimums of the three sets?
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# 8.3 Linear Relationships
### Lesson 1
• I can graph a proportional relationship from a story.
• I can use the constant of proportionality to compare the pace of different animals.
### Lesson 2
• I can graph a proportional relationship from an equation.
• I can tell when two graphs are of the same proportional relationship even if the scales are different.
### Lesson 3
• I can scale and label coordinate axes in order to graph a proportional relationship.
### Lesson 4
• I can compare proportional relationships represented in different ways.
### Lesson 5
• I can find the rate of change of a linear relationship by figuring out the slope of the line representing the relationship.
### Lesson 6
• I can interpret the vertical intercept of a graph of a real-world situation.
• I can match graphs to the real-world situations they represent by identifying the slope and the vertical intercept.
### Lesson 7
• I can use patterns to write a linear equation to represent a situation.
• I can write an equation for the relationship between the total volume in a graduated cylinder and the number of objects added to the graduated cylinder.
### Lesson 8
• I can explain where to find the slope and vertical intercept in both an equation and its graph.
• I can write equations of lines using y=mx+b.
### Lesson 9
• I can give an example of a situation that would have a negative slope when graphed.
• I can look at a graph and tell if the slope is positive or negative and explain how I know.
### Lesson 10
• I can calculate positive and negative slopes given two points on the line.
• I can describe a line precisely enough that another student can draw it.
### Lesson 11
• I can write equations of lines that have a positive or a negative slope.
• I can write equations of vertical and horizontal lines.
### Lesson 12
• I know that the graph of an equation is a visual representation of all the solutions to the equation.
• I understand what the solution to an equation in two variables is.
### Lesson 13
• I can find solutions $(x, y)$ to linear equations given either the $x$- or the $y$-value to start from.
### Lesson 14
• I can write linear equations to reason about real-world situations.
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• Level: GCSE
• Subject: Maths
• Word count: 3708
# Maths Coursework: Number Stairs
Extracts from this document...
Introduction
I have been set the task of working out the relationship between a 3 – step stair total and the position of the stair shape on the prearranged grid and other stair totals.
Aim:
As I work through this task I hope to find a formula that will help me to work out any stair total on a 10 by 10 grid.
Method:
I began by calculating the stair in a vertical sequence ranging from 1 step stair to 8 step stairs. The results of these calculations highlighted a definite pattern. These results were then summarised in a table from which a general formula was found.
Key:
= Base Number
n = Base number
1 Step Stair:
Difference = +1 +1 +1
+1 +1 +1
As we can see, the stair total for a 1-step stair using 1 as the base number on a 10 by 10-number grid is 1 as the total = 1. I can therefore conclude that the general equation for a 1-step stair on a 10 by 10 number grid is 1n + 0 = stair total where n is the base number. (Shaded box).
Total of step containing 1 as its base number – Difference
= 1- 1
= 0
## Formulae: 1n + 0 = Total
Testing the formula:
Total = 1n + 0
Total = 1 * 2 + 0
Total = 2 + 0
Total = 2
2 Step Stairs:
Difference = +3 +3 +3
+3 +3 +3
## 35
As we can see, the stair total for a 2-step stair using 1 as the base number on a 10 by 10-number grid is 14 as 1 + 2 + 11 = 14. I can therefore conclude that the general equation for a 2-step stair on a 10 by 10 number grid is 3n + 11 = stair total where n is the base number. (Shaded box).
Total of step containing 1 as its base number – Difference
= 14 – 3
= 11
Middle
Total = 36n + 924
Total = 36 * 2 + 924
Total = 72 + 924
Total = 996
## Total = 36n + 924
Final Formula:
Section 1:
= First Difference
= Second Difference
## 28
8
36
The data above is summarised in the table below.
n 0 1 2 3 4 5 6 7 8 nth term 0 1 3 6 10 15 21 28 36 1st difference 1 2 3 4 5 6 7 8 2nd difference 1 1 1 1 1 1 1
Observation:
The differences between the nth terms were increasing which meant that I had to use quadratic sequences. The general term of a quadratic sequence is:
un =AN 2 + BN + C
a= the second difference in the sequence over 2.
###### b = is the coefficient of the term in n 2.
c = is the numerical value or constant value.
a = 2nd difference / 2
a = 1 / 2
Counting back, the zeros term is 0. So when n = 0, un = 0. When ‘n’ = 0, both ‘n 2’ and ‘bn’ are equal to 0 so ‘c’ = 0.
C = 0
To find the value of ‘b’, we look at the first term. When ‘n’ = 1, the value of u is 2.
un = n + bn
= 1 + b
2 = 1 + b
b = 1 / 2
1/2s2+ 1/2 s
Section 2:
= First Difference
= Second Difference
= Third Difference
## 616
8
924
The data above is summarized in the table below.
n 0 1 2 3 4 5 6 7 8 nth term 0 0 11 44 110 220 385 616 924 1st difference 0 11 33 66 110 165 231 308 2nd difference 11 22 33 44 55 66 77 3rd difference 11 11 11 11 11 11
Nth term =an 3 + bn 2 + cn + d
When n = 1 13a+12b+1c+d=6 Equation 1: a + b + c + d = 0
When n = 2 23a+22b+2c+d=20 Equation 2: 8a + 4b + 2c + d = 11
When n = 3 33a+32b+3c+d=52 Equation 3: 27a + 9b + 3c + d = 44
When n = 4 43a+42b+4c+d=108 Equation 4: 64a + 16b + 4c + d = 110
Subtracting equation 1 from each of the other 3 gives:
Equation 5 7a+3b+c=11
Equation 6 26a+8b+2c=44
Equation 7 63a+15b+3c=110
Subtracting 2 times equations 5 from equation 6 gives:
Conclusion
an3 + bn2 + cn + d:
##### If n = 1 , 2 , 3 , 4 , 5
a+b+c+d , 8a+4b+2c+d , 27a+9b+3c+d , 64a+164+4c+d , 125a+25b+5c+d , 7a+3b+c , 9a+5b+c , 37a+7b+c , 61a+9b+c, 12a+2b , 18a+2b , 24a+2b
6a , 6a
I am now going to match the numbers in the cubic sequence formula above to the numbers that correspond to their positions in the co-efficient in front of g sequence. 6a =
a = 1/6
12a+2b = 3
12(1/6) + 2b = 3
2 + 2b = 3
2b = 1
b = 0.5
7a+3b+c= 3
7(1/6)+3(1/2)+c = 3
c= 1/3
a+ b + c + d = 1
1/6 + ½ + 21/3 + d = 1.
d = 0
Therefore the equation for this part of the sequence will be :
(1/6 n3 + 1/2n2 + 1/3n) = coefficient of g
Section 3: The Sequence For The Constant
The sequence for the constant will go exactly the same as the sequence for the co-efficient of g as all these values are the same. Therefore the equation for this part of the sequence will be :
(1/6 n3 + 1/2n2 + 1/3n) = the constant
Conclusion
##### Now that we have finished the three parts of the sequence we can put them together in order to make one equation for any size step-stair on any size grid.
(1/2 n2 + 3/2n + 1) x + (1/6 n3 + 1/2n2 + 1/3 n) g + (1/6 n3 + 1/2n2 + 1/3 n) = stair total
x means the number of squares in the stair shape..
g stands for grid size.
n is the equation number. (If you don't know the equation number it is the same number as the number of squares in the 2nd column from the left of the stair shape).
Reference:
• Internet: www.y-maths.co.uk/index.htm
• Higher GCSE for Edexcel
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# Related GCSE Number Stairs, Grids and Sequences essays
1. ## Maths coursework- stair totals. I shall be investigating the total and difference in sets ...
3 star(s)
works for each size of grid it has been proven that it does work. So for the three step total on any size grid the formula is; 6x+4g+4 Different Stair Sizes Next I shall investigate different stair sizes and the formulas for their totals.
2. ## GCSE Maths Sequences Coursework
the sequence so; Nth term for Perimeter = 12N+6 Shaded Stage (N) 1 2 3 4 Sequence 6 12 18 24 1st difference +6 +6 +6 2nd difference I can see a clear pattern here, the sequence for shaded is going up regularly in 6's, and therefore this sequence is a linear sequence and has an Nth term.
1. ## Number Grid Coursework
Generalisation Using this apparent relationship, it can be assumed that, when a 2x2 box is placed anywhere on the grid, the difference of the two products will be z for all possible widths. Therefore, the following equation should be satisfied with any real value of a and any real
2. ## Number stairs
I have highlighted a 3-step stair on the 12 by 12 Number grid The total of the numbers inside the highlighted 3-step stair is: 109 + 97 + 98 + 85 + 86 + 87 = 553 The stair total is all the six numbers in 3-step stair added together
1. ## Investigation of diagonal difference.
I will then calculate the top left corner multiplied by the bottom right corner. Once I have done this I will takeaway both products and find the final formulae. (X - 1)� G (X - 1)� G could be the formula for working out the diagonal difference for any square, size cutout.
2. ## Maths-Number Grid
33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82
1. ## Number Grid Maths Coursework.
Numerical examples 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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48
2. ## For other 3-step stairs, investigate the relationship between the stair total and the position ...
Using any of the algebra equation from the above table, e.g. 6x-36 or 6x-44 we can prove the results for any 3-step grid. The follow exercises in table 11 shows the expected results: Table 11 33 FORMULA = 6x - 36 43 44 1: ( 6 x 53)
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Mathematik Übungen
# Compare fractions (with like numerators or denominators)
3/4>3/7
## Comparing fractions
A fraction is a part of a whole. To find the greater fraction you need to find the fraction that contains a greater part of the whole.
If fractions have the same denominator, you can compare numerators:
\$\$3/4 > 1/4\$\$
It is possible to draw it to see the result:
If there is a common numerator, the fraction with the greater denominator is actually smaller:
\$\$1/3 > 1/4\$\$
You can draw it as well:
If two fractions do not have common numerator or denominator, you can find equivalent fractions with the same denominator.
E.g. compare \$3/8\$ and \$1/3\$:
\$\${3×3}/{8×3}=9/24\;\;\;{1×8}/{3×8}=8/24\;→\;\$\$
\$\$9/24>8/24\$\$
If you compare a fraction with a whole number, you can turn the whole number into a fraction with the same denominator:
\$\$1=1/1=2/2=3/3=4/4=5/5=...\$\$
\$\$4=8/2=12/3=16/4=20/5= ...\$\$
Compare 3 and \$9/4\$:
\$\$3=12/4\;→\;12/4>9/4\$\$
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Home » Maths » Difference Between Log and Ln
# Difference Between Log and Ln
logarithms function are important in the study of Maths , Physics and chemistry We usually express logarithms as log or ln. Both are widely used in problem and an understanding of the difference between them is important for good score also
Lets find the difference between log and ln
1. Log refers to a logarithm to the base 10 and ln refers to a logarithm to the base e(e = 2.718281828)
$log x = log_{10} x = y$ or $10^y= x$
$ln x = log_{e} x = y$ or $e^y= y$
So , log of number x is basically power to which a base of 10 must be raised to obtain a number x
While ln of number of x is basically power to which a base of e must be raised to obtain a number x
Here is the table for common values of log
2. The graph of both are shown below
The Relation between log and ln is given by
$ln x = 2.303 log x$
The ln logarithm graph has exactly the same shape as the base 10 logarithm graph; it is just 2.3 times as tall.
3. log is called the common logarithm while ln is called natural logarithm
4. $log_{10} x$ is usually written as log x . While $log_{e} x$ is usually written as ln x
5. The derivative for ln x is given by
$\frac {d}{dx} [ln x] = \frac {1}{x}$
The derivative for log x is given by
$\frac {d}{dx} [log x] = \frac {1}{x ln 10}$
As per the similarities, the domain of both the log and ln is ${0,-\infty}$ and range is ${\infty,-\infty}$
I hope the difference between log and ln is clear. We need to be very careful while writing these in equation as each denote different thing.
Here we example to practice the values of them
Example
Find the value of
a. $ln e^{3.2}$
b. $ln \frac {1}{e}$
c. $log 10^e$
d. $log 10000$
Solution
a. $ln e^{3.2} = 3.2 ln e = 3.2$
b. $ln \frac {1}{e} = ln e^{-1} = -1 ln e = -1$
c. $log 10^e = 2.303$
d. $log 10000 = log 10^4 = 4$
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Math Central Quandaries & Queries
Question from Kristy, a student: Can you help me with this equation? Find the equation, in standard form of the line perpendicular to 2x-3y=-5 and passing through (3,-2) Write the equation in standard form with all integer coefficient.
Hi Kristy. I can show you how it's done with a similar problem, then you can follow those steps in solving your question.
Example: Find the equation of the line that is perpendicular to 2x = y - 5 and that passes through the point (4, -3). Write the equation in standard form.
Solution:
Step 1: Find the slope of the equation given. I want a line that is perpendicular to it, so I need to find the "negative reciprocal" of the other line's slope. To find the slope of 2x = y - 5, I first put it in "slope-intercept" form (that is, y = mx + b form):
2x = y - 5
y - 5 = 2x
y = 2x + 5
Now I see that the slope of that line is 2 (the number 2 takes the place of m in y = mx + b). So the slope of the new line that is perpendicular to it is the negative reciprocal of 2. So I change the sign: -2. Then I take the reciprocal of that: (-1/2). Now I know the slope of the new line.
To write the equation of a line, all you need is one point it passes through and the slope. That's because the point is (x, y) and the slope is m, so when you substitute these values into y = mx + b, you just have to solve for b. So for my line, I know m = (-1/2) and x = 4 and y = -3.
y = mx + b
-3 = (-1/2)(4) + b
-3 = -2 + b
b = -1
Therefore the equation of the new line is y = (-1/2)x - 1. I have substituted in the slope m and the b value I just found.
The last step is to write this in "standard form" That means the form "Ax + By + C = 0" where A, B, and C are all whole numbers and A is positive. So really, I just move everything to the left side and multiply by 2 to get rid of the fraction:
y = (-1/2)x - 1
(1/2)x + y + 1 = 0
1x + 2y + 2 = 0. <--- standard form
That's the solution for my example. Use the same steps to solve your question, Kristy.
Hope this helps,
Stephen La Rocque.
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
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# Cross multiplication Method
In this section we will discuss cross multiplication method. It is also known as Cramer's rule.
Let the two equations be ,
$a_{1}x + b_{1}y + c_{1}$ = 0
$a_{2}x + b_{2}y + c_{2}$ = 0
be a system of linear equations in two variables x and y such that
$\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$
i.e $a_{1}. b_{2} - a_{2}.b_{1} \neq$ 0
Then the system has a unique solution given by
$x = \frac{b_{1}c_{2} - b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$ and $y = \frac{c_{1}a_{2} - c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}$
## Examples on Cross multiplication method
Example 1 :
Find the solution of the given equations.
2x + 3y = 17 and 3x – 2y = 6
$a_{1}$= 2 , $b_{1}$ = 3 , and $c_{1}$= -17
$a_{2}$= 3 , $b_{1}$ = -2 and $c_{2}$ = -6
$x= \frac{b_{1}c_{2} - b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$ and y = $\frac{c_{1}a_{2} - c_{2}a_{1}}{a_{1}b_{2}-a_{2}b_{1}}$
$x= \frac{3 \times (-6) - (-2) \times (-17)}{2 \times (-2) - 3 \times 3}$ and y = $\frac{(-17) \times 3 - (-6) \times 2}{2 \times (-2) - 3 \times 3}$
$x =\frac{-18 -34}{-4-9}$ and $y = \frac{-51 +12 }{-4-9}$
$x= \frac{-52}{-13}$ and $y =\frac{-39}{-13}$
x = 4 and y = 3
Linear equation in two variables
Solving linear equation by graphical method.
Substitution method.
Solving system of equation by elimination method
Cross multiplication method or Cramer’s rule
Home
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# 1.04 Changing the subject of a formula
Lesson
In certain situations we may need to rearrange an equation or formula to make a different variable the subject. As an example, consider the formula for finding the circumference $C$C of any circle with radius $r$r:
$C=2\pi r$C=2πr
In this formula, $C$C is currently the subject. By rearranging the formula, however, we can make $r$r the subject. This can be done with the same methods we used for solving equations. In this case however, we don't have any values to substitute, so we are working mainly with variables. Although we use equation-solving techniques, we are not actually solving an equation. Instead, we are rearranging the formula or equation into a different form. In this case, the rearranged formula becomes:
$r=\frac{C}{2\pi}$r=C2π
Remember
The subject of an equation or formula is the single variable by itself on one side of the equals sign, usually the left-hand side.
#### Worked examples
##### example 1
For an object that travels a distance $d$d in time $t$t, the average speed of the object is given by the formula $S=\frac{d}{t}$S=dt. Rearrange the formula to change the subject to:
(a) $d$d
Solution
$S$S $=$= $\frac{d}{t}$dt $S\times t$S×t $=$= $\frac{d}{t}\times t$dt×t (Multiply both sides by $t$t) $St$St $=$= $d$d $d$d $=$= $St$St (Swap so that the subject is on the left-hand side)
(b) $t$t
Solution
$S$S $=$= $\frac{d}{t}$dt $S\times t$S×t $=$= $\frac{d}{t}\times t$dt×t (Multiply both sides by $t$t) $St$St $=$= $d$d $\frac{St}{S}$StS $=$= $\frac{d}{S}$dS (Divide both sides by $S$S) $t$t $=$= $\frac{d}{S}$dS
##### Example 2
Rearrange $y=mx+c$y=mx+c to change the subject of the equation to the following:
(a) $c$c
Solution
$y$y $=$= $mx+c$mx+c $y-mx$y−mx $=$= $mx+c-mx$mx+c−mx (Subtract $mx$mx from both sides) $y-mx$y−mx $=$= $c$c $c$c $=$= $y-mx$y−mx (Swap so that the subject is on the left-hand side)
(b) $m$m
Solution
$y$y $=$= $mx+c$mx+c $y-c$y−c $=$= $mx+c-c$mx+c−c (Subtract $c$c from both sides) $y-c$y−c $=$= $mx$mx $\frac{y-c}{m}$y−cm $=$= $\frac{mx}{m}$mxm (Divide both sides by $m$m) $\frac{y-c}{m}$y−cm $=$= $x$x $x$x $=$= $\frac{y-c}{m}$y−cm (Swap so that the subject is on the left-hand side)
##### example 3
The kinetic energy of an object is given by the formula $E=\frac{1}{2}mv^2$E=12mv2. Rearrange to make $v^2$v2 the subject of the formula.
Solution
We begin here by multiplying both sides by $2$2 in order to remove the fraction.
$E$E $=$= $\frac{1}{2}mv^2$12mv2 $E\times2$E×2 $=$= $\frac{1}{2}mv^2\times2$12mv2×2 (Multiply both sides by $2$2) $2E$2E $=$= $mv^2$mv2 $\frac{2E}{m}$2Em $=$= $\frac{mv^2}{m}$mv2m (Divide both sides by $m$m) $\frac{2E}{m}$2Em $=$= $v^2$v2 $v^2$v2 $=$= $\frac{2E}{m}$2Em (Swap so that the subject is on the left-hand side)
##### example 4
Rearrange $ax=b-cx$ax=bcx to make $x$x the subject.
Solution
In this question we have $x$x variables on both sides. We want to get all the $x$x variables on the left, so we begin by adding $cx$cx to both sides. This will give us the expression $ax+cx$ax+cx on the left. We can factorise this expression by taking $x$x out as a common factor. To isolate the $x$x, our final step is to divide both sides by $a+c$a+c.
$ax$ax $=$= $b-cx$b−cx $ax+cx$ax+cx $=$= $b-cx+cx$b−cx+cx (Add $cx$cx to both sides) $ax+cx$ax+cx $=$= $b$b $x\left(a+c\right)$x(a+c) $=$= $b$b (Take out the common factor $x$x) $\frac{x\left(a+c\right)}{a+c}$x(a+c)a+c $=$= $\frac{b}{a+c}$ba+c (Divide both sides by $a+c$a+c) $x$x $=$= $\frac{b}{a+c}$ba+c
#### Practice questions
##### Question 1
Make $m$m the subject of the following equation:
$\frac{m}{y}=gh$my=gh
##### Question 2
Make $m$m the subject of the following equation:
$y=6mx-9$y=6mx9
##### Question 3
Rearrange the formula $r=\frac{k}{k-9}$r=kk9 to make $k$k the subject.
### Outcomes
#### MS11-1
uses algebraic and graphical techniques to compare alternative solutions to contextual problems
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# Parts of Circle
The parts of Circle are Diameter,Radius,Arc and Sector.
Or radius = r = d / 2.
Circumference ( C ) : The length of a boundary of a circle.
C = 2 Πr
Or C = Π D
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Examples
1) Radius = 4 cm, find the diameter.
Solution :
⇒ Diameter = 2 x 4
⇒ Diameter = 8 cm.
________________________________________________________________
2) Diameter = 6.4 cm, find the radius.
Solution :
radius = r = d / 2.
_________________________________________________________________
2)If the radius of circle is 1.5 cm then find the diameter and the circumference.
Solution :
Diameter = D = 2 r
D = 2 x 1.5
D = 3 cm.
Circumference = C = &pie;D
C = 3.14 x 3
C = 9.42 cm
________________________________________________________________
Some more parts of circle are as follows:
Arc : A part of a circle between any two points on the circle.
There are two types of arcs 1) Minor arc 2) Major arc.
Here PQR is the minor arc and PR is a major arc.
As arc is a part of a circle so,
Length of a minor arc = Π rθ / 180
Length of major arc = 2Πr - Π rθ / 180
________________________________________________________________
Examples
1) Find the length of arc of radius 6 cm and angle formed by the two radii is 600.
Solution :
r = 6 cm , θ = 600 and Π = 3.14
Length of a minor arc = Π rθ / 180
Length of a minor arc = ( 3.14 x 6 x 60 ) / 180
Length of arc = 6.28 cm
________________________________________________________________
2) Sector : The region enclosed by two radii and an arc.
Area ( minor sector ) = Πr2θ / 360
Area ( major sector) = Πr2 - Πr2θ / 360
________________________________________________________________
Example : Find the area of sector whose radius is 8 cm , angle θ 300 and Π = 3.14.
Solution :
r = 8 cm , θ = 300 and Π = 3.14
Area of sector = Πr2θ / 360
Area = (3.14 x 82 x 30 ) / 360
Area of sector = 16.75 cm2
Circles
Circles
Parts of Circle
Arc and Chords
Equal Chords of a Circle
Arc and Angles
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# Steps in solving algebra problems
Algebra Balance Scales This virtual manipulative allows you to solve simple linear equations through the use of a balance beam. That means that 35 hot dogs were sold. Rational Inequalities — We continue solving inequalities in this section.
If you are subtracting a number, you want to add on both sides, and if you are adding a number, you want to subtract on each side. As we will see we will need to be very careful with the potential solutions we get as the process used in solving these equations can lead to values that are not, in fact, solutions to the equation.
If you need a review on solving linear equations, feel free to go to Tutorial How many hot dogs were sold and how many sodas were sold? Circles — In this section we discuss graphing circles.
We will also define simplified radical form and show how to rationalize the denominator. This gets rid of fractions. The equations in this tutorial will lead to either a linear or a quadratic equation.
In addition, we will introduce the standard form of the line as well as the point-slope form and slope-intercept form of the line.
If your device is not in landscape mode many of the equations will run off the side of your device should be able to scroll to see them and some of the menu items will be cut off due to the narrow screen width. Polynomial Functions - In this chapter we will take a more detailed look at polynomial functions.
Here are more examples, since solving these is an important foundation of algebra. Even after doing that, there is still a 3 multiplied by the variable, so division will be necessary to eliminate it.
We will also take a quick look at using augmented matrices to solve linear systems of equations. When solving radical equations, extra solutions may come up when you raise both sides to an even power.
Sometimes you start out with two or more radicals in your equation. Rational Exponents — In this section we will define what we mean by a rational exponent and extend the properties from the previous section to rational exponents.
If the top section of the calculator becomes to tall for your viewport, tapping in the "Enter a problem Note that if you are unsure how to enter a problem, tap the stacked dots icon in the upper right-hand corner to see a list of example entries.
Linear Systems with Two Variables — In this section we will solve systems of two equations and two variables. Now we are ready to apply these strategies to solve real world problems! We also define the domain and range of a function. The strategy for getting the variable by itself with a coefficient of 1 involves using opposite operations.
More on the Augmented Matrix — In this section we will revisit the cases of inconsistent and dependent solutions to systems and how to identify them using the augmented matrix method. Integer Exponents — In this section we will start looking at exponents.Practice solving equations that take two steps to solve.
For example, solve = x/4 + 2. Solving Inequalities Practice Problems. Now that you've studied the many examples for solving inequalities, are you ready for some practice problems?. Let's quickly recap some of the steps for solving these practice problems. A two-step equation is as straightforward as it sounds.
You will need to perform two steps in order to solve the equation. One goal in solving an equation is to have only variables on one side of the equal sign and numbers on the other side of the equal sign.
The other goal is to have the number in front of the variable equal to one. The variable does not. Quickly check your math homework with this free online algebra calculator for solving algebra, trigonometry, calculus, or statistics equations.
Free math problem solver answers your algebra homework questions with step-by-step explanations. Learn to solve word problems. This is a collection of word problem solvers that solve your problems and help you understand the solutions. All problems are customizable (meaning that you can change all parameters).
We try to have a comprehensive collection of school algebra problems. The good news is that the steps to solve word problems .
Steps in solving algebra problems
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Open In App
# Linearity of Expectation
Prerequisite: Random Variable This post is about mathematical concepts like expectation, linearity of expectation. It covers one of the required topics to understand Randomized Algorithms
Let us consider the following simple problem.
Problem: Given a fair dice with 6 faces, the dice is thrown n times, find the expected value of the sum of all results. For example, if n = 2, there are total 36 possible outcomes.
```(1, 1), (1, 2), .... (1, 6)
(2, 1), (2, 2), .... (2, 6)
................
................
(6, 1), (6, 2), ..... (6, 6)```
Expected value of a discrete random variable is R defined as following. Suppose R can take value r1 with probability p1, value r2 with probability p2, and so on, up to value rk with probability pk. Then the expectation of this random variable R is defined as:
`E[R] = r1*p1 + r2*p2 + ... rk*pk`
Let us calculate expected value for the above example.
```Expected Value of sum = 2*1/36 + 3*1/36 + .... + 7*1/36 +
of two dice throws 3*1/36 + 4*1/36 + .... + 8*1/36 +
........................
.........................
7*1/36 + 8*1/36 + .... + 12*1/36
= 7```
The above way to solve the problem becomes difficult when there are more dice throws. If we know about the linearity of expectation, then we can quickly solve the above problem for any number of throws.
Linearity of Expectation: Let R1 and R2 be two discrete random variables on some probability space, then
` E[R1 + R2] = E[R1] + E[R2] `
Using the above formula, we can quickly solve the dice problem.
```Expected Value of sum of 2 dice throws = 2*(Expected value of one dice throw)
= 2*(1/6 + 2/6 + .... 6/6)
= 2*7/2
= 7
Expected value of sum for n dice throws is = n * 7/2 = 3.5 * n ```
Some interesting facts about Linearly of Expectation:
1. Linearity of expectation holds for both dependent and independent events. On the other hand the rule E[R1R2] = E[R1]*E[R2] is true only for independent events.
2. Linearity of expectation holds for any number of random variables on some probability space. Let R1, R2, R3, … Rk be k random variables, then E[R1 + R2 + R3 + … + Rk] = E[R1] + E[R2] + E[R3] + … + E[Rk]
Another example that can be easily solved with the linearity of expectation: Hat-Check Problem: Let there be a group of n men where every man has one hat. The hats are redistributed and every man gets a random hat back. What is the expected number of men that get their original hat back? Solution: Let Ri be a random variable, the value of random variable is 1 if i’th man gets the same hat back, otherwise 0.
```So the expected number of men to get the right hat back is
= E[R1] + E[R2] + .. + E[Rn]
= P(R1 = 1) + P(R2 = 1) + .... + P(Rn = 1)
[Here P(Ri = 1) indicates probability that Ri is 1]
= 1/n + 1/n + ... + 1/n
= 1```
So on average 1 person gets the right hat back.
Exercise:
1. Given a fair coin, what is the expected number of heads when coin is tossed n times.
2. Balls and Bins: Suppose we have m balls, labelled i = 1, … , m and n bins, labeled j = 1, .. ,n. Each ball is thrown into one of the bin independently and uniformly at random. a) What is the expected number of balls in every bin b) What is the expected number of empty bins.
3. Coupon Collector: Suppose there are n types of coupons in a lottery and each lot contains one coupon (with probability 1 = n each). How many lots have to be bought (in expectation) until we have at least one coupon of each type.
See following for solution of Coupon Collector.
The linearity of expectation is useful in algorithms. For example, expected time complexity of random algorithms like randomized quick sort is evaluated using linearity of expectation (See this for reference).
References:
This article is contributed by Shubham Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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# Find the Sample Standard Deviation
Along with finding the complete standard deviation of a set of data, we can also find the sample standard deviation if we wished.
The sample standard deviation is a measurement that is very useful for times when you would like to get a good idea of the standard deviation of a perhaps a very large group, but without having to use all the values in the whole group.
Now there was a formula for complete standard deviation, and there is also a formula for sample standard deviation that can be used.
## Sample Standard Deviation Formula
The formula for the sample standard deviation is the following.
Where we have n being the size of the sample of data, and not the size of the whole complete group.
## Standard Deviation andSample Standard Deviation Comparison
To show sample standard deviation in practice, we can look at a random list of 10 numbers.
We will work out the complete standard deviation, then proceed to work out a sample standard deviation from just some of the 10 numbers.
List of 10 numbers: 7 , 11 , 14 , 12 , 9 , 10 , 8 , 9 , 12 , 18
Total amount of values. n = 10
Average/mean ( μ ) = \bf{\frac{7 \space + \space 11 \space + \space 14 \space + \space 12 \space + \space 9 \space + \space 10 \space + \space 8 \space + \space 9 \space + \space 12 \space + \space 18}{10}} = 11
### Standard Deviation:
Standard Deviation = \sqrt{\frac{\sum_{i=1}^n \space (x_i)^2}{n} \space – \space \mu^2}
= \bf{\sqrt{\frac{7^2 \space + \space 11^2 \space + \space 14^2 \space + \space 12^2 \space + \space 9^2 \space + \space 10^2 \space + \space 8^2 \space + \space 9^2 \space + \space 12^2 \space + \space 18^2}{10} \space {\text{–}} \space 11^2}}
= \bf{\sqrt{130.4 \space – \space 121}} = \bf{\sqrt{9.4}} = 3.07 ( to 2 decimal places )
Looking at a range 1 standard deviation away from the mean:
113.07 = 7.93 , 11 + 3.07 = 14.07
So we should expect that the majority of the values in the list of numbers would be between 7.93 and 14.07.
This does turn out to be the case, as only 18 happens to lie outside the range of one standard deviation away from the mean/average.
### Sample Standard Deviation:
Now we will look to take a sample standard deviation, using 5 of the 10 numbers as a sample.
Selection of 5 numbers from the list: 7 , 8 , 9 , 10 , 18
Amount of values. n = 5
Average/mean ( μ ) = \bf{\frac{7 \space + \space 8 \space + \space 9 \space + \space 10 \space + \space 18}{5}} = 10.4
Sample Standard Deviation = \sqrt{\frac{\sum_{i=1}^n \space (x_i)^2 \space {\text{–}} \space {\frac{(\sum_{i=1}^nx_i)^2}{n}}}{n \space {\text{–}} \space 1}}
= \bf{\sqrt{\frac{(49 \space + 64 \space + \space 81 \space + \space 100 \space + \space 324) \space {\text{–}} \space {\frac{2704}{5}}}{4}}} = \bf{\sqrt{19.3}} = 4.39
Again now looking at a range of 1 sample standard deviation away from the sample mean/average.
10.044.39 = 5.65 , 10.04 + 4.39 = 14.43
### Result
These new calculations to find the sample standard deviation, indicate we should expect most of the values in the complete list of 10 numbers to be between 5.65 and 14.43.
Again this does turn out to be the case, as it is still only 18 that lies outside this range in the whole complete list
So the Sample Standard Deviation does still give a fairly accurate result for the larger group.
## Find the Sample Standard DeviationExample
(1.1)
A person who plays golf at a country club with 100 members, would like to know roughly how many rounds each member played in one week.
They want to get a good measure of how spread out the results would likely be.
Instead of asking all 100 members, the person decides to take a random sample of 9 members.
In order to get an idea of what the rounds played results would likely be for the whole 100 members.
The number of rounds played for sample of 7 members were:
1 , 4 , 2 , 2 , 5 , 8 , 3 , 2 , 3
Mean = \bf{\frac{1 \space + \space 4 \space + \space 2 \space + \space 2 \space + \space 5 \space + \space 8 \space + \space 3 \space + \space 2 \space + \space 3}{9}} = \bf{\frac{26}{9}} = 2.89 , n = 9
Now to obtain the Sample Standard Deviation s with these values.
s = \bf{\sqrt{\frac{1^2 \space + \space 2^2 \space + \space 2^2 \space + \space 2^2 \space + \space 3^2 \space + \space 3^2 \space + \space 4^2 \space + \space5^2 \space + \space 8^2 \space \space {\text{–}} \space \space {\frac{(30)^2}{9}}}{9 \space {\text{–}} \space 1}}}
\bf{\sqrt{\frac{136 \space \space {\text{–}} \space \space {\frac{900}{9}}}{8}}} = \bf{\sqrt{\frac{36}{8}}} \bf{\sqrt{4.5}} = 2.12
2.892.12 = 0.77 , 2.89 + 2.12 = 5.01
From the results of the sample of 9 club members.
It’s reasonable to assume that for the whole group of 100 club members:
The average number of rounds played in the one week studied should be around 3.
With the majority of the rounds played results being between 0.77 and 5.01.
So effectively between 1 and 5 rounds played.
1. Home
2. ›
3. Probability and Statistics
4. › Sample Standard Deviation
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# NCERT Solutions for Class 10 Maths Chapter 6 - Triangles
Exercise 6.1
1. Fill in the blanks using the correct word given in 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).
Sol. (i) All circles are similar because they have similar shape but their radii could be different. So they are similar.
(ii) All squares are similar because they have similar shape but not of the same size.
(iii) All equilateral triangles are similar because they have similar shape but not of same size as their side lengths could be different.
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are equal.
(b) their corresponding sides are proportional.
2. Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.
Sol. (i) (a) Pair of circles are similar figures.
(b) Pair of squares are similar figures.
(ii) (a) A triangle and a rectangle form a pair of non-similar figures.
(b) A circle and a square form a pair of non-similar figures.
3. State whether the following quadrilaterals are similar or not :
Sol. The two quadrilaterals shown in figure are not similar because their corresponding angles are not equal. It is clear from the figure that, ∠A is 90° but ∠P is not equal to 90°. (less than 90°)
Exercise 6.2
1. In figures (i) and (ii), DE || BC. Find EC in figure (i) and AD is figure (ii).
Sol. (i) In figure (i), in DABC, DE || BC
$$\therefore\space\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}\\\text{(By basic proportionality theorem)}\\\Rarr\space\frac{1.5}{3}=\frac{1}{\text{EC}}\\{(\because \text{AD =} 1.5 \text{cm, DB = 3 cm and AE = 1 cm)}}\\\Rarr\space\text{EC}=\frac{3}{1.5}=2\text{cm}$$
∴ EC = 2 cm
(ii) In figure (ii), in ΔABC, DE || BC
$$\therefore\space\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}\\\text{(By basic proportionality theorem)}\\\Rarr\space\frac{\text{AD}}{7.2}=\frac{\text{1.8}}{5.4}\\(\because \text{AE = 1.8 cm, EC = 5.4 cm and DB = 7.2 cm)}\\\Rarr\space\text{AD}=\frac{1.8×7.2}{5.4}=2.4\space\text{cm}$$
2. E and F are points on the sides PQ and PR respectively of a DPQR. 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 = 0.36 cm
Sol. (i) In figure,
$$\frac{\text{PE}}{\text{EQ}}=\frac{3.9}{3}=1.3\\\text{and}\space\frac{\text{PF}}{\text{FR}}=\frac{3.6}{2.4}=\frac{3}{2}=1.5\\\Rarr\space\frac{\text{PE}}{\text{EQ}}\neq\frac{\text{PF}}{\text{FR}}$$
⇒ EF is not parallel QR because converse of basic proportionality theorem is not satisfied.
(ii) In figure,
$$\frac{\text{PE}}{\text{EQ}}=\frac{4}{4.5}=\frac{40}{4.5}=\frac{8}{9}\\\text{and}\space\frac{\text{PF}}{\text{FR}}=\frac{8}{9}\\\Rarr\space\frac{\text{PE}}{\text{EQ}}=\frac{\text{PF}}{\text{FR}}$$
⇒ EF || QR because converse of basic proportionality of theorem is satisfied.
(iii) In figure,
EQ = PQ – PE
= 1.28 – 0.18
= 1.10 cm
and FR = PR – PF
= 2.56 – 0.36
= 2.20 cm
$$\frac{\text{PE}}{\text{EQ}}=\frac{0.18}{1.10}=\frac{9}{55}\\\text{and}\space\frac{\text{PF}}{\text{FR}}=\frac{0.36}{2.20}=\frac{9}{55}\\\Rarr\space\frac{\text{PE}}{\text{EQ}}=\frac{\text{PF}}{\text{FR}}$$
⇒ EF || QR because converse of basic proportionality theorem is satisfied.
3. In figure, if LM || CB and LN || CD. Prove that
$$\frac{\textbf{AM}}{\textbf{AB}}=\frac{\textbf{AN}}{\textbf{AD}}$$
Sol. Given : LM || CB and LN || CD
$$\textbf{Proof :}\space\text{In ΔACB,}\\\text{LM|| CB (Given)}\\\text{So, by basic proportionality theorem}\\\Rarr\space\frac{\text{AM}}{\text{MB}}=\frac{\text{AL}}{\text{LC}}\space\text{...(i)}$$
In ΔACD, LN || CD (Given)
By basic proportionality theoerm,
$$\Rarr\space\frac{\text{AN}}{\text{AD}}=\frac{\text{AL}}{\text{LC}}\space\text{...(ii)}$$
From equation (i) and (ii), we get
$$\frac{\text{AM}}{\text{MB}}=\frac{\text{AN}}{\text{ND}}\\\Rarr\space\frac{\text{MB}}{\text{AM}}=\frac{\text{ND}}{\text{AN}}\\\Rarr\space\frac{\text{MB}}{\text{AM}}=\frac{\text{ND}}{\text{AN}}\\\Rarr\space\frac{\text{MB}}{\text{MB}}=\frac{\text{ND}}{\text{AN}}\\\Rarr\space\frac{\text{MB}}{\text{AM}}+1=\frac{\text{ND}}{\text{AN}}+1\\\text{((On adding both sides by 1))}\\\Rarr\space\frac{\text{MB+AM}}{\text{AM}}=\frac{\text{ND+AN}}{\text{AN}}\\\Rarr\space\frac{\text{AM}}{\text{AM+MB}}=\frac{\text{AN}}{\text{AN+ND}}\\\Rarr\space\frac{\text{AM}}{\text{AB}}=\frac{\text{AN}}{\text{AD}}\space\textbf{Hence Proved.}$$
4. In figure, DE || AC and DF || AC. Prove that
$$\frac{\textbf{BF}}{\textbf{FE}}=\frac{\textbf{BE}}{\textbf{EC}}$$
Sol. Given : DE || AC and DE || AC.
$$\textbf{To prove :}\space\frac{\text{BF}}{\text{FE}}=\frac{\text{BE}}{\text{EC}}\\\textbf{Proof}\text{: In} \Delta\text{BAC, DE || AC (Given)}\\\Rarr\space\frac{\text{BF}}{\text{FE}}=\frac{\text{BD}}{\text{DA}}\space\text{...(i)}\\\text{(by basic proportionality theorem)}\\\text{In} \Delta\text{BAE, DF || AE (Given)}\\\Rarr\space\frac{\text{BF}}{\text{FE}}=\frac{\text{BD}}{\text{DA}}\space\text{...(ii)}$$
(by basic proportionality theorem)
From equation (i) and (ii), we get
$$\frac{\text{BF}}{\text{FE}}=\frac{\text{BE}}{\text{EC}}\space\textbf{Hence Proved.}$$
5. In figure, DE || OQ and DF || OR. Show that EF || QR.
Sol. Given : DE || OQ
and DF || OR
To prove : EF || QR
Proof : In ΔPQO, we have
DE || OQ
(by basic proportionality Theorem)
$$\therefore\space\frac{\text{PE}}{\text{EQ}}=\frac{\text{PD}}{\text{DO}}\space\text{...(i)}\\\text{and in} \Delta \text{POR, DF || OR}\\\therefore\space\frac{\text{PF}}{\text{FR}}=\frac{\text{PD}}{\text{DO}}\space\text{...(ii)}$$
(by basic proportionality Theorem)
From equation (i) and (ii), we get
$$\frac{\text{PE}}{\text{EQ}}=\frac{\text{PF}}{\text{FR}}\\\text{Now, in} \Delta\text{PQR, we have proved that}\\\frac{\text{PE}}{\text{EQ}}=\frac{\text{PF}}{\text{FR}}$$
∴ EF || QR Hence Proved.
Play Video about Chapter 6 Triangles in One Shot
6. In figure A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Sol. Given : AB || PQ and AC || PR
To Prove : BC || QR
Proof : In given figure,
∴ AB || PQ (Given)
$$\Rarr\space\frac{\text{OA}}{\text{AP}}=\frac{\text{OB}}{\text{BQ}}\space\text{...(i)}$$
(Basic proportionality theorem)
Also, in figure, AC || PR (Given)
$$\Rarr\space\frac{\text{OA}}{\text{AP}}=\frac{\text{OC}}{\text{CR}}\space\text{...(ii)}$$
(Basic proportionality theorem)
From equations (i) and (ii), w
e get
$$\Rarr\space\frac{\text{OB}}{\text{BQ}}=\frac{\text{OC}}{\text{CR}}$$
∴ BC || QR Hence Proved.
7. Using Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Sol. Given : A ΔABC, in which D is the mid-point of AB and l || BC.
To Prove : E is the mid-point of AC.
Proof : In ΔABC, D is the mid-point of AB.
$$\text{i.e.}\space\frac{\text{AD}}{\text{DB}}=1\space\text{...(i)}$$
As straight line l || BC.
Line l is drawn through D and it meets AC at E.
By basic proportionality theorem, we get
$$\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}\\\Rarr\space\frac{\text{AE}}{\text{EC}}=1\space\text{[From equation (i)]}$$
⇒ AE = EC
$$\Rarr\space\frac{\text{AE}}{\text{EC}}=1$$
⇒ E is the mid-point of AC. Hence Proved.
8. Using 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).
Sol. Given : A DABC in which D and E are the mid-points of sides D and E.
To Prove : DE || BC
Proof : In ΔABC, D and E are mid-points of side AB and AC, respectively.
$$\Rarr\space\frac{\text{AD}}{\text{DB}}=1\text{and}\space\frac{\text{AE}}{\text{EC}}=1\space\text{(Given)}\\\Rarr\space\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$$
∴ DE || BC
(By converse of basic proportionality theorem)
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.
$$\textbf{Show that}\frac{\textbf{AO}}{\textbf{BO}}=\frac{\textbf{CO}}{\textbf{DO}}$$
Construction : Draw EOF || AB
Proof :
In ΔACD, OE || CD
$$\Rarr\space\frac{\text{AE}}{\text{ED}}=\frac{\text{AO}}{\text{OC}}$$
(by basic proportionality theorem)
In ΔABD, OE || BA
$$\Rarr\space\frac{\text{DE}}{\text{ED}}=\frac{\text{DO}}{\text{OB}}$$
(by basic proportionality theorem)
$$\text{or}\space\frac{\text{AE}}{\text{ED}}=\frac{\text{OB}}{\text{OD}}\space\text{...(i)}$$
From equation (i) and (ii), w
e get
$$\text{or}\space\frac{\text{AE}}{\text{ED}}=\frac{\text{OB}}{\text{OD}}\space\text{...(i)}$$
From equation (i) and (ii), w
e get
$$\frac{\text{AO}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}\\\text{i.e.\space}\frac{\text{AO}}{\text{BO}}=\frac{\text{CO}}{\text{DO}}\space\textbf{Hence Proved.}$$
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that
$$\frac{\textbf{AO}}{\textbf{BO}}=\frac{\textbf{CO}}{\textbf{DO}}$$
show that ABCD is a trapezium.
Sol. Given : ABCD is a quadrilateral, in which diagonals intersect at O, such that
$$\frac{\text{AO}}{\text{OB}}=\frac{\text{OC}}{\text{OD}}$$
To Prove : ABCD is a trapezium
Proof : In figure,
$$\frac{\text{AO}}{\text{BO}}=\frac{\text{CO}}{\text{DO}}$$
$$\frac{\text{AO}}{\text{BO}}=\frac{\text{CO}}{\text{DO}}\space\text{(Given)}\\\Rarr\frac{\text{AO}}{\text{OC}}=\frac{\text{BO}}{\text{OD}}\space\text{...(i)}$$
Through O, we draw
OE || BA
OE meets AD at E.
In ΔDAB, EO || AB
$$\Rarr\space\frac{\text{DE}}{\text{EA}}=\frac{\text{DO}}{\text{OB}}\\\text{or}\space\frac{\text{AE}}{\text{ED}}=\frac{\text{BO}}{\text{OD}}\space\text{..(ii)}$$
From equation (i) and (ii), w
e get
$$\frac{\text{AO}}{\text{OC}}=\frac{\text{AE}}{\text{ED}}$$
⇒ OE || CD
(by converse of basic proportionality theorem)
Now, we have
BA|| OE and OE || CD
Therefore, AB || CD
Quadrilateral ABCD is a trapezium.
Hence Proved.
Exercise 6.3
1. State which pairs of triangles in 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 :
Sol. (i) In ΔABC and ΔPQR,
∠A = ∠P = 60°, ∠B = ∠Q = 80°
and ∠C = ∠R = 40°
Here, corresponding angles are equal.
Therefore, ΔABC ~ ΔPQR
(By AAA similarity criterion)
(ii) In DABC and ΔPQR,
$$\frac{\text{AB}}{\text{QR}}=\frac{\text{2}}{\text{4}}=\frac{\text{1}}{\text{2}},\frac{\text{BC}}{\text{RP}}=\frac{\text{2.5}}{\text{5}}=\frac{1}{2}\space\text{and}\space\frac{\text{CA}}{\text{PQ}}=\frac{\text{3}}{\text{6}}=\frac{\text{1}}{\text{2}}$$
Here, ratio of all corresponding sides are equal.
Therefore, ΔABC ~ ΔQRP
(By SSS similarity criterion)
(iii) In ΔLMP and ΔDEF
$$\frac{\text{MP}}{\text{DE}}=\frac{\text{2}}{\text{4}}=\frac{\text{1}}{\text{2}},\frac{\text{BC}}{\text{RP}}=\frac{2.5}{5}=\frac{1}{2}\\\text{and}\space\frac{\text{CA}}{\text{PQ}}=\frac{3}{6}=\frac{1}{2}$$
Here, ratio of all corresponding sides are equal.
Therefore, ΔABC ~ ΔQRP
(By SSS similarity criterion)
(iii) In ΔLMP and ΔDEF
$$\frac{\text{MP}}{\text{DE}}=\frac{2}{4}=\frac{1}{2}\\\text{and}\space\frac{\text{LP}}{\text{DF}}=\frac{\text{3}}{\text{6}}=\frac{\text{1}}{\text{2}}\\\text{and}\space\frac{\text{LM}}{\text{EF}}=\frac{\text{2.7}}{\text{5}}\neq\frac{1}{2}\\\text{i.e.\space}\frac{\text{MP}}{\text{DE}}=\frac{\text{LP}}{\text{DF}}\neq\frac{\text{LM}}{\text{EF}}$$
Here, ratio of all corresponding sides are not equal.
Thus, the two triangles are not similar.
(iv) In ΔLMN and ΔPQR
∠M = ∠Q = 70°
$$\frac{\text{MN}}{\text{PQ}}=\frac{\text{2.5}}{\text{5}}=\frac{\text{1}}{\text{2}}\\\frac{\text{ML}}{\text{QR}}=\frac{5}{10}=\frac{1}{2}\\\text{i.e.\space}\frac{\text{MN}}{\text{PQ}}=\frac{\text{ML}}{\text{QR}}$$
Here, the ratio of two corresponding adjacent sides are equal and one angle is also equal.
Therefore, ΔMNL ~ ΔQPR
(By SAS similarity criterion)
(v) In ΔABC, ∠A is given but the included side AC is not given.
So, can't decide whether ΔABC and ΔDEF are similar or not.
(vi) In ΔDEF and ΔPQR,
∠D = 70°, ∠E = 80° then ∠F = 30°
(∵ In ΔDEF, ∠D + ∠E + ∠F = 180°)
∠Q = 80°, ∠R = 30°, then ∠P = 70°
(∵ In ΔQPR, ∠Q + ∠P + ∠R = 180°)
Here, ∠D = ∠P, ∠E = ∠Q, ∠F = ∠R
Therefore, ΔDEF ~ ΔPQR
(By AAA similarity criterion)
2. In figure, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Sol. Since, DOB is a straight line.
∴ ∠DOC + ∠COB = 180°
⇒ ∠DOC + 125° = 180°
⇒ ∠DCO + 180° – 125° = 55°
In ΔOC,
∠DCO + ∠CDO + ∠DOC = 180°
(Angle sum property)
∠DCO + 70° + 55° = 180°
⇒ ∠DCO + 125° = 180°
⇒ ∠DCO = 180° – 125° = 55°
Now, it is given that,
ΔODC ~ ΔOBA
∴ ∠OCD = ∠OAB
⇒ ∠OAB = ∠OCD = ∠DCO = 55°
i.e., ∠OAB = 55°
Hence, we have ∠DOC = 55°, ∠DCO = 55° and ∠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,
$$\textbf{show that}\space\frac{\textbf{OA}}{\textbf{OC}}=\frac{\textbf{OB}}{\textbf{OD}}.$$
Sol. Given : ABCD is a trapezium with AB || CD and AC and BD are diagonals intersect at O.
$$\textbf{To Prove :}\space\frac{\textbf{OA}}{\textbf{OC}}=\frac{\textbf{OB}}{\textbf{OD}}.$$
Proof : In figure, AB || DC (Given)
⇒ ∠1 = ∠3, ∠2 = ∠4
(Alternate interior angles)
Also,
∠DOC = ∠BOA
(Vertically opposite angles)
∴ ΔOCD ~ ΔOAB
$$\Rarr\space\frac{\text{OC}}{\text{OA}}=\frac{\text{OD}}{\text{OB}}$$
(Ratios of the corresponding sides of the similar triangles are equal)
$$\Rarr\space\frac{\text{OA}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}\space\text{(taking reciprocals)}\\\textbf{Hence Proved.}$$
4. In figure,$$\frac{\textbf{QR}}{\textbf{OC}}=\frac{\textbf{OB}}{\textbf{OD}}\space\textbf{and}\angle\textbf{1}=\angle\textbf{2},\textbf{show that}\Delta\textbf{PQS}∼\Delta\textbf{TQR.}$$
$$\textbf{Sol.}\space\text{Given}:\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PR}}\space\text{and}\space\angle1=\angle2$$
To Proof : ΔPQS ~ ΔTQR
Proof : In figure,
∠1 = ∠2 (Given)
⇒ PQ = PR
(Sides opposite to equal angle are equal)
$$\text{And,}\space\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PR}}\space(\text{given})\\\Rarr\space\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PQ}}\space(\because \text{PQ=PR})\\\text{or}\space\frac{\text{QS}}{\text{QR}}=\frac{\text{PQ}}{\text{QT}}$$
(Taking reciprocals) ...(i)
Now, in ΔPQS and ΔTQR, we have
∠PQS = ∠TQR = ∠Q
$$\text{and}\space\frac{\text{QS}}{\text{QR}}=\frac{\text{PQ}}{\text{QT}}\space\text{[By equation (i)]}$$
Therefore, by SAS similarity criterion, we have ΔPQS ~ ΔTQR. Hence Proved.
5. S and T are points on sides PR and QR of DPQR such that ∠P = ∠RTS. Show that DRPQ ~ DRTS.
Sol. Given : A DRPQ such that S and T are points on sides PR and QR and ∠P = ∠RTS.
To Prove : ΔRPQ ~ ΔRTS
Proof : In figure, we have ΔRPQ and ΔRTS in which
∠RPQ = ∠RTS (Given)
∠PRQ = ∠SRT = ∠R
Then, by AA similarity criterion, we have
ΔRPQ ~ΔRTS Hence Proved.
6. In figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
Sol. Given : ΔABE ≅ ΔACD
To Prove : ΔADE ~ ΔABC
Proof : Since, ΔABE ≅ ΔACD
∴ AB = AC and AE = AD (cpct)
$$\Rarr\space\frac{\text{AB}}{\text{AC}}=1\text{and}\space\frac{\text{AD}}{\text{AE}}=1\\\text{or}\space\frac{\text{AB}}{\text{AC}}=\frac{\text{AD}}{\text{AE}}\space\text{...(i)}$$
Now, in ΔADE and ΔABC, we have
$$\frac{\text{AD}}{\text{AE}}=\frac{\text{AB}}{\text{AC}}\space\text{(from (i))}\\\text{i.e.}\space\frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}$$
and also, ∠DAE = ∠BAC = ∠A
⇒ DADE ~ DABC
(By SAS similarity criterion)
Hence Proved.
7. In figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that :
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC
Sol. (i) Given : Altitudes AD and CE of ΔABC intersect each other at point P.
In ΔAEP and ΔCDP
∠AEP = ∠CDP = 90°
and ∠APE = ∠CPD
(Vertically opposite angles)
∴ ΔAEP ~ ΔCDP
(By AA similarity criterion)
(ii) In ΔABD and ΔCBE
∠ADB = ∠CEB = 90°
and ∠ABD = ∠CBE = ∠B
∴ ΔABD ~ ΔCBE
(By AA similarity criterion)
(iii) In ΔAEP and ΔADB
∠AEP = ∠ADB = 90°
and ∠PAE = ∠DAB (Common angle)
∴ ΔAEP ~ ΔADB
(By AA similarity criterion)
(iv) In ΔPDC and ΔBEC
∠PDC = ∠BEC = 90°
and ∠PCD = ∠BCE (Common angle)
∴ ΔPDC ~ ΔBEC
(By AA similarity criterion)
Hence Proved.
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Sol. Given : A parallelogram ABCD in which a line AD is produced to E and BE is joined.
To Prove : ΔABE ~ ΔCFB
Proof : In parallelogram ABCD,
∠A = ∠C
(opposite angles of parallelogram) ...(i)
Now, in ΔABE and ΔCFB,
we have
∠EAB = ∠BCF [From equation (i)]
∠ABE = ∠BFC
(Pair of alternate angles as AB || FC)
⇒ ΔABE ~ ΔCFB (by AA similarity)
9. In figure, ABC and AMP are two right triangles, right angled at B and M, respectively. Prove that :
(i) ΔABC ~ ΔAMP
$$\textbf{(ii)}\space\frac{\textbf{CA}}{\textbf{PA}}=\frac{\textbf{BC}}{\textbf{MP}}$$
Sol. Given : ΔABC and ΔAMP are two right triangles right angled at B and H.
To Prove : (i) ΔABC ~ ΔAMP
$$\text{(ii)}\space\frac{\text{CA}}{\text{PA}}=\frac{\text{BC}}{\text{MP}}$$
Proof : (i) In ΔABC and ΔAMP
∠ABC = ∠AMP (Each = 90°)
Because in ΔABC and ΔAMP are right angled at B and M, respectively.
Also, ∠BAC = ∠PAM = ∠A
⇒ ΔABC ~ ΔAMP
(By AA similarity criterion)
(ii) Since ΔABC ~ ΔAMP,
$$\therefore\space\frac{\text{AC}}{\text{AP}}=\frac{\text{BC}}{\text{MP}}$$
(Ratio of the corresponding sides of similar triangles are equal)
$$\text{or}\space\frac{\text{CA}}{\text{PA}}=\space\frac{\text{BC}}{\text{MP}}\space\textbf{Hence Proved.}$$
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 :
$$\textbf{(i)}\space\frac{\textbf{CD}}{\textbf{GH}}=\frac{\textbf{AC}}{\textbf{FG}}$$
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Sol. Given : Two ΔABC and ΔEFG and CD and GH are the bisectors of ∠ACB and ∠EGG.
ΔABC ~ ΔFEG
$$\textbf{To Prove : (i)}\frac{\text{CD}}{\text{GH}}=\frac{\text{AC}}{\text{FG}}$$
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF
Proof : (i) In ΔACD and ΔFGH
∠CAD = ∠GFH ...(i)
$$\bigg[\because\Delta \text{ABC}∼\Delta \text{FEG}\therefore\space\angle\text{CAB}=\angle\text{GFE}\bigg]$$
∠ACD = ∠FGH ...(ii)
$$\begin{bmatrix}\because\space\Delta\text{ABC∼}\Delta\text{FEG}\\\therefore\angle\text{ACB}=\angle\text{FGE}\\\Rarr\space\frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{FGE}\space(\text{halves of equal angles})\end{bmatrix}$$
From equation (i) and (ii), we get
ΔACD ~ ΔFGH
(by AA similarity criterion)
$$\therefore\space\frac{\text{CD}}{\text{GH}}=\frac{\text{AC}}{\text{FG}}$$
(∵ Corresponding sides of two similar triangles are proportional)
(ii) In ΔDCB and ΔHGE,
∠DBC = ∠HEF ...(iii)
$$\\\begin{cases}\because\Delta\text{ABC}∼\Delta\text{FEG}\\\therefore \angle\text{ABC}=\angle\text{FEG}\end{cases}$$
and ∠DCB = ∠HGE ...(iv)
$$\\\begin{cases}\because\space\Delta\text{ABC}∼\Delta\text{FEG}\\\therefore\angle\text{ABC}=\angle\text{FGE}\\\Rarr\space\frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{FGE}\space\text{(Halves of equals angles)}\end{cases}$$
From equations (iii) and (iv), we get
ΔDCB ~ ΔHGE
(by AA similarity criterion)
(iii) In ΔCDA and DHGF,
∠DAC = ∠HFG ...(v)
$$\begin{bmatrix}\because\space\Delta\text{ABC}∼\Delta\text{FEG}\\\therefore\space\angle\text{CAB}=\angle\text{GFE}\\\Rarr\space\angle\text{CAD}=\angle\text{GFH}\end{bmatrix}\\\text{∠DCA = ∠HGF ...(vi)}\\\begin{bmatrix}\because\space\Delta\text{ABC}∼\Delta\text{FEG},\space\therefore\angle\text{ACB}=\angle\text{FGE}\\\therefore\frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{FGE}\space\Rarr\space\angle\text{DAC}=\angle\text{HGF} \end{bmatrix}$$
(Halves of equal angles)
From equation (v) and (vi), we get
ΔDCA ~ ΔHGF
(by AA similarity criterion)
Hence Proved.
11. In figure, E is a point on side CB produced of an isosceles ΔABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
Sol. Given : ΔABC is an isosceles triangle in which AB = AC.
AD ⊥ BC and EF ⊥ AC
To Prove : ΔABD ~ ΔECF
Proof : ΔABC is an isosceles triangle.
∴ AB = AC ⇒ ∠B = ∠C ...(i)
For ΔADB and ΔECF,
∠ABD = ∠ECF [From equation (i)]
and ∠ADB = ∠EFC = 90°
ΔABD ~ ΔECF
(by AA similarity criterion)
Hence Proved.
12. Sides AB and BC are median AD of a ΔABC are respectively proportional to sides PQ and QR and median PM of DPQR. Show that ΔABC ~ ΔPQR.
Sol. Given : In ΔABC and ΔPQR, where AD and PM are median of triangles,
$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AD}}{\text{PM}}$$
To prove : ΔABC ~ ΔPQR
Proof : Since, AD and PM are medians of ΔABC and ΔPQR
$$\therefore\space\text{BD}=\frac{1}{2}\text{BC}\space\text{and QN}=\frac{1}{2}\text{QR}\space\text{...(i)}\\\text{Given that}\\\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AD}}{\text{PM}}\space\text{...(ii)}$$
∴ From (i) and (ii),
$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{AB}}{\text{QR}}=\frac{\text{AB}}{\text{PM}}\space\text{...(iii)}\\\text{In} \Delta \text{ABD and} \Delta\text{PQM,}\\\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}=\frac{\text{AD}}{\text{PM}}\space\text{[by (iii)]}$$
∴ By SSS criterion of proportionality
ΔABD ~ ΔPQM
∴ ∠B = ∠Q ...(iv)
In ΔABC and ΔPQR
$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}\space\text{(from (ii))}$$
and ∠B = ∠Q (from iv)
∴ ΔABC ~ ΔPQR (by SAS criterion)
Hence Proved.
13. D is point on the side BC of a ΔABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.
Sol. Given : A ΔABC and D is a point on BC.
Such that ∠ADC = ∠ABC.
To Prove : CA2 = CB . CD
Construction : Join AD
Proof : In ΔABC and ΔDAC, we have
∠BAC = ∠ADC (Given)
and ∠ACB = ∠DCA = ∠C (Common)
∴ ΔABC ~ ΔDAC
(by AA similarity criterion)
$$\therefore\space\frac{\text{AC}}{\text{CB}}=\frac{\text{CD}}{\text{CA}}$$
(Corresponding sides are similar Δs)
$$\Rarr\space\frac{\text{CA}}{\text{CD}}=\frac{\text{CB}}{\text{CA}}$$
⇒ CA × CA = CB × CD
⇒ CA2 = CB × CD Hence Proved.
