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Home | Math | Functions-Types of Function
# Functions-Types of Function
July 20, 2022
written by Azhar Ejaz
## What is Function in Math?
A function is a rule that relates each element of a set X with exactly one element of set Y. It also is defined as the relationship between one variable (the independent variable) and another variable (the dependent variable).
• “…each element…” means that every element in set X is associated with or connected to at least one element in set Y.
(Note that there may be some elements in Y that remain unrelated to any elements in X, and that’s acceptable.)
• “…exactly one…” means that a function is single-valued means that there is only one output of X.
The formula for the area of a circle, A = πr2, gives the dependent variable A (the area) as a function of the independent variable r (the radius).
Functions are fundamental building blocks in mathematics, and they have various applications in science, engineering, and computer science.
### Representation of Functions
Let f be a function from X to Y is represented as
f :X Y
read as “y is a function of x”.
Functions are generally represented as y = f(x).
Let, f(x) = x3.
It is said as f of x is equal to x cube.
Functions can also be represented by g(), t(),… etc.
### The domainof a Function
The domain of a function is the set of all possible inputs for the function.
### The codomain of a Function
The codomain or set of destinations of a function is the set into which all of the output of the function is constrained to fall.
### Rangeof a Function
The range of a function refers to all the possible outputs of a function.
Example:
Let f(x) = x2. Find the domain and range of f.
Solution:
f(x) is defined for every real number x. Further for every real number x, f(x) = x2 is a non-negative real number.
So, Domain f = Set of all real numbers.
Range f = Set of all non-negative real numbers.
Vertical line test
A vertical line test is used to determine whether a curve is a function or not. If any curve cuts a vertical line at more than one point, then the curve is not a function.
## Types of Function
There are various types of functions given as:
• One-to-one Function (Injective Function)
• Onto Function (Surjective Function)
• One to One and Onto function (bijective function)
• Into Function
• Polynomial Function
• Linear Function
• Constant Function
• Identity Function
• Rational Function
• Algebraic Functions
• Cubic Function
• Even and Odd Function
• Periodic Function
• Composite Function
### Into Function
If a function f: A → B is such that Ran f ⊂ B i.e., Ran f ≠ B, then f is said to be a function from A into B.
Fig f is a function. But Ran f ≠ B.
Therefore, f is a function from A to B.
### Onto (Surjective) function
If a function f: A→B is such that
Ran f = B i.e., every element of B is the image of some elements of A, then f is called an onto function or a surjective function.
### One to One and Into (Injective) function
If a function f from A into B is such that second elements of no two of its ordered pairs are equal, then it is called an injective (1 – 1, and into) function. The function shown in Fig is such a function.
### One to One and Onto function (bijective function)
If f is a function from A onto B such that the second element of no two of its ordered pairs are the same, then ƒ is said to be (1-1) function from A onto B. Such a function is also called a (1-1) Correspondence between A and B. It is
also called a bijective function. Fig shows a (1-1) correspondence between the sets A and B.
(a, z), (b, x) and (c, y) are the pairs of corresponding elements i.e., in this case
f= {(a, z), (b, x). (c, y)} which is a bijective function or (1-1) correspondence
between the sets A and B.
### Polynomial Function
A polynomial function is a function that is defined by a polynomial expression in one or more variables. It consists of non-negative integer powers of the variable(s) multiplied by constants.
Example: f(x) = 3x^2 + 2x – 1 is a polynomial function of degree 2.
### Linear Function
A linear function is a polynomial function of degree 1, and its graph is a straight line. It has the form f(x) = mx + b, where m is the slope and b is the y-intercept.
Example: f(x) = 2x + 3 is a linear function.
### Constant Function
A constant function is a function that always returns the same constant value, regardless of the input.
Example: f(x) = 5 is a constant function as it always gives the output 5.
### Identity Function
The identity function is a function that returns the same value as its input. The domain and codomain are the same for the identity function.
Example: f(x) = x is the identity function.
A quadratic function is a polynomial function of degree 2. Its graph is a parabola.
Example: f(x) = x^2 + 2x – 3 is a quadratic function.
Example
A rough sketch of the functions
i) ((x, y) | 3x+y = 2)
ii) ((x, y) l y= ½ x2)
Solution
i) The equation defining the function is 3x + y = 2
⇒ y = -3x + 2
We know that this equation, being linear, represents a straight line. Therefore, for drawing its sketch or graph only two of its points are sufficient.
When
x=0, y = 2.
y = 0, x = 2/3 = 0.6 nearly.
So two points on the line are A (0, 2) and B = (0.6, 0).
Joining A and B and producing the AB bar in both
directions, we obtain the line AB i.e., the graph of the given function.
ii) The equation defining the function is
y =1/2 x2.
Corresponding to the values {…,-3, -2, -1, 0, 1, +2, +3} of x, values of y are 0, .5, 2, 4.5, …
We plot the points
(0, 0), (± 1, .5), (± 2, 2),(± 3, 4.5),… Joining them by means of a smooth
curve and extending it upwards we get the required graph. We notice that:
i) The entire graph lies above the x-axis.
ii) Two equal and opposite values of x corresponding to every value of y (but not vice versa).
iii) 0 As x increases (numerically) y increases and there is no end to their increase. Thus the graph goes infinitely upwards.
### Rational Function
A rational function is a function that can be represented as the ratio of two polynomial functions.
Example: f(x) = (3x^2 + 2x – 1) / (x – 2) is a rational function.
### Algebraic Functions
An algebraic function is a function that can be constructed using algebraic operations (addition, subtraction, multiplication, division) and applying roots and powers to the variable(s).
Example: f(x) = √(x^3) + 2x is an algebraic function.
### Cubic Function
A cubic function is a polynomial function of degree 3. Its graph often has an S-shape.
Example: f(x) = x^3 – 4x^2 + 2x + 1 is a cubic function.
### Even and Odd Function
An even function is a function that satisfies f(x) = f(-x) for all x in its domain.
An odd function satisfies f(x) = -f(-x) for all x in its domain.
Example of an even function: f(x) = x^2
Example of an odd function: f(x) = x^3
### Periodic Function
A periodic function is a function that repeats its values at regular intervals. There exists a positive constant ‘P’ such that f(x + P) = f(x) for all x in the domain.
Example: f(x) = sin(x) is a periodic function with a period of 2π.
### Composite Function
A composite function is formed by applying one function to the output of another function.
Example: If f(x) = 2x and g(x) = x + 3, then the composite function (g ∘ f)(x) = g(f(x)) = g(2x) = 2x + 3.
## Set-Builder Notation for a function
We know that sub-builder notation is more
suitable for infinite sets. So is the case in respect of a function comprising an infinite
number of ordered pairs. Consider, for instance, the function
f = {(1,1), (2, 4), (3, 9), (4, 16)….}
Dom f= {1, 2, 3, 4,…}.
Ran f = {1,4,9, 16, …}
This function may be written as:
f = {(x, y) l y =r², x ∈ N}
For the sake of brevity this function may be written as:
f = function defined by the equation y=x2, x ∈ N.
Or, to be still more brief: The function, x2, x ∈ N.
In algebra and Calculus the domain of most functions is R and if evident from the context it is, generally, omitted.
## The inverse of a function
If a relation or a function is given in the tabular form i.e., as a set of ordered pairs, its inverse is obtained by interchanging the components of each ordered pair.
The inverse of r and f is denoted r-1 and f-1 respectively.
If r or f are given in set-builder notation the inverse of each is obtained by interchanging x and y in the defining equation. The inverse of a function may or may not be a function.
The inverse of the linear function
{(x, y) y= mx+ c) is ((x, y) l x= my+ c) which is also a linear function.
Briefly, we may say that the inverse of a line is a line.
The line y= x is clearly self-inverse. The function defined by this equation i.e., the function (x, y) | y= x) is called the identity function.
Example
Find the inverse of
i) {(1, 1), (2, 4). (3, 9), (4, 16), x ∈ Z+}
ii) {(x, y) l y= 2x+ 3, x ∈ R)
Which of these are functions.
Solution
i) The inverse is:
{(2, 1), (4, 2), (9,3), (16, 4)…}
This is also a function.
Note: Remember that the equation
y= √x, x ≥ 0
defines a function but the equation y2 = x, x ≥ 0 does not define a function.
The function defined by the equation.
y= √x, x ≥ 0
is called the square root function.
The equation y2 = x ⇒ y = ±√x
Therefore, the equation y= x (x ≥ 0) may be regarded as defining the union of the functions defined by
y = √x, x ≥ 0 and y = -√x, x ≥ 0.
ii) The given function is a linear function.
Its inverse is: {(x, y) | x= 2y + 3}
Which is also a linear function.
Points (0, 3), (-1.5, 0) lie on the given line and points (3, 0), (0, -1.5) lie on its inverse.
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# Simplify? 27^(-3/4)
${27}^{- \frac{4}{3}} = \frac{1}{81}$
#### Explanation:
We have ${27}^{- \frac{3}{4}}$. Let's talk about that power for a moment.
There are three things it's asking us to do:
• the numerator (4) is asking us to take the base number to that power
• the denominator (3) is asking us to take the nth root of the base number
• and lastly, because the power is negative, we are to put the base number into a fraction as the denominator (with 1 as the numerator)
This all becomes a little easier by seeing that $27 = {3}^{3}$
Let's take the cube root first:
${27}^{\frac{1}{3}} = {\left({3}^{3}\right)}^{\frac{1}{3}} = {3}^{3 \times \left(\frac{1}{3}\right)} = 3$
Now let's take that to the 4th power:
${3}^{4} = 81$
${81}^{-} 1 = \frac{1}{81}$
And so we can say that ${27}^{- \frac{4}{3}} = \frac{1}{81}$
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# 2003 AIME I Problems/Problem 3
## Problem
Let the set $\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}.$ Susan makes a list as follows: for each two-element subset of $\mathcal{S},$ she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.
## Solution
Order the numbers in the set from greatest to least to reduce error: $\{34, 21, 13, 8, 5, 3, 2, 1\}.$ Each element of the set will appear in $7$ two-element subsets, once with each other number.
• $34$ will be the greater number in $7$ subsets.
• $21$ will be the greater number in $6$ subsets.
• $13$ will be the greater number in $5$ subsets.
• $8$ will be the greater number in $4$ subsets.
• $5$ will be the greater number in $3$ subsets.
• $3$ will be the greater number in $2$ subsets.
• $2$ will be the greater number in $1$ subsets.
• $1$ will be the greater number in $0$ subsets.
Therefore the desired sum is $34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=\boxed{484}$.
Note: Note that $7+6+5+4+3+2+1=\binom{8}{2}$, so we have counted all the possible cases.
~Yiyj1
## Solution
Thinking of this problem algorithmically, one can "sort" the array to give: $${1, 2, 3, 5, 8, 13, 21, 34}$$
Now, notice that when we consider different pairs, we are only going to fixate one element and look at the all of the next elements in the array, basically the whole $j = i + 1$ shebang. Then, we see that if we set the sum of the whole array to $x,$ we get out answer to be
$$(x-1) + (x-3) + (x-6) + (x-11) + (x-19) + (x-32) + (x-53) = 7x - 125$$
Finding $7x$ isn't hard, and we see that it is equal to $609$:
$$609 - 125 = \boxed{484}$$
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# (5/3)=(x/y) If x-y=5
hala718 | High School Teacher | (Level 1) Educator Emeritus
Posted on
5/3 = x/y
First we will cross multiply:
==> 5y = 3x .......(1)
It is given that:
x- y = 5
==> x= y + 5 ........(2)
Now we will substitute with x values in (2)
5y = 3x
==> 5y = 3(y+5)
==> 5y = 3y + 15
==> 2y = 15
==> y= 15/2 = 7/5
==> x= y+5 = 7.5 + 5 = 12.5
Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian
Posted on
(5/3)=(x/y) If x-y=5
Given two equations and two variables we can solve for both:
5/3 = x/y
can be rewritten by cross multiplying.
5y = 3x
x - y = 5
can be written by bringing the y to the other side:
x = 5 + y
Plug in this x value into the other equation:
5y = 3 ( 5 + y )
Now we can solve for y:
5y = 15 + 3y
2y = 15
y = 15/2
Plug y into one of the equations to solve for y:
x = 5 + (15/2) = 12. 5
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
We'll use the proportion rule:
a/b = c/d <=> (a-b)/b = (x-y)/y
We'll apply this rule to the first equation:
(5/3)=(x/y) <=> (5-3)/3 = (x-y)/y (3)
But, from the second equation, we have x-y=5.
We'll substitute the difference x-y by 5, in (3).
(5-3)/3 = 5/y
2/3 = 5/y
We'll cross multiply:
2y = 15
We'll divide by 2:
y = 15/2
y = 7.5
Now, we'll substitute y by 7.5, into the second equation:
x-y=5
x - 7.5 = 5
x = 5 + 7.5
x = 12.5
william1941 | College Teacher | (Level 3) Valedictorian
Posted on
I assume your question is to find the values of x and y for which the conditions (5/3)=(x/y) and x-y=5 are true.
Take x-y=5 first, we get x=5+y
Now replace x in (5/3)=(x/y) by the corresponding value of y.
We get (5/3)=(5+y)/y
=> 5y=3*(5+y)
=>5y= 15+3y
=>5y-3y=15
=>2y=15
=>y=15/2 = 7.5
As x=5+y , it is equal to 7.5+5=12.5.
Therefore x=12.5 and y=7.5
neela | High School Teacher | (Level 3) Valedictorian
Posted on
(5/3)=(x/y) If x-y=5
To solve for x and y.
The 1st equation is 5/3 = x/y . Multiply by 3y and we get:
5y = 3x. ...(1)
Multiply 2nd equation by 3:
3x-3y = 15......(2)
Eliminate 3x from (1) and (2) by adding:
5y +3x-3y = 3x+15
2y = 15
y = 7.5.
Using (1) 5y = 3x, 5(7.5) = 3x. Or x = 5*7.5/3 = 12.5.
So (x,y) = (12.5 , 7.5)
.
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# Adding, Subtracting, Multiplying, and Dividing
## Presentation on theme: "Adding, Subtracting, Multiplying, and Dividing"— Presentation transcript:
Fractions Adding, Subtracting, Multiplying, and Dividing
When adding or subtracting fractions with like denominators, you will keep the denominator and add the numerators together. When adding or subtracting fractions with unlike denominators, you will need to turn them into common denominators first and then add the numerators.
Now you try adding and subtracting like denominators.
Figure out the problem first and then use the mouse key to view the answer.
Some things to remember: The denominators must always be the same before you can add or subtract the numbers together. Determine what the common denominator should be and turn each fraction into a fraction with the common denominator. Then add or subtract. Common denominators are always needed before you can add or subtract fractions.
Converting to like denominators
A quick review of how to put fractions into common denominators First determine what the common denominator is going to be. Convert the fractions to the common denominator by multiplying the numerator and denominator by the factor needed to make the common denominator equal. Follow basic rules to complete the problem.
Example
Remember to always reduce to the lowest terms.
P R A C T I E Work out the problems first, and then click here to see if you got the right answer.
A N S W E R
Multiplication When you are multiplying or dividing fractions, there is no need to have the same denominators. You can cross-cancel like numbers You can multiply straight across and reduce your answer to lowest terms
Examples: You just multiply the numerators together and then the denominators together. The answer is in lowest terms already so I don’t have to reduce. For this problem, I was able to reduce the numbers (working cross-wise) to help reduce the fractions before multiplying. I have also shown that you can just multiply straight across and still get the same answer.
P R A C T I E Work out the problems first, and then click here to see if you got the right answer.
A N S W E R
Division When dividing fractions such as: you will first need to change it to a multiplication problem by rewriting the first fraction, change the division sign to multiplication and “flip” the fraction that follows the division sign.
Changing from division to multiplication
Write down the first fraction as it appears Change the division sign to a multiplication sign “Flip” or take the inverse of the fraction that follows the division sign. Multiply as usual.
P R A C T I E Work out the problems first, and then click here to see if you got the right answer.
A N S W E R
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## RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A.
Question 1.
Solution:
Number of trials = 500 times
Let E be the no. of events in each case, then
∴No. of heads (E1) = 285 times
and no. of tails (E2) = 215 times
∴ Probability in each case will be
∴(i)P(E1) = $$\frac { 285 }{ 500 }$$ = $$\frac { 57 }{ 100 }$$ = 0.57
(ii) P(E2) = $$\frac { 215 }{ 500 }$$ = $$\frac { 43 }{ 100 }$$ = 0.43
Question 2.
Solution:
No. of trials = 400
Let E be the no. of events in each case, then
No. of 2 heads (E1) = 112
No. of one head (E2) = 160 times
and no. of O. head (E3) = 128 times
∴ Probability in each case will be:
∴ (i)P(E1) = $$\frac { 112 }{ 400 }$$ = $$\frac { 28 }{ 100 }$$ = 0.28
(ii)P(E2) = $$\frac { 160 }{ 400 }$$ = $$\frac { 40 }{ 100 }$$= 0.40
(iii) P(E3) = $$\frac { 128 }{ 400 }$$ = $$\frac { 32 }{ 100 }$$ = 0.32 Ans.
Question 3.
Solution:
Number of total trials = 200
Let E be the no. of events in each case, then
No. of three heads (E1) = 39 times
No. of two heads (E2) = 58 times
No. of one head (E3) = 67 times
and no. of no head (E4) = 36 times
∴ Probability in each case will be .
(i) P(E1) = $$\frac { 39 }{ 200 }$$ = 0.195
(ii) P(E3) = $$\frac { 67 }{ 200 }$$ = 0.335
(iii) P(E4) = $$\frac { 36 }{ 200 }$$ = $$\frac { 18 }{ 100 }$$ = 0.18
(iv) P(E2) = $$\frac { 58 }{ 200 }$$ = $$\frac { 29 }{ 100 }$$ = 0.29
Question 4.
Solution:
Solution No. of trials = 300 times
Let E be the no. of events in each case, then
No. of outcome of 1(E1) = 60
No. of outcome of 2(E2) = 72
No. of outcome of 3(E3) = 54
No. of outcome of 4(E4) 42
No. of outcome of 5(E5) = 39
No. of outcome of 6(E6) = 33
The probability of
(i) P(E3) = $$\frac { 54 }{ 300 }$$ = $$\frac { 18 }{ 100 }$$ = 0.18
(ii) P(E6) = $$\frac { 33 }{ 100 }$$ = $$\frac { 11 }{ 100 }$$= 0.11
(iii) P(E5) = $$\frac { 39 }{ 300 }$$ = $$\frac { 13 }{ 100 }$$ = 0.13
(iv) P(E1) = $$\frac { 60 }{ 300 }$$ = $$\frac { 20 }{ 100 }$$= 0.20 Ans.
Question 5.
Solution:
Let E be the no. of events in each case.
No. of ladies who like coffee (E1) = 142
No. of ladies who like coffee (E2) = 58
Probability of
(1) P(E1) = $$\frac { 142 }{ 200 }$$ = $$\frac { 71 }{ 100 }$$ = 0.71
(ii) P(E2) = $$\frac { 58 }{ 200 }$$ = $$\frac { 29 }{ 100 }$$ = 0.29 Ans.
Question 6.
Solution:
Total number of tests = 6
No. of test in which the students get more than 60% mark = 2
Probability will he
P(E) = $$\frac { 2 }{ 6 }$$ = $$\frac { 1 }{ 3 }$$Ans.
Question 7.
Solution:
No. of vehicles of various types = 240
No. of vehicles of two wheelers = 64.
Probability will be P(E) = $$\frac { 84 }{ 240 }$$ = $$\frac { 7 }{ 20 }$$ = 0.35 Ans.
Question 8.
Solution:
No. of phone numbers are one page = 200
Let E be the number of events in each case,
Then (i) P(E5) = $$\frac { 24 }{ 200 }$$ = $$\frac { 12 }{ 100 }$$ = 0.12
(ii) P(E8) = $$\frac { 16 }{ 200 }$$ = $$\frac { 8 }{ 100 }$$ = 0.08 Ans.
Question 9.
Solution:
No. of students whose blood group is checked = 40
Let E be the no. of events in each case,
Then (i) P(E0) = $$\frac { 14 }{ 40 }$$ = $$\frac { 7 }{ 20 }$$ = 0.35
(ii) P(EAB) = $$\frac { 6 }{ 40 }$$ = $$\frac { 3 }{ 20 }$$ = 0.15 Ans.
Question 10.
Solution:
No. of total students = 30.
Let E be the number of elements, this probability will be of interval 21 – 30
P(E) = $$\frac { 6 }{ 30 }$$ = $$\frac { 1 }{ 5 }$$ = 0.2 Ans.
Question 11.
Solution:
Total number of patients of various age group getting medical treatment = 360
Let E be the number of events, then
(i) No. of patient which are 30 years or more but less than 40 years = 60.
P(E) = $$\frac { 60 }{ 360 }$$ = $$\frac { 1 }{ 6 }$$
(ii) 50 years or more but less than 70 years = 50 + 30 = 80
P(E) = $$\frac { 80 }{ 360 }$$ = $$\frac { 2 }{ 9 }$$
(iii) Less than 10 years = zero
P(E) = $$\frac { 0 }{ 360 }$$ = 0
(iv) 10 years or more 90 + 50 + 60 + 80 + 50 + 30 = 360
Hope given RS Aggarwal Class 9 Solutions Chapter 15 Probability Ex 15A are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14F
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F.
Other Exercises
Question 1.
Solution:
Question 2.
Solution:
Question 3.
Solution:
Question 4.
Solution:
Question 5.
Solution:
Mean = 8
Question 6.
Solution:
Mean = 28.25
Question 7.
Solution:
Mean = 16.6
Question 8.
Solution:
Mean = 50
Question 9.
Solution:
Question 10.
Solution:
Let assumed mean = 67
Question 11.
Solution:
Here h = 1, Let assumed mean (A) = 21
Question 12.
Solution:
Here h = 400 and let assumed mean (A) = 1000
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14H
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H.
Other Exercises
Question 1.
Solution:
Arranging the given data in ascending order :
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
We see that 6 occurs in maximum times.
Mode = 6 Ans.
Question 2.
Solution:
Arranging in ascending order, we get:
15, 20, 22, 23, 25, 25, 25, 27, 40
We see that 25 occurs in maximum times.
Mode = 25 Ans.
Question 3.
Solution:
Arranging in ascending order we get:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9
Here, we see that 9 occurs in maximum times.
Mode = 9 Ans.
Question 4.
Solution:
Arranging in ascending order, we get:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
Here, we see that 50 occurs in maximum times.
Modal score = 50 scores Ans.
Question 5.
Solution:
Arranging in ascending order, we get:
10, 10, 11, 11, 12, 12, 13, 14,15, 17
Here, number of terms is 10, which is even
∴ Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (5th term + 6th term)
Question 6.
Solution:
Question 7.
Solution:
Writing its cumulative frequency table
Question 8.
Solution:
Writing its cumulative frequency table
Here, number of items is 40 which is even.
∴ Median = $$\frac { 1 }{ 2 } \left[ \frac { 40 }{ 2 } th\quad term+\left( \frac { 40 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (20th term + 21th term)
= $$\frac { 1 }{ 2 }$$ (30 + 30) = $$\frac { 1 }{ 2 }$$ x 60 = 30
Mean= $$\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } }$$ = $$\frac { 1161 }{ 40 }$$ = 29.025
∴Mode = 3 median – 2 mean = 3 x 30 – 2 x 29.025 = 90 – 58.05 = 31.95
Question 9.
Solution:
Preparing its cumulative frequency table we get:
Here number of terms is 50, which is even
Question 10.
Solution:
Preparing its cumulative frequency table :
Question 11.
Solution:
Preparing its cumulative frequency table we have,
Question 12.
Solution:
Preparing its cumulative frequency table we have,
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G.
Other Exercises
Question 1.
Solution:
Arranging in ascending order, we get:
2,2,3,5,7,9,9,10,11
Here, number of terms is 9 which is odd.
∴ Median = $$\frac { n+1 }{ 2 }$$ th term = $$\frac { 9+1 }{ 2 }$$ th term = 5th term = 7 Ans.
(ii) Arranging in ascending order, we get: 6, 8, 9, 15, 16, 18, 21, 22, 25
Here, number of terms is 9 which is odd.
∴ Median = $$\frac { n+1 }{ 2 }$$ th term = $$\frac { 9+1 }{ 2 }$$ th term = 5th term = 16 Ans.
(iii) Arranging in ascending order, we get: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, number of terms is 11 which is odd.
∴ Median = $$\frac { 11+1 }{ 2 }$$ th term = $$\frac { 12 }{ 2 }$$ th term = 6th term = 16 Ans.
(iv) Arranging in ascending order, we get:
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, number of terms is 13, which is odd.
Median = $$\frac { 13+1 }{ 2 }$$ th term = $$\frac { 14 }{ 2 }$$ th term = 7th term = 4 Ans.
Question 2.
Solution:
Arranging in ascending order, we get 9, 10, 17, 19, 21, 22, 32, 35
Here, number of terms is 8 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [4th term + 5th term] = $$\frac { 1 }{ 2 }$$ (19 + 21) = $$\frac { 1 }{ 2 }$$ x 40 = 20
(ii) Arranging in ascending order, we get:
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here number of terms is 10 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (60 + 63) = $$\frac { 1 }{ 2 }$$ x 123 = 61.5 Ans.
(iii) Arranging in ascending order we get
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here number of terms is 12 which is even.
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 12 }{ 2 } th\quad term+\left( \frac { 12 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (6th term + 7th term) = $$\frac { 1 }{ 2 }$$ (15 + 17)= $$\frac { 1 }{ 2 }$$ x 32
= 16 Ans.
Question 3.
Solution:
Arranging the given data in ascending order, we get :
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
∴ Median = $$\frac { 15+1 }{ 2 }$$ th term = $$\frac { 16 }{ 2 }$$ th term = 8th term = 23
∴ Median score = 23 Ans.
Question 4.
Solution:
Arranging in ascending order, we get:
143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1
Here, number of terms is 9 which is odd.
Median = $$\frac { 9+1 }{ 2 }$$ th term = $$\frac { 10 }{ 2 }$$ th term = 5th term = 147.3 cm
Hence, median height = 147.3 cm Ans.
Question 5.
Solution:
Arranging in ascending order, we get:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here number of terms is 8 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$[4th term + 5th term]
= $$\frac { 1 }{ 2 }$$ (13.4 + 14.3) = $$\frac { 1 }{ 2 }$$ (27.7) = 13.85
∴ Median weight = 13.85 kg. Ans.
Question 6.
Solution:
Arranging in ascending order, we get:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, number of terms is 10 which is even.
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [5th term + 6th term ] = $$\frac { 1 }{ 2 }$$ (40 + 44) = $$\frac { 1 }{ 2 }$$ x 84 = 42 .
∴ Median age = 42 years.
Question 7.
Solution:
The given ten observations are 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41
These are even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ [5th term + 6th term ] = $$\frac { 1 }{ 2 }$$(x + 1 + x + 3) = $$\frac { 1 }{ 2 }$$(2x + 4)
= x + 2
But median is given = 24
∴ x + 2 = 24 => x = 24 – 2 = 22
Hence x = 22.
Question 8.
Solution:
Preparing the cumulative frequency table, we have:
Here, number of terms (n) = 41, which is odd,
Median = $$\frac { 41+1 }{ 2 }$$ th term = $$\frac { 42 }{ 2 }$$ th term = 21st term = 50 (∵ 20th to 28th term = 50)
Hence median weight = 50 kg Ans.
Question 9.
Solution:
Arranging first in ascending order, we get:
Now preparing its cumulative frequency table
Here, number of terms is 37 which is odd.
Median = $$\frac { 37+1 }{ 2 }$$ th term = $$\frac { 38 }{ 2 }$$ th term = 19 th term = 22 (∵18th to 21st = 22)
Hence median – 22 Ans.
Question 10.
Solution:
first arranging in ascending order we get
Now preparing its cumulative frequency table,we find:
Here, number of terms is 43, which if odd.
Median = $$\frac { 43+1 }{ 2 }$$ th term = $$\frac { 44 }{ 2 }$$ th term = 22nd term = 25 25 (∵ 11th to 26th = 25)
Question 11.
Solution:
Arranging in ascending order,we get
Now preparing its cumulative frequency table, we find :
Here, number of terms = 50 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 50 }{ 2 } th\quad term+\left( \frac { 50 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (154 + 155) = $$\frac { 1 }{ 2 }$$ (309) = 154.5 (∵ 22nd to 25th = 154, 26th to 34th= 155)
Question 12.
Solution:
Arranging in ascending order, we get:
Now, preparing its cumulative frequency table.
Here, number of terms is 60 which is even
∴Median = $$\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } th\quad term+\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right]$$
= $$\frac { 1 }{ 2 }$$ (30th term + 31st term)
= $$\frac { 1 }{ 2 }$$ (20 + 23) = $$\frac { 1 }{ 2 }$$ x 43 = 21.5 (∵ 18th to 30th term = 20, 31st term to 34th = 23)
Hence median = 21.5 Ans.
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14E
These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E.
Other Exercises
Question 1.
Solution:
Mean marks of 7 students = 226
∴Total marks of 7 students = 226 x 1 = 1582
Marks obtained by 6 of them are 340, 180, 260, 56, 275 and 307
∴ Sum of marks of these 6 students = 340 + 180 + 260 + 56 + 275 + 307 = 1418
∴ Marks obtained by seventh student = 1582 – 1418 = 164 Ans.
Question 2.
Solution:
Mean weight of 34 students = 46.5 kg.
∴ Total weight of 34 students = 46.5 x 34 kg = 1581 kg
By including the weight of teacher, the mean weight of 35 persons = 500 g + 46.5 kg
= 46.5 + 0.5 = 47.0 kg
∴ Total weight = 47.0 x 35 = 1645 kg
∴ Weight of the teacher = (1645 – 1581) kg = 64 kg Ans.
Question 3.
Solution:
Mean weight of 36 students = 41 kg
∴ Total weight of 36 students = 41 x 36 kg = 1476 kg
After leaving one student, number of students = 36 – 1 =35
Their new mean = 41.0 – 200g = (41.0 – 0.2) kg = 40.8 kg.
∴ Total weight of 35 students = 40.8 x 35 = 1428 kg
∴ Weight of leaving student = (1476 – 1428) kg = 48 kg Ans.
Question 4.
Solution:
Average weight of 39 students = 40 kg
∴ Total weight of 39 students = 40 x 39 = 1560 kg
By admitting of a new student, no. of students = 39 + 1 =40
and new mean = 40 kg – 200 g = 40 kg – 0.2 kg = 39.8 kg
∴Weight of 40 students = 39.8 x 40 kg = 1592 kg
∴Weight of new student = 1592 – 1560 kg = 32 kg Ans.
Question 5.
Solution:
Average salary of 20 workers = Rs. 7650
∴Their total salary = Rs. 7650 x 20 = Rs. 153000
By adding the salary of the manager, their mean salary = Rs. 8200
∴Their total salary = Rs. 8200 x 21 = Rs. 172200
∴Salary of the manager = Rs. 172200 – Rs. 153000 = Rs. 19200 Ans.
Question 6.
Solution:
Average wage of 10 persons = Rs. 9000
Their total wage = Rs. 9000 x 10 = Rs. 90000
Wage of one person among them = Rs. 8100
and wage of new member = Rs. 7200
∴ New total wage = Rs. (90000 – 8100 + 7200) = Rs. 89100
Their new mean wage = Rs.$$\frac { 89100 }{ 10 }$$ = Rs. 8910
Question 7.
Solution:
Mean consumption of petrol for 7 months of a year = 330 litres
Mean consumption of petrol for next 5 months = 270 litres
∴Total consumption for first 7 months = 7 x 330 = 2320 l
and total consumption for next 5 months = 5 x 270 = 1350 l
Total consumption for 7 + 5 = 12 months = 2310 + 1350 = 3660 l
Average consumption = $$\frac { 3660 }{ 12 }$$ = 305 liters per month.
Question 8.
Solution:
Total numbers of numbers = 25
Mean of 15 numbers = 18
∴Total of 15 numbers = 18 x 15 = 270
Mean of remaining 10 numbers = 13
∴Total = 13 x 10 = 130
and total of 25 numbers = 270 + 130 = 400
∴Mean = $$\frac { 400 }{ 25 }$$ = 16 Ans.
Question 9.
Solution:
Mean weight of 60 students = 52.75 kg.
Total weight = 52.75 x 60 = 3165 kg
Mean of 25 out of them = 51 kg.
∴Their total weight = 51 x 25 = 1275 kg
∴ Total weight of remaining 60 – 25 = 35 students = 3165 – 1275 = 1890 kg 1890
Mean weight = $$\frac { 1890 }{ 35 }$$ = 54 kg Ans.
Question 10.
Solution:
Average increase of 10 oarsman = 1.5 kg.
∴ Total increased weight =1.5 x 10 = 15kg
Weight of out going oarsman = 58 kg .
∴ Weight of new oarsman = 58 + 15 = 73 kg Ans.
Question 11.
Solution:
Mean of 8 numbers = 35
∴Total of 8 numbers = 35 x 8 = 280
After excluding one number, mean of 7 numbers = 35 – 3 = 32
Total of 7 numbers = 32 x 7 = 224
Hence excluded number = 280 – 224 = 56 Ans.
Question 12.
Solution:
Mean of 150 items = 60
Total of 150 items = 60 x 150 = 9000
New total = 9000 + 152 + 88 – 52 – 8 = 9000 + 240 – 60 = 9180
∴New mean = $$\frac { 9180 }{ 150 }$$ = 61.2 Ans.
Question 13.
Solution:
Mean of 31 results = 60
∴Total of 31 results = 60 x 31 = 1860
Mean of first 16 results = 58
∴Total of first 16 results = 58 x 16 = 928
and mean of last 16 results = 62
∴Total of last 16 results = 62 x 16 = 992
∴16th result = (928 + 992) – 1860 = 1920 – 1860 = 60 Ans.
Question 14.
Solution:
Mean of 11 numbers = 42
∴Total of 11 numbers = 42 x 11 = 462
Mean of first 6 numbers = 37
∴Total of first 6 numbers = 37 x 6 = 222
Mean of last 6 numbers = 46
∴Total of last 6 numbers = 46 x 6 = 276
∴6th number = (222 + 276) – 462 = 498 – 462 = 36 Ans.
Question 15.
Solution:
Mean weight of 25 students = 52 kg
∴ Total weight of 25 students = 52 x 25 = 1300 kg
Mean weight of first 13 students = 48 kg
∴ Total weight of first 13 students = 48 x 13 kg = 624 kg
Mean of last 13 students = 55 kg
∴Total of last 13 students = 55 x 13 kg = 715 kg
∴Weight of 13th student = (624 + 715) – 1300 = 1339 – 1300 = 39 kg Ans.
Question 16.
Solution:
Mean of 25 observations = 80
Total of 25 observations = 80 x 25 = 2000
Mean of another 55 observations = 60
∴ Total of these 55 observations = 60 x 55 = 3300
Total number of observations = 25 + 55 = 80
and total of 80 numbers = 2000 + 3300 = 5300
Mean of 80 observations = $$\frac { 5300 }{ 80 }$$ = 66.25 Ans.
Question 17.
Solution:
Marks in English = 36
Marks in Hindi = 44
Marks in Mathematics = 75
Marks in Science = x
∴Total number of marks in 4 subjects = 36 + 44 + 75 + x = 155 + x
Average marks in 4 subjects = 50
∴Total marks = 50 x 4 = 200
∴155 + x = 200
=> x = 200 – 155
=> x = 45
Hence, marks in Science = 45 Ans.
Question 18.
Solution:
Mean of monthly salary of 75 workers = Rs. 5680
Total salary of 75 workers = Rs. 5680 x 75 = Rs. 426000
Mean salary of 25 among them = Rs. 5400
Total of 25 workers = Rs. 5400 x 25 = Rs. 135000
Mean salary of 30 among them = Rs. 5700
∴Total of 30 among them = Rs. 5700 x 30 = Rs. 171000
∴ Total salary of 25 + 30 = 55 workers = Rs. 135000 + 171000 = Rs. 306000
∴Total salary of remaining 75 – 55 = 20 workers = Rs. 426000 – 306000 = Rs. 120000
∴ Mean of remaining 20 workers = Rs. $$\frac { 120000 }{ 20 }$$ = Rs. 6000 Ans.
Question 19.
Solution:
Let distance between two places = 60 km
∴Time taken at the speed of 15 km/h = $$\frac { 60 }{ 15 }$$ = 4 hours
and time taken at speed of 10 km/h for coming back = $$\frac { 60 }{ 10 }$$ = 6 hours
Total the taken = 4 + 6 = 10
hours and distance covered = 60 + 60 = 120 km
∴Average speed = $$\frac { 120 }{ 10 }$$ =12 km/hr.
Question 20.
Solution:
No. of total students = 50
No. of boys = 40
∴ No. of girls = 50 – 40 = 10
Average weight of class = 44kg
∴Total weight of 50 students = 44 x 50 = 2200 kg
Average weight of 10 girls = 40 kg
∴Total weight = 40 x 10 = 400 kg
Total weight of 40 boys = 2200 – 400 = 1800 kg
∴Average weight of 40 boys = $$\frac { 1800 }{ 40 }$$kg = 45 kg Ans.
Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.
|
Mathematics 5
Conics
Applications
# Mathematics Conics Ellipse
### 1. Ellipse: Construction:
An ellipse is defined as:
dist (P,F) + dist(P,F') = 2a
P is any point on the curve. F and F' are the two focii of the ellipse; and "dist" is the distance between the two related points.
Let's express the above equation:
sqrt [(c - x)2 + (0 - y)2 ] + [(- c - x)2 + (0 - y)2]1/2 = 2a
[(c - x)2 + y2 ]1/2 + [(- c - x)2 + y2 ]1/2 = 2a
[c2 - 2cx + x2 + y2]1/2 = 2a - [c2 + 2cx + x2 + y2]1/2
c2 - 2cx + x2 + y2 = 4a2 - 4a [c2 + 2cx + x2 + y2]1/2 + c2 + 2cx + x2 + y2
a [c2 + 2cx + x2 + y2]1/2 = a2 + cx
a2 [c2 + 2cx + x2 + y2] = a4 + 2cx a2 + c2x2
a2 [c2 + x2 + y2] = a4 + c2x2
a2 [ x2 + y2] - c2x2 = a2[a2 - c2]
x2[a2 - c2] + a2 y2] = a2[a2 - c2]
x2 b2 + a2 y2 = a2b2
Then:
x2/ a2 + y2/ b2 = 1
x2/ a2 + y2/ b2 = 1
That is the equation of an ellipse.
## 2.1. Eccentricity of an ellipse
The eccentricity "e" of a conic is defined as the ratio :
e = dist(P,F)/dist(P,L)
Applied to the point P = A, and then to the point P = A', yields:
e = dist(A,F)/dist(A,L)
e = (a - c)/(d - a) (1)
e = dist(A',F)/dist(A',L)
e = (a + c)/(d + a) (2)
Solving for "e":
ed - ea = a - c
ed + ea = a + c
2ed = 2a, then: e = a/d, or
Subtracting (2) from (1) gives: -2 ea = - 2c, then: e = c/a
e = c/a , with: 0 < e < 1
## 2.2. Polar equation of en ellipse
Cosine law gives:
PF2 = r2 + c2 - 2rc cosθ
We have :
PL = d - rcosθ
dist(P,F) = e dist(P,L), yields:
sqrt[r2 + c2 - 2r c cosθ] = e(d - r cosθ)
[r2 + c2 - 2r c cosθ]1/2 = e(d - r cosθ)
r2 + c2 - 2r c cosθ = e2d2 + e2 r2 cos2 θ - 2rde2 cosθ
r2[1 - e2 cos2 θ] + 2r cosθ [ de2 - c] + c2 - e2d2 = 0
Since ed = a, ae = c, and b2 = a2 - c2 then:
r2[1 - e2 cos2 θ] = b2
r = b/[1 - e2 cos2 θ]1/2
r = b/[1 - e2 cos2 θ]1/2 =
[(a2 - c2)/(1 - e2 cos2 θ)]1/2
To simplify, lest' consider F = O, that is the focus is at the origin c = 0, therefore:
ro = dist(F,P) = e dist(P,L)
d - c = do
dist(P,L) = do - rocos θo; then:
ro = e[do - rocos θo]
ro[1 + e cos θo] = edo
ro = edo/[1 + e cos θo]
ro = edo/(1 + e cos θo)
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# Integrals Involving sin(x) with Odd Power
Tutorial to find integrals involving odd powers of sin(x). Exercises with answers are at the bottom of the page.
## Examples with Detailed Solutions
In what follows, C is the constant of integration.
### Example 1
Evaluate the integral
sin3(x) dx
Solution to Example 1:
The main idea is to rewrite the power of sin(x) as the product of a term with power 1 and a term with an even power. Example: sin3(x) = sin2(x) sin(x). Hence the given integral may be written as follows:
sin3(x) dx = sin2(x) sin(x) dx
=
(1 - cos2(x)) sin(x) dx
We now let u = cos(x), hence du/dx = -sin(x) or -du = sin(x)dx and substitute in the given intergral to obtain
sin3(x) dx = - (1 - u2) du
sin3(x) dx = (1/3) u3 - u + C
Substitute u by cos(x) to obtain
sin3(x) dx = (1/3)cos3(x) - cos(x) + C
### Example 2
Evaluate the integral
sin5(x) dx
Solution to Example 2:
Rewrite sin5(x) as follows sin5(x) = sin4(x) sin(x). Hence the given integral may be written as follows:
sin5(x) dx = sin4(x) sin(x) dx
We now use the identity sin2(x) = 1 - cos2(x) to rewrite sin4(x) in terms of power of cos(x) and rewrite the given integral as follows:
sin5(x) dx = (1 - cos2(x))2 sin(x) dx
We now let u = cos(x), hence du/dx = -sin(x) or du = -sin(x)dx and substitute in the given intergral to obtain
sin5(x) dx = - (1 - u2)2 du
Expand and calculate the integral on the right
sin5(x) dx = - (u4 - 2u2 + 1) du
= -(1/5)u
5 + (2/3)u3 - u + C
and finally
sin5(x) dx = -(1/5)cos5(x) + (2/3)cos3(x) - cos(x) + C
## Exercises
Evaluate the following integrals.
1.
sin7(x)dx
2.
sin9(x)dx
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# 6.5 Solve proportions and their applications (Page 2/12)
Page 2 / 12
Determine whether each equation is a proportion:
1. $\phantom{\rule{0.2em}{0ex}}\frac{4}{9}=\frac{12}{28}$
2. $\frac{17.5}{37.5}=\frac{7}{15}$
## Solution
To determine if the equation is a proportion, we find the cross products. If they are equal, the equation is a proportion.
ⓐ Find the cross products. $28\cdot 4=112\phantom{\rule{2em}{0ex}}9\cdot 12=108$
Since the cross products are not equal, $28·4\ne 9·12,$ the equation is not a proportion.
ⓑ Find the cross products. $15\cdot 17.5=262.5\phantom{\rule{2em}{0ex}}37.5\cdot 7=262.5$
Since the cross products are equal, $15·17.5=37.5·7,$ the equation is a proportion.
Determine whether each equation is a proportion:
1. $\phantom{\rule{0.2em}{0ex}}\frac{7}{9}=\frac{54}{72}$
2. $\frac{24.5}{45.5}=\frac{7}{13}$
1. ⓐ no
2. ⓑ yes
Determine whether each equation is a proportion:
1. $\phantom{\rule{0.2em}{0ex}}\frac{8}{9}=\frac{56}{73}\phantom{\rule{0.2em}{0ex}}$
2. $\phantom{\rule{0.2em}{0ex}}\frac{28.5}{52.5}=\frac{8}{15}$
1. ⓐ no
2. ⓑ no
## Solve proportions
To solve a proportion containing a variable, we remember that the proportion is an equation. All of the techniques we have used so far to solve equations still apply. In the next example, we will solve a proportion by multiplying by the Least Common Denominator (LCD) using the Multiplication Property of Equality .
Solve: $\frac{x}{63}=\frac{4}{7}.$
## Solution
To isolate $x$ , multiply both sides by the LCD, 63. Simplify. Divide the common factors.
Check: To check our answer, we substitute into the original proportion. Show common factors. Simplify.
Solve the proportion: $\frac{n}{84}=\frac{11}{12}.$
77
Solve the proportion: $\frac{y}{96}=\frac{13}{12}.$
104
When the variable is in a denominator, we’ll use the fact that the cross products of a proportion are equal to solve the proportions.
We can find the cross products of the proportion and then set them equal. Then we solve the resulting equation using our familiar techniques.
Solve: $\frac{144}{a}=\frac{9}{4}.$
## Solution
Notice that the variable is in the denominator, so we will solve by finding the cross products and setting them equal.
Find the cross products and set them equal. Simplify. Divide both sides by 9. Simplify.
Show common factors.. Simplify.
Another method to solve this would be to multiply both sides by the LCD, $4a.$ Try it and verify that you get the same solution.
Solve the proportion: $\frac{91}{b}=\frac{7}{5}.$
65
Solve the proportion: $\frac{39}{c}=\frac{13}{8}.$
24
Solve: $\frac{52}{91}=\frac{-4}{y}.$
## Solution
Find the cross products and set them equal. Simplify. Divide both sides by 52. Simplify.
Check: Show common factors. Simplify.
Solve the proportion: $\frac{84}{98}=\frac{-6}{x}.$
−7
Solve the proportion: $\frac{-7}{y}=\frac{105}{135}.$
−9
## Solve applications using proportions
The strategy for solving applications that we have used earlier in this chapter, also works for proportions, since proportions are equations. When we set up the proportion , we must make sure the units are correct—the units in the numerators match and the units in the denominators match.
When pediatricians prescribe acetaminophen to children, they prescribe $5$ milliliters (ml) of acetaminophen for every $25$ pounds of the child’s weight. If Zoe weighs $80$ pounds, how many milliliters of acetaminophen will her doctor prescribe?
## Solution
Identify what you are asked to find. How many ml of acetaminophen the doctor will prescribe Choose a variable to represent it. Let $a=$ ml of acetaminophen. Write a sentence that gives the information to find it. If 5 ml is prescribed for every 25 pounds, how much will be prescribed for 80 pounds? Translate into a proportion. Substitute given values—be careful of the units. Multiply both sides by 80. Multiply and show common factors. Simplify. Check if the answer is reasonable. Yes. Since 80 is about 3 times 25, the medicine should be about 3 times 5. Write a complete sentence. The pediatrician would prescribe 16 ml of acetaminophen to Zoe.
You could also solve this proportion by setting the cross products equal.
so some one know about replacing silicon atom with phosphorous in semiconductors device?
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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Good
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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# How Many Games Over .500?
A week ago, Brandon McCarthy posted a tweet that sent Twitter into a frenzy.
Many people believed that McCarthy was incorrect. After all, they reasoned, all the media outlets report games over .500 this way. How could all of them possibly be wrong?
Let’s do some math. For the sake of simplicity, we will say that the Dodgers have played one more game than their record indicates so that we have a nice even number. They are currently 80-59. We will add that extra game to the win column, thereby making their hypothetical record 81-59. With that record, how many games have they played in the season? To figure that out, we add the number of wins they have to their losses.
Wins + Losses = Games played so far
81 + 59 = 140
Now, we want to find what the .500 mark is at this point in the season. The way we do that is to divide the number of games played so far by 2. Why by 2? Because that is how you calculate half of a number.
Games played so far / 2 = .500 mark
140/2 = 70
So if a team’s record is .500 at 140 games played, then they have an equal number of wins and losses. Thus, a .500 team has a record of 70-70 at 140 games played.
Now if we have a team that is 81-59, that team is NOT 22 games over .500. Many people are claiming that in order to reach a .500 record, a team at 81-59 has to lose 22 games but what happens when we do that? If we add 22 losses to our 81-59 Dodgers, their new record becomes 81-81. How many games played is that? Going back to our formula, it is 162 games played. But we are not concerned with 162 games played. We are concerned with only the 140 that have been played thus far. So what do we do? Well, we look at the current record of 81-59 and the .500 mark of 70-70. We subtract the number of wins at the .500 mark from the number of wins a team has and that will give us how many games over or under .500 a given team is. For example, our 81-59 Dodgers are 11 games over .500 because
81 – 70 = 11
Now the confusion lies in this comment.
A team with a .500 record is the midway point. And any record over or under .500 is also how many games ahead or back other teams are. In other words, a team that is 11 games over .500 is also 11 games ahead of a .500 team. OR, a .500 team that is eleven games back in the standings also means that the first place team is eleven games over .500.
Math is not an easy subject. I get that. But for the love of god, just because everyone else is spewing incorrect information does not mean that it is right. And really, there is nothing wrong with being wrong. Math is math and the most basic of math is practically indisputable.
Your regularly scheduled Piazza Parlor content will continue tomorrow. The Dodgers take on the Arizona Diamondbacks. Alex Wood will face Robbie Ray. Game time is at 6:40 PM.
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# Sets and Set Notation
In Math, a set is a collection of objects or elements.
With sets and set notation, the common notation for a set, is a collection of elements inside curly brackets, separated by commas.
For example { a , b , c , d }.
The numbers from 11 to 15 can be a set. { 11 , 12 , 13 , 14 , 15 }
Order is not important, { 13 , 11 , 14, 15 , 12 } is the same set as above.
## Sets and Set Notation, Further
Consider 2 sets, A and B.
A = { 1 , 2 , 3 , 4 , 5 } B = { 6 , 7 , 8 , 9 , 10 }
The number 2 is a member of set A. This is denoted as 2A.
= “is a member/element of”.
So it’s also the case that 7B.
However, the number 2 is NOT in set B,
and the number 8 is NOT in set A.
These situations are denoted by 2 ∉ B and 8 ∉ B.
= “is NOT a member/element of”.
It's possible that a set can also have no elements in it at all.
This set would be known as the empty set { }, which is denoted by .
## Set Builder Notation
Sets don’t have to just be finite groups though. They can be infinite.
One can have the set of all numbers larger than 100.
{ 100 , 101, 102 , .... }
There is set builder notation that can make building or representing sets a bit shorter and easier, particularly when the sets are infinite.
A set for all number larger than 5 can be set up as: { x | x > 5 }.
Meaning that we have a set where the letter x can be any single number, but it must be larger than 5.
A good page with a detailed explanation on set builder notation can be seen at the Mathwords website.
## Sets and Set Notation,Subsets and Set Compliment
A set G, could be a "subset" of another set H.
If it's the case that all of the elements in set G, are elements that are also in set H.
Consider 2 sets.
G = { 6, 14 , 56 } H = { 3 , 6 , 78 , 14 , 56 }
In the example above, set G is a subset of set H.
This is denoted G H.
The "complement" of a set, is the elements that are NOT in the set.
For a set labelled A, the complement of the set would be denoted Ac.
Also, with 2 separate sets, there can be another set that's either:
G minus H, or H minus G.
Which is denoted with a backslash between the letters.
G\H = Elements in G, but not in H.
H\G = Elements in H, but not in G.
So for the sets shown above, the set H\G is { 3 , 78 }.
As 3 and 78 are in set H, but NOT in set G.
## Set NotationIntersection and Union
For two sets labelled S and T.
The intersection is the set of elements in both set S and set T.
Denoted ST.
For  2 sets:
S = { 12 , 3 , 5 , 86 , 13 } T = { 7 , 3 , 6 , 86 , 13 }
The set ST = { 3 , 13 , 86 }.
The union of two sets is the set of elements in either set S or in set T.
Denoted ST.
The set ST = { 3 , 5 , 6 , 7 , 12 , 13 , 86 }.
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### math by Jasmeet Nijjar
Have you ever wondered how roller coasters are made? Roller coasters have math behind it. Linear relationships and parabolas are used to make them. We developed our learning from Linear relationships to parabolas to understand our world better. Quadratic comes from the word QUAD, which means square like (x²). The power to 2 is what causes quadratics to have a curve in them. Quadratics is used to measure curves or anything 'u' shaped for example, kicking a soccer ball, bridges, the moon going around the earth and more.
## TABLE OF CONTEXT
vertex form:
• first and second differences
• parts of a parabola/how to read parabolas
• Transformations
• how to make a parabola +step pattern
• finding zeros
Factored form:(what you need to find out to graph factored form)
• finding x intercept
• finding axis of symmetry
• finding optimal value
Standard Form:
• Discriminant
• Axis of symmetry
• Optimal value
• Completing a square
• Factoring to turn into factored form (common factoring, factor by grouping, factor simple trinomials, factor complex trinomial, difference of squares, perfect square)
## First and Second Differences
We use tables of values to determine if the table is linear, quadratic or none. In the picture you have to see the difference in the Y column, 3-2=1, 6-3=3 ect If the difference between the Y in the first column were all the same numbers it was be linear. So we go on to the next column and find the difference between the first difference, ex 3-1=2, 5-3=2, see now the difference keeps coming as 2 which means its quadratic. If there was no common difference even in the second difference column then it wouldn't be linear or quadratic.
Just so you know a parabola is the name for Quadratic when its graphed.
Axis Of Symmetry/Vertex: This is the point that divides the parabola in two and also where it meets. (add the two x int and divide by 2)
X intercept: where the parabola hits the x axis, is also called zeros (plug y as 0)
Y intercept: When the parabola hits the y axis (plug x as 0)
Optimal value: The highest or lowest point (max or min)
Direction of opening: when the graph is smiling its opening upward and when its frowning its opening down
## The original parabola is y=x² every graph is added onto the original
Vertex form is a way to write the equation
y=a(x-h)² +k
a- tells if the parabola stretches up or gets wider
h- tells you if the parabola moved left or right
k- tells you if the parabola moves up or down
ex.
narrow: y=2x²
wider: y=o.5x²
move up: y=x² +3 *for left and right your probably wondering why
move down: y=x²-3 moving to the right actually and move to the left
move right: y=(x-1)² you actually move to the right. This is because
move left: y=(x+1)² you have to use the opposite sign from the bracket
flip: y=-2x (x-1)² is actually x= 1
y=a(x-h)²+k
y= -4 (x-3)² + 4
• narrowed by 4 (a)
• flipped (-a)
• move to the right (opposite of everything in bracket)
• up 4 (k)
## How to make a parabola in the video below
1. find vertex; y=a(x-h)²+k opposite of h and k are your vertex
2. use step pattern: original- over 1 up 1, over 2 up 4 now you do ax2 and ax4
3. AND YOUR PARABOLA IS DONE
in vertex form opp h and k are your vertex on the graph
ex y=a(x-4)²+6
vertex is (4,6)
3.3 More Graphing from Vertex Form
## How to find zeros/x intercepts
In vertex form you are given a, h and k but not x or y. To solve for x you can plug y=0 because x intercepts have a value of 0. You should start bringing everything away from x using samdeb. At the end x should equal to one positve and negative number. solve the last part by using just a plus and the other part using just minus, this should lead to your 2 x intercepts.
x intercepts from vertex form
## factored form
Another form you learn in quadratics is called factored form:
y=a(x-r)(x-s)
This is called factored form because you need to factor to get x intercept.
## step 1 for graphing
Finding your x intercepts from factored form is very easy;
y=a(x-r)(x-s)
1. separate x-r and x-s and put =0
x-r=0 x-s=0
2. now bring your r and s to the other side and x=r x=s switch your signs
x=r x=s
ex.
y=(x-6)(x-4)
x-6=0 x-4=0
x=6 x=4
## step 2 for graphing
Finding axis of symmetry;
you do r plus s and divide the answer by 2, which should give the half of the two roots.
y=a(x-r)(x-s)
r+s/2
ex
(x-6)(x-4) *opposite of whats in bracket
6+4=10/2
aos=5
## step 3 for graphing
Finding the optimal value:
*we already got our roots, and our axis of symmetry
to find optimal value you have to plug the x in the factored form equation with the axis of symmetry you found earlier.
y=(x-6)(x-4)
y=(5-6)(5-4)
y=(-1)(1)
y= -1
now all we have to do is plot the 4 things you got on the graph and use the step pattern from the video how to make a parabola.
## Standard form
The last form in quadratics is standard form;
y=ax²+bx+c
• A formula you learn that can help you find the x intercept for ANY equation is called quadratic formula.
which looks like...
you must be wondering where a b and c came from, well
ex
y=2x²+9x+10
• 2 is a
• 9 is b
• 12 is c
Watch the video below, i will be teaching you how to use the quadratic formula
## discriminant
The discriminate is only the portion inside of the square root from the quadratic formula.
## How to find axis of symmetry using as similar fromula
-b/2a= AOS
using the other equation it will be:
-9/2x2=AOS
-2.25=AOS
## OPTIMAL VALUE
to find the optimal value you have to plug your aos value as x in your original formula
y=2x²+9x+10
y=2(-2.25)²+9(-2.25)+10
y=2(5.0625)-20.25+10
y=10.125-20.25+10
y=10.125-10.25
y=0.125
## now you got your x intercept, aos and optimal value all you have to do is graph it
once againg use the step patter you watched in the how to make a parabola video
## completing the square and turning it into vertex form
The video below demonstrates how to convert from standard to vertex from by completing a square.
1. factor whats common and bring it out from the first two terms and kick the last term out. put the two terms in brackets.
2. make it a perfect square trinominal by getting the second term, dividing that by 2 and power it by 2
3. the answer you got for step #2, you write it twice; first add second subtract
4. keep the 3 numbers in the bracket and kick the last one out, but first you have to multiple it with the common number we found in step #1
Converting Standard to Vertex Form
## Factoring to turn into factored form
• common factoring
• Factoring by grouping
• simple trinomial
• complex trinomial
• perfect square
• difference of square
## Common factoring
Watch the video below, it should take you through the process of common factoring, but common factoring is when you find the factors of numbers and write it in factored form
## Factoring by grouping
This is basically the same thing as common factoring, but it has 4 terms, and we are get grouping but making binomials. After you group you find the common factor and write it in factored form. Watch the video below
## fatoring Simple Trinomial
factor means, all the possible ways to multiply to get that number. It is called simple because the coefficient is one, and it has three terms, which is called a trinomial. Watch the 2 videos below on factoring simple trinomial
## Factoring complex trinomial
Very similar to simple trinomial, only difference it that the coefficient ha can not be one. Watching the video below will help your understanding of complex trinomials
3.9 Complex Trinomial Factoring
## DIfference of squares
Difference of squares(dos) have 2 different symbols so problems for D.O.S are in a²-b² formula, but the end result should look like (a+b) (a-b).
a²-b²
square root a² = a
square root b² = b
and write your answer in brackets, the first one use a '+' and the second one use a '-' sign.
