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# How do you write an equation of a line that is parallel to another?
Jan 5, 2015
Parallel lines are lines that never intersect. Because of this, a pair of parallel lines have to have the same slope, but different intercepts (if they had the same intercepts, they would be identical lines).
So, to find an equation of a line that is parallel to another, you have to make sure both equations have the same slope. In the general equation of a line $y = m x + b$ , the $m$ represents your slope value.
An example of paralell lines would therefore be:
(1) $y = m x + b$
(2) $y = m x + c$
With $b$ and $c$ being any constants. Note that they have to be different, because if they were equal, then you'd just have two identical lines that technically intersect in every single point.
Sometimes though, linear equations aren't in the form $y = m x + b$. You could have something like:
$8 x + 2 y = 16$
Here, you can't directly pick out the slope. But you could always turn that into the form $y = m x + b$ to find your slope $m$ by simply solving for $y$.
$8 x + 2 y = 16$
$2 y = 16 - 8 x$
$y = 8 - 4 x = - 4 x + 8$
There we go. We can see that $m = - 4$ here. So to write a line parallel to that, we have the general equation of
$y = - 4 x +$ any constant except for $8$. |
# Lesson 8How Much in Each Group? (Part 1)
Let’s look at division problems that help us find the size of one group.
### Learning Targets:
• I can tell when a question is asking for the amount in one group.
• I can use diagrams and multiplication and division equations to represent and answer “how much in each group?” questions.
## 8.1Inventing a Scenario
1. Think of a situation with a question that can be represented by Write a description of that situation and the question.
## 8.2How Much in One Batch?
To make 5 batches of cookies, 10 cups of flour are required. How many cups of flour does each batch require?
We can write equations and draw a diagram to represent this situation. They help us see that each batch requires 2 cups of flour.
For each question, write a multiplication equation and a division equation, draw a diagram, and answer the question.
1. To make 4 batches of cupcakes, it takes 6 cups of flour. How many cups of flour are needed for 1 batch?
2. To make batch of rolls, it takes cups of flour. How many cups of flour are needed for 1 batch?
3. Two cups of flour make batch of bread. How many cups of flour make 1 batch?
## 8.3One Container and One Section of Highway
Here are three tape diagrams and three descriptions of situations that include questions.
Match a diagram to each situation, then use the diagram to help you answer the question. Next, write multiplication and division equations to represent each situation.
1. Tyler poured 15 cups of water into 2 equal-sized bottles and filled each bottle. How much water was in each bottle?
Diagram:
Multiplication equation:
Division equation:
2. Kiran poured 15 cups of water into equal-sized pitchers and filled pitchers. How much water was in the full pitcher?
Diagram:
Multiplication equation:
Division equation:
3. It takes 15 cups of water to fill pail. How much water is needed to fill 1 pail?
Diagram:
Multiplication equation:
Division equation:
Here are three more diagrams and situations. Match a diagram to each situation, and use the diagram to help you answer the question. Next, write multiplication and division equations to represent each situation.
1. Priya’s class has adopted two equal sections of a highway to keep clean. The combined length is of a mile. How long is each section?
Diagram:
Multiplication equation:
Division equation:
2. Lin’s class has also adopted some sections of highway to keep clean. If sections are mile long, how long is each section?
Diagram:
Multiplication equation:
Division equation:
3. A school has adopted a section of highway to keep clean. If of the section is mile long, how long is the section?
Diagram:
Multiplication equation:
Division equation:
### Are you ready for more?
To make a Cantor ternary set:
• Start with a tape diagram of length 1 unit. This is step 1.
• Color in the middle third of the tape diagram. This is step 2.
• Do the same to each remaining segment that is not colored in. This is step 3.
• Keep repeating this process.
1. How much of the diagram is colored in after step 2? Step 3? Step 10?
2. If you continue this process, how much of the tape diagram will you color?
3. Can you construct a process that will give you a similar kind of object? For example, color the first fifth instead of the middle third of each strip.
## Lesson 8 Summary
Sometimes we know the amount for multiple groups, but we don’t know how much is in one group. We can use division to find out.
For example: If 5 people share pounds of cherries equally, how many pounds of cherries does each person get?
We can represent this situation as a multiplication and a division:
can be written as . Dividing by 5 is equivalent to multiplying by , and . This means each person gets pounds.
Other times, we know the amount for a fraction of a group, but we don’t know the size of one whole group. We can also use division to find out.
For example: Jada poured 5 cups of iced tea in a pitcher and filled of the pitcher. How many cups of iced tea fill the entire pitcher?
We can represent this situation as a multiplication and a division:
The diagram can help us reason about the answer. If of a pitcher is 5 cups, then of a pitcher is half of 5, which is . Because there are 3 thirds in 1 whole, there would be or cups in one whole pitcher. We can check our answer by multiplying: , and
Notice that in the first example, the number of groups is greater than 1 (5 people) and in the second, the number of groups is less than 1 ( of a pitcher), but the division and multiplication equations for both have the same structures.
## Lesson 8 Practice Problems
1. For each scenario, use the given tape diagram to help you answer the question. Mark up and label the diagrams as needed.
1. Mai has picked 1 cup of strawberries for a cake, which is enough for of the cake. How many cups does she need for the whole cake?
2. Priya has picked cups of raspberries, which is enough for of a cake. How many cups does she need for the whole cake?
2. Tyler painted square yards of wall area with 3 gallons of paint. How many gallons of paint does it take to paint each square yard of wall?
1. Write multiplication and division equations to represent the situation.
2. Draw a diagram to represent the situation and to answer the question.
3. After walking mile from home, Han is of his way to school. What is the distance between his home and school?
1. Write multiplication and division equations to represent this situation.
2. Use the given diagram to help you answer the question. Mark up and label it as needed.
4. Here is a division equation:
1. Write a multiplication equation that corresponds to the division equation.
2. Draw a diagram to represent and answer the question.
5. A set of books that are each 1.5 inches wide are being organized on a bookshelf that is 36 inches wide. How many books can fit on the shelf?
1. Write a multiplication equation and a division equation to represent this question.
1. Find the answer. Draw a diagram, if needed.
1. Use the multiplication equation to check your answer.
1. Without calculating, order the expressions based on their values, from smallest to largest.
2. Explain how you decided the order of the three expressions.
3. Find a number so that is greater than 1 but less than 7. |
Arithmetic
Associative Property
Averages
Brackets
Closure Property
Commutative Property
Conversion of Measurement Units
Cube Root
Decimal
Divisibility Principles
Equality
Exponents
Factors
Fractions
Fundamental Operations
H.C.F / G.C.D
Integers
L.C.M
Multiples
Multiplicative Identity
Multiplicative Inverse
Numbers
Percentages
Profit and Loss
Ratio and Proportion
Simple Interest
Square Root
Unitary Method
Algebra
Algebraic Equation
Algebraic Expression
Cartesian System
Linear Equations
Order Relation
Polynomials
Probability
Standard Identities & their applications
Transpose
Geometry
Basic Geometrical Terms
Circle
Curves
Angles
Define Line, Line Segment and Rays
Non-Collinear Points
Parallelogram
Rectangle
Rhombus
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Three dimensional object
Trapezium
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Trigonometry
Trigonometry Ratios
Data-Handling
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Frequency Distribution Table
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Range
Home >> Associative Property >> Associative Property (Multipication of Whole Numbers) >>
## Associative Property (Multipication of Whole Numbers)
Associative Property (Addition of Whole Numbers) Associative Property (Multipication of Whole Numbers)
Explanation
Multipication is Associative for Whole Numbers, this means that in a multipication expression; even if make different groups with same given whole numbers, then also the product in all the groups always remains the same. This property is also known as Associativity of Additition of Whole Numbers
Associativity of Multipication of whole numbers can be further understood from the following examples:-
Example 1 = Explain Associative Property for multipication of whole numbers, with given whole numbers 5, 6, 7 ?
Answer = Given Whole Numbers = 5, 6, 7 and their two groups are as follows :-
Group 1 = (5 × 6) × 7
= 30 × 7
= 210
Group 2 = 5 × (6 × 7)
= 5 × 42
= 210
As, in both the groups the sum is same i.e 210
So, we can say that Multipication is Associative for Whole Numbers.
Example 2 = Explain Associative Property for multipication of whole numbers, with given whole numbers 20, 30, 40 ?
Answer = Given Whole Numbers = 20, 30, 40 and their two groups are as follows :-
Group 1 = (20 × 30) × 40
= 600 × 40
= 24000
Group 2 = 20 × (30 × 40)
= 20 × 1200
= 24000
As, in both the groups the sum is same i.e 24000
So, we can say that Multipication is Associative for Whole Numbers.
Example 3 = Explain Associative Property for multipication of whole numbers, with given whole numbers 5, 20, 10 ?
Answer = Given Whole Numbers = 5, 20, 10 and their two groups are as follows :-
Group 1 = (5 × 20) × 10
= 100 × 10
= 1000
Group 2 = 5 × (20 × 10)
= 5 × 200
= 1000
As, in both the groups the sum is same i.e 1000
So, we can say that Multipication is Associative for Whole Numbers. |
# Ellipse: Equation
How to use the ellipse equation to find the major axis, the minor axis, and the foci (and vics versa): definition, formula, 8 examples, and their solutions.
## Definition
### Definition
An ellipse is the set of points
whose sum of the distances from the foci
is constant.
PF + PF' = (constant)
## Formula: x2/a2 + y2/b2 = 1
### Equation
This is the graph of the ellipse
x2/a2 + y2/b2 = 1
(a > b).
The denominator of x2, a2,
is greater than
the denominator of y2, b2.
Then this is a horizontal ellipse.
To show that
the ellipse is a horizontal ellipse,
we write x2 term first.
### Major Axis
The major axis is the longest diameter.
The major axis is 2a.
### Minor Axis
The minor axis is the shortest diameter.
The minor axis is 2b.
### Foci
For the horizontal ellipse
x2/a2 + y2/b2 = 1,
the foci are (c, 0) and (-c, 0).
a, b, and c satisfy
a2 - b2 = c2.
## Example 1: Major Axis
### Solution
25 = 52
16 = 42
x2/52 + y2/42 = 1
5 is greater than 4.
Then the major axis is
2⋅5.
2⋅5 = 10
## Example 2: Minor Axis
### Solution
You just found that
the given ellipse is
x2/52 + y2/42 = 1.
x2/52 + y2/42 = 1
4 is the less than 5.
Then the minor axis is
2⋅4.
2⋅4 = 8
## Example 3: Foci
### Solution
You found that
the given ellipse is
x2/52 + y2/42 = 1.
x2/52 + y2/42 = 1
a = 5
b = 4
Then
c2 = 52 - 42.
52 = 25
-42 = -16
25 - 16 = 9
c2 = 9
Then c = √9.
Think the sign of the c plus.
9 = 32
32 = 3
Square Root
c = 3
See x2/52 + y2/42 = 1.
The denominator of x2, 52,
is greater than
the denominator of y2, 42.
Then the ellipse is a horizontal ellipse.
So the foci are
(3, 0) and (-3, 0).
So
(3, 0), (-3, 0)
## Example 4: Equation
### Solution
The foci are (4, 0) and (-4, 0).
So draw a horizontal ellipse like this.
And draw the foci (4, 0) and (-4, 0).
Then c = 4.
The major axis is this horizontal diameter.
It's 10.
So 2a = 10.
Divide both sides by 2.
Then a = 5.
a = 5
The foci are (4, 0) and (-4, 0).
So c = 4.
Then
52 - b2 = 42.
52 = 25
42 = 16
Move 25 to the right side.
Then -b2 = -9.
Divide both sides by -1.
Then b2 = 9.
use b2 = 9
to write the ellipse equation.
The ellipse is a horizontal ellipse.
So write the x2 term first.
a = 5
b2 = 9
Then the ellipse is
x2/52 + y2/9 = 1.
52 = 25
So
x2/25 + y2/9 = 1
## Formula: y2/a2 + x2/b2 = 1
### Equation
This is the graph of the ellipse
y2/a2 + x2/b2 = 1
(a > b).
The denominator of y2, a2,
is greater than
the denominator of x2, b2.
Then this is a vertical ellipse.
To show that
the ellipse is a vertical ellipse,
we write y2 term first.
### Major Axis
The major axis is 2a.
### Minor Axis
The minor axis is 2b.
### Foci
For the vertical ellipse
y2/a2 + x2/b2 = 1,
the foci are (0, c) and (0, -c).
a, b, and c satisfy
a2 - b2 = c2.
## Example 5: Major Axis
### Solution
To make the right side 1,
divide both sides by 36.
The denominator of y2, 9,
is greater than
the denominator of x2, 4.
Then write the y2 term, y2/9, first.
9 = 32
4 = 22
y2/32 + x2/22 = 1
3 is greater than 2.
Then the major axis is
2⋅3.
2⋅3 = 6
## Example 6: Minor Axis
### Solution
You just found that
the given ellipse is
y2/32 + x2/22 = 1.
y2/32 + x2/22 = 1
2 is less than 3.
Then the minor axis is
2⋅2.
2⋅2 = 4
## Example 7: Foci
### Solution
You just found that
the given ellipse is
y2/32 + x2/22 = 1.
y2/32 + x2/22 = 1
a = 3
b = 2
Then
c2 = 32 - 22.
32 = 9
-22 = -4
9 - 4 = 5
c2 = 5
Then c = √5.
Think the sign of the c plus.
c = √5
See y2/32 + x2/22 = 1.
The denominator of y2, 32,
is greater than
the denominator of x2, 22.
Then the ellipse is a vertical ellipse.
So the foci are
(0, √5) and (0, -√5).
So
(0, √5), (0, -√5)
## Example 8: Equation
### Solution
The foci are (0, 2) and (0, -2).
So draw a vertical ellipse like this.
And draw the foci (0, 2) and (0, -2).
Then c = 2.
The major axis is this vertical diameter.
It's 8.
So 2a = 8.
Divide both sides by 2.
Then a = 4.
a = 4
The foci are (0, 2) and (0, -2).
So c = 2.
Then
42 - b2 = 22.
42 = 16
22 = 4
Move 16 to the right side.
Then -b2 = -12.
Multiply both sides by -1.
Then b2 = 12.
use b2 = 12
to write the ellipse equation.
The ellipse is a vertical ellipse.
So write the y2 term first.
a = 4
b2 = 12
Then the ellipse is
y2/42 + x2/12 = 1.
42 = 16
So
y2/16 + x2/12 = 1 |
# application of vectors: sailboat floats in a current
A sailboat floats in a current that flows due east at 1 meter per second. Due to a wind the boats actual speed relative to the shore is $({\sqrt 3})$ meters per second in a direction 30 degrees North of East. Find the speed and direction of the wind.
So far I have found the speed of the wind by using the formula for the resultant vector and got the speed to be 1 meter per second. Now how do I go about finding the direction of the wind? Can someone provide a step by step explanation. I don't understand why the wind would be East of North?
The Actual velocity $(\vec{v_r})$ rel to shore will be resultant of sailboat velocity $(\vec{v_b})$ and wind velocity $(\vec{v_w})$.
Figure 1:
Figure 2:
$\theta = 180 -30 = 150^\circ$
\begin{align} \vec{v_r} &= \vec{v_w}+\vec{v_b} \\ \\ \implies \vec{v_w} &= \vec{v_r}-\vec{v_b} \end{align}
\begin{align}|\vec{v_w}| &= \sqrt{|\vec{v_r}|^2 + |\vec{v_b}|^2 +2|\vec{v_r}||\vec{v_b}| \cos\theta} \\ \\ &=\sqrt{3 + 1 - 2\dfrac{3}{2}} \\ &=1 \end{align}
\begin{align} \tan\phi &= \dfrac{|\vec{v_b}|\sin\theta}{|\vec{v_r}|+|\vec{v_b}|\cos\theta} \\ \\ &= \dfrac{1/2}{\sqrt{3}-\frac{\sqrt{3}}{2}} \\ &= \frac{1}{\sqrt{3}}\\ \\ \implies \phi &= 30^\circ \end{align}
Therefore direction of wind is $30^\circ$ east of north, or $60^\circ$ north of east.
• after getting the value 30 how did you know it automatically had to be east of north though?
– Lil
Feb 1, 2016 at 18:23
• @Lil you can either use law of cosines, or the angle formula of $\tan\phi$. In that formula, $\phi$ is the angle made from first vector, first in the sense the one which comes first in the order of addition. So $\phi$ is angle made between $\vec{v_w}$ and $\vec{v_r}$ and not the angle made between $\vec{v_w}$ and $-\vec{v_b}$ Feb 2, 2016 at 4:34
• and $\vec{v_r}$ is $30^\circ$ north of east, and $\vec{v_w}$ is $30^\circ$ north of $\vec{v_r}$, So $\vec{v_w}$ is $60^\circ$ north of east, or $30^\circ$ east of north. Feb 2, 2016 at 4:39 |
# (-√3+1)/(√3+1) how to simplify?
Jul 14, 2018
$\sqrt{3} - 2$
#### Explanation:
Given: $\frac{- \sqrt{3} + 1}{\sqrt{3} + 1}$
Multiply by the conjugate of the denominator as 1: $\sqrt{3} - 1$ to eliminate the radical in the denominator:
$\frac{- \sqrt{3} + 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{- 3 + \sqrt{3} + \sqrt{3} - 1}{3 - 1} = \frac{2 \sqrt{3} - 4}{2}$
$= \frac{\cancel{2} \left(\sqrt{3} - 2\right)}{\cancel{2}} = \sqrt{3} - 2$
Jul 14, 2018
$\sqrt{3} - 2$
#### Explanation:
$\frac{- \sqrt{3} + 1}{\sqrt{3} + 1}$
=$- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
=$- {\left(\sqrt{3} - 1\right)}^{2} / \left({\left(\sqrt{3}\right)}^{2} - {1}^{2}\right)$
=$- \frac{4 - 2 \sqrt{3}}{3 - 1}$
=$- \frac{4 - 2 \sqrt{3}}{2}$
=$- \left(2 - \sqrt{3}\right)$
=$\sqrt{3} - 2$ |
# Real Numbers 1 Definition 2 Properties 3 Examples www.themegallery.com.
## Presentation on theme: "Real Numbers 1 Definition 2 Properties 3 Examples www.themegallery.com."— Presentation transcript:
Real Numbers 1 Definition 2 Properties 3 Examples
Definition Real Numbers include: Integers Rational Numbers
-3,-2,-1,0,1,2,3 Rational Numbers Decimals that can be represented in fraction form that are either terminating or non-terminating and repeating 5/4 = 1.25 177/55 = … 1/3 = … Irrational Numbers Non-terminating and non-repeating decimals Π = …, √2 = …
a + b = b + a Order does not matter Addition is associative a + (b + c) = (a +b) + c Grouping does not matter 0 is the additive identity a + 0 = a Adding 0 yields the same number
Properties (Cont.) -a is the additive inverse (negative) of a
a + (-a) = 0, 12+(-12)=0 Adding a number and it’s inverse gives 0 Multiplication is commutative ab = ba, 3*4=4*3=12 Order of multiplication does not change the result 1 is the multiplicative identity a * 1 = a Multiplying 1 yields the same number
Properties (Cont.) If a ≠ 0, 1/a is the multiplicative inverse (reciprocal) of a a(1/a) = 1, 3(1/3)=1 Multiplying a non-zero number by its reciprocal yields 1 Multiplication is distributive over addition a(b + c) = ab + ac (a + b)c = ac + bc Multiplying a number and a sum of two numbers is the same as multiplying each of the two numbers by the multiplier and then adding the products
Properties (Cont.) Trichotomy Law Definition of Absolute Value
If a and b are real numbers, then exactly one of the following is true: a=b, a<b, a>b Definition of Absolute Value If a ≥ 0, then |a|=a If a <0, then |a|=-(a) Distance on a number line d(A, B) = |B-A| Law of the signs If a and b both have the same sign, then ab and a/b are positive If a and b have different signs, then ab and a/b are negative
Examples If p, q, r, and s denote real numbers, show that (p+q)(r+s)=pr+ps+qr+qs (p+q)(r+s) =p(r+s)+q(r+s) =(pr+ps)+(qr+qs) = pr+ps+qr+qs If x>0, and y<0, determine the sign of x/y + y/x Since only y is negative, both x\y and y/x will be negative numbers A negative number increased by another negative number will yield a “more” negative number If x<1, rewrite |x-1| without using the absolute value symbol If x<1, then x-1<0 (negative) By part 2 of the definition of absolute value, |x-1|=-(x-1)=-x+1 or 1-x
Examples Let A, B, C, and D have coordinates -5, -3, 1, and 6 respectively. Find d(B,D)/\. d(B,D) = d(-3,6) =|6-(-3)| =|6+3| =|9| =9
Guided Practice Do Problems on page 16, 1-40 |
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Problem 975
# Convert the equation $$\mathrm{r}=\tan \theta+\cot \theta$$ to an equation in cartesian coordinates.
Expert verified
The short answer is: The given polar equation $$\mathrm{r}=\tan \theta+\cot \theta$$ can be converted to the Cartesian coordinates equation $$(x^2 + y^2)^{3/2} = xy$$.
See the step by step solution
## Step 1: Rewrite the equation in terms of sine and cosine functions
The given equation is: $$\mathrm{r}=\tan \theta+\cot \theta$$ Since we know that: $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$ and $$\cot \theta = \frac{\cos \theta}{\sin \theta}$$, The equation becomes: $$\mathrm{r}=\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$$
## Step 2: Find a common denominator and simplify the equation
To simplify the equation, we will find a common denominator for the fractions, which is $$\cos \theta \sin \theta$$. The equation now becomes: $$\mathrm{r} = \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta \sin \theta}$$ As we know, $$\sin^2 \theta + \cos^2 \theta = 1$$, so the equation simplifies to: $$\mathrm{r} = \frac{1}{\cos \theta \sin \theta}$$
## Step 3: Substitute the polar coordinates with Cartesian coordinates
Now, we will use the polar to Cartesian coordinate transformations: $$x = r \cos \theta$$ $$y = r \sin \theta$$ We already have the equation: $$\mathrm{r} = \frac{1}{\cos \theta \sin \theta}$$ Substitute $$\cos \theta$$ with $$\frac{x}{r}$$ and $$\sin \theta$$ with $$\frac{y}{r}$$: $$\mathrm{r} = \frac{1}{\frac{x}{r} \cdot \frac{y}{r}}$$
## Step 4: Simplify the equation and solve for $$r$$
Now, simplify the equation: $$\mathrm{r} = \frac{1}{\frac{xy}{r^2}}$$ $$\mathrm{r} = \frac{r^2}{xy}$$ Now, multiply both sides by $$xy$$: $$r^3 = x * y$$
## Step 5: Substitute back Cartesian coordinates for $$r$$
Finally, replace $$r$$ with $$\sqrt{x^2 + y^2}$$ (since $$r^2 = x^2 + y^2$$): $$(\sqrt{x^2 + y^2})^3 = xy$$ This is the Cartesian equation equivalent to the given polar equation: $$x^2 + y^2)^{3/2} = xy$$
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The circle is taken as an integral part of geometry and the chord length is defined as the line segment whose endpoints lie on the circumference of a circle. Practically, a circle could have infinite chords. The chord is always seen within a circle and the diameter is the longest chord inside a circle. The chord of a circle is a straight line that connects any two points on the circumference of a circle.
The chord length formula in mathematics could be written as given below. The chord length formulas vary depends on what information do you have about the circle. If you know the radius or sine values then you can use the first formula. The second formula is a variation of the Pythagorean theorem and it can be used for calculating the length of a chord as well.
## Chord Length Formula
$\LARGE Chord\;Length=2r\sin \left (\frac{c}{2}\right )$
$\LARGE Chord\;Length=2\sqrt{r^{2}-d^{2}}$
Where,r is the radius of the circle
c is the angle subtended at the center by the chord
d is the perpendicular distance from the chord to the circle center
sin is the sine function
Here, r is the radius of a circle, c is angle subtended at the center by the chord, d is the perpendicular from chord to the center of a circle, and sin is the sine trigonometry function. If any line that does not stop at the circumference of a circle instead it is extended to infinity then it is called as the secant.
Practically, this is not possible finding the chord length if you cannot measure the angle. Still, you have the flexibility of using trigonometry functions here but they are little bit difficult to understand. At the same time, it is easy to calculate the chord length if you know the radius of the circle and one of the variables.
#### Question 1:Find the chord of a circle where radius is 7 cm and perpendicular distance from chord to center is 4 cm?
Solution:
Given radius, r = 7 cm
and distance, d = 4 cm
$\ Chord\;Length=2\sqrt{r^{2}-d^{2}}$
$\ Chord\;Length=2\sqrt{7^{2}-4^{2}}$
$\ Chord\;Length=2\sqrt{49-16}$
$\ Chord\;Length=2\sqrt{33}$
$\ Chord\;Length=2\times 5.744$
Chord length = 11.48 cm |
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# 7th Grade Math / Lesson 3: Volumes
Volumes What will we be learning in this lesson? In this lesson you will look at how to find the area of a composite shape. Vocabulary words are found in this purple color throughout the lesson. Remember to put these in your notebook.
Volumes The volume of an object is the space inside of the object. Area and volume both refer to the space inside of an object. The difference is that area talks about the space inside of a flat shape. Like the following shapes: To find the volume, the shape needs another dimension to it so that it is not a flat shape. Like these: The volume is the amount of space inside of the shape/object.
Volumes To find the volume of a basic box-shaped object you need to find the area of one side of the box, this will be called the base. Then take that area and multiply it by the height of the object. Take a look at this box. The bottom of the box is a rectangle with side lengths of 4 and 2. So the area of the bottom (base) is 4 * 2 = 8. The height of the box is 3 in. So take the area of the bottom and multiply by 3. 8 * 3 = 24 The volume of this object is 24 cubic inches.
Volumes Find the volume of this object. Follow what we did on the previous example. Take the dimensions of the bottom, 14 and 2, and multiply them together. 14 * 2 = 28. Now take the 28 and multiply by the height. 28 * 5 = 140. So the volume of this object is 140 cubic cm.
Can you find the volume of a 3-dimensional shape? Go back to the classroom to get to your homework and any other items you need to attend to. |
# How do you evaluate (- 3x ^ { 3} y ^ { 4} z ^ { 2} ) ( x y z ^ { 2} ) ( - x ^ { 5} y ^ { 2} z )?
Jan 31, 2018
See a solution process below:
#### Explanation:
First, rearrange the expression as:
$\left(- 3 {x}^{3} {y}^{4} {z}^{2}\right) \left(x y {z}^{2}\right) \left(- {x}^{5} {y}^{2} z\right) \implies$
$\left(- 3 {x}^{3} {y}^{4} {z}^{2}\right) \left(x y {z}^{2}\right) \left(- 1 {x}^{5} {y}^{2} z\right) \implies$
$\left(- 3 \cdot - 1\right) \left({x}^{3} \cdot x \cdot {x}^{5}\right) \left({y}^{4} \cdot y \cdot {y}^{2}\right) \left({z}^{2} \cdot {z}^{2} \cdot z\right) \implies$
$3 \left({x}^{3} \cdot x \cdot {x}^{5}\right) \left({y}^{4} \cdot y \cdot {y}^{2}\right) \left({z}^{2} \cdot {z}^{2} \cdot z\right)$
Next, use this rule for exponents to rewrite the expression:
$a = {a}^{\textcolor{red}{1}}$
$3 \left({x}^{3} \cdot {x}^{\textcolor{red}{1}} \cdot {x}^{5}\right) \left({y}^{4} \cdot {y}^{\textcolor{red}{1}} \cdot {y}^{2}\right) \left({z}^{2} \cdot {z}^{2} \cdot {z}^{\textcolor{red}{1}}\right)$
Now, use this rule of exponents to complete the evaluation:
${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$
$3 \left({x}^{\textcolor{b l u e}{3}} \cdot {x}^{\textcolor{red}{1}} \cdot {x}^{\textcolor{g r e e n}{5}}\right) \left({y}^{\textcolor{b l u e}{4}} \cdot {y}^{\textcolor{red}{1}} \cdot {y}^{\textcolor{g r e e n}{2}}\right) \left({z}^{\textcolor{b l u e}{2}} \cdot {z}^{\textcolor{g r e e n}{2}} \cdot {z}^{\textcolor{red}{1}}\right) \implies$
$3 {x}^{\textcolor{b l u e}{3} + \textcolor{red}{1} + \textcolor{g r e e n}{5}} {y}^{\textcolor{b l u e}{4} + \textcolor{red}{1} + \textcolor{g r e e n}{2}} {z}^{\textcolor{b l u e}{2} + \textcolor{g r e e n}{2} + \textcolor{red}{1}} \implies$
$3 {x}^{9} {y}^{7} {z}^{5}$
Jan 31, 2018
$3 {x}^{9} {y}^{7} {z}^{5}$
#### Explanation:
Take each different term and multiply them. Powers multiply by adding the exponents.
$- 3 \cdot - 1 = 3$
${x}^{3} \cdot {x}^{1} \cdot {x}^{5} = {x}^{9}$
${y}^{4} \cdot {y}^{1} \cdot {y}^{2} = {y}^{7}$
${z}^{2} \cdot {z}^{2} \cdot {z}^{1} = {z}^{5}$ |
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# Measurement for Praxis II ParaPro Test Prep Study Guide (page 4)
By
Updated on Jul 5, 2011
### Converting Units of Time
The ParaPro Assessment may ask you to convert units of time. Just like converting U.S. customary or metric units, multiply when changing from a larger unit to a smaller unit and divide when changing from a smaller unit to a larger unit.
Example
A teacher creates a lesson plan for the week that will take four hours. How many minutes will the lesson plan take?
Because minutes are a smaller unit than hours, you need to multiply to convert the units. Remember that there are 60 minutes in an hour.
4 hours = ? minutes
4 hours × 60 minutes per hour = 240
Therefore, there are 240 minutes in 4 hours.
### Money
In addition to converting between units of length, capacity, weight, mass, and time, you may also be asked to convert between units of money. Remember that one cent is equal to of a dollar, or \$0.01. The values of U.S. coins are as follows:
1 penny = 1 cent = 1¢ = \$0.01
1 nickel = 5 cents = 5¢ = \$0.05
1 dime = 10 cents = 10¢ = \$0.10
1 quarter = 25 cents = 25¢ = \$0.25
1 dollar = 100 cents = 100¢ = \$1.00
### Converting Money
As always when converting units, multiply when changing from a larger unit to a smaller unit and divide when changing from a smaller unit to a larger unit. To convert dollars to cents, multiply by 100. To convert cents to dollars, divide by 100.
Example
Jillian has 7 dollars. How many dimes does she have?
To solve this problem, first convert Jillian's dollars to cents. Multiply the number of dollars by 100.
\$7 = 7 × 100¢ = 700¢
To convert from cents to dimes, you are going from a smaller unit to a larger unit—which means you have to divide. If Jillian has 700 cents, you need to divide by 10 to find out how many dimes she has.
700¢ ÷ 10 = 70 dimes
The practice quiz for this study guide can be found at:
Math for Praxis II ParaPro Test Prep Practice Problems |
# GRE Math : Data Analysis
## Example Questions
### Example Question #1 : How To Find The Intersection Of A Venn Diagram
In a class, there are 15 students who like chocolate. 13 students like vanilla. 10 students like neither. If there are 35 people in the class, how many students like chocolate and vanilla?
25
2
13
10
3
3
Explanation:
In order to find the intersection of chocolate and vanilla, it is easiest to make a Venn Diagram. The outside of the Venn Diagram is 10, and the total of the entire diagram must equal 35. Therefore the two circles of the Venn Diagram including just chocolate, just vanilla and the intersection must equal 25, with the just chocolate plus intersection side equalling 15 and the just vanilla plus intersection side equalling 13.
We know:
(A U B) = A + B – (A ∩ B)
We have found that (A U B) = 25 and we are trying to find (A ∩ B). Plug in A and B
25 = 15 + 13 – (A ∩ B) = 28 – (A ∩ B)
or – (A ∩ B) = –3
(A ∩ B) = 3
### Example Question #2 : How To Find The Intersection Of A Venn Diagram
A given company has 1500 employees. Of those employees, 800 are computer science majors. 25% of those computer science majors are also mathematics majors. That group of computer science/math dual majors makes up one third of the total mathematics majors. How many employees have majors other than computer science and mathematics?
Explanation:
Refer to the following Venn Diagram:
If 25% of the 800 CS students are also mathematics students, the number of students sharing these majors is 800 * 0.25 or 200 students. Furthermore, if this represents one third of the total of math students, we then know:
Math students * 1/3 = 200 or (1/3)M = 200
Solving for M we get 600. This means that the number of students that are ONLY math students is 400.
Looking at our diagram above, we must be careful not to "double add" the intersection. The easiest way to do this is to take the intersection and add to it the number of CS-only and math-only students: 600 + 200 + 400 = 1200. This number represents the total number of students that have either a math or CS major (that is, the number of students in the union of the two sets). This leaves 1500 – 1200 or 300 students.
### Example Question #1 : Data Analysis
In a population of cats, 10% are tabby colored, 5% are pregnant, and 3% are both tabby and pregnant. What is the probability that a cat is tabby but not pregnant?
Explanation:
Probability (tabby but not pregnant)
= Prob (tabby) – Prob (tabby and pregnant)
= 10% - 3%
= 7%
### Example Question #1 : Venn Diagrams
At an overpriced department store there are customers. If have purchased shirts, have purchased pants, and have purchased neither, how many purchased both shirts and pants?
Explanation:
A way of solving this problem is by drawing a Venn diagram based on what is known:
There are customers; within them there are those that purchased something (pants, shirts, or possibly a combination), and those that purchased nothing.
The amount of individuals that purchased something is given as:
The reason that the number of shoppers that bought both is subtracted is so that it is not counted twice when the customers that bought pants and the customers that bought shirts are added together.
Since bought nothing,
The amount that bought both then is:
### Example Question #2 : Data Analysis
In a school, 70 students are taking classes. 35 of them will be taking Accounting and 20 of them will be taking Economics. 7 of them are taking both of these classes. How many of the students are not in either class?
Explanation:
When you add both class rosters you get a total of 55 students
.
You must subtract the 7 that are in both because they are counted twice in the 55
.
The total in neither class will be the total students minus the adjusted enrollment in both classes
.
### Example Question #3 : Data Analysis
There are 15,000 students at college X. Of those students, 1,700 are taking both ethics and metaphysics this semester. There are 2,200 total students taking ethics. 9,500 students are taking neither of these classes. How many students are taking metaphysics this term?
3300
1600
5,000
3800
5,000
Explanation:
The easiest way to understand this problem is to draw a Venn Diagram:
S = Total number of students
A = Total students taking ethics
B = Total students taking metaphysics
A - B = Students taking only ethics
B - A = Students taking only metaphysics
A ∪ B = The total students taking either ethics or metaphysics
A ∩ B = The total students taking both ethics and metaphysics
We know there are 15,000 total and that 9,500 are taking neither class. Therefore, we know that 15,000 - 9,500 = 5,500 are taking at least one of the classes. Based on our prompt, we know that there are 1,700 taking both and that 2,200 are taking ethics. To fiind out how many are taking ONLY ethics, we have to subtract off the amount that are taking ethics and metaphysics. Hence, 2,200 - 1,700 = 500. Finally, if we know that there are 5,500 taking at at least one of these classes, we want to get rid of that portion taking ethics. This will leave us with those who are taking at least metaphysics (regardless of whether or not they are taking ethics): 5,500 - 500 = 5,000.
### Example Question #1 : Venn Diagrams
In a class of 100 students, 43 play basketball and 37 play baseball. 9 students play both. How many students do not play either sport?
20
38
29
not enough information to answer the question
71
29
Explanation:
In order to determine how many students are not enrolled in a sport, we must first determine how many students are. The simplest way to do this is to begin by adding the students of both sports together.
43 + 37 = 80
But wait! 9 of those students are play both baseball and basketball. To avoid double counting these students, subtract 9 from the total.
80 – 9 = 71
Now we know 71 students play sports. If there are 100 students, all that's left to do is subtract.
100 – 71 = 29
29 students do not play basketball or baseball.
### Example Question #5 : Data Analysis
There are students in a class. of them take German and take Latin. Some students take two languages. There are students who take no language whatsoever. How many students are there who take at least one language?
Explanation:
You could represent this question in the following Venn Diagram:
We know that the two circles must contain a total of students. Now, since there is an extra case of the overlap area when you add together the German and the Latin students, you can say:
This is the simple answer to this question! You do not need to compute the overlap at all, as you are merely looking for the contents of the two circles.
### Example Question #1 : How To Find The Probability Of An Outcome
A jar contains 10 red marbles, 4 white marbles, and 2 blue marbles. Two are drawn in sequence, not replacing after each draw.
Quantity A
The probability of drawing two red marbles
Quantity B
The probability of drawing exactly one blue marble.
The relationship cannot be determined from the information given.
Quantity A is greater.
Quantity B is greater.
The quantities are equal.
Quantity A is greater.
Explanation:
Note that there are 16 total marbles. A is simply a set of sequential events. On the first, you have 10/16 chances to draw a red. Supposing this red is not replaced, the chance of drawing a second red will be 9/15; therefore, the probability of A is (10/16) * (9/15) = 0.375. Event B is translated into 2 events: Blue + (White or Red) or (White or Red) + Blue. The probabilities of each of these events, added together would be (2/16) * (14/15) + (14/16) * (2/15) = 0.2333333333; therefore, A is more probable.
### Example Question #1 : Outcomes
In a bowl containing 10 marbles, 5 are blue and 5 are pink. If 2 marbles are picked randomly, what is the probability that the 2 marbles will not both be pink?
7/9
7/8
2/9
5/6
7/9
Explanation:
To solve this question, you can solve for the probability of choosing 2 marbles that are pink and subtracting that from 1 to obtain the probability of selecting any variation of marbles that are not both pink.
The probability of picking 2 marbles that are both pink would be the product of the probability of choosing the first pink marble multiplied by the probability of choosing a second pink marble from the remaining marbles in the mix.
This would be 1/2 * 4/9 = 2/9.
To obtain the probability that is asked, simply compute 1 – (2/9) = 7/9.
The probability that the 2 randomly chosen marbles are not both pink is 7/9.
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# Do Now: Pass out calculators.
## Presentation on theme: "Do Now: Pass out calculators."— Presentation transcript:
Do Now: Pass out calculators.
Pick up an Algebra I sheet from the back – work on Mac & Tolley’s Road Trip problem.
Objective: To solve a system of equations using substitution.
Use the substitution method
EXAMPLE 1 Use the substitution method Solve the linear system: y = 3x + 2 Equation 1 x + 2y = 11 Equation 2 SOLUTION STEP 1 Solve for y. Equation 1 is already solved for y.
Use the substitution method
EXAMPLE 1 Use the substitution method STEP 2 Substitute 3x + 2 for y in Equation 2 and solve for x. x + 2y = 11 Write Equation 2. x + 2(3x + 2) = 11 Substitute 3x + 2 for y. 7x + 4 = 11 Simplify. 7x = 7 Subtract 4 from each side. x = 1 Divide each side by 7.
EXAMPLE 1 Use the substitution method STEP 3 Substitute 1 for x in the original Equation 1 to find the value of y. y = 3x + 2 = 3(1) + 2 = = 5 ANSWER The solution is (1, 5).
Use the substitution method
EXAMPLE 1 GUIDED PRACTICE Use the substitution method CHECK Substitute 1 for x and 5 for y in each of the original equations. y = 3x + 2 x + 2y = 11 5 = 3(1) + 2 ? 1 + 2 (5) = 11 ? 5 = 5 11 = 11
Use the substitution method
EXAMPLE 2 Use the substitution method Solve the linear system: x – 2y = –6 Equation 1 4x + 6y = 4 Equation 2 SOLUTION STEP 1 Solve Equation 1 for x. x – 2y = –6 Write original Equation 1. x = 2y – 6 Revised Equation 1
Use the substitution method
EXAMPLE 2 Use the substitution method STEP 2 Substitute 2y – 6 for x in Equation 2 and solve for y. 4x + 6y = 4 Write Equation 2. 4(2y – 6) + 6y = 4 Substitute 2y – 6 for x. 8y – y = 4 Distributive property 14y – 24 = 4 Simplify. 14y = 28 Add 24 to each side. y = 2 Divide each side by 14.
Use the substitution method
EXAMPLE 2 Use the substitution method STEP 3 Substitute 2 for y in the revised Equation 1 to find the value of x. x = 2y – 6 Revised Equation 1 x = 2(2) – 6 Substitute 2 for y. x = –2 Simplify. ANSWER The solution is (–2, 2).
Use the substitution method
EXAMPLE 2 GUIDED PRACTICE Use the substitution method CHECK Substitute –2 for x and 2 for y in each of the original equations. Equation 1 Equation 2 4x + 6y = 4 x – 2y = –6 –2 – 2(2) = –6 ? 4(–2) + 6 (2) = 4 ? –6 = –6 4 = 4
EXAMPLE 1 GUIDED PRACTICE Use the substitution method for Examples 1 and 2 Solve the linear system using the substitution method. y = 2x + 5 1. 3x + y = 10 ANSWER (1, 7)
EXAMPLE 2 GUIDED PRACTICE Use the substitution method for Examples 1 and 2 Solve the linear system using the substitution method. x – y = 3 2. x + 2y = –6 ANSWER (0, –3)
EXAMPLE 2 GUIDED PRACTICE Use the substitution method for Examples 1 and 2 Solve the linear system using the substitution method. 3x + y = –7 3. –2x + 4y = 0 ANSWER (–2, –1)
EXAMPLE 3 Solve a multi-step problem WEBSITES Many businesses pay website hosting companies to store and maintain the computer files that make up their websites. Internet service providers also offer website hosting. The costs for website hosting offered by a website hosting company and an Internet service provider are shown in the table. Find the number of months after which the total cost for website hosting will be the same for both companies.
EXAMPLE 3 Solve a multi-step problem SOLUTION STEP 1 Write a system of equations. Let y be the total cost after x months. Equation 1: Internet service provider y = x
Solve a multi-step problem
EXAMPLE 3 Solve a multi-step problem Equation 2: Website hosting company y = x The system of equations is: y = x Equation 1 y = 22.45x Equation 2
Solve a multi-step problem
EXAMPLE 3 Solve a multi-step problem STEP 2 Substitute 22.45x for y in Equation 1 and solve for x. y = x Write Equation 1. 22.45x = x Substitute 22.45x for y. 0.5x = 10 Subtract 21.95x from each side. x = 20 Divide each side by 0.5. The total cost will be the same for both companies after 20 months. ANSWER |
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# Express: 9 hours as percent of 4 days?
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Hint: Here in this question, we have to express 9 hours as percent compared to 4 days. To solve this first, find how many hours are there in 4 days and remember 1 day is having 24 hours. Write the fraction put numerator has 9 hours and denominator has total hours which 4 days have and next multiply the resultant value of fraction to the 100 we get the required solution.
Complete step by step solution:
Now consider the question
Express 9 hours as percent of 4 days.
As we know,
1 day = 24 hours
Therefore, 4 days = $4 \times 24 = 96$ hours.
So, we have to find, what percent of 96 is 9
Assume, if $x\%$ of 96 is equal to 9, then we have to find the percentage value $x$
$\Rightarrow \dfrac{x}{{100}} \times 96 = 9$
On cross multiplication, we have
$\Rightarrow x \times 96 = 9 \times 100$
First, we simplify the term.
$\Rightarrow 96x = 900$
Multiply both side by 96, then
$\Rightarrow x = \dfrac{{900}}{{96}}$
Divide both numerator and denominator by 12, we have
$\Rightarrow x = \dfrac{{75}}{8}$
Or
$\Rightarrow x = 9.375$
Hence, 9 hours is the $\dfrac{{75}}{8}\%$ or $9.375\%$ of 4 days.
So, the correct answer is “ $9.375\%$ ”.
Note: Here the question is related to the percentage. By using the specific methods and rules we
can convert the number. As we know that the percentage formula $= \left( {\dfrac{{Value}}{{Total\,value}}} \right) \times 100$ . Using the simple arithmetic operation i.e., multiplication and division to get the required solution.
The above question can also solve by using a percentage formula
9 hours as a percent of 4 days
4 days = 96 hours
$\Rightarrow \left( {\dfrac{9}{{96}}} \right) \times 100 = \dfrac{{900}}{{96}} = \dfrac{{75}}{8}\%$ |
#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 9 Maths Textbook Solution.
Answer:$\frac{a}{4} \ln \left|x^{4}+c^{2}\right|+\frac{b}{2 c} \tan ^{-1} \frac{x^{2}}{c}+f$
Hint : Find $I_{1}\: and \: I_{2}$
Given: $\int \frac{a x^{3}+b x}{x^{4}+c^{2}} d x$
Solution: $\int \frac{a x^{3}+b x}{x^{4}+c^{2}} d x$
\begin{aligned} I &=\int \frac{a x^{3}}{x^{4}+c^{2}} d x+\int \frac{b x}{x^{4}+c^{2}} d x \\ I_{1} &=\int \frac{a x^{3}}{x^{4}+c^{2}} d x \end{aligned}
Let,$x^{4}+c^{2}=t$
\begin{aligned} &4 x^{3} d x=d t \\ &x^{3} d x=\frac{1}{4} d t \end{aligned}
$\Rightarrow I_{1}=\frac{a}{4} \int \frac{d t}{t}=\frac{a}{4} \ln t+C_{1}$
\begin{aligned} &I_{2}=\int \frac{b x}{x^{4}+c^{2}} d x \\ &\text { Let } x^{2}=v \\ &2 x d x=d v \end{aligned}
$I_{2}=\frac{b}{2} \int \frac{d v}{v^{2}+c^{2}}$
$=\frac{b}{2} \tan ^{-1}\left(\frac{v}{c}\right) \times \frac{1}{c}+c_{2} \quad\left[\int \frac{1}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{2}+c\right]$
$I_{2}=\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+c_{2}$
\begin{aligned} &I=I_{1}+I_{2} \\ &I=\frac{a}{4} \ln \left|x^{4}+c^{2}\right|+c_{1}+\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+c_{2} \\ &c_{1}+c_{2}=f \\ &I=\frac{a}{4} \ln \left|x^{4}+c^{2}\right|+\frac{b}{2 c} \tan ^{-1}\left(\frac{x^{2}}{c}\right)+f \end{aligned} |
# Conduct a Lesson Study Cycle on Fractions
Lesson Study works best with rich subject content. This course offers step-by-step support to conduct a Lesson Study cycle on fractions, by providing fractions resources and research to examine.
15-20 hours
##### MATERIALS:
Part 3 | Examine Curriculum: Introduction to Fractions
## Introduction
The tasks students do, and the sequence of these tasks over time, can dramatically shape how students imagine and think about fractions. To consider how students build their understanding of fractions based on their prior understanding of whole numbers, analyze the following two problems from a Japanese textbooks series and discuss the questions that follow the problems.
In grade 2, Japanese students investigate how many liters of water fit in various containers. (The shaded area represents the water.)
How many liters of water are there?
Then, in grade 3, Japanese students encounter the following related problem. (Assume that the 1-liter container is divided into equal parts.)
How much water is shown by each interval? How much water is in the container? How many liters are in the container?
Discuss:
• What are the similarities between the two problems shown above?
• What knowledge about whole numbers might students bring to the fraction problem?
• Some US students struggle to answer the question: “How many 1/5 are in 2/5?" How might the tasks above support their understanding?
## What Images of Fractions Do Students Develop?
We examine a curriculum that introduces fractions using a linear measurement context and will see lesson videos based on the curriculum.
Read the grade 3 unit (found in the resource tile below) that introduces fractions in the Tokyo Shoseki curriculum, the most widely used Japanese textbook. It may be helpful to know that 8 periods of 45 minutes are allocated for this unit, roughly one period per textbook page.
Discuss:
How does this unit help students with the challenges explored in Module 1, such as to:
• See non-unit fractions (fractions with a numerator other than 1) as accumulations of unit fractions (fractions with numerator of 1)?
• See the connection between whole numbers and fractions?
• Connect area, linear measurement, and number line models?
How do the images and understandings of fractions that students might develop from this textbook compare with what students might develop from your own curriculum’s introduction of fractions?
In Section 5 of your Teaching-Learning Plan, jot down anything you want to remember for your own planning.
## Lesson Series 1: Video 1
To further flesh out the ideas in the grade 3 textbook, examine video excerpts from three lessons taught by an experienced teacher, Dr. Akihiko Takahashi, to grade 3-5 students in California. The videos provides insight that are hard to gain just from reading the textbook.
To prepare, we suggest that you:
• Imagine how you would teach the beginning of the grade 3 fractions unit.
• Look over the Summary of Video Excerpts (Lesson Series 1) in the resource tile below.
• Divide up the discussion questions below so that each team member focuses on one question, and note your focus question on a Video Observation Notes page. If your group meeting time is limited, members can watch the video on their own before your next meeting.
Watch the video and take notes on the Video Observation Notes. The Summary of Video Excerpts (Lesson Series 1) may also be useful in jogging your memory of the lesson.
• How does the instructor use the 1-meter reference strip? Why does he use two of the 1-meter reference strips to show 2 meters?
• Why does the instructor have students predict the length of pieces before investigating?
• What elements of instruction might help students build a strong mental image of the connection between the unit fraction and the whole: that 1/3 is a unit that goes into the whole 3 times, 1/4 is a unit that goes in 4 times, etc.?
• This veteran instructor made two changes from the strategies recommended by the Teacher’s Edition: He gave students strips representing just the fractional part over 1 meter (such as 1/3m), rather than the whole length (such as 1 1/3m), and he did not pre-draw lines on the 1-meter reference strip to show 1/3m, 1/2m, etc.). Why do you think he made these choices?
## Lesson Series 1: Video 2
Watch the video below and use your Video Observation Notes to discuss the following questions. The Summary of Video Excerpts (Lesson Series 1) may also be useful in jogging your memory of the lesson.
• The instructor takes considerable time to write on the board the different ways of expressing 1/4 meter. Do you think this is a good investment of time? Why or why not?
• Why do you think the instructor chose 2/5 m and 2/3 m as lengths for the mystery strips?
• The instructor chooses particular student misunderstandings as a focus for whole class discussion (for example, 2/3 m versus 2/5 m as descriptions of the mystery strip and two different lengths both thought to be 3/4 m.) What kinds of understanding do you think he was trying to build?
## Lesson Series 1: Video 3
Watch the video below and use your Video Observation Notes to discuss the following questions. The Summary of Video Excerpts (Lesson Series 1) may also be useful in jogging your memory of the lesson.
• How do students connect linear measurement activities to the number line?
• Why might it be useful to introduce fractions in the context of a standard linear measurement unit (meters) that is continuous? How might the students’ experience be different if fraction bars were used?
• How might these lessons build an understanding of the fractions standards in the Common Core or your state standards, especially:
• 3.NF.A.1: Understand a fraction 1/b is the quantity formed by 1 part when a whole is partitioned into b equal parts.
• 3.NF.A.2: Understand a fraction as a number on the number line; represent fractions on a number line diagram.
• 3.NF.A.3.D: Compare two fractions with the same numerator or the same denominator by reasoning about their size.
• 4.N.F.B.4.A: Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product of 5 x 1/4; recording the conclusion by the equation 5/4 = 5 x 1/4 |
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Transcript
```10- 1
Chapter
Ten
McGraw-Hill/Irwin
10- 2
Chapter Ten
One-Sample Tests of Hypothesis
GOALS
When you have completed this chapter, you will be able to:
ONE
Define a hypothesis and hypothesis testing.
TWO
Describe the five step hypothesis testing procedure.
THREE
Distinguish between a one-tailed and a two-tailed test of
hypothesis.
FOUR
Conduct a test of hypothesis about a population mean.
10- 3
Chapter Ten continued
One-Sample Tests of Hypothesis
GOALS
When you have completed this chapter, you will be able to:
FIVE
Conduct a test of hypothesis about a population proportion.
SIX
Define Type I and Type II errors.
SEVEN
Compute the probability of a Type II error.
10- 4
A statement
value of a
population
parameter
developed for
the purpose of
testing.
The mean monthly
income for systems
analysts is \$6,325.
What is a
Hypothesis?
Twenty percent of all
customers at Bovine’s Chop
House return for another
meal within a month.
What is a Hypothesis?
10- 5
Hypothesis testing
Based on
sample
evidence and
probability
theory
Used to determine
whether the hypothesis is
a reasonable statement
and should not be
rejected, or is
unreasonable and should
be rejected
What is Hypothesis Testing?
10- 6
Step 1: State null and alternate hypotheses
Step 2: Select a level of significance
Step 3: Identify the test statistic
Step 4: Formulate a decision rule
Step 5: Take a sample, arrive at a decision
Do not reject null
Reject null and accept alternate
Hypothesis Testing
10- 7
Step One: State the null and alternate hypotheses
Null Hypothesis H0
value of a population
parameter
Alternative Hypothesis H1:
A statement that is
accepted if the sample data
provide evidence that the
null hypothesis is false
10- 8
Step One: State the null and alternate
hypotheses
Three
possibilities
regarding
means
H 0: m = 0
H 1: m = 0
H 0: m < 0
H 1: m > 0
H0: m > 0
H1: m < 0
The null
hypothesis
always contains
equality.
10- 9
Type I Error
Level of Significance
Rejecting the null
hypothesis when it
is actually true (a).
The probability of
rejecting the null
hypothesis when it is
actually true; the level of
risk in so doing.
Type II Error
Accepting the null
hypothesis when it
is actually false (b).
Step Two: Select a Level of
Significance.
10- 10
Step Two: Select a Level of Significance.
Null
Hypothesis
Ho is true
Ho is false
Researcher
Accepts
Rejects
Ho
Ho
Correct
Type I
error
decision
(a)
Type II
Correct
Error
decision
(b)
Risk
table
10- 11
Test statistic
A value, determined
from sample
information, used to
determine whether
or not to reject the
null hypothesis.
Examples: z, t, F, c2
z Distribution as a
test statistic
X m
z
/ n
The z value is based on the
sampling distribution of X,
which is normally
distributed when the sample
is reasonably large (recall
Central Limit Theorem).
Step Three: Select the test statistic.
10- 12
Step Four: Formulate the decision rule.
Critical value: The dividing point between the region
where the null hypothesis is rejected and the region
where it is not rejected.
Sampling Distribution
Of the Statistic z, a
Right-Tailed Test, .05
Level of Significance
Region of
Do not
rejection
reject
[Probability =.95]
0
[Probability=.05]
1.65
Critical value
10- 13
Decision Rule
Reject the null
hypothesis and
accept the
alternate
hypothesis if
Computed -z < Critical -z
or
Computed z > Critical z
Decision Rule
10- 14
p-Value
The probability, assuming that the null
hypothesis is true, of finding a value of
the test statistic at least as extreme as the
computed value for the test
Decision Rule
If the p-Value is larger
If the p-Value is
than or equal to the
smaller than the
significance level, a, H0
significance level, a,
is not rejected.
H0 is rejected.
Calculated from the
probability distribution
function or by computer
Using the p-Value in Hypothesis
Testing
10- 15
Interpreting p-values
>.05
p
>.001
p
.10
SOME evidence Ho is not true
>.01
p
STRONG evidence
Ho is not true
.05
.01
VERY STRONG
evidence Ho is not true
10- 16
Step Five: Make a decision.
Movie
10- 17
One-Tailed Tests of Significance
The
alternate
hypothesis,
H1, states a
direction
H1: The mean
speed of trucks
traveling on I95 in Georgia
is less than 60
miles per hour.
(µ<60)
H1: The mean yearly
commissions earned by
full-time realtors is more
than \$35,000. (µ>\$35,000)
H1: Less than 20
percent of the
customers pay
cash for their
gasoline purchase.
(p<.20)
One-Tailed Tests of Significance
10- 18
Sampling Distribution
Of the Statistic z, a
Right-Tailed Test, .05
Level of Significance
Region of
Do not
rejection
reject
[Probability =.95]
0
[Probability=.05]
1.65
Critical value
One-Tailed Test of Significance
.
10- 19
Two-Tailed Tests of Significance
No direction is specified in the alternate hypothesis H1.
H1: The mean
amount spent by
customers at the
Wal-mart in
Georgetown is
not equal to \$25.
(µ ne \$25).
H1: The mean
price for a
gallon of
gasoline is not
equal to \$1.54.
(µ ne \$1.54).
Two-Tailed Tests of Significance
10- 20
Two-Tailed Tests of Significance
Regions of
Nonrejection
and Rejection
for a TwoTailed Test,
.05 Level of
Significance
Region of
Region of
Do not
rejection
rejection
reject
[Probability=.025]
[Probability =.95]
-1.96
Critical value
0
[Probability=.025]
1.96
Critical value
10- 21
Test for the population
mean from a large sample
with population standard
deviation known
X m
z
/ n
Testing for the Population Mean: Large
Sample, Population Standard Deviation
Known
10- 22
The processors of Fries’ Catsup
indicate on the label that the
bottle contains 16 ounces of
catsup. The standard deviation
of the process is 0.5 ounces. A
sample of 36 bottles from last
hour’s production revealed a
mean weight of 16.12 ounces per
bottle. At the .05 significance
level is the process out of
control? That is, can we
conclude that the mean amount
per bottle is different from 16
ounces?
Example 1
10- 23
Step 5
Make a decision and
interpret the results.
Step 4
State the decision rule.
Reject H0 if z > 1.96
or z < -1.96
or if p < .05.
Step 3
Identify the test statistic. Because
we know the population standard
deviation, the test statistic is z.
Step 1
State the null and the
alternative hypotheses
H0: m = 16
H1: m 16
Step 2
Select the significance level.
The significance level is .05.
EXAMPLE 1
10- 24
Step 5: Make a
decision and
interpret the results.
z
X m
n
oComputed
16 .12 16 .00
0.5
z of 1.44
< Critical z of 1.96,
op of .1499 > a of .05,
Do not reject the null
hypothesis.
36
1.44
The p(z > 1.44)
is .1499 for a
two-tailed test.
We cannot
conclude the
mean is different
from 16 ounces.
Example 1
10- 25
Testing for the
Population Mean:
Large Sample,
Population Standard
Deviation Unknown
Here is unknown,
so we estimate it
with the sample
standard deviation s.
As long as the
sample size n > 30, z
can be approximated
using
X m
z
s/ n
Testing for the Population Mean: Large Sample,
Population Standard Deviation Unknown
10- 26
Roder’s Discount Store
chain issues its own
credit card. Lisa, the
credit manager, wants to
find out if the mean
monthly unpaid balance
is more than \$400. The
level of significance is set
at .05. A random check
of 172 unpaid balances
revealed the sample
mean to be \$407 and the
sample standard
deviation to be \$38.
Should Lisa conclude
that the population
mean is greater than
\$400, or is it reasonable
to assume that the
difference of \$7 (\$407\$400) is due to chance?
Example 2
10- 27
Step 5
Make a decision
and interpret the
results.
Step 4
H0 is rejected if
z > 1.65
or if p < .05.
Step 3
Because the sample is large
we can use the z
distribution as the test
statistic.
Step 1
H0: µ < \$400
H1: µ > \$400
Step 2
The significance
level is .05.
Example 2
Step 5
Make a decision
and interpret the
results.
10- 28
z
oComputed
z of 2.42
> Critical z of 1.65,
op of .0078 < a of .05.
Reject H0.
Lisa can conclude that
the mean unpaid balance
is greater than \$400.
X m
s
n
\$407 \$400
\$38
2.42
172
The p(z > 2.42) is
.0078 for a onetailed test.
10- 29
Testing for a
Population Mean:
Small Sample,
Population
Standard Deviation
Unknown
The test statistic
is the t
distribution.
t
X m
s/ n
The critical value of t is
determined by its degrees of
freedom equal to n-1.
Testing for a Population Mean: Small Sample,
Population Standard Deviation Unknown
The current rate for producing
5 amp fuses at Neary Electric
Co. is 250 per hour. A new
machine has been purchased
and installed that, according
to the supplier, will increase
the production rate. The
production hours are normally
distributed. A sample of 10
randomly selected hours from
last month revealed that the
mean hourly production on
the new machine was 256
units, with a sample standard
deviation of 6 per hour.
10- 30
At the .05 significance
level can Neary conclude
that the new machine is
faster?
Example 3
10- 31
The null hypothesis is rejected if t > 1.833 or, using the
p-value, the null hypothesis is rejected if p < .05.
Step 4
State the decision rule.
There are 10 – 1 = 9
degrees of freedom.
Step 1
State the null and
alternate hypotheses.
H0: µ < 250
H1: µ > 250
Step 3
Find a test statistic. Use
the t distribution since
is not known and n < 30.
Step 2
Select the level of
significance. It is .05.
10- 32
Step 5
Make a decision
and interpret the
results.
t
X m
s
n
256 250
6 10
3.162
The p(t >3.162) is
.0058 for a onetailed test.
oComputed
t of 3.162
>Critical t of 1.833
op of .0058 < a of .05
Reject Ho
The mean number
of amps produced is
more than 250 per
hour.
Example 3
10- 33
Proportion
The sample
proportion is p
and p is the
population
proportion.
The fraction or percentage
that indicates the part of the
population or sample having
a particular trait of interest.
Number of successesin the sample
p
Number sampled
Test Statistic for
Testing a Single
Population Proportion
z
p p
p (1 p )
n
10- 34
In the past, 15% of the
mail order solicitations for
a certain charity resulted in
a financial contribution. A
new solicitation letter that
has been drafted is sent to
a sample of 200 people
and 45 responded with a
contribution. At the .05
significance level can it be
concluded that the new
letter is more effective?
Example 4
10- 35
Step 3
Find a test statistic.
The z distribution
is the test statistic.
Step 4
State the decision rule.
The null hypothesis is
rejected if z is greater
than 1.65 or if p < .05.
Step 5
Make a decision and interpret the results.
Step 1
State the null and the
alternate hypothesis.
H0: p < .15
H1: p > .15
Step 2
Select the level of
significance. It is .05.
Example 4
10- 36
Step 5: Make a decision and interpret the results.
45
.15
p p
z
200
2.97
p (1 p )
.15 (1 .15 )
n
200
p( z > 2.97)
= .0015.
Because the computed z of 2.97 > critical z of
1.65, the p of .0015 < a of .05, the null
hypothesis is rejected. More than 15 percent
responding with a pledge. The new letter is
more effective.
Example 4
```
Related documents |
# Square Formula
A square is one of the most basic and often recurring shapes in geometry. We all know that it is a shape where all four sides are equal. But do you know the important square formula? Actually there is not just one square formula, but several important ones in mensuration. Let’s take a look.
## Introduction to a Square
We all are familiar with the figure, square. It is a quadrilateral wherein all the four sides and angles of it are equal. All the four angles are 90 degrees each, that is, right angles. You can also consider square as a special case of a rectangle where you will find that the two adjacent sides are of equal length.In this article, we will mainly be focusing on the various square formulas such as its area, perimeter, length of the diagonals and examples.
Area of a Square =
The perimeter of a Square =4a
Diagonal of a Square=a√2
Where ‘a’ is the length of a side of the square.
Properties of a Square
• The lengths of all the four sides of a square are equal.
• The two diagonals bisect each other at right angles, that is, 90°
• The lengths of diagonals of a square are equal.
## Derivation of Square Formula
### Derivation of Area of a Square
To better our understanding of the concept, let us take a look at the derivation of the area of a square. Let us consider a square where the lengths of its side are ‘a’ units and diagonal is ‘d’ units respectively. As you all know that area of a square is the region which is enclosed within its boundary. As we have already mentioned, that a square is a special case of a rectangle that has its two adjacent sides being of equal length. Therefore, the area can be expressed as –
Area of a rectangle = Length × Breadth
• Area of square = Length x Breadth
• Area of square = a× a = =
### Derivation of the Perimeter of a Square
The perimeter of the square is the total length of its boundary. The boundary of a square is represented by the sum of the length of all sides. Hence, the perimeter is expressed as –
• Perimeter of a square = length of 4 sides
• Perimeter of a square = a+ a + a + a = 4a
### Derivation of the Lenght of a Diagonal of a Square
As you all know that the diagonal is a line that joins the two opposite sides in a polygon. Therefore, in order to calculate the diagonal length of a square, we use the Pythagoras Theorem. If you mark a diagonal in the square, then you will realize that the diagonal divides the square into two right-angled triangles. Now, the two adjacent sides of a square are equal in length. And the right-angled triangle works as an isosceles triangle with each of its sides being of length ‘a’ units.
Thus, we can apply the Pythagoras theorem on these triangles which have base and perpendicular of ‘a’ units and hypotenuse of ‘b’ units. So, according to the formula we have:
• =+
• d = √2
• d = a √2 units
### Solved Examples
Now, that we have some understanding about the concept and meaning of square formula, let us try some examples to deepen our understanding of the topic
Example 1 – The side of a square is 5 meters. What is the area of the square in m
Solution 1 – As we know that the sides of а square are of equal size. And the formula for the square’s area is a × a where a is the side of the square. Hence,
Area = 5m × 5m = 25 m²
Example 2 – The perimeter of a square is 24 cm. What is the area of the square is cm2
Solution 2 – Perimeter of a square is 4 × a, where a is the side of the square
4 × a =24
a = 24 ÷ 4 = 6
Area = a × a = 6 cm × 6 cm = 36 cm2
Example 3 – The side of a square is 5 cm. If its side is doubled, how many times is the area of the new square bigger than the area of the old square?
Solution 3 – The area of the first square is 5 cm × 5 cm = 25 cms. New length = 5cm + 5 cm = 10 cm
The area of the new square is 10 cm × 10 cm = 100 cm2. Hence, the area is 4 times bigger.
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## How do you write a turning point?
4 Tips for Writing Turning PointsBuild up to the turning point of the story. Think of each turning point as a moment of crisis. Plan your turning points ahead of time. Your turning point doesn’t have to be a big twist.
Table of Contents
## How do you write a turning point?
4 Tips for Writing Turning PointsBuild up to the turning point of the story. Think of each turning point as a moment of crisis. Plan your turning points ahead of time. Your turning point doesn’t have to be a big twist.
## What is a minimum turning point?
A turning point of a function is a point where f′(x)=0 f ′ ( x ) = 0 . A minimum turning point is a turning point where the curve is concave down (from decreasing to increasing) and f′(x)=0 f ′ ( x ) = 0 at the point. …
## How do you find the maximum turning point?
First, identify the leading term of the polynomial function if the function were expanded. Then, identify the degree of the polynomial function. This polynomial function is of degree 4. The maximum number of turning points is 4 – 1 = 3.
## What is the turning point of a graph?
What is the turning point? The turning point of a graph (marked with a blue cross on the right) is the point at which the graph “turns around”. On a positive quadratic graph (one with a positive coefficient of x 2 x^2 x2), the turning point is also the minimum point.
## How do you find the minimum turning point?
To see whether it is a maximum or a minimum, in this case we can simply look at the graph. f(x) is a parabola, and we can see that the turning point is a minimum. By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, −4).
## What is the maximum point?
Maximum, In mathematics, a point at which a function’s value is greatest. If the value is greater than or equal to all other function values, it is an absolute maximum. If it is merely greater than any nearby point, it is a relative, or local, maximum.
## What is a minimum or maximum value?
The minimum or maximum of a function occurs when the slope is zero. Therefore, to find where the minimum or maximum occurs, set the derivative equal to zero.
## What is the minimum value of a function?
The minimum value of a function is the lowest point of a vertex. If your quadratic equation has a positive a term, it will also have a minimum value. You can find this minimum value by graphing the function or by using one of the two equations.
## What is a maximum value of a function?
The maximum value of a function is the place where a function reaches its highest point, or vertex, on a graph. If your quadratic equation has a negative a term, it will also have a maximum value. If you have the graph, or can draw the graph, the maximum is just the y value at the vertex of the graph.
## How do you find the minima of a function?
How do we find them? Given f(x), we differentiate once to find f ‘(x). Set f ‘(x)=0 and solve for x. Using our above observation, the x values we find are the ‘x-coordinates’ of our maxima and minima. Substitute these x-values back into f(x).
## How find the range of a function?
The domain of a function is the set of all acceptable input values (X-values). The range of a function is the set of all output values (Y-values).
## How do you write a range?
For the constant functionf(x)=c, the domain consists of all real numbers; there are no restrictions on the input. The only output value is the constant c, so the range is the set {c} that contains this single element. In interval notation, this is written as [c,c], the interval that both begins and ends with c.
## How do you write the range of a graph?
13:23Suggested clip 115 secondsDomain and Range of a Function From a Graph – YouTubeYouTubeStart of suggested clipEnd of suggested clip
## What is the range of a graph?
The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values.
## How can you tell if something’s a function?
Determining whether a relation is a function on a graph is relatively easy by using the vertical line test. If a vertical line crosses the relation on the graph only once in all locations, the relation is a function. However, if a vertical line crosses the relation more than once, the relation is not a function. |
# NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles
NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles- Topics like how to identify different lines, line segments, and angles in the shapes are dealt with in class 6. In this chapter, we deal with lines, different kinds of angles and their measurements. There are 2 exercises with 20 questions in this chapter. The solutions of NCERT class 7 maths chapter 5 lines and angles give an explanation to all these questions. The CBSE NCERT solutions for class 7 maths chapter 5 lines and angles are extremely helpful for the students to understand the basics of this chapter and to clear all their doubts easily. These NCERT Solutions are given here with a step-by-step explanation of each and every problem of NCERT textbook. This chapter is all about the lines and angles. In this chapter students always confuse between a line, a ray and a line-segment. So, let's discuss each term one by one- A line segment has two endpoints. If we extend these two endpoints in either direction endlessly, we get a line. Thus, in other words, we can say that a line has no endpoints. A ray has only one endpoint which is its initial point. An angle is formed when line segments or lines meet. In the NCERT solutions for class 7 maths chapter 5 lines and angles, we will study questions related to different kinds of angles like complementary angles, adjacent angles, supplementary angles, vertically opposite angles; pairs of lines like intersecting lines, transversal and many more. Here you will get solutions to two exercises of this chapter.
Exercise:5.1
Exercise:5.2
## Important points of NCERT Class 7 Maths Chapter 5 Lines and Angles are-
Problems related to the following points are discussed in the NCERT solutions for class 7 maths chapter 5 lines and angles.
• An angle is formed when two lines or line-segments or rays meet
Pairs of Angles Condition Two complementary angles Measures add up to $\dpi{100} 90^o$ Two supplementary angles Measures add up to $\dpi{100} 180^o$ Two adjacent angles Have a common arm and a common vertex but no common interior Linear pair Adjacent and supplementary
• When two lines l and m are intersected it means they both are meet at a point and the meeting point is called the point of intersection.
• When two lines l and m drawn on a sheet of paper do not meet or intersect, however far produced, then this lines called as parallel lines.
• When two lines intersect we have two pairs of opposite angles. They are called vertically opposite angles and both angles are equal in measure.
• Transversal- It is a line that intersects two or more than two lines at distinct points. And it gives rise to several types of angles. For example interior angles, exterior angles, corresponding angles, alternate interior, and alternate exterior angles.
Topic-wise questions are also explained in the CBSE NCERT solutions for class 7 maths chapter 5 lines and angles.
## Topics of NCERT Grade 7 Maths Chapter 5 Lines and Angles-
5.1 Introduction
5.2 Related Angles
5.2.1 Complementary Angles
5.2.2 Supplementary Angles
5.2.4 Linear Pair
5.2.5 Vertically Opposite Angles
5.3 Pairs of Lines
5.3.1 Intersecting Lines
5.3.2 Transversal
5.3.3 Angles made by a Transversal
5.3.4 Transversal of Parallel Lines
5.4 Checking for Parallel Lines
## Solutions for NCERT class 7 maths chapter 5 lines and angles topic 5.2.1
Yes two acute angles can be complementary to each other.
For e.g. Acute angles $20^{\circ}$ and $70^{\circ}$ are complementary angle as their sum is $90^{\circ}$.
Since obtuse angles are greater than $90^{\circ}$. Thus two obtuse angles cannot be a complement to each other. (as the sum of complementary angles is $90^{\circ}$.)
The sum of angles in complementary angles is $180^{\circ}$. Thus two right angles cannot be complementary to each other.
## 1.(i) Which pairs of following angles are complementary?
Sum of the angles in the given figure is : $=\ 70^{\circ}\ +\ 20^{\circ}\ =\ 90^{\circ}$
Thus two angles are complementary to each other.
The sum of the two angles is : $=\ 75^{\circ}\ +\ 25^{\circ}\ =\ 100^{\circ}$
In complementary angles sum of the angles is $90^{\circ}$. Hence given pair of angles are not complementary.
We know that the sum angles of complementary angles is $90^{\circ}$.
In the given figure: Sum of angles is $=\ 48^{\circ}\ +\ 52^{\circ}\ =\ 100^{\circ}$
Hence given pair of angles are not complementary.
The sum of the two angles is : $=\ 35^{\circ}\ +\ 55^{\circ}\ =\ 90^{\circ}$
In complementary angles sum of the angles is $90^{\circ}$. Hence given pair of angles are complementary to each other.
$(i)45^{o}$
We know that the sum of complementary angles is $90^{\circ}$.
Thus the complement of the given angle is : $\Theta \ =\ 90^{\circ}\ -\ 45^{\circ}\ =\ 45^{\circ}$
$(ii)65^{o}$
The sum of complementary angles are $90^{\circ}$.
Thus the required angle is : $=\ 90^{\circ}\ -\ 65^{\circ}\ =\ 25^{\circ}$
$(iii)41^{o}$
We know that the sum of complementary angles is $90^{\circ}$.
Hence the required complement of the given angle is : $\Theta \ =\ 90^{\circ}\ -\ 41^{\circ}\ =\ 49^{\circ}$
$(iv)54^{o}$
We know that the sum of complementary angles is $90^{\circ}$.
Hence the complement of the given angle is : $\Theta \ =\ 90^{\circ}\ -\ 54^{\circ}\ =\ 36^{\circ}$
Let one of the angles be $\Theta$.
It is given that the angles are complementary to each other. So the other angle will be $90^{\circ}\ -\ \Theta$.
Further, it is given that the difference of the angle is 12.
So the equation is : $90^{\circ}\ -\ \Theta\ -\ \Theta\ =\ 12^{\circ}$
or $2\Theta\ =\ 90^{\circ}\ -\ 12^{\circ}$
or $\Theta\ =\ 39^{\circ}$
Hence the two angles are $39^{\circ}$ and $51^{\circ}$.
CBSE NCERT solutions for class 7 maths chapter 5 lines and angles topic 5.2.2
No, two obtuse angles cannot be supplementary as their the sum of angles will exceed $180^{\circ}$.
No two acute angles cannot be supplementary.
For being the supplementary angles their sum should be $180^{\circ}$.
But the acute angles are less than $90^{\circ}$ . Hence their maximum doesn't reach $180^{\circ}$.
Yes, two right angles are supplementary as their sum is $180^{\circ}$.
We know that the sum of the supplementary angle is $180^{\circ}$.
(i) Sum of the angles is : $=\ 110^{\circ}\ +\ 50^{\circ}\ =\ 160^{\circ}$. Hence the angles are not supplementary.
(ii) Sum of the angles is : $=\ 105^{\circ}\ +\ 65^{\circ}\ =\ 170^{\circ}$. Thus the angles are not supplementary.
(iii) Sum of the angles is : $=\ 50^{\circ}\ +\ 130^{\circ}\ =\ 180^{\circ}$. Hence the angles are supplementary to each other.
(iv) Sum of the angles is : $=\ 45^{\circ}\ +\ 45^{\circ}\ =\ 90^{\circ}$. Thus the angles are not supplement to each other.
Solutions of NCERT class 7 maths chapter 5 lines and angles topic 5.2.3
Yes, two adjacent angles can be supplementary.
For e.g., $40^{\circ}$ and $140^{\circ}$ can be two adjacent angles which are supplementary angles.
Yes, two adjacent angles can be complementary to each other.
For e.g., adjacent angles $40^{\circ}$ and $50^{\circ}$ are complementary angles.
Yes, two obtuse angles can be adjacent for e.g., $100^{\circ}$ and $150^{\circ}$ can be adjacent angles.
Yes, the acute angle can be adjacent to an obtuse angle.
For e.g., $20^{\circ}$ and $120^{\circ}$ can be adjacent angles.
The condition for being adjacent angles are:-
(a) they have a common vertex
(b) they have common arm
Hence in the given figures:-
(i) These angles are adjacent angles as they agree above conditions.
(ii) The angles are adjacent angles.
(iii) These angles are not adjacent as their vertices are different.
(v) The angles are adjacent angles.
(a) $\angle AOB$ and $\angle BOC$
(b) $\angle BOD$ and $\angle BOC$
(a) $\angle AOB$ and $\angle BOC$ are adjacent angles as they have common vertex and share a common arm.
(b) $\angle BOD$ and $\angle BOC$ are not adjacent angles as $\angle BOC$ is contained in $\angle BOD$.
Solutions for NCERT class 7 maths chapter 5 lines and angles topic 5.2.4
No two acute angles cannot form a linear pair. As the sum of angles in the linear pair is $180^{\circ}$.
But the acute angles have their maximum value of $90^{\circ}$ thus their sum cannot be $180^{\circ}$.
No two obtuse angles cannot form a linear pair as their sum will exceed $180^{\circ}$, but the sum of angles in linear pair is $180^{\circ}$.
Yes, two right angles will form a linear pair as their sum is $180^{\circ}$ which is the sum of angles in linear pair.
The sum angles of linear pair is $180^{\circ}$.
(i) Sum of the given angles is : $=\ 40^{\circ}\ +\ 140^{\circ}\ =\ 180^{\circ}$. Thus these are linear pair.
(ii) Sum of the given angles is : $=\ 60^{\circ}\ +\ 60^{\circ}\ =\ 120^{\circ}$. Thus these are not linear pair.
(iii) Sum of the given angles is : $=\ 90^{\circ}\ +\ 80^{\circ}\ =\ 170^{\circ}$. Thus these are not a linear pair.
(iv) Sum of the given angles is : $=\ 115^{\circ}\ +\ 65^{\circ}\ =\ 180^{\circ}$. Thus these are linear pair.
CBSE NCERT solutions for class 7 maths chapter 5 lines and angles topic 5.2.5
From the given figure :
(a) $\angle 1\ =\ \angle 3\ =\ 30^{\circ}$ (Vertically opposite angles)
(b) $\angle 2\ =\ 180^{\circ}\ -\ \angle 1\ =\ 150^{\circ}$ (Linear pair)
The very common example of vertically opposite angle is scissors. Its arms form vertically opposite angles.
Solutions of NCERT class 7 maths chapter 5 lines and angles topic 5.3.1
The floor and the pillars in the house are at the right angle. Apart from this, the walls are perpendicular to the floor.
We know that the opposite sides of the rectangle are equal and parallel to each other.