14. Sides AB and AC and median AD of a DABC are respectively proportional to sides PQ and PR and median PM of another ΔPQR. Show that ΔABC ~ ΔPQR.
Sol. Given : ΔABC and ΔPQR, in which AD and PM are their medians, respectively.
$$\text{Also}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}=\frac{\text{AD}}{\text{PM}}\qquad\text{...(i)}$$
To prove : ΔABC ~ ΔPQR
Construction : Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join BE, CE, QN and RN.
Proof: Quadrilaterials ABEC and PQNR are parallelograms because their diagonals bisect each other at D and M, respectively.
⇒ BE = AC
and QN = PR
$$\Rarr\space\frac{\text{BE}}{\text{QN}}=\frac{\text{AC}}{\text{PR}}\\\Rarr\space\frac{\text{BE}}{\text{QN}}=\frac{\text{AB}}{\text{PQ}}\space\text{[By equation (i)]}\\\text{i.e.\space}\frac{\text{AB}}{\text{PQ}}=\frac{\text{BE}}{\text{QN}}\space\text{...(ii)}\\\text{From equation (i),}\\\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}=\frac{\text{2AD}}{\text{2PM}}=\frac{\text{AE}}{\text{PN}}$$
(∵ Diagonals are bisect each other)
$$\text{i.e.}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AE}}{\text{PN}}\space\text{...(iii)}$$
From equations (ii) and (iii), we have
$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BE}}{\text{QN}}=\frac{\text{AE}}{\text{PN}}\space\text{...(iii)}$$
From equations (ii) and (iii), we have
$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BE}}{\text{QN}}=\frac{\text{AE}}{\text{PN}}$$
∴ ΔABE ~ ΔPQN (by SSS similarity)
⇒ ∠1 = ∠2 ...(iv)
Similarly, we can prove that
ΔACE ~ ΔPRN
∠3 = ∠4 ...(v)
On adding equations (iv) and (v), w
e have
∠1 + ∠3 = ∠2 + ∠4
⇒ ∠A = ∠P
$$\text{and}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}\space\text{(by equation (i))}$$
∴ ΔABC ~ ΔPQR
(SAS similarity criterion)
⇒ ∠1 = ∠2 ...(iv)
Similarly, we can prove that
ΔACE ~ ΔPRN
∠3 = ∠4 ...(v)
On adding equations (iv) and (v), we have
∠1 + ∠3 = ∠2 + ∠4
⇒ ∠A = ∠P
$$\text{and}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}\space\text{(by equation (i))}$$
∴ ΔABC ~ ΔPQR
(SAS similarity criterion)
15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Sol. In figure (i), AB is a pole and behind it a Sun is risen which casts a shadow of length BC = 4 cm and makes an angle q to the horizontal and in figure (ii) PM is a height of the tower and behind a Sun risen which casts a shadow of length,
NM = 28 cm
In ΔACB and ΔPNM,
∠C = ∠N = θ
and ∠ABC = ∠PMN = 90°
∴ ΔABC ~ ΔPMN
(by AA similarity criterion)
$$\Rarr\space\frac{\text{AB}}{\text{PM}}=\frac{\text{BC}}{\text{MN}}\\\Rarr\space\frac{\text{AB}}{\text{BC}}=\frac{\text{PM}}{\text{MN}}\\\Rarr\space\frac{6}{4}=\frac{h}{28}\\\Rarr\space\frac{6×28}{4}=42\space\text{m}$$
16. If AD and PM are medians of ΔABC and ΔPQR, respectively, where ΔABC ~ ΔPQR, prove that
$$\frac{\textbf{AB}}{\textbf{PQ}}=\frac{\textbf{AD}}{\textbf{PM}}.$$
Sol. Given : ΔABC and ΔPQR and AD and PM are the medians of the ΔABC and ΔPQR respectively.
ΔABC ~ ΔPQR
To Prove :
$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}$$
Proof : ΔABC ~ ΔPQR (Given)
$$\text{and}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AC}}{\text{PR}}\space\text{...(i)}$$
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ...(ii)
$$\text{Now,}\space\text{BD = CD =}\frac{1}{2}\text{BC}\space\text{...(iii)}\\\text{and \space QM = RM}=\frac{1}{2}\text{QR}\space\text{...(iv)}$$
(∵ D is mid-point on BC and M is mid-point of QR)
From equation (i),
$$\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}\\=\frac{\text{AB}}{\text{PQ}}=\frac{\text{2BD}}{\text{2QM}}$$
[By equation (iii) & (iv)]
$$\Rarr\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QM}}\\\text{Thus, we have}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}$$
and ∠ABD = ∠PQM (∵∠B = ∠Q)
∴ ΔABD ~ ΔPQM
(by SAS similarity criterion)
$$\Rarr\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}\space\textbf{Hence Proved.}$$
Exercise 6.4
1. Let ΔABC ~ ΔDEF and their areas be respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC
Sol. Given, ΔABC ~ ΔDEF
Since, the ratio of the areas of two similar Δs is equal to the square of the ratio of their corresponding sides.
$$\therefore\space\frac{\text{ar(}\Delta \text{ABC})}{\text{ar(}\Delta \text{DEF})}=\frac{\text{BC}^{2}}{\text{EF}^{2}}\\\Rarr\space\frac{64}{121}=\frac{\text{BC}^{2}}{\text{EF}^{2}}\\\Rarr\space\bigg(\frac{\text{BC}}{\text{EF}}\bigg)^{2}=\bigg(\frac{\text{8}}{\text{11}}\bigg)^{2}\Rarr\frac{\text{BC}}{\text{EF}}=\frac{8}{11}\\\Rarr\space\text{BC}=\frac{8}{11}×\text{EF}\\\Rarr\space\text{BC}=\frac{8}{11}×15.4=11.2\space\text{cm}$$
2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, Find the ratio of the areas of Triangles AOB and Triangle COD.
Sol. ABCD is a trapezium, with AB = 2CD
$$\frac{\text{ar(}\Delta\text{AOB})}{\text{ar(}\Delta\text{COD})}=\frac{\text{AB}^{2}}{\text{CD}^{2}}\\\text{(By property of area of similar triangles)}$$
$$=\frac{(2\text{CD})^{2}}{\text{CD}^{2}}\space\text{(∵ AB = 2CD)}\\=\frac{(2\text{CD})^{2}}{\text{CD}^{2}}=\frac{4}{1}$$
3. In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O,
$$\textbf{show that}\space\frac{\textbf{ar}\textbf{(ABC)}}{\textbf{ar}\textbf{(DBC)}}=\frac{\textbf{AO}}{\textbf{DO}}.$$
Sol. Given : ΔABC and ΔDBC on the same base BC.
AD and BC intersect at O.
$$\textbf{To Prove :}\space\frac{\text{ar(}\Delta\text{ABC})}{\text{ar(}\Delta\text{DBC})}=\frac{\text{AO}}{\text{OD}}\\\textbf{Construction :} \text{Draw AL} \perp \text{BC and DM} \perp \text{BC(See figure)}$$
Proof : In ΔOLA and ΔOMD
∠ALO = ∠DMO = 90°
and ∠AOL = ∠DOM
(Vertically opposite angle)
∴ ΔOLA ~ ΔOMD
(AAA similarity criterion)
$$\Rarr\space\frac{\text{AL}}{\text{DM}}=\frac{\text{AO}}{\text{DO}}\space\text{...(i)}\\\text{Now,}\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{DBC})}=\frac{\frac{1}{2}×\text{(BC)}×\text{(AL)}}{\frac{1}{2}×\text{(BC)}×\text{(DM)}}\\=\frac{\text{AL}}{\text{DM}}=\frac{\text{AO}}{\text{DO}}\\\text{[By equation (i)]}\\\therefore\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{DBC})}=\frac{\text{AO}}{\text{BO}}\space\textbf{Hence Proved.}$$
4. If the areas of two similar triangles are equal, prove that they are congruent.
Sol. Given : ΔABC ~ ΔPQR and ar(ΔABC) = ar(ΔPQR)
To Prove : ΔABC ≅ ΔPQR
Proof : Given,
ar(ΔABC) = ar(ΔPQR)
$$\therefore\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{PQR})}=1\\\Rarr\space\frac{\text{AB}^{2}}{\text{PQ}^{2}}=\frac{\text{BC}^{2}}{\text{QR}^{2}}=\frac{\text{CA}^{2}}{\text{PR}^{2}}=1$$
(By property of area of similar triangles)
⇒ AB = PQ, BC = QR and CA = PR
(by SSS congruency criterion)
ΔABC ≅ ΔPQR.
5. D, E and F are respectively the mid-points of sides AB, BC and CA of DABC. Find the ratio of the areas of ΔDEF and ΔABC.
Sol. Given : A ΔABC in which D, E and F are mid points of sides AB, BC and AC, respectively. Join them.
By Mid-point theorem, we get
$$\text{DF}=\frac{1}{2}\text{BC},\space\text{DE =}\frac{1}{2}\text{CA}\\\text{and}\space\text{EF}=\frac{1}{2}\text{AB}\space\text{...(i)}\\\Rarr\space\frac{\text{DF}}{\text{BC}}=\frac{\text{DE}}{\text{CA}}=\frac{\text{EF}}{\text{AB}}=\frac{1}{2}$$
∴ ΔDEF ~ ΔCAB
(by SSS similarity criterion)
$$\Rarr\space\frac{\text{ar}(\Delta\text{DEF})}{\text{ar}(\Delta\text{CAB})}=\frac{\text{DE}^{2}}{\text{CA}^{2}}\\\text{(by SSS similarity criterion)}\\\frac{\bigg(\frac{1}{2}\text{CA}\bigg)^{2}}{\text{CA}^{2}}=\frac{1}{4}\\\bigg(\because\space\text{DE}=\frac{1}{2}\text{CA}\bigg)\\\therefore\space\frac{\text{ar}(\Delta\text{DEF})}{\text{ar(}\Delta\text{ABC})}=\frac{1}{4}$$
[∵ ar(ΔCAB) = ar(ΔABC)]
Hence, the required ratio is 1 : 4.
6. Prove that the ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding medians.
Sol. Given : AD is a median of ΔABC and PM is a median of ΔPQR. Therefore, D is mid-point of BC and M is mid-point of QR and ΔABC ~ ΔPQR.
$$\textbf{To Prove :}\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{PQR})}=\frac{\text{AD}^{2}}{\text{PM}^{2}}$$
Proof :
∴ ΔABC ~ ΔPQR (given)
⇒ ∠B = ∠Q ...(i)
(Corresponding angles are equal)
$$\text{Also}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}$$
(Ratio of corresponding sides of similar triangles are equal)
$$\Rarr\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{2BD}}{\text{2QM}}$$
(∵D is mid-point of BC and M is mid-point of QR)
$$\Rarr\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}\space\text{...(i)}$$
In ΔABD and ΔPQM,
∠ABD = ∠PQM [By equation (i)]
$$\text{and}\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}\space\text{[By equation (ii)]}$$
⇒ ΔABD ~ ΔPQM
(by SAS similarity criterion)\
$$\therefore\space\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}\space\text{...(iii)}$$
(corresponding sides of similar triangles are proportion)
$$\text{Now,}\space\frac{\text{ar(}\Delta\text{ABC})}{\text{ar(}\Delta\text{PQR})}=\frac{\text{AB}^{2}}{\text{PQ}^{2}}\\\text{[From equation (iii)]}\\\text{(Using property of area of similar triangles)}\\\Rarr\space\frac{\text{ar(}\Delta\text{ABC})}{\text{ar(}\Delta\text{PQR})}=\frac{\text{AD}^{2}}{\text{PM}^{2}}$$
[From equation (iii)]
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.
Sol. Given : A square ABCD having sides of length = a.
Then, the diagonal,
$$\text{BD}=a\sqrt{2}$$
$$\textbf{To Prove :}\space\text{ar}(\Delta\text{PAB})=\frac{1}{2}\text{ar}(\Delta\text{QBD})$$
Proof : Triangles ΔPAB and ΔQBD are equilaterial triangle.
∴ ΔPAB ~ ΔQBD
(Equilateral triangles are similar)
$$\Rarr\space\frac{\text{ar}(\Delta\text{PAB})}{\text{ar}(\Delta\text{QBD})}=\frac{\text{AB}^{2}}{\text{BD}^{2}}$$
(By property of area of similar triangles)
$$=\frac{a^{2}}{(a\sqrt{2})^{2}}=\frac{1}{2}\\\Rarr\space\text{ar}(\Delta\text{PAB})=\frac{1}{2}\text{ar}(\Delta\text{QBD})\space\textbf{Hence Proved.}$$
Thick 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 ΔABC and ΔBDE is :
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4
Sol. (c) 4 : 1
Explanation :
Here, ΔABC is an equilaterial triangle
∴ AB = BC = CA = a (Say)
$$\text{Now,\space BD}=\frac{1}{2}a\\(\because\text{D is mid-point of BC})$$
Now, ΔABC ~ ΔBDE
(∵ Equilaterial triangles are similar)
$$\Rarr\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{BDE})}=\frac{\text{AB}^{2}}{\text{BD}^{2}}\\\text{(By property of area of similar to triangles)}\\=\frac{a^{2}}{\bigg(\frac{1}{2}a\bigg)}=\frac{4}{1}$$
i.e. The ratio is 4 : 1.
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
Sol. (d) 16 : 81
Explanation : Areas of two similar triangles are in the ratio of the square of their corresponding sides
$$=\bigg(\frac{4}{9}\bigg)^{3}=\frac{16}{81}$$
Exercise 6.5
1. 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
Sol. By pythagoras theorem in right triangles, sum of squares of two smalller sides is equal to the square of the third (longest) side (hypotenuse).
(i) Here, (7)2 + (24)2 = 49 + 576 = 625 = (25)2
Therefore, given sides are 7 cm, 24 cm and 25 cm, will make a right triangle and length of its hypotenuse is 25 cm. (longest)
(ii) Here, (3)2 + (6)2 = 9 + 36 = 45 and (8)2 = 64. Both values are not equal.
Therefore, given sides 3 cm, 8 cm and 6 cm does not make a right angled triangle.
(iii) Here, (50)2 + (80)2 = 2500 + 6400 = 8900 and (100)2 = 10000. Both values are not equal.
Therefore, given sides 50 cm, 80 cm and 100 cm does not make a right angled triangle.
(iv) Here, (12)2 + (5)2 = 144 + 25 = 169 = (13)2
Therefore, given sides 13 cm, 12 cm and 5 cm makes a right angled triangle and length of the hypotenuse is 13 cm.
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.
Sol. Given : ΔPQR is a right angled triangle, in which ∠P = 90° and PM QR
To Prove : PM2 = QM × MR
Proof : In ΔPQR and ΔMPQ,
∠1 + ∠2 = ∠2 + ∠4 (Each = 90°)
⇒ ∠1 = ∠4
Similarly, ∠2 = ∠3
and ∠PMR = ∠PMQ = 90° (given)
∴ ΔQPM ~ ΔPRM (by AA criterion)
$$\Rarr\space\frac{\text{ar(}\Delta\text{QPM})}{\text{ar(}\Delta\text{PRM})}=\frac{\text{PM}^{2}}{\text{RM}^{2}}\\\text{(By property of area of similar triangles)}\\\Rarr\space\frac{\frac{1}{2}(\text{QM}×\text{PM})}{\frac{1}{2}(\text{RM}×\text{PM})}=\frac{\text{PM}^{2}}{\text{RM}^{2}}\\\Rarr\space\frac{\text{QM}}{\text{RM}}=\frac{\text{PM}^{2}}{\text{RM}^{2}}$$
⇒ PM2 = QM × RM
or PM2 = QM × MR Hence Proved.
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
Sol. Given : DABD is right angled at A and AC ⊥ BD
To Prove : (i) AB2 = BC.BD
(ii) AC2 = BC.DC
(iii) AD2 = BD.CD
By theorem, that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to whole triangle and to each other.
∴ ΔABC ~ ΔDAC ~ ΔDBA
(i) Now ΔABC ~ ΔDBA
$$\text{Then}\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}(\Delta\text{DBA})}=\frac{\text{AB}^{2}}{\text{DB}^{2}}\\\text{(By property of area of similar triangles)}\\\Rarr\frac{\frac{1}{2}(\text{BC})×(\text{AC})}{\frac{1}{2}(\text{BD})×(\text{AC})}=\frac{\text{AB}^{2}}{\text{DB}^{2}}$$
⇒ AB2 = BC.BD
Hence Proved.
(ii) ΔABC ~ ΔDAC
$$\Rarr\space\frac{\text{ar}(\Delta\text{ABC})}{\text{ar}\Delta(\text{DAC})}=\frac{\text{AC}^{2}}{\text{DC}^{2}}\\\text{(By property of area of similar triangles)}\\\Rarr\space\frac{\frac{1}{2}(\text{BC})×(\text{AC})}{\frac{1}{2}(\text{DC})×\text{(AC)}}=\frac{\text{AC}^{2}}{\text{DC}^{2}}$$
⇒ AC2 = BC. DC
Hence Proved.
(iii) ΔDAC ~ ΔDBA
$$\Rarr\space\frac{\text{ar}(\Delta\text{DAC})}{\text{ar}(\Delta\text{DAC})}=\frac{\text{DA}^{2}}{\text{DB}^{2}}$$
(By property of area of similar triangles)
$$\Rarr\frac{\frac{1}{2}(\text{CD})×(\text{AC})}{\frac{1}{2}(\text{BD})×(\text{AC})}=\frac{\text{AD}^{2}}{\text{BD}^{2}}$$
⇒ AD2 = BD. CD
Hence Proved.
4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
Sol. Given : ΔABC is an isosceles triangle right angled at C.
and AC = BC ...(i)
To Prove : AB2 = 2AC2
Proof : By Pythagoras theorem, in ΔABC,
AB2 = AC2 + BC2
= AC2 + AC2 = 2AC2 [∵ BC = AC]
Hence Proved.
5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Sol. Given : An isosceles DABC with AC = BC and AB2 = 2AC2
To prove : ΔABC is a right angled triangle.
Proof : In ΔABC, we are given that
AC = BC ...(i)
and AB2 = 2AC2 ...(ii)
Now, AC2 + BC2 = AC2 + AC2 [By equation (i)]
= 2AC2 = AB2 [By equation (ii)]
i.e., AC2 + BC2 = AB2
Hence, by the converse of the Pythagoras theorem, we have ΔABC is right angled at C.
6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Sol. Given : An equilateral ΔABC, each of side 2a and AD ⊥ BC.
and AB = AC
(sides of equilateral triangles)
$$\therefore\space\text{BD = CD =}\frac{1}{2}\text{BC}=a\space\text{(c.p.c.t.)}$$
Now, in ΔABD by Pythagoras theorem,
AB2 = AD2 + BD2
⇒ (2a)2 = AD2 + a2
⇒ AD2 = 3a2
$$\Rarr\space\text{AD}=\sqrt{3a}$$
7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Sol. Given : ABCD is a rhombus in which AB = BC = CD = DA = a
Its diagonal AC and BD bisect each other at right angle at O.
To Prove : AB2 + BC2 + CD2 + AD2 = AC2 + BD2
Proof : In ΔOAB,
∠AOB = 90°
$$\text{and OA =}\frac{1}{2}\text{AC}\space\text{and OB =}\frac{1}{2}\text{BD}$$
In ΔAOB, by Pythagoras theorem
OA2 + OB2 = AB2
(∵ ∠AOB = 90°)
$$\Rarr\space\bigg(\frac{1}{2}\text{AC}\bigg)^{2}+\bigg(\frac{1}{2}\text{BD}\bigg)^{2}=\text{AB}^{2}$$
⇒ AC2 + BD2 = 4AB2
or 4AB2 = AC2 + BD2
⇒ AB2 + BC2 + CD2 + DA2 = AC2 + DB2
(∵ AB = BC = CD = DA)
Hence Proved.
8. In figure, 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) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Sol. In ΔABC, from point O join lines OB, OC and OA.
(i) In right angled ΔOFA.
⇒ OA2 = OF2 + AF2
(By Pythagoras theorem)
⇒ OA2 – OF2 = AF2 ...(a)
Similarly, in ΔOBD,
OB2 – OD2 = BD2 ...(b)
and in ΔOCE,
OC2 – OE2 = CE2 ...(c)
On adding equation (a), (b) and (c), we get
OA2 + OB2 + OC2 – OD2 – OE2 – OF2
= AF2 + BD2 + CE2
(ii) From part (i), we get
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ...(d)
Similarly,
OA2 + OB2 + OC2 – OD2 – OE2 – OF2
= BF2 + CD2 + AE2 ...(e)
From equation (d) and (e), we get
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.
Sol. Let A be the position of the window and AC be the length of the ladder.
Then, AB = 8 m (Height of window)
AC = 10 m (Length of ladder)
Let BC = x m be the distance of the foot of the ladder from the base of the wall.
Using Pythagoras theorem in DABC, we get
AC2 = AB2 + BC2
∴ 102 = B2 + x2
⇒ x2 = 100 – 64 = 36
⇒ x = 6 i.e., BC = 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?
Sol. Let AB be the vertical pole of height 18 m. A guy wire is of length AC = 24 m.
Let BC = x m be the distance of the stake from the base of the pole.
Using Pythagoras theorem is DABC, we get
i.e., AC2 = AB2 + BC2
∴ (24)2 = x2 + 182
⇒ x2 = (24)2 – (18)2
= 576 – 324
= 252
$$\Rarr\space x=\sqrt{252}\space \text{m}$$
(∵ we take positive sign becasue cannot be negative)
$$\Rarr\space x=6\sqrt{7}\space \text{m}$$
11. An aeroplane leaves an airport and flies due north at a speed of 1000 kmh–1. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 kmh–1.
$$\textbf{How far apart will be two planes after}\space\textbf{1}\frac{\textbf{1}}{\textbf{2}}?$$
$$\therefore\space\text{Length of BC = 1000}×\frac{3}{2}\text{km}$$
(∵ Distance = speed × time)
= 1500 km
The second plane travels distance BA in the direction of west at a speed of 1200 km/h.
$$\therefore\space\text{Length of BA = 1200}×\frac{3}{2}=1800\space\text{km}$$
In right angled ΔABC,
AC2 = AB2 + BC2
(By Pythagoras theorem)
= (1800)2 + (1500)2
= 3240000 + 2250000
= 5490000
$$\Rarr\space\text{AC}=\sqrt{5490000}\space\text{km}\\\Rarr\space\text{AC}=300\sqrt{61}\space\text{km}\\\text{Hence, the distance between two planes is}\\\space300\sqrt{61}\space\text{km.}$$
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.
Sol. Let BC and AD be the two poles of heights 11 m and 6 m and distance between them is AB = 12 m
Then, CE = BC – BE
= 11 – 6
= 5 m
Now, AB = DE = 12 m
and AD = EB = 6 m
Let distance between tops of two poles DC = x m
By using Pythagoras theorem in ΔDEC, we get
i.e., DC2 = DE2 + CE2
⇒ x2 = (12)2 + (5)2 = 169
⇒ x = 13
Hence, distance between their tops = 13 m.
13. D and E are points on the sides CA and CB respectively of a ΔABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Sol. Given : DABC right angled at C. D and E are points on the sides CA and BC, respectively.
To Prove : AE2 + BD2 = AB2 + DE2
Construction : Join ED, BD and EA.
Proof : In right angled ΔACE,
AE2 = CA2 + CE2 ...(i)
(By Pythagoras theorem)
and in right angled DBCD,
BD2 = BC2 + CD2 ...(ii)
On adding equation (i) and (ii), we get
AE2 + BD2 = (CA2 + CE2) + (BC2 + CD2)
= (BC2 + CA2) + (CD2 + CE2)
(∵ In ΔABC, BA2 = BC2 + CA2 and in ΔECD,
DE2 = CD2 + CE2)
= BA2 + DE2
(By Pythagoras theorem)
∴ AE2 + DB2 = AB2 + DE2 Hence Proved.
14. The perpendicular from A on side BC of a ΔABC intersects BC at D such that DB = 3 CD (see in figure). Prove that 2AB2 = 2AC2 + BC2.
Sol. Given : ΔABD, in which DB = 3CD
To Prove : 2 AB2 = 2AC2 + BC2
Proof : Given, DB = 3CD
$$\Rarr\space\text{CD}=\frac{1}{4}\text{BC}\\\text{and}\space\text{DB}=\frac{3}{4}\text{BC}$$
In ΔABD, by pythagoras theorem
AB2 = DB2 + AD2 ...(ii)
In ΔACD, by pythagoras theorem
AC2 = CD2 + AD2 ...(iii)
On subtracting equations (iii) from equation (ii), we get
AB2 – AC2 = DB2 – CD2
$$=\bigg(\frac{3}{4}\text{BC}\bigg)^{2}-\bigg(\frac{1}{4}\text{BC}\bigg)^{2}\\=\frac{9}{16}\text{BC}^{2}-\frac{1}{16}\text{BC}^{2}=\frac{1}{2}\text{BC}^{2}$$
⇒ 2AB2 – 2AC2 = BC2
⇒ 2AB2 = 2AC2 + BC2 Hence Proved.
15. In an equilateral DABC, D is a point on side BC such that
$$\textbf{BD}=\frac{1}{3}\textbf{BC}.\textbf{Prove that 9AD}^2 = \textbf{7AB}^2.$$
Sol. Given : ΔABC is an equilateral triangle,
$$\text{D is a point on side BC such that BD}=\frac{1}{3}\text{BC.}$$
To Prove : 9AD2 = 7AB2
Consturction : Draw a line AE is perpendicular to BC.
Proof :
AB = BC = CA = a (Say)
(By property of equilateral triangle)
$$\text{and}\space\text{BD}=\frac{1}{3}\text{BC}=\frac{1}{3}a\space\text{(given)}\\\Rarr\space\text{CD}=\frac{2}{3}\text{BC}=\frac{2}{3}\text{a}$$
∵ AE ⊥ BC
∵ In an equilateral triangel, altitude AE is perpendicular bisector of BC.
$$\Rarr\space\text{BE}=\text{EC}=\frac{1}{2}a\\\text{Now,}\space\text{DE = BE – BD =}\frac{1}{2}a-\frac{1}{3}a-\frac{1}{6}a$$
Using Pythagoras theorem in ΔADE,
AD2 = AE2 + DE2 ...(i)
In ΔABE,
AB2 = AE2 + BE2
or
AE2 = AB2 – BE2 ...(ii)
Put the value of (ii) in (i),
AD2 = AB2 – BE2 + DE2
$$=a^{2}-\bigg(\frac{1}{2}a\bigg)^{2}+\bigg(\frac{1}{6}a\bigg)^{2}\\=a^{2}-\frac{1}{4}a^{2}+\frac{1}{36}a{2}\\=\frac{(36-9+1)a^{2}}{36}\\=\frac{28}{36}a^{2}\\=\frac{7}{9}\text{AB}^{2}$$
⇒ 9 AD2 = 7 AB2 Hence Proved.
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.
Sol. Given : An equilateral ΔABC is of side a.
and AD ⊥ BC
To Prove : 3(Side)2 = 4(Altitude)2
Proof :
Let AD = x and AB = BC = AC = a
$$\text{Now,}\space\text{BD = CD =}\frac{1}{2}\text{BC}\frac{1}{2}a$$
(In an equilateral triangle altitude AD is a perpendicular bisector of BC)
In right angled ΔABD,
AB2 = AD2 + BD2
$$\Rarr\space a^{2}=x^{2}+\bigg(\frac{1}{2}a\bigg)^{2}\\\Rarr\space a^{2}=x^{2}+\frac{1}{4}a^{2}$$
⇒ 4a2 = 4x2 + a2 ⇒ 3a2 = 4x2
Hence Proved.
17. Tick the correct answer and justify : In ΔABC,
$$\textbf{AB}=\textbf{6}\sqrt{\textbf{3}},\textbf{AC = 12 cm and BC = 6 cm. The angle B is :}$$
(a) 120° (b) 60° (c) 90° (d) 45°
Sol. (c) 90°
Explanation : Given BC = 6 cm and
$$\text{AB=}\space 6\sqrt{3}\space\text{and AC = 12 cm}$$
$$\text{Now, AB}^{2}+\text{BC}^{2}=(6\sqrt{3})^{2}+(6)^{2}$$
= 108 + 36 = 144
= (12)2 = (AC)2
So by converse of pythagoras theorem,
⇒ ΔABC is right angled at B.
∴ ∠B = 90°
Exercise 6.6 (Optional)
1. In figure, PS is the bisector of ∠QPR of ΔPQR.
$$\textbf{Prove that}\space\frac{\textbf{QS}}{\textbf{SR}}=\frac{\textbf{PQ}}{\textbf{PR}}.$$
Sol. Given : PS is the bisector of ∠QPR of ΔPQR.
$$\textbf{To Prove :}\space\frac{\text{QS}}{\text{SR}}=\frac{\text{PQ}}{\text{PR}}.$$
Construction : Draw RT || SP to meet QP produced in T.
Proof : ∵ RT || SP and transversal PR intersects them
∴ ∠1 = ∠2
(Pair of alternate interior angle) ...(i)
∵ RT || SP and transversal QT intersects them
∴ ∠3 = ∠4
(Pair of corresponding angle) ...(ii)
But ∠1 = ∠3 (As PS bisects ∠QPR)
∴ ∠2 = ∠4
[From equations (i) and (ii)]
∴ PT = PR ...(iii)
(∵ Sides opposite to equal angles of a triangle are equal)
Now, in ΔQRT,
PS || RT (By construction)
$$\frac{\text{QS}}{\text{SR}}=\frac{\text{PQ}}{\text{PT}}\\\text{(By basic proportionality theorem)}\\\Rarr\space\frac{\text{QS}}{\text{SR}}=\frac{\text{PQ}}{\text{PR}}$$
[From equations (iii)]
Hence Proved.
2. In figure, D is a point on hypotenuse AC of DABC, such that BD ⊥ AC, DM ⊥ BC and
DN ⊥ AB. Prove that :
(i) DM2 = DN . MC (ii) DN2 = DM . AN
Sol. Given : Δ is a point of hypotenuse AC of DABC, DM ⊥ BC and DN ⊥ AB.
To Prove : (i) DM2 = DN . MC
(ii) DN2= DM . AN
Construction : Join NM. Let BD and NM intersect at O.
Proof : (i) In ΔDMC and ΔNDM, ∠DMC = ∠NDM = 90° ...(i) Let MCD = ∠1 Then, ∠MDC = 90° – ∠1 (∵ ∠MCD + ∠MDC + ∠DMC = 180°) ∴ ∠ODM = 90° – (90° – ∠1) = ∠1 ⇒ ∠DMN = ∠1 or ∠MCD = ∠DMN ...(ii) From (i) and (ii), ΔDMC ~ ΔNDM (AA similarity criterion) $$\frac{\text{DM}}{\text{ND}}=\frac{\text{MC}}{\text{DM}}$$ (Corresponding sides of the similar triangles are proportional) ⇒ DM2 = MC.ND (ii) In ΔDNM and ΔNAD. ∠NDM = ∠AND (Each equal to 90°) Let ∠NAD = ∠2 Then, ∠NDA = 90° – ∠2 (∵ ∠NDA + ∠DAN + ∠DNA = 180°) ⇒ ∠ODN = 90° – (90° – ∠2) = ∠2 ⇒ ∠ODN = ∠DNM = ∠2 ⇒ ∠DNM = ∠NAD ∠ΔDNM ~ ΔNAD (AA similarity criterion) $$\therefore\space\frac{\text{DN}}{\text{NA}}=\frac{\text{DM}}{\text{ND}}\\\text{(corresponding sides of similar triangles)}\\\Rarr\space\frac{\text{DN}}{\text{AN}}=\frac{\text{DM}}{\text{DN}}$$ ⇒ DN2 = DM × AN
3. In figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC . BD.
Sol. Given : ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced.
To Prove : AC2 = AB2 + BC2 + 2BC.BD
Proof : In ΔADC, ∠D = 90°
∴ AC2 = AD2 + DC2 (By Pythagoras theorem)
= AD2 + (BD + BC)2 [∵ DC = DB + BC]
= (AD2 + DB2) + BC2 + 2 BD.BC
[∵ (a + b)2 = a2 + b2 + 2ab]
= AB2 + BC2 + 2BC.BD
(∵ In right ΔADB with ∠D = 90°, AB2 = AD2 + DB2) Hence Proved.
4. In figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC.BD.
Sol. Given : ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC.
To Prove : AC2 = AB2 + BC2 – 2BC.BD
Proof : In right ΔADC,
∴ ∠D = 90°
∴ AC2 = AD2 + DC2
(by pythagoras theorem)
= AD2 + (BC –BD)2 (∵ BC = BD + DC)
= AD2 + BC2 + BD2 – 2BC.BD
[∵ (a – b)2 = a2 + b2 – 2ab]
= (AD2 + BD2) + BC2 – 2BC.BD
= AB2+ BC2 – 2 BC.BD
{∴ In right DADB with ∠D = 90°, AB2 = AD2 + BD2 } Hence Proved.
5. In figure, AD is a median of a DABC and AM ⊥ BC. Prove that
$$\textbf{(i)\space}\textbf{AC}^{2}=\textbf{AD}^{2}+\textbf{BC}.\textbf{DM}+\bigg(\frac{\textbf{BC}}{\textbf{2}}\bigg)^{2}\\\textbf{(ii)}\space\textbf{AB}^{2}=\textbf{AD}^{2}-\textbf{BC.DM}+\bigg(\frac{\textbf{BC}}{\textbf{2}}\bigg)^{2}\\\textbf{(iii)}\space\textbf{AC}^{2}+\textbf{AB}^{2}=2\textbf{AD}^{2}+\frac{\textbf{1}}{\textbf{2}}\textbf{BC}^{2}$$
Sol. Given that, in figure, AD is a median of a ΔABC and AM ⊥ BC.
Proof :
(i) In right ΔAMC,
∵ ∠M = 90°
∴ AC2 = AM2 + MC2
(By Pythagoras theorem)
= AM2 + (MD + DC)2
(∵ MC = MD + DC)
= (AM2 + MD2) + DC2 + 2 MD.DC
[∵ (a + b)2 = a2 + b2 + 2ab]
= AD2 + DC2 + 2DC.MD
[∴ In right ΔAMD with ∠M = 90°,
AM2 + MD2
= AD2 (By Pythagoras theorem)]
$$=\text{AD}^{2}+\bigg(\frac{\text{BC}}{2}\bigg)^{2}+2\bigg(\frac{\text{BC}}{2}\bigg)\text{DM}\\\text{[}\therefore \text{2DC = BC(AD is a median of DABC)}]\\\therefore\space\text{AC}^{2}=\text{AD}^{2}+\bigg(\frac{\text{BC}}{2}\bigg)^{2}+\text{BC.DM}\\\text{...(i)}$$
(ii) In right ΔAMB,
∵ ∠M = 90°
∴ AB2 = AM2 + MB2
(By Pythagoras theorem)
= AM2 + (BD – MD)2
[∵ BD = BM + MD]
= AM2 + BD2 + MD2 – 2BD.MD
[∵ (a – b)2 = a2 + b2 – 2ab]
= (AM2 + MD2) + BD2 – 2BD.MD
= AD2 + BD2 – 2 BD.MD
[∵ In right ΔAMD with ∠M = 90°,
(By Pythagoras theorem)]
AM2 + MD2 = AD2
$$=\text{AD}^{2}-2\bigg(\frac{\text{BC}}{2}\bigg)\text{DM}+\bigg(\frac{\text{BC}}{2}\bigg)^{2}$$
(∵ 2BD = BC, AD is a median of ΔABC)
$$\therefore\space\text{AB}^{2}=\text{AD}^{2}-\text{BC.DM}+\bigg(\frac{\text{BC}} (iii) On adding Equations (i) and (ii), we get$$\text{AC}^{2}+\text{AB}^{2}=2\text{AD}^{2}+\frac{1}{2}\text{(BC)}^{2}$$6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides. Sol. Given : ABCD is a parallelogram whose diagonals are AC and BD. To Prove : AC2 + BD2 = AB2 + BC2 + CD2 + AD2 Construction : Draw AM ⊥ DC and BN ⊥ DC (produced). Proof : In right ΔAMD and ΔBNC. AD = BC (Opposite sides of a parallelogram) AM = BN (Both are altitudes of the same parallelogram to the same base) ∴ ΔAMD ≅ ΔBNC (RHS Congruency criterion) ∴ MD = NC (CPCT) ...(i) In right ΔBND, ∵ ∠N = 90° ∴ BD2 = BN2 + DN2 (By Pythagoras theorem) = BN2 + (DC + CN)2 (∵ DN = DC + CN) = BN2 + DC2 + CN2 + 2 DC.CN [∵ (a + b)2 = a2 + b2 + 2ab] = (BN2 + CN2) + DC2 + 2DC.CN ⇒ BD2 = BC2 + DC2 + 2 DC.CN ...(ii) (∵ In right ΔBNC with ∠N = 90°) BN2 + CN2 = BC2 (By Pythagoras theorem) In right ΔAMC, ∠M = 90° ∴ ACM2 = AM2 + MC2 (∵ MC = DC – DM) = AM2 + (DC – DM)2 [∵ (a – b)2 = a2 + b2 – 2ab] = AM2 + DC2 + DM2 – 2DC.DM = (AM2 + DM2) + DC2 – 2DC.DM = AD2 + DC2 – 2DC.DM [∵ In right triangle AMD with ∠M = 90°, AD2 = AM2 + DM2 (By Pythagoras theorem)] ⇒ AC2 = AD2 + AB2 – 2DC.CN ...(iii) [∵ DC = AB, opposite sides of parallelogram and MD = NC from equation (i)] On adding equations (iii) and (ii), we get AC2 + BD2 = (AD2 + AB2) + (BC2 + DC2) = AB2 + BC2 + CD2 + DA2 Hence Proved. 7. In figure, two chords AB and CD intersect each other at the point P. Prove that : (i) ΔAPC ~ ΔDPB (ii) AP.PB = CP.DP Sol. Given : Two chords AB and CD intersects each other at the point P. To Prove : (i) ΔAPC ~ ΔDPB (ii) AP.PB = CP.DP Proof : (i) ΔAPC and ΔDPB ∠APC = ∠DPB (Vertically opposite angles) ∠CAP = ∠BDP (Angles in the same segment) ∴ ΔAPC ~ ΔDPB (AA similarity criterion) (ii) ΔAPC ~ ΔDPB [Proved in (i)]$$\therefore\space\frac{\text{AP}}{\text{DP}}=\frac{\text{PC}}{\text{PB}}$$(∵ Corresponding sides of two similar triangles are proportional) ⇒ AP.BP = CP.DP ⇒ AP.PB = CP.DP 8. In figure, two chords AB and CD of a circle intersect each other at the point P(when produced) outside the circle. Prove that (i) ΔPAC ~ ΔPDB (ii) PA.PB = PC.PD Sol. Given : Two chords AB and CD of a a circle intersects each other at the point P (when produced) out the circle. To Prove : (i) ΔPAC ~ ΔPDB (ii) PA.PB = PC.PD Proof : (i) We know that, in a cyclic quadrilaterals, the exterior angle is equal to the interior opposite angles. Therefore, ∠PAC = ∠PDB ...(i) and ∠PCA = ∠PBD ...(ii) By equations (i) and (ii), we get ΔPAC ~ ΔPDB (By AA similarity criterion) (ii) Since, ΔPAC ~ ΔPDB (By AA similarity criterion) (ii) Since, ΔPAC ~ ΔPDB$$\therefore\space\frac{\text{PA}}{\text{PD}}=\frac{\text{PC}}{\text{PB}}$$(∵ Corresponding sides of the similar triangles are proportional) ⇒ PA.PB = PC.PD 9. In figure, D is a point on side BC of DABC such that$$\frac{\textbf{BD}}{\textbf{CD}}=\frac{\textbf{AB}}{\textbf{AC}}.$$Prove that AD is the bisector of ∠BAC. Sol. Given : D is a point on side BC of ΔABC such that$$\frac{\text{BD}}{\text{CD}}=\frac{\text{AB}}{\text{AC}}.$$To Proof : AD is bisector of ∠BAC. Construction : Produce BA to E such that AC = AE. Join CE.$$\textbf{Proof :}\space\frac{\text{BD}}{\text{CD}}=\frac{\text{AB}}{\text{AC}}\space\text{(Given)}\\\Rarr\space\frac{\text{BD}}{\text{CD}}=\frac{\text{AB}}{\text{AE}}
[∵ AC = AE (by construction)]
∴ In ΔBCE,
(By converse of basic proportionality theorem)
∴ ∠BAD = ∠AEC
(Pair of corresponding angle) ...(i)
and
(Pair of alternate interior angle) ...(ii)
∵ AC = AE (By construction)
∴ ∠AEC = ∠ACE ...(iii)
(Angles opposite to equal sides of a triangle are equal)
Using equations (i), (ii) and (iii), we get
i.e., AD is the bisector of ∠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? If she pulls in the string at the rate of 5 cms–1, what will be the horizontal distance of the fly from her after 12 s?
Sol. Let AB be the height of the top of the fishing rod from the water surface. Let BC be the horizontal distance of the fly from the tip of the fishing rod. Then, AC is the length of the string.
In ΔABC, by pythagoras theorem,
AC2 = AB2 + BC2
= (1.8)2 + (2.4)2
= 3.24 + 5.76
= 9
AC = 3 m
Thus, length of string out is 3 m.
Length of string is pulled at the rate of 5 cm/s.
∴ String pulled in 12 sec = 12 × 5 = 60 cm = 0.6 m
Let, fly be at D after 12 seconds. Length of string out after 12 second is AD.
⇒ AD = 3 – 0.6 = 2.4 m
AB2 + BD2 = AD2
⇒ (1.8)2 + BD2 = (2.4)2
⇒ BD2 = 5.76 – 3.24
= 2.52
BD = 1.587 m
Horizontal distance of fly = BD + 1..2 m
Horizontal distance of fly = 1.587 + 1.2
Horizontal distance of fly = 2.79 m
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# JobTestPrep's
Numerical Reasoning
Formulas
Although taking a numerical reasoning test is not the same as taking a maths
exam, in order to succeed on a numerical test you will need to have mastered
some basic maths skills.
## Numerical tests usually target the following mathematic skills:
2. Subtraction
3. Multiplication
4. Division
5. Averages
6. Percentages
7. Ratios
## More advanced calculations, such as averages, percentages and ratios can
become simpler with the use of specific formulas. Such is the case with
algebraic questions that involve rate problems (work/ speed/ distance/ time)
as well as financial-oriented problems.
In this PDF we offer a short guide to basic as well as advanced formulas that
you are expected to be able to apply in your numerical test. We will focus on
the following subjects:
1. Averages
2. Percentages
3. Ratios
4. Rate formulas
5. Finance
## Let's Get Started!
Averages
Average
Definition: A calculated “central” value of a set of numbers.
Sum of items
Average =
Number of items
∑x
̅=
X
n
Weighted average
Definition: A calculated “central” value of a set of numbers, in which each
value or set of values is assigned a different weight.
## Sum of observations × weight
Weighted average =
Sum of weights
∑ x ∗ wi
̅w =
X
∑ wi
Percentages
## Percentages and fractions
Definition: A percentage is a part of a whole, where the whole is defined as
100. A fraction is a part of a whole, where the whole can be any number.
%= fraction ×
## Note that when dealing with percentages it is sometimes easier to convert
them into decimals and use the decimals in percentages calculations. For
example, 50% = 0.5; 120% = 1.2; 11% = 0.11 etc.
Calculating a percentage
%=( )×
��
For example, if you own 20 company shares and the total number of shares is
## 400, this means you own: × = 5% of the shares.
Percentage Increase/Decrease
% Increase:
% Decrease:
## New value = − Decrease × Original amount
For example, if a shirt cost £30 and a week later was offered at a 15%
discount, how much does the shirt cost? − . 5 × = . 5 × = £ 5.5
Calculating Percentage Change
## Definition: Percentage change refers to the relative percent change of an
increase or decrease in the original amount.
% Increase:
ew − igi ew
× = - ×
igi igi
% Decrease:
igi − ew ew
× = - ×
igi igi
For example, if a shirt cost £30 and a week later was offered for the price of
£24, what was the discount on that shirt? × = 20%
## Note: Percentage change is different from absolute change. While
percentage change is calculated in relation to the original amount, absolute
change is calculated as an absolute amount. In other words, it is not divided
by the original amount.
## Calculating Percentage Difference
Definition: Percentage difference refers to the relative percentage change in a
certain amount, when you are not able to determine which amount is the
original one.
## First amount − Second amount
| | ×
First amount + Second amount /
For example, “Molly's designs” gets 200 customers a week while “Best wear”
gets 240 customers. What is the percentage difference in customers between
− − −
the two stores? | |× = | | × = | | × = ��. ��%
+ / /
Reversed Percentages
ew
% Increase: Original amount =
+I e e
ew
% Decrease: Original amount =
−De e e
For example, if a shirt costs £33 after a 20% increase in price, how much did
it cost prior to the price change? = = £27.5
+ . .
Percentage Points
## Definition: Percentage points refer to an increase or decrease of a
percentage. This is an absolute term (in contrast to percentage
change/difference).
## Percentage points difference = New percent − Old percent
Ratios
Definition: The relative size of two or more values. The values are usually
separated by a colon sign.
## a:b is a given ratio.
N is the total sum of items.
## The number of a items = ×N
+
For example, there are 70 red and blue marbles in a jar. The ratio of red to
blue marbles is 3:4. How many red marbles are there?
× = × =
+
Rate Formulas
## What are rate problems?
A rate is a mathematical way of relating two quantities, which are usually
measured in different units. Rate problems usually involve three variables
such as speed/distance/time or product/time/number of workers etc. You are
usually given 2 variables and are required to find the missing variable
according to the data given in the question.
## Work: W =P ×T W= work; P = power; T = time
For example, Jill drove across a 0.3 mile long bridge. The time it took her car
to travel from one side to the other was 20 seconds. How fast was Jill
.
driving? . = V × V = V = . 5 Miles per second
(or 0.9 miles per minute).
Finance
Fixed and variable costs: Fixed costs are set expenses a company has which
never change and variable costs are costs that vary depending on a
company's production volume.
## Total cost = Fixed costs + Variable costs
For example, if the rent a pencil company pays for its offices is £100 per
month, each pencil costs them £0.10 to make, and they make 100 pencils
each month, what is the company's total monthly cost?
�� � = + . × = + =£
as a percentage.
Gain − Cost
ROI = ×
Cost
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# Arc Length Formula
Last updated date: 04th Aug 2024
Total views: 67.8k
Views today: 0.67k
## What is the Arc Length Formula?
To find the length of an arc of a circle, let us understand the arc length formula. An arc is a component of a circle's circumference.
Again, if we want an exact answer when working with π, we use π. We substitute a rounded form of π, such as 3.14, if we want to approximate a response. Also, r refers to the radius of the circle, which is the distance from the center to the circumference of a circle.
(Image will be uploaded soon)
### What is an Arc?
An arc is a connected subset of a separable curve in Euclidean geometry. Depending on whether they are confined or not, arcs of lines are called segments or rays. Two arcs are determined by every pair of distinct points on a circle. If the two points are not directly opposite each other, an angle at the center of the circle that is less than π radians (180 degrees) will be subtended by one of these arcs, the minor arc, and an angle greater than π radians by the other arc, the major arc.
(Image will be uploaded soon)
### What is Arc Length?
Now, let us understand what is arc length. Arc length is the distance along a segment of a curve between two points. Rectification of a curve is often called evaluating the length of an irregular arc section. The advent of infinitesimal calculus led to a general arc angle formula which, in some instances, provides closed-form solutions.
(Image will be uploaded soon)
### Finding Arc Length
Now, let us find out how to find the arc length under different conditions. For finding arc length, there are different arc angle formula for different conditions.
• ## Arc Length Formula Radians
If θ is given in radians, S = θ × r
• ## Arc Length Formula Degrees
If θ is given in degrees S = 2πr(θ/360)
• ## Arc Length Formula Integral Form
Integral form $S = \int_{a}^{b} \sqrt{1 + (\frac{dy}{dx})^{2} dx}$
Where, s: arc length of the circle,
r: radius of the circle,
θ: central angle of a circle.
### Conclusion
The arc formula is used to find the length of an arc in the circle. And as seen above, there are different formulas under different conditions. So, while we calculate the arc length we have to focus on the given conditions.
## FAQs on Arc Length Formula
1. What is the Arc Formula for a Circle?
Solution: Arc length = 2πr(θ/360)
2. Can the Length of Arcs be Negative?
Solution: A curve's arc length can not be negative, just as the distance can not be negative between two points.
3. What are the Two Ways to Make an Arc Begin?
Solution: Scratching and tapping are the two ways to begin an arc.
|
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     Â
Special Features Of Isosceles Triangles
Triangles are the polygons with three sides. We say that the triangle is the closed figure with three sides. Now we need to know that the Triangles are classified as per the measure of their lengths of their sides. We say that the triangles which have all the sides of the same measure, then the triangles are called equilateral triangles.
Now we will talk about isosceles triangles: An Isosceles Triangle has two equal sides and the third side is unequal, which works as the base of the triangle. If we look at the special properties of the isosceles triangle, we say:
As isosceles triangle have two lines of same length
As the two sides of the triangle are same, we say that the angles corresponding to the equal sides are also equal.
Also the Median of the isosceles triangle, drawn from non equal side, is also the perpendicular bisector of the non equal side.
As we know that the sum of the three angles of the triangle are equal, thus if one angle of the triangle is known, we are able to find the remaining two angles of the isosceles triangle. Let us see how: If the triangle has one angle = 70 degree, then if we want to find the measure of two angles of the triangle, which are unequal. Let the measure of those angles = x degrees. Now we say that 70 + x + x = 180,
2x + 70 = 180,
2x = 180- 70,
2x = 110,
X = 55, so other two angles of the triangle are 55 and 55 degrees
Similarly, if we know the two equal angles of the triangle, then we double the measure and then subtract it from 180 degrees to get the third angle of the triangle.
Points Shown above are the Special Features of Isosceles Triangles.
Special Features of Isoceles Triangles.
An Isosceles Triangle With A Specifies Vertex Angle
Triangles are classified according to their length of the line segments and as per their angles. We know that if we have the Triangles classified as per the length of their line segments then the triangles are of following types: 1. Equilateral Triangle 2. Isosceles triangle 3. Scalene triangle. Here we are going to study about an Iso...Read More
An Isosceles triangle with a median
A Median of a triangle is defined as the line segment which is used to join a vertex of a triangle to the midpoint of the opposite sides of a triangle. There are three vertices present in a triangle, so three medians are present in a triangle. Median value depends on the vertices. There are some properties of An Isosceles Triangle with a median which ...Read More
Math Topics
Top Scorers in Geometry Worksheets
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# Mathsight Class 7 Solutions Chapter 13
## Mathsight Class 7 Solutions Chapter 13 Congruence
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students to solve MathSight Class 7 Mathematics Book, Chapter 13, Congruence. Here students can easily get all the exercise questions solution for Chapter 13, Congruence Exercise 13.1, 13.2, 13.3, 13.4 and 13.5
Congruence Exercise 13.1 Solution :
Question no – (1)
Solution :
(a) and (d) are congruent.
AB ≅ HJ,
DC ≅ JK,
BC ≅ HK
Question no – (2)
Solution :
(a) Same measurement
(b) Same length
(d) Same side
(e) Same length and breath
Question no – (3)
Solution :
The measure of the other angle will be 120°
Question no – (4)
Solution :
∠SQR = 90°
So, the angle 90° is congruent to ∠SQR
Question no – (5)
Solution :
(a) – True
(b) – True
(c) – True
(d) – True
(e) – False
Question no – (6)
Solution :
∴ Yes, they are congruent.
Question no – (7)
Solution :
Due to same Diameter, two circular tabletops are congruent.
Congruence Exercise 13.2 Solution :
Question no – (1)
Solution :
Required photo,
(a) ⇒ Congruent angle,
⇒ ∠ABC ≅ ∠MNO
∠BAC ≅ ∠NMO
∠ACB ≅ ∠MON
Congruent sides,
⇒ AB ≅ MN
BC ≅ NO
AC ≅ MO
(d) Photo ⇒ congruent angle
⇒ ∠ABC ≅ ∠FED
∠ACB ≅ ∠EFD
∠CAB ≅ ∠FDE
Congruent sides,
⇒ CB ≅ FE
AC ≅ FD
AB ≅ ED
Question no – (2)
Solution :
Required photo :
(a) ∠P ≅ ∠L
(b) PQ2 = LM
(c) ∠Q = ∠M
(d) QR2 ≅ MN
(e) ∠R2 ≅ ∠N
(f) PR2 ≅ LN
Question no – (3)
Solution :
△ABC ≅ △DEF
Correspondence Side Correspondence Angles AB2 ≅ ED ∠A ≅ ∠D BC2 ≅ EF ∠B ≅ ∠E AC ≅ DF ∠C ≅ ∠F
Question no – (4)
Solution :
Correspondence Side Correspondence Angles IM ≅ RS ∠L ≅ ∠R NM2 ≅ TS ∠N ≅ ∠T IN ≅ RT ∠M ≅ ∠S
Question no – (5)
Solution :
3n – 30° = 57°
Or, 3n = 57 + 30° = 87°
Or, n = 87/3 = 29°
∠B ≅ ∠N
Congruence Exercise 13.3 Solution :
Question no – (1)
Solution :
Common side Corresponding side Congruent triangle A BD ≅ BD AD ≅ BC AB ≅ CD △ABD ≅ △BDC B BC ≅ BC AB ≅ BD AC ≅ CD △ABC ≅ △BDC C AC ≅ FD AB ≅ ED BC ≅ FE △ABC ≅ △DEF
Question no – (2)
Solution :
(a) Not true
(b) QR ≅ SR
(c) QP ≅ ST
(d) PR ≅ TR
△PQR ≅ △TSR
(c) Not True.
Question no – (3)
Solution :
(a) Due to equal of correspondence part △DGF ≅ △JKH
(b) Due to equal of correspondence part △PTQ ≅ △SRT
Question no – (4)
Solution :
DE = GE, DF = GE
EF = EF (common side)
△DEF ≅ △GEF [Proved]
Question no – (5)
Solution :
AB = AC
BO = OC [∵ O, midpoint of BC]
AO = AO [common side]
△AOB ≅ △AOC [Proved]
Question no – (6)
Solution :
(a) EG = EG [Common side]
ED = EF
DG = FG
∴ △DEG ≅ △FEG
(b) Yes, EG is common side of both triangle.
Question no – (7)
Solution :
(a) SQ = SQ, PS = SR, RQ = PQ
(b) Yes, their correspondence part are equal.
(c) Yes, it is common side.
Question no – (8)
Solution :
(a) Given statement is True.
(b) Given statement is False.
Congruence Exercise 13.4 Solution :
Question no – (1)
Solution :
(a) Correspondence part,
Sides Angles EG = EG ∠GED = ∠GEF EF = ED ∠EDG = ∠EFG FG = DG
△EDG ≅ △EFG
(b) Correspondence part,
Sides Angles QR = CB ∠PQR = ∠ABC PR = AC PQ = AB
△PQR ≅ △ABC
(c) Correspondence part,
Sides Angles BD = BD ∠ABD = ∠CBD AB = BC AD = BC
△ABD ≅ △BCD
Question no – (2)
Solution :
(a) True
(b) False. Because they are not fulfilled the condition of S.A.S
(c) False. Because they are not fulfilled the condition of S.A.S
Question no – (3)
Solution :
(a) △PQS ≅ △QSR
(b) △ACB ≅ △FED
Question no – (4)
Solution :
(a) △ABC ≅ △DFE
Question no – (5)
Solution :
We know that,
2a + 30° = a + 50°
Or, 2a – a = 50 – 30° = 20°
Or, a = 20°
Again, 3b – 10° = b + 20°
Or, 3b – b = 20° + 10° = 30°
Or, 2b = 30°
Or, b = 30/2 = 15°
Question no – (6)
Solution :
O is midpoint of YZ.
X bisect ∠X on YZ
YO = OZ
∠XOY = ∠XOZ
XO = XO (Common side)
△YXO ≅ △ZOX
Question no – (7)
Solution :
∠D ≅ ∠M
DF ≅ MO
EF ≅ NO
DE = MN
∠E = ∠N
∠F = ∠O
△DEF ≅ △MNO ….[Proved]
Question no – (8)
Solution :
QR bisect the angle ∠QO.
PR bisect the angle ∠R
PR = SQ, QR = common side
△PQR = △RQS ….[Proved]
Congruence Exercise 13.5 Solution :
Question no – (1)
Solution :
(a) △DEF ≅ △DFG
(b)△ABC ≅ △DEF
(c) △PRQ ≅ △PQS
Question no – (2)
Solution :
(a) False. It does not fulfilled the condition of SAS sides should be shower here.
(b) True.
(c) False. It does not fulfilled the condition of SAS. Angles and side should be properly show here.
Question no – (3)
Solution :
∠BAC = 90° = ∠BDC
AC = BD
BC = BC [common side]
∠ABC = ∠DCB
AB = DC
They are satisfied the condition of SAS.
△ABC ≅ △DCB [P]
Question no – (4)
Solution :
PQ = PQ = common side
∠RQP = ∠SPQ
∠RPQ = ∠SQP = 90°
RP = QS
RQ = PS, ∠R = ∠S
∴ They are satisfied the condition of SAS.
△RPQ ≅ △SPQ [P]
Question no – (5)
Solution :
∠XPY = ∠XOZ = 90°
XP = ∠OZ
XY = XZ
XO = YP
∠Y = ∠X = ∠Z
△XYP ≅ △XZO [P]
Question no – (6)
Solution :
Yes, the information’s are enough.
Question no – (7)
Solution :
No, Because the diaries are not satisfied the condition of SAS. They all are diff height.
Question no – (8)
Solution :
We know that,
12 = 4x
Or, x = 12/4
∴ x = 3 cm.
Previous Chapter Solution :
Updated: May 27, 2023 — 8:32 am
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Multiplying and Factoring Polynomials
# Multiplying and Factoring Polynomials
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## Multiplying and Factoring Polynomials
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##### Presentation Transcript
1. Multiplying and Factoring Polynomials Tuesday, October 29, 2013
2. Remember, to add polynomials, you combine like terms. • To subtract polynomials, you add the opposite. • So, how do you multiply polynomials??? Performing operations
3. When multiplying, multiply the constants, then the variables. • Remember to use the laws of exponents (when multiplying you add the exponents). • Ex: 2x2 * 5x3 = 10x5 How do I multiply monomials?
4. To multiply a monomial and a polynomial, you simply distribute. • Ex: 2x2 (3x3 – 4x2 + x – 5) 6x5 – 8x4 + 2x3 – 10x2 How do I multiply a monomial and a polynomial?
5. FOIL – multiply first, outside, inside, then last (basically distribute) • Box it – draw a box, put the numbers in, multiply and add like terms. How do I multiply a binomial by a binomial?
6. 1) Distribute 2) Line up your like terms 3) Add • Or… Box it Ex: (2x2 – 3x + 4) (x4 + 2x3 – 4x – 3) 2x6 + 4x5 – 8x3 – 6x2 – 3x5 – 6x4 + 12x2+9x + 4x4 + 8x3 – 16x – 12 2x6 + x5 – 2x4 + 6x2 – 7x – 12 How do I multiply a polynomial by a polynomial?
7. 1. 5(x +2) • 2. -2x2 (-2 + 6xy) • 3. (x + 2) (x + 5) • 4. (3x + 10) (2x – 5) • 5. (x + 2) (x2 + 5x + 6) • *Bonus* • (x2 – 2x + 1) (x2 + 5x + 6) Classwork
8. Factoring Fanatic Uncover the mystery of factoring complex trinomials!
9. Tic-Tac-But No ToePart 1: In the following tic tac’s there are four numbers. Find the relationship that the two numbers on the right have with the two numbers on the left. Observations 1. What did you find? 2. Did it follow the pattern every time?
10. Tic-Tac-But No ToePart 2: Use your discoveries from Part 1 to complete the following Tic Tac’s. • Did your discovery work in every case? • Can you give any explanation for this? Observations
11. Finally! Factoring with a Frenzy! • Arrange the expression in descending (or ascending) order. ax2 + bx + c = 0 • Be sure the leading coefficient is positive. • Factor out the GCF, if necessary. • Multiply the coefficients “a” and “c” and put the result in quadrant II of the Tic Tac. • Put the coefficient “b” in quadrant III of the Tic Tac. • Play the game! Just like the previous problems. (Find the relationship!)
12. Once you have completed your Tic Tac– WHERE’S the ANSWER? • Use the “a” coefficient as the numerator of two fractions. Use the results in quadrants I and IV as the two denominators. • Reduce the fractions. • The numerator is your coefficient for x in your binominal and the denominator is the constant term. • EXAMPLE: If you get the fractions ½ and -3/5, your answer would be (x + 2) (3x – 5).
13. EXAMPLES X2 – X - 12 What 2 numbers complete the Tic Tac? Since a = 1, put a 1 in for the numerator in two fractions. You found 3 and -4. These are the denominators for the two fractions. Your fractions are 1/3 and –1/4 Your answer is (x + 3) (x – 4).