(a+b)(a-b)
axa=a²
ax-b=-ab
bxa=ba
bx-b=-b²
-ab and ba cancel out leaving you with a²-b²
## perfect square
Perfect square are the factors of themselves, they use the formula a²+2ab+b² and your answer should end up as (a+b)²
a²+2ab+b²
square root a²=a
square root b²=b
write in a bracket and put '²' sign outside of it
(a+b)²
Check by multiplying inner and outer
expand (a+b)²
(a+b) (a+b)
axa=a²
bxb=b²
and for the middle do 2xaxb=2ab
## If you found my videos on factoring confusing to understand watch this
3.11 Factoring
• All three forms can be graphed
• you can ind all of there Axis of symmetry, optimal value, y and x intercepts, and vertex
• in vertex form you can find the vertex by using opposite of h and original k
• Use step pattern to graph all forms
• Discriminate will tell you in advance how many roots there will be
• x+x/2=aos and -b/2a= aos
• can use quadratic formula to solve any type of x, even complex trinomials
• all forms can be created into eachother
• distributive property works on all factoring to turn into factored form
More Word Problems Using Quadratic Equations - Example 1
a) the question below, in vertex form, h and k are your vertex. (3,19) is my vertex, and its also when the ball reached its highest so i looked at x axis (time) and wrote 3 sec
b) same thing as a vertex is (3,19) so i looked at y axis (height) and wrote 19
c) punted is like something being kicked started above the ground so what we are finiding is how much in the air was it before kicked
i used the original equation h=-2(t-3)²+19 and substituted t as 0 and you should get your answer
i did the same thing here just like the last question
vertex (10,20)
c) i subsitute 1 as d in the original equation and solved it
The question says bryanna sells 15 shirts for \$20, but she starts to decrease money amount by \$1 and she sells 2 more shirts on top of 15
first you make a chart on her revenue, which means her income and price change
a)
20-price change times 15+ 2 times the price change
b)
you have to find the x intercepts, (20-p)(15+2p) is already in factored form so i easily found it.
then you will have to find the axis of symmetry from there which is 6.25 and plug it in (15+2p) and solve
the question is askign for the largest fencing, and finding aos gives you the maximum value
the total area is 500 which is 2x and 2y
we can make the equation more simple
250=x+y
to find y we will bring x over = 250-x=y
plug that in for y times x (lxw for area)
its in factored form so you can find the x intercepts
x=250 x=0
find the axis of symmetry by addign the zeros and dividing it by 2 which is 125
now plug that in as x in the A=(250-x)(x) equation
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2020-03-31
Recently, you've probably seen a lot of graphs that look like this:
The graph above shows something that is growing exponentially: its equation is $$y=kr^x$$, for some constants $$k$$ and $$r$$. The value of the constant $$r$$ is very important, as it tells you how quickly the value is going to grow. Using a graph of some data, it is difficult to get an anywhere-near-accurate approximation of $$r$$.
The following plot shows three different exponentials. It's very difficult to say anything about them except that they grow very quickly above around $$x=15$$.
$$y=2^x$$, $$y=40\times 1.5^x$$, and $$y=0.002\times3^x$$
It would be nice if we could plot these in a way that their important properties—such as the value of the ratio $$r$$—were more clearly evident from the graph. To do this, we start by taking the log of both sides of the equation:
$$\log y=\log(kr^x)$$
Using the laws of logs, this simplifies to:
$$\log y=\log k+x\log r$$
This is now the equation of a straight line, $$\hat{y}=m\hat{x}+c$$, with $$\hat{y}=\log y$$, $$\hat{x}=x$$, $$m=\log r$$ and $$c=\log k$$. So if we plot $$x$$ against $$\log y$$, we should get a straight line with gradient $$\log r$$. If we plot the same three exponentials as above using a log-scaled $$y$$-axis, we get:
$$y=2^x$$, $$y=40\times 1.5^x$$, and $$y=0.002\times3^x$$ with a log-scaled $$y$$-axis
From this picture alone, it is very clear that the blue exponential has the largest value of $$r$$, and we could quickly work out a decent approximation of this value by calculating 10 (or the base of the log used if using a different log) to the power of the gradient.
Log-log plots
Exponential growth isn't the only situation where scaling the axes is beneficial. In my research in finite and boundary element methods, it is common that the error of the solution $$e$$ is given in terms of a grid parameter $$h$$ by a polynomial of the form $$e=ah^k$$, for some constants $$a$$ and $$k$$.
We are often interested in the value of the power $$k$$. If we plot $$e$$ against $$h$$, it's once again difficult to judge the value of $$k$$ from the graph alone. The following graph shows three polynomials.
$$y=x^2$$, $$y=x^{1.5}$$, and $$y=0.5x^3$$
Once again is is difficult to judge any of the important properties of these. To improve this, we once again begin by taking the log of each side of the equation:
$$\log e=\log (ah^k)$$
Applying the laws of logs this time gives:
$$\log e=\log a+k\log h$$
This is now the equation of a straight line, $$\hat{y}=m\hat{x}+c$$, with $$\hat{y}=\log e$$, $$\hat{x}=\log h$$, $$m=k$$ and $$c=\log a$$. So if we plot $$\log x$$ against $$\log y$$, we should get a straight line with gradient $$k$$.
Doing this for the same three curves as above gives the following plot.
$$y=x^2$$, $$y=x^{1.5}$$, and $$y=0.5x^3$$ with log-scaled $$x$$- and $$y$$-axes
It is easy to see that the blue line has the highest value of $$k$$ (as it has the highest gradient, and we could get a decent approximation of this value by finding the line's gradient.
As well as making it easier to get good approximations of important parameters, making curves into straight lines in this way also makes it easier to plot the trend of real data. Drawing accurate exponentials and polynomials is hard at the best of times; and real data will not exactly follow the curve, so drawing an exponential or quadratic of best fit will be an even harder task. By scaling the axes first though, this task simplifies to drawing a straight line through the data; this is much easier.
So next time you're struggling with an awkward curve, why not try turning it into a straight line first.
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Comments in green were written by me. Comments in blue were not written by me.
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# How to Use Number Lines to Identify Equivalent Fractions?
Fractions that have different numerators and denominators but are equal to the same value are equivalent fractions. In this step-by-step guide, you can learn how to use number lines to identify equivalent fractions.
## A Step-by-step guide to using number lines to identify equivalent fractions
Two or more fractions that show the same parts of a whole are called equivalent fractions. Equivalent fractions always represent the same point on the number line. There are two ways to indicate equivalent fractions: 1-Fraction bars, and 2-Number line.
In the Number line method, we draw a number line with whole numbers to represent any number. Then, we divide the space between each whole into equal parts, and the number of equal parts must match the number in the denominator. Then specify the point equal to the number.
The Absolute Best Book for 4th Grade Students
### Using Number Lines to Identify Equivalent Fractions-Example 1:
What is the equivalent fraction of $$\frac{1}{2}$$?
Solution: First, draw a number line of $$0$$ and $$1$$. Then divide the total into two parts equal to the denominator. Then check line $$\frac{1}{2}$$. Then divide each part into two equal parts. Then for the same situation, we can see that the new fraction is $$\frac{2}{4}$$ because there are $$4$$ equal parts and the marking is after two equal parts. So, $$\frac{1}{2}$$ and $$\frac{2}{4}$$ are equivalent fractions.
## Exercises forUsing Number Lines to Identify Equivalent Fractions
### Use a number line to find an equivalent fraction.
1. $$\color{blue}{\frac{1}{4}}$$
2. $$\color{blue}{\frac{2}{4}}$$
A Perfect Book for Grade 4 Math Word Problems!
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# Teaching Ideas for Place Value & Two Freebies
Hi everyone! I want to share some tips about learning place value. Having a deep root in understanding place value plays such an important role in grasping math. This post is geared for teaching place value in first grade, but I put in some ideas for 3-4 digit number too. That way, second and third grade teachers can have some take aways, along with those wanting some ideas to challenge some advanced first graders.
Tip #1: Let students show numbers with 10-blocks and 1-blocks. This will help them to understand that the tens place stands for a group of tens. (For second or third grade, add in numbers with the hundreds place and 100-block.) One idea is to have students draw numbers and show those numbers with the blocks.
(Please excuse the card, it is the wrong one. It should say 65. Each cup has a group of 10 bears.)
Tip #2: Break apart numbers by identifying how many tens and ones. (For second or third grade, have students identify how many hundreds, tens, and ones.) Students can draw a card and write it out on a sheet of paper.
Tip #3: Have students roll or spin a number for each place and merge it into a number. While regular dice only go to 6, you can still have students roll out some numbers. Have students identify the place value of each number and then write the number. (Sorry, no picture for the dice right now. I am hoping to have a dice picture later for this idea. I decided I wanted to use jumbo dice, so I need to go buy some for the picture when I get time).
Tip #4: Give students opportunity to show numbers in expanded form. Students can draw number cards and write out the number in expanded form.
Tip #5: Have students solve mystery number problems. I like this concept because it taps into some more critical thinking skills when students have to do things like figure out a number between 22 and 31 that has a five in the ones place.
Tip #6: Incorporate money with place value learning. Think about it dollar bills are the 100’s place. Dimes are the 10’s place. Pennies are the 1’s place. Students can count the money and identify how many tens and ones.
Tip #7: Count by tens in various forms. Give students plenty of opportunity to count by groups of ten with ten frames, number cards, hundred charts, and whatever else you can dream up. Let them not only tell you how many total, but how many groups of ten.
Tip #8: Compare numbers. Let students compare a mix of numbers. Have them compare numbers that depend on looking at the tens place to determine the bigger number. Have them compare numbers that depend on the ones place. For second and third grade, have them compare numbers that depend on the hundreds place. For an activity, students can draw two cards. Have them write the first number they draw first and the second number they draw second. Then have them use the comparison symbols to show greater than or less than.
Tip #9: Order numbers from either greater to lesser or from lesser to greater. You can have students draw five cards. Let them place the cards in order from lesser to greater or vice versa. Students can record this on a sheet of paper.
Tip #10: Let students work with concepts of one more, one less, ten more, and ten less. (For second and third grade, throw in concepts of a hundred more and a hundred less.) The concept of more than and less than can be tricky to some young learners, so I can’t recommend extra practice with this concept enough. Let students draw a number card. Then let them write ten more or less than as a number sentence. Let them do this for the ones place too. (Hundreds or even thousands place for second, third, and students who need more of a challenge.)
A tip for differentiating for first grade: I like to differentiate by providing students who are emerging with numbers that deal with up to 20 at first. Once they get a footing with that, they can move up to numbers up to 30 or beyond. Of course, you want them working with numbers up to a hundred in first grade, but it’s good to scaffold to get them to where they need to be with their understanding of place value. If they are struggling, breaking it down to lower numbers helps them to grasp that foundational understanding.
If you are wondering where all those fun no prep printables come from, they come from my Place Value No Prep for First Grade resource. Snag it here by clicking on the picture or by clicking here or on the picture.
I have created two resources for FREE with the number cards and posted them over in my TPT store. Snag them by clicking here for the first grade one and here for the 3-4 digit number one. You can also click on the pictures to link to them.
Thanks you so much for stopping by the Candy Class!
Jolene 🙂
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# FREE Equivalent Fraction Pattern Block Task Cards
*This post contains affiliate links from Amazon and other bloggers
Our fun with fractions is continuing this week. My boys are young (4 and 6), so my only goal for them is to understand that a fraction is equal parts, and that the numerator tells them how many parts or groups they have or in some cases don’t have. Though that is what I want for them to master, I am exposing them to other aspects. We are definitely working on what an equivalent fraction is, and how to find equivalent fractions on models (maybe next year we will discuss the algorithm!!) We have even done some adding and subtracting using our pattern blocks.
Today we pulled out our pattern blocks and completed the task cards.
### How to Start
If you have not worked with pattern blocks before you may want to do a short introduction. Simply pull out the hexagons, and have the children fill it up with trapezoids. Ask how many trapezoids were used, and what is the fraction for one trapezoid. This should lead itself into the awesome conversation about equal parts, ect!!! I would then have them place triangles on top of the trapezoids. You can then discuss what one triangle equals and what three triangles equal. Hopefully, your students will see that three triangles equal one trapezoid so therefore 3/6 is equal to 1/2.You can then do the same with the rhombuses or rhombi. (Did you know that both are the correct plural form of a rhombus?)
### Time for Task Cards
Once that background knowledge is laid for them, they are ready for the task cards. There are three different wholes in the sample pack, and five in the complete pack. You will find that the students are finding equivalent fractions for halves, thirds, fourths, and sixths numerous times but in various ways. I feel like this is important because students need to realize that 1/2 is always equal to 2/4 or 3/6 no matter the size of the equal parts.
### Struggles that Led to Better Understanding
There were a couple of things that stumped the boys, and gave us a chance to focus again on a fraction being equal parts and that the denominator tells us how many equal parts it takes to make a whole.
My four-year-old was working on this task card and had followed the directions. He had covered up 2/3 and knew that the numerator was 6. He was struggling to figure out what was the denominator. We went back over that the denominator is how many equal parts it takes to make a whole. He ended up filling the third rhombus with triangles and then was able to tell me that he had 4 out of the 6 triangles, so 2/3 and 4/6 are equivalent fractions.
### If You are Using Equivalent Fraction Patter Block Task Cards in a classroom
As a last minute tip, I highly suggest doing these in small groups where you can monitor. If you are working with higher grades I can possibly see doing this as a whole group with you walking around constantly checking and making note of those who will need to be pulled later for a small group. If you feel that any child is struggling with the concept of fractions, I would do this with them next to you.
These task cards will quickly bring to light any misconceptions, and I know no one wants to miss opportunities to see misconception and work through it.
You’ve Got This
## Or get 7 Activites and Games (with all 20 task cards) designed to help you teach equivalent fractions and allow your students fun, hands – on practice.
This packet contains
1. clip cards
2. interactive notebook
3. 3 games
4. cut and paste fraction book
5. activity for teaching fractions with food
6. task cards using pattern blocks
You’ve Got This
### 11 thoughts on “FREE Equivalent Fraction Pattern Block Task Cards”
1. Mother of 3 says:
What a fun activity! I love teaching math in a fun hands on way.
2. Mother of 3 says:
What a fun activity! I love teaching math in a fun hands on way.
3. Mother of 3 says:
What a fun activity! I love teaching math in a fun hands on way.
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Warm-Up!
1 / 11
# Warm-Up! - PowerPoint PPT Presentation
Warm-Up!. Number Cruncher!. Algebra I – Chapter 1. 1.4 – Translating words into symbols 1.5 – Translating sentences into Equations. Symbols:. You Try!. Four times a number Four less than five times a number Ten times the sum of a number and nine Eleven more than one third of a number.
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Warm-Up!
Number Cruncher!
### Algebra I – Chapter 1
1.4 – Translating words into symbols
1.5 – Translating sentences into Equations
You Try!
• Four times a number
• Four less than five times a number
• Ten times the sum of a number and nine
• Eleven more than one third of a number
You Try!
• Four times a number 4N
• Four less than five times a number 5N -4
• Ten times the sum of a number and nine
10(N + 9)
• Eleven more than one third of a number
⅓N + 11
Formulas:
• Formulas are equations that state rules about relationships.
• Area of a Rectangle:
• Perimeter of a Rectangle:
• Distance Traveled:
• Cost:
Equations:
• When you add the word is, then it becomes an equation. Now it can be solved.
• Ex: Three times a number decreased
by 8 is 10
You Try!
2x
3
x
x
P = 30m
A = 24m
4
P = 10m
Reflection:
• What is a symbol? Give a couple of examples.
• What is a formula?
• When does an expression become an equation?
### Tonight’s Homework:
Page 17 (17-26)
Page 20/21 (1-21 odd)
Quiz next class
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# What are the local extrema of f(x)= sinx on [0,2pi]?
Mar 31, 2016
At $x = \frac{\pi}{2}$ $f ' ' \left(x\right) = - 1$ we have a local maxima and at $x = 3 \frac{\pi}{2}$, $f ' ' \left(x\right) = 1$ we have a local minima.
#### Explanation:
A maxima is a high point to which a function rises and then falls again. As such the slope of the tangent or the value of derivative at that point will be zero.
Further, as the tangents to the left of maxima will be sloping upwards, then flattening and then sloping downwards, slope of the tangent will be continuously decreasing, i.e. the value of second derivative would be negative.
A minima on the other hand is a low point to which a function falls and then rises again. As such the tangent or the value of derivative at minima too will be zero.
But, as the tangents to the left of minima will be sloping downwards, then flattening and then sloping upwards, slope of the tangent will be continuously increasing or the value of second derivative would be positive.
However, these maxima and minima may either be universal i.e. maxima or minima for the entire range or may be localized, i.e. maxima or minima in a limited range.
Let us see this with reference to the function described in the question and for this let us first differentiate $f \left(x\right) = \sin x$.
$f ' \left(x\right) = \cos x$ and on $\left[0 , 2 \pi\right]$ it is $0$ at $x = \frac{\pi}{2}$ and $x = \frac{3 \pi}{2}$.
$f ' ' \left(x\right) = - \sin x$ and while at $x = \frac{\pi}{2}$ $f ' ' \left(x\right) = - 1$ meaning we have a local maxima, at $x = 3 \frac{\pi}{2}$, $f ' ' \left(x\right) = 1$ meaning we have a local minima.
graph{sinx [-1, 7, -1.5, 1.5]}
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components of 1296 are the list of integers that deserve to be evenly divided into 1296. Over there are complete 25 factors of 1296, of i beg your pardon 2, 3 room its prime factors. The element Factorization the 1296 is 24 × 34.
You are watching: Factors of 1296
All components of 1296: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 81, 108, 144, 162, 216, 324, 432, 648 and also 1296 Prime factors of 1296: 2, 3 Prime administrate of 1296: 24 × 34 Sum of factors of 1296: 3751
1 What are the components of 1296? 2 Factors that 1296 by element Factorization 3 Factors the 1296 in Pairs 4 FAQs on determinants of 1296
## What are factors of 1296?
Factors of 1296 are pairs that those numbers who products result in 1296. These components are either prime numbers or composite numbers.
### How to discover the components of 1296?
To discover the determinants of 1296, us will have to discover the perform of numbers that would divide 1296 there is no leaving any remainder.
1296/27 = 48; therefore, 27 is a aspect of 1296 and also 48 is likewise a element of 1296.1296/162 = 8; therefore, 162 is a aspect of 1296 and also 8 is likewise a variable of 1296. Similarly we can uncover other factors. Hence, the factors of 1296 space 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 81, 108, 144, 162, 216, 324, 432, 648, 1296.
## determinants of 1296 by prime Factorization
The number 1296 is composite and therefore that will have actually prime factors. Currently let us learn how to calculate the prime components of 1296. The very first step is to division the number 1296 with the smallest prime factor, here it is 2. We keep dividing until it gives a non-zero remainder. 1296 ÷ 2 = 648 648 ÷ 2 = 324 324 ÷ 2 = 162 162 ÷ 2 = 81
Further dividing 81 by 2 offers a non-zero remainder. For this reason we prevent the procedure and proceed dividing the number 81 by the next smallest element factor. We stop ultimately if the next prime element doesn"t exist or when we can"t divide any type of further.
So, the element factorization the 1296 deserve to be written as 24 × 34 where 2, 3 are prime.
## Factors the 1296 in Pairs
Pair determinants of 1296 are the pairs of numbers that once multiplied give the product 1296. The determinants of 1296 in bag are:
1 × 1296 = (1, 1296) 2 × 648 = (2, 648) 3 × 432 = (3, 432) 4 × 324 = (4, 324) 6 × 216 = (6, 216) 8 × 162 = (8, 162) 9 × 144 = (9, 144) 12 × 108 = (12, 108) 16 × 81 = (16, 81) 18 × 72 = (18, 72) 24 × 54 = (24, 54) 27 × 48 = (27, 48) 36 × 36 = (36, 36)
Negative pair determinants of 1296 are:
-1 × -1296 = (-1, -1296) -2 × -648 = (-2, -648) -3 × -432 = (-3, -432) -4 × -324 = (-4, -324) -6 × -216 = (-6, -216) -8 × -162 = (-8, -162) -9 × -144 = (-9, -144) -12 × -108 = (-12, -108) -16 × -81 = (-16, -81) -18 × -72 = (-18, -72) -24 × -54 = (-24, -54) -27 × -48 = (-27, -48) -36 × -36 = (-36, -36)
NOTE: If (a, b) is a pair factor of a number then (b, a) is additionally a pair factor of the number.
## Factors that 1296 solved Examples
Example 1: How numerous factors are there for 1296?
Solution:
The factors of 1296 space too many, thus if we can uncover the element factorization of 1296, climate the total variety of factors deserve to be calculated using the formula displayed below. If the element factorization that the number is ax × by × cz wherein a, b, c space prime, then the total variety of factors have the right to be provided by (x + 1)(y + 1)(z + 1). prime Factorization that 1296 = 24 × 34 Therefore, the total number of factors space (4 + 1) × (4 + 1) = 5 × 5 = 25
example 3: discover if 4, 6, 18, 54, 108, 144, 324 and 710 are determinants of 1296.
Solution:
as soon as we division 1296 by 710 it pipeline a remainder. Therefore, the number 710 is no a factor of 1296. Every numbers except 710 are determinants of 1296.
instance 4: discover the product of all the prime determinants of 1296.
See more: Urban Slavery Was Well Developed In Spanish America And, Urban Slavery Was Well
Solution:
Since, the prime factors of 1296 space 2, 3. Therefore, the product that prime components = 2 × 3 = 6.
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## FAQs on factors of 1296
### What room the determinants of 1296?
The factors of 1296 room 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 81, 108, 144, 162, 216, 324, 432, 648, 1296 and also its negative factors room -1, -2, -3, -4, -6, -8, -9, -12, -16, -18, -24, -27, -36, -48, -54, -72, -81, -108, -144, -162, -216, -324, -432, -648, -1296.
### What is the sum of the components of 1296?
amount of all components of 1296 = (24 + 1 - 1)/(2 - 1) × (34 + 1 - 1)/(3 - 1) = 3751
### What Numbers space the Prime components of 1296?
The prime determinants of 1296 room 2, 3.
### What is the Greatest typical Factor that 1296 and 204?
The determinants of 1296 and also 204 room 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 81, 108, 144, 162, 216, 324, 432, 648, 1296 and 1, 2, 3, 4, 6, 12, 17, 34, 51, 68, 102, 204 respectively. common factors of 1296 and 204 room <1, 2, 3, 4, 6, 12>. Hence, the Greatest typical Factor (GCF) that 1296 and 204 is 12.
### How many Factors the 1296 are likewise Factors of 546?
Since, the factors of 1296 space 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 36, 48, 54, 72, 81, 108, 144, 162, 216, 324, 432, 648, 1296 and also the factors of 546 space 1, 2, 3, 6, 7, 13, 14, 21, 26, 39, 42, 78, 91, 182, 273, 546. Hence, <1, 2, 3, 6> are the common factors of 1296 and also 546.
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# Class 6 Maths Chapter 1 Exercise 1.3 NCERT Solution – Knowing Our Numbers
NCERT Class 6 Math Exercise 1.3 Solution of Chapter 1 Knowing Our Numbers with explanation. Here we provide Class 6 Maths all Chapters in Hindi for cbse, HBSE, Mp Board, UP Board and some other boards.
Also Read : – Class 6 Maths NCERT Solution
NCERT Class 6 Maths Chapter 1 Knowing our Numbers Exercise 1.3 Solution in english Medium.
## Class 6 Maths Chapter 1 Exercise 1.3 Solution
Q.1. Estimate each of the following using general rule:
A. 730 + 998
Ans. 700 + 1000
= 1700
B. 796 – 314
Ans. 800 – 300
= 500
C. 12,904 + 2,888
Ans. 13,000 + 3000
= 16,000
D. 28,292 – 21496
Ans. 28000 – 21000
= 7,000
Q.2. Give a rough estimate ( by rounding off to nearest hundreds ) and also a closer estimate ( by rounding off to nearest tens ) :
A. 439 + 334 + 4,317
Ans. Estimation by rounding of to nearest
Hundreds:-
= 400 + 300 +4300
= 5,000
Estimation by rounding of to nearest
tens :-
= 440 + 330 + 4320
= 5,090
B. 1,08,734 – 47,599
Ans. Estimation by rounding of to nearest
Hundreds:-
= 1,08,700 – 47,600
= 61,100
Estimation by rounding of to nearest
tens :-
= 1,08,730 – 47,600
= 61,130
C. 8,325 – 491
Ans. Estimation by rounding of to nearest
Hundreds:-
= 8300 – 500
= 7,800
Estimation by rounding of to nearest
tens :-
= 8,330 – 490
= 7,840
D. 4,89,348 – 48,365
Estimation by rounding of to nearest
Hundreds:-
= 4,89,300 – 48,400
= 4,40,900
Estimation by rounding of to nearest
tens :-
= 4,89,350 – 48,370
= 4,40,980
Q.3. Estimate the following products using general rule :
A. 578 x 161
Ans. = 600 x 200
= 1,20,000
B. 5281 x 3491
Ans. = 5,000 x 3000
= 1,50,00,000
C. 1291 x 592
Ans. = 1000 x 600
= 6,00,000
D. 9250 x 29
Ans. = 9000 x 30
= 2,70,000
error:
|
# Vector Spaces and Linear Systems
In document Linear Algebra (Page 137-145)
We will now reconsider linear systems and Gauss’s Method, aided by the tools and terms of this chapter. We will make three points.
For the first, recall the insight from the Chapter One that Gauss’s Method works by taking linear combinations of rows — if two matrices are related by row operationsA−→ · · · −→Bthen each row ofBis a linear combination of
the rows ofA. Therefore, the right setting in which to study row operations in
general, and Gauss’s Method in particular, is the following vector space.
3.1 Definition Therow space of a matrix is the span of the set of its rows. The row rank is the dimension of this space, the number of linearly independent
rows.
3.2 Example If
A= 2 3 4 6
!
then Rowspace(A)is this subspace of the space of two-component row vectors. {c1·(2 3) +c2·(4 6)|c1, c2∈R}
The second row vector is linearly dependent on the first and so we can simplify the above description to{c·(2 3)|c∈R}.
3.3 Lemma If two matricesAandBare related by a row operation A ρi−→↔ρj B or A −→kρi B or A kρi+ρj−→ B
(for i6= jand k6=0) then their row spaces are equal. Hence, row-equivalent
matrices have the same row space and therefore the same row rank.
Proof Corollary One.III.2.4shows that whenA−→Bthen each row ofBis a linear combination of the rows ofA. That is, in the above terminology, each row
ofB is an element of the row space ofA. Then Rowspace(B)⊆Rowspace(A)
follows because a member of the set Rowspace(B)is a linear combination of the
rows ofB, so it is a combination of combinations of the rows ofA, and by the
Linear Combination Lemma is also a member of Rowspace(A).
For the other set containment, recall Lemma One.III.1.5, that row opera- tions are reversible so A−→Bif and only if B−→A. Then Rowspace(A)⊆
Rowspace(B)follows as in the previous paragraph. QED Of course, Gauss’s Method performs the row operations systematically, with the goal of echelon form.
3.4 Lemma The nonzero rows of an echelon form matrix make up a linearly
independent set.
Proof Lemma One.III.2.5says that no nonzero row of an echelon form matrix is a linear combination of the other rows. This result restates that using this
chapter’s terminology. QED
Thus, in the language of this chapter, Gaussian reduction works by eliminating linear dependences among rows, leaving the span unchanged, until no nontrivial linear relationships remain among the nonzero rows. In short, Gauss’s Method produces a basis for the row space.
3.5 Example From any matrix, we can produce a basis for the row space by
performing Gauss’s Method and taking the nonzero rows of the resulting echelon form matrix. For instance,
1 3 1 1 4 1 2 0 5 −ρ1+ρ2 −→ −2ρ1+ρ3 6ρ2+ρ3 −→ 1 3 1 0 1 0 0 0 3
produces the basish(1 3 1),(0 1 0),(0 0 3)ifor the row space. This is a basis for the row space of both the starting and ending matrices, since the two row spaces are equal.
Using this technique, we can also find bases for spans not directly involving row vectors.
3.6 Definition Thecolumn space of a matrix is the span of the set of its columns.
Thecolumn rank is the dimension of the column space, the number of linearly
independent columns.
Our interest in column spaces stems from our study of linear systems. An example is that this system
c1+3c2+7c3=d1 2c1+3c2+8c3=d2 c2+2c3=d3 4c1 +4c3=d4
has a solution if and only if the vector ofd’s is a linear combination of the other
column vectors, c1 1 2 0 4 +c2 3 3 1 0 +c3 7 8 2 4 = d1 d2 d3 d4
meaning that the vector ofd’s is in the column space of the matrix of coefficients. 3.7 Example Given this matrix,
1 3 7 2 3 8 0 1 2 4 0 4
to get a basis for the column space, temporarily turn the columns into rows and reduce. 1 2 0 4 3 3 1 0 7 8 2 4 −3ρ1+ρ2 −→ −7ρ1+ρ3 −2ρ2+ρ3 −→ 1 2 0 4 0 −3 1 −12 0 0 0 0
Now turn the rows back to columns.
h 1 2 0 4 , 0 −3 1 −12 i
3.8 Definition Thetranspose of a matrix is the result of interchanging its rows
and columns, so that columnjof the matrixAis rowjofAT and vice versa. So we can summarize the prior example as “transpose, reduce, and transpose back.”
We can even, at the price of tolerating the as-yet-vague idea of vector spaces being “the same,” use Gauss’s Method to find bases for spans in other types of vector spaces.
3.9 Example To get a basis for the span of {x2+x4, 2x2+3x4,x23x4} in
the spaceP4, think of these three polynomials as “the same” as the row vectors
(0 0 1 0 1),(0 0 2 0 3), and(0 0 −1 0 −3), apply Gauss’s Method
0 0 1 0 1 0 0 2 0 3 0 0 −1 0 −3 −2ρ1+ρ2 −→ ρ1+ρ3 2ρ2+ρ3 −→ 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0
and translate back to get the basishx2+x4, x4i. (As mentioned earlier, we will
make the phrase “the same” precise at the start of the next chapter.)
Thus, the first point for this subsection is that the tools of this chapter give us a more conceptual understanding of Gaussian reduction.
For the second point observe that row operations on a matrix can change its column space. 1 2 2 4 ! −2ρ1+ρ2 −→ 1 2 0 0 !
The column space of the left-hand matrix contains vectors with a second compo- nent that is nonzero but the column space of the right-hand matrix contains only vectors whose second component is zero, so the two spaces are different. This observation makes next result surprising.
3.10 Lemma Row operations do not change the column rank.
Proof Restated, ifAreduces toBthen the column rank ofBequals the column rank ofA.
This proof will be finished if we show that row operations do not affect linear relationships among columns, because the column rank is the size of the largest set of unrelated columns. That is, we will show that a relationship exists among columns (such as that the fifth column is twice the second plus the fourth) if and only if that relationship exists after the row operation. But this is exactly the
first theorem of this book, Theorem One.I.1.5: in a relationship among columns, c1· a1,1 a2,1 ... am,1 +· · ·+cn· a1,n a2,n ... am,n = 0 0 ... 0
row operations leave unchanged the set of solutions(c1, . . . , cn). QED
Another way to make the point that Gauss’s Method has something to say about the column space as well as about the row space is with Gauss-Jordan reduction. It ends with the reduced echelon form of a matrix, as here.
1 3 1 6 2 6 3 16 1 3 1 6 −→ · · · −→ 1 3 0 2 0 0 1 4 0 0 0 0
Consider the row space and the column space of this result.
The first point made earlier in this subsection says that to get a basis for the row space we can just collect the rows with leading entries. However, because this is in reduced echelon form, a basis for the column space is just as easy: collect the columns containing the leading entries,h~e1,~e2i. Thus, for a reduced echelon form matrix we can find bases for the row and column spaces in essentially the same way, by taking the parts of the matrix, the rows or columns, containing the leading entries.
3.11 Theorem For any matrix, the row rank and column rank are equal.
Proof Bring the matrix to reduced echelon form. Then the row rank equals the number of leading entries since that equals the number of nonzero rows. Then also, the number of leading entries equals the column rank because the set of columns containing leading entries consists of some of the~ei’s from a
standard basis, and that set is linearly independent and spans the set of columns. Hence, in the reduced echelon form matrix, the row rank equals the column rank, because each equals the number of leading entries.
But Lemma3.3 and Lemma3.10show that the row rank and column rank are not changed by using row operations to get to reduced echelon form. Thus the row rank and the column rank of the original matrix are also equal. QED
3.12 Definition Therank of a matrix is its row rank or column rank.
So the second point that we have made in this subsection is that the column space and row space of a matrix have the same dimension.
Our final point is that the concepts that we’ve seen arising naturally in the study of vector spaces are exactly the ones that we have studied with linear systems.
3.13 Theorem For linear systems withnunknowns and with matrix of coefficients A, the statements
(1) the rank ofAisr
(2) the vector space of solutions of the associated homogeneous system has dimensionn−r
are equivalent.
So if the system has at least one particular solution then for the set of solutions, the number of parameters equalsn−r, the number of variables minus the rank
of the matrix of coefficients.
Proof The rank ofAis rif and only if Gaussian reduction onAends withr nonzero rows. That’s true if and only if echelon form matrices row equivalent toAhaver-many leading variables. That in turn holds if and only if there are
n−rfree variables. QED
3.14 Corollary Where the matrixAisn×n, these statements (1) the rank ofAisn
(2) Ais nonsingular
(3) the rows ofAform a linearly independent set
(4) the columns ofAform a linearly independent set
(5) any linear system whose matrix of coefficients isAhas one and only one
solution are equivalent.
Proof Clearly (1) ⇐⇒ (2) ⇐⇒ (3) ⇐⇒ (4). The last, (4) ⇐⇒ (5), holds because a set ofncolumn vectors is linearly independent if and only if it is a
basis forRn, but the system
c1 a1,1 a2,1 ... am,1 +· · ·+cn a1,n a2,n ... am,n = d1 d2 ... dm
has a unique solution for all choices ofd1, . . . , dn∈Rif and only if the vectors
ofa’s on the left form a basis. QED
3.15 Remark [Munkres] Sometimes the results of this subsection are mistakenly
system usesn−mparameters. The number of equations is not the relevant
number; rather, what matters is the number of independent equations, the num- ber of equations in a maximal independent set. Where there arerindependent
equations, the general solution involvesn−r parameters. Exercises 3.16 Transpose each. (a) 2 1 3 1 (b) 2 1 1 3 (c) 1 4 3 6 7 8 (d) 0 0 0 (e) (−1 −2)
X3.17 Decide if the vector is in the row space of the matrix. (a) 2 1 3 1 ,(1 0) (b) 0 1 3 −1 0 1 −1 2 7 ,(1 1 1)
X3.18 Decide if the vector is in the column space. (a) 1 1 1 1 , 1 3 (b) 1 3 1 2 0 4 1 −3 −3 , 1 0 0
X3.19 Decide if the vector is in the column space of the matrix. (a) 2 1 2 5 , 1 −3 (b) 4 −8 2 −4 , 0 1 (c) 1 −1 1 1 1 −1 −1 −1 1 , 2 0 0
X3.20 Find a basis for the row space of this matrix. 2 0 3 4 0 1 1 −1 3 1 0 2 1 0 −4 −1
X3.21 Find the rank of each matrix. (a) 2 1 3 1 −1 2 1 0 3 (b) 1 −1 2 3 −3 6 −2 2 −4 (c) 1 3 2 5 1 1 6 4 3 (d) 0 0 0 0 0 0 0 0 0
3.22 Give a basis for the column space of this matrix. Give the matrix’s rank. 1 3 −1 2 2 1 1 0 0 1 1 4
X3.23 Find a basis for the span of each set. (a) {(1 3),(−1 3),(1 4),(2 1)}⊆M1×2 (b) { 1 2 1 , 3 1 −1 , 1 −3 −3 }⊆R3
(c){1+x, 1−x2, 3+2xx2}P 3 (d) { 1 0 1 3 1 −1 , 1 0 3 2 1 4 , −1 0 −5 −1 −1 −9 }⊆M2×3
3.24 Give a basis for the span of each set, in the natural vector space. (a){ 1 1 3 , −1 2 0 , 0 12 6 } (b) {x+x2, 22x, 7, 4+3x+2x2}
3.25 Which matrices have rank zero? Rank one?
X3.26 Givena, b, c∈R, what choice ofdwill cause this matrix to have the rank of one?
a b c d
3.27 Find the column rank of this matrix.
1 3 −1 5 0 4
2 0 1 0 4 1
3.28 Show that a linear system with at least one solution has at most one solution if and only if the matrix of coefficients has rank equal to the number of its columns. X3.29 If a matrix is5×9, which set must be dependent, its set of rows or its set of
columns?
3.30 Give an example to show that, despite that they have the same dimension, the row space and column space of a matrix need not be equal. Are they ever equal? 3.31 Show that the set {(1,−1, 2,−3),(1, 1, 2, 0),(3,−1, 6,−6)} does not have the
same span as{(1, 0, 1, 0),(0, 2, 0, 3)}. What, by the way, is the vector space? X3.32 Show that this set of column vectors
{ d1 d2 d3
|there arex,y, andzsuch that:
3x+2y+4z=d1
x − z=d2
2x+2y+5z=d3
}
is a subspace ofR3. Find a basis.
3.33 Show that the transpose operation is linear:
(rA+sB)T=rAT
+sBT forr, s∈RandA, B∈Mm×n.
X3.34 In this subsection we have shown that Gaussian reduction finds a basis for the row space.
(a)Show that this basis is not unique — different reductions may yield different bases.
(b) Produce matrices with equal row spaces but unequal numbers of rows. (c)Prove that two matrices have equal row spaces if and only if after Gauss-Jordan
reduction they have the same nonzero rows.
3.35 Why is there not a problem with Remark3.15in the case thatris bigger than n?
3.36 Show that the row rank of anm×nmatrix is at most m. Is there a better bound?
3.37 Show that the rank of a matrix equals the rank of its transpose.
3.38 True or false: the column space of a matrix equals the row space of its transpose. X3.39 We have seen that a row operation may change the column space. Must it?
3.40 Prove that a linear system has a solution if and only if that system’s matrix of coefficients has the same rank as its augmented matrix.
3.41 Anm×nmatrix hasfull row rank if its row rank ism, and it hasfull column rank if its column rank isn.
(a) Show that a matrix can have both full row rank and full column rank only if it is square.
(b) Prove that the linear system with matrix of coefficientsAhas a solution for anyd1, . . . ,dn’s on the right side if and only ifAhas full row rank.
(c)Prove that a homogeneous system has a unique solution if and only if its matrix of coefficientsAhas full column rank.
(d) Prove that the statement “if a system with matrix of coefficientsAhas any solution then it has a unique solution” holds if and only ifA has full column rank.
3.42 How would the conclusion of Lemma3.3change if Gauss’s Method were changed to allow multiplying a row by zero?
3.43 What is the relationship between rank(A)and rank(−A)? Between rank(A)
and rank(kA)? What, if any, is the relationship between rank(A), rank(B), and rank(A+B)?
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# Simple Exponent and Logarithm Equations Help (page 2)
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By McGraw-Hill Professional
Updated on Oct 4, 2011
#### Examples
Sketch the graph of the logarithmic functions.
• y = log 2 x
• Rewrite the equation in exponential form, x = 2 y , and let the exponent, y , be the numbers −3, −2, −1, 0, 1, 2, and 3.
Table 9.6
x y −3 −2 −1 1 0 2 1 4 2 8 3
Fig. 9.9
• y = ln x
• Rewritten as an exponent equation, this is x = e y. Let y = −3, −2, −1, 0, 1, 2, and 3.
Table 9.7
x y 0.05 −3 0.14 −2 0.37 −1 1 0 2.72 1 7.39 2 20.09 3
Fig. 9.10
As you can see by these graphs, the domain of the function f ( x ) = log a x is all positive real numbers, (0, ∞).
### Graphing Simple Exponent and Logarithm Equations Practice Problems
#### Practice
Sketch the graph of the logarithmic function.
1. y = log 1.5 x
2. y = log 3 x
1.
Fig. 9.11
2.
Fig. 9.12
## Finding the Domain of a Function
As long as a is larger than 1, all graphs for f ( x ) = log a x look pretty much the same. The larger a is, the flatter the graph is to the right of x = 1. Knowing this and knowing how to graph transformations, we have a good idea of the graphs of many logarithmic functions.
• The graph of f ( x ) = log 2 ( x − 2) is the graph of y = log 2 x shifted to the right 2 units.
• The graph of f ( x ) = − 5 + log 3 x is the graph of y = log 3 x shifted down 5 units.
• log x is the graph of y = log x flattened vertically by a factor of one-third.
The domain of f ( x ) = log a x is all positive numbers. This means that we cannot take the log of 0 or the log of a negative number. The reason is that a is a positive number. Raising a positive number to any power is always another positive number.
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#### Q:
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Mathematics | Random Variables
Random variable is basically a function which maps from the set of sample space to set of real numbers. The purpose is to get an idea about result of a particular situation where we are given probabilities of different outcomes. See below example for more clarity.
Example :
```Suppose that two coins (unbiased) are tossed
X = number of heads. [X is a random variable
or function]
Here, the sample space S = {HH, HT, TH, TT}.
The output of the function will be :
X(HH) = 2
X(HT) = 1
X(TH) = 1
X(TT) = 0
```
Formal definition :
X: S -> R
X = random variable (It is usually denoted using capital letter)
S = set of sample space
R = set of real numbers
Suppose a random variable X takes m different values i.e. sample space X = {x1, x2, x3………xm} with probabilities P(X=xi) = pi; where 1 ≤ i ≤ m. The probabilities must satisfy the following conditions :
1. 0 <= pi <= 1; where 1 <= i <= m
2. p1 + p2 + p3 + ……. + pm = 1 Or we can say 0 ≤ pi ≤ 1 and ∑pi = 1.
Hence possible values for random variable X are 0, 1, 2.
X = {0, 1, 2} where m = 3
P(X=0) = probability that number of heads is 0 = P(TT) = 1/2*1/2 = 1⁄4.
P(X=1) = probability that number of heads is 1 = P(HT | TH) = 1/2*1/2 + 1/2*1/2 = 1⁄2.
P(X=2) = probability that number of heads is 2 = P(HH) = 1/2*1/2 = 1⁄4.
Here, you can observe that
1) 0 ≤ p1, p2, p3 ≤ 1
2) p1 + p2 + p3 = 1/4 + 2/4 + 1/4 = 1
Example :
Suppose a dice is thrown X = outcome of the dice. Here, the sample space S = {1, 2, 3, 4, 5, 6}. The output of the function will be:
1. P(X=1) = 1/6
2. P(X=2) = 1/6
3. P(X=3) = 1/6
4. P(X=4) = 1/6
5. P(X=5) = 1/6
6. P(X=6) = 1/6
See if there is any random variable then there must be some distribution associated with it.
Discrete Random Variable:
A random variable X is said to be discrete if it takes on finite number of values. The probability function associated with it is said to be PMF = Probability mass function.
P(xi) = Probability that X = xi = PMF of X = pi.
1. 0 ≤ pi ≤ 1.
2. ∑pi = 1 where sum is taken over all possible values of x.
The examples given above are discrete random variables.
Example:- Let S = {0, 1, 2}
Find the value of P (X=0):
Sol:- We know that sum of all probabilities is equals to 1.
==> p1 + p2 + p3 = 1
==> p1 + 0.3 + 0.5 = 1
==> p1 = 0.2
Continuous Random Variable:
A random variable X is said to be continuous if it takes on infinite number of values. The probability function associated with it is said to be PDF = Probability density function
PDF: If X is continuous random variable.
P (x < X < x + dx) = f(x)*dx.
1. 0 ≤ f(x) ≤ 1; for all x
2. ∫ f(x) dx = 1 over all values of x
Then P (X) is said to be PDF of the distribution.
Example:- Compute the value of P (1 < X < 2).
```Such that f(x) = k*x^3; 0 ≤ x ≤ 3
= 0; otherwise
f(x) is a density function
```
Solution:- If a function f is said to be density function, then sum of all probabilities is equals to 1. Since it is a continuous random variable Integral value is 1 overall sample space s.
==> K*[x^4]/4 = 1 [Note that [x^4]/4 is integral of x^3]
==> K*[3^4 – 0^4]/4 = 1
==> K = 4/81
The value of P (1 < X < 2) = k*[X^4]/4 = 4/81 * [16-1]/4 = 15/81.
Next Topic :
Linearity of Expectation
Reference:
MIT Video Lecture
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# Math109Fall2019HW3SolutionsRubric.pdf - Homework 3 Partial...
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Homework 3 Partial Solutions Problems IV 1. Prove that if an integer n is the sum of two squares ( n = a 2 + b 2 for some a, b Z ) then n = 4 q or n = 4 q + 1 or n = 4 q + 2 for some q Z . Deduce that 1234567 cannot be wrtten as the sum of two squares. Proof. Suppose n = a 2 + b 2 for some a, b Z . By exercise 15.5, any perfect square is equal to 4 q or 4 q + 1 for some q Z . This gives three cases for a 2 , b 2 : If a 2 = 4 p and b 2 = 4 q for some p, q Z , then n = a 2 + b 2 = 4 p + 4 q = 4( p + q ) . If a 2 = 4 p and b 2 = 4 q + 1 for some p, q Z , then n = a 2 + b 2 = 4 p + 4 q + 1 = 4( p + q ) + 1 . We obtain a similar result when a 2 = 4 p + 1 and b 2 = 4 q for some p, q Z . If a 2 = 4 p + 1 and b 2 = 4 q + 1 for some p, q Z , then n = a 2 + b 2 = 4 p + 1 + 4 q + 1 = 4( p + q ) + 2 . Therefore, n = 4 q or n = 4 q + 1 or n = 4 q + 2 for some q Z . Since 1234567 = 4(308641) + 3, 1234567 is not the sum of two squares. 2. Let a be an integer. Prove that a 2 is divisble by 5 if and only if a is divisible by 5. Proof. ( ) Suppose a 2 is divisible by 5. We may express a in one of the following ways: If a = 5 q for some q Z , then 5 divides a . If a = 5 q + 1 for some q Z , then a 2
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# Graphs of Inequalities in One Variable
## Graph inequalities like y>4 and x<6 on the coordinate plane
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Graphs of Inequalities in One Variable
### Graphs of Inequalities in One Variable
A linear inequality in two variables takes the form y>mx+b\begin{align*}y > mx+b\end{align*} or y<mx+b\begin{align*}y < mx + b\end{align*}. Linear inequalities are closely related to graphs of straight lines; recall that a straight line has the equation y=mx+b\begin{align*}y = mx + b\end{align*}.
When we graph a line in the coordinate plane, we can see that it divides the plane in half:
The solution to a linear inequality includes all the points in one half of the plane. We can tell which half by looking at the inequality sign:
> The solution set is the half plane above the line.
\begin{align*}\ge\end{align*} The solution set is the half plane above the line and also all the points on the line.
< The solution set is the half plane below the line.
\begin{align*}\le\end{align*} The solution set is the half plane below the line and also all the points on the line.
For a strict inequality, we draw a dashed line to show that the points in the line are not part of the solution. For an inequality that includes the equals sign, we draw a solid line to show that the points on the line are part of the solution.
#### Solution Sets
This is a graph of ymx+b\begin{align*}y \ge mx + b\end{align*}; the solution set is the line and the half plane above the line.
This is a graph of y<mx+b\begin{align*}y < mx + b\end{align*}; the solution set is the half plane above the line, not including the line itself.
#### Graph Linear Inequalities in One Variable in the Coordinate Plane
In the last few sections we graphed inequalities in one variable on the number line. We can also graph inequalities in one variable on the coordinate plane. We just need to remember that when we graph an equation of the type x=a\begin{align*}x = a\end{align*} we get a vertical line, and when we graph an equation of the type y=b\begin{align*}y = b\end{align*} we get a horizontal line.
#### Graphing Inequalities
1. Graph the inequality x>4\begin{align*}x > 4\end{align*} on the coordinate plane.
First let’s remember what the solution to x>4\begin{align*}x > 4\end{align*} looks like on the number line.
The solution to this inequality is the set of all real numbers x\begin{align*}x\end{align*} that are bigger than 4, not including 4. The solution is represented by a line.
In two dimensions, the solution still consists of all the points to the right of x=4\begin{align*}x = 4\end{align*}, but for all possible y\begin{align*}y-\end{align*}values as well. This solution is represented by the half plane to the right of x=4\begin{align*}x = 4\end{align*}. (You can think of it as being like the solution graphed on the number line, only stretched out vertically.)
The line x=4\begin{align*}x = 4\end{align*} is dashed because the equals sign is not included in the inequality, meaning that points on the line are not included in the solution.
2. Graph the inequality |y|<5\begin{align*}|y| < 5\end{align*}
The absolute value inequality |y|<5\begin{align*}|y| < 5\end{align*} can be re-written as 5<y<5\begin{align*}-5 < y < 5\end{align*}. This is a compound inequality which can be expressed as
y>5andy<5\begin{align*}y > -5 \quad \text{and} \quad y < 5\end{align*}
In other words, the solution is all the coordinate points for which the value of y\begin{align*}y\end{align*} is larger than -5 and smaller than 5. The solution is represented by the plane between the horizontal lines y=5\begin{align*}y = -5\end{align*} and y=5\begin{align*}y = 5\end{align*}.
Both horizontal lines are dashed because points on the lines are not included in the solution.
### Example
#### Example 1
Graph the inequality |x|2\begin{align*}|x| \ge 2\end{align*}.
The absolute value inequality |x|2\begin{align*}|x| \ge 2\end{align*} can be re-written as a compound inequality:
x2orx2\begin{align*}x \le -2 \quad \text{or} \quad x \ge 2\end{align*}
In other words, the solution is all the coordinate points for which the value of x\begin{align*}x\end{align*} is smaller than or equal to -2 or greater than or equal to 2. The solution is represented by the plane to the left of the vertical line x=2\begin{align*}x = -2\end{align*} and the plane to the right of line x=2\begin{align*}x = 2\end{align*}.
Both vertical lines are solid because points on the lines are included in the solution.
### Review
Graph the following inequalities on the coordinate plane.
1. x<20\begin{align*}x < 20\end{align*}
2. y5\begin{align*}y \ge -5\end{align*}
3. x>0.5\begin{align*}x > 0.5\end{align*}
4. x12\begin{align*}x \le \frac{1}{2}\end{align*}
5. y>23\begin{align*}y > -\frac{2}{3}\end{align*}
6. y<0.2\begin{align*}y < -0.2\end{align*}
7. |x|>10\begin{align*}|x| > 10\end{align*}
8. |y|7\begin{align*}|y| \le 7\end{align*}
9. |y|<13\begin{align*}|y| < \frac{1}{3}\end{align*}
10. |x|10\begin{align*}|x| \ge -10\end{align*}
### Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook® resource, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9616.
### Notes/Highlights Having trouble? Report an issue.
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2. [2]^ License: CC BY-NC 3.0
3. [3]^ License: CC BY-NC 3.0
4. [4]^ License: CC BY-NC 3.0
5. [5]^ License: CC BY-NC 3.0
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7. [7]^ License: CC BY-NC 3.0
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Lesson Worksheet: Variables and Constants Mathematics • 6th Grade
In this worksheet, we will practice using the concepts of constants, variables, and mathematical relations to solve simple real-word problems.
Q1:
The equation shows the relation between the number of socks and the number of shoes Michael owns, where represents the number of socks, and represents the number of shoes. If Michael has 5 socks, how many shoes does he have?
Q2:
The equation shows the number of cakes Michael baked in relation to the number of hours he spent baking them. The variable represents the number of hours spent baking the cakes, and the variable represents the number of cakes baked. How many cakes did Michael make if he spent 7 hours baking?
Q3:
The equation can be used to calculate the total number of CDs Elizabeth had listened to days after she started working as a music producer. How many CDs had Elizabeth listened to one day after she started her job?
Q4:
The equation shows the number of ice cream cones served by an ice cream truck in a number of hours. The variable represents the number of hours, and the variable represents the number of ice cream cones served. How many ice cream cones were served after 2 hours?
Q5:
The equation shows the total number of flowers Olivia has in relation to the number of vases she owns. The variable represents the number of vases she has, and the variable represents the total number of flowers. If Olivia has 5 vases, how many flowers does she have?
Q6:
True or False: In , the variable can have any value.
• AFalse
• BTrue
Q7:
Which one of the following expressions does not belong with the others?
• A
• B
• C
• D
• E
Q8:
What terminology do we use to describe the in ?
• Aa variable
• Ba constant
• Can expression
• Da term
• Ea coefficient
Q9:
The side length of a rhombus is and its perimeter is .
What is the mathematical relation between and ?
• A
• B
• C
• D
What term do we use to describe in ?
• AA constant
• BA variable
Q10:
The side length of an equilateral triangle is and its perimeter is .
What is the mathematical relation between and ?
• A
• B
• C
• D
What is the term used to describe in ?
• AA constant
• BA variable
|
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Nerd For Tech
# Understanding Binary
An Introduction to the 0s and 1s
Even before I learned how to code, I knew there was something called binary and that it was the language computers spoke. I assumed that at some point during my 15 week bootcamp, we would discuss the elusive 0s and 1s but I was wrong. As it turns out, the code I was writing in JavaScript was being compiled and translated into 0s and 1s as instructions for the computer to understand. You may be thinking “ok, so I do not need to know about binary or number systems to code, then why should I bother learning it at all?” Even though understanding number systems is not required to code, it remains essential to our understanding of Computer Science. If you have any interest in taking a deep dive into how computers work, your starting point should be number systems and specifically, binary.
## The numbers we know, aka decimal
Numbers are presented to us in something called place-value notation. This means that both the digit and its position(or place-value) must be known to determine its value.
Let’s look at this in the context of our number system. 5 by itself would be considered in the first position in our number system and therefore to determine its value we multiply it by 1 and get 5. What if the 5 is in the second position? The second positions in our number system multiplies the digit by 10, making its value 50. What if the 5 is in the third position? The third position multiplies the digit by 100, making its value 500.
I am sure most of you are saying “duh” right now. But breaking numbers down like this will help us understand other number systems we encounter. Another way of looking at the above is that each place value is determined by raising 10 to an ever increasing value (multiplying 10 by itself a certain number of times) and multiplying the result by the digit. For these reason, our number system is called base 10 or decimal. In the above examples:
5 in the first position = 5 *10⁰ = 5 * 1 = 5
5 in the second position = 5 * 10¹ = 5 * 10 = 50
5 in the third position = 5 * 10² = 5 * 100 = 500
The number 10 not only determines the order of magnitude for each place-value but also represents the number of possible values for each digit, which is 0–9. The above example only uses a single digit in different positions but what if we have multiple digits in different positions.
2 in the first position = 2*1⁰ = 2* 1 = 2
5 in the second position = 5 * 1¹ = 5 * 10 = 50
6 in the third position = 6* 1² = 6* 100 = 600
600 + 50 + 2 = 652
One thing to note is 0 which represents a position with no value, regardless of the position it is in. 0 times 10¹⁰ (10,000,000,000) is still 0. For this reason, the lowest digit that carries value is 1, not zero.
With only 10 possible digits in each position, what happens when the highest possible digit (9) increases by 1? We add the lowest digit that carries value to the next position and reset the current position to 0 or no value. In other words, 9 becomes 10.
## The numbers computers know, aka binary
Now that we have had a grade school math refresher and introduced some new glossary terms, we will break down binary in the same way we did our numeral system of base 10.
To begin, ask yourself what you know about binary. Odds are your answer will be that each position can only be a 0 or 1. Given what we discussed above, what base would binary be? The answer is 2. Base 2 not only means that each position can only have one of two possible digits but also that the order of magnitude of each place-value increases by powers of 2 instead of 10, as is the case with our number system.
Let us use the model from our base 10 examples to break down binary, using the number 1
1 at the first position (1) = 1 * 2⁰ = 1 * 1 = 1
1 at the second position (10) = 1 * 2¹ = 1 * 2 = 2
1 at the third position (100) =1 * 2² = 1 * 4 = 4
1 at the fourth position (1000) =1 * 2³ = 1 * 8 = 8
1 at the fifth position (10000) =1 * 2⁴ = 1 * 16 = 16
The above demonstrates how we can convert binary numbers to decimal based on 1 being at different positions and having different place-values. If the binary number is 11111, we can add all the above values up to get the total value in decimal
11111 = 1 + 10 + 100 + 1000 + 10000 = 1 + 2 + 4 + 8 + 16 = 31
Another way of looking at this is
11111 is 100000 minus 1 or
(1 * 2⁵) - 1 = 32 - 1 = 31
To make sense of this, we need to do some basic counting in binary. 0 and 1 are the only two possible numbers in each position with 1 being the number with the highest value. Starting with 0, if we add 1, we get 1 (same thing we see in decimal if we were to add 1 to 0). What happens if we add 1 to 1? Just as is the case with 9 in decimal, when we add 1 to 1 in binary, we add the lowest digit of value to the next position and reset the current position.
1 + 1 = 10
Following this pattern of increasing by 1
0
1
10
11
100
101
110
111
1000
Converting 0–100 in binary to decimal:
0= 0 * 2⁰ = 0 * 1 = 0
1= 1* 2⁰ = 1 * 1 = 1
10=(1*2¹) + (0 * 2⁰) = (1 * 2) + (0 * 1) = 2 + 0 = 2
11=(1* 2¹) + (1* 2⁰) = (1 * 2) + (1 * 1) = 2 + 1 = 3
100=(1*2²) + (0 * 2¹) + (0 * 2⁰)= (1 * 4) + (0 * 2) + (0 * 1) = 4 + 0 + 0 =4
This helps us understand binary as a number system but their true value in computers is more than simply representing decimal numbers in a different way.
## Turning binary in building blocks
How does a series of 0s and 1s become every computer application that has ever existed? Just as letters can be grouped to form words, words to form sentences, and sentences to form paragraphs. Binary can be grouped to represent everything from a simple decimal number (as we saw above) to complicated and elaborate instructions dictating the behavior of a character in a computer game.
Above, 0s and 1s were referred to as digits in the context of comparing binary to our number system. However, in the context of the code where binary is the foundation of every computer operation, 0s and 1s are bits. That is to say a 0 by itself is a bit and a 1 by itself is also a bit. When eight bits are grouped together, they form a byte. Finally a familiar term!
Besides numbers, one of the simplest uses of a byte is to represent a letter. This is done with the American Standard Code for Information Interchange, or ASCII, which “is a character encoding standard for electronic communication. ASCII codes represent text in computers”.
ABC: 01000001 01000010 01000011
Hello, world!: 01001000 01100101 01101100 01101100 01101111 00101100 00100000 01110111 01101111 01110010 01101100 01100100 00100001
(Make your own secret messages in binary here)
If one byte can only hold a single letter, you can imagine how many are needed for a song, movie, or game.
