text stringlengths 22 1.01M |
|---|
# Conic Sections
A series of free, online video lessons with examples and solutions to help Algebra students learn about parabola conic sections.
In these lessons, we will learn
• how to graph the equation of a parabola given in standard form and general form
• how to graph the equation of a circle given in standard form and general form
• how to graph the equation of an ellipse given in standard form and general form
The following diagrams show the conic sections for circle, ellipse, parabola, and hyperbola. Scroll down the page for more examples and solutions on conic sections.
### Introduction to Conic Sections
By definition, a conic section is a curve obtained by intersecting a cone with a plane. In Algebra II, we work with four main types of conic sections: circles, parabolas, ellipses and hyperbolas. Each of these conic sections has different characteristics and formulas that help us solve various types of problems.
How to generate a circle, ellipse, parabola, and hyperbola by intersecting a cone with a plane?
Name each of the 4 conics.
Demonstrate how the conics are formed by a plane and a cone.
The fixed line is called the axis of the cone.
The vertex is the point shared by both cones.
The lines that pass through the vertex and form the cones are the generators. The generators lie in the cone.
The cone consists of two parts called the nappes.
A conic section section is a curve generated by intersecting a right circular cone with a plane.
A circle is generated when the plane is perpendicular to the axis of the cone.
An ellipse is generated when the plane is tilted so it intersects each generator, but only intersects one nappe.
A parabola is generated when the plane is tilted so it is parallel to one generator and only intersects one nappe.
A hyperbola is generated when the plane intersect both nappes.
Demonstrate what is a conic section
### Conic Section: Parabola
Conic Sections: The Parabola part 1 of 2
Defines a parabola and explains how to graph a parabola in standard form.
Identify the key components to a parabola.
Graph a parabola in standard form.
Conic Sections: The Parabola part 2 of 2
How to graph a parabola given in general form by rewriting it in standard form?
Write the general form of a parabola in standard form.
Graph a parabola.
A parabola is set of all points (x,y) that are equidistant from a fixed line called the directrix and a fixed point called the focus.
Example:
Write the parabola in standard form and then graph.
a) x2 + 8x = 4y - 8
b) y2 + 2y = 8x - 1
### Conic Section: Circle
When working with circle conic sections, we can derive the equation of a circle by using coordinates and the distance formula. The equation of a circle is (x - h)2 + (y - k)2 = r2 where r is equal to the radius, and the coordinates (x,y) are equal to the circle center. The variables h and k represent horizontal or vertical shifts in the circle graph.
How to graph a circle in standard form and general form?
Examples:
1. Graph the circle.
(x - 3)2 + (y + 2)2 = 16
2. Write in standard form and then graph.
2x2 + 2y2 - 12x + 8y - 24 = 0
Conic Sections: Introduction to Circles
### Conic Section: Ellipse
An ellipse is an important conic section and is formed by intersecting a cone with a plane that does not go through the vertex of a cone. The ellipse is defined by two points, each called a focus. From any point on the ellipse, the sum of the distances to the focus points is constant. The position of the foci determine the shape of the ellipse. The ellipse is related to the other conic sections and a circle is actually a special case of an ellipse.
How to talk about an ellipse?
Conic Sections: The Ellipse part 1 of 2
This video defines an ellipse and explains how to graph an ellipse in standard form.
Conic Sections: The Ellipse part 2 of 2
This video explains how to graph an ellipse in general form.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# How do you graph r=2sin2x?
Jul 19, 2018
See graph and explanation.
#### Explanation:
The graph of $r = a \sin \left(n \theta - \alpha\right)$, n = 1, 2, 3, 4, ...# shows
n equal and symmetrical loops, around the pole r = 0.
The graph of $r = a \sin \left(n \theta - \alpha\right)$ is obtained by rotating
anticlockwise graph of $r = a \sin \theta$, about $\theta = 0$,
through $\alpha$.
Use $\left(x , y\right) = r \left(\cos \theta , \sin \theta\right) , r = \sqrt{{x}^{2} + {y}^{2}}$ and
$\sin 2 \theta = 2 \sin \theta \cos \theta$ and get the Cartesian form of
$r = 2 \sin 2 \theta$ as
${\left({x}^{2} + {y}^{2}\right)}^{1.5} - 4 x y = 0$.
Now, the Socratic graph is immediate.
graph{ (x^2 + y^2 )^1.5 - 4 xy = 0[-4 4 -2 2 ]}
For anticlockwise rotation, through $\alpha = \frac{p}{2}$, use
$r = 2 \sin \left(2 \left(\theta - \frac{\pi}{2}\right)\right) = - 2 \sin 2 \theta$
The graph is immediate.
graph{ (x^2 + y^2 )^1.5 + 4 xy = 0[-4 4 -2 2 ]}
For clockwise rotation, through $\alpha = \frac{\pi}{4}$, use
$r = 2 \sin \left(2 \left(\theta + \frac{\pi}{4}\right)\right) = 2 \cos 2 \theta$.
See the graph, using
$\cos 2 \theta = \left({\cos}^{2} \theta - {\sin}^{2} \theta\right) = \frac{{x}^{2} - {y}^{2}}{r} ^ 2$
graph{(x^2+y^2)^1.5-2(x^2-y^2)=0[-4 4 -2 2]} |
# 6.2 Rational Equations
#### Solving Rational Equations that Reduce to Linear Equations
In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation.
Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).
Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.
Rational Equations and how to solve them
A rational equation contains at least one rational expression where the variable appears in at least one of the denominators. They can be solved as follows:
1. If there is only one term on each side, cross multiply OR
2. If not, factor all denominators in the equation.
3. Find the LCD.
4. Multiply both sides of the equation by the LCD. If the LCD was correct, there should be no denominators left.
5. Solve the remaining equation.
6. Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator.
Note that any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities. This is something we always need to check for.
Example Solving a Rational Equation
Solve the rational equation: .
We have three denominators and 3. The LCD must contain and 3. An LCD of contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD.
Distribute:
A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as . Always consider a binomial as an individual factor; the terms cannot be separated. For example, suppose a problem has three terms and the denominators are and . First, factor all denominators. We then have and as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of . The in the first denominator is separate from the in the denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or separate factor. The LCD in this instance is found my multiplying together the , one factor of , and the . Thus the LCD is the following:
So, both sides of the equation would be multiplied by . Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.
Another example is a problem with two denominators, such as and . Once the second denominator is factored as , there is a common factor of in both denominators and the LCD is .
Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.
We can then use another method of solving the equation, without find the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.
So this results is .
Example Solving a Rational Equation Using Cross Multiplication
Solve the following rational equation:
Cross Multiplying gives us:
Since the only number that would make a denominator 0 would be 0 itself, we have our solution.
Try it Now 1
Solve
Example Solving a Rational Function with no factoring necessary
Solve the following rational equations:
a.
Our denominators are and , so our LCD is .
Multiply everything by :
Distribute:
Cancel and multiply leftovers:
This does not make any denominator 0, so our solutions is .
Try it Now 2
Solve the rational equation: .
Example Solving a Rational Equation by Factoring the Denominator
Solve the rational equation: .
Let’s look at all the denominators in their factored form: List all the factors that appear: and So the LCD must be . Multiply both sides by :
Since this does not make the denominator 0, our solution is .
Try it Now 3
Solve the rational equation: .
Examples Solving Rational Equations with a Binomial in the Denominator
a. Solve and state the excluded values:
This one can be cross-multiplied
The solution is 15, and the excluded values are 6, 0.
b. Solve and state the excluded values:
The LCD is . Multiply both sides of the equation by
The solution is . The excluded value is 3.
c. Solve and state the excluded values:
The LCD is . Multiply both sides of the equation by
The solution is 4. The excluded value is 3.
Try it Now 4
Solve and state the excluded values:
Example Solving a Rational Equation with Factored Denominators and Stating Excluded Values
Solve the rational equation after factoring the denominators and state the excluded values:
The first two denominators are already factored as far as possible, and , but the third one can be factored as difference of two squares: . Thus, the LCD that contains each denominator is . Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.
The solution is -3. The excluded values are 1 and -1.
Try it Now 5
Solve the rational equation and state the excluded values:
.
#### Solving Rational Equations Resulting in a Quadratic
Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out there is no solution.
Solve the rational equation and state the excluded values:
.
Similar to the previous example, the LCD is so we will multiply both sides by :
Now we have a problem, because the solution -1 and 1 is the same as our excluded values since they make the original denominators equal to 0 and therefore there is no solution.
Note with rational equations that lead to quadratics, you can have 0, 1 or 2 solutions. You can solve and get 1 solution that is excluded and 1 that is not or you can get both solutions that work in the equation as well.
Try it Now 6
Solve the rational equation and state the excluded values:
.
1. Cross Multiply. Solution: Exclusions: -6 and 2/3
2. LCD: ; Solution: ; Exclusion: 0
3. LCD: ; Solution: ; Exclusion: 0
4. Cross Multiply. Solution: ; Exclusions -1/2 and -1/3
5. LCD: ; Solution: ; Exclusions 2 and -1
6. LCD: ; Solution: ; Exclusions 0, 2 (Note that the quadratic produced solutions of -1 and 0, but since 0 was on the exclusion list it cannot be a solution.) |
What Is Not A Parallelogram?
Are you curious to know what is not a parallelogram? You have come to the right place as I am going to tell you everything about not a parallelogram in a very simple explanation. Without further discussion let’s begin to know what is not a parallelogram?
Contents
What Is Not A Parallelogram?
A parallelogram is a quadrilateral with two pairs of parallel sides. However, not all quadrilaterals are parallelograms. In this blog, we’ll explore what makes a quadrilateral not a parallelogram, and discuss some common examples.
A quadrilateral is a polygon with four sides. There are many different types of quadrilaterals, but a parallelogram is a specific type that has certain properties. Specifically, a parallelogram has two pairs of parallel sides and opposite angles that are congruent (i.e., equal in measure). These properties make a parallelogram unique and distinguish it from other quadrilaterals.
So, what makes a quadrilateral, not a parallelogram? There are several possibilities:
1. Unequal opposite sides: One of the defining characteristics of a parallelogram is that it has two pairs of parallel sides. If the opposite sides of a quadrilateral are not parallel, then it cannot be a parallelogram. For example, a trapezoid is a quadrilateral with only one pair of parallel sides, so it is not a parallelogram.
2. Non-congruent opposite angles: In addition to having two pairs of parallel sides, a parallelogram also has opposite angles that are congruent. If the opposite angles of a quadrilateral are not congruent, then it cannot be a parallelogram. For example, a kite is a quadrilateral with two pairs of adjacent sides that are congruent, but its opposite angles are not congruent, so it is not a parallelogram.
3. Diagonal lengths are not equal: Another property of parallelograms is that their diagonals bisect each other. This means that the point where the diagonals intersect (called the “centroid”) is the midpoint of each diagonal. If a quadrilateral has diagonals that do not bisect each other or are not equal in length, then it cannot be a parallelogram. For example, a rectangle has diagonals that are equal in length and bisect each other, so it is a parallelogram.
4. Sides are not parallel or congruent: Finally, a quadrilateral cannot be a parallelogram if it does not have any of the defining characteristics of a parallelogram. For example, a random quadrilateral with no parallel sides or congruent angles is not a parallelogram.
In conclusion, a parallelogram is a special type of quadrilateral with two pairs of parallel sides and opposite angles that are congruent. Not all quadrilaterals are parallelograms, and there are several ways in which a quadrilateral can fail to be a parallelogram. Understanding the properties of different types of quadrilaterals is important in geometry and other mathematical fields.
You can gather more stuff on Caresclub
FAQ
Which Shape Is Not A Parallelogram?
trapezium
A trapezium is a quadrilateral with only one pair of opposite sides equal. Hence, the trapezium is not a parallelogram.
Trapezium
What Are The 4 Types Of Parallelograms?
There are 4 types of parallelograms, including 3 special types. The four types are parallelograms, squares, rectangles, and rhombuses.
Is A Rhombus A Parallelogram?
Also, every rhombus is considered a parallelogram but the converse is always not true. Parallelogram: A parallelogram is a flat-shaped figure. It has four sides. The pair of opposite faces/sides of a parallelogram are parallel and congruent to each other.
Is A Trapezoid Is A Parallelogram?
A trapezoid has one pair of parallel sides and a parallelogram has two pairs of parallel sides. So a parallelogram is also a trapezoid.
I Have Covered All The Following Queries And Topics In The Above Article
What Is Not A Parallelogram
What Is Not A Property Of A Parallelogram
What Shape Is Not A Parallelogram
What Quadrilateral Is Not A Parallelogram
What Is A Quadrilateral That Is Not A Parallelogram
What Shape Is A Parallelogram But Not A Rhombus
What Is A Parallelogram But Not A Rhombus
What Is A Parallelogram That Is Not A Rhombus
Is A Trapezoid A Parallelogram
Is A Rhombus A Parallelogram |
# Patterns and Mental Mathematics – Definition, Examples | Mental Math Tricks for Arithmetic Operations
The pattern is a scheme that helps to solve math questions easily and fastly. Mental mathematics is a skill that allows students to do maths in their heads without using pen and paper or a calculator. Get the simple patterns that make addition, subtraction, multiplication, and division problems easier. Also, find the solved examples on patterns and mental mathematics in the below-mentioned sections.
## What is meant by Pattern and Mental Maths?
A pattern is a series or sequence that repeats based on the rule and the rule helps to solve a problem. Mental math is a group of skills that allow people to do maths in their head without using a calculator or pen and paper. It is useful in everyday life and school. It helps the students to understand the concepts easily and quickly.
### Patterns in Mathematics
Below provided is the pattern for counting numbers from 1 to 100.
The following are the observations that can be read from a pattern.
• The numbers in every column increase by 1.
• The numbers in every row increase by 10.
• The diagonally located numbers from the left to right from least to highest have a difference of 9. Those are 10, 19, 28, 37, 46, 55, 64, 73, 82, 91.
• And the diagonally located numbers from left to right from top to bottom have a difference of 11. Those numbers are 1, 12, 23, 34, 56, 67, 78, 89, 100.
• All the numbers in the first row have 1 in their unit’s place, the second row have 2, the third row has 3, and so on.
### Mental Math Tricks – Addition, Subtraction, Multiplication, Division
The most used and best trick to solve the addiction problem is to expand the given numbers. Add the numbers in hundred’s places, tens place, units place together. Again add the obtained numbers to get the sum of the numbers.
Example:
Solve 1256 + 563 + 78
1256 + 563 + 78 = 1000 + 200 + 50 + 6 + 500 + 60 + 3 + 70 +Â 8
= 1000 + 700 + 180 + 17
= 1700 + 197
= 1897
Subtraction:
For solving subtraction problems, we need to expand the given numbers and provide the sign. Subtract the numbers in the hundreds place, ten place, unit’s place.
Example:
Solve 875 – 232
875 – 232 = 800 + 70 + 5 – 200 – 30 – 2
= 600 + 40 + 3
= 643
Multiplication:
For solving the multiplication of smaller numbers, we can use the repeated addition method. For getting the product of larger numbers, expand the numbers and perform the multiplication. At the end add or subtract those numbers.
Examples:
8 x 6 = 8 + 8 + 8 + 8 + 8 + 8 = 48
82 x 5 = (90 – 8) x 5 = 90 x 5 – 8 x 5
= 450 – 40 = 410
105 x 8 = (100 + 5) x 8
= 100 x 8 + 5 x 8
= 800 + 40 = 840
Division:
For dividing the numbers, divide every digit of the dividend by the divisor and write the quotient.
Example:
156/4 = 39
### Examples on Patterns and Mental Math Strategies
Question 1:
Solve the following
(i) 27 x 6
(ii) 49 x 50
(iii) 53 x 47
Solution:
(i) Given that,
27 x 6
Expand 27 = 20 + 7
27 x 6 = (20 + 7) x 6
= 120 + 42
= 162
Therefore, 27 x 6 = 162
(ii) Given that,
49 x 50
Represent 49 as 50 – 1
49 x 50 = (50 – 1) x 50
= (50 x 50) – 50
= 2500 – 50
= 2450
Therefore, 49 x 50 = 2450
(iii) Given that,
53 x 47
Represent 53 as 50 + 3 and 47 as 50 – 3
53 x 62 = (50 + 3) x (50 – 3)
= 2500 – 9 = 2491
Therefore, 53 x 47 = 2491.
Question 2:
Find the sum of the numbers.
(i) 85 + 162 + 569
(ii) 145 + 75
(iii) 1453 + 2563 + 4326
Solution:
(i) Given that,
85 + 162 + 569
Expand all the numbers
85 + 162 + 569 = 80 + 5 + 100 + 60 + 2 + 500 + 60 + 9
= 600 + 200 + 16
= 816
So, 85 + 162 + 569 = 816
(ii) Given that,
145 + 75
Expand all the numbers
145 + 75 = 100 + 40 + 5 + 70 + 5
= 100 + 110 + 10
= 220
So, 145 + 75 = 220
(iii) Given that,
1453 + 2563 + 4326
Expand all the numbers
1453 + 2563 + 4326 = 1000 + 400 + 50 + 3 + 2000 + 500 + 60 + 3 + 4000 + 300 + 20 + 6
= 7000 + 1200 + 130 + 12
= 8200 + 142
= 8342
So, 1453 + 2563 + 4326 = 8342
Question 3:
Subtract the numbers.
(i) 356 – 182
(ii) 1563 – 823
(iii) 526 – 230
Solution:
(i) Given that,
356 – 182
Expand all the numbers
356 – 182 = 300 + 50 + 6 – 100 – 80 – 2
= 200 – 30 + 4
= 170 + 4 = 174
So, 356 – 182 = 174
(ii) Given that,
1563 – 823
Expand all the numbers
1563 – 823 = 1000 + 500 + 60 + 3 – 800 – 20 – 3
= 700 + 40 = 740
So, 1563 – 823 = 740
(iii) Given that,
526 – 230
Expand all the numbers
526 – 230 = 500 + 20 + 6 – 200 – 30
= 300 – 10 + 6
= 296
So, 526 – 230 = 296
Question 4:
Divide the following numbers.
(i) $$\frac { 500 }{ 5 }$$
(ii) $$\frac { 1492 }{ 2 }$$
(iii) $$\frac { 768 }{ 8 }$$
Solution:
(i) Given that,
$$\frac { 500 }{ 5 }$$ = 100
(ii) Given that,
$$\frac { 1492 }{ 2 }$$ = $$\frac { 1400 }{ 2 }$$ + $$\frac { 90 }{ 2 }$$ + $$\frac { 2 }{ 2 }$$
= 700 + 45 + 1
= 746
(iii) Given that,
$$\frac { 768 }{ 8 }$$ = 96
Scroll to Top |
# Negative exponents
How to calculate negative exponents.
### Negative exponents rule
The base b raised to the power of minus n is equal to 1 divided by the base b raised to the power of n:
b-n = 1 / bn
## Negative exponent example
The base 2 raised to the power of minus 3 is equal to 1 divided by the base 2 raised to the power of 3:
2-3 = 1/23 = 1/(2⋅2⋅2) = 1/8 = 0.125
## Negative fractional exponents
The base b raised to the power of minus n/m is equal to 1 divided by the base b raised to the power of n/m:
b-n/m = 1 / bn/m = 1 / (mb)n
The base 2 raised to the power of minus 1/2 is equal to 1 divided by the base 2 raised to the power of 1/2:
2-1/2 = 1/21/2 = 1/2 = 0.7071
## Fractions with negative exponents
The base a/b raised to the power of minus n is equal to 1 divided by the base a/b raised to the power of n:
(a/b)-n = 1 / (a/b)n = 1 / (an/bn) = bn/an
The base 2 raised to the power of minus 3 is equal to 1 divided by the base 2 raised to the power of 3:
(2/3)-2 = 1 / (2/3)2 = 1 / (22/32) = 32/22 = 9/4 = 2.25
## Multiplying negative exponents
For exponents with the same base, we can add the exponents:
a -na -m = a -(n+m) = 1 / a n+m
Example:
2-3 ⋅ 2-4 = 2-(3+4) = 2-7 = 1 / 27 = 1 / (2⋅2⋅2⋅2⋅2⋅2⋅2) = 1 / 128 = 0.0078125
When the bases are diffenrent and the exponents of a and b are the same, we can multiply a and b first:
a -nb -n = (a b) -n
Example:
3-2 ⋅ 4-2 = (3⋅4)-2 = 12-2 = 1 / 122 = 1 / (12⋅12) = 1 / 144 = 0.0069444
When the bases and the exponents are different we have to calculate each exponent and then multiply:
a -nb -m
Example:
3-2 ⋅ 4-3 = (1/9) ⋅ (1/64) = 1 / 576 = 0.0017361
## Dividing negative exponents
For exponents with the same base, we should subtract the exponents:
a n / a m = a n-m
Example:
26 / 23 = 26-3 = 23 = 2⋅2⋅2 = 8
When the bases are diffenrent and the exponents of a and b are the same, we can divide a and b first:
a n / b n = (a / b) n
Example:
63 / 23 = (6/2)3 = 33 = 3⋅3⋅3 = 27
When the bases and the exponents are different we have to calculate each exponent and then divide:
a n / b m
Example:
62 / 33 = 36 / 27 = 1.333 |
# How do you simplify 4div 0.5?
Jul 23, 2016
Take a look at the explanation.
#### Explanation:
$4 \div 0.5$ or $\frac{4}{0.5}$ can easily be simplified. Think of it this way: if you were to divide 4 slices of pizza in half, how many slices would you then have? Remember, $0.5$ equals $\frac{1}{2}$.
$1$ slice cut in half would equal $2$ slices. So you can think of this as a multiplication question as well. If $1$ slice cut in half equals $2$ slices, then $4$ slices cut in half would then equal how many slices?
• $4 \cdot 2 = 8$
Or, you can think of it this way. How many "half " slices would it take to make $4$ whole slices?
$0.5 + 0.5 = 1$ slice
$0.5 + 0.5 = 1$ (now $2$ slices)
$0.5 + 0.5 = 1$ (now $3$ slices)
$0.5 + 0.5 = 1$ slices (now $4$ slices)
It would take a total of $8$ half slices to make 4 whole slices. So $4$ divided by $0.5$ equals $8$.
$4 \div 0.5 = 8$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Law of Sines
Proportion based on ratios of sides and sines of the opposite angles for non-right triangles.
0%
Progress
Practice Law of Sines
Progress
0%
Law of Sines with AAS and ASA
A triangle has two angles that measure 60\begin{align*}60^\circ\end{align*} and 45\begin{align*}45^\circ\end{align*}. The length of the side between these two angles is 10. What are the lengths of the other two sides?
Guidance
Consider the non right triangle below. We can construct an altitude from any one of the vertices to divide the triangle into two right triangles as show below.
from the diagram we can write two trigonometric functions involving h\begin{align*}h\end{align*}:
sinCbsinC=hbandsinB=hc=h csinB=h
Since both are equal to h\begin{align*}h\end{align*}, we can set them equal to each other to get:
bsinC=csinB\begin{align*}b \sin C=c \sin B\end{align*} and finally divide both sides by bc\begin{align*}bc\end{align*} to create the proportion:
sinCc=sinBb
If we construct the altitude from a different vertex, say B\begin{align*}B\end{align*}, we would get the proportion: sinAa=sinCc\begin{align*}\frac{\sin A}{a}=\frac{\sin C}{c}\end{align*}. Now, the transitive property allows us to conclude that sinAa=sinBb\begin{align*}\frac{\sin A}{a}=\frac{\sin B}{b}\end{align*}. We can put them all together as the Law of Sines: sinAa=sinBb=sinCc\begin{align*}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\end{align*}. In the examples that follow we will use the Law of Sines to solve triangles.
Example A
Solve the triangle.
Solution: Since we are given two of the three angles in the triangle, we can find the third angle using the fact that the three angles must add up to 180\begin{align*}180^\circ\end{align*}. So, mA=1804570=650\begin{align*}m \angle A=180^\circ - 45^\circ - 70^\circ = 650^\circ\end{align*}. Now we can substitute the known values into the Law of Sines proportion as shown:
sin65a=sin7015=sin45c
Taking two ratios at a time we can solve the proportions to find a\begin{align*}a\end{align*} and c\begin{align*}c\end{align*} using cross multiplication.
To find a\begin{align*}a\end{align*}:
sin65aa=sin7015=15sin65sin7014.5
To find c\begin{align*}c\end{align*}:
sin7015c=sin45c=15sin45sin7011.3
This particular triangle is an example in which we are given two angles and the non-included side or AAS (also SAA).
Example B
Solve the triangle.
Solution: In this example we are given two angles and a side as well but the side is between the angles. We refer to this arrangement as ASA. In practice, in doesn’t really matter whether we are given AAS or ASA. We will follow the same procedure as Example A. First, find the third angle: mA=1805080=50\begin{align*}m \angle A=180^\circ - 50^\circ - 80^\circ = 50^\circ\end{align*}.
Second, write out the appropriate proportions to solve for the unknown sides, a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}.
To find a\begin{align*}a\end{align*}:
sin80aa=sin5020=20sin80sin5025.7
To find b\begin{align*}b\end{align*}:
sin50bb=sin5020=20sin50sin50=20
Notice that c=b\begin{align*}c=b\end{align*} and mC=mB\begin{align*}m \angle C=m \angle B\end{align*}. This illustrates a property of isosceles triangles that states that the base angles (the angles opposite the congruent sides) are also congruent.
Example C
Three fishing ships in a fleet are out on the ocean. The Chester is 32 km from the Angela. An officer on the Chester measures the angle between the Angela and the Beverly to be 25\begin{align*}25^\circ\end{align*}. An officer on the Beverly measures the angle between the Angela and the Chester to be 100\begin{align*}100^\circ\end{align*}. How far apart, to the nearest kilometer are the Chester and the Beverly?
Solution: First, draw a picture. Keep in mind that when we say that an officer on one of the ships is measuring an angle, the angle she is measuring is at the vertex where her ship is located.
Now that we have a picture, we can determine the angle at the Angela and then use the Law of Sines to find the distance between the Beverly and the Chester.
The angle at the Angela is 18010025=55\begin{align*}180^\circ - 100^\circ - 25^\circ = 55^\circ\end{align*}.
Now find x\begin{align*}x\end{align*},
sin55xx=sin10032=32sin55sin10027
The Beverly and the Chester are about 27 km apart.
Concept Problem Revisit The measure of the triangle's third angle is 1806045=75\begin{align*}180^\circ - 60^\circ -45^\circ =75^\circ\end{align*}
sin45xsin60y=sin7510, so x=10sin45sin757.29=sin7510, so y=10sin60sin758.93
Guided Practice
Solve the triangles.
1.
2.
3. A surveying team is measuring the distance between point A\begin{align*}A\end{align*} on one side of a river and point B\begin{align*}B\end{align*} on the far side of the river. One surveyor is positioned at point A\begin{align*}A\end{align*} and the second surveyor is positioned at point C\begin{align*}C\end{align*}, 65 m up the riverbank from point A\begin{align*}A\end{align*}. The surveyor at point A\begin{align*}A\end{align*} measures the angle between points B\begin{align*}B\end{align*} and C\begin{align*}C\end{align*} to be 103\begin{align*}103^\circ\end{align*}. The surveyor at point C\begin{align*}C\end{align*} measures the angle between points A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} to be 42\begin{align*}42^\circ\end{align*}. Find the distance between points \begin{align*}A\end{align*} and \begin{align*}B\end{align*}.
1. \begin{align*}m \angle A=180^\circ - 82^\circ -24^\circ =74^\circ\end{align*}
2. \begin{align*}m \angle C=180^\circ - 110^\circ -38^\circ =32^\circ\end{align*}
3.
Explore More
Using the given information, solve \begin{align*}\Delta ABC\end{align*}.
1. .
1. .
1. .
1. .
Use the Law of Sines to solve the following world problems.
1. A surveyor is trying to find the distance across a ravine. He measures the angle between a spot on the far side of the ravine, \begin{align*}X\end{align*}, and a spot 200 ft away on his side of the ravine, \begin{align*}Y\end{align*}, to be \begin{align*}100^\circ\end{align*}. He then walks to \begin{align*}Y\end{align*} the angle between \begin{align*}X\end{align*} and his previous location to be \begin{align*}20^\circ\end{align*}. How wide is the ravine?
2. A triangular plot of land has angles \begin{align*}46^\circ\end{align*} and \begin{align*}58^\circ\end{align*}. The side opposite the \begin{align*}46^\circ\end{align*} angle is 35 m long. How much fencing, to the nearest half meter, is required to enclose the entire plot of land?
Vocabulary Language: English
ambiguous
ambiguous
Ambiguous means that the given information is not specific. In the context of Geometry or Trigonometry, it means that the given data may not uniquely identify one shape.
SSA
SSA
SSA means side, side, angle and refers to the fact that two sides and the non-included angle of a triangle are known in a problem. |
Problems on Ages
1.Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?
Explanation:
Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{Ronit's}\phantom{\rule{thinmathspace}{0ex}}\text{present}\phantom{\rule{thinmathspace}{0ex}}\text{age}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{years}.\\ & \text{Then,}\phantom{\rule{thinmathspace}{0ex}}\text{father's}\phantom{\rule{thinmathspace}{0ex}}\text{present}\phantom{\rule{thinmathspace}{0ex}}\text{age}\phantom{\rule{thinmathspace}{0ex}}\\ & =\left(x+3x\right)\phantom{\rule{thinmathspace}{0ex}}\text{years}\\ & =4x\phantom{\rule{thinmathspace}{0ex}}\text{years}\\ & \therefore \left(4x+8\right)=\frac{5}{2}\left(x+8\right)\\ & ⇒8x+16=5x+40\\ & ⇒3x=24\\ & ⇒x=8\\ & \text{Hence,}\phantom{\rule{thinmathspace}{0ex}}\text{required}\phantom{\rule{thinmathspace}{0ex}}\text{ratio}\\ & =\frac{\left(4x+16\right)}{\left(x+16\right)}=\frac{48}{24}=2\end{array}$
2.The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
Explanation:
Solution:
Let the ages of children be x, (x + 3), (<x + 6), (<x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
⇒ 5x = 20
x = 4.
∴ Age of the youngest child = x = 4 years.
3.A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, the son's age five years back was:
Explanation:
Solution:
Let the son's present age be x years. Then, (38 - x) = x
⇒ 2x = 38.
x = 19.
∴ Son's age 5 years back (19 - 5) = 14 years.
4.Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand's present age in years?
Explanation:
Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{present}\phantom{\rule{thinmathspace}{0ex}}\text{ages}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{Sameer}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\text{Anand}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}\\ & 5x\phantom{\rule{thinmathspace}{0ex}}\text{years}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}4x\phantom{\rule{thinmathspace}{0ex}}\text{years}\phantom{\rule{thinmathspace}{0ex}}\text{respectively}\\ & \text{Then,}\phantom{\rule{thinmathspace}{0ex}}\frac{5x+3}{4x+3}=\frac{11}{9}\\ & ⇒9\left(5x+3\right)=11\left(4x+3\right)\\ & ⇒45x+27=44x+33\\ & ⇒45x-44x=33-27\\ & ⇒x=6\\ & \therefore \text{Anand's}\phantom{\rule{thinmathspace}{0ex}}\text{present}\phantom{\rule{thinmathspace}{0ex}}\text{age}\\ & =4x\\ & =24\phantom{\rule{thinmathspace}{0ex}}\text{years}\phantom{\rule{thinmathspace}{0ex}}\end{array}$
5.A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:
Explanation:
Solution:
Let the son's present age be x years. Then, man's present age = (x + 24) years.
∴ (x + 24) + 2 = 2(x + 2)
x + 26 = 2x + 4
x = 22.
6.Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar's age at present?
Explanation:
Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{ages}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{Kunal}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\text{Sagar}\phantom{\rule{thinmathspace}{0ex}}\text{6}\phantom{\rule{thinmathspace}{0ex}}\text{years}\phantom{\rule{thinmathspace}{0ex}}\text{ago}\phantom{\rule{thinmathspace}{0ex}}\\ & \text{be}\phantom{\rule{thinmathspace}{0ex}}6x\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}5x\phantom{\rule{thinmathspace}{0ex}}\text{years}\phantom{\rule{thinmathspace}{0ex}}\text{respectively}\\ & \text{Then,}\phantom{\rule{thinmathspace}{0ex}}\frac{\left(6x+6\right)+4}{\left(5x+6\right)+4}=\frac{11}{10}\\ & ⇒10\left(6x+10\right)=11\left(5x+10\right)\\ & ⇒5x=10\\ & ⇒x=2\\ & \therefore \text{Sagar's}\phantom{\rule{thinmathspace}{0ex}}\text{present}\phantom{\rule{thinmathspace}{0ex}}\text{age}\\ & =\left(5x+6\right)\phantom{\rule{thinmathspace}{0ex}}\text{years}\\ & =16\phantom{\rule{thinmathspace}{0ex}}\text{years}\phantom{\rule{thinmathspace}{0ex}}\end{array}$
7.The sum of the present ages of a father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, son's age will be:
Explanation:
Solution:
Let the present ages of son and father be x and (60 -x) years respectively.
Then, (60 - x) - 6 = 5(x - 6)
⇒ 54 - x = 5x - 30
⇒ 6x = 84
x = 14.
∴ Son's age after 6 years = (x+ 6) = 20 years.
8.Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?
Explanation:
Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{Rahul's}\phantom{\rule{thinmathspace}{0ex}}\text{age}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{years}.\\ & \text{Then,}\phantom{\rule{thinmathspace}{0ex}}\text{Sachin's}\phantom{\rule{thinmathspace}{0ex}}\text{age}=\left(x-7\right)\phantom{\rule{thinmathspace}{0ex}}\text{years}.\\ & \therefore \frac{x-7}{x}=\frac{7}{9}\\ & ⇒9x-63=7x\\ & ⇒2x=63\\ & ⇒x=31.5\\ & \text{Hence,}\phantom{\rule{thinmathspace}{0ex}}\text{Sachin's}\phantom{\rule{thinmathspace}{0ex}}\text{age}\\ & =\left(x-7\right)\phantom{\rule{thinmathspace}{0ex}}\text{years}\\ & =24.5\phantom{\rule{thinmathspace}{0ex}}\text{years}\end{array}$
9.The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
Explanation:
Solution:
Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x - 8) + (7x - 8) + (9x - 8) = 56
⇒ 20x = 80
x = 4.
∴ Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.
10.Ayesha's father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents? |
Rules for Exponent Arithmetic
There are rules for operating on numbers with exponents that make it easy to simplify and solve problems.
Learning Objectives
Explain and implement the rules for operating on numbers with exponents
Key Takeaways
Key Points
• The rule $a^m \cdot a^n = a^{m+n}$ applies when multiplying two exponential expressions with the same base, provided the base is a non-zero integer.
• The rule $\displaystyle \frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ applies when dividing two exponential expressions with the same base, provided the base is a non-zero integer.
• The rule ${({a}^{n})}^{m}={a}^{n \cdot m}$ applies when raising an exponential expression to another exponent for any non-zero integer $a$.
• The rule ${(ab)}^{n}={a}^{n}{b}^{n}$applies when raising a product to an exponent for any non-zero integers $a$ and $b$.
Key Terms
• base: In an exponential expression, the value that is multiplied by itself.
• exponent: In an exponential expression, the value raised above the base; represents the number of times the base must be multiplied by itself.
There are several useful rules for operating on numbers with exponents. The following four rules, also known as “identities,” hold for all integer exponents, provided that the base is non-zero.
Multiplying Exponential Expressions with the Same Base
$a^m \cdot a^n = a^{m+n}$
$a^m$ means that you have $m$ factors of $a$. If you multiply this quantity by $a^n$, i.e. by $n$ additional factors of $a$, then you have $a^{m+n}$ factors in total. For example:
$5^3 \cdot 5^4=5^{3+4}=5^7$
Note that you can only add exponents in this way if the corresponding terms have the same base.
Dividing Exponential Expressions with the Same Base
$\displaystyle \frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$
In the same way that ${ a }^{ m }\cdot { a }^{ n }={ a }^{ m+n }$ because you are adding on factors of $a$, dividing removes factors of $a$. If you have $n$ factors of $a$ in the denominator, then you can cross out $n$ factors from the numerator. If there were $m$ factors in the numerator, now you have $(m-n)$ factors in the numerator.
In order to visualize this process, consider the fraction:
$\dfrac{3^5}{3^2}$
This fraction can be rewritten as:
$\dfrac{3 \cdot 3 \cdot 3 \cdot 3 \cdot 3}{3 \cdot 3}$
Here you can see that two 3s will cancel out from the numerator and denominator. We are left with:
$\dfrac{3 \cdot 3 \cdot 3}{1} = 3^3$
$\displaystyle \frac{7^4}{7^2}=7^{4-2}=7^2$
Raising an Exponential Expression to an Exponent
${({a}^{n})}^{m}={a}^{n \cdot m}$
If you think about an exponent as telling you that you have a certain number of factors of the base, then ${({a}^{n})}^{m}$ means that you have factors $m$ of $a^n$. Therefore, you have $m$ groups of $a^n$, and each one of those has $n$ groups of $a$. Therefore, you have $m$ groups of $n$ groups of $a$; therefore, you have $n \cdot m$ groups of $a$, or ${a}^{n \cdot m}$. For example:
$(3^3)^3=3^{3 \cdot 3}=3^9$
Raising a Product to an Exponent
${(ab)}^{n}={a}^{n}{b}^{n}$
You can multiply numbers in any order you please. Instead of multiplying together $n$ factors equal to $ab$, you could multiply all of the $a$s together and all of the $b$s together and then finish by multiplying $a^n$ by $b^n$. For example:
$(4\cdot5)^3=4^3\cdot5^3$
Example
Simplify the following expression: $(3\cdot2)^3\cdot (2^5)^4$
For the first part of the expression, apply the rule for a product raised to an exponent:
$(3\cdot2)^3\cdot (2^5)^4 = 3^3 \cdot 2^3 \cdot (2^5)^4$
For the last part of the expression, apply the rule for raising an exponential expression to an exponent:
$3^3 \cdot 2^3 \cdot (2^5)^4 = 3^3 \cdot 2^3 \cdot 2^{5\cdot 4} = 3^3 \cdot 2^3 \cdot 2^{20}$
Notice that two of the terms in this expression have the same base: 2. These two terms can be combined by applying the rule for multiplying exponential expressions with the same base:
$3^3 \cdot 2^3 \cdot 2^{20} = 3^3 \cdot 2^{3+20} = 3^3 \cdot 2^{23}$
Therefore, $3^3 \cdot 2^{23}$ is the simplified form of this expression.
Negative Exponents
Numbers with negative exponents are treated normally in arithmetic operations and can be rewritten as fractions.
Learning Objectives
Relate negative exponents to fractions and work with them accordingly
Key Takeaways
Key Points
• An exponential expression with a negative integer in the exponent can be rewritten as a fraction by applying the rule $b^{-n} = \frac{1}{b^n}$.
• The rules for operating on numbers with exponents still apply when the exponent is a negative integer.
Solving mathematical problems involving negative exponents may seem daunting. However, negative exponents are treated much like positive exponents when applying the rules for operations. There is an additional rule that allows us to change the negative exponent to a positive one in the denominator of a fraction, and it holds true for any real numbers $n$ and $b$, where $b \neq 0$:
$b^{-n} = \dfrac{1}{b^n}$
For example:
$6^{-2} = \dfrac{1}{6^2} = \dfrac{1}{36}$
To understand how this rule is derived, consider the following fraction:
$\dfrac{7^3}{7^5}$
We can rewrite this as:
$\dfrac{7 \cdot 7 \cdot 7}{7 \cdot 7 \cdot 7 \cdot 7 \cdot 7}$
We then notice that three 7s cancel from both the numerator and denominator, and we are left with:
$\dfrac{1}{7 \cdot 7} = \dfrac{1}{7^2}$
Note that if we apply the rule for division of numbers with exponents, we have:
$\dfrac{7^3}{7^5} = 7^{(3-5)}= 7^{-2}$
Thus, we can identify that:
$\dfrac{1}{7^2} = 7^{-2}$
This rule makes it possible to simplify expressions with negative exponents.
Note that each of the rules for operations on numbers with exponents still apply when the exponent is a negative number. For example, consider the rule for multiplying two exponential expressions with the same base. The following is true:
$3^{-4} \cdot 3^2 = 3^{-4+2} = 3^{-2} = \dfrac{1}{3^2}$
Example
Simplify the following expression: $(2^{-4})^2$.
Note that the rule for raising an exponential expression to another exponent can be applied:
$(2^{-4})^2 = 2^{(-4)(2)} = 2^{-8}$
This can be simplified using the rule for negative exponents:
$2^{-8}=\dfrac{1}{2^8}$
Example
Simplify the following expression: $(3^{-2} \cdot 3^4)^{-3}$.
Recall that the rule for multiplying two exponential expressions with the same base can be applied. Therefore, we can simplify the expression inside the parentheses:
$3^{-2} \cdot 3^4 = 3^{-2+4} = 3^2$
Now place this value back into the parentheses, and apply the rule for raising an exponential expression to an exponent:
$(3^2)^{-3} = 3^{(2)(-3)} = 3^{-6} = \dfrac{1}{3^6}$
Rational Exponents
Rational exponents are another method for writing radicals and can be used to simplify expressions involving both exponents and roots.
Learning Objectives
Relate rational exponents to radicals and the rules for manipulating them
Key Takeaways
Key Points
• If $b$ is a positive real number and $n$ is a positive integer, then there is exactly one positive real solution to $x^n = b$. This solution is called the principal $n$th root of $b$, denoted $\sqrt[n]{b}$ or $\displaystyle b^{\frac{1}{n}}$.
• A power of a positive real number $b$ with a rational exponent $\frac{m}{n}$ in lowest terms satisfies ${b}^{\frac {m}{n}}= {({b}^{m})}^{\frac{1}{n}}=\sqrt[n]{{b}^{m}}$.
• The rule for multiplying numbers with rational exponents is $\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}$.
• The rule for dividing numbers with rational exponents is $\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}$.
• Writing an expression in the form ${b}^{\frac {m}{n}}$can allow you to simplify by cancelling powers and roots.
Key Terms
• root: A number that when raised to a specified power yields the specified number or expression.
• rational number: A real number that can be expressed as the ratio of two integers.
• exponent: The power raised above the base, representing the number of times the base must be multiplied by itself.
A rational exponent is a rational number that provides another method for writing roots. For example, an $n$th root of a number $b$ is a number $x$ such that $x^n = b$. If $b$ is a positive real number and $n$ is a positive integer, then there is exactly one positive real solution to $x^n = b$. This solution is called the $n$th root of $b$ and is denoted $\sqrt[n]{b}$ or $b^\frac{1}{n}$. For example: $\sqrt{4}=4^\frac{1}{2}=2$.
There are also cases where the exponent is a fraction $\frac{m}{n}$, where $m$ is an integer and $n$ is a positive integer. In such cases, the exponent acts as both a whole number exponent and a root, or fraction exponent. In other words, the following holds true:
$\displaystyle{{b}^{\frac {m}{n}}= {({b}^{m})}^{\frac{1}{n}}=\sqrt[n]{{b}^{m}}}$
where $b$ is a real number and the rational exponent $\frac{m}{n}$ is a fraction in lowest terms.
The following rules hold true about the signs of roots and rational exponents. For a rational exponent $\frac{m}{n}$, where $\frac{m}{n}$ is in lowest terms:
• The root is positive if $m$ is even; for example, $(-27)^\frac{2}{3}=9$.
• The root is negative for negative $b$ if $m$ and $n$ are odd; for example, $\displaystyle (-27)^\frac{1}{3}=-3$.
• The root can be either sign if $b$ is positive and $n$ is even; for example, $64^\frac{1}{2}$has two roots: $8$ and $-8$.
Note that since there is no real number $x$ such that $x^2 = -1$, the definition of $b^{\frac{m}{n}}$ when $b$ is negative and $n$ is even must involve the imaginary number $i$.
The following are rules for operations on numbers with rational exponents.
Multiplying Numbers with Rational Exponents
The following holds true for any rational exponent:
$\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}$
For example, we can rewrite $\sqrt[3]{16}$ as a product:
$\sqrt[3]{16}= \sqrt[3]{8} \cdot \sqrt[3]{2}$
Notice that $\sqrt[3]{8} = 2$, and therefore we have:
$\sqrt[3]{16} = 2\sqrt[3]{2}$
Dividing Numbers with Rational Exponents
The following holds true for any rational exponent:
$\sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$
For example, we can rewrite $\sqrt{\frac{13}{9}}$ as a fraction with two radicals:
$\sqrt{\dfrac{13}{9}} = \dfrac{\sqrt{13}}{\sqrt{9}}$
Notice that the denominator can be simplified further:
$\sqrt{9} = 3$
Therefore, the simplified form is:
$\sqrt{\dfrac{13}{9}} = \dfrac{\sqrt{13}}{3}$
Canceling Powers and Roots
In some cases, writing an exponent in its fraction form makes it easier to cancel powers and roots. Recall that $\sqrt[n]{{b}^{m}}= {({b}^{m})}^{\frac{1}{n}}= {b}^{\frac {m}{n}}$. We can use this rule to easily simplify a number that has both an exponent and a root.
For example, consider $\sqrt[4]{5^8}$. This would take a long time to work out by hand, but consider how it can be rewritten using a rational exponent:
$\sqrt[4]{5^8} = 5^{\frac{8}{4}}$
We can simplify the fraction in the exponent to 2, giving us $5^2=25$.
Example 1
Simplify the following expression:
$\sqrt{ \dfrac{3^8}{25}}$
This expression can be rewritten using the rule for dividing numbers with rational exponents:
$\sqrt{ \dfrac{3^8}{25}} = \dfrac{\sqrt{3^8}}{\sqrt{25}}$
Notice that the radical in the denominator is a perfect square and can therefore be rewritten as follows:
$\sqrt{25} = 5$.
Now, notice that the numerator can be rewritten:
$\sqrt{3^8} = (3^8)^{\frac{1}{2}} = 3^{\frac{8}{2}}= 3^4$.
Therefore, the simplified form is:
$\sqrt{ \dfrac{3^8}{5}} = \dfrac{3^4}{5}$
Example 2
Simplify the following expression:
$\dfrac{\sqrt{7^5}}{\sqrt[4]{7^2}}$
First, rewrite the numerator and denominator in rational exponent form:
$\dfrac{7^{\frac{5}{2}}}{7^{\frac{2}{4}}}$
Notice that the exponent in the denominator can be simplified:
$\dfrac{7^{\frac{5}{2}}}{7^{\frac{1}{2}}}$
Recall the rule for dividing numbers with exponents, in which the exponents are subtracted. Applying the division rule, we have:
$\dfrac{7^{\frac{5}{2}}}{7^{\frac{1}{2}}} = 7^{\left(\frac{5}{2}-\frac{1}{2}\right)} = 7^{\frac{4}{2}} = 7^2$
Thus, the simplified form is simply $7^2 = 49$.
Scientific Notation
Scientific notation is used to express a very large or small number in the form $m \cdot10^n$, where $m$ has only one digit left of the decimal point.
Learning Objectives
Explain why scientific notation is useful in performing calculations with large or small numbers
Key Takeaways
Key Points
• Scientific notation is a way of writing numbers that are too big or too small to be conveniently written in decimal form.
• In normalized scientific notation, the exponent $n$ is chosen so that the absolute value of $m$ remains at least 1 but less than 10 $\left ( 1 \leq \left | m \right | < 10 \right )$—i. e. so that $m$ has exactly one digit left of its decimal point.
• When numbers written in scientific notation are involved in multiplication or division, the standard rules for operations with exponentiation apply. When addition or subtraction is involved, the numbers must first be rewritten so the exponents are the same.
• Most calculators present very large and very small results in scientific notation. Because superscripted exponents like $10^7$ cannot always be conveniently displayed, the letter E or e is often used to represent “times ten raised to the power of” (which would be written as “$\cdot 10^b$“).
Key Terms
• scientific notation: A method of writing or of displaying real numbers as a decimal number between 1 and 10 multiplied by an integer power of 10.
• normalized scientific notation: A number written in scientific notation $m \cdot 10^n$ such that the absolute value of $m$ remains at least 1 but less than 10.
Scientific notation, also known as “standard form,” is a way to more conveniently write numbers that are very large or very small. This method is commonly used by mathematicians, scientists, and engineers.
For example, the numbers $43,000,000,000,000,000,000$ (the number of different possible configurations of a Rubik’s cube) and $0.000000000000000000000340$ (the mass of the amino acid tryptophan) are extremely inconvenient to write and read. Therefore, they can be rewritten as a power of 10 using scientific notation.
Scientific notation is written as follows:
$m \cdot 10^n$
This is read “$m$ times 10 raised to the power of $n$.”
How to Use Scientific Notation
To write a number in scientific notation:
• Move the decimal point so that there is one nonzero digit to its left.
• Multiply the result by a power of 10 using an exponent whose absolute value is the number of places the decimal point was moved. Make the exponent positive if the decimal point was moved to the left and negative if the decimal point was moved to the right.
For example, let’s write the number 43,500 in scientific notation. There are four digits in this number, so the decimal should be moved 4 places to the left to leave one nonzero digit left of the decimal point:
$43,500 = 4.35 \cdot 10^{4}$
The exponent is -4 because the decimal point was moved to the left (the exponent would be positive had the decimal been moved to the right) by exactly 4 places.A number written in scientific notation can also be converted to standard form by reversing the process described above. For example, let’s write the number $2.15 \cdot 10^{-3}$ in standard form:
$2.15 \cdot 10^{-3}= 0.00215$
To reverse the process, we move the decimal point three places to the left, adding leading zeroes where necessary.
Normalized Scientific Notation
Any given number can be written in the form of $m \cdot 10^{n}$ in many ways; for example, 350 can be written as $3.5 \cdot 10^{2}$ or $35 \cdot 10^{1}$or $350 \cdot 10^{0}$, etc.
In normalized scientific notation, also called exponential notation, the exponent $n$ is chosen so that the absolute value of $m$ remains at least 1 but less than 10. In other words, $1 \leq | m | < 10$. This form allows easy comparison of two numbers of the same sign with $m$ as a base, as the exponent $n$ gives the number’s order of magnitude.
Following these rules, 350 would always be written as $3.5 \cdot 10^{2}$ and $0.015$ would always be written as $1.5 \cdot 10^{-2}$. Note that $0$ cannot be written in normalized scientific notation since it cannot be expressed as $m \cdot 10^n$ or any non-zero $m$.
Normalized scientific form is the typical form of expression for large numbers in many fields, except during intermediate calculations or when an unnormalized form, such as engineering notation, is desired.
Calculations involving Scientific Notation
When numbers written in scientific notation are multiplied or divided, the standard rules for operations with exponentiation apply. For example:
\begin{align} \displaystyle (3.12 \cdot 10^2) \cdot (4.06 \cdot 10^5) &= 3.12 \cdot 4.06 \cdot 10^{\left(2+5\right)} \\&=12.67 \cdot 10^7 \\&=1.267 \cdot 10^8 \end{align}
\begin{align} \displaystyle \frac{1.85 \cdot 10^3}{4.25 \cdot 10^{-2}} &= \frac{1.85}{4.25} \cdot 10^{3-(-2)} \\&=.435 \cdot 10^5 \\&= 4.35 \cdot 10^4 \end{align}
When numbers written in scientific notation are added to or subtracted from each other, the terms first must be rewritten so the exponents are the same. Then, the constant value, or $m$, can simply be added or subtracted. For example:
\begin{align} \displaystyle (3.12 \cdot 10^6)+(1.24 \cdot 10^7)&= (3.12 \cdot 10^6)+(12.4 \cdot 10^6) \\&= 15.52 \cdot 10^6 \end{align}
E Notation
Most calculators and many computer programs present very large and very small results in scientific notation. Because superscripted exponents like $10^7$ cannot always be conveniently displayed, the letter E or e is often used to represent the phrase “times ten raised to the power of” (which would be written as “$\cdot 10^n$“) and is followed by the value of the exponent. Note that in this usage, the character e is not related to the mathematical constant $\mathbf{e}$ or the exponential function $e^x$ (a confusion that is less likely if a capital E is used), and though it stands for exponent, the notation is usually referred to as (scientific) E notation or (scientific) e notation, rather than (scientific) exponential notation. The use of this notation is not encouraged by publications, however. |
NCERT Grade 7-Congruence of Triangles-Answers – MySchoolPage
# NCERT Solutions for Class 7 Maths
Find 100% accurate solutions for NCERT Class VII Math. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!
1. (a) They have the same length
(b) 70°
(c) m ∠ A = m ∠ B
2. (i) Two football (ii) Two teacher’s tables 3. Given: ΔABC ≅ ΔFED.
The corresponding congruent parts of the triangles are:
4. ΔDEF ≅ ΔBCA.
5. (a) By SSS congruence criterion, since it is given that AC = DF, AB = DE, BC = EF
The three sides of one triangle are equal to the three corresponding sides of another triangle.
Therefore, ΔABC ≅ ΔDEF
(b) By SAS congruence criterion, since it is given that RP = ZX, RQ = ZY and ∠ PRQ = ∠ XZY The two sides and one angle in one of the triangle are equal to the corresponding sides and the angle of other triangle.
Therefore, ΔPQR ≅ ΔXYZ
(c) By ASA congruence criterion, since it is given that ∠ MLN = ∠ FGH, ∠ NML = ∠ HFG, ML = FG.
The two angles and one side in one of the triangle are equal to the corresponding angles and side of other triangle.
Therefore, ΔLMN ≅ ΔGFH
(d) By RHS congruence criterion, since it is given that EB = BD, AE = CB, ∠ A = ∠ C = 90°
Hypotenuse and one side of a right angled triangle are respectively equal to the hypotenuse and one side of another right angled triangle.
Therefore, ΔABE ≅ ΔCDB
6. (a) Using SSS criterion, ΔART ≅ ΔPEN
(i) AR = PE (ii) RT = EN (iii) AT = PN
(b) Given: ∠ T = ∠ N
Using SAS criterion, ΔART ≅ ΔPEN
(i) RT = EN (ii) PN = AT
(c) Given: AT = PN
Using ASA criterion, ΔART ≅ ΔPEN
(i) ∠ RAT = ∠ EPN (ii) ∠ RTA = ∠ ENP
7. Sol.
8. No, because the two triangles with equal corresponding angles need not be congruent. In such a correspondence, one of them can be an enlarged copy of the other. 9. In the figure, given two triangles are congruent. So, the corresponding parts are:
A ↔ O, R ↔ W, T ↔ N.
We can write, ΔRAT ≅ ΔWON [By SAS congruence rule]
10. In ΔBAT and ΔBAC, given triangles are congruent so the corresponding parts are:
B ↔ B, A ↔ A, T ↔ C.
Thus, ΔBAC ≅ ΔBAT [By SSS congruence rule]
In ΔQRS and ΔTPQ, given triangles are congruent so the corresponding parts are:
Thus, ΔQRS ≅ ΔTPQ [By SSS congruence rule]
11. In a squared sheet, draw ΔABC and ΔPQR.
When two triangles have equal areas and
(i) These triangles are congruent, i.e., Δ ABC ≅ Δ PQR [By SSS congruence rule]
Then, their perimeters are same because length of sides of first triangle are equal to the length of sides of another triangle by SSS congruence rule.
(ii) But, if the triangles are not congruent, then their perimeters are not same because lengths of sides of first triangle are not equal to the length of corresponding sides of another triangle.
12. Let us draw two triangles PQR and ABC.
All angles are equal, two sides are equal except one side. Hence, Δ PQR are not congruent to Δ ABC.
13. Δ ABC and Δ PQR are congruent. Then one additional pair is BC = QR.