14. EXAMPLES 3X2 + 5X = 12 *Remember to re-write in standard form 3X2 + 5X - 12 What 2 numbers complete the Tic Tac? Since a = 3, put a 3 in for the numerator in two fractions. You found 9 and -4. These are the denominators for the two fractions. Your fractions are 3/9 = 1/3 and –3/4 Your answer is (x + 3) (3x – 4).
15. EXAMPLES 2X2 + 8X - 64 *Remember that sometimes a GCF should be factored out before beginning. 2(X2 + 4X – 32) What 2 numbers complete the Tic Tac? Since a = 1, put a 1 in for the numerator in two fractions. You found 8 and -4. These are the denominators for the two fractions. Your fractions are 1/8 and –1/4. Your answer is 2 (x + 8) (x – 4).
16. EXAMPLES 1/2X2 + 1/2X - 6 *Remember that sometimes a GCF should be factored out before beginning. 1/2(X2 + X – 12) What 2 numbers complete the Tic Tac? Since a = 1, put a 1 in for the numerator in two fractions. You found -3 and 4. These are the denominators for the two fractions. Your fractions are –1/3 and 1/4. Your answer is ½ (x – 3) (x + 4).
17. 1. b2 + 8b + 7 • 2. m2 + m – 90 • 3. n2 – 10n + 9 • 4. k2 – 13k + 40 • 5. 2p2 + 2p – 4 • *Bonus* • 7a2 + 53a + 28 Classwork
18. Do you prefer the FOIL or Box method for multiplying binomials? Why? Exit Ticket
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Strand: NUMBER AND OPERATIONS - FRACTIONS (4.NF)
Extend understanding of equivalence and ordering of fractions (Standards 4.NF.1-2). Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers (Standards 4.NF.3-4). Understand decimal notation to the hundredths and compare decimal fractions with denominators of 10 and 100 (Standards 4.NF.5- 7). Denominators for fourth grade are limited to 2, 3, 4, 5, 6, 8, 10, 12, and 100.
Standard 4.NF.6
Use decimal notation for fractions with denominators 10 or 100. For example, rewrite 0.62 as 62/100, describe a length as 0.62 meters; locate 0.62 on a number line diagram.
This problem simply asks students to add the tenths and hundredths together.
• Dimes and Pennies
This task asks: "A dime is 1/10 of a dollar and a penny is 1/100 of a dollar. What fraction of a dollar is 6 dimes and 3 pennies? Write your answer in both fraction and decimal form."
• Expanded Fractions and Decimals
The purpose of this task is for students to show they understand the connection between fraction and decimal notation by writing the same numbers both ways.
• Fraction Conversion 2 (with percents)
When completing this lesson students will understand how to convert fractions, decimals, and percentages.
• Fraction Equivalence
Students are given this task: "Explain why 6/10=60.100. Draw a picture to illustrate your explanation."
• Grade 4 Mathematics Module 6: Decimal Fractions
This 20-day module gives students their first opportunity to explore decimal numbers via their relationship to decimal fractions, expressing a given quantity in both fraction and decimal forms. Utilizing the understanding of fractions developed throughout Module 5, students apply the same reasoning to decimal numbers, building a solid foundation for Grade 5 work with decimal operations.
• Grade 4 Unit 5: Fractions and Decimals (Georgia Standards)
In this unit, students will express fractions with denominators of 10 and 100 as decimals, understand the relationship between decimals and the base ten system, understand decimal notation for fractions, use fractions with denominators of 10 and 100 interchangeably with decimals and express a fraction with a denominator 10 as an equivalent fraction with a denominator 100.
• How Many Tenths and Hundredths?
In this task students are given a list of equations such as "1 tenth + 4 hundredths = ______________ hundredths" and asked to finish the equations to make true statements.
• Number and Operations - Fractions (3.NF) - Third Grade Core Guide
The Utah State Board of Education (USBE) and educators around the state of Utah developed these guides for Third Grade Mathematics - Number and Operations - Fractions (3.NF)
• Number and Operations - Fractions (4.NF) - Fourth Grade Core Guide
The Utah State Board of Education (USBE) and educators around the state of Utah developed these guides for Fourth Grade Mathematics - Number and Operations - Fractions (4.NF)
• Rational Number Project
This portal leads to 28 lesson plans designed to help students understand the four operations with fractions.
• Using Place Value
Each part of this task highlights a slightly different aspect of place value as it relates to decimal notation. More than simply being comfortable with decimal notation, the point is for students to be able to move fluidly between and among the different ways that a single value can be represented and to understand the relative size of the numbers in each place.
http://www.uen.org - in partnership with Utah State Board of Education (USBE) and Utah System of Higher Education (USHE). Send questions or comments to USBE Specialists - Patricia Stephens-French or Molly Basham and see the Mathematics - Elementary website. For general questions about Utah's Core Standards contact the Director - Jennifer Throndsen.
These materials have been produced by and for the teachers of the State of Utah. Copies of these materials may be freely reproduced for teacher and classroom use. When distributing these materials, credit should be given to Utah State Board of Education. These materials may not be published, in whole or part, or in any other format, without the written permission of the Utah State Board of Education, 250 East 500 South, PO Box 144200, Salt Lake City, Utah 84114-4200.
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# Interval in Math – Meaning, Definition, Examples
Welcome, math explorers! At Brighterly, we believe in illuminating the path of mathematical discovery for children. Today, we’re setting off on a journey to understand the concept of “interval” in mathematics. This might seem like a complex term at first, but don’t worry! We’re here to break it down, making it simpler and brighter, just as the name Brighterly implies.
An interval in math is, in essence, a way of expressing a range of numbers. Think about going on a treasure hunt: you know that the treasure lies somewhere between point A and point B. That ‘somewhere’ is essentially what we mean when we talk about intervals. Just as in our treasure hunt, an interval contains all the “points” or numbers between two given numbers.
For example, if we talk about an interval between 3 and 5, it will include numbers like 3.1, 4.5, 4.999, and every possible number you can think of between 3 and 5! Intervals are like the secret bridges in the kingdom of mathematics, connecting different points on the number line and opening up a path for us to understand more complex concepts like continuity, functions, and calculus.
## Interval in Math
An interval in math is a set of real numbers that contains all numbers between any two numbers in the set. For example, if you have the numbers 3 and 5 in your set, all the real numbers between 3 and 5 are also included in the set. This means that 4, 3.5, 4.5, and even numbers like 3.789 are included! Intervals are an important part of mathematics and can be found in various branches such as calculus, algebra, and statistics.
## What is an Interval in Math?
In math, an interval can be thought of as a ‘stretch’ of the number line. It’s the building block for the idea of continuity in calculus, a crucial concept that helps us understand change and motion. In simpler terms, if you imagine a number line, an interval is a segment of that line that includes all the numbers between two given points. Whether those endpoints are included or excluded, depends on the type of interval.
## What is Interval Notation?
Interval notation is a method used in mathematics to specify and simplify the way we represent these intervals. Instead of writing out all the numbers, interval notation uses parentheses and brackets to show which numbers are included in the set. Parentheses, `( )`, are used to indicate that the endpoints are not included, while brackets, `[ ]`, are used when the endpoints are included.
## Different Types Of Intervals
Understanding the different types of intervals is key to mastering interval notation. There are primarily four types: open intervals, closed intervals, half-open/half-closed intervals, and time intervals.
## Interval Notation
When representing these intervals on a number line, we use interval notation. This is a system that uses brackets and parentheses to denote whether endpoints are included in the interval.
## Types of Intervals in Math
To fully grasp this concept, it’s vital to understand the four different types of intervals: open, closed, half-open, and time intervals.
## Open Interval
An open interval `(a, b)` includes all the real numbers between a and b but does not include a and b themselves.
## Closed Interval
A closed interval `[a, b]` includes all the real numbers between a and b, and also includes a and b.
## Half-Open and Half-Closed Interval
A half-open or half-closed interval `(a, b]` or `[a, b)` includes all the real numbers between a and b, but includes only one endpoint (a or b).
## Time Interval
A time interval is a measure of elapsed time between two events. It can also be open, closed, or half-open.
## Notations For Different Types of Intervals
The notations for these different types of intervals are crucial for understanding and solving problems using interval notation.
## Examples on Interval Notation
Let’s look at a few examples:
• The open interval `(2, 5)` includes all real numbers greater than 2 and less than 5. It does not include 2 or 5.
• The closed interval `[2, 5]` includes all real numbers greater than or equal to 2 and less than or equal to 5. It includes 2 and 5.
• The half-open interval `(2, 5]` includes all real numbers greater than 2 and less than or equal to 5. It includes 5 but not 2.
## Practice Questions on Interval Notation
Try these practice questions:
1. Write the interval notation for all real numbers greater than -3 and less than or equal to 7.
2. Write the interval notation for all real numbers less than 5.
3. Write the interval notation for all real numbers greater than or equal to 0.
## Conclusion
Embarking on the journey to understand intervals in mathematics, we’ve uncovered an intricate yet fascinating aspect of the subject. Whether it’s expressing a range of numbers or understanding continuity, intervals prove to be a vital concept. Just like our brand Brighterly aims to illuminate the path of learning for children, intervals light up multiple avenues in mathematics.
Understanding intervals and interval notation is a crucial stepping stone in advancing mathematical knowledge. By simplifying complex ranges into a condensed form, interval notation provides a tool that is both elegant and practical. So, as we explore the kingdom of numbers, know that understanding intervals equips us with a valuable tool, just as a lantern guides us through the darkness. Let’s continue to explore, discover, and learn with Brighterly – where learning math is a bright and delightful journey!
## Frequently Asked Questions on Interval Notation
### What is the interval notation for all real numbers?
• The interval notation for all real numbers is (-∞, ∞). It means the set of all real numbers includes everything from negative infinity to positive infinity.
### What does it mean if an interval is closed?
• A closed interval means that the end numbers (or endpoints) are included in the interval. For example, in the interval [3, 5], both 3 and 5 are part of the interval along with all the numbers between them.
### How do you write an open interval?
• An open interval is written with parentheses. For example, (3, 5) is an open interval. This means that all the numbers between 3 and 5 are included, but 3 and 5 themselves are not included in the interval.
Sources:
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11 / 21 CBSE Class 10 Mathematics NCERT Exemplar Solutions Chapter 8 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS EXERCISE 8.4 1. If cosec θ + cot θ = p, then prove that cos θ = . Sol. cosec θ + cot θ = p [Given] Squaring Both Sides, We get [By using the rule of componendo and dividendo can be written as ] So, by using componendo and dividendo, we have https://www.schoolconnects.in/
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### Transcript of CBSE Class 10 Mathematics NCERT Exemplar Solutions ...
11 / 21
CBSE Class 10 Mathematics
NCERT Exemplar Solutions
Chapter 8 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS
EXERCISE 8.4
1. If cosec θ + cot θ = p, then prove that cos θ = .
Sol. cosec θ + cot θ = p [Given]
Squaring Both Sides, We get
[By using the rule of componendo and dividendo can be written as ]
So, by using componendo and dividendo, we have
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⇒ cos θ =
Hence, proved.
2. Prove that = tan θ + cot θ.
Sol. LHS
= cosec θ sec θ
RHS = tan θ + cot θ
= cosec θ · sec θ = LHS
Hence, proved.
3. The angle of elevation of the top of a tower from a certain point is 30°. If the observer
moves 20 m towards the tower, the angle of elevation of the top increases by 15°. Find
the height of the tower.
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Sol. Consider the height of the vertical tower (TW) = x m (let)
Ist position of observer at A makes angle of elevation at the top of tower is 30°.
Now, angle of elevation of the top of tower is increased by 15°
i.e., it becomes 30° + 15° = 45°.
In ΔTWB, we have
tan 45° =
⇒ 1 =
⇒ x = y (I)
Now, ΔTWA, we have
tan 30° =
⇒ [From (I)]
⇒ = 20 + x
⇒ = 20
⇒ = 20
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⇒ x = 10(1.732 + 1)
⇒ x = 10 × 2.732 = 27.32 m
Hence, the height of the tower is 27.32 m.
4. If 1 + sin2 θ = 3 sin θ cos θ, then prove that tan θ = 1, or
.
Sol. To solve an equation in θ, we have to change it into one trigonometric ratio.
Given trigonometric equation is
1 + sin2 θ = 3 sin θ cos θ
⇒ [Dividing by sin2 θ both sides]
⇒ cosec2 θ + 1 = 3 cot θ
⇒ 1 + cot2 θ + 1 – 3 cot θ = 0
⇒ cot2 θ – 3 cot θ + 2 = 0
⇒ cot2 θ – 2 cot θ – 1 cot θ + 2 = 0
⇒ cot θ(cot θ – 2) –1(cot θ – 2) = 0
⇒ (cot θ – 2) (cot θ – 1) = 0
⇒ cot θ – 2 = 0 or (cot θ – 1) = 0
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⇒ cot θ = 2 or cot θ = 1
⇒ tan θ = or tan θ = 1
Hence, tan θ = , or 1.
5. Given that sin θ + 2 cos θ = 1, then prove that 2 sin θ – cos θ = 2.
Sol. sin θ + 2 cos θ = 1 [Given]
On squaring both sides, we get
(sin θ)2 + (2 cos θ)2 + 2(sin θ) (2 cos θ) = 1
⇒ sin2 θ + 4 cos2 θ + 4 sin θ cos θ = 1
⇒ 1 – cos2 θ + 4 (1 – sin2 θ) + 4 sin θ cos θ = 1
⇒ 1 – cos2 θ + 4 – 4 sin2 θ + 4 sin θ cos θ = 1
⇒ –cos2 θ – 4 sin2 θ + 4 sin θ cos θ = –4
⇒ cos2 θ + 4 sin2 θ – 4 sin θ cos θ = 4
⇒ (cos θ)2 + (2 sin θ)2 – 2(cos θ) (2 sin θ) = 4
⇒ (2 sin θ – cos θ)2 = 22
Taking square root both sides, we have
2 sin θ – cos θ = 2
Hence, proved.
6. The angle of elevation of the top of a tower from two distant points s and t from its
foot are complementary. Prove that the height of tower is .
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Sol. Let the height of the vertical tower (TW) = x m
Points of observation A and B are at distances ‘t’ and ‘s’ from the foot of tower.
The angles of elevation of top of the tower from observation points A and B are (90° – θ) and
θ which are complementary.
In ΔTWB, we have
tan θ = (I)
Now, in ΔTWA, we have
tan (90° – θ) =
⇒ cot θ =
⇒ cot θ · tan θ = [Multiply eq (i) and (ii) ]
⇒ x2 = st
⇒ x =
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Hence, proved.
7. The shadow of a tower standing on a plane level is found to be 50 m. longer when
Sun’s elevation is 30° than when it is 60°. Find the height of the tower.
Sol. Let a tower TW of light x (let) is standing vertically upright on a level plane ABW. A and
B are two positions of observation when angle of elevation changes from 30° to 60°
respectively.
Let BW = y [Given]
AB = 50 m
In ΔTWB, we have
tan 60° =
⇒ x = (I)
Now, in ΔTWA, we have
tan 30° = (II)
⇒ [From (I)]
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⇒ 3y = y + 50
⇒ 3y – y = 50
⇒ 2y = 50
⇒ y = 25 (I)
Now, x =
⇒ x =
⇒ x = m
8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag
staff of height ‘h’. At a point on the plane, the angles of elevation on the bottom and top
of the flag staff are α and β, respectively. Prove that the height of the tower is
.
Sol. Let the height of vertical tower (TW) = x.
And, the height of flag staff (TF) = h (Given)
The angle of elevation at A on ground from the base and top of flag staff are α, β respectively.
Let AW = y
In ΔTWA, we have
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tan α =
⇒ y = (I)
Now, in ΔFWA, we have
tan β =
⇒ y tan β = x + h
⇒ = x+ h [From (I)]
⇒ x tan β = x tan α + h tan α
⇒ x(tab β – tan α) = h tan α
Hence, proved.
9. If tan θ + sec θ = l, then prove that sec θ = .
Sol. [Recall identity sec2 θ – tan2 θ = 1
and now change sec θ + tan θ to sec2 θ – tan2 θ by multiplying and dividing the given
expression to (sec θ – tan θ).
sec θ + tan θ = l [Given]
⇒ (sec θ + tan θ)
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⇒ [∵ 1 + tan2 θ = sec2 θ]
or sec θ – tan θ = (II)
Now, get sec θ by eliminating tan θ from (I) and (II).
It can be obtained by adding (I) and (II).
⇒ 2 sec θ =
⇒ 2 sec θ =
⇒ sec θ =
Hence, proved.
10. If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
Sol. sin θ + cos θ = p
sec θ + cosec θ = q
[ IInd expression can be changed into sin θ, cos θ and eliminate trigonometric ratio from (I)
and (II)]
sec θ + cosec θ = q
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⇒ [Using (I)]
⇒ sin θ cos θ = (III)
sin θ + cos θ = p [From (I)]
⇒ (sin θ + cos θ)2 = p2 [Squaring both sides]
⇒ sin2 θ + cos2 θ + 2 sin θ cos θ = p2
⇒ 1 + 2 · = p2 [∵ sin2 θ + cos2 θ = 1 and using (III)]
⇒ q + 2p = p2q
⇒ 2p = p2q – q
⇒ 2p = q(p2 – 1)
Hence, proved.
11. If a sin θ + b cos θ = c, then prove that a cos θ – b sin θ =
.
Sol. a sin θ + b cos θ = c [Given]
On squaring both sides, we get
(a sin θ)2 + (b cos θ)2 + 2(a sin θ) (b cos θ) = c2
⇒ a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ = c2
⇒ a2(1 – cos2 θ) + b2 (1 – sin2 θ) + 2 ab sin θ cos θ = c2
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⇒ a2 – a2 cos2 θ + b2 – b2 sin2 θ + 2ab sin θ cos θ = c2
⇒ –a2 cos2 θ – b2 sin2 θ + 2ab sin θ cos θ = c2 – a2 – b2
⇒ a2 cos2 θ + b2 sin2 θ – 2ab sin θ cos θ = a2 + b2 – c2
⇒ (a cos θ)2 + (b sin θ)2 – 2(a cos θ) (b sin θ) = a2 + b2 – c2
⇒ (a cos θ – b sin θ)2 = a2 + b2 – c2
⇒ a cos θ – b sin θ = (Taking square root both sides)
Hence, a cos θ – b sin θ =
Hence, proved.
12. Prove that
Sol. Recall identity sec2 θ – tan2θ = 1
LHS
[∵ a2 – b2 = (a – b)(a + b)]
= sec θ – tan θ
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= RHS
Hence, proved.
13. The angle of elevation of the top of a tower 30 m high from the foot of another tower
in the same plane is 60°, and the angle of elevation of the top of second tower from the
foot of first tower is 30°. Find the distance between the two towers and also the height
of the other tower.
Sol. Two vertical tower TW = 30 m and ER = x m (let) are standing on a horizontal plane RW
= y(let). The angle of elevation from R to top 90 m high tower is 60° and the angle of elevation
of second tower from W is 30°.
In ΔERW,
tan 30° =
⇒ y = (I)
Now, In ΔTWR,
tan 60° = (I)
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⇒ [From (I)]
⇒ 3x = 30
⇒ x = 10 m
Now, y =
⇒ y =
⇒ y = 1.732 × 10
⇒ y = 17.32 m
Hence, the distance between the two towers is 17.32 m and the height of the second tower is
10 m.
14. From the top of a tower h m high, the angles of depression of two objects, which are
in line with the foot of the tower are α and β, (β > α). Find the distance between the two
objects.
Sol. Consider a vertical tower TW = h m. Two objects A and B are x m apart in the line joining
A, B and W and BW = y(let).
The angle of depression from top a tower to objects A and B are α and β respectively.
In ΔTWB, we have
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tan β =
⇒ y = (I)
Now, in ΔTWA,
tan α =
⇒ tan α (x + y) = h
⇒ tan α = h [From (I)]
⇒ x tan α + = h
⇒ x tan α = h –
⇒ x tan α =
⇒ x =
⇒ x =
⇒ x =
⇒ x = h[cot α – cot β]
Hence, the distance between the two objects is h(cot α – cot β) m.
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15. A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is
pulled away from the wall through a distance p, so that its upper end slides a distance q
down the wall and then the ladder makes an angle β with horizontal. Show that
.
Sol. Consider a vertical wall WB. Two positions AW and LD of a ladder as shown in figure
such that LA = p, WD = q and LD = AW = z. Angle of inclination of ladder at two positions A
and L are α and β respectively.
Let AB = y and DB = x.
In ΔABW, we have
sin α =
and cos α =
In ΔLBD, we have
Sin β =
and cos β =
Taking RHS =
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= LHS
Hence, proved.
16. The angle of elevation of the top of a vertical tower from a point on the ground is
60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find
the height of the tower.
Sol. Let the height of the vertical tower TW = x m.
It stands on a horizontal plane AW = y.
Also BC = y.
Observation point B is 10 m above the first observation point A.
The angles of elevation from point of observations A and B are 60° and 45° respectively.
TC = x – 10
In right angled triangle TBC, we have
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tan 45° =
⇒ 1 =
⇒ y = x – 10 (I)
Now, in ΔTAW,
tan 60° =
⇒ [From (I)]
⇒ = x
⇒ x =
⇒ x =
⇒ x = 23.66 m
= 100 × 2.366
Hence, the height of the tower = 23.66 m.
17. A window of a house is h m above the ground. From the window, the angles of
elevation and depression of the top and the bottom of another house situated on the
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opposite side of the lane are found to be α and β respectively. Prove that the height of
the other house is h(1 + tan α cot β) m.
Sol. Window W, h m above the ground point A, another house HS = x(m), AS = y m away
from observation window, AS = WN = y (let), NS = h, HN = (x – h).
Angle of elevation and depression of top and bottom of house HS from window W are α, β
respectively.
In right angled ΔWNS,
tan β =
⇒ y = (I)
Now, in right angled ΔHNW,
⇒ tan α =
⇒ y tan α = x – m h
⇒ = x – h [From (I)]
⇒ h tan α = x tan β – h tan β
⇒ x tan β = h tan α + h tan β
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⇒ x =
⇒ x =
⇒ x = h[tan α · cot β + 1]
Hence, the height of the house on the other side of the observer is h[1 + tan α · cot β] m.
18. The lower window of a house is at a height of 2 m above the ground and its upper
window is 4 m vertically above the lower window. At certain instant the angles of
elevation of a balloon from these windows are observed to be 60° and 30°, respectively.
Find the height of the balloon above the ground.
Sol. Let B be a balloon at a height GB = x m.
Let W1 be the window, which is 2 m above the ground H.
∴ W1 H = 2
⇒ AG = 2 m
Let W2 be the second window, which is the 4 m above the window W1.
∴ W2W1 = AC = 4 m
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The angles of elevation of balloon B from W1, W2 are 60° and 30° respectively.
BA = (x – 2) m
BC = x – 2 – 4 = (x – 6) m
In right angled ΔW2CB, we have
tan 30° =
⇒ y =
Now, in right angled ΔW1 AB, (I)
tan 60° =
⇒ = (x – 2) (II)
⇒ (x – 6) = x – 2 [From (I)]
⇒ 3x – 18 = x – 2
⇒ 3x – x = 18 – 2
⇒ 2x = 16
⇒ x = 8 m
Hence, the height of the balloon above the ground is 8 m.
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# Resistors In Parallel
## Introduction
In the previous part we identified how the resistance of a circuit can be calculated when one or more resistors are combined in series. In this section we develop our knowledge of circuit analysis by examining parallel combination of resistors.
## Identifying and Analysing Parallel Circuits
When two or more resistors are connected between identical points in a circuit the resistors are said to be in parallel. In this situation the current has a number of different possible pathways, for example in the circuit below which represents two resistors in parallel there are two possible pathways for the current to flow.
Each parallel current path is called a branch and if additional branches are added then more possible current paths are created eg
In this circuit all the connecting points along the bottom rail are equivalent to (B) and all the connecting points along the top rail equivalent to point (A). The total current flowing (iT) can be calculated by using Kirchhoff's current law.
This states that the sum of the currents into a junction is equal to the sum of the currents out of the junction.
Using the example below
the total current flowing is given by
iT = i1 + i2 + i3+ i4
The voltage dropped across each branch is identical to the source voltage (V), so using Ohms law to replace each current with the voltage and resistance gives
V/RT = V/R1 + V/R2 + V/R3 + V/R4
Cancelling the voltage terms shows how the total resistance can be calculated from the individual components ie
1/RT = 1/R1 + 1/R2 + 1/R3 + 1/R4
Now let us use an example to evaluate the individual currents flowing in each branch of the following circuit.
As we know the total current flowing in the circuit we can work out the value of the voltage source by calculating the total resistance and then using Ohms law to convert this to the applied voltage.
RT = 1/(1/R1 + 1/R2 + 1/R3) = 1/(1/680 + 1/330 + 1/220) = 111 Ω
and so the source voltage (V) is
V = iTRT = 10 x 111 = 1110 V
As we know that each branch has the source voltage dropped across it we are now able to evaluate the current for each branch
i1 = 1110/680 = 1.63 A i2 = 1110/330 = 3.36 A i3 = 1110/220 = 5.05 A
We have now seen how to analyse circuits containing resistors in series or parallel, in many electrochemical measurements we find that the systems can be quantified in terms of combinations of series and parallel components. The circuit analysis approach for these types of problems are set out in the following section
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UPSKILL MATH PLUS
Learn Mathematics through our AI based learning portal with the support of our Academic Experts!
Result $$6$$:
The two direct common tangents drawn to the circles are equal in length.
Explanation:
The two direct common tangents $$AC$$ and $$BD$$ from $$P$$ drawn to the circles are equal in length.
$$\Rightarrow$$ $$AC$$ $$=$$ $$BD$$
Proof of the result:
By the result $$3$$, we have:
The lengths of the two tangents drawn from an exterior point to a circle are equal.
$$PA$$ $$=$$ $$PB$$ and $$PC$$ $$=$$ $$PD$$.
Subtract the above two equations.
$$PA$$ $$-$$ $$PC$$ $$=$$ $$PB$$ $$-$$ $$PD$$
$$AC$$ $$=$$ $$BD$$
Example:
In the above given figure if $$PB$$ $$=$$ $$9$$ $$cm$$ and $$AC$$ $$=$$ $$6$$ $$cm$$, find the length of the tangent $$PD$$.
Solution:
By the result, we know that $$AC$$ $$=$$ $$BD$$.
So, $$BD$$ $$=$$ $$6$$ $$cm$$.
From the figure it is observed that, $$PD$$ $$=$$ $$PB$$ $$-$$ $$BD$$.
$$PD$$ $$=$$ $$9$$ $$cm$$ $$-$$ $$6$$ $$cm$$
$$PD$$ $$=$$ $$3$$ $$cm$$
Therefore, the length of the tangent $$PD$$ $$=$$ $$3$$ $$cm$$.
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32
Q:
# The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team ?
A) 20 years B) 21 years C) 23 years D) 25 years
Explanation:
Let the average age of the whole team be x years.
11x - (26 + 29) = 9 (x - 1)
=> 11x - 9x = 46
=> 2x = 46
=> x = 23.
So, average age of the team is 23 years.
Q:
The average score of boys in an examination in a school is 81 and that of the girls is 83 and the average score of the school is 81.8. Then what will be ratio between the number of girls to the number of boys, in the examination?
A) 3:2 B) 2:3 C) 4:3 D) 3:4
Explanation:
Let the number of boys = b
Let the number of girls = g
From the given data,
81b + 83g = 81.8(b + g)
81.8b - 81b = 83g - 81.8g
0.8b = 1.2g
b/g = 1.2/0.8 = 12/8 = 3/2
=> g : b = 2 : 3
Hence, ratio between the number of girls to the number of boys = 2 : 3.
2 195
Q:
Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, what will be the sum of the highest number and the lowest number?
A) 120 B) 160 C) 80 D) 60
Explanation:
Let the three numbers be x, y, z.
From the gien data,
x = 2y ....(1)
x = z/2 => z = 2(2y) = 4y .....(From 1) ...........(2)
Given average of three numbers = 56
Then,
Now,
x = 2y => x = 2 x 24 = 48
z = 4y = 4 x 24 = 96
Now, the highest number is z = 96 & smallest number is y = 24
Hence, required sum of highest number and smallest number
= z + y
= 96 + 24
= 120.
2 330
Q:
How many window coverings are necessary to span 50 windows, if each window covering is 15 windows long?
A) 4 B) 15 C) 3 D) 50
Explanation:
Given that,
Number of windows = 50
Each window covering covers 15 windows
=> 50 windows requires 50/15 window coverings
= 50/15 = 3.333
Hence, more than 3 window coverings are required. In the options 4 is more than 3.
Hence, 4 window coverings are required to cover 50 windows of each covering covers 15 windows.
1 1304
Q:
Ian has 14 boxes of paper and divides them evenly between 4 coworkers. How many whole boxes did each coworker get?
A) 2 B) 2.5 C) 3 D) 3.5
Explanation:
Given number of boxes = 14
Number of workers = 4
Now, number of whole boxes per worker = 14/4 = 3.5
Hence, number of whole boxes per each coworker = 3
1 4069
Q:
Five boxes of bananas sell for Rs. 30. how many boxes can you buy for Rs.9?
A) 2 B) 1.5 C) 1.25 D) 2.5
Explanation:
Given Five boxes of bananas sell for Rs. 30.
=> 1 Box of Bananas for = 30/5 = Rs. 6
Then, for Rs. 9
=> 9/6 = 3/2 = 1.5
Hence, for Rs. 9, 1.5 box of bananas can buy.
8 2602
Q:
Third proportion of 10 and 20 is
A) 30 B) 40 C) 25 D) 20
Explanation:
The third proportional of two numbers p and q is defined to be that number r such that
p : q = q : r.
Here, required third proportional of 10 & 20, and let it be 'a'
=> 10 : 20 = 20 : a
10a = 20 x 20
=> a = 40
Hence, third proportional of 10 & 20 is 40.
8 1724
Q:
The average weight of 45 passengers in a bus is 52 kg. 5 of them whose average weight is 48 kg leave the bus and other 5 passengers whose average weight is 54 kg join the bus at the same stop. What is the new average weight of the bus?
A) 54.21 kgs B) 51.07 kgs C) 52.66 kgs D) 53.45 kgs
Explanation:
Given total number of passengers in the bus = 45
First average weight of 45 passengers = 52 kgs
Average weight of 5 passengers who leave bus = 48
Average weight of passengers who joined the bus = 54
Therefore, the net average weight of the bus is given by
24 2697
Q:
12 boys decided to constribute Rs. 750 each to an Orphange. Suddenly few of them boys dropped out and consequently the rest had to pay Rs. 150 more. Then the number of boys who dropped out?
A) 4 B) 6 C) 2 D) 3
Explanation:
Total money decided to contribute = 750 x 12 = 9000
Let 'b' boys dropped
The rest paid 150/- more
=> (12 - b) x 900 = 9000
=> b = 2
Hence, the number of boys who dropped out is 2.
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Tuesday , July 5 2022
# NCERT 5th Class (CBSE) Mathematics: Ratio And Proportion
## UNDERSTANDING RATIO
This necklace has been made by Annisha using red and yellow beads.
The ration of red beads to yellow beads is 5 : 4 0r 5/4
5 ( Number of red beads) /4 ( number of yellow beads )
The ration of yellow beads to red beads is 4 : 5 or 4/5
4 ( Number of yellow beads) /5 ( Number of red beads )
The ratio of yellow beads to all the beads is 4 : 9 or 4/9
4 ( Number of yellow beads) / 9 ( Total number of beads in the necklace )
The ration of red beads to all the beads is 5 : 9 or 5/9
5 ( Number of yellow beads) / 9 ( Total number of beads in the necklace )
### Equal Rations
Sunalini is making lime juice. She uses limes and sugar in the ration of 3 : 5 or 3/5
F0r one jug of juice she uses 3 limes and 5 spoons of sugar.
In case she wants to make 3 jugs of lime juice this is what she will do.
One jugs have lime to sugar in the ratio of 3 : 5 or 3/5
Two jugs have lime to sugar in the ratio of 6 : 10 or 6/10
Three jugs have lime to sugar in the ratio of 9 : 15 or 9/15
For three jugs of lime juice she needs 9 limes and 15 spoons of sugar.
She can also use the rules of fractions to find out how many limes and spoons of sugar she needs for three jugs.
For one jug 3 ( limes)/5 ( spoons of sugar ) or 3/5
For two jugs it will be 3 × 2/5 × 2 = 6/10
Similarly, for three jugs it will be 3 × 3/5 × 3 = 9/15
3/5 = 6/10 = 9/15
We can find equal rations by multiplying or dividing the terms by the same number like we do in fractions.
We can use this relationship to find missing terms in equal ratios.
(a) 3 : 7 = ? : 14
3/7 = ?/14 Since 7 × 2 = 14, we should multiply 3 × 2 to get the answer.
3 × 2 / 7 × 2 = 6/14 and 3 : 7 = 6 : 14
(b) 32 : 24 = 4 : ?
32/24 = 4/? Since 32 ÷ 8 = 4, we should also divide 24 by 8 to get the answer.
32 ÷ 8 / 24 ÷ 8 = 4/3 and 32 : 24 = 4 : 3
## चुनौती हिमालय की 5th NCERT CBSE Hindi Rimjhim Ch 18
चुनौती हिमालय की 5th Class NCERT CBSE Hindi Book Rimjhim Chapter 18 प्रश्न: लद्दाख जम्मू-कश्मीर राज्य में …
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# Finding a peak in 2 Dimensional Array
In last post I talked about finding 1-D array peak.
A peak in a 2-D array is an element which has left, right, top and bottom elements lower than it. A simple approach will be to extend the 1-D array approach. Treat each row (or column) as 1-D array, and apply 1-D peak finding algorithm, which we know takes lgN operations. Once we find that we will check if this is 2-D peak as well (a 2-D peak is 1-D peak by default, but not vice versa). If yes, we are good, else continue with next row.
In case we have a N*N 2-D array, above algorithm gives me a time complexity of N*lgN.
Lets improve the above complexity by introducing divide and conquer approach.
1. Divide the matrix (2-D array) into 4 equal parts, divided on mid row and mid column.
2. Find the local peak (highest element) in row and column.
3 a. If local peak is found in horizontal column (we know left and right are small), check if top and bottom are small, if yes current element is 2-D peak, if no, choose the sub matrix which has higher (top or bottom) number than current element.
3 b. If local peak is found in vertical row (we know top and bottom are small), check if right and left are small, if yes current element is 2-D peak, if no, choose the sub matrix which has higher (left or right) number than current element.
4. After 3, we have got a matric of N/4*N/4 size of inital matrix. Repeat from 1 with new matrix.
1 2 3 4 5
1 9 7 5 3
2 3 6 5 3
3 2 4 8 1
1 9 2 3 7
7 is found as local maximum, but 9 on left is larger so we move ahead
Step 2
1 2 3
1 9 7
2 3 6
Repeating steps above give us 2-D peak
Complexity
Step 1: Check N+N elements to find the maximum= T(N/2)+ CN
Step 2: Check N/2+N/2 elements= T(N/4)+C(N/2)+CN
= T(N/8)+C(N/4)+C(N/2)+CN
=T(1)+CN(1+1/2+1/4+1/8+….1/N)
we get a geometric series which tends to 1.
Hence overall complexity O(N)
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# 2016 AMC 8 Problems/Problem 21
A top hat contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?
$\textbf{(A) }\dfrac{3}{10}\qquad\textbf{(B) }\dfrac{2}{5}\qquad\textbf{(C) }\dfrac{1}{2}\qquad\textbf{(D) }\dfrac{3}{5}\qquad \textbf{(E) }\dfrac{7}{10}$
## Solution 1
We put five chips randomly in order, and then pick the chips from the left to the right. To find the number of ways to rearrange the three red chips and two green chips, we solve for $\binom{5}{2} = 10$. However, we notice that whenever the last chip we draw is red, we pick both green chips before we pick the last (red) chip. Similarly, when the last chip is green, we pick all three red chips before the last (green) chip. This means that the last chip must be green in all the situations that work. This means we are left with finding the number of ways to rearrange three red chips and one green chip, which is $\binom{4}{3} = 4$. Because a green chip will be last $4$ out of the $10$ situations, our answer is $\boxed{\textbf{(B) } \frac{2}{5}}$.
## Solution 2
There are two ways of ending the game, either you picked out all the red chips or you picked out all the green chips. We can pick out $3$ red chips, $3$ red chips and $1$ green chip, $2$ green chips, $2$ green chips and $1$ red chip, and $2$ green chips and $2$ red chips. Because order is important in this problem, there are $1+4+1+3+6=15$ ways to pick out the chip. But we noticed that if you pick out the three red chips before you pick out the green chip, the game ends. So we need to subtract cases like that to get the total number of ways a game could end, which $15-5=10$. Out of the 10 ways to end the game, 4 of them ends with a red chip. The answer is $\frac{4}{10} = \frac{2}{5}$, or $\boxed{\textbf{(B) } \frac{2}{5}}$.
## Solution 3
Assume that after you draw the three red chips in a row without drawing both green chips, you continue drawing for the next turn. The last/fifth chip that is drawn must be a green chip because if both green chips were drawn before, we would've already completed the game. So technically, the problem is asking for the probability that the "fifth draw" is a green chip. This probability is symmetric to the probability that the first chip drawn is green, which is $\frac{2}{5}$. Thus the probability is $\boxed{\textbf{(B) } \frac{2}{5}}$.
Note: This problem is almost identical to 2001 AMC 10 #23.
## Video Solution
https://youtu.be/fTtUAtfWKyQ - Happytwin
2016 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 20 Followed byProblem 22 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
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# Domain and Range
A relation may be described by:
• a listed set of ordered pairs
• a graph
• a rule
The set of all first elements of a set of ordered pairs is known as the domain and the set of all second elements of a set of ordered pairs is known as the range.
Alternatively, the domain is the set of independent values and the range is the set of dependent values.
The domain of a relation is the set of values of $x$ in the relation.
The range of a relation is the set of values of $y$ in the relation.
The domain and range of a relation are often described using set notation.
If a relation id described by a rule, it should also specify the domain. For example:
• the relation $\{(x,y):y=x+2,x\in\{1,2,3\}\}$ describes the set of ordered pairs $\{(1,3),(2,4),(3,5)\}$
• the domain is the set $X=\{1,2,3\}$, where is given
• the range is the set $Y=\{3,4,5\}$, and can be found by applying the rule $y=x+2$ to the domain values
If the domain of a relation is not specifically stated, it is assumed to consist of all real numbers for which the rule has meaning. This is referred to as the implied domain of a relation.
• $y=x^2$ has the implied domain $x \in \mathbb{R}$, and implied range $y=x^2 \ge 0$, where $y \in \mathbb{R}$.
• $y=\sqrt{x}$ has the implied domain $x \ge 0$, where $x \in \mathbb{R}$, and implied range $y=\sqrt{x} \ge 0$, where $y \in \mathbb{R}$.
• $y=\dfrac{1}{x}$ has the implied domain $x \ne 0$, where $x \in \mathbb{R}$, and implied range $y=\dfrac{1}{x} \ge 0$, where $y \in \mathbb{R}$.
• $y=\dfrac{1}{\sqrt{x}}$ has the implied domain $x \gt 0$, where $x \in \mathbb{R}$, and implied range $y=\dfrac{1}{x} \gt 0$, where $y \in \mathbb{R}$.
### Example 1
State the domain and range of $\{(2,4),(3,9),(4,14),(5,19)\}$.
### Example 2
State the domain and range of the graph.
### Example 3
State the domain and range of the graph.
### Example 4
State the domain and range of $y=\sqrt{x}$.
### Example 5
State the domain and range of $y=\dfrac{1}{x+1}$.
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Close
## How do you find the maximum percent error?
Calculate the percent error of your measurement.
1. Subtract one value from the other: 2.68 – 2.70 = -0.02.
2. Depending on what you need, you may discard any negative sign (take the absolute value): 0.02.
3. Divide the error by the true value:0.02/2.70 = 0.0074074.
4. Multiply this value by 100% to obtain the percent error:
## Can you have a percent error over 100?
yes, a percent error of over 100% is possible. A percent error of 100% is obtained when the experimental value is twice the value of the true value. In experiments, it is always possible to get values that are way greater or lesser than the true value due to human or experimental errors.
## What is maximum error?
The maximum difference between the point estimate and the actual parameter, which is 1/2 the width of the confidence interval for means and proportions.
## What is a good margin of error?
– An acceptable margin of error used by most survey researchers typically falls between 4% and 8% at the 95% confidence level. It is affected by sample size, population size, and percentage.
## How do you do absolute error?
Here absolute error is expressed as the difference between the expected and actual values. For example, if you know a procedure is supposed to yield 1.0 liters of solution and you obtain 0.9 liters of solution, your absolute error is 1.0 – 0.9 = 0.1 liters.
READ: What is the author purpose for writing the essay how the Grimm brothers saved the fairy tale?
## What is absolute error with example?
Absolute Error is the amount of error in your measurements. It is the difference between the measured value and “true” value. For example, if a scale states 90 pounds but you know your true weight is 89 pounds, then the scale has an absolute error of 90 lbs – 89 lbs = 1 lbs.
## How do you calculate total error?
You must first find the percentage error of each of the values you are testing before you can find the total error value. Find the difference between the estimated result and the actual result. For example, if you estimated a result of 200 and ended up with a result of 214 you would subtract 200 from 214 to get 14.
## How do you find the constant error?
Constant Error: Constant error measures the deviation from the target. The formula for it is: Σ (xi-T)/N, where T is the target and N is the number of shots.
## How can one avoid constant error?
This type of constant error can be eliminated by carrying out your experimental procedure on a reference quantity — for which the accurate result is already known — and applying any necessary correction to unknown quantities.
## How do you get rid of constant error?
One way of reducing the effect of constant error is to increase the sample size until the error is acceptable. until the error is acceptable. Proportional errors decrease or increase in proportion to the size of the sample. A common cause of proportional errors is the presence of interfering contaminants in the sample.
## What is the cause of constant error?
Systematic error due to faulty apparatus causes a constant error. Systematic Error: The error caused due to imperfect measurement technique, defective or imperfect apparatus or some personal reasons is called systematic error.
## What is the constant error?
Constant error is computed as the average positive or negative difference between the observed and actual values along a dimension of interest. For example, if a weight of 1 kg is judged on average to be 1.5 kg, and a weight of 2 kg is judged to be 2.5 kg, the constant error is 500 g.
## What is error and different types of error?
Errors are normally classified in three categories: systematic errors, random errors, and blunders. Systematic Errors. Systematic errors are due to identified causes and can, in principle, be eliminated. Errors of this type result in measured values that are consistently too high or consistently too low.
READ: Why is the sodium-potassium pump considered an active transport which direction are the sodium and potassium being pumped?
## What is random error example?
Random errors in experimental measurements are caused by unknown and unpredictable changes in the experiment. Examples of causes of random errors are: electronic noise in the circuit of an electrical instrument, irregular changes in the heat loss rate from a solar collector due to changes in the wind.
## Can random error be corrected?
It comes from unpredictable changes during an experiment. Systematic error always affects measurements the same amount or by the same proportion, provided that a reading is taken the same way each time. It is predictable. Random errors cannot be eliminated from an experiment, but most systematic errors can be reduced.
## What type of error is human error?
Random errors are natural errors. Systematic errors are due to imprecision or problems with instruments. Human error means you screwed something up, you made a mistake. In a well-designed experiment performed by a competent experimenter, you should not make any mistakes.
## What are sources of error?
Common sources of error include instrumental, environmental, procedural, and human. All of these errors can be either random or systematic depending on how they affect the results. Instrumental error happens when the instruments being used are inaccurate, such as a balance that does not work (SF Fig. 1.4).
three types
## What is a zero error?
zero error Any indication that a measuring system gives a false reading when the true value of a measured quantity is zero, eg the needle on an ammeter failing to return to zero when no current flows. A zero error may result in a systematic uncertainty.
## What is acceptable percent error?
Explanation: In some cases, the measurement may be so difficult that a 10 % error or even higher may be acceptable. In other cases, a 1 % error may be too high. Most high school and introductory university instructors will accept a 5 % error.
## Is a 10 margin of error acceptable?
It depends on how the research will be used. If it is an election poll or census, then margin of error would be expected to be very low; but for most social science studies, margin of error of 3-5 %, sometimes even 10% is fine if you want to deduce trends or infer results in an exploratory manner.
## What does a small percent error mean?
Smaller percent errors indicate that we are close to the accepted or original value. For example, a 1% error indicates that we got very close to the accepted value, while 48% means that we were quite a long way off from the true value.
READ: What are the negative effects of water scarcity?
## Can u have negative percent error?
An individual measurement may be accurate or inaccurate, depending on how close it is to the true value. If the experimental value is less than the accepted value, the error is negative. If the experimental value is larger than the accepted value, the error is positive.
## What does percent error tell you about accuracy?
Percent error is the accuracy of a guess compared to the actual measurement. It’s found by taking the absolute value of their difference and dividing that by actual value. A low percent error means the guess is close to the actual value.
## What does negative percentage mean?
If the answer is a negative number, that means the percentage change is a decrease.
## Can you have a negative absolute error?
Absolute error in measurement When a number is absolute, it is not negative. When we say that the absolute error is the absolute value of the actual value minus the measured value, that means we should subtract the measured value from the actual value, then remove the negative sign (if any).
## How do you find absolute error percentage?
The computation of percentage error involves the use of the absolute error, which is simply the difference between the observed and the true value. The absolute error is then divided by the true value, resulting in the relative error, which is multiplied by 100 to obtain the percentage error.
## What is negative absolute error?
As its name implies, negative MAE is simply the negative of the MAE, which (MAE) is by definition a positive quantity. And since MAE is an error metric, i.e. the lower the better, negative MAE is the opposite: a value of -2.6 is better than a value of -3.0 .
## How do you find the maximum absolute error?
Method 1 of 3: equals the actual value. Subtract the actual value from the measured value. Since absolute error is always positive, take the absolute value of this difference, ignoring any negative signs. This will give you the absolute error.
## What is absolute error in math?
The difference between the measured or inferred value of a quantity and its actual value , given by. (sometimes with the absolute value taken) is called the absolute error. The absolute error of the sum or difference of a number of quantities is less than or equal to the sum of their absolute errors.
2021-05-14
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# 5.6: Determination of Unknown Triangle Measures Given Area
Difficulty Level: At Grade Created by: CK-12
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You are working on creating a mobile for your art class. A mobile is a piece of art that has a rod with different shapes hanging from it, so they can spin.
To create your project, you need to cut a set of triangles that have a variety of sizes. You are about to start cutting triangles, when your friend, who is helping you with the project, comes over. She tells you that each piece needs to have a rod through the side of it to balance the shape in a certain way. She wants you to make a piece that looks like this:
You have already cut a triangle by cutting a piece out of construction paper. You know that one side of your triangle is 6 inches long, but you don't know the length of the other two sides. Can you use the information you have to find the length of in the mobile piece above? (The area of the triangle is , and the interior angle between the six inch side and the side you want to know is ).
### Finding the Measures of a Triangle Given the Area
In this section, we will look at situations where we know the area but need to find another part of the triangle, as well as an application involving a quadrilateral. All of this will involve the use of the Law of Cosines, Law of Sines, and the Alternate Formula for the Area of a Triangle.
1. The jib sail on a sailboat came untied and the rope securing it was lost. If the area of the jib sail is 56.1 square feet, use the figure and information below to find the length of the rope.
Since we know the area, one of the sides, and one angle of the jib sail, we can use the formula to find the side of the jib sail that is attached to the mast. We will call this side .
Now that we know side , we know two sides and the included angle in the triangle formed by the mast, the rope, and the jib sail. We can now use the Law of Cosines to calculate the length of the rope.
The length of the rope is approximately 9.6 feet.
2. In quadrilateral below, the area of , the area of , and . Find the perimeter of .
In order to find the perimeter of , we need to know sides , and . Since we know the area, one side, and one angle in each of the triangles, we can use to figure out and .
Now that we know and , we know two sides and the included angle in each triangle (SAS). This means that we can use the Law of Cosines to find the other two sides, and . First we will find and .
Finally, we can calculate the perimeter since we have found all four sides of the quadrilateral.
3. In is an altitude from to . The area of , and . Find .
First, find by using the Pythagorean Theorem. . Then, using the area and formula , you can find . . .
### Examples
#### Example 1
Earlier, you were asked to find the length of in the mobile piece.
Since you know that the mobile piece is six inches on one side, and that the area of the triangle is , you can use the formula to find the length of the other side:
#### Example 2
Find "h" in the triangle below: Area
Since we know the area, one of the sides (18.15), and one angle of the triangle (), we can use the formula to find the other side of the triangle. We can then use the Pythagorean Theorem to find the height of the triangle.
This gives a result of:
#### Example 3
Find in the triangle below: Area
Since we know the area and the lengths of two of the sides of the triangle, we can use the formula to solve for the included angle, which gives:
#### Example 4
Find the area of below: Area
Area of
### Review
1. The area of the triangle below is . Solve for x, the height.
2. The area of the triangle below is . A height is given on the diagram. Solve for x.
Use the triangle below for questions 3-5. The area of the large triangle is .
1. Solve for x.
2. Find the perimeter of the large triangle.
3. Find the measure of all three angles of the large triangle.
Use the triangle below for questions 6-8. The area of the triangle is .
1. Solve for .
2. Solve for x.
3. Find the perimeter of the triangle.
Use the triangle below for questions 9-11. The area of the triangle is .
1. Solve for x.
2. Solve for y.
3. Find the measure of the other two angles of the triangle.
Use the triangle below for questions 12-15. The area of the large triangle is .
1. Solve for x.
2. Solve for y.
3. Solve for z.
4. Solve for .
To see the Review answers, open this PDF file and look for section 5.6.
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### Vocabulary Language: English
TermDefinition
law of cosines The law of cosines is a rule relating the sides of a triangle to the cosine of one of its angles. The law of cosines states that $c^2=a^2+b^2-2ab\cos C$, where $C$ is the angle across from side $c$.
law of sines The law of sines is a rule applied to triangles stating that the ratio of the sine of an angle to the side opposite that angle is equal to the ratio of the sine of another angle in the triangle to the side opposite that angle.
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## Ways to prove a bijection
You have a function $f : A \to B$ and want to prove it is a bijection. What can you do?
## By the book
A bijection is defined as a function which is both one-to-one and onto. So prove that $f$ is one-to-one, and prove that it is onto.
This is straightforward, and it’s what I would expect the students in my Discrete Math class to do, but in my experience it’s actually not used all that much. One of the following methods usually ends up being easier in practice.
## By size
If $A$ and $B$ are finite and have the same size, it’s enough to prove either that $f$ is one-to-one, or that $f$ is onto. A one-to-one function between two finite sets of the same size must also be onto, and vice versa. (Of course, if $A$ and $B$ don’t have the same size, then there can’t possibly be a bijection between them in the first place.)
Intuitively, this makes sense: on the one hand, in order for $f$ to be onto, it “can’t afford” to send multiple elements of $A$ to the same element of $B$, because then it won’t have enough to cover every element of $B$. So it must be one-to-one. Likewise, in order to be one-to-one, it can’t afford to miss any elements of $B$, because then the elements of $A$ have to “squeeze” into fewer elements of $B$, and some of them are bound to end up mapping to the same element of $B$. So it must be onto.
However, this is actually kind of tricky to formally prove! Note that the definition of “$A$ and $B$ have the same size” is that there exists some bijection $g : A \to B$. A proof has to start with a one-to-one (or onto) function $f$, and some completely unrelated bijection $g$, and somehow prove that $f$ is onto (or one-to-one). Also, a valid proof must somehow account for the fact that this becomes false when $A$ and $B$ are infinite: a one-to-one function between two infinite sets of the same size need not be onto, or vice versa; we saw several examples in my previous post, such as $f : \mathbb{N} \to \mathbb{N}$ defined by $f(n) = 2n$. Although tricky to come up with, the proof is cute and not too hard to understand once you see it; I think I may write about it in another post!
Note that we can even relax the condition on sizes a bit further: for example, it’s enough to prove that $f$ is one-to-one, and the finite size of $A$ is greater than or equal to the finite size of $B$. The point is that $f$ being a one-to-one function implies that the size of $A$ is less than or equal to the size of $B$, so in fact they have equal sizes.
## By inverse
One can also prove that $f : A \to B$ is a bijection by showing that it has an inverse: a function $g : B \to A$ such that $g(f(a)) = a$ and $f(g(b)) = b$ for all $a \in A$ and $b \in B$. As we saw in my last post, these facts imply that $f$ is one-to-one and onto, and hence a bijection. And it really is necessary to prove both $g(f(a)) = a$ and $f(g(b)) = b$: if only one of these hold then $g$ is called a left or right inverse, respectively (more generally, a one-sided inverse), but $f$ needs to have a full-fledged two-sided inverse in order to be a bijection.
…unless $A$ and $B$ are of the same finite size! In that case, it’s enough to show the existence of a one-sided inverse—say, a function $g$ such that $g(f(a)) = a$. Then $f$ is (say) a one-to-one function between finite equal-sized sets, hence it is also onto (and hence $g$ is actually a two-sided inverse).
We must be careful, however: sometimes the reason for constructing a bijection in the first place is in order to show that $A$ and $B$ have the same size! This kind of thing is common in combinatorics. In that case one really must show a two-sided inverse, even when $A$ and $B$ are finite; otherwise you end up assuming what you are trying to prove.
## By mutual injection?
I’ll leave you with one more to ponder. Suppose $f : A \to B$ is one-to-one, and there is another function $g : B \to A$ which is also one-to-one. We don’t assume anything in particular about the relationship between $f$ and $g$. Are $f$ and $g$ necessarily bijections?
Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 7 Responses to Ways to prove a bijection
1. Callan says:
Cheeky example: S: Nat -> Nat is injective so take f = g = S!
• Brent says:
By S do you mean the successor function? That’s a good example, not cheeky at all!
2. I did mean successor, yes!
3. blaisepascal2014 says:
If $f:A\to B$ is one-to-one, then $|A| \geq |B|$. Similarly, if $g:B\to A$, then $|B| \leq |A|$. If both are true, then $|A| = |B|$ and a bijection exists between $A,B$, but there’s no guarantee that either $f,g$ are bijections.
If $A,B$ are finite, then (as you argued in “By Size”), both $f,g$ are bijections. But if they are infinite, then there are $f$ that are one-to-one, but not onto. Proof: Since $B$ is infinite, there exists a $h:B\to B$ that is one-to-one but not onto, so if $f':A\to B$ is one-to-one, then $f:A\to B = h\circ f'$ is one-to-one, but not onto.
This is one of my favorite ways to show $|A| = |B|$. It usually isn’t hard to come up with the necessary injections, even if it is hard to find a bijection. It is easy to show that $f:\mathbb{N}\to\mathbb{N}^2, f(n) = (n,0)$ and $g:\mathbb{N}^2\to\mathbb{N}, g(a,b) = 2^a3^b$ are both one-to-one, It is harder to come up with an easy-to-describe bijection.
• Brent says:
Indeed! I think I will probably write about this in an upcoming post. One fun thing is that the proof of the Schröder-Bernstein theorem is actually constructive, so in theory you can take any two injections and actually use them to construct a bijection. But indeed, there is no guarantee that the resulting bijection is easy to describe. I am still trying to work out how to describe the one generated by your pair of injections between $\mathbb{N}$ and $\mathbb{N}^2$. It’s something like this: any number can be written in the form 2^2^…^m where $m$ is not a power of two, that is, a tower of zero or more 2’s with a final power of $m$ on top. (If $n$ is not a power of two then $m = n$.) If $m = 0$ or $m$ has prime factors other than 2 or 3, then send $n$ to $(n,0)$. Otherwise, if $n$ is a power of two, send it to $\log_2(n)$; finally, if $n = m = 2^j 3^k$, send it to $(j,k)$. It’s very non-obvious that this is a bijection (and I might have even gotten the description wrong)!
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# Horizontal and Vertical Asymptotes
## Guidelines that graphs approach based on zeros and degrees in rational functions.
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Horizontal and Vertical Asymptotes
What if you had a function like \begin{align*}y= \frac {x^2}{2x^2 - 1}\end{align*}? How could you find its vertical and horizontal asymptotes? After completing this Concept, you'll be able to find asymptotes of rational functions like this one.
### Try This
To explore more graphs of rational functions, try the applets available at http://www.analyzemath.com/rational/rational1.html.
### Guidance
We said that a horizontal asymptote is the value of \begin{align*}y\end{align*} that the function approaches for large values of \begin{align*}|x|\end{align*}. When we plug in large values of \begin{align*}x\end{align*} in our function, higher powers of \begin{align*}x\end{align*} get larger much quickly than lower powers of \begin{align*}x\end{align*}. For example, consider:
\begin{align*}y = \frac{2x^2 + x - 1}{3x^2 - 4x + 3}\end{align*}
If we plug in a large value of \begin{align*}x\end{align*}, say \begin{align*}x = 100\end{align*}, we get:
\begin{align*}y = \frac{2(100)^2 + (100) - 1}{3(100)^2 - 4(100) + 3} = \frac{20000 + 100 - 1}{30000 - 400 + 2}\end{align*}
We can see that the beginning terms in the numerator and denominator are much bigger than the other terms in each expression. One way to find the horizontal asymptote of a rational function is to ignore all terms in the numerator and denominator except for the highest powers.
In this example the horizontal asymptote is \begin{align*}y = \frac{2x^2}{3x^2}\end{align*}, which simplifies to \begin{align*}y = \frac{2}{3}\end{align*}.
In the function above, the highest power of \begin{align*}x\end{align*} was the same in the numerator as in the denominator. Now consider a function where the power in the numerator is less than the power in the denominator:
\begin{align*}y = \frac{x}{x^2 + 3}\end{align*}
As before, we ignore all the terms except the highest power of \begin{align*}x\end{align*} in the numerator and the denominator. That gives us \begin{align*}y = \frac{x}{x^2}\end{align*}, which simplifies to \begin{align*}y = \frac{1}{x}\end{align*}.
For large values of \begin{align*}x\end{align*}, the value of \begin{align*}y\end{align*} gets closer and closer to zero. Therefore the horizontal asymptote is \begin{align*}y = 0\end{align*}.
To summarize:
• Find vertical asymptotes by setting the denominator equal to zero and solving for \begin{align*}x\end{align*}.
• For horizontal asymptotes, we must consider several cases:
• If the highest power of \begin{align*}x\end{align*} in the numerator is less than the highest power of \begin{align*}x\end{align*} in the denominator, then the horizontal asymptote is at \begin{align*}y = 0\end{align*}.
• If the highest power of \begin{align*}x\end{align*} in the numerator is the same as the highest power of \begin{align*}x\end{align*} in the denominator, then the horizontal asymptote is at \begin{align*}y = \frac{coefficient \ of \ highest \ power \ of \ x}{coefficient \ of \ highest \ power \ of \ x}\end{align*}.
• If the highest power of \begin{align*}x\end{align*} in the numerator is greater than the highest power of \begin{align*}x\end{align*} in the denominator, then we don’t have a horizontal asymptote; we could have what is called an oblique (slant) asymptote, or no asymptote at all.
#### Example A
Find the vertical and horizontal asymptotes for \begin{align*}y = \frac {1}{x-1}\end{align*}.
Solution
Vertical asymptotes:
Set the denominator equal to zero. \begin{align*}x-1=0\Rightarrow x=1\end{align*} is the vertical asymptote.
Horizontal asymptote:
Keep only the highest powers of \begin{align*}x\end{align*}. \begin{align*}y=\frac {1}{x}\Rightarrow y=0\end{align*} is the horizontal asymptote.
#### Example B
Find the vertical and horizontal asymptotes for \begin{align*}y= \frac {3x}{4x+2}\end{align*}.
Solution
Vertical asymptotes:
Set the denominator equal to zero. \begin{align*}4x+2=0\Rightarrow x=-\frac{1}{2}\end{align*} is the vertical asymptote.
Horizontal asymptote:
Keep only the highest powers of \begin{align*}x\end{align*}. \begin{align*}y=\frac {3x}{4x}\Rightarrow y=\frac{3}{4}\end{align*} is the horizontal asymptote.
#### Example C
Find the vertical and horizontal asymptotes for \begin{align*}y=\frac {x^3}{x^2-3x+2}\end{align*}.
Solution
Vertical asymptotes:
Set the denominator equal to zero: \begin{align*}x^2 - 3x + 2 =0\end{align*}
Factor: \begin{align*}(x - 2)(x - 1) = 0\end{align*}
Solve: \begin{align*}x = 2\end{align*} and \begin{align*}x = 1\end{align*} are the vertical asymptotes.
Horizontal asymptote. There is no horizontal asymptote because the power of the numerator is larger than the power of the denominator.
Notice the function in part d had more than one vertical asymptote. Here’s another function with two vertical asymptotes.
#### Example D
Graph the function \begin{align*}y = \frac{-x^2}{x^2 - 4}\end{align*}.
Solution
\begin{align*}\text{Let's set the denominator equal to zero:} \qquad x^2 - 4 = 0\!\\ \\ \text{Factor:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (x - 2)(x + 2) = 0\!\\ \\ \text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ x = 2, x = -2\end{align*}
We find that the function is undefined for \begin{align*}x = 2\end{align*} and \begin{align*}x = -2\end{align*}, so we know that there are vertical asymptotes at these values of \begin{align*}x\end{align*}.
We can also find the horizontal asymptote by the method we outlined above. It’s at \begin{align*}y = \frac{-x^2}{x^2}\end{align*}, or \begin{align*}y = -1\end{align*}.