This is obviously and extensive topic but as the subtitle of this article says, this is meant to be an “Introduction to the 0s and 1s”, not a definitive guide. I hope this article can serve as a starting point for your own exploration into binary and other Computer Science concepts.
--
--
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Solving rational expressions can be a difficult and time consuming task, but it is a necessary skill in order to solve problems. The process of solving rational expressions requires: Many methods can be employed to solve rational expressions, including: A rational expression is any equation that contains the following symbols: >, , >, and . In order to solve a rational expression, you must first find the roots of the equation by evaluating each term. For example: When evaluating an expression with values on both sides of the equal sign, evaluate both sides before finding the root. For example: When evaluating an expression with only one side of the equal sign, evaluate that side before finding the root. For example: If an expression cannot be simplified by any means, it is said to be irreducible. To solve such an equation, you must factor out all terms until no terms remain. Once all factors are removed from an irreducible expression, you can then find roots using elementary algebra. It is always better to factor out terms before simplifying expressions if possible. Factors are often written in scientific notation; for example: In cases where "a" = "b" or "c" = "d", you can swap the exponents and simplify by dividing by "a". If you have only one pair of exponents, it may make
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# 42% of adults say cashews are their favorite kind of nut. Y
James Dale 2021-12-20 Answered
42% of adults say cashews are their favorite kind of nut. You randomly select 12 adults and ask each to name his or her favorite nut. Find the probability that the number who say cashews are their favorite nut is (a) exactlythree, (b) at least four, and (c) at most two. If convenient, use technology to find the probabilities.
(a) $$\displaystyle{P}{\left({3}\right)}=$$
(b) $$\displaystyle{P}{\left({x}\geq{4}\right)}=$$
(c) $$\displaystyle{P}{\left({x}\le{2}\right)}=$$
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Ronnie Schechter
Step 1
Given Data:
The probability that adults say that cashews are their favorite kind of nut is: $$\displaystyle{p}={0.43}$$
The total number of adults randomly selected is: $$\displaystyle{n}={12}$$.
The probability that the cashew nuts are not their favorite kind of nut is,
$$\displaystyle{q}={1}-{p}$$
Substitute values in the above expression.
$$\displaystyle{q}={1}-{0.43}$$
$$\displaystyle={0.57}$$
The binomial expression to calculate the probability of 'x' adults who says that cashews are their favorite kind of nut is,
$$\displaystyle{P}{\left({x}\right)}=^{{{n}}}{C}_{{{x}}}{p}^{{{x}}}{q}^{{{n}-{x}}}$$
Substitute values in the above expression.
$$\displaystyle{P}{\left({x}\right)}=^{{{12}}}{C}_{{{x}}}\times{\left({0.43}\right)}^{{{x}}}\times{\left({0.57}\right)}^{{{12}-{x}}}$$
Step 2
(a) The expression to calculate the probability that exactly three adults say that the cashews are their favorite kind of nut is,
$$\displaystyle{P}{\left({3}\right)}=^{{{12}}}{C}_{{{3}}}\times{\left({0.43}\right)}^{{{3}}}\times{\left({0.57}\right)}^{{{12}-{3}}}$$
$$\displaystyle={220}\times{\left({0.43}\right)}^{{{3}}}\times{\left({0.57}\right)}^{{{9}}}$$
$$\displaystyle={0.1111}$$
Thus, the probability that exactly three adults say that the cashews are their favorite kind of nut is 0.111.
(b) The expression to calculate the probability that at least four of the adults like cashews nuts.
$$\displaystyle{P}{\left({x}\geq{4}\right)}={1}-{\left({P}{\left({0}\right)}+{P}{\left({1}\right)}+{P}{\left({2}\right)}+{P}{\left({3}\right)}\right)}$$
Substitute values in the above expression.
$$\displaystyle{P}{\left({x}\geq{4}\right)}={1}-{\left(^{\left\lbrace{12}\right\rbrace}{C}_{{{0}}}\times{\left({0.43}\right)}^{{{0}}}\times{\left({0.57}\right)}^{{{12}}}+^{{{12}}}{C}_{{{1}}}\times{0.43}\times{\left({0.57}\right)}^{{{11}}}+^{{{12}}}{C}_{{{2}}}\times{\left({0.43}\right)}^{{{2}}}\times{\left({0.43}\right)}^{{{2}}}\times{\left({0.57}\right)}^{{{10}}}+^{{{12}}}{C}_{{{3}}}\times{\left({0.43}\right)}^{{{3}}}\times{\left({0.57}\right)}^{{{9}}}\right)}$$
$$\displaystyle={1}-{\left({0.0017}+{0.0106}+{0.0441}+{0.1110}\right)}$$
$$\displaystyle={1}-{0.1674}$$
$$\displaystyle={0.8326}$$
Thus, the probability that at least four adults like cashews is0.8326.
Step 3
(c) The expression to calculate the probability that at most two adults' like cashews is,
$$\displaystyle{P}{\left({x}\le{2}\right)}={P}{\left({0}\right)}+{P}{\left({1}\right)}+{P}{\left({2}\right)}$$
Substitute values in the above expression.
$$\displaystyle{P}{\left({x}\le{2}\right)}=^{{{12}}}{C}_{{{0}}}\times{\left({0.43}\right)}^{{{0}}}\times{\left({0.57}\right)}^{{{12}}}+^{{{12}}}{C}_{{{1}}}\times{\left({0.43}\right)}\times{\left({0.57}\right)}^{{{11}}}+^{{{12}}}{C}_{{{2}}}\times{\left({0.43}\right)}^{{{2}}}\times{\left({0.57}\right)}^{{{10}}}$$
$$\displaystyle={0.0017}+{0.0106}+{0.0441}$$
$$\displaystyle={0.0564}$$
Thus, the probability that at most two adults like cashews is 0.0564.
###### Not exactly what you’re looking for?
ambarakaq8
Step 1
The following notation varies slightly from book-to-book
binomial formula:
$$\displaystyle{P}{\left({x}\right)}={\left(\begin{array}{c} {n}\\{x}\end{array}\right)}\times{p}^{{{x}}}\times{\left({1}-{p}\right)}^{{{x}}}$$
where $$\displaystyle{\left(\begin{array}{c} {n}\\{x}\end{array}\right)}={\frac{{{n}!}}{{{x}!{\left({n}-{x}\right)}!}}}$$
Here $$\displaystyle{n}={12},\ {p}={0.42}$$ and $$\displaystyle{1}-{p}={1}-{0.42}={0.58}$$
a) $$\displaystyle{P}{\left({X}={3}\right)}={\frac{{{12}!}}{{{3}!{\left({12}-{3}\right)}!}}}{\left({0.42}^{{{3}}}\right)}{\left({0.58}^{{{12}-{3}}}\right)}$$
$$\displaystyle{P}{\left({X}={3}\right)}={220}{\left({0.42}^{{{3}}}\right)}{\left({0.58}^{{{9}}}\right)}$$
$$\displaystyle{P}{\left({X}={3}\right)}={0.1211}$$
b) $$\displaystyle{P}{\left({X}\geq{4}\right)}={P}{\left({4}\right)}+{P}{\left({5}\right)}+{d}{o}{c}{s}+{P}{\left({12}\right)}$$
$$\displaystyle{P}{\left({X}\geq{4}\right)}={1}-{P}{\left({3}\right)}-{P}{\left({2}\right)}+{P}{\left({1}\right)}-{P}{\left({0}\right)}$$
$$\displaystyle{P}{\left({X}={3}\right)}={\frac{{{12}!}}{{{3}!{\left({12}-{3}\right)}!}}}{\left({0.42}^{{{3}}}\right)}{\left({0.58}^{{{12}-{3}}}\right)}={0.121066}$$
$$\displaystyle{P}{\left({X}={2}\right)}={\frac{{{12}!}}{{{2}!{\left({12}-{2}\right)}!}}}{\left({0.42}^{{{2}}}\right)}{\left({0.58}^{{{12}-{2}}}\right)}={0.050156}$$
$$\displaystyle{P}{\left({X}={1}\right)}={\frac{{{12}!}}{{{1}!{\left({12}-{1}\right)}!}}}{\left({0.42}^{{{1}}}\right)}{\left({0.58}^{{{12}-{1}}}\right)}={0.012593}$$
$$\displaystyle{P}{\left({X}={0}\right)}={\frac{{{12}!}}{{{0}!{\left({12}-{0}\right)}!}}}{\left({0.42}^{{{0}}}\right)}{\left({0.58}^{{{12}-{0}}}\right)}={0.001449}$$
$$\displaystyle{P}{\left({X}\geq{4}\right)}={1}-{0.121006}-{0.050156}-{0.012593}-{0.001449}$$
$$\displaystyle{P}{\left({X}\geq{4}\right)}={0.8147}$$
c) $$\displaystyle{P}{\left({X}\le{2}\right)}={P}{\left({2}\right)}+{P}{\left({1}\right)}+{P}{\left({0}\right)}$$
$$\displaystyle{P}{\left({X}\le{2}\right)}={0.050156}+{0.012593}+{0.001449}$$
$$\displaystyle{P}{\left({X}\le{2}\right)}={0.0642}$$
nick1337
Solution:
Given that
$$p=42\%=0.42$$
$$q=1-P=1-0.42=0.58$$
$$n=12$$
Using binomial probability formula
$$p(x=x)=^{n}C_{x}p^{x}q^{n\times x}$$
a) $$p(x=3)=^{12}C_{3}(0.42)^{3}(0.58)^{12-3}$$
$$=0.121$$
$$P(3)=0.121$$
b) $$P(x\geq4)=1-p(x<4)$$
$$=1-p(x=0)+p(x=1)+p(x=2)+p(x=3)$$
$$=1-0.0014+0.0126+0.0502+0.1211$$
$$=1-0.1853$$
$$P(x\geq4)=0.815$$
c) $$P(x\le2)=p(x=0)+p(x=1)+p(x=2)$$
$$=0.0014+0.0126+0.0502$$
$$p(x\le2)=0.064$$
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# GMAT Quantitative: Symbolism
For the most part, GMAT math is high school math. This is a helpful mantra for many math-phobic test takers, who remind themselves (correctly!) that if they could do the math back when they were sixteen, they can do it today, too. But occasionally, the GMAT will throw something at you that you’ve never seen before, a weird symbol or other doohickey that looks like it came from outer space. Here is an example:
If x☼y = 4xy – x/y, for all and for all y0, then….
Okay, what’s ☼? Certainly nothing you learned in school. But fortunately:
#### Rule 1: The problem will always tell you what the symbols mean.
You aren’t expected to know ☼ at a glance. It’s defined for you. That first part of the sentence, the equation for xy, is that definition. This is tricky, because the equals sign in the definition is the same symbol in every algebraic equation. But in reality, the = in this problem doesn’t mean “equals.”
#### Rule 2: An = in the definition of a symbol should be read as “is defined as.”
In other words, in ‘real’ math, ☼ is meaningless. So the GMAT testmaker has to give that symbol a definition. Paraphrasing the question above into English, we get the following: “When two numbers are connected by the ☼ sign, that operation is defined as four times the product of those two numbers, minus the first over the second.” Or perhaps more simply: “When two numbers are on either side of the ☼, we plug the first number into every x in the definition in the question stem and we plug the second number into every y in that definition.”
So if we see them complete version of the problem:
If x☼y = 4xy – x/y, for all and for all y0, then 6☼3 equals
Then the problem, despite looking intimidating at first, turns out to be quite simple.
#### Rule 3: The vast majority of symbolism problems test no math more complex than simple substitution.
We follow the instructions in the question stem, as we paraphrased them. If it’s easier for you to think of the ☼ as a process, than follow the rules for the process as you paraphrased them. If it’s easier to wrap your head around substitution, then substitute for and y. The final math is the same either way:
6☼3 = 4*6*3 – 6/3 = 72 – 2 = 70
And on test-day, we get the points for a correct answer just like that.
Don’t let yourself get intimidated by strange symbols. These problems are testing your flexibility—your ability to understand new definitions and apply them. And as long as you remember the three rules of symbolism in this article, you’ll be able to answer any weird signs or operations that the GMAT throws at you.
Here is one more chance to try the same skills on a slightly tougher problem. Good luck!
Analyze: What do the answer choices represent? The value of when it is part of the operation described in the stem.
Outline the information in the stem. The operation shows what we do with two numbers, and b.
The stem tells us that when we substitute 3 for and for b, the operation will equal –45.
Task: How do we use the operation in the stem to get a value for x? Just do the substitution and simplify, isolating x.
Approach Strategically:
Replace with 3 and with x, and then solve for x.
(3 + 2)(– 3) = –45
5(– 3) = –45
– 3 = –9
= –6
That gives us (B) as the correct answer.
Confirm: Plug 3 in for and –6 in for in the operation and FOIL: (3 + 2)(–6 – 3) = (–18) – 9 – 12 – 6 = –45.
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# Solve algebra with steps
In this blog post, we will take a look at how to Solve algebra with steps. We will also look at some example problems and how to approach them.
## Solving algebra with steps
When you try to Solve algebra with steps, there are often multiple ways to approach it. Algebra is a branch of mathematics that deals with the operations and relationships between numbers. Algebra is needed to solve many problems in everyday life, such as how to budget your money or how to figure out your taxes. In order to do algebra, you need to know some basic math facts, such as how to add, subtract, multiply, and divide. You will also need to know the rules of algebra. For example, in order to multiply two numbers together, you must multiply them both by 1. Algebra can be very complicated and difficult at first, but with practice and patience it can become easier. There are different types of algebra: algebraic expressions (such as 2x + 2) and linear equations (like x + 3 = 12). Both types of equations can be solved using addition and subtraction (i.e., adding or subtracting one or more). Algebraic expressions are also referred to as equations. Algebraic expressions can have variables (such as x) that represent specific values. These values can range from 0 up to infinity (or any other integer number). The variable represents a value that changes over time. Linear equations are also called linear equations because they all have a constant value on both sides (such as x + 3 = 12).
The side ratios of an equilateral triangle are equal: 1:1:1. The three angles at each vertex are all equal, as well: 90° for each. If "a" is the length of the lateral side, then "b" is the length of the hypotenuse and "c" is the length of one leg (the shorter one). Then, we can write that: A = b = c = 1 Therefore, the side of a triangle can be found by dividing any two sides together and adding 1 to the result. So if you want to find the hypotenuse, you would add 1 to both "a" and "c". If you want to find one of a pair's legs, you would add 1 to both "b" and "c". Another way to solve for a side of a triangle is to use Pythagoras' theorem. This says that in order for two triangles to share a common side, they must have identical altitudes (measured from their highest point). This means that if you want to find the hypotenuse or one leg, you can simply measure them from their top or bottom respectively
Long division is the process of calculating a long number in two or more steps. Long division is useful for calculating a large number that cannot be calculated in one step, such as the area of a shape or the sum of money owed. Long division is also used to calculate change. The steps of long division include: There are several different ways to solve long division. These include: To solve long division by hand, start with the left-most number, then add your divisor and continue to the right; To solve long division by calculator, enter all numbers into the calculator and press the "=" button; To solve long division by computer software, use online calculators or online software programs; To solve long division by machine, use a large-scale calculator that can handle large numbers.
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Assignment 4
Explorations of the Centroid
Drew Wilson
The medians used to construct the centroid divide the triangle into six smaller triangles. Show that the area for each of these triangles are the same. Below is a figure that expresses the triangle ABC and the six smaller triangles formed by the construction of the centroid, G, of triangle ABC.
First, we know the formula for area of a triangle is one half of the base multiplied by the height, or altitude, of the triangle. We also know that points D, E, and F are the midpoints of their respective side. Therefore, we know segment AD=DB, CE=EB, AF=FC.
Say we let the base of each triangle be the side of the triangle that is shared with one of the sides of the original triangle. For example, the base for triangles CGE and EGB will be segments CE and EB which we know are the same. Now if we consider the formula for the area of a triangle we can see that the area of triangle CGE=the area of triangle CGB because they have the same base and the same height as shown to the right. Line segment, a, represents the altitude for both triangles CGE and EGB because the altitude is created by a line segment perpendicular to the base of the triangle through the vertex opposite the base of the triangle. As we can see from the figure, we know the altitude will be the same for both triangles because the vertex opposite the side of the base for both triangles is the point G, and segment, a, is perpendicular to CE and EB. Therefore, since the altitude and the base are the same for each triangle, the area of both triangles are the same.
Similarly if we consider triangles AGF and FGC, we would let segments AF and FC be the base for each triangle and point G would be the vertex opposite the base that would be used to create the altitude for each triangle. Once again, since each triangle have the same base and altitude the area for each triangle will be the same. The same occurs for triangles ADG and DGB if we let segments AD and DB be the base and G be the vertex opposite the base that is used to create the altitude for each triangle.
Now we have three pairs of triangles that share the same area. We need to show that two different pair of triangles have the same area in order to show that all six triangles have the same area. For example, we need to show that triangles CGB and CGF have the same area and triangles AGF and AGD have the same area, and from this we can conclude that all six triangles have the same area.
Let's consider triangles CGE and CGF and compare the area for each of these triangles. Let's define the base for each triangle as the line segment CG. Since each triangle shares this side, we know the base for each triangle is the same. Now, we need to consider the altitude for each triangle and determine if the altitude for each triangle is the same. We know that the altitude for each triangle will be perpendicular to the base, CG, so the altitude for each triangle will be parallel to each other. If we were to construct segments through each vertex of each triangle, F and E, perpendicular to the altitude, then we would have two line segments of equal length. Now, if we were to extend each altitude to the newly constructed segments, then we would have constructed a rectangle. We know this is a rectangle because each side is parallel to the opposite side and perpendicular to the adjacent side. We also know that each angle is 90 degrees because opposite sides are parallel. We also know, based on our construction, that the line segment CG intersects this rectangle in such a way that the rectangle is divided into two equal parts because CG is the base of each triangle and is the basis for the construction of each altitude. Since line segment CG divides the rectangle into two equal parts, we know that the altitudes for each triangle are equivalent. If the altitude for triangle CGE were shorter than the altitude of triangle CGF, then we would have constructed a quadrilateral that is not a rectangle because adjacent sides would not be perpendicular. An example of the rectangle formed by the two altitudes is show at the right. As you can see the altitude for each triangle is given by segment a and the base CG divides the rectangle into two equal parts.
Similarly, if we consider triangles AGF and AGD, and we set segment AG as the base for each triangle, then the altitude for each triangle will be the same as well. We could also construct rectangles from the two altitudes as we did before and show that segment AG divides the rectangle into two equal parts which provides the result that the two altitudes are equivalent. An example is given to the right. Now that we have shown that triangle CGE has the same area as triangle CGF, then we can conclude that triangles EGB=CGE=CFG=FAG all have equal area. Since the area triangle FAG=the area of triangle AGD, we can conclude that triangles FAG=AGD=GDB and since the previous conclusion we know that triangles EGB=CGE=CFG=FAG=AGD=GDB all have the same area. Therefore, each of the six smaller triangles formed by the construction of a centroid have the same area.
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Determine the Discharge of the Channel for which it is to be Designed
Determine the Discharge of the Channel for which it is to be Designed
Mathematical Problem
A channel is to be designed for irrigating 6000 hectares of Kharif crop and 5500 hectares of Rabi crop. The water requirement for the Kharif crop is 70 cm and the water requirement for Rabi is 35 cm. The Kor period for Kharif is 2 weeks and for Rabi is 3 weeks and 4 days. Determine the discharge of the channel for which it is to be designed.
Solution:
Using the relation
∇ = (8.64 × B)/ D
Discharge for Kharif crop
Here,
• ∇ = 70 cm = 0.70 m
• B = 2 weeks = 14 days
Duty (D) = (8.64 × 21)/ 0.70 = 259.2 hectares/cumec.
Area to be irrigated = 6000 hectares.
Required discharge of channel for Kharif crop = (6000/259.2) = 23.15 cumec.
Discharge for Rabi crop
Here,
• ∇ = 35 cm = 0.35 m.
• B = 3 weeks and 4 day = (21 + 4) = 25 days.
Duty (D) = (8.64 × 25)/ 0.35 = 617.14 hectares/cumec.
Area to be irrigated = 5500 hectares.
Required discharge of channel for Kharif crop = (5500/617.14) = 8.91 cumec.
Therefore, the channel is to be designed for the maximum discharge of 23.15 cumec, because this discharge capacity of the channel will be able to supply water to both seasons.
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# 6.2 Solving Nonlinear Equations
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1 6.2. SOLVING NONLINEAR EQUATIONS Solving Nonlinear Equations We begin by introducing a property that will be used extensively in this and future sections. The zero product property. If the product of two or more numbers equals zero, then at least one of the numbers must equal zero. That is, if ab = 0, then a = 0 or b = 0. Let s use the zero product property to solve a few equations. EXAMPLE 1. Solve for x: (x+3)(x 5) = 0 Solve for x: Solution: The product of two factors equals zero. (x 7)(x 2) = 0 (x+3)(x 5) = 0 Hence, at least one of the factors must equal zero. Using the zero product property, set each factor equal to zero, then solve the resulting equations for x. x+3 = 0 or x 5 = 0 x = 3 x = 5 Hence, the solutions are x = 3 and x = 5 Check: Check that each solution satisfies the original equation. Substitute 3 for x: Substitute 5 for x: (x+3)(x 5) = 0 ( 3+3)( 3 5) = 0 (0)( 8) = 0 0 = 0 (x+3)(x 5) = 0 (5+3)(5 5) = 0 (8)(0) = 0 0 = 0 Because each check produces a true statement, both x = 3 and x = 5 are solutions of (x+3)(x 5) = 0. Answer: 7, 2 The zero product property also works equally well if more than two factors are present. For example, if abc = 0, then either a = 0 or b = 0 or c = 0. Let s use this idea in the next example.
2 400 CHAPTER 6. FACTORING Solve for x: 6x(x+4)(5x+1) = 0 EXAMPLE 2. Solve for x: x(2x+9)(3x 5) = 0 Solution: The product of three factors equals zero. x(2x+9)(3x 5) = 0 Using the zero product property, set each factor equal to zero, then solve the resulting equations for x. x = 0 or 2x+9 = 0 or 3x 5 = 0 2x = 9 3x = 5 x = 9 2 x = 5 3 Answer: 0, 4, 1/5 Hence, the solutions are x = 0, x = 9/2, and x = 5/3. We encourage the reader to check the solution. Linear versus Nonlinear All of the equations solved in previous chapters were examples of what are called linear equations. If the highest power of the variable we are solving for is one, then the graphs involved are lines. Hence the term, linear equation. However, if the power on the variable we are solving for exceeds one, then the graphs involved are curves. Hence the term, nonlinear equation. In this chapter we will learn how to solve nonlinear equations involving polynomials. However, let s first make sure we can recognize the difference between a linear and a nonlinear equation. Linear versus Nonlinear. Use the following conditions to determine if an equation is linear or nonlinear. 1. If the highest power of the variable we are solving for is one, then the equation is linear. 2. If the highest power of the variable we are solving for is larger than one, then the equation is nonlinear. Classify the following equation as linear or nonlinear: 2x = x 3 4 EXAMPLE 3. Iftheinstructionis solveforx, classifyeachofthe following equations as linear or nonlinear.
3 6.2. SOLVING NONLINEAR EQUATIONS 401 a) 3x 5 = 4 7x b) x 2 = 8x Solution: Because the instruction is solve for x, to determine whether the equation is linear or nonlinear, we identify the largest power of x present in the equation. a) The highest power of x present in the equation 3x 5 = 4 7x is one. Hence, this equation is linear. b) The equation x 2 = 8x contains a power of x higher than one (it contains an x 2 ). Hence, this equation is nonlinear. Answer: nonlinear Now that we can classify equations as either linear or nonlinear, let s introduce strategies for solving each type, the first of which should already be familiar. Strategy for solving a linear equation. If an equation is linear, start the solution process by moving all terms containing the variable you are solving for to one side of the equation, then move all terms that do not contain the variable you are solving for to the other side of the equation. EXAMPLE 4. Solve for x: 3x 5 = 4 7x Solve for x: 2x+4 = 6x+3 Solution: Because the instruction is solve for x and we note that the largest power of x present is one, the equation 3x 5 = 4 7x is linear. Hence, the strategy is to move all terms containing x to one side of the equation, then move all the remaining terms to the other side of the equation. 3x 5 = 4 7x 3x 5+7x = 4 3x+7x = 4+5 Original equation. Add 7x to both sides. Add 5 to both sides. Note how we havesucceeded in moving all terms containing x to one side of the equation and all terms that do not contain x to the other side of the equation. 10x = 9 x = 9 10 Simplify both sides. Divide both sides by 10. Hence, the solution of 3x 5 = 4 7x is x = 9/10. Readers are encouraged to check this solution. Answer: 1/4
4 402 CHAPTER 6. FACTORING The situation is much different when the equation is nonlinear. Strategy for solving a nonlinear equation. If an equation is nonlinear, first move everything to one side of the equation, making one side of the equation equal to zero. Continue the solution process by factoring and applying the zero product property. Solve for x: x 2 = 5x EXAMPLE 5. Solve for x: x 2 = 8x Solution: Because the instruction is solve for x, and the highest power of x is larger than one, the equation x 2 = 8x is nonlinear. Hence, the strategy requires that we move all terms to one side of the equation, making one side zero. x 2 = 8x x 2 8x = 0 Original equation. Subtract 8x from both sides. Note how we have succeeded in moving all terms to one side of the equation, making one side equal to zero. To finish the solution, we factor out the GCF on the left-hand side. x(x 8) = 0 Factor out the GCF. Note that we now have a product of two factors that equals zero. By the zero product property, either the first factor is zero or the second factor is zero. x = 0 or x 8 = 0 Hence, the solutions are x = 0 and x = 8. x = 8 Check: Check that each solution satisfies the original equation. Substitute 0 for x: Subtitute 8 for x: Answer: 0, 5 x 2 = 8x (0) 2 = 8(0) 0 = 0 x 2 = 8x (8) 2 = 8(8) 64 = 64 Note that both results are true statements, guaranteeing that both x = 0 and x = 8 are solutions of x 2 = 8x.
5 6.2. SOLVING NONLINEAR EQUATIONS 403 Warning! The following is incorrect! Consider what would happen if we divided both sides of the equation x 2 = 8x in Example 5 by x: x 2 = 8x x 2 x = 8x x x = 8 Note that we have lost the second answer found in Example 5, x = 0. This example demonstrates that you should never divide by the variable you are solving for! If you do, and cancellation occurs, you will lose answers. Let s try solving a nonlinear equation that requires factoring by grouping. EXAMPLE 6. Solve for x: 6x 2 +9x 8x 12 = 0 Solve for x: 5x 2 20x 4x+16 = 0 Solution: Because we are solving for x and there is a power of x larger than one, this equation is nonlinear. Hence, the first step is to move everything to one side of the equation, making one side equal to zero. Well that s already done, so let s factor the left-hand side by grouping. Note that we can factor 3x out of the first two terms and 4 out of the second two terms. 6x 2 +9x 8x 12 = 0 3x(2x+3) 4(2x+3) = 0 Factor out the common factor 2x+3. (3x 4)(2x+3) = 0 We now have a product of two factors that equals zero. Use the zero product property to write: 3x 4 = 0 or 2x+3 = 0 3x = 4 2x = 3 x = 4 3 x = 3 2 Hence, the solutions are x = 4/3 and x = 3/2. Check. Let s use the graphing calculator to check the solution x = 4/3. First, store the solution 4/3 in the variable X using the following keystrokes (see the first image in Figure 6.1.
6 404 CHAPTER 6. FACTORING 4 3 STO X,T,θ,n ENTER Next, enter the left-hand side of the equation 6x 2 +9x 8x 12 = 0 using the following keystrokes. Note that the result in the second image in Figure 6.1 indicates that the expression 6x 2 +9x 8x 12 equals zero when x = 4/3. 6 X,T,θ,n X,T,θ,n 8 X,T,θ,n 1 2 ENTER Figure 6.1: Checking the solution x = 4/3. Answer: 4/5, 4 Therefore, the solution x = 4/3 checks. Readers are encouraged to use their graphing calculators to check the second solution, x = 3/2. Using the Graphing Calculator In this section we will employ two different calculator routines to find the solution of a nonlinear equation. Before picking up the calculator, let s first use an algebraic method to solve the equation x 2 = 5x. The equation is nonlinear, so the first step is to move everything to one side of the equation, making one side equal to zero. x 2 = 5x x 2 +5x = 0 x(x+5) = 0 Nonlinear. Make one side zero. Add 5x to both sides. Factor out the GCF. Use the zero product property, setting each factor equal to zero, then solving the resulting equations for x. x = 0 or x+5 = 0 Hence, the solutions are x = 0 and x = 5. x = 5
7 6.2. SOLVING NONLINEAR EQUATIONS 405 We ll now use the calculator to find the solutions of x 2 = 5x. The first technique employs the 5:intersect routine on the calculator s CALC menu. EXAMPLE 7. Use the 5:intersect utility on the graphing calculator to Use the 5:intersect utility solve the equation x 2 = 5x for x. on the graphing calculator to solve the equation x 2 = 4x Solution: Load the left-hand side of x 2 = 5x in Y1 and the right-hand for x. side in Y2 (see Figure 6.2). Selecting 6:ZStandard from the ZOOM menu produces the graphs shown in the image on the right in Figure 6.2. Figure 6.2: Sketch the graphs of each side of the equation x 2 = 5x. Note that the graph of y = x 2 is a parabola that opens upward, with vertex (turning point) at the origin. This graph reveals why the equation x 2 = 5x is called a nonlinear equation (not all the graphs involved are lines). Next, the graph of y = 5x is a line with slope 5 and y-intercept at the origin. The two graphs obviously intersect at the origin, but it also appears that there may be another point of intersection that is off the screen. Let s increase Ymax in an attempt to reveal the second point of intersection. After some experimentation, the settings shown in the first image in Figure 6.3 reveal both points of intersection. Pushing the GRAPH button produces the image on the right in Figure 6.3. Figure 6.3: Adjust the WINDOW parameters to reveal both points of intersection. Tofindthesolutionsoftheequationx 2 = 5x, wemustfindthecoordinates of the points where the graphs of y = x 2 and y = 5x intersect. The x- coordinate of each point of intersection will be a solution of the equation x 2 = 5x.
8 406 CHAPTER 6. FACTORING Start by selecting 5:intersect from the CALC menu. When prompted for the First curve?, press ENTER. When prompted for the Second curve?, press ENTER. When prompted for a Guess, press ENTER. The result is the point (0,0) shown in the image on the left in Figure 6.4. Repeat the process a second time. Select 5:intersect from the CALC menu. When prompted for the First curve?, press ENTER. When prompted for the Second curve?, press ENTER. When prompted for a Guess, use the left-arrow key to move the cursor closer to the leftmost point of intersection, then press ENTER. The result is the point ( 5, 25) shown in the image on the right in Figure 6.4. Figure 6.4: Use the 5:intersect utility to find the points of intersection. Answer: Reporting the solution on your homework: Duplicate the image in your calculator s viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. Label the horizontal and vertical axes with x and y, respectively (see Figure 6.5). PlaceyourWINDOW parametersat the end ofeachaxis(seefigure 6.5). Label each graph with its equation (see Figure 6.5). Drop dashed vertical lines through each point of intersection. Shade and label the x-values of the points where the dashed vertical line crosses the x-axis. Thesearethe solutionsofthe equationx 2 = 5x(seeFigure 6.5). Hence, the solutions of x 2 = 5x are x = 5 and x = 0. Note now these match the solutions found using the algebraic technique. Before demonstrating a second graphing calculator technique for solving nonlinear equations, let s take a moment to recall the definition of a zero of a function, which was first presented in Chapter 5, Section 3. Zeros and x-intercepts. The points where the graph of f crosses the x-axis are called the x-intercepts of the graph of f. The x-value of each x-intercept is called a zero of the function f.
9 6.2. SOLVING NONLINEAR EQUATIONS 407 y = 5x y 50 y = x 2 x Figure 6.5: Reporting your graphical solution on your homework. We ll now employ the 2:zero utility from the CALC menu to find the solutions of the equation x 2 = 5x. EXAMPLE 8. Use the 2:zero utility on the graphing calculator to solve Use the 2:zero utility on the the equation x 2 = 5x for x. graphing calculator to solve Solution: First, make one side of the equation equal to zero. the equation x 2 = 4x for x. x 2 = 5x x 2 +5x = 0 Make one side zero. Add 5x to both sides. produce the image in Figure 6.6. Note that the graphoff hastwo x-intercepts, and the x-values of each of these points are the zeros of the function f. To determine the values of x that make x 2 +5x = 0, we must locate the points where the graph of f(x) = x 2 + 5x crosses the x-axis. These points are the x-intercepts of the graph of f and the x-values of these points are the zeros of the function f. Load the function f(x) = x 2 + 5x in Y1, then select 6:ZStandard to It s often easierto find the solutions of a nonlinear equation by making one side zero and identifying where the graph of the resulting function crosses the x-axis. Figure 6.6: Sketch the graph of p(x) = x 2 +5x. Select 2:zero from the CALC menu (see Figure 6.7).
10 408 CHAPTER 6. FACTORING The calculator responds by asking for a Left Bound? Use the left-arrow key to move the cursor so that it lies to the left of the x-intercept near ( 5,0) (see the second image in Figure 6.7), then press the ENTER key. The calculator responds by asking for a Right Bound? Move the cursor so that is slightly to the right ofthe x-intercept near ( 5,0)(see the third image Figure 6.7), then press the ENTER key. Figure 6.7: Setting the left and right bounds when using the 2:zero utility to find the x-intercepts of the graph of f(x) = x 2 +5x. The calculator responds by asking for a Guess? Note the two triangular marks near the top of the viewing window in the first image in Figure 6.8 that mark the left- and right-bounds. As long as you place the cursor so that the x-value of the cursor location lies between these two marks, you ve made a valid guess. Because the cursor already lies between these two marks, we usually leave it where it is and press the ENTER key. Figure 6.8: Setting the right bound and making a guess. After making your guess and pressing the ENTER key, the calculator proceeds to find an approximation of the x-intercept that lies between the left- and right-bounds previously marked (see the second image in Figure 6.8). Hence, this x-intercept is ( 5,0), making 5 a zero of f(x) = x 2 +5x and a solution of the equation x 2 +5x = 0. We ll leave it to our readers to repeat the 2:zero process to find the second zero at the origin. Reporting the solution on your homework: Duplicate the image in your calculator s viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves.
11 6.2. SOLVING NONLINEAR EQUATIONS 409 Label the horizontal and vertical axes with x and y, respectively (see Figure 6.9). PlaceyourWINDOWparametersat the end ofeachaxis(see Figure 6.9). Label each graph with its equation (see Figure 6.9). Drop dashed vertical lines through each x-intercept. Shade and label the x-values of each x-intercept. These are the solutions of the equation x 2 = 5x (see Figure 6.9). y 10 y = x 2 +5x x Answer: 10 Figure 6.9: Reporting your graphical solution on your homework. Hence, the solutions of x 2 = 5x are x = 5 and x = 0. Note how nicely this agrees with the solutions found using the algebraic technique and the solutions found using the 5:intersect utility in Example 7.
12 410 CHAPTER 6. FACTORING Exercises In Exercises 1-8, solve the given equation for x. 1. (9x+2)(8x+3) = 0 2. (2x 5)(7x 4) = 0 3. x(4x+7)(9x 8) = 0 4. x(9x 8)(3x+1) = x(9x+4) = x(3x 6) = 0 7. (x+1)(x+6) = 0 8. (x 4)(x 1) = 0 In Exercises 9-18, given that you are solving for x, state whether the given equation is linear or nonlinear. Do not solve the equation. 9. x 2 +7x = 9x x 2 +9x = 4x x 2 = 5x x+5 = 6x x 2 = 2x 14. 4x 2 = 7x 15. 3x 2 +8x = x 2 2x = x+6 = x 5 = 3 In Exercises 19-34, solve each of the given equations for x x+8 = x+4 = x 2 = x 22. 6x 2 = 7x 23. 3x+9 = 8x x+5 = 6x x 2 = 2x 26. 8x 2 = 18x 27. 9x+2 = x+2 = x 2 = 6x 30. 6x 2 = 14x 31. 7x 2 = 4x 32. 7x 2 = 9x 33. 7x+2 = 4x x+3 = 2x+8
13 6.2. SOLVING NONLINEAR EQUATIONS 411 In Exercises 35-50, factor by grouping to solve each of the given equations for x x 2 +56x+54x+48 = x 2 +36x+6x+8 = x 2 18x+40x 45 = x 2 35x+54x 45 = x 2 +18x+20x+8 = x 2 +21x+30x+35 = x 2 +10x+4x+40 = x 2 +11x+10x+110 = x 2 +6x 11x 66 = x 2 +6x 2x 12 = x 2 24x+35x 56 = x 2 10x+54x 45 = x 2 +2x+9x+18 = x 2 +8x+4x+32 = x 2 +4x 8x 32 = x 2 +8x 5x 40 = 0 In Exercises 51-54, perform each of the following tasks: i) Use a strictly algebraic technique to solve the given equation. ii) Use the 5:intersect utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown in Example x 2 = 4x 52. x 2 = 6x 53. x 2 = 5x 54. x 2 = 6x In Exercises 55-58, perform each of the following tasks: i) Use a strictly algebraic technique to solve the given equation. ii) Use the 2:zero utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown in Example x 2 +7x = x 2 8x = x 2 3x = x 2 +2x = 0 Answers 1. x = 2 9, x = 0, 7 4, x = 0, x = 1, 6 9. Nonlinear
14 412 CHAPTER 6. FACTORING 11. Linear 35. x = 6 7, Nonlinear 15. Nonlinear 17. Linear x = 0, x = 5 2, x = 4 9, x = 4, x = 11, x = 0, x = 7 3, x = 9, x = 8, x = 0, x = 0, x = 4,0 53. x = 0,5 55. x = 7,0 57. x = 0,3
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# How do you solve for $x$: $\dfrac{12}{x}+\dfrac{3}{4}=\dfrac{3}{2}$?
Last updated date: 29th Feb 2024
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Hint:This is a linear equation in one variable as there is only one variable in an equation. In the given question, the variable is the letter ‘x’, to solve this question we need to get ‘x’ on one side of the “equals” sign, and all the other numbers on the other side. To solve this equation for a given variable ‘x’, we have to undo the mathematical operations such as addition, subtraction, multiplication and division that have been done to the variables.
Complete step by step solution:
We have given that,
$\Rightarrow \dfrac{12}{x}+\dfrac{3}{4}=\dfrac{3}{2}$
Subtracting $\dfrac{3}{4}$ from both the sides of the equation, we get
$\Rightarrow \dfrac{12}{x}+\dfrac{3}{4}-\dfrac{3}{4}=\dfrac{3}{2}-\dfrac{3}{4}$
$\Rightarrow \dfrac{12}{x}=\dfrac{3}{2}-\dfrac{3}{4}$
Taking the LCM of 2 and 4 on the right side of the equation,,
LCM of 2 and 4 is 4
$\Rightarrow \dfrac{12}{x}=\dfrac{3\times 2}{2\times 2}-\dfrac{3}{4}$
$\Rightarrow \dfrac{12}{x}=\dfrac{6}{4}-\dfrac{3}{4}$
$\Rightarrow \dfrac{12}{x}=\dfrac{3}{4}$
Multiplied both the sides by $x$, we get
$\Rightarrow 12=\dfrac{3}{4}x$
Multiplied both the sides of the equation by 4, we get
$\Rightarrow 48=\dfrac{3x}{4}\times 4$
Simplifying the above, we get
$\Rightarrow 48=3x$
Dividing both the sides by, we get
$\Rightarrow x=\dfrac{48}{3}=16$
$\Rightarrow x=16$
Therefore, the value of ‘x’ is equal to 16.
It is the required solution.
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## Course: 8th grade > Unit 1
• Scientific notation word problem: red blood cells
• Scientific notation word problem: U.S. national debt
• Scientific notation word problem: speed of light
• Numbers and operations: FAQ
## Scientific notation word problems
Chapter 6: Polynomials
## 6.3 Scientific Notation (Homework Assignment)
Scientific notation is a convenient notation system used to represent large and small numbers. Examples of these are the mass of the sun or the mass of an electron in kilograms. Simplifying basic operations such as multiplication and division with these numbers requires using exponential properties.
Scientific notation has two parts: a number between one and nine and a power of ten, by which that number is multiplied.
$\text{Scientific notation: }a \times 10^b, \text{ where }1 \le a \le 9$
The exponent tells how many times to multiply by 10. Each multiple of 10 shifts the decimal point one place. To decide which direction to move the decimal (left or right), recall that positive exponents means there is big number (larger than ten) and negative exponents means there is a small number (less than one).
Example 6.3.1
Convert 14,200 to scientific notation.
$\begin{array}{rl} 1.42&\text{Put a decimal after the first nonzero number} \\ \times 10^4 & \text{The exponent is how many times the decimal moved} \\ 1.42 \times 10^4& \text{Combine to yield the solution} \end{array}$
Example 6.3.2
Convert 0.0028 to scientific notation.
$\begin{array}{rl} 2.8&\text{Put a decimal after the first nonzero number} \\ \times 10^{-3}&\text{The exponent is how many times the decimal moved} \\ 2.8\times 10^{-3}&\text{Combine to yield the solution} \end{array}$
Example 6.3.3
Convert 3.21 × 10 5 to standard notation.
Starting with 3.21, Shift the decimal 5 places to the right, or multiply 3.21 by 10 5 .
321,000 is the solution.
Example 6.3.4
Convert 7.4 × 10 −3 to standard notation
Shift the decimal 3 places to the left, or divide 6.4 by 10 3 .
0.0074 is the solution.
Working with scientific notation is easier than working with other exponential notation, since the base of the exponent is always 10. This means that the exponents can be treated separately from any other numbers. For instance:
Example 6.3.5
Multiply (2.1 × 10 −7 )(3.7 × 10 5 ).
First, multiply the numbers 2.1 and 3.7, which equals 7.77.
Second, use the product rule of exponents to simplify the expression 10 −7 × 10 5 , which yields 10 −2 .
Combine these terms to yield the solution 7.77 × 10 −2 .
Example 6.3.6
(4.96 × 10 4 ) ÷ (3.1 × 10 −3 )
First, divide: 4.96 ÷ 3.1 = 1.6
Second, subtract the exponents (it is a division): 10 4− −3 = 10 4 + 3 = 10 7
Combine these to yield the solution 1.6 × 10 7 .
For questions 1 to 6, write each number in scientific notation.
For questions 7 to 12, write each number in standard notation.
• 2.56 × 10 2
For questions 13 to 20, simplify each expression and write each answer in scientific notation.
• (7 × 10 −1 )(2 × 10 −3 )
• (2 × 10 −6 )(8.8 × 10 −5 )
• (5.26 × 10 −5 )(3.16 × 10 −2 )
• (5.1 × 10 6 )(9.84 × 10 −1 )
• $\dfrac{(2.6 \times 10^{-2})(6 \times 10^{-2})}{(4.9 \times 10^1)(2.7 \times 10^{-3})}$
• $\dfrac{(7.4 \times 10^4)(1.7 \times 10^{-4})}{(7.2 \times 10^{-1})(7.32 \times 10^{-1})}$
• $\dfrac{(5.33 \times 10^{-6})(9.62 \times 10^{-2})}{(5.5 \times 10^{-5})^2}$
• $\dfrac{(3.2 \times 10^{-3})(5.02 \times 10^0)}{(9.6 \times 10^3)^{-4}}$
## The Math You Need, When You Need It
math tutorials for students majoring in the earth sciences
## Scientific Notation - Practice Problems
Solving earth science problems with scientific notation, × div[id^='image-'] {position:static}div[id^='image-'] div.hover{position:static} introductory problems.
These problems cover the fundamentals of writing scientific notation and using it to understand relative size of values and scientific prefixes.
Problem 1: The distance to the moon is 238,900 miles. Write this value in scientific notation.
Problem 2: If one mile is 1609.34 meters. What the distance to the moon in meters using scientific notation.
1609.34 m/(mi) xx 238","900 mi = 384,400,000 m
Notice in the above unit conversion the 'mi' units cancel each other out because 'mi' is in the denominator for the first term and the numerator for the second term
Problem 4: The atomic radius of a magnesium atom is approximately 1.6 angstroms, which is equal to 1.6 x 10 -10 meters (m). How do you write this length in standard form?
0.00000000016 m
Fissure A = 40,0000 m Fissure B = 5,0000 m
This shows fissure A is larger (by almost 10 times!). The shortcut to answer a question like this is to look at the exponent. If both coefficients are between 1-10, then the value with the larger exponent is the larger number.
Problem 6: The amount of carbon in the atmosphere is 750 petagrams (pg). One petagram equals 1 x 10 15 grams (g). Write out the amount of carbon in the atmosphere in (i) scientific notation and (ii) standard decimal format.
The exponent is a positive number, so the decimal will move to the right in the next step.
750,000,000,000,000,000 g
Scientific notation is used in solving these earth and space science problems and they are provided to you as an example. Be forewarned that these problems move beyond this module and require some facility with unit conversions, rearranging equations, and algebraic rules for multiplying and dividing exponents. If you can solve these, you've mastered scientific notation!
Problem 7: Calculate the volume of water (in cubic meters and in liters) falling on a 10,000 km 2 watershed from 5 cm of rainfall.
10,000 km^2 = 1 xx 10^4 km^2
5 cm of rainfall = 5 xx 10^0 cm
Let's start with meters as the common unit and convert to liters later. There are 1 x 10 3 m in a km and area is km x km (km 2 ), therefore you need to convert from km to m twice:
1 xx 10^3 m/(km) * 1 xx 10^3 m/(km) = 1 xx 10^6 m^2/(km)^2 1 xx 10 m^2/(km)^2 * 1 xx 10^4 km^2 = 1 xx 10^10 m^2 for the area of the watershed.
For the amount of rainfall, you should convert from centimeters to meters:
5 cm * (1 m)/(100 cm)= 5 xx 10^-2 m
V = A * d
When multiplying terms with exponents, you can multiply the coefficients and add the exponents:
V = 1 xx 10^10 m^2 * 5.08 xx 10^(-2) m = 5.08 xx 10^8 m^3
Given that there are 1 x 10 3 liters in a cubic meter we can make the following conversion:
1 xx 10^3 L * 5.08 xx 10^8 m^3 = 5.08 xx 10^11 L
V = 4/3 pi r^3
Using this equation, plug in the radius (r) of the dust grains.
V = 4/3 pi (2 xx10^(-6))^3m^3
Notice the (-6) exponent is cubed. When you take an exponent to an exponent, you need to multiply the two terms
V = 4/3 pi (8 xx10^(-18)m^3)
Then, multiple the cubed radius times pi and 4/3
V = 3.35 xx 10^(-17) m^3
m = 3.35 xx 10^(-17) m^3 * 3300 (kg)/m^3
Notice in the equation above that the m 3 terms cancel each other out and you are left with kg
m = 1.1 xx 10^(-13) kg
V = 4/3 pi (2.325 xx10^(15) m)^3
V = 5.26 xx10^(46) m^3
Number of dust grains = 5.26 xx10^(46) m^3 xx 0.001 grains/m 3
Number of dust grains = 5.26xx10^43 "grains"
Total mass = 1.1xx10^(-13) (kg)/("grains") * 5.26xx10^43 "grains"
Notice in the equation above the 'grains' terms cancel each other out and you are left with kg
Total mass = 5.79xx10^30 kg
If you feel comfortable with this topic, you can go on to the assessment . Or you can go back to the Scientific Notation explanation page .
« Previous Page Next Page »
## Module 12: Exponents
Problem solving with scientific notation, learning outcomes.
• Solve application problems involving scientific notation
Water Molecule
## Solve application problems
Learning rules for exponents seems pointless without context, so let’s explore some examples of using scientific notation that involve real problems. First, let’s look at an example of how scientific notation can be used to describe real measurements.
Match each length in the table with the appropriate number of meters described in scientific notation below.
Red Blood Cells
One of the most important parts of solving a “real” problem is translating the words into appropriate mathematical terms, and recognizing when a well known formula may help. Here’s an example that requires you to find the density of a cell, given its mass and volume. Cells aren’t visible to the naked eye, so their measurements, as described with scientific notation, involve negative exponents.
Human cells come in a wide variety of shapes and sizes. The mass of an average human cell is about $2\times10^{-11}$ grams [1] Red blood cells are one of the smallest types of cells [2] , clocking in at a volume of approximately $10^{-6}\text{ meters }^3$. [3] Biologists have recently discovered how to use the density of some types of cells to indicate the presence of disorders such as sickle cell anemia or leukemia. [4] Density is calculated as the ratio of $\frac{\text{ mass }}{\text{ volume }}$. Calculate the density of an average human cell.
Read and Understand: We are given an average cellular mass and volume as well as the formula for density. We are looking for the density of an average human cell.
Define and Translate: $m=\text{mass}=2\times10^{-11}$, $v=\text{volume}=10^{-6}\text{ meters}^3$, $\text{density}=\frac{\text{ mass }}{\text{ volume }}$
Write and Solve: Use the quotient rule to simplify the ratio.
$\begin{array}{c}\text{ density }=\frac{2\times10^{-11}\text{ grams }}{10^{-6}\text{ meters }^3}\\\text{ }\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\times10^{-11-\left(-6\right)}\frac{\text{ grams }}{\text{ meters }^3}\\\text{ }\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\times10^{-5}\frac{\text{ grams }}{\text{ meters }^3}\\\end{array}$
If scientists know the density of healthy cells, they can compare the density of a sick person’s cells to that to rule out or test for disorders or diseases that may affect cellular density.
The average density of a human cell is $2\times10^{-5}\frac{\text{ grams }}{\text{ meters }^3}$
The following video provides an example of how to find the number of operations a computer can perform in a very short amount of time.
Light traveling from the sun to the earth.
In the next example, you will use another well known formula, $d=r\cdot{t}$, to find how long it takes light to travel from the sun to the earth. Unlike the previous example, the distance between the earth and the sun is massive, so the numbers you will work with have positive exponents.
The speed of light is $3\times10^{8}\frac{\text{ meters }}{\text{ second }}$. If the sun is $1.5\times10^{11}$ meters from earth, how many seconds does it take for sunlight to reach the earth? Write your answer in scientific notation.
Read and Understand: We are looking for how long—an amount of time. We are given a rate which has units of meters per second and a distance in meters. This is a $d=r\cdot{t}$ problem.
Define and Translate:
$\begin{array}{l}d=1.5\times10^{11}\\r=3\times10^{8}\frac{\text{ meters }}{\text{ second }}\\t=\text{ ? }\end{array}$
Write and Solve: Substitute the values we are given into the $d=r\cdot{t}$ equation. We will work without units to make it easier. Often, scientists will work with units to make sure they have made correct calculations.
$\begin{array}{c}d=r\cdot{t}\\1.5\times10^{11}=3\times10^{8}\cdot{t}\end{array}$
Divide both sides of the equation by $3\times10^{8}$ to isolate t.
$\begin{array}{c}1.5\times10^{11}=3\times10^{8}\cdot{t}\\\text{ }\\\frac{1.5\times10^{11}}{3\times10^{8}}=\frac{3\times10^{8}}{3\times10^{8}}\cdot{t}\end{array}$
On the left side, you will need to use the quotient rule of exponents to simplify, and on the right, you are left with t.
$\begin{array}{c}\frac{1.5\times10^{11}}{3\times10^{8}}=\frac{3\times10^{8}}{3\times10^{8}}\cdot{t}\\\text{ }\\\left(\frac{1.5}{3}\right)\times\left(\frac{10^{11}}{10^{8}}\right)=t\\\text{ }\\\left(0.5\right)\times\left(10^{11-8}\right)=t\\0.5\times10^3=t\end{array}$
This answer is not in scientific notation, so we will move the decimal to the right, which means we need to subtract one factor of $10$.
$0.5\times10^3=5.0\times10^2=t$
The time it takes light to travel from the sun to the earth is $5.0\times10^2=t$ seconds, or in standard notation, $500$ seconds. That’s not bad considering how far it has to travel!
Scientific notation was developed to assist mathematicians, scientists, and others when expressing and working with very large and very small numbers. Scientific notation follows a very specific format in which a number is expressed as the product of a number greater than or equal to one and less than ten, and a power of $10$. The format is written $a\times10^{n}$, where $1\leq{a}<10$ and n is an integer. To multiply or divide numbers in scientific notation, you can use the commutative and associative properties to group the exponential terms together and apply the rules of exponents.
## Contribute!
• Orders of magnitude (mass). (n.d.). Retrieved May 26, 2016, from https://en.wikipedia.org/wiki/Orders_of_magnitude_(mass) ↵
• How Big is a Human Cell? ↵
• How big is a human cell? - Weizmann Institute of Science. (n.d.). Retrieved May 26, 2016, from http://www.weizmann.ac.il/plants/Milo/images/humanCellSize120116Clean.pdf ↵
• Grover, W. H., Bryan, A. K., Diez-Silva, M., Suresh, S., Higgins, J. M., & Manalis, S. R. (2011). Measuring single-cell density. Proceedings of the National Academy of Sciences, 108(27), 10992-10996. doi:10.1073/pnas.1104651108 ↵
• Application of Scientific Notation - Quotient 1 (Number of Times Around the Earth). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/15tw4-v100Y . License : CC BY: Attribution
• Application of Scientific Notation - Quotient 2 (Time for Computer Operations). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/Cbm6ejEbu-o . License : CC BY: Attribution
• Screenshot: water molecule. Provided by : Lumen Learning. License : CC BY: Attribution
• Screenshot: red blood cells. Provided by : Lumen Learning. License : CC BY: Attribution
• Screenshot: light traveling from the sun to the earth. Provided by : Lumen Learning. License : CC BY: Attribution
• Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : http://nrocnetwork.org/resources/downloads/nroc-math-open-textbook-units-1-12-pdf-and-word-formats/ . License : CC BY: Attribution
## Scientific Notation Calculator
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## HiSET: Math : Solve problems using scientific notation
Study concepts, example questions & explanations for hiset: math, all hiset: math resources, example questions, example question #1 : solve problems using scientific notation.
Simplify the following expression using scientific notation.
You can solve this problem in several ways. One way is to convert each number out of scientific notation and write it out fully, then find the sum of the two values and convert the answer back into scientific notation.
Another, potentially faster, way to solve this problem is to convert one answer into the same scientific-notational terms as the other and then sum them.
## Example Question #2 : Solve Problems Using Scientific Notation
Multiply, and express the product in scientific notation:
Convert 7,200,000 to scientific notation as follows:
Move the (implied) decimal point until it is immediately after the first nonzero digit (the 7). This required moving the point six units to the left:
Rearrange and regroup the expressions so that the powers of ten are together:
Multiply the numbers in front. Also, multiply the powers of ten by adding exponents:
In order for the number to be in scientific notation, the number in front must be between 1 and 10. An adjustment must be made by moving the implied decimal point in 36 one unit left. It follows that
the correct response.
## Example Question #3 : Solve Problems Using Scientific Notation
Express the product in scientific notation.
None of the other choices gives the correct response.
Scientific notation refers to a number expressed in the form
Each factor can be rewritten in scientific notation as follows:
Now, substitute:
Apply the Product of Powers Property:
This is in scientific notation and is the correct choice.
## Report an issue with this question
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## Chemical Measurement Unit Plan Mark as Favorite (105 Favorites)
LESSON PLAN in Density , Percent Composition , Accuracy , Dimensional Analysis , Measurements , Scientific Notation , Significant Figures , SI Units , Unlocked Resources , Unit Plans . Last updated September 23, 2020.
The AACT High School Classroom Resource library and multimedia collection has everything you need to put together a unit plan for your classroom: lessons, activities, labs, projects, videos, simulations, and animations. We constructed a unit plan for introducing the concepts needed for students to collect and use chemical measurements:Percent Composition, Metric Units, Accuracy and Precision, Percent Error, Density, Scientific Notation, Significant Figures, and Unit Conversion. These math based topics are very important for your students to master before they dig into other chemistry concepts.This unit is designed to be used at beginning of the school year. You may want to leave activities out, based on the math abilities of your incoming students and the level of the class.
High School
By the end of this unit, students should be able to
• Critically analyze a given problem, and complete appropriate calculations for mass and volume.
• Define percent composition.
• Calculate the percent composition of a substance in a sample.
• Take measurements using metric units.
• Determine the accuracy of different pieces of glassware.
• Accurately use laboratory equipment to gather data.
• Explain why some pieces of glassware produce better accuracy than others.
• Construct and analyze a line graph using Excel.
• Determine a method to measure the mass and volume of an irregular object.
• Calculate density and be able to explain why the density of an object does not change.
• Calculate the density of an irregular object using their data.
• Calculate the percentage error of their results using both the experimentally determined value and the accepted value for density.
• Convert very large and very small values into proper scientific notation.
• Recognize the benefit of using scientific notation to solve large scale problems.
• Use dimensional analysis for mass, length, volume, temperature, and density unit conversion problems.
• Convert between units of measurement using dimensional analysis.
• Understand the purpose of using dimensional analysis for converting between units of measurement.
• Use lab equipment to measure volume, mass, and length to the correct number of significant figures.
## Chemistry Topics
This unit supports students’ understanding of
• Measurement
Percent Composition
• Dimensional Analysis
Metric Units
• Accuracy and Precision
• Percent Error
Scientific Notation
Significant Figures
Unit Conversion
Teacher Preparation : See individual resources.
Lesson : 10 - 15 class periods, depending on class level.
• Refer to the materials list given with each individual activity.
• Refer to the safety instructions given with each individual activity.
## Teacher Notes
• The activities shown below are listed in the order that they should be completed.
• The teacher notes, student handouts, and additional materials can be accessed on the page for each individual activity.
• Please note that most of these resources are AACT member benefits.
## Classroom Resources
• "Keys for Success in Teaching Chemistry: Imagination and Resourcefulness ": Before you begin the unit, take a few minutes to read this article from our March 2016 issue of Chemistry Solutions . This article discusses several labs that the author uses to help teach his students to be creative and resourceful when collecting and using lab data.
• Mineral Investigation: Start your year off with this great introductory lab that familiarizes students with data collection and manipulation while also incorporating engineering principles and guided inquiry.In this lab, students put their problem solving skills to work as a team to determine how many specific samples of ore can be made from a lode equivalent to the size of their classroom using percent composition. This lab is perfect for the start of the school year to engage students in real-life applications of chemistry, as well as essential mathematic and measurement skills. You can then choose from one of these two labs to give your students more practice with percent composition calculations.