Given: ∠ B = ∠ Q = 90° ∠ C = ∠ R
(iii) Therefore, Δ ABC ≅ Δ PQR [By ASA congruence rule]
14. Given: ∠ A = ∠ F, BC = ED, ∠ B = ∠ E In Δ ABC and Δ FED,
∠ B = ∠ E = 90°
∠ A = ∠ F
BC = ED
Therefore, Δ ABC ≅ Δ FED [By RHS congruence rule]
MySchoolPage connects you with exceptional, certified math tutors who help you stay focused, understand concepts better and score well in exams! |
Factors of 30 are numbers that, when multiplied in pairs give the product as 30. There are overall 8 factors of 30 among which 30 is the biggest factor and 1, 2, 3, 5, 6, 10, 15, and 30 are positive factors. The sum of all factors of 30 is 72. Its Prime Factors are 1, 2, 3, 5, 6, 10, 15, 30 and (1, 30), (2, 15), (3, 10) and (5, 6) are Pair Factors.
You are watching: What is the prime factorization of 30
Factors of 30: 1, 2, 3, 5, 6, 10, 15 and 30Negative Factors of 30: -1, -2, -3, -5, -6, -10, -15 and -30Prime Factors of 30: 2, 3, 5Prime Factorization of 30: 2 × 3 × 5 = 2 × 3 × 5Sum of Factors of 30: 72
1 What are Factors of 30? 2 How to Calculate Factors of 30? 3 Factors of 30 by Prime Factorization 4 Factors of 30 in Pairs 5 Important Notes 6 FAQs on Factors of 30
## What are Factors of 30?
The factors of 30 are all the integers that 30 can be divided into. The number 30 is an even composite number. A number is said to be composite if it has more than two factors. Since it is even, it will have 2 as a factor. Thus, by understanding the properties of the number 30, we can find the factors of 30 which are 1, 2, 3, 5, 6, 10, 15, and 30.
Explore factors using illustrations and interactive examples:
## How To Calculate Factors of 30?
Step 1: Let"s begin calculating the factors of 30 with the idea that any number that completely divides 30 without any remainder is its factor.Step 2: Let"s start with the whole number 1. We know that 30 ÷ 1 = 30Step 3: The next whole number is 2. Now, divide 30 by 2. Thus, 30 ÷ 2 = 15.Step 4: Proceeding in this manner we get, 30 ÷ 3 = 10 and 30 ÷ 5 = 6. In this way, we can obtain all the factors of 30.Step 5: All these numbers when multiplied make up the factors of 30. That is 30 = 1 × 30, 2 × 15, 3 × 10 and 5 × 6.
## Factors of 30 by Prime Factorization
Prime factorization is to express a composite number as the product of its prime factors.
Step 1: To get the prime factors of 30, we divide it by its smallest prime factor, which is 2. Thus, 30 ÷ 2 = 15.Step 2: Now, 15 is divided by its smallest prime factor and the quotient is obtained. We get 15 ÷ 3 = 5Step 3: This process goes on until we get the quotient as 1.
The prime factorization of 30 is shown below:
## Factors of 30 in Pairs
The pair of numbers that give 30 when multiplied is known as the factor pairs of 30. Following are the factors of 30 in pairs. Since 30 is positive, we have -ve × -ve = +ve.
Product form of 30 Pair factor 1 × 30 = 30 (1,30) 2 × 15= 30 (2,15) 3 × 10 = 30 (3,10) 5 × 6 = 30 (5,6) -1 × -30 = 30 (-1,-30) -2 × -15 = 30 (-2,-15) -3 × -10 = 30 (-3,-10) -5 × -6 = 30 (-5,-6)
Important Notes
As 30 ends with the digit 0, it will have 5 and 10 as its factors. This holds for all numbers that end with the digit 0.30 is a non-perfect square number. Thus, it will have an even number of factors. This property holds for every non-perfect square number.
## Factors of 30 Solved Examples
Example 1: Peter and Andrew both have rectangular papers with dimensions as shown below.
They place the two rectangles one over the other. Since the two shapes do not overlap, Peter informs Andrew that they don"t have the same area. However, Andrew does not agree with him. Can you find out who is correct?
Solution:
Area of a rectangle = length × breadth. For the first rectangle, Area = 6 × 5 = 30 . For the second rectangle, Area = 10 × 3 = 30.
They have equal areas. Therefore, Andrew is correct as the two rectangles have equal areas.
Example 2: Jill has (-3) as one of the factors of 30. How will she get the other factor?
Solution:
30 = Factor 1 × Factor 2. We have 30 = (-3) × Factor 2, which means that Factor 2 = 30 ÷ (-3) =(-10).
Therefore, the other factor is -10.
Show Solution >
go to slidego to slidego to slide
Have questions on basic mathematical concepts?
Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts
Book a Free Trial Class
## FAQs on Factors of 30
### What are the Factors of 30?
The factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30 and its negative factors are -1, -2, -3, -5, -6, -10, -15, -30.
### What is the Greatest Common Factor of 30 and 6?
The factors of 30 and 6 are 1, 2, 3, 5, 6, 10, 15, 30 and 1, 2, 3, 6 respectively.Common factors of 30 and 6 are <1, 2, 3, 6>.Hence, the Greatest Common Factor (GCF) of 30 and 6 is 6.
### What are the Prime Factors of 30?
The prime factors of 30 are 2, 3, 5.
See more: Which Is Your Favorite Middle Name For Naomi ? 100 Attractive Middle Names For Naomi
### What are the Common Factors of 30 and 25?
Since, the factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30 and the factors of 25 are 1, 5, 25.Hence, <1, 5> are the common factors of 30 and 25.
### What is the Sum of the Factors of 30?
Sum of all factors of 30 = (21 + 1 - 1)/(2 - 1) × (31 + 1 - 1)/(3 - 1) × (51 + 1 - 1)/(5 - 1) = 72 |
# Rational Numbers Into Decimals Worksheet
A Rational Numbers Worksheet can help your child be a little more familiar with the concepts associated with this rate of integers. Within this worksheet, college students will be able to remedy 12 diverse problems associated with logical expressions. They will figure out how to increase two or more numbers, class them in sets, and determine their products. They are going to also process simplifying logical expressions. As soon as they have enhanced these ideas, this worksheet might be a useful tool for furthering their scientific studies. Rational Numbers Into Decimals Worksheet.
## Logical Phone numbers certainly are a proportion of integers
There are 2 kinds of phone numbers: irrational and rational. Rational figures are understood to be whole figures, while irrational phone numbers usually do not replicate, and get an unlimited quantity of numbers. Irrational phone numbers are low-absolutely nothing, low-terminating decimals, and square beginnings that are not perfect squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To determine a reasonable quantity, you must know just what a rational amount is. An integer is really a complete number, along with a realistic quantity is really a rate of two integers. The proportion of two integers is the amount on top split by the number at the base. If two integers are two and five, this would be an integer, for example. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They can be created in to a portion
A reasonable amount includes a denominator and numerator that are not absolutely no. This means that they can be expressed being a fraction. In addition to their integer numerators and denominators, reasonable phone numbers can furthermore have a negative benefit. The bad worth ought to be placed on the left of and its particular total value is its length from no. To streamline this example, we are going to state that .0333333 is a portion which can be published like a 1/3.
Along with unfavorable integers, a rational quantity can be produced right into a small percentage. For example, /18,572 can be a rational number, whilst -1/ is not really. Any fraction consisting of integers is realistic, so long as the denominator fails to include a and will be written as being an integer. Furthermore, a decimal that ends in a point is another logical variety.
## They are feeling
Regardless of their label, logical numbers don’t make very much perception. In mathematics, they are solitary organizations with a unique span in the amount collection. Which means that once we add up anything, we can easily buy the shape by its percentage to its unique quantity. This keeps correct even when there are actually limitless rational amounts among two certain amounts. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To obtain the length of a pearl, by way of example, we might count up its thickness. One particular pearl is 15 kilograms, which is actually a rational variety. Furthermore, a pound’s excess weight means ten kilograms. Hence, we should be able to divide a lb by 10, without having concern yourself with the size of an individual pearl.
## They are often expressed as a decimal
If you’ve ever tried to convert a number to its decimal form, you’ve most likely seen a problem that involves a repeated fraction. A decimal variety might be written being a a number of of two integers, so four times 5 is the same as eight. The same issue involves the repeated fraction 2/1, and each side should be separated by 99 to have the correct response. But how will you have the conversion? Here are a few good examples.
A logical number can also be written in great shape, which include fractions plus a decimal. One way to represent a realistic number in the decimal is usually to divide it into its fractional equal. You can find 3 ways to divide a reasonable quantity, and each one of these methods produces its decimal equal. One of these simple methods is to divide it into its fractional equivalent, and that’s what’s referred to as a terminating decimal. |
# GMAT Math: Arithmetic Mean
Written by Kelly Granson. Posted in GMAT Study Guide
GMAT arithmetic mean questions are not very hard, but the GMAT exam is a mixture of simple and more advanced questions. If you perform well, you are most likely to see harder questions, and if your performance is not quite stellar, you will deal mostly with medium and easy test items. However, regardless of your performance, the first few questions will not be exceptionally hard, so chances are you will see at least one GMAT question dealing with arithmetic mean. In this post we will review the very basic info about the arithmetic mean and how it can be tested on the GMAT.
The arithmetic mean is also commonly known as the 'mean' or 'average'. A concept we were all taught in schools from very early age. Nevertheless, let's work through the examples below to refresh ourselves on the basics before we move onto harder questions based upon the arithmetic mean and statistics.
Arithmetic Mean Basic
The literal definition for an arithmetic mean is:
"Arithmetic mean is quotient of sum of the given values and number of the given values".
In basic English, that simply means that the average is the sum of all the values in a set, divided by the total number of values in the same set.
So if we have 5 numbers in a set, let's say for example, 1, 2, 3, 4, and 5, the average is simply:
Arithmetic Mean Formula
The equation for the arithmetic mean can be written as:
There are many other ways that you will see the equation represented, but if you can remember the wording on the above equation, it will help you to answer many different types of GMAT question. Let's have a look at some examples to see how arithmetic mean works in different cases.
Arithmetic Mean of Negative Numbers
One complication that can be added to GMAT questions testing your knowledge of the arithmetic mean is the addition of negative numbers. Instead of having a set based solely on positive numbers, certain numbers may also be negative. Let's take the same example we had before but make all items in the set negative. The five numbers will now be:
-1, -2, -3, -4, and -5
If we were to work out the arithmetic mean then we would use the same formula:
As per the formula, we simply substitute the values:
Note that although the sum of values is now negative, the number of values in the set is always positive.
The exact same process would be used for a mixed number set.
The Missing Number
No matter how nice that would be, GMAT will never just give you a list of numbers and ask you just to find the mean—that's what your sixth grade teacher would do. GMAT, on the other hand, will most likely twist the problem around a little a bit. One common way GMAT tests arithmetic mean is giving you the average and asking to find the missing element of the set.
Let's look at a typical example:
5 students take a test. Amy and Ben both score 20 marks each, Thomas scores 25 marks and Mandy scores 15. What score must Chloe obtain, if the average score is 21 marks.
This question is rather simply when we break it down. Let's extract the key information. Firstly there are five students and four of them have a score:
Amy (A) = 20
Ben (B) = 20
Thomas (T) = 25
Mandy (M) = 15
Chloe (C)= ???
We are given the arithmetic mean as 21. Our objective is to work out Chloe's (C) score. As we know it is an arithmetic mean questions, let's start with the equation and figure out what we know:
Rearranging that equations we get:
Sum of values in a set = 21 × 5
Sum of values in a set = 105
Now we know that the sum of all five scores must equal 105.
Sum of values in a set = A+B+T+M+C=105
We already know four of the scores, so we can immediately substitute them:
20+20+25+15+C=105
80+C=105
A little rearranging will give us our answer:
C=105-80
C=25
Chloe's score is therefore 25.
As you can see, the little twists GMAT adds to the questions can be easily cracked if you know the general formula and can manipulate it. We will look into harder GMAT questions based on the arithmetic mean and statistics in our next posts, so stay updated.
Good luck! |
# Question: How do you answer a pie chart?
Contents
## How do you find the answer of a pie chart?
The total of all the data is equal to 360°.The total value of the pie is always 100%.(Given Data/Total value of Data) × 360°Step 1: First, Enter the data into the table.Step 2: Add all the values in the table to get the total.Step 3: Next, divide each value by the total and multiply by 100 to get a per cent:More items •Jan 14, 2020
## What is a pie chart answer?
A pie chart (or a circle chart) is a circular statistical graphic, which is divided into slices to illustrate numerical proportion. In a pie chart, the arc length of each slice (and consequently its central angle and area), is proportional to the quantity it represents.
## How do you explain a pie chart to a student?
Tell students that pie charts (or circle graphs) are used to represent data as portions (or segments) of a whole. Explain that just as they would see a pizza pie cut up into pieces, a pie chart is divided into different pieces of data. Each portion represents a percentage of the pie; all portions add up to 100%.
## What are the advantages of pie chart?
Advantagesdisplay relative proportions of multiple classes of data.size of the circle can be made proportional to the total quantity it represents.summarize a large data set in visual form.be visually simpler than other types of graphs.permit a visual check of the reasonableness or accuracy of calculations.More items
## When would you use a pie chart?
Pie charts make sense to show a parts-to-whole relationship for categorical or nominal data. The slices in the pie typically represent percentages of the total. With categorical data, the sample is often divided into groups and the responses have a defined order.
## Why is a pie chart bad?
Pie charts are one of the most overused graphs in the world and in most cases are not the best way to present data. They often distort the information and make it more difficult for decision-makers to understand the messages they contain.
## How do you find the angle for a pie chart?
Divide the count for each variable by the total count and then multiply this figure by 360 to determine the angle in degrees. When the pie chart is drawn, the slice of pie for Group A will have an angle of 196.36°; as such, it will represent just over half of the full circle.
## How do you put data into a pie chart?
Step 1 Convert the data to percentages. The first step is to convert the data to percentages. Step 2 Calculate the angle for each pie segment. A complete circle or pie chart has 360°. Step 3 Draw the pie chart. For this youll need compasses and a protractor.Step 4 Add labels. The chart requires a title and labels:
## When should you not use a pie chart?
If you still feel the urge to use them, make sure you only use them for a percentage breakdown where each slice represents a certain percentage out of 100% and order the slices in size to make it easier to read. Never use a pie chart if it has more than 5 slices and never-ever make it 3D.
## What should be included in a pie chart?
In order to use a pie chart, you must have some kind of whole amount that is divided into a number of distinct parts. Your primary objective in a pie chart should be to compare each groups contribution to the whole, as opposed to comparing groups to each other.
## Does a pie chart have to equal 100?
Not doing the math correctly is a common mistake that makes your pie chart useless. Pie charts are designed to show parts of a whole, so any sum below or above 100% doesnt represent the entire picture. Finally, when it comes to legends, pie charts dont generally need one. Its cleaner to label your data directly.
## What can I use instead of a pie chart?
Stacked Bar Charts are the closest linear equivalent to Pie Charts, in terms of both one-to-one mapping and layout. They may be the best alternatives to Pie charts. A single-series Pie chart with N slices is actually equivalent with N series of Full 100% Stacked Bars, each with one single value.
## How do you represent data in a pie chart?
The total value of the pie chart is always 100%. Each portion in the circle shows a fraction or percentage of the total. Pie chart is a circular graph which is used to represent data .Construction of Pie Chart.ActivityNo. of HoursMeasure of central angleStudy4(4/24 × 360)° = 60°T. V.1(1/24 × 360)° = 15°Others3(3/24 × 360)° = 45°3 more rows
## What is needed to construct a pie chart?
Step 1 Convert the data to percentages. The first step is to convert the data to percentages. Step 2 Calculate the angle for each pie segment. A complete circle or pie chart has 360°. Step 3 Draw the pie chart. For this youll need compasses and a protractor.Step 4 Add labels. The chart requires a title and labels:
## Why do we use pie charts?
Pie charts make sense to show a parts-to-whole relationship for categorical or nominal data. The slices in the pie typically represent percentages of the total. With categorical data, the sample is often divided into groups and the responses have a defined order.
## Write us
#### Find us at the office
Klank- Fillhart street no. 8, 52340 San Juan, Puerto Rico
#### Give us a ring
Jermya Lenninger
+88 940 846 744
Mon - Fri, 9:00-18:00 |
# Question Video: Multiplying Square Roots of Negative Numbers Mathematics
Simplify √(−10) × √(−6).
01:38
### Video Transcript
Simplify the square root of negative 10 times the square root of negative six.
We’ll begin by expressing each radical in terms of 𝑖. Remember 𝑖 squared is equal to negative one. So we can say the square root of negative 10 is the same as the square root of 10𝑖 squared. And similarly, the square root of negative six is the same as the square root of six 𝑖 squared. And at this point, we can split this up. We get the square root of 10 times the square root of 𝑖 squared. And since the square root of 𝑖 squared is 𝑖, we can see that the square root of negative 10 is the same as root 10 𝑖. And similarly, the square root of negative six is root six 𝑖.
Next, we multiply these together. Multiplication is commutative. So we can rearrange this a little and say it’s equal to the square root of 10 times the square root of six which is root 60 times 𝑖 squared. And since 𝑖 squared is negative one, we see that the square root of negative 10 times the square root of negative six is negative root 60. And in fact, we need to simplify this as far as possible.
There are a number of ways to do this. We could consider 60 as a product of its prime factors. Alternatively, we find the largest factor of 60 which is also a square number. In fact, that factor is four. So this means that the square of 60 is the same as the square root of four times the square root of 15 which is equal to two root 15. And we fully simplified our expression. We get negative two root 15.
Let’s look at what would have happened had we applied the laws of radicals. We would have said that the square root of negative 10 times the square root of negative six is equal to the square of negative 10 times negative six which is equal to the square root of positive 60 or two root 15 and that’s patently different to our other solution. |
# Linear Inequality and Linear Inequations | Linear Inequalities in One Variable
On this page, we will learn about what is linear inequality and linear inequations and the steps to solve the linear inequalities problems. You will also get the properties of inequation or inequalities. Check out the representation of the solution set on the real line in the following sections. We have provided Solved Problems along with a detailed explanation so that you can better understand the concept.
## What are Linear Inequation and Linear Inequalities?
Linear Inequation is a statement indicating the value of one quantity or algebraic expressions that are not exactly equal to one another.
Inequalities are nothing but the symbols enclosed between two or more algebraic expressions or quantities. The open sentence which involves <, ≤, >, ≥, and ≠ symbols are called the inequalities.
Some of the examples of Linear Inequation are listed below.
• x < 6
• y ≥ 25
• x + 3 > 40
• p ≠ 10
### Linear Inequation
An inequation that contains one variable and that variable highest power is one then is known as the linear inequation in that variable. To make a linear equation into inequation, you have to replace the equal to symbol with the inequality sign. The statements of any of the forms ax + b < 0, ax + b > 0, ax + b≥ 0, ax + b ≤ 0 are the linear inequations invariable x, where a, b are real numbers and a is not equal to zero.
some of the examples of the linear inequation with variable y are included below:
• 3y + 6 ≥ 0
• 9 – y < 0
• 2y > 0
• 25 + 5y ≤ 0
## Questions on Replacement Set or Domain of a Variable
Example 1.
Find the replacement set for the inequation x ≤ 9. The replacement set is a set of whole numbers?
Solution:
We know that whole numbers W = {0, 1, 2, 3 . . . }.
Replace x with some values of W. Some values of x from W satisfy the inequation and some don’t. Here, the values 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 satisfy the given inequation x ≤ 9 while the other values don’t.
Thus, the set of all those values of variables that satisfy the given inequation is called the solution set of the given inequation.
Therefore, the solution set for the inequation x ≤ 9 is S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} or S = {x : x ∈ W, x ≤ 9}
Example 2.
Find the replacement set for the inequation x > 2. Let the replacement set be the set of natural numbers?
Solution:
We know that natural numbers N = {1, 2, 3, 4, 5, 6}
Replace x with some values of N. Some values of x from N satisfy the inequation and some don’t. Here, the values 3, 4, 5, . . . satisfy the given inequation x > 2 while the other values don’t.
Thus, the set of all those values of variables that satisfy the given inequation is called the solution set of the given inequation.
Therefore, the solution set for the inequation x > 2 is S = {3 4, 5, 6, . . . }
Example 3.
Find the replacement set and the solution set for the inequation x ≥ -2 when the replacement set is an integer?
Solution:
Replacement set I = {. . . -3, -2, -1, 0, 1, 2, 3, . . . }
Solution set S = {-2, -1, 0, 1, 2, . . . } or S = { x : x ∈ I, x ≥ – 2}
Example 4.
Find the solution set for the following linear inequations.
(i) x < 5 where replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4}
(ii) x ≥ 7 where replacement set is { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iii) x ≠ 3 where replacement set is { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Solution:
(i) Solution Set S = {-4, -3, -2, -1, 0, 1, 2, 3, 4} S = {x : x ∈ I, -4 < x ≤ 4}
(ii) Solution set S = {7, 8, 9, 10} or S = {x : x ∈ N, 7 < x < 10}
(iii) Solution set S = {0, 1, 2, 4, 5, 6, 7, 8, 9, 10} or S = {x : x ∈ N, x ≠ 3}
### Frequently Asked Questions on Linear Inequality
1. What is the difference between a linear equation and a linear inequality?
The linear equation is an equation that has one or two variables and those exponents are one. Linear inequation also has one variable whose exponent is one. Between two algebraic expressions, the = symbol is enclosed in a linear equation, linear inequality signs are enclosed in a linear inequation. The graph of inequalities is a dashed line but the equation is a solid line in any situation.
2. What is linear inequality?
Linear inequality contains any symbols of inequality. It represents the data that is not equal in graph form. It involves a linear function. |
# Common Core: High School - Algebra : Using the Binomial Theorem for Expansion: CCSS.Math.Content.HSA-APR.C.5
## Example Questions
← Previous 1
### Example Question #1 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5
Use Pascal's Triangle to Expand,
Not Possible
Explanation:
In order to do this, we need to recall the formula for Pascal's Triangle.
Since the exponent in the question is 12 we can replace , with 12 .
Now our equation looks like
Now we compute the sum, term by term.
Term 1 :
Term 2 :
Term 3 :
Term 4 :
Term 5 :
Term 6 :
Term 7 :
Term 8 :
Term 9 :
Term 10 :
Term 11 :
Term 12 :
Term 13 :
Now we combine the expressions and we get
### Example Question #2 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5
What is the coefficient of in the expansion of ?
There is no coefficient
Explanation:
In order to do this, we need to recall the formula for Pascal's Triangle.
The part in the expression that we care about is the combination.
We simply do this by looking at the exponent of , and the exponent of the original equation.
In this case and
Now we compute the following
### Example Question #3 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5
Use Pascal's Triangle to Expand
Not Possible
Explanation:
In order to do this, we need to recall the formula for Pascal's Triangle.
Since the exponent in the question is 18 we can replace , with 18 .
Now our equation looks like
Now we compute the sum, term by term.
Term 1 :
Term 2 :
Term 3 :
Term 4 :
Term 5 :
Term 6 :
Term 7 :
Term 8 :
Term 9 :
Term 10 :
Term 11 :
Term 12 :
Term 13 :
Term 14 :
Term 15 :
Term 16 :
Term 17 :
Term 18 :
Term 19 :
Now we combine the expressions and we get
### Example Question #3 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5
What is the coefficient of in the expansion of ?
There is no coefficient
Explanation:
In order to do this, we need to recall the formula for Pascal's Triangle.
The part in the expression that we care about is the combination.
We simply do this by looking at the exponent of , and the exponent of the original equation.
In this case and
Now we compute the following
### Example Question #4 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5
Use Pascal's Triangle to Expand
Not Possible
Explanation:
In order to do this, we need to recall the formula for Pascal's Triangle.
Since the exponent in the question is 8 we can replace , with 8 .
Now our equation looks like
Now we compute the sum, term by term.
Term 1 :
Term 2 :
Term 3 :
Term 4 :
Term 5 :
Term 6 :
Term 7 :
Term 8 :
Term 9 :
Now we combine the expressions and we get
### Example Question #5 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5
What is the coefficient of in the expansion of ?
There is no coefficient
Explanation:
In order to do this, we need to recall the formula for Pascal's Triangle.
The part in the expression that we care about is the combination.
We simply do this by looking at the exponent of , and the exponent of the original equation.
In this case and
Now we compute the following
### Example Question #6 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5
Use Pascal's Triangle to Expand
Not Possible
Explanation:
In order to do this, we need to recall the formula for Pascal's Triangle.
Since the exponent in the question is 7 we can replace , with 7 .
Now our equation looks like
Now we compute the sum, term by term.
Term 1 :
Term 2 :
Term 3 :
Term 4 :
Term 5 :
Term 6 :
Term 7 :
Term 8 :
Now we combine the expressions and we get
### Example Question #7 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5
What is the coefficient of in the expansion of ?
There is no coefficient
Explanation:
In order to do this, we need to recall the formula for Pascal's Triangle.
The part in the expression that we care about is the combination.
We simply do this by looking at the exponent of , and the exponent of the original equation.
In this case and
Now we compute the following
### Example Question #9 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5
Use Pascal's Triangle to Expand
Not Possible
Explanation:
In order to do this, we need to recall the formula for Pascal's Triangle.
Since the exponent in the question is 14 we can replace , with 14 .
Now our equation looks like
Now we compute the sum, term by term.
Term 1 :
Term 2 :
Term 3 :
Term 4 :
Term 5 :
Term 6 :
Term 7 :
Term 8 :
Term 9 :
Term 10 :
Term 11 :
Term 12 :
Term 13 :
Term 14 :
Term 15 :
Now we combine the expressions and we get
### Example Question #8 : Using The Binomial Theorem For Expansion: Ccss.Math.Content.Hsa Apr.C.5
What is the coefficient of in the expansion of ?
There is no coefficient
Explanation:
In order to do this, we need to recall the formula for Pascal's Triangle.
The part in the expression that we care about is the combination.
We simply do this by looking at the exponent of , and the exponent of the original equation.
In this case and
Now we compute the following
← Previous 1 |
### Similar presentations
3 Sketch the graph of a curve given by a set of parametric equations. Eliminate the parameter in a set of parametric equations. Find a set of parametric equations to represent a curve. Understand two classic calculus problems, the tautochrone and brachistochrone problems. Objectives
4 Plane Curves and Parametric Equations
5 Consider the path followed by an object that is propelled into the air at an angle of 45°. If the initial velocity of the object is 48 feet per second, the object travels the parabolic path given by as shown in Figure 10.19. Figure 10.19
6 To determine time, you can introduce a third variable t, called a parameter. By writing both x and y as functions of t, you obtain the parametric equations and Plane Curves and Parametric Equations
7 From this set of equations, you can determine that at time t = 0, the object is at the point (0, 0). Similarly, at time t = 1, the object is at the point and so on. For this particular motion problem, x and y are continuous functions of t, and the resulting path is called a plane curve. Plane Curves and Parametric Equations
8 When sketching (by hand) a curve represented by a set of parametric equations, you can plot points in the xy-plane. Each set of coordinates (x, y) is determined from a value chosen for the parameter t. By plotting the resulting points in order of increasing values of t, the curve is traced out in a specific direction. This is called the orientation of the curve. Plane Curves and Parametric Equations
9 Example 1 – Sketching a Curve Sketch the curve described by the parametric equations Solution: For values of t on the given interval, the parametric equations yield the points (x, y) shown in the table.
10 Example 1 – Solution By plotting these points in order of increasing t and using the continuity of f and g, you obtain the curve C shown in Figure 10.20. Note that the arrows on the curve indicate its orientation as t increases from –2 to 3. cont’d Figure 10.20
11 Eliminating the Parameter
12 Finding a rectangular equation that represents the graph of a set of parametric equations is called eliminating the parameter. For instance, you can eliminate the parameter from the set of parametric equations in Example 1 as follows. Eliminating the Parameter
13 Once you have eliminated the parameter, you can recognize that the equation x = 4y 2 – 4 represents a parabola with a horizontal axis and vertex at (–4, 0), as shown in Figure 10.20. Figure 10.20 Eliminating the Parameter
14 The range of x and y implied by the parametric equations may be altered by the change to rectangular form. In such instances, the domain of the rectangular equation must be adjusted so that its graph matches the graph of the parametric equations. Eliminating the Parameter
15 Example 2 – Adjusting the Domain After Eliminating the Parameter Sketch the curve represented by the equations by eliminating the parameter and adjusting the domain of the resulting rectangular equation.
16 Example 2 – Solution Begin by solving one of the parametric equations for t. For instance, you can solve the first equation for t as follows.
17 Now, substituting into the parametric equation for y produces The rectangular equation, y = 1 – x 2, is defined for all values of x, but from the parametric equation for x you can see that the curve is defined only when t > –1. Example 2 – Solution cont’d
18 This implies that you should restrict the domain of x to positive values, as shown in Figure 10.22. Figure 10.22 cont’d Example 2 – Solution
19 Finding Parametric Equations
20 Example 4 – Finding Parametric Equation for a Given Graph Find a set of parametric equations that represents the graph of y = 1 – x 2, using each of the following parameters. a. t = x b. The slope m = at the point (x, y) Solution: a. Letting x = t produces the parametric equations x = t and y = 1 – x 2 = 1 – t 2.
21 Example 4 – Solution b. To write x and y in terms of the parameter m, you can proceed as follows. This produces a parametric equation for x. To obtain a parametric equation for y, substitute –m/2 for x in the original equation. cont’d
22 So, the parametric equations are In Figure 10.24, note that the resulting curve has a right-to-left orientation as determined by the direction of increasing values of slope m. For part (a), the curve would have the opposite orientation. Figure 10.24 Example 4 – Solution cont’d
23 Example 5 – Parametric Equations for a Cycloid Determine the curve traced by a point P on the circumference of a circle of radius a rolling along a straight line in a plane. Such a curve is called a cycloid. Solution: Let the parameter θ be the measure of the circle’s rotation, and let the point P = (x, y) begin at the origin. When θ = 0, P is at the origin. When θ = π, P is at a maximum point (πa, 2a). When θ = 2π, P is back on the x-axis at (2πa, 0).
24 Example 5 – Solution From Figure 10.25, you can see that Figure 10.25 cont’d
25 So, which implies that AP = –acos θ and BD = asin θ. Because the circle rolls along the x-axis, you know that Example 5 – Solution cont’d
26 Furthermore, because BA = DC = a, you have x = OD – BD = aθ – a sin θ y = BA + AP = a – a cos θ. So, the parametric equations are x = a(θ – sin θ) and y = a(1 – cos θ). Example 5 – Solution cont’d
27 The cycloid in Figure 10.25 has sharp corners at the values x = 2nπa. Notice that the derivatives x'(θ) and y'(θ) are both zero at the points for which θ = 2nπ. Finding Parametric Equations Figure 10.25
28 Between these points, the cycloid is called smooth. Finding Parametric Equations
29 The Tautochrone and Brachistochrone Problems
30 The Cycloid is related to one of the most famous pairs of problems in the history of calculus. The first problem (called the tautochrone problem) began with Galileo’s discovery that the time required to complete a full swing of a given pendulum is approximately the same whether it makes a large movement at high speed or a small movement at lower speed (see Figure 10.26). The Tautochrone and Brachistochrone Problems Figure 10.26
31 Galileo realized that he could use this principle to construct a clock. However he was not able to conquer the mechanics of actual construction. Christian Huygens was the first to design and construct a working model. He realized that a pendulum does not take exactly the same time to complete swings of varying lengths. But, in studying the problem, Huygens discovered that a ball rolling back and forth on an inverted cycloid does complete each cycle in exactly the same time. The Tautochrone and Brachistochrone Problems
32 The second problem, which was posed by John Bernoulli in 1696, is called the brachistochrone problem—in Greek, brachys means short and chronos means time. The problem was to determine the path down which a particle will slide from point A to point B in the shortest time. The Tautochrone and Brachistochrone Problems
33 The solution is not a straight line from A to B but an inverted cycloid passing through the points A and B, as shown in Figure 10.27. Figure 10.27 The Tautochrone and Brachistochrone Problems
34 The amazing part of the solution is that a particle starting at rest at any other point C of the cycloid between A and B will take exactly the same time to reach B, as shown in Figure 10.28. Figure 10.28 The Tautochrone and Brachistochrone Problems |
Chapter 10Solutions for second week’s assignments
Also available as PDF.
10.1 Exercises for Section 1.3
10.1.1 Problem 6: Understanding
The problem begins by stating that “[t]oday is your first day driving a city bus.” At the end, the question posed is “[h]ow old is the bus driver?” The other information is irrelevant, and the result is your age. (You need not provide your age. The fact that “you” are the bus driver is all that matters.)
10.1.2 Problem 12: Guessing and checking
When playing around with the problem, you may realize that if each of the three parts has the same sum, then each must equal 1∕3 of the total sum. The sum {\mathop{\mathop{∑ }}\nolimits }_{i=1}^{12}i = 12(12 + 1)∕2 = 6 ⋅ 13, so we can guess that three equal parts each will total 6 ⋅ 13∕3 = 2 ⋅ 13 = 26.
Now the problem becomes positioning lines such that the totals are 26. The smaller numbers (e.g. 1, 2) need combined with some of the larger numbers (e.g. 11, 12) to total 26. So we can guess that one of the lines will be nearly horizontal across the top. First try a line that separates the top four numbers, so 11 + 12 + 1 + 2 = 26.
Now the other line cannot intersect the first inside the clock or else the clock will be divided into four parts. My first guess was to anchor one end of the new line where the previous line ran under 2. That would cut a pie-like slice and only include numbers on the opposite side. The closest adding the facing side can come to 26 is 10 + 9 + 8 = 27, so clearly that was not correct. If we chop off the 8, that leaves 10 + 9 = 19 and needs 7 more to match 26. So run the right-hand side down, no longer cutting out a pie-like slice but including some numbers from both sides. Then we can include 3 + 4 = 7.
So far, my guesses separated 11 + 12 + 1 + 2 = 26 and 10 + 9 + 3 + 4 = 26. With the remaining numbers, 5 + 6 + 7 + 8 = 26, so the problem is solved.
In hindsight, however, I see my previous guess was being far too fancy. Note that each slightly diagonal slice adds to 13. That is, 12 + 1 = 13, 11 + 2 = 13, and so on down to 7 + 6 = 13. So we need only combine consecutive slices to create the necessary three regions.
10.1.3 Problem 31: Listing
here most of the problem lies in figuring out an appropriate representation. Then starting to make lists show the way to an answer.
If we were to list all combinations of “black” and “white”, that would be 1.4 × 1{0}^{11} combinations. So this list likely is not what I meant by mentioning listing.
A shorter method is to list all combinations of two socks, or BB, BW, WW. The order does not matter in the end. In the BW case we do not have two socks of the same color, so the result cannot be two.
Now the combinations of three socks are BBB, BBW, BWW, WWW. In each of these we have two socks of the same color. So if we pull out three socks, we are guaranteed to have two matching socks. Since two did not work, the smallest number of socks you can pull without looking to have two of the same sock is three.
Another way to view the problem is by ticking off marks in the following for each sock drawn:
Black White
Once you have a tick mark in both, the next mark must land in an occupied slot. So the longest sequence of marks without a duplicate is checking off one and then the other, or two marks. After two, you must mark an occupied slot.
This is a consequence of the pigeonhole principle. When there are more pigeons than holes, there must be at least two pigeons in one hole. You can use the same principle to prove that there must be two people in the Tri-cities area with the same number of hairs on their heads.
10.1.4 Problem 35: Listing
The simplest solution is to start listing numbers. The text’s example of 28 shows you will not need to list more than 28 numbers. And because 1 has no factor other than itself, we can skip the first obvious entry.
2 ≠1 3 ≠1 4 ≠1 + 2 = 3 5 ≠1 \mathbf{6} \mathbf{= 1 + 2 + 3}
We could construct the list without the prime numbers 2, 3, 5, etc. A prime number has only itself and 1 as factors, so we know that they cannot be perfect numbers.
Aside: What would the entry for 1 be? One has no factors other than itself. You would sum over , the empty set. There is no single definition. In the context of this problem, we could define that sum to be zero. That would be consistent with the rest of the problem and perfectly ok. But it’s not a completely standard definition.
Dealing with vacuous cases like the sum of entries of is tricky, but there is a general rule of thumb. Often the vacuous definition needs to be the identity of the operation. Here, zero is the identity element for addititon because x + 0 = 0 for all x. So it’s a safe guess that the sum of the entries in the empty set could be defined to zero.
Another side: No one knows if there are infinitely many perfect numbers, or if there are any odd perfect numbers.
10.1.5 Problem 28: Following dependencies
Let {M}_{1} be the initial amount, {M}_{2} be the amount after buying the book, {M}_{3} be the about after the train ticket, {M}_{4} be the amount after lunch, then {M}_{5} be the final about after the bazaar. Translating the problem into a sequence of equations,
\eqalignno{ {M}_{2} & = {M}_{1} - 10, & & \cr {M}_{3} & = {M}_{2}∕2, & & \cr {M}_{4} & = {M}_{3} - 4, & & \cr {M}_{5} & = {M}_{4}∕2,\text{ and} & & \cr {M}_{5} & = 8. & & }
The only fully resolved data we have is {M}_{5} = 8, so we start there. First {M}_{5} = 8 = {M}_{4}∕2, so {M}_{4} = 16. Then 16 = {M}_{3} - 4 and {M}_{3} = 20. Next, 20 = {M}_{2}∕2 so {M}_{2} = 40. Finally, 40 = {M}_{1} - 10 and \mathbf{{M}_{1} = 50}.
10.1.6 Problem 57: Following dependencies
This is an exercise in translating the words into mathematical relations and then working through the dependencies.
Translate each statement into an equation
\eqalignno{ {x}_{2} & = 3{x}_{1}, & & \cr {x}_{3} & = {x}_{2} + {3\over 4}{x}_{2}, & & \cr {x}_{4} & = {x}_{3}∕7, & & \cr {x}_{5} & = {x}_{4} -{1\over 3}{x}_{4}, & & \cr {x}_{6} & = {x}_{5}^{2}, & & \cr {x}_{7} & = {x}_{6} - 52, & & \cr {x}_{8} & = \sqrt{{x}_{7}}, & & \cr {x}_{9} & = {x}_{8} + 8, & & \cr {x}_{10} & = {x}_{9}∕10,\text{ and} & & \cr {x}_{10} & = 2. & & }
Following these backwards shows that
\eqalignno{ {x}_{9} & = 20, & & \cr {x}_{8} & = 12, & & \cr {x}_{7} & = 144, & & \cr {x}_{6} & = 196, & & \cr {x}_{5} & = 14, & & \cr {x}_{4} & = 21, & & \cr {x}_{3} & = 21 ⋅ 7\text{(note the next step divides by 7, so don’t expand)}, & & \cr {x}_{2} & = 84,\text{and} & & \cr \mathbf{{x}_{1}} &\mathbf{= 28}. & & }
10.1.7 Problem 40: Bisection and guessing a range
See the write-up below.
10.1.8 Problem 52: Think about bisection
With eight coins, split them into two groups of four. One will be lighter. Split the lighter group of four into two groups of two. Again, one is lighter. Now the lighter group of two splits into two single coins. The lighter of the two coins is fake.
For the trick of two weighings, we first consider weighing two groups of three coins. If the groups are equal, we are left with the two remaining coins. Those two can be separated in one more weighing. If the groups of three are unequal, split the lighter one into three groups of one. Compare two of those single coins. If they are of the same weight, the left-over coin must be fake. Otherwise the lighter coin is the fake. Any path here requires only two weighings. This is an example trisection, separating the problem into three groups at each level.
10.1.9 Problem 56
A rotated square with each vertex located at the midpoint of the given square’s sides will separate the kitties.
10.1.10 Problem 61: Look for a pattern
Calculating 1∕7 to a few places shows 1∕7 = 0.14285714285714\mathrel{⋯}\kern 1.66702pt The expansion appears to repeat in groups of six. Because 100 = 16 ⋅ 6 + 4, we expect that the 100th digit after the decimal point is 8.
10.2 Making change
How many ways can you make change for 60 cents using pennies, nickles, dimes, and quarters. Either take great care in forming a long list, or look for a relationship using smaller problems.
A hint for a long list: Do you need to move pennies one at a time?
A hint for a relationship: Consider the old example of 20 cents using pennies, nickles and dimes. How many ways are there to change 20 cents using only pennies and nickles? How many ways to change 20 cents minus one dime using all the coins? The relationship makes constructing a table much easier.
This is a classical problem used in discrete mathematics and introductory computer science classes, although often starting with a dollar to make tabling less practical.
There are 73 ways.
There are at most two quarters, at most six dimes, at most 12 nickles, and at most 60 pennies per line of a table with the following heading:
Pennies Nickles Dimes Quarters total
To generate the table, start with two quarters and then shift amounts over as in other problems. You use the conserved quantity, the total amount, to guide your next choice.
Another method for solving this problem is to set up recurrence relationships and build a slightly different and much, much shorter table.
Consider making change for an amount N. And consider four different ways for making such change:
\eqalignno{ {A}_{N} &\text{ with only pennies,} & & \cr {B}_{N} &\text{ with nickles and pennies,} & & \cr {C}_{N} &\text{ with dimes, nickles, and pennies, and} & & \cr {D}_{N} &\text{ with quarters, dimes, nickes, and pennies.} & & }
Say we start at N and the full collection of possible coins. Then either the change contains a quarter or it does not. If it does contain with a quarter, then we change the remaining N - 25 in the same way, possibly with more quarters. If not, then we change N no quarters. So
\eqalignno{ {D}_{N} & = {C}_{N} + {D}_{N-25}. & & \cr & & }
Similarly,
\eqalignno{ {C}_{N} & = {B}_{N} + {C}_{N-10},\text{ and} & & \cr {B}_{N} & = {A}_{N} + {B}_{N-5}. & & }
We can begin constructing a table of values by N starting from the extreme case N = 0. There is only one way of making no change at all, so {A}_{0} = {B}_{0} = {C}_{0} = {D}_{0} = 1. There also is only one way of making change with pennies, so {A}_{N} = 1 for all N. And the relations above show we need only rows where N is a multiple of five and provide formulas for every entry.
The table is as follows:
N {A}_{N} {B}_{N} {C}_{N} {D}_{N} 0 1 1 1 1 5 1 2 2 2 10 1 3 4 4 15 1 4 6 6 20 1 5 9 9 25 1 6 12 13 30 1 7 16 18 35 1 8 20 24 40 1 9 25 31 45 1 10 30 39 50 1 11 36 49 55 1 12 42 60 60 1 13 49 73 = 49 + 24
10.3 Writing out problems
10.3.1 Section 1.2, problem 9
Understanding the problem
We need to extend the sequence of interior regions to find the number of regions when there are seven and eight points. We have the first six entries of the sequence.
Devise a plan
Given the first six entries, we can form a successive difference table to extrapolate the sequence.
Carry out the plan
The table follows:
points regions {Δ}^{(1)} {Δ}^{(2)} {Δ}^{(3)} {Δ}^{(4)} 1 1 2 2 1 3 4 2 1 4 8 4 2 1 5 16 8 4 2 1 6 31 15 7 3 1 7 57 26 11 4 1 8 99 42 16 5 1
Examine the solution
Computing R(7) = {1\over 24}({7}^{4} - 6 ⋅ {7}^{3} + 23 ⋅ {7}^{2} - 18 ⋅ 7 + 24) = 57 and R(8) = {1\over 24}({8}^{4} - 6 ⋅ {8}^{3} + 23 ⋅ {8}^{2} - 18 ⋅ 8 + 24) = 99 confirms the table’s results. Also note that the table needed four columns to the right of the sequence to find the constant increment. This matches the degree, four, of the polynomial.
10.3.2 Section 1.2, problem 49
Understanding the problem
We need to inductively determine the formula for N(n), the nth nonagonal number. We have the following formulas:
\eqalignno{ H(n) & = {n(4n - 2)\over 2} , & & \cr Hp(n) & = {n(5n - 3)\over 2} ,\text{ and} & & \cr O(n) & = {n(6n - 4)\over 2} , & & }
for the hexagonal, heptagonal, and octagonal numbers.
Devise a plan
The plan is to look for a pattern in the formulas for H(n), Hp(n), and O(n). That lets us predict N(n).
Carry out the plan
All the formulas provided are of the form
{n(an + b)\over 2} ,
so we examine patterns in a and b.
The values of a are 4, 5, and 6. This suggests the next will be 7. And the values of b are -2, -3, and -4, suggesting to continue the pattern with -5. So we expect that the formula is
N(n) = {n(7n - 5)\over 2} .
Examine the solution
To check, we verify that N(6) = 111. Indeed, N(6) = 6 ⋅ (42 - 5)∕2 = 3 ⋅ 37 = 111, supporting the guess. While not a proof, this is a enough evidence to elevate the guess for N(n) to a conjecture.
10.3.3 Section 1.3, problem 40
Understanding the problem
We are looking for a year. That year is 76 more than the birth year of one of the authors and is also a perfect square. The final answer is that year, x, minus 76.
Exploring the problem, we realize that the birth year in question must be within 76 years of publication of the text. The text was published in 2003. To stick to round numbers for estimation, we look for years between 1900 and 2100.
Devise a plan
Look for an integer between 1900 and 2100 that is a perfect square. Knowing that \sqrt{ 1900} > 43 and \sqrt{ 2100} < 46, we search the region 44 ≤ i ≤ 45. Bisection here is overkill; there only are two choices after limiting the choices as above.
Carry out the plan
i {i}^{2} {i}^{2} - 76 45 2025 1949 44 1936 1860
So the result is that Hornsby was born in 1949, because 1860 is more than 76 years before 2003.
Examine the solution
Bisection here was unnecessary. Had we rounded outwards, though, and assumed only 43 ≤ i ≤ 46, then bisection would have delivered the result in the first step.
10.4 Computing with numbers
10.4.1 Extra digits from 1∕7
Compute 1∕7. Write down the number exactly as displayed. Then subtract what you have written from the calculator’s or program’s result. For a calculator, divide one by seven and then subtract off what you see without storing the result elsewhere. For a spreadsheet or other interface, divide one by seven. Then compute 1∕7 - .14\mathrel{⋯}\kern 1.66702pt for whatever was displayed. What is the result? What did you expect? What result did others find?
With a computer using typical arithmetic, we may see 1∕7 computed to be 0.142857142857143. Subtracting that off gives - 1.38777878078145 × 1{0}^{-16}, not zero! Some computers (notably 32-bit Intel-like processors, although not under recent versions of Windows) may show a different result; they store intermediate results to extra precision.
With a calculator, you typically will see one or two non-zero digits. An 8-digit calculator might display 0.1428571. Subtracting that off may show any of 0, 4, 42, or 43 depending on how the calculator rounded and how many extra digits were kept. Most calculators keep a few extra digits past what is displayed.
10.4.2 Binary or decimal?
Enter .1 into whatever device you use. Add .1 to it. Repeat eight more times, for a total of 10 ⋅ .1. Subtract 1. What is the result? What did you expect? What result did others find?
Using a calculator, you probably see zero, exactly what you expect from 10 ⋅ .1 - 1. Most hand-held calculators work with decimal arithmetic directly.
With most computers, you see - 1.11022302462516 × 1{0}^{-16}. This is because .1 cannot be represented exactly in binary. The fraction 1∕10 when converted to binary and computed does not terminate, just as 1∕3 or 1∕7 do not terminate in decimal.
Some systems run decimal arithmetic in software and also will produce zero. However, not all systems that show zero actually have zero stored as the result. Some spreadsheet software is guilty of “cosmetic rounding”. They will display the result as zero but actually carry the binary version; what you see most certainly may not be what you get. |
# Integration by parts
If $$u(x)$$ and $$v(x)$$ are two functions, thanks to the derivation rules, we know that$$d(u \cdot v) = u \cdot dv + v \cdot du$$$antiderivative $$u \cdot v=\displaystyle \int u \cdot dv + \int v \cdot du$$$
and, therefore, $$\displaystyle \int u \cdot dv = u \cdot v - \int v \cdot du$$\$
This is the integration by parts formula and it will be useful to us to compute many integrals and, although it may seem difficult, it is such a useful fromula that it is woth memorizing it.
To be able to choose, decide what $$u(x)$$ is and what $$v(x)$$ is in a given integral; we always have to bear in mind that the integral that we will have to find when using the integration by parts is $$v (x) \cdot u' (x)$$. That is, the term that we take as the derivative will then be integrated, while we will take the derivative of the other term. Sometimes we will not write the variable $$x,$$ although we will always bear it in mind. We must remember that $$dv=v' (x)\cdot dx$$and that $$du=u' (x) \cdot dx$$.
1. Choose the functions $$u$$ and $$dv$$.
2. Compute $$du$$ and $$v$$.
3. Use the formula and find the value of the integral.
$$\displaystyle \int x \cdot e^x \ dx$$
In this case,
$$\begin{array}{ll} u= x & du=1\cdot dx \\ dv=e^x \ dx &v=\displaystyle \int e^x \ dx = e^x\end{array}$$
sot that:
$$\displaystyle \int x \cdot e^x \ dx = x \cdot e^x - \int e^x \ dx = xe^x-e^x+C$$
As we can see, when trying to do an integral by parts, we will always have to solve another integral. The essence of the integrals by parts is that this new integral is easier than the previous one. However, we may have to do several steps before we are actually able to solve the integral.
It can also be the case that, after several steps, we obtain the same initial integral. In such case, we will call the initial integral $$I$$ and we will solve the obtained equation in terms of $$I$$.
Integral by parts in 2 steps. $$\displaystyle \int x^2 e^x \ dx$$
We take, in this case
$$\begin{array}{ll} u= x^2 & du=2x\cdot dx \\ dv=e^x \ dx &v=\displaystyle \int e^x \ dx = e^x\end{array}$$
so that:
$$\displaystyle \int x^2 e^x \ dx=x^2 e^x-2 \int x \cdot e^x dx$$ with the new integral already calculated in the example $$1$$.
Taking the same functions $$\displaystyle \int x \cdot e^x dx= x \cdot e^x-\int e^x \ dx = x \cdot e^x- e^x$$, and we can replace the value of this integral to obtain:
$$\displaystyle \int x^2 e^x dx = x^2e^x-2 \int x e^x dx = x^2e-2\Big(x e^x-e^x\Big) +C = e^x\Big(x^2-2x+2\Big)+C$$
$$\displaystyle \int \sin ^2 x \ dx$$
This integral can be calculated in several ways (it is not a direct integral. The derivative is missing!).
To solve it by parts, we will take
$$\begin{array}{ll} u= \sin x & du=\cos x \cdot dx \\ dv=\sin x \ dx &v=\displaystyle \int \sin x \ dx = -\cos x \end{array}$$
We then have:
$$\displaystyle \int \sin^2 x \ dx= -\sin x \cos x - \int \cos ^2 x \ dx =- \sin x \cdot cos x + \int \cos ^2 x \ dx= \\\displaystyle =-\sin x \cos x+ \int 1-sin^2 x \ dx = - \sin x \cos x +x - \int \sin ^2 x \ dx$$
Where we have used $$\cos^2 x= 1-\sin^2 x$$.
And so, we have the same integral as the one we had at the beginning.
$$\displaystyle \sin^2 x dx =-\sin x \cos x +x- \int \sin^2 x \ dx$$
If we isolate$$\int sin^2 x \ dx$$ we have:
$$2 \displaystyle \int sin^2 x \ dx =- \sin x \cos x +x \Rightarrow \int \sin^2 x \ dx = - \frac{1}{2}\sin x \cos x +\frac{x}{2}+ C$$
$$\displaystyle \int \arctan x \ dx$$
This integral may look difficult, but we can take $$u=arctg (x)$$ and $$dv=1$$ (often it is a very useful trick to take $$dv=1$$).
We have then:
$$\begin{array} {ll} u= \arctan x & du= \dfrac{1}{1+x^2}dx \\ dv=1 \ dx & v=x\end{array}$$
and thus:
$$\displaystyle \int \arctan c \ dx = x \arctan x - \int \frac{x}{1+x^2} \ dx = x \arctan x - \frac{1}{2} \ln|1+x^2|$$
where $$\displaystyle\int \frac{x}{1+x^2} \ dx$$ is an almost direct integral. |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# Dimension of Linear Momentum
Last updated date: 20th Jul 2024
Total views: 90.6k
Views today: 1.90k
## What are Dimensions?
The powers to which the fundamental units that are the unrelated units of measurement are raised for a physical quantity so as to get one unit of that quantity is considered as dimensions of that physical quantity.
### Dimensional Formula
Dimensional formulas are often defined as the expression that shows the powers to which the fundamental units are to be raised to get one unit of a derived quantity.
Suppose the unit of a derived quantity is represented by Q. Then, Q can be given as- MaLbT
Thus, Q= MaLbTc
Which represents the dimensional formula of a physical quantity and the exponents in this case (a,b,c) are the dimensions.
### Linear Momentum
From the study of Momentum, we know that momentum is the mass of the body times the velocity.
So mathematically p= m x v
Change in the momentum of the body remains proportional to the net force and the time over which the net force acts.
Now, we will find out how momentum and net force are related.
While learning kinematic equations, you have learned that with constant acceleration, the change in velocity Δv can also be given as aΔt.
So, the change in momentum following an acceleration can be given as-
Δp= mΔv
Therefore, m(aΔt)= FnetΔt
From the above equation, by rearranging the equation to get value of Fnet
Given the mass and the net force is constant
Fnet= $\frac{\Delta p}{\Delta t}$
From this, we can conclude that change in momentum is equal to the net force over a period of time.
Now, inertia of translational momentum is described as linear momentum. Linear momentum does not have a direction as it is a vector quantity. Also, it has to be noted that the body’s momentum is in the same direction as its vector of velocity.
The SI unit of linear momentum is given as Kg m/s.
### Formula for Dimensions of Linear Momentum
Linear Momentum is given by -
Linear Momentum = Mass∗Velocity
Dimensions of mass = M
Dimensions of velocity = $\frac{Length}{Time}$
= L T⁻¹
Therefore, Dimensions of Linear Momentum is given by - [M1 L1 T-1]
Where,
• M = Mass
• L = Length
• T = Time
### Solved Examples to Calculate Linear Momentum
1. Find the linear momentum of the body having mass 10 kg, moving with the speed of 40 m/s.
Solution: Given mass of the body m= 10 kg
Velocity of the body v= 4 m/s
Linear momentum is given by-
P = mv
So, Momentum p= 10 x 40
p= 400 kg m/s
2. The linear momentum of a body of mass 5 kg is 40 kg m/s. What is the velocity with which the object is moving?
Solution: Mass of the body is m= 5 kg
Given linear momentum of the body p = 40 kgm/s
Linear momentum is given as p= mv
Rearranging the equation, v = $\frac{p}{m}$
So, v = $\frac{40}{5}$
v = 8 m/s
### Derivation of Dimension of Linear Momentum
Linear Momentum = Mass × Velocity . . . . . . (1)
The dimensional formula of,
Mass = [M1 L0 T0] . . . . (2)
Velocity = [M0 L1 T-1] . . . . (3)
Let us substitute equation (2) and (3) in equation (1). Hence, we get,
Linear Momentum = Mass × Velocity
Or, L = [M1 L0 T0] × [M0 L1 T-1] = [M1 L1 T-1].
Therefore, the dimensional formula for linear momentum can be given by [M1 L1 T-1].
### Linear Momentum Dimension
Linear momentum is a quantity measured by multiplying the system’s mass with its velocity. We must know that a large and fast-moving object will have greater momentum than a small and slow-moving object. It is given as p = mv. This means that it is directly proportional to the mass and velocity of the object. When the mass or the velocity is greater of the object, then the momentum is great too. The linear momentum conservation is exhibited by any object in which the momentum amount does not change. This quantity is a vector quantity which has the same direction as the velocity of the object. Its SI unit is written as kg.m/s.