So, we start plotting the function by drawing the vertical and horizontal asymptotes on the graph.
Now, let’s make a table of values. Because our function has a lot of detail we must make sure that we pick enough values for our table to determine the behavior of the function accurately. We must make sure especially that we pick values close to the vertical asymptotes.
\begin{align*}x\end{align*} \begin{align*}y = \frac {-x^2}{x^2-4}\end{align*}
\begin{align*}-5\end{align*} \begin{align*}y = \frac {-(-5)^2}{(-5)^2-4} = \frac {-25}{21} = -1.19\end{align*}
-4 \begin{align*}y = \frac {-(-4)^2}{(-4)^2-4} = \frac {-16}{12} = -1.33\end{align*}
-3 \begin{align*}y = \frac {-(-3)^2}{(-3)^2-4} = \frac {-9}{5} = -1.8\end{align*}
-2.5 \begin{align*}y = \frac {-(-2.5)^2}{(-2.5)^2-4} = \frac {-6.25}{2.25} = -2.8\end{align*}
-1.5 \begin{align*}y = \frac {-(-1.5)^2}{(-1.5)^2-4} = \frac {-2.25}{-1.75} = 1.3\end{align*}
-1 \begin{align*}y = \frac {-(-1)^2}{(-1)^2-4} = \frac {-1}{-3} = 0.33\end{align*}
0 \begin{align*}y = \frac {-0^2}{(0)^2-4} = \frac {0}{-4} = 0\end{align*}
1 \begin{align*}y= \frac {-1^2}{(1)^2-4} = \frac {-1}{-3} = 0.33\end{align*}
1.5 \begin{align*}y = \frac {-1.5^2}{(1.5)^2-4} = \frac {-2.25}{-1.75} = 1.3\end{align*}
2.5 \begin{align*}y = \frac {-2.5^2}{(2.5)^2-4} = \frac {-6.25}{2.25} = -2.8\end{align*}
3 \begin{align*}y = \frac {-3^2}{(3)^2-4} = \frac {-9}{5} = -1.8\end{align*}
4 \begin{align*}y = \frac {-4^2}{(4)^2-4} = \frac {-16}{12} = -1.33\end{align*}
5 \begin{align*}y = \frac {-5^2}{(5)^2-4} = \frac {-25}{21} = -1.19\end{align*}
Here is the resulting graph.
Watch this video for help with the Examples above.
### Vocabulary
• Graphs of rational functions are very distinctive, because they get closer and closer to certain values but never reach those values. This behavior is called asymptotic behavior, and we will see that rational functions can have horizontal asymptotes, vertical asymptotes or oblique (or slant) asymptotes.
### Guided Practice
Find the vertical and horizontal asymptotes for \begin{align*}y=\frac {x^2-2}{2x^2+3}\end{align*}.
Solution
Vertical asymptotes:
Set the denominator equal to zero: \begin{align*} 2x^2+3 = 0 \Rightarrow 2x^2 = -3 \Rightarrow x^2 = -\frac{3}{2}\end{align*}. Since there are no solutions to this equation, there is no vertical asymptote.
Horizontal asymptote:
Keep only the highest powers of \begin{align*}x\end{align*}. \begin{align*}y=\frac {x^2}{2x^2} \Rightarrow y= \frac {1}{2}\end{align*} is the horizontal asymptote.
### Practice
Find all the vertical and horizontal asymptotes of the following rational functions.
1. \begin{align*}y=\frac {4}{x+2}\end{align*}
2. \begin{align*}y=\frac {5x-1}{2x-6}\end{align*}
3. \begin{align*}y=\frac {10}{x}\end{align*}
4. \begin{align*}y=\frac {2}{x}-5\end{align*}
5. \begin{align*}y=\frac {x + 1}{x^2}\end{align*}
6. \begin{align*}y=\frac {4x^2}{4x^2+1}\end{align*}
7. \begin{align*}y=\frac {2x}{x^2-9}\end{align*}
8. \begin{align*}y=\frac {3x^2}{x^2-4}\end{align*}
9. \begin{align*}y=\frac {1}{x^2+4x+3}\end{align*}
10. \begin{align*}y=\frac {2x+5}{x^2-2x-8}\end{align*}
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# Solution of Quadratic Equation by factorization
In this method, write the quadratic equation in the standard form as
ax2 + bx + c = 0 (i)
If two numbers r and s can be found for the equation (i) such that
r + s = b and rs = ac
then ax2 + bx + c can be factorized into two linear factors.
### Example
Solve the quadratic equation 3x2 − 6x = x + 20 by factorization.
### Solution:
3x2 − 6x = x + 20 (i)
The standard form of (i) is
3x2 − 7x − 20 = 0 (ii)
Here a = 3, b = −7, c = −20
and ac = 3 × (−20) = −60
As −12 + 5 = −7 and −12 × 5 = −60, so
the equation (ii) can be written as
3x2 − 12x + 5x − 20 = 0
or 3x (x − 4) + 5 (x − 4) = 0
⇒ (x − 4) (3x + 5) = 0
Either x − 4 = 0 or 3x + 5 = 0
x = 4 or 3x = −5 ⇒ x = −5⁄3
∴ x = −5⁄3, 4 are the solutions of the given equation.
Thus, the solution set is {−5⁄3, 4}.
### Activity:
Factorize x2 − x − 2 = 0.
### Solution:
x2 − x − 2 = 0 (i)
Here a = 1, b = −1, c = −2
and ac = 1 × (−2) = −2
As −2 + 1 = −1 and −2 × 1 = −2, so
the equation (i) can be written as
x2 − 2x + x − 2 = 0
or x (x − 2) + 1 (x − 2) = 0
⇒ (x − 2) (x + 1) = 0
(x − 2), (x + 1) are factors of the given equation.
### Example
Solve 5x2 = 30x by factorization.
### Solution:
5x2 = 30x
Remember that cancelling of x on both sides of 5x2 = 30x means the loss of one root, i.e., x = 0
5x2 − 30x = 0 (i)
The equation (i) is factorized as
5x (x − 6) = 0
Either 5x = 0 or x − 6 = 0 ⇒ x = 0 or x = 6
∴ x = 0, 6 are the roots of the given equation.
Thus, the solution set is {0, 6}.
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# Solve Word Problems with Line Plots
Related Topics:
Lesson Plans and Worksheets for Grade 4
Lesson Plans and Worksheets for all Grades
Examples, videos, and solutions to help Grade 4 students learn how to solve word problems with line plots.
Common Core Standards: 4.NF.1, 4.NF.2, 4.NF.3, 4.NBT.6, 4.NF.4a, 4.MD.4
### NYS Common Core Grade 4 Module 5, Lesson 28
Lesson 28 Problem Set
1. The chart to the right shows the distance fourth-graders in Ms. Smith’s class were able to run before stopping for a rest. Create a line plot to display the data in the table.
2. Solve each problem.
a. Who ran a mile farther than Jenny?
b. Who ran a mile less than Jack?
c. Two students ran exactly 2 1/4 miles. Identify the students. How many quarter miles did each student run?
d. What is the difference, in miles, between the longest and shortest distance run?
e. Compare the distances run by Arianna and Morgan using >, <, or =.
f. Ms. Smith ran twice as far as Jenny. How far did Ms. Smith run? Write her distance as a mixed number.
g. Mr. Reynolds ran 1 4/10 miles. Use >, <, or = to compare the distance Mr. Reynolds ran to the distance that Ms. Smith ran. Who ran farther?
3. Using the information in the table and on the line plot, develop and write a question similar to those above. Solve, and then ask your partner to solve. Did you solve in the same way? Did you get the same answer?
Lesson 28 Homework
1. A group of children measured the lengths of their shoes. The measurements are shown in the table. Make a line plot to display the data.
2. Solve each problem. Write an equation and a statement for each. Draw models as needed.
a. Who has a shoe length 1 inch longer than Dickon?
b. Who has a shoe length 1 inch shorter than Susan?
c. How many quarter inches long is Martha’s shoe length?
d. What is the difference, in inches, between Lilias's and Martha’s shoe lengths?
e. Compare the shoe length of Ben and Frances using >, <, or =.
f. How many students had shoes that measured less than 8 inches?
g. How many children measured the length of their shoes?
h. Mr. Jones’s shoe length was 25/2 inches. Use >, <, or = to compare the length of Mr. Jones’s shoe to the length of the longest student shoe length. Who had the longer shoe?
3. Using the information in the table and on the line plot, write a question you could solve by using the line plot. Solve.
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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# Order of Operations
The lesson below is still images taken from the fully animated PowerPoint I created that is available in my Teachers Pay Teachers store page. If the PowerPoint is not one of my freebies, you may also head over to my YouTube channel to see the slideshow fully animated and can pause as needed to be sure you grasp each concept before moving forward (each lesson will always be free there!).
This lesson will introduce how to determine the correct order to perform operations when looking at a problem. We go over why the order of operations is so crucial to getting the correct answer and walk through examples step-by-step to help clear up some of the most common errors students make.
So without further ado, read through the slides below to get a feel for how to correctly use the order of operations!
Phew, that’s a lot to take in. Once you’ve gone over this and found some practice problems to cement the idea in your head what integers are, you may move on to the next lesson below!
Solving Equations by Adding and Subtracting
Just in case the slides aren’t your thing, here is a text outline of the main points of the lessons above!
• Order of operations
• Objectives
• By the end of this lesson you should feel comfortable:
• Recognizing different operations
• Utilizing the order of operations to solve problems
• Where to start?
• Sometimes, we are presented with a problem that requires us to perform more than one calculation, or operation, in order to arrive at an answer.
• For example, let’s say you your two friends are saving up for a trip by mowing lawns. Your dad also says he will give you double what you earn yourself (so you will have three times as much as what you make on your own).
• Let’s say you make \$15, one friend earns \$20, and the other friend earns \$10. How much did you make together?
• Where to start?
• So…what would we do first?
• If we add everything up, we would have 15+20+10 = 45
• Then multiply 45 by 3 to get 135. But is this correct?
• Your dad is only tripling the amount YOU earn, not everyone.
• So you’d have to multiply first.
• 75 and 135 certainly aren’t equal.
• Clearly the ORDER you perform the operations matters.
• Operations
• Before we can decide the correct order, however, first we should know all the different operations that are possible.
• Operations are the different calculation techniques we can do to numbers.
• Still fuzzy? How about a list of examples from the first things we learned.
• All of these have a specific order we perform them in when combined together in a single problem.
• pemdas
• So what order do we follow?
• Cool, so what does that mean?
• It’s an acronym, which means each letter stands for an entire word.
• In fact, you can remember this with a silly saying:
• Please Excuse My Dear Aunt Sally”
• Using Order of Operations
• Alright, so let’s try putting PEMDAS to use and see what it really means:
• Great, so lots of stuff going on. Let’s go down the line.
• Do we see any parentheses? No.
• Do we see any exponents? Yes. Let’s perform that operation first.
• Next, we look for either multiplication or division. Whichever we see first reading from left to right, we perform first. Here we only have multiplication, so let’s do that next.
• Using order of operations
• Finally, we look for addition or subtraction. Again, we see which comes first reading from left to right. This time, subtraction appears first, so:
• Then we have a simple addition problem left.
• Ta da! You’ve solved your first multistep problem using order of operations!
• Notice: In each step, we focus on one operation, then change only that answer in the next line and rewrite the rest of the problem exactly the same so that we can retrace our steps if we somehow make a mistake!
• Using order of operations
• One more problem so that we can introduce a final idea.
• We use parentheses to designate that a specific operation should be done first.
• In our example, we can see multiplication outside the parentheses, which normally would come before addition.
• But remember our order!
• Parentheses are the first thing we check for. Though they are not an operation, they designate which operation, or set of operations, are to be performed first.
• Using order of operations
• With that in mind, let’s do what’s in the parentheses!
• Notice that now we have 5 x 4. This is because a number outside of a set of parentheses is math notation for multiplication.
• Now we can follow along with our order!
• Are there any exponents? No.
• Do we see multiplication or division? Yes, so let’s go left to right.
• And the last step is to simply multiply our final problem.
• Ta da! Another problem finished successfully!
• conclusion
• Operations include:
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## Financial Maths, Part 1
For some of my students, interest calculations are troublesome: you can say that they quickly lose interest in interest.
If I still have your interest after that bad joke, I will continue. The two main types of interest are simple and compound interest. In simple interest, the principal (the amount initially invested) stays the same and interest is calculated on that amount at all times. In compound interest, the principal grows and the value upon which interest is calculated changes.
I have previously talked about percentages and how to take a percentage of a number. Please review that if you do not know how to take a percentage of a number.
As always, in any new topic, there are some definitions to know so that we understand each other. The following are the main definitions with the abbreviations for them that will be used in equations:
Principal (P): the amount invested or borrowed
Interest rate (r): a percentage to be applied to the principal. This can be a percentage (eg. 15%) or its decimal equivalent (0.15).
Interest (I): the dollar amount which results when the interest rate is applied to the principal
Time (t): the amount of time to be used in a problem
Period: the basic amount of time used by the interest rate. For example, 15% per annum (abbreviated p.a.) means that the period is 1 year.
Number of periods (n): The number of periods to be used in a given problem. Note that equations can be in terms of time (t) or number of periods (n).
Let’s start out with a simple interest situation. Suppose I invest $1000 at a simple interest rate of 3% p.a., that is 3% each year. Though I haven’t asked a question yet, let me identify the key items of this set up: P =$1000
r = 3% or 0.03
period = 1 year
So my first question is: how much interest do I earn after 1 year? At the end of each year, if I keep that initial amount 0f $1000 in the investment, I will earn 3% of$1000 in interest. If you remember, to take a percentage “of” something, the “of” means to multiply. So after 1 year:
I = 3% × $1000 = (3/100) × 1000 or 0.03 × 1000 =$30
Note that in equations where you can put the interest rate in directly (the “3”), there will be a “/100” part in the equation. In equations where the decimal equivalent of the interest rate (0.03) is to be used, there will be no “/100” part. So the formulas to find the amount of interest (I) earned in 1 period are:
I = Pr/100, if you like to use the interest rate number directly (the “3”)
I = Pr, if you like to use the decimal equivalent of the interest rate (0.03)
This is why you may see different formulas in different books.
## Percentages, Part 3
So now that we know what a percentage is, how is it used? Let’s look at some sample percentage problems.
Melbourne’s Silvan Reservoir has a capacity of 40,446 ML(megaliters). Currently, it is 88% full. How much water is in the reservoir? In other words, what is 88% of 40,446?
Whenever you see or interpret a problem where you need to take a percentage “of” something, equate the word “of” with “multiply”. So to take 88% of 40,446, we multiply 40, 446 by 88%. If you do this on a calculator, you need to use the decimal equivalent of 88% which is 0.88 (see my previous post). On some calculators, there is a percent key. On these, you can type 88 which the calculator will interpret as 0.88 when you use that key. Regardless of which calculator you use, when you multiply 40, 446 by 88%, you should get 35592.48 ML.
If you do not live in Melbourne, the above problem does not interest you much. As money is of interest to most everyone, let’s look at some typical money related percentage problems.
Since DavidTheMathsTutor has effectively educated the masses on how to use percentages, calculators with percentage keys are no longer in demand. So a store has discounted the normal $24.95 price of these calculators by 30%. What is the new price? This is a two-step problem: first find what the amount of the discount is, then subtract it from the original price. The amount of the discount is 30% × 24.95 = 0.3 × 24.95 =$7.48 (rounded to 2 decimals as we are talking about money). So the store is taking $7.48 off each calculator. So the new price is 24.95 – 7.48 =$17.47.
On the other hand, again because of DavidThe MathsTutor, there is a big demand on fancy calculators that do all sorts of mathematical things like graph equations. So the store decides to markup the normal price of these calculators by 25% to make up for the loss of the percentage calculators. The normal price of these are $149.50 each. What is the new price? Again, a two-step problem, but this time you are find the amount of the price increase, then add it to the original price since this is a markup. The amount of markup is 0.25 × 149.50 =$37.38. So the store is increasing the price by $37.38, so the new price is 149.50 + 37.38 =$186.88.
Many times, you need to calculate the original price. For example, you are looking to buy a car. The sticker says $24,500. The salesperson says that’s a good price because they are only making a 5% profit on it. What is the cost of the car to the store? This is a reverse markup problem: what price plus 5% of that price is$24,500?
In equation form, that is the equivalent maths sentence, this is
x + (0.05 × x) = 24,500.
If you remember my posts on equations, factoring, and the distributive property, this is solved by factoring the left side and then dividing both sides by the number that results from the factoring:
$\begin{array}{l} {{x}\hspace{0.33em}{+}\hspace{0.33em}{(}{0}{.}{05}\hspace{0.33em}{+}\hspace{0.33em}{x}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}{(}{0}{.}{05}\hspace{0.33em}{+}\hspace{0.33em}{1}{)}\hspace{0.33em}{=}\hspace{0.33em}{1}{.}{05}{x}\hspace{0.33em}{=}\hspace{0.33em}{24500}}\\ {\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{24500}{1.05}\hspace{0.33em}{=}\hspace{0.33em}{\}{23333}{.}{33}} \end{array}$
If you knew the cost to the store was $23,333.33 and knew that they marked up that cost by 5%, if you calculated the new price as we did before, you would get$24,500.
Sometimes you need to calculate the actual percentage. What if your salary went up from 88,000 to 99,000? What percentage pay rise is this (so you can brag to your friends)?
The amount of pay increased by 99,000 – 88,000 = 11,000, so we want to know what percentage of 88,000 is 11,000. Note that you always work with the original price or amount when working out a percentage. So the equivalent maths sentence is
88,000 × x% = 11,000
Dividing both sides by 88,000 gives
${x}\hspace{0.33em}{=}\hspace{0.33em}\frac{11000}{88000}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{125}\hspace{0.33em}{=}\hspace{0.33em}{12}{.}{5}{\%}$
You must be very good at your job!
## Percentages, Part 2
So how do you convert percentages to fractions and decimals and vice versa? This post will show examples of each.
1. Convert a percentage to a fraction:
This one is easy as if you remember, a percentage is already a fraction where the numerator is displayed and the denominator is 100. So you just create the fraction and simplify it (see my posts on fractions):
${40}{\%}\hspace{0.33em}{=}\hspace{0.33em}\frac{40}{100}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}{0}\hspace{0.33em}\times\hspace{0.33em}{2}}{{20}\hspace{0.33em}\times\hspace{0.33em}{5}\hspace{0.33em}}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{5}$
2. Convert a percentage to a decimal:
This one is just a matter of moving the decimal point, two places to the left. Keep in mind that the decimal point will not usually show at the end of integer percentage, but you can assume it to be at the end of the number:
37% = 37.% = 0.37
18.5% = 0.185
112% = 1.12
0.15% = 0.0015
Any 0’s at the end of the decimal, can be left off:
40% = 0.40 = 0.4
3. Convert a decimal to a percentage:
This is just the opposite of of the above: you just move the decimal point two places to the right, then add the % symbol:
0.25 = 25% (if an integer results, you can leave the decimal point off)
0.2786 = 27.86%
0.002 = 0.2%
2.345 = 234.5%
4. Convert a fraction to a percentage:
Here you multiply by 100/1, simplify, then multiply numerators together and denominator together. It is advisable to simplify before multiplying:
$\frac{3}{5}\hspace{0.33em}\times\hspace{0.33em}\frac{100}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{\rlap{/}{5}\hspace{0.33em}\times\hspace{0.33em}{1}}\hspace{0.33em}\times\hspace{0.33em}\frac{\rlap{/}{5}\hspace{0.33em}\times\hspace{0.33em}{20}}{1}\hspace{0.33em}{=}\hspace{0.33em}{60}{\%}$
Sometimes though, not as much cancels and you will need to do some division in the end (long or short – see my post on long division):
$\frac{8}{9}\hspace{0.33em}\times\hspace{0.33em}\frac{100}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{800}{9}\hspace{0.33em}{=}\hspace{0.33em}{800}\hspace{0.33em}\div\hspace{0.33em}{9}\hspace{0.33em}{=}\hspace{0.33em}{88}{.}{89}{\%}$
In my next post, I will show how to do some of the more common problems using percentages.
## Percentages, Part 1
I am confident that the chance of you liking this post is 100%. But what does 100% mean? I will talk about this in the next few posts.
First let’s review what a fraction is. The fraction
$\frac{3}{4}$
means that I have 3 pieces (the numerator) of some whole thing that was divided into 4 pieces (the denominator). So the denominator sets the size of the pieces (the bigger the denominator, the smaller the pieces) and the numerator sets how many pieces.
Being creatures that have 10 fingers, we naturally migrate to things that are powers of 10. Way back in the Roman empire days (I remember them fondly), the Romans frequently used fractions that had a denominator of 100. In fact, the word “percentage” comes from the latin (Roman) per centum which means “by the hundred”. That has continued to today, as a percentage is really a fraction where the denominator is fixed at 100. What you see in a percentage is the numerator:
${37}{\%}\hspace{0.33em}{=}\hspace{0.33em}\frac{37}{100}$
So a percentage like 37% means that if I divide something into 100 equal pieces, I have 37 of these pieces. Percentages are also used to indicate a probability. My first sentence in this post used a percentage in this way. If you remember, a probability of an event is a fraction. The probability of event A is expressed as P(A):
${P}{(}{A}{)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{\mathrm{Number}}\hspace{0.33em}{\mathrm{of}}\hspace{0.33em}{\mathrm{times}}\hspace{0.33em}{\mathrm{event}}\hspace{0.33em}{\mathrm{A}}\hspace{0.33em}{\mathrm{occurs}}}{{100}\hspace{0.33em}{\mathrm{trials}}}$
So 100% of you liking this post means that if I take 100 random people who have read this post, 100 of them will like this post. This means all will like this post as that is what 100% means: the whole:
${100}{\%}\hspace{0.33em}{=}\hspace{0.33em}\frac{100}{100}\hspace{0.33em}{=}\hspace{0.33em}{1}$
Now percentages, fractions, and decimals are all ways to express a part of something. In my next post, I will show how to convert percentages to fractions and decimals and vice versa.
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# What is the general solution of the differential equation xy'' = y'+x(y')^2?
Oct 25, 2017
$y = C - \ln | {x}^{2} + B |$
#### Explanation:
We have:
$x y ' ' = y ' + x {\left(y '\right)}^{2}$ ..... [A]
Which is a Second Order Non-Linear ODE. As there is no term in $y$ then let us attempt the following substitution:
$v = y ' \implies v ' = y ' '$
Substituting into the DE [A] we get:
$x v ' = v + x {v}^{2}$ ..... [B]
Which has reduced the initial ODE to a First Order Non-Linear ODE (in this case a Bernoulli Equation), so now we attempt a second substitution of the form:
$u = {v}^{- 1} \implies \frac{\mathrm{du}}{\mathrm{dv}} = - {v}^{- 2}$ and $\frac{\mathrm{dv}}{\mathrm{du}} = - {v}^{2}$
Substituting into the last DE [B], in conjunction with the chain rule we get;
$x \frac{\mathrm{dv}}{\mathrm{dx}} = v + x {v}^{2}$
$\therefore x \frac{\mathrm{dv}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = v + x {v}^{2}$
$\therefore x \left(- {v}^{2}\right) \cdot \frac{\mathrm{du}}{\mathrm{dx}} = v + x {v}^{2}$
$\therefore - \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x v} + 1$
$\therefore - \frac{\mathrm{du}}{\mathrm{dx}} = \frac{u}{x} + 1$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}} + \frac{u}{x} = - 1$ ..... [C]
So the second substitution has reduced the DE into a first order linear differential equation of the form:
$\frac{d \zeta}{\mathrm{dx}} + P \left(x\right) \zeta = Q \left(x\right)$
We solve this using an Integrating Factor
$I = \exp \left(\setminus \int \setminus P \left(x\right) \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(\int \setminus \left(\frac{1}{x}\right) \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(\ln x\right)$
$\setminus \setminus = x$
And if we multiply the last equation [C] by this Integrating Factor, $I$, we will have a perfect product differential;
$\setminus \setminus \setminus x \frac{\mathrm{du}}{\mathrm{dx}} + u = - x$
$\therefore - \frac{d}{\mathrm{dx}} \left(x u\right) = x$
Which is now a trivial separable DE, so we can integrate to get:
$- x u = \int \setminus x \setminus \mathrm{dx}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {x}^{2} / 2 + K$
And restoring the $u$ substitution we have:
$- x \left(\frac{1}{v}\right) = {x}^{2} / 2 + K$
$\therefore - \frac{x}{v} = \frac{{x}^{2} + 2 K}{2}$
$\therefore - \frac{v}{x} = \frac{2}{{x}^{2} + A}$
$\therefore v = - \frac{2 x}{{x}^{2} + A}$
And restoring the $v$ substitution we have::
$y ' = - \frac{2 x}{{x}^{2} + B}$
Which again is a First Order separable ODE, so again we "separate the variable" to get:
$\int \setminus \mathrm{dy} = - \int \setminus \frac{2 x}{{x}^{2} + B} \setminus \mathrm{dx}$
Integrating we get:
$y = - \ln | {x}^{2} + B | + C$
Which is the GS of [A]
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# FINAL_EXAM_REVIEW_QUESTIONS - REVIEW FOR THE FINAL The hope...
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REVIEW FOR THE FINAL The hope is that this will prepare you for the final. If you work out the problems diligently, bring up questions as necessary and review daily, the final will seem like just another exercise. Here we go: Fractions 1) 1/3 + 2/5 * 2/9 – 2/7 = 2) 7/16 divided by 1/8 = Conversion of fractions, decimals and percents Express as a percent: (Use your judgment when you round off) 3) 2.045 4) 2/225 = 5) 876 = 6) .00456 = Express as a decimal: (Round to at least two decimals) 7) 3 /125 = 8) 6/57 = Express as a fraction: 9) 2.075 = 10) .0035% = Order of Operation (Leave your answer as a fraction) 11) (-4) + (-6)(-8) divided by [16 – (-1)] = 12) (-7)(5)(+2) divided by (10 +7) = Evaluation Problems
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13) Given the equation X = Y 3 + Z/2 divided by [K - 3] , Find X when K = (-7), Y = 3 and Z = 6 Solve for "x" : 14) 3x + 7 = 2(4x –5) –2x = 15) 7(y –2) +4 = 3(8y + 7) = Counting Techniques 16) Interpret the following: 12! /7! = 8! /(8 - 4)! = Intersection % Union of Sets, Venn diagrams 17) In the rolling of a fair die, what is the probability of rolling an even number or a number that is less than 3? 18) Given X = {2, 5, 7} and Y = {3, 4, 5} depict these two sets as a union and an intersection, i.e., (x u y) =? and (x ∩ y) =? Depreciation Expense A Widget maker was purchased at a cost of \$225,000. Depreciation will be calculated on the Straight-line method over a period of 20 years. The salvage value of this machine is \$25,000. Answer the following question: 19) What is the annual depreciation expense?
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# QUADRATIC EQUATIONS OMTEX CLASSES www.omtexclasses.
com
EXERCISE - 2.1 1. Which of the following are quadratic equations ?
(i) 11 = – 4x2 – x3 (ii) -¾ y2 = 2y + 7 (iii) (y – 2) (y + 2) = 0 (iv) 3/y – 4 = y [Ans.] [Ans.] [Ans.] [Ans.] [Ans.]
(v) m3 + m + 2 = 4m (vi) n – 3 = 4n (vii) y2 – 4 = 11y [Ans.]
[Ans.]
(viii) z – 7/z = 4z + 5 [Ans.] (ix) 3y2 – 7 = √3 y [Ans.]
(x) (q2 – 4)/q2 = - 3 [Ans.]
2. Write the following quadratic equations in standard form ax2 + bx + c = 0
(i) 7 – 4x –x2 = 0 [Ans] (ii) 3y2 = 10y + 7 [Ans]
(iii) (m + 4) (m – 10) = 0 [Ans] (iv) p(p – 6) = 0 [Ans] (v) (x2/25) – 4 = 0 (vi) n – (7/n) = 4 (vii) y2 – 9 = 13y [Ans] [Ans] [Ans] [Ans]
(viii) 2z – (5/z) = z – 6 (ix) x2 = –7 – √10 x
[Ans] [Ans]
(x) (m2 +5)/m2 = –3
EXERCISE - 2.2 1. In each of the examples given below determine whether the values given against each of the quadratic equation are the roots of the equation or not.
(i) x2 + 3x – 4 = 0, x = 1, –2, – 3 [Ans] (ii) 4m2 – 9 = 0, m = 2, 2/3, 3/2 [Ans] (iii) x2 + 5x – 14 = 0, x = √2 , –7, 3 [Ans] (iv) 2p2 + 5p – 3 = 0, p = 1, ½, –3 [Ans] (v) n2 + 4n = 0, n = 0, – 2, – 4 [Ans]
2. If one root of the quadratic equation x2 – 7x + k = 0 is 4, then find the value of k. [Ans]
3. If one root of the quadratic equation 3y2 – ky + 8 = 0 is 2/3, then find the value of k. [Ans]
4. State whether k is the root of the given equation y2 – (k – 4)y – 4k = 0. [Ans] 5. If one root of the quadratic equation kx2 – 7x + 12 = 0 is 3, then find the value of k. [Ans]
EXERCISE - 2.3 Solve the following quadratic equations by factorization method..
(i) x2 – 5x + 6 = 0 [Ans.] (ii) x2 + 10x + 24 = 0 [Ans.] (iii) x2 – 13x – 30 = 0 [Ans.] (iv) x2 – 17x + 60 = 0 [Ans.]
(v). m2 – 84 = 0 [Ans.]
(vi) x + 20/x – 12 = 0 [Ans.] (vii) x2 = 2(11x – 48) [Ans.] (viii) 21x = 196 – x2 [Ans.]
(x) x2 – x – 132 = 0 [Ans.]
(xi) 5x2 – 22x – 15 = 0 [Ans.]
(xii) 3x2 – x – 10 = 0 [Ans.]
(xiii) 2x2 – 5x – 3 = 0 [Ans]
(xiv) x (2x + 3) = 35 [Ans.]
(xv) 7x2 + 4x – 20 = 0 [Ans]
(xvi) 10x2 + 3x – 4 = 0 [Ans.]
(xvii) 6x2 – 7x – 13 = 0 [Ans.] (xviii) 3x2 + 34x + 11 = 0 [Ans.] (xix) 3x2 – 11x + 6 = 0 [Ans.] (xx) 3x2 – 10x + 8 = 0 [Ans.] (xxi) 2m2 + 19m + 30 = 0 [Ans.] (xxii) 7m2 – 84 = 0 [Ans.] (xxiii) x2 – 3√3 x + 6 = 0 [Ans.]
EXERCISE - 2.4 Solve the following quadratic equations by completing square.
(i) x2 + 8x + 9 = 0 [Ans]
(ii) z2 + 6z – 8 = 0 [Ans]
(iii) m2 – 3m – 1 = 0 [Ans]
(iv) y2 = 3 + 4y [Ans]
(v) p2 – 12p + 32 = 0 [Ans]
(vi) x (x – 1) = 1 [Ans]
(vii) 3y2 + 7y + 1 = 0 [Ans]
(viii) 4p2 + 7 = 12p [Ans] (ix) 6m2 + m = 2 [Ans]
EXERCISE - 2.5 1. Solve the following quadratic equations by using formula.
(i) m2 – 3m – 10 = 0 [Ans] (ii) x2 + 3x – 2 = 0 [Ans] (iii) x2 + (x – 1)/3 = 0 [Ans] (iv) 5m2 – 2m = 2 [Ans.] (v) 7x + 1 = 6x2 [Ans.] (vi) 2x2 – x – 4 = 0 [Ans.] (vii) 3y2 + 7y + 4 = 0 [Ans.] (viii) 2n2 + 5n + 2 = 0 [Ans.] (ix) 7p2 – 5p – 2 = 0 [Ans.] (x) 9s2 – 4 = – 6s [Ans.] (xi) 3q2 = 2q + 8 [Ans.] (xii) 4x2 + 7x + 2 = 0 [Ans.]
EXERCISE - 2.6 1. Find the value of discriminant of each of the following equations :
(i) x2 + 4x + 1 = 0 [Ans] (ii) 3x2 + 2x – 1 = 0 [Ans] (iii) x2 + x + 1 = 0 [Ans] (iv) √3 x2 + 2√2 x – 2√3 = 0 [Ans] (v) 4x2 + kx + 2 = 0 [Ans] (vi) x2 + 4x + k = 0 [Ans]
2. Determine the nature of the roots of the following equations from their discriminants :
(i) y2 – 4y – 1 = 0 [Ans.] (ii) y2 + 6y – 2 = 0 [Ans.] (iii) y2 + 8y + 4 = 0 [Ans.] (iv) 2y2 + 5y – 3 = 0 [Ans.]
(v) 3y2 + 9y + 4 = 0 [Ans.] (vi) 2x2 + 5√3 x + 16 = 0 [Ans.]
3. Find the value of k for which given equation has real and equal roots :
(i) (k – 12)x2 + 2 (k – 12)x + 2 = 0 [Ans.] (ii) k2x2 – 2 (k – 1)x + 4 = 0 [Ans.]
EXERCISE - 2.7
1. If one root of the quadratic equation kx2 – 5x + 2 = 0 is 4 times the other, find k. [Ans.] 2. Find k, if the roots of the quadratic equation x2 + kx + 40 = 0 are in the ratio 2 : 5. [Ans.] 3. Find k, if one of the roots of the quadratic equation kx2 – 7x + 12 = 0 is 3. [Ans.] 4. If the roots of the equation x2 + px + q = 0 differ by 1, prove that p2 = 1 + 4q. [Ans.]
5. Find k, if the sum of the roots of the quadratic equation 4x2 + 8kx + k + 9 = 0 is equal to their product. [Ans.]
6. If α and β are the roots of the equation x2 – 5x + 6 = 0, find[Ans.]
(i) α2+β2 (ii) α/β +β/α
7. If one root of the quadratic equation kx2 – 20x + 34 = 0 is 5 – 2√2 ,
find k.
[Ans.]
EXERCISE - 2.8
1. Form the quadratic equation if its roots are (i) 5 and – 7 [Ans.] [Ans.]
(ii) ½ and – ¾
(iii) - 3 and –11 [Ans.] (iv) -2 and 11/2 [Ans.] (v) ½ and – ½ (vi) 0 and – 4 [Ans.] [Ans.]
2. Form the quadratic equation if one of the root is
(i) 3 – 2√ 5 [Ans.]
(ii) 4 – 3√ 2 [Ans.] (iii) √ 2 + √3 [Ans.] (iv) 2√3 – 4 [Ans.] (v) 2+√5 [Ans.]
(vi) √ 5 - √3 [Ans.]
3. If the sum of the roots of the quadratic is 3 and sum of their cubes is 63, find the quadratic equation. [Ans.] 4. If the difference of the roots of the quadratic equation is 5 and the difference of their cubes is 215, find the quadratic equation.[Ans.]
EXERCISE - 2.9 Solve the following equations.
(i) x4 – 3x2 + 2 = 0 [Ans.] (ii) (x2 + 2x) (x2 + 2x – 11) + 24 = 0 [Ans.] (iii) 2(x2 + 1/x2 ) – 9(x+1/x) + 14 = 0 [Ans.] (iv) 35y2 + 12/y2 = 44 [Ans.] (v) x2 + 12/x2 = 7 [Ans.] (vi) (x2 + x) (x2 + x – 7) + 10 = 0 [Ans.] (vii) 3x4 – 13x2 + 10 = 0 [Ans.] (viii) 2y2 + 15/y2 = 12 [Ans.]
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# Math for Transforming 3D Geometry
Whether you’re writing your own game engine, using Unity or Unreal, or you’re an artist making 3D graphics, a basic understanding of the kinds of math computers use to get graphics on the screen is a huge asset. This article introduces vectors, complex numbers, Euler angles, quaternions, and matrices, and how they are used to move, rotate, and scale 3D models.
# Vectors
The positions of things in graphics are usually represented by vectors. The first number in a vector is called the x-coordinate and the second number is called the y-coordinate. Each number represents a dimension; a 3-dimensional vector has an additional coordinate called z.
Here are two different vectors, a and b. a’s x coordinate is 2 and its y coordinate is 3. Since these vectors have two numbers each, they represent positions in 2-dimensional space.
Vectors can be added together or subtracted. a’s x coordinate is added to b’s x coordinate, and a’s y coordinate is added to b’s y coordinate. Likewise for subtraction.
We can think of a 2D vector as a distance from the origin (0,0) on the x and y axes. Addition and subtraction move one vector by another vector.
When talking about vectors, we call a regular real number a scalar. In this example, a scalar, s, is equal to 5. Multiplying a vector by a scalar multiplies each coordinate by the scalar. However, there is no straightforward multiplication of one vector with another. We won’t need the dot product or cross product in this tutorial, so we won’t cover them here.
The length of a vector is its distance from the origin. If c is a vector, the length of c is notated by |c|. Suppose the units of our space are measured in inches. You could find the length of a vector with a ruler by measuring the distance from the origin to the point represented by the vector. Mathematically, the length is found using the Pythagorean theorem.
A unit vector is a vector with a length of 1. A vector may represent a position, but it also represents a direction. In this example, the vectors m, n, o, p, and q are unit vectors and therefor lie on the unit circle, representing an angle.
# 3D Geometry
Most 3D graphics are made up of triangles. A triangle is defined by three vertices, and a vertex’s position is simply a vector.
This is Suzanne, a simple example model that comes with Blender. By putting together many triangles that share vertices, we can create a 3D model.
In order to transform 3D graphics, we need to transform triangles. We do this by transforming every vertex that makes up the triangle. If we apply the same transform to every vertex, we can move, scale, or rotate the triangle without deforming it. To move around models in a 3D world we need to apply a transformation to every triangle that makes up the model.
When an artist creates a model in a program like Blender, the model is in model space. Character models are usually made with the origin of this space at the character’s feet. This way, if we put the origin point on the ground, the character stands on the ground. The origin here works as a pivot point: everything happened “around” the origin.
The first step of transforming the model is to scale it if necessary. For example, the character may have been modeled too big relative to the scene and we would like to scale it down. Second is rotation: what direction is the player looking in? The last step is moving the model, known as translation. We need to move the model from being near the origin to where in the world we actually want it. A game does these transformations on every object in the scene for every single frame.
So how can we do these transformations? We’ll start by looking at translation. Since the position of vertices are vectors, we can use vector addition and subtraction to move them. Given an offset vector, such as the player’s position in the world, we can move every vertex on the model by this offset in order to move the model.
Scaling is done by multiplying the vertex’s position by a scalar. Multiplying by a number above 1 makes the triangle bigger, a number between 0 and 1 makes it smaller, and multiplying by 1 clearly has no effect on the triangle. Notice how the triangles scale around the origin.
So we have translation and scaling, but what about rotation? Unfortunately there’s no simple vector math for rotating a point around the origin. The rest of this tutorial will be on rotation.
When we rotate a vertex, we want to spin it around the origin. It’s important that if the vertex was 6 inches away from the origin before being rotated, it’s still 6 inches away afterwards. If we rotate every vertex on the model, we’ll have rotated the model.
# Complex Numbers
Before we get to 3D rotation, we’ll first discuss complex numbers. Complex numbers have a real part and an imaginary part. The imaginary part is multiplied by i.
We can think of a complex number like a 2D vector. The real part corresponds to the x coordinate and the imaginary part corresponds to the y coordinate. Most of the time you can think of i as just meaning “the vertical axis.”
We can add and subtract complex numbers just like we can vectors by applying straightforward algebra. This has the same effect of “moving” one complex number by another.
So what’s the deal with i? The only time you have to worry about it is when i is multiplied by another i. i times i (i squared) is equal to -1. That’s all there is to it; a single i means nothing to us besides indicating the vertical axis.
Unlike vectors, complex numbers can be multiplied. Here we see the formula for multiplying the complex numbers C and D.
There are two interesting properties of complex numbers. The first is that their lengths multiply. If the length of C is 4 and the length of D is 6, the length of C times D is 4*6 = 24. This visual shows multiplication of complex numbers with only a real part for simplicity.
The second property is that their angles add together. If A represents a point 30 degrees from the x axis and B represents a point 45 degrees from the x axis, A times B will be 75 degrees from the x axis with the lengths multiplied.
What we’re looking for here is the rotational property. We would actually like to get rid of the part where the lengths multiply. We can do this by making the complex numbers unit complex numbers. If A and B have a length of 1, A times B will also have a length of 1. This way we will only get the effect of adding their angles.
This is actually how we can derive the angle addition formula you may have used in trigonometry.
Let’s say P is a vertex we would like to rotate around the origin and r is a unit complex number that represents a pure rotation. Multiplying P with r will only rotate P without changing its length.
Actually doing the multiplication is mostly straightforward algebra. Put each complex number in parentheses and use the distributive property. Notice how we end up with a term that’s multiplied by i twice. Replace the two i’s by -1, then combine like terms. We end up with another complex number.
# 3D Rotation
So far we have looked at using complex numbers for rotation in 2D space. There are a variety of practical ways to do rotations in 3D space, and we will first look at Euler rotations. In this form we break up a rotation into rotations around the x, y, and z axes. Think of the axis piercing the model. The model spins around this axis like a wheel. By combining all three, we can rotate the model in any possible way.
When a 3D model rotates around an axis, the vertices are rotating on a 2D plane. The coordinate corresponding to the axis it is rotating on is not changed, but the other two are. We can plug the two coordinates we want to change into a complex number and rotate by multiplying by a unit complex number. For example, to rotate around the x axis, we would set the real part of the complex number to y and the imaginary part to z.
There are some drawbacks to the Euler representation, however. One problem is that we can only rotate around the x, y, or z axis, making it difficult to smoothly rotate around any other axis.
# Quaternions
Quaternions are another way of doing rotations in 3D. A quaternion allows us to rotate around any arbitrary axis, not just x, y, or z. It’s based on complex numbers and has two additional imaginary numbers, j and k. w is the real part, and xi, yj, and zk are called the vector part. Like i squared, j squared and k squared equal -1, but i is not equal to j or k. With 4 terms and some confusing rules, deriving quaternion multiplication gets pretty complicated, so we won’t cover it here. Check out the Wikipedia article for more information if you want to give it a shot yourself.
Building a quaternion is pretty easy. We’ll need to know the axis we’ll want to rotate around and the angle indicating how much we want to rotate. The axis is represented by a unit vector. This way, the axis can be any combination of x, y, and z as long as the length is 1. We multiply all three numbers in the vector by the sine of half the angle. w is just the cosine of half the angle.
On the left half of this example, we have a reminder of complex number multiplication. A point p is multiplied by a unit complex number a or b in order to rotate it in a circle around the origin. It can also be multiplied by both a and b to apply the rotations of both. The right half shows the equivalent idea for quaternions. A 3D point v is multiplied by A or B to rotate it around the origin. We can think of v moving on the surface of a sphere. If we multiply by A by B, we get a new quaternion that represents the rotations of both.
# Matrices
Matrices are another of the common ways to represent rotations, as well as translation and scaling. We can multiply a matrix by a vector by thinking of the vector as a matrix with one column. We can then apply matrix multiplication.
A 2x2 matrix can represent 2D rotation. We can take a complex number and directly translate it into matrix form. When this matrix is multiplied with a vector, we will do the exact same math operations as the complex number would. Quaternions can be converted to a 3x3 matrix.
Matrices can come in many sizes. A 3x3 matrix allows us to rotate a 3D vector. A 4x4 matrix is best for 3D graphics, though 4x3 matrices may be used if a bit more efficiency is needed.
The identity matrix contains 1s along the diagonal and 0s everywhere else. It has no effect on a vector when multiplied. Since the matrix has three rows, the vector would need to be 3D. In this example we’re using a 2D vector, so we set the z coordinate to 1.
A small modification to the identity matrix allows us to create a matrix that will scale a vector. By changing the 1s in the matrix to any other number, the vector’s coordinates are scaled by that number. The numbers do not all have to be the same; we could apply different scales to the x and y coordinates of vertices to squash and stretch triangles.
The purpose of using a 3x3 matrix for transforming a 2D vector is so that we can use the matrix for translation. In this example, we want to move the vertex by 5 on the x axis and 8 on the y axis. We take advantage of the 1 at the end of the vector, which is multiplied by the 5 and the 8.
We now know how to represent scaling, rotation, and translation as matrices. By multiplying them with each other, we end up with a single matrix that represents all three. This one matrix can then be applied to a vector to do all of the transformations at once.
# The Graphics Card
So we have a model made up of triangles which are made up of vertices, and we have a matrix we want to use to transform the model and put it where it should be in the virtual world. This is the job of a graphics card, which applies the matrix to every vertex of the model.
# The Camera
One last question is how we decide which models actually get displayed on the screen.
Let’s say you want to take a picture of Mount Everest. What most people would do is get on a plane, fly to Nepal, and set up their camera in front of Mount Everest. But there’s another option: you could set your camera on a floating tri-pod, pick up everything in the universe besides your camera, and rotate the universe so that Mount Everest is in front of your camera. This is the method computer graphics use.
Everything inside the box shown here is what will be rendered to the screen. Our goal, then, is to get the things we want to see inside this box. Applications like the Unity Engine Editor and Blender give you a camera object to intuitively work with. The camera’s position and rotation can be represented by a matrix. But when it comes to implementing the math, rather than moving the camera, the camera stays in place and we move everything else in front of it. This is done by simply applying the inverse of the camera’s matrix to every model in the scene. If you want to take a picture 100 meters to the right of the origin, just move every object 100 meters to the left.
Lastly, we collapse the z-axis to make the triangles 2D. They are then filled in with pixels (the other main responsibility of the graphics card) and the image is transferred to the screen.
Being comfortable with trigonometry is incredibly useful for anyone working with 3D graphics. Dave’s Short Trig Course is a great place to brush up.
For more about vector and matrix math, I recommend this detailed tutorial.
Getting into the math of quaternions is a little more tricky. A good way to get an intuitive feel for them is to play with the w, x, y, and z values in the transform panel in Blender, or write some scripts to tweak them in your favorite engine.
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# What is the slope-intercept form (y=mx+b) of x-y=5?
Feb 26, 2018
$y = x - 5$
#### Explanation:
$5 = x - y$
$y + 5 = x$
$y = x - 5$
The coefficient of x is 1, thus so is the gradient. $b$, or the y intercept, is -5.
Feb 26, 2018
$y = x - 5$
#### Explanation:
$\text{rearrange "x-y=5" into slope-intercept form}$
$\text{subtract x from both sides}$
$\cancel{x} \cancel{- x} - y = 5 - x$
$\Rightarrow - y = 5 - x$
$\text{multiply all terms by } - 1$
$\Rightarrow y = - 5 + x \Rightarrow y = x - 5$
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Difference between revisions of "1997 USAMO Problems/Problem 6"
Problem
Suppose the sequence of nonnegative integers $a_1,a_2,...,a_{1997}$ satisfies
$a_i+a_j \le a_{i+j} \le a_i+a_j+1$
for all $i, j \ge 1$ with $i+j \le 1997$. Show that there exists a real number $x$ such that $a_n=\lfloor{nx}\rfloor$ (the greatest integer $\le nx$) for all $1 \le n \le 1997$.
Solution
Given nonnegative integers $a_1,a_2,\dots, a_{1997}$ satisfying the given inequalities, let $I_n$ be the set of all $x$ such that $a_n=\lfloor nx\rfloor$. Therefore, $$I_n=\{x\,:\,a_n\le nx We can rewrite this as $\[I_n=\left[\frac{a_n}{n},\frac{a_n+1}{n}\right).\tag{1}$$ So we know that each $I_n$ is an interval. We start by proving that if $n$ and $k$ are positive integers, then $I_n\cap I_{k}\ne \emptyset$. To prove this, we need the following lemma.
Lemma 1: If $n$ and $k$ are positive integers, then $$na_k< k(a_n+1)\tag{2}$$
Proof: We prove this by induction. If $k=1$, then we wish to show that $na_1< a_n+1$. By the given lower bound, we know that $$a_n\ge a_{n-1}+a_1\ge (a_{n-2}+a_1)+a_1\ge \cdots \ge na_1.$$ Therefore, $na_1\le a_n, so the hypothesis is true for $k=1$. If $n=1$, then we wish to prove that $a_k\le k(a_1+1)$. By the given upper bound, we know that $$a_k\le a_{k-1}+(a_1+1)\le (a_{k-2}+(a_1+1))+(a_1+1)\le\cdots\le ka_1+(k-1).$$ Therefore, $a_k, so the hypothesis is true for $n=1$.
Now suppose that (2) holds for all $k,n. If $k=n=j$, then (2) is equivalent to $0, which is obviously true. If $n=j>k$, then by the division algorithm, we can write $n=qk+r$ for nonnegative integers $q$ and $r$ with $0\le r. Thus by applying the lower bound repeatedly (with one of the indices equal to $1$), we find $$k(a_n+1)\ge k(qa_k+a_r+1)=kqa_k+k(a_r+1).\tag{3}$$ But then as $k,r, we can apply the inductive hypothesis with $n=r$ and $k=k$ to find $k(a_r+1)> ra_k$. Substituting this into (3), we find $$k(a_n+1)> kqa_k+ra_k=(kq+r)a_k=na_k.$$ This proves the hypothesis for $n=j>k$.
Suppose that $k=j>n$. Then by the division algorithm, we can write $k=qn+r$ for nonnegative integers $q$ and $r$ with $0\le r. Then by applying the upper bound repeatedly (with one of the indices equal to $1$), we find $$na_k\le n(qa_n+a_r+q)=nqa_n+na_r+nq.\tag{4}$$ But as $n,r, we can apply the inductive hypothesis with $n=n$ and $k=r$, finding that $na_r< r(a_n+1)$. Subsituting this into (4), we find $$na_k< nqa_n+r(a_n+1)+nq=(nq+r)(a_n+1)=k(a_n+1).$$ This proves the hypothesis for $k=j>n$.
Therefore, if the hypothesis is true for all $k,n, then it must be true for all $k,n\le j$. Hence by induction, it must be true for all positive integers $k$ and $n$. $\hfill\ensuremath{\square}$
Now suppose for the sake of contradiction that $I_n$ and $I_k$ are disjoint intervals. Without loss of generality, we may assume that $I_n$ precedes $I_k$ on the number line. Hence the upper bound of $I_n$ is less than or equal to the minimum value of $I_k$, or rather $$\frac{a_{n}+1}{n}\le \frac{a_k}{k}.$$ This simplifies to $na_k\ge k(a_n+1)$. But this contradicts the statement of Lemma 1. Therefore, $I_k\cap I_n\ne \emptyset$.
Now we claim that $I_1\cap I_2\cap \cdots \cap I_{1997}\ne \emptyset$. We prove this by induction. By the above, we know that $I_1\cap I_2\ne \emptyset$. Now suppose that $I_1\cap I_2\cap\cdots\cap I_k\ne \emptyset$. The intersection of two overlapping intervals of the form $[a,b)$ and $[c,d)$ is an interval of the form $[e,f)$, where $e=\max\{a,c\}$ and $f=\min\{b,d\}$. Therefore, by induction, we know that if the intersection of $k$ overlapping intervals is nonempty, then it must also be an interval, say $$I_1\cap I_2\cap\cdots\cap I_k=[x_k,y_k).$$ If $I_{k+1}$ does not intersect $[x_k,y_k)$, then as an interval, it must appear either completely before $x_k$ or completely after $y_k$. If $I_{k+1}$ appears completely before $x_k$, then it has a nonempty intersection with each of $I_1, I_2, \dots, I_k$. But we also know that $x_k$ is a lower bound of one of the intervals, hence $I_{k+1}$ cannot intersect that interval, a contradiction. A similar contradiction arises if $I_{k+1}$ appears completely after $y_k$. Therefore, $$I_1\cap I_2\cap\cdots\cap I_{k+1}\ne \emptyset.$$ Thus by induction, $I_1\cap I_2\cap\cdots\cap I_{1997}\ne \emptyset$. So let $x\in I_1\cap I_2\cap\cdots\cap I_{1997}$. Then by the definition of $I_n$, we know that $a_n=\lfloor nx\rfloor$ for all $1\le n\le 1997$, and we are done.
See Also
1997 USAMO (Problems • Resources) Preceded byProblem 5 Followed byLast Question 1 • 2 • 3 • 4 • 5 • 6 All USAMO Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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Class 10 Area of a Triangle
Topic Covered
♦ Area of a Triangle
Area of a Triangle
In your earlier classes, you have studied how to calculate the area of a triangle when
its base and corresponding height (altitude) are given. You have used the formula :
text (Area of a triangle) = 1/2 xx text (base) xx text (altitude )
Now, if the coordinates of the vertices of a triangle are given. Well, you could find the lengths of the three sides using the distance formula and then use Heron’s formula.
But this could be tedious, particularly if the lengths of the sides are irrational numbers. Let us see if there is an easier way out.
Let ABC be any triangle whose vertices are A(x_1, y_1), B(x_2, y_2) and C(x_3, y_3). Draw AP, BQ and CR perpendiculars from A, B and C, respectively, to the x-axis. Clearly ABQP, APRC and BQRC are all trapezia (see Fig. 7.13).
Now, from Fig. 7.13, it is clear that area of
Delta ABC = text ( area of trapezium ) ABQP + text (area of trapezium) APRC – text (area of trapezium) BQRC.
You also know that the
text (area of a trapezium ) = 1/2xx "sum of parallel sides" xx "distance between them"
Therefore, Area of
Delta ABC =1/2 (BQ +AP) QP +1/2 (AP +CR) PR -1/2 (BQ +CR) QR
=1/2 (y_2 +y_1) (x_1 - x_2) +1/2 (y_1 +y_3) (x_3 -x_1) -1/2 (y_2 +y_3) (x_3 -x_2 )
=1/2 [x_1 (y_2 -y_3) +x_2 (y_3 -y_1) + x_3 ( y_1 -y_2) ]
Thus, the area of Delta ABC is the numerical value of the expression
color{red}{= 1/2 [ x_1 (y_2 - y_3) +x_2 (y_3 -y_1) + x_3 (y_1 - y_2)]}
Q 3129780611
Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and
(–3, –5).
Class 10 Chapter 7 Example 11
Solution:
The area of the triangle formed by the vertices A(1, –1), B(– 4, 6) and
C (–3, –5), by using the formula above, is given by
1/2 [ 1 (6+5) + (-4) (-5 +1 ) + (-3) (-1-6) ]
= 1/2 (11 +16 +21 ) = 24
So, the area of the triangle is 24 square units.
Q 3139780612
Find the area of a triangle formed by the points A(5, 2), B(4, 7) and C (7, – 4).
Class 10 Chapter 7 Example 12
Solution:
The area of the triangle formed by the vertices A(5, 2), B(4, 7) and
C (7, – 4) is given by
1/2 [ 5 (7+4) +4 (-4-2) + 7 (2-7) ]
= 1/2 (55-24 -35) = (-4)/2 = -2
Since area is a measure, which cannot be negative, we will take the numerical value
of – 2, i.e., 2. Therefore, the area of the triangle = 2 square units.
Q 3159780614
Find the area of the triangle formed by the points P(–1.5, 3), Q(6, –2)
and R(–3, 4).
Class 10 Chapter 7 Example 13
Solution:
The area of the triangle formed by the given points is equal to
1/2 [-1.5 (-2-4) + 6 (4-3) +(-3) (3+2) ]
=1/2 (9 + 6 -15 ) = 0
Can we have a triangle of area 0 square units? What does this mean?
If the area of a triangle is 0 square units, then its vertices will be collinear.
Q 3169780615
Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are
collinear.
Class 10 Chapter 7 Example 14
Solution:
Since the given points are collinear, the area of the triangle formed by them
must be 0, i.e.,
1/2 [2 ( k+3) +4 (-3-3 ) + 6 (3- k) ] = 0
i.e., 1/2 (-4k) = 0
Therefore , k = 0
area of Delta ABC = 1/2 [2 (0+3 ) +4 ( -3-3 ) + 6 (3 - 0 ) ] = 0
Q 3179780616
If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a
quadrilateral, find the area of the quadrilateral ABCD .
Class 10 Chapter 7 Example 15
Solution:
By joining B to D, you will get two triangles ABD and BCD .
Now the area of Delta ABD = 1/2 [ -5 (-5-5) + (-4) (5-7) + 4 (7+5) ]
= 1/2 (50 + 8 + 48) = 106/2 = 53 square units
Also, the area of Delta BCD = 1/2 [ -4 (-6-5) -1 ( 5+5) +4 (-5 +6) ]
= 1/2 (44 -10 +4) =19 square units
So, the area of quadrilateral ABCD = 53 + 19 = 72 square units.
Note : To find the area of a polygon, we divide it into triangular regions, which have
no common area, and add the areas of these regions.
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Hypotenuse Leg (HL) Congruence Theorem
This lesson will introduce a very long phrase abbreviated CPCTC. It's easy to remember because every other letter is "C," you see? The Hypotenuse Leg or HL Theorem, is not as funny as the Hypotenuse Angle or HA Theorem, but it is useful. This theorem is really a derivation of the Side Angle Side Postulate, just as the HA Theorem is a derivation of the Angle Side Angle Postulate.
Right triangles have exactly one interior angle measuring 90°, and the other two interior angles are acute (because they can only add up to 90°). .
The longest side of a right triangle is called its hypotenuse
CPCTC
CPCTC is an acronym for corresponding parts of congruent triangles are congruent. It is shortened to CPCTC, which is easy to recall because you use three Cs to write it.
Here are two congruent, right triangles, $△PAT$ and $△JOG$. Notice the hash marks for the two acute interior angles. Notice the hash marks for the three sides of each triangle. Notice the squares in the right angles.
Every part of one triangle is congruent to every matching, or corresponding, part of the other triangle. Usually you need only three (or sometimes just two!) parts to be congruent to prove that the triangles are congruent, which saves you a lot of time.
Here are all the congruences:
CPCTC reminds us that, if two triangles are congruent, then every corresponding part of one triangle is congruent to the other.
The converse of this, of course, is that if every corresponding part of two triangles are congruent, then the triangles are congruent. The HL Theorem helps you prove that.
SAS Postulate
Recall the SAS Postulate used to prove congruence of two triangles if you know congruent sides, an included congruent angle, and another congruent pair of sides. The included angle has to be sandwiched between the sides.
HL Theorem
The Hypotenuse Leg Theorem, or HL Theorem, tells us a suspiciously similar story:
The HL Theorem states; If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the triangles are congruent.
Hold on, you say, that so-called theorem only spoke about two legs, and didn't even mention an angle. Aha, have you forgotten about our given right angle? Every right triangle has one, and if we can somehow manage to squeeze that right angle between the hypotenuse and another leg...
Of course you can't, because the hypotenuse of a right triangle is always (always!) opposite the right angle. So we have to be very mathematically clever. We have to enlist the aid of a different type of triangle.
Proving the HL Theorem
We must first prove the HL Theorem. Once proven, it can be used as much as you need. To prove that two right triangles are congruent if their corresponding hypotenuses and one leg are congruent, we start with … an isosceles triangle.
Here we have isosceles $△JAK$. We know by definition that , because they are legs. We are about to turn those legs into hypotenuses of two right triangles. Can you guess how?
Construct an altitude from side $AK$. Recall that the altitude of a triangle is a line perpendicular to the base, passing through the opposite angle. Label its point on $AK$ as .
That altitude, $JC$, complies with the Isosceles Triangle Theorem, which makes the perpendicular bisector of the base the angle bisector of the vertex angle. We have two right angles at , $\angle JCA$ and $\angle JCK$. We have two right triangles, $△JAC$ and $△JCK$, sharing .
We know by the reflexive property that side (it is used in both triangles), and we know that the two hypotenuses, which began our proof as equal-length legs of an isosceles triangle, are congruent. So, we have one leg and a hypotenuse of $△JAC$ congruent to the corresponding leg and hypotenuse of $△JCK$.
Now verify that and all the interior angles are congruent:
• (the altitude of the base of an isosceles triangle bisects the base, since it is by definition the perpendicular bisector)
• (they are both right angles)
• (they were angles opposite to the legs in accordance with the Isosceles Triangle Theorem)
• ( was the angle bisector of original $\angle AJK$)
So, all three interior angles of each right triangle are congruent, and all sides are congruent. CPCTC! How about that, $JACK$?
We originally used the isosceles triangle to find the hypotenuse and a single leg congruent, and from that, we built proof that both triangles are congruent.
So, we have proven the HL Theorem, and can use it confidently now!
HL Theorem Practice Proof
You have two suspicious-looking triangles, $△MOP$ and $△RAG$.
You get out your mathematical detective's magnifying glass and notice that $\angle O$ and $\angle G$ are marked with the tell-tale little squares, $\square$, indicating right angles.
Aha! These are two right triangles, because by definition a right triangle has one right angle.
You also notice, masterful detective that you are, the sides opposite the right angles are congruent:
Finally, you zero in on the little hash marks on sides $OP$ and $AG$, which indicate they are congruent, too. So you have two right triangles, with congruent hypotenuses, and one congruent side. You can whip out the ol' HL Theorem and state without fear of contradiction that these two right triangles are congruent.
That also means, thanks to CPCTC, the two as-yet-unidentified interior angles of one right triangle are congruent to the corresponding interior angles of the other triangle.
Lesson Summary
After working your way through this lesson, you are now able to recall and state the Hypotenuse Leg (HL) Theorem of congruent right triangles, use the HL Theorem to prove congruence in right triangles, and recall what CPCTC means (corresponding parts of congruent triangles are congruent), using as needed.