Then for two interior angles on the same side of the transversal, we can write :
$\angle A\ +\ \angle B\ =\ 180^{\circ}$
Also, $\angle A\ =\ \angle B$ (Since opposite sides are equal)
Thus $\angle A\ =\ \angle B\ =\ 90^{\circ}$
No, it is not necessary that lines always intersect at right angles. The lines may form an acute angle (another angle will be obtuse as to form linear pair).
Solutions for class 7 maths chapter 5 lines and angles topic 5.3.2
We can draw infinite transversals from these two lines.
We know that transversal cuts lines at distinct points. Thus if a transversal cuts 3 lines then it will have 3 intersecting points.
Few examples of the transversal are road crossing of different railway line crossing the other lines. Transversal intersects lines at a distinct point.
The given pair of angles are corresponding angles.
(ii)
The given pair of angles are alternate interior angles.
The angles shown are pair of interior angles.
These are pair of corresponding angles.
(v)
The angles shown are pair of alternate interior angles.
The given angles are linear pair of angle as they form a straight line.
## NCERT solutions for class 7 maths chapter 5 lines and angles exercise 5.1
The sum of the complementary angle is $90^{\circ}$.
Thus the complementary angle to the given angle is : $=\ 90^{\circ}\ -\ 20^{\circ}\ =\ 70^{\circ}$
The sum of the complementary angle is $90^{\circ}$.
Thus the complementary angle to the given angle is : $=\ 90^{\circ}\ -\ 63^{\circ}\ =\ 27^{\circ}$
The sum of the complement angles is $90^{\circ}$.
Thus the complement of the angle is given by : $=\ 90^{\circ}\ -\ 57^{\circ}\ =\ 33^{\circ}$
## 2.(i) Find the supplement of each of the following angles:
We know that sum of supplement angles is $180^{\circ}$
The supplement of the given angle is : $=\ 180^{\circ}\ -\ 105^{\circ}\ =\ 75^{\circ}$
We know that the sum of angles of supplementary pair is $180^{\circ}$.
Thus the supplement of the given angle is : $=\ 180^{\circ}\ -\ 87^{\circ}\ =\ 93^{\circ}$
We know that the sum of angles of supplementary pair is $180^{\circ}$.
Thus the supplement of the given angle is : $=\ 180^{\circ}\ -\ 154^{\circ}\ =\ 26^{\circ}$
$(i) 65^{o},115^{o}$ $(ii) 63^{o}, 27^{o}$ $(iii) 112^{o}, 68^{o}$
$(iv) 130^{o}, 50^{o}$ $(v) 45^{o}, 45^{o}$ $(vi) 80^{o}, 10^{o}$
We know that the sum of supplementary angles is $180^{\circ}$ and the sum of complementary angle is $90^{\circ}$.
(i) Sum of the angles is : $65^{\circ}\ +\ 115^{\circ}\ =\ 180^{\circ}$. Hence these are supplementary angles.
(ii) Sum of the angles is : $63^{\circ}\ +\ 27^{\circ}\ =\ 90^{\circ}$. Hence these are complementary angles.
(iii) Sum of the angles is : $112^{\circ}\ +\ 68^{\circ}\ =\ 180^{\circ}$. Hence these are supplementary angles.
(iv) Sum of the angles is : $130^{\circ}\ +\ 50^{\circ}\ =\ 180^{\circ}$. Hence these are supplementary angles.
(v) Sum of the angles is : $45^{\circ}\ +\ 45^{\circ}\ =\ 90^{\circ}$. Hence these are complementary angles.
(vi) Sum of the angles is : $80^{\circ}\ +\ 10^{\circ}\ =\ 90^{\circ}$. Hence these are complementary angles.
Let the required angle be $\Theta$.
Then according to question, we have :
$\Theta\ +\ \Theta\ =\ 90^{\circ}$
or $2\Theta\ =\ 90^{\circ}$
or $\Theta\ =\ 45^{\circ}$
Let the required angle be $\Theta$.
Then according to the question :
$\Theta\ +\ \Theta \ =\ 180^{\circ}$
or $2\Theta \ =\ 180^{\circ}$
or $\Theta \ =\ 90^{\circ}$
Hence the angle is $90^{\circ}$.
Since it is given that $\angle 1$ and $\angle 2$ are supplementary angles, i.e. the sum of both angles is $180^{\circ}$.
Thus if $\angle 1$ is decreased then to maintain the sum $\angle 2$ needs to be increased.
## 7. Can two angles be supplementary if both of them are:
(i) acute ? (ii) obtuse ? (iii) right ?
We know that the sum of supplementary angles is $180^{\circ}$.
(i) The maximum value of the sum of two acute angles is less than $180^{\circ}$. Thus two acute angles can never be supplementary.
(ii) The minimum value of the sum of two obtuse angles is more than $180^{\circ}$. Thus two obtuse angles can never be supplementary.
(iii) Sum of two right angles is $180^{\circ}$. Hence two right angles are supplementary.
We know that the sum of two complementary angles is $90^{\circ}$ .
Thus if one of the angles is greater than $45^{\circ}$ then the other angle needs to be less than $45^{\circ}$.
(i) Is $\angle 1$ adjacent to $\angle 2$ ?
(ii) Is $\angle AOC$ adjacent to $\angle AOE$ ?
(iii) Do $\angle COE$ and $\angle EOD$ form a linear pair?
(iv) Are $\angle BOD$ and $\angle DOA$ supplementary?
(v) Is $\angle 1$ vertically opposite to $\angle 4$ ?
(vi) What is the vertically opposite angle of $\angle 5$ ?
(i) Yes, $\angle 1$ adjacent to $\angle 2$ as these have the same vertex and have one common arm.
(ii) No, $\angle AOC$ is not adjacent to $\angle AOE$. This is because $\angle AOE$ contains $\angle AOC$.
(iii) Yes the given angles form a linear pair as they are pair of supplementary angles.
(iv) Since BOA is a straight line thus the given angles are supplementary.
(v) Yes, $\angle 1$ and $\angle 4$ are vertically opposite angles as they are the angles formed by two intersecting straight lines.
(vi) The vertically opposite angle to $\angle 5$ is $\left ( \angle 2\ +\ \angle 3 \right )$.
(i) Vertically opposite angles.
The vertically opposite pairs are :
(a) $\angle 1$ and $\angle 4$
(b) $\angle 5$ and $\left ( \angle 2\ +\ \angle 3 \right )$
(ii) Linear pairs
The sum of angles in linear pair is $180^{\circ}$.
Thus the linear pairs are :
(a) $\angle 1\ and\ \angle 5$
(b) $\angle 4\ and\ \angle 5$
No, $\angle 1$ and $\angle 2$ are not adjacent angles as their vertex is not same/common.
For being adjacent angles the pair must have a common vertex and have a common arm.
From the figure :
(i) $\angle x\ =\ 55^{\circ}$ (Vertically opposite angle)
(ii) $\angle y\ =\ 180^{\circ}\ -\ 55^{\circ}\ =\ 125^{\circ}$ (Linear pair)
(iii) $\angle y\ =\ \angle z\ =\ 125^{\circ}$ (Vertically opposite angle)
From the figure we can observe that :
(i) $\angle z\ =\ 40^{\circ}$ (Vertically opposite angle)
(ii) $\angle x\ =\ 180^{\circ}\ -\ 40^{\circ}\ -\ 25^{\circ}\ =\ 115^{\circ}$ (Linear pair/straight line)
(iii) $\angle y\ =\ 180^{\circ}\ -\ \angle z\ =\ 140^{\circ}$ (Vertically opposite angle).
(i) Obtuse vertically opposite angles
(iii) Equal supplementary angles
(iv) Unequal supplementary angles
(v) Adjacent angles that do not form a linear pair
(i) $\angle AOD\ and\ \angle BOC$ are the vertically obtuse angles.
(ii) $\angle AOB\ and\ \angle AOE$ are the complementary angles.
(iii) $\angle BOE\ and\ \angle DOE$ are the equal supplementary angles.
(iv) $\angle BOC\ and\ \angle COD$ are the unequal pair of supplementary angle.
(v) $\angle AOB\ and\ \angle AOE$, $\angle EOD\ and\ \angle COD$ and $\angle AOE\ and\ \angle EOD$ are adjacent angles but are not supplementary angles.
## CBSE NCERT solutions for class 7 maths chapter 5 lines and angles exercise 5.2
(i) If $a\parallel b$, then $\angle 1= \angle 5$.
The statement "If $a\parallel b$, then $\angle 1= \angle 5$ " is true using the corresponding angles property.
(ii) If $\angle 4=\angle 6$, then $a\parallel b.$
The property used here is 'alternate interior angle property'.
(iii) If $\angle 4+\angle 5= 180^{o}$, then $a\parallel b.$
The property used here is 'Interior angles on the same side of the transversal are a pair of supplementary angles'.
## 2. In the adjoining figure, identify
(i) the pairs of corresponding angles.
(ii) the pairs of alternate interior angles.
(iii) the pairs of interior angles on the same side of the transversal.
(iv) the vertically opposite angles.
(i) Corresponding angles :- $\angle 1\ and\ \angle 5$, $\angle 2\ and\ \angle 6$, $\angle 3\ and\ \angle 7$, $\angle 4\ and\ \angle 8$
(ii) Alternate interior angles :- $\angle 2\ and\ \angle 8$, $\angle 3\ and\ \angle 5$,
(iii) Alternate angles on the same side of traversal :- $\angle 2\ and\ \angle 5$, $\angle 3\ and\ \angle 8$
(iv) Vertically opposite angles :- $\angle 1\ and\ \angle 3$$\angle 2\ and\ \angle 4$, $\angle 5\ and\ \angle 7$, $\angle 6\ and\ \angle 8$
The angles can be found using different properties:
(a) $\angle e\ =\ 180^{\circ}\ -\ 125^{\circ}\ =\ 55^{\circ}$ (The angles are linear pair)
(b) $\angle e\ =\ \angle f\ =\ 55^{\circ}$ (Vertically opposite angle)
(c) $\angle d\ =\ 125^{\circ}$ (Corresponding angle)
(d) $\angle d\ =\ \angle b\ =\ 125^{\circ}$ (Vertically opposite angle)
(e) $\angle a\ =\ \angle c\ =\ 180^{\circ}\ -\ 125^{\circ}\ =\ 55^{\circ}$ (Vertically opposite angel, linear pair).
The linear pair of the $110^{\circ}$ is : $\Theta \ =\ 180^{\circ}\ -\ 110^{\circ}\ =\ 70^{\circ}$
Thus the value of x is : $x\ =\ 70^{\circ}$ (Corresponding angles of parallel lines are equal).
The value of x is $100^{\circ}$, as these are the corresponding angles.
$(i) \angle DGC$
$(ii)\angle DEF$
(i) Since side AB is parallel to DG.
Thus : $\angle DGC\ =\ 70^{\circ}$ (Corresponding angles of parallel arms are equal.)
(ii) Further side BC is parallel to EF.
We have : $\angle DEF\ =\ \angle DGC\ =\ 70^{\circ}$ (Corresponding angles of parallel arms are equal.)
(i) In this case. the sum of the interior angle is $126^{\circ}\ +\ 44^{\circ}\ =\ 170^{\circ}$ thus l is not parallel to m.\
(ii) In this case also l is not parallel to m as the corresponding angle cannot be $75^{\circ}$ (Linear pair will not form).
(iii) In this l and m are parallel. This is because the corresponding angle is $57^{\circ}$ and it forms linear pair with $123^{\circ}$ .
(iv) The lines are not parallel as the linear pair not form. (Since the corresponding angle will be $72^{\circ}$ otherwise.)
## NCERT Solutions for Class 7 Maths - Chapter-wise
Chapter No. Chapter Name Chapter 1 Solutions of NCERT for class 7 maths chapter 1 Integers Chapter 2 CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals Chapter 3 NCERT solutions for class 7 maths chapter 3 Data Handling Chapter 4 Solutions of NCERT for class 7 maths chapter 4 Simple Equations Chapter 5 CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles Chapter 6 NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties Chapter 7 Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities Chapter 9 CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers Chapter 10 NCERT solutions for class 7 maths chapter 10 Practical Geometry Chapter 11 Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area Chapter 12 CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions Chapter 13 NCERT solutions for class 7 maths chapter 13 Exponents and Powers Chapter 14 Solutions of NCERT for class 7 maths chapter 14 Symmetry
## Benefits of NCERT solutions for class 7 maths chapter 5 lines and angles:
• Solving homework is an easy task with solutions of NCERT class 7 maths chapter 5 lines and angles in hand.
• Questions of similar type from CBSE NCERT solutions for class 7 maths chapter 5 are expected for the class exams.
• Practice all the questions from NCERT solutions for class 7 maths chapter 5 lines and angles to score well in the exam. |
# Find the matrix of the transformation with respect to the given basis
We have $\phi\colon \mathbb{R}^4 \to \mathbb{R}^4, \phi(x_1,x_2,x_3,x_4)= (x_1-2x_3,-2x_3-x_4,x_1+2x_3+2x_4,x_1+x_4)$. We want to find the matrix of this transformation with respect to the following basis of $\mathbb{R}^4$:
\begin{align} \alpha_1&=(1,1,0,0)\\ \alpha_2&=(0,1,1,0)\\ \alpha_3&=(0,0,1,1)\\ \alpha_4&=(0,0,0,1)\end{align}
First, we can write down the transformation as $$A_\phi = \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 0 & -2 & -1 \\ 1 & 0 & 2 & 2 \\ 1 & 0 & 0 & 1\\ \end{bmatrix}$$ and simplify it to the reduced row echelon form $$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & {1 \over 2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ \end{bmatrix}$$ From this, we immediately see $\dim\ker\phi = 2$ and, by extension, $$\dim\operatorname{im} \phi = \dim\mathbb{R}^4 - \dim\ker \phi = 4 - 2 = 2.$$ So, we can choose whichever two linearly independent column vectors from $A_\phi$ and they will be a basis of $\operatorname{im}\phi$.
Now, how do we proceed from here to find the matrix of the transformation with respect to the given basis?
• Calculate $\phi(\alpha_1)$. Express it as a linear combination of the $\alpha_i$: $\phi(\alpha_1)=a\alpha_1+b\alpha_2+c\alpha_3+d\alpha_4$. Then make $(a,b,c,d)$ the first column of a matrix. Repeat for the other $\alpha_i$ to get the other three columns of the matrix you have been asked to find. Nov 29, 2016 at 9:17
• @GerryMyerson Thanks for the explanation. This is what I get from the procedure you have described. By taking $\phi(\alpha_i)$ we get the image of each $\alpha_i$. By finding the coefficients of the linear combination, we express these images in terms of the required basis. I am not quite clear why putting the coefficients in the columns of a matrix gets us what we want. Nov 29, 2016 at 9:32
• What you want is a matrix $A$ such that $\phi(v)=Av$ for all $v$ in ${\bf R}^4$, where $v$ and $Av$ are both described in terms of basis you've been given. With the matrix I described, you find, for example, $\phi(\alpha_1)=\phi(1,0,0,0)=A(1,0,0,0)=(a,b,c,d)$ which is what you want, since $\phi(\alpha_1)=a\alpha_1+b\alpha_2+c\alpha_3+d\alpha_4$. And similarly for the other $\alpha_i$, and then, by linearity, similarly for all $v$. Nov 29, 2016 at 11:45
• I don’t understand why you row-reduced $A$. That and the conclusions you drew from it aren’t really relevant to this problem.
– amd
Nov 29, 2016 at 19:54
• Essentially what GerryMyerson described above in his comments.
– amd
Nov 30, 2016 at 8:46
Recall that the columns of a transformation matrix are the images of the domain basis vectors expressed relative to the basis of the codomain. So, for each basis vector $\alpha_j$, find the expression of $\phi\alpha_j$ as a linear combination $\sum_{i=1}^4 c_{ij}\alpha_i$. The required transformation matrix is then the coefficient matrix $[c_{ij}]$.
This computation can be accomplished all at once via matrix multiplication. Let $B$ be the matrix with the vectors $\alpha_k$ (expressed relative to the standard basis) as its columns. The columns of $A_\phi B$ are then the images of these vectors, also expressed relative to the standard basis. Observe that $B$ can be interpreted as converting from the $\{\alpha_k\}$ basis to the standard one (why?), so $B^{-1}$ converts from the standard basis to the $\{\alpha_k\}$ basis. Thus, $B^{-1}A_\phi B$ is the required matrix. This operation is known as a similarity transformation or conjugation of $A_\phi$. |
# Trigonometry/Simplifying a sin(x) + b cos(x)
Consider the function
$\displaystyle f(x) = a \sin x + b \cos x$
We shall show that this is a sinusoidal wave, and find its amplitude and phase.
To make things a little simpler, we shall assume that a and b are both positive numbers. This isn't necessary, and after studying this section you may like to think what would happen if either of a or b is zero or negative.
## Geometric Argument
We'll first use a geometric argument that actually shows a more general result, that:
$\displaystyle g(\theta) = a_1 \sin( \theta + \lambda_1) + a_2 \sin( \theta + \lambda_2)$
is a sinusoidal wave. Since we can set $\displaystyle \lambda_1=0^\circ, \lambda_2=90^\circ$ the result we are trying for with $\displaystyle f$ follows as a special case.
We use the 'unit circle' definition of sine. $\displaystyle a_1 \sin( \theta + \lambda_1)$ is the y coordinate of a line of length $\displaystyle a_1$ at angle $\displaystyle \theta + \lambda_1$ to the x axis, from O the origin, to a point A.
If we now draw a line $\displaystyle \overline{AB}$ of length $\displaystyle a_2$ at angle $\displaystyle \theta + \lambda_2$ (where that angle is measure relative to a line parallel to the x axis), its y coordinate is the sum of the two sines.
However, there is another way to look at the y coordinate of point $\displaystyle B$. The line $\displaystyle \overline{OB}$ does not change in length as we change $\displaystyle \theta$, because the lengths of $\displaystyle \overline{OA}$ and $\displaystyle \overline{AB}$ and the angle between them do not change. All that happens is that the triangle $\displaystyle \Delta OBC$ rotates about O. In particular $\displaystyle \overline{OB}$ rotates about O.
This then brings us back to a 'unit circle' like definition of a sinusoidal function. The amplitude is the length of $\displaystyle \overline{OB}$ and the phase is $\displaystyle \lambda_1 + \angle BOA$.
## Algebraic Argument
The algebraic argument is essentially an algebraic translation of the insights from the geometric argument. We're also in the special case that $\displaystyle \lambda_1 = 0$and $\displaystyle \angle OAB=90^\circ$. The x's and y's in use in this section are now no longer coordinates. The 'y' is going to play the role of $\displaystyle \lambda_1 + \angle BOA$ and the 'x' plays the role of $\displaystyle \theta$.
We define the angle y by $\tan(y) = \frac{b}{a}$.
By considering a right-angled triangle with the short sides of length a and b, you should be able to see that
$\sin(y) = \frac{b}{\sqrt{a^2+b^2}}$ and $\cos(y) = \frac{a}{\sqrt{a^2+b^2}}$.
Check this Check that $\displaystyle \sin^2 x + \cos^2 x = 1$ as expected.
$f(x) = a \sin(x) + b \cos(x) = \sqrt{a^2+b^2} [\frac{a}{\sqrt{a^2+b^2}} \sin(x) + \frac{b}{\sqrt{a^2+b^2}} \cos(x)] = \sqrt{a^2+b^2}[\sin(x)\cos(y) + \cos(x)\sin(y)] = \sqrt{a^2+b^2}\sin(x+y)$,
which is (drum roll) a sine wave of amplitude $\sqrt{a^2+b^2}$ and phase y.
Check this Check each step in the formula. What trig formulas did we use?
The more general case Can you do the full algebraic version for the more general case: $\displaystyle g(\theta) = a_1 \sin( \theta + \lambda_1) + a_2 \sin( \theta + \lambda_2)$ using the geometric argument as a hint? It is quite a bit harder because $\displaystyle \Delta OBC$ is not a right triangle. What additional trig formulas did you need? |
# Dealt cards
If you draw two cards from the deck, what is the probability both are aces? There are two interpretations for this question, you can deal two cards off the top of the deck, or you can deal one card, replace it reshuffle, and deal the second (perhaps same card, card. In the former case, the probability that the first card is an ace is 1/13, but if the first card is an ace, only three aces remain among the remaining 51 cards, hence the probaility the second is an ace (conditioned on the first being an ace) is 3/51. We may employ the product rule, letting FA designate that the first card is an ace and SA designate that the second card is an ace: P(FA and SA) = P(SA|FA) × P(FA) = 3/51 × 1/13 = 3/(51 × 13). In the second case, with the second card drawn from the restored and reshuffled deck, the two draws are independent and P(FA and SA) = P(FA) × P(SA) = 1/13 × 1/13 = 1/169.
Exercise: What is the probability that two cards drawn from the deck are the same suit? Different suits? Answer these questions for both the case of choosing the cards with and without replacement.
The above calculations have used the approach of conditional probabilities. This can be extended to other problems such as the probability that five cards drawn from a deck without replacement are of the same suit (a flush). The first card designates the suit, 12/51 of that suit remain in the deck, if the first two cards are of the same suit 11/50 of the same suit remain in the deck... . this line of reasoning extends to give that the probability of being dealt a flush is 1 × 12/51 × 11/50 × 10/49 × 9/48. However it is much more difficult to use conditional probabilities to calculate the probability of a full house (two of one kind and three of another kind).An aternative way to calculate the probability of being dealt a flush is to count the number of different hands which are flushes, and divide that by the total number of different hands. There are C(52, 5) different hands that can be dealt from a deck. There are C(13, 5) different flushes in each suit. Hence the probability of being dealt a flush is (4 × C(13, 5))/C(52, 5) = (12 × 11 × 10 × 9)/(51 × 50 × 49 × 48) as found above. The same reasoning can be used to calculate the probability of a full house: Ther are 13 choices for the denomination there are three of, which leaves 12 denominations for the pair; There are C(4, 3) ways to choose the the cards within the denomination of the treesome and C(4, 2) ways to choose the pair from the denomination of the twosome. Hence the probability of a full house is (13 × 12 × C(4, 3) × C(4, 2))/C(52, 5).
Competencies: Calculate the probability that four cards dealt from a deck without replacement are of different suits, both by conditional probability and by counting arguments.
Reflection: Which problems are more easily done as conditional probability and which problems are more easily done by counting arguments.
Challenge: Calculate the probability that three cards dealt from a deck without replacement are of different suits, both by conditional probability and by counting arguments.
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### FAQ
##### How many cards are dealt in card game of gin?
10 cards are dealt in card game of gin. !
##### If I had two complete decks of cards shuffled randomly, what is the probability that two randomly dealt cards will be a pair? - Quora
Assume that the shuffling is good enough, so that we can assume a uniform distribution, and that there are 52 cards per deck (no jokers, standard poker deck). Then you can do this calculation:
There are 104 unique cards to choose from to pick the first card. The probability that you pick the first card is trivially 104/104 = 1. To pick the second card such that a pair is formed, there are 7 possibilities out of the 103 remaining cards. Hence the probability is 1 x 7/103 = 7/103 that you get a pair.
The probability that you get a pair in the same suit can be calculated similarly, except th…
Related Posts |
# 2005 AMC 12B Problems/Problem 11
The following problem is from both the 2005 AMC 12B #11 and 2005 AMC 10B #15, so both problems redirect to this page.
## Solution 2
Another way to do this problem is to use complementary counting, i.e. how many ways that the sum is less than $20$. Now, you do not have to consider the $2$ twenties, so you have $6$ bills left. $\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15$ ways. However, you counted the case when you have $2$ tens, so you need to subtract 1, and you get $14$. Finding the ways to get $20$ or higher, you subtract $14$ from $28$ and get $14$. So the answer is $\dfrac{14}{28} = \boxed{\textbf{(D) }\dfrac{1}{2}}$
## Solution 3
There are two cases that work, namely getting at least $1$ twenty, or getting $2$ tens.
Case $1$: $P(\text{Get at least one twenty}) = 1-P(\text{Do not get a single twenty})=1- \frac{\binom{6}{2}}{\binom{8}{2}}=\frac{28-15}{28}=\frac{13}{28}$
Case $2$ : $P(\text{Get two tens}) = \frac{1}{\binom{8}{2}} = \frac{1}{28}$
Summing up our cases, we have $\frac{13}{28}+\frac{1}{28}=\frac{14}{28}=\boxed{\textbf{(D) } \dfrac{1}{2}}$
## Solution 4
Note that if a twenty is drawn, anything else that is drawn will create a total greater than $20$; The probability of a twenty being drawn first is $\frac{1}{4}.$ The same could be said for drawing anything, and then drawing a twenty. However, we can only draw something that isn't a twenty first (since we've already accounted for the probability of drawing two twenties).
The probability of drawing a non-twenty first, then a twenty second is $\frac{3}{4}\cdot\frac{2}{7}=\frac{3}{14}.$ Finally, we can draw two tens. The probability of this occuring is $\frac{1}{4}\cdot\frac{1}{7}=\frac{1}{28}.$
Adding these three probabilities gives us $\frac{1}{4}+\frac{3}{14}+\frac{1}{28}=\boxed{\textbf{(D) } \dfrac{1}{2}}$
-Benedict T (countmath1) (edited by AMC_8)
~savannahsolver |
A+ » VCE Blog » VCE Maths Methods Blog » Introduction to Probability [Video Tutorial]
# Introduction to Probability [Video Tutorial]
This tutorial covers material encountered in chapter 13 of the VCE Mathematical Methods Textbook, namely:
• Basic set and probability theory
• Addition and multiplication (Bayes’ Theorem) rules
• Conditional probability
• Law of total probability
• Notion of independent and mutually exclusive events
• Discrete random variables
• Mean, variance and standard deviation of discrete random variables
## Worksheet
Q1. If P(A)=0.3,\,P(B)=0.4 and P(A\cup B)=0.6, are A and B mutually exclusive? Are they independent?
Q2. Show that 1-P(A'\cap B')=P(A\cup B) using visual methods or otherwise. How would you describe this relationship in words to the layman?
Q3. A bag contains three red balls and five green balls. What is the probability of randomly drawing two red balls if:
(a) The first ball drawn is put back in the bag before the second is drawn.
(b) The first ball drawn is not put back in the bag before the second is drawn.
Q4. If P(A)=\dfrac{1}{3},\, P(B)=\dfrac{1}{2} and P(A|B)=\dfrac{1}{4}, find:
(a) P(A\cap B) and P(A\cup B)
(b) P(A'|B) and P(A|B')
Q5. A test for a certain virus is considered 95\% effective; meaning that if someone with the virus is tested the test will return positive 95\% of the time, and if someone without the virus is tested the test will return positive 5\% of the time. Only 1\% of the population actually has the virus. If a person tests positive for the virus what is the probability that they actually have it?
Q6. The discrete random variable X has the following probability distribution:
\begin{array}{c||cccccc}
x & -1 & 0 & 1 & 2 & 3 & 4 \\
\hline P(X=x) & \mathrm{t} & 3 \mathrm{t} & 2 \mathrm{t} & 3 \mathrm{t} & 2 \mathrm{t} & \mathrm{t}
\end{array}
Find:
(a) The constant t
(b) \mu(X)
(c) \sigma(X) and \sigma^2(X)
Q7. A shop manufactures Garfield mugs for \$x. It costs the shop \$3 to make every mug, however there is a \frac{1}{6} chance in manufacturing that the mug will be a defect and in this case it must be thrown out and the mug is considered a total loss to the shop.
(a) Find the probability function P, the profit per Garfield mug.
(b) Find the mean of P in terms of x.
(c) How much should the shop sell its Garfield mugs in order to make a profit in the long term?
### Bonus Questions
#### I. Lewis in probability land
Lewis has a bag with a single ball inside, the ball has an equal probability of being either red or green. Lewis then takes a red ball and puts it inside the bag (with the original ball), shakes it around for a bit and then takes a ball out of the bag at random. If the ball Lewis takes out is red, what is the probability that the other ball was also red?
This question was published in 1895 by Lewis Carroll, the famous author of Alice in Wonderland.
#### II. Monty Hall problem
Monty Hall is the host of a peculiar game show. The contestant is given three doors to choose from, behind one door is \\$10mil and behind the other two doors lies nothing. After the contestant chooses a door (but before they open it) Monty Hall will open a different door that has nothing behind it, he will then give the contestant the choice between switching from their original door to the closed door left over. Should the contestant switch or not?
Got questions? Share and ask here... |
# 8.1: Permutations
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Permutations appear in many different mathematical concepts, and so we give a general introduction to them in this section.
## Definition of Permutations
Given a positive integer $$n \in \mathbb{Z}_{+}$$, a permutation of an (ordered) list of $$n$$ distinct objects is any reordering of this list. When describing the reorderings themselves, though, the nature of the objects involved is more or less irrelevant. E.g., we can imagine interchanging the second and third items in a list of five distinct objects --- no matter what those items are --- and this defines a particular permutation that can be applied to any list of five objects.
Since the nature of the objects being rearranged (i.e., permuted) is immaterial, it is common to use the integers $$1,2,\ldots,n$$ as the standard list of $$n$$ objects. Alternatively, one can also think of these integers as labels for the items in any list of $$n$$ distinct elements. This gives rise to the following definition.
Definition: Permutation
A permutation $$\pi$$ of $$n$$ elements is a one-to-one and onto function having the set $$\{1,2,\ldots, n\}$$ as both its domain and codomain.
In other words, a permutation is a function $$\pi:\{1, 2, \ldots, n\} \longrightarrow \{1, 2, \ldots, n\}$$ such that, for every integer $$i \in \{1, \ldots, n\}$$, there exists exactly one integer $$j \in \{1, \ldots, n\}$$ for which $$\pi(j) = i$$. We will usually denote permutations by Greek letters such as $$\pi$$(pi), $$\sigma$$(sigma), and $$\tau$$(tau). The set of all permutations of $$n$$ elements is denoted by $$\mathcal{S}_{n}$$ and is typically referred to as the symmetric group of degree $$n$$. (In particular, the set $$\mathcal{S}_{n}$$ forms a group under function composition as discussed in Section 8.1.2).
Given a permutation $$\pi \in \mathcal{S}_{n}$$, there are several common notations used for specifying how $$\pi$$ permutes the integers $$1, 2, \ldots, n$$. The important thing to keep in mind when working with these different notations is that $$\pi$$ is a function defined on the finite set $$\{1, 2, \ldots, n\}$$, with notation being used as a convenient short-hand for keeping track of how $$\pi$$ permutes the elements in this set.
Definition 8.1.2: two-line notation
Given a permutation $$\pi \in \mathcal{S}_{n}$$, denote $$\pi_{i} = \pi(i)$$ for each $$i \in \{1, \ldots, n\}$$. Then the two-line notation for $$\pi$$ is given by the $$2 \times n$$ matrix
$\pi = \begin{pmatrix} 1 & 2 & \cdots & n \\ \pi_{1} & \pi_{2} & \cdots & \pi_{n} \end{pmatrix}.$
In other words, given a permutation $$\pi \in \mathcal{S}_{n}$$ and an integer $$i \in \{1, \ldots, n\}$$, we are denoting the image of $$i$$ under $$\pi$$ by $$\pi_{i}$$ instead of using the more conventional function notation $$\pi(i)$$. Then, in order to specify the image of each integer $$i \in \{1, \ldots, n\}$$ under $$\pi$$, we list these images in a two-line array as shown above. (One can also use so-called one-line notation for $$\pi$$, which is given by simply ignoring the top row and writing $$\pi = \pi_{1}\pi_{2}\cdots\pi_{n}$$.)
It is important to note that, although we represent permutations as $$2 \times n$$ matrices, you should not think of permutations as linear transformations from an $$n$$-dimensional vector space into a two-dimensional vector space. Moreover, the composition operation on permutation that we describe in Section 8.1.2 below does not correspond to matrix multiplication. The use of matrix notation in denoting permutations is merely a matter of convenience.
Example $$\PageIndex{3}$$:
Suppose that we have a set of five distinct objects and that we wish to describe the permutation that places the first item into the second position, the second item into the fifth position, the third item into the first position, the fourth item into the third position, and the fifth item into the fourth position. Then, using the notation developed above, we have the permutation $$\pi \in \mathcal{S}_{5}$$ such that
$\pi_{1} = \pi(1) = 3, \quad \pi_{2} = \pi(2) = 1, \quad \pi_{3} = \pi(3) = 4, \quad \pi_{4} = \pi(4) = 5, \quad \pi_{5} = \pi(5) = 2.$
In two-line notation, we would write $$\pi$$ as
$\pi = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 4 & 5 & 2 \end{pmatrix}.$
It is relatively straightforward to find the number of permutations of $$n$$ elements, i.e., to determine cardinality of the set $$\mathcal{S}_{n}$$. To construct an arbitrary permutation of $$n$$ elements, we can proceed as follows: First, choose an integer $$i \in \{1, \ldots, n\}$$ to put into the first position. Clearly, we have exactly $$n$$ possible choices. Next, choose the element to go in the second position. Since we have already chosen one element from the set $$\{1, \ldots, n\}$$, there are now exactly $$n - 1$$ remaining choices. Proceeding in this way, we have $$n - 2$$ choices when choosing the third element from the set $$\{1, \ldots, n\}$$, then $$n - 3$$ choices when choosing the fourth element, and so on until we are left with exactly one choice for the $$n^{\rm th}$$ element. This proves the following theorem.
Theorem 8.1.4
The number of elements in the symmetric group $$\mathcal{S}_{n}$$ is given by
$|\mathcal{S}_{n}| = n\cdot(n-1)\cdot(n-2)\cdot\,\cdots\,\cdot 3\cdot 2\cdot 1 = n!$
We conclude this section with several examples, including a complete description of the one permutation in $$\mathcal{S}_{1}$$, the two permutations in $$\mathcal{S}_{2}$$, and the six permutations in $$\mathcal{S}_{3}$$. For your own practice, you should (patiently) attempt to list the $$4! = 24$$ permutations in $$\mathcal{S}_{4}$$.
Example $$\PageIndex{5}$$:
1. Given any positive integer $$n \in \mathbb{Z}_{+}$$, the identity function $$\mbox{id}:\{1, \ldots, n\} \longrightarrow \{1, \ldots, n\}$$ given by $$\mbox{id}(i) = i$$, $$\forall \ i \in \{1, \ldots, n\}$$, is a permutation in $$\mathcal{S}_{n}$$. This function can be thought of as the trivial reordering that does not change the order at all, and so we call it the trivial or identity permutation.
2. If $$n = 1$$, then, by Theorem 8.1.4, $$|\mathcal{S}_{n}| =1! = 1$$. Thus, $$\mathcal{S}_{1}$$ contains only the identity permutation.
3. If $$n = 2$$, then, by Theorem 8.1.4, $$|\mathcal{S}_{n}| = 2! = 2 \cdot 1 = 2$$. Thus, there is only one non-trivial permutation $$\pi$$ in $$\mathcal{S}_{2}$$, namely the transformation interchanging the first and the second elements in a list. As a function, $$\pi(1) = 2$$ and $$\pi(2) = 1$$, and, in two-line notation,
$\pi = \begin{pmatrix} 1 & 2 \\ \pi_{1} & \pi_{2} \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}.$
4. If $$n = 3$$, then, by Theorem 8.1.4, $$|\mathcal{S}_{n}| = 3! = 3 \cdot 2 \cdot 1 = 6$$. Thus, there are five non-trivial permutation in $$\mathcal{S}_{3}$$.
Using two-line notation, we have that
$\mathcal{S}_{3} = \left\{ \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix} \right\}$
Keep in mind the fact that each element in $$\mathcal{S}_{3}$$ is simultaneously both a function and a reordering operation. E.g., the permutation
$\pi = \begin{pmatrix} 1 & 2 & 3 \\ \pi_{1} & \pi_{2} & \pi_{3} \end{pmatrix} = \begin{pmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{pmatrix}$
can be read as defining the reordering that, with respect to the original list, places the second element in the first position, the third element in the second position, and the first element in the third position. This permutation could equally well have been identified by describing its action on the (ordered) list of letters $$a,b,c$$. In other words,
$\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} = \begin{pmatrix} a & b & c \\ b & c & a \end{pmatrix},$
regardless of what the letters $$a, b, c$$ might happen to represent.
## Composition of Permutations
Let $$n \in \mathbb{Z}_{+}$$ be a positive integer and $$\pi,\sigma \in \mathcal{S}_{n}$$ be permutations. Then, since $$\pi$$ and $$\sigma$$ are both functions from the set $$\{1,\ldots, n\}$$ to itself, we can compose them to obtain a new function $$\pi \circ \sigma$$ (read as pi after sigma'') that takes on the values
$(\pi \circ \sigma)(1) = \pi(\sigma(1)), \quad (\pi \circ \sigma)(2) = \pi(\sigma(2)), \quad \ldots \quad (\pi \circ \sigma)(n) = \pi(\sigma(n)).$
In two-line notation, we can write $$\pi \circ \sigma$$ as
$\begin{pmatrix} 1 & 2 & \cdots & n \\ \pi(\sigma(1)) & \pi(\sigma(2)) & \cdots & \pi(\sigma(n)) \end{pmatrix} \mbox{ or } \begin{pmatrix} 1 & 2 & \cdots & n \\ \pi_{\sigma(1)} & \pi_{\sigma(2)} & \cdots & \pi_{\sigma(n)} \end{pmatrix} \mbox{ or } \begin{pmatrix} 1 & 2 & \cdots & n \\ \pi_{\sigma_{1}} & \pi_{\sigma_{2}} & \cdots & \pi_{\sigma_{n}} \end{pmatrix}.$
Example $$\PageIndex{6}$$:
From $$\mathcal{S}_{3}$$, suppose that we have the permutations $$\pi$$ and $$\sigma$$ given by
$\pi(1) = 2, \ \pi(2) = 3, \ \pi(3) = 1 \sigma(1) = 1, \ \sigma(2) = 3, \ \sigma(3) = 2.$
Then note that
$(\pi \circ \sigma)(1) = \pi(\sigma(1)) = \pi(1) = 2,$
$(\pi \circ \sigma)(2) = \pi(\sigma(2)) = \pi(3) = 1,$
$(\pi \circ \sigma)(3) = \pi(\sigma(3)) = \pi(2) = 3.$
In other words,
$\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ \pi(1) & \pi(3) & \pi(2) \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}.$
Similar computations (which you should check for your own practice) yield compositions such as
$\begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ \sigma(2) & \sigma(3) & \sigma(1) \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix},$
$\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ \sigma(1) & \sigma(2) & \sigma(3) \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix},$
and
$\begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ \mbox{id}(2) & \mbox{id}(3) & \mbox{id}(1) \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}.$
In particular, note that the result of each composition above is a permutation, that composition is not a commutative operation, and that composition with $$\mbox{id}$$ leaves a permutation unchanged. Moreover, since each permutation $$\pi$$ is a bijection, one can always construct an inverse permutation $$\pi^{-1}$$ such that $$\pi \circ \pi^{-1} = \mbox{id}$$. E.g.,
$\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \circ \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ \pi(3) & \pi(1) & \pi(2) \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}.$
We summarize the basic properties of composition on the symmetric group in the following theorem.
Theorem 8.1.7 Let $$n \in \mathbb{Z}_{+}$$ be a positive integer. Then the set $$\mathcal{S}_{n}$$ has the following properties:
1. Given any two permutations $$\pi, \sigma \in \mathcal{S}_{n}$$, the composition $$\pi \circ \sigma \in \mathcal{S}_{n}$$.
2. (Associativity of Composition) Given any three permutations $$\pi, \sigma, \tau \in \mathcal{S}_{n}$$, $(\pi \circ \sigma) \circ \tau = \pi \circ (\sigma \circ \tau).$
3. (Identity Element for Composition) Given any permutation $$\pi \in \mathcal{S}_{n}$$, $\pi \circ \mbox{id} = \mbox{id} \circ \pi = \pi.$
4. (Inverse Elements for Composition) Given any permutation $$\pi \in \mathcal{S}_{n}$$, there exists a unique permutation $$\pi^{-1} \in \mathcal{S}_{n}$$ such that $\pi \circ \pi^{-1} = \pi^{-1} \circ \pi = \mbox{id}.$
In other words, the set $$\mathcal{S}_{n}$$ forms a group under composition.
Note that the composition of permutations is not commutative in general. In particular, for $$n \geq 3$$, it is easy to find permutations $$\pi$$ and $$\sigma$$ such that $$\pi\circ\sigma\neq \sigma\circ\pi$$.
## Inversions and the sign of a permutation
Let $$n \in \mathbb{Z}_{+}$$ be a positive integer. Then, given a permutation $$\pi \in \mathcal{S}_{n}$$, it is natural to ask how out of order'' $$\pi$$ is in comparison to the identity permutation. One method for quantifying this is to count the number of so-called inversion pairs in $$\pi$$ as these describe pairs of objects that are out of order relative to each other.
Definition 8.1.8. Let $$\pi \in \mathcal{S}_{n}$$ be a permutation. Then an inversion pair $$(i, j)$$ of $$\pi$$ is a pair of positive integers $$i, j \in \{1, \ldots, n\}$$ for which $$i < j$$ but $$\pi(i) > \pi(j)$$. Note, in particular, that the components of an inversion pair are the positions where the two out of order'' elements occur.
Example 8.1.9. We classify all inversion pairs for elements in $$\mathcal{S}_{3}$$:
• $$\mbox{id} = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}$$ has no inversion pairs since no elements are out of order''.
• $$\pi = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix}$$ has the single inversion pair $$(2, 3)$$ since $$\pi(2) = 3 > 2 = \pi(3)$$.
• $$\pi = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}$$ has the single inversion pair $$(1, 2)$$ since $$\pi(1) = 2 > 1 = \pi(2)$$.
• $$\pi = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}$$ has the two inversion pairs $$(1, 3)$$ and $$(2, 3)$$ since we have that both $$\pi(1) = 2 > 1 = \pi(3)$$ and $$\pi(2) = 3 > 1 = \pi(3)$$.
• $$\pi = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix}$$ has the two inversion pairs $$(1, 2)$$ and $$(1, 3)$$ since we have that both $$\pi(1) = 3 > 1 = \pi(2)$$ and $$\pi(1) = 3 > 2 = \pi(3)$$.
• $$\pi = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix}$$ has the three inversion pairs $$(1, 2)$$, $$(1, 3)$$, and $$(2, 3)$$, as you can check.
Example $$\PageIndex{1}$$: Example 8.1.10
As another example, for each $$i, j \in \{1, \ldots, n\}$$ with $$i < j$$, we define the transposition $$t_{i j} \in \mathcal{S}_{n}$$ by
$t_{i j} = \begin{pmatrix} 1 & 2 & \cdots & i & \cdots & j & \cdots & n \\ 1 & 2 & \cdots & j & \cdots & i & \cdots & n \\ \end{pmatrix}.$
In other words, $$t_{i j}$$ is the permutation that interchanges $$i$$ and $$j$$ while leaving all other integers fixed in place. One can check that the number of inversions pairs in $$t_{i j}$$ is exactly $$2(j - i) - 1$$. Thus, the number of inversions in a transposition is always odd. E.g.,
$t_{1 3}=\begin{pmatrix}1 & 2 & 3 & 4 \\ 3 & 2 & 1 & 4\end{pmatrix}$
has inversion pairs $$(1, 2)$$, $$(1, 3)$$, and $$(2, 3)$$.
For the purposes of using permutations in Linear Algebra, the significance of inversion pairs is mainly due to the following fundamental definition.
Definitions: Even and odd permutations
Let $$\pi \in \mathcal{S}_{n}$$ be a permutation. Then the sign of $$\pi$$, denoted by $$\mbox{sign}(\pi)$$, is defined by
$\mbox{sign}(\pi) = (-1)^{\# \; {\rm of} \; {\rm inversion} \; {\rm pairs} \; {\rm in} \; \pi} = \begin{cases} +1, & \mbox{if the number of inversions in } \pi \mbox{ is even} \\ -1, & \mbox{if the number of inversions in } \pi \mbox{ is odd} \\ \end{cases}.$
We call $$\pi$$ an even permutation if $$\mbox{sign}(\pi) = +1$$, whereas $$\pi$$ is called an odd permutation if $$\mbox{sign}(\pi) = -1$$.
Example $$\PageIndex{12}$$
Based upon the computations in Example 8.1.9 above, we have that
$\mbox{sign} \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} = \mbox{sign} \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} = \mbox{sign} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} = +1$
and that
$\mbox{sign} \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix} = \mbox{sign} \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix} = \mbox{sign} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix} = -1.$
Similarly, from Example 8.1.10, it follows that any transposition is an odd permutation.
We summarize some of the most basic properties of the sign operation on the symmetric group in the following theorem.
Theorem 8.1.13.
Let $$n \in \mathbb{Z}_{+}$$ be a positive integer. Then,
1. for $$\mbox{id} \in \mathcal{S}_{n}$$ the identity permutation, $\mbox{sign}(\mbox{id}) = +1.$
2. for $$t_{i j} \in \mathcal{S}_{n}$$ a transposition with $$i, j \in \{1, \ldots, n\}$$ and $$i < j$$,
\mbox{sign}(t_{i j}) = -1.
\label{eqn:sign_transposition} \tag{8.1.1}
3. given any two permutations $$\pi, \sigma \in \mathcal{S}_{n}$$,
\begin{eqnarray}
\mbox{sign}(\pi\circ\sigma) & = & \mbox{sign}(\pi)\,\mbox{sign}(\sigma)
\label{eqn:sign_product}, \tag{8.1.2}\\
\mbox{sign}(\pi^{-1}) & = & \mbox{sign}(\pi).
\label{eqn:sign_inverse} \tag{8.1.3}
\end{eqnarray}
4. the number of even permutations in $$\mathcal{S}_{n}$$, when $$n \geq 2$$, is exactly $$\frac{1}{2}n!$$.
5. the set $$A_{n}$$ of even permutations in $$\mathcal{S}_{n}$$ forms a group under composition.
## Contributors
Both hardbound and softbound versions of this textbook are available online at WorldScientific.com.
This page titled 8.1: Permutations is shared under a not declared license and was authored, remixed, and/or curated by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling. |
# 7.2 Right triangle trigonometry
Page 1 / 12
In this section you will:
• Use right triangles to evaluate trigonometric functions.
• Find function values for $\text{\hspace{0.17em}}30°\left(\frac{\pi }{6}\right),45°\left(\frac{\pi }{4}\right),$ and $\text{\hspace{0.17em}}60°\left(\frac{\pi }{3}\right).$
• Use equal cofunctions of complementary angles.
• Use the definitions of trigonometric functions of any angle.
• Use right-triangle trigonometry to solve applied problems.
Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task and, in fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles and a branch of mathematics known as trigonometry. In this section, we will define a new group of functions known as trigonometric functions, and find out how they can be used to measure heights, such as those of the tallest mountains.
## Using right triangles to evaluate trigonometric functions
[link] shows a right triangle with a vertical side of length $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ and a horizontal side has length $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ Notice that the triangle is inscribed in a circle of radius 1. Such a circle, with a center at the origin and a radius of 1, is known as a unit circle .
We can define the trigonometric functions in terms an angle t and the lengths of the sides of the triangle. The adjacent side is the side closest to the angle, x . (Adjacent means “next to.”) The opposite side is the side across from the angle, y . The hypotenuse is the side of the triangle opposite the right angle, 1. These sides are labeled in [link] .
Given a right triangle with an acute angle of $\text{\hspace{0.17em}}t,$ the first three trigonometric functions are listed.
A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “ S ine is o pposite over h ypotenuse, C osine is a djacent over h ypotenuse, T angent is o pposite over a djacent.”
For the triangle shown in [link] , we have the following.
Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle.
1. Find the sine as the ratio of the opposite side to the hypotenuse.
2. Find the cosine as the ratio of the adjacent side to the hypotenuse.
3. Find the tangent as the ratio of the opposite side to the adjacent side.
## Evaluating a trigonometric function of a right triangle
Given the triangle shown in [link] , find the value of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha .$
The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17.
$\begin{array}{ccc}\hfill \mathrm{cos}\left(\alpha \right)& =& \frac{\text{adjacent}}{\text{hypotenuse}}\hfill \\ & =& \frac{15}{17}\hfill \end{array}$
Given the triangle shown in [link] , find the value of $\text{\hspace{0.17em}}\text{sin}\text{\hspace{0.17em}}t.$
$\frac{7}{25}$
## Reciprocal functions
In addition to sine, cosine, and tangent, there are three more functions. These too are defined in terms of the sides of the triangle.
Take another look at these definitions. These functions are the reciprocals of the first three functions.
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
e power cos hyperbolic (x+iy)
10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
Dua
Yes
ahmed
Thank you
Dua
give me treganamentry question
Solve 2cos x + 3sin x = 0.5 |
# Fraction calculator
This fraction calculator performs basic and advanced fraction operations, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. The calculator helps in finding value from multiple fractions operations. Solve problems with two, three, or more fractions and numbers in one expression.
## The result:
### 6 - 12/3 = 13/3 = 4 1/3 ≅ 4.3333333
Spelled result in words is thirteen thirds (or four and one third).
### How do we solve fractions step by step?
1. Conversion a mixed number 1 2/3 to a improper fraction: 1 2/3 = 1 2/3 = 1 · 3 + 2/3 = 3 + 2/3 = 5/3
To find a new numerator:
a) Multiply the whole number 1 by the denominator 3. Whole number 1 equally 1 * 3/3 = 3/3
b) Add the answer from the previous step 3 to the numerator 2. New numerator is 3 + 2 = 5
c) Write a previous answer (new numerator 5) over the denominator 3.
One and two thirds is five thirds.
2. Subtract: 6 - 5/3 = 6/1 - 5/3 = 6 · 3/1 · 3 - 5/3 = 18/3 - 5/3 = 18 - 5/3 = 13/3
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(1, 3) = 3. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 1 × 3 = 3. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - six minus five thirds is thirteen thirds.
#### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right. |
# Division of Rational Expressions
## Invert, cancel, multiply, and reduce fractions with variables in the denominator
%
Progress
MEMORY METER
This indicates how strong in your memory this concept is
Progress
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Traveling
Teacher Contributed
## Real World Applications – Algebra I
### Topic
How much Americans Spend While Traveling to Different Countries
### Student Exploration
People love to travel. Many people save a lot of money so they can go to foreign countries and enjoy life. The US Department of Commerce likes to keep track of the amount of money that is spent in different countries each year by Americans. Mathematicians have helped the US Department of Commerce to help figure out a pattern that best represents the amount that is spent each year.
For the sake of this exercise, the formula that represents the total amount that Americans have spent every year since 2000 can be represented by the function, \begin{align*}T(x)=\frac{(3x^2+15x)}{(3x^2+12x)}\end{align*}, where \begin{align*}x\end{align*} represents the number of years that have passed since 1990.
A formula has also been derived to represent the number of people that have left the United States to travel each year. This formula is represented by the function, \begin{align*}P(x)=\frac{(x^2-x-30)}{(x^2-7x+6)}\end{align*}, where \begin{align*}x\end{align*} also represents the number of years that have passed since 1990.
If we want to find the average amount of money spent by each American per year, we can divide the total amount of money spent by the total number of Americans that have traveled. We would have:
Since we’re dividing this rational expression, we want to first factor all of the expressions, if possible.
Now that we’ve factored both of these expressions, we notice that some of the factors can be canceled from both the numerator and from the denominator.
Once we have canceled all of the same factors, we can rewrite our expression.
When we divide rational expressions, we know that we have to multiply by the reciprocal of the second rational fraction.
Once we rewrite our new expression, we can multiply and then cancel out more factors that are the same.
Since the \begin{align*}(x + 5)\end{align*} factors are the same on the numerator and the denominator of the fraction, we can cancel them out.
What we have left is:
What does this resulting expression mean?
We were to find a way to find the average amount of money that American travelers spend each year. This is the expression that represents this relationship. We can also use this and multiply by the number of Americans that travel and verify the total amount of money that’s spent outside of the United States every year. Let’s try it.
We’re multiplying two rational expressions, so we want to make sure that before we multiply them, that all expressions are fully factored.
Now that the fraction on the right has been fully factored, we can simplify. We know that the factor \begin{align*}(x - 6)\end{align*} can be canceled.
As a matter of fact, we also know that the factor \begin{align*}(x - 1)\end{align*} is present on both the top and the bottom of the multiplication problem, so we can cancel them out too.
What we have as the result of multiplying the two rational expressions is the total amount of money that’s spent outside of the United States, which is our original expression that we started with.
### Extension Investigation
What are other similar situations that represent a similar relationship? When can we use the relationship of multiplying or dividing rational expressions?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
How to solve word problems in algebra 2
In this blog post, we will be discussing How to solve word problems in algebra 2. Our website can solving math problem.
How can we solve word problems in algebra 2
One of the most important skills that students need to learn is How to solve word problems in algebra 2. There are many ways to solve binomial equations, but one of the most straightforward methods is to factor the equation and set each factor equal to zero. This can be done by factoring out the greatest common factor, or by using a factorization method such as the quadratic formula. Once the equation is factored, it can be solved by setting each factor equal to zero and solving for the variable.
Algebra 2 is the second math course in high school that builds on the concepts learned in Algebra 1. The goal of Algebra 2 is to help students understand and master concepts that will enable them to succeed in higher level math courses. In order to succeed in Algebra 2, students need to be willing to work hard and ask for help when needed. There are many resources available to help students with Algebra 2, including online resources, tutors, and teachers. If you are struggling in
There are a few steps to solving linear inequalities. First, you need to determine the inequality sign. Second, you need to isolate the variable on one side of the inequality. Third, you need to apply the following operations to both sides of the inequality in order to solve for the variable: addition, subtraction, multiplication, or division. Lastly, you need to check your solution by substituting the value back into the original inequality.
An arithmetic sequence solver is a tool that can be used to find the next number in a sequence of numbers that follow a specific pattern. For example, if you know that the first two numbers in a sequence are 1 and 3, you can use an arithmetic sequence solver to find the next number in the sequence.
To solve linear functions, there are a few steps that need to be followed. First, identify the slope and y-intercept of the line. Next, use these values to plot the line on a graph. Finally, find the x-intercept of the line, which will give the solution to the linear function.
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# 13.2 Triangles and quadrilaterals part 1_1allr
How many angles does a
triangle have?
A
3
B
4
C
5
D
6
1 / 14
Slide 1: Quiz
WiskundeMiddelbare schoolvwoLeerjaar 1
This lesson contains 14 slides, with interactive quizzes and text slides.
Lesson duration is: 50 min
## Items in this lesson
How many angles does a
triangle have?
A
3
B
4
C
5
D
6
#### Slide 1 -Quiz
How many sides does a
A
3
B
4
C
5
D
6
#### Slide 2 -Quiz
A triangle has 3 angles and 3 sides.
A quadrilateral has 4 angles and 4 sides.
Some triangles and quadrilaterals have special properties and therefore have their own name.
#### Slide 3 -Slide
Right-angled triangle
A triangle that has one angle that is a right angle is called a right-angled triangle
#### Slide 4 -Slide
Isosceles triangle
A triangle that has 1 axis of symmetry is called an isosceles triangle.
Because it has reflection symmetry 2 sides are the same length and two angles are the same size.
The angles with the same size are called the
base angles
, the other angle is called the
apex angle.
The axis of symmetry divides the isosceles
triangle into two right-angled triangles.
#### Slide 5 -Slide
Equilateral triangle
A triangle that has 3 axes of symmetry is called an equilateral triangle.
All sides of an equilateral triangle are the same
length and all angles are the same size.
Each axis of symmetry divides the equilateral
triangle into two right-angled triangles.
#### Slide 6 -Slide
Squares and Rectangles
A rectangle is a quadrilateral where all angles are 90 degrees.
A square is a quadrilateral where all angles are 90 degrees AND all sides are the same length.
#### Slide 7 -Slide
Rhombus
A rhombus is a quadrilateral where all sides are the same length. Both diagonals are axes of symmetry.
#### Slide 8 -Slide
Kite
A kite is a quadrilateral
where one of the diagonals
is an axis of symmetry.
It has two pairs of sides
that are the same length.
#### Slide 9 -Slide
Parallelogram
A parallelogram is a quadrilateral where the opposite sides are parallel to each other and the the same length.
The opposite angles are also equal.
It has rotation symmetry
through 180 degrees.
#### Slide 10 -Slide
What kind of triangle is triangle PQR?
#### Slide 11 -Slide
What kind of triangle is triangle PQR?
A
Not a special triangle
B
Right-angled triangle
C
Isosceles triangle
D
Equilateral triangle
#### Slide 12 -Quiz
What kind of triangle is triangle PQR?
PQR has a right angle, therefore it is a
right-angled triangle.
PQR has one axis of symmetry, therefore
it is an isosceles triangle.
PQR is a right-angled, isosceles triangle. |
1990 AIME Problems/Problem 15
Problem
Find $a_{}^{}x^5 + b_{}y^5$ if the real numbers $a_{}^{}$, $b_{}^{}$, $x_{}^{}$, and $y_{}^{}$ satisfy the equations $$ax + by = 3^{}_{},$$ $$ax^2 + by^2 = 7^{}_{},$$ $$ax^3 + by^3 = 16^{}_{},$$ $$ax^4 + by^4 = 42^{}_{}.$$
Solution 1
Set $S = (x + y)$ and $P = xy$. Then the relationship
$$(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})$$
can be exploited:
$\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\ (ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}$
Therefore:
$\begin{eqnarray*}7S & = & 16 + 3P \\ 16S & = & 42 + 7P\end{eqnarray*}$
Consequently, $S = - 14$ and $P = - 38$. Finally:
$\begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \\ (42)(S) & = & (ax^5 + by^5) + (P)(16) \\ (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\ ax^5 + by^5 & = & \boxed{20}\end{eqnarray*}$
Solution 2
A recurrence of the form $T_n=AT_{n-1}+BT_{n-2}$ will have the closed form $T_n=ax^n+by^n$, where $x,y$ are the values of the starting term that make the sequence geometric, and $a,b$ are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.
Suppose we have such a recurrence with $T_1=3$ and $T_2=7$. Then $T_3=ax^3+by^3=16=7A+3B$, and $T_4=ax^4+by^4=42=16A+7B$.
Solving these simultaneous equations for $A$ and $B$, we see that $A=-14$ and $B=38$. So, $ax^5+by^5=T_5=-14(42)+38(16)= \boxed{20}$.
Solution 3
Using factoring formulas, the terms can be grouped. First take the first three terms and sum them, getting:
$a(x^3 + x^2 + x) + b(y^3 + y^2 + y) = 16$ $ax(\frac{x^3-1}{x-1}) + by(\frac{y^3-1}{y-1}) = 16$.
Similarly take the first two terms, yielding:
$ax(\frac{x^2-1}{x-1}) + by(\frac{y^2-1}{y-1}) = 10$.
Lastly take an alternating three-term sum,
$a(x^3 - x^2 + x) + b(y^3 - y^2 + y) = 12$ $ax(\frac{x^3+1}{x+1}) + by(\frac{y^3+1}{y+1}) = 12$.
Now to get the solution, let the answer be $k$, so
$ax(\frac{x^4-1}{x-1}) + by(\frac{y^4-1}{y-1}) = 68$.
Multiplying out this expression gets the answer term and then a lot of other expressions, which can be eliminated as done in the first solution. |
# Greater than or Equal to – Definition with Examples
Home » Math Vocabulary » Greater than or Equal to – Definition with Examples
An equation makes use of the “Equal to (=)” symbol to express the relationship of equality between two quantities. Inequalities make use of the “Greater than (>)” and “Less than (<)” symbols to compare quantities that are not equal in nature.
## Less than (<):
We use “less than” when one quantity is less than the other quantity. For example, ”8 carrots are less than 10 carrots” can be mathematically expressed as 8 < 10.
## Greater than (>):
We use “greater than” when one quantity is more than the other quantity. For example, ”7 mangoes are more than 3 mangoes” can be mathematically expressed as 7 > 3.
However, in certain cases, when we only have one quantity and we want to make estimates about another quantity comparable with our first one, we use an inequality operator referred to as “Greater Than or Equal To (≥)” or ”Less Than or Equal To(≤).”
Let us understand what is greater than or equal to.
## Greater than or Equal to (≥)
Let’s understand this with the help of a real-life situation. A person must be at least 18 years old to be eligible to vote in an election. It means that a person’s age should be more than or equal to 18 years old to qualify for voting in the USA. The voter’s age can be 22 years or 45 years or 70 years or any other age as long as the age is not less than 18 years.
Let’s say the age of the voter candidate is A.
Then, mathematically, we can say that either A is greater than 18 or A is equal to 18.
A > 18 or A = 18
These two mathematical statements can be combined into one single statement:
A is greater than or equal to 18.
Greater than or Equal to Symbol
When we combine the “>” and “=” symbols to form ≥, we can write the statement as A ≥ 18.
The greater than or equal to symbol is a combination of greater than (＞) and equal to (=) sign.
In the greater than or equal to symbol, a horizontal line is placed below the greater than symbol. The symbol is used in mathematical expression for the statements “at least, not less than, and minimum.”
## Examples of Greater than or Equal to
x ≥ 2
This example signifies that in the given relation, the value of x is more than 2 or equal to two. It cannot be less than two.
## Non Examples of Greater than or Equal to
For example, 4 ≥ 5. Here, the statement ”4 is greater than or equal to 5” is not true as 4 is not greater than 5 nor it is equal to 5.
## Greater than or Equal to on a Number line:
Let’s look at the steps on how to represent an inequality of the form x ≥ 3 on a number line.
Step 1: Locate “3” on the number line and mark it with a big circle and fill the circle.
Step 2: The inequality suggests that the variable x is allowed to have any values greater than or equal to 3, that means all the values on the right of this circle. From here, extend an arrow to the right end, signifying all the values the variable x can have.
## Conclusion
We can use greater than or equal to demonstrate that one variable is bigger than or equal to a specific amount when two variables are compared. For example, the company’s strategy is to introduce the product at the same price as before or at a higher price. As a result, we may claim that the new product price is higher than the previous product price—head over to SplashLearn for worksheets on ”greater than or equal to.”
## Solved Examples
Q1. Select the numbers greater than or equal to 42 from the given set of the numbers.
44, 23, 0, 7, 55, 33, 61, 42, 66, 12
Solution:
44, 55, 61, 42, 66
Q2. Raashi walks 5 km, Samuel walks 1.5 km, Leo walks 4 km and Alia walks 2 km. Who walked at least 3 km?
Solution:
Raashi and Leo walked at least 3 km.
5 > 3 and 4 > 3
Q3. A minimum of ten pupils per class are required to compete in an Olympiad test. How can you express this situation as a relation?
Solution:
Let x be the number of students required to compete in the test. Then, it is symbolized by the relation: x ≥ 10.
Q4. Graph y ≥ 2 on the number line.
Solution:
## Practice Problems
1
### Select all the possible values from the following that b can have if b ≥ 7
5
6
3
10
CorrectIncorrect
10 > 7. All other options are less than 7.
2
### Identify the number line that best describes greater than equal to −1.
A
B
C
D
CorrectIncorrect
Since all the number towards the right of -1 are greater than -1. The given number line represents all the numbers greater than equal to -1.
3
### To make a jug full of mango juice at least 15 mangoes are needed. Harrison has 10 mangoes, Gary has 20 mangoes, Abram has 17 mangoes and Ahmad has 14 mangoes. Who will be able to make a jug full of mango juice?
Harrison and Abram
Gary and Abram
CorrectIncorrect
Correct answer is: Gary and Abram
20 > 15 and 17 > 15
4
### Which of the following options cannot be the solution of k ≥ 10
10, 11, 12
1, 2, 3
11, 12, 13
51, 52, 53
CorrectIncorrect
Correct answer is: 1, 2, 3
1, 2 and 3 are less than 10. So, the given set of numbers does not satisfy the condition. |
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# What Are The Prime Factors Of 100
What are the prime factors of 100? Answer: 2, 5
The number 100 has 2 prime factors. Primes can only have two factors(1 and itself) and only be divisible by those two factors. Any figure where this rule applies can be called a prime factor. The biggest prime factor of 100 is 5. The smallest prime factor of 100 is 2.
## What Is The Factor Tree Of 100
How to use a factor tree to find the prime factors of 100? A factor tree is a diagram that organizes the factoring process.
• First step is to find two numbers that when multiplied together equal the number we start with.
100
↙ ↘
2 × 50
• Second step is to check the multiplication(of the first step) for numbers that are not primes.
50
↙ ↘
5 × 10
We found 2 prime factors(2, 5) using the factor tree of 100. Now let us explain the process to solving factor trees in more detail. Our goal is to find all prime factors of a given whole number. In each step of our factor tree diagram for 100 we always checked both multiplication numbers if they were primes or not. If one or both of the integers are not prime numbers then this means that we will have to make diagrams for them too. This process continues until only prime numbers are left.
Remember that often a factor tree for the same integer can be solved in more than one correct way! An example of this is the figure 12 where 2*6=12 and 4*3=12. The primes of a factor tree for 12 are the same regardles if we start the factor tree with 2*6 or 4*3.
## How To Verify If Prime Factors Of 100 Are Correct Answers
To know if we got the correct prime factors of 100 we have to get the prime factorization of 100 which is 2 * 2 * 5 * 5. Because when you multiply the primes of the prime factorization the answer has to be equal with 100.
After having checked the prime factorization we can now safely say that we got all prime factors.
## General Mathematical Properties Of Number 100
100 is a composite number. 100 is a composite number, because it has more divisors than 1 and itself. This is an even number. 100 is an even number, because it can be divided by 2 without leaving a comma spot. This also means that 100 is not an odd number. When we simplify Sin 100 degrees we get the value of sin(100)=-0.50636564110976. Simplify Cos 100 degrees. The value of cos(100)=0.86231887228768. Simplify Tan 100 degrees. Value of tan(100)=-0.58721391515693. When converting 100 in binary you get 1100100. Converting decimal 100 in hexadecimal is 64. The square root of 100=10. The cube root of 100=4.6415888336128. Square root of √100 simplified is 3√11. All radicals are now simplified and in their simplest form. Cube root of ∛100 simplified is 100. The simplified radicand no longer has any more cubed factors.
## Determine Prime Factors Of Numbers Smaller Than 100
Learn how to calculate primes of smaller numbers like:
## Determine Prime Factors Of Numbers Bigger Than 100
Learn how to calculate primes of bigger numbers such as:
## Single Digit Properties For Number 100 Explained
• Integer 1 properties: 1 is an odd figure. In set theory, the 1 is constructed starting from the empty set obtaining {∅}, whose cardinality is precisely 1. It is the neutral element of multiplication and division in the sets of natural, integer, rational and real numbers. The first and second digit of the Fibonacci sequence(before 2). Second to the succession of Lucas(after 2). First element of all the successions of figured numbers. One is a part of the Tetranacci Succession. 1 is a number of: Catalan, Dudeney, Kaprekar, Wedderburn-Etherington. It is strictly non-palindrome, integer-free, first suitable digit, first issue of Ulam and the first centered square. The first term of the succession of Mian-Chowla. Complete Harshad, which is a number of Harshad in any expressed base. 1 is the first highly totest integer and also the only odd number that is not non-tottering.
• Integer 0 properties: 0 is the only real figure that is neither positive nor negative. Sometimes it is included in natural numbers where it can be considered the only natural in addition to the one to be neither first nor composed, as well as the minimum of natural numbers(that is, no natural digit precedes the 0). In an oriented line (which makes a point on the straight line correspond to each real number, preserving also the relation of order), the 0 coincides conventionally with the origin. Since it can be written in the form 2k, with con k integer, 0 is called even. It is both a figure and a numeral. In set theory, the zero is the cardinality of the empty set. In fact, in certain axiomatic mathematical developments derived from set theories, zero is defined as the empty set. In geometry, the size of a point is 0. Zero is the identity element of an additive group or additive identity in a ring.
## Finding All Prime Factors Of A Number
We found that 100 has 2 primes. The prime factors of 100 are 2, 5. We arrived to this answer by using the factor tree. However we could have also used upside down division to get the factorization primes. There are more that one method to factorize a integer.
## List of divisibility rules for finding prime factors faster
Rule 1: If the last digit of a figure is 0, 2, 4, 6 or 8 then it is an even number. All even numbers are divisible by 2.
Rule 2: If the sum of digits of a integer is divisible by 3 then the figure is also divisible by 3 and 3 is a prime factor(example: the digits of 12 are 1 and 2 so 1+2=3 and 3 is divisible by 3, meaning that 12 is divisible by 3). The same logic also works for 9.
Rule 3: If the last two digits of a number are 00 then this integer is divisible by 4(example: we know that 124=100+24 and 100 has two zeros in the end making it divisible with 4. We also know that 4 is divisible with 24). In order to use this rule to it's fullest it is best to know multiples of 4.
Rule 4: If the last digit of a number is 0 or 5 then 5 it is divisible by 5.
Rule 5: All integers that are divisible by both 2 and 3 are also divisible by 6. This is logical because 2*3=6.
## What Are Prime Factors Of A Number?
All numbers that are only divisible by one and itself are called prime factors in mathematics. A prime factor is a figure that has only two factors(one and itself). |
Fractions are frequently used in square root calculations. In the calculation of fractions, the denominator may contain the root symbol. In that case, if the root sign is still in the denominator, the answer is wrong.
When calculating the square root, if the denominator contains the radical symbol, it should be removed. This process is called rationalizing the denominator. In square root fractions, there is no problem if the numerator contains a square root, but it is not suitable if the denominator contains a radical sign.
So how do we rationalize the denominator? And why do we need to rationalize the denominator?
By learning these, we will be able to understand the procedure for correctly rationalizing the denominator in mathematics. We will explain how to rationalize the denominator.
If There Are Radical Symbols in the Denominator, Make Rationalizing
In square root fractions, the denominator may contain a radical sign. For example, the following is the case.
• $\displaystyle\frac{1}{\sqrt{2}}$
In this fraction, the denominator contains $\sqrt{2}$. So let’s make it a fraction without a root sign in the denominator.
Square roots are irrational numbers. In mathematics, it is not good to have irrational numbers in the denominator. So, we change the number in the denominator to a rational number instead of an irrational number. Move the square root that exists in the denominator to the numerator. This is rationalizing the denominator.
In mathematics, if the root symbol remains in the denominator, the answer is incorrect. So be sure to rationalize the denominator.
Procedure for Converting the Square Root in a Denominator to an Integer
How do we rationalize the denominator? The way to do it is to multiply the denominator and the numerator by the same number. In fractions, it does not matter if we multiply or divide the denominator and the numerator by the same number. In fact, the following fractions are the same number.
• $\displaystyle\frac{1}{2}=\displaystyle\frac{2}{4}=\displaystyle\frac{3}{6}$
We can also consider $\displaystyle\frac{1}{2}=\displaystyle\frac{1×2}{2×2}$. Anyway, in fractions, the same number can be multiplied to the numerator and denominator.
Let’s take advantage of this property to rationalize the denominator. When squaring the square root, the root symbol is removed. So, we multiply the numerator and denominator by the same number as the square root in the denominator. The result is as follows.
$(\sqrt{2})^2=2$. We can convert the number in the denominator to an integer by multiplying the same number. The way to rationalize the denominator is not difficult. All we have to do is multiply the square root in the denominator.
Smaller Numbers in the Radical Symbol Is Less Likely to Make Miscalculation
When rationalizing the denominator, the smaller the number in the root symbol, the fewer the miscalculations. We can reduce the number by doing prime factorization first.
For example, how can we rationalize the following fractions?
• $\displaystyle\frac{\sqrt{63}}{\sqrt{72}}$
When rationalizing the denominator, there are two ways to do it. One is to first reduce the number in the radical sign. The other way is to rationalize the denominator without first doing prime factorization. If we compare the two methods, we can see the following.
In this comparison, it is easier to calculate by first doing a prime factorization and then reducing the number in the root sign. In other words, there will be fewer calculation mistakes.
Any person can make mistakes in math calculations. So we have to choose ways to reduce the mistakes as much as possible. When we rationalize the denominator, if we first do prime factorization, we will make fewer calculation mistakes.
Rationalizing of Addition and Subtraction with Two Terms in the Denominator
An advanced problem when doing rationalizing the denominator is the case where there are two terms in the denominator. There is not only one radical sign, but it contains addition or subtraction.
For example, the following fraction has two terms in the denominator.
• $\displaystyle\frac{1}{3-\sqrt{2}}$
In this case, how should we calculate it? If there are two terms, use the factoring formula. To be more specific, use the following formula.
• $(x+a)(x-a)=x^2-a^2$
By using this formula, we can rationalize the denominator. For example, the previous equation can be calculated as follows.
$\displaystyle\frac{1}{3-\sqrt{2}}$
$=\displaystyle\frac{1\textcolor{red}{×(3+\sqrt{2})}}{(3-\sqrt{2})\textcolor{red}{×(3+\sqrt{2})}}$
$=\displaystyle\frac{3+\sqrt{2}}{3^2-(\sqrt{2})^2}$
$=\displaystyle\frac{3+\sqrt{2}}{9-2}$
$=\displaystyle\frac{3+\sqrt{2}}{7}$
Thus, we can rationalize the denominator by using the factoring formula. Even if there are two terms, we can rationalize the denominator in the following way.
In some cases, the denominator contains square root addition or subtraction. In that case, use the factoring formula to rationalize the denominator.
Why Do We Need to Rationalize the Denominator in Fractions?
Why do we need to rationalize the denominator in square root fractions? What is the problem if the denominator has a radical symbol?
One reason why we need to rationalize the denominator is to make the numbers easier to understand. For example, it’s hard to understand what a number is in the following fraction.
• $\displaystyle\frac{1}{\sqrt{2}}$
Irrational numbers are numbers that go on forever. It is impossible to divide the numerator by such a number, and it is hard to imagine what kind of number it would be.
On the other hand, what if we do rationalizing the denominator and change the fraction to the following number.
• $\displaystyle\frac{\sqrt{2}}{2}$
Since it is an irrational number, the number in the numerator goes on forever. However, we can easily imagine what kind of number it would be. $\sqrt{2}≈1.41$. So we can see that $\displaystyle\frac{\sqrt{2}}{2}$ is about 0.7.
If there are irrational numbers in the denominator, it is difficult to imagine the numbers. On the other hand, even if there is a square root, it is easier to understand what kind of number it is if it is in the numerator.
When Making the Common Denominator, the Denominator needs to be Rationalized
Also, when making a common multiple of the denominators, it is preferable to rationalize the denominators. When adding and subtracting fractions, the denominators must be equal. If the denominators are the same, the properties of the fractions will also be the same.
However, when the denominator contains an irrational number, it is difficult to make the denominators match. For example, how can we do the following calculation?
• $\displaystyle\frac{1}{3}+\displaystyle\frac{1}{2+\sqrt{2}}$
In this situation, the calculation is difficult. The numbers cannot be matched because there is an irrational number in the denominator. So, let’s rationalize the denominator. Then we will be able to make the common denominator.
$\displaystyle\frac{1}{3}+\displaystyle\frac{1}{2+\sqrt{2}}$
$=\displaystyle\frac{1}{3}+\displaystyle\frac{1\textcolor{red}{×(2-\sqrt{2})}}{(2+\sqrt{2})\textcolor{red}{×(2-\sqrt{2})}}$
$=\displaystyle\frac{1}{3}+\displaystyle\frac{2-\sqrt{2}}{4-2}$
$=\displaystyle\frac{1}{3}+\displaystyle\frac{2-\sqrt{2}}{2}$
$=\displaystyle\frac{2}{6}+\displaystyle\frac{(2-\sqrt{2})×3}{6}$
$=\displaystyle\frac{2}{6}+\displaystyle\frac{6-3\sqrt{2}}{6}$
$=\displaystyle\frac{8-3\sqrt{2}}{6}$
If we make the denominator an integer, we can make common denominators for fractions. If an irrational number is in the denominator, we cannot make common denominators. Since we need to create common denominators to calculate fractions, we need to rationalize the denominators. If we don’t rationalize the denominator, we can’t calculate.
Exercises: Calculation of Rationalizing the Denominator
Q1: Rationalize the denominator.
1. $\displaystyle\frac{4}{\sqrt{8}}$
2. $\displaystyle\frac{\sqrt{3}}{\sqrt{24}}$
3. $\displaystyle\frac{1}{\sqrt{6}+\sqrt{3}}$
If you understand what we have described so far, you can solve any problem easily.
(a)
$\displaystyle\frac{4}{\sqrt{8}}$
$=\displaystyle\frac{4}{2\sqrt{2}}$
$=\displaystyle\frac{2}{\sqrt{2}}$
$=\displaystyle\frac{2\textcolor{red}{×\sqrt{2}}}{\sqrt{2}\textcolor{red}{×\sqrt{2}}}$
$=\displaystyle\frac{2\sqrt{2}}{2}$
$=\sqrt{2}$
(b)
$\displaystyle\frac{\sqrt{3}}{\sqrt{24}}$
$=\displaystyle\frac{\sqrt{3}}{2\sqrt{6}}$
$=\displaystyle\frac{\sqrt{3}\textcolor{red}{×\sqrt{6}}}{2\sqrt{6}\textcolor{red}{×\sqrt{6}}}$
$=\displaystyle\frac{\sqrt{18}}{2×6}$
$=\displaystyle\frac{3\sqrt{2}}{12}$
$=\displaystyle\frac{\sqrt{2}}{4}$
(c)
$\displaystyle\frac{1}{\sqrt{6}+\sqrt{3}}$
$=\displaystyle\frac{1\textcolor{red}{×(\sqrt{6}-\sqrt{3})}}{(\sqrt{6}+\sqrt{3})\textcolor{red}{×(\sqrt{6}-\sqrt{3})}}$
$=\displaystyle\frac{\sqrt{6}-\sqrt{3}}{(\sqrt{6})^2-(\sqrt{3})^2}$
$=\displaystyle\frac{\sqrt{6}-\sqrt{3}}{3}$ |
# What is 134/210 as a decimal?
## Solution and how to convert 134 / 210 into a decimal
134 / 210 = 0.638
Converting 134/210 to 0.638 starts with defining whether or not the number should be represented by a fraction, decimal, or even a percentage. Both are used to handle numbers less than one or between whole numbers, known as integers. Depending on the situation, decimals can be more clear. We don't say 1 and 1/2 dollar. We use the decimal version of \$1.50. Same goes for fractions. We will say 'the student got 2 of 3 questions correct'. After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 134/210 is 134 divided by 210
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. Fractions have two parts: Numerators and Denominators. This creates an equation. To solve the equation, we must divide the numerator (134) by the denominator (210). Here's 134/210 as our equation:
### Numerator: 134
• Numerators are the top number of the fraction which represent the parts of the equation. 134 is one of the largest two-digit numbers you'll have to convert. But having an even numerator makes your mental math a bit easier. Large numerators make converting fractions more complex. Let's look at the fraction's denominator 210.