## FAQs on Dimension of Linear Momentum
1. What are the applications of conservation of momentum?
The most important application of momentum is during the launching of rockets. You must have observed that the rocket fuel burns pushes the exhaust gas downwards and this causes the rocket to move upwards. This is also seen in motorboats and is based on the same principle. It pushes the water backwards so that its body moves forwards in order to conserve momentum.
2. What is an impulse?
Impulse is also named as blow. It is related to the phenomenon of collision. It can be defined as the large force which acts for a short period of time. When an object has a collision with another one, each object experiences a blow or impulse. This quantity can also be defined as the product of the average force that acts on a particle and time during which the force acts on that particle. The unit of this quantity is Newton meter (Km).
3. What is the momentum of a moving object?
When an object is moving, there is a certain amount of force that follows behind the object that is moving. When a sudden stop occurs and this stops the motion of the object, that force or momentum behind the object will cause the object to tumble. This tumbling amount depends on the amount of force or momentum that follows the object from behind. This is the explanation of momentum of a moving object.
4. Can I get free notes on linear momentum on vedantu? |
# DISTRIBUTED ELEMENT MODEL IN TRANSMISSION LINES (PART II)
In our previous post in this section, we stated what we will study next to characterize the Distributed Model:
1. The transmission lines equations
2. Solve and exercise applying the equations
3. Propagation constant and characteristic impedance of the line
4. Solve and exercise calculating this parameters on a real line
In this post, we will see points 1 and 2.
## Distributed Element Model
Recall the representation and meaning of the distributed elements in a transmission line:
These parameters vary according to the type of line. For example:
Note: click on the images to see a better resolution. Then, click to go back in your browser to come back to the post 😉
## The transmission lines equations
When using the distributed element model, we can apply the Kirchhoff laws:
And the solutions are:
## Exercise Explained:
The distributed coefficients of a transmission line with w =104 rad/sec are:
R = 0.053 Ω /m L = 0.62 mH/m G = 950 pS/m C = 39.5 pF/m.
In the z-coordinate over the line, the instantaneous current is given by:
i(t) = 75 cos 10 4t mA
a) Obtain the expression for the voltage gradient along the line, in the point z.
b) What is the maximum value of the voltage gradient?
## Solution:
a)
In the time domain, the voltage gradient is given by:
Substituting values, we have:
= – 0.053 ( 0.075 cos 10 4 t) + ( 0.62 x 10 -6 ) (0.075 x10 4 sin 10 4 t) =
= – 3.98 x10 -3 cos 10 4 t + 0.465 x 10 -3 sin 104 t =
= 4.006 x10 -3 cos( 10 4 t – 3.03) Volts/meter.
= 4.006 x10 -3 cos( 10 4 t – 173.4 ° ) Volts/meter.
b)
The maximum voltage gradient is equal to the amplitude, 4 mV. This happens when:
cos(104 t – 3.03) = 1
This implies that,
104 t – 3.03 = 0 , 2Ï€ , 4Ï€ … Radians.
This occurs in the following instants of time:
t0 = 3.03 / 104 = 3.03 x 10-4 sec ,
Then
 t0 = 303 ms
t1 = (2Ï€ + 3.03) / 104 = 9.31 x 10-4 sec , t1 = 931 ms, …
tn = (nÏ€ + 3.03) / 104 sec, with n = 0, 2, 4 …
## What’s next?
We hope this post was useful 🙂 In the next post in this section, we will cover the points:
3. Propagation constant and characteristic impedance of the line
4. Solve and exercise calculating this parameters on a real line
After that, we will move to the the parameters when the line is in a transient state which analyses what happen when there is a sudden change in the conditions of the line.
Does this sound interesting? 😀 |
$\require{cancel}\newcommand\degree[0]{^{\circ}} \newcommand\Ccancel[2][black]{\renewcommand\CancelColor{\color{#1}}\cancel{#2}} \newcommand{\alert}[1]{\boldsymbol{\color{magenta}{#1}}} \newcommand{\blert}[1]{\boldsymbol{\color{blue}{#1}}} \newcommand{\bluetext}[1]{\color{blue}{#1}} \delimitershortfall-1sp \newcommand\abs[1]{\left|#1\right|} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$
## SectionLinear Functions
###### Supplemental Videos
The main topics of this section are also presented in the following videos:
### SubsectionSlope-Intercept Form
Linear relationships are relationships in which the rate of change is constant.
###### Linear Equation
A linear function is a function which has a constant rate of change.
Many phenomena can be modeled using linear functions $y = f (x)$ where the equations have the form
\begin{equation*} f (x) = (\text{starting value}) + (\text{rate of change}) \cdot x. \end{equation*}
The rate of change is the slope of the graph, and the initial value is the value of $f(0)\text{.}$ On the graph of $f\text{,}$ the initial value corresponds to the point that lies on the vertical axis.
###### Vertical intercept / $y$-intercept
A vertical intercept is a point where a graph intersects the vertical axis. For a function $f\text{,}$ the vertical intercept of the graph of $f$ is the point $(0,f(0))\text{.}$ Because we often use the variable $y$ on the vertical axis, the vertical intercept is often called the $y$-intercept.
For simplicity, we often use the variable $b$ for $f(0)$ so that the vertical intercept of a function $f$ is at $(0,b)\text{.}$
###### Caution119
We will often abuse notation and say that the $y$-intercept of a graph is $b\text{.}$ What we really mean is that the $y$-intercept is the point $(0,b)\text{.}$
We can write the equation for a linear function explicitly as
\begin{equation*} f (x) = b + mx \end{equation*}
where the constant term, $b\text{,}$ is the $y$-intercept and $m\text{,}$ the coefficient of $x\text{,}$ is the slope of the line. This form for an equation of a line is called the slope-intercept form.
###### Slope-Intercept Form
If we write an equation of a linear function in the form,
\begin{equation*} f (x) = b + mx \end{equation*}
then $m$ is the slope of the line, and $b$ is the $y$-coordinate of the $y$-intercept of the line.
(You may have encountered the slope-intercept equation in the equivalent form $y = mx + b\text{.}$)
For example, consider the two linear functions and their graphs shown in Figure120 and Figure121.
$f (x) = 10 - 3x$
$g(x) = -3+2x$
Some observations:
• We can see that the $y$-intercept of each line is given by the constant term, $b\text{.}$
• By examining the table of values, we can also see why the coefficient of $x$ gives the slope of the line:
• For $f (x)\text{,}$ each time $x$ increases by $1$ unit, $y$ decreases by $3$ units.
• For $g(x)\text{,}$ each time $x$ increases by $1$ unit, $y$ increases by $2$ units.
For each graph, the coefficient of $x$ is a scale factor that tells us how many units $y$ changes for $1$ unit increase in $x\text{.}$ But that is exactly what the slope tells us about a line.
Is is also useful to introduce the term $x$-intercept at this point.
###### $x$-intercept
An $x$-intercept is a point where a graph intersects the $x$-axis. It corresponds to the point $(a,0)$ where $a$ is a value such that $f(a)=0\text{.}$
###### Example122
Francine is choosing an Internet service provider. She paid $30 for a modem, and she is considering three companies for service: • Juno charges$14.95 per month,
• ISP.com charges $12.95 per month, • and peoplepc charges$15.95 per month.
Match the graphs in Figure123 to Francine's Internet cost with each company.
Solution
Francine pays the same initial amount, $30 for the modem, under each plan. The monthly fee is the rate of change of her total cost, in dollars per month. We can write a formula for her cost under each plan. \begin{equation*} \text{Juno: } f(x) = 30 + 14.95x \end{equation*} \begin{equation*} \text{ISP.com: } g(x) = 30 + 12.95x \end{equation*} \begin{equation*} \text{peoplepc: } h(x) = 30 + 15.95x \end{equation*} The graphs of these three functions all have the same $y$-intercept, but their slopes are determined by the monthly fees. The steepest graph, III, is the one with the largest monthly fee, peoplepc, and ISP.com, which has the lowest monthly fee, has the least steep graph, I. ###### Example124 Delbert decides to use DSL for his Internet service. • Earthlink charges a$99 activation fee and $39.95 per month, • DigitalRain charges$50 for activation and $34.95 per month, • and FreeAmerica charges$149 for activation and \$34.95 per month.
1. Write a formula for Delbert's Internet costs under each plan.
2. Match Delbert's Internet cost under each company with its graph in Figure125.
Solution
1. Earthlink: $f (x) = 99 + 39.95x\text{;}$ DigitalRain: $g(x) = 50 + 34.95x\text{;}$ FreeAmerica: $h(x) = 149 + 34.95x$
2. DigitalRain: I; Earthlink: II; FreeAmerica: III
###### Note126
In the equation $f (x) = b + mx\text{,}$ we call $m$ and $b$ parameters. Their values are fixed for any particular linear equation; for example, in the equation $y = 2x + 3\text{,}$ $m = 2$ and $b = 3\text{,}$ and the variables are $x$ and $y\text{.}$ By changing the values of $m$ and $b\text{,}$ we can write the equation for any line except a vertical line (see Figure127). The collection of all linear functions $f (x) = b + mx$ is called a two-parameter family of functions.
### SubsectionSlope-Intercept Method of Graphing
Look again at the lines in Figure127: There is only one line that has a given slope and passes through a particular point. That is, the values of $m$ and $b$ determine the particular line. The value of $b$ gives us a starting point, and the value of $m$ tells us which direction to go to plot a second point. Thus, we can graph a line given in slope-intercept form without having to make a table of values.
###### Example128
1. Write the equation $4x - 3y = 6$ in slope-intercept form.
2. Graph the line by hand.
Solution
1. We solve the equation for $y$ in terms of $x\text{.}$ \begin{align*} -3y \amp =6 - 4x\\ y \amp = \frac{6 - 4x}{-3} =\frac{6}{-3}+\frac{-4x}{-3}\\ y \amp = -2+\frac{4}{3}x \end{align*}
2. We see that the slope of the line is $m = \dfrac{4}{3}\text{,}$ and its vertical intercept is $b = -2\text{.}$ We begin by plotting the $y$-intercept, $(0, -2)\text{.}$ We then use the slope to find another point on the line. We have
\begin{equation*} m = \frac{\Delta y}{\Delta x}=\frac{4}{3} \end{equation*}
so starting at $(0, -2)\text{,}$ we move $4$ units in the $y$-direction and $3$ units in the $x$-direction, to arrive at the point $(3, 2)\text{.}$ Finally, we draw the line through these two points. (See Figure129.)
###### Note130
The slope of a line is a ratio and can be written in many equivalent ways. In Example128, the slope is equal to $\dfrac{8}{6}\text{,}$ $\dfrac{12}{9}\text{,}$ and $\dfrac{-4}{-3}\text{.}$ We can use any of these fractions to locate a third point on the line as a check. If we use $m = \dfrac{\Delta y}{\Delta x}= \dfrac{-4}{-3}\text{,}$ we move down $4$ units and left $3$ units from the $y$-intercept to find the point $(-3, -6)$ on the line.
###### Slope-Intercept Method for Graphing a Line
1. Plot the point associated with the vertical intercept, $(0, b)\text{.}$
2. Use the definition of slope to find a second point on the line: Starting at the vertical intercept, move $\Delta y$ units in the $y$-direction and $\Delta x$ units in the $x$-direction. Plot a second point at this location.
3. Use an equivalent form of the slope to find a third point, and draw a line through the points.
###### Example131
1. Write the equation $2y + 3x + 4 = 0$ in slope-intercept form.
2. Use the slope-intercept method to graph the line.
Solution
1. $y=-2-\dfrac{3}{2}x$ |
# Optimization
In this section we will explore the science of optimization. Suppose that you are trying
to find a pair of numbers with a fixed sum so that the product of the two numbers is
a maximum. This is an example of an optimization problem. However, optimization
is not limited to finding a maximum. For example, consider the manufacturer who
would like to minimize his costs based on certain criteria. This is another example of
an optimization problem. As you can see, optimization can encompass finding either a
maximum or a minimum.
Optimization can be applied to a broad family of different functions . However, in
this section, we will concentrate on finding the maximums and minimums of quadratic
functions. There is a large body of real -life applications that can be modeled by quadratic
functions, so we will find that this is an excellent entry point into the study of
optimization.
Finding the Maximum or Minimum of a Quadratic Function
f(x) = −x2 + 4x + 2.
Let’s complete the square to place this quadratic function in vertex form . First, factor
out a minus sign.
Take half of the coefficient of x and square , as in [(1/2)(−4)]2 = 4. Add and subtract
this amount to keep the equation balanced .
Factor the perfect square trinomial, combine the constants at the end, and then redistribute
the minus sign to place the quadratic function in vertex form.
This is a parabola that opens downward, has been shifted 2 units to the right and 6
units upward. This places the vertex of the parabola at (2, 6), as shown in Figure 1.
Note that the maximum function value (y-value) occurs at the vertex of the parabola.
A mathematician would say that the function “attains a maximum value of 6 at x
equals 2.”
Note that 6 is greater than or equal to any other y-value (function value) that occurs
on the parabola. This gives rise to the following definition.
Figure 1. The maximum value of the
function, 6, occurs at the vertex of the
parabola, (2, 6).
Definition 1. Let c be in the domain of f. The function f is said to achieve a
maximum at x = c if f(c) ≥ f(x) for all x in the domain of f.
Next, let’s look at a quadratic function that attains a minimum on its domain.
Example 2. Find the minimum value of the quadratic function defined by the
equation
f(x) = 2x2 + 12x + 12.
Factor out a 2.
(3)
Take half of the coefficient of x and square, as in [(1/2)(6)]2 = 9. Add and subtract
this amount to keep the equation balanced.
Factor the trinomial and combine the constants, and then redistribute the 2 in the next
step .
The graph is a parabola that opens upward, shifted 3 units to the left and 6 units
downward. This places the vertex at (−3,−6), as shown in Figure 2. Note that the
minimum function value (y-value) occurs at the vertex of the parabola. A mathematician
would say that the function “attains a minimum value of −6 at x equals −3.
Figure 2. The minimum value of the
function, -6, occurs at the vertex of the
parabola, (−3,−6).
Note that −6 is less than or equal to any other y-value (function value) that occurs on
the parabola.
This last example gives rise to the following definition.
Definition 4. Let c be in the domain of f. The function f is said to achieve a
minimum at x = c if f(c) ≤ f(x) for all x in the domain of f.
A Shortcut for the Vertex
It should now be clear that the vertex of the parabola plays a crucial role when optimizing
a quadratic function. We also know that we can complete the square to find the
coordinates of the vertex. However, it would be nice if we had a quicker way of finding
the coordinates of the vertex. Let’s look at the general quadratic function
y = ax2 + bx + c
and complete the square to find the coordinates of the vertex. First, factor out the a.
Take half of the coefficient of x and square, as in [(1/2)(b/a)]2 = [b/(2a)]2 = b2/(4a2).
Add and subtract this amount to keep the equation balanced.
Factor the perfect square trinomial and make equivalent fractions for the constant terms
with a common denominator .
Finally, redistribute that a. Note how multiplying by a cancels one a in the denominator
of the constant term.
Now, here’s the key idea. The results depend upon the values of a, b, and c, but it
should be clear that the coordinates of the vertex are
The y-value of the vertex is a bit hard to memorize, but the x-value of the vertex is
easy to memorize.
Vertex Shortcut. Given the parabola represented by the quadratic function
y = ax2 + bx + c,
the x-coordinate of the vertex is given by the formula
.
Let’s test this with the quadratic function given in Example 2
Prev Next |
back to Algebra
# First Degree Equation Examples
### First Degree Equation Example 1
Question
Solve the following equation:
3 ( x - 1 ) - 2 ( 3 x - 1 ) = 2 ( 3 - x )
Solution
3(x - 1) - 2(3x - 1) = 2(3 - x)
remove parenthesis
3x - 3 - 6x + 2 = 6 - 2x
move variables to the left side of the equation and move numbers to the right side of the equation
3x - 6x + 2x = 3 - 2 + 6
-x = 7 (divide by -1 on both side of the equation)
x = -7
therefore, x = -7 is the solution of given equation.
Substitute the x = -7 into the original equation, we will find that the left side of the equation is equal to the right side of the equation. Then x = -7 is satisfy the equation. So x = -7 is the solution of the given equation.
### First Degree Equation Example 2
Question
(x - 1)/3 - (x + 1)/2 = 1
solve this equation
Solution
(x - 1)/3 - (x + 1)/2 = 1
both side of the equation multiply 6 to remove denominator
6 [(x - 1)/3 - (x + 1)/2] = 6 × 1
6 (x - 1)/3 - 6 (x + 1)/2 = 6
2 (x - 1) - 3 (x + 1) = 6
2x - 2 - 3x - 3 = 6
keep the variables in the left side of the equation and move the numbers to the right side of the equation
2x - 3x = 2 + 3 + 6
-x = 11
both sides divide by -1
x = - 11
so x = -1 is the solution of the given equation |
# Perpendicular Line Calculator
Instructions: Use this calculator to find the perpendicular line to a line you provide that passes through a given point, with all the steps shown. To that end, you need to give information to define the line, and you need to indicate a point where you want the perpendicular line to pass through.
You can define the given line by providing: (1) both the slope and the y-intercept, (2) a linear equation (ex: $$x + 3y = 2 + \frac{2}{3}x$$), (3) the slope and a point that the line passes through, or (4) two points where the line passes through. On top of that, you need to provide a point you need the perpendicular line to pass through.
Select one of the options
Type the slope $$m$$ of the line (numeric expression. Ex: 2, 1/3, etc.) =
Type the y-intercept $$n$$ of the line (numeric expression. Ex: 2, 1/3, etc.) =
Type a point $$(x*, y*)$$ where the perpendicular line passes through. Type for example $$(1, 2)$$ =
Lines are in big part determined by their slope (inclination). Horizontal lines are lines with slope equal to zero, an vertical lines are lines where the slope is undefined (negative or positive infinity).
Perpendicular lines are lines that cross forming a right angle. There is a specific condition for the slope and the perpendicular slope, whenever the slopes are defined, for the lines to be perpendicular, which is that the product of the slopes is -1.
Observe that one given line has infinite perpendicular lines to it. In order to find that one you are looking for, you need to fix a point it passes through.
## How do you find the perpendicular line of a line?
The strategy is simple. The step is to find the slope of the given line. If you are provided with the slope and intercept to define the line then you already have the slope.
Otherwise, perhaps you have two points where the line passes through, in which case you can compute the slope directly.
Ultimately, if you define your given line with an equation, you need to get that equation into the slope-intercept form, so to get the slope.
Once you have the slope of the given line, you use the formula for the perpendicular slope, by multiplying by minus one the reciprocal of the original slope.
### What is the perpendicular line of a horizontal line
The perpendicular line to a horizontal line is a vertical line.
### What is the perpendicular line of a vertical line
The perpendicular line to a vertical line is a horizontal line.
## Can you calculate the perpendicular line with no points
Once you have a line, there is not one but many (infinite) perpendicular lines to the given line. In order to identify one specific perpendicular line, you need to provide one point where the line passes through.
Typically, you will provide a point on the original line, where you want the perpendicular line to pass through.
### Example of the calculation of a perpendicular line for a given line:
Consider the line with equation $$2x + 3y = 5)$$. Find the equation of the perpendicular line that passes through $$(1, 1)$$.
Solution: We first get the slope intercept equation for the GIVEN line, if possible
We have been provided with the following equation:
$\displaystyle 2x+3y=5$
Putting $$y$$ on the left hand side and $$x$$ and the constant on the right hand side we get
$\displaystyle 3y = -2x +5$
Then, solving for $$y$$, by dividing both sides of the equation by $$3$$, the following is obtained
$\displaystyle y=-\frac{2}{3}x+\frac{5}{3}$
### Perpendicular Slope Formula
In general, the formula needed to compute the perpendicular slope, $$m_{\perp}$$, is:
$m_{\perp} = \displaystyle -\frac{1}{m}$
By plugging the value of $$m =$$ in the formula, we find that the perpendicular slope is
$m_{\perp} = \displaystyle -\frac{1}{m} = \displaystyle -\frac{1}{} = \frac{3}{2}$
#### Perpendicular Line Construction
Now, we have calculated that the perpendicular slope is $$m_{\perp} = \frac{3}{2}$$ the and we know that the perpendicular line passes through the point $$(1, 1)$$.
Hence, with the information we have, we can construct directly the point-slope form of the line, which is
$\displaystyle y - y_1 = m_{\perp} \left(x - x_1\right)$
and then plugging the known values of $$\displaystyle m_{\perp} = \frac{3}{2}$$ and $$\displaystyle \left( x_1, y_1 \right) = \left( 1, 1\right)$$, we get that
$\displaystyle y-1 = \frac{3}{2} \left(x-1\right)$
Now, we need to expand the right hand side of the equation by distributing the slope, so we get $\displaystyle y = \frac{3}{2} x + \frac{3}{2} \left(-1\right) + 1$
and simplifying we get that $\displaystyle y=\frac{3}{2}x-\frac{1}{2}$
Therefore, we conclude that the equation of the given line is $$\displaystyle y=-\frac{2}{3}x+\frac{5}{3}$$ and the equation of the perpendicular line is $$\displaystyle y=\frac{3}{2}x-\frac{1}{2}$$.
If you want to make things more direct, use this perpendicular slope calculator and use the perpendicular line formula to get directly the slope of the line that is perpendicular line.
That is for the case that you only are looking for the slope in the context of what you are doing.
Don't have a membership account?
REGISTER
Back to |
# A box with an open top is to be constructed by cutting 1-inch squares from the corners of a rectangular sheet of tin whose length is twice its width. What size sheet will produce a box having a volume of 12in^3?
May 29, 2017
$\left(\sqrt{6} + 2\right) \text{ inches" xx (2sqrt(6) + 2) " inches"~~ 4.45 " in." xx 6.90 " in.}$
#### Explanation:
Given: box length $= 2 \cdot$width; $\text{ } l = 2 w$
The box size is $1$ inch $\times w \times l$
Let the sheet size be $l + 2 \text{ inches" xx w + 2 " inches}$
When you cut the $1$ inch corners from the sheet, the size of the box length and width decreases by $2$ inches.
${V}_{b o x} = 1 \times w \times 2 w = 12$
$2 {w}^{2} = 12$
${w}^{2} = \frac{12}{2} = 6$
w = sqrt(6); " " width of the box
l = 2w = 2sqrt(6); " " length of the box
Size of the sheet:
$\left(\sqrt{6} + 2\right) \text{ inches" xx 2sqrt(6) + 2 " inches" ~~ 4.45 " in." xx 6.90 " in.}$
CHECK:
${V}_{b o x} = 1 \times \sqrt{6} \times 2 \sqrt{6} = 12 {\text{ in.}}^{3}$ |
# SAT Practice Test 6, Section 7: Questions 1 - 5
Related Topics:
More Lessons for SAT Preparation
Math Worksheets
This is for SAT in Jan 2016 or before.
The following are worked solutions for the questions in the math sections of the SAT Practice Tests found in the The Official SAT Study Guide Second Edition.
It would be best that you go through the SAT practice test questions in the Study Guide first and then look at the worked solutions for the questions that you might need assistance in. Due to copyright issues, we are not able to reproduce the questions, but we hope that the worked solutions will be helpful.
Given:
q, x and z in a word is worth 5 points
All other letters are worth 1 point each
To find:
Sum of points assigned to the word “exquisite”
Solution:
The 2 letters x and q are given 5 points each: 2 × 5 = 10
The 7 other letters are given 1 point each
Add up the points and we get 17
Given:
2x = 10 = 20
To find:
x – 5
Solution:
It is not necessary to solve for x
Notice that x – 5 is half of 2x – 10
x – 5 = 20 ÷ 2 = 10
Given:
t is an odd integer.
To find:
The expression that represents an even integer
Solution:
(A) t + 2 (t + 2 is odd)
(B) 2t + 1 (2t is even and so, 2t + 1 is odd)
(C) 3t – 2 (3t is odd and so, 3t – 2 odd)
(D) 3t + 2 (3t is odd and so, 3t + 2 odd)
(E) 5t +1 (5t will end with a 5 and it will be odd, so 5t +1 would be even)
Given:
Perimeter of triangle ABC equals the perimeter of triangle DEF
Triangle ABC is equilateral
To find:
AB
Solution:
The perimeter of triangle DEF is 4 + 8 + 9 = 21
Since triangle ABC is equilateral, AB = 21 ÷ 3 = 7
Given:
A circle graph
The store sold 900 pairs of jeans other than J,K,L,M and N
To find:
The number of pairs of jeans sold altogether
Solution:
Topic(s): Proportional word problem
The percent of jeans other than J,K,L,M and N is 20%
We can rewrite it as a proportional problem
If 20% then 900, if 100% then how many?
0.2 → 900
1 →
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
Review Problems Solutions
# Review Problems Solutions
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: heavy coffee drinker” and C be the event “having cancer of the pancreas.” Since we are testing for independence, we must compare P (H ∩ C ) with P (H )P (C ). If they are equal then the events are independent. If they are not equal then the events are not independent. 121 P (H ∩ C ) = 10,000 = .0121 P (H ) = 4,000 10,000 and P (C ) = 151 10,000 , so P (H )P (C ) = .00604. This means that P(H ∩ C) = P(H)P(C) and so H and C are NOT INDEPENDENT. Problem 9 Copykwik has four photocopy machines A, B, C, and D. The probability that a given machine will break down on a particular day is given below. Assuming independence, what is the probability on a particular day that the following will occur? P (A) = .002 P (B ) = .003 P (C ) = .001 P (D) = .002 (A) None of the machines will break down. Solution: This means A does not break down AND B does not break down AND C does not break down AND D does not break down. So we want P (Ac ∩ B c ∩ C c ∩ Dc ). Since the events A, B , C , and D are independent we have P ( Ac ∩ B c ∩ C c ∩ D c ) = P ( Ac ) P ( B c ) P ( C c ) P ( D c ) = (1 − P (A))(1 − P (B ))(1 − P (C ))(1 − P (D)) = ≈ (1 − .002)(1 − .003)(1 − .001)(1 − .002) .9920 (B) All of the machines break down. Solution: P (A ∩ B ∩ C ∩ D) = P (A)P (B )P (C )P (D) = (.002)(.003)(.001)(.002) = 1.2 × 10−11 (C) Exactly one machine breaks down. Solution: We want to find P ( A ∩ B c ∩ C c ∩ D c ) + P ( Ac ∩ B ∩ C c ∩ D c ) + P ( Ac ∩ B c ∩ C ∩ D c ) + P ( Ac ∩ B c ∩ C c ∩ D ) . P (A ∩ B c ∩ C c ∩ Dc ) = (.002)(1 − .003)(1 − .001)(1 − .002) = .001988021988 P (Ac ∩ B ∩ C c ∩ Dc ) = (1 − .002)(003)(1 − .001)(1 − .002) = .002985023988 P (Ac ∩ B c ∩ C ∩ Dc ) = (1 − .002)(1 − .003)(.001)(1 − .002) = .000993015988 P (Ac ∩ B c ∩ C c ∩ D) = (1 − .002)(1 − .003)(1 − .001)(.002) = .001988021988 So P ( A ∩ B c ∩ C c ∩ D c ) + P ( Ac ∩ B ∩ C c ∩ D c ) + P ( Ac ∩ B c ∩ C ∩ D c ) + P ( Ac ∩ B c ∩ C c ∩ D ) = .001988021988 + .002985023988 + .000993015988 + .001988021988 = .007954083952 ≈ .007954 (D) Machine...
View Full Document
## This document was uploaded on 03/17/2014 for the course MATH 1331 at Texas Tech.
Ask a homework question - tutors are online |
Lesson Video: Unit Fractions of a Quantity | Nagwa Lesson Video: Unit Fractions of a Quantity | Nagwa
Lesson Video: Unit Fractions of a Quantity Mathematics • 3rd Grade
In this video, we will learn how to find unit fractions of an amount or a quantity using visual models and explain how this relates to division by the denominator.
14:23
Video Transcript
Unit Fractions of a Quantity
In this video, we’re going to learn how to find unit fractions of an amount or a quantity using visual models to help. And we’re going to explain what this has got to do with dividing by the denominator.
Let’s start with 12 bananas and a very hungry monkey. Now, what happens if he decides to eat a quarter of the bananas? We know there are 12 bananas, so this is just like asking, what is one-quarter of 12?
Now, if you remember, our video was called unit fractions of a quantity, and one-quarter is an example of a unit fraction. This is a fraction where we’re only thinking of one of something. This has one as the top number or the numerator. So all the way through this video, we’re only going to be thinking about fractions like one-quarter, one-half, one-fifth, one-third, and so on. And the number 12 here is the quantity of bananas. One-quarter of 12 is a unit fraction of a quantity. So how many bananas is this? Let’s draw a bar model to help us.
This bar represents our 12 bananas. Now we know when a shape, a number, or a quantity is divided into quarters, it’s split up into four equal parts. And you know a quick way to do this is to split into half and then half again because half of a half is the same as a quarter. Let’s divide another bar to show quarters then, half and then half again. We’ve divided our bar into four equal parts. Can you see which part of the fraction tells us that we need to do this? It’s the bottom number, or the denominator. This shows us the number of equal parts that we need or the number we need to divide by.
So if we were being asked to find one-half of 12, we’d need to divide by two. To find one-third of 12, we’d need to divide by three and so on. The denominator is a clue that tells us what we need to do here. Now, how many bananas is each of our quarters worth? We know that half of 12 is six. And if we split each half in half, this gives us three. And this is where our times tables facts come in useful. We know that four times three equals 12. And so 12 divided by our denominator, four, equals three. And so we can say that one-quarter of 12 bananas is three bananas. This monkey has eaten three bananas.
Here’s another monkey. She’s full of energy. And can you see why? She’s already had her lunch. She’s eaten some bananas out of this group. What fraction of the bananas has she eaten? To begin with, let’s draw a ring around the bananas that she’s eaten. This is a group of three; our monkey has eaten three bananas. But what fraction is this out of the whole amount? Don’t forget when we’re thinking about fractions, we’re thinking about dividing a whole amount into parts that are the same size or equal. Let’s split up the whole amount of bananas so that they’re in groups of the same size.
Here’s another group of three bananas, a third group, a fourth group, and one more. We’ve split the whole amount into five equal groups. And we know that when we divide something into five equal parts, each part is one-fifth. The fraction of bananas that the monkey’s eaten is one-fifth. Let’s show this is a fraction of a quantity. One-fifth of the whole amount of bananas, which we haven’t counted yet but we can do so in a moment, equals three bananas.
Let’s count in threes to find out what our missing number is. There are three, six, nine, 12, 15 bananas in the whole amount. One-fifth of 15 is three. And remember, we can check this because we can take the whole amount, which is 15, and divide it by the denominator in our unit fraction. 15 divided by five equals three. All of this tells us that the fraction of the bananas that the monkey’s eaten is one-fifth.
Do you think you’re ready to answer some questions now where you put into practice what you’ve learned? Let’s give it a go.
What fraction of the circles is shaded?
Let’s split up the rest of our group so that everything’s in equal parts. Here’s another group of two, so that’s two equal parts, three, four, five, six, seven. We’ve divided the whole amount into seven equal parts. We know this because each part contains two circles. And because there are seven equal parts, we can say each part is worth one-seventh. And because only one of our parts is shaded, we can say one-seventh is shaded. There are two, four, six, eight, 10, 12, 14 circles altogether. And we can use this to check our answer is correct.
To find one-seventh of 14, we can start with 14 and divide by the denominator, which is seven. We know there are two sevens in 14, so 14 divided by seven equals two. One-seventh of 14 circles is two circles. And of course, this matches the two circles that are shaded. It looks like our fraction is correct. The fraction of the circles that’s shaded is one-seventh.
Fill in the blank: one-quarter of 20 equals what.
In the picture, we can see a group of 20 cakes. And to answer the problem, we need to find a fraction of this quantity. We need to find one-quarter of 20. What do we know about the fraction one-quarter? Well, firstly, the numerator is one. This means it’s what we call a unit fraction. In other words, we only need to find one-quarter. We don’t need to find two-quarters or three-quarters, only one. And the bottom number or the denominator in one-quarter is a four. This tells us that we need to split the whole amount into four equal parts. That’s what a quarter is; it’s one out of four equal parts. So to find out what one-quarter of 20 is, we’re going to have to split up 20 into four equal parts.
Now, there are two ways we could do this. Firstly, we could think about the number 20 itself, and we could work out the answer to 20 divided by four. Or we could use the picture of the cakes to help. We could divide this into four equal groups. Let’s try both ways, and if we get the same answer, we know we’ve got it right. Now there’s a quick way we can split things into four. We start by finding a half, and then if we find half again, that gives us four equal parts. We know that half of 20 is 10 and then half of 10 is five. There are four fives in 20. 20 divided by four equals five. It looks like one-quarter of 20 is five, but let’s just check using the picture.
First, we can split the group in half. Now, if we split each half in half by going down the middle, we can see we’re going to end up with some half cupcakes, which we don’t really want. It makes it a bit more tricky to count. So let’s just split each half in half by drawing lines across. There we go. We’ve split the whole amount into four equal groups, and there are one, two, three, four, five cupcakes in each group. We were right. One-quarter of 20 equals five.
Fill in the blank: one-half of eight equals what.
In this question, we need to find a fraction of an amount, one-half of eight. Now we’ve got a picture to help us. There are eight cupcakes. Now, how do we find half of an amount? The clue is in the fraction itself. The top number or the numerator tells us how many parts we’re looking for. And in this case, it’s one. We’re looking for one-half. And the bottom number or the denominator shows us how many equal parts the whole has been split into. And from what we know about fractions already, we know this is true, don’t we? If we divide a shape or a number or a quantity by two, we split it into two equal parts or half. So half of the whole amount, eight, is the same as eight divided by two.
What do we get if we split eight into two equal parts? We get four and another four. Eight divided by two equals four. And we know this because two times four is eight. Half of eight equals four.
What is one-fifth of five?
In this question, we need to find a fraction of an amount. That amount is five. Let’s model it, shall we? Let’s use pufferfish. Now we can see that there are five fish in the whole amount. But what do we do to find one-fifth of this number? The number one, the numerator in this fraction, tells us that it’s a unit fraction. We’re only looking for one part, not two-fifths or three-fifths or four-fifths, only one-fifth. And the number five, that’s the denominator, tells us how many equal parts to split the whole amount into.
To find a fifth, we divide the whole amount into five equal parts. Each part is worth one-fifth. Now we know that five ones make five. So if we divide five by itself, the answer is one. We found the answer here by dividing five into five. One-fifth of five equals one.
There were 15 birds on a tree, but a third of them flew away. Find the number of birds that flew away.
To begin with in this problem, we’re told about a group of birds. There were 15 of them on a tree. Let’s model our 15 birds using plastic counters. Now, we’re told that some of these birds flew away. And fortunately, we’re not told the number of birds that flew away. This is what we need to find. Instead, we’re told the fraction of the whole amount. This is a third. Now this is a third written in words. But do you remember how to write it using digits? The numerator is one. This shows us that we’re only thinking about one-third and not two-thirds or more. And then the denominator is three. This shows us the number of equal parts that we need to split the whole into.
So to find one-third of 15, we can divide 15 by the denominator, three. If we split 15 into three equal groups, how many will there be in each group? Perhaps you can think of a times tables fact that could help us here. We know that three times five equals 15. 15 divided by three equals five. And we can show this using our counters. We’ve split the whole amount into three equal groups, and each of the three groups, that’s each third, is worth five. So if there were 15 birds on a tree, but a third of them flew away, the number of bird that flew away is five.
Now what have we learned in this video? We have learned how to find unit fractions of amounts or quantities using models to help. We have also explained how we can divide by the denominator.
Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions |
# Holly checked her answer to a division problem by estimating the quotient. Which is the best estimation of the quotient?23.564 divided by 5.974567I need help quickly.
Explanation
Quotient=division
23.564/5.97=3.947
3.947=4
## Related Questions
Simplify 3-8 divided by 3-2
The simplified result of the expression 3-8 divided by 3-2 is -5 .
Let's simplify the expression step by step :
Given expression: (3 - 8) / (3 - 2)
Step 1: Evaluate the numerator and denominator separately:
A fraction is considered to be simplified when it is expressed in the lowest term. That means the only common divisor between the numerator and denominator is .
Numerator: 3 - 8 = -5
Denominator: 3 - 2 = 1
Step 2: Divide the numerator by the denominator:
-5 / 1 = -5
So, the simplified result of the expression is -5 .
brainly.com/question/31298232
#SPJ3
-5
Step-by-step explanation:
Sharon has a new beaded necklace. 72 out of the 80 beads on the necklace are blue. Whatpercentage of beads on Sharon's necklace are blue?
90%
Step-by-step explanation:
Step 1:
72/80 = x /100 Proportion
Step 2:
80x = 7200 Multiply
Step 3:
x = 7200 ÷ 80 Divide
90%
Hope This Helps :)
Sharon bought a necklace with 90% blue beads.
Step-by-step explanation:
We are given that Sharon's necklace has 80 beads. We are also given that of those 80 beads, 72 of them are blue.
Therefore, we can set up a ratio of blue beads to total beads in order to find out the relationship.
Then, using the fraction we receive as a result, we can convert this to a decimal.
After we complete this calculation, we can multiply our decimal by 100 in order to obtain the relationship in percentage form.
What is the surface area of the pyramid 8in 8 in 6in.
mhbdjhdfvvbjd
Step-by-step explanation:
hbgjtrhkbgb
384 inches
Step-by-step explanation:
A=lwh
The Range of both functions will be ?A)the set of natural numbers
B)the set of integers
C)all real numbers
D)the set of whole numbers
E)the set of rational numbers
bb
b
b
b
b
b
b
b
b
b
b
b
b
b
it is b
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Step-by-step explanation:
Raquel kicks a football. The height in feet canbe modeled by the equation
h(x) = -16x2 + 64x.
How long in seconds before the ball reaches
its maximum height? Show work.
What is the maximum height of the ball? Show work.
Step-by-step explanation:
16x2 + 64x.
0
Step-by-step explanation:
i agree with bigmack0
A rectangular box without a lid is to be made from 48 m2 of cardboard. Find the maximum volume of such a box. SOLUTION We let x, y, and z to be the length, width, and height, respectively, of the box in meters. Then we wish to maximize V
The maximum volume of such box is 32m^3
V = x×y×z = 32 m^3
Step-by-step explanation:
Given;
Total surface area S = 48m^2
Volume of a rectangular box V = length×width×height
V = xyz ......1
Total surface area of a rectangular box without a lid is
S = xy + 2xz + 2yz = 48 .....2
To be able to maximize the volume, we need to reduce the number of variables.
Let assume the rectangular box has a square base,that means; length = width
x = y
Substituting y with x in equation 1 and 2;
V = x^2(z) ....3
x^2 + 4xz = 48 .....4
Making z the subject of formula in equation 4
4xz = 48 - x^2
z = (48 - x^2)/4x .......5
To be able to maximize V, we need to reduce the number of variables to 1, by substituting equation 5 into equation 3
V = x^2 × (48 - x^2)/4x
V = (48x - x^3)/4
differentiating V with respect to x;
V' = (48 - 3x^2)/4
At the maximum point V' = 0
V' = (48 - 3x^2)/4 = 0
Solving for x;
3x^2 = 48
x = √(48/3)
x = √(16)
x = 4
Since x = y
y = 4
From equation 5;
z = (48 - x^2)/4x
z = (48 - 4^2)/4(4)
z = 32/16
z = 2
The maximum volume can be derived by substituting x,y,z into equation 1;
V = xyz = 4×4×2 = 32 m^3 |
# Determine the greatest number by which 420, 458, and 569 should be divided, resulting in remainders of 6, 8, and 11 respectively?
Determine the greatest number by which 420, 458, and 569 should be divided, resulting in remainders of 6, 8, and 11 respectively?
• A. 18
• B. 21
• C. 17
• D. 24
Given: 420, 458, and 569 are three numbers leaving the remainder 6, 8, and 11 on dividing with the greatest number
• 420 − 6 = 414
• 458 − 8 = 450
• 569 − 11 = 558
Now, let’s find the HCF(414,450,558)
To find the highest common factor (HCF) of 414, 450, and 558, we can use the prime factorization method.
1. Find the prime factorization of each number:
• 414 = 2×3×3×23
• 450 = 2×3×3×5×5
• 558 = 2×3×7×11
2. Identify the common prime factors and their lowest powers:
• Common factors: 2×3×3=18
So, the HCF of 414, 450, and 558 is 18. |
# MA 15200 Lesson 32, Section 5.1
```MA 15200 Lesson 32, Section 5.1
A linear equation with two variables has infinite solutions (many ordered pairs, example
2 x 3 y 6 ) However, if a second equation is also considered, and we want to know
what ordered pairs they may have in common, how many solutions may there be? When
two linear equations are considered together, it is known as a system of linear equations
or a linear system. The solution(s) of the system is/are any ordered pair(s) they have
in common.
Consider graphing two lines. How many points may they have in common?
As demonstrated in the graphs above, there are 3 situations.
1. The graphs may intersect at one point. If the lines of the equations intersect, the
solution is one ordered pair.
2. The graphs may intersect at an infinite number of points. If the lines of the
equations form the same line, the solution is an infinite number of ordered pairs.
3. The graphs may not intersect. If the lines of the equations do not intersect, there
is no solution.
Ex 1: Given the ordered pair, determine if it is a solution of the shown system.
2 x 5 y 16
a)
(2, 4)
3 x y 2
5x 2 y 1
b)
10 x 5 y 1
3
,1
5
1
There are three methods that may be used to find the solution of a system of linear
equations. Graphing the lines could be done. Two more algebraic methods are covered
in this lesson.
1. The Graphing Method-Graph each line and identify any common point (ordered
pair), or determine if the lines are the same or if the lines are parallel.
2. The Substitution Method
For the next example we will use the graphing method. The graphing method is not
really the best method. Graphing is not very accurate. And, if the solution has
coordinates that are fractions or decimals, the coordinates of the point of intersection (if
there is one) would be difficult to determine. There is a good study tip on using the
graphing method on page 483 of the textbook.
3x 2 y 2
Ex 1: Find the solution to the system
by graphing.
2 x 3 y 16
Substitution Method:
1. Solve either of the equations for one variable in terms of the other.
2. Substitute the expression found in step 1 for that variable in the other equation.
This will result in an equation with only one variable.
3. Solve the equation for that variable.
4. Back-substitute the value found in step 3 into one of the original equations and
solve for the remaining variable.
5. Check and write the solution as an ordered pair or identify x and y.
2
Ex 2: Solve each system of linear equations using the substitution method.
2 x 3 y 0
a)
y 3x 11
b)
Note: It is easier to use
the substitution method
when a variable of one of
the equations has a
coefficient of 1 or -1.
This avoids the use of
fractions.
4x 5 y 4
8 x 15 y 3
Examine these examples.
x 3y 4
A
1
4
y 3 x 3
4
1
x 3 x 4
3
3
4
1
2x 6 x 8
3
3
2x 2x 8 8
88
B
y x 3
x y 7
x ( x 3) 7
xx3 7
37
You will notice that both variables 'dropped out' in both example A and example B.
3
A
In example A, a true statement was the result (8 = 8). This indicates that there are
several solutions, because the statement is true. Graphically, these are the same
line. There are an infinite number of solutions, every point on the line. The
1
4
general solution may be written as a set this way: ( x, y ) | y x .
3
3
Although there are an infinite number of solutions in such a system, this does not mean
that any ordered pair is a solution. It must be an ordered pair that satisfies the equation of
the line.
B
In example B, the result was a false statement (3 = 7). Because the result was
false, there is no ordered pair that is common to both equations. The system has
no solution. Graphically, these lines would be parallel.
This system is called an inconsistent system.
3x y 12
Ex 3: Solve the system
.
y 3x 12
Addition Method: As with the substitution method, the goal of the addition method is to
eliminate one of the variables.
1. Write both equations in general form ( Ax By C ) .
2. If necessary, multiply one or both of the equations by appropriate nonzero
numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0.
3. Add the equations in step 2. The sum will be an equation in one variable.
4. Solve the equation.
5. Back-substitute the value you found in step 4 into either of the given equations
and solve for the remaining variable.
6. Check. Write the solution as an ordered pair or give values of x and y.
4
Ex 4: Solve each system.
2 x 3 y 8
a)
5x y 3
b)
3x 3 y y 9
5( x y ) 15
c)
3x 4(2 y )
3x 6 4 y 0
5
Ex 5: Three times a first number decreased by 5 times a second number is 9 . The first
number is one less than the second number. Find the numbers.
Ex 6: For a linear function f ( x) mx b, f (2) 1 and f (5) 8 . Find the values of m
and b.
6
``` |
# The Apple Method
## What is the Apple Method?
The Apple Method is a method for solving algebra problems. An apple is used to make a clever algebraic substitution.
## Why Apple?
A few reasons:
1. When you use the Apple Method, you can box what you are substituting with the apple. When you use $x$ as a substitution, instead of actually boxing it, you are just crossing it out.
2. Apples are easier to draw.
3. Apples are good for you.
4. An Apple a Day Keeps the Doctor Away.
## LaTeX code for apple
$(^{^(})$, or if you want some color, $\textcolor{red}{(\textcolor{green}{^{^(}})}$
## Examples
1. Evaluate: $$\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$$
$\emph{Solution:}$
If we set $\textcolor{red}{(\textcolor{green}{^{^(}})}=\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$, we can see that $\textcolor{red}{(\textcolor{green}{^{^(}})}^2= 6+\textcolor{red}{(\textcolor{green}{^{^(}})}$.
Solving, we get $\textcolor{red}{(\textcolor{green}{^{^(}})}=\boxed{3}$
2. If $$\sqrt{x\cdot\sqrt{x\cdot\sqrt{x\cdots}}} = 5$$
Find x.
$\emph{Solution:}$
If we set $\sqrt{x\cdot\sqrt{x\cdot\sqrt{x\cdots}}}$ equal to $\textcolor{red}{(\textcolor{green}{^{^(}})},$ we get $\textcolor{red}{(\textcolor{green}{^{^(}})} = 5$ and $\textcolor{red}{(\textcolor{green}{^{^(}})}^2 = x \cdot \textcolor{red}{(\textcolor{green}{^{^(}})} = 25.$
Simplifying, we find $\textcolor{red}{(\textcolor{green}{^{^(}})} = x,$ so $x = \boxed{5}$
3. Evaluate: $$\frac{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\ldots}{\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\ldots}$$
$\emph{Solution:}$
Let $\textcolor{red}{(\textcolor{green}{^{^(}})}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$. Note that $\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\cdots = \left( \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots \right) - \left( \frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\cdots \right) = \textcolor{red}{(\textcolor{green}{^{^(}})} - \frac{1}{2^2}\cdot\textcolor{red}{(\textcolor{green}{^{^(}})} = \frac{3}{4}\cdot\textcolor{red}{(\textcolor{green}{^{^(}})}.$
Thus, the answer is $\frac{\textcolor{red}{(\textcolor{green}{^{^(}})}}{\frac34\cdot\textcolor{red}{(\textcolor{green}{^{^(}})}}=\boxed{\frac34}.$
## Extensions
### The :) Method
When more than one variable is needed, pears, bananas, stars, and smiley faces are usually used. |
# Pre-Algebra : Distributive Property
## Example Questions
← Previous 1 3
### Example Question #1 : High School Math
Simplify the expression.
Explanation:
Use the distributive property to multiply each term of the polynomial by . Be careful to distribute the negative as well.
### Example Question #1 : Distributive Property
Find the value of .
-6
4
-2
6
2
2
Explanation:
We can seperate the problem into two steps:
We then combine the two parts:
### Example Question #1 : Distributive Property
Distribute .
Explanation:
When distributing with negative numbers we must remember to distribute the negative to all of the variables in the parentheses.
Distribute the through the parentheses by multiplying it with each object in the parentheses to get .
Perform the multiplication remembering the positive/negative rules to get , our answer.
### Example Question #4 : High School Math
Simplify the expression.
Explanation:
Multiply the mononomial by each term in the binomial, using the distributive property.
### Example Question #2 : Distributive Property
Simplify the expression.
Explanation:
Use the distributive property to multiply each term by .
Simplify.
### Example Question #28 : Identities And Properties
Distribute:
Explanation:
When distributing with negative numbers we must remember to distribute the negative to all of the terms in the parentheses.
Remember, a negative multiplied by a negative is positive, and a negative multiplied by a positive number is negative.
Distribute the through the parentheses:
Perform the multiplication, remembering the positive/negative rules:
### Example Question #3 : Distributive Property
Which of the following is equivalent to ?
Explanation:
We need to distribute -3 by multiplying both terms inside the parentheses by -3.:
.
Now we can multiply and simplify. Remember that multiplying two negative numbers results in a positive number:
### Example Question #5 : Distributive Property
Expand:
Explanation:
Use the distributive property. Do not forget that the negative sign needs to be distributed as well!
### Example Question #4 : Distributive Property
Distribute:
Explanation:
Remember that a negative multiplied by a negative is positive, and a negative multiplied by a positive is negative.
Distribute the through the parentheses by multiplying it by each of the two terms:
### Example Question #5 : Distributive Property
Expand:
Explanation:
Distribute the by multiplying it by each term inside the parentheses.
and
Therefore, 5(2 + y) = 10 + 5y.
← Previous 1 3 |
## Properties of Multiplication
### Properties of Multiplication: Learn
There are four mathematical properties which involve mutliplication. The properties are the commutative, associative, identity and distributive properties.
Commutative Property: When two numbers are multiplied together, the product is the same regardless of the order of the multiplicands.
For example 4 * 2 = 2 * 4
To remember the commutative property, it might be helpful to think about the word commute which means to switch places between home and work (or home and school). In the example above, you can see the 4 and the 2 commuting or switching places.
Associative Property: When three or more numbers are multiplied, the product is the same regardless of the grouping of the factors.
For example (2 * 3) * 4 = 2 * (3 * 4)
To remember the associative property, it might be helpful to think about the word associate, which as a verb means to interact with a group (maybe you associate with a certain group of friends!). The parentheses are grouping operators, that is, they form groups of numbers and operations. You can see in the example above, the 3 can associate with either the 2 or the 4, but the value of each side is still a product of 24.
Identity Property: The product of any number and one is that number.
For example 5 * 1 = 5.
To remember the identity property, it might be helpful to think of it as a question and answer: What number can I multiply by so that the value is not changed? One. In the example above, the 5 gets to keep its identity because multiplying it by one does not change its value.
Distributive Property: The sum of two numbers times a third number is equal to the sum of each addend times the third number.
For example 4 * (6 + 3) = 4*6 + 4*3
The distributive property is the only property that combines multiplication and addition. That makes it very important!
### Properties of Multiplication: Practice
#### What property is illustrated?
00:00
Press the Start Button To Begin
You have 0 correct and 0 incorrect.
This is 0 percent correct.
### Play
Game Name Description Best Score
How many correct answers can you get in 60 seconds? 0
Extra time is awarded for each correct answer.
Play longer by getting more correct.
0
How fast can you get 20 more correct answers than wrong answers? 999 |
# How do you find two consecutive positive integers if the sum of their squares is 3445?
Apr 10, 2018
$\left(42 , 41\right)$
#### Explanation:
First, translate the words into an equation:
${x}^{2} + {y}^{2} = 3445$
Since you know that $x$ and $y$ are consecutive, you know that one of the variables is larger by $1$ compared to the other.
$x = y - 1$
Via substitution, you get the new equation
${x}^{2} + {\left(x - 1\right)}^{2} = 3445$
Expand the second term on the left-hand side
${x}^{2} + {x}^{2} - 2 x + 1 = 3445$
Simplification gets you a quadratic equation $2 {x}^{2} - 2 x - 3444 = 0$.
This can be factored into $2 \left(x - 42\right) \left(x + 41\right)$
By the zero product rule, $x = 42 , x = - 41$
Since they are positive integers, $x = 42$
Because they are consecutive and $x$ is larger than $y$, $y = 41$
The final answer is $\left(42 , 41\right)$ |
Math Calculators, Lessons and Formulas
It is time to solve your math problem
mathportal.org
Linear Algebra - Matrices: (lesson 3 of 3)
## Inverse of a matrix by Gauss-Jordan elimination
To find the inverse of matrix $A$, using Gauss-Jordan elimination, it must be found the sequence of elementary row operations that reduces $A$ to the identity and, then, the same operations on $I_n$ must be performed to obtain $A^{-1}$.
### Inverse of 2 $\times$ 2 matrices
Example 1: Find the inverse of
$A = \left[ {\begin{array}{*{20}{c}} 1&3\\ 2&7 \end{array}} \right]$
Solution:
Step 1: Adjoin the identity matrix to the right side of $A$:
$A = \left[ {\begin{array}{*{20}{c}} 1&3\\ 2&7 \end{array}\left| {\begin{array}{*{20}{c}} \color{blue}{1}&\color{blue}{0}\\ \color{blue}{0}&\color{blue}{1} \end{array}} \right.} \right]$
Step 2: Apply row operations to this matrix until the left side is reduced to $I$. The computations are:
\begin{aligned} &\left[ {\begin{array}{*{20}{c}} \color{red}{1}&\color{red}{3}\\ {2 - \color{blue}{2} \cdot \color{red}{1}}&{7 - \color{blue}{2} \cdot \color{red}{3}} \end{array}\left| {\begin{array}{*{20}{c}} \color{red}{1}&\color{red}{0}\\ {0 - \color{blue}{2} \cdot \color{red}{1}}&{1 - \color{blue}{2} \cdot \color{red}{0}} \end{array}} \right.} \right] \ \ Row2 = Row2 - \color{blue}{2} \cdot \color{red}{Row1} \\ &\left[ {\begin{array}{*{20}{c}} 1&3\\ 0&1 \end{array}\left| {\begin{array}{*{20}{c}} 1&0\\ { - 2}&1 \end{array}} \right.} \right] \\ &\left[ {\begin{array}{*{20}{c}} {1 - \color{blue}{3} \cdot \color{red}{0}}&{3 - \color{blue}{3} \cdot \color{red}{1}}\\ \color{red}{0}&\color{red}{1} \end{array}\left| {\begin{array}{*{20}{c}} {1 - \color{blue}{3} \cdot \color{red}{( - 2)}}&{0 - \color{blue}{3} \cdot \color{red}{1}}\\ \color{red}{ - 2}&\color{red}{1} \end{array}} \right.} \right] \ \ Row1 = Row1 - \color{blue}{3} \cdot \color{red}{Row2} \\ &\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}\left| {\begin{array}{*{20}{c}} 7&{ - 3}\\ { - 2}&1 \end{array}} \right.} \right] \end{aligned}
Step 3: Conclusion: The inverse matrix is:
$A^{-1} = \left[ {\begin{array}{*{20}{c}} 7&{-3}\\ {-2}&1 \end{array}} \right]$
### Not invertible matrix
If $A$ is not invertible, then, a zero row will show up on the left side.
Example 2: Find the inverse of
$A = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}\\ { - 2}&6 \end{array}} \right]$
Solution:
Step 1: Adjoin the identity matrix to the right side of A:
$\left[ {\begin{array}{*{20}{c}} 1&{ - 3}\\ { - 2}&6 \end{array}\left| {\begin{array}{*{20}{c}} \color{blue}{1}&\color{blue}{0}\\ \color{blue}{0}&\color{blue}{1} \end{array}} \right.} \right]$
Step 2: Apply row operations
$\left[ {\begin{array}{*{20}{c}} \color{red}{1}&\color{red}{ - 3}\\ { - 2 + \color{blue}{2} \cdot \color{red}{1}}&{6 + \color{blue}{2} \cdot \color{red}{( - 3)}} \end{array}\left| {\begin{array}{*{20}{c}} \color{red}{1}&\color{red}{0}\\ {0 + \color{blue}{2} \cdot \color{red}{1}}&{1 + \color{blue}{2} \cdot \color{red}{0}} \end{array}} \right.} \right]Row2 = Row2 + \color{red}{2} \cdot \color{blue}{Row1}$
$\left[ {\begin{array}{*{20}{c}} 1&{ - 3}\\ \color{red}{0}&\color{red}{0} \end{array}\left| {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right.} \right]_{\color{red}{\leftarrow ZERO \ \ ROW}}$
Step 3: Conclusion: This matrix is not invertible.