• Percent Composition of Bubble Gum : In this lab students determine the amount of sweetener in various brands of gum by determining the mass difference of the gum before and after it is chewed.By the end of this lesson, students should be able to define and calculate percent composition.
• Percent Composition : In this lab students calculate the percent composition of sugar in gum and the percent composition of water in popcorn kernels.By the end of this lab, students should be able to calculate the percent composition of a substance in a sample.
• Mysteriously Melodramatic & Maniacal Metric Measurements : Introduce the metric system and units with this activity, which asks students to predict the measurements of objects using metric units. They then take the actual measurements and compare them to their predictions.This activity works well as a competition, assigning points for teams based on how closely their predictions match actual measurements.
Accuracy, Precision, and Percent Error
• Measuring Volume : This simulation shows students an image of a graduated cylinder and asks them to report an accurate volume measurement with the correct number of significant figures. They then are asked to determine the uncertainty value of the graduated cylinder. The simulation includes several different sizes of graduated cylinders, each containing unique markings, so students will be challenged to analyze each individually.
• Glassware Accuracy : Follow the simulation with this laboratory activity, which allows students to further explore the concepts of accuracy, precision, and percent error with this lab.Students use different types of laboratory glassware to measure 50 mL of water and determine the accuracy of each piece of glassware.By the end of this lesson, students should be able to determine the accuracy of different pieces of glassware and calculate percent error.
• The Chemistry Composition Challenge : Your students will be challenged to design a method to solve three chemistry problems with this lab. One problem requires students to determine the thickness of a piece of aluminum foil and compare their value to the actual one. Another has them determine the identity of an unknown metal by calculating its density. This resource includes extensive teacher notes to help you guide your students through this inquiry activity.
• Density Animation : Use this animation to introduce the concept of density and help your students visualize density on the particulate level. There are opportunities to make qualitative and quantitative comparisons between substances.
• Density : You can then use this lab to allow your students to determine the density of several liquids and solids. They then identify an unknown metal by determining its density and calculate the percent error within the class for a specific sample.By the end of this activity, students should be able to calculate the density of a liquid by measuring volume and mass, calculate the density of a solid using the displacement method for finding volume identify an unknown substance by determining its density, calculate percent error, and explain if their results are accurate or precise.
• Graphing Density : Finish up the topic of density with this lab, which requires students to collect data and use graphing to determine the density values of unknown metal samples. This activity will help your students learn to construct a line graph using Excel and analyze a linear equation to help determine density.
• Investigating the Density of an Irregular Solid Object Lab : If you think your students need another activity to solidify their understanding, try out this lab. Students use common laboratory equipment to devise a method to measure the density of several irregular objects. They will then create a formal laboratory report using both their own data and data from the entire class. Read an article about this activity in the September 2016 issue of Chemistry Solutions .
• Use the lab, Colors of the Rainbow to assess your students’ understanding of the concept of density. In this lab, students calculate the density of several unknown liquids and then use their findings to build a density column. There are eight unknown solutions that are grouped nine different ways so that each of the lab groups will have a unique combination of liquids to build their density column.
• "Bringing Real-Life Context to Chemical Math ": Before moving on to the topic of scientific notation, read this article in the March 2016 issue of Chemistry Solutions .
• Scientific Notation: Then use this activity to introduce the topic to your students.Students have a button, which they move like a decimal point, to be actively involved in putting numbers into and taking numbers out of scientific notation format.
• Using Scientific Notation in Chemistry : Follow up with this lesson, which has students solve a variety of real-world problems using scientific notation. Students listen to a convoluted radio conversation about coffee which will relate to a math-based problem that this lesson is developed around. Students begin to recognize the benefits of using scientific notation in their calculations.This lesson includes a formative quiz, summative quiz, slides, and a radio conversation on YouTube
• The Significant Figures and Lab Data : Many students seem to struggle with significant figures, especially when it is taught out of context.This activity allows students to use laboratory equipment of different precision to collect data for several different metals, and then use the data to calculate the density of each.They then compare their calculated densities to accepted values and determine the combination of equipment that leads to the most accurate calculation of density.
• The Measurement Animation allows students to review the fundamentals of measurement in length, mass, and volume. Various units of measurement will be presented for comparison, and several conversion calculations will be demonstrated using dimensional analysis.
• The Temperature Guys : Open your discussion of unit conversion with this video from our Founders of Chemistry series.It tells the story of how temperature as we currently know it evolved. The first thermometers invented in the early 1600s are very different than ones we use today! An activity sheet that includes question for students to answer as they watch the video is available to download.
• Dimensional Analysis and Unit Conversion : Follow the video with this lesson plan to introduce the process of unit conversion.This resource includes a PowerPoint Presentation and student handout with practice problems.
• The activity, Dimensional Analysis with Notecards is a great resource to help students who are struggling with this important concept. This activity allows them to visualize dimensional analysis using pre-made conversion factors on notecards and understand the importance of cancelling units to solve conversion problems.
• Unit Conversion Online Tutorial : Next, use this activity to have your students interact with a web-based tutorial that uses a drag and drop interface in order to learn how to convert between units of measurement using dimensional analysis. The tutorial allows students to learn at their own pace, and also provides feedback while they are solving problems.You may want to read the March 2016 issue of Chemistry Solutions article, "A Student-Centered, Web-Based Approach to Teaching Unit Conversions " before using the activity.
• Math and Measurement : Get your students ready for a unit assessment with this lab, which allows students to practice introductory math skills that will be used in chemistry all year. This includes metric conversion, significant figures, scientific notation, dimensional analysis, density, percent error, accuracy and precision, as well as using lab equipment.
• Nanoscale & Self-Assembly : An option for an advanced culminating lab or extension for the unit could be this lab, which incorporates measurements, and dimensional analysis.In this activity, students determine both the diameter of one single BB and the length of an oleic acid molecule using simple measurements and volume/surface area relationships.
• ChemMatters : If you’d like to increase your students’ scientific literacy while connecting dimensional analysis to the real world, have them read “Recycling Aluminum: A Way of Life or a Lifestyle?” from the April 2012 magazine or “ Drivers, Start Your (Electric) Engines ” from the February 2013 issue.
• The lesson, Captivating Chemistry of Coins will allow students to demonstrate their understanding of physical and chemical properties of matter by comparing the composition of different pennies. This is done by determining the density of pre- and post-1982 pennies which will be compared to the density of different metals. This lab is introduced with the ChemMatters article, The Captivating Chemistry of Coins .
## Complexity=3, Mode=toNormal
Complexity=4, mode=toscientific.
#### IMAGES
1. 30 Scientific Notation Word Problems Worksheet
2. Scientific Notation Word Problem Practice PLUS Spiral Review by Jessica
3. Scientific notation problem solving
4. Lesson 6 Problem Solving Practice
5. Scientific Notation Word Problems
6. Scientific Notation Practice Packet by Maisonet Math
#### VIDEO
1. Intro to Scientific Notation
2. 6) Problem solving skills| Dimensional formulas
3. Chemistry: Math Basics and Scientific Notation
4. 8th Grade Math 2.4b, Multiplying with Scientific Notation
5. Scientific Notation
6. How To Write Scientific Notation #math #mathematics #scientificnotation
1. Scientific notation (practice)
Scientific notation (practice) | Khan Academy. 8th grade. Course: 8th grade > Unit 1. Lesson 10: Scientific notation intro. Scientific notation example: 0.0000000003457. Scientific notation examples. Scientific notation review. Math > 8th grade > Numbers and operations > Scientific notation intro. Google Classroom.
2. Unit 7: Exponents and scientific notation
Learn. Exponent properties with products. Exponent properties with quotients. Negative exponent intuition. Multiplying & dividing powers (integer exponents) Practice. Multiply powers. Start. Practice. Divide powers. Practice. Multiply & divide powers (integer exponents) Get 5 of 7 questions to level up! Practice. Lesson 8: Combining bases. Learn.
3. Scientific notation examples (video)
Scientific notation is a way of writing very large or very small numbers. A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10. For example, 650,000,000 can be written in scientific notation as 6.5 10^8. Created by Sal Khan and CK-12 Foundation. Created by Sal Khan and CK-12 Foundation.
4. Roots, exponents, & scientific notation
Learn. Intro to square roots. Square roots of perfect squares. Intro to cube roots. Worked example: Cube root of a negative number. Square root of decimal. Dimensions of a cube from its volume. Square roots review. Cube roots review. Practice. Up next for you: Square roots Get 5 of 7 questions to level up! Start.
5. PDF SSON 6 Using Scientific Notation
Using Scientific Notation. LESSON 6 . Using Scientific Notation. EXAMPLE A . Multiply: (3.5 103)(2 105). Express the answer in scientific notation. Multiplication is commutative and associative. Rearrange the factors and group the coeficients and powers of 10. (3.5 103)(2 105) 5 (3.5 2.0)(103 105) 2 Multiply the coeficients.
6. Scientific notation word problems (practice)
Google Classroom. You might need: Calculator. Light travels 9.45 \cdot 10^ {15} 9.45 ⋅ 1015 meters in a year. There are about 3.15 \cdot 10^7 3.15 ⋅107 seconds in a year. How far does light travel per second? Write your answer in scientific notation. meters. Show Calculator. Stuck? Review related articles/videos or use a hint. Report a problem. 7.
7. 6.3 Scientific Notation (Homework Assignment)
Scientific notation is a convenient notation system used to represent large and small numbers. Examples of these are the mass of the sun or the mass of an electron in kilograms. Simplifying basic operations such as multiplication and division with these numbers requires using exponential properties.
8. Scientific Notation
Problem 1: The distance to the moon is 238,900 miles. Write this value in scientific notation. Problem 2: If one mile is 1609.34 meters. What the distance to the moon in meters using scientific notation. Problem 3: The age of the Earth is roughly four billion and six hundred million years. Write this number in (i) standard decimal format, and ...
9. Scientific Notation: Practice Problems
In this lesson, review what scientific notation is and examine two practice problems that show how to use this notation easily. Updated: 10/21/2021. Scientific Notation. Do you know what the mass...
10. Problem Solving With Scientific Notation
Summary. Scientific notation was developed to assist mathematicians, scientists, and others when expressing and working with very large and very small numbers. Scientific notation follows a very specific format in which a number is expressed as the product of a number greater than or equal to one and less than ten, and a power of 10 10.
11. PDF Module 1
Answers: 943 = 943 x 1 = 943 x 100 = 9.43 x 102 in scientific notation. 0.00036 = 0.00036 x 100 = 3.6 x 10―4 in scientific notation. When a number is converted to scientific notation, numbers that are • larger than one have positive exponents (zero and above) in scientific notation;
12. Scientific Notation Calculator
How to solve your scientific notation problems. To solve your scientific notation problem, type in your number like 23400. The scientific notation calculator will then show you the steps to help you learn how to convert the number to scientific notation. Scientific Notation Video Lesson. Khan Academy Video: Scientific Notation Examples.
13. Math Practice Problems
1. 2223000000. 2. 0.000007505. Answers. Complexity=3, Mode=toNormal. Convert from scientific notation into normal notation. Complexity=4, Mode=toScientific. Convert from normal notation into scientific notation. Sample answer: 4.5 × 106. Type: 4.5 * 10^6. Learn more about our online math practice software . "MathScore works."
14. Lesson 6 Scientific Notation Classwork Practice.pdf
View Lesson 6 Scientific Notation Classwork Practice.pdf from MATH MISC at Baruch College, CUNY. NAME _ DATE _ PERIOD _ Lesson 6 Scientific Notation Classwork Practice Course 3 • Chapter 1 Real ... Lesson 6 Scientific Notation Classwork Practice.pdf - NAME ... Doc Preview. Pages 2. Total views 19. Baruch College, CUNY. MATH. MATH MISC ...
15. Lesson 6 Homework Practice Scientific Notation
Lesson 6 Homework Practice Scientific Notation Write each number in standard form. 1. 9.03 1022. 7.89 1033. 4.115 1054. 3.201 1065. 5.1 1026. 7.7 1057. 3.85 1048. 1.04 103Write each number in scientific. Fill & Sign Online, Print, Email, Fax, or Download. Get Form. We are not affiliated with any brand or entity on this form.
16. Solve problems using scientific notation
Academic Tutoring. » Solve problems using scientific notation. Simplify the following expression using scientific notation. It follows that 7,200,000 can be rewritten as. By similar reasoning, 5,000,000 can be rewritten as. Express the product in scientific notation. This is in scientific notation and is the correct choice.
17. Scientific Notation: Practice Problems
Quiz. Course. Try it risk-free for 30 days. Instructions: Choose an answer and hit 'next'. You will receive your score and answers at the end. question 1 of 3. What is the value of (2 x 10^6) + (3...
18. Scientific Notation Lesson Plan
Use Study.com's video lesson on scientific notation to practice the rules required to add, subtract, multiply and divide numbers that are written in scientific notation. End the lesson with a game...
19. Scientific Notation
This animated Math Shorts video from UEN explains the term scientific notation and provides several examples in converting extreme numbers to and from scientific notation. It also demonstrates how to multiply numbers using scientific notation. In the accompanying classroom activity, students practice writing numbers in scientific notation and develop real-world problems for each other to solve.
20. Scientific Notation Practice Worksheets & Teaching Resources
. Scientific Notation Practice Worksheets & Teaching Resources | TpT. Browse scientific notation practice resources on Teachers Pay Teachers, a marketplace trusted by millions of teachers for original educational resources. DID YOU KNOW: Seamlessly assign resources as digital activities. Learn how in 5 minutes with a tutorial resource.
21. Lesson 6 Homework Practice Scientific Notation Answer Key
Fill Lesson 6 Homework Practice Scientific Notation Answer Key, Edit online. Sign, fax and printable from PC, iPad, tablet or mobile with pdfFiller Instantly. Try Now! Home; For Business. ... Lesson 6 Homework Practice Scientific Notation Write each number in standard form. 1. 9.03 1022. 7.89 1033. 4.115 1054. 3.201 1065. 5.1 1026. 7.7 1057. 3. ...
22. Classroom Resources
Recognize the benefit of using scientific notation to solve large scale problems. Use dimensional analysis for mass, length, volume, temperature, and density unit conversion problems. Convert between units of measurement using dimensional analysis. Understand the purpose of using dimensional analysis for converting between units of measurement.
23. Scientific Notation Sample Problems
These sample problems below for Scientific Notation were generated by the MathScore.com engine. Sample Problems For Scientific Notation. Complexity=3, Mode=toNormal. Convert from scientific notation into normal notation. 1. 8.39 × 10 6. 2. 5.86 × 10 8. Complexity=4, Mode=toScientific. Convert from normal notation into scientific notation.
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# AREA AND PERIMETER OF SIMILAR FIGURES
## About "Area and perimeter of similar figures"
Area and perimeter of similar figures :
Two figures that have the same shape are said to be similar. When two figures are similar, the ratios of the lengths of their corresponding sides are equal. To determine if the triangles below are similar, compare their corresponding sides.
length of side AB/length of side AD = Length of side EF/Length of side EH
(4/2) = (8/4)
2 : 1 = 2 : 1
Perimeter :
If two similar triangles have a scale factor of a : b, then the ratio of their perimeters is a : b.
Area :
If two similar triangles have a scale factor of a : b, then the ratio of their areas is a2 : b2.
Let us see some examples based on the above concept.
Example 1 :
Find the missing length of following similar shapes
Solution :
Since the above rectangles ABCD and EFGH are similar, the ratio of the corresponding sides will be in the same ratio.
Length of side AB/Length of side AD = Length of EF/Length of EH
4/6 = a/12
4 (12) = 6a
a = 48/6
a = 8
Hence the missing length of the rectangle EFGH is 8 cm.
Example 2 :
Find the missing length of following similar shapes
Solution :
Since the above rectangles ABCD and EFGH are similar, the ratio of the corresponding sides will be in the same ratio.
Length of side AB/Length of side AD = Length of EF/Length of EH
b/1 = 7.5/3
b (3) = 7.5 (1)
b = 7.5/3
b = 2.5
Hence the missing length of the rectangle EFGH is 2.5 cm.
Example 3 :
The perimeters of two similar triangles is in the ratio 3 : 4. The sum of their areas is 75 cm2. Find the area of each triangle.
Solution :
Since the perimeters of two similar triangles is in the ratio 3 : 4,
Let "3x" be the side length of first triangle
Let "4x" be the side length of second triangle
Area of 1st triangle/Area of 2nd triangle = (3x/4x)²
= 9x²/16x²
The sum of their areas = 75 cm2
9x²+ 16x² = 75
25x² = 75
x² = 75/25 = 3
Area of first triangle = 9x² = 3(3) = 9 cm²
Area of second triangle = 16x² = 16(3) = 48 cm²
Example 4 :
The areas of two similar triangles are 45 cm2 and 80 cm2. The sum of their perimeters is 35 cm. Find the perimeter of each triangle.
let the scale factor of the two similar triangles be a : b.
Area of 1st triangle/Area of 2nd triangle = 45/80
a : b is the reduced form of the scale factor. 3 : 4 is then the reduced form of the comparison of the perimeters.
(9/16) = (a/b)2
(a/b) = 3/4
Perimeter of 1st triangle = 3x
Perimeter of 2nd triangle = 4x
Sum of the perimeters of the 1st and 2nd triangle = 35
3x + 4x = 35
7x = 35 ==> x = 5
Perimeter of 1st triangle = 3x = 3(5) = 15 cm
Perimeter of 2nd triangle = 4x = 4(5) = 20 cm
After having gone through the stuff given above, we hope that the students would have understood "Area and perimeter of similar figures".
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In fractions for kids 1, we learnt about the different components of a fraction and how to convert mixed numbers into improper fractions. In this article, we will learn how to add, subtract, multiply and divide fractions and mixed numbers.
## Fractions for kids 2 – Addition and subtraction
Adding or subtracting fractions can be very simple. There is one golden rule that we must all bear in mind whenever we want to add or subtract fractions. The rule is:
Before we add or subtract any 2 fractions, we must make sure that they have the same denominator.
To recap, denominator is the lower number and numerator is the upper number in a fraction.
Credit: iamsy
Once both fractions have the same denominator, adding them up is as easy as adding the numerators while keeping the denominators the same. For example, if we were to solve 1/5 + 2/5, the answer would be (1+2)/5 = 3/5.
Credit: iamsy
The same rule applies for subtraction of fractions. So long we have the same common denominator, subtracting fractions is just subtracting the numerators while keeping the denominator the same. For example, if we were to solve 2/5 - 1/5, the answer would be (2-1)/5 = 1/5
## Fractions for kids 2 – Finding the common denominator
We have seen how easy it is to add or subtract fractions when they have the same denominator. But how do we add or subtract if the fractions do not have the same denominator? What we can do is to change the fractions so that both have the same denominator.
For example, if we were to add ½ and ¼, we can’t add them up directly as they do not share the same denominator. In this case, we would need to change them so they have the same denominators. Recall in fractions for kids 1, we can change a fraction such as 2/4 into its simplest form which is ½. Conversely, we can also change ½ into 2/4.
To solve ½ + ¼, we simply change the fraction ½ into 2/4 and the problem becomes 2/4 + ¼ = ¾
You can do the same thing for subtraction. To solve ½ - 1/4 , just change ½ into 2/4 and we will get 2/4 - ¼ = ¼
### Fractions for kids 2 – Quick way to find common denominator
Somtimes, the common denominator is not obvious. For example, to solve 1/3 + 1/5, we will need to find a common denominator for both of the fractions. One way to quickly identify the common denominator would be to multiply each fraction by the denominator of the other fraction.
## Fractions for kids 2 – adding and subtracting mixed numbers
There are two ways to add or subtract mixed numbers. The first way is to add the mixed fractions directly. For example, to solve 1¼ + 2¼; we add the whole number and the fractions parts of the mixed number separately. In this case, we will add 1 and 2 together to obtain 3 and ¼ and ¼ together to obtain 2/4 or ½. The answer will then be 3½.
The second method is to convert the mixed fractions into improper fractions before adding them together (to recap on improper fraction, read fractions for kids 1). Using the above example, if we were to solve 1¼ + 2¼, we would convert 1¼ into 5/4 and 2¼ into 9/4. Now both fractions have the same denominator and we can add them up, giving an answer of 14/4. 14/4 can then be changed back into the mixed number is 3½.
## Fractions for kids 2 – Multiplying fractions
Multiplying fractions is very simple; there is no need for fractions to have common denominator before we multiply or divide them. To multiply 2 fractions, we simply multiply the numerators of both fractions first, then multiply the denominators of both fractions. For example, to solve ½ x ¼, we multiply both numerators which would give us 1 x 1 = 1 followed by both denominators which would give us 2 x 4 = 8. The answer for ½ x ¼ = 1/8
Credit: iamsy
## Fractions for kids 2 – Dividing fractions
Dividing fractions requires an one more step than multiplying fractions. To divide fractions, we change them into a multiplication first. To do this, we switch the numerator and denominator of the fraction on the right side of the divide sign. For example, to solve ½ ÷ ¼, we change it into a multiplication by swapping the denominator of ¼ . In this case, it would become ½ x 4/1 =4/2 = 2.
Credit: iamsy
## Fractions for kids 2 – Multiplying and dividing fixed numbers
Unlike addition and subtraction of mixed numbers, we cannot multiply or divide mixed numbers directly. Before we multiply or divide a mixed number, we must also change the mixed number into an improper fraction.
After the mixed number is changed into an improper fraction, the multiplication or division can be carried out using the same methods mentioned above.
## Fractions for kids 2 – Summary
We have learnt a couple of things about fractions today.
1. We have learnt about the golden rule of having the same common denominator before adding or subtracting fractions.
2. We have also learnt how to quickly find the common denominator
3. We have learnt how to add and subtract mixed numbers
4. Lastly, we have learnt how to multiply and divide fractions and mixed numbers
### Learn about fractions by cutting up pizzas
Learning Resources Pizza Fraction Fun
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The Pizza Fraction by Learning Resources game lets kids learn about fractions in a fun and engaging way. Learn how to manipulate fractions, add and subtract fractions with 13 double-sided pizzas and 7 different fraction games. Elementary school students or even younger will find it easy to learn fractions by visualizing how they work with the pizzas. This award-winning game is recommended for 2 to 6 players.
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# Adding and subtracting decimals
Here we will learn about adding and subtracting decimals, including calculations with two or more decimals, or with a mixture of decimals and whole numbers.
Students will first learn about adding and subtracting decimals as part of number and operations in base ten in 5th grade.
## What is adding and subtracting decimals?
Adding and subtracting decimals involves the addition and subtraction of decimal numbers by understanding place value.
When adding or subtracting with decimals special care must be taken to ensure that the decimal points line up with each other. This means that each place value should also line up.
For example, let’s look at 12.5 + 6.23.
Decimal numbers are used in real life particularly when using measurements such as money, length, mass, and capacity. Therefore you may find the skill of adding and subtracting decimals useful when you are problem solving or answering word problems in a real-world context.
On this page, we will be focusing on using the standard algorithm to add or subtract decimals to the thousandths place. No calculations will involve negative numbers or recurring decimals. For information on calculating with negative numbers and different types of decimal numbers, you can follow these links.
See also: Adding and subtracting negative numbers
See also: Recurring decimals
## Common Core State Standards
How does this relate to 5th grade math and 6th grade math?
• 5th grade – Numbers and Operations in Base Ten (5.NBT.7)
Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used
• 6th grade – The Number System (6.NS.3)
Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.
## How to add and subtract decimals
In order to add or subtract decimals:
1. Make sure each number has a decimal point and write any \bf{0} placeholders that are required.
2. Stack the numbers, ensuring that the decimal points line up.
3. Use the standard algorithm for addition/subtraction, ensuring the decimal point is also written in the answer.
## Adding and subtracting decimals examples
### Example 1: adding two decimals using the standard algorithm with no regrouping
Calculate 12.3 + 4.5.
1. Make sure each number has a decimal point and write any 0 placeholders that are required.
Each number has a decimal point and one decimal place, so no zero placeholders are required.
2Stack the numbers, ensuring that the decimal points line up.
3Use the standard algorithm for addition/subtraction, ensuring the decimal point is also written in the answer.
Adding the digits in each place value from right to left, we have
Note that the decimal point is placed in the same column in the solution.
So 12.3 + 4.5 = 16.8.
### Example 2: adding a whole number and a decimal using the standard algorithm with no regrouping
Calculate 52 + 31.07.
Make sure each number has a decimal point and write any 0 placeholders that are required.
Stack the numbers, ensuring that the decimal points line up.
Use the standard algorithm for addition/subtraction, ensuring the decimal point is also written in the answer.
### Example 3: adding two decimals using the standard algorithm with regrouping
Calculate 6.7 + 9.31.
Make sure each number has a decimal point and write any 0 placeholders that are required.
Stack the numbers, ensuring that the decimal points line up.
Use the standard algorithm for addition/subtraction, ensuring the decimal point is also written in the answer.
### Example 4: subtracting two decimals using the standard algorithm with no regrouping
Calculate 26.87-14.2.
Make sure each number has a decimal point and write any 0 placeholders that are required.
Stack the numbers, ensuring that the decimal points line up.
Use the standard algorithm for addition/subtraction, ensuring the decimal point is also written in the answer.
### Example 5: subtracting a decimal from a whole number using the standard algorithm with regrouping
Calculate 16-9.4.
Make sure each number has a decimal point and write any 0 placeholders that are required.
Stack the numbers, ensuring that the decimal points line up.
Use the standard algorithm for addition/subtraction, ensuring the decimal point is also written in the answer.
### Example 6: subtracting two decimals using the standard algorithm with regrouping
Calculate 2.04-0.952.
Make sure each number has a decimal point and write any 0 placeholders that are required.
Stack the numbers, ensuring that the decimal points line up.
Use the standard algorithm for addition/subtraction, ensuring the decimal point is also written in the answer.
### Teaching tips for adding and subtracting decimals
• Review place values before beginning to add or subtract decimals, especially decimal place values. Not only will students need to line up the decimal point, but they also need to ensure that each place value is lined up, so their understanding of place values is vital.
• Students may struggle to line up the digits when stacking the numbers, so it may help to provide them with graph paper so they may write the numbers into boxes and keep them aligned. This will also help students see what place values are “missing” a number, and they can add in a zero placeholder into that box.
• Provide students with opportunities to solve real-world word problems involving adding or subtracting decimals, such as money or measurement. This will help them better understand the problems and what the numbers represent in a real-life context.
### Easy mistakes to make
• Lining up decimal numbers in each place value incorrectly
When using the standard algorithm for addition of decimals or subtraction of decimals, students can sometimes line up the numbers incorrectly. This is because younger students are sometimes told to line up the numbers from the right side, but this method only works for whole numbers.
When stacking the decimal numbers, you must line up the decimal points. This will ensure that the digits are in the correct column according to their place value. Using zero placeholders can also help you to avoid making this mistake.
• Subtracting the smaller digit from the larger digit during standard algorithm subtraction even though the smaller digit is above the larger digit
When using the standard algorithm for decimal subtraction, students can find themselves automatically subtracting the smaller digit from the larger digit, rather than taking time to note which digit is supposed to be subtracted from the other. The digit below should always be subtracted from the digit above.
If the digit above is smaller than the digit below then a method often referred to as ‘borrowing’ must take place. For instance, here is the incorrect method for example 5 where the smaller digit which is above has been incorrectly subtracted from the larger digit. Also see the correct method where ‘borrowing’ has taken place.
Incorrect solution:
Correct solution:
• Forgetting to write the decimal point in the solution to a standard algorithm addition/subtraction problem involving decimals
When using the standard algorithm for addition or subtraction of decimals, students can sometimes forget to write the decimal point in their solution. It is important to remember to write this, particularly if the answer is not a whole number. It is good practice to write the decimal point in the solution area before you start to solve. The decimal point should line up with the points above.
Note how the decimal point has been written in the solution area before the calculation has been started.
### Practice adding and subtracting decimals questions
1. Solve 58.1 + 0.46.
62.7
58.56
6.27
5.856
2. Solve 41.3 + 38.
79.3
41.68
41.41
45.1
3. Solve 10.62 + 7.73.
17.23
17.135
18.35
87.92
4. Solve 16.9-3.3.
13.6
16.57
13.87
1.36
5. Solve 27-1.24.
26.24
25.76
14.6
1.46
6. Solve 7.11-6.84.
64.26
1.73
1.95
0.27
## Adding and subtracting decimals word problems
1. This table shows the 4 most recent world records for the men’s 100 meter race.
Usain Bolt holds the current world record for the men’s 100 meter race at 9.58 seconds.
How many seconds did he shave off the previous world record holder’s time?
Show answer
9.74-9.58 = 0.16 seconds
2. Abi, Bobby and Cyrus each have some money.
They want to buy a ball from a local shop costing \$3.60 to play catch with. They decide to put their money together in order to buy the ball. Abi has \$2.30.
Bobby has \$1.25. Cyrus has 9 cents. If they buy the ball, how much change will they get? Show answer 2.30 + 1.25 + 0.09 = 3.64 3.64-3.60 = 0.04 Change is \$0.04 or 4 cents.
3. Ali is harvesting potatoes. He weighs and measures the length of a sample of 10 potatoes. Below is a table showing his results.
(a) Find the difference between the longest potato and the shortest potato in the sample.
(b) What is the total weight of the 3 longest potatoes?
Show answer
(a) Longest shortest = 6.1-2.98 = 3.12 \, cm
(b) Potatoes 1, 2, and 4\text{: } 36.1 + 60.8 + 27.7 = 124.6 \, g
## Adding and subtracting decimals FAQs
What is the first step in adding and subtracting decimals?
The first step is to stack the numbers, lining up the numbers according to place value and lining up the decimal points.
How can you add or subtract decimals if the numbers do not have the same number of digits in the decimal places?
To add or subtract decimals that do not have the same number of digits in the decimal places, you can use zeros as placeholders and then begin to solve.
Where do you place the decimal point in the answer when you add or subtract decimals?
In the answer, the decimal point should line up with the decimal points in the numbers you are adding or subtracting. It may be helpful to place the decimal point in the answer space first before beginning to solve.
## Still stuck?
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#### [FREE] Common Core Practice Tests (3rd to 8th Grade)
Prepare for math tests in your state with these 3rd Grade to 8th Grade practice assessments for Common Core and state equivalents.
Get your 6 multiple choice practice tests with detailed answers to support test prep, created by US math teachers for US math teachers!
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# Probability Theory
Daniel Weibel
Created 8 Nov 2017
$\newcommand{\P}{\text{P}} \newcommand{\Vert}{\,\vert\,}$
# Conditional Probability
## Probability of B Given A
$\P(B\Vert A)$
• If $A$ occured, what is the probability that now also $B$ occurs?
• Percentage of $A$ that is also in $B$.
## Probability of A Given B
$\P(A\Vert B)$
• If $B$ occured, what is the probability that now also $A$ occurs?
• Percentage of $B$ that is also in $A$.
# Probability of A And B
\begin{align*} \P(A \cap B) & = \P(A) \cdot \P(B\Vert A) \\ & = \P(B) \cdot \P(A\Vert B) \end{align*}
# Probability of A Or B
$\P(A \cup B) = \P(A) + \P(B) - \P(A \cap B)$
# Conditional Probability Revisited I
## Probability of B Given A (I)
$\P(B\Vert A) = \frac{\P(A \cap B)}{\P(A)}$
Deduced from this formula for $\P(A \cap B)$.
## Probability of A Given B (I)
$\P(A\Vert B) = \frac{\P(A \cap B)}{\P(B)}$
Deduced from this formula for $\P(A \cap B)$.
# Conditional Probability Revisited II
## Probability of B Given A (II)
$\P(B\Vert A) = \frac{\P(B) \cdot \P(A\Vert B)}{\P(A)}$
In this formula for $\P(B\Vert A)$, replace $\P(A \cap B)$ with this formula for $\P(A \cap B)$.
This equation is called Bayes’ Theorem.
## Probability of A Given B (II)
$\P(A\Vert B) = \frac{\P(A) \cdot \P(B\Vert A)}{\P(B)}$
In this formula for $\P(A\Vert B)$ replace $\P(A \cap B)$ with this formula for $\P(A \cap B)$.
This equation is called Bayes’ Theorem.
# Independent Events
The happening of one event has no effect on the probability of the other event.
For example:
• $A$ = getting head on first toss of a coin
• $B$ = getting head on second toss of a coin
$\P(A \cap B) = \P(A) \cdot \P(B)$
In this formula for $\P(A \cap B)$, replace $\P(B\Vert A)$ with $\P(B)$.
We can do this, because if $A$ occurred, the probability that $B$ occurs is not affected by that. The probability that $B$ occurs is still $\P(B)$.
# Mutual Exclusive Events
The happening of one event prevents the happening of the other event.
For example:
• $A$ = rolling a 1 with a die
• $B$ = rolling a 2 with a die
$\P(A \cup B) = \P(A) + \P(B)$
In this formula for $\P(A \cup B)$, replace $\P(A \cap B)$ with 0.
We can do this, because events $A$ and $B$ cannot occur together (the probability that they occur together is 0).
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1. ## 9th Grade Easy Triangle Question For Most
The three angles of a triangle have measures 12x degrees, 3x degrees, and 7y degrees, where 7y is greater than 60. If x and y are integers, what is the value of X?
2. Originally Posted by 4woods
The three angles of a triangle have measures 12x degrees, 3x degrees, and 7y degrees, where 7y is greater than 60. If x and y are integers, what is the value of X?
Solve 15x + 7y = 180, where 7y > 60 and x and y are whole numbers, by trial and errror.
eg. y = 9 => 7y = 63 satisfies the first condition. But 15x = 180 - 63 = 117 does not have a solution for x that's a whole number. So try again ....
3. Hello, 4woods!
The three angles of a triangle have measures $\displaystyle 12x$ degrees, $\displaystyle 3x$ degrees,
and $\displaystyle 7y$ degrees, where $\displaystyle {\color{red}\rlap{///////}}7y > 60$ .
Unnecessary and confusing . . .
If $\displaystyle x$ and $\displaystyle y$ are integers, what is the value of $\displaystyle x$ ?
The three angles of a triangle always total 180°.
Hence: .$\displaystyle 12x + 3x + 7y \:=\:180 \quad\Rightarrow\quad x \:=\:\frac{180-7y}{15} \:=\:12 - \frac{7y}{15}$
Since $\displaystyle x$ is an integer, $\displaystyle y$ must be a multiple of 15.
$\displaystyle \text{Since }7y < 180\text{, then }\,y < 26.\;\;\text{ Hence: }\:y = 15$.
Therefore: .$\displaystyle x \;=\;12 - \frac{7(15)}{15} \quad\Rightarrow\quad \boxed{x \:=\:5}$
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Is 57 A Prime Number?
• No the number 57 is not a prime number.
• Fifty-seven is a composite number. Because 57 has more divisors than 1 and itself.
Prime Factorization Of 57
• Prime factors of number 57 are: 3, 19
• Equcation of 57 is: 3 * 19
• The smallest common factor of 57 is number 3
• Highest or greatest common factor GCF of 57 is number 19
How To Calculate Prime Number Factors
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
Mathematical Information About Numbers 5 7
• About Number 5. Integers with a last digit as a zero or a five in the decimal system are divisible by five. Five is a prime number. All odd multiples of five border again with the five (all even with zero). The fifth number of the Fibonacci sequence is a five. Five is also the smallest prime number that is the sum of all other primes which are smaller than themselves. The Five is a Fermat prime: 5 = 2 ^ {2 ^ 1} +1 and the smallest Wilson prime. Number five is a bell number (sequence A000110 in OEIS). There are exactly five platonic bodies. There are exactly five tetrominoes.
• About Number 7. Seven is a prime number. It is the lowest natural number that cannot be represented as the sum of the squares of three integers. The corresponding cyclic number is 142857. You can use this feature to calculate the result of the division of natural numbers by 7 without a calculator quickly. A seven-sided shape is a heptagon. One rule for divisibility by 7 leads to a simple algorithm to test the rest loose divisibility of a natural number by 7: Take away the last digit, double it and subtract them from the rest of the digits. If the difference is negative, then you're leaving the minus sign. If the result has more than one digit, so you repeat steps 1 through fourth. Eventually results are 7 or 0, then the number is divisible by 7 and not otherwise.
What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers. Primes can thus be considered the basic building blocks of the natural numbers. There are infinitely many primes, as demonstrated by Euclid around 300 BC. The property of being prime (or not) is called primality.
In number theory, the prime number theorem describes the asymptotic distribution of the prime numbers among the positive integers. It formalizes the intuitive idea that primes become less common as they become larger.
Primes are used in several routines in information technology, such as public-key cryptography, which makes use of properties such as the difficulty of factoring large numbers into their prime factors.
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# What is the vertex form of 3y=-5x^2 - x +7?
##### 1 Answer
Aug 5, 2017
$y = - \frac{5}{3} {\left(x - \left(- \frac{1}{10}\right)\right)}^{2} + \frac{141}{60}$
#### Explanation:
Given:
$3 y = - 5 {x}^{2} - x + 7$
Divide both sides by $3$ to get $y$ on the left hand side, then complete the square...
$y = \frac{1}{3} \left(- 5 {x}^{2} - x + 7\right)$
$\textcolor{w h i t e}{y} = - \frac{5}{3} \left({x}^{2} + \frac{1}{5} x - \frac{7}{5}\right)$
$\textcolor{w h i t e}{y} = - \frac{5}{3} \left({x}^{2} + 2 \left(\frac{1}{10}\right) x + \frac{1}{100} - \frac{141}{100}\right)$
$\textcolor{w h i t e}{y} = - \frac{5}{3} \left({\left(x + \frac{1}{10}\right)}^{2} - \frac{141}{100}\right)$
$\textcolor{w h i t e}{y} = - \frac{5}{3} {\left(x + \frac{1}{10}\right)}^{2} + \frac{141}{60}$
$\textcolor{w h i t e}{y} = - \frac{5}{3} {\left(x - \left(- \frac{1}{10}\right)\right)}^{2} + \frac{141}{60}$
The equation:
$y = - \frac{5}{3} {\left(x - \left(- \frac{1}{10}\right)\right)}^{2} + \frac{141}{60}$
is in the form:
$y = a {\left(x - h\right)}^{2} + k$
which is vertex form for a parabola with vertex $\left(h , k\right) = \left(- \frac{1}{10} , \frac{141}{60}\right)$ and multiplier $a = - \frac{5}{3}$
graph{3y = -5x^2-x+7 [-4.938, 5.06, -1.4, 3.6]}
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6 th Grade Math Focus 2: Rates, including percents
# 6 th Grade Math Focus 2: Rates, including percents
Télécharger la présentation
## 6 th Grade Math Focus 2: Rates, including percents
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##### Presentation Transcript
1. 6th Grade MathFocus 2: Rates, including percents Standards: 6.RP.2, 6.RP.3b, 6.RP.3c Resource: Connected Math Program 2 Bits and Pieces I, Investigation 4.3
2. Bits and Pieces 1- Investigation 4: Working with PercentsMathematical Learning and Problem-Solving Goals • Introduce percents as a part-whole relationship where the whole is not out of 100 but scaled to be “out of 100” (4.1) • Use fraction partitioning and fraction benchmarks to make sense of percents (4.1) • Develop strategies, including percents, to use in comparisons where the whole is less than 100 (4.2) • Understand that comparing situations with different numbers of trials is difficult unless we use percents or some other form of equivalent representation (4.2) • Work with situations where the whole is sometimes greater than 100 and sometimes less than 100 (4.3) • Develop connections between fractions, decimals, and percents (4.3) • Develop strategies for expressing data in percent form (4.3) • Relate fractions, decimals, and percents (4.4) • To move from percents to other representations and from other representations to percents (4.4)
3. Bits and Pieces 1, Investigation 4.3Mathematical Learning Targets Students will be able to work with percents by understanding and completing the following: • Work with situations where the whole is sometimes greater than 100 and sometimes less than 100 • Develop connections between fractions, decimals, and percents • Develop strategies for expressing data in percent form
4. Getting Ready for Problem 4.3
5. Think about the situations we looked at in Problems 4.1 and 4.2 • In Problem 4.1, with Yao and Shaq, we looked at a situation where the whole was GREATER than 100. • In Problem 4.2, the Portland girls’ basketball team problem, we looked at situations where the whole was LESS than 100. This problem asks you to think about fractions, decimals, and percents!
6. Problem 4.3, Parts A-D
7. TEACHER ONLY! Guide to Observations
8. TEACHER ONLY!
9. Problem 4.3, Parts E and F
10. TEACHER ONLY!
11. TEACHER ONLY! SUMMARY!
12. Summary Situation
13. Bits and Pieces 1, Investigation 4.3Mathematical Learning Targets Review Students will be able to work with percents by understanding and completing the following: • How do you work with situations where the whole is sometimes greater than 100 and sometimes less than 100? • What are some connections between fractions, decimals, and percents? • Explain a strategy you used for expressing data in percent form.
14. 6th Grade Math Homework Bits & Pieces 1 ACE #7-15
15. TEACHER ONLY!
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# Difference between revisions of "1985 AIME Problems/Problem 14"
## Problem
In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?
## Solution 1
Let us suppose for convenience that there were $n + 10$ players overall. Among the $n$ players not in the weakest 10 there were $n \choose 2$ games played and thus $n \choose 2$ points earned. By the givens, this means that these $n$ players also earned $n \choose 2$ points against our weakest 10. Now, the 10 weakest players playing amongst themselves played ${10 \choose 2} = 45$ games and so earned 45 points playing each other. Then they also earned 45 points playing against the stronger $n$ players. Since every point earned falls into one of these categories, It follows that the total number of points earned was $2{n \choose 2} + 90 = n^2 - n + 90$. However, there was one point earned per game, and there were a total of ${n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}$ games played and thus $\frac{(n + 10)(n + 9)}{2}$ points earned. So we have $n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}$ so $2n^2 - 2n + 180 = n^2 + 19n + 90$ and $n^2 -21n + 90 = 0$ and $n = 6$ or $n = 15$. Now, note that the top $n$ players got $n(n - 1)$ points in total (by our previous calculation) for an average of $n - 1$, while the bottom 10 got 90 points total, for an average of 9. Thus we must have $n > 10$, so $n = 15$ and the answer is $15 + 10 = \boxed{25}$.
## Solution 2
Suppose that there are $n$ players participating in the tournament. We break this up into a group of the weakest ten, and the other $n-10$ people. Note that the $10$ players who played each other generated a total of $\dbinom{10}{2} = 45$ points playing each other. Thus, they earned $45$ playing the $n-10$ other people. Thus, the $n-10$ people earned a total of $10(n-10)-45 = 10n-145$ points playing vs. this group of 10 people, and also earned a total of $10n-145$ playing against themselves. Since each match gives a total of one point, we must have that $\dbinom{n-10}{2}=10n-145$. Expanding and simplifying gives us $n^2-41n+400=0$. Thus, $n=16$ or $n=25$. Note however that if $n=16$, then the strongest $16$ people get a total of $16*10-145=15$ playing against the weakest $10$ who gained $45$ points vs them, which is a contradiction since it must be larger. Thus, $n=\boxed{25}$.
Solution by GameMaster402
## Solution 3
Note that the total number of points accumulated must sum to ${p \choose 2} = \frac{p(p-1)}{2}$. Say the number of people is $n$. Consider the number of points gained when the 10 lowest scoring people play each other. The problem tells us that each of these 10 people must earn exactly half of the total number of points they will earn during the whole game. This implies that this group of 10 people must accumulate half their total combined points after they (the 10 people) all play each other, meaning they must earn the other half of their points by playing the $n-10$ stronger players. The problem also tells us that the $n-10$ people who aren't part of the losers group will earn half of their points by playing the $10$ losers. Since the $n-10$ group and $10$ losers will earn half their points by playing each other, the sum of the number of points that they gain playing each other must then be half of the total amount of points earned by everyone in the game. Therefore, $\frac{p(p-1)}{4} = 10(p-10)$. This equation is the same as above, and by the same logic, the answer is $n=\boxed{25}$.
## See also
1985 AIME (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
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# Lesson 11
Multiplication Expressions
### Lesson Purpose
The purpose of this lesson is for students to use multiplication expressions to represent equal groups.
### Lesson Narrative
In previous lessons, students represented situations involving equal groups with drawings and tape diagrams. Students were also shown how to represent equal groups as an expression. In this lesson, students connect the structure of drawings, tape diagrams, and multiplication situations to the structure of multiplication expressions (MP7). Students create diagrams and drawings to represent multiplication expressions and ultimately write their own expressions to represent drawings, diagrams, and situations (MP2).
When generating multiplication expressions, consider using the convention of the number of groups as the first factor and the size of the groups as the second factor. However, it is not necessary for students to write the factors in this order. It is important that students connect their expressions to the corresponding situations and representations. They should be able to correctly explain what each factor represents in their expressions. If students ask questions about the idea of commutativity, consider recording the questions publicly for future investigation.
To allow time for students to focus on the meaning of multiplication, it is not an expectation that students find the product of each expression in this lesson. In subsequent lessons, students will work on strategies for finding the product. If students mention the product in today’s lesson, it is okay to note that, but try to maintain focus on the connections between the expression and the diagrams.
• Action and Expression
• MLR2
### Learning Goals
Teacher Facing
• Write multiplication expressions to represent situations involving equal groups and diagrams.
### Student Facing
• Let’s write multiplication expressions.
### Required Materials
Materials to Gather
### Required Preparation
Activity 1:
• Each group of 2 needs 1 card from the card sort in the previous lesson.
• Post these expressions around the room:
• $$3\times5$$
• $$4\times3$$
• $$3\times2$$
• $$2\times10$$
• $$3\times10$$
Addressing
### Lesson Timeline
Warm-up 10 min Activity 1 10 min Activity 2 15 min Activity 3 10 min Lesson Synthesis 10 min Cool-down 5 min
### Teacher Reflection Questions
What did you say, do, or ask during the lesson synthesis that helped students be clear on the learning of the day? How did understanding the cool-down of the lesson before you started teaching today help you synthesize that learning?
### Suggested Centers
• Capture Squares (1–3), Stage 4: Subtract within 20 (Supporting)
• Five in a Row: Addition and Subtraction (1–2), Stage 7: Add within 1,000 without Composing (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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# How do you rationalize (2sqrt5)/ ( 2sqrt5 + 3sqrt2)?
May 11, 2018
$\frac{2 \sqrt{5}}{2 \sqrt{5} + 3 \sqrt{2}} = 10 - 3 \sqrt{10}$
#### Explanation:
Note that the difference of squares identity tells us that:
${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$
Hence we can rationalize the denominator of the given expression by multiplying both numerator and denominator by $2 \sqrt{5} - 3 \sqrt{2}$ ...
$\frac{2 \sqrt{5}}{2 \sqrt{5} + 3 \sqrt{2}} = \frac{2 \sqrt{5} \left(2 \sqrt{5} - 3 \sqrt{2}\right)}{\left(2 \sqrt{5} - 3 \sqrt{2}\right) \left(2 \sqrt{5} + 3 \sqrt{2}\right)}$
$\textcolor{w h i t e}{\frac{2 \sqrt{5}}{2 \sqrt{5} + 3 \sqrt{2}}} = \frac{{\left(2 \sqrt{5}\right)}^{2} - \left(2 \sqrt{5}\right) \left(3 \sqrt{2}\right)}{{\left(2 \sqrt{5}\right)}^{2} - {\left(3 \sqrt{2}\right)}^{2}}$
$\textcolor{w h i t e}{\frac{2 \sqrt{5}}{2 \sqrt{5} + 3 \sqrt{2}}} = \frac{20 - 6 \sqrt{10}}{20 - 18}$
$\textcolor{w h i t e}{\frac{2 \sqrt{5}}{2 \sqrt{5} + 3 \sqrt{2}}} = 10 - 3 \sqrt{10}$
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# How to Write a Decimal From the Shaded Graph
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When students first start learning about decimals, teachers may use shaded graphs to help demonstrate how they work. The whole graph represents the number 1, and it is divided into a number of equal parts. It may be divided into 10 parts, 100 parts or 1,000 parts. Teachers use these graphs to teach place value in decimals. They first show their students a 10-square graph, then a 100-square graph, then a 1,000-square graph. They shade different amounts of the graphs to represent different decimals.
Identify the graph. See if it has 10 squares, 100 squares or 1,000 squares.
Count the number of shaded squares. If the graph has 100 squares, count each fully-shaded row as 10, then count the individual squares in a partially-shaded row. If the graph has 1,000 squares, count each fully-shaded box as 100, then each left-over fully-shaded row as 10, then each left-over individual shaded square.
Count the zeros in the total number of squares (10 has one zero; 100 has two zeros; 1,000 has three zeros). Write down the number of shaded squares, and use the same number of digits as the zeros you just counted. For example, if you counted three shaded squares in a 10 graph, write "3," with only one digit; for three shaded squares in a 100 graph, write "03," with two digits; for three shaded squares in a 1,000 graph, write "003," with three digits.
Put a decimal point on the left of the number, before any zeros you added. For example, for three shaded squares on a 1,000 graph, write .003.
#### Tips
• Note that 1 shaded square on a 10 graph, 10 shaded squares on a 100 graph and 100 shaded squares on a 1,000 graph are all the same size. This is because .1, .10 and .100 are all the same value. One square on a 10 graph is one tenth. One square on a 100 graph is one hundredth. One square on a 1,000 graph is one thousandth.
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# How do you write an equation for a circle given center (0,3) and radius is 7 units?
Jan 10, 2017
${x}^{2} + {\left(y - 3\right)}^{2} = {7}^{2}$, or
${x}^{2} + {y}^{2} - 6 y - 40 = 0$
#### Explanation:
The Cartesian equation of a circle with centre $\left(a , b\right)$ and radius $r$ is:
${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
So a circle with centre $\left(0 , 3\right)$ and radius $7$ would be:
${\left(x - 0\right)}^{2} + {\left(y - 3\right)}^{2} = {7}^{2}$
$\therefore \setminus \setminus \setminus \setminus \setminus {x}^{2} + {\left(y - 3\right)}^{2} = {7}^{2}$
We can multiply out if required to get:
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus {x}^{2} + {y}^{2} - 6 y + 9 = 49$
$\therefore {x}^{2} + {y}^{2} - 6 y - 40 = 0$
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## Using Decimals and Fractions With Circles
### Learning Outcomes
• Find the circumference of circles
• Find the area of circles
The properties of circles have been studied for over $2,000$ years. All circles have exactly the same shape, but their sizes are affected by the length of the radius, a line segment from the center to any point on the circle. A line segment that passes through a circle’s center connecting two points on the circle is called a diameter. The diameter is twice as long as the radius. See the image below.
The size of a circle can be measured in two ways. While the radius is one way to measure the size of a circle, another way to measure the size of a circle is with the circumference which is the distance around a circle.
Archimedes discovered that for circles of all different sizes, dividing the circumference by the diameter always gives the same number. The value of this number is pi, symbolized by Greek letter $\pi$ (pronounced “pie”). However, the exact value of $\pi$ cannot be calculated since the decimal never ends or repeats (we will learn more about numbers like this in The Properties of Real Numbers.)
Doing the Manipulative Mathematics activity Pi Lab will help you develop a better understanding of pi.
If we want the exact circumference or area of a circle, we leave the symbol $\pi$ in the answer. We can get an approximate answer by substituting $3.14$ as the value of $\pi$. We use the symbol $\approx$ to show that the result is approximate, not exact.
### Properties of Circles
$\begin{array}{c}r\text{ is the length of the radius.}\hfill \\ d\text{ is the length of the diameter.}\hfill \end{array}$
$\begin{array}{cccc}\text{The circumference is }2\pi \mathit{\text{r}}.\hfill & & & C=2\pi \mathit{\text{r}}\hfill \\ \text{The area is }\pi{\mathit{\text{r}}}^{2}.\hfill & & & A=\pi{\mathit{\text{r}}}^{2}\hfill \end{array}$
Since the diameter is twice the radius, another way to find the circumference is to use the formula $C=\pi \mathit{\text{d}}$.
Suppose we want to find the exact area of a circle of radius $10$ inches. To calculate the area, we would evaluate the formula for the area when $r=10$ inches and leave the answer in terms of $\pi$.
$\begin{array}{}\\ A=\pi {\mathit{\text{r}}}^{2}\hfill \\ A=\pi \text{(}{10}^{2}\text{)}\hfill \\ A=\pi \cdot 100\hfill \end{array}$
We write $\pi$ after the $100$. So the exact value of the area is $A=100\pi$ square inches.
To approximate the area, we would substitute $\pi \approx 3.14$.
$\begin{array}{ccc}A& =& 100\pi \hfill \\ \\ & \approx & 100\cdot 3.14\hfill \\ & \approx & 314\text{ square inches}\hfill \end{array}$
Remember to use square units, such as square inches, when you calculate the area.
### example
A circle has radius $10$ centimeters. Approximate its circumference and area.
Solution
1. Find the circumference when $r=10$. Write the formula for circumference. $C=2\pi \mathit{\text{r}}$ Substitute $3.14$ for $\pi$ and 10 for $r$ . $C\approx 2\left(3.14\right)\left(10\right)$ Multiply. $C\approx 62.8\text{ centimeters}$
2. Find the area when $r=10$. Write the formula for area. $A=\pi {\mathit{\text{r}}}^{2}$ Substitute $3.14$ for $\pi$ and 10 for $r$ . $A\approx \left(3.14\right){\text{(}10\text{)}}^{2}$ Multiply. $A\approx 314\text{ square centimeters}$
### example
A circle has radius $42.5$ centimeters. Approximate its circumference and area.
### try it
Watch the following video to see another example of how to find the circumference of a circle.
In the next video example, we find the area of a circle.
## Approximate $\pi$ with a Fraction
Convert the fraction ${\Large\frac{22}{7}}$ to a decimal. If you use your calculator, the decimal number will fill up the display and show $3.14285714$. But if we round that number to two decimal places, we get $3.14$, the decimal approximation of $\pi$. When we have a circle with radius given as a fraction, we can substitute ${\Large\frac{22}{7}}$ instead of $3.14$. And, since ${\Large\frac{22}{7}}$ is also an approximation of $\pi$, we will use the $\approx$ symbol to show we have an approximate value, so $\pi\approx{\Large\frac{22}{7}}$.
### example
A circle has radius ${\Large\frac{14}{15}}$ meters. Approximate its circumference and area.
## Contribute!
Did you have an idea for improving this content? We’d love your input.
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A Beginner's Guide to ASCE 7-05 Chapter 7 - W: Wind Loads © 2007, T. Bartlett Quimby
Overview Nature of Wind Method 2 Provisions Example Problems Homework Problems References Report Errors or Make Suggestions
Section 7.4.1
Example Problem 7.1
Two Story Building
Method 2
Last Revised: 11/04/2014
Given: The enclosed office building shown in Figure 7.4.1.1. The building is located in a region with a wind speed (3-sec gust) of 120 mph. The exposure is Exposure C. The building is on flat terrain.
Figure 7.4.1.1
Building Definition
Wanted: The wind pressures applied to the surfaces and the net forces applied to the building.
Solution: To solve this problem, we need to independently look at two different wind directions, but first we will define a few parameters.
Determine critical elevations:
• Mean Roof Height: h = 2*11' + (3/12)*25'/2 = 25.1 ft
• Mean 2nd Floor Height: h = 11' + 11'/2 = 16.5 ft
• Mean 1st Floor Height: h = 11'/2 = 5.5 ft
Compute the Velocity Pressures, qz = .00256 Kz Kzt Kd V2 I (ASCE 7-05 Equation 6-15)
• Kzt = 1 (Flat Terrain)
• Kd = .85 (ASCE 7-05 Table 6-4, Buildings)
• I = 1.0 (ASCE 7-05 Table 6-1, Category II building)
• Kz = varies with elevation = 2.01(max(h,15)/900)(2/9.5), (ASCE 7-05 Tables 6-2 and 6-3)
h Kz qz (ft) (psf) Roof 25.125 0.946 29.7 2nd flr 16.5 0.866 27.1 1st flr 5.5 0.849 26.6
Compute the Internal Pressures, qhGCpi:
• GCpi = + 0.18 (ASCE 7-05 Figure 6-5)
• qh = 29.7 psf
• qhGCpi = + 5.34 psf
Determine the Gust Factor, G
• G = 0.85 (ASCE 7-05, 6.5.8.1)
Wind in the N/S Direction:
For this part of the problem we need to determine pressure coefficients for the locations shown in Figure 7.4.1.2 as well as for the side walls. These coefficients are then combined with the gust factor and velocity pressures to obtain the external pressures in each region.
Figure 7.4.1.2
N/S Building Section
The pressure coefficients for the walls are found in ASCE 7-05 Figure 6-6 (pg 49)
For the Windward wall (P1 & P2), Cp is 0.8 for all elevations.
For the Leeward wall (P5 & P6), Cp is dependent on the ration of L/B. In this case L/B = 50'/90' = 0.556, so Cp = -0.50 for all elevations.
For the sidewalls (not shown in Figure 7.4.1.2), the value of Cp is -0.7 in all cases.
For the roof, the slope angle is 14.0 degrees. This is close to 15 degrees and probably not worth interpolating between the values given in ASCE 7-05 Figure 6-6. We also need to know that h/L = 25.1'/50' = 0.50. From the Figure we get that the values of Cp for the Windward side of the roof is -0.70 and -0.18. These values represent two different load cases. For the Leeward side, Cp is -0.50.
We can now compute the external pressures, qGCp, for each surface. The following table shows the computation results:
Pressure Cp q qGCp (psf) (psf) Windward Wall P1 0.8 26.6 18.1 P2 0.8 27.1 18.5 Windward Roof P3 -0.7 29.7 -17.6 P3 -0.18 29.7 -4.5 Leeward Roof P4 -0.5 29.7 -12.6 Leeward Wall P5 -0.5 29.7 -12.6 P6 -0.5 29.7 -12.6 Side Walls P7 -0.7 29.7 -17.6
Combining with the internal pressures you get the following four load cases where:
• Case I includes the maximum windward pressure (-17.6 psf) and positive internal pressure
• Case II includes the minimum windward pressure (-4.5 psf) and positive internal pressure
• Case III includes the maximum windward pressure (-17.6 psf) and negative internal pressure
• Case IV includes the minimum windward pressure (-4.5 psf) and negative internal pressure
The net forces are found by multiplying the appropriate pressures by the areas over which they act. In this building all but the gable ends are rectangles, making the area calculation easier. Note that we are computing actual surface areas (as opposed to projected areas) in each of the cases below. Also, the sign is important. Negative signs indicate a force that is outward from the surface and a positive sign is inward. All forces are normal to their respective surfaces.