What you'll learn:
Once you work your way through these instructions and the multimedia, you will be able to:
• Recall and state the Hypotenuse Leg (HL) Theorem of congruent right triangles
• Use the HL Theorem to prove congruence in right triangles
• Recall and apply that corresponding parts of congruent triangles are congruent (CPCTC)
Instructor: Malcolm M.
Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.
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# Surface Area of Prisms Cylinders Objectives 1 To
• Slides: 11
Surface Area of Prisms & Cylinders Objectives: 1) To find the surface area of a prism. 2) To find the surface area of a cylinder.
I. Surface Area of a Prism § Prism – Is a polyhedron with exactly 2 , // faces, called bases. § Name it by the shape of its bases. Bases are Rectangles: Lateral Faces – All faces that are not bases. (Sides)
Right Prisms vs Oblique Prisms Right Prism – Edges are Altitudes. Oblique Prism
Lateral Area – The sum of the areas of the lateral faces (sides) • Right Prisms - Lateral Faces are Rectangles A = l • w Base Area – The sum of the areas of the 2 bases • Rectangle: A = l • w • Triangle: A = ½bh • Polygon: A = ½bh Total Surface Area = Lateral Area + Base Area
Ex. 1: Use the net to find the Surface Area of the rectangular Prism. Area of Bases: A = l • w 2 different Lats: A = l • w 4 5 cm 3 cm 4 cm 3 4 3 15 20 15 SA = LA + Area of Bases = 70 cm 2 + 24 cm 2 = 94 cm 2 12 3 20 5 12 3
Ex. 2: Find the total surface area of the following triangular prism. 5 cm LA = l • w (Area of Sides) (5 x 12) = 60 cm 2 5 cm (5 x 12) = 60 cm 2 12 cm (6 x 12) = 72 cm 2 Area of Triangle 6 cm 192 cm 2 BA = ½bh = ½(6)(4) = 12 cm 2 x 2 24 cm 2 a 2 + b 2 = c 2 h 2 + 32 = h=4 52 h 5 SA = LA + BA = 192 cm 2 + 24 cm 2 6 3 = 216 cm 2
Ex. 2: Find the total surface area of the following regular hexagonal prism. LA = l • w (10 x 12) = 120 m 2 12 m x 6 720 m 2 BA = ½ap = ½(8. 7)(60) = 260 m 2 x 2 520 m 2 10 30° a 5 10 m SA = LA + BA Tan 30 = 5/a = 720 m 2 + 520 m 2 . 577 = 5/a = 1240 m 2 a = 8. 7
II. Finding Surface Area of a Cylinder § Has 2 , // bases § Base → Circle r § C = 2 πr § A = π r 2 height r h r
Net of a Cylinder: LA is just a Rectangle! LA = 2 rh Area of a circle BA = r 2 r Circumference of the circle SA = LA + 2 BA
Ex. 4: SA of a right cylinder LA = 2 rh 6 ft 9 ft = 2 (6)(9) Area of Base = 108 ft 2 BA = r 2 = 339. 3 ft 2 = (6)2 = 36 ft 2 x 2 SA = LA + BA = 72 ft 2 = 339. 3 ft 2 + 226. 2 ft 2 = 226. 2 ft 2 = 565. 5 ft 2
What did I learn today? ? § Find the area of the lateral sides first!! § Usually rectangles § Be careful, the rectangles are not always the same size. § Second, find the area of the Base § Rectangle, Triangle, Polygon, or a Circle § There always 2 bases in prisms. § Multiply by 2!
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Practice problems - simplifying fractions. Explained with pictures and diagrams.
# Equivalent Fractions Lesson
### Methods to Simplify Fraction
#### General Steps
Let's examine the fraction 2/4.
• The first step is to determine the largest number that evenly divides the numerator and the denominator (also called the Greatest Common Factor of these numbers). In other words, what's the biggest number that goes into '2' (the numerator) and '4', the denominator?
• Step 2. Once you've answered step 1 and realized that, for our fraction at least, the greatest number that evenly divides 2 and 4 is '2'. Now divide the top and the bottom by 2 to get your simplified fraction which is ½.
If you look at the picture above, you can see that the simplified fraction ½ and the original fraction (2/4) are equal. They both take up the same amount of the pizza.
### Practice how to Simplify Fraction
##### Problem 1
Remember the first step in how to simplify fraction is to identify the largest number that divides both the numerator (3) and the denominator (6).
3 is what we want because it evenly divides both 3 and 6. The final simplified fraction is ½. (see picture for visual)
##### Problem 2
Remember the first thing to do is identify the largest number that divides both the numerator (6) and the denominator (9).
3 is what we want because it evenly divides both 6 and 9. The final simplified fraction is . (see picture for visual)
##### Problem 3
Remember the first step in how to simplify fraction is to identify the largest number that divides both the numerator (2) and the denominator (10).
2 is what we want because it evenly divides both 2 and 10. The final simplified fraction is . (see picture for visual)
##### Problem 4
Remember the first step in how to simplify fraction is to identify the largest number that divides both the numerator (3) and the denominator (15).
3 is what we want because it evenly divides both 3 and 15. The final simplified fraction is .
##### Problem 5
Remember the first step in how to simplify fraction is to identify the largest number that divides both the numerator (6) and the denominator (8).
2 is what we want because it evenly divides both 6 and 8. The final simplified fraction is . (see picture for visual).
##### Problem 6
Remember the first thing to do is identify the largest number that divides both the numerator (12) and the denominator (15).
3 is what we want because it evenly divides both. The answer is . (see picture for visual)
##### Problem 7
Remember the first step in how to simplify fraction is to identify the largest number that divides both the numerator (12) and the denominator (28).
4 is what we want because it evenly divides both. The answer is . (see picture for visual)
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# A Pair of Complementary Angles
One pair of complementary angles
Name a pair of complementary angles, a pair of complementary angles, and a pair of adjacent angles in the figure. Complimentary Angles | Complementary Angle Theorem Within geography we are learning about angles and forms. A square is the shape created by two beams with the same end point. Sometimes the angular pair is used. If two angles come together and create a design, it is called an angular pair.
Different kinds of angle pairs exist, e.g. - straight pair angles, additional angles, complementary angles, vertical opposite angles, etc. The two angles are considered complementary angles if the total of the two angles is 90??. Adding the two angles 180 means that they are referred to as complementary angles.
Complimentary angles are important in geometric design. On this page we will concentrate on complementary angles. The two neighboring angles are called complementary angles if their total is perpendicular. If two angles divide an arms and their total is 90??, then the angles are called complementary angles.
Even an corner of them is called an addition to another. If we consider whether two neighboring angles X and Y should be complementary angles, then ??X + ??Y = 90??. The complementary angles are complementary. Learn to grasp the concepts of complementary angles in this Tutorial and get high-quality mathematical help!
When their angles are added to 90 deg., two angles are referred to as complementary angles. Complimentary angles = angles a + angles a = 90 deg. Two neighboring angles that are 90 deg. in total are referred to as complementary angles. This means that corner "a" is the complementary corner of corner "b" and the other way round.
The two angles make a pair of complementary, complementary and perpendicular angles. Two angles that add up to 90 degrees are known as complementary angles. This can be useful in the determination of other angles. Alternatively, complement shapes with the same angles are matched. They are complementary angles. Stage 2: Let ??A and ??C be complementary angles.
We have from steps 1 and 2, so the complement of the same corner are matching. The two angles are considered complementary if their total is 90 degrees. Complementary angles in the pair, if an angel is degrees greater than degrees, then its complementary dimension is degrees (90 - x).
Complementary angles are seen in many places in reality. Below are just a few samples of complementary angles: Below are some of the issues resolved using complementary angles: Q1: If an arc is 60 degrees, you will find the second arc if the two angles complement each other.
As the two angles are complementary angles, they must be added to 90 degrees. Insert the specified angular value into the above formula, ?? 2 = 30 degrees. Q2: In the given pair of complementary angles, find the value of x and y. solution: Using the complementary angular value we can find the value of x.
Q3: Find x, y, z from the pair of complementary angles. In the following, the practical issues related to complementary angles are presented. Q1: The measurement for angles 1 is 25 and 2 is 65 deg. Determine whether the angles are complementary or not. Q2: If you let the first corner be 20 deg, you will find the second corner if the two angles complement each other.
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# How to factor
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How to factor? Are you scared of factor? Do you face trouble while doing these math’s? They are not as hard as you think and since they are a part of algebra you will understand it once you start practicing. Let us start with knowing what is a factor. You are definitely familiar with the times table such as 3 X 3 = 9 or 4 X 3 = 12. The 3 and 4 are factors and the result is the product (9 and 12). It is a whole number that divides precisely into another number. Here are the whole numbers: 1, 2, 4, 8 and not 3 or 5.
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Five or three are not whole numbers because you cannot divide them equally unlike the other numbers. Suppose, there are four sweets. You can divide them equally between two people. Two for each and you can divide the four sweets among four people and again equally. Therefore, you can say.
2 x 2 = 4
4 x 1 =4
The factors are 2, 1 and 4.
Let us factor second degree polynomials. The format is ax2 + bx + c = 0
Rearrange your equation with the smallest one at the beginning such as 3 + 3x2 + 11x = 0
For example, you have an equation x²+ x + 12
x² + 4x – 3x +12
x(x+4) -3(x+4)
(x+4) (x-3) are factors.
Here is how you do it. First of all, look at the numbers. Arrange them and make sure there is no number on the other side, that means on the other side it should be zero. If the equation was x²+3x+12= -3x, rearrange them and make them as small as possible. It will be x²+ 3x+ 3x+ 12 =O, later add the two 3x and your final equation is x²+6x+12 as you have shortened it out, it is now ready to have its factors.
For the previous equation, if you had a zero at the end with an equal sign, you would have to find the value of the x. Then, the result would be x= -4 and x =3. Here is how you do it.
(x=4)(x-3)
x + 4 = 0 & x-3=0
x = -4 x= 3
Let us see how we got the 4 and 3. We look at the single number of the equation that does not have any x with it. In this case, it is the 12. Go through the times tables and think which numbers that can be multiplied to get 12 and also be added or subtracted to get 1. We know,
6 x 2 = 12
3 x 4 = 12
12 x 1 = 12
Adding or subtracting 6 and 2 will never get you 1 neither will 12 and 1. It is clear that 4 and 3 can be subtracted to get 1.
In case of simple equations, you will be able to figure it out yourself with just looking at it. For example, 4x²+ 4x+1 will result in (2x + 1) (2x +1) when you try to factor it.
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Lesson Objectives
• Learn how to apply a horizontal shift to a function
• Learn how to apply a vertical shift to a function
## How to Apply a Horizontal or Vertical Shift
In this lesson, we want to show how to apply a horizontal and or vertical shift to a function. Essentially, a vertical shift occurs when the action happens outside of the function:
f(x) + k » vertical shift up k units (if k > 0)
f(x) - k » vertical shift down k units (if k> 0)
A horizontal shift occurs when the action happens inside of the function:
f(x + h) » horizontal shift left by h units (if h > 0)
f(x - h) » horizontal shift right by h units (if h > 0)
Let's begin by looking at the absolute value function.
### Absolute Value Function
The absolute value function:
f(x) = |x|
domain: (-∞, ∞)
range: [0, ∞)
Because of the absolute value operation, our graph is a V shape. Recall that opposites have the same absolute value. This means each y-value other than 0 will be associated with two different x-values. Let's look at a table of ordered pairs:
x y (x, y)
-44(-4, 4)
-22(-2, 2)
00(0, 0)
22(2, 2)
44(4, 4)
Notice how the x-values of 4 and -4 both produce a y-value of 4. Similarly, we see the same for the x-values of 2 and -2. They each produce a y-value of 2. Let's plot our ordered pairs and sketch the graph: When we learned how to graph parabolas, we learned about vertical and horizontal shifts. These shifts can be applied to any elementary function. The key is to understand what causes each type of shift. Let's continue to think about our absolute value function.
f(x) = |x|
What happens to the graph if we add something to the end?
f(x) = |x| + 4
This creates a vertical shift. The same x-value will now produce a y-value that is 4 units larger. This will move or shift the graph up by 4 units. Let's look at the graph of both f(x) = |x| and f(x) = |x| + 4 on the same coordinate plane. We will see the graph shifted up by 4 units. We know we have a vertical shift when the action happens "outside of the function". In general for a vertical shift, we will see:
f(x) = |x| + k
is the graph of f(x) = |x|
shifted up k units if k > 0
shifted down k units if k < 0
Let's look at an example.
Example 1: Find the shift based on f(x) = |x|
f(x) = |x| - 14
Based on the rules above, we can say this function is shifted 14 units down since the -14 happens outside of the function.
Additionally, we have horizontal shifts. These will happen "inside of the function". Horizontal shifts are a bit counterintuitive. What would happen to our graph if we add something inside of the absolute value operation?
f(x) = |x + 1|
One might think this produces a shift to the right by one unit, but it actually does the opposite. It will create a shift of 1 unit to the left. Why is this the case? Well if we think about it the same y-value is now produced from an x-value that is one unit less. This creates our shift of 1 unit left. Let's look at the graph of both f(x) = |x| and f(x) = |x + 1| on the same coordinate plane. We will see the graph shifted left by 1 unit. We know we have a horizontal shift when the action happens "inside of the function". In general for a horizontal shift, we will see:
f(x) = |x - h|
is the graph of f(x) = |x|
shifted h units right if h > 0
shifted h units left if h < 0
Let's look at an example.
Example 2: Find the shift based on f(x) = |x|
f(x) = |x - 12|
Based on the rules above, we can say this function is shifted 12 units to the right since the -12 happens inside of the function.
Of course, we can have both a vertical and a horizontal shift. We just look at what's happening inside of the function and outside of the function to determine the impact of each shift. Let's look at an example.
Example 3: Find the shift based on f(x) = |x|
f(x) = |x - 9| + 13
Based on the rules above, we can say this function is shifted 9 units to the right and 13 units up. This is because we are subtracting 9 away inside of the function and adding 13 outside of the function.
### Square Root Function
The square root function: $$f(x)=\sqrt{x}$$ For the square root function, x must be non-negative. Taking the square root of a negative number will not yield a real solution.
domain: [0, ∞)
range: [0, ∞)
Let's gather a few ordered pairs and sketch the graph:
x y (x, y)
00(0, 0)
11(1, 1)
42(4, 2)
93(9, 3)
Let's sketch the graph of our function. This function follows the same process for horizontal and vertical shifts that were demonstrated with the absolute value function.
For a vertical shift: $$f(x)=\sqrt{x}+ k$$ shifts up by k units when k > 0
shifts down by k units when k < 0
For a horizontal shift: $$f(x)=\sqrt{x - h}$$ shifts right by h units when h > 0
shifts left by h units when h < 0
Let's look at an example.
Example 4: Find the shift based on f(x) = sqrt(x) $$f(x)=\sqrt{x - 1}+ 1$$ Based on the rules above, we can say the graph shifts one unit right and one unit up. This is because we are subtracting away 1 inside of the function and adding 1 outside of the function.
### Reciprocal Function
The reciprocal function: $$f(x)=\frac{1}{x}$$ For the reciprocal function, x cannot be 0 since division by 0 is not defined. This also means that y can't be 0 as well.
domain: (-∞, 0) ∪ (0, ∞)
range: (-∞, 0) ∪ (0, ∞)
Let's gather a few ordered pairs and sketch the graph:
x y (x, y)
-1/4-4(-1/4, -4)
-1/2-2(-1/2, -2)
-1-1(-1, -1)
-2-1/2(-2, -1/2)
-4-1/4(-4, -1/4)
1/44(1/4, 4)
1/22(1/2, 2)
11(1, 1)
21/2(2, 1/2)
41/4(4, 1/4)
Let's sketch the graph of our function. This function follows the same process for horizontal and vertical shifts that were demonstrated with the absolute value function.
For a vertical shift: $$f(x)=\frac{1}{x}+ k$$ shifts up by k units when k > 0
shifts down by k units when k < 0
For a horizontal shift: $$f(x)=\frac{1}{x - h}$$ shifts right by h units when h > 0
shifts left by h units when h < 0
Let's look at an example.
Example 5: Find the shift based on f(x) = 1/x $$f(x)=\frac{1}{x - 13}- 8$$ Based on the rules above, we can say the graph shifts 13 units right and 8 units down. This is because we are subtracting away 13 units inside of the function and subtracting 8 units away outside of the function.
### Greatest Integer Function
The greatest integer function, which is also known as the floor function: $$f(x)=[\![x]\!]$$ This function takes an x-value as an input and outputs a y-value that is less than or equal to the x-value. $$[\![9]\!]=9$$ $$[\![9.75]\!]=9$$ $$[\![9.9998]\!]=9$$ domain: (-∞, ∞)
range: {...,-2,-1,0,1,2,...}
The domain consists of all real numbers. We can plug anything we would like in for x. The range is the set of integers since this is the only output that is possible. To make a table of values is a bit different. Now a range of x-values produces the same y-value.
x y
-3 ≤ x < -2-3
-2 ≤ x < -1-2
-1 ≤ x < 0-1
0 ≤ x < 10
1 ≤ x < 21
2 ≤ x < 32
3 ≤ x < 43
Let's sketch the graph of our function. Because the graph looks like a series of steps, this type of function is also called a "step function".
This function follows the same process for horizontal and vertical shifts that were demonstrated with the absolute value function.
For a vertical shift: $$f(x)=[\![x]\!] + k$$ shifts up by k units when k > 0
shifts down by k units when k < 0
For a horizontal shift: $$f(x)=[\![x - h]\!]$$ shifts right by h units when h > 0
shifts left by h units when h < 0
Let's look at an example.
Example 6: Find the shift based on f(x) = [[x]] $$f(x)=[\![x + 5]\!] + 17$$ Based on the rules above, we can say the graph shifts 5 units left and 17 units up. This is because we are adding 5 units inside of the function and adding 17 units outside of the function.
#### Skills Check:
Example #1
Describe the transformation.
Please choose the best answer. $$f(x)=\frac{1}{x}$$ $$g(x)=\frac{1}{x + 1}- 2$$
A
shifts 1 unit right, shifts 2 units up
B
shifts 1 unit right, shifts 2 units down
C
shifts 1 unit left, shifts 2 units down
D
shifts 2 units right, shifts 1 unit down
E
shifts 2 units left, shifts 1 units up
Example #2
Describe the transformation. $$f(x)=x^2$$ $$g(x)=(x + 3)^2 - 3$$
A
shifts 3 units left, shifts 3 units up
B
shifts 3 units right, shifts 3 units up
C
shifts 3 units left, shifts 3 units down
D
shifts 6 units right, shifts 3 units up
E
shifts 3 units left, shifts 6 units down
Example #3
Describe the transformation. $$f(x)=\sqrt{x}$$ $$g(x)=\sqrt{x + 3}+ 2$$
A
shifts 3 units right, shifts 2 units up
B
shifts 3 units left, shifts 2 units up
C
shifts 3 units left, shifts 2 units down
D
shifts 2 units right, shifts 3 units up
E
shifts 2 units right, shifts 3 units down
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# The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80 . Then which one of the following gives possible values a and b? i a = 1, b = 6 ii. a = 3, b = 4 iii. a = 0, b = 7 iv. a = 5, b = 2
Latika Leekha
7 years ago
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80.
Now, using the deifnition of mean we have,
(a + b + 8 + 5 + 10)/ 5 = 6.
This gives a + b = 7. … (1)
Using the definition of variance, we get
[(a-6)2 + (b-6)2 + (8-6)2 + (5-6)2 + (10-6)2] / 5 = 6.8
Hence, we have (a2 + b2 + 36 + 36 – 12a -12b + 4 + 1 + 16) = 34.
(a2 + b2 - 12a -12b) = -59.
From (1), we can use b = 7 – a
Hence, the above equation reduces to
(a2 + (7-a)2 - 12a -12 (7-a)) = -59
Simplifying we get,
2a2 – 14a + 24 = 0
This gives a2 – 7a + 12 = 0
Hence, we have (a-3)(a-4) = 0
‘This yields a = 3 or 4.
If you want you can use these values of a to find the corresponding values of b, but since there is only one option that has the value of a as 3, hence, the correct option is (2).
Thanks & Regards
Latika Leekha
Rishi Sharma
one year ago
Dear Student,
The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80.
Now, using the deifnition of mean we have,
(a + b + 8 + 5 + 10)/ 5 = 6.
This gives a + b = 7. … (1)
Using the definition of variance, we get
[(a-6)2 + (b-6)2 + (8-6)2 + (5-6)2 + (10-6)2] / 5 = 6.8
Hence, we have (a2 + b2 + 36 + 36 – 12a -12b + 4 + 1 + 16) = 34.
(a2 + b2 - 12a -12b) = -59.
From (1), we can use b = 7 – a
Hence, the above equation reduces to
(a2 + (7-a)2 - 12a -12 (7-a)) = -59
Simplifying we get,
2a2 – 14a + 24 = 0
This gives a2 – 7a + 12 = 0
Hence, we have (a-3)(a-4) = 0
‘This yields a = 3 or 4.
If you want you can use these values of a to find the corresponding values of b, but since there is only one option that has the value of a as 3, hence, the correct option is (2).
Thanks and Regards
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# Lesson 3: Circuit Fundamentals 2
We’re finally starting to get armed with some more useful tools to allow us to understand and design some practical circuits.
Today I’m going to talk about another of Kirchoff’s contributions to circuit theory, Kirchoff’s current law. To beign we need to define something called a node within the circuit. Basically a node is where two or more wires connect to each other.
Since our wires have the same voltage along their length, all terminals attached to the same node will be at the same voltage. (We also call voltage as potential, so we could say that all terminals connected to the same node are at the same potential.)
Now what does Kirchoff say about current? Have a look at the next circuit.
We have one source and two loads. Instead of lamps, we’ve used a component called a resistor. You’ll become very familiar with resistors. This circuit looks a little different than previous ones. We know from above that the top terminal of each resistor is at the same potential (voltage) because they’re both connected to node A. Also both bottom terminals are at the same potential because they’re connected to node B. So we can say $V_{R1}=V_{R2}=5V$. This allows us to find $I_{R1}$ and $I_{R2}$:
$I_{R1}=\frac{5V}{20\Omega}=0.25A$
$I_{R2}-\frac{5V}{50\Omega}=0.1A$
Okay, that’s great, but what now? Now we come to Kirchoff’s current law. He says:
All current’s flowing into any node must be equal to all current flowing out of that node.
Here $I_{1}$, $I_{3}$ and $I_{5}$ come into the node, while $I_{2}$ and $I_{4}$ leave the node. So…
$I_{1}+I_{3}+I_{5}=I_{2}+I_{4}$
So what does that mean for our two resistor circuit? Let’s see…
From Kirchoff’s current law we now know that $I_{SRC}=I_{R1}+I_{R2}=0.25+0.1=0.35A$
By the same law we see that $I_{RTN}=I_{R1}+I_{R2}=0.35A$. So $I_{SRC}=I_{RTN}$. This makes sense, all the current leaving the battery returns to the battery.
Now, you may well be asking how will this help us in designing circuits? I know I was wondering such things when I first learned this stuff. Well, it turns out this is a useful tool in understanding our next topic which is a bit more practical.
Basic Circuits: Starting at the beginning
We’re going to discuss something called circuit topology here. Circuit topology is really just a fancy way of talking about how components are connected to from a circuit. We’ve already seen the two basic toplogies we’ll discuss here, series and parallel. Have a look…
It doesn’t take much imagination to figure out why they’re call series and parallel. So why are these two topologies important? That’s what we’re going to look at now.
We’ve seen that we can use Ohm’s law to find unknown voltage, currents, and resistances. Now we’ll use that and Kirchoff’s laws to find something called equivalent resistance. To get an idea of what we call equivalent resistance think of a battery connected to a circuit. Some current flows out of the positive terminal of the battery, I, and that terminal is at some voltage, V. From Ohm’s Law we know that $\frac{V}{I}=R$. So we say that the battery is connected to circuit with an equivalent resistance of R. In other words we could replace the circuit with a resistor of value R and get the same current out of the battery.
Let’s look at one example starting with a series resistance.
Actually we’ve aleady seen this problem. Here the current through each resistor is the same, ICCT, from Kircoff’s Current Law. According to Kirchoff’s Voltage Law (KVL) we know that $V_{R1}+V_{R2}=5V$. Throwing Ohm’s law in the mix we know that:
$V_{R1}=ICCT*20\Omega$
$V_{R2}=ICCT*30\Omega$
which gives:
$ICCT*2O\Omega+ICCT*30\Omega=5V$ —> $20\Omega+30\Omega=\frac{5V}{ICCT}$
Remember we just said that the current out of the battery divided by the current out of the battery is the equivalent resistance, $R_{eq}$. So…
$R_{eq}=20\Omega+30\Omega=50\Omega$
In this case the equivalent resistance is simply the sum of $R_{1}$ and $R_{2}$:
$R_{eq}=R_{1}+R_{2}$
So in general the equivalent resistance of a bunch of resistors in series is just the sum of all the resistor values. Just add ’em all up.
Note that the value of $R_{eq}$ is always greater than any single resistor in the series. (50 > 20, 50 > 30)
Now what about the parallel case? ….
As stated above, $R_{eq}=\frac{5V}{I_{SRC}}$. Using KCL we found that $I_{SRC}=I_{1}+I_{2}$. We also know from KVL that $I_{1}=\frac{5V}{20\Omega}$ and $I_{2}=\frac{5V}{30\Omega}$. We can combine these results giving…
$I_{SRC}=\frac{5V}{20\Omega}+\frac{5V}{30\Omega}=0.35A$
Now all we need to do to find Req is….
$R_{eq}=\frac{5V}{0.35A}=14.28\Omega$
Note that the value of $R_{eq}$ is less than any of the resistor in the circuit. This is always the case.
You’re probably thinking, ‘Hey, I don’t wanna have to solve all those current’s and whatnot just to get an $R_{eq}$ value.’ I don’t either. Fortunately there’s a shortcut.
$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$
Try it with a calculator and the values from our parallel circuit. So if you have three resistors you’d get:
$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$
For four:
$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}$
This is reasonably easy to do on scientific calculators.
Before continuing to a final example I’d just like to clarify one point that may not be obvious. When we refer to equivalent resistance, we’re talking about the resistance that the source ‘sees’ looking into the circuit. In the previous two examples we showed that $R_{eq}$ draws the same current from the source as the original circuit. The source can’t tell the difference between the two, so they are considered equivalent.
Now, here’s one more example to get us home.
All we want is $R_{eq}$, so we don’t care about the voltages or currents. We’ll just use our two formulae.
It’s best to start with the equivalent resistance for just the parallel resistors.
$\frac{1}{R_{eqparallel}}=\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}=\frac{1}{10}+\frac{1}{20}+\frac{1}{50}=0.17$ $R_{eqparallel}=\frac{1}{0.17}=5.88\Omega$
Then $R_{eq}$ must be equal to the sum of $R_{eqparallel}+R_{1}+R_{2}$:
$R_{eq}=12+46+5.88=63.88\Omega$
In most cases I like to reduce any parallel resistances to an $R_{eqparallel}$ and then add the series components. So here the source would ‘see’ an equivalent load of 63.88 Ohms.
So far we’ve developed a quick theoretical foundation to understand how this all boils down to electrons. Then we’ve moved to some basic circuit theory in Kirchoffs laws. We used this theory to gain an understanding of Ohms law and to figure out how to analyze some circuit topologies. Finally we discovered how to figure out equivalent resistance. So far you may still be scratching you’re head about how most or any of this applies to actual circuit designs. I’ll give you the low down on what I use.
In general I don’t think much about atoms or electrons. I don’t really consider electric fields (well not for circuit design anyway). I do use the ideas behind Kirchoff’s laws, not to calculate values, but you need to know that the total voltage around a circuit is zero (all sources balance by all loads). Even more important, I use Ohm’s law all the time and frequently need to calculate an equivalent resistance. You’ll see soon that we come back to these basics time and again.
Next time we’ll start looking at basic components: resistors and diodes.
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Free Essay
# Mth209 Notes
In: Other Topics
Submitted By hanersp
Words 298
Pages 2
Monomials and Polynomials
A monomial is a number, a variable, or a product of numbers and variables raised to natural number powers.
Examples of monomials: [pic]
The degree of monomial is the sum of the exponents of the variables. If the monomial has only one variable, its degree is the exponent of that variable.
The number in a monomial is called the coefficient of the monomial.
Determine whether the expression is a polynomial. If it is, state how many terms and variables the polynomial contains and its degree.
a. 9y2 + 7y + 4
b. 7x4 – 2x3y2 + xy – 4y3 c. [pic]
Solution
a. The expression is a polynomial with three terms and one variable. The term with the highest degree is 9y2, so the polynomial has degree 2.
b. The expression is a polynomial with four terms and two variables. The term with the highest degree is 2x3y2, so the polynomial has degree 5.
c. The expression is not a polynomial because it contains division by the polynomial x + 4.
multiply monomials
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multiply polynomials
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Product of a Sum and Difference
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Division by a Monomial
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Factoring
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Factoring Trinomial
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#### Integrated Chinese
...ADVANCE SAMPLE Integrated Chinese 2nd Edition Level 1 Part 1 Textbook (Simplified Character Ed.) DO NOT DUPLICATE ▲ ▲ 中文聽說讀寫 ▲ © 姓 呢 叫 是 嗎
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# Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2
Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.2 Questions and Answers.
## Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2
Question 1.
Check whether each of the following is p.d.f.
(i) $$f(x)= \begin{cases}x & \text { for } 0 \leq x \leq 1 \\ 2-x & \text { for } 1<x \leq 2\end{cases}$$
Solution:
Given function is
f(x) = x, 0 ≤ x ≤ 1
Each f(x) ≥ 0, as x ≥ 0.
∴ The given function is a p.d.f. of x.
(ii) f(x) = 2 for 0 < x < 1
Solution:
Given function is
f(x) = 2 for 0 < x < 1 Each f(x) > 0,
∴ The given function is not a p.d.f.
Question 2.
The following is the p.d.f. of a r.v. X.
$$f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}$$
Find (i) P(X < 1.5), (ii) P(1 < X < 2), (iii) P(X > 2)
Solution:
Question 3.
It is felt that error in measurement of reaction temperature (in Celsius) in an experiment is a continuous r.v. with p.d.f.
$$f(x)= \begin{cases}\frac{x^{3}}{64} & \text { for } 0 \leq x \leq 4 \\ 0 & \text { otherwise }\end{cases}$$
(i) Verify whether f(x) is a p.d.f.
(ii) Find P(0 < X ≤ 1).
(iii) Find the probability that X is between 1 and 3.
Solution:
(i) f(x) is p.d.f. of r.v. X if
(a) f(x) ≥ 0, ∀ x ∈ R
(b) $$\int_{0}^{4} f(x) d x$$ = 1
Question 4.
Find k, if the following function represents the p.d.f. of a r.v. X.
(i) $$f(x)= \begin{cases}k x & \text { for } 0<x<2 \\ 0 & \text { otherwise }\end{cases}$$
Also find P[$$\frac{1}{4}$$ < X < $$\frac{1}{2}$$]
Solution:
(ii) $$f(x)= \begin{cases}k x(1-x) & \text { for } 0<x<1 \\ 0 & \text { otherwise }\end{cases}$$
Also find (a) P[$$\frac{1}{4}$$ < X < $$\frac{1}{2}$$], (b) P[X < $$\frac{1}{2}$$]
Solution:
We know that
Question 5.
Let X be the amount of time for which a book is taken out of the library by a randomly selected student and suppose that X has p.d.f.
$$f(x)= \begin{cases}0.5 x & \text { for } 0 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}$$
Calculate (i) P(X ≤ 1), (ii) P(0.5 ≤ X ≤ 1.5), (iii) P(X ≥ 1.5).
Solution:
Given p.d.f. of X is f(x) = 0.5x for 0 ≤ x ≤ 2
∴ Its c.d.f. F(x) is given by
(i) P(X < 1) = F(1)
= 0.25(1)2
= 0.25
(ii) P(0.5 < X < 1.5) = F(1.5) – F(0.5)
= 0.25(1.5)2 – 0.25(0.5)2
= 0.25[2.25 – 0.25]
= 0.25(2)
= 0.5
(iii) P(X ≥ 1.5) = 1 – P(X ≤ 1.5)
= 1 – F(1.5)
= 1 – 0.25(1.5)2
= 1 – 0.25(2.25)
= 1 – 0.5625
= 0.4375
Question 6.
Suppose X is the waiting time (in minutes) for a bus and its p.d.f. is given by
$$f(x)=\left\{\begin{array}{cl} \frac{1}{5} & \text { for } 0 \leq x \leq 5 \\ 0 & \text { otherwise } \end{array}\right.$$
Find the probability that (i) waiting time is between 1 and 3 minutes, (ii) waiting time is more than 4 minutes.
Solution:
p.d.f. of r.v. X is given by
f(x) = $$\frac{1}{5}$$ for 0 ≤ x ≤ 5
This is a constant function.
(i) Probability that waiting time X is between 1 and 3 minutes
(ii) Probability that waiting time X is more than 4 minutes
Question 7.
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
$$f(x)= \begin{cases}k\left(4-x^{2}\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}$$
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
Since given f(x) is a p.d.f. of r.v. X
Since -2 ≤ x ≤ 2
∴ x2 ≤ 4
∴ 4 – x2 ≥ 0
∴ k(4 – x2) ≥ 0
∴ k ≥ 0 [∵ f(x) ≥ 0]
Question 8.
Following is the p.d.f. of a continuous r.v. X.
$$f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}$$
(i) Find an expression for the c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7, and 5.
Solution:
The p.d.f. of a continuous r.v. X is
$$f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}$$
(i) c.d.f. of continuous r.v. X is given by
(ii) F(0.5) = $$\frac{(0.5)^{2}}{16}=\frac{0.25}{16}=\frac{1}{64}$$ = 0.015
F(1.7) = $$\frac{(1.7)^{2}}{16}=\frac{2.89}{16}$$ = 0.18
For any of x greater than or equal to 4, F(x) = 1
∴ F(5) = 1
Question 9.
The p.d.f. of a continuous r.v. X is
$$f(x)=\left\{\begin{array}{cl} \frac{3 x^{2}}{8} & \text { for } 0<x<2 \\ 0 & \text { otherwise } \end{array}\right.$$
Determine the c.d.f. of X and hence find (i) P(X < 1), (ii) P(X < -2), (iii) P(X > 0), (iv) P(1 < X < 2).
Solution:
The p.d.f. of a continuous r.v. X is
Question 10.
If a r.v. X has p.d.f.
$$f(x)= \begin{cases}\frac{c}{x} & \text { for } 1<x<3, c>0 \\ 0 & \text { otherwise }\end{cases}$$
Find c, E(X) and V(X). Also find f(x).
Solution:
The p.d.f. of r.v. X is
f(x) = $$\frac{c}{x}$$, 1 < x < 3, c > 0
For p.d.f. of X, we have
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# McGraw Hill Math Grade 8 Lesson 7.1 Answer Key Place Value and Rounding
Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 7.1 Place Value and Rounding to secure good marks & knowledge in the exams.
## McGraw-Hill Math Grade 8 Answer Key Lesson 7.1 Place Value and Rounding
Exercises
ROUND
Round to the nearest whole number.
Question 1.
48.6
49
Explanation:
If the digit in tenth place is greater than 5 add 1 to the nearest whole number.
So, 48.6 is rounded to nearest whole number 49.
Question 2.
98.3
98
Explanation:
If the digit in tenth place is less than 5 round down and keep the nearest whole number.
So, 98.3 is rounded to nearest whole number 98.
Question 3.
156.67
157
Explanation:
If the digit in hundredth place is greater than 5 add 1 to the nearest whole number.
So, 156.67 is rounded to nearest whole number 157.
Question 4.
3026.92
3027
Explanation:
If the digit in hundredth place is greater than 5 add 1 to the nearest whole number.
So, 3026.92 is rounded to nearest whole number 3027.
Question 5.
189.41233
189
Explanation:
If the digit in tenth place is less than 5 round down and keep the nearest whole number.
So, 189.41233 is rounded to nearest whole number 189.
Question 6.
2244.66795
2245
Explanation:
If the digit in tenth place is greater than 5 add 1 to the nearest whole number.
So, 2244.66795 is rounded to nearest whole number 2245.
Question 7.
279.99556
280
Explanation:
If the digit in tenth place is greater than 5 add 1 to the nearest whole number.
So, 279.99556 is rounded to nearest whole number 279.99556.
Question 8.
428.5
429
Explanation:
If the digit in tenth place is greater than 5 add 1 to the nearest whole number.
So, 428.5 is rounded to nearest whole number 4285.
Round to the nearest tenth.
Question 9.
124.5755
124.6
Explanation:
If the digit in hundredth place is greater than 5 add 1 to the nearest tenth place.
So, 124.5755 is rounded to nearest tenth place 124.6
Question 10.
175.5133
175.5
Explanation:
If the digit in hundredth place is less then 5 round down to the tenth place.
So, 175.5133 is rounded to nearest tenth place 175.5
Question 11.
349.49888
349.5
Explanation:
If the digit in hundredth place is greater than 5 add 1 to the nearest tenth place.
So, 349.49888 is rounded to nearest tenth place 349.5
Question 12.
313.35664
313.4
Explanation:
If the digit in hundredth place is greater than 5 add 1 to the nearest tenth place.
So, 313.35664 is rounded to nearest tenth place 313.4
Question 13.
375.77454
375.8
Explanation:
If the digit in hundredth place is greater than 5 add 1 to the nearest tenth place.
So, 375.77454 is rounded to nearest tenth place 375.8
Question 14.
44.00913
44.0
Explanation:
If the digit in hundredth place is less than 5 round down to the nearest tenth place.
So, 44.009133 is rounded to nearest tenth place 44.0
Question 15.
566.9943
567.0
Explanation:
If the digit in hundredth place is greater than 5 add 1 to the nearest tenth place.
So, 566.9943 is rounded to nearest tenth place 567.0
Question 16.
61.15
61.2
Explanation:
If the digit in hundredth place is greater than 5 round down to the nearest tenth place.
So, 61.15 is rounded to nearest tenth place 61.2
Round to the nearest hundredth.
Question 17.
1536.3357
1536.34
Explanation:
If the digit in thousandth place is greater than 5 add 1 to the nearest hundredth place.
So, 1536.3357 is rounded to nearest hundredth place 1536.34
Question 18.
32.4589
32.46
Explanation:
If the digit in thousandth place is greater than 5 add 1 to the nearest hundredth place.
So, 32.4589 is rounded to nearest hundredth place 32.46
Question 19.
118.9977
119.00
Explanation:
If the digit in thousandth place is greater than 5 add 1 to the nearest hundredth place.
So, 118.9977 is rounded to nearest hundredth place 119.00
Question 20.
523.75776
523.76
Explanation:
If the digit in thousandth place is greater than 5 add 1 to the nearest hundredth place.
So, 523.75776 is rounded to nearest hundredth place 523.76
Question 21.
1099.989877
1099.99
Explanation:
If the digit in thousandth place is greater than 5 add 1 to the nearest hundredth place.
So, 1099.989877 is rounded to nearest hundredth place 1099.99
Question 22.
1.11881
1.12
Explanation:
If the digit in thousandth place is greater than 5 add 1 to the nearest hundredth place.
So, 1.11881 is rounded to nearest hundredth place 1.12
Question 23.
33.43718
33.44
Explanation:
If the digit in thousandth place is greater than 5 add 1 to the nearest hundredth place.
So, 33.43718 is rounded to nearest hundredth place 33.44
Question 24.
555.555
555.56
Explanation:
If the digit in thousandth place is greater than 5 add 1 to the nearest hundredth place.
So, 555.555 is rounded to nearest hundredth place 555.56
Round to the nearest thousandth.
Question 25.
729.239788
729.240
Explanation:
If the digit in ten thousandth place is greater than 5 add 1 to the nearest thousandth place.
So, 729.239788 is rounded to nearest thousandth place 729.240
Question 26.
409.13391
409.134
Explanation:
If the digit in ten thousandth place is less than 5 round down to the nearest thousandth place.
So, 409.133391 is rounded to nearest thousandth place 409.134
Question 27.
8056.708035
8056.709
Explanation:
If the digit in ten thousandth place is greater than 5 add 1 to the nearest thousandth place.
So, 8056.708035 is rounded to nearest thousandth place 8056.709
Question 28.
549.594959
549.595
Explanation:
If the digit in ten thousandth place is greater than 5 add 1 to the nearest thousandth place.
So, 549.594959 is rounded to nearest thousandth place 549.595
Question 29.
99.80007
99.800
Explanation:
If the digit in ten thousandth place is less than 5 round down to the nearest thousandth place.
So, 99.80007 is rounded to nearest thousandth place 99.800
Question 30.
177.555901
177.556
Explanation:
If the digit in ten thousandth place is greater than 5 add 1 to the nearest thousandth place.
So, 177.555901 is rounded to nearest thousandth place 177.556
Question 31.
2012.20507
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Triangle
# Solution of Triangle
There are three sides and three angles in a triangle. They are called the six elements of a triangle. Given any three elements (at least one side) in any triangle, remaining three elements can be determined. This process of finding the unknown three elements is called solution of triangle.
If each of the three parts given is an angle, then in this case only the ratios of the sides can be determined. And hence, there exists an infinite number of triangles with the same set of angles.
In solving a triangle, the following different cases arise:
1. Three angles given
2. Three sides given
3. Two angles and one side given
4. Two sides and the included angle given
5. Two sides and an angle opposite to one of them is given
Let $ABC$ be a triangle. Let us denote the angle $BAC$, $CBA$, $ACB$ of the triangle $ABC$ by $A$, $B$, $C$ respectively; and the lengths of the sides $BC$, $CA$, $AB$ by $a$, $b$, $c$ respectively. Let $s$ be the semi perimeter of the triangle $ABC$. $\therefore s=\frac{a+b+c}{2}$
## Case I: Given Three Angles
In this case, no unique solution is possible. We can only determine the ratios of the sides. From sine law, we know, $a:b:c=\sin A:\sin B:\sin C$ Hence, only the shape of the triangle can be determined but not the size.
## Case II: Given Three Sides
Applying any one of the following formulae, we can determine the angle $A$, $\cos A=\frac{b^2+c^2-a^2}{2bc}$ $\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}$ $\sin A=\frac{2}{bc}\sqrt{s(s-a)(s-b)(s-c)}$ $\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}$ $\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$
Once the value of $A$ is obtained, we can now determine the angle $B$ from the relation $\sin A:\sin B=a:b$ Then, angle $C$ may be found from the relation $C=π-(A+B)$
## Case III: Given Two Angles and a Side
Let $A$ and $B$ be two given angles and $a$ be the given side. Then, the angle $C$ can be obtained from the relation $C=π-(A+B)$
Now, by using the sine law, we can determine the other two sides $b$ and $c$ i.e. $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
## Case IV: Given Two Sides and an Included Angle
Let $b$ and $c$ be the two given sides and the angle be $A$. Then, we may find the side $a$ by using the cosine law, $a^2=b^2+c^2-2bc\cos A$
Then, we can determine $B$ from $\frac{a}{\sin A}=\frac{b}{\sin B}$ Finally, we can find $C$ from $C=π-(A+B)$
## Case V: Given Two Sides and an Opposite Angle
Let $b$ and $c$ be the given two sides and $B$ be the given angle. Then, we can determine the side $c$ from the sine law, $\frac{b}{\sin B}=\frac{c}{\sin C}$ Then, we can find $A$ from $A=π-(B+C)$
Once the value of $A$ is obtained, we can find the side $a$ from $\frac{a}{\sin A}=\frac{b}{\sin B}$
Thus, the solution of the triangle depends upon the possibility the determination of $c$ from $\frac{b}{\sin B}=\frac{c}{\sin C}$ $\text{or,}\;\;\sin C=\frac{c\sin B}{b}$
In the determination of $c$, the following three cases arise:
### Case I:
If $c\sin B>b$, then $\sin C>1$ which is not possible and hence no triangle is possible.
### Case II:
If $c\sin B=b$, then $\sin C=1$ or $C=90°$. Then, a right angled triangle is the required solution. Then, $A=\frac{π}{2}-B$ and, we can determine $a$ from Pythagorean theorem, $a=\sqrt{c^2-b^2}$
### Case III:
If $c\sin B<b$, then two supplement values of $C$ are possible, one acute and one obtuse. In this case, the following three sub-cases arise:
#### Sub-case I:
If $c<b$, then $C<B$, hence $C$ must be acute and only one triangle is possible.
#### Sub-case II:
If $c=b$, then $C=B$, hence $C$ can not be obtuse otherwise there will be two obtuse angles in a triangle. So, only one triangle is possible.
#### Sub-case III:
If $c>b$, then the angle $C$ is not restricted and it can have either acute or obtuse value as long as $B$ is less than $90°$. Since $C$ can have either acute or obtuse value, two triangles are possible with given parts.
This is usually known as the Ambiguous Case in the solution of triangle.
Previous: Area of a Triangle
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# FIND THE EQUATION OF A TRANSFORMED EXPONENTIAL FUNCTION FROM A GRAPH
Any transformed exponential can be written in the form
y = abx - h + k
where y = k is the horizontal asymptote.
Note that, due to the shift, the vertical intercept is shifted to (0, a+k)
Find the exponential function represented by the graph given below.
Problem 1 :
Solution :
y = abx + k
Here, the Horizontal asymptote is y = -1. So, k = -1
y = abx - 1 ------(1)
Tracing two points from the graph of the given function given above.
(0, 2) and (-5, -1)
Applying the point (0, 2) in (1), we get
2 = ab0 - 1
2 = a(1) - 1
a = 2 + 1
a = 3
Applying the point (1, 5) in (1), we get
5 = ab1 - 1
6 = 3(b1)
b = 6/3
b = 2
So, the required exponential function is
y = 3(2)x - 1
Problem 2 :
Solution :
y = abx + k
Here, the Horizontal asymptote is y = 5. So, k = 5
y = abx + 5 ------(1)
Tracing two points from the graph of the given function given above.
(0, 2) and (-1, 3)
Applying the point (0, 2) in (1), we get
2 = ab0 + 5
2 - 5 = a(1)
a = -3
Applying the point (-1, 3) in (1), we get
3 = (-3)b-1 + 5
3 - 5 = -3(1/b)
-2 = -3/b
b = 3/2
b = 1.5
So, the required exponential function is
y = -3(1.5)x + 5
Problem 3 :
Solution :
y = abx + k
Here, the Horizontal asymptote is y = 1. So, k = 1
y = abx + 1 ------(1)
Tracing two points from the graph of the given function given above.
(-2, 0) and (0, -3)
Applying the point (-2, 0) in (1), we get
0 = ab-2 + 1
-1 = a/b2
Applying the point (0, -3) in (1), we get
-3 = ab0 + 1
-3 - 1 = a(1)
a = -4
Applying the value of a, we get
-4/b2 = -1
b2 = 4
b = 2
So, the required exponential function is
y = -4(1.5)x + 5
Problem 4 :
Solution :
y = abx + k
Here, the Horizontal asymptote is y = 3. So, k = 3
y = abx + 3 ------(1)
Tracing two points from the graph of the given function given above.
(0, 2) and (1, 1)
Applying the point (0, 2) in (1), we get
2 = ab0 + 3
-1 = a(1)
a = -1
Applying the point (1, 1) in (1), we get
1 = (-1)b1 + 3
1 - 3 = -b
b = 2
So, the required exponential function is
y = -1(2)x + 3
Problem 5 :
Solution :
y = abx + k
Here, the Horizontal asymptote is y = 3. So, k = 3
y = abx + 3 ------(1)
Tracing two points from the graph of the given function given above.
(0, 1) and (-1, -1)
Applying the point (0, 1) in (1), we get
1 = ab0 + 3
-2 = a(1)
a = -2
Applying the point (-1, -1) in (1), we get
-1 = (-2)b-1 + 3
-4/(-2) = 1/b
b = 1/2
So, the required exponential function is
y = -2(1/2)x + 3
Problem 6 :
Solution :
y = abx + k
Here, the Horizontal asymptote is y = -2. So, k = -2
y = abx - 2------(1)
Tracing two points from the graph of the given function given above.
(0, 1) and (-1, 4)
Applying the point (0, 1) in (1), we get
1 = ab0 - 2
3 = a(1)
a = 3
Applying the point (-1, 4) in (1), we get
4 = (2)b-1 - 2
4 + 2 = 2/b
6b = 2
b = 1/3
So, the required exponential function is
y = 3(1/3)x - 2
Problem 7 :
Solution :
y = abx + k
Here, the Horizontal asymptote is y = 7. So, k = 7
y = abx + 7------(1)
Tracing two points from the graph of the given function given above.
(0, 5) and (1, 1)
Applying the point (0, 5) in (1), we get
5 = ab0 + 7
5 - 7 = a(1)
a = -2
Applying the point (1, 1) in (1), we get
1 = (-2)b1 + 7
-6 = -2b
b = 3
So, the required exponential function is
y = -2(3)x + 7
Problem 8 :
Solution :
y = abx + k
Here, the Horizontal asymptote is y = 5. So, k = 5
y = abx + 5------(1)
Tracing two points from the graph of the given function given above.
(0, 3) and (-1, -1)
Applying the point (0, 3) in (1), we get
3 = ab0 + 5
-2 = a(1)
a = -2
Applying the point (-1, -1) in (1), we get
-1 = (-2)b-1 + 5
-6 = -2/b
b = 1/3
So, the required exponential function is
y = -2(1/3)x + 5
## Recent Articles
1. ### Finding Range of Values Inequality Problems
May 21, 24 08:51 PM
Finding Range of Values Inequality Problems
2. ### Solving Two Step Inequality Word Problems
May 21, 24 08:51 AM
Solving Two Step Inequality Word Problems
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#### Need Help?
Get in touch with us
# Rational Exponents
Sep 15, 2022
## Key Concepts
• Define a rational exponent.
• Solve equations with rational exponents using the product of powers property.
• Solve equations with rational exponents using the power of a power property.
• Solve equations with rational exponents using the power of a product property.
• Solve equations with rational exponents using the quotient of powers property.
## Rational Exponents
### Fractions
A part of a whole is called a fraction.
• All fractions can be placed on the number line.
### Decimal numbers
The numbers whose whole number part and fractional part are separated by a decimal point are called decimal points.
### Factors and multiples
A factor is a number or a group of numbers that are multiplied together to make a product.
A multiple is the product of a quantity and a whole number.
### Exponents
Repeated multiplication can be represented in more than one way.
You can use an exponent to write the repeated multiplication of a number.
A number that can be written using exponents is called a power.
We read as 2 raised to the power of 3.
#### Rational exponents
When a number p is raised to power 1/2, we can write them as √p.
The expressions with exponents that are rational numbers are called rational exponents (also called fractional exponents).
#### Laws of exponents
Law: When two terms with the same base are multiplied, the powers are added.
am×an=am+n
Example: Evaluate 24 × 29
Sol: 24 × 29 = 2(4+9)
= 213
= 8192
• Use the product of powers property to solve equations with rational exponents
#### Law of exponents
Law: When raising a power to a new power, multiply the exponents.
(am)n=amn
Example: Evaluate (53)2
Sol: (53)2 = 5(3×2)
= 56
= 15625
Use the power of a power property to solve equations with rational exponents
#### Law of exponents
Law: When multiplying expressions with the same exponent but different bases, multiply the bases and use the same exponent.
am×bm=(a×b)m
Example: Evaluate 62×52
Sol: 62×52 = (6×5)2
= 302
= 900
• Use the power of a product property to solve equations with rational exponents
#### Law of exponents
Law: When dividing two powers with the same base, we subtract the exponents.
• Use the quotient of powers property to solve equations with rational exponents
## Exercise
• Write the radical √14641 using rational exponents.
• What is the value of x in 27(x/2) = 3(x-1)?
• Solve: 3(x/2+1) = 3(-5x/2)
• If the volume of a sphere is V=4/3 πr3 is equal to 392 m3. Find the radius.
• Write the radical √ba using rational exponent.
### What we have learned
• Repeated multiplication can be represented in more than one way.
• You can use an exponent to write the repeated multiplication of a number.
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […]
|
Probability Gambling (Age 7)
The Activities
1. Topic: Logic, Problem Solving: Book: More Stories to Solve by Shannon. This is a book of 15 short stories with surprise endings you can try to figure out. The kids really loved this, and begged to do more stories. We read 5 stories, and were able to solve 4 of them by discussing them together.
2. Topic: Logic. We played Logic Links again. This time I had printed out sheets with the formations needed for the harder puzzles.
A solved logic puzzle.
3. Topic: Probability. Probability race: each kid gets two dice, rolls them, adds them up and colors in the corresponding number on the chart. The first number to get rolled 5 times wins the race.
How did it go?
School is back in session now, so we had all 5 kids. It was a really fun circle with lots of excited contributors, and even some hard thinking.
This time we did puzzles with nine clues. The first puzzle was easy enough that everyone solved it with few problems. The second puzzle was harder, and eventually one kid stumbled across the correct arrangement, and the kids used her paper as reference. This game works pretty well with a large group, though it can be frustrating for the kids who don’t happen to get the solution first.
Probability Race
These kids did a bunch of probability races in past years, but at first only my daughter remembered the activity. As soon as I explained the rules, they started speeding through the races. The kids are MUCH faster now at rolling, adding, and marking their charts. Again, all the kids enjoyed my commentary about how 7, 8, and 9 are tied on Kid X’s paper, etc.
After we finished around 8 races, we stopped and reviewed the winners. 7 won four times, 3 once, 6 twice, and 9 once. I asked the kids what number they would pick if they could win a dollar for guessing right. Most kids said 7.
I asked why 7 had won so many times, and a girl suggested that there are more ways to get 7 than any other number. She counted up “1+6”, “2+5”, “3+4” to get 7. Then she figured out there were 3 ways to get 6 also: “1+5”, “2+4”, “3+3”.
At this point I got a piece of paper, and wrote die A and die B at the top, and we made a chart of possible ways to get 7 based on various rolls of die A and B. We found 6 ways. In contrast there were only 5 ways to get 6. The kids were pretty interested in figuring out how many ways to get various number.
One kid said she thought “3+4” and “4+3” should count as the same. I said that they were different because they were on different dice, but she asked how we could tell the dice apart. Next time, we should use dice of two different colors.
While we were figuring out the ways to roll 5, I asked what if die A rolls a 6? What does die B need to be to add up to 5? At this point, one of the kids said that negative numbers are not allowed at her school. I said no? And she said they never do problems like 5 – 6 at school. I asked everyone if negative numbers are allowed at Math Circle? They all said “Yes!”. I asked, “Can we roll a negative 1 on a die?”, and we all agreed no.
Finally, I let the kids pick 3 numbers. If one of those numbers wins the race, they get a prize from our prize box. They quickly (and wisely) picked 6, 7, and 8. We had a very dramatic race, ending with 6 winning everything, and lots of cheering. We picked prizes, and circle was over.
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Question Video: Solving Trigonometric Equations by Squaring | Nagwa Question Video: Solving Trigonometric Equations by Squaring | Nagwa
# Question Video: Solving Trigonometric Equations by Squaring Mathematics • First Year of Secondary School
## Join Nagwa Classes
By first squaring both sides, or otherwise, solve the equation 4 sin 𝜃 − 4 cos 𝜃 = √3, where 0° < 𝜃 ⩽ 360°. Be careful to remove any extraneous solutions. Give your answers to two decimal places.
06:47
### Video Transcript
By first squaring both sides, or otherwise, solve the equation four sin 𝜃 minus four cos 𝜃 is equal to the square root of three, where 𝜃 is greater than zero degrees and less than or equal to 360 degrees. Be careful to remove any extraneous solutions. Give your answers to two decimal places.
The question advises that we approach the problem by first squaring both sides of the equation. Doing so, we obtain four sin 𝜃 minus four cos 𝜃 all squared is equal to root three squared. Distributing the parentheses and then collecting like terms on the left-hand side gives us 16 sin squared 𝜃 minus 32 sin 𝜃 cos 𝜃 plus 16 cos squared 𝜃. On the right-hand side, root three squared is equal to three.
Next, we recall the Pythagorean identity, which states that sin squared 𝜃 plus cos squared 𝜃 is equal to one. Simplifying the left-hand side further gives us 16 multiplied by sin squared 𝜃 plus cos squared 𝜃 minus 32 sin 𝜃 cos 𝜃. Replacing sin squared 𝜃 plus cos squared 𝜃 with one gives us the equation 16 minus 32 sin 𝜃 cos 𝜃 equals three. Subtracting 16 from both sides of this equation gives us negative 32 sin 𝜃 cos 𝜃 is equal to negative 13. We can then divide through by negative 32 such that sin 𝜃 cos 𝜃 equals 13 over 32.
We now have two equations in the two variables sin 𝜃 and cos 𝜃. This means that the system of equations can be solved simultaneously. Adding four cos 𝜃 to both sides of our original equation, we have four sin 𝜃 is equal to root three plus four cos 𝜃. Dividing both sides of this equation by four, we have sin 𝜃 is equal to root three plus four cos 𝜃 all divided by four.
After clearing some space, we will now consider how we can solve these two simultaneous equations. We will begin by substituting the expression for sin 𝜃 in equation two into equation one. This gives us root three plus four cos 𝜃 over four multiplied by cos 𝜃 is equal to 13 over 32. We can simplify this equation by firstly distributing the parentheses. We can then multiply through by 32, giving us eight root three cos 𝜃 plus 32 cos squared 𝜃 equals 13. Finally, subtracting 13 from both sides of this equation, we have the quadratic equation in terms of cos 𝜃 as shown.
This can be solved using the quadratic formula, where 𝑎 is 32, 𝑏 is eight root three, and 𝑐 is negative 13. Substituting in these values and then simplifying gives us cos of 𝜃 is equal to negative three plus or minus the square root of 29 all divided by eight. Taking the inverse cosine of both sides with positive root 29 gives us 𝜃 is equal to 62.829 and so on. To two decimal places, this is equal to 62.83 degrees. Taking the inverse cosine of our equation with negative root 29 gives us 𝜃 is equal to 152.829 and so on. This rounds to 152.83 degrees to two decimal places.
We were asked to give all solutions that are greater than or equal to zero degrees and less than or equal to 360 degrees. We therefore need to consider the symmetry of the cosine function such that the cos of 𝜃 is equal to the cos of 360 degrees minus 𝜃. Subtracting each of our values from 360 degrees gives us further solutions 297.17 degrees and 207.17 degrees to two decimal places.
We have therefore found four possible solutions to the given equation. However, we were reminded in the question to remove any extraneous solutions, these extra solutions that were created when we squared our original equation. We need to substitute each of our four solutions into the initial equation to check they are valid.
The initial equation was four sin 𝜃 minus four cos 𝜃 equals root three. Substituting 𝜃 equals 62.83 into the left side of our equation gives us an answer of root three. This means that this is a valid solution. However, when we substitute 𝜃 is equal to 152.83 degrees into the left-hand side of our equation, we do not get root three. This means that this is not a valid solution. Repeating this process for 207.17 degrees and 297.17 degrees, we see that 207.17 is a valid solution, whereas the fourth answer of 297.17 is not. We can therefore conclude that there are two solutions that satisfy the equation in the given interval of 𝜃, which are 62.83 and 207.17 degrees.
## Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions
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# CBSE & NCERT Solutions Class 9 Maths Circles Takshila Learning
## Maths NCERT Solutions Class 9 – Chapter 11 – Circle – Part 1
Class 9 Maths Circle – A circle is the locus of a point which moves in a plane in such a way that its distance from a given fixed point is always constant.
The permanent tip is called the centre and the stable distance is called the radius of the circle.
A circle with centre O and radius r is denoted by C (O,r).
For video lectures of Class 9 Maths and NCERT solutions click CBSE Class 9th Maths
Terms Related to a Circle
Radius : A Line segment joining the centre and a point on the circle is called its radius.
In the given figure, OA and OB are radii of circle C (O, r)
Circumference : The perimeter of a circle is called its circumference.
Circumference = 2πr.
Chord-A chord of a circle is a line segment joining any two points on the circle.
In the given figure, PQ, RS, and AOB are the chords of a circle with centre O.
Diameter-A diameter is a chord of a circle passing through the centre of the circle.
Hence AOB is a diameter of a circle with O.
A diameter is the longest chord of a circle.
Secant-A line which intersects a circle in two distinct points is called a scent of the circle.
In the given figure, the line l cuts the circle in two points A and B. So l is a secant of the circle.
Tangent-A line that intersects the circle in exactly one point is called a tangent to the circle. The point at which the tangent meets the circle is called its point of contact.
Clear your doubts with experts, click 9th Class Maths for details.
Position of a point with respect to a circle
A point P is said to lie
• Inside the circle C (O,r), if OP<r
• On the circle C(O,r)if OP=r
• Outside the circle C(O,r)if OP>r
Interior of a circle-The region consisting of all points lying on the circumference of a circle and inside it is called interior of a circle. A point P lies in the interior of a circle C (O,r) if OP<=r.
The exterior of a circle-The region consisting of all points lying outside a circle is called exterior of a circle. A point P lies in the exterior of a circle C (O,r) if OP>r.
The circular region-The region consisting of all points which are either on the circle or lie inside the circle is called the circular region or circular disc.
### Maths NCERT solutions Class 9 designed by experts are available.
Takshila Learning is an online education portal offers live classes for Class 1 to 12th. Animated videos, worksheets, sample papers and doubt sessions are also available. For details www.takshilalearning.com.
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# The Largest Rectangular Area in a Histogram
The Largest Rectangular Area in a Histogram
## Introduction
One of the most feared things by human beings is a failure. The extent to which people push themselves to avoid disappointment is quite surprising, but being prepared is one of the easiest ways to steer clear of it. To quote Abraham Lincoln,
“Give me six hours to chop down a tree, and I will spend the first four sharpening the axe.”