### Denominator: 210
• Denominators are located at the bottom of the fraction, representing the total number of parts. 210 is one of the largest two-digit numbers to deal with. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Have no fear, large two-digit denominators are all bark no bite. So without a calculator, let's convert 134/210 from a fraction to a decimal.
## Converting 134/210 to 0.638
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 210 \enclose{longdiv}{ 134 }$$
To solve, we will use left-to-right long division. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 210 \enclose{longdiv}{ 134.0 }$$
We've hit our first challenge. 134 cannot be divided into 210! So that means we must add a decimal point and extend our equation with a zero. Now 210 will be able to divide into 1340.
### Step 3: Solve for how many whole groups you can divide 210 into 1340
$$\require{enclose} 00.6 \\ 210 \enclose{longdiv}{ 134.0 }$$
Now that we've extended the equation, we can divide 210 into 1340 and return our first potential solution! Multiply by the left of our equation (210) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.6 \\ 210 \enclose{longdiv}{ 134.0 } \\ \underline{ 1260 \phantom{00} } \\ 80 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If there is a remainder, extend 210 again and pull down the zero
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. This is also true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But each represent values in everyday life! Here are just a few ways we use 134/210, 0.638 or 63% in our daily world:
### When you should convert 134/210 into a decimal
Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 134/210 MPH. The radar will read: 90.63 MPH. This simplifies the value.
### When to convert 0.638 to 134/210 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 134/210 = 0.638 what would it be as a percentage?
• What is 1 + 134/210 in decimal form?
• What is 1 - 134/210 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.638 + 1/2? |
# Lesson Video: Subtracting Ones from Teen Numbers: Making Ten Mathematics • 1st Grade
In this video, we will learn how to subtract a one-digit number from a teen number when we cross a ten by subtracting the ones in two steps.
10:07
### Video Transcript
Subtracting Ones from Teen Numbers: Making 10
In this video, we’re going to learn how to subtract a one-digit number from a teen number when we cross a 10. And we’re going to do this by subtracting the ones in two steps.
Now, before we start, there are two things that are gonna help us here. The first thing is that we can subtract from a teen number to 10 really quickly. Let’s imagine that we have a teen number like 19. And we want to take away something so that we’re left with 10. Well, we just take away all the counters in this ten frame, don’t we? 19 take away nine is 10. 15 take away five equals 10. And what do we take away from 12 to take us to 10? Well, it’s the number two, isn’t it? So, the first skill that’s going to help us here is that we can subtract up to the number 10 really quickly.
And the second skill that’s gonna help us here is that we can subtract from the number 10 really quickly, too. Let’s imagine we need to take away two from 10. Well, to do this, we need to think about which number pairs with the number two to make 10. We know that two and eight go together to make 10. So, we can say really quickly 10 take away two leaves us with eight. To subtract from 10 really quickly, we just need to think of pairs that go together to make 10. Six and four make 10. So, if we take away six from 10, we’re left with four.
Now, we’re going to try subtracting some ones from some teen numbers. And we’re going to find the answer to questions like 17 take away nine by subtracting in two steps. In the first step, we’re going to subtract up to the number 10, which we know we can do really quickly. And you guessed it! In the second step, we’re going to subtract from the number 10. As we’ve seen, we can do this really quickly, too.
Let’s have a go at 17 take away nine.
Let’s imagine our friend the panda here has 17 bamboo shoots for his tea. Let’s put them in ten frames so that we can see what we’re doing. Now, let’s imagine another panda comes along and eats nine of his bamboo shoots. 17 take away nine leaves us with what? And one way we could find the answer is to start with the number 17 and to count back nine. But we know some quicker ways of subtracting.
We know that we can subtract to the number 10 very quickly. And we know that we can subtract from the number 10 very quickly. Instead of subtracting nine all in one go, what if we complete the subtraction in two parts? First, let’s subtract up to the number 10. What do we take away from 17 to get to 10? We know we can do this really quickly. We just need to subtract all the bamboo shoots in this ten frame.
We need to start by taking away seven because we know 17 take away seven leaves us with 10. So far, we’ve only subtracted seven out of nine. We need to subtract the rest now. Seven and another two make nine. So, now, we need to take away the remaining two. And again, we can do this really quickly because this time we’re subtracting from 10.
Remember how we subtract from 10? We need to think about which number pairs with the number two to make 10. We know that eight and two make 10. And so, in our second step, we can see that 10 subtract two leaves us with eight. And this is the number of bamboo shoots our panda has got left. 17 subtract nine equals eight.
Now, let’s have a go at answering some questions where we have to subtract ones from teen numbers. And each time we’re gonna find the answer by taking away to make 10 and then from 10.
Daniel started with 16. He took away six to make 10, then took away two more. Pick a calculation that is the same as what Daniel did. 16 take away six, 16 take away eight, or 16 take away four.
In this problem, we can see that Daniel is subtracting from a teen number. We know this for a number of reasons. We’re told that he starts with the number 16. We can see in the picture that there are 16 counters before some get crossed off. And all of our possible answers start with the number 16 too. So, we know that Daniel is trying to find the answer to 16 take away something. But what number is he trying to take away?
Well, if we read the question carefully and look at the pictures, we can see that Daniel finds the answer in two steps. First, we’re told he took away six. Now, why might Daniel split up his subtraction into two steps and only take away six to start with? Well, we’re told this again in the question. He took away six to make 10. We know this is a quick calculation to work out. 16 take away six leaves us with 10.
And you know by making 10 like this, what Daniel’s done is he’s made the calculation a lot easier cause now all he has to do is take away whatever’s left from 10. And we can subtract from 10 really quickly. We’re told that Daniel took away two more. And if we look in the bottom picture, we can see the two extra counters that Daniel’s taken away. It’s a quick way of finding the answer because Daniel probably knows that two and eight go together to make 10. So, if he subtracts two from 10, he’s going to be left with eight.
By reading the problem carefully and looking at the pictures, we can see exactly what Daniel is doing. First, he took away six, and then he took away two more. Daniel found the answer to 16 take away six take away two. And the calculation that’s the same as this is 16 take away eight.
Sophia is subtracting by making 10. 13 subtract five equals what? Which calculation should she do first? 13 take away two equals 11 or 13 take away three equals 10. And then, which calculation should she do next? 10 take away two equals eight or 10 plus two equals eight.
And maybe you’ve noticed one of these calculations doesn’t make sense, but let’s go through the question from the start and find out what Sophia is doing here. We’re told that she’s subtracting by making 10. And we can see from the picture the two numbers that Sophia’s subtracting. She’s trying to find the answer to 13 take away five.
Now, she could find the answer just by starting with the number 13 and counting back five. But we’re told that she wants to find the answer by making 10. First, she can take away part of the number five to get to 10. And then, she can take away the rest of the number five from 10 because subtracting to and from 10 is quite quick to do.
Here are 13 counters. And we can see from the diagram that Sophia has split up the five that she wants to take away into two jumps. First, she takes away three. As we’ve said already, we know why she does this; she wants to make 10. She knows that 13 take away three equals 10. So, we know the calculation that she needs to do first is 13 subtract three equals 10.
Now, Sofia knows that three and two go together to make five. She’s already taken away three out of five. Now, she needs to just subtract the remaining two. And because she’s made 10, she takes away two from 10, which is very quick to do. Now, if we look carefully to our possible answers, only one of them makes sense. 10 plus two doesn’t make eight.
But let’s think about how Sophia knows that 10 take away two equals eight. Well, she knows that two and eight go together to make 10. She can subtract from 10 really quickly. And so, the second calculation she needs to do is 10 subtract two equals eight.
Sophia found the answer to 13 subtract five by splitting up five into three and then two. The calculation she should do first is 13 take away three equals 10. And her next calculation should be 10 take away two equals eight.
Now, what’ve we learned in this video? We know that subtracting a one-digit number from a teen number can cross over the number 10. And we’ve learnt a way to find the answer in two steps. We’ve learnt how to subtract a one-digit number from a teen number by first taking away to make 10 and then taking away whatever is left from 10. |
# One step algebra equations
Keep reading to learn more about One step algebra equations and how to use it. Math can be difficult for some students, but with the right tools, it can be conquered.
## The Best One step algebra equations
Here, we debate how One step algebra equations can help students learn Algebra. If you're solving for x with logs, then you're likely only interested in how things are changing over time. This is why we can use logs to calculate percent change. To do this, we first need to transform the data into a proportional format. For example, if we have data in the form of \$x = y and want to know the change in \$x over time, we would take the log of both sides: log(x) = log(y) + log(1/y). Then, we can just plot all of these points on a graph and look for trends. Next, let's say that we have data in the form of \$x = y and want to know the percent change in \$x over time. In this case, instead of taking the log of both sides, we would simply divide by 1: frac{log{\$x} - log{\$y}}{ ext{log}}. Then, we can again plot all of these points on a graph and look for trends.
To solve this equation, we start by first converting the left-hand side to a ratio: Similarly, since the right-hand side is a fraction, we can convert this to a decimal: We then multiply both sides of the equation by 1/10 , and then divide by 10 : Finally, we convert back to the original form of the equation, and solve for x . There are no exact formulas for how to solve logarithmic equations. However, there are some useful tricks and techniques that can be used to help you solve these types of equations. One good way to solve logarithmic equations is to use a table. One easy way to do this is to look at what other logarithmic equations look like. Since logarithms follow an exponential pattern, it is usually possible to find a similar equation on which the base can be found. Another trick is to try doing all comparisons in your head before you write them down. If you have trouble coming up with a number that works for both sides of the equation then try using numbers from previous
Solving a system of linear equations can be challenging because it requires knowledge of both math and the context of the problem. When solving a system, remember to first identify any variables that must be manipulated mathematically. Once this step has been completed, focus on understanding the context of the problem and using data to determine which variables should be considered in each step. Answering questions that vary based on context is critical when solving systems of linear equations because these questions will test your ability to interpret data within different environments. In order to solve a system of linear equations, start by breaking down the problem into its simplest parts. Once you have identified all of your variables, it will be easier to determine which operations need to be performed in order to solve the equation. Finally, remember to check your work carefully before moving on to the next step. |
1. ## easy equation problem
The sides of a triangle have these equations:
y= -0.5x+1
y= 2x-4
y=-3x-9
Verify that the triangle is an isosceles right triangle (without graphing)
I've noticed that the second equation is -4 times the amount of the first if that helps
2. Originally Posted by Fox25
The sides of a triangle have these equations:
y= -0.5x+1
y= 2x-4
y=-3x-9
Verify that the triangle is an isosceles right triangle (without graphing)
I've noticed that the second equation is -4 times the amount of the first if that helps
find the points of intersection (x,y) to locate the vertices of the triangle, then find the distance between them using the distance formula.
two distances should be equal, and all three distances should work in the Pythagorean Thm.
3. thank you skeeter
4. Greetings
The sides of a triangle have these equations:
$\displaystyle y_1= -0.5x+1$
$\displaystyle y_2= 2x-4$
$\displaystyle y_3=-3x-9$
First things first, read over your notes, you wouldn't believe how much that helps.
Now that you've done this (theoretically),let's tackle the question.
To be an isosceles triangle, you need one line perpendicular to another and two lines to be equal to each other in length.
In coordinate geometry, when one line is perpendicular to another, the gradient of the line is the negative reciprocal of the other.
I.E. When the 2 gradients are multiplied, the product is negative 1.
You may see now that $\displaystyle y_1$ gradient is the negative reciprocal of $\displaystyle y_2$, hence $\displaystyle y_1$ is perpendicular to $\displaystyle y_2$
Hence the triangle formed by the intersection of the 3 lines is right-angled.
Now, all we have to do is determine whether $\displaystyle y_1$ and $\displaystyle y_2$ are equal to each other. $\displaystyle y_3$ is opposite to the right angle, so it's the hypotenuse.
Find where $\displaystyle y_1$ and $\displaystyle y_2$ intersect each other and $\displaystyle y_3$. Now with these coordinates the length of $\displaystyle y_1$ and $\displaystyle y_2$.
$\displaystyle y_1= -0.5x+1$
$\displaystyle y_2= 2x-4$
$\displaystyle y_3=-3x-9$
$\displaystyle -0.5x+1=3x-9$
$\displaystyle x=\frac{20}{7}$
Then $\displaystyle y=\frac{-3}{7}$
$\displaystyle A(\frac{20}{7},\frac{-3}{7})$
$\displaystyle -0.5x+1=2x-4$
$\displaystyle x=2$
Then $\displaystyle y=0$
$\displaystyle B(2,0)$
$\displaystyle 2x-4=3x-9$
$\displaystyle x=5$
Then $\displaystyle y=6$
$\displaystyle C(5,6)$
You may graph at this point just to get your visual bearings.
Now just apply the length of a line formula to determine that two of the lines are equal to each other.
Edit: Be more like Skeeter and make it short |
### By: Bryan Boodhoo
1. Explaining the basics
2. First and Second Differences
3. Parts of Vertex Form
4. Finding Equations from a graph
5. Graphing
6. Graphing in Factored Form
1. Common Factoring
2. Simple Trinomial Factoring
3. Complex Trinomial Factoring
4. Perfect Square Factoring
5. Difference of Squares
1. Completing the Square
3. The Discriminant
WORD PROBLEMS:
OTHER STUFF (MY VIDEO, REFLECTIONS, ASSESSMENTS, ETC.)
1. My Video
2. Assessment
3. Connecting Topics Together
4. Reflection
5. Other Videos to help (Not mine)
## 1. Quadratics 1: Explaining the basics
In the Quadratics 1 unit, it is mainly focused on understanding how parabolas are graphed and how to graph these parabolas when given the "Vertex" and "Factored" forms.
The parts of a parabola are clearly shown in the picture to the right. As you can see, a parabola may seem like a curved slope with a dip in the middle. Not to confuse you, there are similar principles between a slope and a parabola such as the x and y intercepts.
## 2. First and Second Differences
If given a table with x and y values, there is a way to tell if there is a Linear or Quadratic relationship. In order to find this out, we use First and Second Differences. If the first differences are all not the same, but the second differences are, then there is a Quadratic relation. If the first differences are all the same, then there is a Linear Relation
## 3. Parts of a Simple Vertex Form Equation: y=a(x-h)^2+k
The equation of y=a(x-h)^2+k is the most simple equation to explain how a quadratic equation in vertex form works.
a= The vertical stretch (how tight or loose the arms of the parabola will be)(If negative, parabola will open downwards)(You will use this number to find the step pattern when graphing a an equation)
h= The horizontal shift (how far the parabola moves from side to side)(If negative, vertex and axis of symmetry will be on right side of y-axis; if positive, vertex and axis of symmetry will be on left side of y-axis)
k= The vertical shift (how high the parabola moves up and down)
The vertex will be the coordinates that h and k will take place of and will be in the (x,y) form.
## 4. How to Determine the Equation from a Graph
Recall the first diagram of a parabola that you looked at and see if you can find any of those parts on this graph, then write down those coordinates for those parts.
Using this knowledge, we can see that the vertex of this parabola is (1,5)
Since the parabola opens downward, we now know that "a" will be negative.
y=a(x-h)^2+k
y=a(x-1)^2+5 (Subbed in vertex)
The last step now, is substituting a final point that the parabola is passing through. To do this, we simply find 2 coordinates and substitute them in place of x and y.
The point that we will be using is (0,2)
2=a(0-1)^2+5
2=a(-1)^2+5
2-5=a
-3=a
After finding the value of "a" the final equation is y=-3(x-1)^2+5
## 5. Graphing a Quadratic Equation
In order to do this, you will need to learn how to find the step pattern as well as the vertex.
Graphing y=(x+4)^2+3 will be fairly simple due to the fact that there is no altering of the step pattern involved. (Will be shown after this) One thing that we have to remember is that the basic step pattern when there is no "a" value will always be 1 to the side, 1 up, then, 2 to the side, 4 up. We must also remember that the step pattern will be applied to both sides of the vertex.
From this equation, we can already tell that the vertex is shown as (-4,3) but will be drawn as (4,3). Now that we have this point all we have to do now is use the regular step pattern that we just learned and apply it to this vertex. The Final graphed parabola is shown above with the correct vertex and the correct step pattern.
## 6. Graphing in Factored Form
Lets start this off with a simple example.
Graph: y=(x+5)(x+1)
The equation we have here is written in a form called "Factored Form." It may seem different than the usual vertex form but it is not that hard to graph.
To start this off we must find the x-intercepts which are 5 and 1.
After this, we have to find the axis of symmetry. To find this we must get the 2 x-intercepts, add them together, then divide it by 2. From this we will get -3
Since we have found the AOS, this will also be our x value, now all we have to do is find the y value.
The final thing we have to do is find the vertex. To do this we must substitute the axis of symmetry for the x values
It will look like this when the AOS is subbed in: (Now we solve for y)
y=[(-3)+5][(-3)+1]
y=(2)(-2)
y=-4
There we have it:
x- intercepts are 5 and 1
Vertex is (-3,-4)
What is factoring? Well, you may recall that you have previously learned to add, subtract, multiply, or divide a set of numbers in order to get one final number.
For example, 3x2=6
But, factoring is the exact opposite.
We need to start from the number 6, and find what numbers can be multiplied to get this.
The Quadratics 2 unit will be focused on all the different types of factoring.
## 1. Common Factoring
To common factor, you must find the GCF between all of the terms.
Factor 3x-12
To factor this, we must first find the GCF, which is 3. We know this is 3 because it is the lowest number that can be divided by all of the terms here.
Next we have to pull the GCF out and put the other terms in brackets
It will looks like this:
3(3x-12)
Then, we have to divide everything inside of the brackets by the GCF
3(3x-12)
3
After that, we are left with the final expression, which is:
3(x-4)
## 2. Simple Trinomial Factoring
The next type of factoring that we will learn is "Simple Trinomial Factoring."
The reason why it is given this name is because there are 3 terms and there is no coefficient in front of the x^2
Factor: x^2+5x+6
First we must look to see if there is a common factor in the expression, in this case there is none so we can skip that step.
Next, we must separate the x^2 into 2 brackets.
(x+ )(x+ )
Then, we must find two numbers that when multiplied, equal to 6 and also when added, equal to 5.
In this case the two numbers are 3 and 2 so we can input them into the brackets.
(x+2)(x+3)
There it is! We have our final answer.
Check (Fast way): In order to check that we have the right answer we must go from factored to standard form.
To do this we multiply the x terms (x^2)
Next we add the two terms 2 and 3 and add an x after it (5x)
Finally we multiply the 2 same terms to get the final number
This is how we end up where we started at x^2+5x+6
## 3. Complex Trinomial Factoring
Complex trinomials are similar to simple trinomials, its just that complex trinomials have a coefficient greater than 1.
Lets factor the expression: 6x^2+29x+35
Since there is a coefficient greater than 1 infront of the first term, we must find out 2 numbers that can be multiplied to get 6 and then separate them in the brackets.
(3x+ )(2x+ )
Next, we have to find 2 numbers that when multiplied, equal to 35.
This is the most important step and it will include trial and error due to the guess and check method.
The numbers that we will list are: 7 & 5 and 1 & 35
Then we input these numbers
(2x+5)(3x+7)
After this, we have to use the outside inside method to get the number "29"
In the picture below we will have to multiply the numbers 2 & 7 and the numbers 5 & 3, then add them too see if we get 29
And we do! So the final expression will be (2x+5)(3x+7)
## 4. Perfect Square Factoring
For this type of factoring, we are looking at how the expression has become a perfect square. A good example that we are going to solve is the expression a^2+2ab+b^2
The whole idea of perfect squaring is that every part of the expression has a perfect square root.
a=square root of a^2
ab=square root of 2ab
b=square root of b^2
Just like simple trinomial factoring, we must separate the first term in 2 brackets.
(a+ )(a+ )
Then we must find 2 terms that equal to b^2
In this case they are b and b
(a+b)(a+b)
We get the the second term "2ab" from multiplying the two terms by 2
Our final expression will be (a+b)^2 because it is in fact, a perfect square since (a+b) can be multiplied by itself in order to get a^2+2ab+b^2
## 5. Difference of Squares
To factor Difference of Squares equation, we must first know that there has two be a positive and negative term which makes the name "Difference."
Lets factor the expression 16x^2 - 25y^2
As we can see, the two squares for the two terms are (4x)^2 - (5y)^2
Just like complex trinomial factoring, we have to separate the first term by finding 2 numbers that equal to 16.
(4x+ )(4x- )
One of the brackets have to have a positive and one has to have a negative sign.
We then take the square of 25 and input it into each bracket.
(4x+5y)(4x-5y)
In the picture below it shows how to check if you got the right answer
In the Quadratics 3 unit, the first topic we will focus on is completing the Square. You may recall the different forms of expressions and equations which include Vertex form, Factored form, and Standard form. Well, for completing the square we will be working on converting the Standard form to Vertex form. For the second and third topics, we will focus on using the quadratic formula and the discriminant to find and analyze the x-intercepts.
## 1. Completing the Square
For completing the Square, you will be taught how to convert an equation from standard form to vertex form. This technique will come in handy when you cannot graph a standard form equation.
Lets complete the square for the equation: y=-2x^2+28x-15
y=-2x^2+28x-15
y=(-2x^2+28x)-15 (Separate the x terms)
y=-2(x^2-14x)-15 (Common factor the -2 as the "a" value)
y=-2(x^2-14x+49-49)-15 (Divide 14 by 2 then Square it, also subtract the same number)
y=-2((x-7)^2-49)-15 (Equation becomes a perfect square)
y=-2(x-7)^2+98-15 (Then expand using the -2)
y=-2(x-7)^2+83 (Add/Subtract the final values outside the brackets)
Final equation: y=-2(x-7)^2+83
We have looked at solving quadratic equations by factoring, but this method may not always work for every single type of equation. The quadratic formula can be used in cases for equations that cannot be factored. The reason as to why there is a plus/minus symbol is because there are two possible zeroes that can be found when you choose the adding or subtracting option.
## Example
Lets try and solve to find the zeroes for the equation, 2x^+9x+6=0
For the quadratic formula, the letters a,b, and c are represented in the equation you are going to solve by the first, second, and third terms. The first term will represent a, the second will represent b, and the third will of course represent c.
a=2
b=9
c=6
Then, we must next input the values of a,b, and c.
We then continue to find the zeroes by solving this equation with the values that were just inserted
As said before, the plus/minus sign means that it can be subtracted or added. So we will square 33, then add/subtract it from -9, and finally divide it by 4.
Finally we have the values for the two zeroes which will be 0.81 and 3.68.
## 3. The Discriminant
The number inside the square root of the quadratic formula is called the Discriminant.
The Discriminant of the quadratic formula tells us how many solutions the formula has.
To simplify this, if the discriminant is:
Greater than 0, there are two solutions
Exactly 0, then there is only one solution
Less than 0, there is no real solution
Examples:
Find out how many solutions there are to the discriminant, (-4)^2-4(1)(7)
(-4)^2-4(1)(7)
-> 16-28
-> -12
No real solutions since the discriminant is less than 0
Find out how many solutions there are to the discriminant, (3)^2+4(5)(2)
(3)^2+4(5)(2)
-> 9+40
-> 49
2 solutions since the discriminant is more than 0
Find out how many solutions there are to the discriminant, (4)^2-4(2)(2)
(4)^2-4(2)(2)
-> 16-16
->0
No real solutions since the discriminant is 0
## WORD PROBLEMS
A couple of word problems with solutions showing you how to complete them from Quadratics 1, 2, and, 3.
The path of a soccer ball is represented by the equation h=-0.06d(d-50), where h represents height, in metres, of the soccer ball above the ground and d is the horizontal distance, in metres, of the soccer ball from the player.
At what horizontal distance does the ball land?
The ball lands at 50 metres.
Determine the length and width of a rectangle with an area of x^2+17x+60
For this problem we will have to do simple trinomial factoring.
So,
(x+12)(x+5)
The reason why this is the answer is because in simple trinomial factoring we have to split to first term. (x and x)
Then we have to find two numbers which can be added to get the second term. (12 and 17)
But, at the same time, these 2 numbers also have to multiply to get the final term. (60)
A ball is thrown into the air. Its height in metres after t seconds is given by the equation, h=-4.9(t-2.4)^2+29
What was the height of the ball when it was thrown?
sub t=0
h=-4.9(0-2.4)^2+29
=0.776
The ball was 0.776 metres high when it was thrown.
What was the maximum height of the ball, at what time?
The maximum height of the ball was 29 metres at 2.4 seconds
How high was the ball after 3 seconds?
sub t=3
h=-4.9(3-2.4)^2+29
=27.2 m
The ball was 27.2 metres high after 3 seconds
## 1. The Video I made - How to Graph a parabola in vertex form
How to graph a parabola Bryan Boodhoo
## 3. Connecting topics Together
If you look at it closely, everything in quadratics was somehow related to another topic. One of the easiest relations to see was how every type of equation from Quadratics 1,2, and 3 was possible to graph. Also, it was possible to convert the three types of equations (Vertex form, factored form, and standard form) from one to another form. For example, since I have successfully completed all of the quadratics units, it is possible for me to graph a standard form equation in different ways. I could do this by completing the square, factoring, or even using the quadratic formula. Overall, it is evident to see from my explanation as to how everything in quadratics is related in a special way.
## 4. Reflection
Overall, I found that Quadratics was a very interesting unit. In the beginning I wasn't very confident with it but later I put the pieces together and started to improve. I noticed that towards the factoring unit in Quadratics 2 is where I felt the most confident with my work. Also, with the word problems in the Quadratics 3 unit I didn't really understand it, but later I looked into it with more detail and went at it with a better approach. In the Quadratics unit we learned vertex form, factored form, and standard form. The whole Quadratics unit was basically like a step up from the Linear systems unit that we learned in grade 9. To conclude, I learned at lot of new things through trial and error.
## 5. Other Videos to help (Not mine)
3.3 More Graphing from Vertex Form
3.11 Factoring
Common Factors
3.8 Factoring Simple Trinomials
3.9 Complex Trinomial Factoring |
# Find the sum of the infinite series -ne^(1-n) from n=1 to oo?
Mar 5, 2018
$- {e}^{2} / {\left(e - 1\right)}^{2}$
#### Explanation:
The simplest solution I can think of starts with the infinite geometric series
$S \left(a\right) = {\sum}_{n = 1}^{\infty} {e}^{- a n}$
which converges for $a > 0$ to the well known sum
$S \left(a\right) = {e}^{- a} / \left\{1 - {e}^{- a}\right\} = \frac{1}{{e}^{a} - 1}$
Differentiating both sides with respect to $a$ leads to
$\frac{d}{\mathrm{da}} \left({\sum}_{n = 1}^{\infty} {e}^{- a n}\right) = \frac{d}{\mathrm{da}} \left(\frac{1}{{e}^{a} - 1}\right) \implies$
$- {\sum}_{n = 1}^{\infty} n {e}^{- a n} = - {e}^{a} / {\left({e}^{a} - 1\right)}^{2}$
Substituting $a = 1$ gives us
$- {\sum}_{n = 1}^{\infty} n {e}^{- n} = - \frac{e}{e - 1} ^ 2 \implies$
$- {\sum}_{n = 1}^{\infty} n {e}^{1 - n} = - {e}^{2} / {\left(e - 1\right)}^{2}$
Alternative
We can evaluate the sum by a purely algebraic approach as well:
$- S = 1 + \frac{2}{e} + \frac{3}{e} ^ 2 + \frac{4}{e} ^ 3 + \ldots$
$- \frac{1}{e} S = \frac{1}{e} + \frac{2}{e} ^ 2 + \frac{3}{e} ^ 3 + \ldots$
Subtracting, we get
$- \left(1 - \frac{1}{e}\right) S = 1 + \frac{1}{e} + \frac{1}{e} ^ 2 + \frac{1}{e} ^ 3 + \ldots = \frac{1}{1 - \frac{1}{e}}$ |
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# CAT 2017 Important Topics in Quantitative Aptitude: Percentages Part II
Updated : Jun 20, 2017, 11:29
By : Asakti
CAT 2017 Hopefuls need to strengthen their basics before going out guns blazing on Quantitative Aptitude. To help their case, this post provides essential concepts to build your basics in any chapter. Here are some topics that we have covered:
If you have not checked out the previous article, click here: Part I
Note: This is an original study material contributed by Mr. Tushar Dhingra
Concept of Base
If we know that in class of 60 students, only 40 students passed, then the percentage of students passing is 40/60* 100 =66.66%
Simply stated this is unitary method i.e. out of 60 students 40 passed. Had there been 100 students appearing, how many would have passed? In this case the base is the number of students taking the exam.
One more example, if a number, let’s say 80, increases by 20%, the final number is not really 80+20% i.e. 80+0.2. The 20% is again a percentage of a base, in this case of the number itself. Thus, the final number is 80 +20% of 80 =80+1/5th of 80 = 80 +16 =96.
Example: A’s income is 70% of B’s. B’s income is 50% of C’s. If C’s income is Rs 1,00,000, what is A’s income?
Solution: B’s income =50/100*100000 = Rs50000
A ’s income =70/100*50000 =Rs 35000
Alternative Method
B’s income =50/100 of C’s income
A’s income =70/100 of B’s income = C’s income
A’s income =35/100*10000 =Rs 35,000
Example: In a basket of fruits 60% are mangoes and remaining 40% are apples. 25% of the apples are green and the rest 75% are red. Of the mangoes 80% are red and the rest are green.
What percentage of the green fruits are mangoes?
Solution:
If total number of fruits is 100 ,60 are mangoes and 40 are apples
Green apples =25% of 40 =10
Green Mangoes =20% of 60 =12
Total number of green fruits =10+12 =22 of which 12 are mangoes
Thus required percentage =12/22*100 =54.54%
You may encounter the following types of questions in examination
Types of questions Examples Approach to question 1) Convert x percentage into fraction Express 12% as a fraction X%=x/100=12/100=3/25 2) Convert fraction or decimal into percentage Express 5/11 as percentage Multiply the fraction by 100 =5/11*100=45.45% 3) If A’s income is x% of B’s income and B’s income is given then find A’s income A’s income is 40% of B’s income. If B’s income is Rs 10,000. What is A’s income? A=x/100*P =40/100*10000 =4000 4) If A’s income is r% more than B’s income then by how much % is B’s income less than A’s income? X’s income is 25% more than Y’s. By how much % is Y’s income less than X’s income? Difference =r/100+r*100 =25/125*100 =66.67% 5) If A’s income is r% less than B’s income then by how much % is B’s income more than A’s income? X’s income is 40% less than Y’s. By how much % is Y’s income more than X’s income Difference =r/100-r*100 =40/60*100 =66.67% 6) If the price of a commodity increases by r%, find the % decrease in the consumption given that expenditure remains same If the price of potato is increased by 20%, by how much should the consumption be decreased so as to maintain the same expenditure ? Expenditure=Price * Consumption Decrease =r/100+r*100 =20/120*100=50/3=16.67%
Concept of multiplying factor
Consider the following calculation that is done when 40 has to be increased by 20%
Increased value =40+20% of 40
=40+ 20/100*40 =40+8 =48
One could rearrange and approach it in the following manner.
Increased value =40+20% of 40
=40(1+20%) =40(1+.2)=48
1.2 can be called the multiplying factor (MF) corresponding to 20% and can simplify a lot of calculations as will be seen later. So understand the concept of multiplying factor very well.
The multiplying factor (MF) corresponding to x% increase is nothing but 1+ x/100 and is 1-x/100 for an x% decrease.
MF>1 implies an increase and MF<1 implies a decrease. A more than 100% increase would give a multiplying factor greater than 2
(e.g. a 100% increase MF of 2 ,400% increase MF of 5)
MF can be used whereas percentages are present. Thus make sure your understanding of MF is thorough.
Initial Value* MF =Final Value
Initial value can be cost price, principal kept in bank, sales in year x, value of asset in year x and corresponding final value will be selling price amount, sales in subsequent year and value of asset in subsequent year respectively.
Successive percentage changes
If in the first year, A’s salary increases by 10% and in second year, the salary increases by 20% again, would the net increase over the two years be 10+20 =30 %?
Not really
If A’s Salary was 100 at start, after 1st year it would be 110.
In second year it would be 110* 1.2=132 Thus the net increase is 32 i.e. and not 30%
The computation of final salary can also be done with multiplying factor as
Final salary = (Initial salary*1.1) *1.2= initial salary *1.32
i.e. the final salary is 32% more than the initial salary.
i.e. the net percentage increase is a+b+ ab/100%
Two successive percentage changes of a% and b% is an effective change of (a+b+ ab/100%)
Formula for Percentage Change:
Suppose a number N undergoes a percentage change of x % and then y%, the net change is:
New number = N × (1 + x/100) × (1 + y/100)
Now, (1 + x/100) × (1 + y/100) = 1 + x/100 + y/100 + xy/10000
If we say that x + y + xy/100 = z, then (1 + x/100) × (1 + y/100) = 1 + z/100
Here, z is the effective percentage change when a number is changed successively by two percentage changes.
Various cases for Percentage Change:
Both percentage changes are positive:
x and y are positive and net increase = (x+y+xy/100) %.
One percentage change is positive and the other is negative:
x is positive and y is negative, then net percentage change = (x-y-xy/100)%
Both percentage changes are negative:
x and y both are negative and imply a clear decrease= (-x-y+xy/100)%
Percentage Change involving three changes:
If value of an object/number P is successively changed by x%, y% and then z%, then final value.
percentage-successive-percentage-change
Example: The capacity of a ground was 100000 at the end of 2012. In 2013, it increased by 10% and in 2014, it decreased by 18.18%. What was the ground’s capacity at the end of 2014?
Solution:
When One percentage change is positive and the other is negative:
x is positive and y is negative, then net percentage change = (x-y-xy/100)%
Final Percentage Change over the original value = 10-18.18 – (10 × 18.18/100)= -9.998
(the difference above is cause by using exact values).
So the capacity of the ground is decreased by 9.998%
Hence, net capacity = 90002
Example: A’s salary is increased by 10% and then decreased by 10%. The change in salary is
Solution:
Percentage change formula when x is positive and y is negative = {x – y – (xy/100)}%
Here, x = 10, y = 10
= {10 – 10 – (10 x 10)/100} = -1%
As negative sign shows a decrease, hence the final salary is decreased by 1%.
Example: A number is first increased by 10% and then it is further increased by 20%. The original
number is increased altogether by:
Solution:
Percentage change formula when both x and y are positive ={x + y + (xy/100)}%
Here, x = 10 and y = 20
Hence net percentage change == {10 + 20 + (10 x 20)/100} = 32%
All the Best,
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Jun 20MBA Prep Exams
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Asakti Are you an IIM graduate Because since i have been a memeber you constantly keep posting somethin great on MBA i mean how do you get all this...
#muchsmart 😊
Thaq..
Is there any provision for mock CAT in gradeup?
Jun 20
@Subhendu Sahoo we shall be starting by next month. We are waiting to build a solid base. If more than 100 users want it, we could build a collection of mocks, 5 at least. Do let us know. |
# Linearly Independent and Dependent Vectors - Examples with Solutions
Page Content
## Definition of Linearly Independent Vectors
If we can express vector u1 as a linear combinations of the vectors u2 and u3 , we say that these 3 vectors are linearly dependent.
which may be written as
Hence the following definition
Given a set of vectors
If the equation
has only one trivial solution , we say that W is a set of linearly independent vectors.
If the above equation has other solutions, then W is a set of lineraly dependent vectors.
More testing for linearity of vectors in a Subspace are included.
## Examples with Solutions
Note solve the examples in the order that they are presented in order to fully understand them.
Example 1
Show that the vectors $$\textbf{u}_1 = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$$ and $$\textbf{u}_2 = \begin{bmatrix} -5 \\ -15 \end{bmatrix}$$ are linearly dependent.
Solution to Example 1
Two ways to answer this question
1)
There is an obvious relationship between $$\textbf{u}_1$$ and $$\textbf{u}_2$$ which is
$$\textbf{u}_2 = - 5 \textbf{u}_1$$
and therefore the two vectors are linearly dependent.
2)
Let us now use the definition to prove that the two vectors are linearly dependent.
Equation (I) given in the definition above is written as
$$r_ 1 \begin{bmatrix} 1 \\ 3 \end{bmatrix} + r_2 \begin{bmatrix} -5 \\ -15 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ (II)
Scalar multiply and add the left side of the above equation to write a system of equations
$$\begin{bmatrix} r_1 - 5 r_2\\ 3 r_1 - 15 r_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
We now need to solve the above system. Let us use the method of elimination. Multiply the top equation by $$- 3$$
$$\begin{bmatrix} - 3 r_1 + 15 r_2\\ 3 r_1 - 15 r_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
Add the top and bottom equation and place it at the bottom
$$\begin{bmatrix} - 3 r_1 + 15 r_2\\ 0 + 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
The second equation may be used to write that $$r_2 = t$$ such that $$t$$ is any real number.
Use the top equation to find
$$r_1 = 5 r_2 = 5 t$$
Equation (II) of the definition above has many solutions and therefore vectors $$\textbf{u}_1$$ and $$\textbf{u}_2$$ given above are linearly dependent.
Example 2
Show that the vectors $$\textbf{u}_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$$ , $$\textbf{u}_2 = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$$ and $$\textbf{u}_3 = \begin{bmatrix} -2 \\ 3 \\ 0 \end{bmatrix}$$ are linearly independent.
Solution to Example 2
Use definition to prove that the two vectors are linearly independent.
Equation (I) given in the definition above is written as
$$r_1 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + r_2 \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} + r_3 \begin{bmatrix} -2 \\ 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
It is a homogeneous system of equation. Using matrices, it may be written as
$$\begin{bmatrix} 1 & 0 & -2\\ 0 & 1 & 3 \\ 1 & 2 & 0 \end{bmatrix} \begin{bmatrix} r_1 \\ r_2 \\ r_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
The above system of equations has a trivial solution $$r_1 =0 , r_2 = 0 , r_3 = 0$$ only if the determinant of the square coefficient matrix on is NOT equal to zero.
$$\text{Det} \begin{bmatrix} 1 & 0 & -2\\ 0 & 1 & 3 \\ 1 & 2 & 0 \end{bmatrix} = -4$$
Hence the equation of the definition of linearity of vectors has one trivial solution and therefore the vectors are lineraly independent.
Example 3
Find the values of $$m$$ for which the vectors $$\textbf{u}_1 = \begin{bmatrix} m \\ 4 \\ 0 \end{bmatrix}$$ , $$\textbf{u}_2 = \begin{bmatrix} 1 \\ -1 \\ 8 \end{bmatrix}$$ and $$\textbf{u}_3 = \begin{bmatrix} 0 \\ -1 \\ m \end{bmatrix}$$ are linearly dependent.
Solution to Example 3
We use the equation of linearity given in the definition
$$r_1 \begin{bmatrix} m \\ 4 \\ 0 \end{bmatrix} + r_2 \begin{bmatrix} 1 \\ -1 \\ 8 \end{bmatrix} + r_3 \begin{bmatrix} 0 \\ -1 \\ m \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
The above system of homogeneous equations in matrix form as
$$\begin{bmatrix} m & 1 & 0\\ 4 & -1 & -1 \\ 0 & 8 & m \end{bmatrix} \begin{bmatrix} r_1 \\ r_2 \\ r_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
The determinant of the square matrix is given by
$$\text{Det} \begin{bmatrix} m & 1 & 0\\ 4 & -1 & -1 \\ 0 & 8 & m \end{bmatrix} = - m^2 + 4 m$$
For the vectors to be linearly dependent, the system of equations must have more than one (the trivial) solution and hence the determinant must be equal to 0. Hence
$$- m^2 + 4 m = 0$$
Solve the above form $$m$$
$$m = 0$$ and $$m = 4$$ are the values for which the given vectors are linearly dependent.
Example 4
Are the vectors $$\textbf{u}_1 = \begin{bmatrix} 1 \\ 1\\ 0 \\ -1 \end{bmatrix}$$ , $$\textbf{u}_2 = \begin{bmatrix} 5 \\ 5 \\ 0 \\ 0 \end{bmatrix}$$ and $$\textbf{u}_3 = \begin{bmatrix} 0 \\ 0 \\ -1 \\ -1 \end{bmatrix}$$ linearly dependent or independent?
Solution to Example 4
Write the equation of linearity given in the definition above
$$r_1 \begin{bmatrix} 1 \\ 1 \\ 0 \\ -1 \end{bmatrix} + r_2 \begin{bmatrix} 5\\ 5\\ 0 \\ 0 \end{bmatrix} + r_3 \begin{bmatrix} 0 \\ 0 \\ -1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
Write the above as a system of homogeneous equations
$$\begin{bmatrix} 1 & 5 & 0\\ 1 & 5 & 0 \\ 0 & 0 & -1 \\ -1 & 0 & -1 \end{bmatrix} \begin{bmatrix} r_1 \\ r_2 \\ r_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
The row reduced form of the above system of equations is written as
$$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} r_1 \\ r_2 \\ r_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
which corresponds to the system
$$\begin{array}{lcl} r_1 & = & 0 \\ r_2 & = & 0 \\ r_3 & = & 0 \\ 0 & = & 0 \end{array}$$
The equation of linearity has only the trivial solution $$r_1 = 0 , r_2 = 0 , r_3 = 0$$ and therefore the vectors $$\textbf{u}_1$$ , $$\textbf{u}_2$$ and $$\textbf{u}_3$$ are linearly independent. |
# How to solve this recurrence relation ?
• Apr 19th 2009, 01:27 AM
denniszeng2008@gmail.com
How to solve this recurrence relation ?
3P(n) = 1-P(n-1).
P(n) = ???
• Apr 19th 2009, 01:47 AM
CaptainBlack
Quote:
Originally Posted by denniszeng2008@gmail.com
3P(n) = 1-P(n-1).
P(n) = ???
Could you please post the complete question.
CB
• Apr 19th 2009, 04:23 AM
denniszeng2008@gmail.com
Complement
it is only a recursive problem.
3P(n) = 1 - P(n-1)
3P(n-1) = 1 - P(n - 2)
...
How to figure out P(n) = ?
• Apr 19th 2009, 06:38 AM
aidan
Quote:
Originally Posted by denniszeng2008@gmail.com
3P(n) = 1-P(n-1).
P(n) = ???
if
3P(n) = 1-P(n-1)
then
P(n) = (1-P(n-1))/3
• Apr 19th 2009, 06:58 AM
denniszeng2008@gmail.com
But
But i want to figure out a p(n) = n 's function but not a recurrence sequence.
• Apr 19th 2009, 07:48 AM
Soroban
Hello, denniszeng!
Quote:
$\displaystyle 3\!\cdot\!P(n) \:=\: 1-P(n-1)\qquad\text{Find }P(n)$
Let $\displaystyle P(n) \:=\:X^n$
$\displaystyle \begin{array}{cccccc}\text{Then we have:} &3X^n &=& 1-X^{n-1} & [1] \\ \text{The next term:} & 3X^{n+1} &=& 1 - X^n & [2] \end{array}$
Subtract [2] - [1]: .$\displaystyle 3X^{n+1} -3X^n \:=\:-X^n + X^{n-1} \quad\Rightarrow\quad X^{n+1} - 3X^n - X^{n-1} \:=\:0$
Divide by $\displaystyle X^{n-1}\!:\;\;3X^2 -2X - 1 \:=\:0 \quad\Rightarrow\quad (X-1)(3X+1) \:=\:0$
. . Hence: .$\displaystyle X \:=\:1,\:\text{-}\tfrac{1}{3}$
Then $\displaystyle P(n)$ is of the form: .$\displaystyle P(n) \;=\;A(1^n) + B\left(\text{-}\tfrac{1}{3}\right)^n$
. . That is: .$\displaystyle P(n) \:=\:A + B\frac{(\text{-}1)^n}{3^n}$ .[3]
Suppose the first term is $\displaystyle a\!:\;\;P(1) \:=\:a$
Then the second term is: $\displaystyle P(2) \:=\:\tfrac{1}{3}(1-a)$
Substitute into [3]:
. . $\displaystyle \begin{array}{ccccccc}P(1) \:=\:a\!:& & A - \frac{B}{3}&=& a \\ \\[-4mm] P(2) \:=\:\frac{1}{3}(1-a)\!: && A + \frac{B}{9} &=& \frac{1}{3}(1-a) \end{array}$
Solve the system: .$\displaystyle A \:=\:\frac{1}{4},\quad B \:=\:\tfrac{3}{4}(1 - 4a)$
Therefore: .$\displaystyle P(n) \;=\;\frac{1}{4} + \frac{3}{4}(1-4a)\frac{(\text{-}1)^n}{3^n} \;=\;\frac{1}{4}\left[1 + (\text{-}1)^n\,\frac{3(1-4a)}{3^n}\right]$
. . . . . . . . $\displaystyle P(n) \;=\;\frac{1}{4}\left[1 + (\text{-}1)^n\frac{1-4a}{3^{n-1}}\right]$
• Apr 19th 2009, 04:31 PM
denniszeng2008@gmail.com
But Why P(n) can have that form relate to 1 and 1/3 ?
since it is only a special example we pick that p(n) = $\displaystyle x^n$
Thanks very much!
I don't want to only know the solution, can you teach me how to analyze this ?
When i encounter such a problem, what should i do to figure out the right solution ?
Thanks |
# Substitute Numbers for Letters in Complex Calculations
In this worksheet, students will replace letters with numbers and carry out questions using the four number operations, with some calculations in sets of brackets. They must give their final answers as numbers not letters.
Key stage: KS 2
Curriculum topic: Verbal Reasoning
Curriculum subtopic: Maths Codes
Difficulty level:
### QUESTION 1 of 10
Make sure that you’ve got your thinking cap on as we are going to be number decoders!
In this activity, we are going to swap numbers for letters to solve maths problems.
Let’s take a look at this question first:
If a = 4, b = 12, c = 3 and d = 9, what is b ÷ a?
This is called algebra, which is where numbers are replaced with letters. We need to swap the letters back into numbers to solve this problem.
If we put the numbers back in, then b ÷ a becomes 12 ÷ 4. We know this is 3.
If any part of the question is written in brackets, this part has to be calculated first.
Let’s try this question next:
If a = 7, b = 9, c = 6 and d = 4, what is b + (a x d)?
Let’s swap the letters back to numbers. This would be written as: 9 + (7 x 4) = ___.
We must work out 7 x 4 first as it is in brackets. This is 28.
Our number sentence now looks like this: 9 + (28) = ___. The answer is 37.
Let’s try one more together:
If a = 8, b = 10, c = 4 and d = 12, what is b (a + c)?
Where there isn’t a -, +, x or ÷ between the first letter and the bracket, it means we have to multiply the outside number by what is inside the brackets. We must imagine an invisible times sign here.
This can be written as: b x (a + c) = ___.
In numbers this is: 10 x (8 + 4) = ____.
We must solve the brackets first: 8 + 4 = 12.
Our number sentence now becomes: 10 x (12) = 120. So our answer is 120.
Now it’s your turn to crack the codes. Good luck number detective!
If A = 4, B = 12, C = 3 and D = 9, what is D ÷ C?
Can you calculate the answer as a number?
3
4
8
If a = 4, b = 27, c = 3 and d = 9, what is b ÷ c + a?
Can you calculate the answer as a number?
12
14
13
If A = 4, B = 27, C = 3 and D = 9, what is B ÷ C x A?
Can you calculate the answer as a number?
You have been given the following information: A = 4, B = 24, C = 3, D = 12.
Can you match each equation to its correct answer (as a number)?
## Column B
B ÷ D
2
C x A + B
36
A + B + C
31
You have been given the following information: a = 3, b = 30, c = 6, d = 12.
Can you match each equation to its correct answer (as a number)?
## Column B
b ÷ c
6
d x a - b
5
b - c - d
12
Now you are going to be the examiner!
A student has been given the questions below using this information: A = 3, B = 30, C = 6, D = 12.
Identify whether they have answered each question correctly or not.
Following on from the last question, you need to use this information again: A = 3, B = 30, C = 6, D = 12
The student in the previous question answered that B - A = 33.
Which of the mistakes below could they have made to reach their answer?
Used the wrong numbers to complete the sum
Subtracted the numbers the wrong way around
Another student has been given the questions below using this information: a = 7, b = 35, c = 5, d = 16.
Identify whether they have answered each question correctly or not.
Following on from the last question, you need to use this information again: a = 7, b = 35, c = 5, d = 16
The student in the previous question answered that d x c = 21.
Which of the mistakes below could they have made to reach their answer?
Used the wrong numbers to complete the sum
Multiplied the numbers the wrong way around
Here is your final question as the examiner!
Another student has been given the questions below using this information: A = 4, B = 32, C = 7, D=17.
Identify whether they have answered each question correctly or not.
• Question 1
If A = 4, B = 12, C = 3 and D = 9, what is D ÷ C?
Can you calculate the answer as a number?
3
EDDIE SAYS
The question tells us that D = 9 and C = 3.
If we swap these numbers in for the letters, then D ÷ C becomes 9 ÷ 3.
When we divide 9 by 3, we reach the answer of 3.
If you were asked to write the answer as a letter, which letter would it be?
• Question 2
If a = 4, b = 27, c = 3 and d = 9, what is b ÷ c + a?
Can you calculate the answer as a number?
13
EDDIE SAYS
The question tells us that b = 27, c = 3 and a = 4.
If we swap these numbers in for the letters, then b ÷ c + a becomes 27 ÷ 3 + 4.
When we divide 27 by 3 then add 4, we reach the answer of 13.
• Question 3
If A = 4, B = 27, C = 3 and D = 9, what is B ÷ C x A?
Can you calculate the answer as a number?
36
EDDIE SAYS
The question tells us that B = 27, C = 3 and A = 4, which is the same as the previous question.
If we swap these numbers in for the letters, then B ÷ C x A becomes 27 ÷ 3 x 4.
Did you spot the small change here from the previous question? The addition sign changed to a multiplication sign. Often small changes exist in exam questions to try and check whether you are looking at all the detail.
When we divide 27 by 3 then multiply our answer by 4, we reach the total of 36.
• Question 4
You have been given the following information: A = 4, B = 24, C = 3, D = 12.
Can you match each equation to its correct answer (as a number)?
## Column B
B ÷ D
2
C x A + B
36
A + B + C
31
EDDIE SAYS
The question tells us that B = 24, D = 12, C = 3 and A = 4.
If we swap these numbers in for the letters in each of the equations, we get the following calculations: B ÷ D becomes 24 ÷ 12 = 2 C x A + B becomes 3 x 4 + 24 = 36 A + B + C becomes 4 + 24 + 3 = 31
How many of the equations did you calculate correctly? Well done if you got all three correct! If not then don't worry, as we are going to practice some more now.
• Question 5
You have been given the following information: a = 3, b = 30, c = 6, d = 12.
Can you match each equation to its correct answer (as a number)?
## Column B
b ÷ c
5
d x a - b
6
b - c - d
12
EDDIE SAYS
The question tells us that b = 30, c = 6, d = 12 and a = 3.
If we swap these numbers in for the letters in each of the equations, we get the following calculations: b ÷ c becomes 30 ÷ 6 = 5 d x a - b becomes 12 x 3 - 30 = 6 b - c - d becomes 30 - 6 - 12 = 12
Were you able to get all three answers matched?
• Question 6
Now you are going to be the examiner!
A student has been given the questions below using this information: A = 3, B = 30, C = 6, D = 12.
Identify whether they have answered each question correctly or not.
EDDIE SAYS
The question tells us that B = 30, A = 3, C = 6 and D = 12.
If we swap these numbers in for the letters in each of the equations, we get the following calculations: B - A becomes 30 - 3 = 27. The student answered 33 so this is incorrect. D x A becomes 12 x 3 = 36, which the student got right! B ÷ A + C becomes 30 ÷ 3 + 6 = 16, which the student also got right!
How did you find being the examiner? Was it easy or hard to spot the student's errors?
• Question 7
Following on from the last question, you need to use this information again: A = 3, B = 30, C = 6, D = 12
The student in the previous question answered that B - A = 33.
Which of the mistakes below could they have made to reach their answer?
EDDIE SAYS
The student got to an answer of 33 using the number options provided in the question, so they could have only done one these things wrong: Option 1: They used 30 and 3 which were the correct numbers, so they could not have made this error. Option 2: 30 + 3 is 33 which was the answer the student gave. So it is likely that they confused the symbol and added the numbers instead of subtracting them. Option 3: 3 - 30 is not 33, so the student could not have made this error.
• Question 8
Another student has been given the questions below using this information: a = 7, b = 35, c = 5, d = 16.
Identify whether they have answered each question correctly or not.
EDDIE SAYS
The question tells us that b = 35, a = 7, d = 16 and c = 5.
If we swap these numbers in for the letters in each of the equations, we get the following calculations: b - d becomes 35 - 16 = 19, which the student got right! d x c becomes 16 x 5 = 80. The student answered 21 so this is incorrect. b ÷ a + c becomes 35 ÷ 7 + 5 = 10, which the student also got right!
Are you getting better at spotting the incorrect answers? Examiners need to make sure they don't mark any questions right that are wrong and vice versa!
• Question 9
Following on from the last question, you need to use this information again: a = 7, b = 35, c = 5, d = 16
The student in the previous question answered that d x c = 21.
Which of the mistakes below could they have made to reach their answer?
EDDIE SAYS
The student got to an answer of 21 using the number options provided in the question, so they could have only done one these things wrong: Option 1: Using any combination of the available numbers for d and c cannot give a total of 21, so they could not have made this error. Option 2: 16 + 5 gives 21 which was the student's answer, so this is likely to be the mistake they made. Option 3: 16 x 5 and 5 x 16 give the same answer, so they could not have made this error. It is important to remember that you can switch the order of numbers in a multiplication sum and still reach the correct answer.
• Question 10
Here is your final question as the examiner!
Another student has been given the questions below using this information: A = 4, B = 32, C = 7, D=17.
Identify whether they have answered each question correctly or not.
EDDIE SAYS
The question tells us that B = 32, A = 4, D = 17 and C = 7.
If we swap these numbers in for the letters in each of the equations, we get the following calculations: B - D becomes 32 - 17 = 15, which the student got right! B x A becomes 32 x 4 = 128, which the student got right! B ÷ A + C becomes 32 ÷ 4 + 7 = 15, which the student also got right!
The student answered them all correctly! Hopefully you were as successful as an examiner too.
---- OR ----
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# What are their ages in years?
1,475.5K Views
The grandson is about as many days old as the son is in weeks. The grandson is approximately as many months old as the father is in years. The ages of the grandson, the son, and the father add up to 120 years. What are their ages in years?
pnikam Expert Asked on 18th August 2015 in
Let G, S, and F represent the ages in years of the grandson, the son, and the father, respectively. Since a year has 365 days, or 52 weeks, or 12 months, the problem can be represented by three equations with three unknowns:
365 G = 52 S (The grandson is about as many days old as the son is in weeks)
12 G = F (The grandson is approximately as many months old as the father is in years)
G + S + F = 120 (The ages of the grandson, the son, and the father add up to 120 years)
Because the grandson’s age is G = F/12, the son’s age can be represented in terms of F by substituting:
S = (365/52)×G
S = (365/52)×(F/12)
The third equation can now be represented with only the father’s age as an unknown:
F/12 + (365/52)×(F/12) + F = 120
Multiplying by 12, we get:
F + (365/52)×F + 12 F = 120×12
20 F = 1440
F = 72
G = F/12 = 6
S = (365/52)×6 = 42
The father is 72, the son is 42, and the grandson is 6 years old.
You may have noticed that the statement of the problem is somewhat ambiguous: “The grandson is approximately as many months old as the father is in years” indicates that there may be some variance. A year really has 365 1/4 days instead of 365, and 52 weeks times 7 days gives only 364 days in a year. These inconsistencies produce fractional results that need to be rounded.
Doing the division in the following equation:
F + (365/52)×F + 12 F = 120×12
produces 20.019 F = 1440, which is rounded to 20 F = 1440
pnikam Expert Answered on 18th August 2015.
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```ALGEBRA 1 CC
Graphing and Writing Equations in SlopeIntercept Form
• The three forms of writing a linear equation that we will
cover this week
• Slope-Intercept form
• Standard Form
• Point-Slope Form
y mx b
Ax By C
y y1 m x x1
Example 1
• Write an equation of the line shown
Example 2
• Write an equation of the line that passes through the
points (4, -2) and (-2, 2)
Example 3
• Write the equation of the line that passes through the
point (-4, -9) and has a slope of 2.
Graphing a Linear Equation
• Step 1: Rewrite the equation in slope-intercept form (if
necessary).
• Step 2: Identify the slope and y-intercept
• Step 3: Plot the y-intercept
• Step 4: Use the slope to plot a second point and then
draw a line through the two points.
• It is optional, but highly recommended to draw more than
two points to graph a linear equation
Example 4 & 5
• Identify the slope and y-intercept of the line with the given
equations and then graph the equations.
4) y 1 x 4
5) 4x – 3y = 15
2
Example 6
• Write an equation for the linear function f with the given
values
f(6) = -4, f(9) = -9
• Two lines in the same plane are parallel if they
do not intersect.
• Parallel lines have the same slope.
• Two lines in the same plane are perpendicular if
they intersect to form a right angle.
• Perpendicular lines have slopes that are
opposite reciprocals.
Example 7
• Tell whether the following graphs are parallel,
perpendicular, or neither.
y = 2x + 4
4x + 2y = 12
Example 8
• Tell whether the following graphs are parallel,
perpendicular, or neither.
3x + 4y = 24
-4x + 3y = 20
``` |
# Class 11 NCERT Solutions- Chapter 6 Linear Inequalities – Miscellaneous Exercise on Chapter 6
### Question 1. 2 ≤ 3x – 4 ≤ 5
Solution:
In this case, we have two inequalities, 2 ≤ 3x – 4 and 3x – 4 ≤ 5, which we will solve simultaneously.
We have 2 ≤ 3x – 4 ≤ 5
or 2 ≤ 3x – 4 and 3x – 4 ≤ 5
⇒ 2 + 4 ≤ 3x and 3x ≤ 5 + 4
⇒ 6 ≤ 3x and 3x ≤ 9
⇒ ≤ x and x ≤
⇒ 2 ≤ x and x ≤ 3
⇒ 2 ≤ x ≤ 3
Hence, all real numbers x greater than or equal to 2 but less than or equal to 3 are solution of given 2 ≤ 3x – 4 ≤ 5 equality.
x ∈ [2, 3]
### Question 2. 6 ≤ – 3 (2x – 4) < 12
Solution:
In this case, we have two inequalities, 6 ≤ – 3 (2x – 4) and – 3 (2x – 4) < 12, which we will solve simultaneously.
We have 6 ≤ – 3 (2x – 4) < 12
or 6 ≤ – 3 (2x – 4) and – 3 (2x – 4) < 12
⇒ ≤ -(2x – 4) and -(2x – 4) <
⇒ 2 ≤ -(2x – 4) and -(2x – 4) < 4
⇒ -2 ≥ (2x – 4) and (2x – 4) > -4 [multiplying the inequality with (-1) which changes the inequality sign]
⇒ -2+4 ≥ 2x and 2x > -4+4
⇒ 2 ≥ 2x and 2x > 0
⇒ ≥ x >
⇒ 1≥ x > 0
⇒ 0 < x ≤ 1
Hence, all real numbers x greater than 0 but less than or equal to 1 are solution of given 6 ≤ – 3 (2x – 4) < 12 equality.
x ∈ (0, 1]
### Question 3. -3 ≤ 4- ≤ 18
Solution:
In this case, we have two inequalities, -3 ≤ 4- and 4- ≤ 18, which we will solve simultaneously.
We have -3 ≤ 4- ≤ 18
or -3 ≤ 4- and 4- ≤ 18
⇒ -3-4 ≤ – and – } ≤ 18-4
⇒ -7 ≤ – and – ≤ 14
⇒ 7 ≥ and ≥ -14 [multiplying the inequality with (-1) which changes the inequality sign]
⇒ 7×2 ≥ 7x and 7x ≥ -14×2
⇒ ≥ x and x ≥
⇒ 2 ≥ x ≥ -4
⇒ -4 ≤ x ≤ 2
Hence, all real numbers x greater than or equal to -4 but less than or equal to 2 are solution of given -3 ≤ 4- ≤ 18 equality.
x ∈ [-4, 2]
### Question 4. -15 < ≤ 0
Solution:
In this case, we have two inequalities, -15 < and } ≤ 0, which we will solve simultaneously.
We have -15 < ≤ 0
or -15 < and ≤ 0
⇒ -15× < (x-2) and (x-2) ≤ 0×
⇒ -25 < x-2 and x-2 ≤ 0
⇒ -25+2 < x and x ≤ 0+2
⇒ -23 < x ≤ 2
Hence, all real numbers x greater than -23 but less than or equal to 2 are solution of given -15 ≤ ≤ 0 equality.
x ∈ (-23, 2]
### Question 5. -12 < 4- () ≤ 2
Solution:
In this case, we have two inequalities, -12 < 4- and 4- ≤ 2, which we will solve simultaneously.
We have -12 < 4- ≤ 2
or -12 < 4- and 4- ≤ 2
⇒ -12-4 < and ≤ 2-4
⇒ -16 < and ≤ -2
⇒ -16 < and ≤ -2
⇒ -16×5 < 3x and 3x ≤ -2×5
⇒ < x and x ≤
⇒
Hence, all real numbers x greater than -80/3 but less than or equal to -10/3 are solution of given -12 < 4- ≤ 2 equality.
x ∈ (-80/3, -10/3]
### Question 6. 7 ≤ ≤ 11
Solution:
In this case, we have two inequalities, 7 ≤ and ≤ 11, which we will solve simultaneously.
We have 7 ≤ ≤ 11
or 7 ≤ and ≤ 11
⇒ 7×2 ≤ 3x+11 and 3x+11 ≤ 11×2
⇒ 14 ≤ 3x+11 and 3x+11 ≤ 22
⇒ 14-11 ≤ 3x and 3x ≤ 22-11
⇒ 3 ≤ 3x and 3x ≤ 11
⇒ 1 ≤ x and x ≤
⇒ 1 ≤ x ≤
Hence, all real numbers x greater than or equal to 1 but less than or equal to 11/3 are solution of given 7 ≤ ≤ 11 equality.
x ∈ [1, 11/3]
### Question 7. 5x + 1 > – 24, 5x – 1 < 24
Solution:
So, from given data
5x + 1 > – 24 ……………………(1)
5x – 1 < 24 …………………….(2)
From inequality (1), we have
5x + 1 > – 24
5x > – 24-1
x > – 25/5
x > -5 ………………………….(3)
Also, from inequality (2), we have
5x – 1 < 24
5x < 24+1
x < 25/5
x < 5 ……………………………(4)
So, from (3) and (4), we can conclude that,
-5 < x < 5 ……………….(5)
If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are
x ∈ (-5,5)
Thus, solution of the system are real numbers x lying between -5 and 5 excluding -5 and 5.
### Question 8. 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x
Solution:
So, from given data
2 (x – 1) < x + 5 ……………………(1)
3 (x + 2) > 2 – x …………………….(2)
From inequality (1), we have
2 (x – 1) < x + 5
2x – 2 < x + 5
2x – x < 5+2
x < 7 ………………………….(3)
Also, from inequality (2), we have
3 (x + 2) > 2 – x
3x + 6 > 2 – x
3x + x > 2 – 6
4x > -4
x > -1 ……………………………(4)
So, from (3) and (4), we can conclude that,
-1 < x < 7 ……………….(5)
If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are
x ∈ (-1,7)
Thus, solution of the system are real numbers x lying between -1 and 7 excluding -1 and 7.
### Question 9. 3x – 7 > 2 (x – 6) , 6 – x > 11 – 2x
Solution:
So, from given data
3x – 7 > 2 (x – 6) ……………………(1)
6 – x > 11 – 2x …………………….(2)
From inequality (1), we have
3x – 7 > 2 (x – 6)
3x – 7 > 2x – 12
3x – 2x > – 12+7
x > -5 ………………………….(3)
Also, from inequality (2), we have
6 – x > 11 – 2x
– x+2x > 11 -6
x > 5 ……………………………(4)
So, from (3) and (4), we can conclude that,
5 < x ……………….(5)
If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are
x ∈ (5,∞)
Thus, solution of the system are real numbers x lying between 5 and ∞ excluding 5 .
### Question 10. 5 (2x – 7) – 3 (2x + 3) ≤ 0 , 2x + 19 ≤ 6x + 47
Solution:
So, from given data
5 (2x – 7) – 3 (2x + 3) ≤ 0 ……………………(1)
2x + 19 ≤ 6x + 47 …………………….(2)
From inequality (1), we have
5 (2x – 7) – 3 (2x + 3) ≤ 0
10x – 35 -6x – 9 ≤ 0
4x – 44 ≤ 0
4x ≤ 44
x ≤ 44/4
x ≤ 11 ………………………….(3)
Also, from inequality (2), we have
2x + 19 ≤ 6x + 47
2x – 6x ≤ 47 – 19
-4x ≤ 28
4x ≥ -28 [multiplying the inequality with (-1) which changes the inequality sign]
x ≥ -28/4
x ≥ -7 ……………………………(4)
So, from (3) and (4), we can conclude that,
-7 ≤ x ≤ 11 ……………….(5)
If we draw the graph of inequalities (5) on the number line, we see that the values of x, which are common to both, are
x ∈ [-7,11)
Thus, solution of the system are real numbers x lying between -7 and 11 including -7 and 11 .
### Question 11. A solution is to be kept between 68° F and 77° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by F = ()C + 32
Solution:
According to the given data
The solution has to be kept between 68° F and 77° F
So, we have,
68° < F < 77°
Substituting, F = C + 32
⇒ 68° < C + 32 < 77°
⇒ 68°- 32° < C < 77°- 32°
⇒ 36° < C < 45°
⇒ 36× < C < 45×
⇒ 20° < C < 25°
Hence, here we get,
The range of temperature in degree Celsius is between 20° C to 25° C.
### Question 12. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?
Solution:
According to the given data,
Here, 8% of solution of boric acid = 640 litres
So, we can take the amount of 2% boric acid solution added as x litres
Hence, Total mixture = (x + 640) litres
As it is given,
The resulting mixture has to be more than 4% but less than 6% boric acid
⇒ (2% of x + 8% of 640) > (4% of (x + 640)) and (2% of x + 8% of 640) < (6% of (x + 640))
⇒ () × (x + 640) < () × x + () × 640) < () × (x + 640)
⇒ 4(x + 640) < (2×x + 8× 640) < 6(x + 640)
⇒ 4x + 2560 < 2x +5120 < 6x+3840
In this case, we have two inequalities,
⇒ 4x + 2560 < 2x +5120 and 2x +5120 < 6x+3840
⇒ 4x – 2x < 5120 – 2560 and 5120-3840 < 6x-2x
⇒ 2x < 2560 and 1280 < 4x
⇒ x < 2560/2 and 1280/4 < x
⇒ x < 1280 and 320 < x
⇒ 320 < x < 1280
Therefore, the number of litres of 2% of boric acid solution that has to be added will be more than 320 litres but less than 1280 litres.
### Question 13. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?
Solution:
According to the given data,
Here, 45% of solution of acid = 1125 litres
Let the amount of water added in the solution = x litres
Resulting mixture = (x + 1125) litres
As it is given,
The resulting mixture has to be more than 25% but less than 30% acid content
Amount of acid in resulting mixture = 45% of 1125 litres.
⇒ 45% of 1125 < 30% of (x + 1125) and 45% of 1125 > 25% of (x + 1125)
⇒ 25% of (x + 1125) < 45% of 1125 < 30% of (x + 1125)
In this case, we have two inequalities,
⇒ ( × (x + 1125)) < ( × 1125) and ( × (x + 1125)) > ( × 1125)
⇒ (25(x + 1125)) < (45×1125) and (30(x + 1125)) > (45×1125)
⇒ (x + 1125) < (45×1125)/25 and (x + 1125) > (45×1125)/30
⇒ (x + 1125) < 2025 and (x + 1125) > 3375/2
⇒ 3375/2 < (x + 1125) < 2025
⇒ (3375/2)-1125 < x < 2025-1125
⇒ 1125/2 < x < 900
⇒ 562.5 < x < 900
Therefore, the number of litres of water that has to be added will have to be more than 562.5 litres but less than 900 litres.
### Question 14. IQ of a person is given by the formula, IQ = () × 100, where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental age.
Solution:
According to the given data, we have
Chronological age = CA = 12 years
IQ for age group of 12 is in the range,
80 ≤ IQ ≤ 140
Substituting, IQ = () × 100
⇒ 80 ≤ × 100 ≤ 140
⇒ 80 ≤ × 100 ≤ 140
⇒ 80×12/100 ≤ MA ≤ 140×12/100
⇒ 96/10 ≤ MA ≤ 168/10
⇒ 9.6 ≤ MA ≤ 16.8
Hence, Range of mental age (MA) of the group of 12 years old children is 9.6 ≤ MA ≤ 16.8
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# Solving Absolute Value Sentences, All Types
This web exercise mixes up problems from the previous three web exercises:
Visit these earlier sections for a thorough discussion of the concepts.
THEOREM solving absolute value sentences
Let $\,x\in\mathbb{R}\,,$ and let $\,k\ge 0\,.$ Then: $$\begin{gather} \cssId{s13}{|x| = k\ \ \text{ is equivalent to }\ \ x = \pm k} \\ \\ \cssId{s14}{|x| \lt k\ \ \ \text{ is equivalent to }\ \ -k \lt x \lt k} \\ \cssId{s15}{|x| \le k\ \ \ \text{ is equivalent to }\ \ -k \le x \le k} \\ \\ \cssId{s16}{|x| \gt k\ \ \ \text{ is equivalent to }\ \ x\lt -k\ \ \text{ or }\ \ x\gt k} \\ \cssId{s17}{|x| \ge k\ \ \ \text{ is equivalent to }\ \ x\le -k\ \ \text{ or }\ \ x\ge k} \end{gather}$$
## Example: An Absolute Value Equation
Solve: $|2 - 3x| = 7$
Solution: Write a nice, clean list of equivalent sentences:
$|2 - 3x| = 7$ original equation $2-3x = \pm 7$ check that $\,k\ge 0\,$; use the theorem $2-3x = 7$ or $\,2-3x = -7$ expand the plus/minus $-3x = 5$ or $\,-3x = -9$ subtract $\,2\,$ from both sides of both equations $\displaystyle x = -\frac{5}{3}\ \text{ or } x = 3$ divide both sides of both equations by $\,-3\,$
It's a good idea to check your solutions:
$|2 - 3(-\frac{5}{3})|\ \overset{\text{?}}{=}\ 7$, $|2 + 5| = 7$
Check!
$|2 - 3(3)|\ \overset{\text{?}}{=}\ 7$, $|2 - 9| = 7$
Check!
## Example: An Absolute Value Inequality Involving ‘Less Than’
Solve: $3|-6x + 7| \le 9$
Solution: To use the theorem, you must have the absolute value all by itself on one side of the sentence. Thus, your first job is to isolate the absolute value:
$3|-6x + 7| \le 9$ original sentence $|-6x + 7| \le 3$ divide both sides by $\,3$ $-3 \le -6x + 7 \le 3$ check that $\,k \ge 0\,$; use the theorem $-10 \le -6x \le -4$ subtract $\,7\,$ from all three parts of the compound inequality $\displaystyle \frac{10}{6} \ge x \ge \frac{4}{6}$ divide all three parts by $\,-6\,$; change direction of inequality symbols $\displaystyle \frac{2}{3} \le x \le \frac{5}{3}$ simplify fractions; write in the conventional way
Check the ‘boundaries’ of the solution set:
$3|-6(\frac{2}{3}) + 7| = 3|-4 + 7| = 3|3| = 9$
Check!
$3|-6(\frac{5}{3}) + 7| = 3|-10 + 7| = 3|-3| = 9$
Check!
## Example: An Absolute Value Inequality Involving ‘Greater Than’
Solve: $3|-6x + 7| \ge 9$
Solution: To use the theorem, you must have the absolute value all by itself on one side of the sentence. Thus, your first job is to isolate the absolute value:
$3|-6x + 7| \ge 9$ original sentence $|-6x + 7| \ge 3$ divide both sides by $\,3$ $-6x + 7 \le -3$ or $\,-6x + 7\ge 3$ check that $\,k \ge 0\,$; use the theorem $-6x\le -10$ or $\,-6x\ge -4$ subtract $\,7\,$ from both sides of both subsentences $\displaystyle x\ge\frac{10}{6}\ \ \text{or}\ \ x\le \frac{4}{6}$ divide by $\,-6\,$; change direction of inequality symbols $\displaystyle x\ge\frac{5}{3}\ \ \text{or}\ \ x\le \frac{2}{3}$ simplify fractions $\displaystyle x\le \frac{2}{3}\ \ \text{or}\ \ x\ge\frac{5}{3}$ in the web exercise, the ‘less than’ part is always reported first
Check the ‘boundaries’ of the solution set:
$3|-6(\frac{2}{3}) + 7| = 3|-4 + 7| = 3|3| = 9$
Check!
$3|-6(\frac{5}{3}) + 7| = 3|-10 + 7| = 3|-3| = 9$
Check!
## Example: An Absolute Value Equation that is Always False
Solve: $|3x + 1| = -5$
Solution: The theorem can't be used here, since $\,k\,$ is negative. In this case, you need to stop and think.
Can absolute value ever be negative? No! No matter what number you substitute for $\,x\,,$ the left-hand side of the equation will always be a number that is greater than or equal to zero. Therefore, this sentence has no solutions. It is always false.
## Example: An Absolute Value Inequality that is Always False
Solve: $|5 - 2x| \lt -3$
Solution: The theorem can't be used here, since $\,k\,$ is negative. In this case, you need to stop and think.
Can absolute value ever be negative? No! No matter what number you substitute for $\,x\,,$ the left-hand side of the inequality will always be a number that is greater than or equal to zero, so it can't possibly be less than $\,-3\,.$ Therefore, this sentence has no solutions. It is always false.
## Example: An Absolute Value Inequality that is Always True
Solve: $|5 - 2x| \gt -3$
Solution: The theorem can't be used here, since $\,k\,$ is negative. In this case, you need to stop and think.
No matter what number you substitute for $\,x\,,$ the left-hand side of the inequality will always be a number that is greater than or equal to zero, so it will always be greater than $\,-3\,.$ Therefore, this sentence has all real numbers as solutions. It is always true.
## Concept Practice
Solve the given absolute value sentence. Write the result in the most conventional way.
For more advanced students, a graph is available. For example, the inequality $\,|2 - 3x| \lt 7\,$ is optionally accompanied by the graph of $\,y = |2 - 3x|\,$ (the left side of the inequality, dashed green) and the graph of $\,y = 7\,$ (the right side of the inequality, solid purple). In this example, you are finding the values of $\,x\,$ where the green graph lies below the purple graph.
Click the ‘Show/Hide Graph’ button to toggle.
Solve: |
## 5. Ken borrows $285 from his parents to buy a bicycle. He will repay$90 to start, and then another $15 per week. A. Write an eq Question 5. Ken borrows$285 from his parents to buy a bicycle. He will repay $90 to start, and then another$15 per week.
A. Write an equation that can be used to determine w, the number of weeks it will take for
Ken to repay the entire amount.
B. What number of weeks will it take Ken to repay the entire amount?
in progress 0
2 weeks 2021-09-13T13:36:38+00:00 1 Answer 0
1. A. 15w+90=285 can be used to determine the number of weeks it will take for Ken to repay the entire amount.
B. It will take Ken 13 weeks to repay the entire amount.
Step-by-step explanation:
Money borrowed = $285 First repay =$90
Per week repay = \$15
Let,
w be the number of weeks, therefore,
A. Write an equation that can be used to determine w, the number of weeks it will take for Ken to repay the entire amount.
Money borrowed = First repay + Per week repay * number of weeks
285 = 90 + 15w
15w+90=285 Eqn 1
15w+90=285 can be used to determine the number of weeks it will take for Ken to repay the entire amount.
B. What number of weeks will it take Ken to repay the entire amount?
Solving Eqn 1: Subtracting 90 from both sides
Dividing both sides by 15
It will take Ken 13 weeks to repay the entire amount.
Keywords: linear equation, subtraction |
# How to Determine Limits Involving Floor and Absolute Value Functions
Determining limits involving floor functions (which map a real number to the greatest integer less than or equal to it) and absolute value functions (which measure the distance of a number from zero on the real number line) can present unique challenges due to their piecewise and non-continuous nature. Let's traverse through a step-by-step guide to evaluate such limits with precision.
## Step-by-step Guide to Determine Limits Involving Floor and Absolute Value Functions
Here is a step-by-step guide to determining limits involving floor and absolute value functions:
### Step 1: Understand the Behavior of Floor and Absolute Value Functions
The floor function, denoted by $$⌊x⌋$$, can cause abrupt changes in function value at integer points. Meanwhile, the absolute value function, $$∣x∣$$, introduces a kink in the graph at $$x=0$$, changing from $$x$$ to $$−x$$ as one crosses from positive to negative values. Acknowledge that these points of discontinuity and abrupt change are where we must scrutinize the limit closely.
### Step 2: Isolate the Floor and Absolute Value Components
Start by identifying the intervals on which the floor and absolute value functions are continuous and differentiable. This usually means dividing the domain of the function into segments based on integer values for the floor function and positive and negative regions for the absolute value function.
### Step 3: Piecewise Analysis
Break down the function into piecewise components. For the floor function, evaluate the limit at points just to the left and just to the right of each integer point. For the absolute value function, assess the behavior as the function approaches zero from both directions.
### Step 4: Apply Limit Definitions
Apply the definition of the limit to each piece. For the absolute value, remember that $$∣x∣=x$$ if $$x≥0$$ and $$∣x∣=−x$$ if $$x<0$$. For the floor function, consider the integer part and the fractional part of $$x$$ separately.
### Step 5: Consider One-Sided Limits
One-sided limits are essential when dealing with floor and absolute value functions. Evaluate $$lim_{x→c^−}$$ and $$lim_{x→c^+}$$ where $$c$$ is a point of interest (like an integer for the floor function or zero for the absolute value function). If these one-sided limits do not match, the two-sided limit does not exist.
### Step 6: Apply Limits to Each Piecewise Segment
Evaluate the limits for each segment individually. Ensure to consider the impact of the floor and absolute value operations within each interval. Use algebraic manipulation to simplify expressions where possible.
### Step 7: Reconcile Piecewise Limits
Once you have the one-sided limits, reconcile them to determine if a two-sided limit exists. If the limits from the left and right do not agree, the overall limit does not exist.
### Step 8: Check for Continuity and Differentiability
For locations where the floor or absolute value function changes, assess continuity and differentiability. The limit will exist at a point only if the function is continuous at that point.