Net Pressures: Net Force Pressure Case I Case II Case III Case IV Areas Case I Case II Case III Case IV (psf) (psf) (psf) (psf) (ft2) (k) (k) (k) (k) Windward Wall P1 12.8 12.8 23.4 23.4 990 12.62 12.62 23.19 23.19 P2 13.1 13.1 23.8 23.8 990 12.99 12.99 23.55 23.55 Windward Roof P3 -23.0 -9.9 -12.3 0.8 2319 -53.29 -22.90 -28.54 1.86 Leeward Roof P4 -17.9 -17.9 -7.3 -7.3 2319 -41.60 -41.60 -16.85 -16.85 Leeward Wall P5 -17.9 -17.9 -7.3 -7.3 990 -17.76 -17.76 -7.19 -7.19 P6 -17.9 -17.9 -7.3 -7.3 990 -17.76 -17.76 -7.19 -7.19 Side Walls P7 -23.0 -23.0 -12.3 -12.3 1413 -32.46 -32.46 -17.38 -17.38
It is often useful to resolve each force into it's global components so that they can be easily added vectorially. Figure 7.4.1.3 shows the location of each of the resulting forces.
Figure 7.4.1.3
Building Forces for N/S Wind
Case I Case II Case III Case IV Force E/W N/S vert. E/W N/S vert. E/W N/S vert. E/W N/S vert. (k) (k) (k) (k) (k) (k) (k) (k) (k) (k) (k) (k) Windward Wall F1 0.00 12.62 0.00 0.00 12.62 0.00 0.00 23.19 0.00 0.00 23.19 0.00 F2 0.00 12.99 0.00 0.00 12.99 0.00 0.00 23.55 0.00 0.00 23.55 0.00 Windward Roof F3 0.00 -12.93 51.70 0.00 -5.55 22.22 0.00 -6.92 27.69 0.00 0.45 -1.80 Leeward Roof F4 0.00 10.09 40.36 0.00 10.09 40.36 0.00 4.09 16.34 0.00 4.09 16.34 Leeward Wall F5 0.00 17.76 0.00 0.00 17.76 0.00 0.00 7.19 0.00 0.00 7.19 0.00 F6 0.00 17.76 0.00 0.00 17.76 0.00 0.00 7.19 0.00 0.00 7.19 0.00 Side Walls F7a -32.46 0.00 0.00 -32.46 0.00 0.00 -17.38 0.00 0.00 -17.38 0.00 0.00 F7b 32.46 0.00 0.00 32.46 0.00 0.00 17.38 0.00 0.00 17.38 0.00 0.00 Sum 0.00 58.29 92.06 0.00 65.66 62.58 0.00 58.29 44.03 0.00 65.66 14.54
Note that the maximum uplift and maximum horizontal force do not occur in the same load cases! Do not combined the two cases, design for each individually.
You will also notice that the internal pressure has no effect on the net horizontal force.
The net force in the lateral direction is zero since the forces on the side walls will cancel each other.
Wind in the E/W Direction
Figure 7.4.1.4 defines the pressures (with the exception of the lateral/side wall pressures) that need to be computed for wind loading from the E/W direction. In this case we combined all the leeward wall segments into one because they all have the same pressures.
Figure 7.4.1.4
E/W Building Section
The pressure coefficients are taken from ASCE 7-05 Figure 6-6. Note that the coefficient for the leeward wall is obtained by interpolation with an L/B ratio of 1.8.
Pressure Cp q qGCp (psf) (psf) Windward Wall P8 0.8 26.6 18.1 P9 0.8 27.1 18.5 P10 0.8 29.7 20.2 Roof P11 -0.9 29.7 -22.7 P11 -0.18 29.7 -4.5 P12 -0.5 29.7 -12.6 P12 -0.18 29.7 -4.5 P13 -0.3 29.7 -7.6 P13 -0.18 29.7 -4.5 Leeward Wall P14 -0.34 29.7 -8.6 Side Walls P15 -0.7 29.7 -17.6
Note that some of the pressures are applied to differently oriented surfaces. When the same pressure is applied to a different surface, we have chosen to label on as "a" and the other as "b". See Figure 7.4.1.5 for force applications. Four cases are computed, based on combinations of maximum/minimum roof pressures and + internal pressures.
Figure 7.4.1.5
Building Forces for E/W Wind
The net forces on each surface, in terms of direction relative the surface, are as follows:
Pressure Case I Case II Case III Case IV Area Case I Case II Case III Case IV (psf) (psf) (psf) (psf) (ft2) (k) (k) (k) (k) Windward Wall P8 12.8 12.8 23.4 23.4 550 7.01 7.01 12.88 12.88 P9 13.1 13.1 23.8 23.8 550 7.21 7.21 13.09 13.09 P10 14.8 14.8 25.5 25.5 156 2.32 2.32 3.98 3.98 Roof P11a -28.0 -9.9 -17.3 0.8 647 -18.12 -6.39 -11.22 0.52 P11b -28.0 -9.9 -17.3 0.8 647 -18.12 -6.39 -11.22 0.52 P12a -17.9 -9.9 -7.3 0.8 647 -11.60 -6.39 -4.70 0.52 P12b -17.9 -9.9 -7.3 0.8 647 -11.60 -6.39 -4.70 0.52 P13a -12.9 -9.9 -2.2 0.8 1026 -13.23 -10.13 -2.28 0.82 P13b -12.9 -9.9 -2.2 0.8 1026 -13.23 -10.13 -2.28 0.82 Leeward Wall P14 -13.9 -13.9 -3.2 -3.2 1256 -17.47 -17.47 -4.06 -4.06 Side Walls P15a -23.0 -23.0 -12.3 -12.3 1980 -45.50 -45.50 -24.36 -24.36 P15b -23.0 -23.0 -12.3 -12.3 1980 -45.50 -45.50 -24.36 -24.36
Restating the forces in terms of the global coordinate system we get:
Case I Case II Case III Case IV Pressure E/W N/S vert. E/W N/S vert. E/W N/S vert. E/W N/S vert. (k) (k) (k) (k) (k) (k) (k) (k) (k) (k) (k) (k) Windward Wall F8 7.01 0.00 0.00 7.01 0.00 0.00 12.88 0.00 0.00 12.88 0.00 0.00 F9 7.21 0.00 0.00 7.21 0.00 0.00 13.09 0.00 0.00 13.09 0.00 0.00 F10 2.32 0.00 0.00 2.32 0.00 0.00 3.98 0.00 0.00 3.98 0.00 0.00 Roof F11a 0.00 4.40 17.58 0.00 1.55 6.20 0.00 2.72 10.88 0.00 -0.13 -0.50 F11b 0.00 -4.40 17.58 0.00 -1.55 6.20 0.00 -2.72 10.88 0.00 0.13 -0.50 F12a 0.00 2.81 11.26 0.00 1.55 6.20 0.00 1.14 4.56 0.00 -0.13 -0.50 F12b 0.00 -2.81 11.26 0.00 -1.55 6.20 0.00 -1.14 4.56 0.00 0.13 -0.50 F13a 0.00 3.21 12.83 0.00 2.46 9.82 0.00 0.55 2.21 0.00 -0.20 -0.80 F13b 0.00 -3.21 12.83 0.00 -2.46 9.82 0.00 -0.55 2.21 0.00 0.20 -0.80 Leeward Wall F14 17.47 0.00 0.00 17.47 0.00 0.00 4.06 0.00 0.00 4.06 0.00 0.00 Side Walls F15a 0.00 45.50 0.00 0.00 45.50 0.00 0.00 24.36 0.00 0.00 24.36 0.00 F15b 0.00 -45.50 0.00 0.00 -45.50 0.00 0.00 -24.36 0.00 0.00 -24.36 0.00 Sum 34.01 0.00 83.34 34.01 0.00 44.43 34.01 0.00 35.31 34.01 0.00 -3.60
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## GRADE 7 MATH WORKSHEET 15 ANSWERS
Grade 7 math worksheet 15 answers :
Here we are going to see 10 practice questions of mixed topics. We have been preparing this set of questions in order to test how you understood the topics.
## Practice questions of mixed topics in 7th grade math
Question 1 :
In a school the total enrollment of class 8th is 115. If the number of boys exceeds the number of girls by 33, find the number of boys in a class 8th.
(A) 74 (B) 89 (C) 50
Solution :
Let "x" be the number of girls
Number of boys = x + 33
Total number of students in the class = 115
x + x + 33 = 115
2x + 33 = 115
Subtract 33 on both sides
2x = 115 - 33
2x = 82
Divide by 2 on both sides
x = 41
x + 33 = 74
Hence the number of boys in the class is 74.
Question 2 :
One of the angles of a triangles is equal to the sum of the remaining tow angles. If the ratio of these angles is 4:5, find the angles of the given triangle.
(A) 60, 60, 60 (B) 60, 80, 40 (C) 40, 50, 90
Solution :
Let the three angles be "9x", "4x" and "5x"
Sum of three angles = 180°
4x + 5x + 4x + 5x = 180°
8x + 10x = 180°
18x = 180° ==> x = 18°
Hence the angles are 90, 50, 40.
Question 3 :
Two equal sides of a triangle are each 4 m less than three times the third side.Find the dimensions of a triangle,if its perimeter is 55 m.
(A) 15, 15, 8 (B) 23, 23, 9 (C) 13, 13, 8
Solution :
Let "x" be the third side of the triangle
Length of two equal sides = 3x - 4
Perimeter of triangle = 55 m
3x - 4 + 3x - 4 + x = 55
7x - 8 = 55
Add both sides by 8
7x = 63
Divide both sides by 7
x = 9
3x - 4 = 3(9) - 4 ==> 27 - 4 ==> 23
Hence the required angles are 23, 23, and 9
Question 4 :
If a + 1/a = 2, find the value of a3 + (1/a)3
(A) 1 (B) 2 (C) 3
Solution :
To find the value of a3 + (1/a)3, we may use the formula given below.
(a + b)3 = a3 + b3 + 3ab(a + b)
a3 + a1/3 = (a + (1/a))+ 3 a (1/a)(a + (1/a))
= 23 + 3 (2)
= 8 + 6
= 14
Hence the answer is 14.
Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".
Question 5 :
A gardener bought twice the number of Lilly plants that he had in his garden but had to throw 3 bad plants. When the new Lilly plants were planted, he had in all 48 plants in the garden.Find the number of Lilly plants he had originally.
(A) 17 (B) 15 (C) 13
Solution :
Let "x" be the number of lilly plants in his garden
Number of plants he buys = 2x
2x + x - 3 = 48
3x - 3 = 48
Add both sides by 3
3x = 48 + 3
3x = 51
Divide both sides by 3
x = 51/3 = 17
Hence the number of Lilly plants in the garden is 17.
Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".
Question 6 :
Simplify a3 - b3 ÷ (a-b)
(A) a+ ab + b2 (B) a- ab + b2 (C) (a+b) (a+b)
Solution :
a3 - b= (a - b) (a+ ab + b2)
a3 - b3 ÷ (a - b) = (a - b) (a+ ab + b2) / (a - b)
= a+ ab + b2
Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".
Question 7 :
If a + b + c = 9 and ab + bc + ca = 23, find the value of a2 + b2 + c2
(A) 15 (B) 12 (C) 35
Solution :
(a + b + c)2 = a2 + b2 + c+ 2ab + 2bc + 2ca
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
a2 + b2 + c2 = (a + b + c)2 - 2(ab + bc + ca)
= (9)2 - 2 (23)
= 81 - 46 = 35
Hence the answer is 35.
Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".
Question 8 :
The difference between the compound interest and the simple interest on a certain principal for 2 years at 4% per annum is \$150. Find the principal
(A) 1875 (B) 3750 (C) 1263
Solution :
Simple interest = PNR/100
150 = P(2)(4) / 100
150 = 8 P / 100
Multiply both sides by 100
150 (100) = 8 P
Divide both sides by 8
P = 150(100) / 8
= 1875
Hence the principal amount is 1875.
Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".
Question 9 :
The watch is marked at \$1150. During off season a discount given and it is sold for \$1100. Find the discount percent allowed.
(A) 2.14% (B) 4.34% (C) 3.04%
Solution :
Discount percentage
= (Discount amount / Marked price) ⋅ 100
Discount amount = 1150 - 1100 = 50
= (50/1150) ⋅ 100
= 0.0434 (100)
= 4.34%
Let us look into the solution of next problem on "Grade 7 math worksheet 15 answers".
Question 10 :
What must be added to each of the numerator and the denominator of the fraction 7/11 to make it equal to 3/4
(A) 2 (B) 8 (C) 5
Solution :
Let "x" be the required number to be added to both numerator and denominator.
(7 + x) / (11 + x) = 3/4
4 (7 + x) = 3 (11 + x)
28 + 4x = 33 + 3x
4x - 3x = 33 - 28
x = 5
After having gone through the stuff given above, we hope that the students would have understood "7th grade math worksheet 15 answers".
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Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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# Networks and Road Maps, Part I
In the last section, you saw how mathematicians create new ideas byremoving requirements from old ones. In this section, we'll create a new idea by adding onto an old one. Not all connections are created equal. In some situations, electrical circuits, for example, we need a way to assign values to the connections.
What You Should Learn How a graph is transformed into a network by adding a set of numbers. Real world situations that are modeled by networks. A roadmap, with travel distances, is a real world example of a network.
A network is a graph or digraph that also has a function that maps the edges onto the set of real numbers. A network derived from a graph is called an undirected network. A network derived from a digraph is called a directed network. A signed graph is a special case of an undirected network where the functional values are either +1 or -1. All that business about "a function that maps ..." just means that each edge is assigned a number.
Let's take a concrete example and look at it from both perspectives.
Let V = { v1, v2, v3, v4, v5 } and let R = { (v4, v2), (v3, v4), (v3, v1), (v3, v5), (v5, v3) } be a digraph. (This is the same graph that we looked at in the last section.) We can turn this into a graph by assigning a value to each of the edges. f = { (v4 v2, -3.5), (v3 v4, 4), (v3 v1, 5), (v3 v5, 2), (v5 v3, -2) }. Notice that the values can be positive, negative, decimals, etc. I could also have used square roots, multiples of π or any other number that I needed. The picture below illustrates the network visually. The drawing is identical to the one from the previous section except I added numbers to each segment to indicate their values.
Our street map example from the previous section can be quickly converted to a network by assigning to the piece of road between two intersections the distance between the intersections. We'll look at this more in the exercises and come back to it again in a later section as part of a classic example called the traveling salesman problem.
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# What is the rule of 7 14 21 28?
What is the next number in the following sequence 7 14 21 28
Arithmetic sequences
7,14,21,28,35,42,49,56…
What is the missing term in the sequence 7 14 28 blank 112
Answer: 7, 14, 28, 56, 112, 224, 448, 896…. Step-by-step explanation: Each number is the previous number doubled.
What is the rule for the sequence 4 14 29 49
So, the rule can be described as adding a multiple of 5 to the previous term, starting with 10. 5. So, the next term in the sequence is 74. The sequence now looks like this: 4, 14, 29, 49, 74.
What is the next number in the following sequence 9 18 27 36
Arithmetic sequences
9,18,27,36,45,54,63,72,81,90,99…
What are the next three multiples of 7 14 21 28
Multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, … and so on.
What is the 10th term of the sequence 7 14 28
Summary: Given the sequence 7, 14, 28, 56, … The expression for the 10th term is a10 = 7.29.
What is the pattern rule for 7 14 28 56
7 , 14 , 28 , 56 , 112. . . That is, each new term is equal to the previous term multiplied by 2. Sequences which involve multiplication/division by a constant number from term to term are called geometric. Thus, the given sequence is geometric.
What is the missing term in the following number sequence 7 14 21 28 35
Therefore, to get the next term of the sequence, we add -7 to the last term of the sequence, which is -35. We see that the next term in the sequence given is -42.
What is the pattern rule for 1 2 4 7 11
Rule: xn = n(n-1)/2 + 1
Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, …
What is the sequence 7 13 25 49
7 13 25 49 97 194 385.
What number should come next in the following series 21 9 21 11 21 13 21
Download more important topics, notes, lectures and mock test series for LR Exam by signing up for free. Here you can find the meaning of Look at this series: 21, 9, 21, 11, 21, 13, 21, … What number should come next a)14b)15c)21d)23Correct answer is option 'B'.
What are the first five multiple of 7 14 21 28 and 35
Multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, … and so on. How do we get to know a number is multiple of another number If a number is multiple of another number, then it is evenly divisible by the original number.
What is the LCM of 7 14 and 21 by listing their multiples
To find the least common multiple of 7, 14, and 21, we need to find the multiples of 7, 14, and 21 (multiples of 7 = 7, 14, 21, 28, 42 . . . .; multiples of 14 = 14, 28, 42, 56 . . . .; multiples of 21 = 21, 42, 63, 84 . . . .) and choose the smallest multiple that is exactly divisible by 7, 14, and 21, i.e., 42.
What is the sequence 7 10 16 28
The next number in this sequence 7 , 10 , 16 , 28 , 52 ,… will be 100.
What is the answer for pattern 7 21 8 72 9
The next number is 243. 7 ->21 ->8 ->72 ->9 ->243 ->10 ->810… Whatever you want. One example is 9 * 3 * 3 * 3 = 243.
What are the next four terms of the sequence 7 14 21 28 35 42
The nth terms: 7,14,21,28,35,42,49,56,63…
What is the missing number 1 1 2 4 7 11 16
What is the pattern rule for 1 2 4 7 11 16 22
Answer: The number that fits best in the sequence 1, 2, 4, 7, 11, …, 22 is 16. So, the rule boils down to: 1 + 0 , 1 + 1, 2 + 2, 4 + 3, 7 + 4 , 11 + 5, 16 + 6, 22 + 7,
What is the pattern rule for the sequence of numbers 7 14 28 56 112
7 , 14 , 28 , 56 , 112. . . That is, each new term is equal to the previous term multiplied by 2. Sequences which involve multiplication/division by a constant number from term to term are called geometric. Thus, the given sequence is geometric.
What is the missing number in the series 11 12 17 18 23 24
∴ Missing number =23+6=29.
What is the missing number in the pattern 1 2 3 5 _ 13 21
Solution: The Fibonacci series is the series of numbers 1, 1, 2, 3, 5, 8, 13, 21, … Therefore, the next Fibonacci number in the following sequence is 34.
What are the multiples of 7 7 14 21 28 35 42 49 56 63 70
What are the Multiples of 7 up to 100 There are 14 multiples of 7 that are less than 100. The first few multiples of 7 are: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91 and 98.
What is the LCM of 7 14 21 and 28
The least common multiple of 7, 14, 21 and 28 is 84.
What is the least common multiple of 7 8 14 21
168
LCM of 7, 8, 14 and 21 is 168. LCM is the short form for Least Common Multiple or the Lowest common multiple, also known as LCD Least Common Divisor.
What sequence is 7 10 13 16 19
1, 4, 7, 10, 13, 16, 19, 22, 25, … This sequence has a difference of 3 between each number.
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# Solve system of linear equations, using matrix method: $2x+3y+3z=5$ $x-2y+z=-4$ $3x-y-2z=3$
Toolbox:
• A matrix is said to be singular if |A|= 0.
• A matrix is said to be invertible if |A|$\neq 0$.
• If A is a non-singular matrix,AX=B,then $X=A^{-1}B.$
• Using this we can solve the system of equation which has unique solution.
This given system can be written in the form AX=B.
$\begin{bmatrix}1 & -1&1\\2 & 1&-3\\1 &1 & 1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\0\\2\end{bmatrix}$
Where $A=\begin{bmatrix}1 & -1& 1\\2 &1&-3\\1 &1 &1\end{bmatrix}\;X=\begin{bmatrix}x\\y\\z\end{bmatrix}\;B=\begin{bmatrix}4\\0\\2\end{bmatrix}$
Let us now find if A is singular or non-singular.The value of determinant A,|A| can be found by expanding along $R_1$
$|A|=1(1\times 1 -1\times -3)-(-1)(2\times 1-1\times -3)+1(2\times 1-1\times 1)$
$\;\;\;=(1+3)+(2+3)+(2-1)$
$\;\;\;=4+5+1=10\neq 0$
Hence it is a non-singular matrix.
Therefore inverse exists
We know $A^{-1}=\frac{1}{|A|}(adj \;A)$
Let us find the (adj A) by finding the minors and cofactors.
$M_{11}=\begin{vmatrix}1 & -3\\1 & 1\end{vmatrix}=1+3=4$
$M_{12}=\begin{vmatrix}2 &-3\\1 & 1\end{vmatrix}=2+3=5$
$M_{13}=\begin{vmatrix}2 &1\\1 & 1\end{vmatrix}=2-1=1$
$M_{21}=\begin{vmatrix}-1 &1\\1 & 1\end{vmatrix}=-1-1=-2$
$M_{22}=\begin{vmatrix}1 &1\\1 & 1\end{vmatrix}=1-1=0$
$M_{23}=\begin{vmatrix}1 &-1\\1 & 1\end{vmatrix}=1+1=2$
$M_{31}=\begin{vmatrix}-1 &1\\1 & -3\end{vmatrix}=3-1=2$
$M_{32}=\begin{vmatrix}1 &1\\2 & -3\end{vmatrix}=-3-2=-5$
$M_{33}=\begin{vmatrix}1 &-1\\2 & 1\end{vmatrix}=1+2=3$
$A_{11}=(-1)^{1+1}.4=4$
$A_{12}=(-1)^{1+2}.5=-5$
$A_{13}=(-1)^{1+3}.1=1$
$A_{21}=(-1)^{2+1}.-2=2$
$A_{22}=(-1)^{2+2}.0=0$
$A_{23}=(-1)^{2+3}.2=-2$
$A_{31}=(-1)^{3+1}.2=2$
$A_{32}=(-1)^{3+2}.-5=5$
$A_{33}=(-1)^{3+3}.3=3$
Therefore adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}4 & 2 & 2\\-5 & 0 & 5\\1 & -2 & 3\end{bmatrix}$
$A^{-1}=\frac{1}{|A|}(adj \;A)$ We know |A|=10
$A^{-1}=\frac{1}{10}\begin{bmatrix}4 & 2 & 2\\-5 & 0 & 5\\1 & -2 & 3\end{bmatrix}$
We know $X=A^{-1}B$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{10}\begin{bmatrix}4 & 2 & 2\\-5 & 0 & 5\\1 & -2 & 3\end{bmatrix}\begin{bmatrix}4 \\0\\2\end{bmatrix}$
We can do matrix multiplication by multiplying the rows of matrix A by the column of matrix B
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{10}\begin{bmatrix}16 +0+4\\-20+0+10\\4+0+6\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{10}\begin{bmatrix}20\\-10\\10\end{bmatrix}=\begin{bmatrix}2\\-1\\1\end{bmatrix}$
Hence x=2,y=-1,z=1.
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### Tetra Inequalities
Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.
### Staircase
Solving the equation x^3 = 3 is easy but what about solving equations with a 'staircase' of powers?
### Proof Sorter - the Square Root of 2 Is Irrational
Try this interactivity to familiarise yourself with the proof that the square root of 2 is irrational. Sort the steps of the proof into the correct order.
# Be Reasonable
##### Stage: 5 Challenge Level:
Thank you to M. Grender-Jones for this solution.
To show that $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ cannot form part of any arithmetic progression we give a proof by contradiction. Suppose they can, then $$\sqrt 3- \sqrt 2 = px$$ $$\sqrt 5 - \sqrt 2 = qx$$ where $x$ is the common difference of the progression, and $p$ and $q$ are integers.
Eliminating $x$ from these two equations we get $$p(\sqrt 5 - \sqrt 2) = q(\sqrt 3 - \sqrt 2)$$ so $$q\sqrt 3 = p\sqrt 5 + (q-p)\sqrt 2.$$ We know $p$ and $q$ are integers so to simplify this expression write $p-q = s$ where $s$ is an integer: $$q\sqrt 3 = p\sqrt 5 + s\sqrt 2.$$ Squaring this: $$3q^2 = 5p^2 + 2s^2 + 2ps\sqrt 10.$$ Rearranging this expression gives: $$\sqrt 10 = {3q^2 - 5p^2 - 2s^2 \over 2ps}.$$ As $\sqrt 10$ is irrational and all the other terms in this expression are integers this is impossible and we have reached a contradiction. Therefore our assumption was false and $\sqrt 2$, $\sqrt 3$ and $\sqrt 5$ cannot be terms of an AP.
By the same method can you prove that $\sqrt{1}$, $\sqrt{2}$ and $\sqrt{3}$ cannot be terms of ANY arithmetic progression?
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Question Video: Finding the Value of an Unknown in the Equation of a Line in Three Dimensions given the Equation of Another Line Perpendicular to It | Nagwa Question Video: Finding the Value of an Unknown in the Equation of a Line in Three Dimensions given the Equation of Another Line Perpendicular to It | Nagwa
Question Video: Finding the Value of an Unknown in the Equation of a Line in Three Dimensions given the Equation of Another Line Perpendicular to It Mathematics • Third Year of Secondary School
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Given that the lines (π₯ β 8)/3 = (π¦ + 4)/5 = (π§ + 6)/β2 and (π₯ β 10)/β5 = (π¦ + 7)/9 = (π§ β 3)/π are perpendicular, what is π?
02:30
Video Transcript
Given that the lines π₯ minus eight over three is equal to π¦ plus four over five is equal to π§ plus six over negative two and π₯ minus 10 over negative five is equal to π¦ plus seven over nine is equal to π§ minus three over π are perpendicular, what is π?
In this question, weβre given the equations of two lines written in Cartesian form. And we can see in one of the equations of these lines, we have an unknown value of π. We need to determine the value of π such that the two lines are perpendicular. So to answer this question, we should start by recalling how do we determine if two lines are perpendicular. And we know if two lines are perpendicular, then the direction vectors of each line must also be perpendicular. In particular, if we call these direction vectors π sub one and π sub two, then since these vectors are perpendicular, the dot product of the two vectors must be equal to zero.
This means we can determine if two lines are perpendicular by calculating the dot product of their direction vectors. And we can find the direction vectors of these two lines by noting theyβre written in Cartesian form. The components of the direction vector will be equal to the denominator of each part of the equation. We get the vector π sub one is equal to three, five, negative two. And we can do the exact same thing to find the direction vector of the second line. π sub two is the vector negative five, nine, π.
We can now calculate the dot product between these two vectors. And to do this, we recall to find the dot product of two vectors of the same dimension, we need to find the sum of the products of the corresponding components. This gives us three times negative five plus five times nine plus negative two multiplied by π. And remember, we want our lines to be perpendicular. So this dot product needs to be equal to zero. So we can set the left-hand side of this equation equal to zero.
We now want to solve this equation for π. Weβll do this by simplifying and rearranging. First, we can simplify the equation to get zero is equal to negative 15 plus 45 minus two π. We can subtract 30 from both sides of the equation to get negative 30 is equal to negative two π. Then we can solve for π by dividing both sides of the equation through by negative two. We get π is equal to 15, which is our final answer.
Therefore, we were able to show for the lines π₯ minus eight over three is equal to π¦ plus four over five is equal to π§ plus six over negative two and π₯ minus 10 over negative five is equal to π¦ plus seven over nine is equal to π§ minus three over π to be perpendicular, the value of π must be 15.
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# How do you differentiate f(x)=x^2e^(x^2-x) using the product rule?
Apr 7, 2016
$f ' \left(x\right) = x {e}^{{x}^{2} - x} \left(2 {x}^{2} - x + 2\right)$
#### Explanation:
$f \left(x\right) = {x}^{2} {e}^{{x}^{2} - x}$
Differentiating both sides w.r.t 'x'
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{2} {e}^{{x}^{2} - x}\right)$
Taking ${x}^{2}$ as first function and ${e}^{{x}^{2} - x}$ as second function
$f ' \left(x\right) = {x}^{2} \frac{d}{\mathrm{dx}} \left({e}^{{x}^{2} - x}\right) + {e}^{{x}^{2} - x} \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$
$f ' \left(x\right) = {x}^{2} \left({e}^{{x}^{2} - x}\right) \frac{d}{\mathrm{dx}} \left({x}^{2} - x\right) + {e}^{{x}^{2} - x} \left(2 x\right)$
$f ' \left(x\right) = {x}^{2} \left({e}^{{x}^{2} - x}\right) \left(\frac{d}{\mathrm{dx}} \left({x}^{2}\right) \frac{d}{\mathrm{dx}} \left(- x\right)\right) + {e}^{{x}^{2} - x} \left(2 x\right)$
$f ' \left(x\right) = {x}^{2} \left({e}^{{x}^{2} - x}\right) \left(2 x - 1\right) + 2 x {e}^{{x}^{2} - x}$
$f ' \left(x\right) = 2 {x}^{3} {e}^{{x}^{2} - x} - {x}^{2} {e}^{{x}^{2} - x} + 2 x {e}^{{x}^{2} - x}$
$f ' \left(x\right) = {e}^{{x}^{2} - x} \left(2 {x}^{3} - {x}^{2} + 2 x\right)$
$f ' \left(x\right) = x {e}^{{x}^{2} - x} \left(2 {x}^{2} - x + 2\right)$
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In analytic geometry, the direction cosines (or directional cosines) of a vector are the cosines of the angles between the vector and the three coordinate axes. Examples on Direction Cosines and Ratios of Lines Example â 6 Two lines L1 and L2 have direction cosines {l1, m1, n1} { l 1, m 1, n 1 } and {l2, m2, n2} { l 2, m 2, n 2 } respectively. Learn Science with Notes and NCERT Solutions, Chapter 11 Class 12 Three Dimensional Geometry. Direction cosines 1. K=XâX0D,L=YâY0D,M=ZâZ0+dD Taking direction cosines makes it easy to represent the direction of a vector in terms of angles with respect to the reference. Direction cosines and direction ratios are fundamental things to undersrand 3 Dimensional coordinate geometry. DIRECTION COSINES IN PLANAR MOTIONS Now we can also include a rotation at the knee as well as the previous rotation at the ankle. AB = â(2 2 +1 2 +2 2)= 3. Equivalently, they are the contributions of each component of the basis to a unit vector in that direction. Chapter 12Direction Cosines and Vector Geometry12.1IntroductionThe orientations of structural elements are most readily obtained analytically by 3-D vectorgeometry. \ (~~~~~~~~~~\) \ (\frac la \) = \ (\frac {1} {\sqrt {x^2 + y^2 + x^2}} \) Sol: A B = 4 i + 3 j hence the direction ratio is 4 : 3 Now direction cosines are l = 4 2 + 3 2 4 = 5 4 and m = 4 2 + 3 2 3 = 4 3 Law of Cosines: Given three sides. $$x = l| \vec{r} |$$ $$y = m| \vec{r} |$$ $$z = n| \vec{r} |$$ Find the direction cosines of the vector â ð´ = (5, 2, 8). The direction cosines of a line describe the orientation of the unit vector parallelto the line. 2) the direction cosines of a vector a = {ax ; ay ; az} can be found using the following formula. e. 3. direction is taken to be normal to the plane, and a superscript on . The coordinates of the unit vector is equal to its direction cosines. Example 2. Solution: Welcome to OnlineMSchool. Teachoo is free. Example: Find the direction cosines of the line joining the points (2,1,2) and (4,2,0). Ï. QnA , Notes & Videos Solution: Calculate the length of vector a: | a | = â 3 2 + 4 2 = â 9 + 16 = â 25 = 5. BC vector = OC vector - OB vector = (k - j) vector . The following brief MATLAB session reveals that the yaw, pitch, and roll angles for the direction cosine matrix in Example 11.17 are Ï = 109.69°, θ = 17.230°, and Ï = 238.43°. Find the vector A of norm 61 and direction cosines 1 2, â 1 2, â 2 2 . = 3, 2, 8 What this means is that direction cosines do not define how much an object is rotated around the axis of the vector. 54 lessons ⢠10 h 49 m . = 1 ( 2) , 2 4 , 3 ( 5) Solution:-Let r = 6 i ^ + 2 j ^ â 3 k ^ The direction ratios of r are 6, 2, â 3 The direction cosines of the given vector are Teachoo provides the best content available! Then why do we need three Miller indices? Detailed explanation with examples on direction-cosines-and-direction-ratios helps you to understand easily , designed as per NCERT. So, x1 = 2, y1 = 4 , z1 = 5 = , . In our next example, we will demonstrate how to calculate the direction cosines of a vector. Direction Cosines and Ratios Problem Example 1 Watch more videos at https://www.tutorialspoint.com/videotutorials/index.htm Lecture By: Er. On signing up you are confirming that you have read and agree to The coordinates of the point P may also be expressed as the product of the magnitude of the given vector and the cosines of direction on the three axes, i.e. Further Reading. Example: Find the direction cosines of the line joining the points (2,3,-1) and (3,-2,1). He has been teaching from the past 9 years. Thus, the direction cosines of the line joining two points is 3 28,, 77 77 77 â Example 4 Find the direction cosines of x, y and z-axis. The Miller indices prescribe the direction as a vector having a particular length (i.e. This web site owner is mathematician Dovzhyk Mykhailo. If rigid body B (thigh) rotates by angle q 2 from rigid body A (shank or lower leg), and A rotates from N as defined previously, then additional direction cosines may be defined. x axis makes an angle 0 with x axis, 90 with y axis & 90 with z axis. dcm = angle2dcm(rotationAng1,rotationAng2,rotationAng3) calculates the direction cosine matrix given three sets of rotation angles specifying yaw, pitch, and roll. In the case of the plane problem (Fig. I have covered a lot of examples along So, = 90 , = 0 , = 90 Direction cosines are l = cos 90 , m = cos 0 , n = cos ⦠If you want to contact me, probably have some question write me email on support@onlinemschool.com, Direction cosines of a vector - definition, Component form of a vector with initial point and terminal point, Cross product of two vectors (vector product), Linearly dependent and linearly independent vectors. Hence the direction cosines are â
, â
, -â
. Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i.e., 1,0,0. t. i. is represented by . CA vector = OA vector - OC vector = (i - k) vector. Direction cosines and direction ratios of a vector : Consider a vector as shown below on the x-y-z plane. ij. P ( 2, 4, 5) Example, 4 Find the direction cosines of x, y and z-axis. This application of the dot product requires that we be in three dimensional space unlike all the other applications weâve looked at to this point. P ( 2, 4, 5) Q (1, 2, 3) So, x1 = 2, y1 = 4 , z1 = 5 & x2 = 1, y2 = 2 , z2 = 3 Direction ratios = (x2 x1), (y2 y1), (z2 z1) = 1 ( 2) , 2 4 , 3 ( 5) = 1 + 2, 2, 3 + ⦠The angles made by this line with the +ve direactions of the coordinate axes: θx, θy, and θz are used to find the direction cosines of the line: cos θx, cos θy, and cos θz. If the direction cosines of AB are l, ⦠= 1 + 2, 2, 3 + 5 So, = 0 , = 90 , = 90 Direction cosines are l = cos 0 , m = cos 90 , n = cos 90 l = 1 , m = 0, n = 0 Direction cosines of x axis are 1, 0, 0. y axis makes an angle 90 with x axis, 0 with y axis & 90 with z axis. Note that the x, y and z-components of the segment AB are AD, CB and DC respectively. It can easily be shown that the direction cosines will satisfy the following equations: cos2ðx0;xÞþcos2ðx0;yÞþcos2ðx0; zÞ¼1 cos 2ðy0;xÞþcos2ðy0;yÞþcos ðy0;zÞ¼1 cos2ðz0;xÞþcos2ðz0;yÞþcos2ðz0;zÞ¼1 ð2:9Þ As an example, we now assume that stresses are known in the coordinate system (x, y, z), and we would like to find the transformed stresses in the new ⦠t. denotes this normal: 1 1 2 2 3 3. t (e. 3) = + t. e (7.2.1) Each of these components . Transfer of a skew ray from one surface to the next. Algorithm 4.4 (dcm_to_ypr.m in Appendix D.21) is used to determine the yaw, pitch, and roll angles for a given direction cosine matrix. Watch Direction cosines and Direction Ratios in 3D - Definitions & Examples - WORLD ENTERTAINMENT on Dailymotion Answer . Direction cosines = 3 32 + 2 2 + 82 , 2 32 + 2 2 + 82 , 8 32 + 2 2 + 82 Example 3: Finding the Direction Cosines of a Vector. = 3 9 + 4 + 64 , 2 9 + 4 + 64 , 8 9 + 4 + 64 The rotation used in this function is a passive transformation between two coordinate systems. We look at some solved examples based on what we learnt in the last lecture on Direction Cosines and Direction Ratios. Find the size of the largest angle. where l,m,n represent the direction cosines of the given vector on the axes X,Y,Z respectively. Find the angle at which L1 and L2 are inclined to each other respectively. Knowledge of the part of the solutions pertaining to this symmetry applies (with qualifications) to all such problems and it can be factored out of a specific problem at hand, thus reducing its complexity. 1) the direction cosines of a vector a = {ax ; ay} can be found using the following formula, In the case of the spatial problem (Fig. Mathematics Optional For UPSC. Example â 4. Example, 3 Find the direction cosines of the line passing through the two points ( 2, 4, 5) and (1, 2, 3). Let R, Sand Tbe the foots of the perpendiculars drawn from Pto the x, yand zaxes respectively. Login to view more pages. Direction ratios = (x2 x1), (y2 y1), (z2 z1) Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis. Direction cosines and direction ratios - example Point A has coordinates ( 3 , 5 ) , and point B has coordinates ( 7 , 8 ) . Solved Examples - Direction Cosines and Direction Ratios - I. OC vector = 1k vector. Terms of Service. Letâs start with a vector, $$\vec a$$, in three dimensional space. Answer: direction cosines of the vector a is cos α = 0.6, cos β = 0.8. Save. Find the direction cosines of the line segment joining $$A({x_1},\,{y_1},\,{z_1})$$ and $$B({x_2},\,{y_2},\,{z_2})$$ Solution: Refer to Fig - 4. 2. Since the direction cosinesof a line are defined as the differences between the X, Y, Zcoordinates of two points lying on the line divided by the distance between these points, it is clear from Figure 8.8that Figure 8.8. 7.1.1, the . Subscribe to our Youtube Channel - https://you.tube/teachoo. One given SAS and the other given SSS. As we can see from the previous two examples the two projections are different so be careful. Share. & x2 = 1, y2 = 2 , z2 = 3 Topic: Cosine. Solution: Let the points are A(2,1,2) and B(4,2,0). Identify direction ratios and direction cosines when components of a vector is given - example Example:- Find the direction ratios and cosines of the vector 6 i ^ + 2 j ^ â 3 k ^. AB vector = OB vector - OA vector = (j - i) vector. Calculate the direction cosines of the vector a: cos α = a x = 3 = 0.6 | a | 5: cos β = a y = 4 = 0.8 | a | 5: Answer: direction cosines of the vector a is cos α = 0.6, cos β = 0.8. A line is represented by a vector of unit length and a plane by its pole vector orits dip vector. The direction cosines are not independent of each other, they are related by the equation x 2 + y 2 + z 2 = 1, so direction cosines only have two degrees of freedom and can only represent direction and not orientation. C. c 2 = a 2 + b 2 â 2ab cos C = 0.067 â C = 86.2° How to use the cosine rule? Find the direction cosines of the medians. Example: In triangle ABC, a = 9 cm, b = 10 cm and c = 13 cm. Example 1. OB vector = 1j vector. The direction cosines uniquely set the direction of vector. Direction. We can clearly see that lr,mr,nr are in proportion to the direction cosines and these are called as the direction ratios and they are denoted by a,b,c. Q (1, 2, 3) Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Deducing Magnitude and Direction Cosines of a 3D Vector. x 2-x 1 = 4-2 = 2. y 2-y 1 = 2-1 = 1. z 2-z 1 = 0-2 = -2. Find the direction cosines of the vector a = {3; 4}. Author: Rodolfo Fernández de Lara Hadad. Lesson 8 of 54 ⢠14 upvotes ⢠14:52 mins. Solution : Let the given points be A (1, 0, 0) B (0, 1, 0) and C (0, 0, 1) OA vector = 1i vector. 2:39 mins. The direction angles of A are 9 0 â , 9 7 â, and 1 6 5 â. Direction Cosines and Direction Ratios Let Pbe a point in the space with coordinates (x, y, z) and of distance rfrom the origin. Asim Anand. He provides courses for Maths and Science at Teachoo. direction cosines ; This implies that we required only two independent parameters to describe a direction. Find the direction cosines of the line passing through the two points ( 2, 4, 5) and (1, 2, 3). Solution: The largest angle is the one facing the longest side, i.e. Vector Algebra - Vectors are fundamental in the physical sciences.In pure mathematics, a vector is any element of a vector space over some field and is often represented as a co Examples abound in classical mechanics and quantum mechanics. Direction Cosines. dcm = angle2dcm(___,rotationSequence) calculates the direction cosine matrix given three sets of rotation angles. 1. where the first subscript denotes the direction of the normal and the second denotes the direction of the to the component plane. Find the direction cosines of the vector that lies in the positive coordinate plane ð¥ ð§ and makes an angle of 6 0 â with the positive ð§ -axis. Example, 3 For example, in Fig. I designed this web site and wrote all the mathematical theory, online exercises, formulas and calculators. Overview. 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# Help Grasping Algebra
Introduction to help grasping algebra:
Algebra is a branch of the mathematics used to construct the mathematical model of the real-world situations . Algebra includes real numbers, complex numbers, linear equations, vectors etc. In algebra, we are often using the letters for represents the an unknown or variable. In this article we shall discuss about help grasping algebra.
Please express your views of this topic what is pre algebra by commenting on blog.
## Sample problem for help grasping algebra:
Grasping algebra help problem 1:
Theaverage of a list of 5 numbers is 20. If we are reducing a one of the numbers, the average value of the remaining numbers is 27. What is the number that was removed?
Solution:
Step 1: The removed number was might be obtained by the difference between the sum of original 5 numbers and the sum of remaining 4 numbers i.e.
Sum of original 5 numbers – sum of remaining 4 numbers
Step 2: Using the formula
Sum of Terms="Average*number" of Terms
Sum of original 5 numbers = 20 × 5 = 100
sum of remaining 18 numbers = 27× 4 = 108
Step 3: Using the formula from step 1 we get
Number removed = sum of the original 5 numbers – sum of remaining 18 numbers
100 – 108 = -8
Answer: The number removed is -8.
Between, if you have problem on these topics Simplify Algebraic Expressions , please browse expert math related websites for more help on Implicit Differentiation Rules.
Grasping algebra help problem 2:
Solving the given algebraic expression equation and find the value15 + 9(4 + 3)² − 17
Solution:
We have to calculate the values, which is the stand for, we will substitute 4 + 3 with 7:
= 15 + 9 * 7² − 17
Since there is now a just one number, 7, it is not necessary to write parentheses.
Next, evaluate the exponents:
= 15 + 9 * 49 − 17
Now multiply the exponent’s values
= 15 + 441 − 17
Finally, add or subtract, it will not matter. If we add first:
= 456 − 17
= 439.
## Practice problem for help grasping algebra:
• Solve the given algebraic equation 3 + 4(2 + 3)² − 4
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ALGEBRA - Order of Operations
by Stacy
HOW TO WORK OUT ans solve 40-(10-{4-2} )
E evaluate this expression using the PEMDAS order of operations.
P = Parentheses
E = Exponents
M = Multiplication
D = Division
S = Subtraction
Comments for ALGEBRA - Order of Operations
Sep 18, 2011 Evaluate Expression by: Staff ------------------------------------------------------- Part II Step 2: evaluate EXPONENTS PEMDAS order of evaluation for (simplified) overall expression: parentheses, EXPONENTS, multiplication, division, addition, subtraction There are no exponents to evaluate. Step 3: MULTIPLICATION PEMDAS order of evaluation: parentheses, exponents, MULTIPLICATION, division, addition, subtraction There is one multiplication: - (8) = (-1) * (8) = -8 A recap of the PEMDAS simplification process up to this point is shown below: 40 - ( 10 - {4 - 2} ) = 40 - ( 10 - {2} ) = 40 - (10 + (-1) * {2}) = 40 - (10 - 2) = 40 - (8) = 40 + (-1) * (8) = 40 - 8 Step 4: DIVISION PEMDAS order of evaluation: parentheses, exponents, multiplication, DIVISION, addition, subtraction There are no division operations to evaluate. Step 5: evaluate ADDITION & SUBTRACTION PEMDAS order of evaluation: parentheses, exponents, multiplication, division, ADDITION, SUBTRACTION 40 - 8 = 32 After completing the last step, you are finished. The final answer to your question is: 32 40 - ( 10 - {4 - 2} ) = 32 A step-by-step recap of the entire process is shown below: 40 - ( 10 - {4 - 2} ) = 40 - ( 10 - {2} ) = 40 - (10 + (-1) * {2}) = 40 - (10 - 2) = 40 - (8) = 40 + (-1) * (8) = 40 - 8 = 32 Thanks for writing. Staff www.solving-math-problems.com
Sep 18, 2011 Evaluate Expression by: Staff Part IThe question:by Stacy (Trinidad)HOW TO WORK OUT ans solve 40 - ( 10 - {4 - 2} )The answer: You can evaluate this expression using the PEMDAS order of operations. PEMDAS is an acronym. PEMDAS is an abbreviation for a six step process. Each letter stands for a different mathematical procedure:P = Parentheses E = ExponentsM = MultiplicationD = DivisionA = AdditionS = SubtractionThe order of the letters tells you which mathematical operation to complete first, second, . . . etc.Complete 1st = ParenthesesComplete 2nd = ExponentsComplete 3rd = MultiplicationComplete 4th = DivisionComplete 5th = AdditionComplete 6th = SubtractionStep 1: you should evaluate the expression inside each set of PARENTHESESPEMDAS order of evaluation: PARENTHESES, exponents, multiplication, division, addition, subtractionThere are two nested parentheses. Outer parentheses: (10 - { . . . } )andInner parentheses: (. . . {4 - 2})Evaluate the expression inside the inner parentheses (curly brackets in this case) first:{4 - 2}= {2}= 2After simplifying the expression inside the inner parentheses, your original expression looks like this:40 - ( 10 - {4 - 2} )= 40 - ( 10 - {2} )Evaluate the outer PARENTHESES PEMDAS applies inside parentheses . . . The step-by-step procedure for evaluating expressions inside parentheses, square brackets, or curly brackets is exactly the same as the step-by-step procedure for evaluating the overall expression . . .( 10 - {2} )= ( 10 + (-1) * {2} )= ( 10 - 2 )= ( 8 )A recap of the PEMDAS simplification process up to this point is shown below:40 - ( 10 - {4 - 2} )= 40 - ( 10 - {2} )= 40 - (10 + (-1) * {2})= 40 - (10 - 2)= 40 - (8)-------------------------------------------------------
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The present population of a town is 25000.
Question:
The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.
Solution:
Here,
$\mathrm{P}=$ Initial population $=25,000$
$\mathrm{R}_{1}=4 \%$
$\mathrm{R}_{2}=5 \%$
$\mathrm{R}_{3}=8 \%$
$\mathrm{n}=$ Number of years $=3$
$\therefore$ Population after three years $=\mathrm{P}\left(1+\frac{\mathrm{R}_{1}}{100}\right)\left(1+\frac{\mathrm{R}_{2}}{100}\right)\left(1+\frac{\mathrm{R}_{3}}{100}\right)$
$=25,000\left(1+\frac{4}{100}\right)\left(1+\frac{5}{100}\right)\left(1+\frac{8}{100}\right)$
$=25,000(1.04)(1.05)(1.08)$
$=29,484$
Hence, the population after three years will be 29,484 .
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## Tuesday, March 31, 2015
Suppose you want to determine the area bounded by two curves denoted by functions f(x) and g(x).
We find the points of intersection by setting f(x) = g(x). Then we determine which curve is the upper curve in the integral and which is the lower curve in the integral. Suppose the points of intersection are x=1 and x=3.
If f(x) is the upper curve then the integral looks like
Integral (1 to 3) [f(x) - g(x)] dx
If g(x) is the upper curve the the integral looks like
Integral (1 to 3) [g(x) - f(x)] dx
## Saturday, March 28, 2015
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## Saturday, March 21, 2015
Suppose you wish to obtain the exact value of sin(165). We know exact values of the following angles on the unit circle : 0, 30, 45, 60, 90, 120, 135, 150,. 180, 210, 225, 240, 270, 300, 315, 330, 360.
If we pick two angles from the unit circle that add to 165, we can obtain the exact value for sin(165). The reason is we can substitute those values in for this formula.
sin(a + b) = sin(a)cos(b)+cos(a)sin(b)
Notice that 120 + 45 = 165 and both of those values are on the unit circle
sin(120 + 45)= sin(120)cos(45) + cos(120)sin(45)
= (sqrt(3)/2)(sqrt(2)/2) + (-1/2)(sqrt(2)/2)
= sqrt(6)/4 - sqrt(2)/4
= (sqrt(6) - sqrt(2))/4
## Wednesday, March 18, 2015
Here's a great little video on hypothesis testing. It really explains the concept simply, in easy to understand terminology.
The Most Simple Introduction to Hypothesis Testing
## Thursday, March 12, 2015
Remember there is a difference between finding that Z value for a confidence interval and the Z (critical) value for tests and probability.
For example, suppose you want a 95 percent confidence interval for the sample mean. The formula is mean +/- standard error which is Z(alpha/2)standard deviation/square root(n). alpha is 1-.95 = .05. So the Z value is 1.96.
Now if we want an x value that 95% of the data falls below, then we need Z(.05) which is 1.645.
Be careful to understand the difference between the two.
## Friday, March 6, 2015
Notice the graph of a quadratic function f(x) = x^2 + 3x + 8 is a parabola that opens up. What if you know how to find the vertex but forget how to determine whether the vertex is the maximum or minimum point, therefore not knowing that is opens up or down. You can take the second derivative and set equal to zero to determine whether the function is concave up or concave down.
The first derivative f'(x) = 2x + 3
Second derivative f"(x) = 2
Notice the second derivative is positive for all values of x therefore the function is concave up, therefore opening up.
## Tuesday, March 3, 2015
When finding the area of a sector of a circle take the angle of the sector in degrees and divide by 360. Take the result and multiply by Pi times r^2.
For example, suppose the radius of a circle is 6 and the angle of the sector is 45 degrees. Therefore the area of the sector is
(45/360)Pi(6)(6) = (1/9)(36)Pi = 14.14
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# What does the range mean in maths?
What does the range mean in mathematics?
The range is the distinction in between the highest possible and also least expensive worths in a collection of numbers To locate it, deduct the least expensive number in the circulation from the highest possible.
What is the range in mathematics?
Regarding Records. The range is the distinction in between the biggest and also tiniest numbers The midrange is the standard of the biggest and also tiniest number.
What is range clarify? The range is the distinction in between the highest possible and also least expensive worths within a collection of numbers To determine range, deduct the tiniest number from the biggest number in the collection. … To locate the difference in an information collection, deduct each number from the mean, and afterwards square the result.
Just how do you locate the range mean?
Range amounts to optimal worth minus minimum worth which offers us: 12 − 2 = 10. To figure out the worth of the mean, get the overall of all the numbers and afterwards divide by the variety of numbers in the checklist.
## Is setting the highest possible number?
Setting: One of the most constant number– that is, the number that happens the highest possible variety of times Instance: The setting of 4, 2, 4, 3, 2, 2 is 2 since it happens 3 times, which is greater than any type of various other number.
## Exactly How do you do range?
The range is the distinction in between the tiniest and also highest possible numbers in a listing or collection. To locate the range, initially placed all the numbers in order. After that deduct (eliminate) the least expensive number from the highest possible The response offers you the range of the checklist.
## What is range provide instance?
The Range is the distinction in between the least expensive and also highest possible worths Instance: In 4, 6, 9, 3, 7 the least expensive worth is 3, and also the highest possible is 9. So the range is 9 − 3 = 6. It is that basic!
## Why is the range vital?
Range is fairly a valuable indicator of just how expanded the information is, however it has some severe constraints. This is since often information can have outliers that are extensively off the various other information factors. In these situations, the range may not provide a real indicator of the spread of information.
## Just how do we make use of range in day-to-day life?
Utilizing Range In Reality Range is utilized in the real world to make mathematical computations Range can be utilized to determine the quantity of time that has actually passed, like when determining your age. The existing year is 2020, and also you were birthed in 2005.
## Just how do I locate the mean worth?
The mean is the standard of the numbers. It is simple to determine: build up all the numbers, after that divide by the amount of numbers there are. Simply put it is the amount separated by the matter
## Just how do you exercise mean?
The mean is the overall of the numbers separated by the amount of numbers there are To locate the mean, include all the numbers with each other after that separate by the variety of numbers. The mean is 25.
## Which is the number that shows up the frequently?
The setting is the worth that shows up most regularly in an information collection. A collection of information might have one setting, greater than one setting, or no setting in any way.
## Just how do you exercise mean typical and also setting?
The mean indicates typical. To locate it, combine every one of your worths and also divide by the variety of addends The typical is the center variety of your information established when in order from the very least to best. The setting is the number that took place the frequently.
## What if there is no setting?
There is no setting when all observed worths show up the exact same variety of times in an information collection There is greater than one setting when the highest possible regularity was observed for greater than one worth in an information collection. In both of these situations, the setting can not be utilized to find the centre of the circulation.
## Just how setting is determined?
To locate the setting, or modal worth, it is finest to place the numbers in order. After that count the amount of of each number A number that shows up frequently is the setting.
## What are the use range?
Range is commonly utilized to define information spread Nonetheless, because it makes use of just 2 monitorings from the information, it is an inadequate procedure of information diffusion, other than when the example dimension is huge. Keep in mind that the range of Instances 1 and also 2 earlier are both equivalent to 4.
in
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# Surface area and volume of cylinders
1. Examples0/2 watched
1. 0/2
Now Playing:Surface area and volume of cylinders – Example 1a
Examples
0/2 watched
1. Find the surface area and the volume of the following cylinders:
Surface area and volume of prisms
Notes
A cylinder is like stacking up a pile of circle plates. With information such as, height, radius, and circumference, it is easy to calculate the surface area and volume of cylinders.Let's look at how to calculate the surface area and volume of composite solids too.
## Surface area and volume of a cylinder using simple formulas
Cylinders are shaped like a tube, with two circles at both ends that are identical and parallel. In geometry, you'll be required to use formulas in order to find the volume and surface area of this shape. You will encounter problems that may not provide you with all the info you'll need to easily plug in numbers into the formulas, and that's where your previous knowledge of geometry will come into play.
## How to find the surface area of a cylinder
We'll teach you how to find the surface area and the volume of cylinders through an example question. This is not too difficult as we have a nice equation for finding both of these. So for the cylinder in the top left corner below, let's find the surface area first.
To start off, we know that the diameter is equaled to 42 feet and the height is 110 feet. The units are consistent with one another, which is good. But we don't actually need the diameter all that much when it comes to the equation. Instead, we're going to find the radius using the diameter.
If you recall, it's simple to convert between a radius and diameter. In this case, just take the 42 and divide it by 2 and you get the radius which is equaled to 21 feet. Now we can take this and use it in the surface area formula.
## Surface area of a cylinder formula
To begin calculating the surface area of a cylinder, we know that the equation says to take $2 \pi r$-squared, then add $2 \pi r$ with the $h$, which is height.
What is the idea behind this formula? It's actually pretty straightforward. If you were to unwrap a cylinder, you will see that it's a circle on the top with a rectangular shape in the middle and followed by another circle. We've got a total of two circles and one rectangle. A circle's area is equal to $\pi r$ squared, and thus, with two circles, it is $2 \pi r$ squared. The $2 \pi r$ from the second part of the surface area equation is to find the circumference of the circle, which also happens to be the length of the rectangle. It'll help us find the surface area of the rectangle, which is the height multiplied by length, and therefore equal to $2 \pi r h$.
Now that we understand where the equation comes from, let's just plug the numbers that we do know into the equation. What is the cylinder surface area?
You should get:
Surface area
$= 2\pi r^2 + 2\pi rh$
= $2 \pi (21)^2+2 \pi (21)(110)$
=$17285 feet^2$
We've got the radius from the diameter previously (21 feet), and then we just use the height that was given to us, which was 110.
## Volume of a cylinder formula
Before we begin finding the volume, the below shows the volume equation that we'll be using. The volume of a cylinder is simply the area of the base multiplied with the height.
## How to find the volume of a cylinder
Going back to the diagram of the cylinder, the base is circular. The top piece is also circular and identical to the base. We know that the area of a circle is going to be $\pi r$-squared. The height is given, and we know it's once again, 110 feet. What is the cylinder volume?
Let's quickly fill in the information that we have and you should get $152398.7 feet^3$. Remember that volume is in cubes, and that the surface area is in squares. Now you've learned how to calculate the surface area and volume of a cylinder!
To see an interactive example of a cylinder's volume, try out this adjustable figure.
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1. ## word problem
"The sum of the first two digits of a three-digit number exceeds the third digit by 8. If the first two digits are interchanged the new number exceeds the original number by 270. If the number is divided by the sum of the digits the quotient is 31 and the remainder 15. Find the original number."
2. Originally Posted by biva
"The sum of the first two digits of a three-digit number exceeds the third digit by 8. If the first two digits are interchanged the new number exceeds the original number by 270. If the number is divided by the sum of the digits the quotient is 31 and the remainder 15. Find the original number."
Let the number be "xyz." That is to say the number is
$100x + 10y + z$
We know that
$x + y = z + 8$
$100y + 10x + z = (100x + 10y + z) + 270$
$\frac{100x + 10y + z}{x + y + z} = 31 + \frac{15}{x + y + z}$
$z = x + y - 8$
Inserting this into the other two expressions gives:
$100y + 10x + (x + y - 8) = (100x + 10y + (x + y - 8)) + 270$ ==> $101y + 11x - 8 = 101x + 11y + 262$
and
$\frac{100x + 10y + (x + y - 8)}{x + y + (x + y - 8)} = 31 + \frac{15}{x + y + (x + y - 8)}$ ==> $\frac{101x + 11y - 8}{2x + 2y - 8} = 31 + \frac{15}{2x + 2y - 8}$
Let's look at the new top equation:
$101y + 11x - 8 = 101x + 11y + 262$
$-90x + 90y = 270$
$90y = 90x + 270$
$y = x + 3$
Inserting this into the new bottom equation:
$\frac{101x + 11(x + 3) - 8}{2x + 2(x + 3) - 8} = 31 + \frac{15}{2x + 2(x + 3) - 8}$
$\frac{112x + 25}{4x - 2} = 31 + \frac{15}{4x - 2}$<-- Multiply both sides by 4x - 2
$112x + 25 = 31(4x - 2) + 15$
$112x + 25 = 124x - 62 + 15$
$-12x + 25 = - 47$
$-12x = -72$
$x = 6$
So
$y = x + 3$ ==> $y = 9$
and
$z = x + y - 8$ ==> $z = 7$
-Dan
3. Hello, biva!
Another approach . . . with fewer fractions.