Unlike Mr Lincoln, we are not chopping a tree. Instead, we are planting the seeds to a successful career, and to do that, we must be prepared for the placement exams.
In companies like Facebook and Google, a common question in the coding interview is to find the largest rectangular area in Histogram.
Do you know how to find it?
I guess not. So let’s learn how to find the largest rectangular area in Histogram in this article.
## Problem Statement
Before we head on to the solution, we must first correctly understand the largest rectangular area in Histogram.
An array or list “Heights” of length N will be given. The elements of this array are the heights of each bar in a Histogram, and the width of each bar is 1.
For example, if Heights = {3, 2, 6, 5, 1, 2} the Histogram would be as shown below.
For this example, the largest rectangular area in Histogram will be as follows:
Now that we know what question we are solving, we will look into the different methods.
## Brute Force Approach
By brute force, we usually mean the most straightforward method which comes to our mind first. Most of the time, this isn’t the best with respect to time complexity
This method will find the area of all the rectangles possible, starting from each rectangle for our problem.
To understand this better, let us consider our previous example again.
The iterations through all the rectangles to find the largest one is shown below.
To summarise what we learnt, let’s see the algorithm and code for the brute force algorithm.
### Algorithm
```Step 1: Initialise a variable, maxArea, with 0.
Step 2: Create a for loop which runs through all the rectangles.
Step 3: Create a nested loop to find all the possible rectangles, starting from a particular one.
Step 4: Store current index value as both min and max values in minVal and max Val variable, respectively.
Step 4: Find the area of the rectangles formed compare with maxArea and assign maxArea as maxVal if it’s greater.
Step 5: Return maxArea as the largest rectangular area in Histogram.
```
### Python Code
```'''
Time Complexity = O(N^2)
Space Complexity = O(1)
Where N is the number of elements in the given array/list.
'''
def largestRectangle(heights):
n = len(heights)
if n <= 0:
return 0;
maxArea = 0;
for j in range(n):
# Store current index value as both min and max values.
minVal = heights[j]
maxVal = heights[j]
i = j - 1
while i >= 0:
# If the current index value is less than minVal.
if heights[i] < minVal:
minVal = heights[i]
# If the area of the current rectangle is greater than maxVal.
if minVal * (j - i + 1) > maxVal:
maxVal = minVal * (j - i + 1);
# Store the maximum area of the rectangle.
if maxVal > maxArea:
maxArea = maxVal;
i -= 1
# Return area of largest rectangle.
return maxArea
heights = [3,2,6,5,1,2]
print("Maximum area is",largestRectangle(heights))```
Output:
`Largest rectangular area in histogram is 10`
### Complexity Analysis
This method has a time complexity of O(N2) due to the nested loop, where N is the number of elements in the given array/list.
Because of that reason, it isn’t usually used.
Space Complexity = O(1).
## Divide and Conquer
Our age-old favourite technique of divide and conquer can be used to find the largest rectangular area in Histogram too. Let’s see how it is done with the help of our previous example.
In this method, we first have to find the smallest element in the array. Then we will divide the array into two halves, the part to the left of the smallest element and the part to the right of it.
After this, we will calculate the area of two rectangles:
1. The rectangle formed by the elements in the left half
2. The rectangle formed by the elements in the right half
The largest area will be saved. Out of the two halves, the one with the largest area is considered, and the process is repeated till we are left with only one element.
### Algorithm
```The idea is to find the minimum value in the given array/list, which is the lowest histogram bar. Once we have an index of the minimum value, the max area is the maximum of the following three values.
The maximum area on the left side of the minimum value. Here we are trying to find the area of the rectangle towards the left side of the minimum value bar, excluding the min value.
The maximum area on the right side of the minimum value. Here we are trying to find the area of the rectangle towards the right side of the minimum value bar, excluding the min value.
The number of bars is multiplied by the minimum value.
We use a loop with two pointers named “LPTR” and “RPTR”, and at each iteration, we advance the one with a higher neighbour until we reach the end in both directions.
```
### Python Code
```'''
Time Complexity = O(N log N)
Space Complexity = O(log N)
Where N is the number of elements in the given array/list.
'''
def divideAndConquer(heights, l, r):
# If we only have one element then return it.
if l == r:
return heights[l]
mid = (l + r) // 2
# Recur for the left rectangle.
lscore = divideAndConquer(heights, l, mid)
# Recur for the right rectangle.
rscore = divideAndConquer(heights, mid + 1, r)
lptr = mid
rptr = mid
maxArea = 0
minimum = heights[mid]
while lptr >= l and rptr <= r:
'''
Take the minimum of lptr height and
rptr height in comparison to the minimum.
'''
minimum = min(minimum,heights[lptr], heights[rptr])
'''
Take the maximum area for
the rectangle between lptr and rptr.
'''
maxArea = max(maxArea, minimum * (rptr - lptr + 1))
if (lptr > l) and (rptr < r):
if heights[rptr + 1] >= heights[lptr - 1]:
rptr += 1
else:
lptr -= 1
elif lptr > l:
lptr -= 1
elif rptr < r:
rptr += 1
else:
rptr += 1
lptr -= 1
# Return the maximum area.
return max(maxArea,lscore, rscore)
def largestRectangle(heights):
if len(heights) == 0:
return 0
return divideAndConquer(heights, 0, len(heights) - 1)
heights = [3,2,6,5,1,2]
print("Maximum area is",largestRectangle(heights))```
Output:
`Largest rectangular area in histogram is 10`
### Complexity Analysis
Dividing the array into two halves gives us O(log N) complexity, but we repeat that recursively (N-1) times. Therefore, the overall time complexity of the algorithm is O(NlogN), which is better than the complexity of the brute force method,
Space Complexity = O(N), where N is the number of elements in the given array/list.
## Using Next Lower Element
While solving any problem, we aim to solve it using an algorithm with linear time complexity. Let us see how to find the largest rectangular area in Histogram in such a manner.
In this method, we will find the next smaller element to the left and right of it for each element. For example, if we consider element 5, the indices of the next smaller element to the left and right are 1 and 4. respectively.
With these indices, we can find the width of the largest possible rectangle for that element as follows:
Area of the rectangle = Element x (Next smaller element to the right – Next smaller element to the left – 1)
= 5 x (4 – 1 – 1) = 5 x 2 = 10
We can find this area for all the rectangles and compare them to find the largest rectangular area in Histogram.
### Algorithm
```We traverse all histograms from left to the right, maintaining a stack of histograms. Every histogram is pushed to stack once. A histogram is popped from the stack when a histogram of smaller height is seen. When a histogram is popped, we calculate the area with the popped histogram as the smallest histogram. How do we get the left and right indexes of the popped histogram – the current index tells us the ‘RIGHT_INDEX’, and the index of the previous item in the stack is the ‘LEFT_INDEX'.
Create an empty stack.
Start from the first histogram bar, and do the following for every bar ‘heights[i]’ where ‘I’ varies from 0 to ‘N’-1.
If the stack is empty or heights[i] are higher than the bar at the top, push ‘I’ to the stack.
If this bar is smaller than the top of the stack, keep removing the top of the stack while the top is greater. Let the removed bar be heights[tp]. Calculate the area of the rectangle with heights[tp] as the smallest bar. For heights[tp], the ‘left index’ is the previous (previous to tp) item in the stack, and the ‘RIGHT_INDEX' is ‘I’ (current index).
If the stack is not empty, then one by one, remove all bars from the stack and do step 4 for every removed bar.```
### Python Code
```'''
Time Complexity = O(N)
Space Complexity = O(N)
Where N is the number of elements in the given array/list.
'''
def largestRectangle(heights):
n = len(heights)
'''
The stack holds indexes of heights[] array.
The bars stored in stack are always in increasing
order of their heights.
'''
stack = list()
# Initialize max area.
maxArea = 0
# To store the top of the stack.
topOfStack = 0
# To store an area with the top bar as the smallest bar.
areaWithTop = 0
# Run through all bars of a given histogram.
i = 0
while i < n:
'''
If this bar is higher than the bar on top stack,
push it to stack.
'''
if len(stack) == 0 or (heights[stack[-1]] <= heights[i]):
stack.append(i)
i += 1
else:
topOfStack = stack.pop()
'''
Calculate the area with heights[topOfStack]
stack as the smallest bar.
'''
if len(stack) == 0:
areaWithTop = heights[topOfStack] * i
else:
areaWithTop = heights[topOfStack] * (i - stack[-1] - 1)
# Update max area, if needed.
if maxArea < areaWithTop:
maxArea = areaWithTop
'''
Now pop the remaining bars from stack and
calculate area with every popped
bar as the smallest bar.
'''
while len(stack) != 0:
topOfStack = stack.pop()
if len(stack) == 0:
areaWithTop = heights[topOfStack] * i
else:
areaWithTop = heights[topOfStack] * (i - stack[-1] - 1)
if maxArea < areaWithTop:
maxArea = areaWithTop
return maxArea
heights = [3,2,6,5,1,2]
print("Maximum area is",largestRectangle(heights))```
Output:
`Largest rectangular area in histogram is 10`
### Complexity Analysis
As discussed in the beginning, this method has:
Time Complexity = O(N)
Space Complexity = O(N)
Where N is the number of elements in the given array/list.
## Frequently Asked Questions
How do you find the largest area of the rectangle in the Histogram?
The simplest approach is to consider all bars as starting points and calculate the area of the rectangles possible with each bar.
You can also use the concept of the nearest smallest element to left and right to calculate the largest rectangular area in Histogram.
What are stacks in data structures?
Stack is an abstract data type that works on the principle of LIFO (last in, first out).
What is the best time complexity to find the largest rectangular area in Histogram?
The most optimised solution comes up with the time complexity of O(n).
Where can I solve the Next greater element problem?
You can get yourself familiar with the NGE problem here.
## Key Takeaways
In this article, we learnt how to find the largest rectangular area in Histogram. You may, however, try solving it yourself here and get your solution checked too. This will provide you with complete knowledge on a question frequently asked in many interviews.
Apart from this, you can practice other questions commonly asked in the coding rounds of interviews at CodeStudio. You will find practice programming questions there, and the interview experiences of people who were once students like us but are now working in renowned product-based companies.
Happy learning!
By: Neelakshi Lahiri
|
How do you differentiate (3x^2+4) / (sqrt(1+x^2))?
Apr 5, 2015
Hey there! :)
Looks tough when you first see it, right? However, differentiation rules are like super powers.
Here is the short story:
$\frac{d}{\mathrm{dx}} \left(\frac{3 {x}^{2} + 4}{\sqrt{1 + {x}^{2}}}\right) = \frac{6 x}{\sqrt{1 + {x}^{2}}} - \frac{x \left(3 {x}^{2} + 4\right)}{1 + {x}^{2}} ^ \left(\frac{3}{2}\right)$
Now for the far more exciting long story:
The quotient rule says that if we have two differentiable functions, say $f \left(x\right) \mathmr{and} g \left(x\right)$ such that $g \left(x\right) \ne 0$, then we have
$\frac{d}{\mathrm{dx}} \left(f \frac{x}{g} \left(x\right)\right) = \frac{\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) \cdot g \left(x\right) - f \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(g \left(x\right)\right)}{g \left(x\right)} ^ 2$
So now we just apply this!
So,
$\frac{d}{\mathrm{dx}} \left(\frac{3 {x}^{2} + 4}{\sqrt{1 + {x}^{2}}}\right) = \frac{\frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 4\right) \cdot \sqrt{1 + {x}^{2}} - \left(3 {x}^{2} + 4\right) \cdot \frac{d}{\mathrm{dx}} \left(\sqrt{1 + {x}^{2}}\right)}{\sqrt{1 + {x}^{2}}} ^ 2$
Now we have to use a few super powers here. The power rule and chain rule are part of our arsenal.
I hope it is clear that
$\frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 4\right) = 2 \cdot 3 x + 0 = 6 x$
(If not, just shout!) - We just applied the power rule to the first term and the second term is a constant. And constants don't stand a chance against differentiation! They just become zero.
Next, we harness the power of the chain rule, that says for any differentiable functions, say, $u \left(x\right)$ and $h \left(x\right)$,
$\frac{d}{\mathrm{dx}} \left(u \left(h \left(x\right)\right)\right) = \frac{d}{\mathrm{dh} \left(x\right)} \left(u \left(h \left(x\right)\right)\right) \cdot \frac{d}{\mathrm{dx}} \left(h \left(x\right)\right)$
We apply it (here u is the square root function and h(x) = (1+x^2)):
$\frac{d}{\mathrm{dx}} \left(\sqrt{1 + {x}^{2}}\right) = \frac{1}{2} \cdot {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(2 x\right) = x {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}}$
Finally, plugging these things back in their rightful place, we get
$\frac{d}{\mathrm{dx}} \left(\frac{3 {x}^{2} + 4}{\sqrt{1 + {x}^{2}}}\right) = \frac{6 x \cdot \sqrt{1 + {x}^{2}} - \left(3 {x}^{2} + 4\right) \cdot x {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}}}{1 + {x}^{2}}$
$= \frac{6 x \sqrt{1 + {x}^{2}}}{1 + {x}^{2}} - \frac{x \left(3 {x}^{2} + 4\right)}{\left(1 + {x}^{2}\right) \cdot \sqrt{1 + {x}^{2}}}$
$= \frac{6 x}{\sqrt{1 + {x}^{2}}} - \frac{x \left(3 {x}^{2} + 4\right)}{{\left(1 + {x}^{2}\right)}^{\frac{3}{2}}}$
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# Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.3
## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.3
Question 1.
Find the equation of a straight line passing through the mid-point of a line segment joining the points(1, -5) (4, 2) and parallel to
(i) X axis
(ii) Y axis
Solution:
Question 2.
The equation of a straight line is 2(x – y) + 5 = 0. Find its slope, inclination and intercept on the Y axis.
Solution:
2(x – y) + 5 = 0
⇒ 2x – 2y + 5 =
⇒ 2y = 2x + 5
Question 3.
Find the equation of a line whose inclination is 30° and making an intercept – 3 on the Y axis.
Solution:
θ = 30°
Question 4.
Find the slope and y intercept of $$\sqrt{3}$$x + (1 – $$\sqrt{3}$$)y = 3.
Solution:
Question 5.
Find the value of ‘a’, if the line through (-2, 3) and (8, 5) is perpendicular to y = ax = + 2
Solution:
Question 6.
The hill in the form of a right triangle has its foot at (19, 3)The inclination of the hill to the ground is 45°. Find the equation of the hill joining the foot and top.
Solution:
θ = 45°
Coordinate of foot of hill = (19, 3) let equation of line be y = mx + c
m = tan θ = tan 45° = 1
⇒ y = x + c
Substituting y = 3 & x = 19, 3 = 19 + c ⇒ c = -16
Question 7.
Find the equation of a line through the given pair of points
(ii) (2, 3) and (-7, -1)
Solution:
(i) Equation of the line in two point form is
⇒ 9y – 27 = 4x – 8
⇒ 4x – 9y – 8 + 27 = 0
⇒ 4x – 9y + 19 = 0
Question 8.
A cat is located at the point(-6, -4) in xy plane. A bottle of milk is kept at (5, 11). The cat wish to consume the milk traveling through shortest possible distance. Find the equation of the path it needs to take its milk.
Solution:
A = (x1, y1) = (-6, -4)
B = (x2, y2) = (5, 11)
Shortest path between A and B is a line joining A and B.
Question 9.
Find the equation of the median and altitude of ∆ABC through A where the vertices are A(6, 2) B(-5, -1) and C(1, 9)
Solution:
Question 10.
Find the equation of a straight line which -5 has slope $$\frac{-5}{4}$$ and passing through the point (-1, 2).
Solution:
Question 11.
You are downloading a song. The percent y (in decimal form) of mega bytes remaining to get downloaded in x seconds is given by y = -0.1x + 1.
(i) graph the equation.
(ii) find the total MB of the song.
(iii) after how many seconds will 75% of the song gets downloaded?
Solution:
(i) y = -0.1x + 1
when x = 0 ⇒ y = 1
when y = 0 ⇒ y = 10
(ii) Total MB of song can be obtained when time = 0
∴ x = 0
⇒ y = 1 MB
⇒ remaining % = 25% ⇒ y = 0.25
0.25 = -0. 1x + 1
⇒ 0.1x = 0.75
(iv) song will downloaded completely when , remaining % = 0 ⇒ y = 0
⇒ 0 = -0.1x + 1
⇒ x = 10
∴ 10 seconds
Question 12.
Find the equation of a line whose intercepts on the x and y axes are given below.
(i) 4, -6
(ii) $$-5 \frac{3}{4}$$
Solution:
Question 13.
Find the intercepts made by the following lines on the coordinate axes,
(i) 3x – 2y – 6 = 0
(ii) 4x + 3y + 12 = 0
Solution:
(i) The given equation is
3x – 2y – 6 = 0
3x – 2y = 6
Divided by 6
$$\frac { 3x }{ 6 }$$ – $$\frac { 2y }{ 6 }$$ = $$\frac { 6 }{ 6 }$$
$$\frac { x }{ 2 }$$ – $$\frac { y }{ 3 }$$ = 1 ⇒ $$\frac { x }{ 2 }$$ + $$\frac { y }{ -3 }$$ = 1
(Comparing with $$\frac { x }{ a }$$ + $$\frac { y }{ b }$$ = 1)
∴ x intercept = 2; y intercept = -3
(ii) The given equation is
4x + 3y + 12 = 0
4x + 3y = -12
Divided by -12
$$\frac { 4x }{ -12 }$$ + $$\frac { 3y }{ -12 }$$ = $$\frac { -12 }{ -12 }$$
$$\frac { x }{ -3 }$$ + $$\frac { y }{ -4 }$$ = 1
(Comparing with $$\frac { x }{ a }$$ + $$\frac { y }{ b }$$ = 1)
∴ x intercept = -3; y intercept = -4
Question 14.
Find the equation of a straight line
(i) passing through (1, -4) and has intercepts which are in the ratio 2 : 5
(ii) passing through (-8, 4) and making equal intercepts on the coordinate axes
Solution:
(i) ratio of intercept = 2 : 5
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# Derivatives of Non-Polynomial Functions
In this section, we introduce rules for the derivatives of exponential, logarithmic, trigonometric, and inverse trigonometric functions. Although it’s possible to compute each derivative using the difference quotient, it will take a long time to compute derivatives during calculus problems if we have to start from scratch with the difference quotient process every time – so it’s advantageous to remember the derivative rules. The derivative rules are to calculus, what the multiplication table is to arithmetic.
We start with the natural logarithm, which has the derivative $(\ln x)’=\frac{1}{x}$. To see where this formula comes from, we can start by writing and simplifying the difference quotient for $\ln x$.
\begin{align*} (\ln x)' &= \lim\limits_{\Delta x \to 0} \frac{\ln (x+\Delta x) − \ln x}{\Delta x} \\ &= \lim\limits_{\Delta x \to 0} \frac{1}{\Delta x} \left( \ln (x+\Delta x) − \ln x \right) \\ &= \lim\limits_{\Delta x \to 0} \frac{1}{\Delta x} \ln \left( \frac{x+\Delta x}{x} \right) \\ &= \lim\limits_{\Delta x \to 0} \frac{1}{\Delta x} \ln \left( 1+\frac{\Delta x}{x} \right) \\ &= \lim\limits_{\Delta x \to 0} \ln \left[ \left( 1+\frac{\Delta x}{x} \right)^\frac{1}{\Delta x} \right] \\ &= \ln \left[ \lim\limits_{\Delta x \to 0} \left( 1+\frac{\Delta x}{x} \right)^\frac{1}{\Delta x} \right] \end{align*}
Does the limit inside the natural log look familiar? Remember that the constant $e$ can be written as the following limit:
\begin{align*} e = \lim\limits_{u\to 0} (1+u)^\frac{1}{u} \end{align*}
If we substitute $u=\frac{\Delta x}{x}$ and simplify/rearrange, then we can come up with an expression for the limit inside the natural log. (The limit as $\frac{\Delta x}{x} \to 0$ can be thought of as $\Delta x \to 0x$, which is the same as $\Delta x \to 0$.)
\begin{align*} e &= \lim\limits_{\frac{\Delta x}{x} \to 0} \left( 1+\frac{\Delta x}{x} \right)^\frac{1}{\frac{\Delta x}{x} } \\ e &= \lim\limits_{\Delta x \to 0} \left( 1+\frac{\Delta x}{x} \right)^\frac{1}{\frac{\Delta x}{x} } \\ e &= \lim\limits_{\Delta x \to 0} \left( 1+\frac{\Delta x}{x} \right)^\frac{x}{\Delta x} \\ e &= \left[ \lim\limits_{\Delta x \to 0} \left( 1+\frac{\Delta x}{x} \right)^\frac{1}{\Delta x} \right]^x \\ e^\frac{1}{x} &= \lim\limits_{\Delta x \to 0} \left( 1+\frac{\Delta x}{x} \right)^\frac{1}{\Delta x} \end{align*}
Substituting this expression into the natural log, we find that $(\ln x)’ = \frac{1}{x}$.
\begin{align*} (\ln x)' &= \ln \left[ \lim\limits_{\Delta x \to 0} \left( 1+\frac{\Delta x}{x} \right)^\frac{1}{\Delta x} \right] \\ &= \ln \left( e^\frac{1}{x} \right) \\ &= \frac{1}{x} \end{align*}
Knowing this, we can use the chain rule to find the derivative of any natural log function.
\begin{align*} \left( \ln ( 2x^2+x ) \right)' = \frac{1}{2x^2+x} (2x^2+x)' = \frac{4x+1}{2x^2+x} \end{align*}
\begin{align*} \left( (\ln x)^2 \right)' = 2(\ln x)(\ln x)' = 2(\ln x) \left( \frac{1}{x} \right) = \frac{2\ln x}{x} \end{align*}
To differentiate a logarithmic function other than the natural logarithm, we can use the change-of-base formula to rewrite the logarithmic function in terms of natural logarithms. For example, to find the derivative of $\log_2 x$, we can convert it into $\frac{\ln x}{\ln 2}$ and then take the derivative.
\begin{align*} \left( \log_2 x \right)' = \left( \frac{\ln x}{\ln 2} \right)' = \frac{1}{\ln 2} \left( \ln x \right)' = \frac{1}{\ln 2} \left( \frac{1}{x} \right) = \frac{1}{x \ln 2} \end{align*}
In general, performing this procedure on any function of the form $\log_a x$ where $a$ is a constant, we find that $(\log_a x)’ = \frac{1}{x \ln a}$.
Next, we cover exponential functions. The exponential function $e^x$ is very elegant in calculus, because its derivative is simply itself, $(e^x)’=e^x$. To see why this is, we can start with the equation $f(x)=e^x$, then take the logarithm and derivative of both sides, and finally solve for $f’(x)$.
\begin{align*} f(x)&=e^x \\ \ln f(x) &= x \\ \left( \ln f(x) \right)' &= (x)' \\ \frac{1}{f(x)} f'(x) &= 1 \\ f'(x) &= f(x) \\ f'(x) &= e^x \end{align*}
Now that we know the derivative of $e^x$, we can use the chain rule to find the derivative of any exponential function.
\begin{align*} \left( e^{x^2} \right)' = e^{x^2} (x^2)' = e^{x^2} (2x) \\ \left( \sqrt{e^x} \right)' = \left( (e^x)^\frac{1}{2} \right)' = \left( e^{\frac{1}{2}x} \right)' = \frac{1}{2}e^{\frac{1}{2}x} \end{align*}
If we want to take the derivative of an exponential function whose base is not $e$, we can rewrite the exponential function so that its base is $e$, and then differentiate using the chain rule. For example, since $2 = e^{\ln 2}$, we see that
\begin{align*} 2^x = \left( e^{\ln 2} \right)^x = e^{(\ln 2)x}. \end{align*}
Now that we have a function which has base $e$, we can use the chain rule to find the derivative.
\begin{align*} \left( 2^x \right)' = \left( e^{(\ln 2)x} \right)' = e^{(\ln 2)x} \ln 2 \end{align*}
Using the fact that $2^x = e^{(\ln 2)x}$, we can simplify the result a bit to look like the original function.
\begin{align*} \left( 2^x \right)' = e^{(\ln 2)x} \ln 2 = 2^x \ln 2 \end{align*}
In general, performing this procedure on any function of the form $a^x$ where $a$ is a constant, we find that $(a^x)’ = a^x \ln a$.
Now, let’s talk about trig functions. Their derivatives are shown below.
\begin{align*} \begin{matrix} (\sin x)' = \cos x & \hspace{.5cm} & (\cos x)' = -\sin x \\ (\tan x)' = \sec^2 x & & (\cot x)' = -\csc^2 x \\ (\sec x)' = \sec x \tan x & & (\csc x)' = -\csc x \cot x \end{matrix} \end{align*}
To see why the derivative of sine is cosine, consider a section of the unit circle, where $y=\sin \theta$. If we increase $\theta$ by an infinitesimally small amount $d\theta$, the additional arc length $d\theta$ matches the hypotenuse of a triangle which has a leg $dy$ adjacent to an angle $\theta$. In this triangle, we have $\cos \theta = \frac{dy}{d\theta} = (\sin \theta)’$.
Furthermore, we can use the derivative of sine in conjunction with the identities $\cos x = \sin \left( \frac{\pi}{2} - x \right)$ and $\sin x = \cos \left( \frac{\pi}{2} - x \right)$ to compute the derivative of cosine.
\begin{align*} \cos x &= \sin \left( \frac{\pi}{2} − x \right) \\ (\cos x)' &= \left[ \sin \left( \frac{\pi}{2} − x \right) \right]' \\ &= \cos \left( \frac{\pi}{2} − x \right) \left( \frac{\pi}{2} − x \right)' \\ &= \cos \left( \frac{\pi}{2} − x \right) (−1) \\ &= (\sin x)(−1) \\ &= −\sin x \end{align*}
The fundamental trig derivatives are $(\sin x)’=\cos x$ and $(\cos x)’ = -\sin x$; all the other trig derivatives come from using them. For example, to see that $(\sec x)’ = \sec x \tan x$, we express $\sec x$ as $\frac{1}{\cos x}$, take the derivative using the chain rule, and simplify.
\begin{align*} (\sec x)' &= \left( \frac{1}{\cos x} \right)' \\ &= −\frac{1}{\cos^2 x} (\cos x)' \\ &= −\frac{1}{\cos^2 x} (\cos x)' \\ &= −\frac{1}{\cos^2 x} (−\sin x) \\ &= \frac{\sin x}{\cos^2 x} \\ &= \frac{1}{\cos x} \cdot \frac{\sin x }{\cos x} \\ &= \sec x \tan x \end{align*}
However, it will take a long time to compute derivatives if we have to start from scratch with the above process every time, so it’s advantageous to remember the table of trig derivatives.
To make it easier to remember the table, think about three key trends in the table: functions have buddies, “co” functions turn negative, and derivatives of functions other than $\sin$ and $\cos$ have two terms.
More precisely, the functions $\sin$ and $\cos$ are buddies because the derivative of $\sin$ contains $\cos$ and the derivative of $\cos$ contains $\sin$. Likewise, $\sec$ and $\tan$ are buddies because the derivative of $\sec$ contains $\tan$ and the derivative of $\tan$ contains $\sec$, and $\csc$ and $\cot$ are buddies because the derivative of $\csc$ contains $\cot$ and the derivative of $\cot$ contains $\csc$.
“Co” functions include $\cos$, $\csc$, and $\cot$, and each of their derivatives has a negative sign, whereas the other functions do not have a negative sign in their derivatives. Lastly, if we think of squared terms as two terms being multiplied together, then $\sin$ and $\cos$ are the only functions whose derivatives consist of a single term. For example, the derivative of $\sec$ is the product of two terms $\sec$ and $\tan$, and the derivative of $\tan$ is $\sec^2$ which can be interpreted as the product of two terms $\sec$ and $\sec$. On the other hand, the derivative of $\sin$ is just a single term, $\cos$.
Just as we did for exponential and logarithmic derivatives, we can use the chain rule to take the derivative of any trig function.
\begin{align*} \left( \sin ( \ln x ) \right)' = \cos \left( \ln x \right) \left( \ln x \right)' = \frac{\cos \left( \ln x \right)}{x} \\ \left( \sec^2 x \right)' = \left( 2\sec x \right) \left( \sec x \right)' = 2\sec^2 x \tan x \end{align*}
Now that we know the derivatives of trig functions, we can use them to find the derivatives of inverse trig functions, which are shown below.
\begin{align*} (\arcsin x)' = \frac{1}{\sqrt{1-x^2}} \\ (\arccos x)' = -\frac{1}{\sqrt{1-x^2}} \\ (\arctan x)' = \frac{1}{1+x^2} \end{align*}
To see where these derivatives come from, we can proceed in the same way as earlier when we used the logarithmic function to find the derivative of the exponential function. We start with the equation $f(x)=\arcsin x$, then take the $\sin$ and derivative of both sides, and finally solve for $f’(x)$.
\begin{align*} f(x) &= \arcsin x \\ \sin f(x) &= x \\ (\sin f(x))' &= (x)' \\ (\cos f(x)) f'(x) &= 1 \\ f'(x) &= \frac{1}{\cos f(x)} \\ f'(x) &= \frac{1}{\cos (\arcsin x)} \end{align*}
To simplify the denominator, we solve for $\cos \theta$ in the identity $\sin^2 \theta + \cos^2 \theta = 1$ with $\theta = \arcsin x$.
\begin{align*} \sin^2 \theta + \cos^2 \theta &= 1 \\ \cos^2 \theta &= 1−\sin^2 \theta \\ \cos \theta &= \pm \sqrt{1−\sin^2 \theta} \end{align*}
We only need to consider the positive root because $\cos$ is always nonnegative on the range of $\arcsin$, which is $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right]$. Substituting $\theta = \arcsin x$, our expression simplifies.
\begin{align*} \cos \left( \arcsin x \right) &= \sqrt{1−\sin^2 \left( \arcsin x \right)} \\ \cos \left( \arcsin x \right) &= \sqrt{1−x^2} \end{align*}
Substituting the above identity in the denominator of our derivative expression, we obtain the final result.
\begin{align*} f'(x) &= \frac{1}{\cos (\arcsin x)} \\ f'(x) &= \frac{1}{\sqrt{1−x^2}} \end{align*}
The rest of the inverse trig derivatives can be computed by the same process. Now, we can use the chain rule to take the derivative of any inverse trig function.
\begin{align*} \left( \arctan ( e^x ) \right)' = \frac{1}{1+(e^x)^2} (e^x)' = \frac{e^x}{1+e^{2x} } \\ \left( (\arcsin x)^2 \right)' = 2(\arcsin x)(\arcsin x)' = \frac{2 \arcsin x}{\sqrt{1−x^2}} \end{align*}
Practice Problems
Compute the derivative of each function. (You can view the solution by clicking on the problem.)
\begin{align*}1) \hspace{.5cm} f(x)= \frac{1}{\ln x} \end{align*}
Solution:
\begin{align*} f'(x)= -\frac{1}{x(\ln x)^2} \end{align*}
\begin{align*}2) \hspace{.5cm} f(x)= e^{1+\tan x} \end{align*}
Solution:
\begin{align*} f'(x)= \sec^2(x)e^{1+\tan x} \end{align*}
\begin{align*}3) \hspace{.5cm} f(x)= 2^{\sin x}\end{align*}
Solution:
\begin{align*} f'(x)= (\ln 2) 2^{\sin x} \cos x \end{align*}
\begin{align*}4) \hspace{.5cm} f(x)= \arctan \left( \frac{1}{x} \right) \end{align*}
Solution:
\begin{align*} f'(x)= -\frac{1}{x^2+1} \end{align*}
\begin{align*}5) \hspace{.5cm} f(x)= \ln \left( \cos x^2 \right) \end{align*}
Solution:
\begin{align*} f'(x)= -2x \tan x^2 \end{align*}
\begin{align*}6) \hspace{.5cm} f(x)= \log_3 \left( \sin 2x \right) \end{align*}
Solution:
\begin{align*} f'(x)= \frac{2}{\ln 3} \cot 2x \end{align*}
\begin{align*}7) \hspace{.5cm} f(x)= \arccos \left( \log_5 x \right) \end{align*}
Solution:
\begin{align*} f'(x)= - \frac{1}{x \sqrt{ (\ln 5)^2 - (\ln x)^2 }} \end{align*}
\begin{align*}8)\hspace{.5cm} f(x) = \sin^2(x) \cos^3(x) \end{align*}
Solution:
\begin{align*} f'(x)= \left( \sin x \cos^2 x \right) \left( 2\cos^2 x - 3 \sin^2 x \right) \end{align*}
\begin{align*}9) \hspace{.5cm} f(x) = \frac{1+e^x}{1-e^x} \end{align*}
Solution:
\begin{align*} f'(x)= \frac{2e^x}{(1-e^x)^2} \end{align*}
\begin{align*}10) \hspace{.5cm} f(x) = \frac{\arcsin x}{\arccos x} \end{align*}
Solution:
\begin{align*} f'(x) = \frac{\arcsin x + \arccos x}{\sqrt{1-x^2} \arccos^2 x} \end{align*}
\begin{align*}11) \hspace{.5cm} f(x) = \arctan \left( \frac{x}{1+e^x} \right) \end{align*}
Solution:
\begin{align*} f'(x) = \frac{1+e^x-xe^x}{1+x^2+2e^x+e^{2x} } \end{align*}
\begin{align*}12) \hspace{.5cm} f(x) = \arccos(x) \arcsin^2(x) \end{align*}
Solution:
\begin{align*} f'(x) = \frac{ \arcsin(x) (2\arccos x- \arcsin x) }{ \sqrt{1-x^2} } \end{align*}
\begin{align*}13) \hspace{.5cm} f(x) = \frac{e^x}{(\ln x)^2} \end{align*}
Solution:
\begin{align*} f'(x) = \frac{ e^x (x \ln x - 2) }{x (\ln x)^3 } \end{align*}
\begin{align*}14) \hspace{.5cm} f(x) = \frac{\ln (\sin x)}{e^x} \end{align*}
Solution:
\begin{align*} f'(x) = \frac{ \cot x - \ln (\sin x) }{ e^x } \end{align*}
Tags:
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# 10th Maths Chapter 2 Numbers and Sequences Ex 2.1
## 10th Maths Chapter 2 Numbers and Sequences Ex 2.1
You can Download 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations
1.Find all positive integers which when divided by 3 leaves remainder 2.
The positive integers when divided by 3 leaves remainder 2.
By Euclid’s division lemma a = bq + r, 0 ≤ r < b.
Here a = 3q + 2, where 0 ≤ q < 3, a leaves remainder 2 when divided by 3.
∴ 2, 5, 8, 11 ……………..
2.A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over?
Here a = 532, b = 21
Using Euclid’s division algorithm
a = bq + r
532 = 21 × 25 + 7
Number of completed rows = 21
Number of flower pots left over = 7
3.Prove that the product of two consecutive positive integers is divisible by 2.
Solution:
Let n – 1 and n be two consecutive positive integers. Then their product is (n – 1)n.
(n – 1)(n) = n² – n.
We know that any positive integer is of the form 2q or 2q + 1 for some integer q. So, following cases arise.
Case I. When n = 2q.
In this case, we have
n² – n = (2q)² – 2q = 4q² – 2q = 2q(2q – 1)
⇒ n² – n = 2r, where r = q(2q – 1)
⇒ n² – n is divisible by 2.
Case II. When n = 2q + 1
In this case, we have
n² – n = (2q + 1)² – (2q + 1)
= (2q + 1)(2q + 1 – 1) = 2q(2q + 1)
⇒ n² – n = 2r, where r = q (2q + 1).
⇒ n² – n is divisible by 2.
Hence, n² – n is divisible by 2 for every positive integer n.
Hence it is Proved
4.When the positive integers a, b and c are divided by 13, the respective remainders are 9,7 and 10. Show that a + b + c is divisible by 13.
Let the positive integer be a, b, and c
We know that by Euclid’s division lemma
a = bq + r
a = 13q + 9 ….(1)
b = 13q + 7 ….(2)
c = 13q + 10 ….(3)
Add (1) (2) and (3)
a + b + c = 13q + 9 + 13q + 7 + 13q + 10
= 39q + 26
a + b + c = 13 (3q + 2)
This expansion will be divisible by 13
∴ a + b + c is divisible by 13
5.Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.
Let x be any integer.
The square of x is x².
Let x be an even integer.
x = 2q + 0
then x² = 4q² + 0
When x be an odd integer
When x = 2k + 1 for some interger k.
x² = (2k + 1 )²
= 4k² + 4k + 1
= 4k (k + 1) + 1
= 4q + 1
where q = k(k + 1) is some integer.
Hence it is proved.
6.Use Euclid’s Division Algorithm to find the Highest Common Factor (H.C.F) of
(i) 340 and 412
To find the HCF of 340 and 412 using Euclid’s division algorithm. We get
412 = 340 × 1 + 72
The remainder 72 ≠ 0
Again applying Euclid’s division algorithm to the division of 340
340 = 72 × 4 + 52
The remainder 52 ≠ 0
Again applying Euclid’s division algorithm to the division 72 and remainder 52 we get
72 = 52 × 1 + 20
The remainder 20 ≠ 0
Again applying Euclid’s division algorithm
52 = 20 × 2 + 12
The remainder 12 ≠ 0
Again applying Euclid’s division algorithm
20 = 12 × 1 + 8
The remainder 8 ≠ 0
Again applying Euclid’s division algorithm
12 = 8 × 1 + 4
The remainder 4 ≠ 0
Again applying Euclid’s division algorithm
8 = 4 × 2 + 0
The remainder is zero
∴ HCF of 340 and 412 is 4
(ii) 867 and 255
To find the HCF of 867 and 255 using
Euclid’s division algorithm. We get
867 = 255 × 3 + 102
The remainder 102 ≠ 0
Using Euclid’s division algorithm
255 = 102 × 2 + 51
The remainder 51 ≠ 0
Again using Euclid’s division algorithm
102 = 51 × 2 + 0
The remainder is zero
∴ HCF = 51
∴ HCF of 867 and 255 is 51
(iii) 10224 and 9648
Find the HCF of 10224 and 9648 using Euclid’s division algorithm. We get
10224 = 9648 × 1 + 576
The remainder 576 ≠ 0
Again using Euclid’s division algorithm
9648 = 576 × 16 + 432
The remainder 432 ≠ 0
Using Euclid’s division algorithm
576 = 432 × 1 + 144
The remainder 144 ≠ 0
Again using Euclid’s division algorithm
432 = 144 × 3 + 0
The remainder is 0
∴ HCF = 144
The HCF of 10224 and 9648 is 144
(iv) 84,90 and 120
Find the HCF of 84, 90 and 120 using Euclid’s division algorithm
90 = 84 × 1 + 6
The remainder 6 ≠ 0
Using Euclid’s division algorithm
4 = 14 × 6 + 0
The remainder is 0
∴ HCF = 6
The HCF of 84 and 90 is 6
Find the HCF of 6 and 120
120 = 6 × 20 + 0
The remainder is 0
∴ HCF of 120 and 6 is 6
∴ HCF of 84, 90 and 120 is 6
7.Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.
Solution:
The required number is the H.C.F. of the numbers.
1230 – 12 = 1218,
1926 – 12 = 1914
First we find the H.C.F. of 1218 & 1914 by Euclid’s division algorithm.
1914 = 1218 × 1 + 696
The remainder 696 ≠ 0.
Again using Euclid’s algorithm
1218 = 696 × 1 + 522
The remainder 522 ≠ 0.
Again using Euclid’s algorithm.
696 = 522 × 1 + 174
The remainder 174 ≠ 0.
Again by Euclid’s algorithm
522 = 174 × 3 + 0
The remainder is zero.
∴ The H.C.F. of 1218 and 1914 is 174.
∴ The required number is 174.
8.If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.
Find the HCF of 32 and 60
60 = 32 × 1 + 28 ….(1)
The remainder 28 ≠ 0
By applying Euclid’s division lemma
32 = 28 × 1 + 4 ….(2)
The remainder 4 ≠ 0
Again by applying Euclid’s division lemma
28 = 4 × 7 + 0….(3)
The remainder is 0
HCF of 32 and 60 is 4
From (2) we get
32 = 28 × 1 + 4
4 = 32 – 28
= 32 – (60 – 32)
4 = 32 – 60 + 32
4 = 32 × 2 -60
4 = 32 x 2 + (-1) 60
When compare with d = 32x + 60 y
x = 2 and y = -1
The value of x = 2 and y = -1
9.A positive integer when divided by 88 gives the remainder 61. What will be the remainder when the same number is divided by 11?
Solution:
Let a (+ve) integer be x.
x = 88 × y + 61
61 = 11 × 5 + 6 (∵ 88 is multiple of 11)
∴ 6 is the remainder. (When the number is divided by 88 giving the remainder 61 and when divided by 11 giving the remainder 6).
10.Prove that two consecutive positive integers are always coprime.
1. Let the consecutive positive integers be x and x + 1.
2. The two number are co – prime both the numbers are divided by 1.
3. If the two terms are x and x + 1 one is odd and the other one is even.
4. HCF of two consecutive number is always 1.
5. Two consecutive positive integer are always coprime.
Fundamental Theorem of Arithmetic
Every composite number can be written uniquely as the product of power of prime is called fundamental theorem of Arithmetic.
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# How do you do the sum 5+10?
mathsworkmusic | Certified Educator
The meaning of '5+10' is 'what number do you get when you add five and ten together'.
When you add two numbers, write the bits of the number under hundreds, tens and units columns. Write one number below the other, with the bits in the right columns. Then add the numbers up in each of the hundreds (H), tens (T), units (U) columns and write the result under the relevant column.
So for '5+10' write
1 0
+ 5
----------------
= 1 5
When we add up the numbers in the units (U) column here, we have zero plus five. Anything plus zero is the same as it started, so zero plus five (which is the same as five plus zero) is 5. There is only one ten in the tens (T) column so the sum in that column is 1.
The answer to '5+10' is 15, that is one ten plus five units.
aishukul | Student
5+10=15. This means that if you add 10 things to the 5 things, how many do you have in all? For example, suppose you have 10 oranges on your table. Then you add 5 more. How many do you have in total? 15.
`10`
`+ 5`
`= 15`
``The numbers align and add up to 15.
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Stat Trek
Teach yourself statistics
Teach yourself statistics
# F Distribution Probability Calculator
The F distribution calculator makes it easy to find the cumulative probability associated with a specified f statistic. Or you can find the f statistic associated with a specified cumulative probability. For help in using the calculator, read the Frequently-Asked Questions or review the Sample Problems.
To learn more about the F distribution, read Stat Trek's tutorial on the F distribution.
• Enter values for degrees of freedom (v1 and v2).
• Enter a value for one, and only one, of the other textboxes.
• Click Calculate to compute a value for the last textbox.
## Frequently-Asked Questions
Instructions: To find the answer to a frequently-asked question, simply click on the question.
### What are degrees of freedom?
Degrees of freedom can be described as the number of scores that are free to vary. For example, suppose your friend tossed three dice, and the total score added up to 12. If your friend told you that he rolled a 3 on the first die and a 5 on the second, then you know that the third die must be a 4 (otherwise, the total would not add up to 12). In this example, 2 die are free to vary while the third is not. Therefore, there are 2 degrees of freedom.
In many situations, the degrees of freedom are equal to the number of observations minus one. Thus, if the sample size were 20, there would be 20 observations; and the degrees of freedom would be 20 minus 1 or 19.
### What are degrees of freedom v1 and v2?
You can use the following equation to compute an f statistic:
f = [ s1212 ] / [ s2222 ]
where σ1 is the standard deviation of population 1, s1 is the standard deviation of the sample drawn from population 1, σ2 is the standard deviation of population 2, and s1 is the standard deviation of the sample drawn from population 2.
The degrees of freedom (v1) refers to the degrees of freedom associated with the sample standard deviation s1 in the numerator; and the degrees of freedom (v2) refers to the degrees of freedom associated with the sample standard deviation s2 in the denominator.
### What is a cumulative probability?
A cumulative probability is a sum of probabilities. The F distribution calculator computes two cumulative probabilities:
• Probability (F≤f). This is the cumulative probability that an F statistic is less than or equal to the value f.
• Probability (F≥f). This is the cumulative probability that an F statistic is greater than or equal to the value f.
### What is an f statistic?
An f statistic (also known as an f value) is a random variable that has an F distribution. Here are the steps required to compute an f statistic:
• Select a random sample of size n1 from a normal population, having a standard deviation equal to σ1.
• Select an independent random sample of size n2 from a normal population, having a standard deviation equal to σ2.
• The f statistic is the ratio of s1212 and s2222. Thus, f = [ s1212 ] / [ s2222]
### What is a probability?
A probability is a number expressing the chances that a specific event will occur. This number can take on any value from 0 to 1. A probability of 0 means that there is zero chance that the event will occur; a probability of 1 means that the event is certain to occur. Numbers between 0 and 1 quantify the uncertainty associated with the event.
For example, the probability of a coin flip resulting in Heads (rather than Tails) would be 0.50. Fifty percent of the time, the coin flip would result in Heads; and fifty percent of the time, it would result in Tails.
## Sample Problems
1. Suppose we take independent random samples of size n1 = 11 and n2 = 16 from normal populations. Find the f statistic that has the following cumulative probability: P(F≤f)=0.75.
Solution:
We know the following:
• Since the sample size n1 = 11, the degrees of freedom v1 = n1 - 1 = 10.
• And since the sample size n2 = 16, the degrees of freedom v2 = n2 - 1 = 15.
• P(F≤f)=0.75.
Now, we are ready to use the F Distribution Calculator. We enter the degrees of freedom for v1(10), the degrees of freedom for v2 (15), and the cumulative probability (0.75) into the calculator; and hit the Calculate button.
The calculator reports that the f statistic is 1.44907.
1. Suppose we take independent random samples of size n1 = 25 and n2 = 13 from normal populations. What is the probability that the f statistic is less than or equal to 2.51?
Solution:
We know the following:
• Since the sample size n1 = 25, the degrees of freedom v1 = n1 - 1 = 24.
• And since the sample size n2 = 13, the degrees of freedom v2 = n2 - 1 = 12.
• The f statistic of interest is equal to 2.51.
Now, we are ready to use the F Distribution Calculator. We enter the degrees of freedom for v1 (24), the degrees of freedom for v2 (12), and the f statistic (2.51) into the calculator; and hit the Calculate button.
The calculator reports that the probability that f is less than or equal to 2.51 is 0.95032.
## Problem?
Oops! Something went wrong.
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# Excercise 3.3 Data Handling - NCERT Solutions Class 7
Go back to 'Data Handling'
## Chapter 3 Ex.3.3 Question 1
Use the bar graph (see below figure) to answers the following questions.
(i) Which is the most popular pet?
(ii) How many students have dog as a pet?
### Solution
Steps:
From the given bar graph in figure, we have
(i) Cats are the most popular pet among the students.
(ii) 8 students have dog as a pet animal.
## Chapter 3 Ex.3.3 Question 2
Use the bar graph (see below figure) which shows the no. of books sold by a bookstore during $$5$$ consecutive years and answer the following questions:
(i) About how many books were sold in $$1989, 1990, \text &\,\, 1992 ?$$
(ii) In which year was about $$475$$ books sold? About $$225$$ books sold?
(iii) In which years were fewer than $$250$$ books sold?
(iv) Can you explain how you would estimate the number of books sold in $$1989?$$
### Solution
What is known?
Number of books sold during five consecutive years.
Reasoning:
We can answer all the question by observing the bar graph
Steps:
From the given bar graph, we have
(i) In $$1989 ,180$$ books were sold.
• In $$1990, 475$$ books were sold.
• In $$1992, 225$$ books were sold.
(ii) In $$1990$$ about $$475$$ books were sold and in $$1992$$ about $$225$$ books were sold.
• In $$1989$$ and $$1992$$ fewer than $$250$$ books were sold.
• From the graph we can conclude that $$180$$ books were sold in $$1989.$$
## Chapter 3 Ex.3.3 Question 3
Number of children in six different classes is given below. Represent the data on a bar graph.
Class Fifth Sixth Seventh Eighth Ninth Tenth Number of Children $$135$$ $$120$$ $$95$$ $$100$$ $$90$$ $$80$$
1. How would you choose a scale?
(i) Which class has the maximum number of children? And minimum?
(ii) Find the ratio of students of class sixth to the students of class eight?
### Solution
What is known?
Number of children in different classes.
Reasoning:
We have to draw the graph by using the data.
Steps:
a) Scale on y-axis is $$1$$ unit $$= 10$$ children
b)
(i)
• Fifth class has maximum number of children i.e. $$135$$
• Tenth class has the minimum number of children i.e. $$80$$
(ii)
Number of students in class sixth $$=$$ $$120$$
Number of students in class eighth $$=$$ $$100$$
\begin{align} \text{Ratio }&=\frac{\left( \begin{align} & \text{Number of students} \\ & \text{in class sixth} \\ \end{align} \right)}{\left( \begin{align} & \text{Number of students} \\ & \text{in class eighth} \\ \end{align} \right)} \\ & =\frac{120}{100} \\ & =\frac{6}{5} \\ & =6:5 \end{align}
## Chapter 3 Ex.3.3 Question 4
The performance of a student in $$1^{st}$$ term and $$2^{nd}$$ term is given. Draw a double bar graph choosing appropriate scale and answer the following:
Subject English Hindi Maths Science S. Science $$1^{st}$$ Term (M.M. $$100$$) $$67$$ $$72$$ $$88$$ $$81$$ $$73$$ $$2^{nd}$$ Term (M.M. $$100$$) $$70$$ $$65$$ $$95$$ $$85$$ $$75$$
(i) In which subject, has the child improved his performance the most?
(ii) In which subject improvement is the least?
(iii) Has the performance gone down in any subject?
### Solution
What is known?
Marks obtained in different subject
Reasoning:
We have to draw the graph by using the data
Steps:
Difference between the marks of $$1^{st}$$ and $$2^{nd}$$ term
English $$= 70 \, – 67 = 3$$
Hindi $$= 65 \,– 72 = -7$$ (Decrease in marks)
Math $$= 95\, – 88 = 7$$
Science $$= 85\, – 81= 4$$
Social Science $$= 75\, – 73 =2$$
(i) In Math, the performance of the students improved.
(ii) In Social .science, the performance of the students improved is the least.
(iii)Yes, in Hindi the performance of the students has gone down.
## Chapter 3 Ex.3.3 Question 5
Consider this data collected from a survey of a colony.
Favourite Sport Cricket Basketball Swimming Hockey Athletics Watching $$1240$$ $$470$$ $$510$$ $$430$$ $$250$$ Participating $$620$$ $$320$$ $$320$$ $$250$$ $$105$$
(i) Draw a double bar graph choosing an appropriate scale.
What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in the sports?
### Solution
Reasoning:
We have to draw the graph by using the data
Steps:
(i) This bar graph shows the number of persons who are watching and participating in their favorite sports.
(ii) Cricket is the most popular sport.
(iii) Watching different sports is more preferred than participating in the sports.
## Chapter 3 Ex.3.3 Question 6
Take the data giving the maximum and minimum temperatures of various cities given in the beginning of this chapter (in the table below). Draw a double bar graph using the data and answer the following.
(i) Which city has the highest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and maximum temperature?
### Solution
Steps:
Temperature of cities as on $$20.06.2006$$
City Maximum Minimum Ahmedabad $$38^\circ \rm C$$ $$29 ^\circ \rm C$$ Amritsar $$37 ^\circ \rm C$$ $$26 ^\circ \rm C$$ Bangalore $$28 ^\circ \rm C$$ $$21 ^\circ \rm C$$ Chennai $$36 ^\circ \rm C$$ $$27 ^\circ \rm C$$ Delhi $$38^\circ \rm C$$ $$28 ^\circ \rm C$$ Jaipur $$39 ^\circ \rm C$$ $$29 ^\circ \rm C$$ Jammu $$41 ^\circ \rm C$$ $$26 ^\circ \rm C$$ Mumbai $$32 ^\circ \rm C$$ $$27 ^\circ \rm C$$
Difference between the maximum and minimum temperature on $$20.06.2006$$
Ahmedabad $$= (38^\circ \rm C - 29^\circ \rm C) = 9^\circ \rm C$$
Amritsar $$=$$ ($$37 ^\circ \rm C$$ $$-$$ $$26^\circ \rm C$$) $$=$$ $$9 ^\circ \rm C$$
Bangalore $$=$$ ($$28^\circ \rm C$$ $$-$$ $$21^\circ \rm C$$) $$=$$ $$7 ^\circ \rm C$$
Chennai $$=$$ ($$36^\circ \rm C$$ $$-$$ $$27^\circ \rm C$$) $$=$$ $$9 ^\circ \rm C$$
Delhi $$=$$ ($$38 ^\circ \rm C$$ $$-$$ $$28^\circ \rm C$$) $$=$$ $$10 ^\circ \rm C$$
Jaipur $$=$$ ($$39 ^\circ \rm C$$ $$-$$ $$29 ^\circ \rm C$$) $$=$$ $$10 ^\circ \rm C$$
Jammu $$=$$ ($$41 ^\circ \rm C$$ $$-$$ $$26^\circ \rm C$$) $$=$$ $$15 ^\circ \rm C$$
Mumbai $$=$$ ($$32 ^\circ \rm C$$ $$-$$ $$27^\circ \rm C$$) $$=$$ $$5^\circ \rm C$$
(i) As, it is clear from the above calculation, Jammu has the largest difference in the minimum and maximum temperature on the given data i.e. $$=$$ $$15 ^\circ \rm C$$.
(ii) Jammu is the hottest city with $$41^\circ \rm C$$ and Bangalore is the coldest city with $$21^\circ \rm C$$.
Bangalore and Jaipur or Bangalore and Ahmedabad are the two cities where maximum temperature of one was less than the minimum temperature of other.
Mumbai has the least difference between its minimum and maximum temperature.
Maximum temperature $$= 32^\circ \rm C$$
Minimum temperature $$=$$ $$27^\circ \rm C$$
∴ Difference $$=$$ ($$32 ^\circ \rm C$$ $$-$$ $$27^\circ \rm c$$) $$=$$ $$5^\circ \rm C$$
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# Math Worksheets Land
Math Worksheets For All Ages
# Math Worksheets Land
Math Worksheets For All Ages
Home > Topics >
# Subtraction Worksheets
What is Subtraction? We all know that mastering addition is easier than grasping the concept of subtraction. Children can quickly put two things together to obtain a sum but they struggle with taking away one thing from another. Finding differences between values forms the foundation of many advanced mathematical concepts. Therefore, before they hop on complex concepts, it is necessary that they have a firm grip on finding a basic difference. In mathematics, subtraction is defined as taking away one thing from another. After we process this operation, what remains in the group is less than what was present before the calculation was processed. For example, If you take 3 away for 5 you will be left with 2. To express it mathematically, 6 - 4 =2 A simple subtraction problem usually has three parts. The part from which something is being taken away is known as minuend. The part which is being taken away is known as a subtrahend. The part that you are left with is known as difference. For example, 6 - 4 = 2, the number 6 is the minuend, 4 is the subtrahend and the number 2 is the difference. The worksheets and lessons that you will find here are found mostly mixed in with addition operations in our basic math skills sheets. Those operations are direct opposites of one another and it is always good to make that concept concrete for students.
### Tips For Teaching This Skill
Before we get going on this topic let me just remind you of the parts of subtraction. It will be helpful as you read further. Problems follow the model of:
Minuend - Subtrahend = Difference
We normally approach this basic concept with students in first grade. Collective our staff has taught first grade for over 100 years. So we all got together and discussed the best practices for teaching this skill to students at that age. Here is what we find works:
Always Start with Visuals - This is the immediate point that all of us agreed on. You should always approach this skill as a takeaway method. This is where we give students a problem and model it with objects. We do find that students do better when the numbers of the problem are found below the objects. This probably because it is easier to visually isolate the numbers when they are beneath objects rather than above them. Here is an example of a problem: 4 - 2. We will find a fun object that youngsters will be engaged with. We will use cute toy cows in our example. We teach them that your minuend (4) is the number of baby cows you would start with. Your subtrahend (2) is the number cows that you need to take away. We encourage you to use red slashes to serve as the cross out. We also encourage you to always have them cross out the objects that are most left. We find that this is much easier for them to track long term. So the problem 4-2 would be modelled as:
We would then teach them determine the difference by simply counting the number of left-over objects. Some of our teachers prefer starting by using manipulatives and others like visuals such as pictures. Everyone agrees that you will need to use both manipulatives and pictures before you move to anything else. A great manipulative to consider is having students model problems by using their fingers. Remind yourself that you only go as high as 5, but you can put partners together and go to 10. Here is that same problem modelled with fingers.
Number Line - Once students have the basic concept through manipulatives and images the natural progression is to use a number line and teach them to count backwards. The minuend is our starting point. We then count backwards the number of times of the subtrahend.
This really helps them work on the concept that all this skill is composed of is counting in reverse. Once they have this foundation it is pretty easy to build off. You will then progress on to paper and pencil problems. We also encourage students to either by or make math facts cards and practice 5 to 10 minutes everyday until they are memorized. Even then students will need to still look at them once a week to keep it fresh in their memory.
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# IIM CAT Preparation Tips
IIM CAT Preparation Tips: January 2014
## Jan 28, 2014
### Inequalities - Interesting Question
Question
How many positive integer values can x take that satisfy the inequality (x - 8) (x - 10) (x - 12).......(x - 100) < 0?
Explanation
Let us try out a few values to see if that gives us anything.
When x = 8, 10, 12, ....100 this goes to zero. So, these cannot be counted.
When x = 101, 102 or beyond, all the terms are positive, so the product will be positive.
So, straight-away we are down to numbers 1, 2, 3, ...7 and then odd numbers from there to 99.
Let us substitute x =1,
All the individual terms are negative. There are totally 47 terms in this list (How? Figure that out). Product of 47 negative terms will be negative. So, x = 1 works. So, will x =2, 3, 4, 5, 6, and 7.
Remember, product of an odd number of negative terms is negative; product of even number of negative terms is positive. Now, this idea sets up the rest of the question.
When x = 9, there is one positive terms and 46 negative terms. So, the product will be positive.
When x = 11, there are two positive terms and 45 negative terms. So, the product will be negative.
When x = 13, there are three positive terms and 44 negative terms. So, the product will be positive.
and so on.
Essentially, alternate odd numbers need to be counted, starting from 11.
So, the numbers that will work for this inequality are 1, 2, 3, 4, 5, 6, 7...and then 11, 15, 19, 23, 27, 31,..... and so.
What will be the last term on this list?
99, because when x = 99, there are 46 positive terms and 1 negative term.
So, we need to figure out how many terms are there in the list 11, 15, 19,....99. These can be written as
4 * 2 + 3,
4 * 3 + 3,
4 * 4 + 3
4 * 5 + 3
4 * 6 + 3
...
4 * 24 + 3
A set of 23 terms. So, total number of values = 23 + 7 = 30. 30 positive integer values of x exist satisfying the condition.
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# Difference Between Natural and Whole numbers
The key difference between natural number and whole number is the presence of value ‘0’ in whole numbers. A number is an arithmetic value which is used to represent a particular quantity. In Mathematics, the different types of numbers are as follows:
• Integers
• Real Numbers
• Natural Numbers
• Rational Numbers
• Whole numbers
• Irrational Numbers
• Complex Numbers
The different types of numbers are classified based on their characteristic properties. In this article, let us discuss the difference between natural and whole numbers with examples in detail.
## Definition of Natural and Whole Numbers
Natural Numbers: The numbers 1, 2, . . . . , N all are called Natural Numbers. The negative numbers and 0 are not counted as the natural numbers because 1 is considered as the smallest natural number. The examples of natural numbers are 34, 22, 2, 81, 134, 15
Whole Numbers: The numbers 0, 1, 2, . . . . . , N all are called Whole Numbers, i.e. if 0 is included in natural numbers, then it is known as Whole Numbers. Whole numbers are located at the right side of the number line. Zero is the smallest whole number. Also, negative numbers are not whole numbers. For example = 0, 13, 22, 290, 19, 34, 95 all are whole numbers.
## What is the Difference Between Natural and Whole Numbers?
The major difference between natural and whole numbers is that natural number doesn’t include 0 whereas whole numbers include 0 too. Let us discuss the other differences also.
Difference Between Natural and Whole Numbers S.No Natural Numbers Whole Numbers 1 Natural numbers are defined as the basic counting numbers. The natural number in set notation is {1, 2, 3, 4, 5, …..} Whole numbers are defined as the set of natural numbers, and it started with zero. The whole numbers in set notation is {0,1, 2, 3, 4, 5, ….} 2 Natural Numbers are represented using the letter “N” Whole Numbers are represented using the letter “W”. 3 Counting number starts from 1 Counting number starts from 0 4 We can say that all the natural numbers are considered as whole numbers We can say that all the whole numbers are not considered as natural numbers
## Solved Examples
Example 1: Find whole numbers from the given numbers. 13, 0, 5, 89, 4, -4, -3, 23
Sol: Whole Numbers = 13, 0, 5, 89, 4, 23
Example 2: Find whole numbers from 4, 14/5, 0, (0.97), 2, -6, -17, 21
Sol: Whole Numbers = 4, 0, 2, 21.
Example 3: From the given series of numbers, find natural numbers. 41, (0.4), 5, 90, -91, 0, -7.
Sol: Natural Numbers = 41, 5, 90.