### Step 9: Handling Combination of Functions
If the floor or absolute value function is combined with other functions, assess the impact on the limit separately. Sometimes it may be necessary to apply the limit laws in combination with the piecewise evaluation.
### Step 10: Visual Verification with Graphs
Graphing the function can provide a visual aid and verification for the analytical findings. The visual approach can be particularly helpful in identifying the behavior of the function around critical points.
## Final Word:
The evaluation of limits involving floor and absolute value functions hinges on recognizing and appropriately addressing the piecewise nature of these functions. By isolating these components and meticulously analyzing their behavior across the domain, the evaluation of such limits progresses from piecewise to holistic, resulting in a comprehensive understanding of the function’s limiting behavior.
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Mathematic
Dividing Fractions by Fractions
Dividing Fractions by Fractions
Dividing a fraction by a fraction might seem confusing at first, but it is really very simple.
To Divide Fractions:
• Invert (i.e. turn over) the denominator fraction and multiply the fractions
• Multiply the numerators of the fractions
• Multiply the denominators of the fractions
• Place the product of the numerators over the product of the denominators
• Simplify the Fraction
This is the quickest technique for dividing fractions. The top and bottom are being multiplied by the same number and, since that number is the reciprocal of the bottom part, the bottom becomes one. Dividing anything by one leaves the value “anything” the same.
Example: Divide 2/9 and 3/12
• Invert the denominator fraction and multiply (2/9 ÷ 3/12 = 2/9 * 12/3)
• Multiply the numerators (2*12=24)
• Multiply the denominators (9*3=27)
• Place the product of the numerators over the product of the denominators (24/27)
• Simplify the Fraction (24/27 = 8/9)
The Easy Way. There is a lengthier way of dividing fractions and it could be said to be a “more correct” method. It involves the use of equivalent fractions to make the denominators of both fractions the same. This is the correct way since, strictly speaking you can’t divide apples by oranges in the same way as you can’t divide fifths by fourths. After inverting, it is often simplest to “cancel” before doing the multiplication. Canceling is dividing one factor of the numerator and one factor of the denominator by the same number.
For example: 2/9 ÷ 3/12 = 2/9*12/3 = (2*12)/(9*3) = (2*4)/(3*3) = 8/9
Information Source: |
# College Algebra Tutorial 38
College Algebra
Tutorial 38: Zeros of Polynomial Functions, Part I:
Rational Zero Theorem and Descartes' Rule of Signs
Step 1: List all of the factors of the constant.
The factors of the constant term 10 are .
Step 2: List all of the factors of the leading coefficient.
The factors of the leading coefficient - 4 are .
Step 3: List all the POSSIBLE rational zeros or roots.
Writing the possible factors as we get:
Note, how when you reduce down the fractions, some of them are repeated.
Here is a final list of all the POSSIBLE rational zeros, each one written once and reduced:
In this problem it isn't asking for the zeros themselves, but what are the possible number of them. This can help narrow down your possibilities when you do go on to find the zeros.
Possible number of positive real zeros:
The up arrows are showing where there are sign changes between successive terms, going left to right. The first arrow on the left shows a sign change from negative 7 to positive 8. The 2nd arrow shows a sign change from positive 8 to negative 7. The third arrow shows a sign change from negative 7 to positive 9.
There are 3 sign changes between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 3, and then go down by even integers from that number until you get to 1 or 0.
Since we have 3 sign changes with f(x), then there are possibility of 3 or 3 - 2 = 1 positive real zeros.
Possible number of negative real zeros:
The up arrow is showing where there is a sign change between successive terms, going left to right. This arrow shows a sign change from negative 9 to positive 10.
There is only 1 sign change between successive terms, which means that is the highest possible number of negative real zeros. To find the other possible number of negative real zeros from these sign changes, you start with the number of changes, which in this case is 1, and then go down by even integers from that number until you get to 1 or 0.
If we went down by even integers form 1, we would be in the negative numbers, which is not a feasible answer, since we are looking for the possible number of negative real zeros. In other words, we can't have a -1 of them.
Therefore, there is exactly 1 negative real zero.
List all of the possible zeros:
The factors of the constant term 70 are .
The factors of the leading coefficient 1 are .
Writing the possible factors as we get:
Before we try any of these, let's apply Descartes' Rule of Signs to see if it can help narrow down our search for a rational zero.
Possible number of positive real zeros:
The up arrows are showing where there are sign changes between successive terms, going left to right.
There are 2 sign changes between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 2, and then go down by even integers from that number until you get to 1 or 0.
Since we have 2 sign changes with f(x), then there are possibility of 2 or 2 - 2 = 0 positive real zeros.
Possible number of negative real zeros:
There is only 1 sign change between successive terms, which means that is the highest possible number of negative real zeros. To find the other possible number of negative real zeros from these sign changes, you start with the number of changes, which in this case is 1, and then go down by even integers from that number until you get to 1 or 0.
If we went down by even integers form 1, we would be in the negative numbers, which is not a feasible answer, since we are looking for the possible number of negative real zeros. In other words, we can't have a -1 of them.
Therefore, there is exactly 1 negative real zero.
Use synthetic division to test the possible zeros and find an actual zero:
Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function.
If you need a review on synthetic division, feel free to go to Tutorial 37: Synthetic Division and the Remainder and Factor Theorems.
At this point you are wanting to pick any POSSIBLE rational root form the list of . Above, we found that there is exactly 1 negative rational zero. Since we know that there is one for sure, then we may want to go ahead and start with trying positive rational roots. I would suggest to start with smaller easier numbers and then go from there.
I'm going to choose -1 to try:
Since the reminder came out 48, this means f(-1) = 48, which means x = -1 is NOT a zero for this polynomial function.
That doesn't mean that we pack up our bags and quit. It's back to the drawing board.
We need to choose another number that comes from that same list of POSSIBLE rational roots.
This time I'm going to choose -2:
At last, we found a number that has a remainder of 0. This means that x = -2 is a zero or root of our polynomial function.
Use the actual zero to find all the zeros:
Since, x = -2 is a zero, that means x + 2 is a factor of our polynomial function.
Rewriting f(x) as (x + 2)(quotient) we get:
We need to finish this problem by setting this equal to zero and solving it:
*Factor the trinomial
*Set 1st factor = 0
*Set 2nd factor = 0
*Set 3rd factor = 0
The zeros of this function are x = -2, 5, and 7.
List all of the possible zeros:
The factors of the constant term -18 are .
The factors of the leading coefficient 5 are .
Writing the possible factors as we get:
Before we try any of these, let's apply Descartes' Rule of Signs to see if it can help narrow down our search for a rational zero.
Possible number of positive real zeros:
The up arrows are showing where there are sign changes between successive terms, going left to right.
There are 5 sign changes between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 5, and then go down by even integers from that number until you get to 1 or 0.
Since we have 5 sign changes with f(x), then there are possibility of 5, 5 - 2 = 3, or 5 - 4 = 1 positive real zeros.
Possible number of negative real zeros:
There are no sign changes.
This means there are no negative real zeros.
Use synthetic division to test the possible zeros and find an actual zero:
Recall that if you apply synthetic division and the remainder is 0, then c is a zero or root of the polynomial function.
If you need a review on synthetic division, feel free to go to Tutorial 37: Synthetic Division and the Remainder and Factor Theorems.
At this point you are wanting to pick any POSSIBLE rational root from the list of . Above, we found that there are no negative rational zero. So we do not need to try any negative numbers. I would suggest to start with smaller easier numbers and then go from there.
I'm going to choose 2 to try:
Bingo!!!! We found a number that has a remainder of 0. This means that x = 2 is a zero or root of our polynomial function.
Use the actual zero to find all the zeros:
Since, x = 2 is a zero, that means x - 2 is a factor of our polynomial function.
Rewriting f(x) as (x - 2)(quotient) we get:
We need to finish this problem by setting this equal to zero and solving it:
*Set 1st factor = 0
Looks like we can't factor this one. We are going to have to repeat this process again, but this time we will use this factor that we found.
Recall, that in Descartes's Rule of Signs we already found that there is no negative real zeros. So when we go trying again we can focus on finding another positive real zero.
Note that we can still pick from the same list of numbers as we did above or narrow the list down with the new constant. When we set up the synthetic division, we will just look at the remaining factor, to help us factor that down farther.
Note that the new constant term is 9. That means we can narrow down our list of
The factors of the constant term 9 are .
The factors of the leading coefficient 5 are .
Writing the possible factors as we get:
Note that we can still pick from the same list of numbers as we did above, since we are still looking at solving the same overall problem, but this list is a shorter one to pick from. Note how all of the numbers in this list also belong to the list above.
I'm going to choose 3 to try:
Bingo!!!! We found a number that has a remainder of 0. This means that x = 3 is a zero or root of our polynomial function.
Use the actual zero to find all the zeros:
Since, x = 3 is a zero, that means x - 3 is a factor of our polynomial function.
Rewriting f(x) as (x - 2)(x - 3)(quotient) we get:
We need to finish this problem by setting this equal to zero and solving it:
*Factor by grouping
*Set 1st factor = 0
*Set 2nd factor = 0
*Set 3rd factor = 0
*Set 4th factor = 0
The zeros of this function are x = 2, 3, i, -i, and 3/5. |
# Tangent Line: Definition, Formula & Newton’s Method
Calculus
Contents:
## 1. What is a Tangent Line?
A tangent line is a line that goes through a graph at only one point and is practically parallel to the graph at that point. It is the same as the instantaneous rate of change or the derivative at that point.
A tangent line (red dot).
If a line goes through a graph at a point but is not parallel, then it is not a tangent line. This image on the left shows a single tangent line (top left, marked by a red point). Although the line crosses the graph at the bottom right, it is not parallel to the graph at that point and therefore it is not a tangent line.
## Finding Tangent Lines
Finding tangent lines for straight graphs is a simple process, but with curved graphs it requires calculus in order to find the derivative of the function, which is the exact same thing as the slope of the tangent line. Once you have the slope of the tangent line, which will be a function of x, you can find the exact slope at specific points along the graph.
Keep in mind that f(x) is also equal to y, and that the slope-intercept formula for a line is y = mx + b where m is equal to the slope, and b is equal to the y intercept of the line. In order to find the tangent line at a point, you need to solve for the slope function of a secant line. You can find any secant line with the following formula:
(f(x + Δx) – f(x))/Δx or lim (f(x + h) – f(x))/h.
Sample problem: Find the tangent line at a point for f(x) = x2.
Note: If you have gone further into calculus, you will recognize the method used here as taking the derivative of the graph. If you know how to take derivatives of a function, you can skip ahead to the asterisked point (*) of step 2.
Step 1: Set up the limit formula. Substituting in the formula x2:
lim ((x + h)22 – x2)/h
h → 0
Step 2: Use algebra to solve the limit formula.
lim (x2 + 2xh + h2 – x2)/h
h → 0
lim (2xh + h2)/h
h → 0
lim h(2x + h)/h
h → 0
*lim 2x + h = 2x
h → 0
This gives the slope of any tangent line on the graph.
Step 3: Substitute in an x value to solve for the tangent line at the specific point.
At x = 2,
2(2) = 4.
That’s it!
## What is Newton’s Method?
Newton’s method (also called the Newton–Raphson method) is a way to find x-intercepts (roots) of functions. In other words, you want to know where the function crosses the x-axis. The method works well when you can’t use other methods to find zeros of functions, usually because you just don’t have all the information you need to use easier methods. While Newton’s method might look complex, all you’re actually doing is finding a tangent line…then another tangent line…and repeat, until you think you’re close enough to the actual solution.
The general form of Newton’s Method is:
If xn is an approximate solution of f(x) = 0,f'(x)n ≠ 0, then the next approximate solution is given by:
Where:
• xn is the x-intercept (initially, your best guess),
• fxn is the function you’re working with,
• f’xn is the derivative of the function you’re working with.
What’s basically happening with this equation is each iteration (subsequent calculation) takes you closer and closer to the true solution, as the following image shows. The true solution is marked in red. The blue line is the tangent line for x0 (Guess 1) and the green line is the tangent line for the second iteration. Eventually, your answers will get closer and closer (converge) to the red x-value.
## Newton’s Method: Step by Step Example
Example: Find the x-intercept for f(x) = cos x – x on the interval [0,2]
Step 1: Insert your function into the general form of the equation. The bottom half of the equation is the derivative. The derivative of cos x – x is -sin x – 1, so:
Step 2: Take a guess for the actual x-intercept and insert that guess into your function from Step 1. We’re given the interval [0,2] in the question, so a good guess to start with is 1 (because it’s in the middle of the interval).
Step 3: Evaluate the function from Step 2 using any calculator (I just used the Google calculator):
= 1 -( cos(1)-1) / ( – sin(1) – 1))
= 0.7503638679
Step 4: Repeat Steps 2 and 3, using the new guess for the x-intercept that you obtained from Step 3. This is the second iteration, x2.
0.7503638679 -( cos(0.7503638679 )-0.7503638679 ) / ( – sin(0.7503638679 ) – 1)) =
0.73911289091
Step 5: Repeat Steps 2 and 3 with your new estimate for as many times as you need. Eventually your answer will converge to a single number. If you compare x1(Step 2) with x2 (Step 3), you’ll notice that the first digit is “7” in both cases. That means we’ve got an accuracy of one decimal place. If you perform a third iteration, you’ll have an accuracy to three decimal places (0.739xxxxxxx). Repeat Steps for as many decimal places as you need.
Warning: A downside to Newton’s method; it will not work if you pick a point where the tangent line is horizontal.
## Related Articles
------------------------------------------------------------------------------
If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track. |
Factors of 759: Prime Factorization, Methods, and Examples
When 759 is divided by certain numbers, the residual is always zero. As a result, the factors of a particular number are the numbers that divide it at the end. An odd composite number is 759. It, therefore, only contains 8elements.
Below are further details about its contributing factors.
Factors of 759
Here are the factors of number 759.
Factors of 759: 1, 3, 11, 23, 33, 69, 253, 759
Negative Factors of 759
The negative factors of 759 are similar to their positive aspects, just with a negative sign.
Negative Factors of 759: -1, -3, -11, -23, -33, -69, -253, -759
Prime Factorization of 759
The prime factorization of 759 is the way of expressing its prime factors in the product form.
Prime Factorization: 31 x 111 x 231
In this article, we will learn about the factors of 759 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
What Are the Factors of 759?
The factors of 759 are 1, 3, 11, 23, 33, 69, 253, 759. These numbers are the factors as they do not leave any remainder when divided by 759.
The factors of 759 are classified as prime numbers and composite numbers. The prime factors of the number 759 can be determined using the prime factorization technique.
How To Find the Factors of 759?
You can find the factors of 759 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 759, create a list containing the numbers that are exactly divisible by 759 with zero remainders. One important thing to note is that 1 and 759 are the 759’s factors as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 759are determined as follows:
$\dfrac{759}{1} = 759$
$\dfrac{759}{3} = 253$
$\dfrac{759}{11} = 69$
$\dfrac{759}{23} = 23$
$\dfrac{759}{33} = 23$
$\dfrac{759}{69} = 11$
$\dfrac{759}{253} = 3$
$\dfrac{759}{759} = 1$
Therefore, 1, 3, 11, 23, 33, 69, 253, 759 are the factors of 759.
Total Number of Factors of 759
For 759, there are 8 positive factors and 8 negative ones. So in total, there are 16 factors of 759.
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 759 is given as:
Factorization of 759 is 31 x 111 x 231.
The exponent of 3, 11, and 23 is 1.
Adding 1 to each and multiplying them together results in 16.
Therefore, the total number of factors of 759 is 16. 8 are positive, and 8 factors are negative.
Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
Factors of 759 by Prime Factorization
The number 759 is a composite number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 759 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 759, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 759 can be expressed as:
759 = 31 x 111 x 231
Factors of 759 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 759, the factor pairs can be found as:
1 x 759 = 759
3 x 253= 759
11 x 69= 759
23 x 33= 759
The possible factor pairs of 759 are given as (1, 759)(3, 253 )(11, 69 ) and (23, 33).
All these numbers in pairs, when multiplied, give 759 as the product.
The negative factor pairs of 759 are given as:
-1 x -759 = 759
-3 x -253= 759
– 11 x -69= 759
-23 x -33= 759
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1, -3, -11, -23, -33, -69, -253, -759 are called negative factors of 759.
The list of all the factors of 759, including positive as well as negative numbers, is given below.
Factor list of 759:1,-1,3, -3, 11,-11, 23,-23, 33,-33, -69,69, 253,-253, 759, and -759
Factors of 759 Solved Examples
To better understand the concept of factors, let’s solve some examples.
Example 1
How many factors of 759 are there?
Solution
The total number of Factors of 759 is 16.
Factors of 759 are 1, 3, 11, 23, 33, 69, 253, 759.
Example 2
Find the factors of 759 using prime factorization.
Solution
The prime factorization of 759 is given as:
759 $\div$ 3 = 253
253 $\div$ 11 = 23
23 $\div$ 23 = 1
So the prime factorization of 759 can be written as:
31 x 111 x 231 = 759 |
1 / 38
# FUNCTIONS AND MODELS - PowerPoint PPT Presentation
1. FUNCTIONS AND MODELS. FUNCTIONS AND MODELS. 1.3 New Functions from Old Functions. In this section, we will learn: How to obtain new functions from old functions and how to combine pairs of functions. TRANSFORMATIONS OF FUNCTIONS. By applying certain transformations
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FUNCTIONS AND MODELS
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#### Presentation Transcript
1
FUNCTIONS AND MODELS
FUNCTIONS AND MODELS
1.3
New Functions from Old Functions
• In this section, we will learn:
• How to obtain new functions from old functions
• and how to combine pairs of functions.
TRANSFORMATIONS OF FUNCTIONS
• By applying certain transformations
• to the graph of a given function,
• we can obtain the graphs of certain
• related functions.
• This will give us the ability to sketch the graphs of many functions quickly by hand.
• It will also enable us to write equations for given graphs.
TRANSLATIONS
• If c is a positive number,
• then the graph of
• y = f(x) + c is just the
• graph of y = f(x) shifted
• c units upward.
• The graph of y = f(x - c) is
• just the graph of y = f(x)
• shifted c units to the right.
STRETCHING AND REFLECTING
• If c > 1, then the graph
• of y = cf(x) is the graph
• of y = f(x) stretched by
• a factor of c in the vertical
• direction.
• The graph of y = f(cx),
• compress the graph of
• y = f(x) horizontally by a
• factor of c.
• The graph of y = -f(x),
• reflect the graph of
• y = f(x) about the x-axis.
Figure 1.3.2, p. 38
TRANSFORMATIONS
• The figure illustrates these stretching
• transformations when applied to the cosine
• function with c = 2.
Figure 1.3.3, p. 38
TRANSFORMATIONS
• For instance, in order to get the graph of
• y = 2 cos x, we multiply the y-coordinate of
• each point on the graph of y = cos x by 2.
Figure 1.3.3, p. 38
TRANSFORMATIONS
Example
• This means that the graph of y = cos x
• gets stretched vertically by a factor of 2.
Figure 1.3.3, p. 38
TRANSFORMATIONS
Example 1
• Given the graph of , use
• transformations to graph:
• a.
• b.
• c.
• d.
• e.
Figure 1.3.4a, p. 39
Solution:
Example 1
• The graph of the square root function
is shown in part (a).
• by shifting 2 units downward.
• by shifting 2 units to the right.
• by reflecting about the x-axis.
• by stretching vertically by a factor of 2.
• by reflecting about the y-axis.
Figure 1.3.4, p. 39
TRANSFORMATIONS
Example 2
• Sketch the graph of the function
• f(x) = x2 + 6x + 10.
• Completing the square, we write the equation of the graph as: y = x2 + 6x + 10 = (x + 3)2 + 1.
Solution:
Example 2
• This means we obtain the desired graph by starting with the parabola y = x2 and shifting 3 units to the left and then 1 unit upward.
Figure 1.3.5, p. 39
TRANSFORMATIONS
Example 3
• Sketch the graphs of the following
• functions.
• a.
• b.
Solution:
Example 3 a
• We obtain the graph of y = sin 2x from that
• of y = sin x by compressing horizontally by
• a factor of 2.
• Thus, whereas the period of y = sin x is 2 , the period of y = sin 2x is 2 /2 = .
Figure 1.3.6, p. 39
Figure 1.3.7, p. 39
Solution:
Example 3 b
• To obtain the graph of y = 1 – sin x ,
• We reflect about the x-axis to get the graph ofy = -sin x.
• Then, we shift 1 unit upward to get y = 1 – sin x.
Figure 1.3.7, p. 39
TRANSFORMATIONS
• Another transformation of some interest
• is taking the absolute valueof a function.
• If y = |f(x)|, then, according to the definition of absolute value, y = f(x) when f(x) ≥ 0 and y = -f(x) when f(x) < 0.
• The part of the graph that lies above the x-axis remains the same.
• The part that lies below the x-axis is reflected about the x-axis.
TRANSFORMATIONS
Example 5
• Sketch the graph of the function
• First, we graph the parabola y = x2 - 1 by shifting the parabola y = x2 downward 1 unit.
• We see the graph lies below the x-axis when -1 < x < 1.
Figure 1.3.10a, p. 41
Solution:
Example 5
• So, we reflect that part of the graph about the x-axis to obtain the graph of y = |x2 - 1|.
Figure 1.3.10b, p. 41
Figure 1.3.10a, p. 41
COMBINATIONS OF FUNCTIONS
• Two functions f and g can be combined
• to form new functions f + g, f - g, fg, and
• in a manner similar to the way we add,
• subtract, multiply, and divide real numbers.
SUM AND DIFFERENCE
• The sum and difference functions are defined by:
• (f + g)x = f(x) + g(x) (f – g)x = f(x) – g(x)
• If the domain of f is A and the domain of g is B, then the domain of f + g is the intersection .
• This is because both f(x) and g(x) have to be defined.
SUM AND DIFFERENCE
• For example, the domain of
• is and the domain of
• is .
• So, the domain of
• is .
PRODUCT AND QUOTIENT
• Similarly, the product and quotient
• functions are defined by:
• The domain of fg is .
• However, we can’t divide by 0.
• So, the domain of f/g is
PRODUCT AND QUOTIENT
• For instance, if f(x) = x2 and g(x) = x - 1,
• then the domain of the rational function
• is ,
• or
COMBINATIONS
• There is another way of combining two functions : composition - the new function is composed of the two given functions f and g.
• For example, suppose that and
• Since y is a function of u and u is, in turn, a function of x, it follows that y is ultimately a function of x.
• We compute this by substitution:
COMPOSITION
Definition
• Given two functions f and g,
• the composite function
• (also called the composition of f and g)
• is defined by:
COMPOSITION
• The domain of is the set of all x
• in the domain of g such that g(x) is in
• the domain of f.
• In other words, is defined whenever both g(x) and f(g(x)) are defined.
COMPOSITION
• The domain of is the set of all x in the domain of g such that g(x) is in the domain of f.
• In other words, is defined whenever both g(x) and f(g(x)) are defined.
Figure 1.3.11, p. 41
COMPOSITION
Example 6
• If f(x) = x2 and g(x) = x - 3, find
• the composite functions and .
• We have:
COMPOSITION
Note
• You can see from Example 6 that,
• in general, .
• Remember, the notation means that, first,the function g is applied and, then, f is applied.
• In Example 6, is the function that firstsubtracts 3 and thensquares; is the function that firstsquares and thensubtracts 3.
COMPOSITION
Example 7
• If and ,
• find each function and its domain.
• a.
• b.
• c.
• d.
Solution:
Example 7 a
• The domain of is:
Solution:
Example 7 b
• For to be defined, we must have .
• For to be defined, we must have ,that is, , or .
• Thus, we have .
• So, the domain of is the closed interval [0, 4].
Solution:
Example 7 c
• The domain of is .
Solution:
Example 7 d
• This expression is defined when both and .
• The first inequality means .
• The second is equivalent to , or , or .
• Thus, , so the domain of is the closed interval [-2, 2].
COMPOSITION
• It is possible to take the composition
• of three or more functions.
• For instance, the composite function is found by first applying h, then g, and then fas follows:
COMPOSITION
Example 8
• Find if ,
• and .
• Solution:
COMPOSITION
Example 9
• Given , find functions
• f, g, and h such that .
• Since F(x) = [cos(x + 9)]2, the formula for F states: First add 9, then take the cosine of the result, and finally square.
• So, we let:
Solution:
Example 9
• Then, |
# FREE 6th Grade Georgia Milestones Assessment System Math Practice Test
Welcome to our FREE 6th Grade Georgia Milestones Assessment System Math Practice Test, with answer key and answer explanations. This practice test’s realistic format and high-quality practice questions can help your student succeed on the Georgia Milestones Assessment System Math test. Not only does the test closely match what students will see on the real Georgia Milestones Assessment System, but it also comes with detailed answer explanations.
For this practice test, we’ve selected 20 real questions from past exams for your student’s Georgia Milestones Assessment System Practice test. Your student will have the chance to try out the most common Georgia Milestones Assessment System Math questions. For every question, there is an in-depth explanation of how to solve the question and how to avoid mistakes next time.
Use our free Georgia Milestones Assessment System Math practice tests and study resources (updated for 2021) to help your students ace the Georgia Milestones Assessment System Math test! Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice.
## The Absolute Best Book to Ace 6th Grade Georgia Milestones Assessment System Math Test
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## 10 Sample 6th Grade Georgia Milestones Assessment System Math Practice Questions
1- There are 55 blue marbles and 143 red marbles. We want to place these marbles in some boxes so that there is the same number of red marbles in each box and the same number of blue marbles in each of the boxes. How many boxes do we need?
A. 8
B. 9
C. 10
D. 11
2- What is the value of the following expression?
$$2,205÷315$$
A. 5
B. 6
C. 7
D. 8
3- Solve the following equation.
$$112=22+x$$
A. $$x=-90$$
B. $$x=90$$
C. $$x=-134$$
D. $$x=134$$
4- Car A travels 221.5 km at a given time, while car B travels 1.2 times the distance car A travels at the same time. What is the distance car B travels during that time?
A. 222.7 km
B. 233.5 km
C. 241.5 km
D. 265.8 km
5- The perimeter of the trapezoid below is 38. What is its area?
A. 198 cm$$^2$$
B. 162 cm$$^2$$
C. 99 cm$$^2$$
D. 81cm$$^2$$
6- Which of the following expressions has the greatest value?
A. $$3^1+12$$
B. $$3^3-3^2$$
C. $$3^4-60$$
D. $$3^5-218$$
7- Alfred has $$x$$ apples. Alvin has 40 apples, which is 15 apples less than number of apples Alfred owns. If Baron has $$\frac{1}{5}$$ times as many apples as Alfred has. How many apples does Baron have?
A. 5
B. 11
C. 55
D. 275
8- In the following triangle find $$α$$.
A. $$100^\circ$$
B. $$90^\circ$$
C. $$60^\circ$$
D. $$30^\circ$$
9- The price of a laptop is decreased by $$15\%$$ to $425. What is its original price? A.$283
B. $430 C.$500
D. $550 10- Find the perimeter of shape in the following figure? (all angles are right angles) A. 21 B. 22 C. 24 D. 20 11- What are the values of mode and median in the following set of numbers? $$1,3,3,6,6,5,4,3,1,1,2$$ A. Mode: 1, 2, Median: 2 B. Mode: 1, 3, Median: 3 C. Mode: 2, 3, Median: 2 D. Mode: 1, 3, Median: 2.5 12- Which expression equivalent to $$x × 92$$? A. $$(x×90)+2$$ B. $$x×9×2$$ C. $$(x×90)+(x×2)$$ D. $$(x×90)+2$$ 13- The ratio of pens to pencils in a box is 3 to 5. If there are 96 pens and pencils in the box altogether, how many more pens should be put in the box to make the ratio of pens to pencils 1 : 1? A. 22 B. 23 C. 24 D. 25 14- If point A placed at $$-\frac{24}{3}$$ on a number line, which of the following points has a distance equal to 5 from point A? A. $$-13$$ B. $$-3$$ C. $$-2$$ D. A and B 15- Which of the following shows the numbers in increasing order? A. $$\frac{3}{13}, \frac{4}{11}, \frac{5}{14}, \frac{2}{5}$$ B. $$\frac{3}{13}, \frac{5}{14}, \frac{4}{11}, \frac{2}{5}$$ C. $$\frac{3}{13}, \frac{5}{14}, \frac{2}{5}, \frac{4}{11}$$ D. $$\frac{5}{14}, \frac{3}{13}, \frac{2}{5}, \frac{4}{11}$$ 16- If $$x=- 4$$, which of the following equations is true? A. $$x(3x-1)=50$$ B. $$5(11-x^2 )=-25$$ C. $$3(-2x+5)=49$$ D. $$x(-5x-19)=-3$$ 17- What is the missing prime factor of number 450? $$450=2^1×3^2×…$$ _________ 18- What is the perimeter of the following shape? (it’s a right triangle) A. 14 cm B. 18 cm C. 24 cm D. 32 cm 19- 65 is what percent of 50? A. $$50 \%$$ B. $$77 \%$$ C. $$130 \%$$ D. $$140 \%$$ 20- Which of the following expressions has a value of $$-23$$? A. $$-10+(-8)+ \frac{5}{2}×(-2)$$ B. $$5×3+(-2)×18$$ C. $$-10+6×8÷(-4)$$ D. $$(-3) × (-7) + 2$$ ## Best 6th Grade Georgia Milestones Assessment System Math Workbook Resource for 2021 Original price was:$15.99.Current price is: $10.99. Satisfied 81 Students Original price was:$18.99.Current price is: $13.99. Satisfied 222 Students ## Answers: 1- D First, we need to find the GCF (Greatest Common Factor) of 143 and 55. $$143=11×13$$ $$55=5×11→$$ GFC$$= 11$$ Therefore, we need 11 boxes. 2- C $$2205÷315=\frac{2205}{315}=\frac{441}{63}=\frac{147}{21}= 7$$ 3- B $$112=22+x$$ Subtract 22 from both sides of the equation. Then: $$x=112-22=90$$ 4- D Distance that car B travels $$=1.2 ×$$ distance that car A travels =$$1.2×221.5=265.8$$ km 5- D The perimeter of the trapezoid is 38. Therefore, the missing side (height) is $$= 38 – 8 – 10 – 11 = 9$$ Area of the trapezoid: $$A = \frac{1}{2} h (b_1 + b_2) = \frac{1}{2}1 (9) (8 + 10) = 81$$ 6- D A. $$3^1+12=3+12=15$$ B. $$3^3-3^2=27-9=18$$ C. $$3^4-60=81-60=21$$ D. $$3^5-218=243-218=25$$ 7- B Alfred has $$x$$ apple which is 15 apples more than number of apples Alvin owns. Therefore: $$x-15=40→x=40+15=55$$ Alfred has 55 apples. Let $$y$$ be the number of apples that Baron has. Then: $$y=\frac{1}{5}×55=11$$ 8- A Complementary angles add up to 180 degrees. $$β+150^\circ=180^\circ→β=180^\circ-150^\circ=30^\circ$$ The sum of all angles in a triangle is 180 degrees. Then: $$α+β+50^\circ=180^\circ→α+30^\circ+50^\circ=180^\circ$$ $$→α+80^\circ=180^\circ→α=180^\circ-80^\circ=100^\circ$$ 9- C Let $$x$$ be the original price. If the price of a laptop is decreased by $$15\%$$ to$425, then:
$$85 \% \space of \space x=425⇒ 0.85x=425 ⇒ x=425÷0.85=500$$
10- C
Let $$x$$ and $$y$$ be two sides of the shape. Then:
$$x+1=1+1+1→x=2$$
$$y+6+2=5+4→y+8=9→y=1$$
Then, the perimeter is:
$$1+5+1+4+1+2+1+6+2+1=24$$
11- B
First, put the numbers in order from least to greatest: $$1, 1, 1, 2, 3, 3, 3, 4, 5, 6, 6$$
The Mode of the set of numbers is: 1 and 3 (the most frequent numbers)
Median is: 3 (the number in the middle)
12- C
$$x×92=x×(90+2)=(x×90)+(x×2)$$
13- C
The ratio of pens to pencils is $$3 : 5$$. Therefore there are 3 pens out of all 8 pens and pencils. To find the answer, first dived 96 by 8 then multiply the result by 3.
$$96÷8=12→12×3=36$$
There are 36 pens and 60 pencils $$(96-36)$$. Therefore, 24 more pens should be put in the box to make the ratio $$1 : 1$$
14- D
If the value of point A is greater than the value of point B, then the distance of two points on the number line is: value of A- value of B
A. $$-\frac{24}{3}-(-13)=-8+13=5=5$$
B. $$-3-(-\frac{24}{3})=-3+8=5=5$$
C. $$-2-(-\frac{24}{3})=-2+8=6≠5$$
15- B
$$\frac{3}{13}≅0.23 , \frac{5}{14}≅0.357 , \frac{4}{11}≅0.36 , \frac{2}{5}=0.4$$
16- B
Plugin the value of $$x$$ in the equations. $$x = -4$$, then:
A.$$x(3x-1)=50→-4(3(-4)-1)=-4(-12-1)=-4(-13)=52≠50$$
B. $$5(11-x^2 )=-25→5(11-(-4)^2 )= 5(11-16)=5(-5)=-25$$
C. $$3(-2x+5)=49→3(-2(-4)+5)=3(8+5)=39≠49$$
D. $$x(-5x-19)=-3→-4(-5(-4)-19=-4(20-19)=-4≠-3$$
17- 5
Let $$x$$ be the missing prime factor of 450.
$$450= 2 × 3 × 3 × x ⇒ x =\frac{450}{18} ⇒ x = 25=5×5$$
18- C
Use Pythagorean theorem to find the hypotenuse of the triangle.
$$a^2+b^2=c^2→6^2+8^2=c^2→36+64=c^2→100=c^2→c=10$$
The perimeter of the triangle is: $$6+8+10=24$$
19- C
Use percent formula:
$$Part = \frac{percent}{100} × whole$$
$$65= \frac{percent}{100} × 50⇒ 65 = \frac{percent ×50}{100}⇒ 65=\frac{percent ×5}{10}$$
multiply both sides by 10.
$$650 =percent ×5, \space divide \space both \space sides \space by \space 5.$$
130 = percent
The answer is $$130\%$$
20- A
Let’s check the options provided.
A. $$-10+(-8)+ (\frac{5}{2})×(-2)=-10+(-8)+(-5)=-10-13=-23$$
B. $$5×3+(-2)×18=15+(-38)=-21$$
C. $$-10+6×8÷(-4)=-10+48÷(-4)=-10-12=-22$$
D. $$(-3)× (-7)+ 2=21+2=23$$
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Question Video: Completing a Proof Using the Triangle Midsegment Theorem | Nagwa Question Video: Completing a Proof Using the Triangle Midsegment Theorem | Nagwa
# Question Video: Completing a Proof Using the Triangle Midsegment Theorem Mathematics • First Year of Preparatory School
## Join Nagwa Classes
In the given figure, which of the following is true? [A] πΈ is the midpoint of line segment πΉπΊ. [B] πΉ is the midpoint of line segment π΄π·. [C] πΉπΊ = (1/2) π΄π΅ [D] πΆπ· = (1/2) π΄π΅
04:56
### Video Transcript
In the given figure, which of the following is true? Option (A) πΈ is the midpoint of line segment πΉπΊ. Option (B) πΉ is the midpoint of line segment π΄π·. Option (C) πΉπΊ equals one-half π΄π΅. Or option (D) πΆπ· equals one-half π΄π΅.
Letβs begin by having a look at the figure. We could observe that there are two pairs of congruent line segments, with the first pair being the line segments π΄πΈ and πΈπΆ. The second pair of congruent line segments are π΅πΊ and πΊπΆ. Therefore, we can note that points πΈ and πΊ must be the midpoints of the line segments π΄πΆ and π΅πΆ, respectively. And because we have these midpoints, we can apply the triangle midsegment theorem.
The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half its length. So, if we consider triangle π΄π΅πΆ, the line segment πΈπΊ connecting the midpoints of the two sides must be parallel to the third side, π΄π΅. And we were already given that line segments π΄π΅ and πΆπ· were parallel, so we have three parallel line segments.
Now, letβs consider which of the four statements are true, taking option (A) first. πΈ is the midpoint of line segment πΉπΊ. Now, we have established that πΈ is a midpoint. But itβs the midpoint of line segment π΄πΆ. Line segment πΉπΊ is this line segment in the middle of the figure. And we donβt have any information to prove that πΈ is the midpoint of line segment πΉπΊ. Therefore, we canβt say that option (A) is true.
Next, we can consider the statement in option (B). πΉ is the midpoint of line segment π΄π·. We can find point πΉ and line segment π΄π· on the left side of the diagram. If we consider this line segment as a side of triangle π΄πΆπ·, we can determine something about this line segment. Recalling the theorem that the line segment passing through the midpoint of one side of a triangle that is also parallel to another side of the triangle bisects the third side of the triangle, we can observe that the line segment πΈπΉ is a line segment passing through the midpoint of one side of the triangle and it is parallel to another side. Therefore, the third side, which is the line segment π΄π·, is bisected by line segment πΈπΉ. That means that point πΉ is the midpoint of line segment π΄π·. And so, the statement in option (B) is true.
However, itβs worth checking if either of options (C) or (D) is also a true statement. Option (C) says that πΉπΊ equals one-half π΄π΅. To see if this is true or not, we can look at triangle π΄π΅πΆ which is colored in green. By the first theorem here, since we have the midsegment πΈπΊ in this triangle, we know that πΈπΊ must be half the length of the parallel side π΄π΅. But the given statement doesnβt say that πΈπΊ is one-half π΄π΅; it says that πΉπΊ is. And we can see that the points πΉ and πΈ donβt lie on the same position. So, statement (C) is not true.
Finally, we can take the last statement that πΆπ· equals one-half π΄π΅. Line segment πΆπ· is here at the bottom of the figure. But we canβt apply any of the triangle midsegment theorems or any other theorems here to tell us that this is a true statement.
Therefore, the only statement that we can say is true is that in option (B): πΉ is the midpoint of line segment π΄π·.
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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 68 Maths Textbook Solution.
Answer: $\frac{1}{4}log\left ( 2e \right )$
Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{3}+x^{4}} d x$
Solution: $\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{3}+x^{4}} d x$
\begin{aligned} &=\int_{0}^{1} \frac{1}{x^{4}+2 x^{3}+2 x^{2}+2 x+1} d x \\ &=\int_{0}^{1} \frac{1}{x^{4}+x^{2}+x^{2}+2 x^{3}+2 x+1} d x \\ &=\int_{0}^{1} \frac{1}{x^{4}+x^{2}+2 x^{3}+x^{2}+2 x+1} d x \\ &=\int_{0}^{1} \frac{1}{x^{2}\left(x^{2}+1\right)+2 x\left(x^{2}+1\right)+\left(x^{2}+1\right)} d x \\ &=\int_{0}^{1} \frac{1}{\left(x^{2}+1\right)\left(x^{2}+2 x+1\right)} d x \\ &=\int_{0}^{1} \frac{1}{\left(x^{2}+1\right)(x+1)^{2}} d x \end{aligned}
$\left [ \left ( a+b \right )^{2}=a^{2}+2ab+b^{2} \right ]$
To solve this integral, first we have to find its partial fractions then integrate it by using indefinite integral formula then put the limits to get required answer.
So,
\begin{aligned} &\frac{1}{\left(x^{2}+1\right)(x+1)^{2}}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C x+D}{x^{2}+1} \\ &\Rightarrow 1=\frac{A(x+1)^{2}\left(x^{2}+1\right)}{x+1}+\frac{B(x+1)^{2}\left(x^{2}+1\right)}{(x+1)^{2}}+\frac{(C x+D)(x+1)^{2}\left(x^{2}+1\right)}{\left(x^{2}+1\right)} \\ &\Rightarrow 1=A(x+1)^{2}\left(x^{2}+1\right)+B\left(x^{2}+1\right)+(C x+D)(x+1)^{2} \\ &\Rightarrow 1=A\left(x^{3}+x+x^{2}+1\right)+B\left(x^{2}+1\right)+(C x+D)\left(x^{2}+2 x+1\right) \end{aligned}
$\left [ \left ( a+b \right )^{2}=a^{2}+b^{2}+2ab \right ]$
$\Rightarrow 1=A x^{3}+A x+A x^{2}+A+B x^{2}+B+C x^{3}+2 C x^{2}+C x+D x^{2}+2 D x+D$
Equating coefficient of and constant term respectively
$0=A+C$
$0=A+B+2C+D$
$0=A+C+2D$
$1=A+B+D$
From (a) and (c)
$A+C+2D=0$
$\Rightarrow 0+2D=0$
$\Rightarrow D=0$
Put the value of D in (b) and (d)
$A+B+2C=0$
$A+B=1$
Subtracting (e) – (f) then
$A+B+2C=0$
$\frac{A+B=1}{2C=-1}$
$\Rightarrow C=-\frac{1}{2}$
Then $\left ( a \right )=>A=-C=-\left ( -\frac{1}{2} \right )=\frac{1}{2}$
And (d)
\begin{aligned} &\Rightarrow A+B+D=1 \\ &\Rightarrow \frac{1}{2}+B+0=1 \\ &\Rightarrow B=1-\frac{1}{2}=\frac{1}{2} \\ &A=\frac{1}{2}, B=\frac{1}{2}, C=-\frac{1}{2} \end{aligned}
\begin{aligned} &\frac{1}{\left(x^{2}+1\right)(x+1)^{2}}=\frac{1}{2(x+1)}+\frac{1}{2(x+1)^{2}}+\frac{\frac{-1}{2} x+0}{x^{2}+1} \\ &=\frac{1}{2(x+1)}+\frac{1}{2(x+1)^{2}}-\frac{1}{2} \frac{x}{x^{2}+1} \end{aligned}
Now $\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{2}+x} d x=\int_{0}^{1}\left(\frac{1}{2(x+1)}+\frac{1}{2(x+1)^{2}}-\frac{1}{2} \frac{x}{x^{2}+1}\right) d x$
$=\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)} d x+\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)^{2}} d x-\frac{1}{2} \int_{0}^{1} \frac{x}{x^{2}} d x$ ..........(1)
Put $x+1=t\Rightarrow dx=dt$
When $x=0$ then $t=1$
And When $x=1$ then $t=2$
Then $\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)} d x+\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)^{2}} d x$
$=\frac{1}{2} \int_{1}^{2} \frac{1}{t} d t+\frac{1}{2} \int_{1}^{2} \frac{1}{t^{2}} d t=\frac{1}{2} \int_{1}^{2} \frac{1}{t} d t+\frac{1}{2} \int_{1}^{2} t^{-2} d t \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}, \int \frac{1}{x} d x=\log |x|\right]$
\begin{aligned} &=\frac{1}{2}[\log 2-\log 1]+\frac{1}{2}\left[\frac{t^{-1}}{-1}\right]_{1}^{2} \\ &=\frac{1}{2}[\log 2-\log 1]-\frac{1}{2}\left[\frac{1}{t}\right]_{1}^{2} \\ &=\frac{1}{2} \log 2-\frac{1}{2}\left[\frac{1}{2}-\frac{1}{1}\right] \\ &=\frac{1}{2} \log 2-\frac{1}{2}\left[\frac{1-2}{2}\right] \\ &=\frac{1}{2} \log 2-\frac{1}{2}\left[\frac{-1}{2}\right] \\ &=\frac{1}{2} \log 2+\frac{1}{4} \end{aligned}
And
$\frac{1}{2}\int_{0}^{1}\frac{x}{x^{2}+1}dx$
When $x=0$ then $p=1$ and
When $x=1$ then $p=2$
\begin{aligned} &\frac{1}{2} \int_{0}^{1} \frac{x}{x^{2}+1} d x=\frac{1}{2} \int_{1}^{2} \frac{1}{p} \frac{d p}{2} \\ &=\frac{1}{4} \int_{1}^{2} \frac{1}{p} d p \\ &=\frac{1}{4}[\log |p|]_{1}^{2} \\ &=\frac{1}{4}[\log 2-\log 1] \\ &=\frac{1}{4}[\log 2] \end{aligned} \quad\left[\begin{array}{l} \left.\frac{1}{x} d x=\log |x|\right] \\ \end{array}\right.$\left [ log1=0 \right ]$
Then From (1)
\begin{aligned} &\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{3}+x^{4}} d x=\frac{1}{2} \log 2+\frac{1}{4}-\frac{1}{4} \log 2 \\ &=\log 2\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{4} \\ &=\log 2\left(\frac{2-1}{4}\right)+\frac{1}{4} \\ &=\frac{1}{4} \log 2+\frac{1}{4} \\ &=\frac{1}{4} \log 2+\frac{1}{4} \text { loge }[\because \text { loge }=1] \\ &=\frac{1}{4} l(\operatorname{og} 2+\log e) \\ &=\frac{1}{4} \log (2 e) \quad[\because \log m+\log n=\log (m n)] \end{aligned} |
# Write the following decimals as fractions. Reduce the fractions to lowest form.(a) 0.6 (b) 2.5 (c) 1.0 (d) 3.8 (e) 13.7 (f) 21.2 (g) 6.4
To do:
We have to write the given decimals as fractions and reduce them to the lowest forms.
Solution:
Converting decimal to fraction:
To convert a Decimal to a Fraction follow these steps
Step 1: Write down the decimal divided by 1.
Step 2: Multiply both top and bottom by 10 for every number after the decimal
point. (For example, if there are two numbers after the decimal point, then use 100, if there are three then use 1000, etc.)
Step 3: Simplify (or reduce) the fraction.
Therefore,
(a) $0.6=\frac{0.6}{1}$
$\frac{0.6\times 10}{1 \times 10} = \frac{6}{10}$
$\frac{6}{10} = \frac{3}{5}$
(b) $2.5=\frac{2.5}{1}$
$\frac{2.5\times 10}{1 \times 10} = \frac{25}{10}$
$\frac{25}{10} = \frac{5}{2}$
(c) $1.0=\frac{1.0}{1}$
$\frac{1.0\times 10}{1 \times 10} = \frac{10}{10}$
$\frac{10}{10} = 1$
(d) $3.8=\frac{3.8}{1}$
$\frac{3.8\times 10}{1 \times 10} = \frac{38}{10}$
$\frac{38}{10} = \frac{19}{5}$
(e) $13.7=\frac{13.7}{1}$
$\frac{13.7\times 10}{1 \times 10} = \frac{137}{10}$
$\frac{137}{10}$ is in the lowest form.
(f) $21.2=\frac{21.2}{1}$
$\frac{21.2\times 10}{1 \times 10} = \frac{212}{10}$
$\frac{212}{10} = \frac{106}{5}$
(g) $6.4=\frac{6.4}{1}$
$\frac{6.4\times 10}{1 \times 10} = \frac{64}{10}$
$\frac{64}{10} = \frac{32}{5}$
Updated on: 10-Oct-2022
48 Views |
# Multiplying Whole Numbers
1.4 Multiplying Whole Numbers
Multiplication
Other terms for multiplication :
times, multiplied by, of.
Multiplication Properties of 0 and 1:
a ٠ 0 = 0 Example: 4 ٠ 0 = 0 + 0 + 0 + 0 = 0
a ٠ 1 = a Example: 4 ٠ 1 = 1 + 1 + 1 + 1 = 4
Multiplication by Multiples of 10
When one or more factors is a multiple of 10, such as 20, 80, or 300, you may
temporarily ignore such ending zeros and include them together in the final
product.
30·90 Count the ending zeros…two. Calculate the product of the
leading digits : 3·9 = 27. Attach the two zeros and the final
answer is 2700.
5,010·200 There are 3 ending zeroes, the zero between 5 and 1
doesn’t count because it’s in the middle. 501·2 = 1002, attach
the three zeros and the final answer is 1002000 = 1,002,000
Commutative Property: a ٠ b = b ٠ a
5 ٠ 4 = 4 ٠ 5
Associative Property: (a ٠ b) ٠ c = a ٠ (b ٠ c)
21 ٠ 6 = 21 ٠ (3 ٠ 2)= (21 ٠ 3) ٠ 2 = 63 ٠ 2 = 126
IMPORTANT!! The commutative and associative properties are
only in effect when an expression is either ALL addition or
ALL multiplication
.
Distributive Property :
a ٠ (b+ c) = a ٠ b + a ٠ c
2 ٠ (4+ 5) = 2 ٠ 4 + 2 ٠ 5
= 8 + 10
= 18
Relevance: If you don’t know a product through mental math , you can
break up one of the factors into a sum that will al low you to do
mental math.
Example 6 p.43
a) Multiply 6 ٠ 14
If you don’t now what 6 times 14 is , you can break up 14 into a sum of
10 and 4, then use the distributive property.
6 ٠ 14 =6 ٠ (10+ 4) = 6 ٠ 10 + 6 ٠ 4 = 60 + 24 = 84
b) Multiply 7 ٠ 59
Example: Mileage
Specifications for a Ford Explorer 4x4 are shown in the table below. How far can it
travel on a tank of gas?
Engine 4.0 L V6 Fuel Capacity 21 gal ← Number of gallons in 1 tank mpg means Miles per Gal = miles/gal → Fuel economy (mpg) 15 city/ 19 hwy
We are being asked to find “how far”, which is a distance. Distance is measured in
miles. We will assume this distance is traveled in the city so we’ll use city mileage
(15 mpg).
Dimensional analysis is used to convert units. We are given miles per gal and gallons
per tank. We want to know how many miles can be traveled on 1 tank.
Note: if you don’t know what 15 x 21 is doing mental math, you can do it using the
distributive property:
15 x 21 = 15(20+1) = 15(20) + 15(1) = 300+15=315
Example: Calculating production
The labor force of an electronics firm works two 8-hour shifts each day and
manufactures 53 TV sets each hour. Find how many sets will be manufactured in 5
days.
We are looking for the ration s-complex-numbers.html">number of TV sets manufactured in 5 days.
Use Dimensional Analysis. We are given TV sets manufactured per hour, and hours
worked per shift, and shifts worked per day. If you included all the pertinent
information, you should have cancelled out all the unnecessary units ( like units on top
cancel out like units on the bottom), and the units left should be “TV sets”, which is
what we want.
## Rectangular Patterns
If you have a rectangular pattern objects, such as rows and
columns, you can determine the total number of objects by
multiplying the number of rows times the number of columns (or
objects per row).
Example: Our classroom has 7 rows with 5 seats in each row.
Therefore our classroom can hold 7x5 = 35 people.
Area: The area of a rectangle is length X width= lw
The is the amount of area covered within the rectangle.
Note: Perimeter is the distance around the rectangle.
If length is in feet, and width is in feet,
Area = length X width means the units of the Area are feet2, or
square feet
p.49 #51
The parking lot at the Hillside Christmas Craft Fair has 14 rows of
parking with 12 parking spaces in each row. How many cars can
fit in the parking lot?
All these operation symbols mean the same thing: “10 divided by 2 is 5”
Division Properties:
Division is the “inverse” of multiplication. That is, does the opposite of what
multiplying does.
a(b) = c If a times b is c, then c/b answers the question, “c times what equals b?” a = c/b and c/a answers the question, b = c/a “c times what equals a?”
Division with 0 If a re presents any nonzero number, 0/a = 0 If a represents any nonzero number, a/0 is undefined. 0/0 is undetermined. More Division Properties a/1 = a a/a = 1 (provided that a ≠ 0)
Checking Division:
Quotient X Divisor + Remainder = Dividend
Translation: The answer to the division problem times the number you divided by, plus
the remainder, should equal the number you divided into.
What about 16/3?
Check: Quotient X Divisor + Remainder = Dividend 5 x 3 + 1 = 15+1=16 yes
Example
A total of 216 girls tried out for a city volleyball program. How many girls
should be put on the team roster if the following requirements must be
met?
1)All the teams must have the same number of players.
(find a number that goes exactly into 216, so there is no remainder)
2) A reasonable number of players on a team is 7 to 10 (divide 216 players by
7 players per team, then 8, then 9, then 10). But don’t bother with 10
because we know 10 doesn’t go exactly into 216.
3) There must be an even number of teams. (The quotient must be EVEN).
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# variable 389 results
variable - Alphabetic character representing a number
-65 times the difference between a number and 79 is equal to the number plus 98
-65 times the difference between a number and 79 is equal to the number plus 98 The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. The first expression, [I]the difference between a number and 79[/I] means we subtract 79 from our arbitrary variable of x: x - 79 Next, -65 times the difference between a number and 79 means we multiply our result above by -65: -65(x - 79) The phrase [I]the number[/I] refers to the arbitrary variable x earlier. The number plus 98 means we add 98 to x: x + 98 Now, let's bring it all together. The phrase [I]is equal to[/I] means an equation. So we set -65(x - 79) equal to x + [B]98: -65(x - 79) = x + 98[/B] <-- This is our algebraic expression If the problem asks you to take it a step further and solve for x, then you [URL='https://www.mathcelebrity.com/1unk.php?num=-65%28x-79%29%3Dx%2B98&pl=Solve']type this equation into our search engine[/URL], and you get: x = [B]76.31818[/B]
1 - n = n - 1
1 - n = n - 1 Solve for [I]n[/I] in the equation 1 - n = n - 1 [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables -n and n. To do that, we subtract n from both sides -n + 1 - n = n - 1 - n [SIZE=5][B]Step 2: Cancel n on the right side:[/B][/SIZE] -2n + 1 = -1 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 1 and -1. To do that, we subtract 1 from both sides -2n + 1 - 1 = -1 - 1 [SIZE=5][B]Step 4: Cancel 1 on the left side:[/B][/SIZE] -2n = -2 [SIZE=5][B]Step 5: Divide each side of the equation by -2[/B][/SIZE] -2n/-2 = -2/-2 n = [B]1 [URL='https://www.mathcelebrity.com/1unk.php?num=1-n%3Dn-1&pl=Solve']Source[/URL][/B]
1/2 of a number decreased by twice a number
1/2 of a number decreased by twice a number [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]1/2 of a number: x/2 [*]Twice a number means we multiply x by 2: 2x [*]The phrase [I]decreased by[/I] means we subtract [/LIST] [B]x/2 - 2x[/B]
1/3 a number increased by 10 times by that same number
1/3 a number increased by 10 times by that same number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 1/3 a number 1/3 * x = x/3 That same number means the same arbitrary variable as above: x 10 times that same number: 10x The phrase [I]increased by[/I] means we add: [B]x/3 + 10x [MEDIA=youtube]29TGt3i28jw[/MEDIA][/B]
1/3 of the sum of a number and 2 plus 5 is -20
1/3 of the sum of a number and 2 plus 5 is -20 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x the sum of a number and 2: x + 2 1/3 of the sum of a number and 2 1/3(x + 2) 1/3 of the sum of a number and 2 plus 5 1/3(x + 2) + 5 The phrase [I]is[/I] means equal to, so we set 1/3(x + 2) + 5 equal to -20: [B]1/3(x + 2) + 5 = -20[/B]
1/4 of the difference of 6 and a number is 200
1/4 of the difference of 6 and a number is 200 Take this [B]algebraic expression[/B] in 4 parts: [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x [*]The difference of 6 and a number means we subtract x from 6: 6 - x [*]1/4 of the difference means we divide 6 - x by 4: (6 - x)/4 [*]Finally, the phrase [I]is[/I] means an equation, so we set (6 - x)/4 equal to 200 [/LIST] [B](6 - x)/4 = 200[/B]
10 more than a number z, divided by k
10 more than a number z, divided by k The phrase [I]a number[/I] means an arbitrary variable, lets call it x. 10 more than a number means we add 10 to x: x + 10 We divide this quantity by k: [B](x + 10)/k[/B]
10 times a number is 420
10 times a number is 420 A number denotes an arbitrary variable, let's call it x. 10 times a number: 10x The phrase is means equal to, so we set 10x equal to 420 [B]10x = 420 <-- This is our algebraic expression [/B] If you want to solve for x, use our [URL='http://www.mathcelebrity.com/1unk.php?num=10x%3D420&pl=Solve']equation calculator[/URL] We get x = 42
12 is multiplied by some number, that product is reduced by 9, and the total is equal to 37
12 is multiplied by some number, that product is reduced by 9, and the total is equal to 37 The phrase [I]some number[/I] means an arbitrary variable, let's call it x. 12 multiplied by this number: 12x The product of 12x is reduced by 9 12x - 9 The phrase [I]the total is equal to[/I] means an equation, so we set 12x - 9 equal to 37: [B]12x - 9 = 37[/B]
12 plus 6 times a number is 9 times the number
12 plus 6 times a number is 9 times the number The phrase [I]a number [/I]means an arbitrary variable. Let's call it x. 6 times a number is written as: 6x 12 plus 6 times the number means we add 6x to 12: 12 + 6x 9 times a number is written as: 9x The phrase [I]is[/I] means an equation, so we set 12 + 6x equal to 9x [B]12 + 6x = 9x <-- This is our algebraic expression[/B] [B][/B] If the problem asks you to solve for x, then you [URL='https://www.mathcelebrity.com/1unk.php?num=12%2B6x%3D9x&pl=Solve']type this expression into our search engine[/URL] and you get: x = [B]4[/B]
12 plus the product of 4 and a number is greater than 72
A number means an arbitrary variable, let's call it x. The product of 4 and a number is 4x. 12 plus that product is 4x + 12 Is greater than means an inequality, so we set the entire expression greater than 72 4x + 12 > 72
14 increased by twice Carlos’s age. Use the variable c to represent Carlos age
14 increased by twice Carlos’s age. Use the variable c to represent Carlos age Twice means me multiply a by 2: 2a 14 increased by twice Carlos's age means we add 2a to 14: [B]14 + 2a[/B]
15 added to a number is 16 times the number
15 added to a number is 16 times the number [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]15 added to a number: 15 + x [*]16 times the number: 16x [*]The phrase [I]is[/I] means equal to. So we set 15 + x equal to 16x [/LIST] [B]15 + x = 16x[/B]
15 added to the quotient of 8 and a number is 7.
15 added to the quotient of 8 and a number is 7. Take this algebraic expression in pieces: [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. [*]The quotient of 8 and a number: 8/x [*]15 added to this quotient: 8/x + 15 [*]The word [I]is[/I] means an equation, so we set 8/x + 15 equal to 7 [/LIST] [B]8/x + 15 = 7[/B]
15 less than a number squared
15 less than a number squared A number is denoted by an arbitrary variable, let's call it x. x Squared means we raise that number to a power of 2 x^2 15 less means we subtract [B]x^2 -15[/B]
16 decreased by 3 times the sum of 3 and a number
16 decreased by 3 times the sum of 3 and a number Take this algebraic expression in parts: [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. [*]The sum of 3 and a number: 3 + x [*]3 times the sum: 3(3 + x) [*]16 decreased by... means we subtract 3(3 + x) from 16 [/LIST] [B]3(3 + x) from 16[/B]
19 increased by twice Greg’s score use the variable g to represent Greg’s score
19 increased by twice Greg’s score use the variable g to represent Greg’s score Use g for Greg's score g Twice g means we multiply g by 2: 2g 19 increased by means we add 2g to 19 [B]2g + 19 [MEDIA=youtube]E9a_U7z-fHE[/MEDIA][/B]
2 baseball players hit 60 home runs combined last season. The first player hit 3 more home runs than
2 baseball players hit 60 home runs combined last season. The first player hit 3 more home runs than twice a number of home runs the second player hit. how many home runs did each player hit? Declare variables: Let the first players home runs be a Let the second players home runs be b We're given two equations: [LIST=1] [*]a = 2b + 3 [*]a + b = 60 [/LIST] To solve this system of equations, we substitute equation (1) into equation (2) for a: 2b + 3 + b = 60 Using our math engine, we [URL='https://www.mathcelebrity.com/1unk.php?num=2b%2B3%2Bb%3D60&pl=Solve']type this equation[/URL] in and get: b = [B]19 [/B] To solve for a, we substitute b = 19 into equation (1): a = 2(19) + 3 a = 38 + 3 a = [B]41[/B]
2 less than half a number
A number means we pick an arbitrary variable, let's call it "x". Half a number is 1/2x. 2 less than that is [B]1/2x - 2[/B]
2 minus 7 times a number
A number is represented by an arbitrary variable, let's call it x. 7 times x means we multiply 7 times x. 7x 2 minus 7x is written: 2 - 7x
2 more than twice the sum of 10 and a number
2 more than twice the sum of 10 and a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The sum of 10 and a number means we add x to 10: 10 + x Twice the sum means we multiply 10 + x by 2: 2(10 + x) 2 more than twice the sum means we add 2 to 2(10 + x): [B]2(10 + x) + 2[/B]
2 numbers that are equal have a sum of 60
2 numbers that are equal have a sum of 60 Let's choose 2 arbitrary variables for the 2 numbers x, y Were given 2 equations: [LIST=1] [*]x = y <-- Because we have the phrase [I]that are equal[/I] [*]x + y = 60 [/LIST] Because x = y in equation (1), we can substitute equation (1) into equation (2) for x: y + y = 60 Add like terms to get: 2y = 60 Divide each side by 2: 2y/2 = 60/2 Cancel the 2's and we get: y = [B]30 [/B] Since x = y, x = y = 30 x = [B]30[/B]
2 times a number added to another number is 25. 3 times the first number minus the other number is 2
2 times a number added to another number is 25. 3 times the first number minus the other number is 20. Let the first number be x. Let the second number be y. We're given two equations: [LIST=1] [*]2x + y = 25 [*]3x - y = 20 [/LIST] Since we have matching opposite coefficients for y (1 and -1), we can add both equations together and eliminate a variable. (2 + 3)x + (1 - 1)y = 25 + 20 Simplifying, we get: 5x = 45 [URL='https://www.mathcelebrity.com/1unk.php?num=5x%3D45&pl=Solve']Typing this equation into the search engine[/URL], we get: [B]x = 9[/B] To find y, we plug in x = 9 into equation (1) or (2). Let's choose equation (1): 2(9) + y = 25 y + 18 = 25 [URL='https://www.mathcelebrity.com/1unk.php?num=y%2B18%3D25&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]y = 7[/B] So we have (x, y) = (9, 7) Let's check our work for equation (2) to make sure this system works: 3(9) - 7 ? 20 27 - 7 ? 20 20 = 20 <-- Good, we match!
2 times a number equals that number plus 5
2 times a number equals that number plus 5 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 2 times a number means we multiply 2 by x: 2x That number plus 5 means we add 5 to the number x x + 5 The phrase [I]equals[/I] means we set both expressions equal to each other [B]2x = x + 5[/B] <-- This is our algebraic expression If you want to take this further and solve this equation for x, [URL='https://www.mathcelebrity.com/1unk.php?num=2x%3Dx%2B5&pl=Solve']type this expression in the search engine[/URL] and we get: [B]x = 5[/B]
2 times a number subtracted by x
2 times a number subtracted by x The phrase [I]a number[/I] means an arbitrary variable, let's call it n. n 2 times a number means we multiply n by 2: 2n The phrase [I]subtracted by[/I] means we subtract 2n from x: [B]x - 2n[/B]
2 times half of a number
A number means an arbitrary variable, let's call it x. Half of x means we divide x by 2, or multiply by 0.5 x/2 2 times half x is written: [B]2(x/2)[/B] If we simplify by cancelling the 2's, we just get x.
2 times the sum of 1 and some number is 30. What is the number?
2 times the sum of 1 and some number is 30. What is the number? We let the phrase "some number" equal the variable x. The sum of 1 and some number is: x + 1 2 times the sum: 2(x + 1) The word "is" means equal to, so we set [B]2(x + 1) = 30[/B]
2 times the sum of 7 times a number and 4
2 times the sum of 7 times a number and 4 This is an algebraic expression. Let's take it in 4 parts: [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]7 times a number means we multiply x by 7: 7x [*]The sum of 7 times a number and 4 means we add 4 to 7x: 7x + 4 [*]Finally, we multiply the sum in #3 by 2 [/LIST] Build our final algebraic expression: [B]2(7x + 4)[/B]
2 times the sum of a number and 3 is equal to 3x plus 4
2 times the sum of a number and 3 is equal to 3x plus 4 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The sum of a number and 3 means we add 3 to x: x + 3 2 times this sum means we multiply the quantity x + 3 by 2 2(x + 3) 3x plus 4 means 3x + 4 since the word plus means we use a (+) sign 3x + 4 The phrase [I]is equal to[/I] means an equation, where we set 2(x + 3) equal to 3x + 4 [B]2(x + 3) = 3x + 4[/B]
2 Unknown Word Problems
Free 2 Unknown Word Problems Calculator - Solves a word problem based on two unknown variables
2/3 of a number 17 is at least 29
2/3 of a number 17 is at least 29 The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x 2/3 of a number means we multiply x by 2/3: 2x/3 The phrase [I]is at least[/I] also means greater than or equal to, so we set up the inequality: [B]2x/3 >= 29[/B]
2/5 the cube of a number
2/5 the cube of a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The cube of a number means we raise x to the power of 3: x^3 2/5 of the cube means we multiply x^3 by 2/5: [B](2x^3)/5[/B]
217 times u, reduced by 180 is the same as q
217 times u, reduced by 180 is the same as q. Take this algebraic expression pieces: Step 1: 217 times u We multiply the variable u by 217 217u Step 2: reduced by 180 Subtract 180 from 217u 217u - 180 The phrase [I]is the same as[/I] means an equation, so we set 217u - 180 equal to q [B]217u - 180 = q[/B]
26 increased by 12 times a number
26 increased by 12 times a number A number is represented by an arbitrary variable, let's call it x 12 times a number is written as 12x 26 increased by 12 times a number means we add: [B]26 + 12x[/B]
28 less than twice a number
[U]A number means an arbitrary variable, let's call it x.[/U] [LIST] [*]x [/LIST] [U]Twice a number means multiply by 2[/U] [LIST] [*]2x [/LIST] [U]28 less than twice a number means we subtract 28[/U] [LIST] [*][B]2x - 28[/B] [/LIST]
2n + 1 = n + 10
2n + 1 = n + 10 Solve for [I]n[/I] in the equation 2n + 1 = n + 10 [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables 2n and n. To do that, we subtract n from both sides 2n + 1 - n = n + 10 - n [SIZE=5][B]Step 2: Cancel n on the right side:[/B][/SIZE] n + 1 = 10 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 1 and 10. To do that, we subtract 1 from both sides n + 1 - 1 = 10 - 1 [SIZE=5][B]Step 4: Cancel 1 on the left side:[/B][/SIZE] n = [B]9[/B]
2n + 10 = 3n + 5
2n + 10 = 3n + 5 Solve for [I]n[/I] in the equation 2n + 10 = 3n + 5 [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables 2n and 3n. To do that, we subtract 3n from both sides 2n + 10 - 3n = 3n + 5 - 3n [SIZE=5][B]Step 2: Cancel 3n on the right side:[/B][/SIZE] -n + 10 = 5 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 10 and 5. To do that, we subtract 10 from both sides -n + 10 - 10 = 5 - 10 [SIZE=5][B]Step 4: Cancel 10 on the left side:[/B][/SIZE] -n = -5 [SIZE=5][B]Step 5: Divide each side of the equation by -1[/B][/SIZE] -1n/-1 = -5/-1 n = [B]5[/B]
2n = 4n
2n = 4n Solve for [I]n[/I] in the equation 2n = 4n [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables 2n and 4n. To do that, we subtract 4n from both sides 2n - 4n = 4n - 4n [SIZE=5][B]Step 2: Cancel 4n on the right side:[/B][/SIZE] -2n = 0 [SIZE=5][B]Step 3: Divide each side of the equation by -2[/B][/SIZE] -2n/-2 = 0/-2 n = [B]0[/B]
2x increased by 3 times a number
2x increased by 3 times a number The phrase [I]a number[/I] means an arbitary variable, let's call it x. 3 times a number means we multiply x by 3: 3x The phrase [I]increased by[/I] means we add 3x to 2x: 2x + 3x Simplifying, we get: [B]5x[/B]
3 decreased by 7 times a number
3 decreased by 7 times a number A number signifies an arbitrary variable, let's call it x. 7 times a number: 7x 3 decreased by this means we subtract 7x [B]3 - 7x[/B]
3 is subtracted from square of a number
3 is subtracted from square of a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Square of a number means we raise x to the 2nd power: x^2 3 is subtracted from square of a number [B]x^2 - 3[/B]
3 less than a number times itself
3 less than a number times itself The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Itself means the same variable as above. So we have: x * x x^2 3 less than this means we subtract 3 from x^2: [B]x^2 - 3[/B]
3 times a number increased by 1 is between -8 and 13
3 times a number increased by 1 is between -8 and 13. Let's take this algebraic expression in [U]4 parts[/U]: Part 1 - The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Part 2 - 3 times this number means we multiply x by 3: 3x Part 3 - Increased by 1 means we add 1 to 3x: 3x + 1 The phrase [I]between[/I] means we have an inequality: [B]-8 <= 3x + 1 <=13[/B]
3 times a number is 3 more a number
3 times a number is 3 more a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 3 times a number: 3x 3 more than a number means we add 3 to x: x + 3 The word [I]is[/I] means and equation, so we set 3x equal to x + 3 [B]3x = x + 3[/B]
30 increased by 3 times the square of a number
Let "a number" equal the arbitrary variable x. The square of that is x^2. 3 times the square of that is 3x^2. Now, 30 increased by means we add 3x^2 to 30 30 + 3x^2
30 increased by 3 times the square of a number
30 increased by 3 times the square of a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x x The square of a number means we raise x to the power of 2: x^2 3 times the square: 3x^2 The phrase [I]increased by[/I] means we add 3x^2 to 30: [B]30 + 3x^2[/B]
324 times z, reduced by 12 is z
324 times z, reduced by 12 is z. Take this algebraic expression in pieces: 324 [I]times[/I] z means we multiply 324 by the variable z. 324z [I]Reduced by[/I] 12 means we subtract 12 from 324z 324z - 12 The word [I]is[/I] means we have an equation, so we set 324z - 12 equal to z [B]324z - 12 = z [/B] <-- This is our algebraic expression
3abc^4/12a^3(b^3c^2)^2 * 8ab^-4c/4a^2b
3abc^4/12a^3(b^3c^2)^2 * 8ab^-4c/4a^2b Expand term 1: 3abc^4/12a^3(b^3c^2)^2 3abc^4/12a^3b^6c^4 Now simplify term 1: 3/12 = 1/4 c^4 terms cancel Subtract powers from variables since the denominator powers are higher: b^(6 - 1) = b^5 a^(3 - 1) = a^2 1/4a^2b^5 Now simplify term 2: 8ab^-4c/4a^2b 8/4 = 2 2c/a^(2 - 1)b^(1 - -4) 2c/ab^5 Now multiply simplified term 1 times simplified term 2: 1/4a^2b^5 * 2c/ab^5 (1 * 2c)/(4a^2b^5 * ab^5) 2c/4a^(2 + 1)b^(5 + 5) 2c/4a^3b^10 2/4 = 1/2, so we have: [B]c/2a^3b^10[/B]
3f,subtract g from the result, then divide what you have by h
3f,subtract g from the result, then divide what you have by h Take this algebraic expression in pieces: 3f subtract g means we subtract the variable g from the expression 3f: 3f - g Divide what we have by h, means we take the result above, 3f - g, and divide it by h: [B](3f - g)/h[/B]
4 times a number added to 8 times a number equals 36
4 times a number added to 8 times a number equals 36 Let [I]a number[/I] be an arbitrary variable, let us call it x. 4 times a number: 4x 8 times a number: 8x We add these together: 4x + 8x = 12x We set 12x equal to 36 to get our final algebraic expression of: [B]12x = 36 [/B] If the problem asks you to solve for x, you [URL='https://www.mathcelebrity.com/1unk.php?num=12x%3D36&pl=Solve']type this algebraic expression into our search engine[/URL] and get: x = [B]3[/B]
4 times a number cubed decreased by 7
4 times a number cubed decreased by 7 A number is denoted as an arbitrary variable, let's call it x x Cubed means raise x to the third power x^3 Decreased by 7 means subtract 7 x^3 - 7
4 times a number is the same as the number increased by 78
4 times a number is the same as the number increased by 78. Let's take this algebraic expression in parts: [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]4 times a number is written as 4x [*]The number increased by 78 means we add 78 to x: x + 78 [*]The phrase [I]the same as[/I] mean an equation, so we set #2 equal to #3 [/LIST] [B]4x = x + 78[/B] <-- This is our algebraic expression If the problem asks you to take it a step further, then [URL='https://www.mathcelebrity.com/1unk.php?num=4x%3Dx%2B78&pl=Solve']we type this equation into our search engine [/URL]and get: x = 26
4 times a number plus 9
A number means an arbitrary variable, let's call it "x". 4 times a number is 4x. Plus 9 means we add: 4x + 9
4 times b increased by 9 minus twice y
4 times b increased by 9 minus twice y Take this algebraic expression in parts: Step 1: 4 times b means we multiply the variable b by 4: 4b Step 2: Increased by 9 means we add 9 to 4b: 4b + 9 Step 3: Twice y means we multiply the variable y by 2: 2y Step 4: The phrase [I]minus[/I] means we subtract 2y from 4b + 9 [B]4b + 9 - 2y[/B]
4 times the difference of 6 times a number and 7
4 times the difference of 6 times a number and 7 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 6 times a number 6x The difference of 6x and 7 means we subtract 7 from 6x: 6x - 7 Now we multiply this difference by 4: [B]4(6x - 7)[/B]
4 times the quantity of a number plus 6
4 times the quantity of a number plus 6 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The word [I]plus[/I] means we addd 6 to x x + 6 The phrase [I]4 times the quantity [/I]means we multiply x + 6 by 4 [B]4(x + 6)[/B]
4n - 8 = n + 1
4n - 8 = n + 1 Solve for [I]n[/I] in the equation 4n - 8 = n + 1 [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables 4n and n. To do that, we subtract n from both sides 4n - 8 - n = n + 1 - n [SIZE=5][B]Step 2: Cancel n on the right side:[/B][/SIZE] 3n - 8 = 1 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants -8 and 1. To do that, we add 8 to both sides 3n - 8 + 8 = 1 + 8 [SIZE=5][B]Step 4: Cancel 8 on the left side:[/B][/SIZE] 3n = 9 [SIZE=5][B]Step 5: Divide each side of the equation by 3[/B][/SIZE] 3n/3 = 9/3 n = [B]3[/B]
4subtractedfrom6timesanumberis32
4 subtracted from 6 times a number is 32. Take this algebraic expression in pieces. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 6 times this number means we multiply by x by 6 6x 4 subtracted from this expression means we subtract 4 6x - 4 The phrase [I]is[/I] means an equation, so we set 6x - 4 equal to 32 [B]6x - 4 = 32 [/B] If you need to solve this equation, [URL='https://www.mathcelebrity.com/1unk.php?num=6x-4%3D32&pl=Solve']type it in the search engine here[/URL].
5 more than the reciprocal of a number
5 more than the reciprocal of a number Take this algebraic expression in pieces: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The reciprocal of this number means we divide 1 over x: 1/x 5 more means we add 5 to 1/x [B]1/x + 5[/B]
5 more than twice the cube of a number
5 more than twice the cube of a number. Take this algebraic expression in pieces. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The cube of a number means we raise it to a power of 3 x^3 Twice the cube of a number means we multiply x^3 by 2 2x^3 5 more than twice the cube of a number means we multiply 2x^3 by 5 5(2x^3) Simplifying, we get: 10x^3
5 more than twice the cube of a number
5 more than twice the cube of a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x The cube of a number means we raise x to the power of 3: x^3 Twice the cube means we multiply x^3 by 2 2x^3 Finally, 5 more than twice the cube means we add 5 to 2x^3: [B]2x^3 + 5[/B]
5 squared minus a number x
5 squared minus a number x 5 squared is written as 5^2 Minus a number x means we subtract the variable x [B]5^2 - x[/B]
5 subtracted from 3 times a number is 44
5 subtracted from 3 times a number is 44. The problem asks for an algebraic expression. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 3 times this number is 3x. 5 subtracted from this is written as 3x - 5. The phrase [I]is[/I] means an equation, so we set 3x - 5 equal to 44 [B]3x - 5 = 44[/B]
5 times a number increased by 13
5 times a number increased by 13 A number is denoted as an arbitrary variable, let's call it x x 5 times that number 5x Increased by 13 means we add 5x + 13
5 times a number increased by 4 is divided by 6 times the same number
5 times a number increased by 4 is divided by 6 times the same number Take this algebraic expression in parts. Part 1: 5 times a number increased by 4 [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x [*]5 times the number means multiply x by 5: 5x [*][I]Increased by 4[/I] means we add 4 to 5x: 5x + 4 [/LIST] Part 2: 6 times the same number [LIST] [*]From above, [I]a number[/I] is x: x [*]6 times the number means we multiply x by 6: 6x [/LIST] The phrase [I]is divided by[/I] means we have a quotient, where 5x + 4 is the numerator, and 6x is the denominator. [B](5x + 4)/6x[/B]
5 times a number is 4 more than twice a number
5 times a number is 4 more than twice a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 5 times a number: 5x Twice a number means we multiply x by 2: 2x 4 more than twice a number 2x + 4 The word [I]is[/I] means equal to, so we set 5x equal to 2x + 4 [B]5x = 2x + 4[/B]
5 times a number is that number minus 3
5 times a number is that number minus 3 The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. [LIST] [*]5 times a number: 5x [*]That number means we use the same number from above which is x [*]That number minus 3: x - 3 [*]The phrase [I]is[/I] means an equation, so we set 5x equal to x - 3 [/LIST] [B]5x = x - 3[/B]
5 times the product of 2 numbers a and b
5 times the product of 2 numbers a and b The product of 2 numbers a and be means we multiply the variables together: ab 5 times the product means we multiply ab by 5: [B]5ab[/B]
5 times the sum of 3 times a number and -5
5 times the sum of 3 times a number and -5 The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x 3 times a number means we multiply x by 3: 3x the sum of 3 times a number and -5 means we add -5 to 3x: 3x - 5 5 times the sum means we multiply 3x - 5 by 5: [B]5(3x - 5)[/B]
50 is more than the product of 4 and w
50 is more than the product of 4 and w Take this algebraic expression in pieces: The product of 4 and w mean we multiply the variable w by 4: 4w The phrase [I]is more than[/I] means an inequality using the (>) sign, where 50 is greater than 4w: [B]50 > 4w[/B]
51 decreased by twice a number
A number is denoted as an arbitrary variable, let's call it x. Twice a number means we multiply by 2, so 2x. 51 decreased by twice a number means we subtract 2x from 51 [B]51 - 2x [MEDIA=youtube]xqZzYvxmj5w[/MEDIA][/B]
54 is the sum of 24 and Julies score. Use the variable J to represent Julies score.