The sum of the first two digits of a three-digit number exceeds the third digit by 8.
If the first two digits are interchanged, the new number exceeds the original number by 270.
If the number is divided by the sum of the digits, the quotient is 31 and the remainder 15.
Find the original number.
Let the digits be $x,\,y,\,z$.
The three-digit number is: $100x + 10y + z$
The first sentence says: . $x + y \:=\:z + 8\quad\Rightarrow\quad x + y - z \:=\:8$ .[1]
The second says: . $100y + 10x + z \:=\:100x+ 10y + z + 270\quad\Rightarrow\quad x - y \:=\:-3$ .[2]
The third says: . $\frac{100x + 10y + z}{x + y + z} \:=\:31 + \frac{15}{x + y + z}\quad\Rightarrow\quad23x - 7y - 10z \:=\:5$ .[3]
We have a system of equations.
From [2], we have: . $y \:=\:x + 3$ .[4]
Substitute into [1]: . $x + (x + 3) - z \:=\:8\quad\Rightarrow\quad 2x - z\:=\:5$ .[5]
Substitute into [3]: . $23x - 7(x+3) - 10z \:=\:5\quad\Rightarrow\quad 8x - 5z \:=\:13$ .[6]
Multiply [5] by -5: . $\text{-}10x + 5x \:=\:\text{-}25$
. . . . . . .Add [6]: . . $8x - 5x \:=\;13$
. . and we get: . $-2x \:=\:-12\quad\Rightarrow\quad\boxed{x \,=\,6}$
Substitute into [4]: . $y \:=\:6 + 3\quad\Rightarrow\quad\boxed{y \,=\,9}$
Substitute into [5]: . $2(6) - z \:=\:5\quad\Rightarrow\quad\boxed{z \,=\,7}$
Therefore, the original number is
697
.
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# Two charges 9e and 3e are placed at a separation r. The distance of the point where the electric field intensity will be zero, is:\eqalign{ & {\text{A) }}\dfrac{r}{{1 + \sqrt 3 }}{\text{from 9e charge}} \cr & {\text{B) }}\dfrac{r}{{1 + \sqrt {\dfrac{1}{3}} }}{\text{from 9e charge}} \cr & {\text{C) }}\dfrac{r}{{1 - \sqrt 3 }}{\text{from 3e charge}} \cr & {\text{D) }}\dfrac{r}{{1 + \sqrt {\dfrac{1}{3}} }}{\text{from 3e charge}} \cr}
Last updated date: 10th Sep 2024
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Hint: Here in this question we need to find the neutral point , where the force experienced by a unit positive charge is zero , A neutral point is a point at which the resultant magnetic field is zero and it is obtained when horizontal component of earth’s field is balanced by the produced magnet.
Complete step by step solution:
Two charges 9e and 3e separation=r
Let the distance at which electric field intensity from 9e charge = 0, be x and from 3e charge it will be r-x.
Due to a system of two like point charge, the electric field of both the charges at neutral point will be equal
\eqalign{ & \Rightarrow \dfrac{1}{{4\Pi {\varepsilon _o}}} \times \dfrac{{9e \times 1}}{{{a^2}}} = \dfrac{1}{{4\Pi {\varepsilon _o}}} \times \dfrac{{3e}}{{{{(r - x)}^2}}} \cr & \Rightarrow 3{(r - x)^2} = {x^2} \cr & {\text{solving we get,}} \cr & \Rightarrow \sqrt 3 (r - x) = x \cr & or,\sqrt 3 r - \sqrt 3 x = x \cr & \Rightarrow \sqrt 3 r = x(1 + \sqrt 3 ) \cr & \therefore x = \dfrac{{r\sqrt 3 }}{{1 + \sqrt 3 }}{\text{ from 9e charge}} \cr}
Hence, the option B is right.
Additional Information: The electric field at a given distance from a point charge is a vector amount, which points away from the positive charge and towards the negative charge. Its magnitude follows the inverse square law: it is proportional to the charge and inversely proportional to the distance.
The zero field location has to be on a line running between two point charges because it is the only place where field vectors can point in exactly opposite directions. This cannot occur between two opposite charges because field vectors from both charges point towards negative charge. It is to be on one side or the other of the two, where the vectors point in opposite directions.
Note: If both charges are the same, then the resulting potential is not zero at any finite point because such charges will have the same sign on the potentials and can therefore never be added to zero so such a point is only at infinity.
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Short-Question
# What is the remainder of 677 divided by 8?
## What is the remainder of 677 divided by 8?
Using a calculator, if you typed in 677 divided by 8, you’d get 84.625. You could also express 677/8 as a mixed fraction: 84 5/8.
What is the remainder of 8 divided by 2?
Using a calculator, if you typed in 8 divided by 2, you’d get 4. You could also express 8/2 as a mixed fraction: 4 0/2. If you look at the mixed fraction 4 0/2, you’ll see that the numerator is the same as the remainder (0), the denominator is our original divisor (2), and the whole number is our final answer (4).
### What is the remainder of 47 8?
The result of division of 478 is 5 with a remainder of 7 .
How do you do long hand division?
How to Do Long Division?
1. Step 1: Take the first digit of the dividend.
2. Step 2: Then divide it by the divisor and write the answer on top as the quotient.
3. Step 3: Subtract the result from the digit and write the difference below.
4. Step 4: Bring down the next number (if present).
5. Step 5: Repeat the same process.
#### How to calculate the remainder in a calculator?
The procedure to use the remainder calculator is as follows: 1 Enter the dividend and divisor in the respective input field 2 Now click the button “Solve ” to get the remainder 3 Finally, the remainder and quotient will be displayed in the output field
Which is the remainder of the multiple of 6?
The remainder is 2. To work this out, find the largest multiple of 6 that is less than 26. In this case it’s 24. Then subtract the 24 from 26 to get the remainder, which is 2. What is the remainder when 599 is divided by 9?
## What’s the difference between the quotient and the remainder?
What is the quotient and the remainder? The quotient is the number of times a division is completed fully, while the remainder is the amount that is left that doesn’t fully go into the divisor. For example, 127 divided by 3 is 42 R 1, so 42 is the quotient and 1 is the remainder.
What is the remainder when 599 is divided by 9?
What is the remainder when 599 is divided by 9? The remainder is 5. To calculate this, first divide 599 by 9 to get the largest multiple of 9 before 599. 5/9 < 1, so carry the 5 to the tens, 59/9 = 6 r 5, so carry the 5 to the digits. 59/9 = 6 r 5 again, so the largest multiple is 66.
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# Three ways to write expressions
(- (- (+ 4 (* 2 7)) (* (- 1 6) 5)) 1)
is the same as
4 + 2 * 7 - (1 - 6) * 5 - 1
is the same as
4 2 7 * + 1 6 - 5 * - 1 -
Not really. More like prefix, infix, and reverse polish / postfix notation. Basically, they’re all just ways of writing down mathematical concepts.
You can really see it if you start with the general structure that all three share:
That’s what’s called an expression tree. Each node is either a number (the boxes) or a function of two numbers (the circles). To calculate the final value of the tree, you can either think of it by starting at the top or the bottom and working recursively. If you start at the top, whenever you have an operator, branch and work out the value of the two child trees. When you have those, apply the function. If you start at the bottom, repeatedly find branches that have just a leaf on each side and simplify them. Either way, you’ll get the same answer.
From the expression tree, you can easily get the three expressions that we were dealing with earlier. Just take the preorder, inorder, or postorder traversal of the tree (I’ll cover those algorithms at some point as well).
Then to evaluate the tree, you’ll get something like this (using the prefix notation as it’s textual representation is most similar to the tree structure):
(- (- (+ 4 (* 2 7)) (* (- 1 6) 5)) 1)
= (- (- (+ 4 14) (* -5 5)) 1)
= (- (- 18 -25) 1)
= (- 43 1)
43
Basically, you end up repeated end up rewriting expressions from the inside out until you’re left with the final answer. The same sort of thing happens with the infix notation:
4 + 2 * 7 - (1 - 6) * 5 - 1
= 4 + 14 - -5 * 5 - 1
= 4 + 14 - -25 - 1
= 18 - -25 - 1
= 43 - 1
= 42
This time it’s a little more complicated to figure out which expressions are on the “inside”, but on the flip side it’s easier to understand since we’re all used to working with infix expressions, we’ve been doing them since grade school after all.
What about the Reverse Polish Notation?
4 2 7 * + 1 6 - 5 * - 1 -
= 4 14 + -5 5 * - 1 -
= 28 25 - 1 -
= 43 1 -
= 42
Same thing. Just this time, you’re looking for two numbers followed by an operator which you can then reduce.
So what if a computer wanted to do any of these operations?
Note: In order to make this a fair comparison, the input to each will be a list of either numbers or functions that take two numbers and return one. This means that the prefix code at least will be a little more complicated than just calling eval.
(define (car+cdr ls)
(values (car ls) (cdr ls)))
(define (prefix-calc main-expr)
(define (step expr)
(cond
[(number? (car expr))
(values (car expr) (cdr expr))]
[else
(let*-values ([(op rest) (car+cdr expr)]
[(lv rest) (step rest)]
[(rv rest) (step rest)])
(values ((eval op) lv rv) rest))]))
(let-values ([(val rest) (step main-expr)])
(if (null? rest)
val
(error 'prefix-calc
(format "malformed expression ~s: ~s left over" main-expr rest)))))
~ (prefix-calc '(+ * 5 8 + 5 - 1 4))
42
There are a few interesting bits here, so let’s go through them one by one. The first thing is the helper function, step. Essentially, step’s job is to take one “step” in the given expression. If the first thing is a number, just return it. If it’s an operator we need to find the value of the two sub-expressions and then apply it. To do that, I’ve defined step to take in an expression and return a pair of values: the numeric result of the step and any part of the expression that we did not consume.
That’s where the let*-values comes in. Basically, it threads the three states that the code can be in together:
(let*-values ([(op rest) (car+cdr expr)]
[(lv rest) (step rest)]
[(rv rest) (step rest)])
...)
The first line pulls apart the operator and the rest. The second line applies step again, starting at the cdr of the original list. If step does everything that it’s supposed to (the promise of recursion), then we should get back a value from the front of the list (as lv) and any part of the list we didn’t process (as the first value of rest). We can then feed that rest right back into step, resulting in another value (the right side) and anything that didn’t get processed to give us that. We now have an operator and it’s two arguments lv and rv and anything we didn’t use to calculate any of those (the final value of rest), so return those two values.
Easy enough right? Well, it turns out that combining that with a base case that deals with single numbers is all that you need. Recursion takes care of the rest. But all of that power doesn’t come for free. The drawback of doing this is that you need to allocate a stack frame each time you make the recursive call. Worse yet, it’s not tail recursive so you can’t even make that optimization. Still, it’s a neat bit of code.
The next one gets even more interesting. Here, we have to deal with infix notation. It may be the one that we’ve all grown up with, but it’s actually generally considered the hardest of the three to deal with from a computer’s perspective. Mostly because we have to deal with precedence and associativity to really do it correctly.
The way that I’m going to do it is to repeatedly go through the given expression and evaluate things based on their precedence. So I’ll find the leftmost multiplication or division and combine it, repeating until there are none left. If we didn’t find any, try the leftmost addition or subtraction. After that, go back and try multiplication and division again. It’s definitely not an efficient way to do it, but it does work.
First, the code that will find and reduce one expression of a given type:
(define (step ls which)
(cond
[(number? ls) ls]
[(or (null? ls) (null? (cdr ls)) (null? (cddr ls))) ls]
[else
(cons (car ls) (step (cdr ls) which))]))
~ (step '(1 + 2 * 3 - 4) '(* /))
(1 + 6 - 4)
Then, a function that will repeatedly step, taking precedence into account. This is the slow part. 😄
(define (step-all ls)
(let ([step1 (step ls '(* /))])
(if (equal? ls step1)
(let ([step2 (step step1 '(+ -))])
(if (equal? step1 step2)
(car step2)
(step-all step2)))
(step-all step1))))
~ (step-all '(1 + 2 * 3 - 4))
3
And finally, the glue that holds the parts together:
(define (infix-calc expr)
(define (step ls which) ...)
(define (step-all ls) ...)
(step-all expr))
~ (infix-calc '(4 + 2 * 7 - 1 * 5 + 6 * 5 - 1))
42
(Yes, I had to distribute the 5 at one point. Without doing so, there’s no way to force the subtraction to go first and the expression doesn’t have the same meaning.)
Whew. That was a fun one. Luckily, I’ve saved the easiest one for last. Here, we have the Reverse Polish Notation. As a side note, you have may have guessed that since the other two were prefix and infix notation, another name for RPN is postfix notation.
The algorithm for this one is neat because all you need to keep track of the operation is a stack of values. Whenever you see a number, you push it onto the stack. When you see a function, you pop two values, apply the function, and push the result back onto the stack. Easy as pie. And as an added benefit, you can easily write it in an iterative / tail-recursive manner that avoids allocated stack frames so all the extra memory you need is in the value stack.
(define (postfix-calc expr)
(define (step expr stack)
(cond
[(null? expr) (car stack)]
[(number? (car expr))
(step (cdr expr)
(cons (car expr) stack))]
[else
(step (cdr expr)
(cons ((eval (car expr))
(car stack))
(cddr stack)))]))
(step expr '()))
~ (postfix-calc '(4 2 7 * + 1 6 - 5 * - 1 -))
42
Pretty much the only thing you have to watch out for in this code is the actual evaluation:
((eval (car expr)) (cadr stack) (car stack))
Because you pushed the first argument onto the stack first, you have to reverse them after popping them (thus cadr before car). This doesn’t matter for + or *, so it’s easy to miss, but it will really mess up your day when you’re working with - or /.
So there you have it. Three ways to write expressions and at least one way a computer could deal with each of them.
Before I go, I bet you were wondering why postfix notation is more commonly called Reverse Polish Notation. It’s mostly because of Jan Łukasiewicz, a Polish logician and philosopher from first half of the 1900s. Essentially, he described (prefix) Polish notation in the 1920s but for whatever reason it stuck on the postfix notation, ergo reverse. And now you know.
The goal tomorrow is to create a web-based simulation of each of these three algorithms that shows the internal works, step by step. We’ll see how that goes.
If you’d like to download today’s source, you can do so here: source code
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Become a math whiz with AI Tutoring, Practice Questions & more.
HotmathMath Homework. Do It Faster, Learn It Better.
Quadratic regression is a statistical method used to model a relationship between variables with a parabolic best-fit curve, rather than a straight line. It's ideal when the data relationship appears curvilinear. The goal is to fit a quadratic equation $y=a{x}^{2}+bx+c$ to the observed data, providing a nuanced model of the relationship. Contrary to historical or biological connotations, "regression" in this mathematical context refers to advancing our understanding of complex relationships among variables, particularly when data follows a curvilinear pattern.
We have already covered linear regression, which we use to predict the value of a variable based on the value of another variable. This gives us the "line of best fit" and allows us to make accurate predictions about our data.
The same general concept applies to quadratic regression. The only difference is that instead of trying to find the line of best fit, we're trying to find the parabola of best fit.
So how exactly does this work?
Instead of getting a simple linear equation, quadratic regression leaves us with this:
If you already know about the "least squares method," you can use the exact same process with quadratic regression.
We need to find the values of $a$ , $b$ , and $c$ so that the squared vertical distance between each point $\left({x}_{i},{y}_{i}\right)$ and the quadratic curve $y=a{x}^{2}+bx+c$ is minimal.
We can use a matrix equation to represent our quadratic curve:
We can also use these formulas to find the values of $a$ , $b$ , and $c$ :
In these equations:
• $x$ and $y$ are our variables
• $a$ , $b$ , and $c$ are our coefficients for our quadratic equation
• $n=$ the number of elements
• $\sum x=$ the sum of $x$ values
• $\sum y=$ = the sum of $y$ values
• $\sum {x}^{2}=$ the sum of the squares of $x$ values
• $\sum {x}^{3}=$ the sum of the cubes of $x$ values
• $\sum {x}^{4}=$ the sum of the fourth powers of the $x$ values
These calculations can be represented by a system of equations in a matrix form, and the coefficients a, b, and c can be determined using these equations. It's important to note that the equations and methods used to calculate these coefficients can be complex, involving several sums and squares of various combinations of the data points. Thus, these computations are typically done with the assistance of software.
Note that the relative predictive power of a quadratic model is denoted by R2. Our predictive power tells us how accurate our predictions truly are. It is the power of a scientific theory to generate testable predictions. In the case of quadratic regression, this applies to the predictions related to our parabola.
The formula for our predictive power is as follows:
${R}^{2}=1-\frac{\mathrm{SSE}}{\mathrm{SST}}$
In this formula:
$\mathrm{SSE}=\sum _{i=1}^{n}{\left({y}_{i}-a-b{x}_{i}-c{{x}_{i}}^{2}\right)}^{2}$
$\mathrm{SST}=\sum _{i=1}^{n}{\left({y}_{i}-\stackrel{̄}{y}\right)}^{2}$
Our relative predictive power ${R}^{2}$ lies anywhere between 0 and 1. The closer it is to 1, the more accurate our model is.
We should also note that the "bar" above the y value in the SST equation represents the average of all y values.
As we might have guessed, these calculations can become quite complex and tedious. We have just gone over a few very detailed formulas, but the truth is that we can handle these calculations with a graphing calculator. This saves us from having to go through so many steps -- but we still must understand the core concepts at play.
Let's try a practice problem that includes quadratic regression. Consider the following set of data:
$\left\{\left(-3,7.5\right),\left(-2,3\right),\left(-1,0.5\right),\left(0,1\right),\left(1,3\right),\left(2,6\right),\left(3,14\right)\right\}$
Can we determine the quadratic regression for this set?
Our first step is to enter our x-coordinates and y-coordinates into our graphing calculator. We can then carry out our operation for a quadratic equation. This will give us the equation of the parabola that best approximates the points:
$y=1.1071{x}^{2}+x+0.5714$
Great! Now all we need to do is plot our graph. We should be left with something like this:
We also know that our relative predictive power ( ${R}^{2}$ ) is 0.9942. That's pretty accurate -- and it tells us that our calculations for quadratic regression worked!
## Flashcards covering the Quadratic Regression
Statistics Flashcards
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## Quotient and Remainder calculator with steps
Qutient and Remainder
Use this calculator to find the remainder by dividing any two integers. In this article, I will guide you through the parts of Division and how to get quotient and remainder.
## Parts of Division
This is what a typical division equation looks like.
a ¸ b = q R x
### Dividend
This refers to the number that is being divided in a division equation. For example, a is the dividend from the above equation.
The formula for finding the dividend is: Dividend = Divisor × Quotient + Remainder
### Divisor
This refers to the number you divide the dividend by. In the example above, b is the dividend.
### Quotient
It refers to the answer you get from dividing the dividend by the divisor, rounded off to the nearest integer.
Formula: Quotient = Dividend ¸ Divisor
### Remainder
Refers to the remainder
### Example 1
Find the remainder of the dividend is 90, the divisor is 6, and the quotient is 15.
In this example, we will use the dividend, divisor, quotient, and remainder formula to find the remainder
Dividend = Divisor × Quotient + Remainder
90= 6 × 15 + r
90 = 90 + r
R = 90 – 90
r = 0
The remainder is 0
### Example 2
Find the Quotient. If the Dividend is 13, the divisor is 2, and the remainder is 1
In this example, we will use long division to find the quotient since we have both the dividend and the divisor.
## How to calculate the remainder step by step
1. Write down the problem
For example Divide 351 by 4
2. Distinguish the divisor and the dividend
Dividend = 351
Divisor = 4
3. Perform division and write down your answer. Include the decimals too if any.
351 ¸ 4 = 87.75
4. Round down the number to get the quotient 87
5. Multiply the answer in step 4 by the divisor
87 × 4 = 348
6. Subtract the number in step 5 from the dividend to get the remainder
351 – 348 = 3
You can save time by using our quotient remainder calculator.
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# Question Video: Converting Units of Atmospheric Pressure Physics
If the pressure of a contained gas in a tank is 2.3 atm, what is its pressure in centimeters of mercury? Let π_(π) = 76 cm Hg.
02:48
### Video Transcript
If the pressure of a contained gas in a tank is 2.3 atmospheres, what is its pressure in centimeters of mercury? Let π sub π equal 76 centimeters of mercury. (A) 174.8 centimeters of mercury, (B) 33.0 centimeters of mercury, (C) 141.8 centimeters of mercury, or (D) 93.1 centimeters of mercury.
To answer this question, we need to convert the value of the gas pressure from units of atmospheres to units of centimeters of mercury. We are told that the value of π sub π, which stands for atmospheric pressure, is equal to 76 centimeters of mercury. This means that in normal atmospheric pressure, which is equal to one atmosphere, the mercury in a barometer will be at a height of 76 centimeters. But if the barometer were placed in a higher-pressure environment, the mercury would rise higher up the tube. This is because there is a greater pressure pushing down on the mercury in the dish and pushing it up the tube. So, a higher pressure corresponds to a greater number of centimeters of mercury.
To answer this question, we need to work out how many centimeters of mercury is equivalent to 2.3 atmospheres.
Letβs start by writing down what we already know. At atmospheric pressure, the pressure is equal to one atmosphere, which is equivalent to 76 centimeters of mercury. In the gas tank, the pressure is equal to 2.3 atmospheres. In other words, the pressure in the gas tank is 2.3 times higher than the atmospheric pressure. So, the value of the pressure, when given in units of centimeters of mercury, must also be 2.3 times higher than at atmospheric pressure.
So, in units of centimeters of mercury, the pressure of the gas in the tank is simply equal to 76 centimeters of mercury multiplied by 2.3. This gives us a pressure of 174.8 centimeters of mercury. So, the correct answer to this question is option (A).
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# 7-5 Dilations Warm Up Problem of the Day Lesson Presentation
## Presentation on theme: "7-5 Dilations Warm Up Problem of the Day Lesson Presentation"— Presentation transcript:
7-5 Dilations Warm Up Problem of the Day Lesson Presentation
Pre-Algebra
7-5 Dilations Warm Up Multiply. 1. 4 2. 12 3. 24 4. –36 3 9 18
Pre-Algebra 7-5 Dilations Warm Up Multiply. 1. 4 3. 24 –36 3 4 3 4 3 9 3 4 3 4 18 –27 5. 4 2.5 10 30
Problem of the Day Every day, a plant grows to three times its size. Every night, it shrinks to half its size. After three days and nights, it is 6.75 in. tall. How tall was the plant at the start? 2 in.
Learn to identify and create dilations of plane figures.
Vocabulary dilation scale factor center of dilation
Your pupils are the black areas in the center of your eyes
Your pupils are the black areas in the center of your eyes. When you go to the eye doctor, the doctor may dilate your pupils, which makes them larger.
Translations, reflections, and rotations are transformations that do not change the size or shape of a figure. A dilation is a transformation that changes the size, but not the shape, of a figure. A dilation can enlarge or reduce a figure.
A scale factor describes how much a figure is enlarged or reduced
A scale factor describes how much a figure is enlarged or reduced. A scale factor can be expressed as a decimal, fraction, or percent. A 10% increase is a scale factor of 1.1, and a 10% decrease is a scale factor of 0.9.
A scale factor between 0 and 1 reduces a figure
A scale factor between 0 and 1 reduces a figure. A scale factor greater than 1 enlarges it. Helpful Hint
Additional Example 1A & 1B: Identifying Dilations
Tell whether each transformation is a dilation. A. B. The transformation is not a dilation. The figure is distorted. The transformation is a dilation.
Additional Example 1C & 1D: Identifying Dilations
Tell whether each transformation is a dilation. D. C. The transformation is not a dilation. The figure is distorted. The transformation is a dilation.
Tell whether each transformation is a dilation.
Try This: Example 1A & 1B Tell whether each transformation is a dilation. B A C A. A B C A' B' C' B. A' B' C' The transformation is a dilation. The transformation is not a dilation. The figure is distorted.
Tell whether each transformation is a dilation.
Try This: Example 1C & 1D Tell whether each transformation is a dilation. D. A' B' C' A B C C. A' B' C' A B C The transformation is not a dilation. The figure is distorted. The transformation is a dilation.
Every dilation has a fixed point that is the center of dilation
Every dilation has a fixed point that is the center of dilation. To find the center of dilation, draw a line that connects each pair of corresponding vertices. The lines intersect at one point. This point is the center of dilation.
Additional Example 2: Dilating a Figure
Dilate the figure by a scale factor of 1.5 with P as the center of dilation. Multiply each side by 1.5.
Try This: Example 2 Dilate the figure by a scale factor of 0.5 with G as the center of dilation. G F H 2 cm F’ H’ 1 cm G 2 cm 2 cm F H 2 cm Multiply each side by 0.5.
Additional Example 3A: Using the Origin as the Center of Dilation
Dilate the figure in Example 3A on page 363 by a scale factor of 2. What are the vertices of the image? Multiply the coordinates by 2 to find the vertices of the image. A’B’C’ ABC A(4, 8) A’(4 2, 8 2) A’(8, 16) B(3, 2) B’(3 2, 2 2) B’(6, 4) C(5, 2) C’(5 2, 2 2) C’(10, 4) The vertices of the image are A’(8, 16), B’(6, 4), and C’(10, 4).
Additional Example 3B: Using the Origin as the Center of Dilation
Dilate the figure in Example 3B by a scale factor of . What are the vertices of the image? 1 3 Multiply the coordinates by to find the vertices of the image. 1 3 A’B’C’ ABC A(3, 9) A’(3 , 9 ) A’(1, 3) 1 3 B(9, 6) B’(9 , 6 ) B’(3, 2) 1 3 C(6, 3) C’(6 , 3 ) C’(2, 1) 1 3 The vertices of the image are A’(1, 3), B’(3, 2), and C’(2, 1).
Try This: Example 3A Dilate the figure by a scale factor of 2. What are the vertices of the image? 10 8 6 C 4 2 A B 2 4 6 8 10
Try This: Example 3A Continued
A’B’C’ ABC A(2, 2) A’(2 2, 2 2) A’(4, 4) B(4, 2) B’(4 2, 2 2) B’(8, 4) C(2, 4) C’(2 2, 4 2) C’(4, 8) The vertices of the image are A’(4, 4), B’(8, 4), and C’(4, 8).
Try This: Example 3A Continued
10 B’ C’ A’ 8 6 C 4 2 A B 2 4 6 8 10
Try This: Example 3B Dilate the figure by a scale factor of What are the vertices of the image? 10 C 8 6 A B 4 2 2 4 6 8 10
Try This: Example 3B Continued
A’B’C’ ABC A(4, 5) A’(4 0.5, 5 0.5) A’(2, 2.5) B(8, 5) B’(8 0.5, 5 0.5) B’(4, 2.5) C(4, 9) C’(4 0.5, 9 0.5) C’(2, 4.5) The vertices of the image are A’(2, 2.5), B’(4, 2.5), and C’(2, 4.5).
Try This: Example 3B Continued
10 C 8 6 B’ C’ A’ A B 4 2 2 4 6 8 10
1. Tell whether the transformation is a dilation.
Lesson Quiz 1. Tell whether the transformation is a dilation. A(0, 4) B(5,5) C(3,3) A’(0, 8) B’(10, 10) C’(6, 6) 2 4 6 2. Dilate the figure by a scale factor of 1.5 with P as the center of dilation. P A B C -2 3. Dilate the figure by a scale factor of 2 with the origin as the center of dilation. What are the coordinates of the image? A(2,4) B(5,6) C(6,1) -4 -6
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# Lesson Notes By Weeks and Term - Junior Secondary School 3
BINARY NUMBERS (BASE 2 NUMBERS)
SUBJECT: MATHEMATICS
CLASS: JSS 3
DATE:
TERM: 1st TERM
REFERENCE BOOKS
• New General Mathematics by M. F Macrae et al bk 3
• Essential Maths by AJS OluwasanmiBk 3
WEEK THREE
TOPIC: BINARY NUMBERS (BASE 2 NUMBERS)
• Addition in base 2
• Subtraction in base 2
• Multiplication & Division in base 2
ADDITION IN BASE TWO
We can add binary numbers in the same way as we separate with ordinary base 10 numbers.
The identities to remember are:-
0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 10, 1 + 1 + 1 = 11, 1 + 1 + 1 + 1 = 100
Worked Examples
Simplify the following
1. 1110 + 1001 2. 1111 + 1101 + 101
Solutions:
1. 1110
+ 1001
10111
1. 1111
+ 1101
101
100001
Note: 11 take 1 carry 1
10 take 0 carry1
100 take 0 carry 10
EVALUATION
1. Simplify the following 101 + 101 +111
2. 10101 + 111
ADDITION IN BICIMALS
In bicimals, the binary point are placed underneath each other exactly the same way like ordinary decimals.
Example:
1. 1.1011two + 10.1001two + 10.01
2. 10.001two+ 101.111
Solution:
1. 1 . 1011
10 . 1001
10 .0100
1. 1000two
1. 101.111
10.001
1000.000
SUBTRACTION IN BASE TWO
The identities to remember on subtraction are: 0 - 0 = 0, 1 - 0 = 1, 10 - 1 = 1, 11 - 1 = 10, 100 - 1 = 11
Worked Examples
Simplify the following:-
(a) 1110 - 1001 (b) 101010 - 111
Solutions:
(a) 1110
- 1001
101
(b) 101010
- 111
1110
SUBTRACTION IN BICIMAL
Example
101.101two – 11.011two
101.101
11.011
10.010two
EVALUATION
1. 10111÷110
2. 10001 x 11
READING ASSIGNMENT
New Gen Maths Book 3, chapter 1 Exercise 1e pg 18 Nos 1-12
Essential Mathematics for J.S.S.3 Pg 8-10
WEEKEND ASSIGNMENT
1. Express 3426 as a number in base 10. (a) 342 (b) 3420 (c) 134
2. Change the number 10010 to base 10 (a) 18 (b) 34 (c) 40
3. Express in base two the square of 11 (a) 1001 (b) 1010 (c) 1011
4. Find the value of (101)2 in base two (a) 1010 (b) 1111 (c) 1001
5. Multiply 1000012 by 11 (a) 1001 (b) 1100011 (c) 10111
THEORY
1. Calculate 1102 x (10112 + 10012 – 1012)
2. Convert 110111 to base five
© Lesson Notes All Rights Reserved 2023
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# 2.15: Graphical Solutions to Absolute Value Inequalities
Difficulty Level: Advanced Created by: CK-12
Estimated6 minsto complete
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Progress
Practice Graphs of Absolute Value Inequalities
Progress
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Solve the following inequality and graph the solution on a number line.
|x+2|3\begin{align*}|x+2|\le 3\end{align*}
### Guidance
Recall that you can graph linear inequalities on number lines. For x>5\begin{align*}x > 5\end{align*}, the graph can be shown as:
Notice that there is only one solution set and therefore one section of the number line has the region shown in red.
What do you think would happen with absolute value linear inequalities? With absolute value linear inequalities, there are two inequalities to solve. Therefore there can be two sections of the number line showing solutions.
For |t|>5\begin{align*}|t|>5\end{align*}, you would actually solve for t>5\begin{align*}t > 5\end{align*} and t<5\begin{align*}t <-5\end{align*}. If you were to graph this solution on a number line it would look like the following:
The solution is t>5\begin{align*} t>5\end{align*} OR t<5\begin{align*}t<-5\end{align*}.
For |t|<5\begin{align*}|t|<5\end{align*}, you would actually solve for t<5\begin{align*}t < 5\end{align*} and t>5\begin{align*}t >-5\end{align*}. If you were to graph this solution on a number line it would look like the following:
The solution is 5<t<5\begin{align*} -5. This is the same as t<5\begin{align*}t<5\end{align*} AND t>5\begin{align*}t>-5\end{align*}.
Graphing the solution set to an absolute value linear inequality gives you the same visual representation as you had when graphing the solution set to linear inequalities. The same rules apply when graphing absolute values of linear inequalities on a real number line. Once the solution is found, the open circle is used for absolute value inequalities containing the symbols > and <. The closed circle is used for absolute value inequalities containing the symbols \begin{align*}\le\end{align*} and \begin{align*}\ge\end{align*}.
#### Example A
Represent the solution set to the following inequality on a number line: |2x|6\begin{align*}|2x|\ge 6\end{align*}.
Solution: First solve the inequality. Then, represent your solution on a number line.
|2x|2x2x2x2x2x2x66623OR6623(Divide by 2 to isolate and solve for the variable)(Simplify)(Divide by 2 to isolate and solve for the variable)(Simplify)\begin{align*}|2x| &\ge 6\\ 2x & \ge 6\\ \frac{2x}{{\color{red}2}} & \ge \frac{6}{{\color{red}2}} && (\text{Divide by 2 to isolate and solve for the variable})\\ x & \ge 3 && (\text{Simplify})\\ & OR\\ 2x & \le -6\\ \frac{2x}{{\color{red}2}} & \le \frac{-6}{{\color{red}2}} && (\text{Divide by 2 to isolate and solve for the variable})\\ x & \le -3 && (\text{Simplify})\end{align*}
The solution sets are x3\begin{align*}x \ge 3\end{align*} OR x3\begin{align*} x \le -3\end{align*}.
#### Example B
Solve the following inequality and graph the solution on a number line: |x+1|>3\begin{align*}|x+1|>3\end{align*}
Solution: First solve the inequality. Then, represent your solution on a number line.
|x+1|x+1x+11xx+1x+11x>3>3>31>2OR<3<31<4(Divide both sides by 2 to solve for the variable)(Subtract 1 from both sides of the inequality sign)(Subtract 1 from both sides of the inequality sign)\begin{align*}|x+1| &> 3 && (\text{Divide both sides by 2 to solve for the variable})\\ x+1 &>3\\ x+1{\color{red}-1} &> 3{\color{red}-1} && (\text{Subtract 1 from both sides of the inequality sign})\\ x & > 2\\ & OR\\ x+1 &< -3\\ x+1{\color{red}-1} &< -3{\color{red}-1} && (\text{Subtract 1 from both sides of the inequality sign})\\ x & < -4\end{align*}
The solution sets are x>2\begin{align*}x>2\end{align*}, OR x<4\begin{align*}x<-4\end{align*}.
#### Example C
Solve the following inequality and graph the solution on a number line: x52<1\begin{align*}\bigg |x-\frac{5}{2} \bigg | < 1\end{align*}
Solution: First solve the inequality. Then, represent your solution on a number line.
x52x52(22)x522x2522x52x5+52x2x2xx52(22)x522x2522x52x5+52x2x2x<1<1<(22)1<22<2<2+5<7<72<72OR>1>(22)(1)<22>2>2+5>3>32>32(Simplify)(Add 5 to isolate the variable)(Simplify)(Multiply to get common denominator (LCD=2))(Simplify)(Simplify)(Add 5 to isolate the variable)(Simplify)(Divide both sides by 2 to solve for the variable)\begin{align*}\bigg |x-\frac{5}{2} \bigg | &< 1\\ x-\frac{5}{2} &< 1\\ \left({\color{red}\frac{2}{2}}\right)x-\frac{5}{2} &< \left(\frac{{\color{red}2}}{{\color{red}2}}\right)1\\ \frac{2x}{2}-\frac{5}{2}&<\frac{2}{2}\\ 2x-5&<2 && (\text{Simplify})\\ 2x-5{\color{red}+5} &< 2{\color{red}+5} && ( \text{Add 5 to isolate the variable})\\ 2x &< 7&& (\text{Simplify})\\ \frac{2x}{{\color{red}2}} &< \frac{7}{{\color{red}2}}\\ x &< \frac{7}{2}\\ & OR\\ x-\frac{5}{2} &> -1\\ \left({\color{red}\frac{2}{2}}\right)x-\frac{5}{2} &> \left(\frac{{\color{red}2}}{{\color{red}2}}\right)(-1) && (\text{Multiply to get common denominator (LCD} = 2))\\ \frac{2x}{2}-\frac{5}{2}&<\frac{-2}{2}&& (\text{Simplify})\\ 2x-5&>-2 && (\text{Simplify})\\ 2x-5{\color{red}+5} &> -2{\color{red}+5} && (\text{Add 5 to isolate the variable})\\ 2x &> 3&& (\text{Simplify})\\ \frac{2x}{{\color{red}2}} &> \frac{3}{{\color{red}2}} && (\text{Divide both sides by 2 to solve for the variable})\\ x &> \frac{3}{2}\end{align*}
The solution is 32<x<72\begin{align*}\frac{3}{2}.
#### Concept Problem Revisited
Solve the following inequality and graph the solution on a number line.
|x+2|3\begin{align*}|x+2|\le 3\end{align*}
First solve the inequality:
x+2x+22xx+2x+22x3321OR3325Subtract 2 from both sides to isolate the variableSimplifySubtract 2 from both sides to isolate the variableSimplify\begin{align*}x+2 &\le 3\\ x+2{\color{red}-2} &\le 3{\color{red}-2} && \text{Subtract 2 from both sides to isolate the variable}\\ x &\le 1 && \text{Simplify}\\ & OR\\ x+2 &\ge -3\\ x+2{\color{red}-2} &\ge -3{\color{red}-2} && \text{Subtract 2 from both sides to isolate the variable}\\ x &\ge -5 && \text{Simplify}\end{align*}
The solution is 5x1\begin{align*}-5 \le x \le 1\end{align*}.
Representing on a number line:
### Vocabulary
Absolute Value Linear Inequality
Absolute Value Linear inequalities can have one of four forms: |ax+b|>c,|ax+b|<c,|ax+b|c\begin{align*}|ax + b| > c, |ax + b| < c, |ax + b| \ge c\end{align*}, or |ax+b|c\begin{align*}|ax + b| \le c\end{align*}. Absolute value linear inequalities have two related inequalities. For example for |ax+b|>c\begin{align*}|ax+b|>c\end{align*}, the two related inequalities are ax+b>c\begin{align*}ax + b > c\end{align*} and ax+b<c\begin{align*}ax + b < -c\end{align*}.
Number Line
A number line is a line that matches a set of points and a set of numbers one to one.
### Guided Practice
1. Represent the solution set to the inequality |2x+3|>5\begin{align*}|2x+3|>5\end{align*} on a number line.
2. Represent the solution set to the inequality |32x16|32\begin{align*}|32x-16| \ge 32\end{align*} on a number line.
3. Represent the solution set to the inequality |x21.5|>12.5\begin{align*}|x-21.5|>12.5\end{align*} on a number line.
1. |2x+3|>5\begin{align*}|2x+3| >5\end{align*}
2x+32x+332x2x2x2x+32x+332x2x2x>5>53>2>22>1OR<5<53<8>82<4(Subtract 3 from both sides of the inequality sign)(Simplify)(Divide by 2 to solve for the variable)(Subtract 3 from both sides of the inequality sign)(Simplify)(Divide by 2 to solve for the variable)\begin{align*}2x+3&>5\\ 2x+3{\color{red}-3}&>5{\color{red}-3} && (\text{Subtract 3 from both sides of the inequality sign})\\ 2x &> 2 && (\text{Simplify})\\ \frac{2x}{{\color{red}2}}&>\frac{2}{{\color{red}2}} && (\text{Divide by 2 to solve for the variable})\\ x &>1\\ & OR\\ 2x+3&<-5\\ 2x+3{\color{red}-3}&<-5{\color{red}-3} && (\text{Subtract 3 from both sides of the inequality sign})\\ 2x &< -8 && (\text{Simplify})\\ \frac{2x}{{\color{red}2}}&>\frac{-8}{{\color{red}2}} && ( \text{Divide by 2 to solve for the variable})\\ x &<-4\end{align*}
The solution sets are x>1\begin{align*}x>1\end{align*} or x<4\begin{align*}x<-4\end{align*}.
2. |32x16|32\begin{align*}|32x-16| \ge 32\end{align*}
32x1632x16+1632x32x32x32x1632x16+1632x32x32x3232+1648483232OR3232+1616163212(Add 16 to both sides of the inequality sign)(Simplify)(Divide by 32 to solve for the variable)(Add 16 to both sides of the inequality sign)(Simplify)(Divide by 32 to solve for the variable)\begin{align*}32x-16 &\ge 32\\ 32x-16{\color{red}+16}&\ge 32{\color{red}+16} && (\text{Add 16 to both sides of the inequality sign})\\ 32x &\ge 48&& (\text{Simplify})\\ \frac{32x}{{\color{red}32}} &\ge \frac{48}{{\color{red}32}}&& (\text{Divide by 32 to solve for the variable})\\ x &\ge \frac{3}{2}\\ & OR \\ 32x-16 &\le -32\\ 32x-16{\color{red}+16}&\le -32{\color{red}+16} && (\text{Add 16 to both sides of the inequality sign})\\ 32x &\le -16&& (\text{Simplify})\\ \frac{32x}{{\color{red}32}} &\le \frac{-16}{{\color{red}32}}&& (\text{Divide by 32 to solve for the variable})\\ x &\le -\frac{1}{2}\end{align*}
The solution sets are x32\begin{align*}x \ge \frac{3}{2}\end{align*} or x12\begin{align*} x \le -\frac{1}{2}\end{align*}.
3. |x21.5|>12.5\begin{align*}|x-21.5|>12.5\end{align*}
x21.5x21.5+21.5xx21.5x21.5+21.5x>12.5>12.5+21.5>34OR<12.5<12.5+21.5<9(Add 21.5 to both sides to isolate the variable)(Simplify)(Add 21.5 to both sides to isolate the variable)(Simplify)\begin{align*}x-21.5 &> 12.5\\ x-21.5{\color{red}+21.5}&>12.5{\color{red}+21.5} && (\text{Add 21.5 to both sides to isolate the variable})\\ x &>34 &&(\text{Simplify})\\ & OR\\ x-21.5 &< -12.5\\ x-21.5{\color{red}+21.5}&<-12.5{\color{red}+21.5} && (\text{Add 21.5 to both sides to isolate the variable})\\ x &<9 &&(\text{Simplify})\end{align*}
The solution sets are x<9\begin{align*}x<9\end{align*} or x>34\begin{align*}x>34\end{align*}.
### Practice
Represent the solution sets to each absolute value inequality on a number line.
1. |32x|<3\begin{align*}|3-2x|<3\end{align*}
2. \begin{align*}2\big|\frac{2x}{3}+1\big|\ge 4\end{align*}
3. \begin{align*}\big|\frac{2g-9}{4}\big|<1\end{align*}
4. \begin{align*}\big|\frac{4}{3}x-5\big|\ge 7\end{align*}
5. \begin{align*}|2x+5|+4 \ge 7\end{align*}
6. \begin{align*}|p-16|>10\end{align*}
7. \begin{align*}|r+2|<5\end{align*}
8. \begin{align*}|3-2k|\ge 1\end{align*}
9. \begin{align*}|8-y|>5\end{align*}
10. \begin{align*}8 \ge |5d-2|\end{align*}
11. \begin{align*}|s+2|-5>8\end{align*}
12. \begin{align*}|10+8w|-2<16\end{align*}
13. \begin{align*}|2q+1|-5 \le 7\end{align*}
14. \begin{align*}\big |\frac{1}{3}(g-2) \big |<4\end{align*}
15. \begin{align*}|-2(e+4)|>17\end{align*}
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
Absolute Value Linear inequalities
Absolute value linear inequalities can have one of four forms $|ax + b| > c, |ax + b| < c, |ax + b| \ge c$, or $|ax + b| \le c$. Absolute value linear inequalities have two related inequalities. For example for $|ax+b|>c$, the two related inequalities are $ax + b > c$ and $ax + b < -c$.
number line
A number line is a line on which numbers are marked at intervals. Number lines are often used in mathematics to show mathematical computations.
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NSW Year 8 - 2020 Edition
2.02 Percentage increase and decrease
Lesson
A percentage increase or decrease is a measure of how much some value has changed compared to its original amount. If a value is greater than its original amount, it has increased. If that value is less than its original amount, it has decreased.
### Adding or subtracting percentage changes
There are two main parts to any percentage increase or decrease; the original amount and the percentage change.
The original amount is equal to $100%$100% of itself and has not been increased or decreased yet.
The percentage change is the amount that we are either increasing or decreasing the original amount by. We can calculate this as some percentage of the original amount.
To find the result of a percentage increase, we add the percentage change to the original amount.
For example: if we increase $40$40 by $25%$25% then we are adding $25%$25% of $40$40 to the original $40$40. The final result will be equal to the expression:
$40+25%\times40$40+25%×40
Since $25%$25% of $40$40 is equal to $10$10, we find that the result of the percentage increase is $50$50.
Similarly, we can calculate a percentage decrease by subtracting the percentage change from the original amount. As such, if we decrease $40$40 by $25%$25% the result will be $30$30.
#### Practice question
##### Question 1
Bob wants to decrease $110$110 by $60%$60%, so he calculates $110-\left(60%\times110\right)$110(60%×110).
1. What was his result?
### Finding percentage changes directly
It is mentioned above that the original amount is equal to $100%$100% of itself. This fact is particularly useful if we want to find the result of a percentage increase or decrease directly.
To increase $40$40 by $25%$25% using the addition method, we are effectively adding $25%$25% of $40$40 to $100%$100% of $40$40. This is the same as finding $125%$125% of $40$40, which we can see gives the same result:
$125%\times40=50$125%×40=50
We can do the same for percentage decreases. To decrease $40$40 by $25%$25%, we take $25%$25% away from the original $100%$100%, leaving only $75%$75% of $40$40. This gives us:
$75%\times40=30$75%×40=30
By adding or subtracting at the percentage level, we can more clearly see how the original amount changes.
#### Practice question
##### Question 2
Sandy starts with the number $110$110, and then calculates $110\times160%$110×160%.
1. What was her result?
2. Which is the best description of her result?
She increased $110$110 by $60%$60%.
A
She decreased $110$110 by $60%$60%.
B
She decreased $110$110 by $160%$160%.
C
She increased $110$110 by $160%$160%.
D
She increased $110$110 by $60%$60%.
A
She decreased $110$110 by $60%$60%.
B
She decreased $110$110 by $160%$160%.
C
She increased $110$110 by $160%$160%.
D
### What percentage change was required?
Now that we are able to calculate the result of percentage increases and decreases, we can also do the reverse.
Suppose we want to increase an amount by $70%$70%, by what percentage would we multiply the original price?
Since we want to increase the amount, we will be adding to the original $100%$100%. Since we want to increase by $70%$70%, that is how much we will be adding. As such, to increase the original amount by $70%$70%, we multiply the original amount by $170%$170%.
We can do the same for flat changes by adding a couple of extra steps.
#### Worked Example
What percentage do we need to multiply $220$220 by to get $187$187?
Think: The flat change will be the difference between the original and desired amounts. The percentage change will be this difference as a percentage of the original amount. Since the amount has decreased, we will want to subtract this percentage change from the original $100%$100%.
Do: The flat change is the difference between $220$220 and $187$187, which we can calculate to be $33$33.
By simplifying the expression $\frac{33}{220}\times100%$33220×100% we find that the percentage change required is $15%$15%.
This means that, to get $187$187, we need to decrease $220$220 by $15%$15%. In other words, we can multiply it by $85%$85%.
Reflect: Consider that $\frac{187}{220}\times100%=85%$187220×100%=85%. Since this result is $15%$15% less than $100%$100%, we need to decrease $220$220 by $15%$15% to get $187$187. By reversing the multiplication required we notice that we can also deduce required percentage changes by dividing the desired amount by the original amount.
#### Practice question
##### Question 3
In training for her next marathon, Sally increased her practise route from $7000$7000 metres to $7910$7910 metres.
By what percentage has Sally increased the distance of her practice route?
Write the answer as a percentage value.
### Outcomes
#### MA4-5NA
operates with fractions, decimals and percentages
#### MA4-6NA
solves financial problems involving purchasing goods
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Lesson Objectives
• Demonstrate an understanding of how to Factor Trinomials
• Learn how to write a Quadratic Equation in Standard Form
• Learn about the Zero-Product Property
• Learn how to solve a Quadratic Equation using Factoring
## How to Solve Quadratic Equations Using Factoring
Over the course of the last few lessons, we have learned to factor quadratic expressions. A quadratic expression contains a squared variable and no term with a higher degree. We will expand on this knowledge and learn how to solve a quadratic equation using factoring. A quadratic equation is an equation that contains a squared variable and no other term with a higher degree. Generally, we think about a quadratic equation in standard form:
ax2 + bx + c = 0
a ≠ 0 (since we must have a variable squared)
a, b, and c are any real numbers (a can't be zero)
Some examples of a quadratic equation are:
4x2 + 7x - 15 = 0
5x2 + 18x + 9 = 0
### Zero-Product Property
Up to this point, we have not attempted to solve an equation in which the exponent on a variable was not 1. For these types of problems, obtaining a solution can be a bit more work than what we have seen so far. When a quadratic equation is in standard form and the left side can be factored, we can solve the quadratic equation using factoring. This works based on the zero-product property (also known as the zero-factor property). The zero product property tells us if the product of two numbers is zero, then at least one of them must be zero:
xy = 0
x could be 0, y could be a non-zero number
y could be 0, x could be a non-zero number
x and y could both be zero
We can apply this to more advanced examples. Suppose we saw the following:
(x - 2)(x + 3) = 0
In this case, we have a quantity (x - 2) multiplied by another quantity (x + 3). The result of this multiplication is zero. This means we can use our zero-product property. To do this, we set each factor equal to zero and solve:
(x - 2) = 0
(x + 3) = 0
x - 2 = 0
x = 2
x + 3 = 0
x = -3
Essentially, x could be 2 or x could be -3. In either scenario, the equation would be true:
Let's check x = 2:
(x - 2)(x + 3) = 0
(2 - 2)(2 + 3) = 0
0(5) = 0
0 = 0
Let's check x = -3:
(-3 - 2)(-3 + 3) = 0
(-5)(0) = 0
0 = 0
### Solving a Quadratic Equation using Factoring
• Place the quadratic equation in standard form
• Factor the left side
• Use the zero-product property and set each factor with a variable equal to zero
• Check the result
Let's look at a few examples.
Example 1: Solve each quadratic equation using factoring.
x2 + 3x = 18
Step 1) Write the quadratic equation in standard form. We want to subtract 18 away from each side of the equation:
x2 + 3x - 18 = 0
Step 2) Factor the left side:
x2 + 3x - 18 » (x + 6)(x - 3)
(x + 6)(x - 3) = 0
Step 3) Use the zero-product property and set each factor with a variable equal to zero:
x + 6 = 0
x - 3 = 0
x = -6
x = 3
We can say that x = -6, 3
This means x can be -6 or x can be 3. Either will work as a solution.
Step 4) Check the result:
Plug in a -6 for x:
(-6)2 + 3(-6) - 18 = 0
36 - 18 - 18 = 0
36 - 36 = 0
0 = 0
Plug in a 3 for x:
(3)2 + 3(3) - 18 = 0
9 + 9 - 18 = 0
18 - 18 = 0
0 = 0
Example 2: Solve each quadratic equation using factoring.
3x2 - 5 = -14x
Step 1) Write the quadratic equation in standard form. We want to add 14x to both sides of the equation:
3x2 + 14x - 5 = 0
Step 2) Factor the left side:
3x2 + 14x - 5 » (3x - 1)(x + 5)
(3x - 1)(x + 5) = 0
Step 3) Use the zero-product property and set each factor with a variable equal to zero:
3x - 1 = 0
x + 5 = 0
x = 1/3
x = -5
We can say that x = -5, 1/3
This means x can be -5 or x can be 1/3. Either will work as a solution.
Step 4) Check the result:
Plug in a -5 for x:
3(-5)2 + 14(-5) - 5 = 0
3(25) - 70 - 5 = 0
75 - 75 = 0
0 = 0
Plug in a 1/3 for x:
3(1/3)2 + 14(1/3) - 5 = 0
3(1/9) + (14/3) - 5 = 0
1/3 + 14/3 - 5 = 0
15/3 - 5 = 0
5 - 5 = 0
0 = 0
#### Skills Check:
Example #1
Solve each equation. $$x^{2}+ 2x - 8=0$$
A
$$x=5, -1$$
B
$$x=-4, 2$$
C
$$x=4, -2$$
D
$$x=5, -5$$
E
$$x=\frac{1}{2}, -3$$
Example #2
Solve each equation. $$x^{2}- 5x=6$$
A
$$x=-1, 6$$
B
$$x=-4, -7$$
C
$$x=1, 4$$
D
$$x=4, 2$$
E
$$x=-\frac{2}{5}, 6$$
Example #3
Solve each equation. $$3x^{2}+ 26x + 17=1$$
A
$$x=-\frac{1}{3}, 4$$
B
$$x=\frac{2}{3}, 8$$
C
$$x=-\frac{2}{3}, -8$$
D
$$x=-\frac{6}{5}, -2$$
E
$$x=-4, \frac{1}{3}$$
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363_pdfsam_math 54 differential equation solutions odd
# 363_pdfsam_math 54 differential equation solutions odd -...
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Unformatted text preview: Exercises 6.2 We find the remaining roots by using the quadratic formula. Thus, we obtain the roots r = 1, 2 ± 3i. From the root r = 1, we obtain the solution y = x. From the roots r = 2 ± 3i, by applying the hint given in the problem, we see that a solution is given by y (x) = x2+3i = x2 {cos(3 ln x) + i sin(3 ln x)} . Therefore, by Lemma 2 on page 172 of the text, we find that two real-valued solutions to this differential equation are y (x) = x2 cos(3 ln x) and y (x) = x2 sin(3 ln x). Since these functions and the function y (x) = x are linearly independent, we obtain the fundamental solution set x, x2 cos(3 ln x), x2 sin(3 ln x) . 33. With suggested values of parameters m1 = m2 = 1, k1 = 3, and k2 = 2, the system (34)–(35) becomes x + 5x − 2y = 0, y − 2x + 2y = 0. (6.7) (a) Expressing y = (x + 5x) /2 from the first equation and substituting this expression into the second equation, we obtain 1 (x + 5x) − 2x + (x + 5x) = 0 2 ⇒ x(4) + 5x − 4x + 2 (x + 5x) = 0 ⇒ as it is stated in (36). (b) The characteristic equation corresponding to (6.8) is r 4 + 7r 2 + 6 = 0. This equation is of quadratic type. Substitution s = r 2 yields s2 + 7 s + 6 = 0 Thus √ r = ± −1 = ±i ⇒ s = −1, −6. √ √ r = ± −6 = ±i 6 , x(4) + 7x + 6x = 0, (6.8) and and a general solution to (6.8) is given by √ √ x(t) = c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t . 359 ...
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# Exercise : 1
1. 12 men take 18 days to complete a job whereas 12 women in 18 days can complete 34 of the same job. How many days will 10 men and 8 women together take to complete the same job?
(a) 6
(b) 1312
(c) 12
(e) None of these
#### View Ans & Explanation
Ans.b
12 M × 18 = 12 W × 18 × 43
\ w = 3/4 M
10M + 8W = 10M + 8 × 34M = 16 M
\16 men can complete the same work
in $\frac{12 \times 18}{16} = \frac{27}{2} = 13\tfrac{1}{2} \; days$
2. Seven men and four boys can complete a work in 6 days. A man completes double the work than a boy. In how many days will 5 men and 4 boys complete the work?
(a) 5
(b) 4
(c) 6
(d) Cannot be determined
(e) None of these
#### View Ans & Explanation
Ans.e
M = 2B
∴ 7M + 4B = 14B + 4B = 18B
5M + 4B = 10B + 4B = 14B
∴ 18 boys complete the work in 6 days.
∴ 14 boys complete the work in
$\frac{6 \times 18}{14} = 7\tfrac{5}{7} \; days$
Note: 7 men and 4 boys complete the work in 6 days. We have to find out the no. of days in which 5 men and 4 boys complete the work. Here, we see that 4 boys are common in both the cases, therefore, 5 men will take more time to complete the work, i.e., more than 6 days, which is not given in any options. Therefore, without calculating we can say that our answer is(e).
3. The work done by a woman in 8 hours is equal to the work done by a man in 6 hours and by a boy in 12 hours. If working 6 hours per day, 9 men can complete a work in 6 days then in how many days can 12 men, 12 women and 12 boys together finish the same work by working 8 hours per day?
(a) 113 days
(b) 323 days
(c) 3 days
(d) 112 days
(e) None of these
#### View Ans & Explanation
Ans.d
8W = 6M = 12B
12M + 12W + 12B Þ 12M + 9M + 6M = 27M
\9 men can complete the work by working 1 hour per day in 6 × 6 days
\27 men working 8 hours per day = $\frac{6 \times 6 \times 9}{27 \times 8} = 1\tfrac{1}{2} \; days$
4. Tap ‘A’ can fill a water tank in 25 minutes, tap ‘B’ can fill the same tank in 40 minutes and tap ‘C’ can empty that tank in 30 minutes. If all the three taps are opened together, in how many minutes will the tank be completely filled up or emptied?
(a) 3213
(b) 15513
(c) 8213
(d) 311119
(e) None of these
#### View Ans & Explanation
Ans.d
Tank filled in 1 minute = 125 + 140 - 130 part
= $\frac{24 + 15 - 20}{600} = \frac{19}{600}$ part
∴ tank will be filled complete in minutes
= 60019 = 311119
5. Machine A can print one lakh books in 8 hours. Machine B can do the same job in 10 hours. Machine C can do the same job in 12 hours. All the three machines start job at 9.00 am. A breaks down at 11.00 am and the other two machines finish the job. Approximately at what time will the job be finished?
(a) 12.00 noon
(b) 1.30 pm
(c) 1.00 pm
(d) 11.30 am
(e) None of these
#### View Ans & Explanation
Ans.c
Part of print done by A, B and C in 2 hours = 2(18 + 110 + 112) = 3760
Remaining = 1 - 3760 = 2360
If B and C print together, then they can print in $\frac{10 \times 12}{10 + 12}$
Therefore, remaining part can be printed by
B and C in $\frac{10 \times 12}{22} \times \frac{23}{60}$ ≈ 2 hrs.
Hence, the job will be finished at
9 am + 2 + 2 = 1.00 p.m.
6. Suresh can complete a job in 15 hours. Ashutosh alone can complete the same job in 10 hours. Suresh works for 9 hours and then the remaining job is completed by Ashutosh. How many hours will it take Ashutosh to complete the remaining job alone?
(a) 4
(b) 5
(c) 6
(d) 12
(e) None of these
#### View Ans & Explanation
Ans.a
The part of job that Suresh completes in 9 hours
= 915 = 35
Remaining job = 1 - 35 = 25
Remaining job can be done by Ashutosh in 25 × 10 = 4 hours
7. 10 men and 15 women finish a work in 6 days. One man alone finishes that work in 100 days. In how many days will a woman finish the work?