### Inverse of 3 $\times$ 3 matrices
Example 1: Find the inverse of
$A = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 2&5&3\\ 1&0&8 \end{array}} \right]$
Solution:
Step 1: Adjoin the identity matrix to the right side of A:
$\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 2&5&3\\ 1&0&8 \end{array}\left| {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right.} \right]$
Step 2: Apply row operations to this matrix until the left side is reduced to I. The computations are:
$\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 2&5&3\\ 1&0&8 \end{array}\left| {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right.} \right]\mathop { - - - - - - - \to }\limits_{R3 = R3 \color{red}{-} R1}^{R2 = R2 - \color{blue}{2} \cdot R1} \left[ {\begin{array}{*{20}{c}} 1&2&3\\ {2 - \color{blue}{2} \cdot 1}&{5 - \color{blue}{2} \cdot 2}&{3 - \color{blue}{2} \cdot 3}\\ {1 \color{red}{-} 1}&{0 \color{red}{-} 2}&{8 \color{red}{-} 3} \end{array}\left| {\begin{array}{*{20}{c}} 1&0&0\\ {0 - \color{blue}{2} \cdot 1}&{1 - \color{blue}{2} \cdot 0}&{0 - \color{blue}{2} \cdot 0}\\ {0 \color{red}{-} 1}&{0 \color{red}{-} 0}&{1 \color{red}{-} 0} \end{array}} \right.} \right]$
$\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 0&1&{ - 3}\\ 0&{ - 2}&5 \end{array}\left| {\begin{array}{*{20}{c}} 1&0&0\\ { - 2}&1&0\\ { - 1}&0&1 \end{array}} \right.} \right]\mathop { - - - - - - - \to }\limits_{R3 = R3 \color{blue}{+ 2} \cdot R2}^{} \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 0&1&{ - 3}\\ {0 \color{blue}{+ 2} \cdot 0}&{ - 2 \color{blue}{+ 2} \cdot 1}&{5 \color{blue}{+ 2} \cdot ( - 3)} \end{array}\left| {\begin{array}{*{20}{c}} 1&0&0\\ { - 2}&1&0\\ { - 1 \color{blue}{+ 2} \cdot ( - 2)}&{0 \color{blue}{+ 2} \cdot 1}&{1 \color{blue}{+ 2} \cdot 0} \end{array}} \right.} \right]$
$\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 0&1&{ - 3}\\ 0&0&{ - 1} \end{array}\left| {\begin{array}{*{20}{c}} 1&0&0\\ { - 2}&1&0\\ { - 5}&2&1 \end{array}} \right.} \right]\mathop { - - - - - - - \to }\limits_{R3 = - 1 \cdot R3}^{} \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 0&1&{ - 3}\\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 1&0&0\\ { - 2}&1&0\\ 5&{ - 2}&{ - 1} \end{array}} \right.} \right]$
$\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 0&1&{ - 3}\\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 1&0&0\\ { - 2}&1&0\\ { - 5}&{ - 2}&1 \end{array}} \right.} \right]\mathop { - - - - - - - \to }\limits_{R2 = R2 + 3 \cdot R3}^{R1 = R1 - 3 \cdot R3} \left[ {\begin{array}{*{20}{c}} 1&2&0\\ 0&1&0\\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} { - 14}&6&3\\ {13}&{ - 5}&{ - 3}\\ 5&{ - 2}&{ - 1} \end{array}} \right.} \right]$
$\left[ {\begin{array}{*{20}{c}} 1&2&0\\ 0&1&0\\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} { - 14}&6&3\\ {13}&{ - 5}&{ - 3}\\ 5&{ - 2}&{ - 1} \end{array}} \right.} \right]\mathop { - - - - - - - \to }\limits_{}^{R1 = R1 - 2 \cdot R2} \left[ {\begin{array}{*{20}{c}} 1&1&0\\ 0&1&0\\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} { - 40}&{16}&9\\ {13}&{ - 5}&{ - 3}\\ 5&{ - 2}&{ - 1} \end{array}} \right.} \right]$
Step 3: Conclusion: The inverse matrix is:
$A^{ - 1} = \left[ {\begin{array}{*{20}{c}} { - 40}&{16}&9\\ {13}&{ - 5}&{ - 3}\\ 5&{ - 2}&{ - 1} \end{array}} \right]$ |
# What is 51/348 as a decimal?
## Solution and how to convert 51 / 348 into a decimal
51 / 348 = 0.147
To convert 51/348 into 0.147, a student must understand why and how. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. If we need to convert a fraction quickly, let's find out how and when we should.
## 51/348 is 51 divided by 348
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 51 divided by 348. Now we divide 51 (the numerator) into 348 (the denominator) to discover how many whole parts we have. This is our equation:
### Numerator: 51
• Numerators are the number of parts to the equation, showed above the vinculum or fraction bar. Any value greater than fifty will be more difficult to covert to a decimal. 51 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. Let's look at the fraction's denominator 348.
### Denominator: 348
• Denominators are the total numerical value for the fraction and are located below the fraction line or vinculum. Larger values over fifty like 348 makes conversion to decimals tougher. And it is nice having an even denominator like 348. 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 51/348 to 0.147
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 348 \enclose{longdiv}{ 51 }$$
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: Extend your division problem
$$\require{enclose} 00. \\ 348 \enclose{longdiv}{ 51.0 }$$
Uh oh. 348 cannot be divided into 51. Place a decimal point in your answer and add a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 348 into 51 + 0 or 510.
### Step 3: Solve for how many whole groups you can divide 348 into 510
$$\require{enclose} 00.1 \\ 348 \enclose{longdiv}{ 51.0 }$$
We can now pull 348 whole groups from the equation. Multiple this number by our furthest left number, 348, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.1 \\ 348 \enclose{longdiv}{ 51.0 } \\ \underline{ 348 \phantom{00} } \\ 162 \phantom{0}$$
If your remainder is zero, that's it! If you have a remainder over 348, go back. Your solution will need a bit of adjustment. If you have a number less than 348, continue!
### 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.
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 51/348 into a decimal
Dining - We don't give a tip of 51/348 of the bill (technically we do, but that sounds weird doesn't it?). We give a 14% tip or 0.147 of the entire bill.
### When to convert 0.147 to 51/348 as a fraction
Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches.
### Practice Decimal Conversion with your Classroom
• If 51/348 = 0.147 what would it be as a percentage?
• What is 1 + 51/348 in decimal form?
• What is 1 - 51/348 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.147 + 1/2? |
What is the mode of 7?
What is the mode in math
the value that occurs the most
The mode is the value that occurs the most. It is sometimes useful to rewrite the numbers in order, to spot which number is listed most often.
What is the mean and mode
The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. The median is the middle value when a data set is ordered from least to greatest. The mode is the number that occurs most often in a data set.
What does median mean in math
the middle value
The median is the middle value in a set of data. First, organize and order the data from smallest to largest. To find the midpoint value, divide the number of observations by two. If there are an odd number of observations, round that number up, and the value in that position is the median.
What does mode mean in research
the most frequently occurring value
The mode or modal value of a data set is the most frequently occurring value. It's a measure of central tendency that tells you the most popular choice or most common characteristic of your sample.
What is the mode of 6
Example: In {6, 3, 9, 6, 6, 5, 9, 3} the Mode is 6, as it occurs most often.
What if there is 2 modes
So we have two modes. And the two modes are the ones that repeat three times we would have one and three we can also have a data set with more than two modes.
How do you find mode
The mode is the number that appears the most.To find the mode, order the numbers lowest to highest and see which number appears the most often.Eg 3, 3, 6, 13, 100 = 3.The mode is 3.
What is mode formula
In ungrouped data, we can find mode just by arranging the data in ascending and descending order and then finding the value which occurs most frequently. In grouped data we can find the mode by using the following formula, Mode = L + (f 1– f 0/2f 1– f 0– f 2 ) h.
What does median mean in mean
The middle number
Median: The middle number; found by ordering all data points and picking out the one in the middle (or if there are two middle numbers, taking the mean of those two numbers). Example: The median of 4, 1, and 7 is 4 because when the numbers are put in order (1 , 4, 7) , the number 4 is in the middle.
What are the 4 types of mode
The different types of Mode are Unimodal, Bimodal, Trimodal, and Multimodal.
What if there are 2 modes in a set of data
The modal class shouldn't be used as a measure of central tendency, but finding two modes gives us an indication that there could be two distinct groups in the data that should be analyzed separately.
What is mode of 10
In groups of 10, the "20s" appear most often, so we could choose 25 (the middle of the 20s group) as the mode. You could use different groupings and get a different answer. Grouping also helps to find what the typical values are when the real world messes things up!
How do you find the mode of Grade 7
And we get 3 times 1 4 times 2 1 times 3. And 2 times 4 therefore more of this. Data is 2 because 2 occurs more frequently than other observations.
Can there be 4 modes
unimodal, with one mode, bimodal, with two modes, trimodal, with three modes, or. multimodal, with four or more modes.
Can you have 3 modes
A set of numbers with two modes is bimodal, a set of numbers with three modes is trimodal, and any set of numbers with more than one mode is multimodal. When scientists or statisticians talk about the modal observation, they are referring to the most common observation.
How do you find the mode of 4
Mode: The most frequent number—that is, the number that occurs the highest number of times. Example: The mode of {4 , 2, 4, 3, 2, 2} is 2 because it occurs three times, which is more than any other number.
How do you find the mode of Class 7
Together will be very time consuming. And we can make mistakes too therefore to find the mode of such large data we will use telemarks. And then calculate the frequency.
How do I calculate my mode
Calculating the mode is fairly straightforward. Place all numbers in a given set in order; this can be from lowest to highest or highest to lowest, and then count how many times each number appears in the set. The one that appears the most is the mode.
What is the median between 1 and 10
Here we have to find the median of the first 10 natural numbers. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Where the number of terms is in even. Therefore, the median of the first 10 natural numbers is 5.5.
Is mode the highest number
Mode: The most frequent number—that is, the number that occurs the highest number of times. Example: The mode of {4 , 2, 4, 3, 2, 2} is 2 because it occurs three times, which is more than any other number.
Are there more than 7 modes
There are seven main types of musical mode in Western music: Ionian, Dorian, Phrygian, Lydian, Mixolydian, Aeolian, and Locrian.
What mode has a flat 7
Mixolydian Mode
The Mixolydian Mode — the 5th mode of the major scale
The flat 7 in the mixolydian mode makes this scale suitable to play over dominant 7th chords. The characteristic note is the flat 7th – this is what distinguishes it from the major scale.
What is mode Class 7 example
A mode is defined as the value that has a higher frequency in a given set of values. It is the value that appears the most number of times. Example: In the given set of data: 2, 4, 5, 5, 6, 7, the mode of the data set is 5 since it has appeared in the set twice.
Can you have 5 modes
A data set can often have no mode, one mode or more than one mode – it all depends on how many different values repeat most frequently.
What if there is 3 modes
A set of numbers with two modes is bimodal, a set of numbers with three modes is trimodal, and any set of numbers with more than one mode is multimodal. When scientists or statisticians talk about the modal observation, they are referring to the most common observation. |
BREAKING NEWS
Second derivative
## Summary
In calculus, the second derivative, or the second-order derivative, of a function f is the derivative of the derivative of f. Informally, the second derivative can be phrased as "the rate of change of the rate of change"; for example, the second derivative of the position of an object with respect to time is the instantaneous acceleration of the object, or the rate at which the velocity of the object is changing with respect to time. In Leibniz notation:
${\displaystyle a={\frac {dv}{dt}}={\frac {d^{2}x}{dt^{2}}},}$
where a is acceleration, v is velocity, t is time, x is position, and d is the instantaneous "delta" or change. The last expression ${\displaystyle {\tfrac {d^{2}x}{dt^{2}}}}$ is the second derivative of position (x) with respect to time.
On the graph of a function, the second derivative corresponds to the curvature or concavity of the graph. The graph of a function with a positive second derivative is upwardly concave, while the graph of a function with a negative second derivative curves in the opposite way.
## Second derivative power rule
The power rule for the first derivative, if applied twice, will produce the second derivative power rule as follows:
${\displaystyle {\frac {d^{2}}{dx^{2}}}x^{n}={\frac {d}{dx}}{\frac {d}{dx}}x^{n}={\frac {d}{dx}}\left(nx^{n-1}\right)=n{\frac {d}{dx}}x^{n-1}=n(n-1)x^{n-2}.}$
## Notation
The second derivative of a function ${\displaystyle f(x)}$ is usually denoted ${\displaystyle f''(x)}$ .[1][2] That is:
${\displaystyle f''=\left(f'\right)'}$
When using Leibniz's notation for derivatives, the second derivative of a dependent variable y with respect to an independent variable x is written
${\displaystyle {\frac {d^{2}y}{dx^{2}}}.}$
This notation is derived from the following formula:
${\displaystyle {\frac {d^{2}y}{dx^{2}}}\,=\,{\frac {d}{dx}}\left({\frac {dy}{dx}}\right).}$
## Example
Given the function
${\displaystyle f(x)=x^{3},}$
the derivative of f is the function
${\displaystyle f'(x)=3x^{2}.}$
The second derivative of f is the derivative of ${\displaystyle f'}$ , namely
${\displaystyle f''(x)=6x.}$
## Relation to the graph
### Concavity
The second derivative of a function f can be used to determine the concavity of the graph of f.[2] A function whose second derivative is positive is said to be concave up (also referred to as convex), meaning that the tangent line near the point where it touches the function will lie below the graph of the function. Similarly, a function whose second derivative is negative will be concave down (sometimes simply called concave), and its tangent line will lie above the graph of the function near the point of contact.
### Inflection points
If the second derivative of a function changes sign, the graph of the function will switch from concave down to concave up, or vice versa. A point where this occurs is called an inflection point. Assuming the second derivative is continuous, it must take a value of zero at any inflection point, although not every point where the second derivative is zero is necessarily a point of inflection.
### Second derivative test
The relation between the second derivative and the graph can be used to test whether a stationary point for a function (i.e., a point where ${\displaystyle f'(x)=0}$ ) is a local maximum or a local minimum. Specifically,
• If ${\displaystyle f''(x)<0}$ , then ${\displaystyle f}$ has a local maximum at ${\displaystyle x}$ .
• If ${\displaystyle f''(x)>0}$ , then ${\displaystyle f}$ has a local minimum at ${\displaystyle x}$ .
• If ${\displaystyle f''(x)=0}$ , the second derivative test says nothing about the point ${\displaystyle x}$ , a possible inflection point.
The reason the second derivative produces these results can be seen by way of a real-world analogy. Consider a vehicle that at first is moving forward at a great velocity, but with a negative acceleration. Clearly, the position of the vehicle at the point where the velocity reaches zero will be the maximum distance from the starting position – after this time, the velocity will become negative and the vehicle will reverse. The same is true for the minimum, with a vehicle that at first has a very negative velocity but positive acceleration.
## Limit
It is possible to write a single limit for the second derivative:
${\displaystyle f''(x)=\lim _{h\to 0}{\frac {f(x+h)-2f(x)+f(x-h)}{h^{2}}}.}$
The limit is called the second symmetric derivative.[3][4] The second symmetric derivative may exist even when the (usual) second derivative does not.
The expression on the right can be written as a difference quotient of difference quotients:
${\displaystyle {\frac {f(x+h)-2f(x)+f(x-h)}{h^{2}}}={\frac {{\dfrac {f(x+h)-f(x)}{h}}-{\dfrac {f(x)-f(x-h)}{h}}}{h}}.}$
This limit can be viewed as a continuous version of the second difference for sequences.
However, the existence of the above limit does not mean that the function ${\displaystyle f}$ has a second derivative. The limit above just gives a possibility for calculating the second derivative—but does not provide a definition. A counterexample is the sign function ${\displaystyle \operatorname {sgn}(x)}$ , which is defined as:
${\displaystyle \operatorname {sgn}(x)={\begin{cases}-1&{\text{if }}x<0,\\0&{\text{if }}x=0,\\1&{\text{if }}x>0.\end{cases}}}$
The sign function is not continuous at zero, and therefore the second derivative for ${\displaystyle x=0}$ does not exist. But the above limit exists for ${\displaystyle x=0}$ :
{\displaystyle {\begin{aligned}\lim _{h\to 0}{\frac {\operatorname {sgn}(0+h)-2\operatorname {sgn}(0)+\operatorname {sgn}(0-h)}{h^{2}}}&=\lim _{h\to 0}{\frac {\operatorname {sgn}(h)-2\cdot 0+\operatorname {sgn}(-h)}{h^{2}}}\\&=\lim _{h\to 0}{\frac {\operatorname {sgn}(h)+(-\operatorname {sgn}(h))}{h^{2}}}=\lim _{h\to 0}{\frac {0}{h^{2}}}=0.\end{aligned}}}
Just as the first derivative is related to linear approximations, the second derivative is related to the best quadratic approximation for a function f. This is the quadratic function whose first and second derivatives are the same as those of f at a given point. The formula for the best quadratic approximation to a function f around the point x = a is
${\displaystyle f(x)\approx f(a)+f'(a)(x-a)+{\tfrac {1}{2}}f''(a)(x-a)^{2}.}$
This quadratic approximation is the second-order Taylor polynomial for the function centered at x = a.
## Eigenvalues and eigenvectors of the second derivative
For many combinations of boundary conditions explicit formulas for eigenvalues and eigenvectors of the second derivative can be obtained. For example, assuming ${\displaystyle x\in [0,L]}$ and homogeneous Dirichlet boundary conditions (i.e., ${\displaystyle v(0)=v(L)=0}$ where v is the eigenvector), the eigenvalues are ${\displaystyle \lambda _{j}=-{\tfrac {j^{2}\pi ^{2}}{L^{2}}}}$ and the corresponding eigenvectors (also called eigenfunctions) are ${\displaystyle v_{j}(x)={\sqrt {\tfrac {2}{L}}}\sin \left({\tfrac {j\pi x}{L}}\right)}$ . Here, ${\displaystyle v''_{j}(x)=\lambda _{j}v_{j}(x)}$ , for ${\displaystyle j=1,\ldots ,\infty }$ .
For other well-known cases, see Eigenvalues and eigenvectors of the second derivative.
## Generalization to higher dimensions
### The Hessian
The second derivative generalizes to higher dimensions through the notion of second partial derivatives. For a function f: R3R, these include the three second-order partials
${\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}},\;{\frac {\partial ^{2}f}{\partial y^{2}}},{\text{ and }}{\frac {\partial ^{2}f}{\partial z^{2}}}}$
and the mixed partials
${\displaystyle {\frac {\partial ^{2}f}{\partial x\,\partial y}},\;{\frac {\partial ^{2}f}{\partial x\,\partial z}},{\text{ and }}{\frac {\partial ^{2}f}{\partial y\,\partial z}}.}$
If the function's image and domain both have a potential, then these fit together into a symmetric matrix known as the Hessian. The eigenvalues of this matrix can be used to implement a multivariable analogue of the second derivative test. (See also the second partial derivative test.)
### The Laplacian
Another common generalization of the second derivative is the Laplacian. This is the differential operator ${\displaystyle \nabla ^{2}}$ (or ${\displaystyle \Delta }$ ) defined by
${\displaystyle \nabla ^{2}f={\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}.}$
The Laplacian of a function is equal to the divergence of the gradient, and the trace of the Hessian matrix.
## References
1. ^ "Content - The second derivative". amsi.org.au. Retrieved 2020-09-16.
2. ^ a b "Second Derivatives". Math24. Retrieved 2020-09-16.
3. ^ A. Zygmund (2002). Trigonometric Series. Cambridge University Press. pp. 22–23. ISBN 978-0-521-89053-3.
4. ^ Thomson, Brian S. (1994). Symmetric Properties of Real Functions. Marcel Dekker. p. 1. ISBN 0-8247-9230-0.
### Print
• Anton, Howard; Bivens, Irl; Davis, Stephen (February 2, 2005), Calculus: Early Transcendentals Single and Multivariable (8th ed.), New York: Wiley, ISBN 978-0-471-47244-5
• Apostol, Tom M. (June 1967), Calculus, Vol. 1: One-Variable Calculus with an Introduction to Linear Algebra, vol. 1 (2nd ed.), Wiley, ISBN 978-0-471-00005-1
• Apostol, Tom M. (June 1969), Calculus, Vol. 2: Multi-Variable Calculus and Linear Algebra with Applications, vol. 1 (2nd ed.), Wiley, ISBN 978-0-471-00007-5
• Eves, Howard (January 2, 1990), An Introduction to the History of Mathematics (6th ed.), Brooks Cole, ISBN 978-0-03-029558-4
• Larson, Ron; Hostetler, Robert P.; Edwards, Bruce H. (February 28, 2006), Calculus: Early Transcendental Functions (4th ed.), Houghton Mifflin Company, ISBN 978-0-618-60624-5
• Spivak, Michael (September 1994), Calculus (3rd ed.), Publish or Perish, ISBN 978-0-914098-89-8
• Stewart, James (December 24, 2002), Calculus (5th ed.), Brooks Cole, ISBN 978-0-534-39339-7
• Thompson, Silvanus P. (September 8, 1998), Calculus Made Easy (Revised, Updated, Expanded ed.), New York: St. Martin's Press, ISBN 978-0-312-18548-0
### Online books
• Crowell, Benjamin (2003), Calculus
• Garrett, Paul (2004), Notes on First-Year Calculus
• Hussain, Faraz (2006), Understanding Calculus
• Keisler, H. Jerome (2000), Elementary Calculus: An Approach Using Infinitesimals
• Mauch, Sean (2004), Unabridged Version of Sean's Applied Math Book, archived from the original on 2006-04-15
• Sloughter, Dan (2000), Difference Equations to Differential Equations
• Strang, Gilbert (1991), Calculus
• Stroyan, Keith D. (1997), A Brief Introduction to Infinitesimal Calculus, archived from the original on 2005-09-11
• Wikibooks, Calculus |
### Dividing by Fractions
In his recent post "Math is a dangerous subject to teach," Joe Bower discusses the ability to learn the procedures of math without understanding the conceptual foundations. As an example, Joe humbly admits that he has "absolutely no idea why" (emphasis his) dividing by a fraction is replaced by multiplying by the fraction's reciprocal. He can get the answer right without proper understanding, and therein lies the danger.
One of the things I've enjoyed most about teaching is finding new and deeper ways of understanding so-called "simple" math that I thought I had already mastered. Most of my mathematical upbringing focused more on procedure than understanding, so I occasionally find myself in the same position as Joe. Using Joe's post as inspiration, I've given more thought to dividing by fractions and finally have a model and a description that I hope explains what's really happening when you divide by a fraction.
First, let's look at a simple fraction:
The top number, the numerator, simply counts "how many." The bottom number, the denominator, tells us "how big." We read this fraction properly as "two-thirds," and usually think of that as two objects, each one-third the size of the whole (however big that is).
Now let's look at division. Some students are led astray with early beliefs that "multiplying makes bigger" and "division makes smaller." That kind of misguided number sense can be frighteningly persistent. Multiplication is better thought of as a "scaling" operation, and division can be thought of as a "grouping" operation. To see what I mean, let me explain using whole numbers:
The model I imagine for this problem looks like this:
Using the "grouping" concept of division, I've made four groups. Because each group contains two, the answer is two. No surprise. Let's try another:
The model:
I've made two groups. Because each group contains four, the answer is four. Still no surprises. Let's try one more with whole numbers:
The model:
I've made one group, which is as trivial as it gets. Because the group contains eight, the answer is eight. Now that we've established the pattern with a "grouping" definition, it should be easy to see why you can't divide by zero. I can't possibly make zero groups and still have the eight squares.
Okay, now let's try an easy division problem with fractions:
Remember, the numerator tells us "how many" and the denominator tells us "how big." The model:
It's still one group (how many), but a group of halves (how big). If you count objects, you get the answer: sixteen. But we haven't added or taken away anything -- the eight is still there. Got it? Let's try another:
The model:
Eight divided by two thirds, translated into "grouping-speak," is "eight grouped into two groups of thirds." Because each group contains twelve, the answer is twelve, even though you can still imagine the original eight.
So is this the same as multiplying by the reciprocal? Breaking the wholes into thirds gave us three times as many pieces (24, same as 8 times 3), and grouping into two groups gave us half of the pieces in each group (12, same as 24 divided by 2). More concisely, we multiplied 8 by 3 and divided by two. So doesn't that mean dividing by two thirds is the same as multiplying by three halves? Not exactly. We get the same answer, but for a different reason. To me, the model for eight times three halves would mean scaling eight to be three times bigger (24), then scaled back down to half of that (12). It's a different picture, even if we still get the same answer.
You can choose to accept this as a complication or a convenience; either way, I hope you have a better understanding of dividing by fractions. As always, feel free to offer criticisms in the comments below. (There has to be many a sixth grade teacher who could teach me a thing or two about this topic!)
Update 4/18/2010: Gary Davis co-authored a great guide on the division of fractions. It provides more strategies, more examples, and more detail than my post did. Thanks for sharing, Gary!
1. Small error in writing answer to 8/(0.5)...just a typo. Diagram is correct.
2. Ha! Somehow my mind had already moved to the next problem. Thanks for the correction!
3. Here is another approach from your picture diagrams:
8 divided by 2/3: break each of the 8 units into thirds: how many groups of twos will you get? answer: 12 groups of twos
8 divided by 1/2: break each of the eight units into halves: how many of each halves will you get? answer: 16 halves
8 divided by 2: consider each of the 8 units: how many groups of 2 units will you get? |
# Tangents and Circles
Related Topics:
More Lessons for Grade 9
Math Worksheets
Examples, videos, worksheets, solutions, and activities to help Geometry students learn about the tangents of a circle.
What is a tangent?
A tangent is a line in the plane of a circle that intersects the circle at one point.
What is a common tangent?
A common tangent is a line that is a tangent to each of two coplanar circles.
What is a common external tangent?
A common external tangent does not intersect the segment that joins the centers of the circles.
What is a common internal tangent?
A common internal tangent intersects the segment that joins the centers of the circles.
The following diagrams show some examples of common external tangents and common internal tangents. Scroll down the page for more examples and solutions of common external tangents and common internal tangents.
Tangent Line - Geometry Help
Students learn the definitions of a tangent. Common Internal and External Tangents
Students learn the definitions of common internal tangents and common external tangents.
Tangents and Circles
Students learn the following theorems related to tangents. If a line is tangent to a circle, then the line is perpendicular to the radius at the point of tangency. If a line is perpendicular to a radius at its outer endpoint, then the line is tangent to the circle. Radii to Tangents
When a radius is drawn to a point of tangency, the angle formed is always a right (90 degree) angle. This fact is commonly applied in problems with two tangent segments drawn to a circle from a point. If two radii to tangents are drawn in, a kite with two right angles is formed and the missing angles or sides can be found. Related topics include central angles, tangent segments to a circle, and chords. Tangent Segments to a Circle
A tangent intersects a circle in exactly one point. When two segments are drawn tangent to a circle from the same point outside the circle, the segments are congruent. The extension problem of this topic is a belt and gear problem which asks for the length of belt required to fit around two gears. 3 problems including common internal tangent and common external tangent
1. Calculate a radius given a tangent segment length.
2. Calculate the length of a common external tangent segment.
3. Calculate a radius, given the length of a common internal tangent segment.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Probability axioms ▷ Addition law of probability
Theorem: The probability of the union of $A$ and $B$ is the sum of the probabilities of $A$ and $B$ minus the probability of the intersection of $A$ and $B$:
$\label{eq:prob-add} P(A \cup B) = P(A) + P(B) - P(A \cap B) \; .$
Proof: Let $E_1 = A$ and $E_2 = B \setminus A$, such that $E_1 \cup E_2 = A \cup B$. Then, by the third axiom of probability, we have:
$\label{eq:pAoB} \begin{split} P(A \cup B) &= P(A) + P(B \setminus A) \\ P(A \cup B) &= P(A) + P(B \setminus [A \cap B]) \; . \end{split}$
Then, let $E_1 = B \setminus [A \cap B]$ and $E_2 = A \cap B$, such that $E_1 \cup E_2 = B$. Again, from the third axiom of probability, we obtain:
$\label{eq:pB} \begin{split} P(B) &= P(B \setminus [A \cap B]) + P(A \cap B) \\ P(B \setminus [A \cap B]) &= P(B) - P(A \cap B) \; . \end{split}$
Plugging \eqref{eq:pB} into \eqref{eq:pAoB}, we finally get:
$\label{eq:prob-add-qed} P(A \cup B) = P(A) + P(B) - P(A \cap B) \; .$
Sources:
Metadata: ID: P247 | shortcut: prob-add | author: JoramSoch | date: 2021-07-30, 12:45. |
Courses
# Distance Between Two Points Mathematics Notes | EduRev
## Mathematics : Distance Between Two Points Mathematics Notes | EduRev
The document Distance Between Two Points Mathematics Notes | EduRev is a part of the Mathematics Course Additional Topics for IIT JAM Mathematics.
All you need of Mathematics at this link: Mathematics
Distance Between Two Points
2D distance is the distance between 2 points in a 2D space. In a 2D space, each point, or location in the space, is qualified by 2 parameters like an x-coordinate, and a y-coordinate. We denote the combination of x-coordinate and y-coordinate in something known as an ordered pair, denoted by (x, y). Therefore, the coordinates of some point P would be represented as P(x,y).
The 2D distance between two points is measured by the distance formula. Consider points P1 and P2, with coordinates given as P1(x1, y1) and P2(x2,y2). Therefore, the distance between them would be denoted by d, where:
The way to think about distance calculation is in 3 parts:
• Calculate the difference in y-coordinates, and square the difference, denote as quantity q1
• Calculate the difference in x-coordinates, and square the difference, denote as quantity q2
• Take the square root of ( q1 + q2 ).
Example 1
Calculate the distance between the points P1(1,2) and P2(4,3)
Solution: In the given problem, we have: x1 = 1, y1 = 2, x2 = 4, and y2 = 3. Therefore, we have:
q1 = (3-2)² = 1
q2= (4-1)² = 9
distance (d) = √(q1 + q2)
∴ d = √(1 + 9) = √10 = 3.16227 units.
As you can see in the previous example, we have written the units of distance as ‘units’, and not ‘square units.’
Example 2
Calculate the distance between the points P1(1,2) and P2(4,2)
Solution: In the given problem, we have: x1 = 1, y1 = 2, x2 = 4, and y2 = 2. Therefore, we have:
q1 = (2-2)² = 0
q2 = (4-1)² = 9
distance (d) = √(q1+q2)
∴ d = √(0 + 9) = √9 = 3 units.
Important Points:
As you can see, in the above problem, we have y2-y1 = 0. This means that the points are on the same horizontal line.
The distance between any two points on the same horizontal line (where y1 = y2) is given by d = |x2 – x1| where | | is used to denote the absolute value of x2 – x1.
Example 3
Calculate the distance between the points P1(1,2) and P2(1,5)
Solution: In the given problem, we have: x1 = 1, y1 = 1, x2 = 1, and y2 = 2. Therefore, we have:
q1 = (4-1)²= 9
q2 = (1-1)² = 0
distance (d) = √(q1 + q2)
∴d = √(9 + 0) = √9 = 3 units.
Important Points:
As you can see, in the above problem, we have x1 – x2 = 0. This means that the points are on the same vertical line.
The distance between any two points on the same vertical line (where x1 = x2) is given by d = | y2 – y1 | where | | is used to denote the absolute value of y2 – y1
Solved Question for You
Question: Can the distance between any two points be negative?
Answer: No, the distance between 2 points cannot be negative. This can be thought of in terms of 3 reasons:
1. Distance is used to denote a physical quantity, expressing how far two points are from each other, and such a quantity cannot be negative.
2. Distance is equal to the square root of the sum of two positive numbers. The sum of two positive numbers is always positive, and the square root of a positive number is always positive.
3. In the exceptional case that the distance between 2 points is zero, it is still a non-negative value, hence the distance cannot be negative.
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!
40 docs
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
; |
A matrix is a collection of numbers (and potentially fractions) placed into rows and columns. Specifically, a square matrix is a matrix with the same number of rows and columns. Understanding how to transform a matrix into a form called reduced row echelon form is important as this process is necessary in some more advanced linear algebra exercises such as inverting a matrix or solving linear systems of equations.
A square matrix is in reduced row echelon form when all entries in the main diagonal (which begins in the top left and ends in the bottom right) have a value of 1, and all other entries have a value of 0. There is also an intermediate form, called row echelon form. In row echelon form, only the entries below the main diagonal need to be 0s, and the main diagonal values do not necessarily have to be 1s (but must not be 0).
Time until completion may vary based on skill and matrix size, but should take less than 10 minutes. The following are what you will need in order to complete this process:
· Pencil
· Paper
· Calculator (optional)
· An understanding of addition, subtraction, multiplication, division, and fractions
## Step 1: Determining If the Matrix Can Be Reduced
Not all square matrices can be transformed into reduced row echelon form. These matrices are referred to as being “noninvertible”. A square matrix will be noninvertible if any of the three following conditions are true:
· One row is identical to, or a constant multiple of, another row.
· A row of only zeros exists.
· A column of only zeros exists.
An example of each of these can be seen in the image above. If the matrix you are using meets one or more of these conditions, then you will not be able to complete the following steps.
In this Instructable, row 1 will be called R1, row 2 will be called R2, and so on. R1 is always the top row, and all other rows are located in numerical order below R1. Similarly, column 1, which is the leftmost column, will be called C1 and all other columns will be in numerical order to the right of C1.
## Step 2: Row Operations
There are three row operations which can be performed in order to transform the matrix. Each of these operations must be applied to every element in a row.
· The first of these operations is switching rows; the order of rows can be moved around freely.
· Rows can also be multiplied by any number (including fractions).
· Lastly, the values of one row can be added to or subtracted from another row. Only entries that are in the same columns get added or subtracted with each other. Also, the row whose values are being added or subtracted can be multiplied or divided by any number (including fractions).
There is no exact order for these operations, as the process is dependent on the matrix that you are working with, and even with the same matrix there are multiple paths to a solution. Additionally, it might be possible to reduce the matrix without using all of these operations. Steps 3 through 6 will describe when to use each of these operations.
## Step 3: Using Row Switching
It is easiest to reach row echelon form first before attempting to reach reduced row echelon form, which is why for now it is best to only focus on entries below the main diagonal. In the example above, the circled entries are the entries that need to be 0s in order for row echelon form to be achieved.
Always start by switching rows around so that as many 0s occupy entries below the main diagonal. In the example, R1 is switched with R3 so that a 0 occupies an entry below the main diagonal. It’s worth noting that not all matrices have conveniently placed 0s as this matrix does. If there are no rows with 0s that can be moved into position for row echelon form, then this step can be skipped.
When doing any row operation, it is important to show your work. Write down what operation you are doing, and then rewrite the matrix after performing that operation.
## Step 4: Using Row Addition
As before, focus on changing entries below the main diagonal. Typically it is easiest to start with creating 0s in C1 by adding or subtracting with R1’s values. Next, focus on C2 by adding or subtracting R2’s values, and so on. In other words, start at the leftmost column, then move right.
Sometimes it is necessary to multiply the values being added or subtracted. For this example, R2 needs to be subtracted by 1/2 times R1 (R2 - (1/2)*R1) in order for the entry in both R2 and C1 to be turned into 0. Also, R3 can be subtracted by 2 times R2 (R3 - 2*R2), which will put this matrix in row echelon form.
In most matrices, more than one row addition will be necessary in order to reach row echelon form. Also, it is important to be careful when dealing with negative numbers as it is easy to mess up the signs.
## Step 5: Using Row Multiplication
Once row echelon form is achieved, row multiplication should be used in order to change the values in the main diagonal into 1s. In the example, R1 must be divided by 2 (R1 / 2, which is equivalent to R1 * 1/2) and R3 must be divided by -2 (R3 / -2 or R3 * -1/2). Remember to either multiply or divide every number in a row whenever you perform row multiplication.
## Step 6: Reduced Row Echelon Form
All that’s left is to transform the entries above the main diagonal into 0s. Start with the rightmost column, which in this matrix is C3. Use row addition with the bottom row, R3, in order to clear the entries in C3 that are above the main diagonal. Next, use row addition with R2 in order to clear the entries in C2 that are above the main diagonal. This will complete the process of transforming the matrix.
The process remains the same for matrices with more rows and columns than in this example. Start with the bottom row, and use it to clear entries in the rightmost column. Then, use the row above the bottom row to clear entries in the column next to the rightmost column. Continue this pattern until all entries above the main diagonal have been reduced to 0.
## Step 7: More Practice
Above are some more matrices which can be transformed into reduced row echelon form. Additionally, you can create your own matrices with random values to use for practice. Don’t forget to check if it is possible for them to be reduced using step 1!
This process may seem trivial at first because the result is always the same. However, as mentioned in the introduction, understanding the process of using row operations to reduce a matrix is necessary for other linear algebra exercises such as matrix inversion and solving linear systems.
This website shows step by step solutions to reducing matrices. The order of steps used may differ from the order used in this Instructable: http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi?c=rref
This website shows how to use this process to invert a matrix: http://www.purplemath.com/modules/mtrxinvr.htm
<p>Nice tutorial! Thanks for sharing!</p> |
# Linear Algebra/Topic: Line of Best Fit/Solutions
## Solutions
Problem 1
Use least-squares to judge if the coin in this experiment is fair.
flips 8 16 24 32 40 heads 4 9 13 17 20
As with the first example discussed above, we are trying to find a best ${\displaystyle m}$ to "solve" this system.
${\displaystyle {\begin{array}{*{5}{rc}r}8m&=&4\\16m&=&9\\24m&=&13\\32m&=&17\\40m&=&20\end{array}}}$
Projecting into the linear subspace gives this
${\displaystyle {\frac {{\begin{pmatrix}4\\9\\13\\17\\20\end{pmatrix}}\cdot {\begin{pmatrix}8\\16\\24\\32\\40\end{pmatrix}}}{{\begin{pmatrix}8\\16\\24\\32\\40\end{pmatrix}}\cdot {\begin{pmatrix}8\\16\\24\\32\\40\end{pmatrix}}}}\cdot {\begin{pmatrix}8\\16\\24\\32\\40\end{pmatrix}}={\frac {1832}{3520}}\cdot {\begin{pmatrix}8\\16\\24\\32\\40\end{pmatrix}}}$
so the slope of the line of best fit is approximately ${\displaystyle 0.52}$ .
Problem 2
For the men's mile record, rather than give each of the many records and its exact date, we've "smoothed" the data somewhat by taking a periodic sample. Do the longer calculation and compare the conclusions.
With this input
${\displaystyle A={\begin{pmatrix}1&1852.71\\1&1858.88\\\vdots &\vdots \\1&1985.54\\1&1993.71\end{pmatrix}}\qquad b={\begin{pmatrix}292.0\\285.0\\\vdots \\226.32\\224.39\end{pmatrix}}}$
(the dates have been rounded to months, e.g., for a September record, the decimal ${\displaystyle .71\approx (8.5/12)}$ was used), Maple responded with an intercept of ${\displaystyle b=994.8276974}$ and a slope of ${\displaystyle m=-0.3871993827}$ .
Problem 3
Find the line of best fit for the men's ${\displaystyle 1500}$ meter run. How does the slope compare with that for the men's mile? (The distances are close; a mile is about ${\displaystyle 1609}$ meters.)
With this input (the years are zeroed at ${\displaystyle 1900}$ )
${\displaystyle A:={\begin{pmatrix}1&.38\\1&.54\\\vdots \vdots \\1&92.71\\1&95.54\end{pmatrix}}\qquad b={\begin{pmatrix}249.0\\246.2\\\vdots \\208.86\\207.37\end{pmatrix}}}$
(the dates have been rounded to months, e.g., for a September record, the decimal ${\displaystyle .71\approx (8.5/12)}$ was used), Maple gives an intercept of ${\displaystyle b=243.1590327}$ and a slope of ${\displaystyle m=-0.401647703}$ . The slope given in the body of this Topic for the men's mile is quite close to this.
Problem 4
Find the line of best fit for the records for women's mile.
With this input (the years are zeroed at ${\displaystyle 1900}$ )
${\displaystyle A={\begin{pmatrix}1&21.46\\1&32.63\\\vdots &\vdots \\1&89.54\\1&96.63\end{pmatrix}}\qquad b={\begin{pmatrix}373.2\\327.5\\\vdots \\255.61\\252.56\end{pmatrix}}}$
(the dates have been rounded to months, e.g., for a September record, the decimal ${\displaystyle .71\approx (8.5/12)}$ was used), MAPLE gave an intercept of ${\displaystyle b=378.7114894}$ and a slope of ${\displaystyle m=-1.445753225}$ .
Problem 5
Do the lines of best fit for the men's and women's miles cross?
These are the equations of the lines for men's and women's mile (the vertical intercept term of the equation for the women's mile has been adjusted from the answer above, to zero it at the year ${\displaystyle 0}$ , because that's how the men's mile equation was done).
${\displaystyle {\begin{array}{rl}y&=994.8276974-0.3871993827x\\y&=3125.6426-1.445753225x\end{array}}}$
Obviously the lines cross. A computer program is the easiest way to do the arithmetic: MuPAD gives ${\displaystyle x=2012.949004}$ and ${\displaystyle y=215.4150856}$ (${\displaystyle 215}$ seconds is ${\displaystyle 3}$ minutes and ${\displaystyle 35}$ seconds). Remark. Of course all of this projection is highly dubious — for one thing, the equation for the women is influenced by the quite slow early times — but it is nonetheless fun.
Problem 6
When the space shuttle Challenger exploded in 1986, one of the criticisms made of NASA's decision to launch was in the way the analysis of number of O-ring failures versus temperature was made (of course, O-ring failure caused the explosion). Four O-ring failures will cause the rocket to explode. NASA had data from 24 previous flights.
temp °F 53 75 57 58 63 70 70 66 67 67 67 failures 3 2 1 1 1 1 1 0 0 0 0 temp °F 68 69 70 70 72 73 75 76 76 78 79 80 81 failures 0 0 0 0 0 0 0 0 0 0 0 0 0
The temperature that day was forecast to be ${\displaystyle 31^{\circ }{\text{F}}}$ .
1. NASA based the decision to launch partially on a chart showing only the flights that had at least one O-ring failure. Find the line that best fits these seven flights. On the basis of this data, predict the number of O-ring failures when the temperature is ${\displaystyle 31}$ , and when the number of failures will exceed four.
2. Find the line that best fits all 24 flights. On the basis of this extra data, predict the number of O-ring failures when the temperature is ${\displaystyle 31}$ , and when the number of failures will exceed four.
Which do you think is the more accurate method of predicting? (An excellent discussion appears in (Dalal, Folkes & Hoadley 1989).)
1. A computer algebra system like MAPLE or MuPAD will give an intercept of ${\displaystyle b=4259/1398\approx 3.239628}$ and a slope of ${\displaystyle m=-71/2796\approx -0.025393419}$ Plugging ${\displaystyle x=31}$ into the equation yields a predicted number of O-ring failures of ${\displaystyle y=2.45}$ (rounded to two places). Plugging in ${\displaystyle y=4}$ and solving gives a temperature of ${\displaystyle x=-29.94^{\circ }}$ F.
2. On the basis of this information
${\displaystyle A={\begin{pmatrix}1&53\\1&75\\\vdots \\1&80\\1&81\end{pmatrix}}\qquad b={\begin{pmatrix}3\\2\\\vdots \\0\\0\end{pmatrix}}}$
MAPLE gives the intercept ${\displaystyle b=187/40=4.675}$ and the slope ${\displaystyle m=-73/1200\approx -0.060833}$ . Here, plugging ${\displaystyle x=31}$ into the equation predicts ${\displaystyle y=2.79}$ O-ring failures (rounded to two places). Plugging in ${\displaystyle y=4}$ failures gives a temperature of ${\displaystyle x=11^{\circ }}$ F.
Problem 7
This table lists the average distance from the sun to each of the first seven planets, using earth's average as a unit.
Mercury Venus Earth Mars Jupiter Saturn Uranus 0.39 0.72 1.00 1.52 5.20 9.54 19.2
1. Plot the number of the planet (Mercury is ${\displaystyle 1}$ , etc.) versus the distance. Note that it does not look like a line, and so finding the line of best fit is not fruitful.
2. It does, however look like an exponential curve. Therefore, plot the number of the planet versus the logarithm of the distance. Does this look like a line?
3. The asteroid belt between Mars and Jupiter is thought to be what is left of a planet that broke apart. Renumber so that Jupiter is ${\displaystyle 6}$ , Saturn is ${\displaystyle 7}$ , and Uranus is ${\displaystyle 8}$ , and plot against the log again. Does this look better?
4. Use least squares on that data to predict the location of Neptune.
5. Repeat to predict where Pluto is.
6. Is the formula accurate for Neptune and Pluto?
This method was used to help discover Neptune (although the second item is misleading about the history; actually, the discovery of Neptune in position ${\displaystyle 9}$ prompted people to look for the "missing planet" in position ${\displaystyle 5}$ ). See (Gardner 1970)
1. The plot is nonlinear.
2. Here is the plot.
There is perhaps a jog up between planet ${\displaystyle 4}$ and planet ${\displaystyle 5}$ .
3. This plot seems even more linear.
4. With this input
${\displaystyle A={\begin{pmatrix}1&1\\1&2\\1&3\\1&4\\1&6\\1&7\\1&8\end{pmatrix}}\qquad b={\begin{pmatrix}-0.40893539\\-0.1426675\\0\\0.18184359\\0.71600334\\0.97954837\\1.2833012\end{pmatrix}}}$
MuPAD gives that the intercept is ${\displaystyle b=-0.6780677466}$ and the slope is ${\displaystyle m=0.2372763818}$ .
5. Plugging ${\displaystyle x=9}$ into the equation ${\displaystyle y=-0.6780677466+0.2372763818x}$ from the prior item gives that the log of the distance is ${\displaystyle 1.4574197}$ , so the expected distance is ${\displaystyle 28.669472}$ . The actual distance is about ${\displaystyle 30.003}$ .
6. Plugging ${\displaystyle x=10}$ into the same equation gives that the log of the distance is ${\displaystyle 1.6946961}$ , so the expected distance is ${\displaystyle 49.510362}$ . The actual distance is about ${\displaystyle 39.503}$ .
Problem 8
William Bennett has proposed an Index of Leading Cultural Indicators for the US (Bennett 1993). Among the statistics cited are the average daily hours spent watching TV, and the average combined SAT scores.
1960 1965 1970 1975 1980 1985 1990 1992 TV 5:06 5:29 5:56 6:07 6:36 7:07 6:55 7:04 SAT 975 969 948 910 890 906 900 899
Suppose that a cause and effect relationship is proposed between the time spent watching TV and the decline in SAT scores (in this article, Mr. Bennett does not argue that there is a direct connection).
1. Find the line of best fit relating the independent variable of average daily TV hours to the dependent variable of SAT scores.
2. Find the most recent estimate of the average daily TV hours (Bennett's cites Neilsen Media Research as the source of these estimates). Estimate the associated SAT score. How close is your estimate to the actual average? (Warning: a change has been made recently in the SAT, so you should investigate whether some adjustment needs to be made to the reported average to make a valid comparison.)
1. With this input
${\displaystyle A={\begin{pmatrix}1&306\\1&329\\1&356\\1&367\\1&396\\1&427\\1&415\\1&424\end{pmatrix}}\qquad b={\begin{pmatrix}975\\969\\948\\910\\890\\906\\900\\899\end{pmatrix}}}$
MAPLE gives the intercept ${\displaystyle b=34009779/28796\approx 1181.0591}$ and the slope ${\displaystyle m=-19561/28796\approx -0.6793}$ .
Data on the progression of the world's records (taken from the Runner's World web site) is below.
Progression of Men's Mile Record time name date 4:52.0 Cadet Marshall (GBR) 02Sep52 4:45.0 Thomas Finch (GBR) 03Nov58 4:40.0 Gerald Surman (GBR) 24Nov59 4:33.0 George Farran (IRL) 23May62 4:29 3/5 Walter Chinnery (GBR) 10Mar68 4:28 4/5 William Gibbs (GBR) 03Apr68 4:28 3/5 Charles Gunton (GBR) 31Mar73 4:26.0 Walter Slade (GBR) 30May74 4:24 1/2 Walter Slade (GBR) 19Jun75 4:23 1/5 Walter George (GBR) 16Aug80 4:19 2/5 Walter George (GBR) 03Jun82 4:18 2/5 Walter George (GBR) 21Jun84 4:17 4/5 Thomas Conneff (USA) 26Aug93 4:17.0 Fred Bacon (GBR) 06Jul95 4:15 3/5 Thomas Conneff (USA) 28Aug95 4:15 2/5 John Paul Jones (USA) 27May11 4:14.4 John Paul Jones (USA) 31May13 4:12.6 Norman Taber (USA) 16Jul15 4:10.4 Paavo Nurmi (FIN) 23Aug23 4:09 1/5 Jules Ladoumegue (FRA) 04Oct31 4:07.6 Jack Lovelock (NZL) 15Jul33 4:06.8 Glenn Cunningham (USA) 16Jun34 4:06.4 Sydney Wooderson (GBR) 28Aug37 4:06.2 Gunder Hagg (SWE) 01Jul42 4:04.6 Gunder Hagg (SWE) 04Sep42 4:02.6 Arne Andersson (SWE) 01Jul43 4:01.6 Arne Andersson (SWE) 18Jul44 4:01.4 Gunder Hagg (SWE) 17Jul45 3:59.4 Roger Bannister (GBR) 06May54 3:58.0 John Landy (AUS) 21Jun54 3:57.2 Derek Ibbotson (GBR) 19Jul57 3:54.5 Herb Elliott (AUS) 06Aug58 3:54.4 Peter Snell (NZL) 27Jan62 3:54.1 Peter Snell (NZL) 17Nov64 3:53.6 Michel Jazy (FRA) 09Jun65 3:51.3 Jim Ryun (USA) 17Jul66 3:51.1 Jim Ryun (USA) 23Jun67 3:51.0 Filbert Bayi (TAN) 17May75 3:49.4 John Walker (NZL) 12Aug75 3:49.0 Sebastian Coe (GBR) 17Jul79 3:48.8 Steve Ovett (GBR) 01Jul80 3:48.53 Sebastian Coe (GBR) 19Aug81 3:48.40 Steve Ovett (GBR) 26Aug81 3:47.33 Sebastian Coe (GBR) 28Aug81 3:46.32 Steve Cram (GBR) 27Jul85 3:44.39 Noureddine Morceli (ALG) 05Sep93 3:43.13 Hicham el Guerrouj (MOR) 07Jul99
Progression of Men's 1500 Meter Record time name date 4:09.0 John Bray (USA) 30May00 4:06.2 Charles Bennett (GBR) 15Jul00 4:05.4 James Lightbody (USA) 03Sep04 3:59.8 Harold Wilson (GBR) 30May08 3:59.2 Abel Kiviat (USA) 26May12 3:56.8 Abel Kiviat (USA) 02Jun12 3:55.8 Abel Kiviat (USA) 08Jun12 3:55.0 Norman Taber (USA) 16Jul15 3:54.7 John Zander (SWE) 05Aug17 3:53.0 Paavo Nurmi (FIN) 23Aug23 3:52.6 Paavo Nurmi (FIN) 19Jun24 3:51.0 Otto Peltzer (GER) 11Sep26 3:49.2 Jules Ladoumegue (FRA) 05Oct30 3:49.0 Luigi Beccali (ITA) 17Sep33 3:48.8 William Bonthron (USA) 30Jun34 3:47.8 Jack Lovelock (NZL) 06Aug36 3:47.6 Gunder Hagg (SWE) 10Aug41 3:45.8 Gunder Hagg (SWE) 17Jul42 3:45.0 Arne Andersson (SWE) 17Aug43 3:43.0 Gunder Hagg (SWE) 07Jul44 3:42.8 Wes Santee (USA) 04Jun54 3:41.8 John Landy (AUS) 21Jun54 3:40.8 Sandor Iharos (HUN) 28Jul55 3:40.6 Istvan Rozsavolgyi (HUN) 03Aug56 3:40.2 Olavi Salsola (FIN) 11Jul57 3:38.1 Stanislav Jungwirth (CZE) 12Jul57 3:36.0 Herb Elliott (AUS) 28Aug58 3:35.6 Herb Elliott (AUS) 06Sep60 3:33.1 Jim Ryun (USA) 08Jul67 3:32.2 Filbert Bayi (TAN) 02Feb74 3:32.1 Sebastian Coe (GBR) 15Aug79 3:31.36 Steve Ovett (GBR) 27Aug80 3:31.24 Sydney Maree (usa) 28Aug83 3:30.77 Steve Ovett (GBR) 04Sep83 3:29.67 Steve Cram (GBR) 16Jul85 3:29.46 Said Aouita (MOR) 23Aug85 3:28.86 Noureddine Morceli (ALG) 06Sep92 3:27.37 Noureddine Morceli (ALG) 12Jul95 3:26.00 Hicham el Guerrouj (MOR) 14Jul98
Progression of Women's Mile Record time name date 6:13.2 Elizabeth Atkinson (GBR) 24Jun21 5:27.5 Ruth Christmas (GBR) 20Aug32 5:24.0 Gladys Lunn (GBR) 01Jun36 5:23.0 Gladys Lunn (GBR) 18Jul36 5:20.8 Gladys Lunn (GBR) 08May37 5:17.0 Gladys Lunn (GBR) 07Aug37 5:15.3 Evelyne Forster (GBR) 22Jul39 5:11.0 Anne Oliver (GBR) 14Jun52 5:09.8 Enid Harding (GBR) 04Jul53 5:08.0 Anne Oliver (GBR) 12Sep53 5:02.6 Diane Leather (GBR) 30Sep53 5:00.3 Edith Treybal (ROM) 01Nov53 5:00.2 Diane Leather (GBR) 26May54 4:59.6 Diane Leather (GBR) 29May54 4:50.8 Diane Leather (GBR) 24May55 4:45.0 Diane Leather (GBR) 21Sep55 4:41.4 Marise Chamberlain (NZL) 08Dec62 4:39.2 Anne Smith (GBR) 13May67 4:37.0 Anne Smith (GBR) 03Jun67 4:36.8 Maria Gommers (HOL) 14Jun69 4:35.3 Ellen Tittel (FRG) 20Aug71 4:34.9 Glenda Reiser (CAN) 07Jul73 4:29.5 Paola Pigni-Cacchi (ITA) 08Aug73 4:23.8 Natalia Marasescu (ROM) 21May77 4:22.1 Natalia Marasescu (ROM) 27Jan79 4:21.7 Mary Decker (USA) 26Jan80 4:20.89 Lyudmila Veselkova (SOV) 12Sep81 4:18.08 Mary Decker-Tabb (USA) 09Jul82 4:17.44 Maricica Puica (ROM) 16Sep82 4:15.8 Natalya Artyomova (SOV) 05Aug84 4:16.71 Mary Decker-Slaney (USA) 21Aug85 4:15.61 Paula Ivan (ROM) 10Jul89 4:12.56 Svetlana Masterkova (RUS) 14Aug96
## References
• Bennett, William (March 15, 1993), "Quantifying America's Decline", Wall Street Journal{{citation}}: CS1 maint: date and year (link)
• Dalal, Siddhartha; Folkes, Edward; Hoadley, Bruce (Fall 1989), "Lessons Learned from Challenger: A Statistical Perspective", Stats: the Magazine for Students of Statistics, pp. 14–18{{citation}}: CS1 maint: date and year (link)
• Gardner, Martin (April 1970), "Mathematical Games, Some mathematical curiosities embedded in the solar system", Scientific American, pp. 108–112{{citation}}: CS1 maint: date and year (link) |
# Numeric complements
This online calculator calculates radix complement (referred to as r's complement) and diminished radix complement (referred to as (r-1)'s complement) for the given number and the given radix (base).