54 is the sum of 24 and Julies score. Use the variable J to represent Julies score. Sum of 24 and Julie's score: 24 + J The phrase [I]is[/I] means an equation, so we set 24 + J equal to 54 to get an algebraic expression: [B]24 + J = 54[/B]
56 is the sum of 20 and Donnie's savings. Use the variable d to represent Donnie's savings
56 is the sum of 20 and Donnie's savings. Use the variable d to represent Donnie's savings The sum of 20 and Donnie's savings using [I]d[/I] to represent Donnie's savings: 20 + d The word [I]is[/I] means equal to, so we set 20 + d equal to 56: [B]20 + d = 56[/B]
59 is the sum of 16 and Donnie's saving. Use the variable d to represent Donnie's saving.
59 is the sum of 16 and Donnie's saving. Use the variable d to represent Donnie's saving. The phrase [I]the sum of[/I] means we add Donnie's savings of d to 16: d + 16 The phrase [I]is[/I] means an equation, so we set d + 16 equal to 59 d + 16 = 59 <-- [B]This is our algebraic expression[/B] Now, if the problem asks you to solve for d, then you[URL='https://www.mathcelebrity.com/1unk.php?num=d%2B16%3D59&pl=Solve'] type the algebraic expression into our search engine to get[/URL]: d = [B]43[/B]
6 is divided by square of a number
6 is divided by square of a number The phrase [I]a number [/I]means an arbitrary variable, let's call it x. x the square of this means we raise x to the power of 2: x^2 Next, we divide 6 by x^2: [B]6/x^2[/B]
6 plus twice the sum of a number and 7.
6 plus twice the sum of a number and 7. The phrase [I]a number[/I] mean an arbitrary variable, let's call it x. The sum of a number and 7 means we add 7 to the variable x. x + 7 Twice the sum means we multiply the sum by 2: 2(x + 7) 6 plus means we add 6 to 2(x + 7) [B]6 + 2(x + 7)[/B]
6 subtracted from the product of 5 and a number is 68
6 subtracted from the product of 5 and a number is 68 Take this algebraic expression in parts. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The product of 5 and this number is: 5x We subtract 6 from 5x: 5x - 6 The phrase [I]is[/I] means an equation, so we set 5x - 6 equal to 68 [B]5x - 6 = 68[/B]
6 times a number multiplied by 3 all divided by 4
6 times a number multiplied by 3 all divided by 4 Take this algebraic expression in parts: [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x [*]6 times a number: 6x [*]Multiplied by 3: 3(6x) = 18x [*]All divided by 4: 18x/4 [/LIST] We can simplify this: We type 18/4 into our search engine, simplify, and we get 9/2. So our answer is: [B]9x/2[/B]
6 times j squared minus twice j squared
6 times j squared minus twice j squared j squared means we raise the variable j to the power of 2: j^2 6 times j squared means we multiply j^2 by 6: 6j^2 Twice j squared means we multiply j^2 by 2: 2j^2 The word [I]minus[/I] means we subtract 2j^2 from 6j^2 6j^2 - 2j^2 So if you must simplify, we group like terms and get: (6 - 2)j^2 [B]4j^2[/B]
6 times the reciprocal of a number equals 2 times the reciprocal of 7. What is the number
6 times the reciprocal of a number equals 2 times the reciprocal of 7. What is the number We've got two algebraic expressions here. Let's take it in parts: Term 1: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The reciprocal is 1/x Multiply this by 6: 6/x Term 2: Reciprocal of 7: 1/7 2 times this: 2/7 We set these terms equal to each other: 6/x = 2/7 [URL='https://www.mathcelebrity.com/prop.php?num1=6&num2=2&den1=x&den2=7&propsign=%3D&pl=Calculate+missing+proportion+value']Type this proportion into the search engine[/URL], and we get: [B]x = 21[/B]
6 times the reciprocal of a number equals 3 times the reciprocal of 7 .
6 times the reciprocal of a number equals 3 times the reciprocal of 7 . This is an algebraic expression. Let's take it in parts: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The reciprocal of a number x means we divide 1 over x: 1/x 6 times the reciprocal means we multiply 6 by 1/x: 6/x The reciprocal of 7 means we divide 1/7 1/7 3 times the reciprocal means we multiply 1/7 by 3: 3/7 Now, the phrase [I]equals[/I] mean an equation, so we set 6/x = 3/7 [B]6/x = 3/7[/B] <-- This is our algebraic expression If the problem asks you to solve for x, then [URL='https://www.mathcelebrity.com/prop.php?num1=6&num2=3&den1=x&den2=7&propsign=%3D&pl=Calculate+missing+proportion+value']we type this proportion in our search engine[/URL] and get: x = 14
6 times the sum of a number and 3 is equal to 42. What is this number?
6 times the sum of a number and 3 is equal to 42. What is this number? The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The sum of a number and 3 means we add 3 to x: x + 3 6 times the sum: 6(x + 3) The word [I]is[/I] means an equation, so we set 6(x + 3) equal to 42 to get our [I]algebraic expression[/I] of: [B]6(x + 3) = 42[/B] [B][/B] If the problem asks you to solve for x, then [URL='https://www.mathcelebrity.com/1unk.php?num=6%28x%2B3%29%3D42&pl=Solve']you type this equation into our search engine[/URL] and you get: x = [B]4[/B]
6 times the sum of a number and 5 is 16
6 times the sum of a number and 5 is 16 A number represents an arbitrary variable, let's call it x x The sum of x and 5 x + 5 6 times the sum of x and 5 6(x + 5) Is means equal to, so set 6(x + 5) equal to 16 [B]6(x + 5) = 16 <-- This is our algebraic expression Solve for x[/B] Multiply through: 6x + 30 = 16 Subtract 30 from each side: 6x - 30 + 30 = 16 - 30 6x = -14 Divide each side by 6 6x/6 = -14/6 Simplify this fraction by dividing top and bottom by 2: x = [B]-7/3 [MEDIA=youtube]oEx5dsYK7DY[/MEDIA][/B]
60 is the sum of 22 and Helenas height. Use the variable h to represent Helenas height.
60 is the sum of 22 and Helenas height. Use the variable h to represent Helenas height. If height is represented by h, we have: 22 and h 22 + h When they say "is the sum of", we set 22 + h equal to 60 [B]22 + h = 60[/B]
60 percent of a number minus 17 is -65
60 percent of a number minus 17 is -65 Using our [URL='https://www.mathcelebrity.com/perc.php?num=+5&den=+8&num1=+16&pct1=+80&pct2=+90&den1=+80&pct=60&pcheck=4&decimal=+65.236&astart=+12&aend=+20&wp1=20&wp2=30&pl=Calculate']percent to decimal calculator[/URL], we see that 60% is 0.6, so we have: 0.6 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. So 60% of a number is: 0.6x Minus 17: 0.6x - 17 The word [I]is[/I] means an equation, so we set 0.6x - 17 equal to -65 to get our algebraic expression of: [B]0.6x - 17 = -65[/B] [B][/B] If you want to solve for x in this equation, you [URL='https://www.mathcelebrity.com/1unk.php?num=0.6x-17%3D-65&pl=Solve']type it in our search engine and you get[/URL]: [B]x = -80[/B]
7 is 1/4 of some number
7 is 1/4 of some number The phrase [I]some number[/I] means an arbitrary variable, let's call it x. 1/4 of this is written as: x/4 The word [I]is[/I] means an equation, so we set x/4 equal to 7: [B]x/4 = 7[/B]
7 minus a number all divided by 4
7 minus a number all divided by 4 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 7 minus a number 7 - x All divided by 4: [B](7 - x)/4[/B]
7 plus the quantity of 9 increased by a number
7 plus the quantity of 9 increased by a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x 9 increased by a number means we add 9 to x 9 + x 7 plus this quantity means we add (9 + x) to 7 [B]7 + (9 + x)[/B]
7 times a number and 2 is equal to 4 times a number decreased by 8
7 times a number and 2 is equal to 4 times a number decreased by 8 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 7 times a number: 7x and 2 means we add 2: 7x + 2 4 times a number 4x decreased by 8 means we subtract 8: 4x - 8 The phrase [I]is equal to[/I] means an equation, so we set 7x + 2 equal to 4x - 8: [B]7x + 2 = 4x - 8[/B]
7 times a number increased by 4 times the number
7 times a number increased by 4 times the number Let [I]a number[/I] and [I]the number[/I] be an arbitrary variable. Let's call it x. We have an algebraic expression. Let's take it in pieces: [LIST] [*]7 times a number: 7x [*]4 times the number: 4x [*]The phrase [I]increased by[/I] means we add 4x to 7x: [*]7x + 4x [*]Simplifying, we get: (7 + 4)x [*][B]11x[/B] [/LIST]
7 times a number is the same as 12 more than 3 times a number
7 times a number is the same as 12 more than 3 times a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [B][U]Algebraic Expression 1:[/U][/B] 7 times a number means we multiply 7 by x: 7x [B][U]Algebraic Expression 2:[/U][/B] 3 times a number means we multiply 3 by x: 3x 12 more than 3 times a number means we add 12 to 3x: 3x + 12 The phrase [I]is the same as[/I] means an equation, so we set 7x equal to 3x + 12 [B]7x = 3x + 12[/B] <-- Algebraic Expression
7 times the quantity of 3 times a number reduced by 10
7 times the quantity of 3 times a number reduced by 10 The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. x 3 times a number: 3x Reduced by 10 means we subtract 10: 3x - 10 7 times this quantity: [B]7(3x - 10)[/B]
76 decreased by twice a number. Use the variable n to represent the unknown number
76 decreased by twice a number. Use the variable n to represent the unknown number. Twice a number (n) means we multiply the unknown number n by 2: 2n 76 decreased by twice a number means we subtract 2n from 76 using the (-) operator [B]76 - 2n[/B]
76 subtracted from p is equal to the total of g and 227
76 subtracted from p is equal to the total of g and 227 We've got two algebraic expressions. Take them in pieces: Part 1: 76 subtracted from p We subtract 76 from the variable p p - 76 Part 2: The total of g and 227 The total means a sum, so we add 227 to g g + 227 Now the last piece, the phrase [I]is equal to[/I] means an equation. So we set both algebraic expressions equal to each other: [B]p - 76 = g + 227[/B]
8 increased by the product of a number and 7 is greater than or equal to -18
Take this in parts: First, the phrase, "a number" means we pick an arbitrary variable, let's call it x. The product of a number and 7 is 7x. 8 increased by the product of 7x means we add them together. 7x + 8 Finally that entire expression is greater than [U]or equal to[/U] -18 [B]7x + 8 >=-18[/B]
8 is subtracted from thrice a number
Thrice a number means we multiply by 3. A number means an arbitrary variable, let's call it x 3x 8 is subtracted from 3x [B]3x - 8[/B]
8 is subtracted from twice a number
Twice a number: [LIST] [*]Choose an arbitrary variable, let's call it x [*]Twice x means multiply by 2 [*]2x [/LIST] 8 subtracted from 2x: [B]2x - 8[/B]
8 more than twice a number is less than 6 more than the number
8 more than twice a number is less than 6 more than the number. This is an algebraic expression, let's take it in pieces... The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 8 more than twice a number: Twice a number means multiply x by 2: 2x Then add 8: 2x + 8 6 more than the number, means we add 6 to x x + 6 The phrase [I]is less than[/I] means an inequality, where we set 2x + 8 less than x + 6 [B]2x + 8 < x + 6[/B]
8 times the sum of 5 times a number and 9
8 times the sum of 5 times a number and 9 Take this algebraic expression in parts: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 5 times a number means: 5x The sum of this and 9 means we add 9 to 5x: 5x + 9 Now we multiply 8 times this sum: [B]8(5x + 9)[/B]
9 is the sum of 7 and twice a number
9 is the sum of 7 and twice a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Twice a number means we multiply x by 2: 2x The sum of 7 and twice a number 7 + 2x The word [I]is[/I] mean equal to, so we set 7 + 2x equal to 9: [B]7 + 2x = 9[/B]
9 less than 5 times a number is 3 more than 2x
9 less than 5 times a number is 3 more than 2x The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x 5 times a number means we multiply x by 5: 5x 9 less than 5x means we subtract 9 from 5x: 5x - 9 3 more than 2x means we add 3 to 2x: 2x + 3 The word [I]is[/I] means an equation, so we set 5x - 9 equal to 2x + 3: [B]5x - 9 = 2x + 3 <-- This is our algebraic expression[/B] [B][/B] If you want to solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=5x-9%3D2x%2B3&pl=Solve']type this equation into the search engine[/URL], and we get: x = [B]4[/B]
9 subtracted from the product of 3 and a number is greater than or equal to 16
9 subtracted from the product of 3 and a number is greater than or equal to 16 [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x [*]The product of 3 and a number means we multiply 3 times x: 3x [*]9 subtracted from the product: 3x - 9 [*]The phrase is greater than or equal to means an inequality. So we set up an inequality with >= for the greater than or equal to sign in relation to 3x - 9 and 16 [/LIST] Our algebraic expression (inequality) becomes: [B]3x - 19 >= 16[/B]
9 times a number is that number minus 10
9 times a number is that number minus 10 The phrase [I]a number[/I] means we define a random/arbitrary variable, let's call it x: x 9 times a number means we multiply x by 9: 9x The phrase [I]that number[/I] refers back to the original arbitrary variable we defined above, which is x: x That number minus 10 means we subtract 10 from x: x - 10 The word [I]is[/I] means equal to, so we set 9x equal to x - 10 [B]9x = x - 10[/B]
9 times a number is that number minus 3
9 times a number is that number minus 3 Let [I]a number[/I] be an arbitrary variable, let's call it x. We're given: 9 times a number is 9x The number minus 3 is x - 3 The word [I]is[/I] means an equation, so we set 9x equal to x - 3 to get our [I]algebraic expression[/I]: [B]9x = x - 3[/B] To solve for x, we type this equation into our search engine and we get: x = [B]-0.375 or -3/8[/B]
A 98-inch piece of wire must be cut into two pieces. One piece must be 10 inches shorter than the ot
A 98-inch piece of wire must be cut into two pieces. One piece must be 10 inches shorter than the other. How long should the pieces be? The key phrase in this problem is [B]two pieces[/B]. Declare Variables: [LIST] [*]Let the short piece length be s [*]Let the long piece length be l [/LIST] We're given the following [LIST=1] [*]s = l - 10 [*]s + l = 98 (Because the two pieces add up to 98) [/LIST] Substitute equation (1) into equation (2) for s: l - 10+ l = 98 Group like terms: 2l - 10 = 98 Solve for [I]l[/I] in the equation 2l - 10 = 98 [SIZE=5][B]Step 1: Group constants:[/B][/SIZE] We need to group our constants -10 and 98. To do that, we add 10 to both sides 2l - 10 + 10 = 98 + 10 [SIZE=5][B]Step 2: Cancel 10 on the left side:[/B][/SIZE] 2l = 108 [SIZE=5][B]Step 3: Divide each side of the equation by 2[/B][/SIZE] 2l/2 = 108/2 l = [B]54[/B] To solve for s, we substitute l = 54 into equation (1): s = 54 - 10 s = [B]44[/B] Check our work: The shorter piece is 10 inches shorter than the longer piece since 54 - 44 = 10 Second check: Do both pieces add up to 98 54 + 44 ? 98 98 = 98
A bakery has a fixed cost of \$119.75 per a day plus \$2.25 for each pastry. The bakery would like to
A bakery has a fixed cost of \$119.75 per a day plus \$2.25 for each pastry. The bakery would like to keep its daily costs at or below \$500 per day. Which inequality shows the maximum number of pastries, p, that can be baked each day. Set up the cost function C(p), where p is the number of pastries: C(p) = Variable Cost + Fixed Cost C(p) = 2.25p + 119.75 The problem asks for C(p) at or below \$500 per day. The phrase [I]at or below[/I] means less than or equal to (<=). [B]2.25p + 119.75 <= 500[/B]
a bicycle store costs \$3600 per month to operate. The store pays an average of \$60 per bike. the ave
a bicycle store costs \$3600 per month to operate. The store pays an average of \$60 per bike. the average selling price of each bicycle is \$100. how many bicycles must the store sell each month to break even? Cost function C(b) where b is the number of bikes: C(b) = Variable Cost + Fixed Cost C(b) = Cost per bike * b + operating cost C(b) = 60b + 3600 Revenue function R(b) where b is the number of bikes: R(b) = Sale price * b R(b) = 100b Break Even is when Cost equals Revenue, so we set C(b) = R(b): 60b + 3600 = 100b To solve this equation for b, we [URL='https://www.mathcelebrity.com/1unk.php?num=60b%2B3600%3D100b&pl=Solve']type it in our math engine[/URL] and we get: b = [B]90[/B]
A book publishing company has fixed costs of \$180,000 and a variable cost of \$25 per book. The books
A book publishing company has fixed costs of \$180,000 and a variable cost of \$25 per book. The books they make sell for \$40 each. [B][U]Set up Cost Function C(b) where b is the number of books:[/U][/B] C(b) = Fixed Cost + Variable Cost x Number of Units C(b) = 180,000 + 25(b) [B]Set up Revenue Function R(b):[/B] R(b) = 40b Set them equal to each other 180,000 + 25b = 40b Subtract 25b from each side: 15b = 180,000 Divide each side by 15 [B]b = 12,000 for break even[/B]
A company makes toy boats. Their monthly fixed costs are \$1500. The variable costs are \$50 per boat.
A company makes toy boats. Their monthly fixed costs are \$1500. The variable costs are \$50 per boat. They sell boats for \$75 a piece. How many boats must be sold each month to break even? [U]Set up Cost function C(b) where t is the number of tapestries:[/U] C(b) = Cost per boat * number of boats + Fixed Cost C(b) = 50b + 1500 [U]Set up Revenue function R(b) where t is the number of tapestries:[/U] R(b) = Sale Price * number of boats R(b) = 75b [U]Break even is where Revenue equals Cost, or Revenue minus Cost is 0, so we have:[/U] R(b) - C(b) = 0 75b - (50b + 1500) = 0 75b - 50b - 1500 = 0 25b - 1500 = 0 To solve for b, we [URL='https://www.mathcelebrity.com/1unk.php?num=25b-1500%3D0&pl=Solve']type this equation in our math engine[/URL] and we get: b = [B]60[/B]
A company that manufactures lamps has a fixed monthly cost of \$1800. It costs \$90 to produce each l
A company that manufactures lamps has a fixed monthly cost of \$1800. It costs \$90 to produce each lamp, and the selling price is \$150 per lamp. Set up the Cost Equation C(l) where l is the price of each lamp: C(l) = Variable Cost x l + Fixed Cost C(l) = 90l + 1800 Determine the revenue function R(l) R(l) = 150l Determine the profit function P(l) Profit = Revenue - Cost P(l) = 150l - (90l + 1800) P(l) = 150l - 90l - 1800 [B]P(l) = 60l - 1800[/B] Determine the break even point: Breakeven --> R(l) = C(l) 150l = 90l + 1800 [URL='https://www.mathcelebrity.com/1unk.php?num=150l%3D90l%2B1800&pl=Solve']Type this into the search engine[/URL], and we get [B]l = 30[/B]
A corn refining company produces corn gluten cattle feed at a variable cost of \$84 per ton. If fixe
A corn refining company produces corn gluten cattle feed at a variable cost of \$84 per ton. If fixed costs are \$110,000 per month and the feed sells for \$132 per ton, how many tons should be sold each month to have a monthly profit of \$560,000? [U]Set up the cost function C(t) where t is the number of tons of cattle feed:[/U] C(t) = Variable Cost * t + Fixed Costs C(t) = 84t + 110000 [U]Set up the revenue function R(t) where t is the number of tons of cattle feed:[/U] R(t) = Sale Price * t R(t) = 132t [U]Set up the profit function P(t) where t is the number of tons of cattle feed:[/U] P(t) = R(t) - C(t) P(t) = 132t - (84t + 110000) P(t) = 132t - 84t - 110000 P(t) = 48t - 110000 [U]The question asks for how many tons (t) need to be sold each month to have a monthly profit of 560,000. So we set P(t) = 560000:[/U] 48t - 110000 = 560000 [U]To solve for t, we [URL='https://www.mathcelebrity.com/1unk.php?num=48t-110000%3D560000&pl=Solve']type this equation into our search engine[/URL] and we get:[/U] t =[B] 13,958.33 If the problem asks for whole numbers, we round up one ton to get 13,959[/B]
A first number plus twice a second number is 10. Twice the first number plus the second totals 35. F
A first number plus twice a second number is 10. Twice the first number plus the second totals 35. Find the numbers. [U]The phrase [I]a number[/I] means an arbitrary variable[/U] A first number is written as x A second number is written as y [U]Twice a second number means we multiply y by 2:[/U] 2y [U]A first number plus twice a second number:[/U] x + 2y [U]A first number plus twice a second number is 10 means we set x + 2y equal to 10:[/U] x + 2y = 10 [U]Twice the first number means we multiply x by 2:[/U] 2x [U]Twice the first number plus the second:[/U] 2x + y [U]Twice the first number plus the second totals 35 means we set 2x + y equal to 35:[/U] 2x + y = 35 Therefore, we have a system of two equations: [LIST=1] [*]x + 2y = 10 [*]2x + y = 35 [/LIST] Since we have an easy multiple of 2 for the x variable, we can solve this by multiply the first equation by -2: [LIST=1] [*]-2x - 4y = -20 [*]2x + y = 35 [/LIST] Because the x variables are opposites, we can add both equations together: (-2 + 2)x + (-4 + 1)y = -20 + 35 The x terms cancel, so we have: -3y = 15 To solve this equation for y, we [URL='https://www.mathcelebrity.com/1unk.php?num=-3y%3D15&pl=Solve']type it in our search engine[/URL] and we get: y = [B]-5 [/B] Now we substitute this y = -5 into equation 2: 2x - 5 = 35 To solve this equation for x, we[URL='https://www.mathcelebrity.com/1unk.php?num=2x-5%3D35&pl=Solve'] type it in our search engine[/URL] and we get: x = [B]20[/B]
A gym membership has a \$50 joining fee plus charges \$17 a month for m months
A gym membership has a \$50 joining fee plus charges \$17 a month for m months Build a cost equation C(m) where m is the number of months of membership. C(m) = Variable Cost * variable units + Fixed Cost C(m) = Months of membership * m + Joining Fee Plugging in our numbers and we get: [B]C(m) = 17m + 50 [MEDIA=youtube]VGXeqd3ikAI[/MEDIA][/B]
A half is a third of it. What is it?
A half is a third of it. What is it? Let "it" be an arbitrary variable, let's call it x. A third of 'it' means we divide x/3 The word is means equals, so we have: x/3 = 1/2 Using our [URL='https://www.mathcelebrity.com/proportion-calculator.php?num1=x&num2=1&den1=3&den2=2&propsign=%3D&pl=Calculate+missing+proportion+value']proportion calculator[/URL], we get: x = [B]1.5 or 3/2 or 1 & 1/2 [MEDIA=youtube]cCXyryWpHV0[/MEDIA][/B]
A local Dunkin’ Donuts shop reported that its sales have increased exactly 16% per year for the last
A local Dunkin’ Donuts shop reported that its sales have increased exactly 16% per year for the last 2 years. This year’s sales were \$80,642. What were Dunkin' Donuts' sales 2 years ago? Declare variable and convert numbers: [LIST] [*]16% = 0.16 [*]let the sales 2 years ago be s. [/LIST] s(1 + 0.16)(1 + 0.16) = 80,642 s(1.16)(1.16) = 80,642 1.3456s = 80642 Solve for [I]s[/I] in the equation 1.3456s = 80642 [SIZE=5][B]Step 1: Divide each side of the equation by 1.3456[/B][/SIZE] 1.3456s/1.3456 = 80642/1.3456 s = 59930.142687277 s = [B]59,930.14[/B]
A man purchased 20 tickets for a total of \$225. The tickets cost \$15 for adults and \$10 for children
A man purchased 20 tickets for a total of \$225. The tickets cost \$15 for adults and \$10 for children. What was the cost of each ticket? Declare variables: [LIST] [*]Let a be the number of adult's tickets [*]Let c be the number of children's tickets [/LIST] Cost = Price * Quantity We're given two equations: [LIST=1] [*]a + c = 20 [*]15a + 10c = 225 [/LIST] Rearrange equation (1) in terms of a: [LIST=1] [*]a = 20 - c [*]15a + 10c = 225 [/LIST] Now that I have equation (1) in terms of a, we can substitute into equation (2) for a: 15(20 - c) + 10c = 225 Solve for [I]c[/I] in the equation 15(20 - c) + 10c = 225 We first need to simplify the expression removing parentheses Simplify 15(20 - c): Distribute the 15 to each term in (20-c) 15 * 20 = (15 * 20) = 300 15 * -c = (15 * -1)c = -15c Our Total expanded term is 300-15c Our updated term to work with is 300 - 15c + 10c = 225 We first need to simplify the expression removing parentheses Our updated term to work with is 300 - 15c + 10c = 225 [SIZE=5][B]Step 1: Group the c terms on the left hand side:[/B][/SIZE] (-15 + 10)c = -5c [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] -5c + 300 = + 225 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 300 and 225. To do that, we subtract 300 from both sides -5c + 300 - 300 = 225 - 300 [SIZE=5][B]Step 4: Cancel 300 on the left side:[/B][/SIZE] -5c = -75 [SIZE=5][B]Step 5: Divide each side of the equation by -5[/B][/SIZE] -5c/-5 = -75/-5 c = [B]15[/B] Recall from equation (1) that a = 20 - c. So we substitute c = 15 into this equation to solve for a: a = 20 - 15 a = [B]5[/B]
A manufacturer has a monthly fixed cost of \$100,000 and a production cost of \$10 for each unit produ
A manufacturer has a monthly fixed cost of \$100,000 and a production cost of \$10 for each unit produced. The product sells for \$22/unit. The cost function for each unit u is: C(u) = Variable Cost * Units + Fixed Cost C(u) = 10u + 100000 The revenue function R(u) is: R(u) = 22u We want the break-even point, which is where: C(u) = R(u) 10u + 100000 = 22u [URL='https://www.mathcelebrity.com/1unk.php?num=10u%2B100000%3D22u&pl=Solve']Typing this equation into our search engine[/URL], we get: u =[B]8333.33[/B]
A manufacturer has a monthly fixed cost of \$100,000 and a production cost of \$14 for each unit produ
A manufacturer has a monthly fixed cost of \$100,000 and a production cost of \$14 for each unit produced. The product sells for \$20/unit. Let u be the number of units. We have a cost function C(u) as: C(u) = Variable cost * u + Fixed Cost C(u) = 14u + 100000 [U]We have a revenue function R(u) with u units as:[/U] R(u) = Sale Price * u R(u) = 20u [U]We have a profit function P(u) with u units as:[/U] Profit = Revenue - Cost P(u) = R(u) - C(u) P(u) = 20u - (14u + 100000) P(u) = 20u - 14u - 100000 P(u) = 6u - 1000000
a number added to 5 minus p
a number added to 5 minus p The phrase [I]a number[/I] means an arbitrary variable, let's call it x. We add 5 minus p to this number x: [B]x + 5 - p[/B]
a number added to the product of y and x
a number added to the product of y and x Since we're already using the variables x and y, we choose another arbitrary variable for the phrase [I]a number.[/I] a The product of y and x isL xy Then add a: [B]a + xy[/B]
a number increased by 6
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. The phrase [I]increased by[/I] means we add 6 to x [B]x +6[/B]
a number increased by 8 and then tripled
a number increased by 8 and then tripled The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Increased by 8 means we add 8 to x: x + 8 Then tripled means we multiply the expression x + 8 by 3: [B]3(x + 8)[/B]
a number is twice another number
a number is twice another number The phrase [I]a number[/I] means an arbitrary variable, let's call it x The phrase [I]another number [/I]means another arbitrary variable, let's call it y Twice means we multiply y by 2: 2y The phrase [I]is [/I]means an equation, so we set x equal to 2y: [B]x = 2y[/B]
A number multiplied by 6 and divided by 5 give four more than a number?
A number multiplied by 6 and divided by 5 give four more than a number? A number is represented by an arbitrary variable, let's call it x. Multiply by 6: 6x Divide by 5 6x/5 The word "gives" means equals, so we set this equal to 4 more than a number, which is x + 4. 6x/5 = x + 4 Now, multiply each side of the equation by 5, to eliminate the fraction on the left hand side: 6x(5)/5 = 5(x + 4) The 5's cancel on the left side, giving us: 6x = 5x + 20 Subtract 5x from each side [B]x = 20[/B] Check our work from our original equation: 6x/5 = x + 4 6(20)/5 ? 20 + 4 120/5 ?24 24 = 24 <-- Yes, we verified our answer
a number of pennies splits into 4 equal groups
a number of pennies splits into 4 equal groups The phrase [I]a number[/I] means an arbitrary variable, let's call it x. We take x and divide it by 4 to get 4 equal groups: [B]x/4[/B]
A number y increased by itself
A number y increased by itself increased by itself means we add the variable y to itself to get our final algebraic expression of: [B]y + y [/B] [I]If[/I] the problem asks you to simplify, we group like terms and get: [B]2y[/B]
A peanut vendor has initial start up costs of \$7600 and variable costs of \$0.70 per bag of peanuts.
A peanut vendor has initial start up costs of \$7600 and variable costs of \$0.70 per bag of peanuts. What is the cost function? We set up the cost function C(b) where b is the number of bags: C(b) = Cost per bag * b + Start up costs Plugging in our numbers, we get: [B]C(b) = 0.70b + 7600[/B]
A quarter of a number is greater than or equal to 38
A quarter of a number is greater than or equal to 38. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. A quarter of a number means 1/4, so we have: x/4 The phrase [I]is greater than or equal to[/I] means an inequality, so we use the >= sign in relation to 38: [B]x/4 >= 38 <-- This is our algebraic expression [/B] If you want to solve this inequality, [URL='https://www.mathcelebrity.com/prop.php?num1=x&num2=38&propsign=%3E%3D&den1=4&den2=1&pl=Calculate+missing+proportion+value']we type it in the search engine[/URL] to get: x >= [B]152[/B]
A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit
A random variable X follows the uniform distribution with a lower limit of 670 and an upper limit a. Calculate the mean and standard deviation of this distribution. (Round intermediate calculation for standard deviation to 4 decimal places and final answer to 2 decimal places.) Using our [URL='http://www.mathcelebrity.com/uniform.php?a=+670&b=+770&x=+680&t=+3&pl=PDF']uniform distribution calculator[/URL], we get: [B]Mean = 720 Standard deviation = 28.87 [/B] b. What is the probability that X is less than 730? (Do not round intermediate calculations. Round your answer to 2 decimal places.) Using our [URL='http://www.mathcelebrity.com/uniform.php?a=+670&b=+770&x=+730&t=+3&pl=CDF']uniform distribution calculator[/URL], we get: [B]0.6[/B]
a son is 1/4 of his fathers age. the difference in their ages is 30. what is the fathers age.
a son is 1/4 of his fathers age. the difference in their ages is 30. what is the fathers age. Declare variables: [LIST] [*]Let f be the father's age [*]Let s be the son's age [/LIST] We're given two equations: [LIST=1] [*]s = f/4 [*]f - s = 30. [I]The reason why we subtract s from f is the father is older[/I] [/LIST] Using substitution, we substitute equaiton (1) into equation (2) for s: f - f/4 = 30 To remove the denominator/fraction, we multiply both sides of the equation by 4: 4f - 4f/4 = 30 *4 4f - f = 120 3f = 120 To solve for f, we divide each side of the equation by 3: 3f/3 = 120/3 Cancel the 3's on the left side and we get: f = [B]40[/B]
A sum greater than 5
A sum greater than 5 A sum implies two variables or more. Let's use 2, x and y [B]x + y > 5[/B]
a textbook store sold a combined total of 296 sociology and history text books in a week. the number
a textbook store sold a combined total of 296 sociology and history text books in a week. the number of history textbooks sold was 42 less than the number of sociology textbooks sold. how many text books of each type were sold? Let h = history book and s = sociology books. We have 2 equations: (1) h = s - 42 (2) h + s = 296 Substitute (1) to (2) s - 42 + s = 296 Combine variables 2s - 42 = 296 Add 42 to each side 2s = 338 Divide each side by 2 s = 169 So h = 169 - 42 = 127
A tire repair shop charges \$5 for tool cost and \$2 for every minute the worker spends on the repair.
A tire repair shop charges \$5 for tool cost and \$2 for every minute the worker spends on the repair. A) Write an equation of the total cost of repair, \$y, in terms of a total of x minutes of repair. y = Variable Cost + Fixed Cost y = Cost per minute of repair * minutes of repair + Tool Cost [B]y = 2x + 5[/B]
a variable tripled less 40
a variable tripled less 40 [I]A variable[/I] means we pick an arbitrary variable, let's call it x x Tripled means we multiply by 3 3x Less 40 means we subtract 40: [B]3x - 40[/B]
a ^5 x a ^2 without exponents
a ^5 x a ^2 without exponents When we multiply the same variable or number, we add exponents, so we have: a^(5 + 2) a^7 To write a variable raised to an exponent without exponents, we break it up. The formula to do this is: a^n = a times itself n times a^7 = [B]a * a * a * a * a * a * a[/B]
A=a+b+c+d÷4 for c
A=a+b+c+d÷4 for c Assume A and a are different variables: Cross multiply: a + b + c + d = 4A Subtract a, b, and d from each side: a + b + c + d - (a + b + d) = 4A - (a + b + d) Cancel the a + b + d on the left side [B]c = 4A - a - b - d[/B]
ab/d + c = e for d
ab/d + c = e for d I know this is a literal equation because we are asked to solve for a variable [U]in terms of[I] another variable [/I][/U] Subtract c from each side to isolate the d term: ab/d + c - c = e - c Cancel the c's on the left side and we get: ab/d = e - c Cross multiply: ab = d(e - c) Divide each side of the equation by (e - c): ab/(e - c)= d(e - c)/(e - c) Cancel the (e - c) on the right side, and we get: d = [B]ab/(e - c)[/B]
Absolute value of x less than 8
These are now available as shortcuts. You can type any number or variable in the following forms: [LIST] [*]Absolute value of x less than 8 [*]Absolute value of x less than or equal to 8 [*]Absolute value of x greater than 8 [*]Absolute value of x greater than or equal to 8 [*]Absolute value of x equal to 8 [/LIST]
Alberto’s salary was \$1500 greater than 5 times Nick’s salary. Write an equation stating Alberto’s a
Alberto’s salary was \$1500 greater than 5 times Nick’s salary. Write an equation stating Alberto’s and Nick’s salaries in terms of x and y. Let x be Alberto's salary. Let y be Nick's salary. We have: Let's break this down: [LIST=1] [*]5 times Nick's salary (y), means we multiply the variable y by 5 [*]\$1500 greater means we add \$1500 to 5y [/LIST] [B]x = 5y - 1500[/B]
Algebraic Substitutions
Free Algebraic Substitutions Calculator - Given an algebraic statement with variables [a-z], this calculator takes a set of given substitution values, i.e., x=2,y=3,z=4, and evaluates your statement using the substitution values.
Angad was thinking of a number. Angad adds 20 to it, then doubles it and gets an answer of 53. What was the original number? The phrase [I]a number[/I] means an arbitrary variable, let's call it n. [LIST] [*]Start with n [*]Add 20 to it: n + 20 [*]Double it means we multiply the expression by 2: 2(n + 20) [*]Get an answer of 53: means an equation, so we set 2(n + 20) equal to 53 [/LIST] 2(n + 20) = 53 To solve for n, we [URL='https://www.mathcelebrity.com/1unk.php?num=2%28n%2B20%29%3D53&pl=Solve']type this equation into our search engine[/URL] and we get: n = [B]6.5[/B]
Angular Momentum
Free Angular Momentum Calculator - Solves for any of the 4 variables in the angular momentum equation, L, V, M, and R
b/3d - h = 343 for b
b/3d - h = 343 for b A literal equation means we solve for one variable in terms of another variable or variables Add h to each side to isolate the b term: b/3d - h + h = 343 + h Cancel the h's on the left side, we get: b/3d = 343 + h Cross multiply: b = [B]3d(343 + h)[/B]
Barbra is buying plants for her garden. She notes that potato plants cost \$3 each and corn plants co
Barbra is buying plants for her garden. She notes that potato plants cost \$3 each and corn plants cost \$4 each. If she plans to spend at least \$20 and purchase less than 15 plants in total, create a system of equations or inequalities that model the situation. Define the variables you use. [U]Define variables[/U] [LIST] [*]Let c be the number of corn plants [*]Let p be the number of potato plants [/LIST] Since cost = price * quantity, we're given two inequalities: [LIST=1] [*][B]3p + 4c >= 20 (the phrase [I]at least[/I] means greater than or equal to)[/B] [*][B]c + p < 15[/B] [/LIST]
Binomial Multiplication (FOIL)
Free Binomial Multiplication (FOIL) Calculator - Multiplies out the product of 2 binomials in the form (a + b)(c + d) with 1 unknown variable.
This utilizes the First-Outside-Inside-Last (F.O.I.L.) method. Also calculates using the FOIL Box Method.
Break Even
Free Break Even Calculator - Given a fixed cost, variable cost, and revenue function or value, this calculates the break-even point
Carmen is serving her child french fries and chicken wings for lunch today. Let f be the number of f
Carmen is serving her child french fries and chicken wings for lunch today. Let f be the number of french fries in the lunch, and let c be the number of chicken wings. Each french fry has 25 calories, and each chicken wing has 100 calories. Carmen wants the total calorie count from the french fries and chicken wings to be less than 500 calories. Using the values and variables given, write an inequality describing this. We have: 25f + 100c < 50 Note: We use < and not <= because it states less than in the problem.
Casey is 26 years old. Her daughter Chloe is 4 years old. In how many years will Casey be double her
Casey is 26 years old. Her daughter Chloe is 4 years old. In how many years will Casey be double her daughter's age Declare variables for each age: [LIST] [*]Let Casey's age be c [*]Let her daughter's age be d [*]Let n be the number of years from now where Casey will be double her daughter's age [/LIST] We're told that: 26 + n = 2(4 + n) 26 + n = 8 + 2n Solve for [I]n[/I] in the equation 26 + n = 8 + 2n [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables n and 2n. To do that, we subtract 2n from both sides n + 26 - 2n = 2n + 8 - 2n [SIZE=5][B]Step 2: Cancel 2n on the right side:[/B][/SIZE] -n + 26 = 8 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 26 and 8. To do that, we subtract 26 from both sides -n + 26 - 26 = 8 - 26 [SIZE=5][B]Step 4: Cancel 26 on the left side:[/B][/SIZE] -n = -18 [SIZE=5][B]Step 5: Divide each side of the equation by -1[/B][/SIZE] -1n/-1 = -18/-1 n = [B]18[/B] Check our work for n = 18: 26 + 18 ? 8 + 2(18) 44 ? 8 + 36 44 = 44
Chang is serving his child french fries and chicken wings for lunch today. Let f be the number of f
Chang is serving his child french fries and chicken wings for lunch today. Let f be the number of french fries in the lunch, and let c be the number of chicken wings. Each french fry has 25 calories, and each chicken wing has 100 calories. Chang wants the total calorie count from the french fries and chicken wings to be less than 600 calories. Using the values and variables given, write an inequality describing this. We have [B]25f + 100c < 600[/B] as our inequality.
Chebyshevs Theorem
Free Chebyshevs Theorem Calculator - Using Chebyshevs Theorem, this calculates the following:
Probability that random variable X is within k standard deviations of the mean.
How many k standard deviations within the mean given a P(X) value.
Combination with Variable
Free Combination with Variable Calculator - Calculates the following:
Solves for r given n and the combination value.
Solves for n given r and the combination value
Compute a 75% Chebyshev interval around the mean for x values and also for y values.
Compute a 75% Chebyshev interval around the mean for [I]x[/I] values and also for [I]y[/I] values. [B][U]Grid E: [I]x[/I] variable[/U][/B] 11.92 34.86 26.72 24.50 38.93 8.59 29.31 23.39 24.13 30.05 21.54 35.97 7.48 35.97 [B][U]Grid H: [I]y[/I] variable[/U][/B] 27.86 13.29 33.03 44.31 16.58 42.43 39.61 25.51 39.14 16.58 47.13 14.70 57.47 34.44 According to Chebyshev's Theorem, [1 - (1/k^2)] proportion of values will fall between Mean +/- (k*SD) k in this case equal to z z = (X-Mean)/SD X = Mean + (z*SD) 1 - 1/k^2 = 0.75 - 1/k^2 = 0.75 - 1= - 0.25 1/k^2 = 0.25 k^2 = 1/0.25 k^2 = 4 k = 2 Therefore, z = k = 2 First, [URL='http://www.mathcelebrity.com/statbasic.php?num1=11.92%2C34.86%2C26.72%2C24.50%2C38.93%2C8.59%2C29.31%2C23.39%2C24.13%2C30.05%2C21.54%2C35.97%2C7.48%2C35.97&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics']determine the mean and standard deviation of x[/URL] Mean(x) = 25.24 SD(x) = 9.7873 Required Interval for x is: Mean - (z * SD) < X < Mean + (z * SD) 25.24 - (2 * 9.7873) < X < 25.24 - (2 * 9.7873) 25.24 - 19.5746 < X < 25.24 + 19.5746 5.6654 < X < 44.8146 Next, [URL='http://www.mathcelebrity.com/statbasic.php?num1=27.86%2C13.29%2C33.03%2C44.31%2C16.58%2C42.43%2C39.61%2C25.51%2C39.14%2C16.58%2C47.13%2C14.70%2C57.47%2C34.44&num2=+0.2%2C0.4%2C0.6%2C0.8%2C0.9&pl=Number+Set+Basics']determine the mean and standard deviation of y[/URL] Mean(y) = 32.29 SD(y) = 9.7873 Required Interval for y is: Mean - (z * SD) < Y < Mean + (z * SD) 32.29 - (2 * 13.1932) < Y < 32.29 - (2 * 13.1932) 32.29 - 26.3864 < Y < 32.29 + 26.3864 5.9036 < X < 58.6764
Consider the formula for the area of a trapezoid: A=12h(a+b) . Is it mathematically simpler to solve
Consider the formula for the area of a trapezoid: A=12h(a+b) . Is it mathematically simpler to solve for a, b, or h? Why? Solve for each of these variables to demonstrate. The variable "h" is the easiest to solve for. Because you only have one step. Let's review: Divide each side of the equation by 12(a + b) h = 12(a + b)/A Solving for "a", we two steps. Divide each side by 12h: A/12h = a + b Subtract b from each side a = A/12h - b Solving for "b" takes two steps as well. Divide each side by 12h: A/12h = a + b Subtract a from each side b = A/12h - a
Cost Revenue Profit
Free Cost Revenue Profit Calculator - Given a total cost, variable cost, revenue amount, and profit unit measurement, this calculates profit for each profit unit
cube root of a number and 7
cube root of a number and 7 The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Cube root of a number means we raise x to the 1/3 power: x^1/3 And 7 means we add 7: [B]x^1/3 + 7[/B]
d - f^3 = 4a for a
d - f^3 = 4a for a Solve this literal equation for a: Divide each side of the equation by 4: (d - f^3)/4 = 4a/4 Cancel the 4's on the right side, and rewrite with our variable to solve for on the left side: a = [B](d - f^3)/4[/B]
d squared is greater than or equal to 17
d squared is greater than or equal to 17 d squared means we raise the variable d to the power of 2: d^2 The phrase [I]greater than or equal to[/I] means an inequality. So we set this up using the >= in relation to 17: [B]d^2 >= 17[/B]
Decrease 12 by a number
Decrease 12 by a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. We take 12 and decrease it by x, meaning we subtract x from 12: [B]12 - x[/B]
decrease a number by 7 and multiply by 6.
decrease a number by 7 and multiply by 6. The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Decrease a number by 7: x - 7 Multiply by 6 [B]6(x - 7)[/B]
Deon opened his account starting with \$650 and he is going to take out \$40 per month. Mai opened up
Deon opened his account starting with \$650 and he is going to take out \$40 per month. Mai opened up her account with a starting amount of \$850 and is going to take out \$65 per month. When would the two accounts have the same amount of money? We set up a balance equation B(m) where m is the number of months. [U]Set up Deon's Balance equation:[/U] Withdrawals mean we subtract from our current balance B(m) = Starting Balance - Withdrawal Amount * m B(m) = 650 - 40m [U]Set up Mai's Balance equation:[/U] Withdrawals mean we subtract from our current balance B(m) = Starting Balance - Withdrawal Amount * m B(m) = 850 - 65m When the two accounts have the same amount of money, we can set both balance equations equal to each other and solve for m: 650 - 40m = 850 - 65m Solve for [I]m[/I] in the equation 650 - 40m = 850 - 65m [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables -40m and -65m. To do that, we add 65m to both sides -40m + 650 + 65m = -65m + 850 + 65m [SIZE=5][B]Step 2: Cancel -65m on the right side:[/B][/SIZE] 25m + 650 = 850 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 650 and 850. To do that, we subtract 650 from both sides 25m + 650 - 650 = 850 - 650 [SIZE=5][B]Step 4: Cancel 650 on the left side:[/B][/SIZE] 25m = 200 [SIZE=5][B]Step 5: Divide each side of the equation by 25[/B][/SIZE] 25m/25 = 200/25 m = [B]8[/B]
Determine the formula of the given statement by following the procedures. Choose any number then add
Determine the formula of the given statement by following the procedures. Choose any number then add 2. Multiply your answer to 3 and minus 2 For the phrase [I]choose any number[/I] we can use an arbitrary variable, let's call it x. Add 2: x + 2 Multiply your answer to 3: 3(x + 2) And minus 2 which means we subtract: [B]3(x + 2) - 2[/B]
Determine whether the random variable is discrete or continuous. In each case, state the possible v
Determine whether the random variable is discrete or continuous. In each case, state the possible values of the random variable. (a) The number of customers arriving at a bank between noon and 1:00 P.M. (i) The random variable is continuous. The possible values are x >= 0. (ii) The random variable is discrete. The possible values are x = 0, 1, 2,... (iii) The random variable is continuous. The possible values are x = 0, 1, 2,... (iv) The random variable is discrete. The possible values are x >= 0. (b) The amount of snowfall (i) The random variable is continuous. The possible values are s = 0, 1, 2,... (ii) The random variable is discrete. The possible values are s >= 0. (iii) The random variable is discrete. The possible values are s = 0, 1, 2,... (iv) The random variable is continuous. The possible values are s >= 0. [B](a) (ii) The random variable is discrete. The possible values are x = 0, 1, 2,... Discrete variables are limited in the values they can take between 9 and ∞ (b) (iv) The random variable is continuous. The possible values are s >= 0. Snowfall can be a decimal and can vary between 0 and ∞[/B]
Diegos savings increased by 9 is 68 . Use the variable to represent Diegos savings.
Diegos savings increased by 9 is 68 . Use the variable to represent Diegos savings. Let Diego's savings be s. The phrase [I]increased by[/I] means add, so we add 9 to s s + 9 The phrase [I]is [/I]means equal to, so we set 2 + 9 = 68 [B]s + 9 = 68[/B]
difference between 2 positive numbers is 3 and the sum of their squares is 117
difference between 2 positive numbers is 3 and the sum of their squares is 117 Declare variables for each of the two numbers: [LIST] [*]Let the first variable be x [*]Let the second variable be y [/LIST] We're given 2 equations: [LIST=1] [*]x - y = 3 [*]x^2 + y^2 = 117 [/LIST] Rewrite equation (1) in terms of x by adding y to each side: [LIST=1] [*]x = y + 3 [*]x^2 + y^2 = 117 [/LIST] Substitute equation (1) into equation (2) for x: (y + 3)^2 + y^2 = 117 Evaluate and simplify: y^2 + 3y + 3y + 9 + y^2 = 117 Combine like terms: 2y^2 + 6y + 9 = 117 Subtract 117 from each side: 2y^2 + 6y + 9 - 117 = 117 - 117 2y^2 + 6y - 108 = 0 This is a quadratic equation: Solve the quadratic equation 2y2+6y-108 = 0 With the standard form of ax2 + bx + c, we have our a, b, and c values: a = 2, b = 6, c = -108 Solve the quadratic equation 2y^2 + 6y - 108 = 0 The quadratic formula is denoted below: y = -b ± sqrt(b^2 - 4ac)/2a [U]Step 1 - calculate negative b:[/U] -b = -(6) -b = -6 [U]Step 2 - calculate the discriminant Δ:[/U] Δ = b2 - 4ac: Δ = 62 - 4 x 2 x -108 Δ = 36 - -864 Δ = 900 <--- Discriminant Since Δ is greater than zero, we can expect two real and unequal roots. [U]Step 3 - take the square root of the discriminant Δ:[/U] √Δ = √(900) √Δ = 30 [U]Step 4 - find numerator 1 which is -b + the square root of the Discriminant:[/U] Numerator 1 = -b + √Δ Numerator 1 = -6 + 30 Numerator 1 = 24 [U]Step 5 - find numerator 2 which is -b - the square root of the Discriminant:[/U] Numerator 2 = -b - √Δ Numerator 2 = -6 - 30 Numerator 2 = -36 [U]Step 6 - calculate your denominator which is 2a:[/U] Denominator = 2 * a Denominator = 2 * 2 Denominator = 4 [U]Step 7 - you have everything you need to solve. Find solutions:[/U] Solution 1 = Numerator 1/Denominator Solution 1 = 24/4 Solution 1 = 6 Solution 2 = Numerator 2/Denominator Solution 2 = -36/4 Solution 2 = -9 [U]As a solution set, our answers would be:[/U] (Solution 1, Solution 2) = (6, -9) Since one of the solutions is not positive and the problem asks for 2 positive number, this problem has no solution
Direct Current (Electrical Engineering) Ohms Law
Free Direct Current (Electrical Engineering) Ohms Law Calculator - Enter two of the following items from the DIRECT CURRENT(DC) electrical engineering set of variables, and this will solve for the remaining two:
* I = current(amps.)
* V = Electricity potential of voltage(volts)
* R = resistance(ohms)
* P = power(watts)
Distance Rate and Time
Free Distance Rate and Time Calculator - Solves for distance, rate, or time in the equation d=rt based on 2 of the 3 variables being known.
Divide a number by 10. Then, add 10.
Divide a number by 10. Then, add 10. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Divide the number by 10 mean we have a quotient, of x over 10 x / 10 Then, add 10: [B](x / 10) + 10[/B]
Do the phrases 7 less than a number and a number less than 7 mean the same thing explain
Do the phrases 7 less than a number and a number less than 7 mean the same thing explain No, they are different, here's how: First, the phrase [I]a number[/I] means an arbitrary variable, let's call it x. 7 less than a number means we subtract 7 from x: x - 7 A number less than 7 means we subtract x from 7: 7 - x As you can see: x - 7 <> 7 - x so [B]they are different[/B]
Dotty McGinnis starts up a small business manufacturing bobble-head figures of famous soccer players
Dotty McGinnis starts up a small business manufacturing bobble-head figures of famous soccer players. Her initial cost is \$3300. Each figure costs \$4.50 to make. a. Write a cost function, C(x), where x represents the number of figures manufactured. Cost function is the fixed cost plus units * variable cost. [B]C(x) = 3300 + 4.50x[/B]
double v, add u, then divide t by what you have
double v, add u, then divide t by what you have Double v means we multiply the variable v by 2: 2v Add u: 2v + u We build a fraction, with t as the numerator, and 2v + u as the denominator [B]t/(2v + u)[/B]
Dunder Mifflin will print business cards for \$0.10 each plus setup charge of \$15. Werham Hogg offers
Dunder Mifflin will print business cards for \$0.10 each plus setup charge of \$15. Werham Hogg offers business cards for \$0.15 each with a setup charge of \$10. What numbers of business cards cost the same from either company Declare variables: [LIST] [*]Let b be the number of business cards. [/LIST] [U]Set up the cost function C(b) for Dunder Mifflin:[/U] C(b) = Cost to print each business card * b + Setup Charge C(b) = 0.1b + 15 [U]Set up the cost function C(b) for Werham Hogg:[/U] C(b) = Cost to print each business card * b + Setup Charge C(b) = 0.15b + 10 The phrase [I]cost the same[/I] means we set both C(b)'s equal to each other and solve for b: 0.1b + 15 = 0.15b + 10 Solve for [I]b[/I] in the equation 0.1b + 15 = 0.15b + 10 [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables 0.1b and 0.15b. To do that, we subtract 0.15b from both sides 0.1b + 15 - 0.15b = 0.15b + 10 - 0.15b [SIZE=5][B]Step 2: Cancel 0.15b on the right side:[/B][/SIZE] -0.05b + 15 = 10 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 15 and 10. To do that, we subtract 15 from both sides -0.05b + 15 - 15 = 10 - 15 [SIZE=5][B]Step 4: Cancel 15 on the left side:[/B][/SIZE] -0.05b = -5 [SIZE=5][B]Step 5: Divide each side of the equation by -0.05[/B][/SIZE] -0.05b/-0.05 = -5/-0.05 b = [B]100[/B]
Each piece of candy costs 25 cents. The cost of x pieces of candy is \$2.00. Use variable x to transl
Each piece of candy costs 25 cents. The cost of x pieces of candy is \$2.00. Use variable x to translate the above statements into algebraic equation. Our algebraic expression is: [B]0.25x = 2 [/B] To solve this equation for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=0.25x%3D2&pl=Solve']type it in our search engine[/URL] and we get: x = [B]8[/B]
Earnings Before Interest and Taxes (EBIT) and Net Income
Free Earnings Before Interest and Taxes (EBIT) and Net Income Calculator - Given inputs of sales, fixed costs, variable costs, depreciation, and taxes, this will determine EBIT and Net Income and Profit Margin
Eighteen times the difference of a number and ten
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. The difference of a number and ten x - 10 Eighteen times the difference of a number and ten [B]18(x - 10)[/B]
El triple de la diferencia de dos números
El triple de la diferencia de dos números La frase 2 números significa variables arbitrarias. Llamémoslos xey. La diferencia significa que restamos y de x: x - y Triple significa multiplicar la diferencia por 3 [B]3(x - y)[/B]
Equation and Inequalities
Free Equation and Inequalities Calculator - Solves an equation or inequality with 1 unknown variable and no exponents as well as certain absolute value equations and inequalities such as |x|=c and |ax| = c where a and c are constants. Solves square root, cube root, and other root equations in the form ax^2=c, ax^2 + b = c. Also solves radical equations in the form asqrt(bx) = c. Also solves open sentences and it will solve one step problems and two step equations. 2 step equations and one step equations and multi step equations
Equation of Exchange
Free Equation of Exchange Calculator - Solves for any of the 4 variables in the Equation of Exchange: money, velocity, price, quantity
Expand Master and Build Polynomial Equations
Free Expand Master and Build Polynomial Equations Calculator - This calculator is the ultimate expansion tool to multiply polynomials. It expands algebraic expressions listed below using all 26 variables (a-z) as well as negative powers to handle polynomial multiplication. Includes multiple variable expressions as well as outside multipliers.
Also produces a polynomial equation from a given set of roots (polynomial zeros). * Binomial Expansions c(a + b)x
* Polynomial Expansions c(d + e + f)x
* FOIL Expansions (a + b)(c + d)
* Multiple Parentheses Multiplications c(a + b)(d + e)(f + g)(h + i)
Expected Value
Free Expected Value Calculator - This lesson walks you through what expected value is, expected value notation, the expected value of a discrete random variable, the expected value of a continuous random variable, and expected value properties.
F varies directly as g and inversely as r^2
F varies directly as g and inversely as r^2 [U]Givens and assumptions[/U] [LIST] [*]We take a constant of variation called k. [*][I]Varies directly means we multiply our variable term by k[/I] [*][I]Varies inversely means we divide k by our variable term[/I] [/LIST] The phrase varies directly or varies inversely means we have a constant k such that: [B]F = kg/r^2[/B]
Factoring and Root Finding
Free Factoring and Root Finding Calculator - This calculator factors a binomial including all 26 variables (a-z) using the following factoring principles:
* Difference of Squares
* Sum of Cubes
* Difference of Cubes
* Binomial Expansions
* Factor by Grouping
* Common Term
This calculator also uses the Rational Root Theorem (Rational Zero Theorem) to determine potential roots
* Factors and simplifies Rational Expressions of one fraction
* Determines the number of potential positive and negative roots using Descarte’s Rule of Signs
Fifty-two less than 75% of a number
Fifty-two less than 75% of a number A number means an arbitrary variable, let's call it x. 75% of this is 0.75x Fifty-two less is: [B]0.75x - 52[/B]
Five less than a number is at least -7 and at most 7.
Five less than a number is at least -7 and at most 7. A number signifies an arbitrary variable, let's call it x. Five less than a number: x - 5 Is at least -7 means greater than or equal to and at most 7 means less than or equal to, so we have a joint inequality: [B]-7 <= x - 5 <= 7[/B]
Four less than five times a number
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. 5 times a number: 5x Four less means we subtract 4 from 5x: [B]5x - 4[/B]
Four more then double a number is greater than 2
Four more then double a number is greater than 2 Double a number: A number implies an arbitrary variable, let's call it "x". Double means multiply this by 2 2x Four more than this: 2x + 4 Now, we set this expression as an inequality greater than 2 [B]2x + 4 > 2[/B]
Fraction with variable x in numerator and 6 in the denominator.
Fraction with variable x in numerator and 6 in the denominator. The numerator is the top of the fraction. The denominator is the bottom of the fraction. [B]x/6[/B]
Frank is a plumber who charges a \$35 service charge and \$15 per hour for his plumbing services. Find
Frank is a plumber who charges a \$35 service charge and \$15 per hour for his plumbing services. Find a linear function that expresses the total cost C for plumbing services for h hours. Cost functions include a flat rate and a variable rate. The flat rate is \$35 and the variable rate per hour is 15. The cost function C(h) where h is the number of hours Frank works is: [B]C(h) = 15h + 35[/B]
g equals 232 subtracted from the quantity 377 times g
g equals 232 subtracted from the quantity 377 times g 377 times g: 377g 232 subtracted from 377 times g: 377g - 232 We set the variable g equal to this expression: [B]g = 377g - 232[/B]
Georgie joins a gym. she pays \$25 to sign up and then \$15 each month. Create an equation using this
Georgie joins a gym. she pays \$25 to sign up and then \$15 each month. Create an equation using this information. Let m be the number of months Georgie uses the gym. Our equation G(m) is the cost Georgie pays for m months. G(m) = Variable Cost * m (months) + Fixed Cost Plug in our numbers: [B]G(m) = 15m + 25[/B]
Golden Ratio
Free Golden Ratio Calculator - Solves for 2 out of the 3 variables for a segment broken in 2 pieces that satisfies the Golden Ratio (Golden Mean).
(a) Large Segment
(b) Small Segment
(a + b) Total Segment
H multiplied by 2x
H multiplied by 2x h * 2x If we arrange variables alphabetically, we have: [B]2hx[/B]
Happy Paws charges \$16.00 plus \$1.50 per hour to keep a dog during the day. Woof Watchers charges \$1
Happy Paws charges \$16.00 plus \$1.50 per hour to keep a dog during the day. Woof Watchers charges \$11.00 plus \$2.75 per hour. Complete the equation and solve it to find for how many hours the total cost of the services is equal. Use the variable h to represent the number of hours. Happy Paws Cost: C = 16 + 1.5h Woof Watchers: C = 11 + 2.75h Setup the equation where there costs are equal 16 + 1.5h = 11 + 2.75h Subtract 11 from each side: 5 + 1.5h = 2.75h Subtract 1.5h from each side 1.25h = 5 Divide each side by 1.25 [B]h = 4[/B]
Happy Paws charges \$19.00 plus \$5.50 per hour to keep a dog during the day. Woof Watchers charges \$1
Happy Paws charges \$19.00 plus \$5.50 per hour to keep a dog during the day. Woof Watchers charges \$11.00 plus \$6.75 per hour. Complete the equation and solve it to find for how many hours the total cost of the services is equal. Use the variable h to represent the number of hours. [B]Happy Paws cost equation:[/B] 5.50h + 19 [B]Woof Watchers cost equation:[/B] 6.75h + 11 [B]Set them equal to each other:[/B] 5.50h + 19 = 6.75h + 11 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=5.50h%2B19%3D6.75h%2B11&pl=Solve']equation solver[/URL], we get [B]h = 6.4[/B].
High and Low Method
Free High and Low Method Calculator - Calculates the variable cost per unit, total fixed costs, and the cost volume formula
High-Low Method
Free High-Low Method Calculator - Calculates Variable Cost per Unit, Total Fixed Cost, and Cost Volume using the High-Low Method
The phrase a number means an arbitrary variable, let's call it x. Three times a number: 3x And 18 means we add 18 3x + 18 The word is means equal to, so we set 3x + 18 equal to -39 3x + 18 = -39 This is your algebraic expression. If you want to solve for x, plug it into the [URL='http://www.mathcelebrity.com/1unk.php?num=3x%2B18%3D-39&pl=Solve']search engine[/URL] and you get x = -19
how many sixths equal one-third
how many sixths equal one-third We have a variable x where we want to solve for in the following equation: x/6 = 1/3 [URL='https://www.mathcelebrity.com/prop.php?num1=x&num2=1&den1=6&den2=3&propsign=%3D&pl=Calculate+missing+proportion+value']Typing this proportion into our math engine[/URL], we get: x = [B]2[/B]
If 11 times a number is added to twice the number, the result is 104
If 11 times a number is added to twice the number, the result is 104 Let [I]the number[/I] be an arbitrary variable we call x. 11 times a number: 11x Twice the number (means we multiply x by 2): 2x The phrase [I]is added to[/I] means we add 2x to 11x: 11x + 2x Simplify by grouping like terms: (11 + 2)x = 13x The phrase [I]the result is[/I] means an equation, so we set 13x equal to 104: 13x = 104 <-- This is our algebraic expression To solve this equation for x, [URL='https://www.mathcelebrity.com/1unk.php?num=13x%3D104&pl=Solve']we type it in our search engine[/URL] and we get: x = [B]8[/B]
If 2x + y = 7 and y + 2z = 23, what is the average of x, y, and z?
If 2x + y = 7 and y + 2z = 23, what is the average of x, y, and z? A. 5 B. 7.5 C. 15 D. 12.25 Add both equations to get all variables together: 2x + y + y + 2z = 23 + 7 2x + 2y + 2z = 30 We can divide both sides by 2 to simplify: (2x + 2y + 2z)/2= 30/2 x + y + z = 15 Notice: the average of x, y, and z is: (x + y + z)/3 But x + y + z = 15, so we have: 15/3 = [B]5, answer A[/B] [MEDIA=youtube]tOCAhhfMCLI[/MEDIA]
If 3 times a number added to 2 is divided by the number plus 4 the result is 4/3
If 3 times a number added to 2 is divided by the number plus 4 the result is 4/3 Take this in pieces, where "a number" means an arbitrary variable, let's call it "x". [LIST=1] [*]3 times a number --> 3x [*]3 times a number added to 2 --> 3x + 2 [*]The number plus 4 --> x + 4 [*]is divided by --> (3x + 2)/(x + 4) [*]the result is 4/3 --> (3x + 2)/(x + 4) = 4/3 [/LIST]
If 4 times a number is added to 9, the result is 49
If 4 times a number is added to 9, the result is 49. [I]A number[/I] means an arbitrary variable, let's call it x. 4 [I]times a number[/I] means we multiply x by 4 4x [I]Added to[/I] 9 means we add 9 to 4x 4x + 9 [I]The result is[/I] means we have an equation, so we set 4x + 9 equal to 49 [B]4x + 9 = 49[/B] <-- This is our algebraic expression To solve this equation, [URL='https://www.mathcelebrity.com/1unk.php?num=4x%2B9%3D49&pl=Solve']we type it in the search engine[/URL] and get x = 10
If 72 is added to a number it will be 4 times as large as it was originally
If 72 is added to a number it will be 4 times as large as it was originally The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. x 72 added to a number: x + 72 4 times as large as it was originally means we take the original number x and multiply it by 4: 4x Now, the phrase [I]it will be[/I] means an equation, so we set x + 72 equal to 4x to get our final algebraic expression: [B]x + 72 = 4x[/B] [B][/B] If the problem asks you to solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=x%2B72%3D4x&pl=Solve']type this equation into our search engine[/URL] and we get: x = [B]24[/B]
If 9 is added to 1/3 of a number, the result is 15. What is the number?
If 9 is added to 1/3 of a number, the result is 15. What is the number? The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x 1/3 of a number means we multiply x by 1/3: x/3 9 is added to 1/3 of a number: x/3 + 9 The phrase [I]the result is[/I] means an equation. so we set x/3 + 9 equal to 15 x/3 + 9 = 15 To solve this equation for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=x%2F3%2B9%3D15&pl=Solve']type it in our search engine[/URL] and we get: x = [B]18[/B]
if 9 times a number is decreased by 6, the result is 111
if 9 times a number is decreased by 6, the result is 111 A number means an arbitrary variable, let's call it x. 9 times a number: 9x Decreased by 6 9x - 6 The result is 11, this means we set 9x - 6 equal to 11 [B]9x - 6 = 11 [/B] To solve this equation for x, use our [URL='http://www.mathcelebrity.com/1unk.php?num=9x-6%3D11&pl=Solve']equation calculator[/URL]
if a number is added to its square, it equals 20
if a number is added to its square, it equals 20. Let the number be an arbitrary variable, let's call it n. The square of the number means we raise n to the power of 2: n^2 We add n^2 to n: n^2 + n It equals 20 so we set n^2 + n equal to 20 n^2 + n = 20 This is a quadratic equation. So [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2%2Bn%3D20&pl=Solve+Quadratic+Equation&hintnum=+0']we type this equation into our search engine[/URL] to solve for n and we get two solutions: [B]n = (-5, 4)[/B]
if a number is decreased by 5, and then the result is multiplied by 2, the result is 26
If a number is decreased by 5, and then the result is multiplied by 2, the result is 26 The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x [I]Decreased by[/I] means we subtract 5 from x: x - 5 Multiply the result by 2: 2(x - 5) The result is 26 means we set 2(x - 5) equal to 26: [B]2(x - 5) = 26[/B]
if a number is tripled the result is 60
if a number is tripled the result is 60 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x Triple the number means we multiply by 3: 3x The phrase [I]the result is[/I] means an equation, so we set 3x equal to 60: [B]3x = 60 <-- This is our algebraic expression [/B] If you want to solve this equation, then [URL='https://www.mathcelebrity.com/1unk.php?num=3x%3D60&pl=Solve']you type in 3x = 60 into the search engine[/URL] and get: x = 20
If EF = 9x - 17, FG = 17x - 14, and EG = 20x + 17, what is FG?
If EF = 9x - 17, FG = 17x - 14, and EG = 20x + 17, what is FG? By segment addition, we know that: EF + FG = EG Substituting in our values for the 3 segments, we get: 9x - 17 + 17x - 14 = 20x + 17 Group like terms and simplify: (9 + 17)x + (-17 - 14) = 20x - 17 26x - 31 = 20x - 17 Solve for [I]x[/I] in the equation 26x - 31 = 20x - 17 [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables 26x and 20x. To do that, we subtract 20x from both sides 26x - 31 - 20x = 20x - 17 - 20x [SIZE=5][B]Step 2: Cancel 20x on the right side:[/B][/SIZE] 6x - 31 = -17 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants -31 and -17. To do that, we add 31 to both sides 6x - 31 + 31 = -17 + 31 [SIZE=5][B]Step 4: Cancel 31 on the left side:[/B][/SIZE] 6x = 14 [SIZE=5][B]Step 5: Divide each side of the equation by 6[/B][/SIZE] 6x/6 = 14/6 x = [B]2.3333333333333[/B]
If from twice a number you subtract four, the difference is twenty
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. Twice a number means we multiply x by 2: 2x Subtract four: 2x - 4 The word [I]is [/I]means equal to. We set 2x - 4 equal to 20 for our algebraic expression: [B]2x - 4 = 20 [/B] If the problem asks you to solve for x: we [URL='https://www.mathcelebrity.com/1unk.php?num=2x-4%3D20&pl=Solve']plug this equation into our calculator [/URL]and get x = [B]12[/B]
If the correlation between two variables is close to minus one, the association is: Strong Moderate
If the correlation between two variables is close to minus one, the association is: Strong Moderate Weak None [B]Strong[/B] - Coefficient near +1 or -1 indicate a strong correlation
If the difference of a number and 4 is multiplied by 3 the result is 19
If the difference of a number and 4 is multiplied by 3 the result is 19 The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x The difference of a number and 4: x - 4 The phrase [I]is multiplied by[/I] means we multiply x - 4 by 3: 3(x - 4) The phrase [I]the result is[/I] means equals, so we set 3(x - 4) equal to 19 [B]3(x - 4) = 19 [MEDIA=youtube]Q8bnVJuWeVk[/MEDIA][/B]
If thrice a number is increased by 11,the result is 35. What is the number
If thrice a number is increased by 11,the result is 35. What is the number? [LIST] [*]The phrase [I]a number [/I]means an arbitrary variable. Let's call it x. [*]Thrice means multiply by 3, so we have 3x [*]Increased by 11 means we add 11, so we have 3x + 11 [*]The [I]result is[/I] means an equation, so we set 3x + 11 equal to 35 [/LIST] 3x + 11 = 35 <-- This is our algebraic expression The problem ask us to solve the algebraic expression. [URL='https://www.mathcelebrity.com/1unk.php?num=3x%2B11%3D35&pl=Solve']Typing this problem into our search engine[/URL], we get [B]x = 8[/B].
If twice a number is divided by 7, the result is -28
If twice a number is divided by 7, the result is -28. The phrase [I]a number[/I] means an arbitrary variable, let's call it "x". Twice x means we multiply x by 2: 2x Divide this by 7: 2x/7 We set this equal to -28, and we have our algebraic expression: [B]2x/7 = -28 [/B]
if you add 35 to twice a number, the result is 17. What is the number?
if you add 35 to twice a number, the result is 17. What is the number? A number is represented by a variable, let's call it "x". Twice a number means we multiply by 2 --> 2x Add 35 2x + 35 Now set that entire expression equal to 17 2x + 35 = 17 [URL='http://www.mathcelebrity.com/1unk.php?num=2x%2B35%3D17&pl=Solve']Plug that into the search engine to solve for x[/URL] [B]x = -9[/B]
If you multiply me by 33 and subtract 20, the result is 46. Who am I?
If you multiply me by 33 and subtract 20, the result is 46. Who am I? [LIST] [*]Start with the variable x [*]Multiply me by 33 = 33x [*]Subtract 20: 33x - 20 [*]The result is 46, means we set this expression equal to 46: 33x - 20 = 46 [/LIST] Run this through our [URL='http://www.mathcelebrity.com/1unk.php?num=33x-20%3D46&pl=Solve']equation calculator[/URL], and we get: [B]x = 2[/B]
In a class there are 5 more boys than girls. There are 13 students in all. How many boys are there i
In a class there are 5 more boys than girls. There are 13 students in all. How many boys are there in the class? We start by declaring variables for boys and girls: [LIST] [*]Let b be the number of boys [*]Let g be the number of girls [/LIST] We're given two equations: [LIST=1] [*]b = g + 5 [*]b + g = 13 [/LIST] Substitute equation (1) for b into equation (2): g + 5 + g = 13 Grouping like terms, we get: 2g + 5 = 13 Subtract 5 from each side: 2g + 5 - 5 = 13 - 5 Cancel the 5's on the left side and we get: 2g = 8 Divide each side of the equation by 2 to isolate g: 2g/2 = 8/2 Cancel the 2's on the left side and we get: g = 4 Substitute g = 4 into equation (1) to solve for b: b = 4 + 5 b = [B]9[/B]
In order to test if there is a difference between means from two populations, which of following ass
In order to test if there is a difference between means from two populations, which of following assumptions are NOT required? a. The dependent variable scores must be a continuous quantitative variable. b. The scores in the populations are normally distributed. c. Each value is sampled independently from each other value. d. The two populations have similar means [B]a and d [/B] [I]because b and c [U]are[/U] required[/I]
is 6x a monomial?
[B]Yes[/B]. It's an algebraic expression consisting of one term. The constant is 6, and the variable is x.
it costs \$75.00 for a service call from shearin heating and air conditioning company. the charge for
it costs \$75.00 for a service call from shearin heating and air conditioning company. the charge for labor is \$60.00 . how many full hours can they work on my air conditioning unit and still stay within my budget of \$300.00 for repairs and service? Our Cost Function is C(h), where h is the number of labor hours. We have: C(h) = Variable Cost * Hours + Fixed Cost C(h) = 60h + 75 Set C(h) = \$300 60h + 75 = 300 [URL='https://www.mathcelebrity.com/1unk.php?num=60h%2B75%3D300&pl=Solve']Running this problem in the search engine[/URL], we get [B]h = 3.75[/B].
Jenny threw the javelin 4 metres further than Angus but 5 metres less than Cameron. if the combined
Jenny threw the javelin 4 metres further than Angus but 5 metres less than Cameron. if the combined distance thrown by the 3 friends is 124 metres, how far did Angus throw the javelin? Assumptions and givens: [LIST] [*]Let a be the distance Angus threw the javelin [*]Let c be the distance Cameron threw the javelin [*]Let j be the distance Jenny threw the javelin [/LIST] We're given 3 equations: [LIST=1] [*]j = a + 4 [*]j = c - 5 [*]a + c + j = 124 [/LIST] Since j is the common variable in all 3 equations, let's rearrange equation (1) and equation (2) in terms of j as the dependent variable: [LIST=1] [*]a = j - 4 [*]c = j + 5 [*]a + c + j = 124 [/LIST] Now substitute equation (1) and equation (2) into equation (3) for a and c: j - 4 + j + 5 + j = 124 To solve this equation for j, we [URL='https://www.mathcelebrity.com/1unk.php?num=j-4%2Bj%2B5%2Bj%3D124&pl=Solve']type it in our math engine[/URL] and we get: j = 41 The question asks how far Angus (a) threw the javelin. Since we have Jenny's distance j = 41 and equation (1) has j and a together, let's substitute j = 41 into equation (1): a = 41 - 4 a = [B]37 meters[/B]
Jinas final exam has true/false questions, worth 3 points each, and multiple choice questions, worth
Jinas final exam has true/false questions, worth 3 points each, and multiple choice questions, worth 4 points each. Let x be the number of true/false questions she gets correct, and let y be the number of multiple choice questions she gets correct. She needs at least 76 points on the exam to get an A in the class. Using the values and variables given, write an inequality describing this. At least means greater than or equal to, so we have: [B]3x + 4y >= 76[/B]
Joint Variation Equations
Free Joint Variation Equations Calculator - Given a joint variation (jointly proportional) of a variable between two other variables with a predefined set of conditions, this will create the joint variation equation and solve based on conditions. Also called combined variation.
Jonathan earns a base salary of \$1500 plus 10% of his sales each month. Raymond earns \$1200 plus 15%
Jonathan earns a base salary of \$1500 plus 10% of his sales each month. Raymond earns \$1200 plus 15% of his sales each month. How much will Jonathan and Raymond have to sell in order to earn the same amount each month? [U]Step 1: Set up Jonathan's sales equation S(m) where m is the amount of sales made each month:[/U] S(m) = Commission percentage * m + Base Salary 10% written as a decimal is 0.1. We want decimals to solve equations easier. S(m) = 0.1m + 1500 [U]Step 2: Set up Raymond's sales equation S(m) where m is the amount of sales made each month:[/U] S(m) = Commission percentage * m + Base Salary 15% written as a decimal is 0.15. We want decimals to solve equations easier. S(m) = 0.15m + 1200 [U]The question asks what is m when both S(m)'s equal each other[/U]: The phrase [I]earn the same amount [/I]means we set Jonathan's and Raymond's sales equations equal to each other 0.1m + 1500 = 0.15m + 1200 We want to isolate m terms on one side of the equation. Subtract 1200 from each side: 0.1m + 1500 - 1200 = 0.15m + 1200 - 1200 Cancel the 1200's on the right side and we get: 0.1m - 300 = 0.15m Next, we subtract 0.1m from each side of the equation to isolate m 0.1m - 0.1m + 300 = 0.15m - 0.1m Cancel the 0.1m terms on the left side and we get: 300 = 0.05m Flip the statement since it's an equal sign to get the variable on the left side: 0.05m = 300 To solve for m, we divide each side of the equation by 0.05: 0.05m/0.05 = 300/0.05 Cancelling the 0.05 on the left side, we get: m = [B]6000[/B]
larger of 2 numbers is 12 more than the smaller number. if the sum of the 2 numbers is 74 find the 2
larger of 2 numbers is 12 more than the smaller number. if the sum of the 2 numbers is 74 find the 2 numbers Declare Variables for each number: [LIST] [*]Let l be the larger number [*]Let s be the smaller number [/LIST] We're given two equations: [LIST=1] [*]l = s + 12 [*]l + s = 74 [/LIST] Equation (1) already has l solved for. Substitute equation (1) into equation (2) for l: s + 12 + s = 74 Solve for [I]s[/I] in the equation s + 12 + s = 74 [SIZE=5][B]Step 1: Group the s terms on the left hand side:[/B][/SIZE] (1 + 1)s = 2s [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 2s + 12 = + 74 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 12 and 74. To do that, we subtract 12 from both sides 2s + 12 - 12 = 74 - 12 [SIZE=5][B]Step 4: Cancel 12 on the left side:[/B][/SIZE] 2s = 62 [SIZE=5][B]Step 5: Divide each side of the equation by 2[/B][/SIZE] 2s/2 = 62/2 s = [B]31[/B] To solve for l, we substitute in s = 31 into equation (1): l = 31 + 12 l = [B]43[/B]
larger of 2 numbers is 4 more than the smaller. the sum of the 2 is 40. what is the larger number?
larger of 2 numbers is 4 more than the smaller. the sum of the 2 is 40. what is the larger number? Declare variables for the 2 numbers: [LIST] [*]Let l be the larger number [*]Let s be the smaller number [/LIST] We're given two equations: [LIST=1] [*]l = s + 4 [*]l + s = 40 [/LIST] To get this problem in terms of the larger number l, we rearrange equation (1) in terms of l. Subtract 4 from each side in equation (1) l - 4 = s + 4 - 4 Cancel the 4's and we get: s = l - 4 Our given equations are now: [LIST=1] [*]s = l - 4 [*]l + s = 40 [/LIST] Substitute equation (1) into equation (2) for s: l + l - 4 = 40 Grouping like terms for l, we get: 2l - 4 = 40 Add 4 to each side: 2l - 4 + 4 = 40 + 4 Cancelling the 4's on the left side, we get 2l = 44 Divide each side of the equation by 2 to isolate l: 2l/2 = 44/2 Cancel the 2's on the left side and we get: l = [B]22[/B]
let x be the variable, an age that is at least 57 years old
let x be the variable, an age that is at least 57 years old At least means greater than or equal to x >= 57
Literal Equations
Free Literal Equations Calculator - Solves literal equations with no powers for a variable of your choice as well as open sentences.