(a) 125 days
(b) 150 days
(c) 90 days
(d) 225 days
(e) None of these
#### View Ans & Explanation
Ans.d
15 women's work of a day = 16 - 110115 part
∴ for 1 whole part a woman will take
= 15 × 15 = 225 days.
8. A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
(a) 35 hours
(b) 25 hours
(c) 20 hours
(d) Cannot be determined
(e) None of these
#### View Ans & Explanation
Ans.a
Here ratio of efficiencies of pipes A, B and C are as follows:
$\frac{\begin{matrix} C \;\;\; & B \;\;\; & A\\ 2 \;\;\; & 1 \;\;\; & \\ & 2 \;\;\; & 1 \end{matrix}} {\begin{matrix} 4 & : & 2 & : & 1 \end{matrix}}$
Suppose the efficiencies of pipes C, B and A are 4K, 2K and K.
Since, the tank is filled in 5 hours by the three pipes having combined efficiency equal to 7K, the time required to fill the tank by A alone = $\frac{7K \times 5}{K} = 35 \; hours$
9. 24 men working 8 hours a day can finish a work in 10 days. Working at the rate of 10 hours a day, the number of men required to finish the same work in 6 days is :
(a) 30
(b) 32
(c) 34
(d) 36
(e) None of these
#### View Ans & Explanation
Ans.b
m1 × d1 × t1 × w2 = m2 × d2 × t2 × w1
24 × 10 × 8 × 1 = m2 × 6 × 10 × 1
⇒ m2 = $\frac{24 \times 10 \times 8}{6 \times 10} = 32 \; men$
10. A water tank is 25th full. Pipe A can fill the tank in 10 minutes and the pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?
(a) 6 minutes to empty
(b) 6 minutes to fill
(c) 9 minutes to empty
(d) 9 minutes to fill
(e) None of these
#### View Ans & Explanation
Ans.a
∵ Pipe A in 1 minute fills 110 part and Pipe Bin 1 min.empties 16 part
\ Pipe A + B in 1 min = 110 - 16 = -115
115 part gets emptied in 1 min
\ 25 part is emptied in 15 × 25 min = 6 min.
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# How do you find the limit of the cubic root of x^2+1 all mines x ? Thank you.
##### 1 Answer
Nov 15, 2016
The question does not say what number $x$ is approaching.
#### Explanation:
The properties of limits assure us that for any number $a$,
${\lim}_{x \rightarrow a} \left(\sqrt[3]{{x}^{2} + 1} - x\right)$ can be found by substituting $a$ in place of $x$ and doing the arithmetic.
For any $a$, the subtraction property of limits gets us:
${\lim}_{x \rightarrow a} \left(\sqrt[3]{{x}^{2} + 1} - x\right) = {\lim}_{x \rightarrow a} \sqrt[3]{{x}^{2} + 1} - {\lim}_{x \rightarrow a} x$
The power (or root) property of limits lets us continue:
$= \sqrt[3]{{\lim}_{x \rightarrow a} \left({x}^{2} + 1\right)} - {\lim}_{x \rightarrow a} x$
Using the addition property:
$= \sqrt[3]{{\lim}_{x \rightarrow a} \left({x}^{2}\right) + {\lim}_{x \rightarrow a} 1} - {\lim}_{x \rightarrow a} x$
And the power property again
= root(3)((lim_(xrarra)x)^2)+lim_(xrarra)1)-lim_(xrarra)x
Now use the fact that ${\lim}_{x \rightarrow a} x = a$ to write
$= \sqrt[3]{{a}^{2} + 1} - a$
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# How do you find (d^2y)/(dx^2) given x=tany?
Sep 5, 2016
$\forall y \in \mathbb{R} , \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- 2 x}{1 + {x}^{2}} ^ 2$.
#### Explanation:
$x = \tan y$.
Case : 1 $\text{Suppose that,} y \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right) \Rightarrow y = a r c \tan x , x \in \mathbb{R}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}$.
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{d}{\mathrm{dx}} \left(\frac{1}{1 + {x}^{2}}\right)$
Since, $\frac{d}{\mathrm{dt}} \left(\frac{1}{t}\right) = - \frac{1}{t} ^ 2$, we have, by the Chain Rule,
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{1}{{\left(1 + {x}^{2}\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(1 + {x}^{2}\right) = \frac{- 2 x}{{\left(1 + {x}^{2}\right)}^{2}}$
Case : 2 $\text{Next, let } y \in \left(\frac{\pi}{2} , 3 \frac{\pi}{2}\right) \Rightarrow \frac{\pi}{2} < , y < , 3 \frac{\pi}{2}$
$\Rightarrow \frac{\pi}{2} - \pi < , y - \pi < , 3 \frac{\pi}{2} - \pi$
rArr -pi/2<,y-pi<,pi/2, &, tan(y-pi)=-tan(pi-y)=-(-tany)=tany=x.
Thus, in this Case, $x = \tan \left(y - \pi\right) , w h e r e , \left(y - \pi\right) \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$
$\therefore \text{ by defn. of arc tan, } y - \pi = a r c \tan x , \mathmr{and} , y = \pi + a r c \tan x$
$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- 2 x}{1 + {x}^{2}} ^ 2$, as in Case : 1
Thus, $\forall y \in \mathbb{R} , \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{- 2 x}{1 + {x}^{2}} ^ 2$.
Sep 5, 2016
To differentiate without using the inverse tangent, see below.
#### Explanation:
$\tan y = x$
Differentiate both sides with respect to $x$.
$\frac{d}{\mathrm{dx}} \left(\tan x\right) = \frac{d}{\mathrm{dx}} \left(x\right)$
${\sec}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} = 1$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\cos}^{2} y$
Differentiate again w.r.t. $x$
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 2 \cos y \left(- \sin y \frac{\mathrm{dy}}{\mathrm{dx}}\right)$
Now rfeplace $\frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 2 \sin y {\cos}^{3} y$
To see that this is the same as the other answer
$\tan y = x$ $\Rightarrow$ $\cos y = \frac{1}{1 + {x}^{2}}$
(To see this draw and label a right triangle with angle $y$, opposite side $x$ and adjacent side $1$. So the hypotenuse is $1 + {x}^{2}$.)
(Or use $\tan y = x$ $\Rightarrow$ ${\tan}^{2} y = {x}^{2}$, so $1 + {\tan}^{2} y = 1 + {x}^{2}$ and ${\sec}^{y} = 1 + {x}^{2}$ so that $\cos y = \frac{1}{1 + {x}^{2}}$
Now,
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - 2 \sin y {\cos}^{3} y$
$= - 2 \sin \frac{y}{\cos} y {\cos}^{4} y$
$= - 2 \tan y {\left({\cos}^{2} y\right)}^{2}$
$= - 2 x \frac{1}{1 + {x}^{2}} ^ 2$
$= \frac{- 2 x}{1 + {x}^{2}} ^ 2$.
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# 2018 AMC 12B Problems/Problem 15
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
How many odd positive $3$-digit integers are divisible by $3$ but do not contain the digit $3$?
$\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120$
## Solution 1
Let $\underline{ABC}$ be one such odd positive $3$-digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$
As $A\in\{1,2,4,5,6,7,8,9\}$ and $C\in\{1,5,7,9\},$ there are $8$ possibilities for $A$ and $4$ possibilities for $C.$ Note that each ordered pair $(A,C)$ determines the value of $B$ modulo $3,$ so $B$ can be any element in one of the sets $\{0,6,9\},\{1,4,7\},$ or $\{2,5,8\}.$ We conclude that there are always $3$ possibilities for $B.$
By the Multiplication Principle, the answer is $8\cdot4\cdot3=\boxed{\textbf{(A) } 96}.$
~Plasma_Vortex ~MRENTHUSIASM
## Solution 2
Let $\underline{ABC}$ be one such odd positive $3$-digit integer with hundreds digit $A,$ tens digit $B,$ and ones digit $C.$ Since $\underline{ABC}\equiv0\pmod3,$ we need $A+B+C\equiv0\pmod3$ by the divisibility rule for $3.$
As $A\in\{1,2,4,5,6,7,8,9\},B\in\{0,1,2,4,5,6,7,8,9\},$ and $C\in\{1,5,7,9\},$ note that:
1. There are $2$ possibilities for $A\equiv0\pmod3,$ namely $A=6,9.$
There are $3$ possibilities for $A\equiv1\pmod3,$ namely $A=1,4,7.$
There are $3$ possibilities for $A\equiv2\pmod3,$ namely $A=2,5,8.$
2. There are $3$ possibilities for $B\equiv0\pmod3,$ namely $B=0,6,9.$
There are $3$ possibilities for $B\equiv1\pmod3,$ namely $B=1,4,7.$
There are $3$ possibilities for $B\equiv2\pmod3,$ namely $B=2,5,8.$
3. There are $1$ possibility for $C\equiv0\pmod3,$ namely $C=9.$
There are $2$ possibilities for $C\equiv1\pmod3,$ namely $C=1,7.$
There are $1$ possibility for $C\equiv2\pmod3,$ namely $C=5.$
We apply casework to $A+B+C\equiv0\pmod3:$ $$\begin{array}{c|c|c||l} & & & \\ [-2.5ex] \boldsymbol{A\operatorname{mod}3} & \boldsymbol{B\operatorname{mod}3} & \boldsymbol{C\operatorname{mod}3} & \multicolumn{1}{c}{\textbf{Count}} \\ [0.5ex] \hline & & & \\ [-2ex] 0 & 0 & 0 & 2\cdot3\cdot1=6 \\ 0 & 1 & 2 & 2\cdot3\cdot1=6 \\ 0 & 2 & 1 & 2\cdot3\cdot2=12 \\ 1 & 0 & 2 & 3\cdot3\cdot1=9 \\ 1 & 1 & 1 & 3\cdot3\cdot2=18 \\ 1 & 2 & 0 & 3\cdot3\cdot1=9 \\ 2 & 0 & 1 & 3\cdot3\cdot2=18 \\ 2 & 1 & 0 & 3\cdot3\cdot1=9 \\ 2 & 2 & 2 & 3\cdot3\cdot1=9 \end{array}$$ Together, the answer is $6+6+12+9+18+9+18+9+9=\boxed{\textbf{(A) } 96}.$
~MRENTHUSIASM
## Solution 3
Analyze that the three-digit integers divisible by $3$ start from $102.$ In the $200$'s, it starts from $201.$ In the $300$'s, it starts from $300.$ We see that the units digits is $0, 1,$ and $2.$
Write out the $1$- and $2$-digit multiples of $3$ starting from $0, 1,$ and $2.$ Count up the ones that meet the conditions. Then, add up and multiply by $3,$ since there are three sets of three from $1$ to $9.$ Then, subtract the amount that started from $0,$ since the $300$'s ll contain the digit $3.$
Together, the answer is $3(12+12+12)-12=\boxed{\textbf{(A) } 96}.$
## Solution 4
Consider the number of $2$-digit numbers that do not contain the digit $3,$ which is $90-18=72.$ For any of these $2$-digit numbers, we can append $1,5,7,$ or $9$ to reach a desirable $3$-digit number. However, we have $7 \equiv 1\pmod{3},$ and thus we need to count any $2$-digit number $\equiv 2\pmod{3}$ twice. There are $(98-11)/3+1=30$ total such numbers that have remainder $2,$ but $6$ of them $(23,32,35,38,53,83)$ contain $3,$ so the number we want is $30-6=24.$ Therefore, the final answer is $72+24= \boxed{\textbf{(A) } 96}.$
## Solution 5
We need to take care of all restrictions. Ranging from $101$ to $999,$ there are $450$ odd $3$-digit numbers. Exactly $\frac{1}{3}$ of these numbers are divisible by $3,$ which is $450\cdot\frac{1}{3}=150.$ Of these $150$ numbers, $\frac{4}{5}$ $\textbf{do not}$ have $3$ in their ones (units) digit, $\frac{9}{10}$ $\textbf{do not}$ have $3$ in their tens digit, and $\frac{8}{9}$ $\textbf{do not}$ have $3$ in their hundreds digit. Thus, the total number of $3$-digit integers is $$900\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{4}{5}\cdot\frac{9}{10}\cdot\frac{8}{9}=\boxed{\textbf{(A) } 96}.$$
~mathpro12345
## Solution 6
We will start with the numbers that could work. This numbers include _ _ $1$, _ _ $5$, _ _ $7$, _ _ $9$. Let's work case by case.
Case $1$: _ _ $1$: The two blanks could be any number that is $2$ mod $3$ that does not include $3$. We have $24$ cases for this case (we could count every case).
Case $2$: _ _ $5$: The $2$ blanks could be any number that is $1$ mod $3$ that does not include $3$. But we could see that this case has exactly the same solutions to case $1$ because we have a $1-1$ correspondence. We can do the exact same for case $3$.
Cases $4$: _ _ $9$: We need the blanks to be a multiple of $3$, but does not contain 3. We have $(12, 15, 18, 21, 24, 27, 42, 45, 48, 51, 54, 57, 60, 66, 69, 72, 75, 78, 81, 84, 87, 90, 96, 99)$ which also contains $24$ numbers. Therefore, we have $24 \cdot 4$ which is equal to $\boxed{\textbf{(A) } 96}.$
~Arcticturn
## Solution 7
This problem is solvable by inclusion exclusion principle. There are $\frac{999-105}{6} + 1 = 150$ odd $3$-digit numbers divisible by $3$. We consider the number of $3$-digit numbers divisible by $3$ that contain either $1, 2$ or $3$ digits of $3$.
For $\underline{AB3}$, $AB$ is any $2$-digit number divisible by $3$, which gives us $\frac{99-12}{3} + 1 = 30$. For $\underline{A3B}$, for each odd $B$, we have $3$ values of $A$ that give a valid case, thus we have $5(3) = 15$ cases. For $\underline{3AB}$, we also have $15$ cases, but when $B=3, 9$, $A$ can equal $0$, so we have $17$ cases.
For $\underline{A33}$, we have $3$ cases. For $\underline{3A3}$, we have $4$ cases. For $\underline{33A}$, we have $2$ cases. Finally, there is just one case for $\underline{333}$.
By inclusion exclusion principle, we get $150 - 30 - 15 - 17 + 3 + 4 + 2 - 1 = \boxed{\textbf{(A) } 96}$ numbers.
~Zeric
~ pi_is_3.14
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# Arithmetic Sequence|Definition & Meaning
## Definition
A list of numbers where subtracting two consecutive entries results in a constant difference is an arithmetic sequence. This difference is called the common difference of the sequence. In that sense, the set of odd integers is an arithmetic sequence with a common difference of 2 since 3 minus 1, 5 minus 3, and any other consecutive pair’s difference is equal to 2.
## What Is an Arithmetic Sequence?
The arithmetic sequence is a sequence in which the expected difference between any two successive terms remains the same. Let us go back and review what a sequence is. A set of numbers arranged in a specific order is called a sequence.
For instance, the sequence 1, 7, 12, and 17 is an example of an arithmetic sequence since there is a sequence where each number is created by adding 6 to the phrase that came before it.
The following steps form an arithmetic sequence since each term is obtained by adding a constant number four to the phrase that came before it.
Figure 1: Graphical representation of the arithmetic sequence.
There are two different formulas for arithmetic sequences.
• A mathematical expression that locates the nth element of a sequence.
• A mathematical expression that finds the sum of an arithmetic sequence’s first n terms.
We can use the formula for the arithmetic sequence to locate any term in the arithmetic sequence if that is what we need to do. To achieve a deeper comprehension of the topic, let us first examine the definition of the arithmetic sequence and then proceed to study the formulae, which derivations and several further examples will accompany.
There are two ways to define what is known as an arithmetic sequence. “sequence where the difference between every two succeeding terms is identical” is one definition of it. In contrast, “in an arithmetic sequence, “every term is formed by adding a specific number (positive or negative or zero) to its preceding” another.
## Example of Arithmetic Sequence
Consider the sequence: 4, 7, 10, 13, 16. This is an arithmetic sequence since each phrase in the sequence can be created by adding a constant number (3) to the value of the preceding item in the sequence.
The initial number, notated as ‘a’, is 4, so a = 4. The common difference, d, is calculated as follows:
(7 – 4) = (10 – 7) = (13 – 10) = (16 – 13) = … = 3
As a result, the notation for an arithmetic sequence looks like this: a, a plus d, a plus 2d, a plus 3d, and so on.
## Calculating the Nth Term of an Arithmetic Sequence
The formula for determining the nth element of an arithmetic series, such as a1, a2, a3, …, is a = a1 + (n – 1) d. The arithmetic sequence can also be referred to by its broad name, the general term. This is a logical conclusion that may be drawn directly from the realization that the arithmetic series a1, a2, a3, … = a1, a1 + d, a1 + 2d, a1 + 3d, …
Figure 2: Illustration of nth term of arithmetic sequence.
## The Formula for the Arithmetic Sequence
The letter a denotes the beginning of an arithmetic sequence, and the letter d denotes the sequence’s common difference. The number n denotes the total number of terms. The AP can be written in any generic form: a + d, a + 2d, a + 3d, etc., up to n words.
A specific arithmetic sequence can be used to compute the nth term, the summation of n terms of an AP, or the common difference of a given arithmetic sequence using one of several possible formulae related to the series.
## The Recursive Formula for the Arithmetic Sequence
When the value of ‘a1 and ‘d’ are known, the formula just presented for determining the nth term of an arithmetic series can be used to determine any term of the sequence. The “recursive formula of an arithmetic series” is another formula that may be used to discover the nth term.
This formula is applied to find a term (an) of the sequence when its preceding term (an – 1) and ‘d’ are known. It says a = an – 1 + d
This formula adheres to the requirements outlined in the concept of the arithmetic sequence.
Figure 3: Representation of recursive formula of arithmetic sequence.
## Properties of Arithmetic Sequence
The common difference, represented by the letter dd, is the difference maintained between all and all pairs of numbers that are consecutive or subsequent in a sequence. We will employ the common difference to get from one phrase to another. How?
The following term can be arrived at by taking the present term and adding the common difference to it, and so on. The terms in the sequence are created in such a way as to do this.
• If the average difference between two consecutive terms is positive, then we state that the sequence is rising.
• Conversely, we conclude that the sequence declines when the difference between the two values is negative.
• The difference in value between any two numbers that are sequential in an arithmetic sequence is always the same.
• An arithmetic series, a1, a2, a3,… have a common difference that can be expressed as the formula d = a2 – a1 = a3 – a2 = …
• An arithmetic series has the following formula for its nth term: a = a1 + (n – 1)d.
• Arithmetic sequence
• s’ differences can be positive, negative, or zero. All three outcomes are possible.
## Examples of Arithmetic Sequence Operations
### Example 1
In the following arithmetic sequence, figure out the next three terms:
9, 18, 27, 36.
### Solution
First, determine the common differences:
18 – 9 = 9
27 – 18 = 9
36 – 27 = 9
So d = 9
Now add the common difference to the last term, then to the result, then as often as required.
36 + 9 = 45
45 + 9 = 54
54 + 9 = 63
### Example 2
Find the 28th term of the given sequence
2, 7, 12, 17, 22, 27, 32, …
### Solution
a = 2
d = 5
a28 = 2 + (28 – 1)5
2 + (27)5
2 + 135 = 137
All images/graphs are created using GeoGebra.
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# IBPS Specialist Officers Exam: Quantitative Aptitude-Number Series-Concept & Sample Questions
Dec 24, 2014 16:17 IST
Institute of Banking and Personnel Selection (IBPS) will conduct common written examination for recruitment of Specialist Officers. The IBPS SO Exam 2015 will be conducted in the month of February 2015.
The banking team of jagranjosh.com has brought concept and sample questions for Alphabet Test (Reasoning). The concept provided by us will help you to understand the topic. The sample questions offered by us are framed by keeping in view need of the question paper.
Type 1 To Find the Missing Term
Directions (Q. Nos. 1 to 2): Find the missing term in each oy the following number series.
Example 1 13, 18, 16, 21, 19, 24,?
Solution: Clearly, the given series follows the pattern +5, -2, +5, -2, …….
So, Missing term = 24 – 2 = 22
Example 2 3333, 3233, 3143, 3063,?
Solution: Clearly, the given series follows the pattern -100, -90, -80, -70, ………
So, Missing term = 2993
Triangular Pattern Series
In such a series, the difference between the consecutive terms of a series, again form a series. The difference between the consecutive terms of the new series so formed again forms a series. This pattern continues till we attain a constant difference between the consecutive terms of the series.
Example 3 4, 18, 48, 100, 180, 294,?
Solution:
So, the missing term = 34 + 6 = 40
Then, 40 + 114 = 154 and 294 + 154 = 448
Type 2 To Pick the Odd One
Directions (Q. Nos. 5 to 6): In each of the following number series, one of the given numbers is wrong. Find out the number.
Example 4 2, 7, 14, 23, 36, 47, 62
Solution: Clearly, the correct sequence is 22 - 2, 32 - 2, 42 - 2, 52 - 2, 62 - 2, 72 - 2, 82 - 2.
So, 36 is wrong and should be replaced by (62 -2) =34
Example 5 1, 27, 125, 343, 729, 1321
Solution: Clearly, the correct sequence is 13, 33, 53, 73, 93, 113
So, 1321 is wrong and should be replaced by 113 = 1331
Sample Questions with Solutions and Explanations
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# Number Bases
## ADDITION AND SUBTRACTION IN NUMBER BASES.
Most students have been finding it so difficult to add or subtract in number bases. The method used on tackling the problems is not different from what we have done here.
### Example 1: (Number Bases)
If you are given 264, 86, 475 all in base nine to be added.
Do not hesitate to convert those numbers first to base ten.
##### For 264nine
= 2 x 9² + 6 x 9¹ + 4 x 9° = 162 + 54 + 4
= 220ten
##### For 86nine
= 8 x 9¹ + 6 x 9° = 72 + 6
= 78ten
##### For 475nine
= 4 x 9² + 7 x 9¹ + 5 x 9° = 324 + 63 + 5
= 392ten
#### Step two: Add up all those numbers in base ten.
220 + 78 + 392 = 690ten.
Final step is to convert it back to base nine.
Learn how to convert from base ten to any required base here.
Therefore, (264 + 86 + 475)nine
= 846nine.
### Example 2. (Number Bases)
Subtract 265 from 317 in base eight.
Solution:
First convert to base ten.
For 265eight
= 2 x 8² + 6 x 8¹ + 5 x 8° = 128 + 48 + 5
= 181ten
For 317eight
= 3 x 8² + 1 x 8¹ + 7 x ° = 192 x 8 x 7
= 207ten
Step two: Now subtract
207 – 181 = 26ten.
Final step is to convert it back to base eight.
Thus, (317 – 265)eight = 32eight.
### Example 3. (Number Bases)
If 123 + 1302 + xxxx in base four is 10122four, find the missing figure.
##### Solution:
First convert the known figures to base ten.
For 123four
= 1 x 4² + 2 x 4¹ + 3 x 4° = 16 + 8 + 3
= 27ten
For 1302four
= 1 x 4³ + 3 x 4² + 0 x 4¹ + 2 x 4°
= 64 + 48 + 0 + 2
= 114ten
For 10122four
= 1 x 4⁴ + 0 x 4³ + 1 x 4² + 2 x 4¹ + 2 x 4°
= 256 + 0 + 16 + 8 + 2
= 282ten
Step 2: Subtract (27 + 114)ten from 282ten
= 282 – 141 = 141ten
Step 3: Convert the 141ten back to the required base .
Therefore, xxxx = 2031.
## MULTIPLICATION AND DIVISION OF NUMBER BASES
This is not different from the one we have done when adding or subtracting number bases. As far as you know what multiplication, division, addition and subtraction are all about. All you have to do first is to convert to base ten, then return the answer gotten to the required base.
### Example 1: (Number Bases)
Evaluate 321six X 25six
Solution:
Step 1: Convert to base ten first.
For 321six
= 3 x 6² + 2 x 6¹ + 1 x 6°
= 108 + 12 + 1
= 121ten
For 25six
= 2 x 6¹ + 5 x 6°
= 12 + 5
= 17ten
Step 2: Multiply 121 by 17
= 121 x 17
= 2057ten
Step 3: Convert back to base six
321six X 25six = 13305six.
### Example 2: (Number Bases)
Find the product of 1011two and 1101two.
Solution:
Step 1: Convert to base ten
For 1011two
= 1 x 2³ + 0 x 2² + 1 x 2¹ + 1 x 2°
= 8 + 0 + 2 + 1
= 11ten
For 1101two
= 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2°
= 8 + 4 + 0 + 1
= 13ten
Step 2: Multiply the two.
11 x 13 = 143ten
Step 3: Convert it back to base two.
1011two X 1101two = 10001111two
### Example 3:
If 231 base four X 10101 base two = “M” base six, Find M.
Solution:
Do you notice that they are of different bases?. Do not panic because you can still solve it😊😊
Step1: Convert the two numbers to base ten
For 231 base four
= 2 x 4² + 3 x 4¹ + 1 x 4°
= 32 + 12 + 1
= 45 base ten
For 10101 base two
= 1 x 2⁴ + 0 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2°
= 16 + 0 + 4 + 0 + 1
= 21 base ten
Step 2: Multiply the two.
45 x 21 = 945 base ten
Step 3: Convert to the required base.
231 base four X 10101 base two = 4213 base six
Therefore, M = 4213
### Example 4:
Evaluate 2115 base seven ÷ 12 base seven
Solution:
Step 1: Convert to base ten
For 2115 base seven
= 2 x 7³ + 1 x 7² + 1 x 7¹ + 5 x 7°
= 686 + 49 + 7 + 5
= 747 base ten
For 12 base seven
= 1 x 7¹ + 2 x 7°
= 7 + 2 = 9 base ten
Step 2: Now divide.
747/9 = 83 base ten
Step 3: Convert back to base seven.
83 base ten = 146 base seven.
So, (2115 ÷ 12) base ten = 146 base seven.
### Example 5:
Divide 1423 by 24 in base five.
Solution:
Step 1: Convert to base ten
For 1423 base five
= 1 x 5³ + 4 x 5² + 2 x 5¹ + 3 x 5°
= 125 + 100 + 10 + 3
= 238 base ten
For 24 base five
= 2 x 5¹ + 4 x 5°
= 10 + 4 = 14 base ten
Step 2: Divide
238/14 = 17 base ten
Step 3: Convert back to base five
Therefore:
1423 base five ÷ 24 base five = 32 base five
## OTHER CASES OF NUMBER BASES.
Number base sometimes have unknown that must be made known.
### Example 1:
Calculate “y” if 134 base “y” = 54 base eight
Step 1: Convert both sides to base ten
For 134 base “y”
= 1 x y² + 3 x y¹ + 4 x y°
= (y² + 3y + 4) base ten
For 54 base eight
= 5 x 8¹ + 4 x 8°
= 40 + 4 = 44 base ten
##### Step 2: Equate the two terms.
(y² + 3y + 4) = 44
y² + 3y + 4 = 44
y² + 3y + 4 -44 = 0
y² + 3y – 40 = 0
y² + 8y – 5y – 40 = 0
y(y + 8) – 5(y + 8) = 0
(y – 5)(y + 8) = 0
Either (y – 5) = 0
or (y + 8) = 0
When y – 5 = 0
y = 5
And when y + 8 = 0
y = – 8
Number base cannot take a negative value,
so y = 5.
### Example 2:
If 251 base “b” = 100 base two, find the value of “b”.
Solution:
Similar thing we did in example 1 above will be done here too.
Step 1: Convert to base ten.
For 251 base “b”
= 2 x b² + 5 x b¹ + 1 x b°
= (2b² + 5b + 1) base ten
For 100 base two
= 1 x 2² + 0 x 2¹ + 0 x 2°
= 4 + 0 + 0 = 4 base ten
##### Equating both of them
(2b² + 5b + 1) = 4
2b² + 5b + 1 = 4
2b² + 5b + 1 – 4 = 0
2b² + 5b – 3 = 0
2b² + 6b – b – 3 = 0
2b(b + 3) – 1(b + 3) = 0
(2b – 1)(b + 3) = 0
So, either 2b – 1 = 0 or
b + 3 = 0
If 2b – 1 = 0
b = 1/2
And if b + 3 = 0
b = -3
Base cannot have negative sign, hence b = 1/2
## QUESTIONS AND SOLUTIONS (Number Bases):
### Q1. Evaluate 3112 base five X 34 base five.
Solution:
Step 1: Convert to base ten
For 3112 base five
= 3 x 5³ + 1 x 5² + 1 x 5¹ + 2 x 5°
= 375 + 25 + 5 + 2 = 407 base ten
For 34 base five
= 3 x 5¹ + 4 x 5°
= 15 + 4 = 19 base ten
Step 2: Multiply both
407 x 19 = 7733 base ten
Step 3: Convert back to base five
(3112 x 34) base five = 221413 base five
### Q2. Divide 2173 by 16 in base nine
Solution:
Convert to base ten.
For 2173 base nine
= 2 x 9³ + 1 x 9² + 7 x 9¹ + 3 x 9°
= 1458 + 81 + 63 + 3
= 1605 base ten
For 16 base nine
= 1 x 9¹ + 6 x 9°
= 9 + 6 = 15 base ten
Divide the numbers
1605/15 = 107 base ten
2173 base nine ÷ 16 base nine = 128 base nine.
### Q3. 111 x 101 in base two
Solution:
Convert to base ten
For 111 base two
= 1 x 2² + 1 x 2¹ + 1 x 2°
= 4 + 2 + 1 = 7 base ten
For 101 base two
= 1 x 2² + 0 x 2¹ + 1 x 2°
= 4 + 0 + 1 = 5 base ten
Multiply the both
7 x 5 = 35 base ten
Return the number base ten to base two
111 x 101 in base two = 100011 base two
### Q4. 3132 base eight + 6 base eight
Solution:
Convert to base ten
For 3132 base eight
= 3 x 8³ + 1 x 8² + 3 x 8¹ + 2 x 8°
= 1536 + 64 + 24 + 2
= 1626 base ten
For 6 base eight
= 6 x 8° = 6 base ten
1626 + 6 = 1632 base ten
Return it to base eight
3140 base eight.
### Q5. If 25 x 14 = 374, find the number base used.
Solution:
Let the number base used be “y”
(25 base y) X (14 base y) = 374 base y
Convert all to base ten
For 25 base y
= 2 x y¹ + 5 x y°
= (2y + 5) base ten
For 14 base y
= 1 x y¹ + 4 x y°
= (y + 4) base ten
For 374 base y
= 3 x y² + 7 x y¹ + 4 x y°
= (3y² + 7y + 4) base ten
If (25 base y) X (14 base y) = 374 base y
##### Then,
(2y + 5)(y + 4) = 3y² + 7y + 4
Expand the equation
2y² + 8y + 5y + 20 = 3y² + 7y + 4
Collection of like terms
2y² – 3y² + 13y – 7y + 20 – 4 = 0
-y² + 6y + 16 = 0
y² – 6y – 16 = 0
y² – 8y + 2y – 16 = 0
y(y – 8) + 2(y – 8) = 0
(y + 2)(y – 8) = 0
y = – 2 or 8
Therefore y = 8
### Q6. Given that 132 base five = P base six, find P.
Solution:
Convert to base ten
For 132 base five
= 1 x 5² + 3 x 5¹ + 2 x 5°
= 25 + 15 + 2
= 42 base ten
For P base six
= P x 6° = P base ten
Equating both sides
P = 42 base ten
Convert back to base six
P = 110 base six
### Q7. If 244 base n = 1022 base four, find n.
Solution:
Convert to base ten
For 244 base n
= 2 x n² + 4 x n¹ + 4 x n °
= (2n² + 4n + 4) base ten
For 1022 base four
= 1 x 4³ + 0 x 4² + 2 x 4² + 2 x 4°
= 64 + 0 + 8 + 2
= 74 base ten
##### Equating both of them
(2n² + 4n + 4) = 74
2n² + 4n – 70 = 0
n² + 2n – 35 = 0
(n – 5)(n + 7) = 0
n = 5 or – 7
Therefore n = 5
because we do not have negative base
### Q8. A number is written as 52 base x. Four times the number is written as 301 base x. What is x?Solution:
A number = 52 base x
Four times the number = 4 X 52 base x = 301 base x
Convert to base ten
For 4
= 4 X x° = 4 base ten
#### For 52 base x
= 5 X x¹ + 2 X x°
= (5x + 2) base ten
For 301 base x
= 3 X x² + 0 X x¹ + 1 X x °
= 3x² + 0 + 1
= (3x² + 1) base ten
4(5x + 2) = 3x² + 1
20x + 8 = 3x² + 1
3x² – 20x – 7 = 0
Let us use quadratic formula to find the value of “x”
x = 7
### Q9. Arrange the following in ascending order.
1101 base two, 26 base eight, 113 base four and 1024 base six.
Solution:
Convert all to base ten.
For 1101 base two
= 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2°
= 8 + 4 + 0 + 1
= 13 base ten
For 26 base eight
= 2 x 8¹ + 6 x 8°
= 16 + 6 = 22 base ten
##### For 113 base four
= 1 x 4² + 1 x 4¹ + 3 x 4°
= 16 + 4 + 3 = 23 base ten
For 1024 base six
= 1 x 6³ + 0 x 6² + 2 x 6¹ + 4 x 6°
= 216 + 0 + 12 + 4
= 232 base ten
Therefore, the ascending order of magnitude
= 1101, 26, 113 and 1024 to their respective bases.
### Q10. If 321 base n = 232 base seven. Find the value of n.
Solution:
Convert all to base ten first.
For 321 base n
= 3 x n² + 2 x n¹ + 1 x n°
= (3n² + 2n + 1) base ten
For 232 base seven
= 2 x 7² + 3 x 7¹ + 2 x 7°
= 98 + 21 + 2
= 121 base ten
##### Equate the two.
3n² + 2n + 1 = 121
3n² + 2n + 1 – 121 = 0
3n² + 2n – 120 = 0
Using quadratic method to find the value of “n”
n = 6
### Q11. Simplify the following binary numbers.
(a). 101 + 111
Solution:
Convert to base ten
For 101
= 1 x 2² + 0 x 2¹ + 1 x 2°
= 4 + 0 + 1 = 5 base ten
For 111
= 1 x 2² + 1 x 2¹ + 1 x 2°
= 4 + 2 + 1 = 7 base ten
101 + 111 = 5 base ten + 7 base ten
= 12 base ten
Convert 12 base ten back to binary.
Therefore 101 + 111 = 1100
(b). 11001 + 1111 + 10110
Solution:
For 11001
= 1 x 2⁴ + 1 x 2³ + 0 x 2² + 0 x 2¹ + 1 x 2°
= 16 + 8 + 0 + 0 + 1
= 25 base ten
For 1111
= 1 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2°
= 8 + 4 + 2 + 1
= 15 base ten
For 10110
= 1 x 2⁴ + 0 x 2³ + 1 x 2² + 1 x 2¹ + 0 x 2°
= 16 + 0 + 4 + 2 + 0
= 22 base ten
25 + 15 + 22 = 62 base ten
Convert it back to binary
Therefore 11001 + 1111 + 10110 = 111110
(c). 11111 – 1010
Solution:
For 11111
= 1 x 2⁴ + 1 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2°
= 16 + 8 + 4 + 2 + 1
= 31 base ten
For 1010
= 1 x 2³ + 0 x 2² + 1 x 2¹ + 0 x 2°
= 8 + 0 + 2 + 0
= 10 base ten
Subtract 10 base ten from 31 base ten
= 21 base ten.
Convert it back to binary
So, 11111 – 1010 = 10101
(d). 1111 x 110
Solution:
Convert both to base ten
For 1111
= 1 x 2³ + 1 x 2² + 1 x 2¹ + 1 x 2°
= 8 + 4 + 2 +1
= 15 base ten
For 110
= 1 x 2² + 1 x 2¹ + 0 x 2°
= 4 + 2 + 0
= 6 base ten
15 x 6 = 90 base ten
Return 90 base ten to binary
Thus, 1111 x 110 = 1011010
### Q12. Simplify 342 base five + 134 base five + 223 base five and leave your answer in base
Solution:
Always convert to base ten first before any other thing.
For 342 base five
= 3 x 5² + 4 x 5¹ + 2 x 5°
= 75 + 20 + 2
= 97 base ten
For 134 base five
= 1 x 5² + 3 x 5¹ + 4 x 5°
= 25 + 15 + 4
= 44 base ten
For 223 base five
= 2 x 5² + 2 x 5¹ + 3 x 5°
= 50 + 10 + 3
= 63 base ten
Then, 97 + 44 – 63 = 78 base ten.
Convert 78 base ten to the required base, which is base five.
So, 342 base five + 134 base five + 223 base five = 303 base five.
### Q13. Multiply (12012) base three by (201) base three.
Solution:
For 12012 base three
= 1 x 3⁴ + 2 x 3³ + 0 x 3² + 1 x 3¹ + 2 x 3°
= 81 + 54 + 0 + 3 + 2
= 140 base ten
For 201 base three
= 2 x 3² + 0 x 3¹ + 1 x 3°
= 18 + 0 + 1
= 19 base ten
140 by 19 = 140 x 19 = 2660 base ten
Convert the 2660 base ten back to base three.
Therefore, (12012) base three by (201) base three = 10122112 base three.
Q14. Divide (100001) base two by (11) base two
Solution:
Convert to base ten.
For 100001 base two
1 x 2^5 + 0 x 2⁴ + 0 x 2³ + 0 x 2² + 0 x 2¹ + 1 x 2°
= 32 + 0 + 0 + 0 + 0 + 1
= 33 base ten
For 11 base two
= 1 x 2¹ + 1 x 2°
= 2 + 1 = 3 base ten
Now divide
33 ÷ 3 = 11 base ten
Convert it back to base two.
Therefore, 100001 base two divided by 11 base two = 1011 base two.
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# What is the integer of 285?
## numere-prime.ro
### Division of the numbers
If it is: 12/4 = 3, remainder 0 and 15/4 = 3, remainder 3, we say 12 divides by 4 but 15 doesn't divide by 4. So 4 is the divisor of 12 but not from 15.
### A natural number "p" shares another number "q" if there is a third natural number "r" with "q = r × p". It is also said that "p" is a factor of "q".
0 is divisible by any number. Any number a, indefinitely from 0, is divisible by 1 and with itself - these are called improper divisors.
### Partial rules
The number 84 is divided by 4 and 3, and can also be divided by 4 * 3 = 12. This is not true if the two factors between them are prime. In general, if a is divisible by m and n and gcd (m, n) = 1, then this is also divisible by m × n.
Establishing the divisors, i.e. identifying that one number is divisible by another, is used a great deal to simplify fractions.
The established rules for finding the divisors are based on the case that the numbers are written in the decimal system. The ten multiples divide by 2 and 5 because 10 divides by 2 and 5; multiples of 100 are divisible by 4 and 25, because 100 is divided by 4 and 25; the multiples of 1,000 divide by 8 because 1,000 divides by 8. All powers of 10, when divided by 3 and 9 have the remainder equal to 1.
Because of the rules with remainder operations, when dividing by 3 and 9 we have the following remainder: 600 has a remainder equal to 6 = 1 × 6 (1 for every hundred); 240 = 2 × 100 + 4 × 10, then the remainder is equal to 2 × 1 + 4 × 1 = 6. When dividing a number at 3 or 9 the remainder is equal to that obtained by dividing the total numbers by their number 3 or 9. 7,309 has the sum of the numbers 7 + 3 + 0 + 9 = 19, which divides by both 3 and 9 without a remainder. So 7.309 is not divided by both 3 and 9.
All even powers of 10, 100, 10,000, 1,000,000 etc, when divided by 11 have an equal remainder with 1, and the odd powers of 10, when divided by 11 have an equal remainder with 10 or 10 - 11 = -1 . In this case, the alternate sum of the numbers has the same remainder as the number. How the alternate sum is calculated is shown in the example below.
Example. 85.976: 8 + 9 + 6 = 23 + 5 + 7 = 12, alternating the sum of the numbers. 23 - 12 = 11. So 85.976 is divided by 11.
A number is divided by:
• 2 if the last number is divisible by 2
• 4 if the last two numbers form a number divisible by 4;
• 8 if the last three numbers form a number divisible by 8;
• 5 if the last number is divisible by 5, i.e. 5 and 0;
• 25 if the last two numbers form a number divisible by 25;
• 3 if the sum of the numbers divides by 3;
• 9 if the sum of the numbers divides by 9;
• 11 if the alternating sum of the numbers is divided by 11.
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# Laws of Exponents (Index Laws)
$\textbf{Laws of Exponents (Index Laws)}$
$$\large a^x \times a^y = a^{x+y}$$
To $\textit{multiply}$ numbers with the $\textit{same base}$, keep the base and $\textit{add}$ the exponents.
$$\large \dfrac{a^x}{a^y} = a^x \div a^y = a^{x-y}$$
To $\textit{divide}$ numbers with the $\textit{same base}$, keep the base and $\textit{substract}$ the exponents.
$$\large (a^x)^y = a^{x \times y}$$
When $\textit{raising a power to a power}$, keep the base and $\textit{multiply}$ the exponents.
$$\large (ab)^x = a^xb^x$$
The power of a product is the product of the powers.
$$\large \Big(\dfrac{a}{b}\Big)^x = \dfrac{a^x}{b^y}, b \ne 0$$
The power of a quotient is the quotient of the powers.
$$\large a^0 = 1, a \ne 0$$
Any non-zero number raised to the power of zero is 1.
$$\large a^{-x} = \frac{1}{a^x} \text{ and } \frac{1}{a^{-x}} = a^x, a \ne 0$$
### Example 1
Simplify $a^5 \times a^6$ using Laws of Exponents (Index Laws).
\begin{align} \displaystyle a^5 \times a^6 &= a^{5+6} \\ &= a^{11} \end{align}
### Example 2
Simplify $\dfrac{a^7}{a^3}$.
\begin{align} \displaystyle \dfrac{a^7}{a^3} &= a^{7-3} \\ &= a^4 \end{align}
### Example 3
Simplify $(a^3)^4$.
\begin{align} \displaystyle (a^3)^4 &= a^{3 \times 4} \\ &= a^{12} \end{align}
### Example 4
Simplify $(a^2b^3)^4$.
\begin{align} \displaystyle (a^2b^3)^4 &= a^{2 \times 4} b^{3 \times 4} \\ &= a^8 b^{12} \end{align}
### Example 5
Write $\dfrac{a^{-2}}{b^{-3}}$ without negative exponents.
\begin{align} \displaystyle \dfrac{a^{-2}}{b^{-3}} &= \dfrac{b^3}{a^2} \end{align}
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## Rates of Change
Calculating rates of change is an important part of the GCSE Maths curriculum for students studying the higher paper.
The graph below shows the cost of three different mobile phone tariffs.
Line A shows a direct proportion. The gradient of the line represent the rate of change.
The formula is therefore the change in the y axis divided by the change in the x axis.
In this example that equals 10 ÷ 40 = 0.25. This represents a charge of 25p per minute and shows a constant proportion.
Line B shows a fixed charge of £5 regardless of how many calls are made. The gradient is 20 ÷ 100 = 0.2 or 20p per minute.
Line C shows a fixed charge of £12.50, with no call charges up to 60 minutes. After 60 minutes the gradient is 3 ÷ 60 = 0.05 0r 5p a minute.
You can use the graph to see which tariff represents best value for you. After about 55 minutes this would be tariff C, if you use fewer minutes than this it would be tariff A. Tariff B is therefore never the best value option.
Repetitive rate of change
With repetitive rates of change the percentage change is applied more than once. You therefore have to calculate one step at a time.
For example, if a business buys a computer for £1,000. In the first year it depreciates by 25%, the next year it loses 20% of its value and then 10% every year after that.
You may then be asked how much is the computer worth after 3 years? To calculate this you can use one of two methods.
Method 1: Step by Step:
Lose 25% so worth 75% after year one.
75% of 1,000 = 0.75 x 1000 = £750
Then 80% of 750 = 0.80 x 750 = £600
Then 90% of 600 = 0.90 x 600 = £540
So the computer is worth £540 after 3 years.
Method 2: Using multipliers:
Lose 25% so worth 75%
Lose 20% so worth 80%
Lose 10% so worth 90%
0.75 x 0.80 x 0.90 = 0.54
Then 0.54 x 1000 = £540
|
Related Articles
# Union of Sets
The representation of similar types of data is called the set. Union of a set is the basic operation on the sets which is used to find all the entries of the given sets. It is one of the operators on the set which is used to solve the set theory problems. Union of two sets means finding a set that contains all the values in both the given sets. It is denoted using the symbol ‘∪’ and is read as the union, i.e.
If A = {1,3.5.7} and B = {2,4,6,8} then A∪B is read as A union B and its value is,
A∪B = {1,2,3,4,5,6,7,8}
Thus, from the above example, it is clear that a set that contains all the elements of set A and set B is called the union of set A and B.
In this article, we will learn about the union of sets, its definition, its properties, and others.
## What is Union of Sets?
The union of any two or more sets is a set that contains all the elements of the previous sets. The union of two sets is equivalent to the logical operation OR and its means any of the given values, for example, if we take a set A = {a, e, i, o, u} and set B = {a, b, c, d, e} then OR operations signifies any of the value of set A and set B and this can be written as A∪B and its value is equal to,
A∪ B = {a, b, c, d, e, i, o, u}
Here, the union set contains all the values either it is in set A or in set B.
In general, for two sets, set A and set B we represent the union of sets in set builder form as,
A ∪ B = {x: x ∈ A or x ∈ B}
## Finding a Union of Sets
We can easily find the union of two sets by taking all the elements of both sets and removing the common elements. Let’s learn this concept through an example.
Example: Find the union of the sets, set A = {p, q, r, s, t, u} and set B = {s, t, u, v, w,}.
Solution:
The union of set A and set B is found by taking all the elements of set A and set B and taking the common element only once.
A∪ B = {p, q, r, s, t, u, v, w}
Here, all the elements of set A and set B are taken and the elements which appear twice (s,t,u) are taken only once.
### Notation of Union of Sets
Union of the sets is represented using the symbol “∪”. It is placed between two sets whose union is to be found. We read this symbol as “union”. Example A∪B is read as A union B, furthermore we can also find the union of two or more sets as the union of set A, set B, and set C is represented as, A∪B∪C and is read as A union B union C.
Note: We can find the union for any number of finite or countable infinite sets.
## A Union B Formula
As we already discussed set A union B contains all the elements of set A as well as set B, but there are some formulas related to the A U B operation that helps us calculate many things. One such formula involving union of two sets, is discussed as follows:
### Formula for Number of Elements in A union B
To find the number of elements in the set of A union B, we can use the following formula:
n(A U B) = n(A) + n(B) – n(A ∩ B)
Where,
• n(A U B) is the number elements in A U B,
• n(A) is the number of elements in A,
• n(B) is the number of elements in B, and
• n(A ∩ B) is the number of elements that are common to both A and B.
Note: n(A) or |A| is called the cardinality of the set A i.e., the number of elements set A contains.
Also Check:
## Venn Diagram of Union of Sets
The union of the set can also be represented using the Venn Diagrams. For example, if we have set A and set B which have some values in common then their Venn diagram is represented in the image below,
We can represent any set using the Venn diagram in the Venn diagram explained above the rectangle represents the Universal set, and set A and set B are represented using the circles. The common area of the two sets represents the intersection of the two sets and both the circles combined along with the common area represent the union of the set.
## Properties of Union of Sets
The intersection of set has various properties. The table below discusses the properties of the union of the set.
Properties of Union
Notation
Commutative Property A∪ B = B ∪ A
Associative Property (A ∪ B) ∪ C = A ∪ (B ∪ C)
Identity Law (Property of Ⲫ) A ∪ ∅ = A
Property of Universal Set A ∪ U = U
Idempotent Property A ∪ A = A
Now let’s learn about these properties in detail.
### Commutative Property
The commutative property of the union of the set explains that the order in which the union of two sets is taken is not important. For example, if take the union of two sets, set A and set B then the value of A ∪ B is equal to the B ∪ A. We can write this property as,
A ∪ B = B ∪ A
Example: Take two sets, set A = {1, 3, 5, 7}, and set B = {a, b, c, d} and find their union.
Solution:
Given sets,
A = {1,3,5,7}
B = {a,b,c,d}
Now, for proving the commutative property.
A ∪ B = {1,3,5,7} ∪ {a,b,c,d}
⇒ A ∪ B = {1,3,5,7,a,b,c,d}…(i)
Similarly,
B ∪ A = {a,b,c,d} ∪ {1,3,5,7} = {a,b,c,d,1,3,5,7}
As we know the order of elements is not important in sets so,
B ∪ A = {a, b, c, d, 1, 3, 5, 7}
⇒ B ∪ A = {1, 3, 5, 7, a, b, c, d}…(ii)
Thus from (i) and (ii) we say that
A ∪ B = B ∪ A,
Thus, commutative property for union of sets can be varified.
### Associative Property
The associative property of the union of the set explains that the order in which the two sets are grouped for finding the union of two or more sets is not important. For example, if take the union of three finite sets, set A, set B, and set C then,
(A ∪ B) ∪ C = A ∪ (B ∪ C)
Example: Take three sets, set P = {1, 3, 5, 7}, set Q = {a, b, c, d}, and set R = {p, q, r, s}. Verify Associative property.
Solution:
Given sets,
P = {1, 3, 5, 7}
Q = {a, b, c, d}
R = {p, q, r, s}
Now, for proving the associative property.
P ∪ Q = {1,3,5,7} ∪ {a,b,c,d} = {1,3,5,7,a,b,c,d}
⇒ (P ∪ Q) ∪ R = {1, 3, 5, 7, a, b, c, d} ∪ {p, q, r, s}
⇒ (P ∪ Q) ∪ R = {1, 3, 5, 7, a, b, c, d, p, q, r, s}…(i)
Similarly,
Q ∪ R = {a,b,c,d} ∪ {p,q,r,s} = {a,b,c,d,p,q,r,s}
⇒ P ∪ (Q ∪ R) = {1, 3, 5, 7} ∪ {a, b, c, d, p, q, r, s}
⇒ (P ∪ Q) ∪ R= {1, 3, 5, 7, a, b, c, d, p, q, r, s}…(ii)
Thus from (i) and (ii) we say that
(P ∪ Q) ∪ R = P ∪ (Q ∪ R)
Thus, the associative property of the union of the set is verified.
### Identity Law (Property of Ⲫ)
The Identity Law of the union of the sets states that the union of any set with an identity element will result in the same set. It can be represented as
A ∪ Ⲫ = A
where Ⲫ is the identity set or null set. This is also called the Property of Ⲫ or the Property of identity set.
Example: If A = {1,2,3,4,5,6} prove A ∪ Ⲫ = A
Solution:
Given,
A ∪ Ⲫ = {1, 2, 3, 4, 5, 6} ∪ { } = {1, 2, 3, 4, 5, 6}
⇒ A ∪ Ⲫ = A
Thus, Identity Law is verified.
### Property of Universal Set
Property of the Universal Set of the union of the sets states that the union of any set with the universal set will result in the Universal set. It can be represented as
A ∪ U = U
Note: This property is sometimes referred to as Domination Law.
Example: If A = {1,2,3} and U = {1,2,3,4,5,6,7,8} then prove A ∪ U = U
Solution:
Given,
A ∪ U = {1, 2, 3} ∪ {1, 2, 3, 4, 5, 6, 7, 8}
⇒ A ∪ U = {1, 2, 3, 4, 5, 6, 7, 8}
⇒ A ∪ U = U
Thus, Property of Universal set is verified.
### Idempotent Property
Idempotent property of the union of the sets states that the union of any set with itself will result in the same set. It can be represented as
A ∪ A = A
Example: If A = {1, 2, 3, 4, 5, 6} then verify the idempotent property.
Solution:
Given,
A ∪ A = {1, 2, 3, 4, 5, 6} ∪ {1, 2, 3, 4, 5, 6}
⇒ A ∪ A = {1,2,3,4,5,6}
⇒ A ∪ A = A
Thus, Idempotent Property is verified.
Also Check:
## Union of Sets Examples
Example 1: Find the Union of the sets,
• A = {1, 2, 3, 4, 5, 6}
• B = {5, 6, 7, 8, 9}
Solution:
Given set,
Set A = {1, 2, 3, 4, 5, 6}
Set B = {5, 6, 7, 8, 9}
Union of sets
A∪ B = {1, 2, 3, 4, 5, 6} ∪ {5, 6, 7, 8, 9}
⇒ A∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Example 2: Find the Union of the sets given below,
• P = {a, e, i, o, u}
• Q = {p, q, r, s, t}
• R = {j, k, l, m, n}
Solution:
Given set,
P = {a, e, i, o, u}
Q = {p, q, r, s, t}
R = {j, k, l, m, n}
Thus, P∪ Q∪ R = {a, e, i, o, u} ∪ {p, q, r, s, t} ∪ {p, q, r, s, t}
⇒ P∪ Q∪ R = {a, e, i, o, u, p, q, r, s, t, j, k, l, m, n}
Example 3: Find the union of sets P and Q, if P = {1, 2, 3, 4, 5} and Q = Ⲫ.
Solution:
Given,
Set P = {1,2,3,4,5}
Set Q = Ⲫ
We know that,
P ∪ Ⲫ = P
⇒ P ∪ Q = {1,2,3,4,5} ∪ Ⲫ
⇒ P ∪ Q = {1,2,3,4,5} = P
Example 4: Find the union of Q = Sets of Rational Nimbers and Qo = Set of Irrational Numbers
Solution:
We know that,
Set of Rational Numbers, Q = {p/q where p, q ∈ z, q ≠0}
Set of Irrational numbers, Qo = {x where x is not a rational number}
Union of these two sets is Q ∪ Qo we know that,
Q ∪ Qo = R {Real Numbers}
Thus, the union of the set of rational numbers and the set of irrational numbers is Real Numbers.
## FAQs on Union of Sets
### 1. What is the Union of Sets in Maths?
In maths, union of sets is the operation on the set which finds a set such that it has all the elements of all the sets of which the union is taken (duplicity of the elements is not allowed.)
### 2. What is the Symbol for Union of Sets?
The symbol which is used to represent the union of the set is ‘∪’. If we have to find the union of set A and set B we write A ∪ B and it is read as A union B.
### 3. What is the difference between Intersection and Union of Sets?
In the union of the set all the unique elements of both sets are taken whereas, in the intersection of the set, only the common elements of the set are taken.
### 4. What is Union of Two Sets?
The union of the two sets is the set which contains all the elements of set A and set B but the duplicity of the element is not allowed.
### 5. How to find Union of Sets?
To find the union of the set follow the steps given below.
Step 1: Compare all the elements of the given set
Step 2: List all the elements of the first set.
Step 3: List all those elements of the second set which are not in the first set.
Step 4: Similarly, repeat step 3 for all the given sets.
Step 5: The resultant set so obtained is the union of all the given sets.
### 6. What is the definition of Union in Sets?
In set theory, the union of two or more sets refers to the combination of all distinct elements present in any of the given sets, resulting in a new set containing those unique elements.
### 7. What are the Applications of Sets?
Sets have diverse applications:
• Mathematics: Fundamental for defining relationships.
• Computer Science: Data structures, databases, and algorithms.
• Statistics: Probability and data analysis.
• Social Sciences: Modeling networks and populations.
• Engineering: Properties and systems.
• Programming: Managing and manipulating data.
• Physics and Biology: Modeling complex systems.
• Finance and Medicine: Risk assessment and data classification.
• Operations Research: Optimization problems.
### 8. What does ∩ and ∪ mean in Math?
In maths, ∩ means intersection of two sets and ∪ means union of sets.
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# If a[1]=4 and r=2, what are the first four terms of the geometric sequence?
Mar 13, 2016
$4 , 8 , 16 , 32$
#### Explanation:
Recall that the geometric sequence formula is written as:
$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{n} = a {r}^{n - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where:
${t}_{n} =$term number
$a =$first term
$r =$common ratio
$n =$number of terms
Determining the First Four Terms
$1$. Since the value of $a$ has already been given, the first term is $4$.
$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{1} = 4 \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$\textcolor{red}{\Rightarrow}$Sequence thus far: $4 , \ldots$
$2$. Using the geometric sequence formula, substitute your known values to determine the second term.
${t}_{n} = a {r}^{n - 1}$
${t}_{2} = 4 {\left(2\right)}^{2 - 1}$
${t}_{2} = 4 {\left(2\right)}^{1}$
$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{2} = 8 \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$\textcolor{red}{\Rightarrow}$Sequence thus far: $4 , 8 , \ldots$
$3$. Repeat for the third term.
${t}_{n} = a {r}^{n - 1}$
${t}_{3} = 4 {\left(2\right)}^{3 - 1}$
${t}_{3} = 4 {\left(2\right)}^{2}$
${t}_{3} = 4 \left(4\right)$
$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{3} = 16 \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$\textcolor{red}{\Rightarrow}$Sequence thus far: $4 , 8 , 16 , \ldots$
$4$. Repeat for the fourth term.
${t}_{n} = a {r}^{n - 1}$
${t}_{4} = 4 {\left(2\right)}^{4 - 1}$
${t}_{4} = 4 {\left(2\right)}^{3}$
${t}_{4} = 4 \left(8\right)$
$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{4} = 32 \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$\textcolor{red}{\Rightarrow}$Sequence thus far: $4 , 8 , 16 , 32$
$\therefore$, the first four terms of the sequence are $4 , 8 , 16 , \text{and}$ $32$.
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# Class 5 Maths Shapes And Angles | Important Questions
IMPORTANT QUESTIONS FOR CBSE EXAMINATION | CLASS 5 MATHEMATICS
SHAPES AND ANGLES – Lesson 2
Fill in the blanks (1 mark):
1. Angle made by two right angles is a———————-
2. The measure of a straight angle is ——————
3. Half of a right angle is ——————-
4. One-third of a right angle is —————-
5. Sum of the angles of any triangle is —————-
6. When two lines meet each other ———– angles are formed.
7. Angle indicated in the clock time 3’O Clock is —————–
8. An angle whose measure is less than 90 degree is known as —————–
9. An angle whose measure is greater than 90 degree is known as ————–
10. An angle whose measure is 90 degree is known as —————–
11. The sum of 2 angles in a triangle is 100 degree. What is the measure of the third angle?
12. Write the name of the angle formed when the clock time is
a) 7.15pm
b) 9’ O Clock
c) 11’ O Clock
13. Classify the following angles as acute, obtuse and right angle.
a) 120
b) 50
c) 90
d)30
e) 150
f) 2 x 45
14. What is the measure of the angle formed by the hands of the clock at
a) 1’ O Clock
b) 2’ O Clock
c) 3’ O Clock
15. Write five English alphabets where you see a right angle
1. Straight angle
2. 180 degree
3. 45 degree
4. 30 degree
5. 180 degree
6. Four
7. Right angle
8. Acute angle
9. Obtuse angle
10. Right angle
11. Sum of 2 angles = 100 degree
Measure of third angle = 180 – 100 = 80 degree
12. a) Obtuse angle
b) Right angle
c) Acute angle
13. a) 120 – Obtuse angle
b) 50 – Acute angle
c) 90 – Right angle
d) 30 – Acute angle
e) 150 – Obtuse angle
f) 2 x 45 – Right angle
14. a) 30 degree
b) 60 degree
c) 90 degree
15. E, F, H, L, T
### 8 Responses
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This is a very helpful page
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• Ankur Singh says:
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|
Difference between revisions of "2007 AMC 12A Problems/Problem 16"
Problem
How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
$\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256$
Solution 1
We can find the number of increasing arithmetic sequences of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.