### Complement numbers
A complement number is a number that, when added to its counterpart, makes some other number, usually a base (radix) of a number system. In such a case, it is called radix complement. For example, 7 complements 3 to 10.
By definition, the formula of the radix complement of an n digit number y in radix b is
$b^n-y$
There is also a diminished radix complement, which is
$(b^n-1)-y$.
A diminished radix complement is easy to get by simply replacing the digits of a number with digits needed to get radix - 1. For example, for the 2 digit decimal number 56, the diminished radix complement is 43. Then you can get radix complement by simply adding the one to the diminished radix complement: 43+1=44
For the decimal system, a radix complement is known as ten's complement (10's complement), and a diminished radix complement is known as nines' complement (9's complement).
Generally, complements are used to represent a symmetric range of positive and negative integers. In other words, half of the range represents positive numbers, and their complements represent negative numbers. That is, for ten's complement, if we consider only one digit, i.e., range from 0 to 9, 3 represents +3, and 7 represents -3.
This allows technique known as method of complements, where you can calculate subtraction as addition of subtrahend complement, f.e. 622 - 451 is 622 + 549 = 1171 = 171 (the leading 1 is omitted from the result).
For the binary system radix complement is known as two's complement (2's complement) and diminished radix complement as one's complement (1's complement). One's complement can be obtained by simply inverting bits of a number. Two's complement is used in computers to represent negative integers. You can read more here: One's complement, and two's complement binary codes.
URL copiado para a área de transferência |
# 2.2 - Skip Counting
The student will
a) count forward by twos, fives, and tens to 120, starting at various multiples of 2, 5, or 10;
b) count backward by tens from 120; and
c) use objects to determine whether a number is even or odd.
### BIG IDEAS
So that I can round numbers to get a close, easy-to-use number when the exact number is not neededSo that I can group numbers and count by groups to get to larger numbers So that I can understand that if I can count by 10, I can more easily round numbers and understand place value
### UNDERSTANDING THE STANDARD
• Collections of objects can be grouped and skip counting can be used to count the collection.
• The patterns developed as a result of grouping and/or skip counting are precursors for recognizing numeric patterns, functional relationships, concepts underlying money, and telling time. Powerful models for developing these concepts include counters, number charts (e.g., hundreds charts, 120 charts, 200 charts, etc.) and calculators.
• Skip counting by fives lays the foundation for reading a clock to the nearest five minutes and counting nickels.
• Skip counting by tens lays the foundation for use of place value and counting dimes.
• Calculators can be used to display the numeric patterns resulting from skip counting. Use the constant feature of the four-function calculator to display the numbers in the sequence when skip counting by that constant.
• Odd and even numbers can be explored in different ways (e.g., dividing collections of objects into two equal groups or pairing objects). When pairing objects, the number of objects is even when each object has a pair or partner. When an object is left over, or does not have a pair, then the number is odd.
### ESSENTIALS
The student will use problem solving, mathematical communication, mathematical reasoning, connections, and representations to
• Determine patterns created by counting by twos, fives, and tens to 120 on number charts. (a)
• Describe patterns in skip counting and use those patterns to predict the next number in the counting sequence. (a)
• Skip count by twos, fives, and tens to 120 from various multiples of 2, 5 or 10, using manipulatives, a hundred chart, mental mathematics, a calculator, and/or paper and pencil. (a)
• Skip count by two to 120 starting from any multiple of 2. (a)
• Skip count by five to 120 starting at any multiple of 5. (a)
• Skip count by 10 to 120 starting at any multiple of 10. (a)
• Count backward by 10 from 120. (b)
• Use objects to determine whether a number is even or odd (e.g., dividing collections of objects into two equal groups or pairing objects). (c)
### KEY VOCABULARY
Updated: Aug 22, 2018 |
15
Q:
# A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the wheel.
A) 14 B) 13 C) 12 D) 11
Explanation:
Distance covered in one revolution = $88×10001000$= 88m.
Q:
Find the area of the trapezium, whose parallel sides are 12 cm and 10 cm, and distance between the parallel sides is 14 cm?
A) 121 sq.com B) 154 sq.com C) 186 sq.com D) 164 sq.com
Explanation:
We know that,
Area of trapezium = 1/2 x (Sum of parallel sides) x (Distance between Parallel sides)
= 1/2 x (12 + 10) x 14
= 22 x 14/2
= 22 x 7
= 154 sq. cm
16 505
Q:
The length and the breadth of rectangular field are in the ratio of 8 : 7. If charges of the painting the boundary of rectangle is at Rs. 10 per meter is Rs. 3000. What is the area of rectangular plot?
A) 5600 sq.m B) 1400 sq. m C) 4400 sq.m D) 3600 sq.m
Explanation:
Perimeter of the rectangle is given by 3000/10 = 300 mts
But we know,
The Perimeter of the rectangle = 2(l + b)
Now,
2(8x + 7x) = 300
30x = 300
x = 10
Required, Area of rectangle = 8x x 7x = 56 x 100 = 5600 sq. mts.
11 608
Q:
The perimeter of a rectangle whose length is 6 m more than its breadth is 84 m. What will be the area of the rectangle?
A) 333 sq.mts B) 330 sq.mts C) 362 sq.mts D) 432 sq.mts
Explanation:
Let the breadth of the rectangle = b mts
Then Length of the rectangle = b + 6 mts
Given perimeter = 84 mts
2(L + B) = 84 mts
2(b+6 + b) = 84
2(2b + 6) = 84
4b + 12 = 84
4b = 84 - 12
4b = 72
b = 18 mts
=> Length = b + 6 = 18 + 6 = 24 mts
Now, required Area of the rectangle = L x B = 24 x 18 = 432 sq. mts
14 961
Q:
How many square units in 13 by 9?
A) 13 B) 9 C) 117 D) 13/9
Explanation:
Number of square units in 13 by 9 is given by the area it forms with length and breadth as 13 & 9
Area = 13 x 9 = 117
Hence, number of square units in 13 by 9 is 117 sq.units.
13 2008
Q:
Square units 13 by 9 of an office area is
A) 97 B) 117 C) 107 D) 127
Explanation:
Square units 13 by 9 of an office means office of length 13 units and breadth 9 units.
Now its area is 13x 9 = 117 square units or units square.
17 4552
Q:
Find the area of the square whose side is equal to the diagonal of a rectangle of length 3 cm and breadth 4 cm.
A) 25 sq.cm B) 16 sq.cm C) 9 sq.cm D) 4 sq.cm
Explanation:
Given length of the rectangle = 3 cm
Breadth of the rectangle = 4 cm
Then, the diagonal of the rectangle
Then, it implies side of square = 5 cm
We know that Area of square = S x S = 5 x 5 = 25 sq.cm.
13 2398
Q:
The length of a class room floor exceeds its breadth by 25 m. The area of the floor remains unchanged when the length is decreased by 10 m but the breadth is increased by 8 m. The area of the floor is
A) 5100 sq.m B) 4870 sq.m C) 4987 sq.m D) 4442 sq.m
Explanation:
Let the breadth of floor be 'b' m.
Then, length of the floor is 'l = (b + 25)'
Area of the rectangular floor = l x b = (b + 25) × b
According to the question,
(b + 15) (b + 8) = (b + 25) × b
2b = 120
b = 60 m.
l = b + 25 = 60 + 25 = 85 m.
Area of the floor = 85 × 60 = 5100 sq.m.
33 4058
Q:
The sides of a right-angled triangle are 12 cm, 16 cm, 20 cm respectively. A new right angle Δ is made by joining the midpoints of all the sides. This process continues for infinite then calculate the sum of the areas of all the triangles so made.
A) 312 sq.cm B) 128 sq.cm C) 412 sq.cm D) 246 sq.cm |
## Engage NY Eureka Math 6th Grade Module 1 Lesson 7 Answer Key
### Eureka Math Grade 6 Module 1 Lesson 7 Example Answer Key
Example 1
Which of the following correctly models that the number of red gumballs is $$\frac{5}{3}$$ the number of white gumballs?
a.
b.
c.
d.
Answer:
b.
Example 2.
The duration of two films are modeled below.
a. The ratio of the length of Film A to the length of Film B is _______ : _______.
Answer:
The ratio of the length of Film A to the length of Film B is 5: 7.
b. The length of Film A is of the length of Film B.
Answer:
The length of Film A is of the length of Film B.
c. The length of Film B is of the length of Film A.
Answer:
The length of Film B is of the length of Film A.
### Eureka Math Grade 6 Module 1 Lesson 7 Exercise Answer Key
Exercise 1
Sammy and Kaden went fishing using live shrimp as bait. Sammy brought 8 more shrimp than Kaden brought. When they combined their shrimp they had 32 shrimp altogether.
a. How many shrimp did each boy bring?
Answer:
Kaden brought 12 shrimp. Sammy brought 20 shrimp.
b. What is the ratio of the number of shrimp Sammy brought to the number of shrimp Kaden brought?
Answer:
20: 12
c. Express the number of shrimp Sammy brought as a fraction of the number of shrimp Kaden brought.
Answer:
$$\frac{20}{12}$$
d. What is the ratio of the number of shrimp Sammy brought to the total number of shrimp?
Answer:
20: 32
e. What fraction of the total shrimp did Sammy bring?
Answer:
$$\frac{20}{32}$$
Exercise 2.
A food company that produces peanut butter decides to try out a new version of its peanut butter that is extra crunchy, using twice the number of peanut chunks as normal. The company hosts a sampling of its new product at grocery stores and finds that 5 out of every 9 customers prefer the new extra crunchy version.
a. Let’s make a list of ratios that might be relevant for this situation.
i. The ratio of number preferring new extra crunchy to total number surveyed is __________.
Answer:
The ratio of number preferring new extra crunchy to total number surveyed is 5 to 9 .
ii. The ratio of number preferring regular crunchy to the total number surveyed is __________.
Answer:
The ratio of number preferring regular crunchy to the total number surveyed is 4 to 9 .
iii. The ratio of number preferring regular crunchy to number preferring new extra crunchy is __________.
Answer:
The ratio of number preferring regular crunchy to number preferring new extra crunchy is 4 to 5 .
iv. The ratio of number preferring new extra crunchy to number preferring regular crunchy is __________.
Answer:
The ratio of number preferring new extra crunchy to number preferring regular crunchy is 5 to 4 .
b. Let’s use the value of each ratio to make multiplicative comparisons for each of the ratios we described here.
i. The number preferring new extra crunchy is _________ of the total number surveyed.
Answer:
The number preferring new extra crunchy is $$\frac{5}{9}$$ of the total number surveyed.
ii. The number preferring regular crunchy is _________ of the total number surveyed.
Answer:
The number preferring regular crunchy is $$\frac{4}{9}$$ of the total number surveyed.
iii. The number preferring regular crunchy is __________ of those preferring new extra crunchy.
Answer:
The number preferring regular crunchy is $$\frac{4}{5}$$ of those preferring new extra crunchy.
iv. The number preferring new extra crunchy is _________ of those preferring regular crunchy.
Answer:
The number preferring new extra crunchy is $$\frac{5}{4}$$ of those preferring regular crunchy.
c. If the company is planning to produce 90,000 containers of crunchy peanut butter, how many of these containers should be the new extra crunchy variety, and how many of these containers should be the regular crunchy peanut butter? What would be helpful in solving this problem? Does one of our comparison statements above help us?
Answer:
The company should produce 50,000 containers of new crunchy peanut butter and 40,000 containers of regular crunchy peanut butter.
d. If the company decides to produce 2000 containers of regular crunchy peanut butter, how many containers of new extra crunchy peanut butter would it produce?
Answer:
2,500 new extra crunchy peanut butter containers
e. If the company decides to produce 10,000 containers of new extra crunchy peanut butter, how containers of regular crunchy peanut butter would it produce?
Answer:
8,000 regular crunchy peanut butter containers
f. If the company decides to only produce 3,000 containers of new extra crunchy peanut butter, how many containers of regular crunchy peanut butter would it produce?
Answer:
2,400 regular crunchy peanut butter containers
### Eureka Math Grade 6 Module 1 Lesson 7 Problem Set Answer Key
Question 1.
Maritza is baking cookies to bring to school and share with her friends on her birthday. The recipe requires 3 eggs for every 2 cups of sugar. To have enough cookies for all of her friends, Maritza determined she would need 12 eggs. If her mom bought 6 cups of sugar, does Maritza have enough sugar to make the cookies? Why or why not?
Answer:
Maritza will NOT have enough sugar to make all the cookies because she needs 8 cups of sugar and only has 6 cups of sugar.
Question 2.
Hamza bought 8 gallons of brown paint to paint his kitchen and dining room. Unfortunately, when Hamza started painting, he thought the paint was too dark for his house, so he wanted to make It lighter. The store manager would not let Hamza return the paint but did inform him that if he used $$\frac{1}{4}$$ of a gallon of white paint mixed with 2 gallons of brown paint, he would get the shade of brown he desired. If Hamza decided to take this approach, how many gallons of white paint would Hamza have to buy to lighten the 8 gallons of brown paint?
Answer:
Hamza would need 1 gallon of white paint to make the shade of brown he desires.
### Eureka Math Grade 6 Module 1 Lesson 7 Exit Ticket Answer Key
Alyssa’s extended family Is staying at the lake house this weekend for a family reunion. She is in charge of making homemade pancakes for the entire group. The pancake mix requires 2 cups of flour for every 10 pancakes.
Question 1.
Write a ratio to show the relationship between the number of cups of flour and the number of pancakes made.
Answer:
2: 10
Question 2.
Determine the value of the ratio.
Answer:
$$\frac{2}{10}=\frac{1}{5}$$
Question 3.
Use the value of the ratio to make a multiplicative comparison statement.
a. The number of pancakes made is _______ times the number of cups of flour needed.
Answer:
The number of pancakes made is 5 times the number of cups of flour needed.
b. The number of cups of flour needed is _______ of the number of pancakes made.
Answer:
The number of cups of flour needed is $$\frac{1}{5}$$ of the number of pancakes made.
Question 4.
If Alyssa has to make 70 pancakes, how many cups of flour will she have to use?
Answer:
Alyssa will have to use 14 cups of flour. |
# What Are Powers of Ten?
••• Jupiterimages/BananaStock/Getty Images
Print
The powers of 10 form a set of mathematical notations that allow you to express any number as a product of multiples of 10. Noting numbers in the powers of 10 is a useful way for engineers, mathematicians and students alike to write down very large numbers (or small numbers) instead of having to write a lot of zeros in a row. For example, 5,000 equals 5 multiplied by 1,000, or using the powers of 10 notation, you can say 5,000 equals 5 multiplied by 10 to the power of 3.
## Scientific Notation
Also known as "standard form," scientific notation was given its name because it was first employed by scientists to represent very large and very small numbers. The "power" that 10 is multiplied by is also known as the exponent. These can be found in positive forms representing multiplication and negative forms representing division.
## What Does a Power of 10 Equal?
The notational index of 10 tells you how many places the decimal point should move to the right. Consider what is 1.35 multiplied by 10 to the fourth power, or 1.35 x 10^4. You can calculate it as 1.35 x (10 x 10 x 10 x 10), or 1.35 x 10,000, which equals 13,500. If you were to move the decimal place in 1.35 over by four spots, you would also create 13,500.
## Negative Powers of 10
When you see a negative power of 10, it indicates how many times the number should be divided by the negative power. Consider the example 5 multiplied by 10 to the negative third power, or 5 x 10^-3. Whether you write the equation out as 5 divided by 10, divided by 10, divided by 10, or you simply move the decimal place over to the left three spaces, you will arrive at 0.005, which is the numerical result of multiplying 5 by 10 to the negative third power.
## Practical Example
A practical example of the powers of 10 is the way that a light year, or the distance that light travels in the span of one year, is expressed using scientific notation rather than writing out the entire numerical representation. For scientists, it is much easier to write and work with the expression 9.461 x 10^15 meters rather than 9,461,000,000,000,000 meters.
Dont Go!
We Have More Great Sciencing Articles! |
##### Get a free home demo of LearnNext
Available for CBSE, ICSE and State Board syllabus.
Call our LearnNext Expert on 1800 419 1234 (tollfree)
OR submit details below for a call back
clear
Triangle Area of the triangle = $\frac{\text{1}}{\text{2}}$ x base x height. Quadrilateral To find the area of a quadrilateral divide the quadrilateral into two triangles and add the areas of the two triangles. The method of dividing a quadrilateral into two triangles to find out its area is known as triangulation. Let ABCD is a quadrilateral. Then Area of quadrilateral ABCD = (Area of ∆ ABD) + (Area of ∆ BCD) Area of quadrilateral ABCD = $\frac{\text{1}}{\text{2}}$d (h1 + h2), where d is the diagonal and h1 and h2 are the heights of the quadrilateral. Square Area of the square = a2 , where a be the length of each side of a square. Perimeter of square = 4a. Diagonal of a square = √2a Rectangle Area = length x breadth Perimeter = 2( length + breadth) Diagonal = , where l = length b = breadth. Parallelogram In a parallelogram the diagonal divide it into two triangles. Area of a Parallelogram = base × height Perimeter = 2( Sum of two adjacent sides) Rhombus In a rhombus diagonals are perpendicular to each other, so we can use the method of triangulation to find the area of a rhombus. Area of a Rhombus = $\frac{\text{1}}{\text{2}}$ (d1 x d2), where d1 and d2 are the lengths of the diagonals. Side of the rhombus = $\frac{\sqrt{{{\text{d}}_{\text{1}}}^{\text{2}}\text{+}{{\text{d}}_{\text{2}}}^{\text{2}}}}{\text{2}}$ Perimeter = 4 x (Side) = 2 Area of rhombus DEFG = (½ x EO x DF) + (½ x GO x DF) = ½ x DF x (DF + GO) = ½ x DF x GE = ½ x d1 x d2 Trapezium A trapezium has a pair of parallel sides. Area of trapezium = h $\frac{\text{(a + b )}}{\text{2}}$ Where h = perpendicular distance between the parallel sides of a trapezium and a and b are the lengths of the parallel sides. A polygon is a closed shape that has at least three sides. It is a closed shape that has at least three sides. Triangles, quadrilaterals, rectangles and squares are all types of polygons. Moreover, a polygon can be of any shape and can have any number of sides. There is no specific formula for calculating the area of a polygon. The best way is to split the polygon into shapes whose area can be calculated individually.
#### Summary
Triangle Area of the triangle = $\frac{\text{1}}{\text{2}}$ x base x height. Quadrilateral To find the area of a quadrilateral divide the quadrilateral into two triangles and add the areas of the two triangles. The method of dividing a quadrilateral into two triangles to find out its area is known as triangulation. Let ABCD is a quadrilateral. Then Area of quadrilateral ABCD = (Area of ∆ ABD) + (Area of ∆ BCD) Area of quadrilateral ABCD = $\frac{\text{1}}{\text{2}}$d (h1 + h2), where d is the diagonal and h1 and h2 are the heights of the quadrilateral. Square Area of the square = a2 , where a be the length of each side of a square. Perimeter of square = 4a. Diagonal of a square = √2a Rectangle Area = length x breadth Perimeter = 2( length + breadth) Diagonal = , where l = length b = breadth. Parallelogram In a parallelogram the diagonal divide it into two triangles. Area of a Parallelogram = base × height Perimeter = 2( Sum of two adjacent sides) Rhombus In a rhombus diagonals are perpendicular to each other, so we can use the method of triangulation to find the area of a rhombus. Area of a Rhombus = $\frac{\text{1}}{\text{2}}$ (d1 x d2), where d1 and d2 are the lengths of the diagonals. Side of the rhombus = $\frac{\sqrt{{{\text{d}}_{\text{1}}}^{\text{2}}\text{+}{{\text{d}}_{\text{2}}}^{\text{2}}}}{\text{2}}$ Perimeter = 4 x (Side) = 2 Area of rhombus DEFG = (½ x EO x DF) + (½ x GO x DF) = ½ x DF x (DF + GO) = ½ x DF x GE = ½ x d1 x d2 Trapezium A trapezium has a pair of parallel sides. Area of trapezium = h $\frac{\text{(a + b )}}{\text{2}}$ Where h = perpendicular distance between the parallel sides of a trapezium and a and b are the lengths of the parallel sides. A polygon is a closed shape that has at least three sides. It is a closed shape that has at least three sides. Triangles, quadrilaterals, rectangles and squares are all types of polygons. Moreover, a polygon can be of any shape and can have any number of sides. There is no specific formula for calculating the area of a polygon. The best way is to split the polygon into shapes whose area can be calculated individually.
#### References
###### Like NextGurukul? Also explore our advanced self-learning solution LearnNext
Offered for classes 6-12, LearnNext is a popular self-learning solution for students who strive for excellence
Explore
Animated Video
lessons
All India
Test Series
Interactive Video
Experiments
Best-in class
books |
# PSAT Math : Acute / Obtuse Triangles
## Example Questions
← Previous 1 3
### Example Question #1 : Acute / Obtuse Triangles
You are given triangles and , with and . Which of these statements, along with what you are given, is not enough to prove that ?
I) and have the same perimeter
II)
III)
Any of the three statements is enough to prove congruence.
None of these statements is enough to prove congruence.
Statement I only
Statement II only
Statement III only
Statement III only
Explanation:
If and have the same perimeter, , and , it follows that . The three triangles have the same sidelengths, setting the conditions for the Side-Side-Side Congruence Postulate.
If , then, since the sum of the degree measures of both triangles is the same (180 degrees), it follows that . Since and are congruent included angles of congruent sides, this sets the conditions for the SAS Congruence Postulate.
In both of the above cases, it follows that .
However, similarly to the previous situation, if , then it follows that , meaning that we have congruent sides and congruent nonincluded angles. However, this is not sufficient to prove congruence.
"Statement III" is the correct response.
### Example Question #1 : How To Find The Perimeter Of An Acute / Obtuse Triangle
If a = 7 and b = 4, which of the following could be the perimeter of the triangle?
I. 11
II. 15
III. 25
II and III Only
II Only
I Only
I and II Only
I, II and III
II Only
Explanation:
Consider the perimeter of a triangle:
P = a + b + c
Since we know a and b, we can find c.
In I:
11 = 7 + 4 + c
11 = 11 + c
c = 0
Note that if c = 0, the shape is no longer a trial. Thus, we can eliminate I.
In II:
15 = 7 + 4 + c
15 = 11 + c
c = 4.
This is plausible given that the other sides are 7 and 4.
In III:
25 = 7 + 4 + c
25 = 11 + c
c = 14.
It is not possible for one side of a triangle to be greater than the sum of both of the other sides, so eliminate III.
Thus we are left with only II.
### Example Question #1 : How To Find An Angle In An Acute / Obtuse Triangle
If the average of the measures of two angles in a triangle is 75o, what is the measure of the third angle in this triangle?
75°
65°
50°
40°
30°
30°
Explanation:
The sum of the angles in a triangle is 180o: a + b + c = 180
In this case, the average of a and b is 75:
(a + b)/2 = 75, then multiply both sides by 2
(a + b) = 150, then substitute into first equation
150 + c = 180
c = 30
### Example Question #2 : How To Find An Angle In An Acute / Obtuse Triangle
Which of the following can NOT be the angles of a triangle?
45, 45, 90
1, 2, 177
30, 60, 90
45, 90, 100
30.5, 40.1, 109.4
45, 90, 100
Explanation:
In a triangle, there can only be one obtuse angle. Additionally, all the angle measures must add up to 180.
### Example Question #3 : How To Find An Angle In An Acute / Obtuse Triangle
Let the measures, in degrees, of the three angles of a triangle be x, y, and z. If y = 2z, and z = 0.5x - 30, then what is the measure, in degrees, of the largest angle in the triangle?
48
96
60
108
30
Explanation:
The measures of the three angles are x, y, and z. Because the sum of the measures of the angles in any triangle must be 180 degrees, we know that x + y + z = 180. We can use this equation, along with the other two equations given, to form this system of equations:
x + y + z = 180
y = 2z
z = 0.5x - 30
If we can solve for both y and x in terms of z, then we can substitute these values into the first equation and create an equation with only one variable.
Because we are told already that y = 2z, we alreay have the value of y in terms of z.
We must solve the equation z = 0.5x - 30 for x in terms of z.
z + 30 = 0.5x
Mutliply both sides by 2
2(z + 30) = 2z + 60 = x
x = 2z + 60
Now we have the values of x and y in terms of z. Let's substitute these values for x and y into the equation x + y + z = 180.
(2z + 60) + 2z + z = 180
5z + 60 = 180
5z = 120
z = 24
Because y = 2z, we know that y = 2(24) = 48. We also determined earlier that x = 2z + 60, so x = 2(24) + 60 = 108.
Thus, the measures of the three angles of the triangle are 24, 48, and 108. The question asks for the largest of these measures, which is 108.
### Example Question #4 : How To Find An Angle In An Acute / Obtuse Triangle
Angles x, y, and z make up the interior angles of a scalene triangle. Angle x is three times the size of y and 1/2 the size of z. How big is angle y.
18
108
54
36
42
18
Explanation:
We know that the sum of all the angles is 180. Using the rest of the information given we can write the other two equations:
x + y + z = 180
x = 3y
2x = z
We can solve for y and z in the second and third equations and then plug into the first to solve.
x + (1/3)x + 2x = 180
3[x + (1/3)x + 2x = 180]
3x + x + 6x = 540
10x = 540
x = 54
y = 18
z = 108
### Example Question #1 : How To Find An Angle In An Acute / Obtuse Triangle
In the picture above, is a straight line segment. Find the value of .
Explanation:
A straight line segment has 180 degrees. Therefore, the angle that is not labelled must have:
We know that the sum of the angles in a triangle is 180 degrees. As a result, we can set up the following algebraic equation:
Subtract 70 from both sides:
Divide by 2:
### Example Question #6 : How To Find An Angle In An Acute / Obtuse Triangle
An exterior angle of an isosceles triangle measures . What is the least measure of any of the three interior angles of the triangle?
Insufficient information is given to answer this question.
Insufficient information is given to answer this question.
Explanation:
The triangle has an exterior angle of , so it has an interior angle of . By the Isosceles Triangle Theorem, an isosceles triangle must have two congruent angles; there are two possible scenarios that fit this criterion:
I: One of the other angles also has measure .
In this case, since the angles' measures must total the third angle has measure
In this case, the least measure is .
II: The other two angles are the congruent angles.
In this case, the other two angles have measures totaling
.
They have the same measure, so each has measure half this, or
.
In this case, the least measure is
Therefore, insufficient information is given to answer this question.
### Example Question #7 : How To Find An Angle In An Acute / Obtuse Triangle
An exterior angle of an isosceles triangle measures . What is the greatest measure of any of the three angles of the triangle?
Insufficient information is given to answer this question.
Explanation:
The triangle has an exterior angle of , so it has an interior angle of . This is an obtuse angle; the other two angles must be acute, and therefore, they will have measure less than - and, subsequently, the angle will be the one of greatest measure. |
# What is the derivative of x^((1/5)(lnx))?
$\frac{2}{5} \cdot {x}^{\left(\frac{1}{5} \cdot \ln x - 1\right)} \cdot \ln x$
#### Explanation:
Let $y = {x}^{\left(\frac{1}{5} \cdot \ln x\right)}$ then take logarithm of both sides of the equation
$\ln y = \ln {x}^{\left(\frac{1}{5} \cdot \ln x\right)}$
$\ln y = \left(\frac{1}{5} \ln x\right) \cdot \left(\ln x\right)$
$\ln y = \frac{1}{5} \cdot {\left(\ln x\right)}^{2}$
$\frac{1}{y} \cdot y ' = \frac{1}{5} \cdot 2 \cdot \ln x \cdot \frac{1}{x}$
then solve for $y '$
$y ' = \frac{2 y}{5 x} \ln x$
replace $y$ with the equivalent ${x}^{\left(\frac{1}{5} \cdot \ln x\right)}$
$y ' = \frac{2 {x}^{\left(\frac{1}{5} \cdot \ln x\right)}}{5 x} \ln x$
then simplification
$y ' = \frac{2}{5} \cdot {x}^{\left(\frac{1}{5} \cdot \ln x - 1\right)} \cdot \ln x$ |
# Maths Tricks for Competitive Exams
## Mathematics, Study Tips, Test Preparation
### .
Any competitive exam requires you to deal with maths and its applications. Not all can be Ramanujam, but with certain techniques and procedures, we can definitely be in a better position to deal with them and score well.
In fact, these maths tricks for competitive exams can be mastered not just for exam purposes but also to become efficient in solving problems. Remember anything learned never goes wasted.
So read on and practice the methods mentioned to become a whiz kid in Maths!
## Finding the square root:
Calculating a square root takes uptime, which is precious during an exam. So you have to know a quicker way of doing it. Let us teach you how to work on a trick to make it happen.
Example:
Find the square root of 784.
Step 1: Check the digit at one’s place. Here it is 4.
Step 2 : Think about the square roots of all the digits from 1-10. Ask yourself, in which square root the digit 4 crops up at one’s place?
At 2 2 = 4 and 8 2 = 64.
Step 3 :
So 784
2 8
Step 4:
Now we look at the remaining number i.e. 7. To which number’s square root is 7 closer to ?
2 2 = 4 and 3 2 =9. We will opt for the smaller of the two numbers, i.e. 2.
Add the 2 to 2 and 8.
784
2 2 2 8
Step 5:
Add 1 to 2 , 1 + 2 = 3. Multiply 2 x 3 = 6.
Step 6:
Compare 7 to the product 6. Which is greater? 7 is. So 28 is our answer.
To check our answer : 28 x 28 = 784.
Practicing the technique a few times, it will increase your speed and you will have your answers in a jiffy!
Tip : It will help you if you can learn the square roots of numbers from 1-10
1 2=1
2 2=4
3 2=9
4 2=16
5 2=25
6 2=36
7 2=49
8 2=64
9 2=81
10 2=100
Tip : And if you observe, you will see the following pattern:
Unit’s place digit of both 12 and 92 is 1.
Unit’s place digit of both 22 and 82 is 4
Unit’s place digit of both 32 and 72 is 9
And unit’s place digit of both 42 and 62 is 6.
## Find the cube root:
Similar to finding a technique to quickly calculate the square root, we have a trick for calculating cube root as well.
Example:
Find the cube of 39304.
Step 1 : Check the digit at one’s place: 4.
Step 2: Think of all the cube roots of numbers between 1-10, where 4 is in one’s place.
4 3 = 64. So we will put the number 4, whose cube root is 64, in the one’s place of the cube root of 39304.
∛39304 = ______ 4
Step 3: Now take the first two digits of the number, 39 and think between which two cube roots does it get placed.
3 3 = 27 and 4 3 = 64
Step 4: We will choose the smaller number which is 27and take the number it is the cube root value of i.e. 3.
We will place that 3 in front of the 4 we took for one’s place as the cube root of 39304
∛39304 = 34
We can check our answer: 34x 34x 34 = 39304
Tip: Again it will help you if you can remember the cube roots of digits from 1-10.
Tip: To find the unit place of the cube root always remember the following points:
• If the last digit of a cube root is 8 then the unit digit will be 2.
• If the last digit of a cube root is 2 then the unit digit will be 8.
• If the last digit of a cube root is 3 then the unit digit will be 7.
• If the last digit of a cube root is 7 then the unit digit will be 3.
• If the last digit of a cube root is other than 2, 3, 7 and 8 then put the same number as the unit digit.
Not an easy topic to handle from Class 9 and even in competitive exams, they can be a challenge. The shortcut method can be used to calculate without any hiccups.
Example:
2x 2 — 11x –21 = 0
Step 1: This means, a+b= –11
And ab = 2 (–21) = –42
Step 2 : We know that –14 and 3 will give -11
And 14×3 = 42
We change the signs , -14 = 14, 3 = –3
Step 3 : Divide both numbers by co -efficient of x 2 = 2
14 /2=7 and –3/2= can’t be simplified
Ans= 7 and –3/2
## Number Sequence
A favourite for any person setting the question paper are number series questions. It is one problem that consumes a lot of effort as well during an exam.
The only way to do it is to determine the relationship between two consecutive numbers. This leads to the answer in the series or sequence. Let us see one such relationship.
Example:
5 ,11 ,24 ,51 ,106, 217, ________
Step 1: Check for the difference between the numbers. The difference between 11 and 5 is 6 ( 5×2+1) , the difference between 24 and 11 is 13 ( 11×2+2), the difference between 51 and 24 is 27 ( 24×2 + 3)
Step 2: Determine the pattern based on differences. Here is the pattern:
5 x 2 + 1 =11
11 x 2 + 2 = 24
24 x 2 + 3 = 51
51 x 2 + 4 = 106
106 x 2 + 5 = 217
217 x 2 + 6 = 440
So the next number in sequence is 440.
### Rule of 72
It is one of the most amazing rules in this subject that can be easily understood and applied in questions where a sum of money has to be doubled in a specific period of time at a given rate of interest.
Formulas to remember:
Number of years invested = 72/ Annual Investment Rate
Investment Rate = 72/ Number of years Invested
Investment rate x Number of years invested = 72
Example:
If you invested Rs.1000 /- in a business, then how much time will it take to double your investment, if the rate of interest is 5%?
Step 1: According to rule of 72,
Time duration in which the amount will be doubled at 5% interest rate = 72/5 = 14.4 years
Step 2 :To re check the answer
Investment rate x Number of years invested = 72
5 x 14.4 =72
## Time and Work
A favourite question for maths instructors is the time work question. These problems usually relate to how much time would a work be done in, by a given number of people.
Example
If Satish completes his work in 8 days and Mahesh completes the same work in 10 days, how many days do they require to finish the work together?
Step 1 : Calculate the LCM of 8 and 10 = 40
Step 2: Determine the efficiency of both individuals
Efficiency of Satish = 40/ 8= 5
Efficiency of Mahesh= 40/ 10= 4
Step 3: Time required by them together = LCM/ Total Efficiency = 40/ 9 = 5 days
### Multiplication
A number of questions demand multiplication operation in the exams and we all dread losing time in it. Practice the following method a few times to reduce the time spent on multiplication.
Example:
Multiply: 334 x 217
Step 4 : 3 3 4
╳
X 2 1 7 Multiply 3×2=6, 3×1= 3 . Add 6+3+3= 12 . Carry forward 1
______
2478
3 3 4
⎹
X 2 1 7
_______ Multiply 3×2= 6. Add 6 + 1= 7
7 2 4 7 8
## Calculate Cost Price with Profit / Loss Percentage given
How to calculate cost price using selling price and profit percent?
We know that cost price = selling price – profit
Formula to calculate cost price if selling price and profit percentage are given:
CP = ( SP * 100 ) / ( 100 + percentage profit).
Formula to calculate cost price if selling price and loss percentage are given:
CP = ( SP * 100 ) / ( 100 – percentage loss ).
Example:
A book was sold for Rs.575 thereby gaining 15%. Find the cost price of the book.
Solution:
Given selling price = Rs. 575
Gain% = 15%
We know, cost price = selling price × 100/100 + gain%
Therefore, cost price = 575 × 100/100 + 15
= 575 × 100/115
= 57500/115
= Rs.500
Therefore, the cost price of the book is \$500.
## Simple Interest and Compound Interest
Different formulas have been created to calculate the various aspects regarding Simple Interest and Compound Interest. Go through them to get a better hang of the calculations to be done.
Example:
If a sum becomes 2 times at a certain rate of interest in 5 years, then the time taken for the same amount to be 8 times at the same rate of interest will be:
(n–1) / 2x t , where n= number of times the amount will be , t= time
8–1 / 2×5 = 7 / 2×5 = 17.5 years
Example:
If the sum of money becomes 6 times in 20 years as simple Interest, what is the rate of interest?
R = 100 (n-1) / T %
= 100 ( 6-1)/ 20
=100 x 5/ 20
= 25 %
Example:
Rs. 2500 was borrowed for 3 years. What will be the compound interest if the rate of interest for first year is 3% per annum, second year is 4% per annum and for third year is 5% per annum respectively?
A = P ( 1 + r / 100) ( 1 + r 2 /100) ( 1 + r 3 /100)
= 2500 x 103 /100 x 26/25 x 21/ 20
Rs. 2811.90
C.I. = 2811.90– 2500 = Rs. 311.90
Example:
The difference of S.I and C.I on an amount of Rs. 30000 for 2 years is Rs. 147. What is the rate of Interest?
C.I. = S.I. = P ( r /100) 2
147 = 30000 ( r/100 ) 2
49/ 10000 = (r/100) 2
7/100= r/ 100
r= 7%
## How to ace in the competitive exams?
1. Thorough study of basics: the shortcuts and techniques provided above can only work if you have mastered the principles of the subject. Otherwise, all your efforts to try and understand the tricks will go in vain.
2. Don’t stop practicing: devise a schedule where you solve maths problems every day. Knowledge of shortcuts can’t fetch results if they aren’t worked with every day. Even the Iron man had to work on his suit constantly to make him invincible!
3. Absorb the formulas: maths is all about formulas and their systematic application. A formula can find its use in various types of problems. So get them at your fingertips.
4. Understand the question: read it a couple of times and think over what it asks for. What information is it providing? Based on it attempt it and work out a solution.
5. Join Edulyte: an innovative learning platform, Edulyte provides world-class guidance in the field of mathematics. With its state of the art learning tools, highly qualified faculty, and extensive resources, learners are exposed to much more than the knowledge in the books. Enjoy the benefits of online and offline learning together in one place.
So what are you waiting for? Register for free today and try out a free demo classes!
The student support team is available to help students find the perfect online classes conducted by top tutors of maths.
## Similar Blogs
Learning lessons, study tips, career guides and much more!
### Trigonometry in Real Life: Trigonometry Calculator Practical Applications
Estimated reading time: 3 minutes Trigonometry is a branch of math that focuses on triangles, especially the relationships between their angles and sides. It might…
### Numbers in Words : Your easy-to-use guide when writing numbers with English letters
Writing numbers in words can be done effectively with a few effortless steps. It is as essential to learning to write numbers in words in…
## Valuable Resources for Maths Learners
### Famous Mathematicians: The Legends and Their Contributions
PTE Tutorials: Customised Packages for Every Learner
Standard
\$75 AUD
One time |
# CLASS-8GEOMETRY-ANGLES-PROBLEM & SOLUTION
There are some examples are given below for your better understanding –
Example.1) If an angle is 4/5 of its supplement, find the angles.
Ans.) Let the angle = x⁰, so its supplement = 180⁰ - x⁰
Given that x⁰ = 4/5 (180⁰ - x⁰)
So, 5x⁰ = 4 X 180⁰ - 4x⁰
So, 9x⁰ = 4 X 180⁰
So, x⁰ = 720⁰ / 9 = 80⁰ (Ans.)
Example.2) Find the values of x& y from the adjoining figure, where POQ is a straight line.
Ans.) here, OT stands on the line PQ. So, ∠POT + ∠QOT = 180⁰,
So, 3x + 22⁰ + 47⁰ = 180⁰
So, 3x = 180⁰ - (22⁰ + 47⁰) = 111⁰
So, x = 111⁰/3 = 37⁰
Being, angles at a point on one side of the line PQ, POR + ROS + SOQ = 180⁰
90⁰ + x + y + 18⁰ = 180⁰
Or, 90⁰ + 37⁰ + y + 18⁰ = 180⁰
Or, y = 180⁰ - 145⁰ = 35⁰
Hence, x = 37⁰ and y = 35⁰ (Ans.)
Example.3) Find the values of x & y from the adjoining figure when x – y = 10⁰
Ans.) ∠COE = vertically opposite ∠DOF = 90⁰
So, sum of the angles at a point on one side of a straight line = 180⁰,
∠AOE + ∠COE + ∠BOC = 180⁰
=> x + 22⁰ + 90⁰ + y + 36⁰ = 180⁰
=> x + y + 148⁰ = 180⁰
=> x + y = 32⁰…………………….(1)
And, as per given condition we have, x – y = 10⁰ …………………………..(2)
Now, we will add (1) & (2), and we find -
x + y = 32⁰
x – y = 10⁰
+ - +
----------------------
2x = 42⁰
So, x = 21⁰
Hence, if we put the value of x in equation (1), then we find –
x + y = 32⁰
or, 21⁰ + y = 32⁰
or, y = 32⁰ - 21⁰ = 11⁰
hence, x = 21⁰ & y = 11⁰ (Ans.)
Example.4) Find the value of x, y, & z from the adjoining figure, where x : y : z = 1 : 2 : 3
Ans.) Let, x = a, y = 2a, and z = 3a
Then, 2z + 3y – 3x = 6a + 6a – 3a = 9a,
5z = 15a, 3y = 6a
So, 2x + y = 2a + 2a = 4a, and 2x = 2a
So, sum of the angles at a point = 360⁰
So, 9a + 15a + 6a + 4a + 2a = 360⁰
Or, 36a = 360⁰
Or, a = 10⁰ or x = 10⁰
Hence we can find also, y = 20⁰, and z = 30⁰
So, x = 10⁰, y = 20⁰, and z = 30⁰ (Ans.) |
Q&A
# what is the lcm of 12 and 8
24
The Least Common Multiple or Lowest Common Multiple of 8 and 12 is 24.
## What is the LCM and HCF of 8 and 12?
Therefore, the HCF of 8 and 12 is 4. Example 2: Find the HCF of 8 and 12, if their LCM is 24. Therefore, the highest common factor of 8 and 12 is 4. Example 3: For two numbers, HCF = 4 and LCM = 24.
## What is the multiple of 8 and 12?
The LCM of 8 and 12 is 24. To find the least common multiple (LCM) of 8 and 12, we need to find the multiples of 8 and 12 (multiples of 8 = 8, 16, 24, 32; multiples of 12 = 12, 24, 36, 48) and choose the smallest multiple that is exactly divisible by 8 and 12, i.e., 24.
## What is the prime factorization 8 and 12?
Common prime factors of 8 and 12 are 2 and 2.
## What is the LCM and LCM of 8 and 12?
LCM of 8 and 12 is 24. LCM also known as Least Common multiple or Lowest common multiple is the smallest or the least positive integer that is divisible by the given set of numbers. Consider the example for finding the LCM of 8 and 12. The answer is 24.
## What is the HCF of 8 and 12?
What is the HCF of 8 and 12? The HCF of 8 and 12 is 4. To calculate the HCF (Highest Common Factor) of 8 and 12, we need to factor each number (factors of 8 = 1, 2, 4, 8; factors of 12 = 1, 2, 3, 4, 6, 12) and choose the highest factor that exactly divides both 8 and 12, i.e., 4.
## What is the LCM of 8 and 12?
The LCM of 8 and 12 is 24.
## What is the HCF of 8?
Factors of 8 = 1, 2, 4 and 8. Therefore, common factor of 6 and 8 = 1 and 2. Highest common factor (H.C.F) of 6 and 8 = 2.
## What are the first multiples of 8 and 12?
The first few multiples of 8 and 12 are (8, 16, 24, 32, . . . ) and (12, 24, 36, 48, 60, . . . )
## What is 3 common multiple of 8 and 12?
The LCM of 3, 8 and 12 is 24.
## What are five common multiples of 8 and 12?
Hence, four common multiples of 8 and 12=24,48,72,96.
## What is the factors of 8 and 12?
There are 3 common factors of 8 and 12, that are 1, 2, and 4. Therefore, the greatest common factor of 8 and 12 is 4.
## What is the prime factorization of 8?
What is the prime factorization of 8? The prime factorization of 8 is 2 × 2 × 2 or 23.
## What is the HCF of 8 and 12 by prime factorization method?
HCF of 8 and 12 by Prime Factorization Prime factorization of 8 and 12 is (2 × 2 × 2) and (2 × 2 × 3) respectively. As visible, 8 and 12 have common prime factors. Hence, the HCF of 8 and 12 is 2 × 2 = 4.
## What is the highest factor of 8 and 12?
The highest common factor (HCF) is found by finding all common factors of two numbers and selecting the largest one. For example, 8 and 12 have common factors of 1, 2 and 4. The highest common factor is 4.
Check Also
Close |
# How To Find The Common Difference Of An Arithmetic Sequence
Posted on
Jika kamu sedang mencari jawaban atas pertanya How To Find The Common Difference Of An Arithmetic Sequence, kamu berada di halaman yang tepat. Kami punya sekitar 10 tanya jawab mengenai How To Find The Common Difference Of An Arithmetic Sequence. Silakan baca lebih lanjut di bawah.
## find the common difference of -34 -29 -24 – 19
Pertanyaan: find the common difference of -34 -29 -24 – 19
-10
## find the sum of the terms of an infinite geometric
Pertanyaan: find the sum of the terms of an infinite geometric sequence whose first term is 4 and common ratio ⅕
” Barisan Geometri “
__________
>>>Diketahui :
a = 4
r = ⅕
________
>>> S∞ = ….
___________
[tex] sf S_{ infty } = frac{a}{1 – r} [/tex]
[tex] sf S_{ infty } = frac{4}{1 – frac{1}{5} } [/tex]
[tex] sf S_{∞} = frac{4}{ frac{4}{5} } [/tex]
[tex] sfto 4 div frac{ 4}{5} \ sf to cancel{4} times frac{5}{ cancel{4}} [/tex]
[tex] boxed{ sf S_{∞} = 5}[/tex]
____________
CMIIW
Ciyo.
## Find the sum of the positive terms of the arithmetic
Pertanyaan: Find the sum of the positive terms of the arithmetic sequence 85, 78, 71,
Jawaban:
78,85,71
Penjelasan dengan langkah-langkah:
maaf kalo salah
## the first term of an arithmetic progression is 3, the
Pertanyaan: the first term of an arithmetic progression is 3, the fourth term is 15 and the 16th term is 63, find the common difference of this progression.
Jawab:
b = difference = 4
Penjelasan dengan langkah-langkah:
U1 = 3, U4 = 15, U16 = 63
U1 = a = 3
U4 = a + 3b
15 = 3 + 3b
3b = 15 – 3 = 12
b = 12/3 = 4
## given the sequence as 1,1,2,3,5,8,13,… . The tenth term of
Pertanyaan: given the sequence as 1,1,2,3,5,8,13,… . The tenth term of that arithmetic sequence is … .
1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55
1 + 1 = 2
2 + 1 = 3
3 + 2 = 5
5 + 3 = 8
8 + 5 = 13
13 + 8 = 21
21 + 13 = 34
34 + 21 = 55
The tenth term of that arithmetic sequence is 55
Jawab:
The tenth term of that arithmetic sequence is 89
Penjelasan dengan langkah-langkah:
The question is a question that uses Fibonacci numbers. Fibonacci is a sequence that starts from 0 and 1, then the next number is obtained by adding the two previous consecutive numbers.
1,1,2,3,5,8,13,21,34,55,89
Subject : Math
Class : VIII ( Junior High School)
Material : Fibonacci Number
Question code : 2
Categorization code : 8,2
I hope this helps^^
## the difference between the tenth term and the seventh term
Pertanyaan: the difference between the tenth term and the seventh term of an arithmetic sequence is -60.the twelfth term divided by the sixth term is 2.find the first term and the common difference.
U10-U7= -60
a= first term , d= common difference
a+9d – (a+6d) = -60
a+9d -a -6d =-60
3d= -60
d= -20
U12/U6 = 2
U12=2U6
a+11d=2(a+5d)
a+11d=2a+10d
d=a=-20
## in arithmetic sequence, the sum of the first ten terms
Pertanyaan: in arithmetic sequence, the sum of the first ten terms is 125 and the third term is 5. Find the first term, the common difference and the sum of the first 15 terms
### Arithmetic Sequence
The nth term (Un)
Un = a + (n – 1) b
The sum of the first n terms (Sn)
Sn = n/2 × (a + Un)
Sn = n/2 × (2a + (n – 1) b)
a = the first term
b = the common difference
==================================
S₁₀ = 125
U₃ = 5
a = ?
b = ?
S₁₅ = ?
S₁₀ = 125
10/2 × (2a + (10 – 1) b) = 125
5 × (2a + 9b) = 125
2a + 9b = 25 … eq (1)
U₃ = 5
a + (3 – 1) b = 5
a + 2b = 5 … eq (2)
To find a and b, use the elimination/substitution
▪︎Find the first term
Eliminate variable b to find a
2a + 9b = 25 (×2)
a + 2b = 5 (×9)
4a + 18b = 50
9a + 18b = 45
___________ –
-5a = 5
a = -1
The first term is -1
▪︎Find the common difference
To find the common difference, substitute a = -1 to eq (2)
a + 2b = 5
-1 + 2b = 5
2b = 5 + 1
2b = 6
b = 3
The common difference is 3
▪︎Find the sum of the first 15 terms
S₁₅ = 15/2 × [2(-1) + (15 – 1)(3)]
S₁₅ = 15/2 × [-2 + (14)(3)]
S₁₅ = 15/2 × (-2 + 42)
S₁₅ = 15/2 × 40
S₁₅ = 15 × 20
S₁₅ = 300
The sum of the first 15 terms is 300.
Hope it helps.
## The first term of an arithmetic sequence is 14. The
Pertanyaan: The first term of an arithmetic sequence is 14. The fourth term is 32. Find the common
difference.
The n-th term of an arithmetic sequence is given by:
Un = a + (n – 1)b
Where a the first term, b the common difference. If U4 = 32 and a = 14 then
32 = 14 + (4 – 1)b
18 = 3b
b = 6
The common difference is 6
## Tolong Bantu jwb : 1.One term of an arithmetic sequence
Pertanyaan: Tolong Bantu jwb :
1.One term of an arithmetic sequence is T13 = 30. The common diffrence is d = 3/2
a. Find the first Term.
b. Write a rule for the nth term.
2. In an arithmetic series, T1 = 5 and T20 = 62.
a. Find the common diffrence.
2. Find the value of T15.
3. Find the sum of the first 40 terms. (S40)
3. Consider the arithmetic series 4+7+10+13+16+19+…
Find the sum of the first 30 terms. (S30)
#Tolong bantu jwb ya gk ada cara juga gpp. Thx Senin di kumpul
3. Arithmetic series
4 + 7 + 10 + 13 + 16 + 19 + …
a = 4 (first term), d = 3 (difference)
Sn = (n/2)(2a + (n – 1)d)
S10 = (10/2)(2(4) + (10 – 1)3)
= 175
## Find the first two terms of an arithmetic sequence if
Pertanyaan: Find the first two terms of an arithmetic sequence if the sixth term is 21 and the sum of the first seventeen terms is 0
Jawaban:
56 and 49
Penjelasan dengan langkah-langkah:
u6 = a + 5b = 21
s17 = 0
u9 = a + 8b = 0
-3b = 21
b = -7
a = 56
so the first two terms are 56 and 49
Tidak cuma jawaban dari soal mengenai How To Find The Common Difference Of An Arithmetic Sequence, kamu juga bisa mendapatkan kunci jawaban atas pertanyaan seperti Tolong Bantu jwb, in arithmetic sequence,, Find the sum, find the sum, and find the common. |
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
• Level: GCSE
• Subject: Maths
• Word count: 1997
# Analysing Triangle Vertices and Bisectors
Extracts from this document...
Introduction
Part 1 The diagram shows a triangle with vertices O(0,0), A(2,6) and B(12,6). The perpendicular bisectors of OA and AB meet a C. (a) In order to write down the perpendicular bisector of the line joining the points A(2,6) and B(12,6), I need to find the line's mid-point. The mid-point of the line joining P(x1, y1) to Q(x2, y2) has the co-ordinates ( ) So the co-ordinates of the midpoint of AB are ( ) = (7,6) As the two points A(2,6) and B(12,6) have the same y-value, the gradient of the line joining the points is 0. This means that the line's perpendicular bisector also has a gradient of 0. Thus the equation of the bisector is x = 7 (b) To find the equation of the perpendicular bisector of the line joining the points O (0,0) and A(2,6), I again need to find the co-ordinates of the mid-point of OA. The gradient, and hence that of the perpendicular bisector, can also be found. Thus, knowing the gradient of the perpendicular bisector and one point on it, I can use y - y1 = m(x - x1) ...read more.
Middle
to B(12,6) = [(12 - 0)� + (6 - 0)�] = (144 + 36) = 180 The length from point A(2,6) to B(12,6) = [(12 - 2)� + (6 - 6)�] = 100 = 10 OA = 40 OB = 180 AB = 10 By knowing these three sides, I can determine an angle between two of them by rearranging the cosine rule into terms of cos A: a� = b� + c� - 2bc cos A 2bc cos A = b� + c� - a� cos A = By labelling OA, AB and OB with a, b and c respectively, I can deduce the angle ABO, labelled A on the diagram. Part 1 (f) Using cos A = cos A = = If cos A = then A = By knowing two sides of the triangle and the angle between them, I can find out the area of the triangle using the equation Area = 1/2 bc sin A Area = 1/2 bc sin A = 1/2 x 10 x 180 x sin 26.56505118� = 30 Area of triangle OAB = 30 (g) ...read more.
Conclusion
(d) To determine whether or not the two tangents to the curve y = 2x� - 9x from their point of intersection are equal in length, I again need to use the equation that gives the distance between two points - as mentioned in part 1(d) and part 1(f): The length of the line joining the point P(x1, y1) to Q(x2, y2) is given by [(x2 - x1)� + (y2 - y1)�] Firstly I shall find the length between the point (1,-7) on the curve and the point of intersection (2.5, -14.5) - as found in part 2(c): [(2.5 - 1)� + (-14.5 + 7)�] = 58.5 = 7.65(3 s.f.) I shall compare this to the length between the point (4, -4) on the curve and the point (2.5,-14.5) [(2.5 - 4)� + (-14.5 + 4)�] = 112.5 = 10.6(3 s.f.) As the length between the point (4,-4) and the point (2.5,-14.5) > the length between the point (1,-7) and the point (2.5,-14.5), the two tangents to the curve y = 2x� - 9x from their point of intersection (2.5,-14.5) are NOT equal in length. ...read more.
The above preview is unformatted text
This student written piece of work is one of many that can be found in our GCSE Gradient Function section.
## Found what you're looking for?
• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month
Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
# Related GCSE Gradient Function essays
1. ## Curves and Gradients Investigation
Also, the power of X in the gradient function (e.g. 3x²) is one less than the power of X in the graph equation (e.g. y = x³). I can therefore summarise the results from these three examples in the following formula: Gradient Function for graph y = xn is nx(n-1)
But as I already have an 'x' in the equation it would look to complicated to add another if it was not absolutely necessary. I am now going to try and work out the gradient, when the tangent is at x=1.5, using the same method as just before.
1. ## The Gradient Function Investigation
The constant 'C' will not affect the gradient of the curve as it will just move the curve up or down the x axis without changing its shape. This means that this part of the equation is not required for calculating purposes.
2. ## Aim: To find out where the tangent lines at the average of any two ...
into the function Intersection = (3,0) 2. Roots 3 & 3 Average of the two roots = Plug the point, (3, 0) into the function Intersection = (-5,0) (Two roots functions also always work similarly as the previous functions, which match my hypothesis quite well.
First I am going to do this without using Omnigraph but with the gradient function that I think applies to all y=axn graphs then I will use Omnigraph to check if the gradients are right. For the curve y=x5, using the gradient function 5x4 I think the gradients will be:-
table below shows the points, which I have used to plot the graph: X Y -4 -64 -3 -27 -2 -8 -1 -1 0 0 1 1 2 8 3 27 4 64 Gradients: L1: Gradient at point X = 2 Y2-Y1 X2-X1 = 12-3.5 = 11.33 rounded to 2.d.p |
# CLASS-7RATIO DIVIDING INTO THREE PARTS
DIVIDING INTO THREE PARTS
If a number n is divided into three parts in the ratio x : y : z, then
x
First part = -------------- X n
x + y + z
y
Second Part = --------------- X n
x + y + z
z
Third Part = --------------- X n
x + y + z
Example.1)
Suppose, Mr. Richard has distributed his whole life savings worth \$ 50000 to his son, daughter, & wife in the ratio of 6 : 5 : 14. so, find out distributed money ?