Example 4: From the given series of numbers, find natural numbers. 4, 4/15, 0, (0.87), 2, -7, -17, 21.
Sol: Natural Numbers = 4, 2, 21.
## Frequently Asked Questions – FAQs
### Are whole numbers and natural numbers same?
Whole numbers include all the natural numbers along with zero. It starts from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and so on. But all whole numbers are not natural numbers.
### What value is not common between natural and whole numbers?
The only difference between natural and whole numbers is the value of 0. The whole number includes 0, whereas natural numbers do not.
### Is 0 a natural number?
Zero (0) is not a natural number but a whole number. Natural numbers start from 1, 2, 3, 4, 5,….and so on.
### Are integers whole numbers or natural numbers?
Integers are the numbers which are positive, negative and zero. It includes all the whole numbers, but it cannot be a fraction.
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PRIMITIVE, DEFINITE INTEGRAL
Def. Primitive of a function. Given a function f(x), the primitive of f(x) is a function F(x) which has f(x) as its derivative i.e. it is a function F(x) such that F'(x) = f(x). If F(x) is a primitive of f(x) then F(x) + c is also a primitive of f(x), where c is any constant. Thus if a function f(x) has one primitive F(x), then along with this one it has an entire family of primitives, infinite in number, of the form F(x) + c. And this family of primitives exhausts the whole set of primitives for f(x) — there are no others.
Syn. Anti-derivative, indefinite integral.
Example. Find the primitive of the function 8x3.
Solution. We ask ourselves the question: What function has as its derivative 8x3? To answer the question we think in terms of our memorized differentiation rules and formulas and work backward, trying to figure out what function, when differentiated, would give 8x3. The answer is 2x4 + c where c is any constant.
Definite integral. The definite integral is the most basic, fundamental concept of integral calculus.
Consider the function y = f(x) shown in Fig. 1. We wish to find the area A bounded by the curve y = f(x), the x-axis and the lines x = a and x = b. To find this area we proceed as follows: We divide the interval [a, b] into n sub-intervals, not necessarily of equal length. We denote the length of the first subinterval by Δx1, of the second by Δx2, and so forth up to the final subinterval Δxn. In each subinterval Δxi we arbitrarily choose a point ξi and set up the sum
1) An = f(ξ1)Δx1 + f(ξ2)Δx2 + ... + f(ξn)Δxn .
An is thus equal to the sum of the areas of the rectangles shown in Fig. 1 and represents an approximation to the area A that we wish to compute. We now define the area A to be the following sum:
or
in which we require that the largest of the sub-intervals, max Δxi, approach zero as the number of sub-intervals n → ∞. We thus make an infinitely fine net on the interval [a, b] in which the largest sub-interval is required to approach zero.
We have assumed that f(x) 0. If f(x) changes sign, then the limit 2) will give us the algebraic sum of the areas of the segments lying between the curve y = f(x) and the x-axis, where the segments above the x-axis are taken with a plus sign and those below with a minus sign. See Fig. 2.
The need to calculate the limit 2) arises in many problems. For example, suppose that a point is moving along a straight line with variable speed s = s(t). How do we determine the distance d covered by the point in the time from t = a to t = b?
Let us assume that the function s(t) is continuous. Let us divide the interval [a, b] up into n sub-intervals, of length Δt1, Δt2,... , Δtn. To calculate the approximate value for the distance covered in each subinterval Δti, let us suppose that the speed during the period of time is constant i.e. it is equal throughout the subinterval to its actual value at some intermediate point ξ1. The whole distance covered will then be expressed approximately by the sum
and the exact value of the distance d covered in the time from a to b, will be the limit of such sums for finer and finer subdivisions; that is, it will be the limit
The example just given is only one of many that could be given. Many practical problems lead to the calculation of such a limit.
The limit 2) above is called the definite integral of the function f(x) taken over the interval [a, b].
Note. There is a natural question that a reader might ask. Why such a complicated definition? Why not just define the sub-intervals to be equal with the ξ’s placed at the midpoints? I can’t give an answer to that. The definition we have given appears to be the standard, universally used definition.
Def. Definite (Riemann) integral. The definite integral of a function f(x) taken over the interval [a, b] is denoted by
and is defined as
where f(x)dx is called the integrand and a and b are the limits of integration; a is the lower limit, b is the upper limit.
Note the source of the notation for an integral by comparing the left and right members of 3) above: dx corresponds to Δx, f(x) corresponds to f(ξi), and the integral sign ∫ represents the letter S, meaning “sum”.
The intuitive interpretation of a definite integral
is that of the area bounded by the function f(x), the x-axis and the lines x = a and a = b.
The connection between differential and integral calculus. One of the fundamental contributions of Newton and Leibnitz was that they clarified the profound connection that exists between differential and integral calculus, a connection which provides us with a general method of calculating definite integrals for an extremely wide class of functions.
To explain this connection let us give the following example from mechanics: Let us suppose that a point is moving along a straight line with a speed s = s(t), where t is time. We have already shown that the distance Δd covered by our point in the time between t = t1 and t = t2 is given by the definite integral
Now let us assume that we know the function d = d(t) giving the distance traveled as a function of time as calculated from some initial point A on the straight line. The distance covered in the interval of time [t1, t2] is then given by
5) Δd = d(t2) - d(t1)
Thus, from 4) and 5), we have
Now the function d(t) is a primitive of s(t) i.e.
Thus 6) provides us with the connection between differential and integral calculus. The relationship 6) tells us that we can compute the definite integral
by finding its primitive d(t), its value being given by d(t2) - d(t1).
From this example from mechanics we are led to the well-known formula of Newton and Leibnitz, the so-called Fundamental Theorem of Integral Calculus:
Fundamental Theorem of Integral Calculus. If f(x) is continuous and F(x) is any arbitrary primitive for f(x) i.e. any function such that F'(x) = f(x), then
This is often written as
where
is a shorthand notation for
F(b) - F(a).
Thus, by this theorem, the problem of calculating the definite integral of a function is reduced to the problem of finding a primitive of the function.
Area function, A(x). The definite integral
represents the area under the curve f(x) between lines x = a and x = b where a and b are regarded as fixed numbers. The integral represents some number, namely the measure of that area. Now let us consider the upper limit b of the integral to be variable. Let us replace b with x. The integral then becomes
and represents a variable area A(x) extending from x = a to a variable point x. In other words,
where A(x) is the area shown in Fig. 3. This area function A(x) is a primitive of the function f(x) i.e. dA/dx = f(x). That this is true is quite obvious. It is quite easy to see that
dA = f(x)dx
and consequently that dA/dx = f(x). A small increase in x of Δx will produce an increase ΔA in
area of approximately f(x)Δx. See Fig. 4. Thus
ΔA f(x)Δx
which we can rewrite as
and
Now let F(x) represent some particular primitive in the family of primitives of f(x). Then, since A(x) is also a primitive,
10) A(x) = F(x) + c
where c is a constant. We can evaluate c by substituting in known values: the value of A(x) corresponding to the initial point x = a is zero. Hence, substituting into 10),
0 = F(a) + c
or
c = - F(a) .
Replacing c by - F(a) in 10) we have the expression for the area from x = a to any other point x :
11) A(x) = F(x) - F(a)
or
Finally, the area under the curve from x = a to x = b is found by replacing x by b so
or
A = F(b) - F(a) .
We see that 12) is the same equation as 7) above and we have derived the Fundamental Theorem of Integral Calculus.
The area function A(x) of some function f(x) is conceived of as the measure of an area generated by an ordinate of variable length which starts at x = a and moves to the right, its upper end always on the curve. The area generated when the moving ordinate has reached any point x is a quantity which depends on x; i.e. it is a function of x.
Problem. Given an arbitrary function f(x), one not necessarily expressed in terms of formulas or analytically, perhaps one in tabular or graphical form. Does a primitive for it exist? If so, exhibit it.
Solution. The area function A(x) representing the area under the curve extending from an arbitrary starting point “a” up to the value x constitutes a primitive for it. It can indeed be proved that if a function f(x) is continuous (and even if it is discontinuous, but Lebesgue-summable) a primitive F(x) satisfying 7) above exists.
Properties of definite integrals. If f(x) and g(x) are continuous on the interval of integration a x b the following hold:
6. The First Mean Value Theorem:
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# Common Core: 8th Grade Math : Understand that the Solution of a System of Two Linear Equations is the Intersection of their Lines: CCSS.Math.Content.8.EE.C.8a
## Example Questions
← Previous 1
### Example Question #1 : Understand That The Solution Of A System Of Two Linear Equations Is The Intersection Of Their Lines: Ccss.Math.Content.8.Ee.C.8a
Identify the point of intersection by solving for the solution of the system of equations in the provided figure.
Explanation:
The graph displays a system of two linear equations. The point where these two lines intersect is the solution to the system of the equations because that coordinate point is the point that both lines have in common, or pass through.
In this case, the solution to the two linear equations that are displayed in the graph is the following point:
### Example Question #1 : Understand That The Solution Of A System Of Two Linear Equations Is The Intersection Of Their Lines: Ccss.Math.Content.8.Ee.C.8a
Identify the point of intersection by solving for the solution of the system of equations in the provided figure.
Explanation:
The graph displays a system of two linear equations. The point where these two lines intersect is the solution to the system of the equations because that coordinate point is the point that both lines have in common, or pass through.
In this case, the solution to the two linear equations that are displayed in the graph is the following point:
### Example Question #3 : Understand That The Solution Of A System Of Two Linear Equations Is The Intersection Of Their Lines: Ccss.Math.Content.8.Ee.C.8a
Identify the point of intersection by solving for the solution of the system of equations in the provided figure.
Explanation:
The graph displays a system of two linear equations. The point where these two lines intersect is the solution to the system of the equations because that coordinate point is the point that both lines have in common, or pass through.
In this case, the solution to the two linear equations that are displayed in the graph is the following point:
### Example Question #2 : Understand That The Solution Of A System Of Two Linear Equations Is The Intersection Of Their Lines: Ccss.Math.Content.8.Ee.C.8a
Identify the point of intersection by solving for the solution of the system of equations in the provided figure.
Explanation:
The graph displays a system of two linear equations. The point where these two lines intersect is the solution to the system of the equations because that coordinate point is the point that both lines have in common, or pass through.
In this case, the solution to the two linear equations that are displayed in the graph is the following point:
### Example Question #5 : Understand That The Solution Of A System Of Two Linear Equations Is The Intersection Of Their Lines: Ccss.Math.Content.8.Ee.C.8a
Identify the point of intersection by solving for the solution of the system of equations in the provided figure.
Explanation:
The graph displays a system of two linear equations. The point where these two lines intersect is the solution to the system of the equations because that coordinate point is the point that both lines have in common, or pass through.
In this case, the solution to the two linear equations that are displayed in the graph is the following point:
### Example Question #6 : Understand That The Solution Of A System Of Two Linear Equations Is The Intersection Of Their Lines: Ccss.Math.Content.8.Ee.C.8a
Identify the point of intersection by solving for the solution of the system of equations in the provided figure.
Explanation:
The graph displays a system of two linear equations. The point where these two lines intersect is the solution to the system of the equations because that coordinate point is the point that both lines have in common, or pass through.
In this case, the solution to the two linear equations that are displayed in the graph is the following point:
### Example Question #7 : Understand That The Solution Of A System Of Two Linear Equations Is The Intersection Of Their Lines: Ccss.Math.Content.8.Ee.C.8a
Identify the point of intersection by solving for the solution of the system of equations in the provided figure.
Explanation:
The graph displays a system of two linear equations. The point where these two lines intersect is the solution to the system of the equations because that coordinate point is the point that both lines have in common, or pass through.
In this case, the solution to the two linear equations that are displayed in the graph is the following point:
### Example Question #8 : Understand That The Solution Of A System Of Two Linear Equations Is The Intersection Of Their Lines: Ccss.Math.Content.8.Ee.C.8a
Identify the point of intersection by solving for the solution of the system of equations in the provided figure.
Explanation:
The graph displays a system of two linear equations. The point where these two lines intersect is the solution to the system of the equations because that coordinate point is the point that both lines have in common, or pass through.
In this case, the solution to the two linear equations that are displayed in the graph is the following point:
### Example Question #3 : Understand That The Solution Of A System Of Two Linear Equations Is The Intersection Of Their Lines: Ccss.Math.Content.8.Ee.C.8a
Identify the point of intersection by solving for the solution of the system of equations in the provided figure.
Explanation:
The graph displays a system of two linear equations. The point where these two lines intersect is the solution to the system of the equations because that coordinate point is the point that both lines have in common, or pass through.
In this case, the solution to the two linear equations that are displayed in the graph is the following point:
### Example Question #10 : Understand That The Solution Of A System Of Two Linear Equations Is The Intersection Of Their Lines: Ccss.Math.Content.8.Ee.C.8a
Identify the point of intersection by solving for the solution of the system of equations in the provided figure.
|
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# 5.4: Dividing Polynomials
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Learning Objectives
By the end of this section, you will be able to:
• Dividing monomials
• Dividing a polynomial by a monomial
• Dividing polynomials using long division
• Dividing polynomials using synthetic division
• Dividing polynomial functions
• Use the remainder and factor theorems
Before you get started, take this readiness quiz.
1. Add: $$\frac{3}{d}+\frac{x}{d}$$.
If you missed this problem, review [link].
2. Simplify: $$\frac{30xy}{35xy}$$.
If you missed this problem, review [link].
3. Combine like terms: $$8a^2+12a+1+3a^2−5a+4$$.
If you missed this problem, review [link].
## Dividing Monomials
We are now familiar with all the properties of exponents and used them to multiply polynomials. Next, we’ll use these properties to divide monomials and polynomials.
Example $$\PageIndex{1}$$
Find the quotient: $$54a^2b^3÷ (−6ab^5)$$.
When we divide monomials with more than one variable, we write one fraction for each variable.
$$\begin{array} {ll} {} &{54a^2b^3÷(−6ab^5)} \\ {\text{Rewrite as a fraction.}} &{\frac{54a^2b^3}{−6ab^5}} \\ {\text{Use fraction multiplication.}} &{\frac{54}{−6}·\frac{a^2}{a}·\frac{b^3}{b^5}} \\ {\text{Simplify and use the Quotient Property.}} &{−9·a·\frac{1}{b^2}} \\ {\text{Multiply.}} &{−\frac{9a}{b^2}} \\ \end{array}$$
Example $$\PageIndex{2}$$
Find the quotient: $$−72a^7b^3÷(8a^{12}b^4)$$.
$$−\frac{9}{a^5b}$$
Example $$\PageIndex{3}$$
Find the quotient: $$−63c^8d^3÷(7c^{12}d^2)$$.
$$\frac{−9d}{c^4}$$
Once you become familiar with the process and have practiced it step by step several times, you may be able to simplify a fraction in one step.
Example $$\PageIndex{4}$$
Find the quotient: $$\frac{14x^7y^{12}}{21x^{11}y^6}$$.
Be very careful to simplify $$\frac{14}{21}$$ by dividing out a common factor, and to simplify the variables by subtracting their exponents.
$$\begin{array} {ll} {} &{\frac{14x^7y^{12}}{21x^{11}y^6}} \\ {\text{Simplify and use the Quotient Property.}} &{\frac{2y^6}{3x^4}} \\ \end{array}$$
Example $$\PageIndex{5}$$
Find the quotient: $$\frac{28x^5y^{14}}{49x^9y^{12}}$$.
$$\frac{4y^2}{7x^4}$$
Example $$\PageIndex{6}$$
Find the quotient: $$\frac{30m^5n^{11}}{48m^{10}n^{14}}$$.
$$\frac{5}{8m^5n^3}$$
## Divide a Polynomial by a Monomial
Now that we know how to divide a monomial by a monomial, the next procedure is to divide a polynomial of two or more terms by a monomial. The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. So we’ll start with an example to review fraction addition. The sum $$\frac{y}{5}+\frac{2}{5}$$ simplifies to $$\frac{y+2}{5}$$. Now we will do this in reverse to split a single fraction into separate fractions. For example, $$\frac{y+2}{5}$$ can be written $$\frac{y}{5}+\frac{2}{5}$$.
This is the “reverse” of fraction addition and it states that if a, b, and c are numbers where $$c\neq 0$$, then $$\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$$. We will use this to divide polynomials by monomials.
definition: DIVISION OF A POLYNOMIAL BY A MONOMIAL
To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.
Example $$\PageIndex{7}$$
Find the quotient: $$(18x^3y−36xy^2)÷(−3xy)$$.
$$\begin{array} {ll} {} &{(18x^3y−36xy^2)÷(−3xy)} \\ {\text{Rewrite as a fraction.}} &{\frac{18x^3y−36xy^2}{−3xy}} \\ {\text{Divide each term by the divisor. Be careful with the signs!}} &{\frac{18x^3y}{−3xy}−\frac{36xy^2}{−3xy}} \\ {\text{Simplify.}} &{−6x^2+12y} \\ \end{array}$$
Example $$\PageIndex{8}$$
Find the quotient: $$(32a^2b−16ab^2)÷(−8ab)$$.
$$−4a+2b$$
Example $$\PageIndex{9}$$
Find the quotient: $$(−48a^8b^4−36a^6b^5)÷(−6a^3b^3)$$.
$$8a^5b+6a^3b^2$$
## Divide Polynomials Using Long Division
Divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully the steps we take when we divide a 3-digit number, 875, by a 2-digit number, 25.
We check division by multiplying the quotient by the divisor. If we did the division correctly, the product should equal the dividend.
$\begin{array} {l} {35·25} \\ {875\checkmark} \\ \nonumber \end{array}$
Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.
Example $$\PageIndex{10}$$
Find the quotient: $$(x^2+9x+20)÷(x+5)$$.
Write it as a long division problem. Be sure the dividend is in standard form. Divide $$x^2$$ by $$x$$. It may help to ask yourself, “What do I need to multiply $$x$$ by to get $$x^2$$?” Put the answer, $$x$$, in the quotient over the $$x$$ term. Multiply $$x$$ times $$x+5$$. Line up the like terms under the dividend. Subtract $$x^2+5x$$ from $$x^2+9x$$. You may find it easier to change the signs and then add. Then bring down the last term, 20. Divide $$4x$$ by $$x$$. It may help to ask yourself, “What do I need to multiply xx by to get $$4x$$?” Put the answer, $$4$$, in the quotient over the constant term. Multiply 4 times $$x+5$$. Subtract $$4x+20$$ from $$4x+20$$. Check: $$\begin{array} {ll} {\text{Multiply the quotient by the divisor.}} &{(x+4)(x+5)} \\ {\text{You should get the dividend.}} &{x^2+9x+20\checkmark}\\ \end{array}$$
Example $$\PageIndex{11}$$
Find the quotient: $$(y^2+10y+21)÷(y+3)$$.
$$y+7$$
Example $$\PageIndex{12}$$
Find the quotient: $$(m^2+9m+20)÷(m+4)$$.
$$m+5$$
When we divided 875 by 25, we had no remainder. But sometimes division of numbers does leave a remainder. The same is true when we divide polynomials. In the next example, we’ll have a division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.
Look back at the dividends in previous examples. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in this example will be $$x^4−x^2+5x−6$$. It is missing an $$x^3$$ term. We will add in $$0x^3$$ as a placeholder.
Example $$\PageIndex{13}$$
Find the quotient: $$(x^4−x^2+5x−6)÷(x+2)$$.
Notice that there is no $$x^3$$ term in the dividend. We will add $$0x^3$$ as a placeholder.
Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms. Divide $$x^4$$ by $$x$$. Put the answer, $$x^3$$, in the quotient over the $$x^3$$ term. Multiply $$x^3$$ times $$x+2$$. Line up the like terms. Subtract and then bring down the next term. Divide $$−2x^3$$ by $$x$$. Put the answer, $$−2x^2$$, in the quotient over the $$x^2$$ term. Multiply $$−2x^2$$ times $$x+1$$. Line up the like terms Subtract and bring down the next term. Divide $$3x^2$$ by $$x$$. Put the answer, $$3x$$, in the quotient over the $$x$$ term. Multiply $$3x$$ times $$x+1$$. Line up the like terms. Subtract and bring down the next term. Divide $$−x$$ by $$x$$. Put the answer, $$−1$$, in the quotient over the constant term. Multiply $$−1$$ times $$x+1$$. Line up the like terms. Change the signs, add. Write the remainder as a fraction with the divisor as the denominator. To check, multiply $$(x+2)(x^3−2x^2+3x−1−4x+2)$$. The result should be $$x^4−x^2+5x−6$$.
Example $$\PageIndex{14}$$
Find the quotient: $$(x^4−7x^2+7x+6)÷(x+3)$$.
$$x^3−3x^2+2x+1+3x+3$$
Example $$\PageIndex{15}$$
Find the quotient: $$(x^4−11x^2−7x−6)÷(x+3)$$.
$$x^3−3x^2−2x−1−3x+3$$
In the next example, we will divide by $$2a−3$$. As we divide, we will have to consider the constants as well as the variables.
Example $$\PageIndex{16}$$
Find the quotient: $$(8a^3+27)÷(2a+3)$$.
This time we will show the division all in one step. We need to add two placeholders in order to divide.
To check, multiply $$(2a+3)(4a^2−6a+9)$$.
The result should be $$8a^3+27$$.
Example $$\PageIndex{17}$$
Find the quotient: $$(x^3−64)÷(x−4)$$.
$$x^2+4x+16$$
Example $$\PageIndex{18}$$
Find the quotient: $$(125x^3−8)÷(5x−2)$$.
$$25x^2+10x+4$$
## Divide Polynomials using Synthetic Division
As we have mentioned before, mathematicians like to find patterns to make their work easier. Since long division can be tedious, let’s look back at the long division we did in Example and look for some patterns. We will use this as a basis for what is called synthetic division. The same problem in the synthetic division format is shown next.
Synthetic division basically just removes unnecessary repeated variables and numbers. Here all the $$x$$ and $$x^2$$ are removed. as well as the $$−x^2$$ and $$−4x$$ as they are opposite the term above.
• The first row of the synthetic division is the coefficients of the dividend. The $$−5$$ is the opposite of the 5 in the divisor.
• The second row of the synthetic division are the numbers shown in red in the division problem.
• The third row of the synthetic division are the numbers shown in blue in the division problem.
Notice the quotient and remainder are shown in the third row.
$\text{Synthetic division only works when the divisor is of the form }x−c. \nonumber$
The following example will explain the process.
Example $$\PageIndex{19}$$
Use synthetic division to find the quotient and remainder when $$2x^3+3x^2+x+8$$ is divided by $$x+2$$.
Write the dividend with decreasing powers of $$x$$. Write the coefficients of the terms as the first row of the synthetic division. Write the divisor as $$x−c$$ and place c in the synthetic division in the divisor box. Bring down the first coefficient to the third row. Multiply that coefficient by the divisor and place the result in the second row under the second coefficient. Add the second column, putting the result in the third row. Multiply that result by the divisor and place the result in the second row under the third coefficient. Add the third column, putting the result in the third row. Multiply that result by the divisor and place the result in the third row under the third coefficient. Add the final column, putting the result in the third row. The quotient is $$2x^2−1x+3$$ and the remainder is 2.
The division is complete. The numbers in the third row give us the result. The $$2\space\space\space−1\space\space\space3$$ are the coefficients of the quotient. The quotient is $$2x^2−1x+3$$. The 2 in the box in the third row is the remainder.
Check:
\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3+3x^2+x+8 &= 2x^3+3x^2+x+8\checkmark \nonumber \end{align}
Example $$\PageIndex{20}$$
Use synthetic division to find the quotient and remainder when $$3x^3+10x^2+6x−2$$ is divided by $$x+2$$.
$$3x^2+4x−2;\space 2$$
Example $$\PageIndex{21}$$
Use synthetic division to find the quotient and remainder when $$4x^3+5x^2−5x+3$$ is divided by $$x+2$$.
$$4x^2−3x+1; 1$$
In the next example, we will do all the steps together.
Example $$\PageIndex{22}$$
Use synthetic division to find the quotient and remainder when $$x^4−16x^2+3x+12$$ is divided by $$x+4$$.
The polynomial $$x^4−16x^2+3x+12$$ has its term in order with descending degree but we notice there is no $$x^3$$ term. We will add a 0 as a placeholder for the $$x^3$$ term. In $$x−c$$ form, the divisor is $$x−(−4)$$.
We divided a $$4^{\text{th}}$$ degree polynomial by a $$1^{\text{st}}$$ degree polynomial so the quotient will be a $$3^{\text{rd}}$$ degree polynomial.
Reading from the third row, the quotient has the coefficients $$1\space\space\space−4\space\space\space0\space\space\space3$$, which is $$x^3−4x^2+3$$. The remainder
is 0.
Example $$\PageIndex{23}$$
Use synthetic division to find the quotient and remainder when $$x^4−16x^2+5x+20$$ is divided by $$x+4$$.
$$x^3−4x^2+5;\space 0$$
Example $$\PageIndex{24}$$
Use synthetic division to find the quotient and remainder when $$x^4−9x^2+2x+6$$ is divided by $$x+3$$.
$$x^3−3x^2+2;\space 0$$
## Divide Polynomial Functions
Just as polynomials can be divided, polynomial functions can also be divided.
definition: DIVISION OF POLYNOMIAL FUNCTIONS
For functions $$f(x)$$ and $$g(x)$$, where $$g(x)\neq 0$$,
$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)} \nonumber$
Example $$\PageIndex{25}$$
For functions $$f(x)=x^2−5x−14$$ and $$g(x)=x+2$$, find:
1. $$(\frac{f}{g})(x)$$
2. $$(\frac{f}{g})(−4)$$.
$$\begin{array} {ll} {\text{Substitute for }f(x)\text{ and }g(x).} &{\left(\frac{f}{g}\right)(x)=\frac{x^2−5x−14}{x+2}} \\ {\text{Divide the polynomials.}} &{\left(\frac{f}{g}\right)(x)=x−7} \\ \end{array}$$
ⓑ In part ⓐ we found $$(\frac{f}{g})(x)$$ and now are asked to find $$(\frac{f}{g})(−4)$$.
$$\begin{array} {ll} {} &{\left(\frac{f}{g}\right)(x)=x−7} \\ {\text{To find }\left(\frac{f}{g}\right)(−4), \text{ substitute }x=−4.} &{\left(\frac{f}{g}\right)(−4)=−4−7} \\ {} &{\left(\frac{f}{g}\right)(−4)=−11} \\ \end{array}$$
Example $$\PageIndex{26}$$
For functions $$f(x)=x^2−5x−24$$ and $$g(x)=x+3$$, find:
1. $$(\frac{f}{g})(x)$$
2. $$(\frac{f}{g})(−3)$$.
$$\left(\frac{f}{g}\right)(x)=x−8$$
$$\left(\frac{f}{g}\right)(−3)=−11$$
Example $$\PageIndex{27}$$
For functions $$f(x)=x2−5x−36$$ and $$g(x)=x+4$$, find:
1. $$\left(\frac{f}{g}\right)(x)$$
2. $$\left(\frac{f}{g}\right)(−5)$$.
$$\left(\frac{f}{g}\right)(x)=x−9$$
$$\left(\frac{f}{g}\right)(x)=x−9$$
## Use the Remainder and Factor Theorem
Let’s look at the division problems we have just worked that ended up with a remainder. They are summarized in the chart below. If we take the dividend from each division problem and use it to define a function, we get the functions shown in the chart. When the divisor is written as $$x−c$$, the value of the function at $$c$$, $$f(c)$$, is the same as the remainder from the division problem.
Dividend Divisor $$x−c$$ Remainder Function $$f(c)$$
$$x^4−x^2+5x−6$$ $$x−(−2)$$ $$−4$$ $$f(x)=x^4−x^2+5x−6$$ $$−4$$
$$3x^3−2x^2−10x+8$$ $$x−2$$ 4 $$f(x)=3x^3−2x^2−10x+8$$ 4
$$x^4−16x^2+3x+15$$ $$x−(−4)$$ 3 $$f(x)=x^4−16x^2+3x+15$$ 3
To see this more generally, we realize we can check a division problem by multiplying the quotient times the divisor and add the remainder. In function notation we could say, to get the dividend $$f(x)$$, we multiply the quotient, $$q(x)$$ times the divisor, $$x−c$$, and add the remainder, $$r$$.
If we evaluate this at c,c, we get:
This leads us to the Remainder Theorem.
definition: REMAINDER THEOREM
If the polynomial function $$f(x)$$ is divided by $$x−c$$, then the remainder is $$f(c)$$.
Example $$\PageIndex{2}$$
Use the Remainder Theorem to find the remainder when $$f(x)=x^3+3x+19$$ is divided by $$x+2$$.
To use the Remainder Theorem, we must use the divisor in the $$x−c$$ form. We can write the divisor $$x+2$$ as $$x−(−2)$$. So, our $$c$$ is $$−2$$.
To find the remainder, we evaluate $$f(c)$$ which is $$f(−2)$$.
To evaluate $$f(−2)$$, substitute $$x=−2$$. Simplify. The remainder is 5 when $$f(x)=x^3+3x+19$$ is divided by $$x+2$$. Check: Use synthetic division to check. The remainder is 5.
Example $$\PageIndex{29}$$
Use the Remainder Theorem to find the remainder when $$f(x)=x^3+4x+15$$ is divided by $$x+2$$.
$$−1$$
Example $$\PageIndex{30}$$
Use the Remainder Theorem to find the remainder when $$f(x)=x^3−7x+12$$ is divided by $$x+3$$.
$$6$$
When we divided $$8a^3+27$$ by $$2a+3$$ in Example the result was $$4a^2−6a+9$$. To check our work, we multiply $$4a2−6a+9$$ by $$2a+3$$ to get $$8a^3+27$$.
$(4a^2−6a+9)(2a+3)=8a^3+27 \nonumber$
Written this way, we can see that $$4a^2−6a+9$$ and $$2a+3$$ are factors of $$8a^3+27$$. When we did the division, the remainder was zero.
Whenever a divisor, $$x−c$$, divides a polynomial function, $$f(x)$$, and resulting in a remainder of zero, we say $$x−c$$ is a factor of $$f(x)$$.
The reverse is also true. If $$x−c$$ is a factor of $$f(x)$$ then $$x−c$$ will divide the polynomial function resulting in a remainder of zero.
We will state this in the Factor Theorem.
definition: FACTOR THEOREM
For any polynomial function $$f(x)$$,
• if $$x−c$$ is a factor of $$f(x)$$, then $$f(c)=0$$
• if $$f(c)=0$$, then $$x−c$$ is a factor of $$f(x)$$
Example $$\PageIndex{31}$$
Use the Remainder Theorem to determine if $$x−4$$ is a factor of $$f(x)=x^3−64$$.
The Factor Theorem tells us that $$x−4$$ is a factor of $$f(x)=x^3−64$$ if $$f(4)=0$$.
$$\begin{array} {ll} {} &{f(x)=x^3−64} \\ {\text{To evaluate }f(4) \text{ substitute } x=4.} &{f(4)=4^3−64} \\ {\text{Simplify.}} &{f(4)=64−64} \\ {\text{Subtract.}} &{f(4)=0} \\ \end{array}$$
Since $$f(4)=0, x−4$$ is a factor of $$f(x)=x^3−64$$.
Example $$\PageIndex{32}$$
Use the Factor Theorem to determine if $$x−5$$ is a factor of $$f(x)=x^3−125$$.
yes
Example $$\PageIndex{33}$$
Use the Factor Theorem to determine if $$x−6$$ is a factor of $$f(x)=x^3−216$$.
yes
Access these online resources for additional instruction and practice with dividing polynomials.
## Key Concepts
• Division of a Polynomial by a Monomial
• To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.
• Division of Polynomial Functions
• For functions $$f(x)$$ and $$g(x)$$, where $$g(x)\neq 0$$,
$$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$$
• Remainder Theorem
• If the polynomial function $$f(x)$$ is divided by $$x−c$$, then the remainder is $$f(c)$$.
• Factor Theorem: For any polynomial function $$f(x)$$,
• if $$x−c$$ is a factor of $$f(x)$$, then $$f(c)=0$$
• if $$f(c)=0$$, then $$x−c$$ is a factor of $$f(x)$$
## Section Exercises
### Practice Makes Perfect
Divide Monomials
In the following exercises, divide the monomials.
$$15r^4s^9÷(15r^4s^9)$$
$$20m^8n^4÷(30m^5n^9)$$
$$\frac{2m^3}{3n^5}$$
$$\frac{18a^4b^8}{−27a^9b^5}$$
$$\frac{45x^5y^9}{−60x^8y^6}$$
$$\frac{−3y^3}{4x^3}$$
$$\frac{(10m^5n^4)(5m^3n^6)}{25m^7n^5}$$
$$\frac{(−18p^4q^7)(−6p^3q^8)}{−36p^{12}q^{10}}$$
$$\frac{−3q^5}{p^5}$$
$$\frac{(6a^4b^3)(4ab^5)}{(12a^2b)(a^3b)}$$
$$\frac{(4u^2v^5)(15u^3v)}{(12u^3v)(u^4v)}$$
$$\frac{5v^4}{u^2}$$
Divide a Polynomial by a Monomial
In the following exercises, divide each polynomial by the monomial.
$$(9n^4+6n^3)÷3n$$
$$(8x^3+6x^2)÷2x$$
$$4x^2+3x$$
$$(63m^4−42m^3)÷(−7m^2)$$
$$(48y^4−24y^3)÷(−8y^2)$$
$$−6y^2+3y$$
$$\frac{66x^3y^2−110x^2y^3−44x^4y^3}{11x^2y^2}$$
$$\frac{72r^5s^2+132r^4s^3−96r^3s^5}{12r^2s^2}$$
$$6r^3+11r^2s−8rs^3$$
$$10x^2+5x−4−5x$$
$$20y^2+12y−1−4y$$
$$−5y−3+\frac{1}{4y}$$
Divide Polynomials using Long Division
In the following exercises, divide each polynomial by the binomial.
$$(y^2+7y+12)÷(y+3)$$
$$(a^2−2a−35)÷(a+5)$$
$$a−7$$
$$(6m^2−19m−20)÷(m−4)$$
$$(4x^2−17x−15)÷(x−5)$$
$$4x+3$$
$$(q^2+2q+20)÷(q+6)$$
$$(p^2+11p+16)÷(p+8)$$
$$p+3−\frac{8}{p+8}$$
$$(3b^3+b^2+4)÷(b+1)$$
$$(2n^3−10n+28)÷(n+3)$$
$$\frac{2n^2−6n+8+4}{n+3}$$
$$(z^3+1)÷(z+1)$$
$$(m^3+1000)÷(m+10)$$
$$m^2−10m+100$$
$$(64x^3−27)÷(4x−3)$$
$$(125y^3−64)÷(5y−4)$$
$$25y^2+20x+16$$
Divide Polynomials using Synthetic Division
In the following exercises, use synthetic Division to find the quotient and remainder.
$$x^3−6x^2+5x+14$$ is divided by $$x+1$$
$$x^3−3x^2−4x+12$$ is divided by $$x+2$$
$$x^2−5x+6; \space 0$$
$$2x^3−11x^2+11x+12$$ is divided by $$x−3$$
$$2x^3−11x^2+16x−12$$ is divided by $$x−4$$
$$2x^2−3x+4; \space 4$$
$$x^4-5x^2+2+13x+3$$ is divided by $$x+3$$
$$x^4+x^2+6x−10$$ is divided by $$x+2$$
$$x^3−2x^2+5x−4; \space −2$$
$$2x^4−9x^3+5x^2−3x−6$$ is divided by $$x−4$$
$$3x^4−11x^3+2x^2+10x+6$$ is divided by $$x−3$$
$$3x^3−2x^2−4x−2;\space 0$$
Divide Polynomial Functions
In the following exercises, divide.
For functions $$f(x)=x^2−13x+36$$ and $$g(x)=x−4$$, find ⓐ $$\left(\frac{f}{g}\right)(x)$$ ⓑ $$\left(\frac{f}{g}\right)(−1)$$
For functions $$f(x)=x^2−15x+54$$ and $$g(x)=x−9$$, find ⓐ $$\left(\frac{f}{g}\right)(x)$$ ⓑ $$\left(\frac{f}{g}\right)(−5)$$
ⓐ $$\left(\frac{f}{g}\right)(x)=x−6$$
ⓑ $$\left(\frac{f}{g}\right)(−5)=−11$$
For functions $$f(x)=x^3+x^2−7x+2$$ and $$g(x)=x−2$$, find ⓐ $$\left(\frac{f}{g}\right)(x)$$ ⓑ $$\left(\frac{f}{g}\right)(2)$$
For functions $$f(x)=x^3+2x^2−19x+12$$ and $$g(x)=x−3$$, find ⓐ $$\left(\frac{f}{g}\right)(x)$$ ⓑ $$\left(\frac{f}{g}\right)(0)$$
ⓐ $$\left(\frac{f}{g}\right)(x)=x^2+5x−4$$
ⓑ $$\left(\frac{f}{g}\right)(0)=−4$$
For functions $$f(x)=x^2−5x+2$$ and $$g(x)=x^2−3x−1$$, find ⓐ $$(f·g)(x)$$ ⓑ $$(f·g)(−1)$$
For functions $$f(x)=x^2+4x−3$$ and $$g(x)=x^2+2x+4$$, find ⓐ $$(f·g)(x)$$ ⓑ $$(f·g)(1)$$
ⓐ $$(f·g)(x)=x^4+6x^3+9x^2+10x−12$$; ⓑ $$(f·g)(1)=14$$
Use the Remainder and Factor Theorem
In the following exercises, use the Remainder Theorem to find the remainder.
$$f(x)=x^3−8x+7$$ is divided by $$x+3$$
$$f(x)=x^3−4x−9$$ is divided by $$x+2$$
$$−9$$
$$f(x)=2x^3−6x−24$$ divided by $$x−3$$
$$f(x)=7x^2−5x−8$$ divided by $$x−1$$
$$−6$$
In the following exercises, use the Factor Theorem to determine if x−cx−c is a factor of the polynomial function.
Determine whether $$x+3$$ a factor of $$x^3+8x^2+21x+18$$
Determine whether $$x+4$$ a factor of $$x^3+x^2−14x+8$$
no
Determine whether $$x−2$$ a factor of $$x^3−7x^2+7x−6$$
Determine whether $$x−3$$ a factor of $$x^3−7x^2+11x+3$$
yes
### Writing Exercises
James divides $$48y+6$$ by 6 this way: $$\frac{48y+6}{6}=48y$$. What is wrong with his reasoning?
Divide $$\frac{10x^2+x−12}{2x}$$ and explain with words how you get each term of the quotient.
Explain when you can use synthetic division.
In your own words, write the steps for synthetic division for $$x^2+5x+6$$ divided by $$x−2$$.
### Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section
ⓑ On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?
## Chapter Review Exercises
Determine the Degree of Polynomials
In the following exercises, determine the type of polynomial.
$$16x^2−40x−25$$
$$5m+9$$
binomial
$$−15$$
$$y^2+6y^3+9y^4$$
other polynomial
In the following exercises, add or subtract the polynomials.
$$4p+11p$$
$$−8y^3−5y^3$$
$$−13y^3$$
$$(4a^2+9a−11)+(6a^2−5a+10)$$
$$(8m^2+12m−5)−(2m^2−7m−1)$$
$$6m^2+19m−4$$
$$(y^2−3y+12)+(5y^2−9)$$
$$(5u^2+8u)−(4u−7)$$
$$5u^2+4u+7$$
Find the sum of $$8q^3−27$$ and $$q^2+6q−2$$.
Find the difference of $$x^2+6x+8$$ and $$x^2−8x+15$$.
$$2x^2−2x+23$$
In the following exercises, simplify.
$$17mn^2−(−9mn^2)+3mn^2$$
$$18a−7b−21a$$
$$−7b−3a$$
$$2pq^2−5p−3q^2$$
$$(6a^2+7)+(2a^2−5a−9)$$
$$8a^2−5a−2$$
$$(3p^2−4p−9)+(5p^2+14)$$
$$(7m^2−2m−5)−(4m^2+m−8)$$
$$−3m+3$$
$$(7b^2−4b+3)−(8b^2−5b−7)$$
Subtract $$(8y^2−y+9)$$ from (11y^2−9y−5)\)
$$3y^2−8y−14$$
Find the difference of $$(z^2−4z−12)$$ and $$(3z^2+2z−11)$$
$$(x^3−x^2y)−(4xy^2−y^3)+(3x^2y−xy^2)$$
$$x^3+2x^2y−4xy^2$$
$$(x^3−2x^2y)−(xy^2−3y^3)−(x^2y−4xy^2)$$
Evaluate a Polynomial Function for a Given Value of the Variable
In the following exercises, find the function values for each polynomial function.
For the function $$f(x)=7x^2−3x+5$$ find:
ⓐ $$f(5)$$ ⓑ $$f(−2)$$ ⓒ $$f(0)$$
ⓐ 165 ⓑ 39 ⓒ 5
For the function $$g(x)=15−16x^2$$, find:
ⓐ $$g(−1)$$ ⓑ $$g(0)$$ ⓒ $$g(2)$$
A pair of glasses is dropped off a bridge 640 feet above a river. The polynomial function $$h(t)=−16t^2+640$$ gives the height of the glasses t seconds after they were dropped. Find the height of the glasses when $$t=6$$.
The height is 64 feet.
A manufacturer of the latest soccer shoes has found that the revenue received from selling the shoes at a cost of $$p$$ dollars each is given by the polynomial $$R(p)=−5p^2+360p$$. Find the revenue received when $$p=110$$ dollars.
In the following exercises, find ⓐ $$(f + g)(x)$$ ⓑ $$(f + g)(3)$$ ⓒ $$(fg)(x)$$ ⓓ $$(fg)(−2)$$
$$f(x)=2x^2−4x−7$$ and $$g(x)=2x^2−x+5$$
ⓐ $$(f+g)(x)=4x^2−5x−2$$ ⓑ $$(f+g)(3)=19$$
ⓒ $$(f−g)(x)=−3x−12$$
ⓓ $$(f−g)(−2)=−6$$
$$f(x)=4x^3−3x^2+x−1$$ and $$g(x)=8x^3−1$$
### Properties of Exponents and Scientific Notation
Simplify Expressions Using the Properties for Exponents
In the following exercises, simplify each expression using the properties for exponents.
$$p^3·p^{10}$$
$$p^{13}$$
$$2·2^6$$
$$a·a^2·a^3$$
$$a^6$$
$$x·x^8$$
$$y^a·y^b$$
$$y^{a+b}$$
$$\frac{2^8}{2^2}$$
$$\frac{a^6}{a}$$
$$a^5$$
$$\frac{n^3}{n^{12}}$$
$$\frac{1}{x^5}$$
$$\frac{1}{x^4}$$
$$3^0$$
$$y^0$$
$$1$$
$$(14t)^0$$
$$12a^0−15b^0$$
$$−3$$
Use the Definition of a Negative Exponent
In the following exercises, simplify each expression.
$$6^{−2}$$
$$(−10)^{−3}$$
$$−\frac{1}{1000}$$
$$5·2^{−4}$$
$$(8n)^{−1}$$
$$\frac{1}{8n}$$
$$y^{−5}$$
$$10^{−3}$$
$$\frac{1}{1000}$$
$$\frac{1}{a^{−4}}$$
$$\frac{1}{6^{−2}}$$
$$36$$
$$−5^{−3}$$
$$(−\frac{1}{5})^{−3}$$
$$−\frac{1}{25}$$
$$−(12)^{−3}$$
$$(−5)^{−3}$$
$$−\frac{1}{125}$$
$$\left(\frac{5}{9}\right)^{−2}$$
$$\left(−\frac{3}{x}\right)^{−3}$$
$$\frac{x^3}{27}$$
In the following exercises, simplify each expression using the Product Property.
$$(y^4)^3$$
$$(3^2)^5$$
$$3^{10}$$
$$(a^{10})^y$$
$$x^{−3}·x^9$$
$$x^5$$
$$r^{−5}·r^{−4}$$
$$(uv^{−3})(u^{−4}v^{−2})$$
$$\frac{1}{u^3v^5}$$
$$(m^5)^{−1}$$
$$p^5·p^{−2}·p^{−4}$$
$$\frac{1}{m^5}$$
In the following exercises, simplify each expression using the Power Property.
$$(k−2)^{−3}$$
$$\frac{q^4}{q^{20}}$$
$$\frac{1}{q^{16}}$$
$$\frac{b8}{b^{−2}}$$
$$\frac{n^{−3}}{n^{−5}}$$
$$n^2$$
In the following exercises, simplify each expression using the Product to a Power Property.
$$(−5ab)^3$$
$$(−4pq)^0$$
$$1$$
$$(−6x^3)^{−2}$$
$$(3y^{−4})^2$$
$$\frac{9}{y^8}$$
In the following exercises, simplify each expression using the Quotient to a Power Property.
$$\left(\frac{3}{5x}\right)^{−2}$$
$$\left(\frac{3xy^2}{z}\right)^4$$
$$\frac{81x^4y^8}{z^4}$$
$$(4p−3q^2)^2$$
In the following exercises, simplify each expression by applying several properties.
$$(x^2y)^2(3xy^5)^3$$
$$27x^7y^{17}$$
$$(−3a^{−2})^4(2a^4)^2(−6a^2)^3$$
$$\left(\frac{3xy^3}{4x^4y^{−2}}\right)^2\left(\frac{6xy^4}{8x^3y^{−2}}\right)^{−1}$$
$$\frac{3y^4}{4x^4}$$
In the following exercises, write each number in scientific notation.
$$2.568$$
$$5,300,000$$
$$5.3×10^6$$
$$0.00814$$
In the following exercises, convert each number to decimal form.
$$2.9×10^4$$
$$29,000$$
$$3.75×10^{−1}$$
$$9.413×10^{−5}$$
$$0.00009413$$
In the following exercises, multiply or divide as indicated. Write your answer in decimal form.
$$(3×10^7)(2×10^{−4})$$
$$(1.5×10^{−3})(4.8×10^{−1})$$
$$0.00072$$
$$\frac{6×10^9}{2×10^{−1}}$$
$$\frac{9×10^{−3}}{1×10^{−6}}$$
$$9,000$$
### Multiply Polynomials
Multiply Monomials
In the following exercises, multiply the monomials.
$$(−6p^4)(9p)$$
$$(\frac{1}{3}c^2)(30c^8)$$
$$10c^{10}$$
$$(8x^2y^5)(7xy^6)$$
$$(\frac{2}{3}m^3n^6)(\frac{1}{6}m^4n^4)$$
$$\frac{m^7n^{10}}{9}$$
Multiply a Polynomial by a Monomial
In the following exercises, multiply.
$$7(10−x)$$
$$a^2(a^2−9a−36)$$
$$a^4−9a^3−36a^2$$
$$−5y(125y^3−1)$$
$$(4n−5)(2n^3)$$
$$8n^4−10n^3$$
Multiply a Binomial by a Binomial
In the following exercises, multiply the binomials using:
ⓐ the Distributive Property ⓑ the FOIL method ⓒ the Vertical Method.
$$(a+5)(a+2)$$
$$(y−4)(y+12)$$
$$y^2+8y−48$$
$$(3x+1)(2x−7)$$
$$(6p−11)(3p−10)$$
$$18p^2−93p+110$$
In the following exercises, multiply the binomials. Use any method.
$$(n+8)(n+1)$$
$$(k+6)(k−9)$$
$$k^2−3k−54$$
$$(5u−3)(u+8)$$
$$(2y−9)(5y−7)$$
$$10y^2−59y+63$$
$$(p+4)(p+7)$$
$$(x−8)(x+9)$$
$$x^2+x−72$$
$$(3c+1)(9c−4)$$
$$(10a−1)(3a−3)$$
$$30a^2−33a+3$$
Multiply a Polynomial by a Polynomial
In the following exercises, multiply using ⓐ the Distributive Property ⓑ the Vertical Method.
$$(x+1)(x^2−3x−21)$$
$$(5b−2)(3b^2+b−9)$$
$$15b^3−b^2−47b+18$$
In the following exercises, multiply. Use either method.
$$(m+6)(m^2−7m−30)$$
$$(4y−1)(6y^2−12y+5)$$
$$24y^2−54y^2+32y−5$$
Multiply Special Products
In the following exercises, square each binomial using the Binomial Squares Pattern.
$$(2x−y)^2$$
$$(x+\frac{3}{4})^2$$
$$x^2+\frac{3}{2}x+\frac{9}{16}$$
$$(8p^3−3)^2$$
$$(5p+7q)^2$$
$$25p^2+70pq+49q^2$$
In the following exercises, multiply each pair of conjugates using the Product of Conjugates.
$$(3y+5)(3y−5)$$
$$(6x+y)(6x−y)$$
$$36x^2−y^2$$
$$(a+\frac{2}3b)(a−\frac{2}{3}b)$$
$$(12x^3−7y^2)(12x^3+7y^2)$$
$$144x^6−49y^4$$
$$(13a^2−8b4)(13a^2+8b^4)$$
### Divide Monomials
Divide Monomials
In the following exercises, divide the monomials.
$$72p^{12}÷8p^3$$
$$9p^9$$
$$−26a^8÷(2a^2)$$
$$\frac{45y^6}{−15y^{10}}$$
$$−3y^4$$
$$\frac{−30x^8}{−36x^9}$$
$$\frac{28a^9b}{7a^4b^3}$$
$$\frac{4a^5}{b^2}$$
$$\frac{11u^6v^3}{55u^2v^8}$$
$$\frac{(5m^9n^3)(8m^3n^2)}{(10mn^4)(m^2n^5)}$$
$$\frac{4m^9}{n^4}$$
$$\frac{(42r^2s^4)(54rs^2)}{(6rs^3)(9s)}$$
Divide a Polynomial by a Monomial
In the following exercises, divide each polynomial by the monomial
$$(54y^4−24y^3)÷(−6y^2)$$
$$−9y^2+4y$$
$$\frac{63x^3y^2−99x^2y^3−45x^4y^3}{9x^2y^2}$$
$$\frac{12x^2+4x−3}{−4x}$$
$$−3x−1+\frac{3}{4x}$$
Divide Polynomials using Long Division
In the following exercises, divide each polynomial by the binomial.
$$(4x^2−21x−18)÷(x−6)$$
$$(y^2+2y+18)÷(y+5)$$
$$y−3+\frac{33}{q+6}$$
$$(n^3−2n^2−6n+27)÷(n+3)$$
$$(a^3−1)÷(a+1)$$
$$a^2+a+1$$
Divide Polynomials using Synthetic Division
In the following exercises, use synthetic Division to find the quotient and remainder.
x3−3x2−4x+12x3−3x2−4x+12 is divided by x+2x+2
$$x^3−3x^2−4x+12$$ is divided by $$x+2$$
$$2x^3−11x^2+11x+12$$ is divided by $$x−3$$
$$2x^2−5x−4;\space0$$
$$x^4+x^2+6x−10$$ is divided by $$x+2$$
Divide Polynomial Functions
In the following exercises, divide.
For functions $$f(x)=x^2−15x+45$$ and $$g(x)=x−9$$, find ⓐ $$\left(\frac{f}{g}\right)(x)$$
ⓑ $$\left(\frac{f}{g}\right)(−2)$$
ⓐ $$\left(\frac{f}{g}\right)(x)=x−6$$
ⓑ $$\left(\frac{f}{g}\right)(−2)=−8$$
For functions $$f(x)=x^3+x^2−7x+2$$ and $$g(x)=x−2$$, find ⓐ $$\left(\frac{f}{g}\right)(x)$$
ⓑ $$\left(\frac{f}{g}\right)(3)$$
Use the Remainder and Factor Theorem
In the following exercises, use the Remainder Theorem to find the remainder.
$$f(x)=x^3−4x−9$$ is divided by $$x+2$$
$$−9$$
$$f(x)=2x^3−6x−24$$ divided by $$x−3$$
In the following exercises, use the Factor Theorem to determine if $$x−c$$ is a factor of the polynomial function.
Determine whether $$x−2$$ is a factor of $$x^3−7x^2+7x−6$$
no
Determine whether $$x−3$$ is a factor of $$x^3−7x^2+11x+3$$
## Chapter Practice Test
For the polynomial $$8y^4−3y^2+1$$
ⓐ Is it a monomial, binomial, or trinomial? ⓑ What is its degree?
ⓐ trinomial ⓑ 4
$$(5a^2+2a−12)(9a^2+8a−4)$$
$$(10x^2−3x+5)−(4x^2−6)$$
$$6x^2−3x+11$$
$$\left(−\frac{3}{4}\right)^3$$
$$x^{−3}x^4$$
$$x$$
$$5^65^8$$
$$(47a^{18}b^{23}c^5)^0$$
$$1$$
$$4^{−1}$$
$$(2y)^{−3}$$
$$\frac{1}{8y^3}$$
$$p^{−3}·p^{−8}$$
$$\frac{x^4}{x^{−5}}$$
$$x^9$$
$$(3x^{−3})^2$$
$$\frac{24r^3s}{6r^2s^7}$$
$$\frac{4r}{s^6}$$
$$(x4y9x−3)2$$
$$(8xy^3)(−6x^4y^6)$$
$$−48x^5y^9$$
$$4u(u^2−9u+1)$$
$$(m+3)(7m−2)$$
$$21m^2−19m−6$$
$$(n−8)(n^2−4n+11)$$
$$(4x−3)^2$$
$$16x^2−24x+9$$
$$(5x+2y)(5x−2y)$$
$$(15xy^3−35x^2y)÷5xy$$
$$3y^2−7x$$
$$(3x^3−10x^2+7x+10)÷(3x+2)$$
Use the Factor Theorem to determine if $$x+3$$ a factor of $$x^3+8x^2+21x+18$$.
yes
ⓐ Convert 112,000 to scientific notation. ⓑ Convert $$5.25×10^{−4}$$ to decimal form.
In the following exercises, simplify and write your answer in decimal form.
$$(2.4×10^8)(2×10^{−5})$$
$$4.4×10^3$$
$$\frac{9×10^4}{3×10^{−1}}$$
For the function $$f(x)=6x^2−3x−9$$ find:
ⓐ $$f(3)$$ ⓑ $$f(−2)$$ ⓒ $$f(0)$$
ⓐ $$36$$ ⓑ $$21$$ ⓒ $$-9$$
For $$f(x)=2x^2−3x−5$$ and $$g(x)=3x^2−4x+1$$, find
ⓐ $$(f+g)(x)$$ ⓑ $$(f+g)(1)$$
ⓒ $$(f−g)(x)$$ ⓓ $$(f−g)(−2)$$
For functions
$$f(x)=3x^2−23x−36$$ and
$$g(x)=x−9$$, find
ⓐ $$\left(\frac{f}{g}\right)(x)$$ ⓑ $$\left(\frac{f}{g}\right)(3)$$
ⓐ $$\left(\frac{f}{g}\right)(x)=3x+4$$
ⓑ $$\left(\frac{f}{g}\right)(3)=13$$
A hiker drops a pebble from a bridge 240 feet above a canyon. The function $$h(t)=−16t^2+240$$ gives the height of the pebble $$t$$ seconds after it was dropped. Find the height when $$t=3$$.
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# How to Interpret a Student's T-Test Results
••• Creatas/Creatas/Getty Images
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Mastering statistical techniques can help us to better understand the world around us, and learning to handle data correctly can prove useful in a variety of careers. T-Tests can help to determine whether or not the difference between an expected set of values and a given set of values is significant. While this procedure may look difficult at first, it can be simple to use with a little bit of practice. This process is vital to interpreting statistics and data, as it tells us whether or not the data is useful.
## Procedure
State the hypothesis. Determine whether the data warrants a one-tailed or two-tailed test. For one-tailed tests, the null hypothesis will be in the form of μ > x if you want to test for a sample mean that is too small, or μ < x if you want to test for a sample mean that is too large. The alternative hypothesis is in the form of μ = x. For two-tailed tests, the alternative hypothesis is still μ = x, but the null hypothesis changes to μ ≠ x.
Determine a significance level appropriate for your study. This will be the value you compare your final result to. Generally, significance values are at α = .05 or α = .01, depending on your preference and how accurate you want your results to be.
Calculate the sample data. Use the formula (x - μ)/SE, where the standard error (SE) is the standard deviation of the square root of the population (SE = s/√n). After determining the t-statistic, calculate degrees of freedom through the formula n-1. Enter the t-statistic, degrees of freedom, and significance level into the t-test function on a graphing calculator to determine the P-value. If you are working with a two-tailed T-Test, double the P-value.
Interpret the results. Compare the P-value to the α significance level stated earlier. If it is less than α, reject the null hypothesis. If the result is greater than α, fail to reject the null hypothesis. If you reject the null hypothesis, this implies that your alternative hypothesis is correct, and that the data is significant. If you fail to reject the null hypothesis, this implies that there is no significant difference between the sample data and the given data.
#### Tips
• Always double check your calculations.
#### Warnings
• T-Test results are subjective to the significance level you choose to compare your results to. Although results are accurate most of the time, it is still possible to misinterpret the data.
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# How do you write the solution in interval notation, and graph -5/3x<=-10?
Feb 17, 2018
Interval notation: [6,∞)
See the graph below.
#### Explanation:
First, solve for x:
$- \frac{5}{3} x \le - 10$
When dividing or multiplying by a negative, then flip the inequality sign.
$\left(- \frac{5}{3} \textcolor{red}{\cdot - \frac{3}{5}}\right) x \le - 10 \left(\textcolor{red}{- \frac{3}{5}}\right)$
$x \ge 6$
To write interval notation, use brackets $\left[\right]$ and parenthesis $\left(\right)$. Brackets are used when the answer is included, and parenthesis are used when the answer is excluded. Interval notation goes from least to greatest.
In this case, the answer is included. The answer also goes up to infinity, which will always have a parenthesis, as you cannot reach infinity:
[6,∞)
This means that any number from $6$ to ∞ is an answer, including $6$ and excluding ∞.
A graph would look like this:
The dot at 6 is shaded in as 6 is included in the answer.
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Lesson 3
Compose Three-digit Numbers
Lesson Purpose
The purpose of this lesson is for students to use base-ten representations to build an understanding of the digits in three-digit numbers.
Lesson Narrative
In previous units, students used base-ten blocks and diagrams to compose and decompose tens when adding and subtracting by place. In previous lessons, students learned that a hundred is a unit that is made up of 10 tens and used base-ten blocks to show composing a hundred with 10 tens.
In this lesson, students represent three-digit numbers that include an amount of hundreds, tens, and ones. In the first activity, students take inventory of the units represented by a collection of base-ten blocks. They use their understanding of the units of hundred and ten to determine how to represent the total value with the fewest number of blocks possible. In the second activity, students use base-ten diagrams to represent values using the fewest number of each unit possible and connect these representations to the meaning of each digit in a three-digit numeral. In both activities, look for the different ways students represent and record the value of their blocks for reference in the activity syntheses and in future lessons.
• Engagement
Learning Goals
Teacher Facing
• Compose three-digit numbers using place value understanding.
Student Facing
• Let’s compose three-digit numbers.
Required Materials
Materials to Gather
Required Preparation
Activity 1:
• Each group of 34 students will need a container with 2 hundreds, 28 tens, and 15 ones.
• Each group of 34 students will need access to additional base-ten blocks (hundred blocks and ten blocks).
Addressing
Lesson Timeline
Warm-up 10 min Activity 1 20 min Activity 2 15 min Lesson Synthesis 10 min Cool-down 5 min
Teacher Reflection Questions
In grade 1, students developed an understanding of the digits in a two-digit number. How did the work of this lesson reinforce that understanding? How did it build on that understanding?
Suggested Centers
• Greatest of Them All (1–5), Stage 1: Two-digit Numbers (Supporting)
• Mystery Number (1–4), Stage 1: Two-digit Numbers (Supporting)
Print Formatted Materials
Teachers with a valid work email address can click here to register or sign in for free access to Cool Down, Teacher Guide, and PowerPoint materials.
Student Task Statements pdf docx Lesson Cover Page pdf docx Cool Down Log In Teacher Guide Log In Teacher Presentation Materials pdf docx
Additional Resources
Google Slides Log In PowerPoint Slides Log In
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# Math Homework: 704520
Math Homework
Question 1
A quadratic function has a series of varying positive gradients on one part and a series of varying negative gradients on the other part. The positive and negative gradients meet at a point where the gradient is zero hence forming a maximum point or a minimum point. The model in part C has a negative gradient in the range . That is,
Additionally, the model has a series of varying positive gradients in the range .
The rates of change, can be represented using a schematic diagram resembling that of a quadratic function as shown in figure 1.1.
Figure 1.1: The schematic model
Therefore, table C can be modelled as a quadratic function.
Question 2
part a
Using the quadratic regression function in Desmos graphing calculator we obtain the result shown in figure 1
describes the direction the quadratic model opens (up or down). That is, the trend of the number of worms as the number of days’ increases. In this case, the negative value of a denotes a decrease in the number of worms with the increase in the number of days. describes the coefficient of the linear term of the model and shows how the number of worms is linearly to the number of days. On the other hand, describes the initial number of worms.
Part c
Solving for x using the quadratic formula,
We ignore since the number of days can’t be negative.
Hence, it takes for all the worms to die.
Question 3
Part a
Given that, . The graph of is shown in figure 3.1.
Figure 3.1: A graph of
At the vertex,
When
Hence, the vertex is
Part b
The vertex form of the equation is in the form where is the vertex.
But we know that,
Therefore, the vertex form is,
Part c
The years when
We substitute 10 into the quadratic model and solve for t using the quadratic formula.
The number of crimes committed was equal to 10million in the years and . That is, 1983.5316 and 1998.8774.
That is, approximately during the years 1983 and 1998.