Common difference Sequences possible Number of sequences 1 $012, \ldots, 789$ 8 2 $024, \ldots, 579$ 6 3 $036, \ldots, 369$ 4 4 $048, \ldots, 159$ 2
This gives us a total of $2 + 4 + 6 + 8 = 20$ sequences. There are $3! = 6$ to permute these, for a total of $120$.
However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are $2! \cdot 4 = 8$ numbers which start with zero, so our answer is $120 - 8 = 112 \Longrightarrow \mathrm{(C)}$.
Solution 2
Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit gives $3! = 6$ possible 3-digit numbers; otherwise, $4$ possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not.
Case 1: 0 is not in the number. Then there are $\binom{5}{2} + \binom{4}{2} = 16$ ways to choose two nonzero digits of the same parity, and each choice generates $3! = 6$ 3-digit numbers, giving $16 \times 6 = 96$ numbers.
Case 2: 0 is in the number. Then there are $4$ ways to choose the largest digit (2, 4, 6, or 8), and each choice generates $4$ 3-digit numbers, giving $4 \times 4 = 16$ numbers.
Thus the total is $96 + 16 = 112 \Longrightarrow \mathrm{(C)}$. (by scrabbler94)
~ pi_is_3.14
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# Writing down Variable Expressions Involving Two Operations
In this guide, the process of writing variable expressions that include two operations will be explained. By comprehending the use of two operations, composing variable expressions becomes effortless.
## A step-by-step guide toWriting down Variable Expressions Involving Two Operations
Variables are considered letters whose values are unknown.
For instance, w is the variable in the expression: 2w$$+$$42
Coefficients are numerical values used along with a variable.
For instance, 2 is the coefficient in the expression 2w$$+$$42.
To identify the operations, you have to write down a mathematical expression and find the keywords.
To convert the description into an expression, you have to use the keywords to do one operation at a time.
Here is a step-by-step guide to writing down variable expressions involving two operations:
1. Identify the operations to be performed: Determine the two operations that need to be performed on the variables in the expression. For example, the operations could be addition and multiplication, or subtraction and division.
2. Write down the expression using variables: Use variables to represent the values in the expression. For example, if you are given the expression “3 + 4x,” you would write “x + 4x.”
3. Use proper mathematical symbols for the operations: Use the appropriate mathematical symbols for the operations identified in step 1. For example, if the operations are addition and multiplication, use the “+” symbol for addition and the “x” symbol for multiplication.
4. Simplify the expression: If possible, simplify the expression by combining like terms and using the distributive property.
5. Check your work: Make sure the expression makes sense and that the answer is in the correct form.
By following these steps, you will be able to write down variable expressions involving two operations.
Remember, the key is to clearly identify the operations to be performed, use variables to represent the values, and use proper mathematical symbols for the operations.
### Writing down Variable Expressions Involving Two Operations – Example 1
Write an expression for this sequence of operations.
Multiply f by 9, then multiply w by the result.
Solution:
Step 1: Find the keywords. “Multiply” is a keyword. Now write the expression.
Step 2: Change the first keyword: Multiply f by $$9→ f×9$$
Step 3: Change the second keyword: Multiply w by the result. w$$×$$ (result)
Step 4: So, w$$(f×9)$$
### Writing down Variable Expressions Involving Two Operations – Example 2
Write an expression for this sequence of operations.
Divide 4 by x, then subtract 12 from the result.
Solution:
Step 1: Find the keywords. Divide and subtract the keywords.
Step 2: Change the first keyword: Divide 4 by x→ $$\frac{4}{x}$$
Step 3: Change the second keyword: Subtract 12 from the result→ result $$-12$$
Step 4: So, $$\frac{4}{x}-12$$
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HomeTren&dThe Orthocenter of a Triangle Formula: Explained and Illustrated
The Orthocenter of a Triangle Formula: Explained and Illustrated
Triangles are fundamental shapes in geometry, and understanding their properties is essential for various mathematical applications. One of the key concepts related to triangles is the orthocenter, which plays a significant role in triangle analysis and construction. In this article, we will explore the orthocenter of a triangle formula, its significance, and how it can be calculated. We will also provide examples and case studies to illustrate its practical applications.
What is the Orthocenter of a Triangle?
The orthocenter of a triangle is the point where the altitudes of the triangle intersect. An altitude is a line segment drawn from a vertex of the triangle perpendicular to the opposite side. Each triangle has its unique orthocenter, which can be inside, outside, or on the triangle itself.
The orthocenter is denoted by the letter H, and it is a crucial point in triangle analysis. It has several interesting properties and applications in various fields, including mathematics, physics, and engineering.
How to Calculate the Orthocenter of a Triangle
Calculating the orthocenter of a triangle requires knowledge of the triangle’s vertices. There are several methods to determine the orthocenter, including algebraic, geometric, and trigonometric approaches. In this article, we will focus on the geometric method, which is widely used and relatively straightforward.
Geometric Method:
To calculate the orthocenter of a triangle using the geometric method, follow these steps:
1. Draw the triangle and label its vertices as A, B, and C.
2. Construct the altitudes of the triangle by drawing perpendicular lines from each vertex to the opposite side.
3. Extend the altitudes until they intersect. The point of intersection is the orthocenter of the triangle.
It is important to note that not all triangles have an orthocenter. For example, an equilateral triangle, where all sides are equal, does not have a unique orthocenter. In such cases, the altitudes coincide with the medians and the circumcenter.
Practical Applications of the Orthocenter
The orthocenter of a triangle has various practical applications in different fields. Let’s explore a few examples to understand its significance:
Architecture and Engineering:
In architecture and engineering, the orthocenter is used to determine the stability and strength of structures. By analyzing the orthocenter, engineers can identify potential weak points in a structure and make necessary adjustments to ensure its stability.
For example, when designing a bridge, engineers consider the orthocenter to determine the optimal placement of support columns. By placing the columns at or near the orthocenter, they can distribute the load evenly and minimize stress on the structure.
The orthocenter is also used in navigation and surveying to calculate distances and angles. By knowing the orthocenter of a triangle formed by three landmarks, surveyors can determine their relative positions accurately.
For instance, in geodesy, the science of measuring the Earth’s shape and size, the orthocenter is used to calculate the geodetic coordinates of specific points on the Earth’s surface. This information is crucial for mapping, satellite positioning systems, and other geospatial applications.
Examples of Calculating the Orthocenter
Let’s explore a couple of examples to illustrate how to calculate the orthocenter of a triangle:
Example 1:
Consider a triangle with vertices A(2, 4), B(6, 2), and C(8, 6). To find the orthocenter, we can follow the geometric method:
1. Plot the triangle on a coordinate plane.
2. Construct the altitudes by drawing perpendicular lines from each vertex to the opposite side.
3. Extend the altitudes until they intersect. The point of intersection is the orthocenter.
After following these steps, we find that the orthocenter of the triangle is H(6, 4).
Example 2:
Let’s consider another triangle with vertices A(0, 0), B(4, 0), and C(2, 6). Using the same geometric method, we can calculate the orthocenter:
1. Plot the triangle on a coordinate plane.
2. Construct the altitudes by drawing perpendicular lines from each vertex to the opposite side.
3. Extend the altitudes until they intersect. The point of intersection is the orthocenter.
After following these steps, we find that the orthocenter of the triangle is H(2, 0).
Summary
The orthocenter of a triangle is a significant point that plays a crucial role in triangle analysis and construction. It is the point where the altitudes of a triangle intersect. Calculating the orthocenter can be done using various methods, including the geometric approach. The orthocenter has practical applications in fields such as architecture, engineering, navigation, and surveying.
Understanding the orthocenter and its properties allows us to analyze and manipulate triangles effectively. By leveraging the orthocenter, we can design stable structures, accurately determine positions, and solve complex geometric problems. So, the next time you encounter a triangle, remember the orthocenter and its formula to unlock its hidden potential.
Q&A
Q1: Can all triangles have an orthocenter?
A1: No, not all triangles have an orthocenter. An equilateral triangle, for example, does not have a unique orthocenter. In such cases, the altitudes coincide with the medians and the circumcenter.
A2: The orthocenter, centroid, and circumcenter are three important points associated with a triangle. The centroid is the point where the medians of a triangle intersect, while the circumcenter is the point where the perpendicular bisectors of the sides intersect. In some cases, these three points coincide, but in general, they are distinct.
Q3: Can the orthocenter be located outside the triangle?
A3: Yes, the orthocenter can be located outside the triangle. This occurs when the triangle is obtuse, meaning it has an angle greater than 90 degrees. In such cases, the orthocenter lies outside the triangle.
Q4: Are there any other methods to calculate
Riya Sharma
Riya Sharma is a tеch bloggеr and UX/UI dеsignеr spеcializing in usеr еxpеriеncе dеsign and usability tеsting. With еxpеrtisе in usеr-cеntric dеsign principlеs, Riya has contributеd to crafting intuitivе and visually appеaling intеrfacеs.
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# Minimize the absolute difference of sum of two subsets
Last Updated : 08 Jun, 2022
Given a number n, divide first n natural numbers (1, 2, …n) into two subsets such that difference between sums of two subsets is minimum.
Examples:
```Input : n = 4
Output : First subset sum = 5,
Second subset sum = 5.
Difference = 0
Explanation:
Subset 1: 1 4
Subset 2: 2 3
Input : n = 6
Output: First subset sum = 10,
Second subset sum = 11.
Difference = 1
Explanation :
Subset 1: 1 3 6
Subset 2: 2 4 5 ```
Approach:
The approach is based on the fact that any four consecutive numbers can be divided into two groups by putting middle two elements in one group and corner elements in other group. So, if n is a multiple of 4 then their difference will be 0, hence the summation of one set will be half of the summation of N elements which can be calculated by using sum = n*(n+1)/2
There are three other cases to consider in which we cannot divide into groups of 4, which will leave a remainder of 1, 2 and 3:
a) If it leaves a remainder of 1, then all other n-1 elements are clubbed into group of 4 hence their sum will be int(sum/2) and the other half sum will be int(sum/2+1) and their difference will always be 1.
b) Above mentioned steps will be followed in case of n%4 == 2 also. Here we form groups of size 4 for elements from 3 onward. Remaining elements would be 1 and 2. 1 goes in one group and 2 goes in other group.
c) When n%4 == 3, then club n-3 elements into groups of 4. The left out elements will be 1, 2 and 3, in which 1 and 2 will go to one set and 3 to the other set which eventually makes the difference to be 0 and summation of each set to be sum/2.
Below is the implementation of the above approach:
## CPP
`// CPP program to Minimize the absolute ` `// difference of sum of two subsets` `#include ` `using` `namespace` `std;` `// function to print difference` `void` `subsetDifference(``int` `n)` `{` ` ``// summation of n elements` ` ``int` `s = n * (n + 1) / 2;` ` ``// if divisible by 4` ` ``if` `(n % 4 == 0) {` ` ``cout << ``"First subset sum = "` ` ``<< s / 2;` ` ``cout << ``"\nSecond subset sum = "` ` ``<< s / 2;` ` ``cout << ``"\nDifference = "` `<< 0;` ` ``}` ` ``else` `{` ` ``// if remainder 1 or 2. In case of remainder` ` ``// 2, we divide elements from 3 to n in groups` ` ``// of size 4 and put 1 in one group and 2 in ` ` ``// group. This also makes difference 1.` ` ``if` `(n % 4 == 1 || n % 4 == 2) {` ` ``cout << ``"First subset sum = "` ` ``<< s / 2;` ` ``cout << ``"\nSecond subset sum = "` ` ``<< s / 2 + 1;` ` ``cout << ``"\nDifference = "` `<< 1;` ` ``}` ` ``// We put elements from 4 to n in groups of` ` ``// size 4. Remaining elements 1, 2 and 3 can` ` ``// be divided as (1, 2) and (3).` ` ``else` ` ``{` ` ``cout << ``"First subset sum = "` ` ``<< s / 2;` ` ``cout << ``"\nSecond subset sum = "` ` ``<< s / 2;` ` ``cout << ``"\nDifference = "` `<< 0;` ` ``}` ` ``}` `}` `// driver program to test the above function` `int` `main()` `{` ` ``int` `n = 6;` ` ``subsetDifference(n);` ` ``return` `0;` `}`
## Java
`// Java program for Minimize the absolute ` `// difference of sum of two subsets` `import` `java.util.*;` `class` `GFG {` ` ` ` ``// function to print difference` ` ``static` `void` `subsetDifference(``int` `n)` ` ``{` ` ``// summation of n elements` ` ``int` `s = n * (n + ``1``) / ``2``;` ` ` ` ``// if divisible by 4` ` ``if` `(n % ``4` `== ``0``) {` ` ``System.out.println(``"First subset sum = "` `+ s / ``2``);` ` ``System.out.println(``"Second subset sum = "` `+ s / ``2``);` ` ``System.out.println(``"Difference = "` `+ ``0``);` ` ``}` ` ``else` `{` ` ` ` ``// if remainder 1 or 2. In case of remainder` ` ``// 2, we divide elements from 3 to n in groups` ` ``// of size 4 and put 1 in one group and 2 in ` ` ``// group. This also makes difference 1.` ` ``if` `(n % ``4` `== ``1` `|| n % ``4` `== ``2``) {` ` ` ` ``System.out.println(``"First subset sum = "` `+ s / ``2``);` ` ``System.out.println(``"Second subset sum = "` `+ ((s / ``2``) + ``1``));` ` ``System.out.println(``"Difference = "` `+ ``1``);` ` ``}` ` ` ` ``// We put elements from 4 to n in groups of` ` ``// size 4. Remaining elements 1, 2 and 3 can` ` ``// be divided as (1, 2) and (3).` ` ``else` ` ``{` ` ``System.out.println(``"First subset sum = "` `+ s / ``2``);` ` ``System.out.println(``"Second subset sum = "` `+ s / ``2``);` ` ``System.out.println(``"Difference = "` `+ ``0``);` ` ``}` ` ``}` ` ``}` ` ` ` ``/* Driver program to test above function */` ` ``public` `static` `void` `main(String[] args) ` ` ``{` ` ``int` `n = ``6``;` ` ``subsetDifference(n);` ` ``}` `}` ` ` `// This code is contributed by Arnav Kr. Mandal.`
## Python3
`# Python3 code to Minimize the absolute ` `# difference of sum of two subsets` `# function to print difference` `def` `subsetDifference( n ):` ` ``# summation of n elements` ` ``s ``=` `int``(n ``*` `(n ``+` `1``) ``/` `2``)` ` ` ` ``# if divisible by 4` ` ``if` `n ``%` `4` `=``=` `0``:` ` ``print``(``"First subset sum = "``, ``int``(s ``/` `2``))` ` ``print``(``"Second subset sum = "``,``int``(s ``/` `2``))` ` ``print``(``"Difference = "` `, ``0``)` ` ``else``:` ` ``# if remainder 1 or 2. In case of remainder` ` ``# 2, we divide elements from 3 to n in groups` ` ``# of size 4 and put 1 in one group and 2 in` ` ``# group. This also makes difference 1.` ` ``if` `n ``%` `4` `=``=` `1` `or` `n ``%` `4` `=``=` `2``:` ` ``print``(``"First subset sum = "``,``int``(s``/``2``))` ` ``print``(``"Second subset sum = "``,``int``(s``/``2``)``+``1``)` ` ``print``(``"Difference = "``, ``1``)` ` ` ` ``# We put elements from 4 to n in groups of` ` ``# size 4. Remaining elements 1, 2 and 3 can` ` ``# be divided as (1, 2) and (3).` ` ``else``:` ` ``print``(``"First subset sum = "``, ``int``(s ``/` `2``))` ` ``print``(``"Second subset sum = "``,``int``(s ``/` `2``))` ` ``print``(``"Difference = "` `, ``0``)` ` ` `# driver code to test the above function` `n ``=` `6` `subsetDifference(n)` `# This code is contributed by "Sharad_Bhardwaj".`
## C#
`// C# program for Minimize the absolute ` `// difference of sum of two subsets` `using` `System;` `class` `GFG {` ` ` ` ``// function to print difference` ` ``static` `void` `subsetDifference(``int` `n)` ` ``{` ` ` ` ``// summation of n elements` ` ``int` `s = n * (n + 1) / 2;` ` ` ` ``// if divisible by 4` ` ``if` `(n % 4 == 0) {` ` ``Console.WriteLine(``"First "` ` ``+ ``"subset sum = "` `+ s / 2);` ` ` ` ``Console.WriteLine(``"Second "` ` ``+ ``"subset sum = "` `+ s / 2);` ` ` ` ``Console.WriteLine(``"Difference"` ` ``+ ``" = "` `+ 0);` ` ``}` ` ``else` `{` ` ` ` ``// if remainder 1 or 2. In case ` ` ``// of remainder 2, we divide ` ` ``// elements from 3 to n in groups` ` ``// of size 4 and put 1 in one ` ` ``// group and 2 in group. This` ` ``// also makes difference 1.` ` ``if` `(n % 4 == 1 || n % 4 == 2) {` ` ` ` ``Console.WriteLine(``"First "` ` ``+ ``"subset sum = "` `+ s / 2);` ` ` ` ``Console.WriteLine(``"Second "` ` ``+ ``"subset sum = "` `+ ((s / 2) ` ` ``+ 1));` ` ` ` ``Console.WriteLine(``"Difference"` ` ``+ ``" = "` `+ 1);` ` ``}` ` ` ` ``// We put elements from 4 to n ` ` ``// in groups of size 4. Remaining ` ` ``// elements 1, 2 and 3 can` ` ``// be divided as (1, 2) and (3).` ` ``else` ` ``{` ` ``Console.WriteLine(``"First "` ` ``+ ``"subset sum = "` `+ s / 2);` ` ` ` ``Console.WriteLine(``"Second "` ` ``+ ``"subset sum = "` `+ s / 2);` ` ` ` ``Console.WriteLine(``"Difference"` ` ``+ ``" = "` `+ 0);` ` ``}` ` ``}` ` ``}` ` ` ` ``/* Driver program to test above ` ` ``function */` ` ``public` `static` `void` `Main() ` ` ``{` ` ``int` `n = 6;` ` ` ` ``subsetDifference(n);` ` ``}` `}` ` ` `// This code is contributed by vt_m.`
## PHP
``
## Javascript
``
Output
```First subset sum = 10
Second subset sum = 11
Difference = 1```
Time Complexity: O(1)
Auxiliary Space: O(1)
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# RD Sharma Class 7 ex 14.1 Solutions Chapter 14 Lines and Angles
In this chapter, we provide RD Sharma Class 7 ex 14.1 Solutions Chapter 14 Lines and Angles for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 14.1 Solutions Chapter 14 Lines and Angles Maths pdf, Now you will get step by step solution to each question.
# Chapter 14: Lines and Angles Exercise – 14.1
### Question: 1
Write down each pair of adjacent angles shown in Figure
### Solution:
The angles that have common vertex and a common arm are known as adjacent angles
∠DOC and ∠BOC
∠COB and ∠BOA
### Question: 2
In figure, name all the pairs of adjacent angles.
### Solution:
In fig (i), the adjacent angles are
∠EBA and ∠ABC
∠ACB and ∠BCF
In fig (ii), the adjacent angles are
∠BDA and ∠CDA
### Question: 3
In fig , write down
(i) each linear pair
(ii) each pair of vertically opposite angles.
### Solution:
(i) The two adjacent angles are said to form a linear pair of angles if their non – common arms are two opposite rays.
∠1 and ∠3
∠1 and ∠2
∠4 and ∠3
∠4 and ∠2
∠5 and ∠6
∠5 and ∠7
∠6 and ∠8
∠7 and ∠8
(ii) The two angles formed by two intersecting lines and have no common arms are called vertically opposite angles.
∠1 and ∠4
∠2 and ∠3
∠5 and ∠8
∠6 and ∠7
### Question: 4
Are the angles 1 and 2 in figure adjacent angles?
### Solution:
No, because they do not have common vertex.
### Question: 5
Find the complement of each of the following angles:
(i) 35°
(ii) 72°
(iii) 45°
(iv) 85°
### Solution:
The two angles are said to be complementary angles if the sum of those angles is 90°
Complementary angles for the following angles are:
(i) 90° – 35° = 55°
(ii) 90° – 72° = 18°
(iii) 90° – 45° = 45°
(iv) 90° – 85° = 5°
### Question: 6
Find the supplement of each of the following angles:
(i) 70°
(ii) 120°
(iii) 135°
(iv) 90°
### Solution:
The two angles are said to be supplementary angles if the sum of those angles is 180°
(i) 180° – 70° = 110°
(ii) 180° – 120° = 60°
(iii) 180° – 135° = 45°
(iv) 180° – 90° = 90°
### Question: 7
Identify the complementary and supplementary pairs of angles from the following pairs
(i) 25°, 65°
(ii) 120°, 60°
(iii) 63°, 27°
(iv) 100°, 80°
### Solution:
(i) 25° + 65° = 90° so, this is a complementary pair of angle.
(ii) 120° + 60° = 180° so, this is a supplementary pair of angle.
(iii) 63° + 27° = 90° so, this is a complementary pair of angle.
(iv) 100° + 80° = 180° so, this is a supplementary pair of angle.
Here, (i) and (iii) are complementary pair of angles and (ii) and (iv) are supplementary pair of angles.
### Question: 8
Can two obtuse angles be supplementary, if both of them be
(i) obtuse?
(ii) right?
(iii) acute?
### Solution:
(i) No, two obtuse angles cannot be supplementary
Because, the sum of two angles is greater than 90 degrees so their sum will be greater than 180degrees.
(ii) Yes, two right angles can be supplementary
Because, 90° + 90° = 180°
(iii) No, two acute angle cannot be supplementary
Because, the sum of two angles is less than 90 degrees so their sum will also be less tha 90 degrees.
### Question: 9
Name the four pairs of supplementary angles shown in Fig.
### Solution:
The supplementary angles are
∠AOC and ∠COB
∠BOC and ∠DOB
∠BOD and ∠DOA
∠AOC and ∠DOA
### Question: 10
In Figure, A, B, C are collinear points and ∠DBA = ∠EBA.
(i) Name two linear pairs.
(ii) Name two pairs of supplementary angles.
### Solution:
(i) Linear pairs
∠ABD and ∠DBC
∠ABE and ∠EBC
Because every linear pair forms supplementary angles, these angles are
∠ABD and ∠DBC
∠ABE and ∠EBC
### Question: 11
If two supplementary angles have equal measure, what is the measure of each angle?
### Solution:
Let p and q be the two supplementary angles that are equal
∠p = ∠q
So,
∠p + ∠q = 180°
=> ∠p + ∠p = 180°
=> 2∠p = 180°
=> ∠p = 180°/2
=> ∠p = 90°
Therefore, ∠p = ∠q = 90°
### Question: 12
If the complement of an angle is 28°, then find the supplement of the angle.
### Solution:
Here, let p be the complement of the given angle 28°
Therefore, ∠p + 28° = 90°
=> ∠p = 90° – 28°
= 62°
So, the supplement of the angle = 180° – 62°
= 118°
### Question: 13
In Figure, name each linear pair and each pair of vertically opposite angles.
### Solution:
Two adjacent angles are said to be linear pair of angles, if their non-common arms are two opposite rays.
∠1 and ∠2
∠2 and ∠3
∠3 and ∠4
∠1 and ∠4
∠5 and ∠6
∠6 and ∠7
∠7 and ∠8
∠8 and ∠5
∠9 and ∠10
∠10 and ∠11
∠11 and ∠12
∠12 and ∠9
The two angles are said to be vertically opposite angles if the two intersecting lines have no common arms.
∠1 and ∠3
∠4 and ∠2
∠5 and ∠7
∠6 and ∠8
∠9 and ∠11
∠10 and ∠12
### Question: 14
In Figure, OE is the bisector of ∠BOD. If ∠1 = 70°, Find the magnitude of ∠2, ∠3, ∠4
### Solution:
Given,
∠1 = 70°
∠3 = 2(∠1)
= 2(70°)
= 140°
∠3 = ∠4
As, OE is the angle bisector,
∠DOB = 2(∠1)
= 2(70°)
= 140°
∠DOB + ∠AOC + ∠COB +∠DOB = 360°
=> 140° + 140° + 2(∠COB) = 360°
Since, ∠COB = ∠AOD
=> 2(∠COB) = 360° – 280°
=> 2(∠COB) = 80°
=> ∠COB = 80°/2
=> ∠COB = 40°
Therefore, ∠COB = ∠AOB = 40°
The angles are,
∠1 = 70°,
∠2 = 40°,
∠3 = 140°,
∠4 = 40°
### Question: 15
One of the angles forming a linear pair is a right angle. What can you say about its other angle?
### Solution:
One of the Angle of a linear pair is the right angle (90°)
Therefore, the other angle is
=> 180° – 90° = 90°
### Question: 16
One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?
### Solution:
One of the Angles of a linear pair is obtuse, then the other angle should be acute, only then their sum will be 180°.
### Question: 17
One of the angles forming a linear pair is an acute angle. What kind of angle is the other?
### Solution:
One of the Angles of a linear pair is acute, then the other angle should be obtuse, only then their sum will be 180°.
### Question: 18
Can two acute angles form a linear pair?
### Solution:
No, two acute angles cannot form a linear pair because their sum is always less than 180°.
### Question: 19
If the supplement of an angle is 65°, then find its complement.
### Solution:
Let x be the required angle
So,
=> x + 65° = 180°
=> x = 180° – 65°
= 115°
But the complement of the angle cannot be determined.
### Question: 20
Find the value of x in each of the following figures
### Solution:
(i) Since, ∠BOA + ∠BOC = 180°
Linear pair:
=> 60° + x° = 180°
=> x° = 180° – 60°
=> x° = 120°
(ii) Linear pair:
=> 3x° + 2x° = 180°
=> 5x° = 180°
=> x° = 180°/5
=> x° = 36°
(iii) Linear pair,
Since, 35° + x° + 60° = 180°
=> x° = 180° – 35° – 60°
=> x° = 180° – 95°
=> x° = 85°
(iv) Linear pair,
83° + 92° + 47° + 75° + x° = 360°
=> x° + 297° = 360°
=> x° = 360° – 297°
=> x° = 63°
(v) Linear pair,
3x° + 2x° + x° + 2x° = 360°
=> 8x° = 360°
=> x° = 360°/8
=> x∘ = 45°
(vi) Linear pair:
3x° = 105°
=> x° = 105°/3
=> x° = 45°
### Question:
In Fig. 22, it being given that ∠1 = 65°, find all the other angles.
### Solution:
Given,
∠1 = ∠3 are the vertically opposite angles
Therefore, ∠3 = 65°
Here, ∠1 + ∠2 = 180° are the linear pair
Therefore, ∠2 = 180° – 65°
= 115°
∠2 = ∠4 are the vertically opposite angles
Therefore, ∠2 = ∠4 = 115°
And ∠3 = 65°
### Question: 22
In Fig. 23 OA and OB are the opposite rays:
(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?
### Solution:
∠AOC + ∠BOC = 180° – Linear pair
=> 2y + 5 + 3x = 180°
=> 3x + 2y = 175°
(i) If x = 25°, then
=> 3(25°) + 2y = 175°
=> 75° + 2y = 175°
=> 2y = 175° – 75°
=> 2y = 100°
=> y = 100°/2
=> y = 50°
(ii) If y = 35°, then
3x + 2(35°) = 175°
=> 3x + 70° = 175°
=> 3x = 175° – 70°
=> 3x = 105°
=> x = 105°/3
=> x = 35°
### Question: 24
In Figure, write all pairs of adjacent angles and all the linear pairs.
∠DOA and ∠DOC
∠BOC and ∠COD
∠AOD and ∠BOD
∠AOC and ∠BOC
Linear pairs:
∠AOD and ∠BOD
∠AOC and ∠BOC
### Question: 25
In Figure, find ∠x. Further find ∠BOC, ∠COD, ∠AOD
### Solution:
(x + 10)° + x° + (x + 20) ° = 180°
=> 3x° + 30° = 180°
=> 3x° = 180° – 30°
=> 3x° = 150°
=> x° = 150°/3
=> x° = 50°
Here,
∠BOC = (x + 20)°
= (50 + 20)°
= 70°
∠COD = 50°
∠AOD = (x + 10)°
= (50 + 10)°
= 60°
### Question: 25
How many pairs of adjacent angles are formed when two lines intersect in a point?
### Solution:
If the two lines intersect at a point, then four adjacent pairs are formed and those are linear.
### Question: 26
How many pairs of adjacent angles, in all, can you name in Figure?
∠EOD and ∠DOC
∠COD and ∠BOC
∠COB and ∠BOA
∠AOB and ∠BOD
∠BOC and ∠COE
∠COD and ∠COA
∠DOE and ∠DOB
∠EOD and ∠DOA
∠EOC and ∠AOC
∠AOB and ∠BOE
### Question: 27
In Figure, determine the value of x.
### Solution:
Linear pair:
∠COB + ∠AOB = 180°
=> 3x° + 3x° = 180°
=> 6x° = 180°
=> x° = 180°/6
=> x° = 30°
### Question: 28
In Figure, AOC is a line, find x.
### Solution:
∠AOB + ∠BOC = 180°
Linear pair
=> 2x + 70° = 180°
=> 2x = 180° – 70°
=> 2x = 110°
=> x = 110°/2
=> x = 55°
### Question: 29
In Figure, POS is a line, find x.
### Solution:
Angles of a straight line,
∠QOP + ∠QOR + ∠ROS = 108°
=> 60° + 4x + 40° = 180°
=> 100° + 4x = 180°
=> 4x = 180° – 100°
=> 4x = 80°
=> x = 80°/4
=> x = 20°
### Question: 30
In Figure, lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u.
### Solution:
Given that,
∠x = 45°
∠x = ∠z = 45°
∠y = ∠u
∠x + ∠y + ∠z + ∠u = 360°
=> 45° + 45° + ∠y + ∠u = 360°
=> 90° + ∠y + ∠u = 360°
=> ∠y + ∠u = 360° – 90°
=> ∠y + ∠u = 270°
=> ∠y + ∠z = 270°
=> 2∠z = 270°
=> ∠z = 135°
Therefore, ∠y = ∠u = 135°
So, ∠x = 45°,
∠y = 135°,
∠z = 45°,
∠u = 135°
### Question: 31
In Figure, three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u
### Solution:
Given that,
∠x + ∠y + ∠z + ∠u + 50° + 90° = 360°
Linear pair,
∠x + 50° + 90° = 180°
=> ∠x + 140° = 180°
=> ∠x = 180° – 140°
=> ∠x = 40°
∠x = ∠u = 40° are vertically opposite angles
=> ∠z = 90° is a vertically opposite angle
=> ∠y = 50° is a vertically opposite angle
Therefore, ∠x = 40°,
∠y = 50°,
∠z = 90°,
∠u = 40°
### Question: 32
In Figure, find the values of x, y and z
### Solution:
∠y = 25° vertically opposite angle
∠x = ∠y are vertically opposite angles
∠x + ∠y + ∠z + 25° = 360°
=> ∠x + ∠z + 25° + 25° = 360°
=> ∠x + ∠z + 50° = 360°
=> ∠x + ∠z = 360° – 50°
=> 2∠x = 310°
=> ∠x = 155°
And, ∠x = ∠z = 155°
Therefore, ∠x = 155°,
∠y = 25°,
∠z = 155°
All Chapter RD Sharma Solutions For Class 7 Maths
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# How many seismographs are needed to locate the epicenter of an earthquake?
A seismogram from at least three seismic stations is required for scientists to pinpoint the location of an earthquake's epicentre. In order to locate an epicentre, there are three processes to follow: Each of the three locations is visited by the scientists, who measure the difference between the arrival timings of the main and secondary waves.
three
### Second, where exactly are the three seismographs that were utilised to determine the epicentre of this earthquake placed today?
Minneapolis, Detroit, and Charleston, to name a few. Explanation: As seen in the image, three seismographs are stationed (situated) in the cities of Minneapolis, Detroit, and Charleston, and the epicentre of the earthquake may be determined with the aid of these three seismographs.
### Also, understand why three seismographs are required to find an epicentre.
Triangulation is the process of determining the precise arrival time of the P and S waves generated by an earthquake for at least three separate seismograph sites, which is known as triangulation. Because P waves move almost twice as quickly as S waves, the greater the distance between the earthquake's epicentre and the two arrival timings, the bigger the time gap between the two arrival times.
### What is the best method for determining the epicentre of an earthquake?
The estimated epicentre of the earthquake is located at the location where all of the circles come together.
Calculate the distance between the first P wave and the first S wave by using a waveform analyzer.
Find the point on the left side of the chart below that corresponds to 24 seconds and make a note of it.
Take note of the amplitude of the most powerful wave.
### How can you tell when an earthquake is on its way?
Seismograms may also be useful for finding earthquakes, and being able to distinguish between the P wave and the S wave is critical for this. While studying P and S waves, you discovered that they cause various types of ground shaking depending on their speed and direction. P waves travel at a quicker rate than S waves, and it is this characteristic that enables us to determine the location of an earthquake.
### What is the maximum distance that an earthquake of magnitude 7.0 may be felt?
Incredibly, deep focus earthquakes may occur hundreds of miles below the surface and be felt practically half a globe away, according to the scientific literature. With an epicentre 609 kilometres (380 miles) under the ocean's surface off the coast of Russia, just north of Japan, the 2013 Okhotsk Sea Earthquake caused significant shaking that was reported as far away as Moscow.
### Is there a cyclical pattern to earthquakes?
cyclic, although not in the same way as periodic. This process is cyclic because it involves the accumulation of tension over a period of time, the release of that stress (as an earthquake), and the re-accumulation of stress for the following series of earthquakes.
### What is the path taken by S waves?
An S wave travels at a slower rate than a P wave and can only go through solid rock; it cannot travel through any other kind of material. S waves move rock particles up and down, or side-to-side, in a direction that is perpendicular to the direction in which the wave is going. S waves are also known as S-waves (the direction of wave propagation). To view a S wave in action, please visit this page.
### What factors influence the location of epicentres?
Scientists draw circles around the seismograph sites in order to estimate the direction in which each wave travelled. The radius of each circle is the same as the known distance from the epicentre of the circle. The epicentre is the point at which these three circles come together.
### Where do the vast majority of earthquakes occur?
Around the planet, earthquakes occur on a regular basis, both along the borders of plate boundaries and along fault lines. Near the margins of the oceanic and continental plates, the majority of earthquakes occur. The earth's crust (the planet's outermost layer) is made up of numerous parts, collectively referred to as plates.
### What are some of the regions in the United States where the danger of earthquakes is high?
Alaska, Arkansas, California, Hawaii, Idaho, Illinois, Kentucky, Missouri, Montana, Nevada, Oregon, South Carolina, Tennessee, Utah, Washington, and Wyoming are the states with the greatest risk of earthquakes.
### What is the significance of determining the epicentre of an earthquake?
The most important reason for finding the epicentre is so that the fault that broke and caused the earthquake may be located and repaired if necessary. However, if the fault is a well-known fault, the earthquake may be utilised to boost confidence in the hazard modelling for the surrounding region.
### What is the location of seismographs?
A seismograph is a device that measures earthquake (seismic) waves and their amplitudes. Both the bedrock and a concrete base serve as a highly sturdy foundation for these structures. The seismometer itself is made up of a frame and a mass that may move in relation to the frame and vice versa.
### What is the rationale of having a bare minimum of three stations?
A seismogram from at least three seismic stations is required for scientists to pinpoint the location of an earthquake's epicentre. The more distant the epicentre is from the centre, the bigger the disparity in time between the two points. A circle is drawn around each station, with the radius of the circle matching to the distance between the station and the epicentre.
### What is the proper way to interpret a seismogram?
Like a book, the seismogram should be "read" from left to right and from top to bottom (this is the direction that time increases). In a book, the right end of any horizontal line "connects" with the left end of the line below it in the same manner. Each line represents 15 minutes of data; four lines per hour represent one hour of data.
### What is the location of an earthquake's epicentre?
It is the point on the earth's surface that lies vertically above the hypocenter (or focus), the point in the crust where a seismic rupture originates, and is located vertically above the epicentre. The Epicenter and the Hypocenter. (
### What is the frequency at which earthquakes occur?
[3] We know from seismometric data that earthquakes often transmit seismic waves with frequencies ranging from 0.01 to 10 Hz, even though they are capable of generating waves with greater frequencies.
### What method do you use to determine the distance to the epicentre?
It is possible to determine the difference in arrival timings between the first shear (s) wave and the first compressional (p) wave using the seismogram. Calculate the distance between the seismograph station and the epicentre by multiplying the difference by 8.4 to get an estimate of the distance in kilometres.
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# APEX Calculus: for University of Lethbridge
## Section12.1Vector-Valued Functions
We are very familiar with real valued functions, that is, functions whose output is a real number. This section introduces vector-valued functions — functions whose output is a vector.
### Definition12.1.2.Vector-Valued Functions.
A vector-valued function is a function of the form
\begin{equation*} \vec r(t) = \la\, f(t),g(t)\,\ra \text{ or } \vec r(t) = \la \,f(t),g(t),h(t)\,\ra\text{,} \end{equation*}
where $$f\text{,}$$ $$g$$ and $$h$$ are real valued functions.
The domain of $$\vec r$$ is the set of all values of $$t$$ for which $$\vec r(t)$$ is defined. The range of $$\vec r$$ is the set of all possible output vectors $$\vec r(t)\text{.}$$
### Subsection12.1.1Evaluating and Graphing Vector-Valued Functions
Evaluating a vector-valued function at a specific value of $$t$$ is straightforward; simply evaluate each component function at that value of $$t\text{.}$$ For instance, if $$\vec r(t) = \la t^2,t^2+t-1\ra\text{,}$$ then $$\vec r(-2) = \la 4,1\ra\text{.}$$ We can sketch this vector, as is done in Figure 12.1.3.(a). Plotting lots of vectors is cumbersome, though, so generally we do not sketch the whole vector but just the terminal point. The graph of a vector-valued function is the set of all terminal points of $$\vec r(t)\text{,}$$ where the initial point of each vector is always the origin. In Figure 12.1.3.(b) we sketch the graph of $$\vec r\text{;}$$ we can indicate individual points on the graph with their respective vector, as shown.
Vector-valued functions are closely related to parametric equations of graphs. While in both methods we plot points $$\big(x(t), y(t)\big)$$ or $$\big(x(t),y(t),z(t)\big)$$ to produce a graph, in the context of vector-valued functions each such point represents a vector. The implications of this will be more fully realized in the next section as we apply calculus ideas to these functions.
#### Example12.1.4.Graphing vector-valued functions.
Graph $$\ds \vec r(t) = \la t^3-t, \frac{1}{t^2+1}\ra\text{,}$$ for $$-2\leq t\leq 2\text{.}$$ Sketch $$\vec r(-1)$$ and $$\vec r(2)\text{.}$$
Solution 1.
We start by making a table of $$t\text{,}$$ $$x$$ and $$y$$ values as shown in Figure 12.1.5.(a). Plotting these points gives an indication of what the graph looks like. In Figure 12.1.5.(b), we indicate these points and sketch the full graph. We also highlight $$\vec r(-1)$$ and $$\vec r(2)$$ on the graph.
Solution 2. Video solution
#### Example12.1.6.Graphing vector-valued functions.
Graph $$\vec r(t) = \la \cos(t) ,\sin(t) ,t\ra$$ for $$0\leq t\leq 4\pi\text{.}$$
Solution 1.
We can again plot points, but careful consideration of this function is very revealing. Momentarily ignoring the third component, we see the $$x$$ and $$y$$ components trace out a circle of radius 1 centered at the origin. Noticing that the $$z$$ component is $$t\text{,}$$ we see that as the graph winds around the $$z$$-axis, it is also increasing at a constant rate in the positive $$z$$ direction, forming a spiral. This is graphed in Figure 12.1.7. In the graph $$\vec r(7\pi/4)\approx (0.707,-0.707,5.498)$$ is highlighted to help us understand the graph.
Solution 2. Video solution
### Subsection12.1.2Algebra of Vector-Valued Functions
#### Definition12.1.8.Operations on Vector-Valued Functions.
Let $$\vec r_1(t)=\la f_1(t),g_1(t)\ra$$ and $$\vec r_2(t)=\la f_2(t),g_2(t)\ra$$ be vector-valued functions in $$\mathbb{R}^2$$ and let $$c$$ be a scalar. Then:
1. $$\vec r_1(t) \pm \vec r_2(t) = \la\, f_1(t)\pm f_2(t),g_1(t)\pm g_2(t)\,\ra\text{.}$$
2. $$c\vec r_1(t) = \la\, cf_1(t),cg_1(t)\,\ra\text{.}$$
A similar definition holds for vector-valued functions in $$\mathbb{R}^3\text{.}$$
This definition states that we add, subtract and scale vector-valued functions component-wise. Combining vector-valued functions in this way can be very useful (as well as create interesting graphs).
#### Example12.1.10.Adding and scaling vector-valued functions.
Let $$\vec r_1(t) = \la\,0.2t,0.3t\,\ra\text{,}$$ $$\vec r_2(t) = \la\,\cos(t) ,\sin(t) \,\ra$$ and $$\vec r(t) = \vec r_1(t)+\vec r_2(t)\text{.}$$ Graph $$\vec r_1(t)\text{,}$$ $$\vec r_2(t)\text{,}$$ $$\vec r(t)$$ and $$5\vec r(t)$$ on $$-10\leq t\leq10\text{.}$$
Solution 1.
We can graph $$\vec r_1$$ and $$\vec r_2$$ easily by plotting points (or just using technology). Let’s think about each for a moment to better understand how vector-valued functions work.
We can rewrite $$\vec r_1(t) = \la\, 0.2t,0.3t\,\ra$$ as $$\vec r_1(t) = t\la 0.2,0.3\ra\text{.}$$ That is, the function $$\vec r_1$$ scales the vector $$\la 0.2,0.3\ra$$ by $$t\text{.}$$ This scaling of a vector produces a line in the direction of $$\la 0.2,0.3\ra\text{.}$$
We are familiar with $$\vec r_2(t) = \la\, \cos(t) ,\sin(t) \,\ra\text{;}$$ it traces out a circle, centered at the origin, of radius 1. Figure 12.1.11.(a) graphs $$\vec r_1(t)$$ and $$\vec r_2(t)\text{.}$$
Adding $$\vec r_1(t)$$ to $$\vec r_2(t)$$ produces $$\vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra\text{,}$$ graphed in Figure 12.1.11.(b). The linear movement of the line combines with the circle to create loops that move in the direction of $$\la 0.2,0.3\ra\text{.}$$ (We encourage the reader to experiment by changing $$\vec r_1(t)$$ to $$\la 2t,3t\ra\text{,}$$ etc., and observe the effects on the loops.)
Multiplying $$\vec r(t)$$ by 5 scales the function by 5, producing $$5\vec r(t) = \langle 5\cos(t) +t,5\sin(t) +1.5t\rangle\text{,}$$ which is graphed in Figure 12.1.11.(c) along with $$\vec r(t)\text{.}$$ The new function is “5 times bigger” than $$\vec r(t)\text{.}$$ Note how the graph of $$5\vec r(t)$$ in Figure 12.1.11.(c) looks identical to the graph of $$\vec r(t)$$ in Figure 12.1.11.(b). This is due to the fact that the $$x$$ and $$y$$ bounds of the plot in Figure 12.1.11.(c) are exactly 5 times larger than the bounds in Figure 12.1.11.(b).
Solution 2. Video solution
#### Example12.1.12.Adding and scaling vector-valued functions.
A cycloid is a graph traced by a point $$p$$ on a rolling circle, as shown in Figure 12.1.13. Find an equation describing the cycloid, where the circle has radius 1.
Solution 1.
This problem is not very difficult if we approach it in a clever way. We start by letting $$\vec p(t)$$ describe the position of the point $$p$$ on the circle, where the circle is centered at the origin and only rotates clockwise (i.e., it does not roll). This is relatively simple given our previous experiences with parametric equations; $$\vec p(t) = \la \cos(t) , -\sin(t) \ra\text{.}$$
We now want the circle to roll. We represent this by letting $$\vec c(t)$$ represent the location of the center of the circle. It should be clear that the $$y$$ component of $$\vec c(t)$$ should be 1; the center of the circle is always going to be 1 if it rolls on a horizontal surface.
The $$x$$ component of $$\vec c(t)$$ is a linear function of $$t\text{:}$$ $$f(t) = mt$$ for some scalar $$m\text{.}$$ When $$t=0\text{,}$$ $$f(t) = 0$$ (the circle starts centered on the $$y$$-axis). When $$t=2\pi\text{,}$$ the circle has made one complete revolution, traveling a distance equal to its circumference, which is also $$2\pi\text{.}$$ This gives us a point on our line $$f(t) = mt\text{,}$$ the point $$(2\pi, 2\pi)\text{.}$$ It should be clear that $$m=1$$ and $$f(t) = t\text{.}$$ So $$\vec c(t) = \la t, 1\ra\text{.}$$
We now combine $$\vec p$$ and $$\vec c$$ together to form the equation of the cycloid: $$\vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra\text{,}$$ which is graphed in Figure 12.1.14.
Solution 2. Video solution
### Subsection12.1.3Displacement
A vector-valued function $$\vec r(t)$$ is often used to describe the position of a moving object at time $$t\text{.}$$ At $$t=t_0\text{,}$$ the object is at $$\vec r(t_0)\text{;}$$ at $$t=t_1\text{,}$$ the object is at $$\vec r(t_1)\text{.}$$ Knowing the locations $$\vec r(t_0)$$ and $$\vec r(t_1)$$ give no indication of the path taken between them, but often we only care about the difference of the locations, $$\vec r(t_1)-\vec r(t_0)\text{,}$$ the displacement.
#### Definition12.1.15.Displacement.
Let $$\vec r(t)$$ be a vector-valued function and let $$t_0\lt t_1$$ be values in the domain. The displacement $$\vec d$$ of $$\vec r\text{,}$$ from $$t=t_0$$ to $$t=t_1\text{,}$$ is
\begin{equation*} \vec d=\vec r(t_1)-\vec r(t_0)\text{.} \end{equation*}
When the displacement vector is drawn with initial point at $$\vec r(t_0)\text{,}$$ its terminal point is $$\vec r(t_1)\text{.}$$ We think of it as the vector which points from a starting position to an ending position.
#### Example12.1.17.Finding and graphing displacement vectors.
Let $$\vec r(t) = \la \cos(\frac{\pi}{2}t),\sin(\frac{\pi}2 t)\ra\text{.}$$ Graph $$\vec r(t)$$ on $$-1\leq t\leq 1\text{,}$$ and find the displacement of $$\vec r(t)$$ on this interval.
Solution 1.
The function $$\vec r(t)$$ traces out the unit circle, though at a different rate than the “usual” $$\la \cos(t) ,\sin(t) \ra$$ parametrization. At $$t_0=-1\text{,}$$ we have $$\vec r(t_0) = \la 0,-1\ra\text{;}$$ at $$t_1=1\text{,}$$ we have $$\vec r(t_1) = \la 0,1\ra\text{.}$$ The displacement of $$\vec r(t)$$ on $$[-1,1]$$ is thus $$\vec d = \la 0,1\ra - \la 0,-1\ra = \la 0,2\ra\text{.}$$
A graph of $$\vec r(t)$$ on $$[-1,1]$$ is given in Figure 12.1.18, along with the displacement vector $$\vec d$$ on this interval.
Solution 2. Video solution
Measuring displacement makes us contemplate related, yet very different, concepts. Considering the semi-circular path the object in Example 12.1.17 took, we can quickly verify that the object ended up a distance of 2 units from its initial location. That is, we can compute $$\vnorm{d} = 2\text{.}$$ However, measuring distance from the starting point is different from measuring distance traveled. Being a semi-circle, we can measure the distance traveled by this object as $$\pi\approx 3.14$$ units. Knowing distance from the starting point allows us to compute average rate of change.
#### Definition12.1.19.Average Rate of Change.
Let $$\vec r(t)$$ be a vector-valued function, where each of its component functions is continuous on its domain, and let $$t_0\lt t_1\text{.}$$ The average rate of change of $$\vec r(t)$$ on $$[t_0,t_1]$$ is
\begin{equation*} \text{ average rate of change } = \frac{\vec r(t_1) - \vec r(t_0)}{t_1-t_0}\text{.} \end{equation*}
#### Example12.1.20.Average rate of change.
Let $$\vec r(t) = \la \cos(\frac{\pi}2t),\sin(\frac{\pi}2t)\ra$$ as in Example 12.1.17. Find the average rate of change of $$\vec r(t)$$ on $$[-1,1]$$ and on $$[-1,5]\text{.}$$
Solution 1.
We computed in Example 12.1.17 that the displacement of $$\vec r(t)$$ on $$[-1,1]$$ was $$\vec d = \la 0,2\ra\text{.}$$ Thus the average rate of change of $$\vec r(t)$$ on $$[-1,1]$$ is:
\begin{equation*} \frac{\vec r(1) -\vec r(-1)}{1-(-1)} = \frac{\la 0,2\ra}{2} = \la 0,1\ra\text{.} \end{equation*}
We interpret this as follows: the object followed a semi-circular path, meaning it moved towards the right then moved back to the left, while climbing slowly, then quickly, then slowly again. On average, however, it progressed straight up at a constant rate of $$\la 0,1\ra$$ per unit of time.
We can quickly see that the displacement on $$[-1,5]$$ is the same as on $$[-1,1]\text{,}$$ so $$\vec d = \la 0,2\ra\text{.}$$ The average rate of change is different, though:
\begin{equation*} \frac{\vec r(5)-\vec r(-1)}{5-(-1)} = \frac{\la 0,2\ra}{6} = \la 0,1/3\ra\text{.} \end{equation*}
As it took “3 times as long” to arrive at the same place, this average rate of change on $$[-1,5]$$ is $$1/3$$ the average rate of change on $$[-1,1]\text{.}$$
Solution 2. Video solution
We considered average rates of change in Sections 1.1 and 2.1 as we studied limits and derivatives. The same is true here; in the following section we apply calculus concepts to vector-valued functions as we find limits, derivatives, and integrals. Understanding the average rate of change will give us an understanding of the derivative; displacement gives us one application of integration.
### Exercises12.1.4Exercises
#### Terms and Concepts
##### 1.
Vector-valued functions are closely related to of graphs.
##### 2.
When sketching vector-valued functions, technically one isn’t graphing points, but rather .
##### 3.
It can be useful to think of as a vector that points from a starting position to an ending position.
##### 4.
In the context of vector-valued functions, average rate of change is divided by time.
#### Problems
##### Exercise Group.
In the following exercises, sketch the vector-valued function on the given interval.
###### 5.
$$\vec r(t) = \la t^2,t^2-1\ra\text{,}$$ for $$-2\leq t\leq 2\text{.}$$
###### 6.
$$\vec r(t) = \la t^2,t^3\ra\text{,}$$ for $$-2\leq t\leq 2\text{.}$$
###### 7.
$$\vec r(t) = \la 1/t,1/t^2\ra\text{,}$$ for $$-2\leq t\leq 2\text{.}$$
###### 8.
$$\vec r(t) = \la \frac1{10}t^2,\sin(t) \ra\text{,}$$ for $$-2\pi\leq t\leq 2\pi\text{.}$$
###### 9.
$$\vec r(t) = \la \frac1{10}t^2,\sin(t) \ra\text{,}$$ for $$-2\pi\leq t\leq 2\pi\text{.}$$
###### 10.
$$\vec r(t) = \la 3\sin(\pi t),2\cos(\pi t)\ra\text{,}$$ on $$[0,2]\text{.}$$
###### 11.
$$\vec r(t) = \la 3\cos(t) ,2\sin(2 t)\ra\text{,}$$ on $$[0,2\pi]\text{.}$$
###### 12.
$$\vec r(t) = \la 2\sec(t) ,\tan(t) \ra\text{,}$$ on $$[-\pi,\pi]\text{.}$$
##### Exercise Group.
In the following exercises, sketch the vector-valued function on the given interval in $$\mathbb{R}^3\text{.}$$ Technology may be useful in creating the sketch.
###### 13.
$$\vec r(t) = \la 2\cos(t) , t, 2\sin(t) \ra\text{,}$$ on $$[0,2\pi]\text{.}$$
###### 14.
$$\vec r(t) = \la 3\cos(t) , \sin(t) , t/\pi\ra$$ on $$[0,2\pi]\text{.}$$
###### 15.
$$\vec r(t) = \la \cos(t) , \sin(t) ,\sin(t) \ra$$ on $$[0,2\pi]\text{.}$$
###### 16.
$$\vec r(t) = \la \cos(t) , \sin(t) ,\sin(2t)\ra$$ on $$[0,2\pi]\text{.}$$
##### Exercise Group.
In the following exercises, find $$\norm{\vec r(t)}\text{.}$$
###### 17.
If $$\vec r(t) = \la t,t^2\ra\text{,}$$ then $$\norm{\vec r(t)}=$$.
###### 18.
$$\vec r(t) = \la 5\cos(t) ,3\sin(t) \ra\text{.}$$
###### 19.
If $$\vec r(t) = \la 2\cos(t) ,2\sin(t) ,t\ra\text{,}$$ then $$\norm{\vec r(t)}=$$.
###### 20.
$$\vec r(t) = \la \cos(t) ,t,t^2\ra\text{.}$$
##### Exercise Group.
Create a vector-valued function whose graph matches the given description.
###### 21.
A circle of radius $$2\text{,}$$ centered at $$(1,2)\text{,}$$ traced counter-clockwise once at constant speed on $$[0,2\pi)\text{.}$$
###### 22.
A circle of radius 3, centered at $$(5,5)\text{,}$$ traced clockwise once on $$[0,2\pi]\text{.}$$
###### 23.
An ellipse, centered at $$(0,0)$$ with vertical major axis of length 10 and minor axis of length 3, traced once counter-clockwise on $$[0,2\pi]\text{.}$$
###### 24.
An ellipse, centered at $$(3,-2)$$ with horizontal major axis of length 6 and minor axis of length 4, traced once clockwise on $$[0,2\pi]\text{.}$$
###### 25.
A line through $$(2,3)$$ with a slope of $$5\text{.}$$
###### 26.
A line through $$(1,5)$$ with a slope of $$-1/2\text{.}$$
###### 27.
The line through points $$(1,2,3)$$ and $$(4,5,6)\text{,}$$ where
$$\vec r(0) = \la 1,2,3\ra$$ and $$\vec r(1) = \la 4,5,6\ra\text{.}$$
###### 28.
The line through points $$(1,2)$$ and $$(4,4)\text{,}$$ where
$$\vec r(0) = \la 1,2\ra$$ and $$\vec r(1) = \la 4,4\ra\text{.}$$
###### 29.
A vertically oriented helix with radius of $$2$$ that starts at $$(2,0,0)$$ and ends at $$(2,0,4\pi)$$ after one revolution on $$[0,2\pi]\text{.}$$
###### 30.
A vertically oriented helix with radius of 3 that starts at $$(3,0,0)$$ and ends at $$(3,0,3)$$ after 2 revolutions on $$[0,1]\text{.}$$
##### Exercise Group.
Find the average rate of change of $$\vec r(t)$$ on the given interval.
###### 31.
$$\vec r(t) = \la t,t^2\ra$$ on $$[-2,2]\text{.}$$
###### 32.
$$\vec r(t) = \la t,t+\sin(t) \ra$$ on $$[0,2\pi]\text{.}$$
###### 33.
$$\vec r(t) = \la 3\cos(t) ,2\sin(t) ,t\ra$$ on $$[0,2\pi]\text{.}$$
###### 34.
$$\vec r(t) = \la t,t^2,t^3\ra$$ on $$[-1,3]\text{.}$$
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# If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d), Prove that A: B = C: D. - ICSE Class 10 - Mathematics
ConceptComponendo and Dividendo Properties
#### Question
If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d), prove that a: b = c: d.
#### Solution
Given (7a + 8b)/(7c – 8d) = (7a – 8b)/(7c + 8d)
Applying componendo and dividendo
(7a + 8b + 7a - 8b)/(7a + 8b - 7 + 8b) = (7c + 8d + 7c - 8d)/(7c + 8d - 7c - 8d)
=> (14a)/(16b) = (14c)/(16d)
=> a/b = c/d
Hence a : b = c : d
Is there an error in this question or solution?
#### APPEARS IN
Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2018-2019) (2017 to Current)
Chapter 7: Ratio and Proportion (Including Properties and Uses)
Ex.7C | Q: 5
#### Video TutorialsVIEW ALL [1]
Solution If (7a + 8b) (7c – 8d) = (7a – 8b) (7c + 8d), Prove that A: B = C: D. Concept: Componendo and Dividendo Properties.
S
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# What is 3/4 divided by 4/5?
#### Understand the Problem
The question is asking for the result of dividing the fraction 3/4 by the fraction 4/5. To solve this, we can use the rule that dividing by a fraction is equivalent to multiplying by its reciprocal.
$\frac{15}{16}$
The result of dividing the fraction $\frac{3}{4}$ by the fraction $\frac{4}{5}$ is $\frac{15}{16}$.
#### Steps to Solve
1. Identify the fractions We need to divide the fraction $\frac{3}{4}$ by the fraction $\frac{4}{5}$.
2. Reciprocal of the second fraction To divide by a fraction, we multiply by its reciprocal. The reciprocal of $\frac{4}{5}$ is $\frac{5}{4}$.
3. Multiply the fractions Now we can rewrite the division as multiplication: $$\frac{3}{4} \div \frac{4}{5} = \frac{3}{4} \times \frac{5}{4}$$
4. Multiply the numerators and denominators Now multiply the numerators and denominators: $$\frac{3 \times 5}{4 \times 4} = \frac{15}{16}$$
The result of dividing the fraction $\frac{3}{4}$ by the fraction $\frac{4}{5}$ is $\frac{15}{16}$.
• Not simplifying the final answer if possible. However, in this case, $\frac{15}{16}$ cannot be simplified further.
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