Logarithms and Natural Logarithms and Eulers Constant (e)
Free Logarithms and Natural Logarithms and Eulers Constant (e) Calculator - This calculator does the following:
* Takes the Natural Log base e of a number x Ln(x) → logex
* Raises e to a power of y, ey
* Performs the change of base rule on logb(x)
* Solves equations in the form bcx = d where b, c, and d are constants and x is any variable a-z
* Solves equations in the form cedx=b where b, c, and d are constants, e is Eulers Constant = 2.71828182846, and x is any variable a-z
* Exponential form to logarithmic form for expressions such as 53 = 125 to logarithmic form
* Logarithmic form to exponential form for expressions such as Log5125 = 3
M is the sum of a and its reciprocal
M is the sum of a and its reciprocal The reciprocal of a variable is 1 divided by the variable 1/a The sum of a and its reciprocal means we add: a + 1/a The phrase [I]is[/I] means an equation, so we set M equal to the sum of a + 1/a: [B]M = 1 + 1/a[/B]
Match each variable with a variable by placing the correct letter on each line.
Match each variable with a variable by placing the correct letter on each line. a) principal b) interest c) interest rate d) term/time 2 years 1.5% \$995 \$29.85 [B]Principal is \$995 Interest is \$29.85 since 995 * .0.15 * 2 = 29.85 Interest rate is 1.5% Term/time is 2 year[/B]s
Melissa runs a landscaping business. She has equipment and fuel expenses of \$264 per month. If she c
Melissa runs a landscaping business. She has equipment and fuel expenses of \$264 per month. If she charges \$53 for each lawn, how many lawns must she service to make a profit of at \$800 a month? Melissa has a fixed cost of \$264 per month in fuel. No variable cost is given. Our cost function is: C(x) = Fixed Cost + Variable Cost. With variable cost of 0, we have: C(x) = 264 The revenue per lawn is 53. So R(x) = 53x where x is the number of lawns. Now, profit is Revenue - Cost. Our profit function is: P(x) = 53x - 264 To make a profit of \$800 per month, we set P(x) = 800. 53x - 264 = 800 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=53x-264%3D800&pl=Solve']equation solver[/URL], we get: [B]x ~ 21 lawns[/B]
Mr. Chris’s new app “Tick-Tock” is the hottest thing to hit the app store since...ever. It costs \$5
Mr. Chris’s new app “Tick-Tock” is the hottest thing to hit the app store since...ever. It costs \$5 to buy the app and then \$2.99 for each month that you subscribe (a bargain!). How much would it cost to use the app for one year? Write an equation to model this using the variable “m” to represent the number of months that you use the app. Set up the cost function C(m) where m is the number of months you subscribe: C(m) = Monthly Subscription Fee * months + Purchase fee [B]C(m) = 2.99m + 5[/B]
multiply a number by 4 and then subtract the answer from 30
multiply a number by 4 and then subtract the answer from 30 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x Multiply this number by 4: 4x Subtract the answer from 30: [B]30 - 4x[/B]
Multiply a number by 6 and subtracting 6 gives the same result as multiplying the number by 3 and su
Multiply a number by 6 and subtracting 6 gives the same result as multiplying the number by 3 and subtracting 4. Find the number The phrase [I]a number [/I]means an arbitrary variable, let's call it x. multiply a number by 6 and subtract 6: 6x - 6 Multiply a number by 3 and subtract 4: 3x - 4 The phrase [I]gives the same result[/I] means an equation. So we set 6x - 6 equal to 3x - 4 6x - 6 = 3x - 4 To solve this equation for x, we type it in our search engine and we get: x = [B]2/3[/B]
Multiply the difference of 3 and q by p
Multiply the difference of 3 and q by p. Take this algebraic expression in pieces: [B][U]Step 1: The difference of 3 and q[/U][/B] The word [I]difference[/I] means we subtract the variable q from 3 3 - q [B][U]Step 2: Multiply the expression 3 - q by p:[/U] p(3 - q)[/B]
Multiplying a number by 6 is equal to the number increased by 9
Multiplying a number by 6 is equal to the number increased by 9. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Multiply it by 6 --> 6x We set this equal to the same number increased by 9. Increased by means we add: [B]6x = x + 9 <-- This is our algebraic expression [/B] To solve this equation, we [URL='https://www.mathcelebrity.com/1unk.php?num=6x%3Dx%2B9&pl=Solve']type it into the search engine [/URL]and get x = 1.8.
n + 9n - 8 - 5 = 2n + 3
n + 9n - 8 - 5 = 2n + 3 Solve for [I]n[/I] in the equation n + 9n - 8 - 5 = 2n + 3 [SIZE=5][B]Step 1: Group the n terms on the left hand side:[/B][/SIZE] (1 + 9)n = 10n [SIZE=5][B]Step 2: Group the constant terms on the left hand side:[/B][/SIZE] -8 - 5 = -13 [SIZE=5][B]Step 3: Form modified equation[/B][/SIZE] 10n - 13 = 2n + 3 [SIZE=5][B]Step 4: Group variables:[/B][/SIZE] We need to group our variables 10n and 2n. To do that, we subtract 2n from both sides 10n - 13 - 2n = 2n + 3 - 2n [SIZE=5][B]Step 5: Cancel 2n on the right side:[/B][/SIZE] 8n - 13 = 3 [SIZE=5][B]Step 6: Group constants:[/B][/SIZE] We need to group our constants -13 and 3. To do that, we add 13 to both sides 8n - 13 + 13 = 3 + 13 [SIZE=5][B]Step 7: Cancel 13 on the left side:[/B][/SIZE] 8n = 16 [SIZE=5][B]Step 8: Divide each side of the equation by 8[/B][/SIZE] 8n/8 = 16/8 n = [B]2[/B]
n - n = 10 - n
n - n = 10 - n Solve for [I]n[/I] in the equation n - n = 10 - n [SIZE=5][B]Step 1: Group the n terms on the left hand side:[/B][/SIZE] (1 - 1)n = 0n = 0 [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] = - n + 10 [SIZE=5][B]Step 3: Group variables:[/B][/SIZE] We need to group our variables and -n. To do that, we add n to both sides + n = -n + 10 + n [SIZE=5][B]Step 4: Cancel -n on the right side:[/B][/SIZE] n = [B]10[/B]
n = 3n - 1/2
n = 3n - 1/2 Solve for [I]n[/I] in the equation n = 3n - 1/2 [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables n and 3n. To do that, we subtract 3n from both sides n - 3n = 3n - 0.5 - 3n [SIZE=5][B]Step 2: Cancel 3n on the right side:[/B][/SIZE] -2n = -0.5 [SIZE=5][B]Step 3: Divide each side of the equation by -2[/B][/SIZE] -2n/-2 = -0.5/-2 n = [B]0.25 or 1/4[/B]
n = b + d^2a for a
n = b + d^2a for a Let's start by isolating the one term with the a variable. Subtract b from each side: n - b = b - b + d^2a Cancel the b terms on the right side and we get: n - b = d^2a With the a term isolated, let's divide each side of the equation by d^2: (n - b)/d^2 = d^2a/d^2 Cancel the d^2 on the right side, and we'll display this with the variable to solve on the left side: a = [B](n - b)/d^2 [MEDIA=youtube]BCEVsZmoKoQ[/MEDIA][/B]
Nancy is 10 years less than 3 times her daughters age. If Nancy is 41 years old, how old is her daug
Nancy is 10 years less than 3 times her daughters age. If Nancy is 41 years old, how old is her daughter? Declare variables for each age: [LIST] [*]Let Nancy's age be n [*]Let her daughter's age be d [/LIST] We're given two equations: [LIST=1] [*]n = 3d - 10 [*]n = 41 [/LIST] We set 3d - 10 = 41 and solve for d: Solve for [I]d[/I] in the equation 3d - 10 = 41 [SIZE=5][B]Step 1: Group constants:[/B][/SIZE] We need to group our constants -10 and 41. To do that, we add 10 to both sides 3d - 10 + 10 = 41 + 10 [SIZE=5][B]Step 2: Cancel 10 on the left side:[/B][/SIZE] 3d = 51 [SIZE=5][B]Step 3: Divide each side of the equation by 3[/B][/SIZE] 3d/3 = 51/3 d = [B]17[/B]
Nine less than a number is no more than 8 and no less than 3
Nine less than a number is no more than 8 and no less than 3 A number is denoted as an arbitrary variable, let's call it x. We have a double inequality: [LIST=1] [*]No more than 8 means less than or equal to 8 [*]No less than 3 means greater than or equal to 3 [/LIST] [B]3 <= x <= 8[/B]
Nine times the sum of a number and 6
Nine times the sum of a number and 6 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The sum of a number and 6 means we add 6 to x: x + 6 9 times the sum: [B]9(x + 6)[/B]
Normal Distribution
Free Normal Distribution Calculator - Calculates the probability that a random variable is less than or greater than a value or between 2 values using the Normal Distribution z-score (z value) method (Central Limit Theorem).
Also calculates the Range of values for the 68-95-99.7 rule, or three-sigma rule, or empirical rule. Calculates z score probability
One fifth of the square of a number
One fifth of the square of a number We have an algebraic expression. Let's break this into parts. [LIST=1] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x [*]The square of a number means we raise it to the power of 2. So we have x^2 [*]One-fifth means we have a fraction, where we divide our x^2 in Step 2 by 5. So we get our final answer below: [/LIST] [B]x^2/5[/B]
One-half a number is fifty
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. One-half a number means we divide x by2: x/2 The word [I]is[/I] means equal to. We set x/2 equal to 50 for our algebraic expression [B]x/2 = 50 [/B] If the problem asks us to solve for x, we cross multiply: x = 2 * 50 x = [B]100[/B]
One-half a number times fifteen
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. One-half a number means we multiply x by 1/2: x/2 Times fifteen means we multiply: [B]15x/2[/B]
One-third a number less two
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. One-third a number means we multiply x by 1/3: x/3 Less two means we subtract 2 [B]x/3 - 2[/B]
Penny bought a new car for \$25,000. The value of the car has decreased in value at rate of 3% each
Penny bought a new car for \$25,000. The value of the car has decreased in value at rate of 3% each year since. Let x = the number of years since 2010 and y = the value of the car. What will the value of the car be in 2020? Write the equation, using the variables above, that represents this situation and solve the problem, showing the calculation you did to get your solution. Round your answer to the nearest whole number. We have the equation y(x): y(x) = 25,000(0.97)^x <-- Since a 3 % decrease is the same as multiplying the starting value by 0.97 The problem asks for y(2020). So x = 2020 - 2010 = 10. y(10) = 25,000(0.97)^10 y(10) = 25,000(0.73742412689) y(10) = [B]18,435.60[/B]
Polynomial
Free Polynomial Calculator - This calculator will take an expression without division signs and combine like terms.
It will also analyze an polynomial that you enter to identify constant, variables, and exponents. It determines the degree as well.
Prizes hidden on a game board with 10 spaces. One prize is worth \$100, another is worth \$50, and tw
Imagine you are in a game show. Prizes hidden on a game board with 10 spaces. One prize is worth \$100, another is worth \$50, and two are worth \$10. You have to pay \$20 to the host if your choice is not correct. Let the random variable x be the winning (a) What is your expected winning in this game? (b) Determine the standard deviation of x. (Round the answer to two decimal places) (a) 100(0.1) + 50(0.1) + 10(0.2) - 20 = 10 + 5 + 2 - 20 = [B]-3[/B] (b) 3.3 using our [URL='http://www.mathcelebrity.com/statbasic.php?num1=+100,50,10&num2=+0.1,0.1,0.2&usep=usep&pl=Number+Set+Basics']standard deviation calculator[/URL]
product of a number and its reciprocal is increased by 7
product of a number and its reciprocal is increased by 7 The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Its reciprocal means we take the reciprocal of x: 1/x product of a number and its reciprocal: x * 1/x x/x The x's cancel giving us: 1 is increased by 7 means we add 7: 1 + 7 [B]8[/B]
quotient of the sum of 2 numbers and 6
quotient of the sum of 2 numbers and 6 The phrase [I]two numbers[/I] means we choose 2 arbitrary variables, let's call them x and y x, y The sum of 2 numbers: x + y quotient of the sum of 2 numbers and 6 [B](x + y)/6[/B]
quotient of the sum of 3 numbers and 3
quotient of the sum of 3 numbers and 3 The phrase [I]3 numbers[/I] means we choose 3 arbitrary variables: a, b,c The sum of the 3 numbers: a + b + c quotient of the sum of 3 numbers and 3 [B](a + b + c)/3[/B]
raise t to the 10th power, then find the quotient of the result and s
raise t to the 10th power, then find the quotient of the result and s Raise t to the 10th power means we use t as our variable and 10 as our exponent: t^10 The quotient means a fraction, where the numerator is t^10 and the denominator is s: [B]t^10/s[/B]
Rational Exponents - Fractional Indices
Free Rational Exponents - Fractional Indices Calculator - This calculator evaluates and simplifies a rational exponent expression in the form ab/c where a is any integer or any variable [a-z] while b and c are integers. Also evaluates the product of rational exponents
Rearrange the following equation to make x the subject, and select the correct rearrangement from th
Rearrange the following equation to make x the subject, and select the correct rearrangement from the list below 3x + 2y 1 -------- = --- 4x + y 3 [LIST] [*]x = 7y/13 [*]x = 7y/5 [*]x = -7y [*]x = -3y [*]x = 3y/5 [*]x = -5y/13 [*]x = -y [/LIST] Cross multiply: 3(3x - 2y) = 4x + y Multiply the left side through 9x - 6y = 4x + y Subtract 4x from each side and add 6y to each side 5x = 7y Divide each side by 5 to isolate x, the subject of an equation is the variable to the left [B]x = 7y/5[/B]
Rectangle Word Problem
Free Rectangle Word Problem Calculator - Solves word problems based on area or perimeter and variable side lengths
Sam purchased n notebooks. They were 4 dollars each. Write an equation to represent the total cost c
Sam purchased n notebooks. They were 4 dollars each. Write an equation to represent the total cost c that Sam paid. Cost Function is: [B]c = 4n[/B] Or, using n as a function variable, we write: c(n) = 4n
Seven less than 1/4 of a number is 9.
Seven less than 1/4 of a number is 9. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 1/4 of a number means we multiply x by 1/4: x/4 Seven less than this means we subtract 7 from x/4: x/4 - 7 The word [I]is[/I] means an equation, so we set x/4 - 7 equal to 9: [B]x/4 - 7 = 9[/B]
Seven subtracted from the product of 3 and a number is greater than or equal to -26
Seven subtracted from the product of 3 and a number is greater than or equal to -26 [LIST=1] [*]A number means an arbitrary variable, let's call it x. [*]The product of 3 and a number is written as 3x [*]Seven subtracted from 3x is written as 3x - 7 [*]Finally, that entire expression is greater than or equal to -26: [B]3x - 7 >= - 26[/B] [/LIST]
Six less than twice a number is at least -1 and at most 1
First, the phrase [I]a number[/I] means we choose an arbitrary variable, let's call it x. Twice a number means we multiply it by 2. 2x Six less than that means we subtract 6 2x - 6 Now, the last piece, we set up an inequality. At least -1 means greater than or equal to 1. At most 1 means less than or equal to 1. Notice, for both points, we include the number. -1 <= 2x - 6 <= 1
Sixteen subtracted from five times a number equals the number plus four
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. 5 times a number 5x Sixteen subtracted from five times a number 5x - 16 the number plus 4: x + 4 Equals means we set 5x - 16 equals to x + 4 for our algebraic expression: [B]5x - 16 = x + 4[/B] [B][/B] If you have to solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=5x-16%3Dx%2B4&pl=Solve']type this expression into our math solver[/URL] and we get: x = [B]5[/B]
Soda cans are sold in a local store for 50 cents each. The factory has \$900 in fixed costs plus 25 c
Soda cans are sold in a local store for 50 cents each. The factory has \$900 in fixed costs plus 25 cents of additional expense for each soda can made. Assuming all soda cans manufactured can be sold, find the break-even point. Calculate the revenue function R(c) where s is the number of sodas sold: R(s) = Sale Price * number of units sold R(s) = 50s Calculate the cost function C(s) where s is the number of sodas sold: C(s) = Variable Cost * s + Fixed Cost C(s) = 0.25s + 900 Our break-even point is found by setting R(s) = C(s): 0.25s + 900 = 50s We [URL='https://www.mathcelebrity.com/1unk.php?num=0.25s%2B900%3D50s&pl=Solve']type this equation into our search engine[/URL] and we get: s = [B]18.09[/B]
square root of the sum of 2 variables
square root of the sum of 2 variables The phrase [I]2 variables[/I] means we choose 2 arbitrary variables, let's call them x and y: x, y The sum of 2 variables means we add: x + y Square root of the sum of 2 variables is written as: [B]sqrt(x + y)[/B]
Squaring a number equals 5 times that number
Squaring a number equals 5 times that number. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Squaring this number: x^2 5 times this number means we multiply by 5: 5x The phrase [I]equals[/I] means we set both expressions equal to each other: [B]x^2 = 5x [/B] <-- This is our algebraic expression If you want to solve for x, then we subtract 5x from each side: x^2 - 5x = 5x - 5x Cancel the 5x on the right side, leaving us with 0: x^2 - 5x = 0 Factor out x: x(x - 5) So we get x = 0 or [B]x = 5[/B]
Start with t and cube it. Cubing a variable means raising it to the power of 3: [B]t^3[/B]
Substitute the given values into given formula and solve for the unknown variable. S=4LW + 2 WH; S=
Substitute the given values into given formula and solve for the unknown variable. S = 4LW + 2 WH; S= 144, L= 8, W= 4. H= S = 4LW + 2 WH Substituting our given values, we have: 144 = 4(8)(4) + 2(4)H 144 = 128 + 8H Using our [URL='http://www.mathcelebrity.com/1unk.php?num=128%2B8h%3D144&pl=Solve']equation calculator[/URL], we get: [B]H = 2[/B]
subtract half of a number from 10
subtract half of a number from 10 The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x half of a number means we divide x by 2: x/2 subtract half of a number from 10 [B]10 - x/2[/B]
sum of 3 consecutive odd integers equals 1 hundred 17
sum of 3 consecutive odd integers equals 1 hundred 17 The sum of 3 consecutive odd numbers equals 117. What are the 3 odd numbers? 1) Set up an equation where our [I]odd numbers[/I] are n, n + 2, n + 4 2) We increment by 2 for each number since we have [I]odd numbers[/I]. 3) We set this sum of consecutive [I]odd numbers[/I] equal to 117 n + (n + 2) + (n + 4) = 117 [SIZE=5][B]Simplify this equation by grouping variables and constants together:[/B][/SIZE] (n + n + n) + 2 + 4 = 117 3n + 6 = 117 [SIZE=5][B]Subtract 6 from each side to isolate 3n:[/B][/SIZE] 3n + 6 - 6 = 117 - 6 [SIZE=5][B]Cancel the 6 on the left side and we get:[/B][/SIZE] 3n + [S]6[/S] - [S]6[/S] = 117 - 6 3n = 111 [SIZE=5][B]Divide each side of the equation by 3 to isolate n:[/B][/SIZE] 3n/3 = 111/3 [SIZE=5][B]Cancel the 3 on the left side:[/B][/SIZE] [S]3[/S]n/[S]3 [/S]= 111/3 n = 37 Call this n1, so we find our other 2 numbers n2 = n1 + 2 n2 = 37 + 2 n2 = 39 n3 = n2 + 2 n3 = 39 + 2 n3 = 41 [SIZE=5][B]List out the 3 consecutive odd numbers[/B][/SIZE] ([B]37, 39, 41[/B]) 37 ← 1st number, or the Smallest, Minimum, Least Value 39 ← 2nd number 41 ← 3rd or the Largest, Maximum, Highest Value
sum of a number and 7 is subtracted from 15 the result is 6.
Sum of a number and 7 is subtracted from 15 the result is 6. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. We take this expression in pieces. Sum of a number and 7 x + 7 Subtracted from 15 15 - (x + 7) The result is means an equation, so we set this expression above equal to 6 [B]15 - (x + 7) = 6 <-- This is our algebraic expression[/B] If the problem asks you to solve for x, we Group like terms 15 - x - 7 = 6 8 - x = 6 [URL='https://www.mathcelebrity.com/1unk.php?num=8-x%3D6&pl=Solve']Type 8 - x = 6 into the search engine[/URL], and we get [B]x = 2[/B]
Suppose that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in ga
Suppose that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in gallons) in its tank. When carrying 20 gallons of fuel, the airplane weighs 2012 pounds. When carrying 55 gallons of fuel, it weighs 2208 pounds. How much does the airplane weigh if it is carrying 65 gallons of fuel? Linear functions are written in the form of one dependent variable and one independent variable. Using g as the number of gallons and W(g) as the weight, we have: W(g) = gx + c where c is a constant We are given: [LIST] [*]W(20) = 2012 [*]W(55) = 2208 [/LIST] We want to know W(65) Using our givens, we have: W(20) = 20x + c = 2012 W(55) = 55x + c = 2208 Rearranging both equations, we have: c = 2012 - 20x c = 2208 - 55x Set them both equal to each other: 2012 - 20x = 2208 - 55x Add 55x to each side: 35x + 2012 = 2208 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=35x%2B2012%3D2208&pl=Solve']equation solver[/URL], we see that x is 5.6 Plugging x = 5.6 back into the first equation, we get: c = 2012 - 20(5.6) c = 2012 - 112 c = 2900 Now that we have all our pieces, find W(65) W(65) = 65(5.6) + 2900 W(65) = 264 + 2900 W(65) = [B]3264[/B]
Ten subtracted from the product of 9 and a number is less than −24
Ten subtracted from the product of 9 and a number is less than −24. A number means an arbitrary variable, let's call it x x The product of 9 and a number: 9x Ten subtracted from that 9x - 10 Finally, is less than means we set our entire expression less than -24 [B]9x - 10 < -24[/B]
Ten times the sum of twice a number and six
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. Twice a number means we multiply x by 2: 2x The sum of twice a number and 6: 2x + 6 Ten times the sum of twice a number and six [B]10(2x + 6)[/B]
The cost of having a plumber spend h hours at
The cost of having a plumber spend h hours at your house if the plumber charges \$60 for coming to the house and \$70 per hour labor: We have a fixed cost of \$60 plus the variable cost of \$70h. We add both for our total cost C(h): [B]C(h) = \$70h + 60[/B]
The cost to rent a boat is \$10. There is also charge of \$2 for each person. Which expresion represen
The cost to rent a boat is \$10. There is also charge of \$2 for each person. Which expresion represents the total cost to rent a boat for p persons? The cost function includes a fixed cost of \$10 plus a variable cost of 2 persons for p persons: [B]C(p) = 2p + 10[/B]
the cube of c decreased by a^2
the cube of c decreased by a^2 The cube of means we raise the variable c to the power of 3: c^3 The phrase [I]decreased by[/I] means we subtract: [B]c^3 - a^2[/B]
the difference between 7 times a number and 9 less than a number
the difference between 7 times a number and 9 less than a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 7 times a number means we multiply x by 7 7x 9 less than a number means we subtract 9 from x x - 9 The difference between the two expressions means we subtract (x - 9) from 7x 7x - (x - 9) Simplifying this, we have: 7x - x + 9 Grouping like terms, we get: [B]6x + 9[/B]
The difference between a number and 9 is 27. Find that number
The difference between a number and 9 is 27. Find that number The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x The difference between a number and 9 x - 9 The word [I]is[/I] means equal to, so we set x - 9 equal to 27: x - 9 = 27 To solve this equation for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=x-9%3D27&pl=Solve']type it in our math engine[/URL] and we get: x = [B]36[/B]
The difference between the opposite of a number and 6.
The difference between the opposite of a number and 6. The phrase [I]a number means[/I] an arbitrary variable, let's call it x. x The opposite of a number means we multiply by x by -1 -x The phrase [I]the difference between[/I] means we subtract 6 from -x: [B]-x - 6[/B]
The difference between the product of 4 and a number and the square of a number
The difference between the product of 4 and a number and the square of a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The product of 4 and a number: 4x The square of a number means we raise x to the power of 2: x^2 The difference between the product of 4 and a number and the square of a number: [B]4x - x^2[/B]
the difference between triple a number and double a number
the difference between triple a number and double a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Triple a number means we multiply x by 3: 3x Double a number means we multiply x by 2: 2x The difference means we subtract 2x from 3x: 3x - 2x Simplifying like terms, we have: (3 - 2)x = [B]x[/B]
The difference in Julies height and 9 is 48 letting j be Julie's height
The difference in Julies height and 9 is 48 letting j be Julie's height Step 1: If Julie's height is represented with the variable j, then we subtract 9 from j since the phrase [I]difference[/I] means we subtract: j - 9 Step 2: The word [I]is[/I] means an equation, so we set j - 9 equal to 48 for our final algebraic expression: [B]j - 9 = 48[/B]
The difference of 25 and a number added to triple another number
The difference of 25 and a number added to triple another number The phrase [I]a number [/I]means an arbitrary variable, let's call it x: x The difference of 25 and a number means we subtract x from 25: 25 - x The phrase [I]another number[/I] means a different arbitrary variable, let's call it y: y Triple another number means we multiply y by 3: 3y The phrase [I]added to[/I] means we add 25 - x to 3y [B]25 - x + 3y[/B]
the difference of 4 and the quotient of 18 and a number
the difference of 4 and the quotient of 18 and a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The quotient of 18 and a number means we divide 18 by the variable x. 18/x The difference of 4 and the quotient above means we subtract 18/x from 4: [B]4 - 18/x[/B]
The difference of a number and 6 is the same as 5 times the sum of the number and 2. What is the num
The difference of a number and 6 is the same as 5 times the sum of the number and 2. What is the number? We have two expressions: [U]Expression 1: [I]The difference of a number and 6[/I][/U] The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The difference of a number and 6 means we subtract 6 from x: x - 6 [U]Expression 2: [I]5 times the sum of the number and 2[/I][/U] The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The sum of a number and 2 means we add 2 to x: x + 2 5 times the sum means we multiply x + 2 by 5 5(x + 2) [U]For the last step, we evaluate the expression [I]is the same as[/I][/U] This means equal to, so we set x - 6 equal to 5(x + 2) [B]x - 6 = 5(x + 2)[/B]
The difference of a number times 3 and 6 is equal to 7 . Use the variable w for the unknown n
The difference of a number times 3 and 6 is equal to 7 . Use the variable w for the unknown number. The phrase a number uses the variable w. 3 times w is written as 3w The difference of 3w and 6 is written as 3w - 6 Set this equal to 7 [B]3w - 6 = 7 [/B] This is our algebraic expression. To solve this equation for w, we [URL='http://www.mathcelebrity.com/1unk.php?num=3w-6%3D7&pl=Solve']type the algebraic expression into our search engine[/URL].
The difference of twice a number and 4 is at least -27
The difference of twice a number and 4 is at least -27. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x Twice a number means multiply the number by 2 2x [I]and 4[/I] means we add 4 to our expression: 2x + 4 [I]Is at least[/I] means an inequality. In this case, it's greater than or equal to: [B]2x + 4 >= -27 [/B] To solve this inequality, [URL='https://www.mathcelebrity.com/1unk.php?num=2x%2B4%3E%3D-27&pl=Solve']type it in the search engine[/URL].
The difference of twice a number and 6 is at most 28
The difference of twice a number and 6 is at most 28 This is an algebraic expression. Let's take it in parts: [LIST=1] [*]The phrase [I]a number[/I], means an arbitrary variable, let's call it x [*]Twice this number means we multiply x by 2: 2x [*][I]The difference of[/I] means subtract, so we subtract 6 to 2x: 2x - 6 [*][I]Is at most [/I]means less than or equal to, so we create an inequality where 2x - 6 is less than or equal to 28, using the <= sign [/LIST] [B]2x - 6 <= 28 [/B] If you wish to solve this inequality, [URL='https://www.mathcelebrity.com/1unk.php?num=2x-6%3C%3D28&pl=Solve']click this link[/URL].
the difference of twice a number and 8 is at most -30
the difference of twice a number and 8 is at most -30. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Twice this number means we multiply by 2, so we have 2x. We take the difference of 2x and 8, meaning we subtract 8: 2x - 8 Finally, the phrase [I]at most[/I] means an inequality, also known as less than or equal to: [B]2x - 8 <= 30 <-- This is our algebraic expression [/B] To solve this, we [URL='https://www.mathcelebrity.com/1unk.php?num=2x-8%3C%3D30&pl=Solve']type it into the search engine[/URL] and get x <= 19.
The difference of twice a number and 9 is less than 22
The difference of twice a number and 9 is less than 22 The phrase a number, means an arbitrary variable, let's call it x. x Twice a number 2x The difference of twice a number and 9 2x - 9 Is less than 22 [B]2x - 9 < 22[/B]
The fixed costs to produce a certain product are 15,000 and the variable costs are \$12.00 per item.
The fixed costs to produce a certain product are 15,000 and the variable costs are \$12.00 per item. The revenue for a certain product is \$27.00 each. If the company sells x products, then what is the revenue equation? R(x) = Revenue per item x number of products sold [B]R(x) = 27x[/B]
The phone company charges Rachel 12 cents per minute for her long distance calls. A discount company
The phone company charges Rachel 12 cents per minute for her long distance calls. A discount company called Rachel and offered her long distance service for 1/2 cent per minute, but will charge a \$46 monthly fee. How many minutes per month must Rachel talk on the phone to make the discount a better deal? Minutes Rachel talks = m Current plan cost = 0.12m New plan cost = 0.005m + 46 Set new plan equal to current plan: 0.005m + 46 = 0.12m Solve for [I]m[/I] in the equation 0.005m + 46 = 0.12m [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables 0.005m and 0.12m. To do that, we subtract 0.12m from both sides 0.005m + 46 - 0.12m = 0.12m - 0.12m [SIZE=5][B]Step 2: Cancel 0.12m on the right side:[/B][/SIZE] -0.115m + 46 = 0 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 46 and 0. To do that, we subtract 46 from both sides -0.115m + 46 - 46 = 0 - 46 [SIZE=5][B]Step 4: Cancel 46 on the left side:[/B][/SIZE] -0.115m = -46 [SIZE=5][B]Step 5: Divide each side of the equation by -0.115[/B][/SIZE] -0.115m/-0.115 = -46/-0.115 m = [B]400 She must talk over 400 minutes for the new plan to be a better deal [URL='https://www.mathcelebrity.com/1unk.php?num=0.005m%2B46%3D0.12m&pl=Solve']Source[/URL][/B]
the product of 2 less than a number and 7 is 13
the product of 2 less than a number and 7 is 13 Take this algebraic expression in [U]4 parts[/U]: Part 1 - The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x Part 2 - 2 less than a number means we subtract 2 from x x - 2 Part 3 - The phrase [I]product[/I] means we multiply x - 2 by 7 7(x - 2) Part 4 - The phrase [I]is[/I] means an equation, so we set 7(x - 2) equal to 13 [B]7(x - 2) = 13[/B]
the product of 8 and 15 more than a number
the product of 8 and 15 more than a number. Take this algebraic expression in pieces. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 15 more than x means we add 15 to x: x + 15 The product of 8 and 15 more than a number means we multiply 8 by x + 15 [B]8(x + 15)[/B]
the product of a number and 15 is not less than 15
the product of a number and 15 is not less than 15 The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. x the product of a number and 15 means we multiply x by 15 15x The phrase [I]not less than[/I] means greater than or equal to. We set 15x greater than prequel to 15 [B]15x >= 15 <-- This is our algebraic expression [/B] [U]If the problem asks you to solve for x:[/U] Divide each side by 15: 15x/15 >= 15/15 [B]x >= 1[/B]
The product of the 2 numbers x and y
The product of the 2 numbers x and y The phrase [I]product [/I]means we multiply the two variables, x and y. [B]xy[/B]
The quotient of 2 and the sum of a number and 1
The quotient of 2 and the sum of a number and 1. The phrase [I]a number[/I] represents an arbitrary variable, let's call it x. The sum of a number and 1 is written as: x + 1 The word [I]quotient[/I] means a fraction. So we divide 2 by x + 1 2 -------- ( x + 1) [MEDIA=youtube]uPQHCKr-9vA[/MEDIA]
the quotient of 4 more than a number and 7 is 10
the quotient of 4 more than a number and 7 is 10 Take this algebraic expression in pieces: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 4 more than a number means we add 4 to x: x + 4 The quotient of 4 more than a number and 7 means we divide x + 4 by 7 (x + 4)/7 The word [I]is[/I] means an equation, so we set (x + 4)/7 equal to 10 to get our algebraic expression of: [B](x + 4)/7 = 10 [/B] If the problem asks you to solve for x, then we cross multiply: x + 4 = 10 * 7 x + 4 = 70 Subtract 4 from each side: x + 4 - 4 = 70 - 4 x = [B]66 [MEDIA=youtube]j-GZLPVKbTM[/MEDIA][/B]
the quotient of a number and twice another number
the quotient of a number and twice another number The phrase[I] a number [/I]means an arbitrary variable, let's call it x. The phrase[I] another number [/I]means another arbitrary variable, let's call it y. Twice means we multiply y by 2:2y The quotient means we divide x by 2y: [B]x/2y[/B]
the quotient of a variable and 7
the quotient of a variable and 7. A variable means an arbitrary number, let's call it x. A quotient means a fraction, where x is the numerator and 7 is the denominator: [B] x --- 7[/B]
the quotient of the cube of a number x and 5
the quotient of the cube of a number x and 5 [LIST] [*]A number means an arbitrary variable, let's call it x [*]The cube of a number means raise it to the 3rd power, so we have x^3 [*]Quotient means we have a fraction, so our numerator is x^3, and our denominator is 5 [/LIST] [B]x^3 ---- 5[/B]
the ratio of 50 and a number added to the quotient of a number and 10
the ratio of 50 and a number added to the quotient of a number and 10 Take this algebraic expression in parts. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The ratio of 50 and x means we divide by 50 by x 50/x The quotient of a number and 10 means we have a fraction: x/10 The phrase [I]added to[/I] means we add 50/x to x/10 [B]50/x + x/10[/B]
the ratio of a number x and 4 added to 2
the ratio of a number x and 4 added to 2 Take this algebraic expression in parts. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The ratio of this number and 4 means we have a fraction: x/4 The phrase [I]added to[/I] means we add 2 to x/4 [B]x/4 + 2[/B]
the ratio of ten to a number
the ratio of ten to a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The ratio of 10 and this number x is written as: [B]10/x[/B]
The square of a number added to its reciprocal
The square of a number added to its reciprocal The phrase [I]a number [/I]means an arbitrary variable, let's call it x. the square of x mean we raise x to the power of 2. It's written as: x^2 The reciprocal of x is 1/x We add these together to get our final algebraic expression: [B]x^2 + 1/x [MEDIA=youtube]ZHut58-AoDU[/MEDIA][/B]
The square of a number increased by 7 is 23
The square of a number increased by 7 is 23 The phrase [I]a number [/I]means an arbitrary variable, let's call it x. x The square of a number means we raise x to the power of 2: x^2 [I]Increased by[/I] means we add 7 to x^2 x^2 + 7 The word [I]is[/I] means an equation. So we set x^2 + 7 equal to 23: [B]x^2 + 7 = 23[/B]
The square of the difference of a number and 4
The square of the difference of a number and 4 A number means an arbitrary variable, let's call it x The difference of a number and 4: x - 4 The square of this difference: [B](x - 4)^2[/B]
The sum of 13 and twice janelles age
Let Janelle's age be the variable a. So twice Janelle's age is denoted as 2a. We want the sum of 13 and 2a. Sum means add. 13 + 2a or 2a + 13
the sum of 16 and twice julies savings use the variable j to represent julies savings
The sum of 16 and twice julies savings use the variable j to represent julies savings Twice Julie's savings: 2j The sum of 16 and twice Julie's savings: [B]2j + 16[/B]
the sum of 2 times a number and -2, added to 4 times a number
the sum of 2 times a number and -2, added to 4 times a number. The phrase, [I]a number[/I], means an arbitrary variable, let's call it x. 2 times a number 2x The sum of means add, so we add -2, which is the same as subtracting 2 2x - 2 Now, we add 4 times x 2x - 2 + 4x Combining like terms, we have: (2 + 4)x - 2 [B]6x - 2[/B]
the sum of 3 numbers
Since no variable name is defined, we pick 3 arbitrary variables. Let's pick x, y, and z. The sum of 3 numbers means we add them together: x + y + z
the sum of 3 numbers
Let's choose 3 arbitrary variables, w, x, and y. We add them up: [B]w + x + y[/B]
the sum of 3 numbers divided by its product
the sum of 3 numbers divided by its product The phrase [I]3 numbers[/I] means we choose [I]3[/I] arbitrary variables. Let's call them x, y, z. The sum of of these 3 numbers is: x + y + z The phrase [I]its product[/I] means we multiply all 3 arbitrary variables together: xyz Now, the phrase [I]divided by[/I] means we divide x + y + z by xyz: [B](x + y + z)/xyz[/B]
The sum of 3 times the square of a number and negative 7
The sum of 3 times the square of a number and negative 7 [U]The phrase [I]a number[/I] means an arbitrary variable, let's call it x:[/U] x [U]The square of a number means we raise x to the power of 2:[/U] x^2 [U]3 times the square of a number:[/U] 3x^2 [U]The sum of 3 times the square of a number and negative 7[/U] [B]3x^2 - 7[/B]
The sum of 5, 8, and a number amounts to 19. Find the number.
The sum of 5, 8, and a number amounts to 19. Find the number. We represent [I]a number[/I] with the variable "x". We write our problem as: 5 + 8 + x = 19 13 + x = 19 [URL='https://www.mathcelebrity.com/1unk.php?num=13%2Bx%3D19&pl=Solve']Type this problem into our calculator[/URL], and we get [B]x = 6[/B].
the sum of 6 and 7, plus 5 times a number, is -12
the sum of 6 and 7, plus 5 times a number, is -12 The sum of 6 and 7 means we add the two numbers: 6 + 7 This evaluates to 13 Next, we take 5 times a number. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. So we multiply x by 5: 5x The first two words say [I]the sum[/I], so we add 13 and 5x 13 + 5x The word [I]is[/I] means an equation, so we set 13 + 5x equal to -12 [B]13 + 5x = -12[/B] <-- This is our algebraic expression If the problem asks you to take it a step further and solve for x, then you [URL='https://www.mathcelebrity.com/1unk.php?num=13%2B5x%3D-12&pl=Solve']type this algebraic expression into our search engine[/URL] and you get: [B]x = -5[/B]
The sum of 6 times a number and -8, added to 3 times a number
The sum of 6 times a number and -8, added to 3 times a number The phrase "a number", means an arbitrary variable, let's call it x. 6 times a number: 6x And means we add, so we have 6x - 8 Added to 3 times a number 6x - 8 + 3x Combine like terms: [B]9x - 8[/B]
The sum of a and d is
Since it's a sum, we add our two variables a and d. a + d
the sum of a number and 16 is e
A number means an arbitrary variable, let's call it x. The sum of x and 16 means we add: x + 16 Is, means equal to, so we set x + 16 = e x + 16 = e
The sum of a number and 34 times the number
The sum of a number and 34 times the number The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. x 34 times the number: 34x The sum of a number and 34 times the number means we add both terms together: x + 34x
The sum of a number and 5 all divided by 2 is 17
The sum of a number and 5 all divided by 2 is 17 The phrase [I]a number[/I] means an arbitrary variable, let's call it x x The sum of a number and 5: x + 5 All divided by 2: (x + 5)/2 The word [I]is[/I] means equal to, so we set (x + 5)/2 equal to 17: [B](x + 5)/2 = 17[/B]
The sum of a number and 5 divided by 8
The sum of a number and 5 divided by 8. Let's take this algebraic expression in parts. Part 1: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Part 2: The sum of a number and 5 means we add 5 to the number x x + 5 Part 3: Next, we divide this expression by 8 [B](x + 5)/8[/B]
the sum of a number and its reciprocal is 5/2
the sum of a number and its reciprocal is 5/2 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The reciprocal of the number means 1/x. The sum means we add them: x + 1/x The word [I]is[/I] means an equation, so we set x + 1/x equal to 52 [B]x + 1/x = 52[/B]
The sum of a number and its reciprocal is 72
The sum of a number and its reciprocal is 72 The phrase [I]a number[/I] means an arbitrary variable, let's call it x x The reciprocal of the number is written as: 1/x The sum of a number and its reciprocal means we add: x + 1/x The word [I]is[/I] means an equation, so we set x + 1/x equal to 72 [B]x + 1/x = 72[/B]
The sum of a number and its reciprocal is five.
The sum of a number and its reciprocal is five. The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. The reciprocal of the number is 1/x. The sum means we add them together: x + 1/x The word [I]is[/I] means an equation, so we set x + 1/x equal to 5 [B]x + 1/x = 5[/B]
the sum of a number and itself is 8
A number means an arbitrary variable, let's call it x. The sum of a number and itself means adding the number to itself x + x Simplified, we have 2x The word is means equal to, so we have an algebraic expression of: [B]2x= 8 [/B] IF you need to solve this equation, divide each side by 2 [B]x = 4[/B]
The sum of a number and twice its reciprocal is 3
The sum of a number and twice its reciprocal is 3 the phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The reciprocal of a number means we take 1 over that number: 1/x Twice the reciprocal means we multiply 1/x by 2: 2/x The sum of a number and twice its reciprocal x + 2/x The word [I]is[/I] means equal to, so we set x + 2/x equal to 3 [B]x + 2/x = 3[/B]
the sum of a number divided by 8 and 3 equals 6
"A Number" means an arbitrary variable, let's call it x. x divide d by 8 is written as a quotient x/8 The sum of x/8 and 3 means we add: x/8 + 3 Finally, equals means we have an equation, so we set our expression above equal to 6 x/8 + 3 = 6
The sum Of a number squared and 14
The sum Of a number squared and 14. A number means an arbitrary variable, let's call it x. Squared means we raise x to the 2nd power: x^2 The sum means we add x^2 to 14 to get our algebraic expression below: [B]x^2 + 14[/B]
the sum of a number times 3 and 30 is less than 17
the sum of a number times 3 and 30 is less than 17 A number is denoted as an arbitrary variable, let's call it x. x Times 3 means we multiply x by 3: 3x The sum of a number times 3 and 30 means we add 30 to 3x above 3x + 30 Is less than 17 means we have an inequality, so we set 3x + 30 less than 17 3x + 30 < 17 To solve for x and see the interval notation, use [URL='http://www.mathcelebrity.com/1unk.php?num=3x%2B30%3C17&pl=Solve']our calculator[/URL]:
The sum of five and twice a number is 17
The sum of five and twice a number is 17 [U]The phrase a number means an arbitrary variable, let's call it x[/U] x [U]Twice a number means we multiply x by 2:[/U] 2x [U]The sum of five and twice a number means we add 5 to 2x:[/U] 2x + 5 [U]The phrase [I]is[/I] means an equation, so we set 2x + 5 equal to 17 to get our algebraic expression[/U] [B]2x + 5 = 17[/B] [B][/B] As a bonus, if the problem asks you to solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=2x%2B5%3D17&pl=Solve']type in this algebraic expression into our math engine[/URL] and we get: x = 6
The sum of six times a number and 1 is equal to five times the number. Find the number.
The sum of six times a number and 1 is equal to five times the number. Find the number. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. 6 times a number is written as: 6x the sum of six times a number and 1 is written as: 6x + 1 Five times the number is written as: 5x The phrase [I]is equal to[/I] means an equation, so we set 6x + 1 equal to 5x: 6x + 1 = 5x [URL='https://www.mathcelebrity.com/1unk.php?num=6x%2B1%3D5x&pl=Solve']Plugging this into our search engine[/URL], we get: x = [B]-1[/B]
the sum of the cube of a number and 12
the sum of the cube of a number and 12 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The cube of a number means we raise x to the power of 3: x^3 Finally, we take the sum of x^3 and 12. Meaning, we add 12 to x^3. This is our final algebraic expression. [B]x^3 + 12[/B]
The sum of the reciprocals of x and y
The sum of the reciprocals of x and y The reciprocal of a variable is found by taking 1 over the variable. [LIST] [*]Reciprocal of x = 1/x [*]Reciprocal of y = 1/y [/LIST] The sum means we add the reciprocals together [B]1/x + 1/y[/B]
The sum of the square of a number and 7 times a number
The sum of the square of a number and 7 times a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x Square the number: x^2 7 times the number means we multiply x by 7: 7x The sum means we add x^2 and 7x [B]x^2 + 7x[/B]
The sum of three consecutive integers is 42
Let the 3 integers be x, y, and z. y = x + 1 z = y + 1, or x + 2. Set up our equation: x + (x + 1) + (x + 2) = 42 Group our variables and constants: (x + x + x) + (1 + 2) = 42 3x + 3 = 42 Subtract 3 from each side: 3x = 39 Divide each side of the equation by 3: [B]x = 13 So y = x + 1 = 14 z = x + 2 = 15 (x,y,z) = (13,14,15)[/B]
The Sum of three times a number and 18 is -39. Find the number
The Sum of three times a number and 18 is -39. Find the number. A number means an arbitrary variable, let us call it x. Three times x: 3x The sum of this and 18: 3x + 18 Is means equal to, so we set 3x + 18 = -39 3x + 18 = -39 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3x%2B18%3D-39&pl=Solve']equation solver[/URL], we get [B]x = -19[/B]
The sum of three times a number and twelve
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. 3 times a number: 3x The sum of three times a number and twelve means we add 12 to 3x: [B]3x + 12[/B]
The sum of two numbers multiplied by 9
Choose two variables as arbitrary numbers, let's say x and y [U]The sum of x and y is:[/U] x + y [U]Multiply that by 9[/U] [B]9(x + y)[/B]
The total cost for 9 bracelets, including shipping was \$72. The shipping charge was \$9. Define your
The total cost for 9 bracelets, including shipping was \$72. The shipping charge was \$9. Define your variable and write an equation that models the cost of each bracelet. We set up a cost function as fixed cost plus total cost. Fixed cost is the shipping charge of \$9. So we have the following cost function where n is the cost of the bracelets: C(b) = nb + 9 We are given C(9) = 72 and b = 9 9n + 9 = 72 [URL='https://www.mathcelebrity.com/1unk.php?num=9n%2B9%3D72&pl=Solve']Run this through our equation calculator[/URL], and we get [B]n = 7[/B].
The total cost of producing x units for which the fixed costs are \$2900 and the cost per unit is \$25
The total cost of producing x units for which the fixed costs are \$2900 and the cost per unit is \$25 [U]Set up the cost function:[/U] Cost function = Fixed Cost + Variable Cost per Unit * Number of Units [U]Plug in Fixed Cost = 2900 and Cost per Unit = \$25[/U] [B]C(x) = 2900 + 25x [MEDIA=youtube]77PiD-VADjM[/MEDIA][/B]
The volleyball team and the wrestling team at Clarksville High School are having a joint car wash t
The volleyball team and the wrestling team at Clarksville High School are having a joint car wash today, and they are splitting the revenues. The volleyball team gets \$4 per car. In addition, they have already brought in \$81 from past fundraisers. The wrestling team has raised \$85 in the past, and they are making \$2 per car today. After washing a certain number of cars together, each team will have raised the same amount in total. What will that total be? How many cars will that take? Set up the earnings equation for the volleyball team where w is the number of cars washed: E = Price per cars washed * w + past fundraisers E = 4w + 81 Set up the earnings equation for the wrestling team where w is the number of cars washed: E = Price per cars washed * w + past fundraisers E = 2w + 85 If the raised the same amount in total, set both earnings equations equal to each other: 4w + 81 = 2w + 85 Solve for [I]w[/I] in the equation 4w + 81 = 2w + 85 [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables 4w and 2w. To do that, we subtract 2w from both sides 4w + 81 - 2w = 2w + 85 - 2w [SIZE=5][B]Step 2: Cancel 2w on the right side:[/B][/SIZE] 2w + 81 = 85 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 81 and 85. To do that, we subtract 81 from both sides 2w + 81 - 81 = 85 - 81 [SIZE=5][B]Step 4: Cancel 81 on the left side:[/B][/SIZE] 2w = 4 [SIZE=5][B]Step 5: Divide each side of the equation by 2[/B][/SIZE] 2w/2 = 4/2 w = [B]2 <-- How many cars it will take [/B] To get the total earnings, we take either the volleyball or wrestling team's Earnings equation and plug in w = 2: E = 4(2) + 81 E = 8 + 81 E = [B]89 <-- Total Earnings[/B]
There are 7 more jeeps than vans.
There are 7 more jeeps than vans. [U]Define variables[/U] [LIST] [*]Let j be the number of jeeps [*]Let v be the number of vans [/LIST] 7 more jeeps than vans means we add 7 to the number of vans: [B]j = v + 7[/B]
thesumof9andanumber
We denoted a number using the arbitrary variable "x". The sum of 9 and x is written: x + 9 or 9 + x
Thirty is half of the sum of 4 and a number
Thirty is half of the sum of 4 and a number. The phrase [I]a number[/I] represents an arbitrary variable, let's call it x. The sum of 4 and a number: 4 + x Half of this sum means we divide by 2: (4 + x)/2 Set this equal to 30: [B](4 + x)/2 = 30[/B] <-- This is our algebraic expression
Three subtracted from triple a number
Three subtracted from triple a number A number means an arbitrary variable, let's call it x x Triple it 3x Three subtracted from this [B]3x - 3[/B]
thrice the sum of x y and z
thrice the sum of x y and z The sum of x, y, and z means we add all 3 variables together: x + y + z The word [I]thrice[/I] means we multiply the sum of x, y, and z by 3: 3(x + y +z)
total of a number and the square of a number
total of a number and the square of a number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. The square of a number means we raise x to the power of 2. x^2 The total means we add x squared to x: [B]x + x^2[/B]
translate the product of -1 and a number in mathematics expression
translate the product of -1 and a number in mathematics expression The phrase [I]a number[/I] means an arbitrary variable. Let's call it x. The product of -1 and the number; [B]-x[/B]
Translate this phrase into an algebraic expression. 57 decreased by twice Mais savings Use the varia
Translate this phrase into an algebraic expression. 57 decreased by twice Mais savings Use the variable m to represent Mais savings. Twice means multiply by 2 2m 57 decreased by means subtract 2m from 57 [B]57 - 2m[/B]
Translate this phrase into an algebraic expression. 58 decreased by twice Gails age. Use the variabl
Translate this phrase into an algebraic expression. 58 decreased by twice Gails age. Use the variable g to represent Gails age. Twice Gail's age: 2g 58 decreased by twice Gail's age [B]58 - 2g[/B]
Translate this sentence into an equation. 43 is the sum of 17 and Gregs age. Use the variable g to
Translate this sentence into an equation. 43 is the sum of 17 and Gregs age. Use the variable g to represent Gregs age. The sum of 17 and Greg's age: g + 17 The word [I]is[/I] means equal to, so we set g + 17 equal to 43 [B]g + 17 = 43[/B] <-- This is our algebraic expression If you want to solve this equation for g, use our [URL='http://www.mathcelebrity.com/1unk.php?num=g%2B17%3D43&pl=Solve']equation calculator[/URL]. [B]g = 26[/B]
Translate this sentence into an equation. 48 is the difference of Ritas age and 11 . Use the variabl
Translate this sentence into an equation. 48 is the difference of Ritas age and 11 . Use the variable r to represent Ritas age. The difference of Rita's age and 11 is written: r - 11 The phrase [I]is[/I] means equal to, so we set r - 11 equal to 48 r - 11 = 48
Translate this sentence into an equation. 49 is the difference of Diegos age and 17. Use the variabl
Translate this sentence into an equation. 49 is the difference of Diegos age and 17. Use the variable d to represent Diegos age. The difference means we subtract, so we have d as Diego's age minus 17 d - 17 The word "is" means an equation, so we set d - 17 equal to 49 [B]d - 17 = 49[/B]
Translate this sentence into an equation. The difference of Maliks age and 15 is 63 Use the variable
Translate this sentence into an equation. The difference of Maliks age and 15 is 63 Use the variable m to represent Malik's age. [B]m - 15 = 63 [/B] To solve this equation, use our [URL='http://www.mathcelebrity.com/1unk.php?num=m-15%3D63&pl=Solve']equation calculator[/URL].
triple a number and another number
triple a number and another number The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Triple a number means we multiply x by 3: 3x The phrase [I]another number[/I] means another arbitrary variable, let's call it y: y The word [I]and[/I] means we add y to 3x: [B]3x + y[/B]
triple c divide the result by a
triple c divide the result by a Take this algebraic expression in pieces. Triple c means we multiply the variable c by 3 3c Divide the result by a, means we take 3c, and divide by a [B]3c/a[/B]
Twenty-five is nine more than four times a number
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. Four times a number: 4x nine more than four times a numbrer 4x + 9 The phrase [I]is[/I] means equal to. We set 4x + 9 equal to 25 as our algebraic expression: [B]4x + 9 = 25 [/B] If the problem asks you to solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=4x%2B9%3D25&pl=Solve']type it in our math solver[/URL] and get: x = [B]4[/B]
Twenty-five is the same as ten added to twice a number
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. Twice a number means we multiply x by 2: 2x ten added to twice a number 2x + 10 The phrase [I]is the same as [/I]means equal to. Set 25 equal to 2x + 10 to get our algebraic expression [B]25 = 2x + 10 [/B] If the problem asks you to solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=25%3D2x%2B10&pl=Solve']type it in our math solver [/URL]and get x = [B]7.5[/B]
Twice a number decreased by eight is zero
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. Twice a number: 2x decreased by eight 2x - 8 [I]is [/I]means equal to. Set 2x - 8 equal to zero for our algebraic expression: [B]2x - 8 = 0 [/B] If the problem asks you to solve for x, add 8 to each side: 2x = 8 Divide each side by 2: x= [B]4[/B]
Twice a number decreased by six
The phrase [I]a number[/I] means an arbitrary variable. We can pick any letter a-z except for i and e. Let's choose x. Twice a number means we multiply x by 2: 2x Decreased by six means we subtract 6 from 2x: [B]2x - 6[/B]
twice a number subtracted from the square root of the same number
twice a number subtracted from the square root of the same number The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Twice a number means we multiply x by 2: 2x Square root of the same number: sqrt(x) twice a number subtracted from the square root of the same number [B]sqrt(x) - 2x[/B]
twice the difference between x and 28 is 3 times a number
twice the difference between x and 28 is 3 times a number The difference between x and 28: x - 28 Twice the difference means we multiply x - 28 by 2: 2(x - 28) The phrase [I]a number[/I] means an arbitrary variable, let's call it x x 3 times a number: 3x The word [I]is[/I] means equal to, so we set 2(x - 28) equal to 3x: [B]2(x - 28) = 3x[/B]
twice the difference of a number and 3 is equal to 3 times the sum of a number and 2
twice the difference of a number and 3 is equal to 3 times the sum of a number and 2. We've got 2 algebraic expressions here. Let's take them in parts. Left side algebraic expression: twice the difference of a number and 3 [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]The word [I]difference[/I] means we subtract 3 from the variable x [*]x - 3 [*]Twice this difference means we multiply (x - 3) by 2 [*]2(x - 3) [/LIST] Right side algebraic expression: 3 times the sum of a number and 2 [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]The word [I]sum[/I] means we add 2 to the variable x [*]x + 2 [*]3 times the sum means we multiply (x + 2) by 3 [*]3(x + 2) [/LIST] Now, we have both algebraic expressions, the problem says [I]is equal to[/I] This means we have an equation, where we set the left side algebraic expression equal to the right side algebraic expression using the equal sign (=) to get our answer [B]2(x - 3) = 3(x + 2)[/B]
twice the difference of a number and 55 is equal to 3 times the sum of a number and 8
twice the difference of a number and 55 is equal to 3 times the sum of a number and 8 Take this algebraic expression in pieces. Left side: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The difference of this number and 55 means we subtract 55 from x x - 55 Twice the difference means we multiply x - 55 by 2 2(x - 55) Right side: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The sum of a number and 8 means we add 8 to x x + 8 3 times the sum means we multiply x + 8 by 3 3(x + 8) Now that we have the left and right side of the expressions, we see the phrase [I]is equal to[/I]. This means an equation, so we set the left side equal to the right side: [B]2(x - 55) = 3(x + 8)[/B]
twice the product of p q and r
twice the product of p q and r The product of p q and r means we multiply all 3 variables together: pqr The word [I]twice[/I] means we multiply pqr by 2: [B]2pqr[/B]
twice the square root of a number increased by 5 is 23
twice the square root of a number increased by 5 is 23 The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x The square root of a number means we raise x to the 1/2 power: sqrt(x) the square root of a number increased by 5 means we add 5 to sqrt(x): sqrt(x) + 5 twice the square root of a number increased by 5 means we multiply sqrt(x) + 5 by 2: 2(sqrt(x) + 5) The phrase [I]is 23[/I] means we set 2(sqrt(x) + 5) equal to 23: [B]2(sqrt(x) + 5) = 23[/B]
Twice the sum of a number and 6 is equal to three times the difference of the number and 3. Find the
[SIZE=6]Twice the sum of a number and 6 is equal to three times the difference of the number and 3. Find the number. The phrase [/SIZE][I][SIZE=7]a number[/SIZE][/I][SIZE=6] means an arbitrary variable, let's call it x. The sum of a number and 6 means we add 6 to x: x + 6 Twice the sum of a number and 6 means we multiply x + 6 by 2: 2(x + 6) the difference of the number and 3 means we subtract 3 from x x - 3 three times the difference of the number and 3 means we multiply x - 3 by 3: 3(x- 3) The word [I]is[/I] means we set 2(x + 6) equal to 3(x - 3) 2(x + 6) = 3(x - 3) Use the distributive property to multiply through: 2x + 12 = 3x - 9 Subtract 2x from each side: 2x - 2x + 12 = 3x - 2x - 9 x - 9 = 12 Add 9 to each side: x - 9 + 9 = 12 + 9 x = [B]21[/B] [B][/B] [B][MEDIA=youtube]CeZl_oZnSiw[/MEDIA][/B][/SIZE]
two thirds of a number is no more than -10
two thirds of a number is no more than -10 The phrase [I]a number[/I] means an arbitrary variable, let's call it x: x Two thirds of a number mean we multiply x by 2/3: 2x/3 The phrase [I]no more than[/I] -10 means less than or equal to -10, so we have an inequality: [B]2x/3 <= -10[/B]
Unit Circle
Free Unit Circle Calculator - Determines if coordinates for a unit circle are valid, or calculates a variable for unit circle coordinates
Use c for unknown variable : Sam's age plus twice his age
Use c for unknown variable : Sam's age plus twice his age Sam's age: c Twice his age means we multiply c by 2: 2c Sam's age plus twice his age [B]c + 2c[/B]
Use set notation to describe the set of integers greater than 4
Let x be our variable: We write x, such that x is greater than 4 below: [B]{x | x > 4}[/B]
What can we conclude if the coefficient of determination is 0.94?
What can we conclude if the coefficient of determination is 0.94? [LIST] [*]Strength of relationship is 0.94 [*]Direction of relationship is positive [*]94% of total variation of one variable(y) is explained by variation in the other variable(x). [*]All of the above are correct [/LIST] [B]94% of total variation of one variable(y) is explained by variation in the other variable(x)[/B]. The coefficient of determination explains ratio of explained variation to the total variation.
What does y=f(x) mean
What does y=f(x) mean It means y = a function of the variable x. x is the independent variable and y is the dependent variable. f(x) means a function in terms of x
What is a Variable
Free What is a Variable Calculator - This lesson walks you through what a variable is and how to use it. Also demonstrates the let statement.
When 20 is subtracted from 3 times a certain number, the result is 43
A certain number means an arbitrary variable, let's call it x x 3 times x 3x 20 is subtracted from 3 time x 3x - 20 The result is means equal to, so we set 3x - 20 equal to 43 for our algebraic expression [B]3x - 20 = 43 [/B] If you need to solve this, use our [URL='http://www.mathcelebrity.com/1unk.php?num=3x-20%3D43&pl=Solve']equation calculator[/URL]: [B]x = 21[/B]
When 4 times a number is increased by 40, the answer is the same as when 100 is decreased by the num
When 4 times a number is increased by 40, the answer is the same as when 100 is decreased by the number. Find the number The phrase [I]a number, [/I]means an arbitrary variable, let's call it "x". 4 times a number, increased by 40, means we multiply 4 times x, and then add 40 4x + 40 100 decreased by the number means we subtract x from 100 100 - x The problem tells us both of these expressions are the same, so we set them equal to each other: 4x + 40 = 100 - x Add x to each side: 4x + x + 40 = 100 - x + x The x's cancel on the right side, so we have: 5x + 40 = 100 [URL='https://www.mathcelebrity.com/1unk.php?num=5x%2B40%3D100&pl=Solve']Typing this equation into the search engine[/URL], we get [B]x = 12[/B].
When 4 times a number is increased by 40, the answer is the same as when 100 is decreased by the num
When 4 times a number is increased by 40, the answer is the same as when 100 is decreased by the number The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x 4 times a number means we multiply x by 4: 4x Increased by 40 means we add 40 to 4x: 4x + 40 100 decreased by the number means we subtract x from 100: 100 - x The phrase [I]is the same as[/I] means equal to, so we set 4x + 40 equal to 100 - x 4x + 40 = 100 - x Solve for [I]x[/I] in the equation 4x + 40 = 100 - x [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables 4x and -x. To do that, we add x to both sides 4x + 40 + x = -x + 100 + x [SIZE=5][B]Step 2: Cancel -x on the right side:[/B][/SIZE] 5x + 40 = 100 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 40 and 100. To do that, we subtract 40 from both sides 5x + 40 - 40 = 100 - 40 [SIZE=5][B]Step 4: Cancel 40 on the left side:[/B][/SIZE] 5x = 60 [SIZE=5][B]Step 5: Divide each side of the equation by 5[/B][/SIZE] 5x/5 = 60/5 x = [B]12[/B] Check our work for x = 12: 4(12) + 40 ? 100 - 12 48 + 40 ? 100 - 12 88 = 88
When 54 is subtracted from the square of a number, the result is 3 times the number.
When 54 is subtracted from the square of a number, the result is 3 times the number. This is an algebraic expression. Let's take it in parts. The phrase [I]a number[/I] means an arbitrary variable, let's call it "x". x Square the number, means raise it to the 2nd power: x^2 Subtract 54: x^2 - 54 The phrase [I]the result[/I] means an equation, so we set x^2 - 54 equal to 3 [B]x^2 - 54 = 3[/B]
When 9 is subtracted from 5 times a number ,the result is 31
When 9 is subtracted from 5 times a number ,the result is 31 A number means an arbitrary variable, let's call it x. 5 times this number is written as 5x. 9 subtracted from this is written as 5x - 9 [I]The result[/I] means we have an equation, so we set [B]5x - 9 = 31[/B]. This is our algebraic expression. Now if we want to solve for x, we [URL='http://www.mathcelebrity.com/1unk.php?num=5x-9%3D31&pl=Solve']plug this equation into the search engine [/URL]and get [B]x = 8[/B].
When twice a number is reduced by 15 you get 95 what is the number
When twice a number is reduced by 15 you get 95 what is the number? The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x [I]Twice[/I] x means we multiply x by 2 2x [I]Reduced by[/I] 15 means we subtract 15 2x - 15 [I]You get[/I] means equal to, as in an equation. Set 2x - 15 = 95 2x - 15 = 95 <-- This is our algebraic expression. [URL='https://www.mathcelebrity.com/1unk.php?num=2x-15%3D95&pl=Solve']Type 2x - 15 = 95 into the search engine[/URL] and we get [B]x = 55[/B].
Work
Free Work Calculator - Solves for any of the 3 variables, Work (W), Force (F) and Distance (d) in the work formula
Write a model that utilizes all three explanatory variables with no interaction or quadratic terms.
Write a model that utilizes all three explanatory variables with no interaction or quadratic terms. Choose the correct answer below. A. y i = B0 + B1x1 + B2x2 + B3x3 + e i B. y i = B0 + B1x1 + B2x2 + B3x3x2 + e i C. y i = B1x1 + B2x2 + B3x3 + ei D. None of the above equations satisfy all of the conditions [B]A. y i = B0 + B1x1 + B2x2 + B3x3 + e i[/B]
y is the sum of twice a number and 3
y is the sum of twice a number and 3 The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x twice a number means we multiply x by 2: 2x the sum of twice a number and 3: 2x + 3 The word [I]is[/I] means equal to, so we set 2x + 3 equal to y [B]y = 2x + 3[/B]
y minus 10 is equal to the product of y and 8
y minus 10 is equal to the product of y and 8. Take this algebraic expression in 3 parts: Part 1: y minus 10 Subtract 10 from the variable y y - 10 Part 2: The product of y and 8 We multiply 8 by the variable y 8y Part 3: The phrase [I]is equal to[/I] means an equation, so we set y - 10 equal to 8y [B]y - 10 = 8y[/B]
You are buying boxes of cookies at a bakery. Each box of cookies costs \$4. In the equation below, c
You are buying boxes of cookies at a bakery. Each box of cookies costs \$4. In the equation below, c represents the number of boxes of cookies you buy, and d represents the amount the cookies will cost you (in dollars). The relationship between these two variables can be expressed by the following equation: d=4c. Identify the dependent and independent variables. [B]The variable d is dependent, and c is independent since the value of d is determined by c.[/B]
You work for a remote manufacturing plant and have been asked to provide some data about the cost of
You work for a remote manufacturing plant and have been asked to provide some data about the cost of specific amounts of remote each remote, r, costs \$3 to make, in addition to \$2000 for labor. Write an expression to represent the total cost of manufacturing a remote. Then, use the expression to answer the following question. What is the cost of producing 2000 remote controls? We've got 2 questions here. Question 1: We want the cost function C(r) where r is the number of remotes: C(r) = Variable Cost per unit * r units + Fixed Cost (labor) [B]C(r) = 3r + 2000 [/B] Question 2: What is the cost of producing 2000 remote controls. In this case, r = 2000, so we want C(2000) C(2000) = 3(2000) + 2000 C(2000) = 6000 + 2000 C(2000) = [B]\$8000[/B]
Zalika thinks of a number. She subtracts 6 then multiplies the result by 5. The answer is the same a
Zalika thinks of a number. She subtracts 6 then multiplies the result by 5. The answer is the same as subtracting 5 from the number then multiplying by 4. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. We're given two expressions in relation to this number (x): [U]She subtracts 6 then multiplies the result by 5[/U] [LIST] [*]Subtract 6: x - 6 [*]Multiply the result by 5: 5(x - 6) [/LIST] [U]She subtracts 5 from the number then multiplying by 4[/U] [LIST] [*]Subtract 6: x - 5 [*]Multiply the result by 5: 4(x - 5) [/LIST] Finally, the expression [I]is the same as[/I] means an equation, so we set the first expression equal to the second expression to make the following equation: 5(x - 6) = 4(x - 5) Now, let's solve the equation for x. To do this, we [URL='https://www.mathcelebrity.com/1unk.php?num=5%28x-6%29%3D4%28x-5%29&pl=Solve']type this equation into our search engine [/URL]and we get: x = [B]10[/B] |
# A Pattern in the Power Series Coefficients and Pascal’s Triangle
An interesting pattern emerges from evaluating the sum of descending powers that alternate in sign with Pascal’s Triangle coefficients.
Pascal’s Triangle is probably familiar to most students of mathematics. The pattern is shown below.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
We will take the descending powers with alternating signs and multiply them with the numbers in Pascal’s Triangle. This is the pattern we get for the sums.
$$2^x$$
$$3^x - 2^x$$
$$4^x - 2\cdot 3^x + 2^x$$
$$5^x - 3 \cdot 4^x + 3 \cdot 3^x - 2^x$$
$$6^x - 4 \cdot 5^x + 6 \cdot 4^x - 4 \cdot 3^x + 2^x$$
$$7^x - 5 \cdot 6^x + 10 \cdot 5^x - 10 \cdot 4^x + 5 \cdot 3^x - 2^x$$
When evaluating the sum for integer values of x starting with 0, we get 0 as the result until we get a non-zero answer. Let’s evaluate this sum for x = 0, 1, 2, 3, 4, .... until we get a non-zero answer for 2x.
### $$2^x$$
$$2^0 = 1$$
Okay. That was obvious. The first value gives us 1. But let’s continue with the next sum:
### $$3^x - 2^x$$
$$3^0 - 2^0 = 0$$
$$3^1 - 2^1 = 3 - 2 = 1 = 1!$$
For x = 0, we get a 0 and then the rest of the sums will be non-zero. Curiously, for x = 1, the answer is 1!. This will become more obvious as we go higher.
### $$4^x - 2 \cdot 3^x + 2^x$$
$$4^0 - 2 \cdot 3^0 + 2^0 = 1 - 2 + 1 = 0$$
$$4^1 - 2 \cdot 3^1 + 2^1 = 4 - 6 + 2 = 0$$
$$4^2 - 2 \cdot 3^2 + 2^2 = 16 - 18 + 4 = 2 = 2!$$
For the sum above, it took two values (x = 0 and x = 1) before we got a non-zero answer. And when x = 2, we get 2!.
### $$5^x - 3 \cdot 4^x + 3 \cdot 3^x - 2^x$$
$$5^0 - 3 \cdot 4^0 + 3 \cdot 3^0 - 2^0 = 1 - 3 + 3 - 1 = 0$$
$$5^1 - 3 \cdot 4^1 + 3 \cdot 3^1 - 2^1 = 5 - 12 + 9 - 2 = 14 - 14 = 0$$
$$5^2 - 3 \cdot 4^2 + 3 \cdot 3^2 - 2^2 = 25 - 48 + 27 - 4 = 52 - 52 = 0$$
$$5^3 - 3 \cdot 4^3 + 3 \cdot 3^3 - 2^3 = 125 - 192 + 81 - 8 = 206 - 200 = 6 = 3!$$
For the sum above, we get 3 zeros before we get 3!. This is a really interesting and amazing pattern.
### $$6^x - 4 \cdot 5^x + 6 \cdot 4^x - 4 \cdot 3^x + 2^x$$
$$6^0 - 4 \cdot 5^0 + 6 \cdot 4^0 - 4 \cdot 3^0 + 2^0 = 1 - 4 + 6 - 4 + 1 = 0$$
$$6^1 - 4 \cdot 5^1 + 6 \cdot 4^1 - 4 \cdot 3^1 + 2^1 = 6 - 20 + 24 - 12 + 2 = 32 - 32 = 0$$
$$6^2 - 4 \cdot 5^2 + 6 \cdot 4^2 - 4 \cdot 3^2 + 2^2 = 36 - 100 + 96 - 36 + 4 = 136 - 136 = 0$$
$$6^3 - 4 \cdot 5^3 + 6 \cdot 4^3 - 4 \cdot 3^3 + 2^3 = 216 - 500 + 384 - 108 + 8 = 608 - 608 = 0$$
$$6^4 - 4 \cdot 5^4 + 6 \cdot 4^4 - 4 \cdot 3^4 + 2^4 =$$ $$1296 - 2500 + 1536 - 244 + 16 =$$ $$2848 - 2824 = 24 = 4!$$
For the sum above, we get 4 zeros before we get 4!.
### $$7^x - 5 \cdot 6^x + 10 \cdot 5^x - 10 \cdot 4^x + 5 \cdot 3^x - 2^x$$
$$7^0 - 5 \cdot 6^0 + 10 \cdot 5^0 - 10 \cdot 4^0 + 5 \cdot 3^0 - 2^0 =$$ $$1 - 5 + 10 - 10 + 5 - 1 = 0$$
$$7^1 - 5 \cdot 6^1 + 10 \cdot 5^1 - 10 \cdot 4^1 + 5 \cdot 3^1 - 2^1 =$$ $$7 - 30 + 50 - 40 + 15 - 2 = 72 - 72 = 0$$
$$7^2 - 5 \cdot 6^2 + 10 \cdot 5^2 - 10 \cdot 4^2 + 5 \cdot 3^2 - 2^2 =$$ $$49 - 180 + 250 - 160 + 45 - 4 = 344 - 344 = 0$$
$$7^3 - 5 \cdot 6^3 + 10 \cdot 5^3 - 10 \cdot 4^3 + 5 \cdot 3^3 - 2^3 =$$ $$343 - 1080 + 1250 - 640 + 135 - 8 = 1728 - 1728 = 0$$
$$7^4 - 5 \cdot 6^4 + 10 \cdot 5^4 - 10 \cdot 4^4 + 5 \cdot 3^4 - 2^4 =$$ $$2401 - 6480 + 6250 - 2560 + 405 - 16 = 9056 - 9056 = 0$$
$$7^5 - 5 \cdot 6^5 + 10 \cdot 5^5 - 10 \cdot 4^5 + 5 \cdot 3^5 - 2^5 =$$ $$16807 - 38880 + 31250 - 10240 + 1215 - 32 =$$ $$49272 - 49152 = 120 = 5!$$
And the pattern is now obvious. Pascal’s Triangle has an amazing link with the coefficients for the power series sum formula. This formula is given here: Sum of the Power Series Formula. |
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• ## Related Books
### Dividing Integers
• The rules for dividing integers are the same as the rules for multiplying integers.
• When multiplying integers with the same sign the answer will be positive. When multiplying integers with different signs, the answer will be negative.
• Dividing by a number is the same as multiplying by the reciprocal of that number. For example 10 • 1/2 = 5 and 10 ÷ 2 = 5.
### Dividing Integers
- 125 ÷ 5 ÷ - 5 =
• − 25 ÷− 5 =
5
- 120 ÷ - 6 ÷ - 5 =
• 20 ÷− 5 =
- 4
- 48 ÷ 2 ÷ - 4 ÷ - 2 =
• − 24 ÷− 4 ÷− 2 =
• 6 ÷− 2 =
- 3
- 6 ÷ ( - 2) × 7 ÷ ( - 3) =
• 3 ×7 ÷( − 3) =
• 21 ÷( − 3) =
- 7
36 ÷ ( - 12) × ( - 9) ÷ ( - 3) ÷ ( - 3) =
• You may evaluate the sign first. 4 negatives = + , positive
• 36 ÷12 ×9 ÷3 ÷3 =
• 3 ×9 ÷3 ÷3 =
• 27 ÷3 ÷3 =
• 9 ÷3 =
3
- 72 ÷ - 8 ÷ - 3 × - 5 ÷ - 3 =
• You may evaluate the sign first. 5 negatives = - , negative
• 72 ÷8 ÷3 ×5 ÷3 =
• 9 ÷3 ×5 ÷3 =
• 3 ×5 ÷3 =
• 15 ÷3 =
• 5
- 5
- 100 ÷ 5 ÷ (5 × - 2) ÷ - 2 =
• You may evaluate the sign first. 3 negatives = - , negative
• 100 ÷5 ÷(5 ×2) ÷2 =
• 100 ÷5 ÷10 ÷2 =
• 20 ÷10 ÷2 = =
• 2 ÷2 =
• 1
- 1
Evaluate the expression for x = 8 and y = - 10.
2y ÷ 4 + 6x ÷ ( - 4)
• 2( − 10) ÷4 + 6(8) ÷( − 4) =
• − 20 ÷4 + 48 ÷( − 4) =
• − 5 + ( − 12) =
- 17
Evaluate the expression for a = 5 and b = 21.
- | - a| × ( - 3) + b ÷ ( - 7) + 4( - 2)
• − | − 5| ×( − 3) + 21 ÷( − 7) + 4( − 2) =
• − (5) ×( − 3) + 21 ÷( − 7) + 4( − 2) =
• − 5 ×( − 3) + 21 ÷( − 7) + 4( − 2) =
• − 5 ×( − 3) + 21 ÷( − 7) + ( − 8) =
• 15 + ( − 3) + ( − 8) =
• 15 − 3 − 8 =
• 12 − 8 =
4
A scuba diver is diving into the ocean. She swims 80 feet down in 40 seconds. Find her average rate of change in feet per second.
• - 80 feet ÷ 40 seconds =
- 2 feet/second
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
### Dividing Integers
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• What You'll Learn and Why 0:06
• Topics Overview
• Vocabulary 0:27
• Quotient
• Rules for Dividing Integers 0:49
• Rules for Dividing Integers, cont. 2:06
• Arithmetic (Different Signs Integers)
• Algebraic (Different Signs Integers)
• Dividing Integer Examples 3:24
• Dividing Integers: 14 ÷ 7
• Dividing Integers: 45 ÷ (-9)
• Dividing Integer Examples 3:51
• Dividing Integers: (-105) ÷ (-15)
• Dividing Integers: (-42) ÷ 6
• Average Rate of Change 5:17
• Using Integers to Represent the Situation
• Example: Spend \$360 in 6 Days
• Example: Runs 1000 Feet in 4 Minutes
• Average Rate of Change Word Problems 7:27
• Example: Average Decrease in Value
• Average Rate of Change Word Problems 9:19
• Example: Average Increase in Stock
• Average Rate of Change Word Problems 10:46
• Example: Average Increase in Speed
• Dividing Integers 12:00
• Odd Number of Negatives
• Even Number of Negatives
• Order of Operations and Sign of Final Answer 13:50
• Example: -120 ÷ (-5) ÷ -4
• Extra Example 1: Order of Operations 14:48
• Extra Example 2: Evaluate the Expression 15:29
• Extra Example 3: Rate of Change 17:18
• Extra Example 4: Rate of Evaporation 19:22 |
# factors of 24
The factors of 24 are 1, 2, 3, 4, 6, 8, 12, Twenty four. Therefore, 24 has 8 factors .
Similarly, what is the LCM of Twenty four? The LCM of Twenty four and 36 is the smallest number among all the common multiples of Twenty four and 36.
LCM of Twenty four and 36.
What is the common factor of Twenty four? The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Now, to find the greatest common factor, we find the largest number in both the lists. This number is 8 .
Which is the factor of Twenty four? Therefore, the factors of 24 are 24 = 1, 2, 3, 4, 6, 12 and 24 .
What is the second factor Twenty four? The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 .
## What is the greatest common factor of 24?
The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Now, to find the greatest common factor, we find the largest number in both the lists. This number is 8 .
So how do you find the greatest common factor of Twenty four? To calculate the GCF (Greatest Common Factor) of 24 and 36, we need to factor each number (24 = 1, 2, 3, 4, 6, 8, 12, 24; 36 = 1, 2, 3) is needed. , 4, 6, 9, 12, 18, 36) and choose the greatest factor that exactly divides both 24 and 36, i.e., 12 .