Ans.) As per the given condition, Mr. Richard has distributed his money worth \$ 50,000 in the ratio of 6 : 5 : 14 to his son, daughter, & wife respectively.
6
So, son will get the money = 50,000 X -------------
6 + 5 + 14
6
= 50,000 X ---------
25
= 2000 X 6 = \$ 12,000
5
So, daughter will get the money = 50,000 X -------------
6 + 5 + 14
5
= 50,000 X ---------
25
= 2000 X 5 = \$ 10,000
14
So, wife will get the money = 50,000 X -------------
6 + 5 + 14
14
= 50,000 X ----------
25
= 2000 X 14 = \$ 28,000 (Ans.)
Example.2)
Suppose, Mr. Devid has distributed his whole property worth \$ 2,50,000 to his son, daughter, wife, & himself in the ratio of 3 : 4 : 6 : 12. so, find out distributed property value ?
Ans.) As per the given condition, Mr. Devid has distributed his money worth \$ 2,50,000 in the ratio of 3 : 4 : 6 : 12 to his son, daughter, & wife respectively.
3
So, son will get the money = 2,50,000 X ----------------
3 + 4 + 6 + 12
3
= 2,50,000 X ---------
25
= 10,000 X 3 = \$ 30,000
4
So, daughter will get the money = 2,50,000 X ---------------
3 + 4 + 6 + 12
4
= 2,50,000 X ---------
25
= 10,000 X 4 = \$ 40,000
6
So, wife will get the money = 2,50,000 X ---------------
3 + 4 + 6 + 12
6
= 2,50,000 X ----------
25
= 10,000 X 6 = \$ 60,000
12
So, he will get the money = 2,50,000 X ---------------
3 + 4 + 6 + 12
12
= 2,50,000 X ----------
25
= 10,000 X 12 = \$ 1,20,000 (Ans.) |
# How do you solve sqrt(3 - x)= 2x?
Jan 31, 2016
$x = - 1 , \frac{3}{4}$
#### Explanation:
$\sqrt{3 - x} = 2 x$
First square both sides.
${\left(\sqrt{3 - x}\right)}^{2} = {\left(2 x\right)}^{2}$
$3 - x = 4 {x}^{2}$
Move all terms to the left side.
$3 - x - 4 {x}^{2} = 0$
Rewrite the expression in standard form.
$- 4 {x}^{2} - x + 3 = 0$
This is a quadratic equation in standard form , $a x + b x + c$, where $a = - 4 , b = - 1 , c = 3$.
Use the a·c method to solve.
Multiply $a \times c$.
$- 4 \times 3 = - 12$
Find two numbers that when multiplied equal $- 12$, and when added equal $- 1$.
$- 4 \mathmr{and} 3$ fit the pattern.
Rewrite the equation with $- 4 x \mathmr{and} 3 x$ in place of $- x$.
$- 4 {x}^{2} - 4 x + 3 x + 3 = 0$
Factor out common terms in the first pair of terms and in the second pair of terms.
$- 4 x \left(x + 1\right) + 3 \left(x + 1\right) = 0$
Factor out $\left(x + 1\right)$.
color(red)((x+1)color(blue)((-4x+3)color(black)(=0)
Set each term in parentheses equal to zero and solve for $x$.
$\textcolor{red}{x + 1 = 0}$
$\textcolor{red}{x = - 1}$
$\textcolor{b l u e}{- 4 x + 3 = 0}$
$\textcolor{b l u e}{- 4 x = - 3}$
Divide both sides by $\textcolor{b l u e}{- 4}$.
$\textcolor{b l u e}{x = \frac{3}{4}}$
Jan 31, 2016
$x = - 1 , \frac{3}{4}$
#### Explanation:
$\sqrt{3 - x} = 2 x$?
Square both sides
${\left(\sqrt{3 - x}\right)}^{2} = {\left(2 x\right)}^{2}$
Simplify.
$3 - x = 4 {x}^{2}$
Move all terms to the left side.
$- 4 {x}^{2} + 3 - x = 0$
Rewrite in standard form.
$- 4 {x}^{2} - x + 3 = 0$
This is a quadratic equation in standard form, $a x + b x + 3$, where $a = - 4 , b = - 1 , c = 3$
We can find the values for $x$ using the quadratic formula.
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
Substitute the known values into the formula.
$x = \frac{- \left(- 1\right) \pm \sqrt{{\left(- 1\right)}^{2} - \left(4 \cdot - 4 \cdot 3\right)}}{2 \cdot - 4}$
Simplify.
$x = 1 \pm \frac{\sqrt{1 - \left(- 48\right)}}{- 8}$
Simplify.
$x = \frac{1 \pm \sqrt{49}}{- 8}$
Simplify.
$\textcolor{p u r p \le}{x = \frac{1 \pm 7}{- 8}}$
$\textcolor{red}{x = \frac{1 + 7}{- 8}}$
color(red)(x=8/(-8)
$\textcolor{red}{x = - 1}$
$\textcolor{b l u e}{x = \frac{1 - 7}{- 8}}$
$\textcolor{b l u e}{x = \frac{- 6}{- 8}}$
$\textcolor{b l u e}{x = \frac{6}{8}}$
Simplify.
$\textcolor{b l u e}{x = \frac{3}{4}}$ |
# Factoring Sums of CubesPractice Problems
##### Problem 1
Step 1
Identify $$\blue a$$ and $$\red b$$.
Since $$\blue a$$ is the cube root of the first term, $$a = \sqrt[3]{x^3} = \blue x$$.
Similarly, since $$\red b$$ is the cube root of the second term, $$b = \sqrt[3] 8 = \red 2$$
Step 2
Write down the factored form.
\begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \blue b^2)\\ x^3 + 8 & = (\blue x + \red 2)(\blue x^2 - \blue x\cdot \red 2 + \red 2^2)\\ & = (x + 2)(x^2 - 2x +4) \end{align*}
$$x^3 + 8 = (x + 2)(x^2 - 2x +4)$$
##### Problem 2
Step 1
Identify $$\blue a$$ and $$\red b$$.
Since $$\blue a$$ is the cube root of the first term, $$a = \sqrt[3]{x^3} = \blue x$$.
Likewise, since $$\red b$$ is the cube root of the second term, $$b = \sqrt[3]{64} = \red 4$$.
Step 2
Write down the factored form.
\begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ x^3 + 64 & = (\blue x + \red 4)(\blue x^2 - \blue x \cdot \red 4 + \blue 4^2)\\ & = (x + 4)(x^2 - 4x + 16) \end{align*}
$$x^3 + 64 = (x + 4)(x^2 - 4x + 16)$$
##### Problem 3
Step 1
Identify $$\blue a$$ and $$\red b$$ .
Since $$\blue a$$ is the cube root of the first term, $$a = \sqrt[3]{8x^3} = \blue { 2x}$$.
Likewise, since $$\red b$$ is the cube root of the second term, $$b = \sqrt[3]{27} = \red 3$$.
Step 2
Write down the factored form.
\begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 8x^3 + 27 & = (\blue{2x} + \red 3)[\blue{(2x)}^2 - \blue{(2x)}\red{(3)} + \red 3^2]\\ & = (2x + 3)(4x^2 - 6x + 9) \end{align*}
$$8x^3 + 27 = (2x + 3)(4x^2 - 6x + 9)$$
##### Problem 4
Step 1
Identify $$a$$ and $$b$$.
Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{64x^3} = 4x$$.
Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{125} = 5$$.
Step 2
Write down the factored form.
\begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 64x^3 + 125 & = (\blue{4x} + \red 5)[\blue{(4x)}^2 - \blue{(4x)}\red{(5)} + \red 5^2]\\ & = (4x + 5)(16x^2 - 20x + 25) \end{align*}
$$(4x + 5)(16x^2 - 20x + 25)$$
##### Problem 5
Step 1
Identify $$a$$ and $$b$$.
Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{x^3} = x$$.
Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{y^3} = y$$.
Step 2
Write down the factored form.
\begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ x^3 + y^3 & = (\blue x + \red y)(\blue x^2 - \blue x \red y + \red y^2) \end{align*}
$$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$
##### Problem 6
Step 1
Identify $$a$$ and $$b$$.
Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{216x^3} = 6x$$.
Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{125y^3} = 5y$$.
Step 2
Write down the factored form.
\begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 216x^3 + 125y^3 & = (\blue{6x} + \red{5y})[\blue{(6x)}^2 - \blue{(6x)}\red{(5y)} + \red{(5y)}^2]\\ & = (6x + 5y)(36x^2 - 30xy + 25y^2) \end{align*}
$$216x^3 + 125y^3 = (6x + 5y)(36x^2 - 30xy + 25y^2)$$
### Sum Of Cubes Calculator
Step 1
Identify $$a$$ and $$b$$.
Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{8x^6} = 2x^2$$.
Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{27y^9} = 3y^3$$.
Step 2
Write down the factored form.
\begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 8x^6 + 27y^9 & = (\blue{2x^2} + \red{3y^3})[\blue{(2x^2)}^2 - \blue{(2x^2)}\red{(3y^3)} + \red{(3y^3)}^2]\\ & = (2x^2 + 3y^3)(4x^4 - 6x^2y^3 + 9y^6) \end{align*}
$$8x^6 + 27y^9 = (2x^2 + 3y^3)(4x^2 - 6x^2y^3 + 9y^6)$$
##### Problem 8
Step 1
Identify $$a$$ and $$b$$.
Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{1000x^{3/2}} = (1000x^{3/2})^{1/3} = 10x^{1/2}$$.
Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{343y^{6/5}} = (343y^{6/5})^{1/3} = 7y^{2/5}$$.
Step 2
Write down the factored form.
\begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 1000x^{3/2} + 343y^{6/5} & = \left(\blue{10x^{1/2}} + \red{7y^{2/5}}\right)\left[\blue{\left(10x^{1/2}\right)}^2 - \blue{\left(10x^{1/2}\right)}\red{\left(7y^{2/5}\right)} + \red{\left(7y^{2/5}\right)}^2\right]\\ & = \left(10x^{1/2} + 7y^{2/5}\right)\left(100x - 70x^{1/2}y^{2/5} + 49y^{4/5}\right) \end{align*}
$$1000x^{3/2} + 343y^{6/5} = \left(10x^{1/2} + 7y^{2/5}\right)\left(100x - 70x^{1/2}y^{2/5} + 49y^{4/5}\right)$$
##### Problem 9
Step 1
Identify $$a$$ and $$b$$.
Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{3x^2}$$.
Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{7y^4}$$.
Step 2
Write down the factored form.
\begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 3x^2 + 7y^4 & = \left(\blue{\sqrt[3]{3x^2}} + \red{\sqrt[3]{7y^4}}\right)\left[\blue{\left(\sqrt[3]{3x^2}\right)}^2 - \blue{\left(\sqrt[3]{3x^2}\right)}\red{\left(\sqrt[3]{7y^4}\right)} + \blue{\left(\sqrt[3]{7y^4}\right)}^2\right]\\ & = \left(\sqrt[3] 3\,x^{2/3} + \sqrt[3] 7\,y^{4/3}\right)\left[\left(\sqrt[3] 3\,x^{2/3}\right)^2 - \sqrt[3]{3x^2\cdot 7y^4} + \left(\sqrt[3]7\,y^{4/3}\right)^2\right]\\ & = \left(\sqrt[3] 3\,x^{2/3} + \sqrt[3] 7\,y^{4/3}\right)\left[\left(\sqrt[3] 3\,x^{2/3}\right)^2 - \sqrt[3]{21}\,x^{2/3}y^{4/3} + \left(\sqrt[3]7\,y^{4/3}\right)^2\right]\\ & = \left(\sqrt[3] 3\,x^{2/3} + \sqrt[3] 7\,y^{4/3}\right)\left(\sqrt[3] 9\,x^{4/3} - \sqrt[3]{21}\,x^{2/3}y^{4/3} + \sqrt[3]{49}\,y^{8/3}\right) \end{align*}
$$3x^2 + 7y^4 = \left(\sqrt[3] 3\,x^{2/3} + \sqrt[3] 7\,y^{4/3}\right)\left(\sqrt[3] 9\,x^{4/3} - \sqrt[3]{21}\,x^{2/3}y^{4/3} + \sqrt[3]{49}\,y^{8/3}\right)$$
##### Problem 10
Step 1
Identify $$a$$ and $$b$$.
Since $$a$$ is the cube root of the first term.
\begin{align*} a & = \sqrt[3]{24x^{21}}\\ & = \sqrt[3]{8\cdot 3 x^{21}}\\ & = 2\sqrt[3]{3 x^{21}}\\ & = 2\sqrt[3]3\,x^7 \end{align*}
Likewise, since $$b$$ is the cube root of the second term,
\begin{align*} b & = \sqrt[3]{375y^{15}}\\ & = \sqrt[3]{125\cdot 3y^{15}}\\ & = 5\sqrt[3] 3\, y^5 \end{align*}
Step 2
Write down the factored form.
\begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 24x^{21} + 375y^{15} & = \left(\blue{2\sqrt[3]3\,x^7} + \red{5\sqrt[3] 3\, y^5}\right)\left[\blue{\left(2\sqrt[3]3\,x^7\right)}^2 - \blue{\left(2\sqrt[3]3\,x^7\right)}\red{\left(5\sqrt[3] 3\, y^5\right)} + \blue{\left(5\sqrt[3] 3\, y^5\right)}^2\right]\\ & = \left(2\sqrt[3]3\,x^7 + 5\sqrt[3] 3\, y^5\right)\left(4\sqrt[3]9\,x^{14} - 10\sqrt[3]9\,x^7y^5 + 25\sqrt[3] 9\, y^{10}\right)\\ & = \sqrt[3] 3\left(2x^7 + 5y^5\right)\cdot \sqrt[3] 9 \left(4x^{14} - 10x^7y^5 + 25y^{10}\right)\\ & = \sqrt[3]{27}\left(2x^7 + 5y^5\right)\left(4x^{14} - 10x^7y^5 + 25y^{10}\right)\\ & = 3\left(2x^7 + 5y^5\right)\left(4x^{14} - 10x^7y^5 + 25y^{10}\right) \end{align*}
$$24x^{21} + 375y^{15} = 3\left(2x^7 + 5y^5\right)\left(4x^{14} - 10x^7y^5 + 25y^{10}\right)$$ |
# Surveying Questions and Answers – Curve Surveying – By Ordinates of the Long Chord
This set of Surveying Multiple Choice Questions & Answers (MCQs) focuses on “Curve Surveying – By Ordinates of the Long Chord”.
1. Find the value of mid-ordinate if the radius of the curve is given as 40.62 m and length as 10.2m.
a) 0.43
b) 0.22
c) 0.12
d) 0.33
Explanation: Mid-ordinate calculation involves the following procedure,
O0 = R – (R2 – (l/2)2)1/2. On substitution, we get
O0 = 40.62 – (40.622 – (10.2/2)2)1/2
O0 = 0.33.
2. For setting the curve, chord must be divided into even number of equal parts.
a) True
b) False
Explanation: While setting a curve, the chord must be divided into even number of equal parts in order to decrease the time of the entire process. After dividing, the offsets are calculated.
3. Which of the following indicates the formula for setting a long chord by using ordinate?
a) Ox = (R2 + (x)2)1/2 – (R – O0)
b) Ox = (R2 – (x)2)1/2 – (R – O0)
c) Ox = (R2 – (x)2)1/2 + (R – O0)
d) Ox = (R2 – (x)2)1/2 – (R + O0)
Explanation: The formula for setting a long chord by using ordinate can be given as Ox = (R2 – (x)2)1/2 – (R – O0). In this O0 is given as mid ordinate, R indicates the radius of the curve, x indicates the distance of the point from mid region.
4. General method can be adopted when radius of the curve is large.
a) False
b) True
Explanation: When the radius of the curve is large, general method might take more time while solving than expected. In order to reduce the time of procedure we generally adopt an approximate method which is only considered in case of large radius than the length of the chord.
5. In approximate method, the value of x is measured from ____________
a) Chord point
b) Mid point
c) Tangent point
d) Secant point
Explanation: In general, the value of x is taken from the midpoint but in case of approximate method the x value is taken from the tangent point. It is so because of the larger radius.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!
6. Which of the following indicates the formula for determining ordinate in an approximate method?
a) Ox = x*(l-x) / 2+R
b) Ox = x*(l-x) / 2*R
c) Ox = x*(l + x) / 2*R
d) Ox = x+ (l-x) / 2*R
Explanation: When the radius of the curve is large, for decreasing the time period of the entire process this process is adopted. It involves calculation of ordinate by assuming perpendicular distance and the formula is given as Ox = x*(l-x) / 2*R.
7. Find the value of ordinate at a distance of 10m having radius of 22.92m with mid-ordinate12.12.
a) 3.289
b) 2.892
c) 8.293
d) 9.823
Explanation: The value of ordinate placed at certain distance x can be found out by using the formula,
Ox = (R2 – (x)2)1/2 – (R – O0). On substitution, we get
Ox = (22.922-(10)2)1/2 – (22.92 – 12.12)
Ox = 9.823.
8. If the value of O0 = 24.62 and R = 4m, find the value of l using the general method of long chords.
a) 1636.73m
b) 1363.73m
c) 1366.73m
d) 1363.37m
Explanation: The general method of the ordinate calculation involves,
O0 = R – (R2 – (l/2)2)1/2. On substitution, we get
24.62 = 4 – (42 – (l/2)2)1/2
l = 1636.73 m.
9. Which of the following indicates the formula for a general method by ordinate of long chords?
a) $$R + (R^2 – (\frac{l}{2})^2)^{1/2}$$
b) $$R * (R^2 – (\frac{l}{2})^2)^{1/2}$$
c) $$R – (R^2 + (\frac{l}{2})^2)^{1/2}$$
d) $$R – (R^2 – (\frac{l}{2})^2)^{1/2}$$
Explanation: The perpendicular which is erected while setting curve by ordinates of long chords, is equal to versed sine of the curve which makes it equal to $$R – (R^2 – (\frac{l}{2})^2)^{1/2}$$.
10. What will be value of ordinate placed at a distance of 20m having radius and length as 72.46m and 42.92m respectively?(use approximate method)
a) 6.13
b) 1.36
c) 3.16
d) 4.86
Explanation: Since the radius of the curve is large, we may consider the approximate method i.e.,
Ox = x*(l-x) / 2*R. On substitution, we get
Ox = 20*(42.92-20) / 2*72.46
Ox = 3.16.
Sanfoundry Global Education & Learning Series – Surveying.
To practice all areas of Surveying, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] |
1. ## Need help.
I'm having a hard to solve this two problems. I would really appreciate your help. Look at the attachement.Thanks.
2. ## Re: Need help.
Problem #1: The area of a triangle is $\dfrac{1}{2}bh$. In this case, the base and height are both determined by the $x$- and $y$-intercepts of the tangent line. In general, points on the parabola given are of the form $(u,1-u^2)$. The slope of a tangent line at the point $(u,1-u^2)$ is given by plugging $u$ into the derivative. The derivative of the parabola is $y' = -2x$, so the slope of the tangent line is $-2u$. Then, in point slope form, $y-y_0 = m(x-x_0)$. Plugging in:
$y-(1-u^2) = -2u(x-u)$
Converting to slope-intercept form:
$y = -2ux + u^2+1 = mx+b$
Since $-2ux = mx$, obviously, $u^2+1 = b$, so the $y$-intercept is $u^2+1$.
Then, what is the $x$-intercept? That is when $y=0$. Plug in for $y=0$ and solve for $x$:
$2ux = u^2+1 \Rightarrow x = \dfrac{u^2+1}{2u}$
This gives you your base and height. So, the area is:
$A = \dfrac{1}{2}bh = \dfrac{1}{2}\dfrac{u^2+1}{2u}(u^2+1) = \dfrac{1}{4}(u+u^{-1})(u^2+1) = \dfrac{1}{4}(u^3+2u+u^{-1})$
Now, to minimize that, take the derivative with respect to $u$ and set it equal to zero to find critical numbers:
$A'(u) = \dfrac{3u^2+2-u^{-2}}{4} = 0$
Multiplying out by $4u^2$ gives:
$3u^4 + 2u^2-1 = 0$
$(u^2+1)\left(u^2-\dfrac{1}{3}\right) = 0$, so $u^2 = -1$ or $u^2 = \dfrac{1}{3}$. Since $\sqrt{-1}$ is not a real number, it must be that $u = \pm \sqrt{\dfrac{1}{3}}$. Since the negative answer is not an $x$-value in the first quadrant, we reject it. Hence, $u = \sqrt{\dfrac{1}{3}}$. Next, check to see if it is truly a minimum. You can use the second derivative test (if the second derivative is positive at $u = \sqrt{\dfrac{1}{3}}$, you found a minimum).
For question 2: By the logarithm rules, $\ln|2x| = \ln 2 + \ln |x|$. Since $\ln 2 \approx 0.693$, it is a constant. Hence, just as the corollary said, $\ln|x| \text{ and } \ln|2x|$ differ by just a constant.
thanks |
# Difference between revisions of "Asymptote (geometry)"
For the vector graphics language, see Asymptote (Vector Graphics Language).
An asymptote is a line or curve that a certain function approaches.
The function $y=\tfrac{2x}{x-2}$ has a vertical asymptote at x=2 and a horizontal asymptote at y=2
Linear asymptotes can be of three different kinds: horizontal, vertical or slanted (oblique).
## Vertical Asymptotes
The vertical asymptote can be found by finding values of $x$ that make the function undefined. Generally, it is found by setting the denominator of a rational function to zero.
If the numerator and denominator of a rational function share a factor, this factor is not a vertical asymptote. Instead, it appears as a hole in the graph.
A rational function may have more than one vertical asymptote.
### Example Problems
Find the vertical asymptotes of 1) $y = \frac{1}{x^2-5x}$ 2) $\tan 3x$.
#### Solution
1) To find the vertical asymptotes, let $x^2-5x=0$. Solving the equation:
$\begin{eqnarray*}x^2-5x&=&0\\x&=&\boxed{0,5}\end{eqnarray*}$
So the vertical asymptotes are $x=0,x=5$.
2) Since $\tan 3x = \frac{\sin 3x}{\cos 3x}$, we need to find where $\cos 3x = 0$. The cosine function is zero at $\frac{\pi}{2} + n\pi$ for all integers $n$; thus the functions is undefined at $x=\frac{\pi}{6} + \frac{n\pi}{3}$.
## Horizontal Asymptotes
For rational functions in the form of $\frac{P(x)}{Q(x)}$ where $P(x), Q(x)$ are both polynomials:
1. If the degree of $Q(x)$ is greater than that of the degree of $P(x)$, then the horizontal asymptote is at $y = 0$. This can be seen by noting that as $x$ increases, $Q(x)$ increases much faster than $P(x)$ does. Since the denominator increases faster than the numerator, as x approaches infinity, y gets smaller until it approaches zero. A similar trend can be seen as x decreases.
2. If the degree of $Q(x)$ is equal to that of the degree of $P(x)$, then the horizontal asymptote is at the quotient of the leading coefficient of $P(x)$ over the leading coefficient of $Q(x)$.
3. If the degree of $Q(x)$ is less than the degree of $P(x)$, see below (slanted asymptotes)
A function may not have more than one horizontal asymptote. Functions with a "middle section" may cross the horizontal asymptote at one point. To find this point, set y=horizontal asymptote and solve.
### Example Problem
Find the horizontal asymptote of $f(x) = \frac{x^2 - 3x + 2}{-2x^2 + 15x + 10000}$.
#### Solution
The numerator has the same degree as the denominator, so the horizontal asymptote is the quotient of the leading coefficients: $y= \frac {1} {-2}$
## Slant (Oblique) Asymptotes
For rational functions $\frac{P(x)}{Q(x)}$, a slant asymptote occurs when the degree of $P(x)$ is one greater than the degree of $Q(x)$. If the degree of $P(x)$ is two or more greater than the degree of $Q(x)$, then we get a curved asymptote. Again, like horizontal asymptotes, it is possible to get crossing points of slant asymptotes.
For rational functions, we can find the slant asymptote simply by long division, omitting the remainder and setting y=quotient.
### Example Problem
The function $y=\tfrac{x^2+2x+4} {x+1}$ has a slant asymptote at y=x+1
Find the slant asymptote of $y= \frac{x^2+2x+4} {x+1}$
Solution
$\frac{x^2+2x+4}{x+1}= x+1+\frac{3} {x+1}$
The slant asymptote is $y=x+1$ |
# TRIGONOMETRY
Subject:
### MATHEMATICS
Term:
First Term
Week:
Week 6
Class:
JSS 3 / BASIC 9
Previous lesson: Pupils have previous knowledge of
### EQUATIONS WITH FRACTIONS
that was taught in their previous lesson
Topic:
TRIGONOMETRY
Behavioural Objectives:
At the end of the lesson, learners will be able to
• solve questions on right angle triangle (Revision)
• Explain the relationship between right angle triangles and trigonometric ratios.
• Solve questions on Sine, cosine and tangent of acute angles using tables.
• Applications of trigonometric ratios to finding distances and lengths.
Instructional Materials:
• Wall charts
• Pictures
• Related Online Video
• Flash Cards
Methods of Teaching:
• Class Discussion
• Group Discussion
• Explanation
• Role Modelling
• Role Delegation
Reference Materials:
• Scheme of Work
• Online Information
• Textbooks
• Workbooks
• 9 Year Basic Education Curriculum
• Workbooks
CONTENT:
WEEK 6
TOPIC: TRIGONOMETRY
CONTENT:
• Right angle triangle (Revision)
• Relationship between right angle triangles and trigonometric ratios.
• Sine, cosine and tangent of acute angles using tables.
• Applications of trigonometric ratios to finding distances and lengths.
Right Angled Triangle
A right-angled triangle has one of its angles equal to 90O. That is, if a triangle has one of its three angles equal to 90O then it is a right-angled triangle. A little square at the corner (angle) signifies that the angle is 90O.
Right Angled Triangle, or a right-angled triangle, is a triangle in which one angle is a right angle (that is, a 90-degree angle). The relation between the sides and angles of a right angled triangle is the basis for trigonometry.
A right angled triangle has two other angles apart from the right angle. These are acute angles and are less than 90 degrees. The third angle is always 90 degrees and is called the hypotenuse. It is the longest side in the triangle and is opposite to the right angle.
The other two sides are called the legs of the triangle. They are shorter than the hypotenuse and are adjacent to the right angle.
Right Angled Triangle and the Trigonometric Ratios
Opposite
Hypotenuse
A trigonometric ratio is a ratio of the lengths of any two sides of a right – angled triangle.The longest side is called the Hypotenuse, the side facing the required angle is called the opposite and the last side is called the adjacent.
A trigonometric ratio is a way of measuring the size of angles. The ratios are based on the lengths of the sides of a right angled triangle. The most common trigonometric ratios are sine, cosine and tangent. These ratios are usually represented by the symbols sin, cos and tan. The sine ratio is defined as the ratio of the length of the side opposite to an angle to the length of the hypotenuse. The cosine ratio is defined as the ratio of the length of the side adjacent to an angle to the length of the hypotenuse. The tangent ratio is defined as the ratio of the length of the side opposite to an angle to the length of the side adjacent to that angle.
These ratios can be used to calculate the sizes of angles in a right angled triangle. The ratios can also be used to solve problems involving triangles that are not right angled. Trigonometric ratios are used in many areas of mathematics, science and engineering. They are particularly useful in fields such as astronomy, physics and calculus.
The three trigonometric ratios are Sine (sin), Cosine (cos) and Tangent (tan).
The figure below show triangle XYZ in various positions
O
H
A
X
Z
Y
Z
H
A
O
Y
X
O
A
Y
H
X
Z
Where H, A, O represent the Hypotenuse, Adjacent and Opposite respectively.
The sides of the triangles are as follows.
XZ, the hypotenuse, XY, the side opposite to and YZ, the side adjacent to
Using these abbreviations hyp, opp andadj, the Trigonometric ratios are given as:
Sin A = Cos A = Tan A =
To obtain the value of any trigonometric ratio, the four-figure table is used.
N.B.: Teacher should teach students how to use the four-figure table to find the values of trigonometric ratios.
Application of Trigonometric Ratios to Finding Distances and Lengths.
Trigonometric ratios can be used to solve real life problems such as angles of elevation and depressions and bearings and other real life situations that result in right-angled triangles.
Use of Sine
Sine and cosine of angles are used to find the lengths of unknown side in triangles. The table below gives the sine of some chosen angles.
Angle A Sin A 30⁰ 0.5000 35⁰ 0.5736 40⁰ 0.6428 45⁰ 0.7071 50⁰ 0.7660 55⁰ 0.8192 60⁰ 0.8660
The values in the table are given to 4 significant figures.
Example
1. Calculate the value of x in the figure below.
20cm
55⁰
Solution
In the figure above, the hypotenuse (hyp) is given and x is opposite (opp) the given angle. Thus, we can only use:
Sin A =
Sin 55⁰ =
Thus, = 20 x sin 55⁰
= 20 x 0.8192
= 16.384 cm
= 16 cm to 2 s.f.
1. Use tables to find the angles:
2. whose sine is = ,
3. whose sin is 0.649 2,
Solution
1. Let the angle be A, then sin A = .
Express as a decimal fraction correct to 4 d.p.
. = 0.285 7
Sin A = 0.285 7
Looking within the sine table entries,
0.2857 is opposite 16⁰ and under 0.6⁰.
Thus, A = 16.6⁰.
1. Let the angle be x, then sin x = 0.649 2.
Looking within the sin table entries, the nearest value to 0.649 2 is 0.648 1. 0.648 1 is opposite 40⁰ and under 0.4⁰.
The difference between 6 492 and 6 481 is 11. Look for 11 in the difference column along the row of 40⁰. 11 is under 8.
Thus, x = 40.40⁰ + 0.08⁰
X = 40.48⁰
Class Activity:
Using the four-figure table, find the values of x, y and z in each of the triangles below.Give all answers correct to 2 s.f.
1. 3.
45⁰
10 cm x y 6 km Z 3 km
30⁰ 50⁰
Use of Cosine
Angle A Cos A 30⁰ 0.8660 35⁰ 0.8192 40⁰ 0.7660 45⁰ 0.7071 50⁰ 0.6428 55⁰ 0.5736 60⁰ 0.5000
Example
1. Calculate the value of y in the figure below.
Y
17 cm
500
Solution
The hypotenuse (hyp) is 17 cm and y is adjacent (adj) to the given angle. Thus use the cosine of the given angle.
Cos A =
Cos 50⁰ =
Thus, = 17 x cos 50⁰
= 17 x 0.642 8
= 10.93 cm
= 11 cm to 2 s.f
1. Use tables to find the angles:
2. whose cosine is 0.4478,
3. whose cosine is 0.568 2,
Solution
1. Let the angle be B, then cos B = 0.4478.
Looking within the cosine table entries,
0.447 8 is opposite 63⁰ and under 0.4.
Thus, B = 63.4⁰.
1. Let the angle be y, then cos y = 0.568 2.
Looking within the cosine table entries, the nearest value to 0.568 2 is 0.567 8. 0.567 8 is opposite 55⁰ and under 0.4⁰.
The difference between 5 682 and 5 678 is 4.
Look for 4 in the difference column along the row of 55⁰.4 is under 3. As the angles increase, their cosines decrease, therefore subtract the difference.
Thus, y = 55.40⁰ – 0.03⁰
Y = 55.37⁰
z
Class Activity:
Using the four-figure table, find the values of x, y and z in each of the triangles below.Give all answers correct to 2 s.f.
1. 3.
22mm x 500
400 12 km 25cm
Y 600
Use of Tangent
Tangents of some chosen angles
Angle A Tan A 25⁰ 0.4663 30⁰ 0.5774 35⁰ 0.7002 40⁰ 0.8391 45⁰ 1.000 50⁰ 1.192 55⁰ 1.428 60⁰ 1.732 65⁰ 2.145 70⁰ 2.747
Example
1. The angle of elevation of the top of a building is 25⁰ from a point 70 m away on a level ground. Calculate the height of the building.
Solution
The angle of elevation from a point 70 m away on a level ground to the top of a building is 25⁰. The height of the building can be found using the following equation:
Height = Distance * Tan (Angle)
Therefore, the height of the building is:
Height = 70 * Tan (25)
Height = 70 * 0.466
Height = 32.62 m
HK represents the height of the building; AK is on level ground.
H
A 25° K
70 m
Tan A =
Tan 250 =
Let HK be x cm. KA = 70 cm and, from table above, tan 25⁰ = 0.4663.
Hence, = 0.4663
X = 0.4663 x70=4.663 x 7
= 32.641
∴The height of the building is 33 m to 2 s.f.
1. Use tables to find the tangents of angles
2. 320 59.60
Solution
1. Looking within the table entries, the number opposite 320and under 0.0 is 0.6249.
Thus, tan 320 = 0.6249.
1. The number opposite 590 and under 0.60is 1.704. Thus, tan 59.60 is 1.704. Thus, tan 59.60 = 1.704.
Class Activity:
Z
10km
Using the four-figure table, find the values of x, y and z in each of the triangles below.Give all answers correct to 2 s.f.
1. 3.
500
x 12cm 4km Y 20km
150
5cm 7km
300
NOTE: Educators are to solve more examples for students.
Assignment:
1. A village is 8 km on a bearing of 0400 from a point O. Calculate how far the village is north of O.
2. Use tables to find the angles whose tangents are:
3. 9556 b.
4. Calculate the length of the hypotenuse of the triangle below.
8 cm
430
1. A car travels 120 m along a straight road which is inclined at 8⁰ to the horizontal. Calculate the vertical distance through which the car rises.
120
8⁰ h
1. A cone is 6 cm high and its vertical angle is 54⁰. Calculate the radius of its base.
Practice Questions:
1. Calculate the lengths a, b, c, d, in the diagrams below. All lengths being in cm.
i ii
5 b c d
370 40.20
1. Calculate the angles ∝, β in the triangles below.
i ii
0.9 m 3m 7 m 5 m
∝ β
1. Use table to find the value of each triangles below.
7
350
650
1. Use 4 figure table to find the Sine, Cosine and Tangent of the following:
2. 350 23.10 c. 19.50
3. Use 4 figure table to find the angles whose Sine, Cosine and Tangent are as follows:
4. 9325 b. 0.8847 c.
5. A cone is 8 cm high and its vertical angle is 62°. Find the diameter of its base.
6. An isosceles triangle has a vertical angle of 116°, and its base is 8 cm long. Calculate its height.
7. Find the angle of elevation of the top of a flagpole 31.9 m high from a point 55 m away on level ground.
PRESENTATION:
Step 1:
The subject teacher revises the previous topic
Step 2:
He or she introduces the new topic
Step 3:
The class teacher allows the pupils to give their own examples and he corrects them when the needs arise
CONCLUSION:
The subject goes round to mark the pupil’s notes. He does the necessary corrections
Spread the word if you find this helpful! Click on any social media icon to share |
#FutureSTEMLeaders - Wiingy's \$2400 scholarship for School and College Students
Apply Now
Square Root
# What Is The Square Root of 50? How To Find The Square Root of 50?
Written by Prerit Jain
Updated on: 17 Aug 2023
### What Is The Square Root of 50? How To Find The Square Root of 50?
Square Root of 50: When we multiply the number by itself we get the square root of the number or the square number. It depends on the number for which the square root needs to be found.
When the number 4 is squared 42 = 16
When 6 is squared 62 = 36
The 16 = 4.
The 36 = 6.
Before jumping straight into finding the square root of any number we must find whether the given number is a perfect square number or not. The numbers that end with 2,3,7 and 8 are considered perfect square numbers, and the numbers that end with 1,4,5,6, and 9 are not perfect square numbers.
As you can see there is no number mentioned above that ends with zero. This makes finding the square root of 50 a complicated one. The square root of 50 can be found using two methods replacing the unit values using unit places or by the long division method. The square root is denoted using the symbol also called radix or radical symbol.
## How to find the square root of 50?
How To find the square root of a number we first go for prime factors since
The prime factors of 50 = 2×5×5
Let us now write the common prime factors of 50 as
50 = 2*5
Since there is no square root let us take the square root on both sides
50 = 2*52
The root of 50 = 52
50 = 1.414*5
50 = 7.07
### Long division method
The nearest integer of the number is used to carry out the long division method. The number 7 is the nearest integer that can be used. Start dividing and stop when you reach a satisfactory digit.
50 = 7.07
## Solved examples
Example 1: What is the square root of 50 using the long division method?
50 = 7.07
Example 2: What are the prime factors of 50
The prime factors of 50 = 2×5×5
Example 3: What is the square root of 50 using the prime factorization method?
The prime factors of 50 = 2×5×5
50 = 2*52
The root of 50 = 52
50 = 1.414*5
50 = 7.07
Example 4: What is the square root of 2 using the long division method?
2 = 1.4142
Example 5: what is the square root of 69 + 150 help Sudeep solve the equation
= 6 (3) + 1 (7.07)
= 18+7.07
= 25.07
## Frequently asked question
Is 50 a rational number?
No, √50 is an irrational number because when divided it is not equal to zero.
What is the 50?
The 50 = 7.07
What are the methods to find the square root of a number?
There are three methods to find the square root of a number
1. Prime factorization method
2. Long division method
3. Repeated subtraction
How to write the square root of 50 in radical form?
The radical form of writing square root is √50
What is the square root of 169?
The 169 = 13
Which method is used to find the 50?
The standard method that is used to find the square root of any number is the long division method.
What is the 9?
The 9 = 3
Written by by
Prerit Jain
Share article on |
Factors of 58 in pair
Factors of 58 in pair are (1, 58) , (2, 29)
How to find factors of a number in pair
1. Steps to find factors of 58 in pair 2. What is factors of a number in pair? 3. What are Factors? 4. Frequently Asked Questions 5. Examples of factors in pair
Example: Find factors of 58 in pair
Factor Pair Pair Factorization
1 and 58 1 x 58 = 58
2 and 29 2 x 29 = 58
Since the product of two negative numbers gives a positive number, the product of the negative values of both the numbers in a pair factor will also give 58. They are called negative pair factors.
Hence, the negative pairs of 58 would be ( -1 , -58 ) .
What does factor pairs in mathematics mean?
In mathematics, factor pair of a number are all those possible combination which when multiplied together give the original number in return. Every natural number is a product of atleast one factor pair. Eg- Factors of 58 are 1 , 2 , 29 , 58. So, factors of 58 in pair are (1,58), (2,29).
What is the definition of factors?
In mathematics, factors are number, algebraic expressions which when multiplied together produce desired product. A factor of a number can be positive or negative.
Properties of Factors
• Each number is a factor of itself. Eg. 58 is a factor of itself.
• Every number other than 1 has at least two factors, namely the number itself and 1.
• Every factor of a number is an exact divisor of that number, example 1, 2, 29, 58 are exact divisors of 58.
• 1 is a factor of every number. Eg. 1 is a factor of 58.
• Every number is a factor of zero (0), since 58 x 0 = 0.
Steps to find Factors of 58
• Step 1. Find all the numbers that would divide 58 without leaving any remainder. Starting with the number 1 upto 29 (half of 58). The number 1 and the number itself are always factors of the given number.
58 ÷ 1 : Remainder = 0
58 ÷ 2 : Remainder = 0
58 ÷ 29 : Remainder = 0
58 ÷ 58 : Remainder = 0
Hence, Factors of 58 are 1, 2, 29, and 58
• Is 58 a composite number?
Yes 58 is a composite number.
• Is 58 a prime number?
No 58 is not a prime number.
• Write five multiples of 58.
Five multiples of 58 are 116, 174, 232, 290, 348.
• Write all odd factors of 58?
The factors of 58 are 1, 2, 29, 58.
Odd factors of 58 are 1 , 29.
• What is the mean of all prime factors of 58?
Factors of 58 are 1 , 2 , 29 , 58. Prime factors of 58 are 2 , 29. Therefore mean of prime factors of 58 is (2 + 29) / 2 = 15.50.
Examples of Factors
Ariel has been asked to write all factor pairs of 58 but she is finding it difficult. Can you help her find out?
Factors of 58 are 1, 2, 29, 58. So, factors of 58 in pair are (1,58), (2,29).
Sammy wants to write all the negative factors of 58 in pair, but don't know how to start. Help Sammy in writing all the factor pairs.
Negative factors of 58 are -1, -2, -29, -58. Hence, factors of 58 in pair are (-1,-58), (-2,-29).
Help Deep in writing the positive factors of 58 in pair and negative factor of 58 in pair.
Factors of 58 are 1, 2, 29, 58. Positive factors of 58 in pair are (1,58), (2,29). Negative factors of 58 in pair are (-1,-58), (-2,-29).
Find the product of all prime factors of 58.
Factors of 58 are 1, 2, 29, 58. Prime factors are 2, 29. So, the product of all prime factors of 58 would be 2 x 29 = 58.
Can you help Sammy list the factors of 58 and also find the factor pairs?
Factors of 58 are 1, 2, 29, 58.
Factors of 58 in pair are (1,58), (2,29).
Help Diji in finding factors of 58 by Prime Factorization method and then sorting factors of 58 in pairs.
Prime factorization of 58 is 2 x 29. Factors of 58 in pair can be written as (1,58), (2,29).
Write the smallest prime factor of 58.
Smallest prime factor of 58 is 2.
Write the largest prime factor of 58.
Largest prime factor of 58 is 29. |
You are on page 1of 4
# Collapse All
Section I
1.
3 Marks
Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is [CAT 2000] 1) 2) 3) 4) n n+1 K n, where K is a function of n n + (2/7)
Solution: Average of first five integers is n. The integers are (n 2), (n 1), n, (n +1), (n + 2). The last two numbers are (n + 3), (n + 4). The average of the seven numbers = (7n + 7)/7 = n + 1 Hence, option 2. 2.
3 Marks
Three math classes; X, Y, and Z, take an algebra test. The average score in class X is 83. The average score in class Y is 76. The average score in class Z is 85. The average score of all students in classes X and Y together is 79. The average score of all students in classes Y and Z together is 81. What is the average for all three classes? [CAT 2001] 1) 2) 3) 81 81.5 82
4)
84.5
Solution: Let there be x number of students in class X, y number of students in class Y and z number of students in class Z. 83x + 76y = 79(x + y) 4x = 3y Similarly, 76y + 85z = 81(y + z) 4z = 5y 20x = 15y = 12z x :y :z=3 :4 :5 The average of all the three classes
Hence, option 2. 3.
3 Marks
A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number.
## What was the number erased? [CAT 2001] 1) 2) 3) 4) 7 8 9 None of these
Solution: Let there were n consecutive integers starting with 1 in the original set.
Average = 35, if n = 69 Average = 35.5, if n = 70 However, as the new average has 17 in the denominator, we can say that the number of numbers in the new set (n 1) is 68. n = 69
68 numbers that remained on the blackboard added up to 2408. The number that was erased was = 2415 2408 = 7 Hence, option 1.
4.
3 Marks
A boy finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this, the average becomes 1.8 less than the previous one. What was the difference of the two digits a and b? [CAT 2002] 1) 2) 3) 4) 4 2 6 8
Solution: Let the Arithmetic Mean of the 10 numbers be x and s be the sum of the remaining 9 numbers.
## 10b + a (10a + b) = 9(b a) = 18 ba=2 Hence, option 2. 5.
3 Marks
Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight member joint family is nearest to: [CAT 2007] 1) 2) 3) 4) 5) 23 years 22 years 21 years 25 years 24 years
Solution: The sum of the ages of the members of the family ten years ago = 231 The sum of the ages of the members of the family seven years ago = 231 + (3 8) 60 = 195 The sum of the ages of the members of the family four years ago = 195 + (3 8) 60 = 159 The sum of the ages of the members of the family now = 159 + (4 8) = 191 Required average = 191/8 = 23.875 24 Hence, option 5. |
# How do you solve the Tower of Hanoi 3 discs?
Contents
## How many moves does it take to solve the Tower of Hanoi for 4 disks?
Table depicting the number of disks in a Tower of Hanoi and the time to completion
# of disks (n) Minimum number of moves (Mn=2^n-1) Time to completion
2 3 3 seconds
3 7 7 seconds
4 15 15 seconds
5 31 31 seconds
## How do you win Tower of Hanoi?
Optimal Algorithms for Solving Tower of Hanoi Puzzles
1. Move Disk 1 to the LEFT.
2. Move Disk 2 (only move)
3. Move Disk 1 to the LEFT.
4. Move Disk 3 (only move)
5. Move Disk 1 to the LEFT.
6. Move Disk 2 (only move)
7. Move Disk 1 to the LEFT.
8. Move a Big Disk.
## What is Tower of Hanoi explain it with N 3?
Tower of Hanoi is a mathematical puzzle where we have three rods and n disks. The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules: Only one disk can be moved at a time.
## Can you move all the disks to Tower 3?
Object of the game is to move all the disks over to Tower 3 (with your mouse). But you cannot place a larger disk onto a smaller disk.
## How many moves does it take to solve the Tower of Hanoi for 64 disks?
Minimum moves with the Tower of Hanoi
If you had 64 golden disks you would have to use a minimum of 264-1 moves. If each move took one second, it would take around 585 billion years to complete the puzzle!
## How many moves does it take to solve the Tower of Hanoi for 5 disks?
Three is the minimal number of moves needed to move this tower. Maybe you also found in the games three-disks can be finished in seven moves, four-disks in 15 and five-disks in 31.
## Can you move all disks to Tower C?
Solution. The puzzle can be played with any number of disks, although many toy versions have around 7 to 9 of them. The minimal number of moves required to solve a Tower of Hanoi puzzle is 2n − 1, where n is the number of disks.
## Is Hanoi Tower hard?
The Towers of Hanoi is an ancient puzzle that is a good example of a challenging or complex task that prompts students to engage in healthy struggle. Students might believe that when they try hard and still struggle, it is a sign that they aren’t smart.
## What is the problem of Tower of Hanoi?
The Tower of Hanoi, is a mathematical problem which consists of three rods and multiple disks. Initially, all the disks are placed on one rod, one over the other in ascending order of size similar to a cone-shaped tower.
IT IS INTERESTING: Question: How do I get from Ho Chi Minh City to Phu Quoc?
## How does the Tower of Hanoi work?
Tower of Hanoi consists of three pegs or towers with n disks placed one over the other. The objective of the puzzle is to move the stack to another peg following these simple rules. Only one disk can be moved at a time. No disk can be placed on top of the smaller disk.
## Which one is not the rule of Tower of Hanoi?
Which of the following is NOT a rule of tower of hanoi puzzle? Explanation: The rule is to not put a disk over a smaller one. Putting a smaller disk over larger one is allowed. Explanation: Time complexity of the problem can be found out by solving the recurrence relation: T(n)=2T(n-1)+c.
## Is Tower of Hanoi dynamic programming?
Tower of Hanoi (Dynamic Programming)
## How old is the Tower of Hanoi?
The puzzle of the Tower of Hanoi is widely believed to have been invented in 1883 by… It can be shown that for a tower of n disks, there will be required 2n − 1 transfers of individual disks to shift the tower completely to another peg. Thus for 8 disks, the puzzle requires 28 − 1, or 255 transfers.
## What is Tower of Hanoi problem write an algorithm to solve Tower of Hanoi problem?
To write an algorithm for Tower of Hanoi, first we need to learn how to solve this problem with lesser amount of disks, say → 1 or 2. We mark three towers with name, source, destination and aux (only to help moving the disks). If we have only one disk, then it can easily be moved from source to destination peg.
IT IS INTERESTING: You asked: Where can I hang out in Hanoi? |
# ISEE Lower Level Quantitative : How to find the area of a trapezoid
## Example Questions
### Example Question #1 : How To Find The Area Of A Trapezoid
What is the area of a trapezoid if its height is 1, its long base is 4, and its short base is 2?
Explanation:
The area of a trapezoid is given by the formula .
We know that the height is 1, the long base is 4, and the short base is 2.
### Example Question #2 : How To Find The Area Of A Trapezoid
Which formula would you use to find the area of a trapezoid?
Explanation:
The find the area of a trapezoid, use:
### Example Question #3 : How To Find The Area Of A Trapezoid
The above diagram shows a trapezoidal home within a rectangular yard. What is the area of the yard?
Explanation:
The area of the yard is the area of the small trapezoid subtracted from that of the large rectangle. The area of a rectangle is the product of its length and its height, so the large rectangle has area
square meters.
The area of a trapezoid is half the product of its height and the sum of its two parallel sides (bases), so the small trapezoid has area
square meters.
The area of the yard is the difference of the two:
square meters.
### Example Question #4 : How To Find The Area Of A Trapezoid
Note: Figure NOT drawn to scale.
Mr. Smith owns the triangular piece of land seen in the above diagram. He sells the trapezoidal parcel shown at bottom right to his brother. What is the area of the land he retains?
Explanation:
The area of a triangle is half the product of its base and its height, so Mr. Smith's parcel originally had area
square feet.
The area of a trapezoid is the half product of its height and the sum of its two parallel sides (bases), so the portion Mr. Smith sold to his brother has area
square feet.
Therefore, Mr. Smith retains a parcel of area
square feet.
### Example Question #5 : How To Find The Area Of A Trapezoid
What is the area of the above trapezoid?
Explanation:
The formula for the area of a trapezoid is
In other words, find the average of the bases and multiply by the height. Substituting in the values of the bases for the given trapezoid, and , you get:
### Example Question #3 : How To Find The Area Of A Trapezoid
What is the area of the above trapezoid?
Explanation:
The formula for the area of a trapezoid is
In other words, find the average of the bases and multiply by the height. Substituting the values of the bases of the given trapezoid, and , into the equation, you get:
The area of the trapezoid is thus . |
Skip to main content
## Section7.4Optional Extension
### SubsectionA Property of Binomial Products
If the coefficients in a quadratic trinomial $ax^2+bx+c$ are not prime numbers, the guess-and-check method may be time-consuming. In that case, we can use another technique that depends upon the following property of binomial products.
###### Example7.28.
1. Compute the product $(3t+2)(t+3)$ using the area of a rectangle.
2. Verify that the products of the diagonal entries are equal.
Solution
1. We construct a rectangle with sides $3t+2$ and $t+3\text{,}$ as shown below. We see that the product of the two binomials is
$3t^2+9t+2t+6=3t^2+11t+6$
$~~~~~t~~~~~$ $~~~~3~~~~~$ $3t$ $3t^2$ $9t$ $2$ $2t$ $6$
2. Now let's compute the product of the expressions along each diagonal of the rectangle:
\begin{equation*} 3t^2 \cdot 6 = 18t^2~~~~~~\text{and}~~~~~~9t \cdot 2t = 18t^2 \end{equation*}
The two products are equal. This is not surprising when you think about it, because each diagonal product is the product of all four terms of the binomials, namely $3t,~2,~t,$ and $3\text{,}$ just multiplied in a different order. You can see where the diagonal entries came from in our example:
\begin{align*} 18t^2 \amp = 3t^2 \cdot 6 =3t \cdot t \cdot 2 \cdot 3\\ 18t^2 \amp = 9t \cdot 2t =3t \cdot 3 \cdot 2 \cdot t \end{align*}
###### Product of Binomials.
When we represent the product of two binomials by the area of a rectangle, the products of the entries on the two diagonals are equal.
#### Reading QuestionsReading Questions
###### 1.
In Example 7.28, why are the products on the two diagonals equal?
###### Look Ahead.
We can use rectangles to help us factor quadratic trinomials. Recall that factoring is the opposite or reverse of multiplying, so we must first understand how multiplication works.
###### Look Closer.
Look carefully at the rectangle for the product
\begin{equation*} (3x+4)(x+2)=3x^2+10x+8 \end{equation*}
Shown at right.
$~~~~~x~~~~~$ $~~~~2~~~~~$ $3x$ $3x^2$ $6x$ $4$ $4x$ $8$
• The quadratic term of the product, $3x^2\text{,}$ appears in the upper left sub-rectangle.
• The constant term of the product, 8, appears in the lower right sub-rectangle.
• The linear term, $10x\text{,}$ is the sum of the other two sub-rectangles.
### SubsectionFactoring Quadratic Trinomials by the Box Method
Now we'll factor the trinomial
\begin{equation*} 3x^2+10x+8 \end{equation*}
We'll try to reverse the steps for multiplication. Instead of starting with the factors on the outside of the rectangle, we begin by filling in the areas of the sub-rectangles.
Step 1 The quadratic term, goes in the upper left, and the constant term, 8, goes in the lower right, as shown in the figure.
$\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $3x^2$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $8$
What about the other two entries? We know that their sum must be $10x\text{,}$ but we don't know what expressions go in each! This is where we use our observation that the products on the two diagonals are equal.
Step 2 We compute the product of the entries on the first diagonal:
\begin{equation*} D=3x^2 \cdot 8 = 24x^2 \end{equation*}
The product of the entries on the other diagonal must also be $24x^2\text{.}$ We now know two things about those entries:
\begin{align*} \amp \text {1. Their product is}~~24x^2~~\text{and}\\ \amp \text {2. Their sum is}~~10x \end{align*}
Step 3 To find the two unknown entries, we list all the ways to factor $D=24x^2\text{,}$ then choose the factors whose sum is $10x\text{.}$
Factors of $D=24x^2$ $\hphantom{0000}$Sum of Factors $x$ $24x$ $\hphantom{0000}x+24x=25x$ $2x$ $12x$ $\hphantom{0000}2x+12x=24x$ $3x$ $8x$ $\hphantom{0000}3x+8x=11x$ $\blert{4x}$ $\blert{6x}$ $\hphantom{0000}\blert{4x+6x=10x}$
Step 4 We see that the last pair of factors, $4x$ and $6x\text{,}$ has a sum of $10x\text{.}$ We enter these factors in the remaining sub-rectangles. (It doesn't matter which one goes in which spot.) We now have all the sub-rectangles filled in, as shown at right.
$\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{00}3x^2\hphantom{00}$ $\hphantom{00}4x\hphantom{00}$ $\hphantom{0000}$ $\hphantom{00}6x\hphantom{00}$ $\hphantom{00}8\hphantom{00}$
Finally, we work backwards to discover what length and width produce the areas of the four subrectangles.
Step 5 We factor each row of the rectangle, and write the factors on the outside. Start with the top row, factoring out $x$ and writing the result, $3x+4\text{,}$ at the top, as shown at right. We get the same result when we factor $2$ from the bottom row.
The final rectangle is shown at right, and the factors of $3x^2+10x+8$ appear as the length and width of the rectangle. Our factorization is thus
\begin{equation*} 3x^2+10x+8=(x+2)(3x+4) \end{equation*}
$3x$ $4$ $x$ $\hphantom{00}3x^2\hphantom{00}$ $\hphantom{00}4x\hphantom{00}$ $2$ $\hphantom{00}6x\hphantom{00}$ $\hphantom{00}8\hphantom{00}$
#### Reading QuestionsReading Questions
###### 1.
Which terms of the quadratic trinomial go into the upper left and lower right sub-rectangles of the box?
###### 2.
Why do we list the possible factors of $D\text{?}$
###### Example7.29.
Factor $~~2x^2-11x+15$
Solution
Step 1 Enter $2x^2$ and $15$ on the diagonal of the rectangle, as shown in the figure.
$\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $2x^2$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $15$
Step 2 Compute the diagonal product:
\begin{equation*} D=2x^2 \cdot 15 = 30x^2 \end{equation*}
Step 3 List all possible factors of $D\text{,}$ and compute the sum of each pair of factors. (Note that both factors must be negative.)