Question 4
Part a
An example of a quadratic model in the form of is . It has a vertical intercept of +12, decreases initially, and then increases. The sketch of the model is shown in figure 4.1.
Figure 4.1: A graph of the model,
Part b
An example of a quadratic model in the form of that has maximum value and a vertical intercept at is as shown in the sketch, figure 4.2.
Figure 4.2: A graph of
Question 5
The quadratic model, is concave up. To ascertain that, we determine the rate of change of As we can see,
, the rate of change of with respect to x is always positive and increases with the increase in the value of x Hence, it is concave up.
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# Basics of Simplification with Important Questions, Formula, Rules and Concepts for Bank Exams
By Jyoti Bisht|Updated : October 31st, 2021
We all know the importance of fast and speedy calculations in bank and insurance exams. Simplification is one of the topics in the quant section that requires a speedy calculation to solve maximum questions on it in less time. It is one of the most widely asked topics in competitive exams, especially in the prelims stage. So today, we are discussing some key points to deal with Simplification topics to score well in this section.
Simplification generally means finding an answer for the complex calculation that may involve numbers on division, multiplication, square roots, cube roots, plus and minus.
Simplification questions are asked in the Bank and Insurance exam to check the ability of an aspirant to deal with numbers which can be in one of the following two types.
• Sometimes, a calculation is given and one of the numbers is missing from the calculation. To find out the missing number, we have to approximate the given numbers or do the basic operations.
• Sometimes all the numbers are given with some operations between them & we have to simplify the calculation.
Attempt Quantitative Aptitude Quizzes
#### Rule-(I) Replace ‘of’ by ‘Multiplication’ & ‘/’ by ‘Division’.
Explanation: Whenever we find ‘of’ in a simplification problem, we can replace that with ‘multiplication(*)’. Similarly ‘/’ can be replaced by ‘÷’.
Example: Find ¼ of 20
Solution: (¼) x 20 = 20÷4 = 5
Rule-(II) Always keep in mind the “BODMAS” rule. These operations have priorities in the same order as mentioned.
Explanation: Whenever we have more than one operation in the given calculation, we have to do the operations according to the priority specified by ‘BODMAS’
• B-Bracket
• O-Of (means multiplication)
• D-Division
• M-Multiplication
• S-Subtraction
Example: Simplify: (2+3)*30
Solution: In this question, we have two things-Bracket & Multiplication. According to the BODMAS rule, we have to solve the bracket first and not multiplication. So now coming to the bracket, we have only one operation-Addition, so we will do addition.
(2+3)*30 = 5*30
Now we have only one operation to do – Multiplication
5*30 = 150
Example: Simplify: (2+5) of 80
Solution: In this question, we have three things – bracket, addition & of. Replacing ‘of’ by ‘multiplication’.
(2+5) of 30 = (2+5)*80
Now we have three things – bracket, addition & Multiplication. According to the BODMAS rule, we have to solve brackets first and not multiplication. So now coming to the bracket, we have only one operation-Addition, we will do addition.
(2+5)*80 = 7*80
Now we will do multiplication.
7*80 = 560
Rule-(III) Multiplication & Division have the same priority(Do that operation first which is on left)
Explanation: Though division has more priority than multiplication according to ‘BODMAS’, we can perform any two operations first if multiplication is on the left.
Example: 8*30/15
8*30 ÷ 15
Solution: In this question, we have two things – Multiplication & Division. Multiplication is on the left So we can perform that first.
Doing Multiplication first:
240 ÷ 15
16
Dividing first:
8*2
16
Rule-(IV) Addition & Subtraction have the same priority.
Explanation: Though addition has more priority than division according to ‘BODMAS’, we can perform any two operations first.
Example: 30+40-15
Solution: In this question, we have two things – Addition & Subtraction. So we can perform any operation first as they have the same priority.
70 – 15
55
Doing Subtraction first:
30 + 25
55
Rule-(V) Don’t hesitate in rounding the numbers to the nearest integers.
Explanation: Most of the time the numbers are given in such a way that you can round them quickly and get the answer (Rounding should be done or not, It can be realized by looking at the given options).
Example: (324.5*15)/(5.01*24.98)
Solution: (325*15)/(5*25)
=13*3
=39
## Previous Years Questions Asked in Bank Exams from Simplification
Now let us see some of the previous year questions asked from 'Simplification' & try to apply the rules learnt so far.
Q. 1) (17 -13)4 - 174 – 134 – [- 52(17)3 – 68(133)] = (?)*221
Using formula: (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3+ b4
⇒ (a – b)4 - a4 + 4a3b + 4ab3 - b4 = + 6a2b2
(17 -13)4 - 174 – 134 – [- 52(17)3 – 68(133)] = (?)*221
Here, a = 17 and b = 13
⇒ (?)= (6 (17)2 (13)2)/221
⇒ (?) = (6 × 289 × 169)/221
⇒ (?) = 1326
Q.2) Simplify: 127.001 * 7.998 + 6.05 * 4.001
1. 1000
2. 1020
3. 1040
4. 1080
5. None of these
Solution: Using the rounding concept
127 * 8 + 6 * 4
Using the BODMAS rule
1016 + 24
1040 (Option 3)
Q.3) What will come at place of ?: 9876 ÷ 24.96 + 215.005 - ? = 309.99
1. 270
2. 280
3. 290
4. 300
5. 310
Solution: Using the rounding concept
9875 ÷ 25 + 215 - ? = 310
Using the BODMAS rule
395 + 215 - ? = 310
610 - ? = 310
? = 300 (Option 4)
Q.4) What will come at place of a: (128 ÷ 16 x a – 7*2)/(72-8*6+a2) = 1
1. 1
2. 5
3. 9
4. 13
5. 17
Solution: Using the BODMAS rule
(8*a – 14)/(49-48+a2) = 1
(8*a – 14)/(1 + a2) = 1
8a – 14 = 1 + a2
a2 – 8a + 15 = 0
a=3 or 5 (Option 2)
Q.5) What will come at place of ?: 85.147 + 34.192*6.2 + ? = 802.293
1. 400
2. 450
3. 550
4. 600
5. 500
Solution: Using the rounding concept
85 + 35*6 + ? = 803
Using the BODMAS rule
85 + 210 + ? = 803
295 + ? = 803
? = 508 [approx. = 500] (Option 5)
Q.6) What will come at place of ? : (3/8 of 168)*15 ÷ 5 + ? = 549 ÷ 9 + 235
1. 189
2. 107
3. 174
4. 296
5. None of these
Solution: Using the BODMAS rule
(3*168÷8)*15 ÷ 5 + ? = 549 ÷ 9 + 235
(504÷8)*3 + ? = 61 + 235
63*3 + ? = 296
189 + ? = 296
? = 107 (Option 2)
## Key Points to Remember while Solving Simplification Question
• Replace ‘of’ by ‘Multiplication’
• Replace ‘/’ by ‘Division’
• Always do the operations in priority according to ‘BODMAS’
• Division & Multiplication have the same priority (Start from left)
• Addition & Subtraction have the same priority
• Rounding can be done to simplify problems
• When the given options are very close then rounding doesn’t help much
• Always look at the options before doing simplification that can help in the elimination of options
Related Articles to Study: BBK Full Form IIBI Full Form BOA Full Form BMB Full Form TGB Full Form AmEx Full Form KVB Full Form CUB Full Form APGVB Full Form IVB Full Form
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1/2÷1/2of1/2=? (1/2 or 2)
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# How do you simplify sqrt(2205)?
Jan 25, 2016
$21 \sqrt{5}$
#### Explanation:
The key to this solution is to represent $2205$ as a product of some numbers that are squares of some other numbers and to use a property of square roots that allows to do the following for any two non-negative real numbers $X$ and $Y$:
$\sqrt{X \cdot Y} = \sqrt{X} \cdot \sqrt{Y}$
Using this, we can perform the following steps:
$\sqrt{2205} = \sqrt{9 \cdot 245} =$
$= \sqrt{9} \cdot \sqrt{245} =$
$= \sqrt{9} \cdot \sqrt{49 \cdot 5} =$
$= \sqrt{9} \cdot \sqrt{49} \cdot \sqrt{5} =$
$= 3 \cdot 7 \cdot \sqrt{5} = 21 \sqrt{5}$
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# Divisibility Rules
A divisibility rule is a shortcut you can use to see if an integer is evenly divisible by another integer, without doing the actual division. Most divisibility rules involve looking at the digits of the number.
Here are some easy divisibility rules for 1 through 11:
## 1
All integers are divisible by 1.
## 2
Last digit is even.
Examples:
• Divisible: 2556, because the last digit (6) is an even number
• Not Divisible: 2655, because the last digit (5) is not an even number
## 3
Sum of digits is divisible by 3.
Examples:
• Divisible: 2334, because 2 + 3 + 3 + 4 = 12 is divisible by 3
• Not Divisible: 2443, because 2 + 4 + 4 + 3 = 13 is not divisible by 3
## 4
Last two digits form a number divisible by 4.
Examples:
• Divisible: 2512, because the number formed by the last two digits (12) is divisible by 4
• Not Divisible: 2242, because the number formed by the last two digits (42) is not divisible by 4
## 5
Last digit is 0 or 5.
Examples:
• Divisible: 2185, because the last digit is a 5
• Not Divisible: 2953, because the last digit (3) is not a 0 or 5
## 6
Divisible by both 2 and 3.
Examples:
• Divisible: 5322, because the last digit (2) is even and the sum of the digits (5 + 3 + 2 + 2 = 12) is divisible by 3
• Not Divisible: 4994, because although the last digit (4) is even, the sum of the digits (4 + 9 + 9 + 4 = 26) is not divisible by 3
## 7
Subtracting double the last digit from the number formed by the remaining digits gives a result that is divisible by 7.
Examples:
• Divisible: 532, because 53 – (2 x 2) = 49 is divisible by 7
• Not Divisible: 270, because 27 – (0 x 2) = 27 is not divisible by 7
## 8
Last three digits form a number divisible by 8.
Examples:
• Divisible: 36136, because the number formed by the last 3 digits (136) is divisible by 8
• Not Divisible: 20238, because the number formed by the last 3 digits (238) is not divisible by 8
## 9
Sum of digits is divisible by 9.
Examples:
• Divisible: 1431720, because 1 + 4 + 3 + 1 + 7 + 2 + 0 = 18 is divisible by 9
• Not Divisible: 2299, because 2 + 2 + 9 + 9 = 22 is not divisible by 9
## 10
Last digit is 0.
Examples:
• Divisible: 35480, because the last digit is 0
• Not Divisible: 30005, because the last digit is not 0
## 11
Alternating sum of digits is divisible by 11. To get the alternating sum, add every other digit starting from the left, and subtract all the other digits.
Examples:
• Divisible: 3729, because 3 – 7 + 2 – 9 = -11, which is divisible by 11
• Not Divisible: 4311, because 4 – 3 + 1 – 1 = 1, which is not divisible by 11
## Larger Divisors
Some larger composite numbers also have simple divisibility rules. For example, a number is divisible by 99 if it is both divisible by 9 and divisible by 11.
## Relevant Brainteasers and Puzzles
Some seemingly difficult brainteasers can be solved by using divisibility rule shortcuts:
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# Illustrative Mathematics Grade 8, Unit 1, Lesson 7: No Bending or Stretching
Learning Targets:
• I can describe the effects of a rigid transformation on the lengths and angles in a polygon.
Related Pages
Illustrative Math
#### Lesson 7: No Bending or Stretching
Let’s compare measurements before and after translations, rotations, and reflections.
Illustrative Math Unit 8.1, Lesson 7 (printable worksheets)
#### Lesson 7 Summary
The following diagram shows the effects of a rigid transformation on the lengths and angles in a polygon..
#### Lesson 7.1 Measuring Segments
For each question, the unit is represented by the large tick marks with whole numbers.
1. Find the length of this segment to the nearest 1/8 of a unit.
2. Find the length of this segment to the nearest 0.1 of a unit.
3. Estimate the length of this segment to the nearest 1/8 of a unit.
4. Estimate the length of the segment in the prior question to the nearest 0.1 of a unit.
#### Lesson 7.2 Sides and Angles
1. Translate Polygon A so point P goes to point P'. In the image, write in the length of each side, in grid units, next to the side using the draw tool.
Open Applet
2. Rotate Triangle B, 90° clockwise using R as the center of rotation. In the image, write the measure of each angle in its interior using the draw tool.
Open Applet
3. Reflect Pentagon C across line l.
a. In the image, write the length of each side, in grid units, next to the side.
b. In the image, write the measure of each angle in the interior.
Open Applet
#### Lesson 7.3 Which One?
Here is a grid showing triangle ABC and two other triangles.
Open Applet
You can use a rigid transformation to take triangle ABC to one of the other triangles.
1. Which one? Explain how you know.
2. Describe a rigid transformation that takes ABC to the triangle you selected.
#### Are you ready for more?
A square is made up of an L-shaped region and three transformations of the region. If the perimeter of the square is 40 units, what is the perimeter of each L-shaped region?
#### Lesson 7 Practice Problems
1. Is there a rigid transformation taking Rhombus P to Rhombus Q? Explain how you know.
2. Describe a rigid transformation that takes Triangle A to Triangle B.
3. Is there a rigid transformation taking Rectangle A to Rectangle B? Explain how you know.
4. For each shape, draw its image after performing the transformation. If you get stuck, consider using tracing paper.
a. Translate the shape so that A goes to A'.
b. Rotate the shape 180 degrees counterclockwise around B.
The Open Up Resources math curriculum is free to download from the Open Up Resources website and is also available from Illustrative Mathematics.
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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# The slope of the line? positive negative zero undefined
Question: The slope of the line? positive negative zero undefined
The slope of a line is a measure of how steeply it rises or falls as we move along the x-axis. The slope can be positive, negative, zero, or undefined depending on the direction and shape of the line. In this blog post, we will explore each of these cases and how to find the slope using the formula m = (y2 - y1) / (x2 - x1), where m is the slope and (x1, y1) and (x2, y2) are two points on the line.
A line has a positive slope if it goes up from left to right. This means that as we increase the value of x, the value of y also increases. For example, the line y = 2x + 5 has a positive slope of 2. To find the slope, we can pick any two points on the line and plug them into the formula. For instance, if we choose (0, 5) and (1, 7), we get m = (7 - 5) / (1 - 0) = 2 / 1 = 2.
A line has a negative slope if it goes down from left to right. This means that as we increase the value of x, the value of y decreases. For example, the line y = -3x + 2 has a negative slope of -3. To find the slope, we can use the same formula as before. For instance, if we choose (0, 2) and (1, -1), we get m = (-1 - 2) / (1 - 0) = -3 / 1 = -3.
A line has a zero slope if it is horizontal. This means that the value of y does not change as we move along the x-axis. For example, the line y = 4 has a zero slope. To find the slope, we can use any two points on the line and plug them into the formula. We will always get m = 0 / (x2 - x1) = 0 for any value of x.
A line has an undefined slope if it is vertical. This means that the value of x does not change as we move along the y-axis. For example, the line x = -2 has an undefined slope. To find the slope, we cannot use the formula because it would involve dividing by zero, which is not possible. We can say that m is undefined or does not exist for a vertical line.
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# Into Math Grade 3 Module 13 Lesson 4 Answer Key Represent and Name Fractions on a Number Line
We included HMH Into Math Grade 3 Answer Key PDF Module 13 Lesson 4 Represent and Name Fractions on a Number Line to make students experts in learning maths.
## HMH Into Math Grade 3 Module 13 Lesson 4 Answer Key Represent and Name Fractions on a Number Line
I Can identify, describe, represent, and locate fractions on a number line.
Michaela works for the Park Service. Her job is to place trail markers on the trail from the Ranger Station to Baldy Mountain. The markers should be placed at locations that mark $$\frac{1}{4}$$, $$\frac{2}{4}$$, $$\frac{3}{4}$$, and $$\frac{4}{4}$$ of the distance from the Ranger Station to Baldy Mountain.
Show each of Michaela’s markers on the number line. Label the distance of each marker from the Ranger Station.
First, we have to know the representation of fractions on a number line.
Representing fractions on a number line means that we can plot fractions on a number line, which is similar to plotting whole numbers and integers. Fractions represent parts of a whole. So, fractions on the number line are represented by making equal parts of a whole i.e. 0 to 1, and the number of those equal parts would be the same as the number written in the denominator of the fraction.
Turn and Talk Trace the number line on a sheet of paper and fold the paper to show the trail marker locations. How do the folds on your paper number line compare to the labels on your drawing?
Build Understanding
Question 1.
On the way to school, Leo stops at Blackberry Pond. What fraction of the 1-mile distance to school has Leo gone?
Complete the fractions on the number line.
A. Into how many equal lengths is the whole mile divided?
To plot fractions on the number line, follow the steps given below:
Step 1: Draw a number line of a suitable length.
Step 2: check the fractions once. If the given fraction is a proper fraction then mark points 0 and 1 on the number line. If it is an improper fraction, then we have to convert it into a mixed fraction, and then mark two integers between those the given fraction lies.
Step 3: Draw an equal number of parts of the numbers marked in step 2 which will be equal to the denominator of the fraction.
Step 4: Starting from the left point, count forward the number of parts shown by the numerator.
Step 5: Mark the point on the line.
In the above problem, Leo walked up to the blackberry pond and he stopped there. If he walks about 2 more steps then he reached the school.
– If we add 1/6 to every next number he reaches the school within one mile.
B. Which part of a fraction shows the number of equal lengths in the whole, the numerator or the denominator? Which part of the fraction shows the number of equal lengths you are counting?
The last part will show the number of equal lengths in the whole.
1 is represented as a whole which is 6/6.
1/6 shows the number of equal lengths.
1 is numerator and 6 is the denominator.
When I was counting the fraction shows the equal length is 1/6 every time.
C. Locate and draw a point on the number line to show the distance from Leo’s home to Blackberry Pond.
To plot fractions on the number line, follow the steps given below:
Step 1: Draw a number line of a suitable length.
Step 2: check the fractions once. If the given fraction is a proper fraction then mark points 0 and 1 on the number line. If it is an improper fraction, then we have to convert it into a mixed fraction, and then mark two integers between those the given fraction lies.
Step 3: Draw an equal number of parts of the numbers marked in step 2 which will be equal to the denominator of the fraction.
Step 4: Starting from the left point, count forward the number of parts shown by the numerator.
Step 5: Mark the point on the line.
D. What fraction represents 1 equal length of the whole? What fraction names the distance from Leo’s home to Blackberry Pond?
1 is represented as a whole infraction.
the fraction 6/6 represents 1 equal length of the whole.
the fraction names the distance from Leo’s home to Blackberry pond
The starting point is 0 and from there one-sixths, two-sixths, and up to 4/6.
For every step, we are adding 1/6.
Question 2.
What fraction does point A show?
A. Into how many equal lengths is the distance from 0 to 1 divided?
Answer: Representing fractions on a number line means that we can plot fractions on a number line, which is similar to plotting whole numbers and integers. Fractions represent parts of a whole. So, fractions on the number line are represented by making equal parts of a whole i.e. 0 to 1, and the number of those equal parts would be the same as the number written in the denominator of the fraction.
The distance from 0 to 1 is divided into 4 equal increments.
The third mark represents 3/4 because it is 3/4 of the distance from 0 to 1.
The point represents the fraction 3/4.
B. How many equal lengths are being counted?
Answer: There are four equal lengths.
Start from 0 to 1
0/4, 1/4, 2/4, 3/4, 4/4.
C. What is the fraction?
Answer: A fraction represents a numerical value, which defines the parts of a whole. Generally, the fraction can be a portion of any quantity out of the whole thing and the whole can be any specific things or value. The basics of fractions explain the top and bottom numbers of a fraction. The top number represents the number of selected or shaded parts of a whole whereas the bottom number represents the total number of parts.
Properties of fraction:
Similar to real numbers and whole numbers, a fractional number also holds some of the important properties. They are:
– Commutative and associative properties hold true for fractional addition and multiplication.
– The identity element of fractional addition is 0, and fractional multiplication is 1.
– The multiplicative inverse of a/b is b/a, where a and b should be non zero elements.
– Fractional numbers obey the distributive property of multiplication over addition.
Question 3.
Enya and Kat walk from home to the local history museum. When they are $$\frac{1}{3}$$ of the way to the museum, they stop at the library. They stop again to read their library books at the park when they are $$\frac{2}{3}$$ of the way to the museum. Complete the number line to show how to represent these distances.
To plot fractions on the number line, follow the steps given below:
Step 1: Draw a number line of a suitable length.
Step 2: check the fractions once. If the given fraction is a proper fraction then mark points 0 and 1 on the number line. If it is an improper fraction, then we have to convert it into a mixed fraction, and then mark two integers between those the given fraction lies.
Step 3: Draw an equal number of parts of the numbers marked in step 2 which will be equal to the denominator of the fraction.
Step 4: Starting from the left point, count forward the number of parts shown by the numerator.
Step 5: Mark the point on the line.
The above-given question Enya and Kat walk from home to the local history museum.
On the way to the museum is 1/3. And from there they stopped at the library and read library books at the park that is 2/3. We need to represent these ways on the number line. The number line is drawn below:
Turn and Talk On a number line, what does the length between each whole number represent? What does the length between each mark represent? How do you know?
The length between each whole number represents is 3/3 is nothing but 1.
The length between each mark represents 1/3.
Check Understanding
Question 1.
Complete the number line. Label the fractions. Locate and draw a point on the number line to show $$\frac{5}{8}$$.
To plot fractions on the number line, follow the steps given below:
Step 1: Draw a number line of a suitable length.
Step 2: check the fractions once. If the given fraction is a proper fraction then mark points 0 and 1 on the number line. If it is an improper fraction, then we have to convert it into a mixed fraction, and then mark two integers between those the given fraction lies.
Step 3: Draw an equal number of parts of the numbers marked in step 2 which will be equal to the denominator of the fraction.
Step 4: Starting from the left point, count forward the number of parts shown by the numerator.
Step 5: Mark the point on the line.
Write the fractions that name points E and F
Question 2.
point E
To plot fractions on the number line, follow the steps given below:
Step 1: Draw a number line of a suitable length.
Step 2: check the fractions once. If the given fraction is a proper fraction then mark points 0 and 1 on the number line. If it is an improper fraction, then we have to convert it into a mixed fraction, and then mark two integers between those the given fraction lies.
Step 3: Draw an equal number of parts of the numbers marked in step 2 which will be equal to the denominator of the fraction.
Step 4: Starting from the left point, count forward the number of parts shown by the numerator.
Step 5: Mark the point on the line.
By applying all those steps we can continue the number line by pointing E and F.
The complete number line is shown below:
The point E represents 3/6.
Question 3.
point F
The point F represents 4/6.
By using the above number line we can write the point F fraction.
Question 4.
Use Tools Reese’s poster is $$\frac{2}{4}$$ yard wide. Complete the number line. Label the fractions. Locate and draw a point on the number line to show the width of Reese’s poster.
To plot fractions on the number line, follow the steps given below:
Step 1: Draw a number line of a suitable length.
Step 2: check the fractions once. If the given fraction is a proper fraction then mark points 0 and 1 on the number line. If it is an improper fraction, then we have to convert it into a mixed fraction, and then mark two integers between those the given fraction lies.
Step 3: Draw an equal number of parts of the numbers marked in step 2 which will be equal to the denominator of the fraction.
Step 4: Starting from the left point, count forward the number of parts shown by the numerator.
Step 5: Mark the point on the line.
The whole length is 4/4
The length in each mark represents 1/4 to next fraction.
Question 5.
Abe marks points H and l on the number line. Write the fraction that names each point.
To plot fractions on the number line, follow the steps given below:
Step 1: Draw a number line of a suitable length.
Step 2: check the fractions once. If the given fraction is a proper fraction then mark points 0 and 1 on the number line. If it is an improper fraction, then we have to convert it into a mixed fraction, and then mark two integers between those the given fraction lies.
Step 3: Draw an equal number of parts of the numbers marked in step 2 which will be equal to the denominator of the fraction.
Step 4: Starting from the left point, count forward the number of parts shown by the numerator.
Step 5: Mark the point on the line.
Question 6.
Reason Izzy marks a number line to show eighths. What is the fraction that is located halfway between $$\frac{6}{8}$$ and $$\frac{8}{8}$$? Explain.
To plot fractions on the number line, follow the steps given below:
Step 1: Draw a number line of a suitable length.
Step 2: check the fractions once. If the given fraction is a proper fraction then mark points 0 and 1 on the number line. If it is an improper fraction, then we have to convert it into a mixed fraction, and then mark two integers between those the given fraction lies.
Step 3: Draw an equal number of parts of the numbers marked in step 2 which will be equal to the denominator of the fraction.
Step 4: Starting from the left point, count forward the number of parts shown by the numerator.
Step 5: Mark the point on the line.
the whole length of the fraction is 8/8 nothing but 1.
The length between each mark represents 1/8.
I’m in a Learning Mindset!
Would drawing my own diagrams help me better understand how to place fractions on a number line?
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# PSAT Math : How to find the endpoints of a line segment
## Example Questions
### Example Question #1 : How To Find The Endpoints Of A Line Segment
The midpoint of line segment AB is (2, -5). If the coordinates of point A are (4, 4), what are the coordinates of B?
(0, -13)
(0, -14)
(6, 14)
(6, 13)
(3, -0.5)
(0, -14)
Explanation:
The fastest way to find the missing endpoint is to determine the distance from the known endpoint to the midpoint and then performing the same transformation on the midpoint. In this case, the x-coordinate moves from 4 to 2, or down by 2, so the new x-coordinate must be 2-2 = 0. The y-coordinate moves from 4 to -5, or down by 9, so the new y-coordinate must be -5-9 = -14.
An alternate solution would be to substitute (4,4) for (x1,y1) and (2,-5) for (x,y) into the midpoint formula:
x=(x1+x2)/2
y=(y1+y2)/2
Solving each equation for (x2,y2) yields the solution (0,-14).
### Example Question #1 : Midpoint Formula
Point A is (5, 7). Point B is (x, y). The midpoint of AB is (17, –4). What is the value of B?
(8.5, –2)
(29, –15)
(12, –11)
(22, –9)
(29, –15)
Explanation:
Point A is (5, 7). Point B is (x, y). The midpoint of AB is (17, –4). What is the value of B?
We need to use our generalized midpoint formula:
MP = ( (5 + x)/2, (7 + y)/2 )
Solve each separately:
(5 + x)/2 = 17 → 5 + x = 34 → x = 29
(7 + y)/2 = –4 → 7 + y = –8 → y = –15
Therefore, B is (29, –15).
### Example Question #1 : Midpoint Formula
Line segment AB has an endpoint, A, located at , and a midpoint at . What are the coordinates for point B of segment AB?
The second endpoint cannot exist
Explanation:
With an endpoint A located at (10,-1), and a midpoint at (10,0), we want to add the length from A to the midpoint onto the other side of the segment to find point B. The total length of the segment must be twice the distance from A to the midpoint.
A is located exactly one unit below the midpoint along the y-axis, for a total displacement of (0,1). To find point B, we add (10+0, 0+1), and get the coordinates for B: (10,1).
### Example Question #1 : How To Find The Endpoints Of A Line Segment
One endpoint of a line segment on the coordinate plane is . The segment has length 10; the other endpoint has -coordinate 10 and is in Quadrant I. Give the -coordinate of the other endpoint. (Nearest tenth if applicable).
Explanation:
Let be the -coordinate of the other endpoint.
The segment has endpoints and .
Apply the distance formula
setting ,
Therefore, there are two values of which fit the distance criterion.
One is .
However, since the endpoint is in Quadrant I, it must have a positive coordinate, so this is eliminated as a choice.
The other is , which is correct since it is positive.
### Example Question #2 : How To Find The Endpoints Of A Line Segment
The midpoint of a line segment is the point . One endpoint is ; give the -coordinate of the other.
Explanation:
Let the endpoints of a line segment be
Then the midpoint of the segment will be
The -coordinate of the midpoint is , and the -coordinate of the known endpoint is , so we can solve for , the -coordinate of the unknown endpoint, in the following equation:
### Example Question #1 : How To Find The Endpoints Of A Line Segment
The midpoint of a line segment is the point . One endpoint is ; give the -coordinate of the other.
Explanation:
Let the endpoints of a line segment be
Then the midpoint of the segment will be
The -coordinate of the midpoint is , and the -coordinate of the known endpoint is , so we can solve for , the -coordinate of the unknown endpoint, in the following equation:
### Example Question #4 : How To Find The Endpoints Of A Line Segment
The midpoint of a line segment is the point . One endpoint is ; give the -coordinate of the other.
Explanation:
Let the endpoints of a line segment be
Then the midpoint of the segment will be
The -coordinate of the midpoint is , and the -coordinate of the known endpoint is
so we can solve for , the -coordinate of the unknown endpoint, in the following equation:
### Example Question #5 : How To Find The Endpoints Of A Line Segment
The midpoint of a line segment is the point . One endpoint is ; give the -coordinate of the other.
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# Examples of Recursive Problems
## Recursion Algorithms - Introduction to Recursion - Examples of Recursive Problems
### Introduction
Recursion is a fundamental concept in computer science and programming. It refers to solving a problem by breaking it down into smaller, similar subproblems until a base case is reached. A recursive algorithm calls itself to solve these subproblems, ultimately providing a solution to the original problem.
Recursion can be a powerful technique that simplifies problem-solving, especially when dealing with complex and repetitive tasks. However, it requires a good understanding of the core principles and careful consideration of base cases to avoid infinite loops.
### Basic Principles
To understand recursion better, let's discuss its basic principles:
1. Base Case: A base case defines an exit condition that stops the recursive calls and returns a result. It represents the simplest form of the problem that doesn't require further recursion. Without a proper base case, a recursive algorithm may run indefinitely, leading to a stack overflow or other errors.
2. Recursive Step: The recursive step defines the problem in terms of smaller subproblems that can be solved in a similar manner. It involves calling the same function with new parameters, bringing the problem closer to the base case.
### Examples of Recursive Problems
Now that we have grasped the core concept of recursion, let's explore a few examples of recursive problems.
#### Factorial Function
The factorial of a non-negative integer `n`, denoted as `n!`, is defined as the product of all positive integers less than or equal to `n`. The factorial function can be implemented recursively as follows:
``````def factorial(n):
if n == 0:
return 1
else:
return n * factorial(n-1)
``````
In this example, the base case is when `n` equals 0, where we return 1 as per the definition of factorial. Otherwise, we calculate and return the product of `n` and the factorial of `n-1`, which reduces the problem size at each recursive call.
#### Fibonacci Sequence
The Fibonacci sequence is a classic example of recursion. Each number in the sequence is the sum of the two preceding numbers, starting with 0 and 1. We can define a recursive function to compute the nth Fibonacci number:
``````def fibonacci(n):
if n <= 1:
return n
else:
return fibonacci(n-1) + fibonacci(n-2)
``````
Here, the base case is when `n` is less than or equal to 1, where we return `n` itself. Otherwise, we recursively compute the Fibonacci number by adding the two numbers preceding it (`n-1` and `n-2`).
#### Binary Search
Binary search is a widely used search algorithm that efficiently finds a target value within a sorted array. It works by repeatedly dividing the search space in half. A recursive implementation of binary search can be achieved as follows:
``````def binary_search(arr, target, low, high):
if low > high:
return -1
else:
mid = (low + high) // 2
if arr[mid] == target:
return mid
elif arr[mid] < target:
return binary_search(arr, target, mid+1, high)
else:
return binary_search(arr, target, low, mid-1)
``````
In this example, the base case is when `low` becomes greater than `high`, indicating that the target value is not present in the array. Otherwise, we calculate the middle index `mid` and compare the element at `arr[mid]` with the target value. Based on the comparison, we either recursively search in the left half or the right half of the array.
### Conclusion
Recursion is a powerful technique that allows programmers to solve complex problems by breaking them down into smaller, manageable subproblems. Understanding the basic principles, defining appropriate base cases, and structuring recursive steps are crucial for writing recursive algorithms.
In this post, we explored the concept of recursion, its basic principles, and examined examples of recursive problems such as the factorial function, Fibonacci sequence, and binary search. These examples demonstrate how recursion can simplify problem-solving and provide elegant solutions.
To summarize, recursion is an essential tool in a programmer's arsenal, and mastering it opens up a world of possibilities in algorithm design and implementation.
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Åk 6–9
English/Soomaali
3.1 Writing Algebraic Expressions
If you have gone through part 2 in this section, you have seen that you can express a problem in many different ways; in words, figures, tables, and finally algebraic expressions. You should even have learned moer about how to write algebraic expressions.
Jasmine works in the summer as a landscaper. One day she is asked to plant flowers in ten rows. In every row there is to be three flowers. How many flowers does Jasmine plant?
This is how the picture of the flowers looks:
If we express in words, this is what it looks like:
3 times the number of rows gives us the total number of flowers.
It can even be written as an algebraic expression:
Total number of flowers: f
total rows: r
number of flowers per row: 3
How do you think the algebraic expression for the number of flowers for the new planting would have looked?
We begin by looking at how many flowers there are in every row.
There are 4 flowers.
So the formula must be: f = 4 · r
f stands for the total number of flowers.
r stands for the number of rows.
The number of rows is 8. We replace the letter r with the number 8.
The number of flowers that are to be planted then becomes:
f = 4 · 8
f = 32
As you can see f and b each take different values, f and b are variables.
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## Revision Notes of Chapter 12 Heron's Formula Class 9th Math
Topics in the Chapter
• Basics Revision
• Area of Triangle
• Heron's Formula
• Area of equilateral Triangle
Basics Revision
• Area of a triangle = 1/2 × Base × Height
Heron’s formula: For sides of a triangle being a, b, c:
Area of Triangle =
where s is semi-perimeter and s = (a+b+c)/2
• For finding area of a quadrilateral we divide it into various triangles. Then we use Heron’s formula to find the area of the triangles.
Area of Triangles
We have already learnt that area of a triangle
= 1/2 × (side) × [altitude corresponding to that side (or height)]
= 1/2 × BC × AD
= 1/2 × a × h
Thus, area of a triangle = 1/2 × Base × Height
Note:
Unit of measurement for area of any plane figure is taken as square metre (m2) or square centimetre (cm2), etc.
Area of Triangle using Heron's Formula
When it is not possible to find the height of the triangle easily and measures of all the three sides are known then we use Heron’s formula, which is given by:
Area of a triangle =
where a, b and c are the sides of the triangle and s = semi-perimeter, i.e. half the perimeter of the triangle = (a+b+c)/2
Area of an Equilateral Triangle
Let the side of the equilateral triangle be ‘a’.
⇒ Area of the triangle
Thus, area of an equilateral triangle = √3/4 × side2
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# Sarah is 25 years older than her son Gavin. In ten years, Sarah will be twice Gavin age. How old are Sarah and Gavin?
Feb 25, 2018
Sarah is 40 and Gavin is 15
#### Explanation:
Let (G+25) be Sarah's age and G be Gavin's age
$S = G + 25$
$G + 25 = 2 \left(G + 10\right)$
$G = 15$
$S = 40$
Feb 25, 2018
Below.
#### Explanation:
Let Garvin's age be $= x$ and Sarah's age $= x + 25$
After $10$ years
Garvin's age $= x + 10$
Sarah's age $= 2 \left(x + 10\right) \mathmr{and} x + 35$
According to the question
$2 \left(x + 10\right) = x + 35$
$2 x + 20 = x + 35$
$2 x - x = 35 - 20$
$x = 15$
Therefore, Gavin's age is $15$ years and Sarah's age is $40$ years.
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# How do you find (d^2y)/(dx^2) for x^2+4y^2=5?
Nov 6, 2016
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \left[{x}^{2} + 4 {y}^{2} = 5\right] = \frac{- 5}{16 {y}^{3}}$
#### Explanation:
First find $\frac{\mathrm{dy}}{\mathrm{dx}}$ by implicitly differentiating ${x}^{2} + 4 {y}^{2} = 5$:
${x}^{2} + 4 {y}^{2} = 5$
$2 x + 8 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(8 y\right) = - 2 x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x}{8 y}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- x}{4 y}$
Now implicitly differentiate $\frac{\mathrm{dy}}{\mathrm{dx}}$:
Use the quotient rule:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\left(4 y\right) \left(- 1\right) - \left(- x\right) \left(4 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right)}{{\left(4 y\right)}^{2}}$
Simplify:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 4 y + 4 x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{16 {y}^{2}}$
Substitute $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- x}{4 y}$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- \cancel{4} y + \cancel{4} x \left(\frac{- x}{4 y}\right)}{4 \cancel{16} {y}^{2}}$
Simplify:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left[\frac{- y}{4 {y}^{2}}\right] - \left[\frac{{x}^{2}}{\left(4 y\right) \left(4 {y}^{2}\right)}\right]$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \left[\frac{- 1}{4 y}\right] - \left[\frac{{x}^{2}}{16 {y}^{3}}\right]$
Make a common denominator to combine into one fraction:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 4 {y}^{2} - {x}^{2}}{16 {y}^{3}}$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- \left(4 {y}^{2} + {x}^{2}\right)}{16 {y}^{3}}$
Recall that the original equation states that $4 {y}^{2} + {x}^{2} = 5$, so:
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 5}{16 {y}^{3}}$
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A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm,
Question:
A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85cm (See figure 18.5). The thickness of the plank is 5cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2. Find the total expenses required for polishing and painting the surface of the bookshelf.
Solution:
External length of book shelf = 85 cm
Height = 110 cm
External surface area of shelf while leaving front face of shelf
lh + 2(lb + bh)
[85 * 110 + 2(85 * 25 + 25 * 110)]
$19100 \mathrm{~cm}^{2}$
Area of Front face $=\left[85^{*} 110-75^{*} 100+2\left(75^{*} 5\right)\right] \mathrm{cm}^{2}$
$=1850+750 \mathrm{~cm}^{2}$
$=2600 \mathrm{~cm}^{2}$
Area to be polished $=19100+2600 \mathrm{~cm}^{2}$
$=21700 \mathrm{~cm}^{2}$
Cost of polishing $1 \mathrm{~cm}^{2}$ area $=$ Rs. $0.20$
Cost of polishing $21700 \mathrm{~cm}^{2}$ area $=21700 * 0.20$
= Rs. 4340
Now, Length (l), breadth (b), height ( $h$ ) of each row of book shelf is $75 \mathrm{~cm}, 20 \mathrm{~cm}$ and $30 \mathrm{~cm}=\left(\frac{110-20}{3}\right)$
respectively.
Area to be painted in 1 row = 2(l + h)b + lh
$[2(75+30) * 20+75 * 30)] \mathrm{cm}^{2}$
$(4200+2250) \mathrm{cm}^{2}$
$6450 \mathrm{~cm}^{2}$
Area to be painted in 3 rows = 3 * 6450
$=$ Rs. $19350 \mathrm{~cm}^{2}$
Cost of painting $1 \mathrm{~cm}^{2}$ area $=$ Rs. $0.10$
Cost of painting $19350 \mathrm{~cm}^{2}$ area $=19350 * 0.10$
= Rs.1935
Total expense required for polishing and painting the surface of the bookshelf = 4340 + 1935
= Rs.6275
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# Square Root By Prime Factorization
Created by: Team Maths - Examples.com, Last Updated: May 10, 2024
## Square Root By Prime Factorization
Square root prime factorization is a method used in mathematics to break down a number into its prime factors to easily find its square and square root. This technique is particularly useful in algebra and number theory, where understanding the properties of integers and their relationships is crucial. The process also intersects with concepts of rational and irrational numbers, especially when determining if a square root is a perfect square or not. Additionally, the method ties into statistical analysis and data fitting through the least squares method, which minimizes the discrepancies between observed values and those predicted by a model. This approach is foundational in various branches of mathematics, including algebra, statistics, and arithmetic involving square and square roots.
## What is Square Root By Prime Factorization?
Square Root Prime Factorization is a mathematical process used to determine the square root of a number by first breaking it down into its prime factors. This method simplifies the extraction of roots, particularly when dealing with large numbers, by utilizing the fundamental properties of primes.
## How to Find Square Root By Prime Factorisation?
Finding the square root of a number through prime factorization is a systematic method that breaks down a number into its basic prime components. This process is especially useful when dealing with large numbers or when simplifying square roots in algebra. Below are detailed steps on how to find the square root by prime factorization:
### Step 1: Factorize the Number into Prime Factors
Begin by dividing the number by the smallest prime number (usually 2) that can exactly divide it without leaving a remainder. Continue this process with the quotient until you reach a prime number. This will give you the prime factors of the number.
### Example
To factorize 72:
• 72÷2 = 36
• 36÷2 = 18
• 18÷2 = 9
• 9÷3 = 3
• 3÷3 = 1
• Prime factors of 72 are: 2³×3²
### Step 2: Organize the Prime Factors into Pairs
Arrange the prime factors in a visible format (like a list) and pair identical factors. If a factor does not have a pair, it remains single.
### Step 3: Multiply One Element from Each Pair
For each pair of identical primes, select one prime from the pair and multiply them. If a prime remains unpaired, it stays under the square root.
### Example
From the prime factors 2³×3²:
• Pairs are: (2,2),(2),(3,3)
• Multiply one element from each pair: 2×3
### Step 4: Calculate the Square Root
Multiply the results of the previous step together, and multiply this product by the square root of any leftover unpaired primes to find the square root of the original number.
## Example
• Paired result is 2×3 = 6
• Unpaired prime is 2
• The square root of 72 is 6×√2 or 6√2.
## Step 1: Prime Factorization
First, factorize 144 into its prime components:
• Divide by 2: 144÷2 = 72
• Divide by 2: 72÷2 = 36
• Divide by 2: 36÷2 = 18
• Divide by 2: 18÷2 = 9
• Divide by 3: 9÷3 = 3
• Divide by 3: 3÷3 = 1
• Prime factors are 2⁴×3².
## Step 2: Pair the Prime Factors
• Pairs: (2,2),(2,2),(3,3)
## Step 3: Multiply the Paired Factors
• Multiply one from each pair: 2×2×3 = 12
## Step 4: Formulate the Square Root
Since all factors are paired, the square root is:
• √144 = 12
## Example 2: Square Root of 200
### Step 1: Prime Factorization
Factorize 200:
• Divide by 2: 200÷2 = 100
• Divide by 2: 100÷2 = 50
• Divide by 2: 50÷2 = 25
• Divide by 5: 25÷5 = 5
• 5 is a prime number.
Prime factors are 2²×5².
### Step 2: Pair the Prime Factors
• Pairs: (2,2),(5,5)
• Unpaired: 2
### Step 3: Multiply the Paired Factors
• Multiply one from each pair: 2×5 = 10
### Step 4: Formulate the Square Root
• Since there is an unpaired factor, the square root is:
• √200=10√2.
## What is the prime factorization method for finding square roots?
The prime factorization method involves breaking down a number into its prime factors, pairing them, and using these pairs to simplify the extraction of the square root. This method is especially useful for large numbers and helps in determining whether the square root is a perfect square or involves irrational numbers.
## How do you determine if a number is a perfect square using prime factorization?
A number is a perfect square if all its prime factors can be paired without any remainder. During the factorization process, if every prime factor can be grouped into identical pairs, then the number is a perfect square. If any prime factor remains unpaired, the number is not a perfect square.
## Can the prime factorization method be used for any number?
Yes, the prime factorization method can be used for any positive integer. However, the method is most effective for numbers where prime factors can be easily determined. For very large numbers or numbers involving large prime factors, the process can be time-consuming without the aid of computational tools.
## Can this method be applied to algebraic expressions?
Yes, the prime factorization method can also be applied to algebraic expressions involving square roots. Factors in polynomial expressions that are perfect squares can be treated similarly to numerical prime factors, allowing for simplification of square roots in algebra.
## What do you do with unpaired prime factors when finding square roots?
Unpaired prime factors remain under the square root symbol in the final expression. If a prime factor does not have a pair, it indicates that the square root of the number will be an irrational number, and the unpaired prime factor will be part of the radical expression.
Text prompt
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## Presentation on theme: "Adding and Subtracting Integers."— Presentation transcript:
Absolute Value -7 +7 -5 5 10 -10 A number line has many functions. Previously, we learned that numbers to the right of zero are positive and numbers to the left of zero are negative. By putting points on the number line, we can graph values. If one were to start at zero and move seven places to the right, this would represent a value of positive seven. If one were to start at zero and move seven places to the left, this would represent a value of negative seven.
Absolute Value -7 +7 -5 5 10 -10 Both of these numbers, positive seven (+7) and negative seven (-7), represent a point that is seven units away from the origin. The absolute value of a number is the distance between that number and zero on a number line. Absolute value is shown by placing two vertical bars around the number as follows: 5 The absolute value of five is five. -3 The absolute value of negative three is three.
What is 5 + 7? We can show how to do this by using algebra tiles. + = 5 + 7 = 12 What is –5 + -7? = + -12 -5 + -7 =
We can show this same idea using a number line. 9 -5 5 10 -10 What is 5 + 4? Move five (5) units to the right from zero. Now move four more units to the right. The final point is at 9 on the number line. Therefore, = 9.
-9 -5 5 10 -10 What is -5 + (-4)? Move five (5) units to the left from zero. Now move four more units to the left. The final point is at -9 on the number line. Therefore, -5 + (-4) = -9.
To add integers with the same sign, add the absolute values of the integers. Give the answer the same sign as the integers. Examples Solution
Additive Inverse What is (-7) + 7? -5 5 10 -10
10 -10 To show this, we can begin at zero and move seven units to the left. Now, move seven units to the right. Notice, we are back at zero (0). For every positive integer on the number line, there is a corresponding negative integer. These integer pairs are opposites or additive inverses. Additive Inverse Property – For every number a, a + (-a) = 0
Additive Inverse When using algebra tiles, the additive inverses make what is called a zero pair. For example, the following is a zero pair. 1 + (-1) = 0. This also represents a zero pair. x + (-x) = 0
Add the following integers: (-4) + 7. -5 5 10 -10 Start at zero and move four units to the left. Now move seven units to the right. The final position is at 3. Therefore, (-4) + 7 = 3.
Add the following integers: (-4) + 7. -5 5 10 -10 Notice that seven minus four also equals three. In our example, the number with the larger absolute value was positive and our solution was positive. Let’s try another one.
-5 5 10 -10 Start at zero and move nine places to the left. Now move three places to the right. The final position is at negative six, (-6). Therefore, (-9) + 3 = -6.
-5 5 10 -10 In this example, the number with the larger absolute value is negative. The number with the smaller absolute value is positive. We know that 9-3 = 6. However, (-9) + 3 = -6. 6 and –6 are opposites. Comparing these two examples shows us that the answer will have the same sign as the number with the larger absolute value.
To add integers with different signs determine the absolute value of the two numbers. Subtract the smaller absolute value from the larger absolute value. The solution will have the same sign as the number with the larger absolute value. Example Subtract Solution
Subtracting Integers We can show the addition of numbers with opposite signs by using algebra tiles. For example, 3 + (-5) would look like this: + Group the zero pairs. Remove the zero pairs
Subtracting Integers We can show the addition of numbers with opposite signs by using algebra tiles. For example, 3 + (-5) would look like this: + = Group the zero pairs. Remove the zero pairs The remainder is the solution. Therefore, 3 + (-5) = -2
Subtracting Integers Let’s see how subtraction works using algebra tiles for Begin with three algebra tiles. Now, take away or remove five tiles from these three. It can’t be done. However, we can add zero pairs until we have five ones because adding zero doesn’t change the value of the number.
Subtracting Integers Let’s see how subtraction works using algebra tiles for Begin with three algebra tiles. Now remove the five ones.
Subtracting Integers Let’s see how subtraction works using algebra tiles for Begin with three algebra tiles. Now remove the five ones. The remainder is negative two (-2). Therefore, 3 – 5 = -2. This is the same as 3 + (-5). To subtract a number, add its additive inverse. For any numbers a and b, a – b = a + (-b).
You Try It Find each sum or difference.
-24 – (-40) – (-14) – Simplify each expression. r – 27r c – (-12c) x + 45x Evaluate each expression. 89 -14 -7 c + 5 if c = -19
Solutions
Solutions
Solutions 13. c + 5 if c = -19
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How do you substitute values in algebraic expressions?
To substitute a number into an algebraic expression, all you need to do is re-write the expression in exactly the same way, except replacing the variable (letter) with the number. It always makes it clearer to put the number in brackets too. Then you can simplify your new expression and you have your answer!
What is substituting into an expression?
Substitution means putting numbers in place of letters to calculate the value of an expression .
How do you translate an expression into an algebraic expression?
When translating phrases into algebraic expressions, you need to identify keywords and phrases which specifically refer to a mathematical operation (addition, subtraction, multiplication, and division). Usually, you can write out the algebraic expression of the verbal description in the order that it is said.
What is an evaluating expression?
To evaluate an algebraic expression means to find the value of the expression when the variable is replaced by a given number. To evaluate an expression, we substitute the given number for the variable in the expression and then simplify the expression using the order of operations.
Do you need parentheses when substituting?
If substituting one variable for another, then parentheses are never needed (the order of operations is clearly identical before and after). If substituting a whole number, then only the situation of juxtaposed numbers after the coefficient can apply.
How do you substitute expressions?
To use the substitution method, use one equation to find an expression for one of the variables in terms of the other variable. Then substitute that expression in place of that variable in the second equation. You can then solve this equation as it will now have only one variable.
How do you write and evaluate algebraic expressions?
To evaluate an algebraic expression, you have to substitute a number for each variable and perform the arithmetic operations. In the example above, the variable x is equal to 6 since 6 + 6 = 12. If we know the value of our variables, we can replace the variables with their values and then evaluate the expression.
Does adding parentheses change the value of the expression?
As you can see, adding parentheses can really change the outcome of our expression! Since this first equation tells us the value of x, we can actually substitute this for x in the second equation using parentheses. So now, instead of 3x−7y, we have 3(2y–3)–7y=−14.
What is this algebraic expression worksheet?
This worksheet also features negative numbers, brackets and powers. A worksheet for pupils to practise creating an algebraic expression and then substituting into it. Can be used as a standalone worksheet or as an extension task for the worksheet above.
How to simplify algebraic expressions?
20 Simplifying Algebraic Expressions by Distributing and Combining Like Terms 1.a 3(x + 6y – 7) 1.b. -7(2x – 4) Distribute first…. then catch and combine like terms.
How do you use distributive property in algebraic expressions?
13 Use the Distributive Property to expand each expression. Ex. -4(2n + 5) a. 3(4x + 2) b. -5( y – 7) c. -2(n + 9) d. 3(-1 – 5c) e. x(3 + 4y) f. -1(a – 1) g. ½(4x – 6) h. − 1 4 (8x + 12) i. 3 4 (8x – 12) x P 14 Expand the expressions that require the distributive property.
What is a substitute worksheet?
A worksheet for practising substituting numbers into an algebraic expression. This worksheet also features negative numbers, brackets and powers. A worksheet for pupils to practise creating an algebraic expression and then substituting into it. Can be used as a standalone worksheet or as an extension task for the worksheet above.
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# Find the points which divide the line segment joining $(8,12)$ and $(12,8)$ into four equal parts.
Last updated date: 11th Aug 2024
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Views today: 8.10k
Answer
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410.1k+ views
Hint: Draw a rough sketch of a line joining the coordinates $(8,12)$ and $(12,8)$ . Then mark the four points on the line segment, assuming that they divide the line segment into four equal parts. Then use the property of the coordinate of midpoint, to find the coordinate of all the three points.
Complete step-by-step answer:
Observe the diagram
Let $A(8,12)$ and $B(12,8)$ be the two given points.
Let $C(a,b),D(b,c),E(p,q)$ be the three points that divide the line AB into four equal parts.
By observing the diagram, we can conclude that D must be the midpoint AB.
Then by the property of mid-point, the coordinates of D can be given as
$c = \dfrac{{8 + 12}}{2}$ and $d = \dfrac{{12 + 8}}{2}$
$\Rightarrow c = \dfrac{{20}}{2} = 10$ and $d = \dfrac{{20}}{2} = 10$
Thus, the coordinate of D are $(10,10)$
Again, by observing the diagram, we can say that, C is the midpoint of AD
Then by the property of mid-point, the coordinates of C can be given as
$a = \dfrac{{8 + 10}}{2}$ and $b = \dfrac{{12 + 10}}{2}$
$\Rightarrow a = \dfrac{{18}}{2} = 9$ and $b = \dfrac{{22}}{2} = 11$
Thus, the coordinates of C are $(9,11)$
Now, by again observing the diagram, we can say that E is the midpoint of DB.
Then by the property of mid-point, the coordinates of E can be given as
$p = \dfrac{{10 + 12}}{2}$ and $q = \dfrac{{10 + 8}}{2}$
$\Rightarrow p = \dfrac{{22}}{2} = 11$ and $q = \dfrac{{18}}{2} = 9$
Thus, the coordinates of E are $(11,9)$
Hence, the points which divide the line segment $(8,12)$ and $(12,8)$ are $(9,11),(10,10),(11,9)$
So, the correct answer is “(9,11),(10,10),(11,9)”.
Note: If it doesn’t click to use the mid-point theorem then this question can also be solved by using, section formula. We can say that the point C divides AB in the ratio $1:4$ . Then find the value of C using section formula. And then repeat the same concept for finding other points as well. But knowing mid-point property will help solve this question easily and in a less number of steps.
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# GMAT Math : Exponents
## Example Questions
← Previous 1 3 4 5 6 7 8 9
### Example Question #1 : Exponents
cannot be determined
Explanation:
Putting these together,
Explanation:
Then,
### Example Question #1 : Exponents
Explanation:
Then putting them together,
### Example Question #1 : Understanding Exponents
Which of the following expressions is equivalent to this expression?
You may assume that .
Explanation:
### Example Question #1 : Exponents
Simplify the following expression without a calculator:
Explanation:
The easiest way to simplify is to work from the inside out. We should first get rid of the negatives in the exponents. Remember that variables with negative exponents are equal to the inverse of the expression with the opposite sign. For example, So using this, we simplify:
Now when we multiply variables with exponents, to combine them, we add the exponents together. For example,
Doing this to our expression we get it simplified to .
The next step is taking the inside expression and exponentiating it. When taking an exponent of a variable with an exponent, we actually multiply the exponents. For example, . The other rule we must know that is an exponent of one half is the same as taking the square root. So for the So using these rules,
### Example Question #2 : Exponents
Rewrite as a single logarithmic expression:
Explanation:
First, write each expression as a base 3 logarithm:
since
Rewrite the expression accordingly, and apply the logarithm sum and difference rules:
### Example Question #1 : Exponents
If , what is in terms of ?
Explanation:
We have .
So , and .
### Example Question #2 : Understanding Exponents
What are the last two digits, in order, of ?
Explanation:
Inspect the first few powers of 6; a pattern emerges.
As you can see, the last two digits repeat in a cycle of 5.
789 divided by 5 yields a remainder of 4; the pattern that becomes apparent in the above list is that if the exponent divided by 5 yields a remainder of 4, then the power ends in the diigts 96.
### Example Question #1 : Exponents
Which of the following expressions is equal to the expression
?
Explanation:
Use the properties of exponents as follows:
Simplify:
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# Class 10 Maths Chapter 8: Introduction to Trigonometry Ex 8.1
In this page we have NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry for EXERCISE 8.1 . Hope you like them and do not forget to like , social_share and comment at the end of the page.
Question 1
In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) Sin A, cos A
(ii) Sin C, cos C
Solution
In Δ ABC, right-angled at B ,using Pythagoras theorem we have
AC= AB2 +BC2 = 576 + 49 = 625
Or AC=25 ( taking positive value only)
Now
(i) In a right angle triangle ABC where B=90° ,
Sin A = BC/AC = 7/25
CosA = AB/AC = 24/25
(ii)
Sin C = AB/AC = 24/25
Cos C = BC/AC =7/25
Question 2
In below find tan P – cot R
Solution
Again consider the above figure . Now by Pythagoras theorem
PQ2 + QR2 =PR2
QR=5
Now
tan P = Perp/Base = 5/12
Cot R = Base/Perm = 5/12
So tan P – cot R=0
Question 3. If $sin A= \frac{3}{4}$. Calculate cos A and tan A.
Solution
Given $sin A= \frac{3}{4}$
Or P/H=3/4
Let P=3k and H=4k
Now By Pythagoras theorem
P2 + B2 =H2
9k2 + B2 =16k2
Or $B =+ k \sqrt 7$
Now $\cos A = {B \over H} = {{\sqrt 7 } \over 4}$
Now $\tan A = {{\sin A} \over {\cos A}} = {3 \over {\sqrt 7 }}$
Question 4
Given 15 cot A = 8, find sin A and sec A.
Solution 4
$Cot A =\frac{8}{15}$
Or
${B \over P} = {8 \over {15}}$
Let B=8K and P=15k
So in a right angle triangle with angle A
P2 + B2 =H2
Or H=17K
Sin A = P/H = 15/17
Sec A = H/B = 17/8
Question 5
Given sec θ =13/12. Calculate all other trigonometric ratios.
Solution 5
Given sec θ=13/12
Or
H/B=13/12
let H=13K ,B=12K
So in a right angle triangle with angle A
P2 + B2 =H2
P=5k
sin θ = P/H = 5/13
Cos θ = B/H = 12/13
Tan θ = P/B = 5/12
Cosec θ= 1/sin θ = 13/5
cot θ = 1/tan θ = 12/5
Question 6
If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.
Solution 6
In a triangle
Cos A =cos B
AC/AB= BC/AB
⇒AC=BC
⇒Angle A and Angle B
Question 7
If cot θ =7/8 evaluate
(i) ${{\left( {1 + \sin\theta } \right)\left( {1 - \sin\theta } \right)} \over {\left( {1 + \cos\theta } \right)\left( {1 - \cos\theta } \right)}}$
(ii) ${\cot ^2}\theta$
Solution
Given
Given cot θ=7/8
Or
B/P=7/8
let B=7K ,P=8K
So in a right angle triangle with angle θ
P2 + B2 =H2
$H = k\sqrt {113}$
$\sin \theta = {P \over H} = {8 \over {\sqrt {113} }}$
$\cos \theta = {B \over H} = {7 \over {\sqrt {113} }}$
(i) ${{\left( {1 + sin\theta } \right)\left( {1 - sin\theta } \right)} \over {\left( {1 + cos\theta } \right)\left( {1 - cos\theta } \right)}}$
$= {{1 - \sin^2 \theta } \over {1 - \cos^2 \theta }}$
${{1 - {{64} \over {113}}} \over {1 - {{49} \over {113}}}} = {{49} \over {64}}$
(ii)Cot2 θ=(cot θ)2= 49/64
Question 8
If 3 cot A = 4, check whether below is true or not
${{1 - \tan^2 A} \over {1 + \tan^2 A}} = \cos^2 A - \sin^2 A$
Solution 8
Given
Cot A=4/3
Or
B/P=4/3
Let B=4k and P=3k
So in a right angle triangle with angle A
P2 + B2 =H2
H=5k
Now tan A=1/cot A=3/4
Cos A=B/H=4/5
Sin A=P/H=3/5
Let us take the LHS
${{1 - ta{n^2}A} \over {1 + ta{n^2}A}} = {{1 - {{\left( {{3 \over 4}} \right)}^2}} \over {1 + {{\left( {{3 \over 4}} \right)}^2}}} = {7 \over {25}}$
RHS=cos2 A - sin2A= 7/25
So LHS=RHS,so the statement is true
Question 9
In triangle ABC, right-angled at B, if $\tan A = {1 \over {\sqrt 3 }}$
Find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution
$\tan A = {1 \over {\sqrt 3 }}$
${P \over B} = {1 \over {\sqrt 3 }}$
Let $P = k$ and $B = k\sqrt 3$
Now by Pythagoras theorem
P2 + B2 =H2
H=2k
(i)
$\sin A \cos C + \cos A \sin C = \left( {{P \over H}} \right)\left( {{B \over H}} \right) + \left( {{B \over H}} \right)\left( {{P \over H}} \right)$
$= \left( {{{BC} \over {AC}}} \right)\left( {{{BC} \over {AC}}} \right) + \left( {{{AB} \over {AC}}} \right)\left( {{{AB} \over {AC}}} \right)$
$= {{{k^2}} \over {4{k^2}}} + {{3{k^2}} \over {4{k^2}}} = 1$
(ii)
$\cos A \cos C - \sin A \sin C = \left( {{P \over H}} \right)\left( {{P \over H}} \right) + \left( {{B \over H}} \right)\left( {{B \over H}} \right)$
$= \left( {{{BC} \over {AC}}} \right)\left( {{{AB} \over {AC}}} \right) - \left( {{{AB} \over {AC}}} \right)\left( {{{BC} \over {AC}}} \right)$
$=0$
Question 10
In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of
sin P, cos P and tan P.
Solution
Let QR=x andPR=y
Then x+y=25
y=25-x
Now by Pythagorus theorem
x2 + 25= y2
x2 + 25==(25-x)2
Solving it ,we get
X=12 cm
Then y=25-12=13 cm
Now Sin P= 12/13
Cos P=5/13
Tan P=12/5
Question 11
(i) The value of tan A is always less than 1.
(ii) sec A =12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ =4/3 for some angle
Solution
1. False. The value of tan A increase from 0 to ∞.
2. True. The value of sec A increase from 1 to ∞.
3. False .Cos A is the abbreviation used for the cosine of angle A
4. False .cot A is one symbol. We cannot separate it
5. False. The value of sin θ always lies between 0 and 0 and 4/3 > 1
Download Class 10 Maths Chapter 8: Introduction to Trigonometry Ex 8.1 as pdf
Go back to Class 10 Main Page using below links
### Practice Question
Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20
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