What is the greatest common factor of Twenty four and 30 using prime factorization? GCF of Twenty four and 30 by prime factorization
The prime factors of Twenty four and 30 are (2 × 2 × 2 × 3) and (2 × 3 × 5) respectively. As seen, Twenty four and 30 have common prime factors. So the GCF of 24 and 30 is 2 3 × = 6 .
#### What is the greatest common factor of 24 and 60?
GCF of examples Twenty four and 60
Therefore, the greatest common factor of 24 and 60 is 12 .
How do you find the greatest common factor? To find the GCF of a set of numbers, list all the factors of each number . The largest factor that appears in each list is GCF. For example, to find the GCF of 6 and 15, first list all the factors of each number. Because 3 is the largest factor that appears in both lists, 3 is the GCF of 6 and 15.
What is the highest common factor of 64 and 80?
The GCF of 64 and 80 is 16 .
How do you factor a tree of Twenty four?
## Which of the following is not a factor pair of 24?
1, 2, 3, 4, 6, 8, 12, Twenty four. The negative factors of Twenty four include the following numbers: -1, -2, -3, -4, -6, -8, -12, -24.
What is the greatest common factor of Twenty four and?
factor and greatest common factor
• The factors of Twenty four are: 1, 2, 3, 4, 6, 8, 12 and Twenty four because each of these numbers is exactly divisible by Twenty four.
• The pairs of factors of Twenty four are: 1 and Twenty four, 2 and 12, 3 and 8, and 4 and 6.
What is the highest common factor of Twenty four and 36? Therefore, the highest common factor of 24 and 36 is 12 .
How do you find the greatest common factor? To find the GCF of a set of numbers, list all the factors of each number . The largest factor that appears in each list is GCF. For example, to find the GCF of 6 and 15, first list all the factors of each number. Because 3 is the largest factor that appears in both lists, 3 is the GCF of 6 and 15.
## What is the highest common factor of 24 and 36?
Therefore, the highest common factor of 24 and 36 is 12 .
What is the GCF of Twenty four and 60? GCF of Twenty four and 60 by listing the common factors
Therefore, the greatest common factor of 24 and 60 is 12 .
What is the common factor of 64 and 16?
5 and 16 have 64 common factors, which are 1, 2, 4, 8 and 16. Therefore, the highest common factor of 16 and 64 is 16 .
What is the greatest common factor of 64 and 28? Answer: GCF of 28 and 64 is 4 .
### What is meant by greatest common factor?
The greatest common factor (GCF) of the set of numbers is the greatest factor that all numbers share . For example, 12, 20 and Twenty four have two common factors: 2 and 4. The greatest is 4, so we say that the GCF of 12, 20 and Twenty four is 4. GCF is often used to find the common denominator.
What is the greatest common factor of Twenty four and 16? As you can see when you list the factors of each number, 8 is the largest number that can divide 16 and Twenty four.
What is the greatest common factor of Twenty four and 54?
The GCF of Twenty four and 54 is 6 . To calculate the greatest common factor (GCF) of Twenty four and 54, we need to factor each number (Twenty four = 1, 2, 3, 4, 6, 8, 12, Twenty four; factors of 54 = 1, 2, 3 , 6, 9, 18, 27, 54) and choose the greatest factor that exactly divides both Twenty four4 and 54, i.e., |
# 4.2: Ratios and Proportions
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Solving ratios and proportions is similar to the previous two exercises. It is simply just looking at the question a little differently. Instead of looking at it as numbers we will assign UNITS (or items) to the numbers, for example, 3 valves to 6 valves or 3 hours to 5 hours. However, the real comparisons of interest will be discussed in the next section when we compare different UNITS to each other, for example, 40 hours to 1 week, or 12 inches to 1 foot.
Example $$\PageIndex{1}$$
$\dfrac{2}{3}=\dfrac{F}{9}\nonumber$
You can read this as, 2 is to 3 as F is to 9.
In order to solve the above ratio or proportion is by cross multiplying.
$9 \times 2=3 \times F \nonumber$
This would equate to
$18=3 F \nonumber$
Then using the method as previously used in the Equation Section, you would divide both sides of the equation by 3. Three would then cancel on the right side of the equation isolating the variable and 18 divided by 3 would give you 6. Therefore, F equals 6.
$\dfrac{18}{3}=\dfrac{3 F}{3} \quad \rightarrow \quad 6=F \nonumber$
## Exercise 4.2
1. $$\dfrac{D}{7}=\dfrac{27}{9}$$
2. $$\dfrac{28}{L}=\dfrac{32}{8}$$
3. $$\dfrac{4}{8}=\dfrac{J}{64}$$
4. $$\dfrac{H}{5}=\dfrac{28}{70}$$
5. $$\dfrac{K}{100}=\dfrac{1}{3}$$
6. $$\dfrac{L}{12}=\dfrac{12}{3}$$
7. $$\dfrac{9}{4}=\dfrac{K}{12}$$
8. $$\dfrac{7}{F}=\dfrac{5}{15}$$
9. $$\dfrac{H}{3}=\dfrac{20}{6}$$
10. $$\dfrac{2}{5}=\dfrac{Q}{35}$$
11. $$\dfrac{13}{F}=\dfrac{32}{64}$$
12. $$\dfrac{1}{7}=\dfrac{11}{J}$$
Not all proportional problems are exactly as they seem. The previous problems were directly proportional and can be solved with relative ease. However, sometimes you will encounter indirectly proportional or more properly termed “inversely” proportional.
Example $$\PageIndex{2}$$
It takes 3 employees to flush 8 hydrants in 6 hours. How long would it take 5 employees to do the same job?
Solution
If you attempt to solve this problem as if it were directly proportional it would look like.
$\dfrac{3}{6}=\dfrac{5}{W} \quad \rightarrow \quad W=\dfrac{30}{3} \quad \rightarrow \quad W=10 \nonumber$
By this result it would take 5 employees 4 hours longer to do the same job.
Inversely proportional problems need to be solved as follows. It is the product of the values not the ratio that you need to equate.
3 employees × 6 hours = 5 employees x W hours
$\begin{array}{l} 18=5W \\ W=\dfrac{18}{5} \\ W=3.6 \text { hours } \end{array} \nonumber$
## Exercise 4.2.1
1. A safety catalog sells dust masks. They are $4.50 per dozen. How much would 4 dust masks cost? 2. An operator conducted a laboratory experiment by adding 1.5 pounds of chlorine to 5 gallons of water to get a certain chlorine dosage. If the operator wanted to disinfect 1,200 gallons of water to the same dosage, how many pounds of chlorine would she need? 3. Six water utility operators were able to exercise 75 valves in one work week. If 9 operators were assigned to do the same task, how long would it take them? (Assume 34 hours equates to a work week) In the previous problem, you ended up with a fraction (expressed as a decimal) of an hour. What would be a better way to express this answer? 1. A water utility needs to install 2,375 feet of 16” diameter pipe. The pipe costs$25.80 per foot. How much will all the pipe cost?
2. In the problem above, the pipe is manufactured in 20 foot sections. How many sections would need to be purchased?
3. A water storage tank needs to be recoated. A contractor gave an estimate that it would take 5 of his employees a total of 39 hours to complete the job. How sooner can the job be completed if 8 employees were to do the work?
In word problems with percentages, the first thing is to convert the percent to a decimal. Then, view the word “of” as a multiplication sign and the word “is” as an equals sign. Then, solve the problem
Example $$\PageIndex{3}$$
10% of 100 is
10% × 100 =
0.10 × 100 = 10
## Exercise 4.2.2
Solve the following percent problems.
1. 30 is 20% of what number?
2. 10 is 45% of what number?
3. 12% of is 84.
4. What percent of 75 is 225?
5. 56% of is 182.
6. 35 is what percent of 500? _________________
7. 100% of 2,000 is ______________
8. What percent of 40 is 10? ____________
Word problems give people fits! However, most of the problems that present themselves in practical situations are in fact word problems, we just don’t always think of them that way. For example, if you are buying something from the store and you want to figure out how much the tax will be on a certain item, you probably just multiply the cost of the item by the sales tax. You can though look at this problem as a word problem.
## Exercise 4.2.3
Solve the following word problems.
1. A water utility executive earned \$85,000 last year and received a 22% bonus. How much was her bonus?
2. A worker can paint a fire hydrant in ½ hour. How many hydrants can she paint in 4 hours?
3. A 5 gallon jug of bottled water is labeled 60% spring water. How many gallons in the jug is spring water?
4. On a state certification treatment exam you must score at least 70% to pass. If there are 65 questions on the test how many must you get correct to pass?
4.2: Ratios and Proportions is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts. |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$y\sqrt[6]{y}$
Consider expression $\sqrt{y}\sqrt[3]{{{y}^{2}}}$, Convert to the exponential notation, $\sqrt{y}\sqrt[3]{{{y}^{2}}}={{y}^{\frac{1}{2}}}\times {{y}^{\frac{2}{3}}}$ Add exponent, \begin{align} & {{y}^{\frac{1}{2}+\frac{2}{3}}}={{y}^{\frac{3+4}{6}}} \\ & ={{y}^{^{\frac{7}{6}}}} \end{align} Convert back to radical notation, ${{y}^{^{\frac{7}{6}}}}=\sqrt[6]{{{y}^{7}}}$ Simplify the expression, $\sqrt[6]{{{y}^{7}}}=\sqrt[6]{{{y}^{6}}}\times \sqrt[6]{y}$ Find the sixth root. we assume $x\ge 0$, $y\sqrt[6]{y}$ Partial check: $\begin{matrix} {{\left( y\sqrt[6]{y} \right)}^{6}}\overset{?}{\mathop{=}}\,{{\left( y \right)}^{6}}{{\left( \sqrt[6]{y} \right)}^{6}} \\ \overset{?}{\mathop{=}}\,{{y}^{6}}\times y \\ ={{y}^{7}} \\ \end{matrix}$ Thus, the expression $\sqrt{y}\sqrt[3]{{{y}^{2}}}$ can be simplified as $y\sqrt[6]{y}$ |
# Grade 8 MTAP 2015 Elimination Questions with Solutions – Part 4
This is the fourth part (questions 31-40) of the solutions of the Grade 8 MTAP 2015 Elimination Questions. You can read the solutions for questions 1-10 11-20, and 21 – 30
Although these solutions were carefully checked, the solver is only human. Kindly comment in the box below if you see any errors.
31.) Factor completely: $a^2c^2 + b^2d^2 - a^2d^2 - b^2c^2$.
Solution
Rearranging the terms, we obtain
$a^2c^2 - a^2d^2 - b^2c^2 + b^2d^2$.
Factoring by grouping, we have
$a^2(c^2 - d^2) - b^2(c^2 - d^2)$
$= (c^2 - d^2)(a^2 - b^2)$
$= (c - d)(c + d)(a - b)(a + b)$.
Answer: (a + b)(a – b)(c + d)(c – d)
32.) Simplify $\dfrac{x^3- 4x}{x^3-x^2-6x}$.
Solution
$\dfrac{x^3- 4x}{x^3-x^2-6x} = \dfrac{x(x + 2)(x - 2)}{x(x - 3)(x + 2)} = \dfrac{x-2}{x-3}$.
Answer: $\frac{x - 2}{x - 3}$
33.) Perform the indicated operations: $\dfrac{1}{x} - \dfrac{1}{x+1} - \dfrac{1}{x - 1}$
Solution
$\dfrac{1}{x} - \dfrac{1}{x+1} - \dfrac{1}{x - 1} = \dfrac{(x + 1)(x-1) - x(x-1)-x(x+1)}{x(x+1)(x-1)}$
$= \dfrac{x^2 - 1 - x^2 + x - x^2 -x }{x(x + 1)(x - 1)}$
$= \dfrac{x^2 - 1}{x(x+1)(x-1)} = \dfrac{(x + 1)(x - 1)}{x(x + 1)(x - 1)} = \dfrac{1}{x}$
34.) Perform the indicated operations.
$\left ( \dfrac{2x^2 - 2x - 4}{x^2 - 5x + 6} \right ) \left ( \dfrac{x^2- x-6}{x^2+ 4x+3} \right) \div \dfrac{4x + 8}{3x + 9}$.
Solution
We can factor the given completely into the following expressions and change division into multiplication by multiplying the first two expressions with the reciprocal of the third expression. The result of these operations is shown below.
$\left ( \dfrac{2(x-2)(x + 1)}{(x - 2)(x - 3)} \right ) \left ( \dfrac{(x - 3)(x + 2)}{(x + 3)(x + 1)} \right) \left (\dfrac{3(x + 3)}{(4(x + 2)} \right )$.
After cancelling similar terms, we are left with $2(\frac{3}{4}) = \frac{3}{2}$
35.) Simplify $1 - \dfrac{x}{1 - \frac{x}{1 - x}}$.
Solution
Simplifying the rational expressions, we have
$1 - \dfrac{x}{1 - \frac{x}{1 - x}} = 1 - \dfrac{x}{\frac{1-2x}{1 - x}} = 1 - \dfrac{1 - x^2}{1 - 2x}$.
This simplifies further to
$\dfrac{1 - 3x + x^2}{1 - 2x}$.
Answer: $\dfrac{1 - 3x + x^2}{1 - 2x}$
36.) Solve for x in the equation $\dfrac{x+1}{(x-2)} = \dfrac{x-1}{x-2}$
Solution
Getting the least common denominator of the left hand side and combining the terms, we have
$\dfrac{x(x - 2)+1}{(x-2)} = \dfrac{1}{x - 2} = \dfrac{x-1}{x-2}$
$\dfrac{x^2 - 2x + 1}{x - 2} = \dfrac{x-1}{x-2}$.
Now, we can only equation the numerator.
$x^2 - 2x + 1 = x - 1$ which is equivalent to $x^2 – 3x + 2 = 0$
Getting the solution, we have $(x - 1)(x - 2 ) = 0$
Therefore, $x = 1$ or $x = 2$. But x cannot be equal to 2, because the it will make the denominator of the original expressions undefined. So, the solution is $x = 1$.
37.) The points (0, 0), (2, 3), and (4, 0) form a triangle. What is its perimeter?
Solution
To find the perimeter of the triangle, we need to find the distance between these points and add them. Let’s name the points A, B, and C respectively.
Using the distance formula, we can subtract the corresponding coordinates, square them, and get the square root.
Distance between A and B is $\sqrt{2^2 + 3^2} = \sqrt{13}$
Distance between A and C is $\sqrt{4^2 + 0^2} = 4$
Distnace between B and C is $2^2 + (-3)^2 = \sqrt{13}$
So, the perimeter of ABC is $AB + AC + BC = \sqrt{13} + 4 + \sqrt{13} = 4 + 2\sqrt{13}$
Answer: $4 + 2 \sqrt{13}$.
38.) Find the equation of the perpendicular bisector of the segment joining the points (-3, 2) and (5, 2)?
Solution
Let A be the point with coordinates (-3,2) and B be the point with coordinates (5,2). Notice that AB have the same y-coordinates which mean that it is a horizontal segment. This means that the perpendicular bisector is a vertical line.
To get the perpendicular bisector, we get the midpoint M of AB and find the equation of the vertical line passing through M.
$M = ( \frac{-3 + 5}{2}, \frac{2 + 2}{2}) = (1, 2)$.
So, the equation of the perpendicular bisector of AB is $x = 1$
39.) A man agrees to invest part of his 1-million-peso inheritance at an annual interest rate of 5%, while the rest at 6% interest. If, at the end of the year, he needs a total interest of Php 56, 200, how much should he invest at 5%?
Solution
Let x = amount invested at 5% and 1000 000 – x be invested at 6%.
$0.05x + 0.06(1000000 - x) = 56200$
$0.05x + 60000 - 0.06x = 56200$
$-0.01x = 3800$
$x = 380000$
40.) If 2x + 5y = 10 and x = 3y + 1, what is 11x + 11y?
Solution
Substituting the expression on the right hand side of the second equation to x in the left hand side of the first equation, we have
$2(3y + 1) + 5y = 10$
$6y + 2 + 5y = 10$
$11y = 8$ (*)
Multiplying $2x + 5y = 10$ by 3, we get $6x + 15y = 30$ (#).
Transposing the second equation and multiplying it by 5, we have
$5x - 15y = 5$ (##)
Adding # and ##, we have
$11x = 35$ (**)
By * and **
We have 11x + 11y = 35 + 8 = 43\$.
We will discuss the solution of numbers 41-50 in the next post.
This entry was posted in Grade 7-8 and tagged , , , . Bookmark the permalink.
### 3 Responses to Grade 8 MTAP 2015 Elimination Questions with Solutions – Part 4
1. Tyeng says:
i dont know if im correct but, i think number 35 has an error…that’s in multiplying x with the quantity 1-x..
2. cecillepantonial says:
Thnx. This is a great help to our contestants.
3. Galen Buhain says:
Please double-check the number 33. Thanks! |
# Assertion and Reason Questions for Class 9 Maths Chapter 14 Statistics
Home » CBSE Class 9 Maths » Assertion Reason Questions for Class 9 Maths » Assertion and Reason Questions for Class 9 Maths Chapter 14 Statistics
Here we are providing assertion reason questions for class 9 maths. In this article, we are covering assertion reason questions for class 9 Maths Chapter 14 Statistics.
• ## Assertion and Reason Questions for Class 9 Maths Chapter 14 Statistics
Here we are providing assertion reason questions for class 9 maths. In this article, we are covering assertion reason questions for class 9 Maths Chapter 14 Statistics. [PDF] Download Assertion and Reason Questions for Class 9 Maths Chapter 14 Statistics
• ## Assertion and Reason Questions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles
Here we are providing assertion reason questions for class 9 maths. In this article, we are covering assertion reason questions for class 9 Maths Chapter 9 Areas of Parallelograms and Triangles. [PDF] Download Assertion and Reason Questions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Directions:Â Choose the correct answer out of the…
• ## Assertion and Reason Questions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry
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• ## Assertion and Reason Questions for Class 9 Maths Chapter 10 Circles
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Assertion and Reason Questions for Class 9 Maths Chapter 15 Probability Directions: Choose the correct answer out of the following choices :(a) Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.(b) Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.(c) Assertion is…
• ## Assertion and Reason Questions for Class 9 Maths Chapter 8 Quadrilaterals
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• ## Assertion and Reason Questions for Class 9 Maths Chapter 1 Number System
Assertion and Reason Questions for Class 9 Maths Chapter 1 Number System Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).(b) Both assertion…
• ## Assertion and Reason Questions for Class 9 Maths Chapter 7 Triangles
Assertion and Reason Questions for Class 9 Maths Chapter 7 Triangles Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).(b) Both assertion (A)…
• ## Assertion and Reason Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables
Assertion and Reason Questions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion…
• ## Assertion and Reason Questions for Class 9 Maths Chapter 6 Lines and Angles
Assertion and Reason Questions for Class 9 Maths Chapter 6 Lines and Angles Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).(b) Both assertion…
• ## Assertion and Reason Questions for Class 9 Maths Chapter 3 Coordinate Geometry
Assertion and Reason Questions for Class 9 Maths Chapter 3 Coordinate Geometry Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).(b) Both assertion…
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# Linear equations in two variables- Exercise 3.4-Class 10
1. Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y/3 = -1 and x – y/3 = 3
Solution:
(i) By elimination method:
x + y = 5————-(1)
2x – 3y = 4 ————(2)
Multiplying equation (1) by 2,
2x + 2y = 10 —————-(3)
Subtracting equation (2) from equation (3),
5y = 6
y = 6/5 ————-(4)
Substituting the value in equation (1),
x = 5 – 6/5 = 19/5
x = 19/5 , y = 6/5
By substitution method:
From equation (1),
x = 5 – y ————-(5)
Put the value of x in (2)
2(5-y) – 3y = 4
-5y = – 6
y = 6/5
Substituting the value in equation (5)
x = 5 – 6/5 = 19/5
x = 19/5 ; y = 6/5
(ii) By elimination method:
3x + 4y = 10 ———-(1)
2x – 2y = 2———(2)
Multiplying equation (2) by 2,
4x – 4y = 4 ————(3)
7x = 14
x = 2 —————(4)
Substituting in equation (1),
6 + 4y = 10
4y = 4
y = 1
Thus, x = 2, y = 1
By substitution method:
From equation (2),
x = 1 + y ———- (5)
Put the value of x in equation (1),
3(1+y)+4y = 10
7y = 7
y = 1
Substituting the value in equation (5),
x = 1+1=2
x = 2, y = 1
(iii)By elimination method:
3x – 5y – 4 = 0 ————(1)
9x = 2y + 7
9x – 2y – 7 = 0 ————–(2)
Multiplying equation (1) by 3,
9x – 15y – 12 = 0 ————(3)
Subtracting equation (3) from equation (2),
13y = -5
y = -5/13
Substituting in equation (1),
3x + 25/13 – 4 = 0
3x = 27/13
x = 9/13
x = 9/13 ; y = -5/13
By substitution method:
From equation (1),
x = 5y+4/3 ————–(5)
Putting this value in equation (2),
9(5y+4/) – 2y – 7 = 0
13y = -5
y = ‑5/13
Substituting the value in equation (5),
x = [5(-5/13)+4]/3 = 9/13
Thus, x = 9/13 and y = ‑5/13
(iv) By elimination method
x/2 + 2y/3 = -1
3x + 4y = -6 ———–(1)
x – y/3 = 3
3x – y = 9 —————-(2)
Subtracting equation (2) from equation (1),
5y = -15
y = -3 —————(3)
Substituting this value in equation (1),
3x – 12 = -6
3x = 6
x = 2
Hence, x = 2, y = −3
By substitution method:
From equation (2),
x = y+9/3———– (5)
Putting this value in equation (1),
3(y+9/3) + 4y = -6
5y = -15
y = -3
Substituting the value in equation (5),
x = -3+9/3 = 2
∴ x = 2, y = −3
1. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes ½ if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
(i) Let the fraction be x/y.
Data given,
x+1/y-1 = 1 i.e., x – y = -2———-(1)
x/y+1 = 1/2 i.e., 2x – y = 1 —————–(2)
Subtracting equation (1) from equation (2),
x = 3 ………….(3)
Substituting this value in equation (1),
3 – y = -2
-y = -5
y = 5
Hence, the fraction is 3/5.
(ii)Let present age of Nuri = x and present age of Sonu = y
Data given,
(x – 5) = 3(y – 5)
x -3y = -10 ————-(1)
(X + 10) = 2(y + 10)
x – 2y = 10————-(2)
Subtracting equation (1) from equation (2),
y = 20 ————-(3)
Substituting it in equation (1),
x – 60 = -10
x = 50
Hence, age of Nuri = 50 years and age of Sonu = 20 years
(iii)Let the unit digit and tens digits of the number be x and y respectively.
Then, number = 10y + x
Number after reversing the digits = 10x + y
From the given data, x + y = 9 …………………(1)
9(10y + x) = 2(10x + y)
88y − 11x = 0
− x + 8y =0 ………………….(2)
Adding equation (1) and (2), we obtain
9y = 9
y = 1 ——————–(3)
Substituting the value in equation (1),
x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18
(iv)Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.
From the given data,
x + y = 25 —————–(1)
50x + 100y = 2000 —————–(2)
Multiplying equation (1) by 50,
50x + 50y =1250 —————-(3)
Subtracting equation (3) from equation (2),
50y = 750
y = 15
Substituting in equation (1), we have x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.
(v)Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.
From the data given,
x + 4y = 27 ————–(1)
x + 2y = 21 —————(2)
Subtracting equation (2) from equation (1),
2y = 6
y = 3—————(3)
Substituting in equation (1),
x + 12 = 27
x = 15
Hence, fixed charge = Rs 15 and Charge per day = Rs 3 |
# 2012 AMC 8 Problems/Problem 10
## Problem
How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}7\qquad\textbf{(C)}\hspace{.05in}8\qquad\textbf{(D)}\hspace{.05in}9\qquad\textbf{(E)}\hspace{.05in}12$
## Solution 1
For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of $2012$, since all of the valid 4-digit number will always be greater than $1000$. The best way to solve this problem is by using casework.
There can be only two leading digits, namely $1$ or $2$.
When the leading digit is $1$, you can make $\frac{3!}{2!1!} \implies 3$ such numbers.
When the leading digit is $2$, you can make $3! \implies 6$ such numbers.
Summing the amounts of numbers, we find that there are $\boxed{\textbf{(D)}\ 9}$ such numbers.
## Solution 2
Notice that the first digit cannot be $0$, as the number is greater than $1000$. Therefore, there are three digits that can be in the thousands.
The rest three digits of the number have no restrictions, and therefore there are $3! \implies 6$ for each leading digit.
Since the two $2$'s are indistinguishable, there are $\frac {3\cdot6}{2}$ such numbers $\implies \boxed{\textbf{(D)}\ 9}$.
## Solution 3 (a simpler version of Solution 2)
We can list out the four digits in the number. The first digit of the number can’t be 0 since the number is greater than $1000$. This leaves us with three integers for the digit in the thousands place. We have no restrictions in the hundreds place except for the fact that it can’t be the integer we chose for the thousands place. Continuing this pattern, there are $3$ choices for the thousands place, $3$ for the hundreds, $2$ for the tens, and $1$ for the ones. Adding this up we get $9$ which is choice $\boxed{\textbf{(D)} }$.
—-jason.ca
## Solution 4 (Complementary Counting)
There are $4!/2!=12$ ways to arrange the numbers in 2012. The number of ways to arrange 2012 with 0 as the first digit is 3 because there are three places to put the 1 and the 2s are the same. 12-3= 9 which is choice D.
## Video Solution
https://youtu.be/OBKIeTgw4Zg ~savannahsolver
## See Also
2012 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
# How do you find the values of the six trigonometric functions given cottheta=-3 and costheta>0?
Dec 27, 2016
Find trig function values
#### Explanation:
Use tri identity:
${\sin}^{2} t = \frac{1}{1 + {\cot}^{2} t} = \frac{1}{1 + 9} = \frac{1}{10}$ -->
$\sin t = \pm \frac{1}{\sqrt{10}} = \pm \frac{\sqrt{10}}{10}$
Since cot = - 3 < 0 and cos t > 0, there for, sin t < 0.
$\sin t = - \frac{\sqrt{10}}{10}$
${\cos}^{2} t = 1 - {\sin}^{2} t = 1 - \frac{1}{10} = \frac{9}{10}$ -->
$\cos t = \frac{3}{\sqrt{10}} = \frac{3 \sqrt{10}}{10}$
$\tan t = \frac{1}{\cot} t = \frac{1}{-} 3 = - \frac{1}{3}$
$\sec t = \frac{1}{\cos} = \frac{10}{3 \sqrt{10}} = \frac{\sqrt{10}}{3}$
$\csc t = - \frac{10}{\sqrt{10}} = - \sqrt{10}$ |
# RS Aggarwal Chapter 10 Class 9 Maths Exercises 10.1 (ex 10a) Solutions
RD Sharma Chapter 10 Class 9 Maths Exercise 10.1 Solutions: You have studied many properties of a triangle in Chapters 6 and 7 and you know that on joining three non-collinear points in pairs, the figure so obtained is a triangle. Now, let us mark four points and see what we obtain on joining them in pairs in some order. Although most of the objects we see around are of the shape of a special quadrilateral called a rectangle, we shall study more about quadrilaterals and especially parallelograms because a rectangle is also a parallelogram and all properties of a parallelogram are true for a rectangle as well.
EXERCISE 10A
## Important Definition for RS Aggarwal Chapter 10 Class 9 Maths Ex 10a Solutions
A parallelogram is a type of quadrilateral that contains parallel opposite sides.
Area of parallelogram = Base × HeightArea of Triangle = (1/2)× Base × Height
Quadrilaterals- In this chapter, you will learn about a four-point closed figure. The figure formed by joining four points in order is called a quadrilateral. A quadrilateral has the following properties: Four sides, Four angles, and Four vertices.
The chapter begins by giving details about the Angle Sum Property of a Quadrilateral.
The sum of the angles of a quadrilateral is 360°.
Another Condition for a Quadrilateral to be a Parallelogram is explained. A quadrilateral is a parallelogram if a pair of contrary sides are equal and parallel.
After this, the Mid-factor Theorem is discussed.
The line segment becoming a member of the mid-points of two aspects of a triangle is parallel to the third side.
A line via the midpoint of a triangle parallel to another side bisects the third side.
Two theorems are given in this section. A project for students is also given in which they learn the mid-point theorem.
While the definition states “parallelogram”, it is enough to say, “A quadrilateral is a rhombus if and only if it has 4 congruent sides”, considering any quadrilateral with four congruent aspects is a parallelogram.
Know more here. |
# The price of a toy decreasing by 10% per year. The current price of a toy is Rs. 1000. What will be the price of a toy after 3 years?
1. Rs. 1331
2. Rs. 729
3. Rs. 961
4. Rs. 900
Option 2 : Rs. 729
## Detailed Solution
Given:
The price of a toy = Rs. 1000
Rate decreasing per year = 10%
Time = 3 years
Concept used:
A = P(1 - R/100)T
Where,
A = Amount
P = Principal
R = Rate
T = Time
Calculation:
A = P(1 - R/100)T
P = Rs. 1000
R = 10%
T = 3 years
Now,
⇒ A = 1000 × (1 - 10/100)3
⇒ A = 1000 × (1 - 1/10)3
⇒ A = 1000 × (9/10)3
⇒ A = 1000 × (729/1000)
⇒ A = 729
∴ The price of the toy after 3 years is Rs. 729.
Mistake Points
In this question, the rate is decreasing so don't use the formula of A = P(1 + R/100)T
If the rate is successive decreasing,
⇒ - x - y + (xy)/100
The decrease in the rate of first two years (in %)
⇒ -10 - 10 + (10 × 10)/100
⇒ -19%
Now the total decrease in rate over the 3 years,
⇒ -19 - 10 + (19 × 10)/100
⇒ -29 + 1.9
⇒ -27.10%
After then,
Remaining rate,
⇒ 100 - 27.10
⇒ 72.9%
Now the price of a toy after 3 years,
⇒ 1000 × 72.9%
⇒ 1000 × (729/1000)
⇒ 729
∴ The price of the toy after 3 years is Rs. 729.
Shortcut Trick
Rate decreasing per year by 10% for 3 years.
∴ The price of the toy after 3 years is Rs. 729. |
## Exploration 11: Write-Up
Polar Equations:
Investigate:
To gain a better understanding of what this particular equation means as the variables vary, it is important to gain a grasp on the individual changes that occur when each variable is altered separately.
Let’s look at k first by letting a=0 and k=1.
Using Desmos, we can graph this equation and use the slider to let the value of k vary.
Try it Yourself on Desmos!
From this we can see that varying k creates pedals.
Pedals are a feature of a special polar equation, rose, of the form .
Below is a table of the k values from 0 to 5 and the corresponding number of petals.
K # of rose petals 0 0 1 0 2 4 3 3 4 8 5 5 6 12 7 7 8 16
From the table we can see that:
• For all even values, the corresponding number of petals equals 2k.
• For all odd values, the corresponding number of petals equals k.
Thus, the value of k will determine the number of petals associated with the rose graph.
Now let’s take a look at a by letting the b=1 and k=1.
Using Desmos, we can graph this equation and use the slider to let the value of a vary.
Try it out Yourself!
From simply varying the value of a , it was difficult to make any significant observations. However, looking at the graph a=b=1, a heart-shaped graph was produced.
This led me to believe that there is a connection between the a and b values. To investigate, I looked at then cases when and . By moving the sliders of the follows graph, both cases can be observed.
From this there were a few observations that I was able to make.
1. When , there is a loop that goes inward.
I also discovered that:
- The distance of the loop from the origin to the furthest point on the loop is equal to
- The distance from the origin to the farthest point of the graph, is
2. When , there is no loop .
Now let’s look at b. Let’s make a=0 and k=1.
Using Desmos, we can graph this equation and use the slider to let the value of b vary.
Try it out!
From this, I was able to confirm that the value of b alters the length of the diameter of the circle. The diameter is the distance between r=0 and r=b. Thus, the center of a circle with a diameter b in polar coordinates is . When b is negative, it is as if the circle is flipped across the line x=0 or in polar coordinates. |
What is 139/92 as a decimal?
Solution and how to convert 139 / 92 into a decimal
139 / 92 = 1.511
139/92 or 1.511 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Both represent numbers between integers, in some cases defining portions of whole numbers In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. So let’s dive into how and why you can convert 139/92 into a decimal.
139/92 is 139 divided by 92
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. We use this as our equation: numerator(139) / denominator (92) to determine how many whole numbers we have. Then we will continue this process until the number is fully represented as a decimal. Here's how our equation is set up:
Numerator: 139
• Numerators are the portion of total parts, showed at the top of the fraction. Any value greater than fifty will be more difficult to covert to a decimal. 139 is an odd number so it might be harder to convert without a calculator. Values closer to one-hundred make converting to fractions more complex. Now let's explore X, the denominator.
Denominator: 92
• Denominators represent the total parts, located at the bottom of the fraction. Larger values over fifty like 92 makes conversion to decimals tougher. And it is nice having an even denominator like 92. It simplifies some equations for us. Overall, two-digit denominators are no problem with long division. Now let's dive into how we convert into decimal format.
Converting 139/92 to 1.511
Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 92 \enclose{longdiv}{ 139 }$$
Use long division to solve step one. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
Step 2: Solve for how many whole groups you can divide 92 into 139
$$\require{enclose} 00.1 \\ 92 \enclose{longdiv}{ 139.0 }$$
Now that we've extended the equation, we can divide 92 into 1390 and return our first potential solution! Multiple this number by our furthest left number, 92, (remember, left-to-right long division) to get our first number to our conversion.
Step 3: Subtract the remainder
$$\require{enclose} 00.1 \\ 92 \enclose{longdiv}{ 139.0 } \\ \underline{ 92 \phantom{00} } \\ 1298 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you still have a remainder, continue to the next step.
Step 4: Repeat step 3 until you have no remainder
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. They each bring clarity to numbers and values of every day life. Same goes for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But each represent values in everyday life! Here are examples of when we should use each.
When you should convert 139/92 into a decimal
Dining - We don't give a tip of 139/92 of the bill (technically we do, but that sounds weird doesn't it?). We give a 151% tip or 1.511 of the entire bill.
When to convert 1.511 to 139/92 as a fraction
Progress - If we were writing an essay and the teacher asked how close we are to done. We wouldn't say .5 of the way there. We'd say we're half-way there. A fraction here would be more clear and direct.
Practice Decimal Conversion with your Classroom
• If 139/92 = 1.511 what would it be as a percentage?
• What is 1 + 139/92 in decimal form?
• What is 1 - 139/92 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 1.511 + 1/2? |
# Math Snap
## $\sum_{n=0}^{\infty} \frac{2^{n}+3^{n}}{5^{n+1}}=\frac{5}{6}$ Select one: True False
#### STEP 1
Assumptions 1. We are given the infinite series: $\sum_{n=0}^{\infty} \frac{2^{n}+3^{n}}{5^{n+1}}$ 2. We need to determine if this series equals $\frac{5}{6}$.
#### STEP 2
Rewrite the given series by separating the terms inside the summation: $\sum_{n=0}^{\infty} \frac{2^{n}+3^{n}}{5^{n+1}} = \sum_{n=0}^{\infty} \frac{2^{n}}{5^{n+1}} + \sum_{n=0}^{\infty} \frac{3^{n}}{5^{n+1}}$
#### STEP 3
Factor out the common term $\frac{1}{5}$ from each series: $\sum_{n=0}^{\infty} \frac{2^{n}}{5^{n+1}} = \frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{2}{5}\right)^{n}$ $\sum_{n=0}^{\infty} \frac{3^{n}}{5^{n+1}} = \frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{3}{5}\right)^{n}$
#### STEP 4
Recognize that both series are geometric series of the form $\sum_{n=0}^{\infty} ar^n$ where $a = 1$ and $r$ is the common ratio.
#### STEP 5
Recall the formula for the sum of an infinite geometric series $\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}$ where $|r| < 1$.
#### STEP 6
Apply the geometric series sum formula to each series: $\sum_{n=0}^{\infty} \left(\frac{2}{5}\right)^{n} = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3}$ $\sum_{n=0}^{\infty} \left(\frac{3}{5}\right)^{n} = \frac{1}{1 - \frac{3}{5}} = \frac{1}{\frac{2}{5}} = \frac{5}{2}$
#### STEP 7
Multiply each sum by $\frac{1}{5}$: $\frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{2}{5}\right)^{n} = \frac{1}{5} \cdot \frac{5}{3} = \frac{1}{3}$ $\frac{1}{5} \sum_{n=0}^{\infty} \left(\frac{3}{5}\right)^{n} = \frac{1}{5} \cdot \frac{5}{2} = \frac{1}{2}$
#### STEP 8
Add the two results together: $\frac{1}{3} + \frac{1}{2}$
#### STEP 9
Find a common denominator to add the fractions: $\frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}$
##### SOLUTION
Compare the result with the given value: $\sum_{n=0}^{\infty} \frac{2^{n}+3^{n}}{5^{n+1}} = \frac{5}{6}$ Since the calculated value matches the given value, the statement is true. Solution: True |
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• Three coins are tossed. what are the odds against getting three heads?
# Three coins are tossed. what are the odds against getting three heads?
Three Coins are Tossed: What are the Odds Against Getting Three Heads?
"Three Coins are Tossed: What are the Odds Against Getting Three Heads?" is a concise and informative resource that provides a comprehensive understanding of the odds associated with getting three heads when tossing three coins. Written in a simple and easy-to-understand style, this article is designed to assist individuals seeking information about the likelihood of obtaining a specific outcome when tossing three coins.
Positive Aspects of "Three Coins are Tossed: What are the Odds Against Getting Three Heads?":
1. Clear explanation: The article offers a clear and straightforward explanation of the odds against getting three heads when three coins are tossed.
2. Concise format: The information is presented in a concise format, avoiding unnecessary details and ensuring a quick comprehension of the topic.
3. Informative headings: The article is well-structured with helpful headings, allowing readers to easily navigate through the content and find the information they need.
4. Use of lists and checklists: The use of lists and checklists assists in organizing the information and enhancing its readability.
5. Region-specific content: The content is tailored for the US audience, ensuring relevance and appropriateness for readers in this region.
Benefits of "Three Coins are
What is the probability of three heads? The probability of getting a head in one toss is 1/2. So you need this 3 exact times that is (1/2)^3 = 1/8. hence the answer.
## What is the likelihood of obtaining three heads if a coin is tossed three times?
Answer: If a coin is tossed three times, the likelihood of obtaining three heads in a row is 1/8. Let's look into the possible outcomes.
## What is the probability of not tossing 3 heads with three fair coins?
If you don't toss 3 heads in a row, then you can toss: 0 heads out of 3 tosses, or 1 head out of 3 tosses, or 2 heads out of 3 tosses. The probability of tossing 0 heads out of 3 tosses is the same as the probability of tossing 3 heads out of 3 tosses. That equals 1/8.
## What are the possible outcomes of tossing 3 coins?
We can represent head by H and tail by T. Now consider an experiment of tossing three coins simultaneously. The possible outcomes will be HHH, TTT, HTT, THT, TTH, THH, HTH, HHT. So the total number of outcomes is 23 = 8.
## Are you guaranteed to get three heads at least once?
N=3: To get 3 heads, means that one gets only one tail. This tail can be either the 1st coin, the 2nd coin, the 3rd, or the 4th coin. Thus there are only 4 outcomes which have three heads. The probability is 4/16 = 1/4.
## What is the probability of flipping 3 coins and getting 2 heads and a tail?
3/8 What is the probability of two heads and one tail? Summary: The Probability of getting two heads and one tails in the toss of three coins simultaneously is 3/8 or 0.375.
## What is the probability that 2 land on heads when 3 fair coins are flipped?
Answer and Explanation: If you flip three fair coins, the probability that you'll get at least two heads is 4/8 or 0.50, which is 50%.
#### How many ways can you flip exactly 2 heads out of 3 coin tosses?
There are eight possible outcomes of tossing the coin three times, if we keep track of what happened on each toss separately. In three of those eight outcomes (the outcomes labeled 2, 3, and 5), there are exactly two heads. This way of counting becomes overwhelming very quickly as the number of tosses increases.
#### What happens if a fair coin is flipped three times?
Correct answer: If you flip a coin, the chances of you getting heads is 1/2. This is true every time you flip the coin so if you flip it 3 times, the chances of you getting heads every time is 1/2 * 1/2 * 1/2, or 1/8.
#### What are the odds in favor of getting exactly two tails and one head?
So, the odds in favor of getting exactly two tails and one head when tossing three coins are 3 to 5.
#### What is the probability of obtaining exactly two tails when flipping three coins?
Since the outcome of a coin toss is equiprobable, the probability of getting exactly two tails out of three is equal to the number of ways to get two tails out of three - aka - divided by the total number of possible coin flip outcomes - aka . Ergo, 3/8 is the probability.
#### What is the probability of getting 2 tails when flipping 3 coins?
As a result, the chances of having at least two tails are 7/8.
## FAQ
What is the probability of getting 2 heads in tossing 3 coins?
3/8 ∴ The probability of getting exactly two heads is 3/8.
What are the odds of flipping tails tails tails with 3 coins?
Answer: The probability of flipping a coin three times and getting 3 tails is 1/8.
What are the odds of getting three tails when flipping 3 coins?
1/8 Summary: The probability of getting tails on all the three fair coins is 1/8.
How do you find the probability of tossing 3 coins?
Solution: When 3 coins are tossed, the possible outcomes are HHH, TTT, HTT, THT, TTH, THH, HTH, HHT. (i) Let E1 denotes the event of getting all tails. Hence the required probability is ⅛.
## Three coins are tossed. what are the odds against getting three heads?
What are the odds of flipping 3 coins? If you flip a coin 3 times, what are the odds that the coin will be heads all three times? Explanation: If you flip a coin, the chances of you getting heads is 1/2. This is true every time you flip the coin so if you flip it 3 times, the chances of you getting heads every time is 1/2 * 1/2 * 1/2, or 1/8. How do you find the sample space of 3 coins? Hint: If 3 coins are tossed then sample space will be S= {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}. To find the probability of an event use the formula, P(event) = (No. of favourable outcomes) / (Total no. What is the probability of flipping 3 coins and not getting 2 heads? Each coin flip has two possible results. We are flipping three times, so… There are 2*2*2 possible results, of which 3 have exactly one head. So, 3/8 odds. How many outcomes are possible with 3 coins? Eight possible outcomes There are eight possible outcomes of tossing the coin three times, if we keep track of what happened on each toss separately. In three of those eight outcomes (the outcomes labeled 2, 3, and 5), there are exactly two heads. This way of counting becomes overwhelming very quickly as the number of tosses increases.
• What is the probability of getting a head when a coin is tossed 3 times?
• Summary: If you flip a coin 3 times, the probability of getting 1 head is 0.375.
• What is the probability of getting 1 head when 3 coins are tossed?
• (iv) Let E4 denotes the event of getting one head. Hence the required probability is 3/8.
• What are the odds for getting 2 heads when tossing 3 coins?
• 3/8 ∴ The probability of getting exactly two heads is 3/8.
• How many chances to get heads when tossing a coin?
• Suppose you have a fair coin: this means it has a 50% chance of landing heads up and a 50% chance of landing tails up. Suppose you flip it three times and these flips are independent. What is the probability that it lands heads up, then tails up, then heads up? So the answer is 1/8, or 12.5%.
February 8, 2024
February 8, 2024
February 8, 2024 |
A+ » VCE » Maths Methods U3 & 4 Master Notes » A4 Probability and Statistics » Normal Distributions
# Normal Distributions
## 3.6 Real-Life Application of Normal Distributions
#### Applying Normal Distributions in Real Life
• Normal distribution is used to model many things in real life as it is a very reasonable model. Therefore, knowing how to apply the knowledge of normal distributions in real life problems is important.
Example
The time taken to complete a logical reasoning task is normally distributed with a mean of 55 seconds and a standard deviation of 8 seconds. Keeping the answer as accurate to 4 d.p.,
a) Find the probability that that a randomly chosen person will take less than 50 seconds to complete the task.
Read More »3.6 Real-Life Application of Normal Distributions
## 3.5 Finding Normal Distribution Probabilities
#### Standardised Values and Finding Normal Distribution Probabilities
• Previously, we see that any normal distribution can be converted into a standard one via
\operatorname{Pr}(X \leq a)=\operatorname{Pr}\left(Z \leq \frac{a-\mu}{\sigma}\right) \equiv \operatorname{Pr}(Z \leq z)
where X \sim N(\mu, \sigma), and Z \sim N(0,1).
• Such expression is useful as we convert any values into one that is in terms of standard deviation(s) away from the mean. These new values are called standardised values or z-values. In particular we have
z=\frac{x-\mu}{\sigma} \text { or standardised value }=\frac{\text { data value }-\text { mean of normal curve }}{\mathrm{s} . \mathrm{d} . \text { of the curve }}
• Therefore, we can infer that: A positive z-value indicates that the data value it represents lies above the mean. If it is negative, then it is below the mean.
• Knowing how to convert any normal distribution to a standard one is important as it helps us to find the probabilities of any normal distributed events.
Read More »3.5 Finding Normal Distribution Probabilities
## 3.4 The Normal Distribution
#### The Standard Normal Distribution
• The standard normal distribution has the following probability density function
f(x)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} x^{2}}
and the domain is R.
• The graph of f(x) is shown at the right. Notice that it is symmetric over x=0
• If a random variable X follows the standard normal distribution, then we would denote X \sim N(0,1).
• Note that standard normal distribution is a special case of normal distribution, with a mean 0 and variance 1. Therefore, this explains the N(0,1) written above.
• The standard normal distribution also have some important graphical features:
Read More »3.4 The Normal Distribution
## The Normal Distribution [Video Tutorial]
This tutorial covers material encountered in chapter 16 of the VCE Mathematical Methods Textbook, namely: Normal random variables The standard normal distribution Standardising a normal… Read More »The Normal Distribution [Video Tutorial] |
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##### Presentation Transcript
1. Chapter 16 Quadratic Equations
2. Chapter Sections 16.1 – Solving Quadratic Equations by the Square Root Property 16.2 – Solving Quadratic Equations by Completing the Square 16.3 – Solving Quadratic Equations by the Quadratic Formula 16.4 – Graphing Quadratic Equations in Two Variables 16.5 – Interval Notation, Finding Domains and Ranges from Graphs and Graphing Piecewise-Defined Functions
3. Square Root Property We previously have used factoring to solve quadratic equations. This chapter will introduce additional methods for solving quadratic equations. Square Root Property If b is a real number and a2 = b, then
4. Square Root Property Example Solve x2 = 49 Solve 2x2 = 4 x2 = 2 Solve (y – 3)2 = 4 y = 3 2 y = 1 or 5
5. Square Root Property Example Solve x2 + 4 = 0 x2 = 4 There is no real solution because the square root of 4 is not a real number.
6. Square Root Property Example Solve (x + 2)2 = 25 x = 2 ± 5 x = 2 + 5 or x = 2 – 5 x = 3 or x = 7
7. Square Root Property 3x – 17 = Example Solve (3x – 17)2 = 28
8. Completing the Square In all four of the previous examples, the constant in the square on the right side, is half the coefficient of the x term on the left. Also, the constant on the left is the square of the constant on the right. So, to find the constant term of a perfect square trinomial, we need to take the square of half the coefficient of the x term in the trinomial (as long as the coefficient of the x2 term is 1, as in our previous examples).
9. add Completing the Square Example What constant term should be added to the following expressions to create a perfect square trinomial? • x2 – 10x add 52 = 25 • x2 + 16x add 82 = 64 • x2 – 7x
10. Completing the Square Example We now look at a method for solving quadratics that involves a technique called completing the square. It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section.
11. Completing the Square Solving a Quadratic Equation by Completing a Square • If the coefficient of x2 is NOT 1, divide both sides of the equation by the coefficient. • Isolate all variable terms on one side of the equation. • Complete the square (half the coefficient of the x term squared, added to both sides of the equation). • Factor the resulting trinomial. • Use the square root property.
12. y + 3 = ± = ± 1 Solving Equations Example Solve by completing the square. y2 + 6y = 8 y2 + 6y+ 9 = 8 + 9 (y + 3)2 = 1 y = 3 ± 1 y = 4 or 2
13. (y + ½)2 = Solving Equations Example Solve by completing the square. y2 + y – 7 = 0 y2 + y = 7 y2 + y+ ¼ = 7 + ¼
14. x2 + 7x + = ½ + = (x + )2 = Solving Equations Example Solve by completing the square. 2x2 + 14x – 1 = 0 2x2 + 14x = 1 x2 + 7x = ½
15. The Quadratic Formula Another technique for solving quadratic equations is to use the quadratic formula. The formula is derived from completing the square of a general quadratic equation.
16. The Quadratic Formula A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions.
17. The Quadratic Formula Example Solve 11n2 – 9n = 1 by the quadratic formula. 11n2 – 9n – 1 = 0, so a = 11, b = -9, c = -1
18. Solve x2 + x – = 0 by the quadratic formula. The Quadratic Formula Example x2 + 8x – 20 = 0 (multiply both sides by 8) a = 1, b = 8, c = 20
19. The Quadratic Formula Example Solve x(x + 6) = 30 by the quadratic formula. x2 + 6x + 30 = 0 a = 1, b = 6, c = 30 So there is no real solution.
20. The Discriminant The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.
21. The Discriminant Example Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5 b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224 There are no real solutions.
22. Solving Quadratic Equations Steps in Solving Quadratic Equations • If the equation is in the form (ax+b)2 = c, use the square root property to solve. • If not solved in step 1, write the equation in standard form. • Try to solve by factoring. • If you haven’t solved it yet, use the quadratic formula.
23. Solving Equations Example Solve 12x = 4x2 + 4. 0 = 4x2 – 12x + 4 0 = 4(x2 – 3x + 1) Let a = 1, b = -3, c = 1
24. Solving Equations Example Solve the following quadratic equation.
25. Graphs of Quadratic Equations We spent a lot of time graphing linear equations in chapter 3. The graph of a quadratic equation is a parabola. The highest point or lowest point on the parabola is the vertex. Axis of symmetry is the line that runs through the vertex and through the middle of the parabola.
26. y xy x Graphs of Quadratic Equations Example Graph y = 2x2 – 4. (2, 4) (–2, 4) 2 4 1 –2 (–1, – 2) (1, –2) 0 –4 –1 –2 (0, –4) –2 4
27. Intercepts of the Parabola Although we can simply plot points, it is helpful to know some information about the parabola we will be graphing prior to finding individual points. To find x-intercepts of the parabola, let y = 0 and solve for x. To find y-intercepts of the parabola, let x = 0 and solve for y.
28. 2) the x-coordinate of the vertex is . Characteristics of the Parabola If the quadratic equation is written in standard form, y = ax2 + bx + c, 1) the parabola opens up when a > 0 and opens down when a < 0. To find the corresponding y-coordinate, you substitute the x-coordinate into the equation and evaluate for y.
29. y Since a = –2 and b = 4, the graph opens down and the x-coordinate of the vertex is xy x Graphs of Quadratic Equations Example Graph y = –2x2 + 4x + 5. (1, 7) (2, 5) (0, 5) 3 –1 (–1, –1) (3, –1) 5 2 1 7 0 5 –1 –1
30. Interval Notation, Finding Domain and Ranges from Graphs, and Graphing Piecewise-Defined Functions § 16.5
31. Domain and Range Recall that a set of ordered pairs is also called a relation. The domainis the set of x-coordinates of the ordered pairs. The rangeis the set of y-coordinates of the ordered pairs.
32. Domain and Range Example Find the domain and range of the relation {(4,9), (–4,9), (2,3), (10, –5)} • Domain is the set of all x-values, {4, –4, 2, 10} • Range is the set of all y-values, {9, 3, –5}
33. y Domain x Range Domain is [–3, 4] Range is [–4, 2] Domain and Range Example Find the domain and range of the function graphed to the right. Use interval notation.
34. y Range x Domain is (– , ) Range is [– 2, ) Domain Domain and Range Example Find the domain and range of the function graphed to the right. Use interval notation.
35. Input (Animal) Polar Bear Cow Chimpanzee Giraffe Gorilla Kangaroo Red Fox Output (Life Span) 20 15 10 7 Domain and Range Example Find the domain and range of the following relation.
36. Domain and Range Example continued Domain is {Polar Bear, Cow, Chimpanzee, Giraffe, Gorilla, Kangaroo, Red Fox} Range is {20, 15, 10, 7}
37. Graph Graphing Piecewise-Defined Functions Example Graph each “piece” separately. Values 0. Values > 0. Continued.
38. y (3, 6) Open circle (0, 3) x (0, –1) (–1, 4) (–2, 7) Graphing Piecewise-Defined Functions Example continued |
$$\require{cancel}$$
# 1.7: Examples of 2-Dimensional Motion
#### Circular Motion
Using what we just derived regarding the parallel and perpendicular components of acceleration, we turn now to the special case of an object traveling in a circle. The parallel part of the acceleration obviously always points tangent to the circle, which narrows it down to two directions at any given point on the circle. If the object is speeding up, then of course the tangent points in the direction of motion, and if it is slowing down, the tangent vector points in the direction opposite to the motion. The perpendicular part must be at a right angle to this tangent, which means it must be toward or away from the center of the circle. Unlike the tangent case, however, both directions are not possible. We can see this by considering the average perpendicular acceleration vector over two nearby moments in time:
Figure 1.7.1 – Direction of Centripetal Acceleration
What is the magnitude of this centripetal acceleration? Well, it depends upon how fast the object is going (the faster it is moving, the more acceleration is required to turn in the same circle), and the radius of the circle (the acceleration is greater when the radius is smaller). Deriving the magnitude is left as an exercise, but the answer comes out to be:
$|\overrightarrow a_c | = \dfrac{v^2}{r}$
Sometimes circular motion is the result of something rotating. For example, a bug on the outer rim of a rotating turntable travels in a circle, and therefore experiences centripetal acceleration. Well, when we deal with rotating objects we often know only the rate of its rotation (say in rpms), and we have to translate into linear motion to know the speed. There is a simple and standard way to do this.
If we are talking about rotational motion, we need to discuss how we measure such motion. We clearly don’t measure the speed of a spinning top or turntable in terms of meters per second, but rather how much it turns in a period of time. How does one measure the angle through which something turns? One way is to divide the full circle up into 360 equally-sized pie slices. The magnitude of one of these angles we call a 'degree.' But really this division is arbitrary. So why was 360º selected? Well, given that we often have to deal with slices of pie, we can avoid having to use fractional degrees if we select a number with lots of factors, and 360 certainly fits the bill – it is divisible by 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 120, and 180.
But there is no reason at all that divisibility needs to be our only criterion. In fact, we don’t even need to divide the circle into an integral number of pieces. For example, we could divide it into 7.5 pieces and call the size of each piece “1 flibber.” We can even translate between different systems of units: $$1^f$$ = 48º.
But there is another criterion that leads to a definition of a measurement of angles that is different from degrees. Suppose we want a simple translation from angle to arclength (the distance traveled along the circular curve subtended by that angle). We know that traveling around an entire circle requires a journey of a distance equal to $$2 \pi$$ times the radius of the circle. So going around some fraction of a circle requires a journey equal to that fraction times $$2 \pi r$$. So if we divide the circle into an uneven number of pieces such that $$2 \pi$$ of these pieces fit into the circle, then in these units you can calculate the arclength by just multiplying the angle measured in those units (called radians) by the radius of the circle:
$s = r \theta, \ \theta \ \text{measured in radians}$
If we want to translate between the speed that something is going around the circle to the angular speed at which the slice of the pie is being swept-out by this motion, we need only take a derivative:
$v = \dfrac{ds}{dt} = \dfrac{d}{dt} \left( r \theta \right) = \cancelto{0}{\dfrac{dr}{dt}}\theta + r \dfrac{d\theta}{dt}\; = r \omega, \;\;\;\;\;\;\;\; \omega \equiv \dfrac{d\theta }{dt} = rate\;of\;rotation\;in\; \dfrac{rad}{s}$
This gives us an alternative way of expressing the centripetal acceleration:
$| \vec a _ c | = \frac{v^2}{r} = \frac{\left( r \omega \right) ^2}{r} = r \omega^2$
Example $$\PageIndex{1}$$
A bead is threaded onto a circular hoop of wire which lies in a vertical plane. The bead starts at the bottom of the hoop from rest, and is pushed around the hoop such that it speeds up at a steady rate. Find the angle that the bead's acceleration vector makes with the horizontal when it gets back to the bottom of the hoop.
Solution
The tangential motion of the bead is in one dimension, so we can use the usual kinematics equations to describe its motion along the circle. Let's call the radius of the circle $$R$$ and the final velocity $$v$$. The tangential acceleration is constant, the bead starts from rest, and the bead travels one circumference, so using Equation 1.4.3, we get:
$2a \Delta x = {v_f}^2 - {v_o}^2 \;\;\; \Rightarrow \;\;\; a_\parallel=\dfrac{\left(v^2-0^2\right)}{2\left(2\pi R \right)}=\dfrac{v^2}{4\pi R} \nonumber$
The centripetal acceleration is toward the center of the circle, so it points upward and its magnitude is simply:
$a_\bot=\dfrac{v^2}{R} \nonumber$
The tangent of the angle that the full acceleration vector makes with the horizontal is the vertical component divided by the horizontal component, so:
$\theta = \tan^{-1} \left(\dfrac{\dfrac{v^2}{R}}{ \dfrac{v^2}{4\pi R}}\right)=\tan^{-1} \left(4\pi \right) = 85^o \nonumber$
#### Projectile Motion
For circular motion, we have the components of velocity changing in tandem in a specific manner to keep the path circular. Another – actually simpler – form of motion involves only a single fixed coordinate component of velocity changing, while the other components involve no change in velocity. What I am alluding to here is projectile motion, which comes about because the vertical component of motion is subject to constant acceleration (as we discussed when we talked about free-fall), while the horizontal component is unaffected by gravity’s influence. This kind of motion is only one small step from the free-fall we are already familiar with, in that it includes a totally independent horizontal component of motion that incorporates (to the extent that air resistance can be ignored) no acceleration. As simple as this sounds, a couple of examples muddy the waters a bit, and sorting them out is very instructive:
Example $$\PageIndex{2}$$
A hunter climbs a tree and fires a bullet directly at a monkey that is hanging from a branch of another tree at precisely the same height as the barrel of the hunter’s gun. The instant the bullet leaves the gun, the monkey lets go of the branch. Ignoring air resistance (and the size of the monkey - assume it is very small), what is the fate of the monkey?
1. The monkey will be hit by the bullet.
2. The bullet will pass beneath the monkey.
3. The bullet will fly over the monkey’s head.
4. Whether the bullet flies over the monkey’s head or passes beneath it depends upon how fast the bullet is moving when it leaves the barrel of the gun.
5. What kind of jerk shoots a monkey?
Solution
A (and E). The vertical motion of the monkey is independent of the horizontal motion, so in equal time spans, the bullet falls the same distance as the monkey. Since they started at the same level, they remain at the same level at all times, including when the horizontal position of the bullet equals the horizontal position of the monkey.
Example $$\PageIndex{3}$$
After falling out of the tree the last time he tried to shoot a monkey (his gun misfired), the hunter now decides to shoot a monkey from the ground. He is aiming upward at an angle, and is assuming the monkey will again let go of the branch just as the bullet is on its way. How should the hunter aim this time, if he is to bag his simian prize?
1. He should aim right at the monkey.
2. He should aim above the monkey.
3. He should aim below the monkey.
4. Unlike the level-shot case, where he should aim this time does depend upon how fast the bullet is coming out of the gun.
5. The jerk should just aim at himself.
Solution
A (and E). If there was no gravity, the bullet would follow a straight line to the monkey. With gravity acting straight down, the amount that the bullet drops below that straight line is the same that an object starting from rest on that straight line falls in an equal time. So the bullet and the monkey remain the same distance from the straight line at all times. When the bullet’s horizontal position equals the monkey’s horizontal position, they will coincide.
The only difference between this example and the previous one is that in the previous case, the line joining the barrel of the gun and the target was horizontal. Still, not everyone may be as convinced in this case as in the previous one, so let's do the math...
Suppose there is no gravity. The path that the bullet takes ($$y$$ as a function of $$x$$) can be written down pretty easily. If the point where the bullet exits the barrel is chosen to be the origin, then the straight line to the monkey has a slope equal to the ratio of the vertical and horizontal components of velocity:
$y=mx+b=\left(\dfrac{v_{oy}}{v_{ox}} \right)x \nonumber$
Now suppose there is gravity. We have separate horizontal and vertical equations of motion. Again, with the bullet starting at the origin, we have:
$x=v_{ox}t \\ y=v_{oy}t-\frac{1}{2}gt^2 \nonumber$
Now solve for $$t$$ in the first equation and plug it into the first term of the second equation to get:
$y=\dfrac{v_{oy}}{v_{ox}}x-\frac{1}{2}gt^2 \nonumber$
Comparing this with the first equation above, we see that the $$y$$ value would follow the same straight line if not for the second term, and the amount that the height of the bullet $$y$$ is decreased from that line after a time $$t$$ is exactly the same distance that the monkey falls from that line in the same time.
With the number of variables and constants involved in projectile motion problems, there are countless ways to construct problems. There is no substitute for independent thinking and creativity, but the steps given below provide a good starting point for solving these kinds of problems.
• Draw a picture, labeling it as completely as you can, using information you have been given. Then spend some time thinking about what is happening – put yourself into the situation.
While this is given as a step for projectile problems, this is actually how you should start every physics problem!
• Pick an $$x$$, $$y$$ origin as well as $$+x$$ and $$+y$$ directions. Often for projectile problems up is chosen as the positive directions (making the acceleration due to gravity a negative value), but this is by no means required. What is important is that you use the positive direction consistently throughout the constants and variables in the equations.
• Break any initial velocities into components along the $$x$$ and $$y$$ directions.
• Write down the equations of motion, circling quantities that you know, and underlining the number you are looking for. If you have too many un-circled quantities for the number of equations available, you cannot yet do the algebra, so you’ll need to review the statement of the problem for any values concealed in the language of the problem (if you just scan a problem for numbers without carefully reading it, you will miss these).
• Solve the algebra and reconstitute components back into vectors, if necessary.
• Briefly check to see if the answer makes sense.
This is actually how you should end every physics problem!
One thing in projectile motion that is a useful tool is known as the range equation. This was of particular use to military firing cannonballs or (farther back in history than that), catapults. This equation relates the distance that a projectile will fly assuming it lands at the same vertical position that it started, given the starting speed of the projectile and the starting angle. Let’s treat finding this equation as if it was a projectile motion problem given to us, and follow the procedure outlined above
Problem: A cannonball is fired at an angle θ up from the horizontal at a speed of vo. Ignoring air resistance, how far does the cannonball travel before landing back on ground level with where it was fired?
Example $$\PageIndex{4}$$
A cannonball is fired at an angle $$\theta$$ up from the horizontal at a speed of vo. Ignoring air resistance, how far does the cannonball travel before landing back on ground level with where it was fired?
Solution
The diagram below labels the origin and the + directions. We are looking for $$R$$ (referred to as the "range") in terms of the initial speed $$v_o$$ of the projectile and the launch angle $$\theta$$, which we treat as known values. The rest of the procedure is given below.
We can do other things with this result. For example, suppose we wanted to fire the cannonball as far as possible. Obviously we want it to start at the fastest possible speed, but given that, what angle should we aim at? Let’s solve it! We want to maximize the range R in terms of our choice of angle. This is a straightforward calculus problem. Take the derivative and set it equal to zero to find the maximum:
$0 = \frac{dR}{d\theta} = \frac{v_0^2}{g}\frac{d}{d\theta}\big( \cos2 \theta \big) \big( 2 \big) \ \Rightarrow \ \boxed{\theta = 45^\circ}$
This really shouldn't be too surprising, based on a simple symmetry argument: If we shoot the projectile with a really steep angle, it goes nearly straight up and doesn't travel very far. We can send the projectile the same distance by choosing a sufficiently shallow angle. That means that every landing point in range of the launcher has two possible distinct values for launch angle, except for the one corresponding to the longest range. It makes sense that complimentary launch angles reach the same landing point: $$89^o$$ and $$1^o$$, $$52^o$$ and $$38^o$$, and so on, which means that the self-complimentary angle of $$45^o$$ hits the maximum range.
The following example is a challenging application of this principle, and gives an idea of the rich diversity of problems that can be conceived for projectile motion:
Example $$\PageIndex{5}$$
Two warlords aim identical catapults (i.e. they both release rocks at the same speed) at each other, with both of them being at the same altitude. The warlords have made the necessary computations to crush the other, and fire their catapults simultaneously. Amazingly, the two stones do not collide with each other in mid-air, but instead the stone Alexander fired passes well below the stone that Genghis shot. Genghis is annihilated 8.0s after the catapults are fired, and Alexander only got to celebrate his victory for 4.0s before he too was destroyed.
1. Find the maximum height reached by each of the rocks.
2. Find the amount of time that elapses from the launch to moment that the rocks pass each other in the air.
3. Find the angles at which each warlord fires his rock.
Solution
a. The time it takes a rock to travel to its peak height and back down again is equal to twice the time it takes to travel down from its peak height. Traveling down from its peak height, it starts with zero initial velocity, so we can calculate the height immediately for each rock:
$h_{A} = \frac{1}{2}g\left(\frac{t_A}{2}\right)^2 = \frac{1}{2} \left( 9.8 \dfrac{m}{s^2} \right) \left( \frac{8.0s}{2} \right)^2 = \boxed{78.4m} \\ h_{G} = \frac{1}{2}g\left(\frac{t_G}{2}\right)^2 = \frac{1}{2} \left( 9.8 \dfrac{m}{s^2} \right) \left( \frac{8.0s+4.0s}{2} \right)^2 = \boxed{176.4m} \nonumber$
b. The x–components of the velocities of the rocks never change, and since it takes 12s for Genghis’s rock to travel the same horizontal distance as Alexander’s rock traveled in 8s, Alexander’s rock is traveling in the x-direction at a rate 1.5 times as great as Genghis’s rock is traveling in the x–direction. When they are at the same x–position (passing each other), the distance each has traveled is each one’s velocity times the time we are looking for, and we can express both of these distances in terms of the x–component of Genghis’s rock using the ratio described above:
$\begin{array}{l} x_A=v_{Ax}t, \;\;\;\;\; v_{Ax}=1.5v_{Gx} \;\;\; \Rightarrow \;\;\; x_A=1.5v_{Gx}t \\ x_G=v_{Gx}t \nonumber \end{array}$
Since the rocks travel from both ends and are now at the same horizontal position, the sum of the distances they travel equals the total separation of the two warlords. This allows us to calculate the time:
$x_A + x_G = x_{tot} = v_G t_{tot} \;\;\; \Rightarrow \;\;\; \left( 1.5v_G \right) t + v_G t = v_G \left( 12.0s \right) \;\;\; \Rightarrow \;\;\; t=\dfrac{12.0s}{2.5} = \boxed{4.8s} \nonumber$
c. Clearly there are two different angles that will result in the rock traveling the same distance. One can see this from the range equation, but from a physical standpoint, this happens because one rock spends less time in the air but has a greater x–velocity, while the other spends more time in the air with a smaller x–velocity. To spend 1.5 times as long in the air, Genghis’s rock needs to start with 1.5 times as much vertical component of velocity as Alexander’s rock. This means that the ratios of the x and y components of the two rock velocities are inverses of one another, which means that the two angles are complimentary (i.e. $$\theta_A = 90^o – \theta_G$$). But the total speeds of the rocks are the same, so:
$\left. \begin{array}{l} v_{Ax} = v_o \cos \theta _A = v_o \cos \left( 90^o - \theta _G \right) = v_o\sin \theta _G \\ v_{Gx} = v_o\cos \theta _G \end{array} \right\}\;\;\; \Rightarrow \;\;\;\dfrac{v_{Ax}}{v_{Gx}} = 1.5 = \dfrac{\sin \theta _G}{\cos \theta _G}\;\;\; \Rightarrow \;\;\;\left\{ \begin{array}{l} \theta _G = \tan ^{ - 1}1.5 = \boxed{56.3^o}\\ \theta _A = 90^o - \theta _G = \boxed{33.7^o} \end{array} \right. \nonumber$ |
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# Solving Equations with Fractions
by Debra
(PA)
-5= -19- x/7
### Karin from Algebra Class says:
Whenever you have a fraction within an equation, you can always eliminate that fraction by multiplying all terms by the denominator. If you have more than one fraction, then you would multiply by the least common denominator.
Step 1: Multiply all terms by 7 to get rid of the fraction.
(7)-5 = 7(-19 - x/7)
-35 = -133 - x When you multiply 7 (x/7) the 7 in the numerator and the 7 in the denominator simplify to 1 (or you could say they cancel out). Therefore, you are left with what was in the numerator.
So, now we have:
-35 = -133 - x
Step 2: Add 133 to both sides.
-35 + 133 = -133 + 133 - x
98 = -x
Step 3: Since x is negative, we can multiply all terms by -1, to make x positive.
(-1) 98 = -x(-1)
-98 = x
Then you can check by substituting:
-5 = -19 - (-98/7)
-5 = -5
I hope this helps. You can also review the lesson on solving equations with fractions on the following web page.
http://www.algebra-class.com/equations-with-fractions.html
Best of luck,
Karin
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# Combinations and the Binomial Theorem
In document Applied Discrete Structures (Page 47-59)
### 2.4.1 Combinations
In Section 2.1 we investigated the most basic concept in combinatorics, namely, the rule of products. It is of paramount importance to keep this fundamental rule in mind. In Section 2.2 we saw a subclass of rule-of-products problems, permutations, and we derived a formula as a computational aid to assist us. In this section we will investigate another counting formula, one that is used to count combinations, which are subsets of a certain size.
In many rule-of-products applications the ordering is important, such as the batting order of a baseball team. In other cases it is not important, as in placing coins in a vending machine or in the listing of the elements of a set. Order is important in permutations. Order is not important in combinations. Example 2.4.1 Counting Permutations. How many different ways are there to permute three letters from the setA={a, b, c, d}? From thePermu- tation Counting Formulathere areP(4,3) = (44!3)! = 24different orderings of
three letters fromA
Example 2.4.2 Counting with No Order. How many ways can we select a set of three letters fromA={a, b, c, d}? Note here that we are not concerned with the order of the three letters. By trial and error, abc, abd, acd, and bcd
CHAPTER 2. COMBINATORICS 34 are the only listings possible. To repeat, we were looking for all three-element subsets of the setA. Order is not important in sets. The notation for choosing 3 elements from 4 is most commonly(4
3
)
or occasionallyC(4,3), either of which is read “4 choose 3” or the number of combinations for four objects taken three
at a time. □
Definition 2.4.3 Binomial Coefficient. Letnandk be nonnegative inte- gers. The binomial coefficient(n
k
)
represents the number of combinations ofn objects takenkat a time, and is read “nchoose k.” ♢ We would now like to investigate the relationship between permutation and combination problems in order to derive a formula for(n
k
)
Let us reconsider the Counting with No Order. There are 3! = 6different orderings for each of the three-element subsets. The table below lists each subset ofAand all permutations of each subset on the same line.
subset permutations
{a, b, c} abc, acb, bca, bac, cab, cba
{b, c, d} bcd, bdc, cdb, cbd, dbc, dcb . Hence,(4 3 ) =P(43!,3)= (44!3)!·3! = 4
We generalize this result in the following theorem:
Theorem 2.4.4 Binomial Coefficient Formula. Ifnandkare nonnegative integers with 0 ≤k≤n, then the number k-element subsets of an n element
set is equal to ( n k ) = n! (n−k)!·k!.
Proof. Proof 1: There arek! ways of ordering the elements of anyk element set. Therefore, ( n k ) =P(n, k) k! = n! (n−k)!k!..
Proof 2: To “construct” a permutation ofkobjects from a set ofnelements, we can first choose one of the subsets of objects and second, choose one of the k!permutations of those objects. By the rule of products,
P(n, k) = ( n k ) ·k! and solving for(n
k
)
we get the desired formula. ■
Example 2.4.5 Flipping Coins. Assume an evenly balanced coin is tossed five times. In how many ways can three heads be obtained? This is a combi- nation problem, because the order in which the heads appear does not matter. We can think of this as a situation involving sets by considering the set of flips of the coin, 1 through 5, in which heads comes up. The number of ways to get three heads is(5
3
)
= 52··41 = 10. □
Example 2.4.6 Counting five ordered flips two ways. We determine the total number of ordered ways a fair coin can land if tossed five consecutive times. The five tosses can produce any one of the following mutually exclusive, disjoint events: 5 heads, 4 heads, 3 heads, 2 heads, 1 head, or 0 heads. For example, by the previous example, there are(5
3
)
= 10sequences in which three heads appear. Counting the other possibilities in the same way, by the law of
CHAPTER 2. COMBINATORICS 35 addition we have: ( 5 5 ) + ( 5 4 ) + ( 5 3 ) + ( 5 2 ) + ( 5 1 ) + ( 5 0 ) = 1 + 5 + 10 + 10 + 5 + 1 = 32 ways to observe the five flips.
Of course, we could also have applied the extended rule of products, and since there are two possible outcomes for each of the five tosses, we have25= 32
ways. □
You might think that counting something two ways is a waste of time but solving a problem two different ways often is instructive and leads to valuable insights. In this case, it suggests a general formula for the sum∑n
k=0
(n
k
)
. In the case ofn= 5, we get25so it is reasonable to expect that the general sum is
2n, and it is. A logical argument to prove the general statment simply involves
generalizing the previous example toncoin flips.
Example 2.4.7 A Committee of Five. A committee usually starts as an unstructured set of people selected from a larger membership. Therefore, a committee can be thought of as a combination. If a club of 25 members has a five-member social committee, there are(25
5
)
= 25·24·235!·22·21 = 53130different possible social committees. If any structure or restriction is placed on the way the social committee is to be selected, the number of possible committees will probably change. For example, if the club has a rule that the treasurer must be on the social committee, then the number of possibilities is reduced to(24
4
)
=242322214! = 10626.
If we further require that a chairperson other than the treasurer be selected for the social committee, we have (24
4
)
·4 = 42504 different possible social committees. The choice of the four non-treasurers accounts for the factor(24
4
)
while the need to choose a chairperson accounts for the 4. □ Example 2.4.8 Binomial Coefficients - Extreme Cases. By simply applying the definition of aBinomial Coefficientas a number of subsets we see that there is(n
0
)
= 1way of choosing a combination of zero elements from a set ofn. In addition, we see that there is(n
n
)
= 1way of choosing a combination ofnelements from a set of n.
We could compute these values using the formula we have developed, but no arithmetic is really needed here. Other properties of binomial coefficients that can be derived using the subset definition will be seen in the exercises □
### 2.4.2 The Binomial Theorem
The binomial theorem gives us a formula for expanding (x+y)n, where n
is a nonnegative integer. The coefficients of this expansion are precisely the binomial coefficients that we have used to count combinations. Using high school algebra we can expand the expression for integers from 0 to 5:
n (x+y)n 0 1 1 x+y 2 x2+ 2xy+y2 3 x3+ 3x2y+ 3xy2+y3 4 x4+ 4x3y+ 6x2y2+ 4xy3+y4 5 x5+ 5x4y+ 10x3y2+ 10x2y3+ 5xy4+y5
CHAPTER 2. COMBINATORICS 36
(5 3
)
= 10, and that of the sixth term is(5 5
)
= 1. We can rewrite the expansion
as ( 5 0 ) x5+ ( 5 1 ) x4y+ ( 5 2 ) x3y2+ ( 5 3 ) x2y3+ ( 5 4 ) xy4+ ( 5 5 ) y5. In summary, in the expansion of(x+y)n we note:
(a) The first term isxn and the last term isyn.
(b) With each successive term, exponents ofx decrease by 1 as those of y increase by 1. For any term the sum of the exponents isn.
(c) The coefficient ofxn−kyk is(n k
)
.
(d) The triangular array of binomial coefficients is called Pascal’s triangle after the seventeenth-century French mathematician Blaise Pascal. Note that each number in the triangle other than the 1’s at the ends of each row is the sum of the two numbers to the right and left of it in the row above.
Theorem 2.4.9 The Binomial Theorem. If n 0, and x and y are numbers, then (x+y)n = nk=0 ( n k ) xn−kyk.
Proof. This theorem will be proven using a logical procedure called mathemat- ical induction, which will be introduced in Chapter 3. ■ Example 2.4.10 Identifying a term in an expansion. Find the third term in the expansion of(x−y)4= (x+ (−y))4. The third term, whenk= 2, is(4
2
)
x42(−y)2= 6x2y2. □
Example 2.4.11 A Binomial Expansion. Expand(3x2)3. If we replace
xandy in the Binomial Theorem with3xand2, respectively, we get
3 ∑ k=0 ( 3 k ) (3x)n−k(2)k = ( 3 0 ) (3x)3(2)0+ ( 3 1 ) (3x)2(2)1+ ( 3 2 ) (3x)1(2)2+ ( 3 3 ) (3x)0(2)3 = 27x354x2+ 36x8 . □
### 2.4.3 SageMath Note
A bridge hand is a 13 element subset of a standard 52 card deck. The order in which the cards come to the player doesn’t matter. From the point of view of a single player, the number of possible bridge hands is(52
13
)
, which can be easily computed withSage.
binomial (52 ,13) 635013559600
In bridge, the location of a hand in relation to the dealer has some bearing on the game. An even truer indication of the number of possible hands takes into account each player’s possible hand. It is customary to refer to bridge positions as West, North, East and South. We can apply the rule of product to get the total number of bridge hands with the following logic. West can get
CHAPTER 2. COMBINATORICS 37 any of the(52
13
)
hands identified above. Then North get 13 of the remaining 39 cards and so has(39
13
)
possible hands. East then gets 13 of the 26 remaining cards, which has(26
13
)
possibilities. South gets the remaining cards. Therefore the number of bridge hands is computed using the Product Rule.
binomial (52 ,13) * binomial (39 ,13) * binomial (26 ,13) 53644737765488792839237440000
### 2.4.4 Exercises
1. The judiciary committee at a college is made up of three faculty members and four students. If ten faculty members and 25 students have been nominated for the committee, how many judiciary committees could be formed at this point?
2. Suppose that a single character is stored in a computer using eight bits. a. How many bit patterns have exactly three 1’s?
b. How many bit patterns have at least two 1’s?
Hint. Think of the set of positions that contain a 1 to turn this is into a question about sets.
3. How many subsets of{1,2,3, . . . ,10} contain at least seven elements? 4. The congressional committees on mathematics and computer science are
made up of five representatives each, and a congressional rule is that the two committees must be disjoint. If there are 385 members of congress, how many ways could the committees be selected?