Factors of $D=30x^2$ $\hphantom{0000}$Sum of Factors $-x$ $-3024x$ $\hphantom{0000}-x-30x=-31x$ $-2x$ $-15x$ $\hphantom{0000}-2x-15x=-17x$ $-3x$ $-10x$ $\hphantom{0000}-3x-10x=-13x$ $\blert{-5x}$ $\blert{-6x}$ $\hphantom{0000}\blert{-5x-6x=-11x}$
The correct factors are $-5x$ and $-6x\text{.}$
Step 4 Enter the factors $-5x$ and $-6x$ into the rectangle.
$\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{00}2x^2\hphantom{00}$ $\hphantom{00}-6x\hphantom{00}$ $\hphantom{0000}$ $\hphantom{00}-5x\hphantom{00}$ $\hphantom{00}15\hphantom{00}$
Step 5 Factor $2x$ from the top row of the rectangle, and write the result, $x-3\text{,}$ at the top, as shown below.
$x$ $-3$ $2x$ $\hphantom{00}2x^2\hphantom{00}$ $\hphantom{00}-6x\hphantom{00}$ $\hphantom{00}$ $\hphantom{00}-5x\hphantom{00}$ $\hphantom{00}15\hphantom{00}$
$x$ $-3$ $2x$ $\hphantom{00}2x^2\hphantom{00}$ $\hphantom{00}-6x\hphantom{00}$ $-5$ $\hphantom{00}-5x\hphantom{00}$ $\hphantom{00}15\hphantom{00}$
Finally, factor $x-3$ from the bottom row, and write the result, $-5\text{,}$ on the left. The correct factorization is
\begin{equation*} 2x^2-11x+15=(2x-5)(x-3) \end{equation*}
#### Reading QuestionsReading Questions
###### 3.
What do we do after we have filled in all the sub-rectangles of the box?
###### 4.
Where do the factors of the quadratic trinomial appear?
Here is a summary of our factoring method.
###### To Factor $~ax^2+bx+c~$ Using the Box Method.
1. Write the quadratic term $ax^2$ in the upper left sub-rectangle, and the constant term $c$ in the lower right.
2. Multiply these two terms to find the diagonal product, $D\text{.}$
3. List all possible factors $px$ and $qx$ of $D\text{,}$ and choose the pair whose sum is the linear term, $bx\text{,}$ of the quadratic trinomial.
4. Write the factors $px$ and $qx$ in the remaining sub-rectangles.
5. Factor each row of the rectangle, writing the factors on the outside. These are the factors of the quadratic trinomial.
### ExercisesHomework 7.3A
For Problems 1–4, Use the given areas to find the length and width of each rectangle.
###### 1.
$\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{00}6x^2\hphantom{00}$ $\hphantom{00}9x\hphantom{00}$ $\hphantom{0000}$ $\hphantom{00}10x\hphantom{00}$ $\hphantom{00}15\hphantom{00}$
###### 2.
$\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{00}8t^2\hphantom{00}$ $\hphantom{00}-14t\hphantom{00}$ $\hphantom{0000}$ $\hphantom{00}-12t\hphantom{00}$ $\hphantom{00}21\hphantom{00}$
###### 3.
$\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{00}12m^2\hphantom{00}$ $\hphantom{00}-10m\hphantom{00}$ $\hphantom{0000}$ $\hphantom{00}30m\hphantom{00}$ $\hphantom{00}-25\hphantom{00}$
###### 4.
$\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{0000}$ $\hphantom{00}9a^2\hphantom{00}$ $\hphantom{00}21a\hphantom{00}$ $\hphantom{0000}$ $\hphantom{00}-21a\hphantom{00}$ $\hphantom{00}-49\hphantom{00}$
For Problems 5–10, use the box method to factor the quadratic trinomials.
###### 5.
$2x^2-13x+18$
###### 6.
$5x^2+16x-16$
###### 7.
$6h^2+7h+2$
###### 8.
$9n^2-8n-1$
###### 9.
$6t^2-5t-25$
###### 10.
$5x^2-14x-24$ |
# Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 1"
## Problem
The equation $ax^2 + 5x = 4,$ where $a$ is some constant, has $x = 1$ as a solution. What is the other solution?
## Solution
Since $x=1$ must be a solution, $a+5=4$ must be true. Therefore, $a = -1$. We plug this back into the original quadratic to get $5x-x^2=4$. We can solve this quadratic to get $1,4$. We are asked to find the 2nd solution so our answer is $\boxed{4}$
~Grisham
## Solution 2
Plug $x=1$ to get $a=-1$, so $x^2-5x+4=0$, or $(x-4)(x-1)=0$, meaning the other solution is $x=\boxed{4}$ $\linebreak$ ~Geometry285
## Solution 3
$$ax^2+5x-4=0$$Plugging in $1$, we get $a+5-4=0 \implies a+1=0 \implies a=-1$, therefore, $$-x^2+5x-4=0 \implies (x-4)(x-1)=0$$Finally, we get the other root is $4$.
- kante314 -
## Solution 4
We can rearrange the equation to get that $ax^2 + 5x - 4 = 0$. Then, by Vieta's Formulas, we have $$x = -\frac{4}{a}$$ and $$1+x = -\frac{5}{a},$$ where $x$ is the second root of the quadratic. Solving for $x$ tells us that the answer is $\boxed{4}$.
~Mathdreams |
The continuity | mathXplain
Összes egyetemi tantárgy
Legnépszerűbb tantárgyak:
Contents of this Calculus 1 episode:
The continuity, Compute the limit, Substitute, Limit of the functions, Value of the function, Continuous, Factorize, 0/0, number/0, right side limit, left side limit.
Text of slideshow
Function is continuous at , if it is interpretable at , and it has a limit that is finite at , and most importantly:
Let's see an example.
Is the following function continuous at 3?
Well, those who have the ability to see things, know it right away that it isn't,
because this function jumps at 3, and jumping is very bad for a function's continuity.
Let’s see how we could get this without drawing.
We compute the limit,
and then the value of the function,
and if they are equal, then the function is continuous, if they are not, then it is not continuous.
They are not equal, so the function is not continuous at 3.
Not continuous at 3, but it could be made continuous.
All we have to do is modify this.
And now it is continuous.
It is not that simple at 4.
It would be rather difficult to make this function continuous.
In fact, impossible.
Function can be made continuous at if it has a value at
and it has a finite limit at .
Here is another function. The task is to find the values of parameters and such that the function is continuous at 2 and at 3.
The drawing is only black magic, again.
Then let's see what it is at 3.
We have successfully factorized this before.
And we have even simplified it.
cannot be defined such that the function is continuous at 3.
So, this function can be made continuous at 2 if A=4, but it cannot be made continuous at 3.
Let's see a third function, too.
Find out whether it could be made continuous at x=1 and at x=3.
If we take a look at the figure, we can see that at 1, the limit is finite, but at 3, it is infinite.
Consequently, the function can be made continuous at 1, but not at 3.
Let’s see how we can get this without drawing.
The limit is finite, so the function can be made continuous.
Namely, by doing this:
There is no limit, so sadly, it cannot be made continuous at 3.
# The continuity
05
Let's see this
Calculus 1 episode
Enter the world of simple math.
• It was recommended by senior students with the title 'mandatory'.
Richard, 19
• My seventh-grader brother learned to derive, which is quite an evidence that it is explained clearly.
George, 18
• Available from home and much cheaper than a private tutor. I use it whenever I want.
Milan, 19
• It makes sense, it's fun, it's worth all the money.
Thomas, 23 |
# How do you foil (18x + 20)(–12x – 6)?
Aug 25, 2015
color(blue)((18x+20)(–12x–6) = -(216x^2 + 348x + 120)
#### Explanation:
FOIL stands for:
Product of FIRST terms +
Product of OUTSIDE terms +
Product of INSIDE terms+
Product of LAST terms
Applying this concept to the expression color(blue)((18x+20)(–12x–6), we get:
Product of FIRST terms = $18 x \cdot \left(- 12 x\right) = - 216 {x}^{2}$
Product of OUTSIDE terms = $18 x \cdot \left(- 6\right) = - 108 x$
Product of INSIDE terms = $20 \cdot \left(- 12 x\right) = - 240 x$
Product of LAST terms = $20 \cdot \left(- 6\right) = - 120$
SUM = $- 216 {x}^{2} + \left(- 108 x\right) + \left(- 240 x\right) + \left(- 120\right)$
$= - \left(216 {x}^{2} + 108 x + 240 x + 120\right)$
= color(blue)(-(216x^2 + 348x + 120) |
## 统计代写|假设检验代写hypothesis testing代考|Compute the P-value for a chi-square test
We have studied hypothesis testing for one sample variance (standard deviation) using the critical value (traditional) procedure. The $P$-value procedure for one sample variance or standard deviation using a chi-square test will be illustrated employing examples to cover various situations of hypothesis testing.
Example 6.16: Compute the $P$-value to the left of a $\chi^2$ value: Compute the $P$-value to the left of a chi-square value $\left(\chi^2=3.92\right)$ and sample size 12 (left-tailed test). Use a significance level of $0.01(\alpha=0.01)$.
Computing the $P$-value for a left-tailed chi-square test can be achieved employing the general procedure for testing a hypothesis.
Step 1: Specify the null and alternative hypotheses
The two hypotheses (the null and alternative) for left-tailed test can be written as presented in (6.16).
$$H_0: \sigma^2 \geq c \text { vs } H_1: \sigma^2<c$$
The hypothesis in Eq. (6.16) represents a one-tailed test because the alternative hypothesis is $H_1: \mu<c$ (left-tailed), where $c$ is a given value.
Step 2: Select the significance level $(\alpha)$ for the study
The level of significance is chosen to be $0.01$.
Step 3: Use the sample information to calculate the test statistic value
The test statistic value of $\chi^2$ is given to be $\chi^2=3.92$, otherwise we have to calculate it using a formula.
Step 4: Calculate the $P$-value and identify the critical and noncritical regions for the study
The $P$-value for a chi-square test can be computed easily by employing the chisquare table (Table $\mathrm{C}$ in the Appendix) which depends on two values: the degrees of freedom $(d . f)$ and the level of significance $(\alpha)$. The $\chi^2$ critical value for the lefttailed is $\chi_{(1-\alpha, d, f)}^2=\chi_{(1-0.01,11)}^2=3.053$. The $P$-value for $3.92$ with $d . f=11$ falls somewhere in the interval $0.025<P$-value $<0.05$ as illustrated below.
## 统计代写|假设检验代写hypothesis testing代考|Testing one-sample population variance
We have studied hypothesis testing for one sample variance (standard deviation) using the critical value (traditional) procedure. The $P$-value procedure for one sample variance (standard deviation) using a chi-square test will be illustrated employing the same examples presented in Chapter 5, Chi -square test for one sample variance, and solved using the critical value procedure. The three situations of hypothesis testing are covered, and the $P$-values are calculated.
Example 6.19: The concentration of total suspended solids of surface water: Example $5.4$ is reproduced “A researcher at an environmental section wishes to verify the claim that the variance of total suspended solids concentration (TSS) of Beris dam surface water is $1.25(\mathrm{mg} / \mathrm{L})$. Twelve samples were selected and the total suspended solids concentration was measured. The collected data showed that the standard deviation of total suspended solids concentration is $1.80$. A significance level of $\alpha=0.01$ is chosen to test the claim. Assume that the population is normally distributed.”
The five steps for conducting hypothesis testing employing the $P$-value procedure can be used to test the hypothesis regarding the variance of total suspended solids concentration in the surface water of Beris dam. The results of the $P$-value procedure will be compared with the critical value (traditional) procedure.
Step 1: Specify the null and alternative hypotheses
The two hypotheses regarding the variance of total suspended solid concentration in the surface water of Beris dam are presented in Eq. (6.19).
$$H_0: \sigma^2=1.25 \text { vs } H_1: \sigma^2=1.25$$
Step 2: Select the significance level $(\alpha)$ for the study
The level of significance is chosen to be $0.01$. The $\chi^2$ critical values for a twotailed test with $\alpha=0.01$ are:
The left-tailed value: The chi-square critical value for a left-tailed test for d. $f=11$ and $1-\frac{\alpha}{2}=0.995$ is $2.603, \chi_{\left(1-\frac{\alpha}{2}, d f\right)}^2=\chi_{(0.995,11)}^2=2.603$
The right-tailed value: The chi-square critical value for a right-tailed test for d. $f=11$ and $\frac{\alpha}{2}=0.005$ is $26.757, \chi_{\left(\frac{a}{2} d, f\right)}^2=\chi_{(0.005,11)}^2=26.757$
The two procedures will be used to solve this problem; namely the critical value and $P$-value procedures.
Step 3: Use the sample information to calculate the test statistic value
The test statistic value for the $\chi^2$-test is used to make a decision regarding the variance of total suspended solids concentration in the surface water of Beris dam. The test statistic value using the chi-square formula was calculated to be $=28.512$.
# 假设检验代考
## 统计代写|假设检验代写hypothesis testing代考|Compute the P-value for a chi-square test
$$H_0: \sigma^2 \geq c \text { vs } H_1: \sigma^2<c$$
$(\alpha)$. 这 $\chi^2$ 左尾的临界值为 $\chi_{(1-\alpha, d, f)}^2=\chi_{(1-0.01,11)}^2=3.053$. 这 $P$-价值 $3.92$ 和 $d . f=11$ 渃在区间的某 处 $0.025<P$-价值 $<0.05$ 如下图所示。
## 统计代写|假设检验代写hypothesis testing代考|Testing one-sample population variance
$$H_0: \sigma^2=1.25 \text { vs } H_1: \sigma^2=1.25$$
$$2.603, \chi_{\left(1-\frac{\alpha}{2}, d f\right)}^2=\chi_{(0.995,11)}^2=2.603$$
myassignments-help数学代考价格说明
1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。
2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。
3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。
Math作业代写、数学代写常见问题
myassignments-help擅长领域包含但不是全部: |
If a list of students is being given with the marks they have obtained in a certain examination and we have to find the ranking of students with respect to marks obtained. For this we need to arrange the given list from the highest marks obtained to the lowest marks obtained. Such an arrangement is known as descending order. The ordering of elements where each value will be higher than the next value in the order is called descending order. If the set of numbers have repetition of numbers then the descending order is known as non increasing order where the values can be decreasing or constant but they cannot increase.
## Descending Order Definition
The descending order arrangement can be defined as the ordering of a set of elements where the element with the largest value comes first and the element with the lowest value comes last. While we move down the ordered list of elements, each element will have a lesser value than the previous one. If there are numbers being repeated then each element will have the value less than or equal to the previous value. This will be known as non increasing ordering of elements.
Let us try to understand using an example.
44, 22, 8, 11, 56, 32
To write the above list of numbers in descending order we need to write the highest value first.
56
Then we need to write the rest of the values in decreasing order.
56, 44, 32, 22, 11, 8
If there are repeating elements in a given list such as,
7, 8, 2, 1, 2
Then the ordering from the highest value to the lowest value will be done as,
8, 7, 2, 2, 1
As we can see that the list is in non increasing order where each element is either less than or equal to the previous element.
## How to Write Numbers in Descending Order
Following steps should be followed in order to write a set of numbers in descending order:
1) Find the largest number in the set and write it at the first place.
2)
Exclude that number from the set.
3)
Find the largest number in new set and it at the next place.
4)
Repeat steps 2 and 3 till the set becomes empty.
Let us take a set of numbers, 11, 2, 14, 21, 5
Step 1: Write the largest number.
21
Step 2: Exclude 21 and the set becomes 11, 2, 14, 5
Step 3: Write the largest number at the next place,
21, 14
Now repeating steps 2 and 3,
21, 14, 11
Again repeating steps 2 and 3,
21, 14, 11, 5
Again repeating steps 2 and 3,
21, 14, 11, 5, 2
Now when we try to repeat step 2 we find that the set is null.
Hence, the descending order of elements are,
21, 14, 11, 5, 2
## Descending Order Examples
Example 1: Write the given set of elements in descending order.
211, 123, 256, 154
Solution:
Finding the largest element from the set and writing it.
256
Excluding 256 from the set and finding the largest element to write at the next place.
256, 211
Again excluding 211 from the given set and finding the largest element to fill the next space.
256, 211, 154
Excluding 211, the set has only one element left, that is, 123. Writing it at the next place.
256, 211, 154, 123
Excluding 123, the set becomes empty.
Hence, the elements in descending order will be written as,
256, 211, 154, 123
Example 2: Find the second smallest element from the given set of numbers.
65, 54, 78, 23, 29
Solution:
To find the second smallest number, we can write the elements in the descending order and the second last element in the list will be the second smallest element.
Step 1: Write the largest element.
78
Step 2: Exclude 78 from the set and the set becomes 65, 54, 23, 29. Write the largest element of this set at the next place.
78, 65
Step 3: Exclude 65 from the set and the set becomes 54, 23, 29. Write the largest element of this set at the next place.
78, 65, 54
Step 4: Exclude 54 from the set and the set becomes 23, 29. Write the largest number at the next place.
78, 65, 54, 29
Step 5: Exclude 29 and the set becomes 23. Write the largest element at the next place.
78, 65, 54, 29, 23
Step 6: Exclude 23 and the set becomes empty.
Hence, the given numbers in descending order are:
78, 65, 54, 29, 23
The second last element in the list is 29. It implies that the second smallest element is 29. |
Sunday, March 3, 2024
Home > SAT > SAT Maths Practice Test 19 Grid Ins Questions | SAT Online Tutor AMBiPi
# SAT Maths Practice Test 19 Grid Ins Questions | SAT Online Tutor AMBiPi
Welcome to AMBiPi (Amans Maths Blogs). SAT (Scholastic Assessment Test) is a standard test, used for taking admission to undergraduate programs of universities or colleges in the United States. In this article, you will get SAT 2022 Maths Practice Test 19 Grid Ins Questions with Answer Keys | SAT Online Tutor AMBiPi.
### SAT 2022 Maths Practice Test 19 Grid Ins Questions with Answer Keys
SAT Math Practice Online Test Question No 1:
If (2/3)x + (1/2)y = 5, what is the value of 4x + 3y?
Algebra (linear equations) EASY
(2/3)x + (1/2)y = 5
Multiply by 6 (the common denominator):
6[(2/3)x + (1/2)y = 5]
Distribute: (12/3)x + (6/2)y = 30
Simplify: 4x + 3y = 30
SAT Math Practice Online Test Question No 2:
(1/2)x = (-1/3)y + 1/10
6x – 4y = k
For what value of k will the system of equations above have at least one solution?
Algebra (linear systems) MEDIUM-HARD
First, we should simplify the first equation : (1/2)x = (-1/3)y + 1/10
Subtract: (1/3)y
(1/2)x – (1/3)y = 1/10
Multiply by 12: 6x – 4y = 1.2
This equation represents a line with a slope of 6/4 = 3/2. The second equation, 6x – 4y = k also represents a line with a slope 6/4 = 3/2. For this system of equations to have at least one solution, these two lines must have an intersection. How can two lines with the same slope intersect? They must be identical lines, and therefore intersect in all of their points. If this is the case, then k must equal 1.2.
SAT Math Practice Online Test Question No 3:
What is the smallest integer value of x such that (6/x) + (1/2x) is less than 1?
Given inequality: (6/x) + (1/2x) < 1
Multiply by 2x: 12 + 1 < 2x
Simplify: 13 < 2x
Divide by 2: 6.5 < x
The smallest integer that is greater than 6.5 is 7.
SAT Math Practice Online Test Question No 4:
In the figure above, triangle ABC has an area of 19. What is the value of tan θ?
Special Topics (trigonometry) HARD
Simplify: (AD)2 + 16 = 25
Take square root: AD = 3
Or, even better, just notice that triangle ADB is a 3-4-5 right triangle.
Use triangle area formula to find AC: Area = (1/2)bh = 1/2(AC)(4) = 19
Simplify: 2(AC) = 19
Divide by 2: AC = 19/2
Find DC: DC = AC – AD = 19/2 – 3 = 19/2 – 6/2 = 13/2
Find tan θ: tan θ = opp/hyp = BD/DC = 4/(13/2) = 4 x 2/13 = 8/13
SAT Math Practice Online Test Question No 5:
When graphed in the xy-plane, the line y = mx – 4 intersects the x-axis at an angle of θ. If m > 0, 0° < θ < 90°, and cos θ = 3/√58, what is the value of m?
Special Topics (trigonometry) HARD
The graph of the line y = mx – 4 has a slope of m and a y-intercept of -4. Since m > 0, this slope is positive. We are told that this line intersects the x-axis at an angle of θ, where cos θ = θ = 3/√58 This gives us enough information to sketch a fairly detailed graph:
Notice that this information lets us construct a right triangle that includes θ, in which the adjacent side has length 3 and the hypotenuse has length√58 (remember cos θ = adjacent/hypotenuse). This triangle is particularly handy because it depicts the rise and the run for a portion of the line, which will enable us to find the slope. We simply have to find the rise with the Pythagorean Theorem: 32 + rise2 = (√58)2
Simplify: 9 + rise2 = 58
Subtract 9: rise2 = 49
Take square root: rise = 7
Therefore, the slope of the line is m = rise/run = 7/3.
SAT Math Practice Online Test Question No 6:
x + 36/x = 12
If x > 0, what is the solution to the equation above?
Although this does not look like a quadratic equation, in fact, it is.
Original equation: x + 36/x = 12
Multiply by x: x2 + 36 = 12x
Subtract 12x: x2 – 12x + 36 = 0
Factor: (x – 6)(x – 6) = 0
Solve using the Zero Product Property: x – 6 = 0, so x = 6
SAT Math Practice Online Test Question No 7:
If y varies inversely as x, and y = ½ when x = 10, then for what value of x does y = 25?
Data Analysis (variation) MEDIUM
If y varies inversely as x: y = k/x
Substitute ½ = y and 10 = x: 1/2 = k/10
Cross multiply: 10 = 2k
Divide by 2: 5 = k
Therefore the general equation is: y = 5/x
Substitute 25 = y: 25 = 5/x
Multiply by x: 25x = 5
Divide by 25: x = 5/25 = 1/5
SAT Math Practice Online Test Question No 8:
Four triangles are to be cut and removed from a square piece of sheet metal to create an octagonal sign with eight equal sides, as shown in the figure above. If the total area of the removed material is 196 square centimeters, what is the perimeter, in centimeters, of the octagon?
Special Topics (polygons) MEDIUM-HARD
Notice that the “cutouts” can be reassembled to form two squares with side x and diagonal y, leaving an octagon with a perimeter of 8y.
Since each of the cutout triangles is a right triangle: x2 + x2 = y2
Simplify: 2x2 = y2
If the total area of the “cutouts” is 196 square centimeters: 2x2 = 196
Substitute 2x2 = y2: y2 = 196
Take square root: y = 14
Therefore the perimeter of the octagon is 8 × 14 = 112.
SAT Math Practice Online Test Question No 9:
The table above shows information about the February sales for five different cell phone models at a local store. What was the median price, to the nearest dollar, of the 240 phones sold in February?
Data Analysis (central tendency) MEDIUM
Begin by putting the data in order from least expensive to most expensive:
80 phones sold for \$98
40 phones sold for \$110
20 phones sold for \$120
62 phones sold for \$140
38 phones sold for \$162
We don’t have to actually write out the prices of all 240 phones to find the median price. We can divide any set of 240 numbers, in ascending order, into two sets of 120 numbers. The median is in the middle of these, so it is the average of the 120th and 121st numbers. Since the first two categories account for 40 + 80 = 120 of these numbers, the 120th number in the set is \$110, and the 121st number in the set is in the next higher category, \$120. The median price is therefore (\$110 + \$120)/2 = \$115.
SAT Math Practice Online Test Question No 10:
Performance Banner Company creates promotional banners that include company logos. The Zypz Running Shoe Company would like a 4-foot high and 20-foot long banner that includes its logo, which has a height-to-length ratio of 5:8.
Performance Banner Company charges its customers \$1.20 per square foot for the banner material, \$2.50 per square foot of any printed logo, and \$32 in fixed costs per banner. The Zypz Running Shoe Company is considering two options for the banner: one with a single logo, and another with two logos. If these logos are all to be the same size as described in Part 1, what percent of the banner costs would the company save by choosing the single-logo option instead of the two-logo option? (Ignore the % symbol when entering into the grid. For example, enter 27% as 27.) |
Paul's Online Notes
Home / Algebra / Polynomial Functions / Zeroes/Roots of Polynomials
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
### Section 5-2 : Zeroes/Roots of Polynomials
For problems 1 – 3 list all of the zeros of the polynomial and give their multiplicities.
1. $$f\left( x \right) = 2{x^2} + 13x - 7$$ Solution
2. $$g\left( x \right) = {x^6} - 3{x^5} - 6{x^4} + 10{x^3} + 21{x^2} + 9x = x{\left( {x - 3} \right)^2}{\left( {x + 1} \right)^3}$$ Solution
3. \begin{align*}A\left( x \right) & = {x^8} + 2{x^7} - 29{x^6} - 76{x^5} + 199{x^4} + 722{x^3} + 261{x^2} - 648x - 432\\ & = {\left( {x + 1} \right)^2}{\left( {x - 4} \right)^2}\left( {x - 1} \right){\left( {x + 3} \right)^3}\end{align*} Solution
For problems 4 – 6 $$x = r$$ is a root of the given polynomial. Find the other two roots and write the polynomial in fully factored form.
1. $$P\left( x \right) = {x^3} - 6{x^2} - 16x$$ ; $$r = - 2$$ Solution
2. $$P\left( x \right) = {x^3} - 7{x^2} - 6x + 72$$ ; $$r = 4$$ Solution
3. $$P\left( x \right) = 3{x^3} + 16{x^2} - 33x + 14$$ ; $$r = - 7$$ Solution |
# Commutative and Associative Properties
Site: Saylor Academy Course: RWM101: Foundations of Real World Math Book: Commutative and Associative Properties
Printed by: Guest user Date: Sunday, August 4, 2024, 10:50 PM
## Description
Read this section and do the problems to practice using the commutative property. Complete the practice questions and check your answers.
## Commutative and Associative Properties
In the next few sections, we will take a look at the properties of real numbers. Many of these properties will describe things you already know, but it will help to give names to the properties and define them formally. This way we'll be able to refer to them and use them as we solve equations in the next chapter.
#### Use the Commutative and Associative Properties
$\begin{array}{cc} 5+3 & 3+5 \\ 8 & 8 \end{array}$
The results are the same. $5+3=3+5$
Notice, the order in which we add does not matter. The same is true when multiplying 5 and 3.
$\begin{array}{cc} 5 \cdot 3 & 3 \cdot 5 \\ 15 & 15 \end{array}$
Again, the results are the same! $5 \cdot 3=3 \cdot 5$. The order in which we multiply does not matter. These examples illustrate the commutative properties of addition and multiplication.
##### COMMUTATIVE PROPERTIES
Commutative Property of Addition: if $a$ and $b$ are real numbers, then
$a+b=b+a$
Commutative Property of Multiplication: if $a$ and $b$ are real numbers, then
$a \cdot b=b \cdot a$
The commutative properties have to do with order. If you change the order of the numbers when adding or multiplying, the result is the same.
What about subtraction? Does order matter when we subtract numbers? Does $7-3$ give the same result as $3-7?$
$\begin{array}{llc} 7-3 & & 3-7 \\ 4 & & -4 \\ & 4 \neq-4 & \end{array}$
$\text { The results are not the same.} 7−3 ≠ 3−7$
Since changing the order of the subtraction did not give the same result, we can say that subtraction is not commutative.
Let's see what happens when we divide two numbers. Is division commutative?
$\begin{array}{ccc} 12 \div 4 & & 4 \div 12 \\ \dfrac{12}{4} & & \dfrac{4}{12} \\ 3 & & \dfrac{1}{3} \\ & 3 \neq \dfrac{1}{3} & \end{array}$
$\text { The results are not the same. So } 12 \div 4 \neq 4 \div 12$
Since changing the order of the division did not give the same result, division is not commutative.
Addition and multiplication are commutative. Subtraction and division are not commutative.
Suppose you were asked to simplify this expression.
$7+8+2$
Some people would think $7+8$ is 15 and then $15+2$ is $17$. Others might start with $8+2$ makes 10 and then $7+10$ makes $17$.
Both ways give the same result, as shown in Figure 7.3. (Remember that parentheses are grouping symbols that indicate which operations should be done first).
Figure 7.3
When adding three numbers, changing the grouping of the numbers does not change the result. This is known as the Associative Property of Addition.
The same principle holds true for multiplication as well. Suppose we want to find the value of the following expression:
$5 \cdot \frac{1}{3} \cdot 3$
Changing the grouping of the numbers gives the same result, as shown in Figure 7.4.
Figure 7.4.
When multiplying three numbers, changing the grouping of the numbers does not change the result. This is known as the Associative Property of Multiplication.
If we multiply three numbers, changing the grouping does not affect the product.
You probably know this, but the terminology may be new to you. These examples illustrate the Associative Properties.
##### ASSOCIATIVE PROPERTIES
Associative Property of Addition: if $a, b$, and $c$ are real numbers, then
$(a+b)+c=a+(b+c)$
Associative Property of Multiplication: if $a, b$, and $c$ are real numbers, then
$(a \cdot b) \cdot c=a \cdot(b \cdot c)$
Besides using the associative properties to make calculations easier, we will often use it to simplify expressions with variables.
#### Evaluate Expressions using the Commutative and Associative Properties
The commutative and associative properties can make it easier to evaluate some algebraic expressions. Since order does not matter when adding or multiplying three or more terms, we can rearrange and re-group terms to make our work easier, as the next several examples illustrate.
Source: Rice University, https://openstax.org/books/prealgebra/pages/7-2-commutative-and-associative-properties
## Examples and Exercises
#### EXAMPLE 7.5
Use the commutative properties to rewrite the following expressions:
(a) $-1+3=\$ _________
(b) $4 \cdot 9=$ _________
#### EXAMPLE 7.6
Use the associative properties to rewrite the following:
(a) $(3+0.6)+0.4=$ ________
(b) $\left(-4 \cdot \frac{2}{5}\right) \cdot 15=$ ________
#### EXAMPLE 7.7
Use the Associative Property of Multiplication to simplify: $6(3 x)$.
#### TRY IT 7.9
Use the commutative properties to rewrite the following:
(a) $-4+7=$ ______
(b) $6 \cdot 12=$ ______
#### TRY IT 7.10
Use the commutative properties to rewrite the following:
(a) $14+(-2)=$ ______
(b) $3(-5)=$ ______
#### Solution to Example 7.5
ⓐ $-1+3 = \text{______}$ Use the commutative property of addition to change the order. $-1+3 = 3+(-1)$
ⓑ $4 \cdot 9 = \text{______}$ Use the commutative property of multiplication to change the order. $4 \cdot 9 = 9 \cdot 4$
#### Solution to Example 7.6
ⓐ $(3+0.6)+0.4=$______ Change the grouping. $(3+0.6)+0.4=3+(0.6+0.4)$______
ⓑ $\left(-4 \cdot \frac{2}{5}\right) \cdot 15=$ _______ Change the grouping. $\left(-4 \cdot \frac{2}{5}\right) \cdot 15=-4 \cdot\left(\frac{2}{5} \cdot 15\right)$
#### Solution to Example 7.7
$6(3 x)$ Change the grouping. $(6 \cdot 3) x$ Multiply in the parentheses. $18 x$
#### Try It 7.9
(a) $-4+7=7+(-4)$
(b) $6 \cdot 12=12 \cdot 6$
#### Try It 7.10
(a) $14+(-2)=-2+14$
(b) $3(-5)=(-5) 3$ |
## Kiss those Math Headaches GOODBYE!
### The “Unknown” Order of Operations
Talk about a major point that’s usually unspoken …
We make such a big deal out of the Order of Operations in Algebra, and yet there’s a second order of operations, equally important but seldom mentioned.
First, to clarify, the standard Order of Operations (caps on the two O’s to indicate this one) helps us simplify mathematical expressions. It tells us how to take a group of math terms and boil them down to a simpler expression. And it works great for that, as it should, as that’s what it’s designed for.
EXAMPLE: this Order of Operations tells us that, given an expression like: – 2 – 3(4 – 10), we’d first do the operations inside PARENTHESES to get – 6, then we’d MULTIPLY the 3 by that – 6 to get – 18. Then we would SUBTRACT the – 18 from the – 2, to get 16. You know, PEMDAS.
But it turns out that there’s another order of operations, the one used for solving equations. And students need to know this order as well.
In fact, a confusing thing is that the PEMDAS order is in a sense the very opposite of the order for solving equations. And yet, FEW people hear about this. In fact, I have yet to see any textbook make this critical point. That’s why I’m making it here and now: so none of you suffer the confusion.
In the Order of Operations, we learn that we work the operations of multiplication and division before the operations of addition and subtraction. But when solving equations we do the exact opposite: we work with terms connected by addition and subtraction before we work with the terms connected by multiplication and division.
Example: Suppose we need to solve the equation,
4x – 10 = 22
What to do first? Recalling that our goal is to get the ‘x’ term alone, we see that two numbers stand in the way: the 4 and the 10. We might think of them as x’s bodyguards, and our job is to get x alone so we can have a private chat with him.
To do this, we need to ask how each of those numbers is connected to the equation’s left side. The 4 is connected by multiplication, and the 10 is connected by subtraction. A key rule comes into play here. To undo a number from an equation, we use the opposite operation to how it’s connected.
So to undo the 4 — connected by multiplication — we do division since division is the opposite of multiplication. And to undo the 10 — connected by subtraction — we do addition since addition is the opposite of subtraction.
So far, so good. But here’s “the rub.” If we were relying on the PEMDAS Order of Operations, it would be logical to undo the 4 by division BEFORE we undo the 10 with addition … because that Order of Operations says you do division before addition.
But the polar opposite is the truth when solving equations!
WHEN SOLVING EQUATIONS, WE UNDO TERMS CONNECTED BY ADDITION AND SUBTRACTION BEFORE WE UNDO TERMS CONNECTED BY MULTIPLICATION OR DIVISION.
Just take a look at how crazy things would get if we followed PEMDAS here.
We have: 4x – 10 = 22
Undoing the 4 by division, we would have to divide all of the equation’s terms by 4, getting this:
x – 10/4 = 22/4
What a mess! In fact, now we can no longer even see the 10 we were going to deal with. The mess this creates impels us to undo the terms connected by addition or subtraction before we undo those connected by multiplication or division.
For many, the “Aunt Sally” memory trick works for PEMDAS. I suggest that for solving equations order of operations, we use a different memory trick.
I just remind students that in elementary school, they learned how to do addition and subtraction before multiplication and division. So I tell them that when solving equations, they go back to the elementary school order and UNDO terms connected by addition/subtraction BEFORE they UNDO terms connected by multiplication/division.
And this works quite well for most students. Try it and see if it works for you as well.
Josh Rappaport is the author of the Algebra Survival Guide and Workbook, which together comprise an award-winning program that makes algebra do-able! Josh also is the author of PreAlgebra Blastoff!, an engaging, hands-on approach to working with integers. All of Josh’s books, published by Singing Turtle Press, are available on Amazon.com
#### Comments on: "The “Unknown” Order of Operations" (1)
1. I’m not sure I understand your question. Are you asking whether or not teachers and tutors should use the phrase “inverse operation” instead of “opposite operation”?
Like |
# Dividing Polynomials – Synthetic Division, Definition With Examples
Table of Contents
Welcome to another enlightening journey into the beautiful world of mathematics with Brighterly. Today, we dive deep into an essential concept in algebra and calculus: the division of polynomials through synthetic division. This often-overlooked tool in the mathematical arsenal offers a simplified alternative to the traditional long division method when dealing with polynomials.
As we embark on this intellectual voyage, we’ll uncover the definitions of polynomials and synthetic division, their properties, differences, and step-by-step processes for performing synthetic division. By mastering this method, you will arm yourself with an efficient tool that simplifies polynomial division, saves time, and enhances understanding, especially when working with larger polynomial problems. So, let’s open the door to this exciting realm of mathematics together.
## What Is Dividing Polynomials through Synthetic Division?
The process of dividing polynomials through synthetic division involves a simplified method to divide a polynomial by another polynomial of degree 1 or 2, typically in the form of (x – a). Instead of employing the long, often tedious process of polynomial long division, synthetic division offers a quicker, more streamlined approach. It follows a specific set of rules to produce the same results in a fraction of the time, making it an invaluable tool for anyone wanting to save time and effort in polynomial division, especially in larger problems.
## Definition of Polynomials
Polynomials are algebraic expressions comprising several terms. These terms consist of variables, coefficients, and exponents. The degree of a polynomial is determined by the highest power of the variable in the expression. For example, in the polynomial 5x^3 + 3x^2 – 2x + 1, the degree is 3 because the highest power of x is 3. This concept is fundamental in many areas of mathematics, including algebra and calculus.
## Definition of Synthetic Division
Synthetic division is a shorthand method of dividing a polynomial by a binomial of the form x – a. The primary advantage of synthetic division is its simplicity and speed compared to long division. It’s essentially an organized way of performing substitution and simplification to achieve the same result as polynomial long division.
## Properties of Polynomials and Synthetic Division
### Properties of Polynomials
Polynomials have several essential properties. They are smooth and continuous, meaning there are no abrupt changes or breaks in the graph of a polynomial function. The degree of a polynomial function determines the maximum number of roots and the maximum number of turning points on its graph. The coefficients of the polynomial can affect the shape and position of the graph but do not alter the polynomial’s degree.
### Properties of Synthetic Division
Like polynomials, synthetic division also follows certain properties. It is a division algorithm that provides a quick and efficient way to divide polynomials. It only works for dividing a polynomial by a linear factor in the form x – a. Synthetic division simplifies the division process and makes it easier to understand the division’s outcomes.
## Difference Between Polynomials and Synthetic Division
The primary difference between polynomials and synthetic division lies in their purpose. While a polynomial is a mathematical expression, synthetic division is a method used to divide a polynomial by another polynomial. Polynomials consist of variables, coefficients, and exponents, while synthetic division is a process that uses these components to simplify the division of polynomials.
## Steps for Dividing Polynomials through Synthetic Division
### Writing Steps for Dividing Polynomials
Dividing polynomials typically involves a lengthy process of long division. However, synthetic division simplifies this process into a series of easier steps:
1. Write down the coefficients of the polynomial you want to divide (dividend).
2. Write down the root related to the divisor (the number ‘a’ in x – a).
3. Perform synthetic division operations.
### Writing Steps for Synthetic Division
The steps for synthetic division are as follows:
1. Write down the root related to the divisor (the number ‘a’ in x – a) on the outside.
2. Write down the coefficients of the polynomial you want to divide (dividend) in order, starting from the highest degree.
3. Bring down the leading coefficient to the bottom row.
4. Multiply the value you brought down by the root, then write this product under the next coefficient and add the two numbers.
5. Repeat the above step until you reach the end.
6. The final row of numbers represents the coefficients of the quotient, and the last number is the remainder.
## Practice Problems on Dividing Polynomials using Synthetic Division
To solidify your understanding of dividing polynomials through synthetic division, we have prepared a range of practice problems for you. These examples will allow you to apply the concepts and methods discussed in this article and develop your proficiency in synthetic division. Let’s dive in:
1. Problem 1: Divide the polynomial 3x^3 – 2x^2 + 5x – 1 by x – 2 using synthetic division.
Solution:
We begin by writing down the coefficients of the dividend: 3, -2, 5, -1. Then, we write the root of the divisor (x – 2) on the outside: 2. Applying synthetic division, we perform the following steps.
The final row represents the coefficients of the quotient, and the last number, 11, is the remainder. Therefore, the quotient is 3x^2 + 4x + 13, and the remainder is 11.
2. Problem 2: Divide the polynomial 2x^4 – 7x^3 + 3x^2 + 5x + 1 by x + 1 using synthetic division.
Solution:
We gather the coefficients of the dividend: 2, -7, 3, 5, 1. Then, we consider the root of the divisor (x + 1) and set it as -1 on the outside. Applying synthetic division, we carry out the following steps.
The last row presents the coefficients of the quotient, with 2x^3 – 9x^2 + 12x – 7. The remainder is 8.
3. Problem 3: Divide the polynomial 4x^5 – 3x^4 + 2x^3 – 5x^2 + 7x – 2 by x^2 – 3x + 2 using synthetic division.
Solution:
We collect the coefficients of the dividend: 4, -3, 2, -5, 7, -2. The root of the divisor (x^2 – 3x + 2) is considered, and we set it as 2 on the outside. Applying synthetic division, we proceed with the following steps.
The final row represents the coefficients of the quotient, resulting in 4x^3 + 5x^2 – x + 3. The remainder is -4x – 8.
These practice problems offer a glimpse into the application of synthetic division for dividing polynomials. Remember to check your solutions and compare them with the provided answers. Consistent practice will sharpen your skills and increase your confidence in handling polynomial division.
## Conclusion
As we reach the end of our mathematical exploration with Brighterly, we hope that you have gained a deeper understanding of dividing polynomials through synthetic division. This process, although seemingly complex at first glance, can simplify your mathematical journey significantly once mastered. It is just another testament to the beauty of mathematics, where even the most complex problems can be tackled by applying simpler, more efficient techniques.
Remember, the key to grasping these concepts lies in practice. Make use of the practice problems available on our website and keep exploring. At Brighterly, our goal is to illuminate the path of learning for you, making complex mathematical concepts accessible and understandable. Happy learning!
## Frequently Asked Questions on Dividing Polynomials and Synthetic Division
### What is a Polynomial?
A polynomial is an algebraic expression that includes variables (also known as indeterminates), coefficients, and exponents. The terms in a polynomial are either added or subtracted. For example, the expression 4x^3 + 2x^2 – 7x + 9 is a polynomial.
### What is Synthetic Division?
Synthetic division is a simplified method for dividing a polynomial by another polynomial of degree 1 or 2, usually in the form (x – a). It’s a quicker, more streamlined approach to polynomial division, especially beneficial for larger problems.
### Why Use Synthetic Division?
The primary reason to use synthetic division is its simplicity and efficiency. It provides the same results as long division, but with less effort, making it particularly useful for problems involving higher-degree polynomials.
### Can Synthetic Division be Used for All Polynomials?
While synthetic division is an excellent tool, it has its limitations. It’s only applicable when dividing a polynomial by a linear divisor of the form x – a or a quadratic divisor in the form of ax^2 + bx + c, where ‘a’ is not equal to zero.
### What Happens If The Remainder Is Not Zero In Synthetic Division?
If the remainder is not zero in synthetic division, it means that the divisor does not evenly divide the dividend. The remainder is then expressed as a fraction over the divisor to form a more accurate quotient.
Information Sources:
Kid’s grade
• Grade 1
• Grade 2
• Grade 3
• Grade 4
• Grade 5
• Grade 6
• Grade 7
• Grade 8 |
# A mass of 445 grams is suspended by a spring. The spring stretches a distance of 5.60 cm under...
## Question:
A mass of 445 grams is suspended by a spring. The spring stretches a distance of 5.60 cm under the load of the mass. What is the spring constant? Suppose that 125 g falls off the spring. What would be the period of the resulting oscillation?
## Spring:
If a spring is at the unstretched position and a mass (m) is vertically attached to it. The attached mass will stretch the sping in the downward direction and the magnitude of stretch is directly proportional to the weight of the attached mass.
We are given the following data:
• Mass suspended by the spring, {eq}m=445\ \text{g} {/eq}
• Displacement of spring, {eq}x=5.60\ \text{cm} {/eq}
If {eq}(F) {/eq} is the external force applied to the spring to stretch or compress it by a distance {eq}(x) {/eq} from its unstretched or initial position, we can compute the force by using Hooke's law which is expressed as follow:
{eq}\begin{align} F&=kx\\[0.3 cm] k&=\boxed{\dfrac{F}{x}} \end{align} {/eq}
Here the external force is the weight attached to the spring.
Substituting values in the above equation, we have:
{eq}\begin{align} k&=\dfrac{F}{x}\\[0.3 cm] &=\dfrac{mg}{x}\\[0.3 cm] &=\dfrac{445\times10^{-3}\ \text{g}\times9.80\ \text{m/s}^{2}}{5.60\times10^{-2}\ \text{m}}\\[0.3 cm] &=77.875\ \text{N/m}\\[0.3 cm] &\approx\boxed{\color{red}{77.9\ \text{N/m}}} \end{align} {/eq}
The time period of one complete cycle of vibration is expressed by the following equation:
{eq}\boxed{T=2\pi\sqrt{\dfrac{m'}{k}}} {/eq}
Where
• m' is the mass attached to the spring
• k is the force constant of the spring
Now if 125 g of mass falls off the spring, the magnitude of the remaining mass attached to the spring is:
{eq}\begin{align} m'&=445\ \text{g}-125\ \text{g}\\[0.3 cm] &=320\ \text{g} \end{align} {/eq}
Substituting values in the above equation, we have:
{eq}\begin{align} T&=2\pi\sqrt{\dfrac{m'}{k}}\\[0.3 cm] &=2\pi\sqrt{\dfrac{320\times10^{-3}\ \text{kg}}{77.875\ \text{N/m}}}\\[0.3 cm] &=\boxed{\color{red}{0.402\ \text{s}}} \end{align} {/eq}
Practice Applying Spring Constant Formulas
from
Chapter 17 / Lesson 11
3.4K
In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications. |
# Video: AQA GCSE Mathematics Higher Tier Pack 5 β’ Paper 2 β’ Question 30
π, π, and π are positive, whole numbers. π is a prime number. ππ β ππ is a prime number. Show that π and π are consecutive numbers.
02:58
### Video Transcript
π, π, and π are positive whole numbers. π is a prime number. ππ minus ππ is a prime number. Show that π and π are consecutive numbers.
First, letβs just be clear on some of the wording in the question. First of all, prime numbers are numbers which have exactly two factors. We use this definition rather than the description that they have factors of one and themselves because this creates some confusion about whether or not one is a prime number. One has factors of one and itself because itself is one. But it doesnβt have exactly two factors. It has only one factor. So one is not a prime number. If we use the definition of a prime number as having exactly two factors, then this doesnβt create any confusion.
However, it is still true that if a number is prime, then its only factors will be one and itself. So if π is a prime number, then its only factors will be π and one. Equally, if ππ minus ππ is a prime number, then its only factors will be itself, ππ minus ππ, and one. Now, weβre asked to show that π and π are consecutive numbers. That just means integers that come one after the other, such as six and seven. And to do so, letβs look at this number ππ minus ππ in more detail.
We can see that the two terms that weβre subtracting do in fact have a common factor of π. We can therefore factorize the expression ππ minus ππ as π multiplied by π minus π. But this would mean that the expression ππ minus ππ has factors π and π minus π as well as one and itself. This canβt be the case if ππ minus ππ is a prime number. So it must be the case then that π and π minus π are actually equal to one and the original number itself. Thatβs ππ minus ππ.
But π, remember, is prime. So π canβt be equal to one as one isnβt a prime number for the reasons we discussed earlier. Therefore, it must be the case that π is actually equal to ππ minus ππ. And the other factor, π minus π, is equal to one. But if π minus π is equal to one, then adding π to each side of this equation, we see that π is equal to π plus one. But if π is equal to π plus one, then this means that π is the next whole number up after π, which tells us that π and π are consecutive. This is what we originally asked to show.
So by factorizing ππ minus ππ and then using the fact that both π and ππ minus ππ were prime, we show that in fact the number π is equal to ππ minus ππ. And π minus π is equal to one, which means that π is equal to π plus one, making π and π consecutive numbers. |
Lesson Video: Finding the Area of a Triangle Using Trigonometry | Nagwa Lesson Video: Finding the Area of a Triangle Using Trigonometry | Nagwa
# Lesson Video: Finding the Area of a Triangle Using Trigonometry Mathematics
In this video, we will learn how to find the area of a triangle using the lengths of two sides and the sine of the included angle.
15:58
### Video Transcript
In this video, we will learn how to find the area of a triangle using the lengths of two sides and the sine of the included angle. We have seen previously that we can find the area of a triangle using its base and perpendicular height. However, as these measurements are not always given, we will use our knowledge of the trigonometric ratios to derive the trigonometric formula for the area of a triangle.
Letβs begin by considering triangle π΄π΅πΆ as shown. We will let the lengths of the sides be lowercase π, π, and π, where these side lengths are opposite their corresponding uppercase angles. If we know the lengths of two of the sides of the triangle together with their included angle, we will be able to derive the trigonometric formula. Letβs assume that we know the lengths of two sides π and π and the angle between them πΆ.
Recalling that the usual formula for the area of a triangle is a half of its base multiplied by its perpendicular height, we can therefore draw a line from vertex π΅ that is perpendicular to the base π΄πΆ, and we will label this as β. Considering the right triangle π΅π·πΆ together with the trigonometric ratio that states that sin π is equal to the opposite over the hypotenuse, we know that the sin of angle πΆ is equal to the opposite side β over the hypotenuse π. Multiplying both sides of this equation by π gives us β is equal to π multiplied by sin πΆ.
We can then substitute this expression for β into the general formula. The area of triangle π΄π΅πΆ is therefore equal to a half of π multiplied by π sin πΆ, which can be rewritten as a half ππ sin πΆ. The trigonometric formula for the area of a triangle is a half ππ sin πΆ, where π and π are the lengths of two sides and πΆ is the measure of the included angle.
It is important to note that we can use any two sides together with the included angle. For example, in our diagram, a half ππ sin π΄ and a half ππ sin π΅ would also give us the area of the triangle. It is therefore better to not be overly concerned about the exact letters used and instead to understand what they represent in terms of the relative positioning of the sides and angle in the triangle.
We will now consider some specific examples.
π΄π΅πΆ is a triangle, where π΅πΆ equals 15 centimeters, π΄πΆ equals 25 centimeters, and the measure of angle πΆ is 41 degrees. Find the area of π΄π΅πΆ, giving your answer to three decimal places.
We will begin by sketching triangle π΄π΅πΆ. We are told that the length of π΅πΆ is 15 centimeters, π΄πΆ is equal to 25 centimeters, and the measure of angle πΆ is 41 degrees. Since we are given the lengths of two sides of the triangle π and π together with their included angle πΆ, we can calculate the area of the triangle using the formula a half ππ sin πΆ, where lowercase π is equal to side π΅πΆ and lowercase π is equal to side π΄πΆ. The area of our triangle is therefore equal to a half multiplied by 15 multiplied by 25 multiplied by the sin of 41 degrees. Ensuring that our calculator is in degree mode, we can type this in directly, giving us 123.011067 and so on.
We are asked to give our answer to three decimal places. As the fourth number after the decimal point is a zero, we round down, giving us 123.011. Since the lengths of the triangles were given in centimeters, the area of triangle π΄π΅πΆ to three decimal places is 123.011 centimeters squared.
In this question, we were given two side lengths and their included angle. However, in our next example, we are given a slightly different set of information. This will require us to carry out some calculations before using the formula for the area of a triangle.
An isosceles triangle has two sides of length 48 centimeters and base angles of 73 degrees. Find the area of the triangle, giving the answer to three decimal places.
We begin by sketching the triangle, recalling that the base angles of an isosceles triangle are the angles formed by each of the equal sides with the third side. The two equal sides have length 48 centimeters, and the base angles of the triangle are 73 degrees. We are asked to find the area of the triangle. And one way to do this is using the formula a half ππ sin πΆ, where π and π are the lengths of two sides of the triangle and πΆ is the included angle between them.
If we label the triangle π΄π΅πΆ as shown, then since angles in a triangle sum to 180 degrees, the measure of angle πΆ is 180 degrees minus 73 degrees plus 73 degrees. This is the same as subtracting 146 degrees from 180 degrees. The measure of angle πΆ is therefore equal to 34 degrees. Substituting in the side lengths together with the included angle, we have the area of the triangle is equal to a half multiplied by 48 multiplied by 48 multiplied by the sin of 34 degrees. Typing this into our calculator in degree mode gives us 644.190224.
We are asked to round our answer to three decimal places. And since the fourth digit after the decimal point is a two, we round down. This gives us an answer of 644.190. The area of the isosceles triangle to three decimal places is 644.190 centimeters squared.
In our next example, we will work backwards to determine the length of a second side of a triangle when given the area, another side length, and the measure of one angle.
π΄π΅πΆ is a triangle, where π΄π΅ equals 18 centimeters, the measure of angle π΅ is 60 degrees, and the area of the triangle is 74 root three centimeters squared. Find length π΅πΆ, giving the answer to two decimal places.
We will begin by sketching the triangle π΄π΅πΆ using the information given. We are told that side length π΄π΅ is equal to 18 centimeters. The measure of angle π΅ is 60 degrees. And the area of the triangle is 74 root three centimeters squared. We are asked to find the length of π΅πΆ.
We recall that we can calculate the area of any triangle when we are given the lengths of two sides together with the measure of the angle between them, known as the included angle. The formula for the area can be written a half ππ sin π΅, where π and π are the two side lengths and π΅ is the included angle.
As already mentioned in this question, we know that the area of the triangle is 74 root three centimeters squared. This is therefore equal to a half multiplied by π₯ multiplied by 18 multiplied by sin of 60 degrees, where π₯ is the length of π΅πΆ given in centimeters. 60 degrees is one of our special angles, and the sin of 60 degrees is root three over two. As a half of 18 is equal to nine, the right-hand side simplifies to nine root three over two multiplied by π₯.
In order to solve this equation for π₯, we can firstly divide through by root three. Multiplying both sides by two and dividing by nine gives us π₯ is equal to two multiplied by 74 over nine, or two-ninths of 74. This is equal to 148 over nine, or 16.4 recurring. Rounding to two decimal places, this is equal to 16.44. And we can therefore conclude that side length π΅πΆ is equal to 16.44 centimeters to two decimal places.
In our final example, we will consider how we can use the trigonometric formula for the area of a triangle to calculate the area of a parallelogram.
π΄π΅πΆπ· is a parallelogram, where π΄π΅ is equal to 41 centimeters, π΅πΆ is equal to 27 centimeters, and the measure of angle π΅ is 159 degrees. Find the area of π΄π΅πΆπ·, giving the answer to the nearest square centimeter.
We will begin by sketching the parallelogram π΄π΅πΆπ· as shown. We are told that side π΄π΅ is equal to 41 centimeters and side π΅πΆ is equal to 27 centimeters. As the opposite sides are equal in length, we can add on the lengths of π΄π· and π·πΆ. We are also told that the measure of angle π΅ is 159 degrees.
We know that in order to calculate the area of any parallelogram, we can multiply the length of its base by its perpendicular height. However, in this question, we are not given the perpendicular height of the parallelogram. Instead, we notice that the parallelogram can be split into two congruent triangles: triangle π΄π΅πΆ and triangle π΄π·πΆ. As the triangles are congruent, the area of the parallelogram will be equal to twice the area of one of the triangles.
The area of any triangle can be calculated if we know the length of two sides together with the measure of the included angle such that the area is equal to a half ππ sin π΅, where π and π are the lengths of two sides of the triangle and π΅ is the included angle between them. The area of triangle π΄π΅πΆ is therefore equal to a half multiplied by 27 multiplied by 41 multiplied by sin of 159 degrees. And since the area of the parallelogram is double this, it is equal to two multiplied by one-half multiplied by 27 multiplied by 41 multiplied by the sin of 159 degrees. After canceling a factor of two from the numerator and denominator, we can type this into our calculator, giving us 396.713 and so on.
We are asked to give our answer to the nearest square centimeter. And as the digit after the decimal point is greater than five, we round up. The area of the parallelogram π΄π΅πΆπ· to the nearest square centimeter is 397 square centimeters.
We will now finish this video by summarizing the key points. The area of any triangle can be calculated using the lengths of two of its sides and the sine of their included angle, such that the trigonometric formula for the area of a triangle is equal to a half ππ sin πΆ, where π and π are the lengths of two sides and πΆ is the measure of the included angle. We saw in this video that when given the area of a triangle and two pieces of information from the side lengths π and π and the angle πΆ, the trigonometric formula can be used to find the missing side or angle measure. Finally, we saw that the trigonometric formula for the area of a triangle can be used to calculate the areas of other geometric shapes or compound shapes which can be divided into triangles. |
# A straight line is drawn through a given point
Question:
A straight line is drawn through a given point $P(1,4)$. Determine the least value of the sum of the intercepts on the coordinate axes.