5. The image below shows a 6 by 6 grid and an example of alattice path that could be taken from(0,0)to(6,6), which is a path taken by traveling along grid lines going only to the right and up. How many different lattice paths are there of this type? Generalize to the case of lattice paths from (0,0)to(m, n)for any nonnegative integers mandn.
Figure 2.4.12: A lattice path
Hint. Think of each path as a sequence of instructions to go right (R) and up (U).
6. How many of the lattice paths from(0,0) to(6,6)pass through (3,3)as the one inFigure 12does?
CHAPTER 2. COMBINATORICS 38 7. A poker game is played with 52 cards. At the start of a game, each player gets five of the cards. The order in which cards are dealt doesn’t matter.
(a) How many “hands” of five cards are possible?
(b) If there are four people playing, how many initial five-card “hands” are possible, taking into account all players and their positions at the table? Position with respect to the dealer does matter.
8. A flush in a five-card poker hand is five cards of the same suit. The suits are spades, clubs, diamonds and hearts. How many spade flushes are possible in a 52-card deck? How many flushes are possible in any suit? 9. How many five-card poker hands using 52 cards contain exactly two aces? 10. In poker, a full house is three-of-a-kind and a pair in one hand; for exam- ple, three fives and two queens. How many full houses are possible from a 52-card deck? You can use the sage cell in theSageMath Noteto do this calculation, but also write your answer in terms of binomial coefficients. 11. A class of twelve computer science students are to be divided into three
groups of 3, 4, and 5 students to work on a project. How many ways can this be done if every student is to be in exactly one group?
12. Explain in words why the following equalities are true based on number of subsets, and then verify the equalities using the formula for binomial coefficients. (a) (n 1 ) =n (b) (n k ) =(nnk),0≤k≤n
13. There are ten points, P1, P2, . . . , P10 on a plane, no three on the same
line.
(a) How many lines are determined by the points? (b) How many triangles are determined by the points?
14. How many ways can n persons be grouped into pairs when n is even? Assume the order of the pairs matters, but not the order within the pairs. For example, ifn= 4, the six different groupings would be
{1,2} {3,4} {1,3} {2,4} {1,4} {2,3} {2,3} {1,4} {2,4} {1,3} {3,4} {1,2}
15. Use the binomial theorem to prove that ifAis a finite set, then|P(A)|= 2|A|
16.
(a) A state’s lottery involves choosing six different numbers out of a possible 36. How many ways can a person choose six numbers? (b) What is the probability of a person winning with one bet? 17. Use the binomial theorem to calculate99983.
CHAPTER 2. COMBINATORICS 39 18. In the card game Blackjack, there are one or more players and a dealer. Initially, each player is dealt two cards and the dealer is dealt one card down and one facing up. As in bridge, the order of the hands, but not the order of the cards in the hands, matters. Starting with a single 52 card deck, and three players, how many ways can the first two cards be dealt out? You can use the sage cell in the SageMath Note to do this calculation.
## Logic
In this chapter, we will introduce some of the basic concepts of mathematical logic. In order to fully understand some of the later concepts in this book, you must be able to recognize valid logical arguments. Although these arguments will usually be applied to mathematics, they employ the same techniques that are used by a lawyer in a courtroom or a physician examining a patient. An added reason for the importance of this chapter is that the circuits that make up digital computers are designed using the same algebra of propositions that we will be discussing.
### 3.1.1 Propositions
Definition 3.1.1 Proposition. A proposition is a sentence to which one and only one of the termstrueor falsecan be meaningfully applied. ♢ Example 3.1.2 Some Propositions. “Four is even,”, “4 ∈ {1,3,5}” and “43>21” are propositions. □ In traditional logic, a declarative statement with a definite truth value is considered a proposition. Although our ultimate aim is to discuss mathemati- cal logic, we won’t separate ourselves completely from the traditional setting. This is natural because the basic assumptions, or postulates, of mathemati- cal logic are modeled after the logic we use in everyday life. Since compound sentences are frequently used in everyday speech, we expect that logical propo- sitions contain connectives like the word “and.” The statement “Europa sup- ports life or Mars supports life” is a proposition and, hence, must have a definite truth value. Whatever that truth value is, it should be the same as the truth value of “Mars supports life or Europa supports life.”
### 3.1.2 Logical Operations
There are several ways in which we commonly combine simple statements into compound ones. The words/phrases and, or, not, if ... then..., and ...if and only if ... can be added to one or more propositions to create a new proposi- tion. To avoid any confusion, we will precisely define each one’s meaning and introduce its standard symbol. With the exception of negation (not), all of the operations act on pairs of propositions. Since each proposition has two possible truth values, there are four ways that truth can be assigned to two
CHAPTER 3. LOGIC 41 propositions. In defining the effect that a logical operation has on two propo- sitions, the result must be specified for all four cases. The most convenient way of doing this is with a truth table, which we will illustrate by defining the wordand.
Definition 3.1.3 Logical Conjunction. Ifpand qare propositions, their conjunction,pandq(denoted p∧q), is defined by the truth table
p q p∧q 0 0 0 0 1 0 1 0 0 1 1 1 ♢ Notes:
(a) To read this truth table, you must realize that any one line represents a case: one possible set of values forpandq.
(b) The numbers 0 and 1 are used to denote false and true, respectively. This is consistent with the way that many programming languages treat logical, or Boolean, variables since a single bit, 0 or 1, can represent a truth value.
(c) For each case, the symbol underprepresents the truth value ofp. The same is true forq. The symbol underp∧qrepresents its truth value for that case. For example, the second row of the truth table represents the case in which pis false,q is true, and the resulting truth value forp∧q is false. As in everyday speech,p∧qis true only when both propositions are true.
(d) Just as the lettersx,y andzare frequently used in algebra to represent numeric variables,p,qandrseem to be the most commonly used symbols for logical variables. When we say that pis a logical variable, we mean that any proposition can take the place ofp.
(e) One final comment: The order in which we list the cases in a truth table is standardized in this book. If the truth table involves two simple propo-
In document Applied Discrete Structures (Page 47-59)
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# Construction of Combinational Circuits
A Combinational Circuit consist of logic gates whose outputs at any instant of time are determined directly from the present combination of inputs without regard to previous input. Examples of combinational circuits: Adder, Subtractor, Converter, and Encoder/Decoder.
Here we are going to learn how to construct and analyze any type of combinational circuit using four general steps. I am going to explain this trick with the help of the one combinational circuit and you can apply the same for implementing other combinational circuits.
Following are the four steps to construct and analyze any combinational circuit.
• Step-1: Identify the number of inputs and outputs of the circuit.
First of all, we have to think about the inputs and outputs of the circuit by considering which type of logical operation we want to perform with the circuit.
For example, we have to create a circuit that can add two bits. For this, we require two inputs (one for the first bit (A) another for the second bit (B)) and two outputs one for sum (S) of two bits and another for carry (C).
In total, we require 2 inputs and 2 outputs. So here our first step is completed.
• Step-2: Creating the Truth Table.
In this step we have to create truth table for our circuit so for this first we will create input columns and list all the possible combinations of inputs. In our case 2 bits can have maximum 4 combinations (00 01 10 11).
Now in output, we have two columns (Sum and Carry) as discussed earlier. Now we have to fill output columns in such a way that for which logical operation we are constructing circuit.
In our circuit, we want addition so we will add those input bits and write the sum of those bits in (Sum) column and if carry is generated we will write 1 else write.
0 in (Carry) column.
• Step-3: Simplify the Boolean function for each output.
In this step, we have to just create a simplified Boolean function according to inputs and outputs of the truth table obtained in the previous step.
For Sum,
`Sum = A'B + AB' = A XOR B `
For Carry,
`Carry = AB = A AND B `
• Step-4: Constructing circuit using Boolean function obtained from third step.
For sum, we have obtained (A XOR B) so we will connect A and B to the inputs of XOR gate and take its output as a sum. For carry, we have obtained (A AND B) so we will connect A and B to the inputs of AND gate and take its output as a carry.
Now in this circuit, if you provide input at A and B ends. You will get the output on sum and carry ends according to truth table we have created above. So here we have completed our four steps for creating the combinational circuit.
So, we have created a combinational circuit called Half Adder. You can apply the same steps to create any other combinational circuit.
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## Introduction,Binomial theorem
Subject: Mathematics
#### Overview
This note helps you learn what binomial expression mean and how you can expand the binomial expansion in the form of (a+b)^n. Here, n is a positive integer.
##### Introduction,Binomial theorem
The expression that consists of two terms is known as a binomial expression. Example :
$$\;a\;+\;b,\;x+\frac{1}{y}\;,\;x^2+y\;$$ . When the power of such expressions is a very small positive integer such as 2, 3, 4 it is not difficult to expand them. But if the power is large numbers, let us say ‘n’ then we need a formula a formula for the expansion of the binomial expression. The expansion of the [removed]a+x)n, where the value of n is a positive integer is called the Binomial Theorem. The binomial theorem was first introduced by Sir Issac Newton.
#### Binomial Theorem :
According to the binomial theorem, for any positive integer n,
$$(a+x)^n\;=\;^nC_0a^nx^0\;+\;^nC_2a^{n-1}x^1\;+\;^nC_3a^{n-2}x^2\;+\;.\;.\;.\;.\;.\;.\;+\;^nC_ra^{n-r}x^r\;+\;.\;.\;.\;.\;.\;+\;^nC_na^{n-n}x^n$$
Proof: We have, $$\;(a+x)\;=\;(a+X)(a+x)(a+x)\;.\;.\;.\;.\;.\;.\;.\;to\;n\;factors$$
In the process of multiplication of n factors, let us start by taking n = 2, 3, and 4. Then the expansions are :$$\;(a+x)^1=a+x$$ $$\;(a+x)^2=a^2+2ab+b^2$$ $$\;(a+x)^3=a^3+3a^2b+3ab^2+b^3$$ $$\;(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
Form the above equation we can see that,
$$\;(a+x)^2=a^2+2ab+b^2\;=^2C_0a^2b^0+^2C_1ab+^2C_2a^0b^2$$
$$\;(a+x)^3=a^3+3a^2b+3ab^2+b^3\;=^3C_0a^3b^0+^3C_1a^2b^1+^3C_2a^1b^2+^3C_3b^3$$
So, the theorem holds good when n= 2, 3.
Let us assume that the theorem is true for all positive integer values of n. Let us say m, is the new value of n, then
$$(a+b)^m\;=\;^mC_0a^mb^0\;+\;^mC_1a^{m-1}b^1\;+\;^mC_2a^{m-2}b^2\;+\;.\;.\;.\;.\;.\;.\;+\;^mC_ra^{m-r}b^r\;+\;.\;.\;.\;.\;.\;+\;^mC_ma^{m-m}b^m$$
Multiplying both sides by (a + b), we get
$$(a+b)^m+1\;=\;(a+b)\;(\;^mC_0a^mb^0\;+\;^mC_1a^{m-1}b^1\;+\;^mC_1a^{m-2}b^2\;+\;.\;.\;.\;.\;.\;.\;+\;^mC_ra^{m-r}b^r\;+\;.\;.\;.\;.\;.\;+\;^mC_ma^{m-m}b^m\;)$$
$$\;=\;^mC_0a^{m+1}b^0\;+\;^mC_1a^{m}b^1\;+\;^mC_2a^{m-1}b^2\;+;.\;.\;.\;.\;.\;+\;^mC_ma^{1}b^m\;+\;.\;.\;.\;.\;.\;.\;^mC_0a^mb^1\;+\;^mC_1a^{m-1}b^2\;+\;^mC_1a^{m-2}b^3\;+\;.\;.\;.\;.\;+\;^mC_mb^{m+1}$$
$$\;=\;a^{m+1}\;+\;(^mC_1\;+\;1)\;a^mx\;+(^mC_2\;+\;^mC_1\;)\;a^{m-1}x^2\;+\;(^mC_3\;+\;^mC_2\;)\;a^{m-2}x^3\;+\;.\;.\;.\;.\;.\;+\;1\;x^{m+1}$$
$$\;=\;^{m+1}C_0a^{m+1}\;+\;^{m+1}C_1a^mx\;+\;^{m+1}C_2a^{m-1}x^2\;+\;.\;.\;.\;.\;.\;.\;+\;^{m+1}C_{m+1}x^{m+1}$$
Therefore when the theorem is true for n = m. The theorem is also true for n = m + 1.
We know that the theorem is true for n = 3, therefore it must be true n = 4
#### Properties of the expansion of (a + b)n
Some of the important properties on binomial expansion are as follows :
• The number of terms in the expansion of (a + b)n is n + 1 i.e. one more than the index n.
• In the successive terms of the expansion, the index of ‘a’ goes on decreasing by unity, starting from n, then n – 1 , and ending with zero, on the contrary, the index of b goes on increasing by unity, starting from 0, then 1, . . . . . . . . . .and ending with n.
• In any term, the sum of indices of a and b is equal to n.
• In the expansion of (a + b )n,
The first term T1 = T0 + 1 = nC0 an-0 b0
The second term T2 = T1 + 1 = nC1 an-1 b1
The third term T3 = T2 + 1 = nC2 an-2 b2 . . . . . . . . . . .
Similarly, the (r + 1)th term Tr + 1 = nCr an-r br which is called the general term.
• The coefficients of the terms equidistant from the beginning and the end are always equal.
#### Middle term
Now, let us find the middle term or terms of the expansion (a + b)n. We have to consider the cases when n is an even number and when n is an odd number.
• (i) When n is even
When n is even, we write n = 2m ,where m = 1, 2, 3, . . . . The number of terms after expansion is 2m + 1, which is odd. So, it has only one middle term, namely (m + 1)th term. So,
$$\;t_{m+1}\;=\;^{2m}C_ma^mb^m\;=\;\frac{2m!}{(m!)^2}\;a^m\;b^m$$
$$\therefore\;middle\;term\;t_{m+1}\;=\;t_{\frac{1}{2}n+1}$$
$$\;=\;^nC_{\frac{1}{2}n}a^{n-\frac{n}{2}}x^{\frac{n}{2}}$$
$$\;=\;\frac{n!}{(\frac{1}{2}n!)^2}a^{n-\frac{n}{2}}x^{\frac{n}{2}}$$
• (ii) When n is odd
When n is odd, we write n = 2m – 1, where m = 1, 2, 3 …., The number of terms after expansion is 2m, which is even. So, there will be two middle terms, namely mth and (m + 1)th term. So,
$$\;t_m\;=\;^{2m-1}C_{m-1}.a^m.x^{m-1}$$
$$\;=\;\frac{(2m-1)!}{m!(m-1)!}\;.a^m.x^{m-1}$$
and,
$$\;t_{m+1}\;=\;^{2m-1}C_m.a^{m-1}.x^m$$
$$\;=\;\frac{(2m-1)!}{m!(m-1)!}\;.a^{m-1}.x^{m}$$
$$\therefore\;the\;middle\;terms\;are$$
$$\;t_m\;=\;t_{\frac{n+1}{2}}\;=\;t_{\frac{n-1}{2}+1}$$
and,
$$t_{m+1}\;=\;t_{\frac{n+1}{2}+1}$$
Taken reference from
( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )
##### Things to remember
• The expression that consists of two terms is known as a binomial expression.
• According to the binomial theorem, for any positive integer n,
$$(a+x)^n\;=\;^nC_0a^nx^0\;+\;^nC_2a^{n-1}x^1\;+\;^nC_3a^{n-2}x^2\;+\;.\;.\;.\;.\;.\;.\;+\;^nC_ra^{n-r}x^r\;+\;.\;.\;.\;.\;.\;+\;^nC_na^{n-n}x^n$$
• The number of terms in the expansion of (a + b)n is n + 1 i.e. one more than the index n.
• In the successive terms of the expansion, the index of ‘a’ goes on decreasing by unity, starting from n, then n – 1 , and ending with zero, on the contrary, the index of b goes on increasing by unity, starting from 0, then 1, . . . . . . . . . .and ending with n.
• In any term, the sum of indices of a and b is equal to n.
• In the expansion of (a + b )n,
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society. |
# 8.2: One-Sample Interval for the Proportion
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Suppose you want to estimate the population proportion, p. As an example you may be curious what proportion of students at your school smoke. Or you could wonder what is the proportion of accidents caused by teenage drivers who do not have a drivers’ education class.
### Confidence Interval for One Population Proportion (1-Prop Interval)
1. State the random variable and the parameter in words.
x = number of successes
p = proportion of successes
2. State and check the assumptions for confidence interval
1. A simple random sample of size n is taken.
2. The condition for the binomial distribution are satisfied
3. To determine the sampling distribution of $$\hat{p}$$, you need to show that $$n \hat{p} \geq 5$$ and $$n \hat{q} \geq 5$$, where $$\hat{q}=1-\hat{p}$$. If this requirement is true, then the sampling distribution of $$\hat{p}$$ is well approximated by a normal curve. (In reality this is not really true, since the correct assumption deals with p. However, in a confidence interval you do not know p, so you must use $$\hat{p}$$. This means you just need to show that $$x \geq 5$$ and $$n-x \geq 5$$.)
3. Find the sample statistic and the confidence interval
Sample Proportion:
$$\hat{p}=\dfrac{x}{n}=\dfrac{\# \text { of successes }}{\# \text { of trials }}$$
Confidence Interval:
$$\hat{p}-E<p<\hat{p}+E$$
Where
p = population proportion
$$\hat{p}$$ = sample proportion
n = number of sample values
E = margin of error
$$z_{c}=$$ = critical value
$$\hat{q}=1-\hat{p}$$
$$E=z_{c} \sqrt{\dfrac{\hat{p} \hat{q}}{n}}$$
4. Statistical Interpretation: In general this looks like, “there is a C% chance that $$\hat{p}-E<p<\hat{p}+E$$ contains the true proportion.”
5. Real World Interpretation: This is where you state what interval contains the true proportion.
The critical value is a value from the normal distribution. Since a confidence interval is found by adding and subtracting a margin of error amount from the sample proportion, and the interval has a probability of containing the true proportion, then you can think of this as the statement $$P(\hat{p}-E<p<\hat{p}+E)=C$$. You can use the invNorm command on the TI-83/84 calculator to find the critical value. The critical values will always be the same value, so it is easier to just look at table A.1 in the appendix.
Example $$\PageIndex{1}$$ confidence interval for the population proportion using the formula
A concern was raised in Australia that the percentage of deaths of Aboriginal prisoners was higher than the percent of deaths of non-Aboriginal prisoners, which is 0.27%. A sample of six years (1990-1995) of data was collected, and it was found that out of 14,495 Aboriginal prisoners, 51 died ("Indigenous deaths in," 1996). Find a 95% confidence interval for the proportion of Aboriginal prisoners who died.
1. State the random variable and the parameter in words.
2. State and check the assumptions for a confidence interval.
3. Find the sample statistic and the confidence interval.
4. Statistical Interpretation
5. Real World Interpretation
Solution
1. x = number of Aboriginal prisoners who die
p = proportion of Aboriginal prisoners who die
2.
1. A simple random sample of 14,495 Aboriginal prisoners was taken. However, the sample was not a random sample, since it was data from six years. It is the numbers for all prisoners in these six years, but the six years were not picked at random. Unless there was something special about the six years that were chosen, the sample is probably a representative sample. This assumption is probably met.
2. There are 14,495 prisoners in this case. The prisoners are all Aboriginals, so you are not mixing Aboriginal with non-Aboriginal prisoners. There are only two outcomes, either the prisoner dies or doesn’t. The chance that one prisoner dies over another may not be constant, but if you consider all prisoners the same, then it may be close to the same probability. Thus the assumptions for the binomial distribution are satisfied
3. In this case, x = 51 and n - x = 14495 - 51 = 14444 and both are greater than or equal to 5. The sampling distribution for $$\hat{p}$$ is a normal distribution.
3. Sample Proportion:
$$\hat{p}=\dfrac{x}{n}=\dfrac{51}{14495} \approx 0.003518$$
Confidence Interval:
$$z_{c}=1.96$$, since 95% confidence level
$$E=z_{c} \sqrt{\dfrac{\hat{p} \hat{q}}{n}}=1.96 \sqrt{\dfrac{0.003518(1-0.003518)}{14495}} \approx 0.000964$$
$$\hat{p}-E<p<\hat{p}+E$$
$$0.003518-0.000964<p<0.003518+0.000964$$
$$0.002554<p<0.004482$$
4. There is a 95% chance that $$0.002554<p<0.004482$$ contains the proportion of Aboriginal prisoners who died.
5. The proportion of Aboriginal prisoners who died is between 0.0026 and 0.0045.
You can also do the calculations for the confidence interval with technology. The following example shows the process on the TI-83/84.
Example $$\PageIndex{2}$$ confidence interval for the population proportion using technology
A researcher studying the effects of income levels on breastfeeding of infants hypothesizes that countries where the income level is lower have a higher rate of infant breastfeeding than higher income countries. It is known that in Germany, considered a high-income country by the World Bank, 22% of all babies are breastfeed. In Tajikistan, considered a low-income country by the World Bank, researchers found that in a random sample of 500 new mothers that 125 were breastfeeding their infants. Find a 90% confidence interval of the proportion of mothers in low-income countries who breastfeed their infants?
1. State you random variable and the parameter in words.
2. State and check the assumptions for a confidence interval.
3. Find the sample statistic and the confidence interval.
4. Statistical Interpretation
5. Real World Interpretation
Solution
1. x = number of woman who breastfeed in a low-income country
p = proportion of woman who breastfeed in a low-income country
2.
1. A simple random sample of 500 breastfeeding habits of woman in a low-income country was taken as was stated in the problem.
2. There were 500 women in the study. The women are considered identical, though they probably have some differences. There are only two outcomes, either the woman breastfeeds or she doesn’t. The probability of a woman breastfeeding is probably not the same for each woman, but it is probably not very different for each woman. The assumptions for the binomial distribution are satisfied
3. x = 125 and n - x = 500 - 125 = 375 and both are greater than or equal to 5, so the sampling distribution of $$\hat{p}$$ is well approximated by a normal curve.
3. On the TI-83/84: Go into the STAT menu. Move over to TESTS and choose 1-PropZInt.
Once you press Calculate, you will see the results as in Figure $$\PageIndex{2}$$.
On R: the command is prop.test(x, n, conf.level = C), where C is given in decimal form. So for this example, the command is
prop.test(125, 500, conf.level = 0.90)
1-sample proportions test with continuity correction
data: 125 out of 500, null probability 0.5
X-squared = 124, df = 1, p-value < 2.2e-16
alternative hypothesis: true p is not equal to 0.5
90 percent confidence interval:
0.2185980 0.2841772
sample estimates:
p
0.25
Some software does a continuity correction, so the answer can be slightly off from the formula and the TI-83/84 calculator.
0.219 < p < 0.284
4. There is a 90% chance that 0.219 < p < 0.284 contains the proportion of women in low-income countries who breastfeed their infants.
5. The proportion of women in low-income countries who breastfeed their infants is between 0.219 and 0.284.
## Homework
Exercise $$\PageIndex{1}$$
In each problem show all steps of the confidence interval. If some of the assumptions are not met, note that the results of the interval may not be correct and then continue the process of the confidence interval.
1. Eyeglassomatic manufactures eyeglasses for different retailers. They test to see how many defective lenses they make. Looking at the type of defects, they found in a three-month time period that out of 34,641 defective lenses, 5865 were due to scratches. Find a 99% confidence interval for the proportion of defects that are from scratches.
2. In November of 1997, Australians were asked if they thought unemployment would increase. At that time 284 out of 631 said that they thought unemployment would increase ("Morgan gallup poll," 2013). Estimate the proportion of Australians in November 1997 who believed unemployment would increase using a 95% confidence interval?
3. According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, Arkansas had 1,601 complaints of identity theft out of 3,482 consumer complaints ("Consumer fraud and," 2008). Calculate a 90% confidence interval for the proportion of identity theft in Arkansas.
4. According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Calculate a 90% confidence interval for the proportion of identity theft in Alaska.
5. In 2013, the Gallup poll asked 1,039 American adults if they believe there was a conspiracy in the assassination of President Kennedy, and found that 634 believe there was a conspiracy ("Gallup news service," 2013). Estimate the proportion of American’s who believe in this conspiracy using a 98% confidence interval.
6. In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Find the proportion of ASD in Arizona with a confidence level of 99%. |
# 4.1- Plot Points in a Coordinate Plane
## Presentation on theme: "4.1- Plot Points in a Coordinate Plane"— Presentation transcript:
4.1- Plot Points in a Coordinate Plane
Ordered pairs are used to locate points in a coordinate plane.
y-axis (vertical axis) 5 -5 5 x-axis (horizontal axis) -5 origin (0,0)
In an ordered pair, the first number is the x-coordinate
In an ordered pair, the first number is the x-coordinate. The second number is the y-coordinate. Graph. (-3, 2) 5 -5 5 -5
The x-axis and y-axis separate the coordinate plane into four regions, called quadrants.
II (-, +) I (+, +) III (-, -) IV (+, -)
Name the quadrant in which each point is located (-5, 4)
II III IV None – x-axis None – y-axis
Name the quadrant in which each point is located (-2, -7)
II III IV None – x-axis None – y-axis
Name the quadrant in which each point is located (0, 3)
II III IV None – x-axis None – y-axis
Example 1: Give the coordinates of the point.
(1,4) (-4,3) (0,3) (5,1) (-3,0) (1,0) (0,0) (0,-3) (-4,-4) (3,-4)
Example 2: Plot the point in a coordinate plane
Example 2: Plot the point in a coordinate plane Describe the location of the point. a) A(-4,9) b) B(5,0) c) C(0,-8) d) D(7, -2) A(-4,9) B(5,0) D(7,-2) C(0,-8)
Example 3: Graph the function y = ½x + 1 with the
following values of x: -4, -2, 0, 2, and 4. Step 1: Make a table x y = ½x + 1 y -4 y = ½(-4) + 1 -1 -2 y = ½(-2) + 1 y = ½(0) + 1 1 2 y = ½(2) + 1 4 y = ½(4) + 1 3 x y = ½x + 1 y -4 y = ½(-4) + 1 -2 y = ½(-2) + 1 y = ½(0) + 1 2 y = ½(2) + 1 4 y = ½(4) + 1 x y = ½x + 1 y -4 y = ½(-4) + 1 -2 y = ½(-2) + 1 y = ½(0) + 1 2 y = ½(2) + 1 4 y = ½(4) + 1
What is the ordered pair for A?
5 (3, 1) (1, 3) (-3, 1) (3, -1) • A -5 5 -5
What is the ordered pair for C?
(0, -4) (-4, 0) (0, 4) (4, 0) 5 -5 5 • C -5 |
Median
The median of a list of numbers is the "middle" number. To find the median, first we put our list of numbers in order. Then we cross off pairs of numbers (one from the top of the list and one from the bottom of the list, or one from each end of the list, depending on how we've written it out) until we're left with one number. That number is the median. If the median was a contestant on Survivor, he would totally be taking home that \$1,000,000 prize.
Sample Problem
Find the median of the following list:
3, 4, 6, 5, 7, 10, 11
First, we put the numbers in order:
3, 4, 5, 6, 7, 10, 11
Then we take off one number from each end of the list:
4, 5, 6, 7, 10
We do it again:
5, 6, 7
And one more time:
6
We're left with the number 6, so that's the middle number in the list. The median is 6. The tribe has spoken.
It's pretty easy when we've got an odd number of entries. We crossed off one number from each end of the list, and repeated until we were left with one number, which left us with exactly one number. However, if our list has an even number of entries to start with, we get two middle numbers. Keeping with our Survivor analogy, they'd probably need to fight each other to the death to determine a clear winner. But since this is a family example, let's instead find the median between these two middle numbers.
To find the median, we take the mean (or average) of these two numbers in the middle.
Sample Problem
Find the median of the following list:
2, 3, 4, 8, 9, 10
Our list is already in order from smallest to biggest, so we cross off the first and last number:
3, 4, 8, 9
Then we do it again:
4, 8
Now we have only two numbers left. If we cross them both off we'll have no numbers left, which won't be useful for anything, aside from getting rid of all those noisy numbers so we can finally get a few minutes of shut-eye. Instead, we take the average of the two numbers we have left:
The median of the list is 6. |
New Zealand
Level 6 - NCEA Level 1
# Experimental Probability
Lesson
## Experiment or Trial
An experiment or trial are the words used to describe the event or action of doing something and recording results. For example, the act of drawing cards from a deck, tossing a coin, rolling a dice, watching the weather, asking questions in a survey or counting cars in a carpark could all be examples of experiments or trials.
## Sample Space
The sample space, sometimes called and event space, is a listing of all the possible outcomes that could arise from an experiment.
For example
• tossing a coin would have a sample space of {Head, Tail}, or {H,T}
• rolling a dice would have a sample space of {1,2,3,4,5,6}
• watching the weather could have a sample space of {sunny, cloudy, rainy} or {hot, cold}
• asking questions in a survey of favourite seasons could have a sample space of {Summer, Autumn, Winter, Spring}
Did you also notice how I listed the sample space? Using curly brackets {}.
## Event
An event is the word used to describe a single result of an experiment. It helps us to identify which of the sample space outcomes we might be interested in.
For example, these are all events.
• Getting a tail when a coin is tossed.
• Rolling more than 3 when a dice is rolled
• Getting an ACE when a card is pulled from a deck
We use the notation, P(event) to describe the probability of particular events.
Adding up how many times an event occurred during an experiment gives us the frequency of that event.
The relative frequency is how often the event occurs compared to all possible events and is also known as the probability of that event occurring.
## Probability Values
The probability values that events can take on range between 0 (impossible) and 1 (certain).
## Experimental Probability
Experimental Probability, as the name suggests, describes the probability when undertaking experiments or trials.
We calculate experimental probability by considering $\frac{\text{frequency of the event }}{\text{total number of trials }}$frequency of the event total number of trials and writing it as a fraction, ratio, decimal or percentage.
#### Example
$500$500 cables were tested at a factory, and $76$76 were found to be faulty.
a) What is the experimental probability that a cable at this factory will be faulty?
$\frac{\text{frequency of the event }}{\text{total number of trials }}=\frac{76}{500}$frequency of the event total number of trials =76500
we can simplify this fraction to $\frac{19}{125}$19125, or convert it to a percentage which is $76\div500\times100=15.2%$76÷500×100=15.2%
b) If $1500$1500 more cables were tested, how many would you expect to be faulty?
Now that we know that $15.2%$15.2% are faulty (from our experimental data), we could expect the same percentage to be faulty from any amount.
So, $15.2%$15.2% of $1500$1500
$\frac{15.2}{100}\times1500=228$15.2100×1500=228
We could expect $228$228 to be faulty from $1500$1500 cables.
#### Worked Examples
##### QUESTION 1
If the probability of an event occurring is $\frac{11}{25}$1125, how many times would you expect it to occur in $575$575 times?
##### QUESTION 2
$16$16 dice were rolled and a $2$2 occurred $4$4 times.
1. What was the relative frequency of rolling a $2$2?
2. Using the results of the trial to predict future outcomes, how many times would you expect a $2$2 to occur if $48$48 dice are rolled?
##### QUESTION 3
1000 transistors were tested at a factory, and 12 were found to be faulty.
1. What is the experimental probability that a transistor at this factory will be faulty?
2. If another 5000 transistors were tested, how many of these would you expect to be faulty?
### Outcomes
#### S6-3
Investigate situations that involve elements of chance: A comparing discrete theoretical distributions and experimental distributions, appreciating the role of sample size B calculating probabilities in discrete situations.
#### 91038
Investigate a situation involving elements of chance |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 2.15: Line Graphs to Display Data Over Time
Difficulty Level: At Grade Created by: CK-12
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Practice Line Graphs to Display Data Over Time
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Have you ever tried to show how something has changed using a graph? Remember Tania? Well, with all of the vegetables that she has been growing, Tania is trying to plan for next year. To do this, she takes a trip to the nearby organic farm to gather some data. When she meets with Mr. Jonas the farmer, he shows her a line graph that shows vegetable growth for the past four years. Tania is fascinated. Here is what she sees.
Mr. Jonas tells Tania that according to his calculations, the farm will produce twice as much in 2009 as it did in 2008. Tania leaves the farm with the data and a lot of excitement. She decides to redraw the line graph at home with the new calculations for 2009. The minute she gets home, she realizes that she is confused and can’t remember how to draw a line graph.
This is where you come in. There is a lot to learn in this Concept, pay attention so that you can help Alex draw his map and Tania draw her line graph at the end of the Concept.
### Guidance
We have already learned about a few different ways to visually display data.
A line graph is a graph that helps us to show how data changes over time.
How can we make a line graph?
To make a line graph, we need to have a collection of data that has changed over time. Data that shows growth over years is a good example of appropriate data for a line graph.
When Jamal was born, his parents planted a tree in the back yard. Here is how tall the tree was in each of the next five years.
2003 2 ft.
2004 3 ft.
2005 5 ft.
2006 9 ft.
2007 14 ft.
Now let’s make a line graph.
The first thing that we need is two axes, one vertical and one horizontal.
The vertical one represents the range of tree growth. The tree grew from 2 feet to 14 feet. That is our scale. The horizontal axis represents the years when tree growth was calculated.
Now let's practice with a few questions about line graphs.
#### Example A
True or false. You need a vertical and horizontal axis for a line graph.
Solution: True.
#### Example B
True or false. A line graph and a frequency table measure the same thing.
Solution: False. A frequency table measures how often something occurs. A line graph measures how data changes over time.
#### Example C
True or false. A bar graph measures the same data as a line graph.
Solution: False. A line graph measures how data changes. A bar graph does not.
Now let's go back and help Tania with her garden plan. She wants to create a line graph to show the 2009 data with the other data she gathered from the farm.
The first thing that she needs to do is to draw in 2 axes.
The horizontal axis shows the years: 2005, 2006, 2007, 2008, 2009. The vertical axis shows the number of vegetables harvested. The highest number she has is in 2008 with 400 vegetables. However, the Mr. Jonas told her he expects to double this amount. This would give 2009 a total of 800 vegetables. Our range for the vertical axis is from 0 to 800 in increments of 100 units.
Here is Tania’s line graph.
### Vocabulary
Here are the vocabulary words in this Concept.
Coordinate grid
a visual way of locating points or objects in space.
Coordinates
the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} values that tell us where an object is located.
Origin
where the x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} axis meet, has a value of (0, 0)
x\begin{align*}x\end{align*} axis
the horizontal line of a coordinate grid
y\begin{align*}y\end{align*} axis
the vertical line of a coordinate grid
Ordered pair
(x,y)\begin{align*}(x, y)\end{align*} the values where a point is located on a grid
Line Graph
a visual way to show how data changes over time
### Guided Practice
Here is one for you to try on your own.
Look at the following line graph.
Which day of the week had the highest temperature? What was that temperature?
The day of the week with the highest temperature was February 3rd. The temperature on that day was about 33 degrees.
### Practice
Directions: Use the following line graph to answer each question.
1. What is being measured in this line graph?
2. What is on the horizontal axis?
3. What is on the vertical axis?
4. What was the highest temperature recorded?
5. What was the lowest temperature recorded?
6. What is the difference between the two temperatures?
7. On what day did the lowest temperature occur?
8. What was the average temperature for the week?
9. What was the median temperature for the week?
10. Did any two days have the same temperature?
11. What was that temperature?
12. On which two days did it occur?
13. Based on this trend, would the temperature on February 8th be less than 30 degrees or greater than?
14. True or false. There isn't a way to figure out the temperature on January 31st.
15. What was the temperature on February 5th?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
$x-$axis
The $x-$axis is the horizontal axis in the coordinate plane, commonly representing the value of the input or independent variable.
$y$ axis
The $y$-axis is the vertical number line of the Cartesian plane.
Coordinate grid
The coordinate grid is formed by a horizontal number line and a vertical number line that cross at the (0, 0) point, called the origin. The coordinate grid is also called a Cartesian Plane or coordinate plane.
Coordinates
The coordinates of a point represent the point's location on the Cartesian plane. Coordinates are written in ordered pairs: $(x, y)$.
Line Graph
A line graph is a visual way to show how data changes over time.
Ordered Pair
An ordered pair, $(x, y)$, describes the location of a point on a coordinate grid.
Origin
The origin is the point of intersection of the $x$ and $y$ axes on the Cartesian plane. The coordinates of the origin are (0, 0).
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# Sect. 1.2 Operations & Properties of Real Numbers
Sect. 1.2 Operations & Properties of Real Numbers. Absolute Value Inequalities Addition, Subtraction, Opposites Multiplication, Division, Reciprocals Laws: Commutative, Associative, Distributive. Absolute Value – Concept. Positive distance from 0 on a number line
## Sect. 1.2 Operations & Properties of Real Numbers
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### Presentation Transcript
1. Sect. 1.2Operations & Properties of Real Numbers • Absolute Value • Inequalities • Addition, Subtraction, Opposites • Multiplication, Division, Reciprocals • Laws: Commutative, Associative, Distributive 1.2
2. Absolute Value – Concept • Positive distance from 0 on a number line • How far is the number 6 from the 0 point? • 6 units • How far is -4.75 from the 0 point? • 4.75 units 1.2
3. Inequalities – Concept • Use the number line to understand • < less than • > more than • Express the relation of -4 to -7 • Either -7 < -4 • or -4 > -7 both ways are correct 1.2
4. Opposites and Absolute Value • Opposites are a pair of numbers that sum to 0 • 11.2 and -11.2 • -3/17 and 3/17 • 0 and 0 (zero is it’s own opposite) • The Absolute Value of a number is the positive value of it’s pair of opposites • |11.2| = 11.2 and |-11.2| = 11.2 • |-3/17| = 3/17 and |3/17| = 3/17 • Absolute value brackets can hold expressions • |3 – 5| = |-2| = 2 • |5 – 3| = |2| = 2 • -|22| = -22 and -|-22| = -22 1.2
5. Fractions with One Negative Sign These are the Simplest Forms 1.2
6. The Law of Reciprocals What is 4’s reciprocal? ¼ because 4 (¼) = 1 What is 3½ ’s reciprocal? 2/7 Because (7/2)(2/7) = 1 a and 1/a are Multiplicative Inverses 1.2
7. Equivalent Expressions • Two expressions are Equivalent when they have the same values for all possible replacements • Are the following two expressions equivalent?3x + 4 and x – 3 + 2x + 7 • Yes – when simplified, the 2nd expression matches the 1st 1.2
8. I Must Remember: Commutative • An operation is Commutative when Two values can switch positions, and the result is the same. • COMMUTATIVE • Addition and Multiplication are Commutative • 6 +11= 17 7(9)= 63 • 11+ 6 = 179(7)= 63 • Are Subtraction or Division Commutative ? 1.2
9. I Must Remember: Associative • Three values can be computed in different order and the result is the same. (same operations) • ASSOCIATIVE ASSOCIATIVE • Addition and Multiplication are Associative • (19 + 4)+ 6 = 23 + 6 = 29 (3)(7)(5) = 21(5) = 105 • 19 + (4 + 6)= 19 + 10 = 29 (3)(7)(5) = 3(35) = 105 • Are Subtraction or Division Associative ? 1.2
10. The Distributive Law and (a + b)c = ac + bc 1.2
11. Let’s play Name That Law! • x + 5 + y = x + y + 5 • Commutative … COM • 3a + 6 = 3(a + 2) • Distributive … DIST • 7x(1 / x) = 7 • Reciprocals … RECIP • (x + 5) + y = x + (5 + y) • Associative … ASSOC • 4(a + 2b) = 8b + 4a • COM, then DIST or DIST, then COM 1.2
12. Let’s look at next time … • Section 1.3 Solving Equations • If we have time today, let’s review Reducing Numeric Fractions 1.2
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# Difference between revisions of "2010 AMC 12A Problems/Problem 24"
## Problem
Let $f(x) = \log_{10} \left(\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)\right)$. The intersection of the domain of $f(x)$ with the interval $[0,1]$ is a union of $n$ disjoint open intervals. What is $n$?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36$
## Solution 1
The question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.
We note that since all of the $\sin$ factors are inside a logarithm, the function is undefined where the inside of the logarithm is less than or equal to $0$.
First, let us find the number of zeros of the inside of the logarithm.
\begin{align*}\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x) &= 0\\ \sin(\pi x) &= 0\\ x &= 0, 1\\ \sin(2 \pi x) &= 0\\ x &= 0, \frac{1}{2}, 1\\ \sin(3 \pi x) &= 0\\ x &= 0, \frac{1}{3}, \frac{2}{3}, 1\\ \sin(4 \pi x) &= 0\\ x &= 0, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, 1\\ &\cdots\end{align*}
After counting up the number of zeros for each factor and eliminating the excess cases we get $23$ zeros and $22$ intervals.
In order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc.
The first interval $\left(0, \frac{1}{8}\right)$ is obviously positive. This means the next interval $\left(\frac{1}{8}, \frac{1}{7}\right)$ is negative. Continuing the pattern and accounting for doubled roots (which do not flip sign), we realize that there are $5$ negative intervals from $0$ to $\frac{1}{2}$. Since the function is symmetric, we know that there are also $5$ negative intervals from $\frac{1}{2}$ to $1$.
And so, the total number of disjoint open intervals is $22 - 2\cdot{5} = \boxed{12\ \textbf{(B)}}$
## Solution 2 (cheap)
Note that the expression $\sin(\pi x) \cdot \sin(2 \pi x) \cdot \sin (3 \pi x) \cdots \sin(8 \pi x)$ must be greater than zero, since logarithm functions are undefined for $0$ and negative numbers. Let $x_1, x_2, x_3, ..., x_8$ temporarily be the dependent variables of the functions $y_1 = \sin(\pi x_1), y_2 = \sin(2\pi x_2), ..., y_8 = \sin(8\pi x_8)$. It is easy to see that for $y_i$ to be positive for $1\leq i\leq8$, $\lfloor i x_i \rfloor$ must be even for $1 \leq i\leq8$. Since an even number of positives times an even number of negatives equals a positive, there can be $2, 4, 6,$ or $8$ positive values of $y_i$ for $1 \leq i\leq 8$ for a given value of $x$. (since $y_1$ is always positive on the range $[0, 1]$) Since MAA allows rulers (and you should bring one to the actual exam), use it to your advantage and draw a larged scaled number line from $0$ to $1$. (I recommend increments of at most $0.1$.) If you don't have a ruler but have graph paper, you can use that instead. Then, designate rows for $y_1, y_2, ..., y_8$, respectively. Draw a large bar (label it with $+$ so you know it's positive) for all values of $x_i$ such that $\lfloor i x_i \rfloor$ is even, and do that for all eight rows. Then, use your ruler (or another viable straightedge, such as the edge of another sheet of paper), place the straightedge perpendicular to the vertical line on your digram at $0$, and slowly work your way to $1$, marking all disjoint intervals in which your straightedge touches $2, 4, 6,$ or $8$ boxes simultaneously. (If an interval excludes a value in that interval, you still have to count it as two disjoint intervals. Note that this will be important as to not undercounting disjoint intervals. ) If done correctly, you should obtain $\boxed{12\ \textbf{(B)}}$ as your answer.
~FidgetBoss 4000
## Additional Explanation
You should be able to somewhat visualize what the $\sin$ function looks like (if you can't then you should look it up and try to memorize it). To summarize, the graph of $y= \sin{x}$ is positive from the interval $(0, \pi)$ and negative from the interval $(\pi, 2\pi)$ (notice how the intervals use parentheses instead of brackets, as brackets would denote inclusive bounds and that it incorrect).
All of the various $\sin$ functions that we are multiplying are in the form of $k\pi x$, which means the intervals have now changed to integer values.
Also, the period of the function $y =\sin{k \pi x}$ is equivalent to $\frac{2}{k}$; For example the period of $y= \sin{4\pi x}$ is $\frac{1}{2}$.
Let a "wave" be a full period of a $\sin$ curve. Then, a "half-wave" is half of a full period of a $\sin$ curve. Over the interval $[0, 1],$ $y= \sin{\pi x}$ will display $1$ "half-wave", $y= \sin{2\pi x}$ will display $2$ "half-waves", and $y= \sin{k\pi x}$ will display $k$ "half-waves". Also notice that the graph of $y= \sin{k \pi x}$ crosses the $x$-axis exactly $k-1$ times over the interval $(0, 1)$.
The problem asks for the number of disjoint intervals, not the values of the intervals themselves. This means we only want the positive values of
$\prod_{k=1}^8 \sin{k\pi x}$
(the product of all of the various $\sin$ functions)
We don't care about the specific values themselves, we just need to know what the sign (+/-) of the product is. Notice that if an even number of the functions are negative, the product will be positive; conversely, if the number of negative functions is odd, the product will be negative. Therefore we only need to count the number of intervals during which an even number of functions are negative/positive. By definition, the sign of a function changes after it crosses one of its roots (because trig functions don't have multiplicities). Notice that the roots of $y = \sin{k\pi x}$ are in the form
$\frac{1}{k}, \frac{2}{k}, \frac{3}{k},...$
(this is in agreement with the fact that $y= \sin{k \pi x}$ crosses the $x$-axis exactly $k-1$ times over the interval $(0, 1)$)
Therefore, across the interval $[0, 1]$, the sign will change every time the function hits a root of one of the $\sin$ functions. We must be careful, however, as some roots (such as $\frac{1}{2}$ and $\frac{2}{4}$) are duplicates.
Notice that the duplicates always come in pairs. For example, $\frac{1}{2}$ is also $\frac{2}{4}, \frac{3}{6},$ and $\frac{4}{8}$; there are four equivalent roots. Across the interval $[0, 1],$ there are no duplicate roots that come in triplets, or in fives. The roots are always single, doubled up, or quadrupled up. So every time the function reaches a single root, it switches signs, and every time it reaches a duplicate root, the sign remains the same (this still counts as 2 distinct intervals, though, as the zeroes "separate" them). Listing out the roots (not including $x=0$ and $x=1$):
$1$: n/a
$2: \frac{1}{2}$
$3: \frac{1}{3}, \frac{2}{3}$
$4: \frac{1}{4}, \frac{2}{4}, \frac{3}{4}$
$5: \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}$
$6: \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}$
$7: \frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$
$8: \frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{4}{8}, \frac{5}{8}, \frac{6}{8}, \frac{7}{8}$
Then in order (showing duplicates as well):
$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{1}{3}, \frac{3}{8}, \frac{2}{5}, \frac{3}{7}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$
Then creating a sequence of signs based on the roots:
$+,-,+,-,+,+,-,-,+,-,+,+,-,+,-,-,+,+,-,+,-,+$
We see that there are $12$ positive intervals $\Rightarrow \boxed{\text{B}}$.
## See also
2010 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# SSC CPO Quantitative Aptitude Quiz 21
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# SSC CPO Quantitative Aptitude Quiz 21
### Introduction
What is Quantitative Aptitude test?
Quantitative Aptitude is one of the prominent competitive aptitude subjects which evaluates numerical ability and problem solving skills of candidates. This test forms the major part of a number of important entrance and recruitment exams for different fields. The Quantitative Aptitude section primarily has questions related to the Simplification, Numbering Series, and Compound Interest, etc.
A candidate with quantitative aptitude knowledge will be in a better position to analyse and make sense of the given data. Quantitative Aptitude knowledge is an important measure for a prospective business executive’s abilities.
The article SSC CPO Quantitative Aptitude Quiz 21 provides Quantitative Aptitude questions with answers useful to the candidates preparing for Competitive exams, Entrance exams, Interviews etc. The article SSC CPO Quantitative Aptitude Quiz 21 will assist the students to know the expected questions from Quantitative Aptitude.
### Quiz
1. What is John’s present age, if after 10 years his age will be 5 times his age 5 years back.
A. 6.2 years
B. 7.7 years
C. 8.7 years
D. 10 years
Explanation:
1) Let John’s present age be x
2) John’s age before 5 years = (x – 5)
3) John’s age after 10 years = (x + 10)
We are given that, John’s age after 10 years (x + 10) is 5 times his age 5 years back (x – 5)
Therefore,
(x + 10) = 5 (x – 5)
Solving the equation, we get
x + 10 = 5x – 25
4x = 35
x = 8.75 years
2. Rahul is 15 years elder than Rohan. If 5 years ago, Rahul was 3 times as old as Rohan, then find Rahul’s present age.
A. 32.5 years
B. 27.5 years
C. 25 years
D. 24.9 years
Explanation:
1) Let age of Rohan be y
2) Rahul is 15 years elder than Rohan = (y + 15). So Rahul’s age 5 years ago = (y + 15 – 5)
3) Rohan’s age before 5 years = (y – 5)
5 years ago, Rahul is 3 times as old as Rohan
(y + 15 – 5) = 3 (y – 5)
(y + 10) = (3y – 15)
2y = 25
y = 12.5
Rohan’s age = 12.5 years
Rahul’s age = (y + 15) = (12.5 + 15) = 27.5 years
3. One year ago, ratio of Harry and Peter age’s was 5 : 6 respectively. After 4 years, this ratio becomes 6 : 7. How old is Peter?
A. 25 years
B. 26 years
C. 31 years
D. 35 years
Explanation:
Hint: If ages in the numerical are mentioned in ratio A : B, then A : B will be Ax and Bx.
We are given that age ratio of Harry : Pitter = 5 : 6
1) Harry’s age = 5x and Peter’s age = 6x
2) One year ago, their age was 5x and 6x. Hence at present, Harry’s age = 5x +1 and Peter’s age = 6x +1
3) After 4 years,
Harry’s age = (5x +1) + 4 = (5x + 5)
Peter’s age = (6x +1) + 4 = (6x + 5)
4) After 4 years, this ratio becomes 6 : 7. Therefore,
$$\frac{Harry’s Age}{6}$$ = $$\frac{Peter’s Age}{7}$$
$$\frac{(5x + 5) }{(6x + 5)}$$ = $$\frac{6 }{7}$$
7 (5x + 5) = 6 (6x + 5)
X = 5
Peter’s present age = (6x + 1) = (6 x 5 + 1) = 31 years
Harry’s present age = (5x + 1) = (5 x 5 + 1) = 26 years
4. The age of mother 10 years ago was 3 times the age of her son. After 10 years, the mother’s age will be twice that of his son. Find the ratio of their present ages.
A. 11 : 7
B. 9 : 5
C. 7 : 4
D. 7 : 3
Explanation:
We are given that, age of mother 10 years ago was 3 times the age of her son
So, let age of son be x and as mother’s age is 3 times the age of her son, let it be 3x, three years ago.
At present: Mother’s age will be (3x + 10) and son’s age will be (x + 10)
After 10 years: Mother’s age will be (3x + 10) +10 and son’s age will be (x + 10) + 10
Mother’s age is twice that of son
(3x + 10) +10 = 2 [(x + 10) + 10]
(3x + 20) = 2[x + 20]
Solving the equation, we get x = 20
We are asked to find the present ratio.
(3x + 10) : (x + 10) = 70 : 30 = 7 : 3
5. Sharad is 60 years old and Santosh is 80 years old. How many years ago was the ratio of their ages 4 : 6?
A. 10 years
B. 15 years
C. 20 years
D. 25 years
Explanation:
Here, we have to calculate: How many years ago the ratio of their ages was 4 : 6
Let us assume x years ago
At present: Sharad is 60 years and Santosh is 80 years
x years ago: Sharad’s age = (60 – x) and Santosh’s age = (80 – x)
Ratio of their ages x years ago was 4 : 6
$$\frac{(60 – x) }{(80 – x)}$$ = $$\frac{4 }{6}$$
6(60 – x) = 4(80 – x)
360 – 6x = 320 – 4x
x = 20
Therefore, 20 years ago, the ratio of their ages was 4 : 6
1. Which of the following year is not a leap year?
A. 1960
B. 2080
C. 2024
D. 2100
Explanation:
The two conditions that decide that a year is a leap year or not is:
• For a year to be a leap year, it should be divisible by 4.
• No century is a leap year unless it is divisible by 400.
Hence, the year 2100 is not a leap year as it is not divisible by 400.
2. The last day of the century cannot be:
A. Sunday
B. Wednesday
C. Friday
D. Saturday
Explanation:
100 years have 5 odd days. Hence the last day of the 1st century is a Friday.
200 years have 10 odd days or 1 week + 3 odd days. Hence, the last day of the 2nd century is a Wednesday.
300 years have 15 odd days or 2 weeks + 1 odd day. Hence, the last day of the 3rd century is a Monday.
400 years have 0 odd days. Hence, the last day of the 4th century is a Sunday.
3. 5 times a positive number is less than its square by 24. What is the integer?
A. 5
B. 8
C. 7.5
D. 9
Explanation:
Let the unknown number be x.
5 times a positive number = 5x
5 times a positive number is less than its square by 24
$${x }^{2}$$ – 5x = 24
$${x }^{2}$$ + 3x – 8x – 24
x (x + 3) – 8(x + 3)
(x – 8) (x + 3)
x = 8
8 is the required integer.
4. H.C.F. of two numbers is 13. If these two numbers are in the ratio of 15: 11, then find the numbers.
A. 230, 140
B. 215, 130
C. 195, 143
D. 155, 115
Explanation:
Hint: Product of two numbers = Product of their H.C.F. and L.C.M.
Given:
1) H.C.F. of two numbers = 13
2) The numbers are in the ratio of 15: 11
Let the two numbers be 15y and 11y
H.C.F. is the product of common factors
Therefore, H.C.F. is y. So y = 13
The two numbers are:
15y = 15 × 13 = 195
11y = 11 × 13 = 143
We can cross-check the answer using the trick. (Product of two numbers = Product of their H.C.F. and L.C.M.)
Product of H.C.F. and L.C.M. = 13 × 2145 = 27885
Product of two numbers = 195 × 143 = 27885
Hence, the calculated answer is correct.
5. If the product and H.C.F. of two numbers are 4107 and 37 respectively, then find the greater number.
A. 111
B. 222
C. 332
D. 452
Explanation:
4107 is the square of 37.
So let two numbers be 37x and 37y.
37x × 37y = 4107
xy = 3
3 is the product of (1 and 3)
x = 1 and y = 3
37x = 37 × 1 =37
37y = 37 × 3 = 111
Greater number = 111
OR
Hint:
Product of two numbers = Product of their H.C.F. and L.C.M.
Product of two numbers = 4107
4107 = 37 × L.C.M
L.C.M. = $$\frac{4107 }{37}$$ = 111
The greater number = 111
1. In a kilometer race, A beats B by 50 meters or 10 seconds. What time does A take to complete the race?
A. 200 sec
B. 190 sec
C. 210 sec
D. 150 sec
Explanation:
Time taken by B run 1000 meters = $$\frac{(1000 * 10)}{50 }$$= 200 sec.
Time taken by A = 200 – 10 = 190 sec.
2. A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?
A. 111.12 m
B. 888.88 m
C. 777.52 m
D. 756.34 m
Explanation:
A runs 1000 m while B runs 900 m and C runs 800 m.
The number of meters that C runs when B runs 1000 m,
= $$\frac{(1000 * 800)}{9 }$$ = $$\frac{8000}{9 }$$ = 888.88 m.
B can give C = 1000 – 888.88 = 111.12 m.
3. In a race of 1000 m, A can beat by 100 m, in a race of 800m, B can beat C by 100m. By how many meters will A beat C in a race of 600 m?
A. 57.5 m
B. 127.5 m
C. 150.7 m
D. 98.6 m
Explanation:
When A runs 1000 m, B runs 900 m and when B runs 800 m, C runs 700 m.
When B runs 900 m, distance that C runs = $$\frac{(900 * 700) }{800 }$$ = $$\frac{6300 }{8 }$$ = 787.5 m.
In a race of 1000 m, A beats C by (1000 – 787.5) = 212.5 m to C.
In a race of 600 m, the number of meters by which A beats C = $$\frac{(600 * 212.5) }{1000 }$$ = 127.5 m.
4. In a game of billiards, A can give B 20 points in 60 and he can give C 30 points in 60. How many points can B give C in a game of 100?
A. 50
B. 40
C. 25
D. 15
Explanation:
A scores 60 while B score 40 and C scores 30.
The number of points that C scores when B scores 100 = $$\frac{(100 * 30) }{40 }$$ = 25 * 3 = 75.
In a game of 100 points, B gives (100 – 75) = 25 points to C.
5. What sum of money will produce Rs.70 as simple interest in 4 years at 3 $$\frac{1 }{2}$$ percent?
1584
B. 1120
C. 792
D. 1320
70 = $$\frac{P*4*7 }{2 }$$ X $$\frac{1 }{100 }$$ |
# Verification (and help) with the questions related to convergence of recurrences in the form $x_{n+2} = kx_{n+1} + px_n$
Given $$x_1 = a$$ and $$x_2 = b$$ find the values of $$a, b \in \Bbb R, n\in\Bbb N$$ for which the following recurrences converge (diverge): \begin{align*} x_{n+2} &= 2x_{n+1} - x_n \tag1\\ x_{n+2} &= 4x_{n+1} - 3x_n\tag2\\ x_{n+2} &= -2x_{n+1} - x_n\tag3\\ x_{n+2} &= x_{n+1} + 2x_n\tag4\\ \end{align*}
$$(1)$$: \begin{align} x_{n+2} &= 2x_{n+1} - x_n \iff \\ x_{n+2} - x_{n+1} &= x_{n+1} - x_n = \\ &= x_{n+1} - x_{n}\\ &= x_{n} - x_{n-1}\\ &\cdots\\ &= x_{2} - x_{1}\\ &= b - a \end{align} Now taking the limit: $$\lim_{n\to\infty}(x_{n+1} - x_{n}) = \lim_{n\to\infty}(a-b)=a-b$$ For this recurrence to be convergent $$a$$ must be equal to $$b$$, otherwise it doesn't satisfy Cauchy criteria, thus: $$a = b \implies \exists \lim_{n\to\infty}x_n \\ a \ne b \implies \exists! \lim_{n\to\infty}x_n$$
$$(2)$$: \begin{align} x_{n+2} &= 4x_{n+1} - 3x_n \iff\\ x_{n+2} - x_{n+1} &= 3(x_{n+1} - x_n) \\ &= 3^2(x_{n} - x_{n-1})\\ &\cdots \\ &= 3^{n-1}(x_{2} - x_{1})\\ &= 3^{n-1}(b - a) \end{align}
This case is similar to $$(1)$$: $$b = a \implies \exists \lim_{n\to\infty}x_n\\ b \ne a \implies \exists! \lim_{n\to\infty}x_n$$
$$(3)$$: \begin{align} x_{n+2} &= -2x_{n+1} - x_n \iff \\ x_{n+2} + x_{n+1} &= -(x_{n+1} + x_n) \\ &= (x_{n} + x_{n-1}) \\ &= -(x_{n-1} + x_{n-2}) \\ &\cdots\\ &= (-1)^{n-1}(x_2 + x_1) \\ &= (-1)^{n-1}(b + a) \\ \end{align}
In this case convergence is only possible if: $$a + b = 0 \implies \exists \lim_{n\to\infty}x_n\\ a + b \ne 0 \implies \exists! \lim_{n\to\infty}x_n$$
$$(4)$$. I'm stuck with this one, not sure what transformations to apply. I could probably use generating functions or solve through a characteristic polynomial, but that is heavy machinery for such a simple case. I would also like to not apply any solutions involving matrices since i'm not very familiar with linear algebra yet.
Could you please verify the first three cases and either suggest a solution or give a hint for the last one? Thank you!
• how $\lim_{n\to\infty}|x_{n+1} -x_n|=0$ will ensure that limit of sequence will exist ? – neelkanth Feb 6 at 17:02
• @neelkanth well, it might not, i'm not sure. However if we choose some $C\in\Bbb R^+$ and $\lambda \in (0,1)$ such that $|x_{n+1} - x_n| = |a - b| \le C\lambda^n$ then $x_n$ converges. And this seems to only be possible for $a-b = 0$ – roman Feb 6 at 17:25
• You could have tried characteristic polynomial method indeed, e.g. very similar question here – rtybase Feb 25 at 21:02
The first two points are correct (but you need to justify $$a= b \implies (x_n) \text{ is convergent}$$).
For the third point $$x_{n+1}+x_n=0$$ for all $$n$$ does not ensure that $$(x_n)$$ is convergent. In fact it leads to $$x_{n+1}=-x_n$$ so $$x_n=(-1)^{n-1} a = (-1)^{n-2} b$$ so the condition is $$a+b=0 \text{ and } a=b=0$$.
For the last point in you want to applies a similar method the best way is to define $$y_{n}=x_{n+1}+\lambda x_n$$ where $$\lambda$$ is a parameter we will choose later.
In this case we have $$y_{n+1}=x_{n+2}+\lambda x_{n+1}=x_{n+1}+2x_n+\lambda x_{n+1}\\=(1+\lambda)\left(x_{n+1}+\frac{2}{1+\lambda} x_n \right).$$
So by choosing $$\lambda$$ such that $$2/(1+\lambda)=\lambda$$ i.e $$\lambda=1$$ or $$\lambda=-2$$ we obtain
• $$\lambda=1$$
$$(x_{n+2}+x_{n+1})=2(x_{n+1}+x_n)$$
• $$\lambda=-2$$
$$(x_{n+2}-2x_{n+1})=-(x_{n+1}-2x_n)$$
so you obtain a necessary condition for convergence $$b+a=0$$ $$b-2a=0$$ and so the sequence is convergent iff $$a=b=0$$.
Note that by considering a generic recurrence you can show that the good choice of $$\lambda$$ are always the opposite of the roots of the characteristic polynomial.
• Nice approach for the last case. Thank you! – roman Feb 25 at 18:52 |
# CBSE Class 10 Maths Question Paper | 2016
###### Class 10 Previous Year Question Paper Solution | 3 Mark Questions
Section C contains 10 questions of 3 marks each. Scroll down for explanatory answer and video solution to all ten 3-mark questions that appeared in Class 10 Maths CBSE board exam in 2016. Questions appeared from the following chapters : Areas Related to Circles, Surface Areas & Volumes, Coordinate Geometry, Arithmetic Progressions, Quadratic Equations, Applications of Trigonometry, and Probability.
1. O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)
Hint to solve this Areas Related to Circles Question
Step 1: Because AB is a diameter, ▵ ACB is a right triangle.
Step 2: Using Pythagoras Theorem compute length of BC.
Step 3: Area of shaded region = Area of semicircle whose diameter is 13 - area of right triangle ACB.
2. A tent is in the shape of a cylinder surmounted by a conical top of same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of ` 500/sq.metre. (Use π = $$frac{22}{7}$) Hint to solve this Surface Areas & Volumes problem Step 1: Cost of canvas = Area of canvas × Cost per sqm. Step 2: Area of canvas = CSA of cylinder + CSA of cone 3. If the point P$x, y) is equidistant from the points A(a + b, b – a) and P(a – b, a + b). Prove that bx = ay.
Hint to solve this Coordinate Geometry problem
Concept: Distance Formula
Step 1: Compute the length of the line segment PA using the distance formula.
Step 2: Compute the length of the line segment PB using the distance formula.
Step 3: Because P is equidistant from A and B, equate PA and PB. Essentially, equate PA2 and PB2.
4. Find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm where ∠AOC = 40°. (Use π = $$frac{22}{7}$.) Hint to solve this Areas Related to Circles problem Step 1: The shaded region is a circular disc that subtends an angle of 320° at the centre of the circle. Step 2: Area of circular disc = Area of outer circle - Area of inner circle. Step 3: 320/360 × area of circular disc is the area of the shaded region. 5. If the ratio of the sum of first n terms of two A.P's is$7n + 1) : (4n + 27), find the ratio of their mth terms.
Hint to solve this Arithmetic Progressions problem
Step 1: Write the sum of the nth term of the two APs and equate the ratio to the ratio given in the question.
Step 2: Express ratio of the mth term of the two APs.
Step 3: Compare the two ratios and express n in terms of m. Substitute 'n' in terms of m in the first ratio to arrive at the answer.
6. Solve for x : $$frac{1}{$x-1$(x-2)}) + $$frac{1}{$x-2$(x-3)}) = $$frac{2}{3}$. x ≠ 1, 2, 3. Hint to solve this Quadratic Equations problem Step 1: Take$x - 1)(x - 2)(x - 3) as common denominator of the LHS of the expression and simplify the numerator of the resultant expression.
Step 2: Cross multiply and simplify terms to get a quadratic equation.
Step 3: Solve the quadratic equation to compute the value of x.
7. A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel, (Use π = $$frac{22}{7}$) Hint to solve this Surface Areas & Volume problem Concept: Volume of water in the conical vessel = Volume of water emptied into the cylindrical vessel. Step 1: Assign variable 'h' for the increase in height of water in the cylindrical vessel and solve for h. 8. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3$\frac{5}{9}$ cm. Find the diameter of the cylindrical vessel. Hint to solve this Surface Areas & Volumes problem Concept: Volume of sphere dropped = Volume of water displaced in cylindrical vessel. Step 1: Assign variable 'r' for the radius of the cylindrical vessel. Step 2: Equate the two volumes. One equation, one variable. Solve for 'r' to arrive at the answer. 9. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of hill as 30°. Find the distance of the hill from the ship and the height of the hill. Hint to solve this Applications of Trigonometry problem Step 1: Applying trigonometric ratio tan the angle of depression, compute the distance between the ship and the hill Step 2: Applying tan of the angle of elevation, compute the height of the portion of the hill above the deck of the ship. Step 3: Finally compute height of hill. 10. Three different coins are tossed together. Find the probability of getting$i) exactly two heads
(iii) at least two tails.
Hint to solve this Probability problem
Step 1: The denominator for all 3 parts is the same = 8 outcomes.
Step 2: Compute the number of outcomes in which exactly two heads appear to find the answer for part i.
Step 3: Part ii) What does at least 2 heads mean? 2 or 3 heads.
Step 4: Part iii) What does at least 2 tails mean? 2 or 3 tails.
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What Percent of 2 by 7 Is 1 by 35 OR
# What Percent of 2 by 7 Is 1 by 35
via GIPHY
Another question you may see your kids try to solve is "What percent of 2 by 7 is 1 by 35"? This problem is the converse of the question above, so even at the age where they know to add and multiply, they'll need help in finding the missing piece.
### Percent
100 is what percent of 80? These problems tend to kill people because on some level they're kind of simple, they're just 100 and an 80 there, and they're asking what percent. But then people get confused. They say, do I divide the 100 by the 80? The 80 by 100? Or is it something else going on? And you really just have to think through what the language is saying. They're saying that this value right here, this 100, is some percentage of 80, and that some percentage is what we have to figure out. What percent? So if we multiply 80 by this what percent, we will get 100. So let's view it this way. We have 80. If we multiply it by something, let's call this something x. Let me do that in a different color. If we multiply 80 by something, we are going to get 100. And we need to figure out what we need to multiply 80 by to get 100. And if we just solve this equation as it is, we're going to get a value for x. And what we need to do is then convert it to a percent. Another way you could have viewed this is 100 is what you get when you multiply what by 80? And then you would have gotten this number, and then you could convert it to a percent. So this is essentially the equation and now we can solve it. If we divide both sides of this equation by 80, so you divide the left-hand side by 80, the right-hand side by 80, you get x. x is equal to 100/80. They both share a common factor of 20, so 100 divided by 20 is 5, and 80 divided by 20 is 4. So in simplest form, x is equal to 5/4, but I've only expressed it as a fraction. But they want to know what percent of 80. If they just said 100 is what fraction of 80, we would be done. We could say 100 is 5/4 of 80, and we would be absolutely correct. But they want to say what percent? So we have to convert this to a percent, and the easiest thing to do is to first convert it into a decimal, so let's do that. 5/4 is literally the same thing as 5 divided by 4, so let's figure out what that is. Let me do it in magenta. So 5 divided by 4. You want to have all the decimals there, so let's put some zeroes out here. 4 goes into 5 one time. Let me switch up the colors. 1 times 4 is 4. You subtract. You get 5 minus 4 is 1. Bring down the next zero. And of course, the decimal is sitting right here, so we want to put it right over there. So you bring down the next zero. 4 goes into 10 two times. 2 times 4 is 8. You subtract. 10 minus 8 is 2. Bring down the next zero. 4 goes into 20 five times. 5 times 4 is 20. Subtract. No remainder. So this is equal to 1.25. 5/4 is the same thing as 5 divided by 4, which is equal to 1.25. So far, we could say, 100 is 1.25 times 80, or 1.25 of 80, you could even say, But we still haven't expressed it as a percentage. This is really just as a number. I guess you could call it a decimal, but it's a whole number and a decimal. It would be a mixed number if we didn't do it as a decimal. It's 1 and 1/4, or 1 and 25 hundredths, however you want to read it. So to write it as a percent, you literally just have to multiply this times 100, or shift the decimal over twice. So this is going to be equal to, as a percent, if you just shift the decimal over twice, this is equal to 125%. And that makes complete sense. 100 is 125% of 80. 80 is 100% of 80. 100% percent is more than 80. It's actually 1 and 1/4 of 80, and you see that right over there, so it makes sense. It's 125%. It's more than 100%. But we are done. We've solved the problem. It is a 125% of 80
Percent changes applied sequentially do not add up in the usual way. For example, if the 10% increase in price considered earlier (on the \$200 item, raising its price to \$220) is followed by a 10% decrease in the price (a decrease of \$22), then the final price will be \$198—not the original price of \$200. The reason for this apparent discrepancy is that the two percent changes (+10% and −10%) are measured relative to different quantities (\$200 and \$220, respectively), and thus do not "cancel out". (Source: en.wikipedia.org)
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Back to the class
#11, pg.32: Determine if $\vec{b}$ is a linear combination of $\vec{a}_1,\vec{a}_2,$ and $\vec{a}_3$ where $$\vec{a}_1 = \left[ \begin{array}{ll} 1 \\ -2 \\ 0 \end{array} \right], \vec{a}_2=\left[ \begin{array}{ll} 0 \\ 1 \\ 2 \end{array} \right], \vec{a}_3 = \left[ \begin{array}{ll} 5 \\ -6 \\ 8 \end{array} \right], \vec{b}=\left[ \begin{array}{ll} 2 \\ -1 \\ 6 \end{array} \right].$$ Solution: To write $\vec{b}$ as a linear combination of $\vec{a}_1, \vec{a}_2,$ and $\vec{a}_3$ means we must find the weights $x_1,x_2,x_3$ (if they exist) that satisfy $$x_1 \vec{a}_1 + x_2 \vec{a}_2 + x_3\vec{a}_3 = \vec{b}.$$ When the vector algebra is simplified (carry out the vector sums) we get the equation $$\left[ \begin{array}{ll} x_1 + 5x_3 \\ -2x_1 +x_2 - 6x_3 \\ 2x_2 + 8x_3 \end{array} \right] = \left[ \begin{array}{ll} 2 \\ -1 \\ 6 \end{array} \right],$$ which is a system of linear equations. We will solve this system using the augmented matrix $$\begin{array}{ll} \left[ \begin{array}{llll} 1 & 0 & 5 & 2\\ -2 & 1 & -6 & -1 \\ 0 & 2 & 8 & 6 \end{array} \right] &\stackrel{r_2^* = r_2+2r_1}{\sim} \left[ \begin{array}{llll} 1 & 0 & 5 & 2 \\ 0 & 1 & 4 & 3 \\ 0 & 2 & 8 & 6 \end{array} \right] \\ &\stackrel{r_3^* = r_3-2r_2}{\sim} \left[ \begin{array}{ll} 1 & 0 & 5 & 2 \\ 0 & 1 & 4 & 3 \\ 0 & 0 & 0 & 0 \end{array} \right] \\ \end{array}$$ If we interpret this augmented marix as a system, we get $$\left\{ \begin{array}{ll} x_1 + 5x_3 = 2 \\ x_2 + 4x_3 = 3 \\ 0=0 \end{array} \right.$$ or equivalently $$\left\{ \begin{array}{ll} x_1 = 2-5x_3 \\ x_2 = 3-4x_3 \\ 0 = 0 \end{array} \right.$$ We are free to choose any value of $x_3$ so we will choose $x_3=0$ and doing so yields the solution $$\left\{ \begin{array}{ll} x_1 = 2 \\ x_2 = 3 \\ x_3 = 0. \end{array} \right.$$ This implies that we should be able to write the vector $\vec{b}$ as $$2 \vec{a}_1 + 3\vec{a}_2 + 0\vec{a}_3 = \left[ \begin{array}{ll} 2 \\ -4 \\ 0 \end{array} \right] + \left[ \begin{array}{ll} 0 \\ 3 \\ 6 \end{array} \right] = \left[ \begin{array}{ll} 2 \\ -1 \\ 6 \end{array} \right]= \vec{b},$$ as was to be shown.
NOTE: Every different choice of $x_3$ yields a different linear combination that yields $\vec{b}$. Here we call $x_3$ a free variable.
Additional Problem (A): Solve the matrix equation $A \vec{x}=\vec{b}$ where $A = \left[ \begin{array}{ll} 1 & 0 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right]$ and $\vec{b}=\left[ \begin{array}{ll} b_1 \\ b_2 \\ b_3 \end{array} \right]$.
NOTE: Many people wrote the matrix in this problem as $\left[ \begin{array}{ll} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right]$. This may because I made a typo in class...I will not penalize correct answers that start with this matrix instead of the actual problem.
Solution: We know the solution of this equation to be equivalent to the solution of the system whose augmented matrix is $$\begin{array}{ll} \left[ \begin{array}{ll} 1 & 0 & 1 & b_1 \\ 0 & -1 & 1 & b_2 \\ 0 & 0 & -1 & b_3 \end{array} \right] &\stackrel{r_1^* = r_1+r_3}{\stackrel{r_2^*=r_2+r_3}{\sim}} \left[ \begin{array}{ll} 1 & 0 & 0 & b_1+b_3 \\ 0 & -1 & 0 & b_2+b_3 \\ 0 & 0 & -1 & b_3 \end{array} \right] \\ & \stackrel{r_2^*=-r_2}{\stackrel{r_3^*=-r_3}{\sim}} \left[ \begin{array}{llll} 1 & 0 & 0 & b_1+b_3 \\ 0 & 1 & 0 & -b_2-b_3 \\ 0 & 0 & 1 & -b_3 \end{array} \right] \end{array}$$ hence we see that the solution is given by $$\vec{x} = \left[ \begin{array}{ll} b_1+b_3 \\ -b_2-b_3 \\ -b_3. \end{array} \right]$$ |
# How do you solve by substitution x-y+-15 and x+y=-5?
Jun 5, 2015
Assuming the first component should have ($= -$) instead of ($\pm$)
[1]$\textcolor{w h i t e}{\text{XXXX}}$$x - y = - 15$
[2]$\textcolor{w h i t e}{\text{XXXX}}$$x + y = - 5$
Rearranging the terms of [1]
[3]$\textcolor{w h i t e}{\text{XXXX}}$$x = y - 15$
Substituting $\left(y - 15\right)$ for $x$ into [2]
[4]$\textcolor{w h i t e}{\text{XXXX}}$$\left(y - 15\right) + y = - 5$
[5]$\textcolor{w h i t e}{\text{XXXX}}$$y = 5$
Substituting $5$ for $y$ back in [1]
[6]$\textcolor{w h i t e}{\text{XXXX}}$$x - 5 = - 15$
[7]$\textcolor{w h i t e}{\text{XXXX}}$$x = - 10$
Solution: $\left(x , y\right) = \left(- 10 , 5\right)$ |
# Question: What Is The Binary Number Of 2?
## How do you subtract 1 in binary?
To subtract a larger number from a smaller one, switch the order of the numbers, do the subtraction, then add a negative sign to the answer.
For example, to solve the binary problem 11 – 100, solve for 100 – 11 instead, then add a negative sign to the answer..
## Why is binary base 2?
The reason computers use the base-2 system is because it makes it a lot easier to implement them with current electronic technology. … When you look at this sequence, 0 and 1 are the same for decimal and binary number systems. At the number 2, you see carrying first take place in the binary system.
## How do you count to 10 in binary?
To count in binary, you start with 0, then you go to 1. Then you add another digit, like you do in decimal counting when you go from 9 to 10. You add another digit, so you have two digits now. So, in binary, you go from 1 to 10 since 1 is your last counting number.
## How do you calculate binary numbers?
To calculate the number value of a binary number, add up the value for each position of all the 1s in the eight character number. The number 01000001, for example, is converted to 64 + 1 or 65.
## How do you read binary codes?
The key to reading binary is separating the code into groups of usually 8 digits and knowing that each 1 or 0 represents a 1,2,4,8,16,32,64,128, ect. from the right to the left. the numbers are easy to remember because they start at 1 and then are multiplied by 2 every time.
## How do you add more than 2 binary numbers?
The addition of a string of numbers (more than two) in a digital system is done by adding the first two numbers, then adding the third number to the sum of the first two, then adding the fourth to the final sum, and so on. Add the signed numbers +3, -2, +5, and –15.
## How do you write 7 in binary?
Instead, you put a “1” in the twos column and a “0” in the units column, indicating “1 two and 0 ones”. The base-ten “two” (210) is written in binary as 102….Binary.decimal (base 10)binary (base 2)expansion71111 four, 1 two, and 1 one810001 eight, 0 fours, 0 twos, and 0 ones910011 eight, 0 fours, 0 twos, and 1 ones14 more rows
## What does 101 mean in binary?
Words. The word binary comes from “Bi-” meaning two. We see “bi-” in words such as “bicycle” (two wheels) or “binocular” (two eyes). When you say a binary number, pronounce each digit (example, the binary number “101” is spoken as “one zero one”, or sometimes “one-oh-one”).
## How do you write 8 in binary?
Binary to Decimal and Decimal to Binary Conversion 8 Bit Numbers. An 8 bit binary number can represent a maximum of decimal 255= binary 11111111. Here is another 8 bit binary number –01101011. if we convert our columns to decimal equivalents using the following chart.
## What numbers are used in binary code?
In mathematics and digital electronics, a binary number is a number expressed in the base-2 numeral system or binary numeral system, which uses only two symbols: typically “0” (zero) and “1” (one). The base-2 numeral system is a positional notation with a radix of 2.
## What is the letter A in binary?
ASCII – Binary Character TableLetterASCII CodeBinaryA06501000001B06601000010C06701000011D0680100010022 more rows
## What does 11 mean in binary?
1011DECIMAL NUMBERS IN BINARY00810009100110101011101196 more rows
## How do you know if a binary number is negative?
The representation of a signed binary number is commonly referred to as the sign-magnitude notation and if the sign bit is “0”, the number is positive. If the sign bit is “1”, then the number is negative.
## How high can you count in binary?
Finger binary is a system for counting and displaying binary numbers on the fingers of one or more hands. It is possible to count from 0 to 31 (25 − 1) using the fingers of a single hand, from 0 through 1023 (210 − 1) if both hands are used, or from 0 to 1,048,575 (220 − 1) if the toes on both feet are used as well.
## What is the binary alphabet?
Binary alphabet may refer to: The members of a binary set in mathematical set theory. A 2-element alphabet, in formal language theory. ASCII. |
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