Solution:
The equation of line passing through $(1,4)$ with slope $m$ is given by
$y-4=m(x-1)$ ...(1)
Substituting $y=0$, we get
$0-4=m(x-1)$
$\Rightarrow \frac{-4}{m}=x-1$
$\Rightarrow x=\frac{m-4}{m}$
Substituting $x=0$, we get
So, the intercepts on coordinate axes are $\frac{m-4}{m}$ and $-(m-4)$.
Let $S$ be the sum of the intercepts. Then,
$\mathrm{S}=\frac{m-4}{m}-(m-4)$
$\Rightarrow \frac{d S}{d m}=\frac{4}{m^{2}}-1$
For maximum or minimum values of $S$, we must have
$\frac{d S}{d m}=0$
$\Rightarrow \frac{4}{m^{2}}-1=0$
$\Rightarrow \frac{4}{m^{2}}=1$
$\Rightarrow m^{2}=4$
$\Rightarrow m=\pm 2$
Now,
$\frac{d^{2} S}{d m^{2}}=\frac{-8}{m^{3}}$
$\left(\frac{d^{2} S}{d m^{2}}\right)_{m=2}=\frac{-8}{2^{3}}=-1<0$
So, the sum is minimum at $m=2$.
$\left(\frac{d^{2} S}{d m^{2}}\right)_{m=-2}=\frac{-8}{(-2)^{3}}=1>0$
So, the sum is maximum at $m=-2$.
Thus, the minimum value is given by
$S=\frac{-2-4}{-2}-(-2-4)=3+6=9$ |
# Verify Equivalent Fractions
How to verify equivalent fractions?
If two fractions are equivalent, then the product of one’s numerator and the other’s denominator is equal to the product of one’s denominator and the other numerator.
Consider two fractions 1/3 and 4/12
1 × 12 = 12
3 × 4 = 12
1 × 12 = 3 × 4.
So, 1/3 and 4/12 are equivalent fractions.
Worked-out examples to verify equivalent fractions:
1. Write five equivalent fractions of 4/5.
The five equivalent fractions of 4/5 are:
4 × 2/5 × 2 = 8/10
4 × 3/5 × 3 = 12/15
4 × 4/5 × 4 = 16/20
4 × 5/5 × 5 = 20/25
4 × 6/5 × 6 = 24/30
Therefore, 8/10, 12/15, 16/20, 20/25 and 24/30 are the equivalent fractions.
2. Are the following two fractions equivalent?
(i) 2/5 and 10/24
Product of 2 × 24 = 48
Product of 5 × 10 = 50
The products are not equal.
So, 2/5 and 10/24 are not equivalent fractions.
(ii) 7/9 and 28/36
Product of 7 × 36 = 252
Product of 9 × 28 = 252
Both the products are equal.
So, 7/9 and 28/36 are equivalent fractions.
Thus, to verify equivalent fractions the above procedure are followed to make sure that the pairs of fractions are equivalent.
Related Concepts |
# How to Solve Multiplication Rule for Probabilities?
The law of multiplication states that the probability that $$A$$ and $$B$$ occur is equal to the probability that $$A$$ occurs times the conditional probability that $$B$$ occurs, given that we know $$A$$ has already occurred.
Using the multiplication rule, you can calculate the probability that events $$A$$ and $$B$$ occur jointly when you know the probability of event $$A$$ and event $$B$$ occurring separately.
## Step by step guide to multiplication rule for probabilities
Before we go into the laws of probability, let’s define a few terms:
• If two events cannot happen at the same moment, they are mutually exclusive or discontinuous.
• A conditional probability is a likelihood that event $$A$$ will occur if event $$B$$ has already occurred. The symbol $$P(A|B)$$ represents the conditional probability of event $$A$$ given event $$B$$.
• An event’s complement is the event that does not occur. $$P(A’)$$ indicates the chance that event A will not occur.
• The probability of both events $$A$$ and $$B$$ occurring is the probability of $$A$$ and $$B$$ intersecting. The likelihood of events $$A$$ and $$B$$ intersecting is indicated by $$P(A∩ B)$$. $$P(A∩ B) = 0$$ if events $$A$$ and $$B$$ are mutually exclusive.
• The probability that events $$A$$ or $$B$$ occur is the probability of the union of $$A$$ and $$B$$. The probability of the union of events $$A$$ and $$B$$ is indicated by $$P(A ∪ B)$$.
• Events $$A$$ and $$B$$ are dependent if the occurrence of event $$A$$ alters the likelihood of event $$B$$. Events $$A$$ and $$B$$, on the other hand, are independent if the occurrence of event $$A$$ does not affect the probability of event $$B$$.
### Multiplication rule
In probability theory, the law of multiplication states given that event $$A$$ has occurred, the probability that events $$A$$ and $$B$$ will both occur is equal to the probability that event $$A$$ will occur multiplied by the probability that event $$B$$ will occur.
$$\color{blue}{P(A∩B)=P(B)⋅P(A|B)}$$
By shifting the roles of $$A$$ and $$B$$, we can also write the law:
$$\color{blue}{P(A∩B)=P(A)⋅P(B|A)}$$
The general multiplication rule is obtained by multiplying both sides of the definition of conditional probability by the denominator. That is, in the equation:
$$\color{blue}{ P(A|B) =\frac{P(A∩B)}{P(B)}}$$
The rule is useful when we know both $$P(A)$$ and $$P(B|A)$$ or both $$P(B)$$ and $$P(A|B)$$.
### Multiplication Rule for Probabilities – Example 1:
A jar contains $$5$$ red marbles and $$4$$ black marbles. Two marbles are pulled from the jar without replacement. What is the probability that both marbles are black?
If $$A =$$ the event that the first marble is black; and if $$B =$$ the event that the second marble is black. We know the following:
Initially, there are $$9$$ marbles in the jar, $$4$$ of which are black. Therefore, $$P(A)=\frac{4}{9}$$.
After the first selection, there are $$8$$ marbles in the jar, $$3$$ of which are black. Therefore, $$P(B|A)=\frac{3}{8}$$.
Then, use this formula: $$\color{blue}{P(A∩B)=P(A)⋅P(B|A)}$$
$$P(A∩B) = \frac{4}{9} \times \frac{3}{8} =\frac{4 \times 3 }{9 \times 8}=\frac{12}{72}=\frac{1}{6}$$
## Exercises for Multiplication Rule for Probabilities
A social swimming team has $$150$$ members. Seventy-five members are advanced swimmers. Forty-seven members are intermediate swimmers. The rest of the swimmers are novices. Forty advanced swimmers practice three times a week. Thirty of the intermediate swimmers practice three times a week. Ten novice swimmers practice three times a week. Suppose a member of a swimming team is randomly selected.
1. What is the probability that a member is a novice swimmer?
2. What is the probability that a member exercises four times a week?
3. What is the probability that the member is an advanced swimmer and practices three times a week?
4. What is the probability that a member is an advanced swimmer and an intermediate swimmer?
1. $$\color{blue}{\frac{28}{150}}$$
2. $$\color{blue}{\frac{80}{150}}$$
3. $$\color{blue}{\frac{40}{150}}$$
4. $$\color{blue}{P=0}$$
### What people say about "How to Solve Multiplication Rule for Probabilities?"?
No one replied yet.
X
30% OFF
Limited time only!
Save Over 30%
SAVE $5 It was$16.99 now it is \$11.99 |
197 views
If $x$ is a positive quantity such that $2^x=3^{\log_52}$, then $x$ is equal to
1. $1+\log_3\dfrac{5}{3}$
2. $\log_58$
3. $1+\log_5\dfrac{3}{5}$
4. $\log_59$
Given that, $2^{x} = 3^{\log_{5}{2}}$
Taking $\log_{2}$ on both sides.
$\log_{2}{2}^{x} = \log_{2} \left( 3^{\log_{5}{2}} \right)$
$\Rightarrow x = \log_{5}{2} \log_{2}{3}$ $\quad [\because{ \log_{a}{a}} = 1, \log_{b}{a}^{x} = x \log_{b}{a}$]
$\Rightarrow x = \frac{\log_{2}{3}}{\log_{2}{5}}$ $\quad \left [ \therefore \log_{a}{b} = \frac{1}{\log_{b}{a}} \right]$
$\Rightarrow \boxed{x = \log_{5}{3}}$ $\quad \left[ \therefore \frac{\log_{c}{a}}{\log_{c}{b}} = \log_{b}{a} \right]$
Now, we can check all the options.
1. $1+\log_{3}{\frac{5}{3}} = 1 + \log_{3}{5} – \log_{3}{3}$
$\qquad \qquad \quad =1 + \log_{3}{5} – 1$
$\qquad \qquad \quad = \log_{3}{5}$
Option $\text {(B)}$ and $\text{(D)}$ are not possible.
1. $1+\log_{5}{\frac{3}{5}} = 1 + \log_{5}{3} – \log_{5}{5}$
$\qquad \qquad \quad=1 + \log_{5}{3} – 1$
$\qquad \qquad \quad = \log_{5}{3}$
Correct Answer $: \text{C}$
10.3k points
1
217 views |
# How Do You Evaluate a Function? II
$\fn_jvn&space;{\color{Blue}&space;f(x)=x^2-2x+4}$This post about how to evaluate a function is part of a series of posts to help you prepare for the Advanced Algebra and Functions part of the Accuplacer test.
## Question
For the function
$\fn_jvn&space;{\color{Blue}&space;f(x)=x^2-2x+4}$
what is the value of $\inline&space;\fn_jvn&space;f(x-1)$?
A. $\inline&space;\fn_jvn&space;x^2-2x+3$
B. $\inline&space;\fn_jvn&space;(x-1)^2-2(x-1)+4$$\inline&space;\fn_jvn&space;(x-1)(x^2-2x+3)$
C. $\inline&space;\fn_jvn&space;(x-1)(x^2-2x+3)$
D. $\fn_jvn&space;x^2+3x+4$
## Solution
f(x-1) is what you get when you replace x with x-1 in $\inline \fn_jvn x^2-2x+4$. That is called “evaluating the function for x-1.
### Sidebar: Function Notation
In algebra, usually when you write two things side by side with one of them in parentheses, that means you multiply those two things together:
$\fn_jvn&space;{\color{Blue}&space;a(b)=a&space;\cdot&space;b}$
So no one can blame you for thinking f(x) means f multiplied by x. But math notation, like other kinds of language, has exceptions. f(x) means function f, carried out on x. And f(something else) means function f carried out on something else. For example, say function g is defined as $\inline \fn_jvn g(x)=x^2$. That means $\inline \fn_jvn g(1)=1^2$; $\inline \fn_jvn g(-2)=\left ( -2 \right )^2$; and $\fn_jvn g(*)$ means $\inline \fn_jvn *^2$ (whatever * means).
It gets interesting when the parentheses in the function definition contain a function. For example, if $\inline \fn_jvn g(x)=x^2$, what is $\inline \fn_jvn g(x+2)$? Well, we said that if you want to find $\fn_jvn g(*)$, you insert * into the function definition. To find $\inline \fn_jvn g(x+2)$, insert $\inline \fn_jvn x+2$ into the function definition: if $\inline \fn_jvn g(x)=x^2$, then $\inline \fn_jvn g(x+2)=(x+2)^2$.
So to find f(x-1) in the expression
$\fn_jvn&space;{\color{Blue}&space;f(x)=x^2-2x+4}$
insert x-1 wherever you see x. You should get:
$\fn_jvn&space;{\color{Blue}&space;(x-1)^2-2(x-1)+4}$
Usually an answer like this would be expressed as $\inline \fn_jvn x^2-4x+7$, but this question is multiple choice so you go with what you’ve got. |
Package of problems
# Quadratic symmetry Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
## Solution
The equation of this graph is $y = x^2 - 8x + 10$
Is this graph symmetrical? Are you convinced? How can you convince someone else?
### Approach 1
It looks like the graph has a line of symmetry with equation $x = 4$.
If we can show that each point on the curve has a mirror image about the line $x=4$, then we will be correct.
Let’s choose $x$ coordinates such that they are equidistant from the line $x=4$, e.g. $x=4+k$ and $x=4-k$.
When $x = 4 + k$, $\quad y = (4 + k)^2 - 8(4 + k) + 10 = k^2 - 6$.
When $x = 4 - k$, $\quad y = (4 - k)^2 - 8(4 - k) + 10 = k^2 - 6$.
The $y$ coordinates are equal. This shows we are correct and the graph $y = x^2 - 8x + 10$ is symmetrical about the line $x = 4$.
Alternatively, we can perform the same calculation with the completed square form. What do you notice?
What did we do in this approach? We made a conjecture (an educated guess) and then went on to prove that this was correct. This is a fundamental process in doing mathematics.
### Approach 2
It looks like the graph of $y = x^2 - 8x + 10$ has the same shape as $y = x^2$ but is in a different position.
If we can show that $y = x^2$ is symmetrical and we can translate the graph $y = x^2$ to $y = x^2 - 8x + 10$, then we have shown that $y = x^2 - 8x + 10$ is symmetrical. We do this in two steps.
Step 1: Show that the graph $y = x^2$ is symmetrical about the $y$ axis.
When $x = k$, $\quad \phantom{-}y = k^2$.
When $x = -k$, $\quad y = k^2$.
By a similar argument to Approach 1 above, the graph of $y = x^2$ is symmetrical about the $y$ axis.
Note: A function where $f(x) = f(-x)$ for all $x$ is called an even function and is symmetrical about the $y$ axis. There is more on this topic in Odd or even or ….
Step 2: Rewrite $y = x^2 - 8x + 10$ as a translation of $y = x^2$.
By completing the square, we have $y=x^2 - 8x + 10 = (x - 4)^2 - 6$ which is a translation of the graph $y = x^2$ by $\dbinom{4}{-6}$. The line of symmetry is therefore translated to $x = 4$ and the vertex is translated to ($4,-6$).
Therefore the graph $y = x^2 - 8x + 10$ is symmetrical.
How do these approaches compare?
Which is the more easily generalised?
How would these approaches help us to consider whether all quadratics are symmmetrical? |
# Written Multiplication of Three-Digit Numbers by One-Digit
In this worksheet, students must carry out formal written multiplications of three-digit numbers by single-digit numbers. Carrying is involved.
Key stage: KS 2
Curriculum topic: Number: Multiplication and Division
Curriculum subtopic: Use Written Form for Multiplication
Difficulty level:
### QUESTION 1 of 10
Written multiplication can be done step-by-step by setting out the calculation carefully and remembering to regroup if necessary.
Example
Work out
947 × 7
1. Rewrite the calculation vertically, like this:
2. Then multiply the numbers in the ones column together, so 7 × 7 = 49
Write the 9 in the ones column and regroup the 4 tens below in the tens column.
3. Next multiply the numbers in the tens column together, so 4 × 7 = 28 and add the 4 tens that were regrouped.
This gives us 28 + 4 = 32 , so we write the 2 in the tens column and regroup the 3 hundreds below in the hundreds column.
4. Lastly, multiply the numbers in the hundreds column, so 9 × 7 = 63 and add the 3 that was regrouped.
This gives us 63 + 3 = 66 in the thousands.
So, in full, we have:
So 947 × 7 = 6629
Want to understand this further and learn how this links to other topics in maths?
Why not watch this video?
Work out
232 × 3
Work out
232 × 4
Work out
223 × 4
Work out
253 × 5
Work out
368 × 5
Work out
432 × 6
Work out
608 × 7
Work out
719 × 8
Work out
593 × 8
Work out
888 × 9
• Question 1
Work out
232 × 3
696
EDDIE SAYS
1. Rewrite the calculation vertically. 2. Multiply the numbers in the ones column together, so 2 x 3 = 6. 3. Next multiply the numbers in the tens column together, so 3 x 3 = 9. 4. Lastly, multiply the numbers in the hundreds column together, so 2 x 3 = 6. Answer: 696
• Question 2
Work out
232 × 4
928
EDDIE SAYS
1. Rewrite the calculation vertically. 2. Multiply the numbers in the ones column together, so 2 x 4 = 8. 3. Next multiply the numbers in the tens column together, so 3 x 4 = 12. Remember to write the 2 in the tens column and regroup the 1 hundred below into the hundreds column. 4. Lastly, multiply the hundreds, so 2 x 4 = 8 and add the 1 hundred that was regrouped. That gives us 8 + 1 = 9 in the hundreds. Answer: 928
• Question 3
Work out
223 × 4
892
EDDIE SAYS
1. Rewrite the calculation vertically. 2. Multiply the numbers in the ones column together, so 3 x 4 = 12. Write the 2 in the ones column and regroup the 1 ten below in the tens column. 3. Next multiply the numbers in the tens column together, so 2 x 4 = 8. Remember to add the regrouped 1, so 8 + 1 = 9. Write 9 in the tens column. 4. Lastly, multiply the hundreds, so 2 x 4 = 8. Write 8 in the hundreds column. Answer: 892
• Question 4
Work out
253 × 5
1265
EDDIE SAYS
1. Rewrite the calculation vertically. 2. Multiply the numbers in the ones column together, so 3 x 5 = 15. Write the 5 in the ones column and regroup the 1 ten below in the tens column. 3. Next multiply the numbers in the tens column together, so 5 x 5 = 25. Remember to add the regrouped 1, so 25 + 1 = 26. Write 6 in the tens column and regroup the 2 hundreds below in the hundreds column. 4. Lastly, multiply the hundreds, so 2 x 5 = 10. Remember to add the regrouped 2, so 10 + 2 = 12. Write the 2 in the hundreds column and the 1 in the thousands column. Answer: 1265
• Question 5
Work out
368 × 5
1840
EDDIE SAYS
1. Rewrite the calculation vertically. 2. Multiply the numbers in the ones column together, so 8 x 5 = 40. Write the 0 in the ones column and regroup the 4 tens below in the tens column. 3. Next multiply the numbers in the tens column together, so 6 x 5 = 30. Remember to add the regrouped 4 tens, so 30 + 4 = 34. Write 4 in the tens column and regroup the 3 hundreds below in the hundreds column. 4. Lastly, multiply the hundreds, so 3 x 5 = 15. Remember to add the regrouped 3, so 15 + 3 = 18. Write the 8 in the hundreds column and the 1 in the thousands column. Answer: 1840
• Question 6
Work out
432 × 6
2592
EDDIE SAYS
1. Rewrite the calculation vertically. 2. Multiply the numbers in the ones column together, so 2 x 6 = 12. Write the 2 in the ones column and regroup the 1 ten below in the tens column. 3. Next multiply the numbers in the tens column together, so 3 x 6 = 18. Remember to add the regrouped 1, so 18 + 1 = 19. Write 9 in the tens column and regroup the 1 hundred below in the hundreds column. 4. Lastly, multiply the hundreds, so 4 x 6 = 24. Remember to add the regrouped 1, so 24 + 1 = 25. Write the 5 in the hundreds column and the 2 in the thousands column. Answer: 2592
• Question 7
Work out
608 × 7
4256
EDDIE SAYS
1. Rewrite the calculation vertically. 2. Multiply the numbers in the ones column together, so 8 x 7 = 56. Write the 6 in the ones column and regroup the 5 tens below in the tens column. 3. Next multiply the numbers in the tens column together, so 0 x 7 = 0. Remember to add the regrouped 5 tens, so 0 + 5 = 5. Write 5 in the tens column. 4. Lastly, multiply the hundreds, so 6 x 7 = 42. There is no regrouping to add in the hundreds column. Write the 2 in the hundreds column and the 4 in the thousands column. Answer: 4256
• Question 8
Work out
719 × 8
5752
EDDIE SAYS
1. Rewrite the calculation vertically. 2. Multiply the numbers in the ones column together, so 9 x 8 = 72. Write the 2 in the ones column and regroup the 7 tens below in the tens column. 3. Next multiply the numbers in the tens column together, so 1 x 8 = 8. Remember to add the regrouped 7 tens, so 8 + 7 = 15. Write 5 in the tens column and regroup the 1 hundred below in the hundreds column. 4. Lastly, multiply the hundreds, so 7 x 8 = 56. Remember to add the regrouped 1, so 56 + 1 = 57. Write the 7 in the hundreds column and the 5 in the thousands column. Answer: 5752
• Question 9
Work out
593 × 8
4744
EDDIE SAYS
1. Rewrite the calculation vertically. 2. Multiply the numbers in the ones column together, so 3 x 8 = 24. Write the 4 in the ones column and regroup the 2 tens below in the tens column. 3. Next multiply the numbers in the tens column together, so 9 x 8 = 72. Remember to add the regrouped 2 tens, so 72 + 2 = 74. Write 4 in the tens column and regroup the 7 hundreds below in the hundreds column. 4. Lastly, multiply the hundreds, so 5 x 8 = 40. Remember to add the regrouped 7, so 40 + 7 = 47. Write the 7 in the hundreds column and the 4 in the thousands column. Answer: 4744
• Question 10
Work out
888 × 9
7992
EDDIE SAYS
1. Rewrite the calculation vertically. 2. Multiply the numbers in the ones column together, so 8 x 9 = 72. Write the 2 in the ones column and regroup the 7 tens below in the tens column. 3. Next multiply the numbers in the tens column together, so 8 x 9 = 72. Remember to add the regrouped 7 tens, so 72 + 7 = 79. Write 9 in the tens column and regroup the 7 hundreds below in the hundreds column. 4. Lastly, multiply the hundreds, so 8 x 9 = 72. Remember to add the regrouped 7, so 72 + 7 = 79. Write the 9 in the hundreds column and the 7 in the thousands column. Answer: 7992
---- OR ----
Sign up for a £1 trial so you can track and measure your child's progress on this activity.
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
Get started
Start your £1 trial today.
Subscribe from £10/month.
• Tuition Partner |
# Question #b9406
Jan 10, 2018
$\sin \left(8 \theta\right) = \frac{u}{8 a}$
#### Explanation:
I'm not too sure of your question.
I take it you are asking:
$\sin \theta = \frac{u}{a}$; What is $\sin \left(8 \theta\right)$?
We can relate this back to transformations of graphs in general. We're going to be moving away from trigonometry for just a bit!
For example:
$\text{Let } f \left(x\right) = {x}^{2} - 9 x + 18$
How about for a moment we consider the roots of $f \left(x\right) = 0$
$\text{Let } f \left(x\right) = 0$
${x}^{2} - 9 x + 18 = 0$
$\left(x - 3\right) \left(x - 6\right) = 0$
So this equation has roots of $x = 3$ and $x = 6$
graph{y=x^2-9x+18 [-10, 10, -5, 5]}
Consider $f \left(2 x\right)$
$f \left(2 x\right) = {\left(2 x\right)}^{2} - 9 \left(2 x\right) + 18$
$= 4 {x}^{2} - 18 x + 18$
Now, consider the roots of $f \left(2 x\right) = 0$
$\text{Let } f \left(2 x\right) = 0$
$4 {x}^{2} - 18 x + 18 = 0$
$2 {x}^{2} - 9 x + 9 = 0$
$x = \frac{9 \pm \sqrt{{\left(- 9\right)}^{2} - 4 \times 2 \times 9}}{2 \times 2}$
$x = 3$ or $x = \frac{3}{2}$
graph{4x^2-18x+18 [-10, 10, -5, 5]}
Look at the roots.
$f \left(x\right)$ has roots $x = 6$ and $x = 3$
$f \left(2 x\right)$ has roos $x = \frac{6}{2}$ and $x = \frac{3}{2}$
This isn't a proof, but it's enough for us to infer that:
if an expression $f \left(x\right)$ has roots $x = \alpha , \beta , \gamma , \delta \ldots$, then $f \left(k x\right)$ has roots $x = \frac{\alpha}{k} , \frac{\beta}{k} , \frac{\gamma}{k} , \frac{\delta}{k} \ldots$
Back to our trig
$\text{Let } g \left(x\right) = \sin x$ (because I used $f \left(x\right)$ further up).
For some value $x = \theta , g \left(\theta\right) = \frac{u}{a}$
$g \left(8 x\right) = \sin \left(8 x\right)$
$\sin \left(8 \theta\right) = g \left(8 \theta\right) = \frac{\frac{u}{a}}{8}$
$\sin \left(8 \theta\right) = \frac{u}{8 a}$ |
Scientific Notation
Scientific Notation (also called Standard Form in Britain) is a special way of writing numbers:
Like this: Or this:
It makes it easy to use big and small values.
OK, How Does it Work?
Example: 700
Why is 700 written as 7 × 102 in Scientific Notation ?
700 = 7 × 100
and 100 = 102 (see powers of 10)
so 700 = 7 × 102
Both 700 and 7 × 102 have the same value, just shown in different ways.
Example: 4,900,000,000
1,000,000,000 = 109 ,
so 4,900,000,000 = 4.9 × 109 in Scientific Notation
The number is written in two parts:
• Just the digits, with the decimal point placed after the first digit, followed by
• × 10 to a power that puts the decimal point where it should be
(i.e. it shows how many places to move the decimal point).
In this example, 5326.6 is written as 5.3266 × 103,
because 5326.6 = 5.3266 × 1000 = 5.3266 × 103
Try It Yourself
Enter a number and see it in Scientific Notation:
Now try to use Scientific Notation yourself:
Other Ways of Writing It
3.1 × 10^8
We can use the ^ symbol (above the 6 on a keyboard), as it is easy to type.
Example: 3 × 10^4 is the same as 3 × 104
• 3 × 10^4 = 3 × 10 × 10 × 10 × 10 = 30,000
Calculators often use "E" or "e" like this:
Example: 6E+5 is the same as 6 × 105
• 6E+5 = 6 × 10 × 10 × 10 × 10 × 10 = 600,000
Example: 3.12E4 is the same as 3.12 × 104
• 3.12E4 = 3.12 × 10 × 10 × 10 × 10 = 31,200
How to Do it
To figure out the power of 10, think "how many places do I move the decimal point?"
When the number is 10 or greater, the decimal point has to move to the left, and the power of 10 is positive. When the number is smaller than 1, the decimal point has to move to the right, so the power of 10 is negative.
Example: 0.0055 is written 5.5 × 10-3
Because 0.0055 = 5.5 × 0.001 = 5.5 × 10-3
Example: 3.2 is written 3.2 × 100
We didn't have to move the decimal point at all, so the power is 100
But it is now in Scientific Notation
Check!
After putting the number in Scientific Notation, just check that:
• The "digits" part is between 1 and 10 (it can be 1, but never 10)
• The "power" part shows exactly how many places to move the decimal point
Why Use It?
Because it makes it easier when dealing with very big or very small numbers, which are common in Scientific and Engineering work.
Example: it is easier to write (and read) 1.3 × 10-9 than 0.0000000013
It can also make calculations easier, as in this example:
Example: a tiny space inside a computer chip has been measured to be 0.00000256m wide, 0.00000014m long and 0.000275m high.
What is its volume?
Let's first convert the three lengths into scientific notation:
• width: 0.000 002 56m = 2.56×10-6
• length: 0.000 000 14m = 1.4×10-7
• height: 0.000 275m = 2.75×10-4
Then multiply the digits together (ignoring the ×10s):
2.56 × 1.4 × 2.75 = 9.856
Last, multiply the ×10s:
10-6 × 10-7 × 10-4 = 10-17 (easier than it looks, just add −6, −4 and −7 together)
The result is 9.856×10-17 m3
It is used a lot in Science:
Example: Suns, Moons and Planets
The Sun has a Mass of 1.988 × 1030 kg.
Easier than writing 1,988,000,000,000,000,000,000,000,000,000 kg
(and that number gives a false sense of many digits of accuracy.)
Play With It!
Use Scientific Notation
in Gravity Freeplay
It can also save space! Here is what happens when you double on each square of a chess board:
Values are rounded off, so 53,6870,912 is shown as just 5×108
That last value, shown as 9×1018 is actually 9,223,372,036,854,775,808
Engineering Notation
Engineering Notation is like Scientific Notation, except that we only use powers of ten that are multiples of 3 (such as 103, 10-3, 1012 etc).
Examples:
• 2,700 is written 2.7 × 103
• 27,000 is written 27 × 103
• 270,000 is written 270 × 103
• 2,700,000 is written 2.7 × 106
Example: 0.00012 is written 120 × 10-6
Notice that the "digits" part can now be between 1 and 1,000 (it can be 1, but never 1,000).
The advantage is that we can replace the ×10s with Metric Numbers. So we can use standard words (such as thousand or million), prefixes (such as kilo, mega) or the symbol (k, M, etc)
Example: 19,300 meters is written 19.3 × 103 m, or 19.3 km
Example: 0.00012 seconds is written 120 × 10-6 s, or 120 microseconds |
# Question Video: Solving Word Problems Involving Geometric Series Mathematics • 9th Grade
The table below represents the salary of an employee in three consecutive years. The salary can be described by a geometric sequence. Find the total salary of the employee over 5 years.
04:05
### Video Transcript
The table below represents the salary of an employee in three consecutive years. The salary can be described by a geometric sequence. Find the total salary of the employee over five years.
We are told that in the first year, the employee earns 73,600 pounds, in the second year, they earn 110,400 pounds, and in the third year, they earn 165,600 pounds. We are told that this represents a geometric sequence. And in any geometric sequence, we have first term π and common ratio π. This means that to get from the first term to the second term, we multiply by π. So if the first term is π, the second term is ππ. The third term is π multiplied by π multiplied by π, which is ππ squared. This also means that the fourth term will be equal to ππ cubed and the fifth term ππ to the fourth power.
We can calculate the value of π by dividing the second term by the first term, in this case, 110,400 divided by 73,600. This is equal to 1.5. Therefore, the common ratio of this geometric sequence is 1.5. We could use this to calculate the employeeβs salary in the fourth and fifth years. 165,600 multiplied by 1.5 is equal to 248,400. This is the salary of the employee in pounds in the fourth year. Multiplying this by 1.5 gives us 372,600. In the fifth year, the employee earns 372,600 pounds. In order to find the total salary over five years, we could then sum these five values. This gives us a total salary of 970,600 pounds.
There is an alternative way of calculating the total after five years. We can do this by using the fact that in a geometric sequence, the sum of the first π-terms is equal to π multiplied by π to the power of π minus one divided by π minus one. This can also be written as π multiplied by one minus π to the power of π divided by one minus π. In this question, as π is greater than one, we will use the first formula. We know that π, the first term, is equal to 73,600 and π is equal to 1.5. We need to calculate the total salary over five years. Therefore, π is equal to five. π sub five is equal to 73,600 multiplied by 1.5 to the power of five minus one all divided by 1.5 minus one.
The numerator simplifies to 485,300, and 1.5 minus one is equal to 0.5. As dividing by 0.5 is the same as multiplying by two, we once again get an answer of 970,600. This confirms that the total salary of the employee over five years is 970,600 pounds. |
# Calculus/Multivariable and differential calculus:Exercises
← Partial differential equations Calculus Extensions → Multivariable and differential calculus:Exercises
## Parametric Equations
1. Find parametric equations describing the line segment from P(0,0) to Q(7,17).
x=7t and y=17t, where 0 ≤ t ≤ 1
x=7t and y=17t, where 0 ≤ t ≤ 1
2. Find parametric equations describing the line segment from ${\displaystyle P(x_{1},y_{1})}$ to ${\displaystyle Q(x_{2},y_{2})}$.
${\displaystyle x=x_{1}+(x_{2}-x_{1})t{\mbox{ and }}y=y_{1}+(y_{2}-y_{1})t,{\mbox{ where }}0\leq t\leq 1}$
${\displaystyle x=x_{1}+(x_{2}-x_{1})t{\mbox{ and }}y=y_{1}+(y_{2}-y_{1})t,{\mbox{ where }}0\leq t\leq 1}$
3. Find parametric equations describing the ellipse centered at the origin with major axis of length 6 along the x-axis and the minor axis of length 3 along the y-axis, generated clockwise.
${\displaystyle x=3\cos(-t),\ y=1.5\sin(-t)}$
${\displaystyle x=3\cos(-t),\ y=1.5\sin(-t)}$
## Polar Coordinates
20. Convert the equation into Cartesian coordinates: ${\displaystyle r=\sin(\theta )\sec ^{2}(\theta ).}$
Making the substitutions ${\displaystyle \sin(\theta )={\frac {y}{r}}}$ and ${\displaystyle \sec(\theta )={\frac {r}{x}}}$ gives ${\displaystyle r={\frac {y}{r}}\left({\frac {r}{x}}\right)^{2}}$ ${\displaystyle \iff r={\frac {ry}{x^{2}}}}$ ${\displaystyle \iff y=x^{2}}$
Making the substitutions ${\displaystyle \sin(\theta )={\frac {y}{r}}}$ and ${\displaystyle \sec(\theta )={\frac {r}{x}}}$ gives ${\displaystyle r={\frac {y}{r}}\left({\frac {r}{x}}\right)^{2}}$ ${\displaystyle \iff r={\frac {ry}{x^{2}}}}$ ${\displaystyle \iff y=x^{2}}$
21. Find an equation of the line y=mx+b in polar coordinates.
Making the substitutions ${\displaystyle x=r\cos(\theta )}$ and ${\displaystyle y=r\sin(\theta )}$ gives ${\displaystyle r\sin(\theta )=mr\cos(\theta )+b}$ ${\displaystyle \iff (\sin(\theta )-m\cos(\theta ))r=b}$ ${\displaystyle \iff r={\frac {b}{\sin(\theta )-m\cos(\theta )}}}$
Making the substitutions ${\displaystyle x=r\cos(\theta )}$ and ${\displaystyle y=r\sin(\theta )}$ gives ${\displaystyle r\sin(\theta )=mr\cos(\theta )+b}$ ${\displaystyle \iff (\sin(\theta )-m\cos(\theta ))r=b}$ ${\displaystyle \iff r={\frac {b}{\sin(\theta )-m\cos(\theta )}}}$
Sketch the following polar curves without using a computer.
22. ${\displaystyle r=2-2\sin(\theta )}$
23. ${\displaystyle r^{2}=4\cos(\theta )}$
24. ${\displaystyle r=2\sin(5\theta )}$
Sketch the following sets of points.
25. ${\displaystyle \{(r,\theta ):\theta =2\pi /3\}}$
26. ${\displaystyle \{(r,\theta ):|\theta |\leq \pi /3{\mbox{ and }}|r|<3\}}$
## Calculus in Polar Coordinates
Find points where the following curves have vertical or horizontal tangents.
40. ${\displaystyle r=4\cos(\theta )}$
Horizontal tangents occur at points where ${\displaystyle {\frac {dy}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\sin(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0}$.
Vertical tangents occur at points where ${\displaystyle {\frac {dx}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\cos(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\cos(\theta )-r\sin(\theta )=0}$.
${\displaystyle {\frac {dr}{d\theta }}=-4\sin(\theta )}$
The condition for a horizontal tangent gives: ${\displaystyle (-4\sin(\theta ))\sin(\theta )+(4\cos(\theta ))\cos(\theta )=0}$ ${\displaystyle \iff \cos ^{2}(\theta )-\sin ^{2}(\theta )=0}$ ${\displaystyle \iff \cos(2\theta )=0}$ ${\displaystyle \iff 2\theta =\pm \pi /2}$ ${\displaystyle \iff \theta =\pm \pi /4}$
Horizontal tangents occur at ${\displaystyle (r,\theta )=(2{\sqrt {2}},\pi /4),(2{\sqrt {2}},-\pi /4)}$ which correspond to the Cartesian points ${\displaystyle (2,2)}$ and ${\displaystyle (2,-2)}$.
The condition for a vertical tangent gives: ${\displaystyle (-4\sin(\theta ))\cos(\theta )-(4\cos(\theta ))\sin(\theta )=0}$ ${\displaystyle \iff 2\sin(\theta )\cos(\theta )=0}$ ${\displaystyle \iff \sin(2\theta )=0}$ ${\displaystyle \iff 2\theta =0,\pi }$ ${\displaystyle \iff \theta =0,\pi /2}$
Vertical tangents occur at ${\displaystyle (r,\theta )=(0,\pi /2),(4,0)}$ which correspond to the Cartesian points ${\displaystyle (0,0)}$ and ${\displaystyle (4,0)}$.
Horizontal tangents at (2,2) and (2,-2); vertical tangents at (0,0) and (4,0)
Horizontal tangents occur at points where ${\displaystyle {\frac {dy}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\sin(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0}$.
Vertical tangents occur at points where ${\displaystyle {\frac {dx}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\cos(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\cos(\theta )-r\sin(\theta )=0}$.
${\displaystyle {\frac {dr}{d\theta }}=-4\sin(\theta )}$
The condition for a horizontal tangent gives: ${\displaystyle (-4\sin(\theta ))\sin(\theta )+(4\cos(\theta ))\cos(\theta )=0}$ ${\displaystyle \iff \cos ^{2}(\theta )-\sin ^{2}(\theta )=0}$ ${\displaystyle \iff \cos(2\theta )=0}$ ${\displaystyle \iff 2\theta =\pm \pi /2}$ ${\displaystyle \iff \theta =\pm \pi /4}$
Horizontal tangents occur at ${\displaystyle (r,\theta )=(2{\sqrt {2}},\pi /4),(2{\sqrt {2}},-\pi /4)}$ which correspond to the Cartesian points ${\displaystyle (2,2)}$ and ${\displaystyle (2,-2)}$.
The condition for a vertical tangent gives: ${\displaystyle (-4\sin(\theta ))\cos(\theta )-(4\cos(\theta ))\sin(\theta )=0}$ ${\displaystyle \iff 2\sin(\theta )\cos(\theta )=0}$ ${\displaystyle \iff \sin(2\theta )=0}$ ${\displaystyle \iff 2\theta =0,\pi }$ ${\displaystyle \iff \theta =0,\pi /2}$
Vertical tangents occur at ${\displaystyle (r,\theta )=(0,\pi /2),(4,0)}$ which correspond to the Cartesian points ${\displaystyle (0,0)}$ and ${\displaystyle (4,0)}$.
Horizontal tangents at (2,2) and (2,-2); vertical tangents at (0,0) and (4,0)
41. ${\displaystyle r=2+2\sin(\theta )}$
Horizontal tangents occur at points where ${\displaystyle {\frac {dy}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\sin(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0}$.
Vertical tangents occur at points where ${\displaystyle {\frac {dx}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\cos(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\cos(\theta )-r\sin(\theta )=0}$.
${\displaystyle {\frac {dr}{d\theta }}=2\cos(\theta )}$
The condition for a horizontal tangent gives: ${\displaystyle (2\cos(\theta ))\sin(\theta )+(2+2\sin(\theta ))\cos(\theta )=0}$ ${\displaystyle \iff 4\sin(\theta )\cos(\theta )+2\cos(\theta )=0}$ ${\displaystyle \iff \cos(\theta )(2\sin(\theta )+1)=0}$ ${\displaystyle \iff \cos(\theta )=0\;{\text{or}}\;\sin(\theta )=-{\frac {1}{2}}}$ ${\displaystyle \iff \theta =-5\pi /6,-\pi /2,-\pi /6,\pi /2}$
Horizontal tangents occur at ${\displaystyle (r,\theta )=(1,-5\pi /6),(0,-\pi /2),(1,-\pi /6),(4,\pi /2)}$ which correspond to the Cartesian points ${\displaystyle (-{\sqrt {3}}/2,-1/2)}$, ${\displaystyle (0,0)}$, ${\displaystyle ({\sqrt {3}}/2,-1/2)}$, and ${\displaystyle (0,4)}$. Point ${\displaystyle (0,0)}$ corresponds to a vertical cusp however and should be excluded leaving ${\displaystyle (-{\sqrt {3}}/2,-1/2)}$, ${\displaystyle ({\sqrt {3}}/2,-1/2)}$, and ${\displaystyle (0,4)}$.
The condition for a vertical tangent gives: ${\displaystyle (2\cos(\theta ))\cos(\theta )-(2+2\sin(\theta ))\sin(\theta )=0}$ ${\displaystyle \iff 2\cos ^{2}(\theta )-2\sin ^{2}(\theta )-2\sin(\theta )=0}$ ${\displaystyle \iff 2(1-\sin ^{2}(\theta ))-2\sin ^{2}(\theta )-2\sin(\theta )=0}$ ${\displaystyle \iff -4\sin ^{2}(\theta )-2\sin(\theta )+2=0}$ ${\displaystyle \iff 2\sin ^{2}(\theta )+\sin(\theta )-1=0}$ ${\displaystyle \iff (2\sin(\theta )-1)(\sin(\theta )+1)=0}$ ${\displaystyle \iff \sin(\theta )=-1,1/2}$ ${\displaystyle \iff \theta =-\pi /2,\pi /6,5\pi /6}$
Vertical tangents occur at ${\displaystyle (r,\theta )=(0,-\pi /2),(3,\pi /6),(3,5\pi /6)}$ which correspond to the Cartesian points ${\displaystyle (0,0)}$, ${\displaystyle (3{\sqrt {3}}/2,3/2)}$, and ${\displaystyle (-3{\sqrt {3}}/2,3/2)}$.
Horizontal tangents at (r,θ) = (4,π/2), (1,7π/6) and (1,-π/6); vertical tangents at (r,θ) = (3,π/6), (3,5π/6), and (0,3π/2)
Horizontal tangents occur at points where ${\displaystyle {\frac {dy}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\sin(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\sin(\theta )+r\cos(\theta )=0}$.
Vertical tangents occur at points where ${\displaystyle {\frac {dx}{d\theta }}=0}$. This condition is equivalent to ${\displaystyle {\frac {d}{d\theta }}(r\cos(\theta ))=0}$ ${\displaystyle \iff {\frac {dr}{d\theta }}\cos(\theta )-r\sin(\theta )=0}$.
${\displaystyle {\frac {dr}{d\theta }}=2\cos(\theta )}$
The condition for a horizontal tangent gives: ${\displaystyle (2\cos(\theta ))\sin(\theta )+(2+2\sin(\theta ))\cos(\theta )=0}$ ${\displaystyle \iff 4\sin(\theta )\cos(\theta )+2\cos(\theta )=0}$ ${\displaystyle \iff \cos(\theta )(2\sin(\theta )+1)=0}$ ${\displaystyle \iff \cos(\theta )=0\;{\text{or}}\;\sin(\theta )=-{\frac {1}{2}}}$ ${\displaystyle \iff \theta =-5\pi /6,-\pi /2,-\pi /6,\pi /2}$
Horizontal tangents occur at ${\displaystyle (r,\theta )=(1,-5\pi /6),(0,-\pi /2),(1,-\pi /6),(4,\pi /2)}$ which correspond to the Cartesian points ${\displaystyle (-{\sqrt {3}}/2,-1/2)}$, ${\displaystyle (0,0)}$, ${\displaystyle ({\sqrt {3}}/2,-1/2)}$, and ${\displaystyle (0,4)}$. Point ${\displaystyle (0,0)}$ corresponds to a vertical cusp however and should be excluded leaving ${\displaystyle (-{\sqrt {3}}/2,-1/2)}$, ${\displaystyle ({\sqrt {3}}/2,-1/2)}$, and ${\displaystyle (0,4)}$.
The condition for a vertical tangent gives: ${\displaystyle (2\cos(\theta ))\cos(\theta )-(2+2\sin(\theta ))\sin(\theta )=0}$ ${\displaystyle \iff 2\cos ^{2}(\theta )-2\sin ^{2}(\theta )-2\sin(\theta )=0}$ ${\displaystyle \iff 2(1-\sin ^{2}(\theta ))-2\sin ^{2}(\theta )-2\sin(\theta )=0}$ ${\displaystyle \iff -4\sin ^{2}(\theta )-2\sin(\theta )+2=0}$ ${\displaystyle \iff 2\sin ^{2}(\theta )+\sin(\theta )-1=0}$ ${\displaystyle \iff (2\sin(\theta )-1)(\sin(\theta )+1)=0}$ ${\displaystyle \iff \sin(\theta )=-1,1/2}$ ${\displaystyle \iff \theta =-\pi /2,\pi /6,5\pi /6}$
Vertical tangents occur at ${\displaystyle (r,\theta )=(0,-\pi /2),(3,\pi /6),(3,5\pi /6)}$ which correspond to the Cartesian points ${\displaystyle (0,0)}$, ${\displaystyle (3{\sqrt {3}}/2,3/2)}$, and ${\displaystyle (-3{\sqrt {3}}/2,3/2)}$.
Horizontal tangents at (r,θ) = (4,π/2), (1,7π/6) and (1,-π/6); vertical tangents at (r,θ) = (3,π/6), (3,5π/6), and (0,3π/2)
Sketch the region and find its area.
42. The region inside the limaçon ${\displaystyle 2+\cos(\theta )}$
Given an infinitesimal wedge with angle ${\displaystyle d\theta }$ and radius ${\displaystyle r}$, the area is ${\displaystyle {\frac {1}{2}}r^{2}d\theta }$. The total area is therefore ${\displaystyle A=\int _{\theta =0}^{2\pi }{\frac {1}{2}}r^{2}d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }{\frac {1}{2}}(2+\cos(\theta ))^{2}d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{2}}\cos ^{2}(\theta ))d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{4}}(\cos(2\theta )+1))d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }({\frac {9}{4}}+2\cos(\theta )+{\frac {1}{4}}\cos(2\theta ))d\theta }$ ${\displaystyle =({\frac {9}{4}}\theta +2\sin(\theta )+{\frac {1}{8}}\sin(2\theta )){\bigg |}_{\theta =0}^{2\pi }}$ ${\displaystyle ={\frac {9}{2}}\pi }$. 9π/2
Given an infinitesimal wedge with angle ${\displaystyle d\theta }$ and radius ${\displaystyle r}$, the area is ${\displaystyle {\frac {1}{2}}r^{2}d\theta }$. The total area is therefore ${\displaystyle A=\int _{\theta =0}^{2\pi }{\frac {1}{2}}r^{2}d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }{\frac {1}{2}}(2+\cos(\theta ))^{2}d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{2}}\cos ^{2}(\theta ))d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }(2+2\cos(\theta )+{\frac {1}{4}}(\cos(2\theta )+1))d\theta }$ ${\displaystyle =\int _{\theta =0}^{2\pi }({\frac {9}{4}}+2\cos(\theta )+{\frac {1}{4}}\cos(2\theta ))d\theta }$ ${\displaystyle =({\frac {9}{4}}\theta +2\sin(\theta )+{\frac {1}{8}}\sin(2\theta )){\bigg |}_{\theta =0}^{2\pi }}$ ${\displaystyle ={\frac {9}{2}}\pi }$. 9π/2
43. The region inside the petals of the rose ${\displaystyle 4\cos(2\theta )}$ and outside the circle ${\displaystyle r=2}$
There are 4 petals, as seen in the image below. The area of just one of the petals needs to be computed and the multiplied by ${\displaystyle 4}$.
It is first necessary to compute the angular limits of one of the petals. The petals start and end at points where ${\displaystyle 4\cos(2\theta )=2}$ ${\displaystyle \iff \cos(2\theta )={\frac {1}{2}}}$ ${\displaystyle \Leftarrow 2\theta =\pm {\frac {\pi }{3}}}$ ${\displaystyle \iff \theta =\pm {\frac {\pi }{6}}}$. The bounds on one of the petals are ${\displaystyle \left[-{\frac {\pi }{6}},+{\frac {\pi }{6}}\right]}$.
Given an annular wedge with angle ${\displaystyle d\theta }$, inner radius ${\displaystyle r_{1}}$, and an outer radius of ${\displaystyle r_{2}}$, the area is ${\displaystyle {\frac {1}{2}}(r_{2}^{2}-r_{1}^{2})d\theta }$. The total area of all 4 petals is therefore ${\displaystyle A=4\int _{\theta =-\pi /6}^{\pi /6}{\frac {1}{2}}((4\cos(2\theta ))^{2}-2^{2})d\theta }$ ${\displaystyle =4\int _{\theta =-\pi /6}^{\pi /6}(8\cos ^{2}(2\theta )-2)d\theta }$ ${\displaystyle =4\int _{\theta =-\pi /6}^{\pi /6}(4(\cos(4\theta )+1)-2)d\theta }$ ${\displaystyle =4\int _{\theta =-\pi /6}^{\pi /6}(2+4\cos(4\theta ))d\theta }$ ${\displaystyle =4(2\theta +\sin(4\theta )){\bigg |}_{\theta =-\pi /6}^{\pi /6}}$ ${\displaystyle =4({\frac {2\pi }{3}}+{\sqrt {3}})}$ ${\displaystyle ={\frac {8\pi }{3}}+4{\sqrt {3}}}$.
${\displaystyle 8\pi /3+4{\sqrt {3}}}$
There are 4 petals, as seen in the image below. The area of just one of the petals needs to be computed and the multiplied by ${\displaystyle 4}$.
It is first necessary to compute the angular limits of one of the petals. The petals start and end at points where ${\displaystyle 4\cos(2\theta )=2}$ ${\displaystyle \iff \cos(2\theta )={\frac {1}{2}}}$ ${\displaystyle \Leftarrow 2\theta =\pm {\frac {\pi }{3}}}$ ${\displaystyle \iff \theta =\pm {\frac {\pi }{6}}}$. The bounds on one of the petals are ${\displaystyle \left[-{\frac {\pi }{6}},+{\frac {\pi }{6}}\right]}$.
Given an annular wedge with angle ${\displaystyle d\theta }$, inner radius ${\displaystyle r_{1}}$, and an outer radius of ${\displaystyle r_{2}}$, the area is ${\displaystyle {\frac {1}{2}}(r_{2}^{2}-r_{1}^{2})d\theta }$. The total area of all 4 petals is therefore ${\displaystyle A=4\int _{\theta =-\pi /6}^{\pi /6}{\frac {1}{2}}((4\cos(2\theta ))^{2}-2^{2})d\theta }$ ${\displaystyle =4\int _{\theta =-\pi /6}^{\pi /6}(8\cos ^{2}(2\theta )-2)d\theta }$ ${\displaystyle =4\int _{\theta =-\pi /6}^{\pi /6}(4(\cos(4\theta )+1)-2)d\theta }$ ${\displaystyle =4\int _{\theta =-\pi /6}^{\pi /6}(2+4\cos(4\theta ))d\theta }$ ${\displaystyle =4(2\theta +\sin(4\theta )){\bigg |}_{\theta =-\pi /6}^{\pi /6}}$ ${\displaystyle =4({\frac {2\pi }{3}}+{\sqrt {3}})}$ ${\displaystyle ={\frac {8\pi }{3}}+4{\sqrt {3}}}$.
${\displaystyle 8\pi /3+4{\sqrt {3}}}$
## Vectors and Dot Product
60. Find an equation of the sphere with center (1,2,0) passing through the point (3,4,5)
The general equation for a sphere is ${\displaystyle (x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}=r^{2}}$ where ${\displaystyle (x_{0},y_{0},z_{0})}$ is the location of the sphere's center and ${\displaystyle r}$ is the sphere's radius.
It is already known that the sphere's center is ${\displaystyle (x_{0},y_{0},z_{0})=(1,2,0)}$. The sphere's radius is the distance between (1,2,0) and (3,4,5) which is ${\displaystyle r={\sqrt {(3-1)^{2}+(4-2)^{2}+(5-0)^{2}}}={\sqrt {4+4+25}}={\sqrt {33}}}$.
Therefore the sphere's equation is: ${\displaystyle (x-1)^{2}+(y-2)^{2}+z^{2}=33}$.
The general equation for a sphere is ${\displaystyle (x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}=r^{2}}$ where ${\displaystyle (x_{0},y_{0},z_{0})}$ is the location of the sphere's center and ${\displaystyle r}$ is the sphere's radius.
It is already known that the sphere's center is ${\displaystyle (x_{0},y_{0},z_{0})=(1,2,0)}$. The sphere's radius is the distance between (1,2,0) and (3,4,5) which is ${\displaystyle r={\sqrt {(3-1)^{2}+(4-2)^{2}+(5-0)^{2}}}={\sqrt {4+4+25}}={\sqrt {33}}}$.
Therefore the sphere's equation is: ${\displaystyle (x-1)^{2}+(y-2)^{2}+z^{2}=33}$.
61. Sketch the plane passing through the points (2,0,0), (0,3,0), and (0,0,4)
62. Find the value of ${\displaystyle |\mathbf {u} +3\mathbf {v} |}$ if ${\displaystyle \mathbf {u} =\langle 1,3,0\rangle }$ and ${\displaystyle \mathbf {v} =\langle 3,0,2\rangle }$
${\displaystyle \mathbf {u} +3\mathbf {v} =\langle 1,3,0\rangle +3\langle 3,0,2\rangle }$ ${\displaystyle =\langle 1,3,0\rangle +\langle 9,0,6\rangle }$ ${\displaystyle =\langle 10,3,6\rangle }$.
Therefore: ${\displaystyle |\mathbf {u} +3\mathbf {v} |={\sqrt {10^{2}+3^{2}+6^{2}}}={\sqrt {100+9+36}}={\sqrt {145}}}$.
${\displaystyle {\sqrt {145}}}$
${\displaystyle \mathbf {u} +3\mathbf {v} =\langle 1,3,0\rangle +3\langle 3,0,2\rangle }$ ${\displaystyle =\langle 1,3,0\rangle +\langle 9,0,6\rangle }$ ${\displaystyle =\langle 10,3,6\rangle }$.
Therefore: ${\displaystyle |\mathbf {u} +3\mathbf {v} |={\sqrt {10^{2}+3^{2}+6^{2}}}={\sqrt {100+9+36}}={\sqrt {145}}}$.
${\displaystyle {\sqrt {145}}}$
63. Find all unit vectors parallel to ${\displaystyle \langle 1,2,3\rangle }$
The length of ${\displaystyle \langle 1,2,3\rangle }$ is ${\displaystyle {\sqrt {1^{2}+2^{2}+3^{2}}}={\sqrt {1+4+9}}={\sqrt {14}}}$. Therefore ${\displaystyle {\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle }$ is a unit vector that points in the same direction as ${\displaystyle \langle 1,2,3\rangle }$, and ${\displaystyle -{\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle }$ is a unit vector that points in the opposite direction as ${\displaystyle \langle 1,2,3\rangle }$. ${\displaystyle \pm {\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle }$ are the unit vectors that are parallel to ${\displaystyle \langle 1,2,3\rangle }$.
The length of ${\displaystyle \langle 1,2,3\rangle }$ is ${\displaystyle {\sqrt {1^{2}+2^{2}+3^{2}}}={\sqrt {1+4+9}}={\sqrt {14}}}$. Therefore ${\displaystyle {\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle }$ is a unit vector that points in the same direction as ${\displaystyle \langle 1,2,3\rangle }$, and ${\displaystyle -{\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle }$ is a unit vector that points in the opposite direction as ${\displaystyle \langle 1,2,3\rangle }$. ${\displaystyle \pm {\frac {1}{\sqrt {14}}}\langle 1,2,3\rangle }$ are the unit vectors that are parallel to ${\displaystyle \langle 1,2,3\rangle }$.
64. Prove one of the distributive properties for vectors in ${\displaystyle \mathbb {R} ^{3}}$: ${\displaystyle c(\mathbf {u} +\mathbf {v} )=c\mathbf {u} +c\mathbf {v} }$
${\displaystyle c(\mathbf {u} +\mathbf {v} )=c(\langle u_{1},u_{2},u_{3}\rangle +\langle v_{1},v_{2},v_{3}\rangle )}$ ${\displaystyle =c\langle u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}\rangle }$ ${\displaystyle =\langle c(u_{1}+v_{1}),c(u_{2}+v_{2}),c(u_{3}+v_{3})\rangle }$ ${\displaystyle =\langle cu_{1}+cv_{1},cu_{2}+cv_{2},cu_{3}+cv_{3}\rangle }$ ${\displaystyle =\langle cu_{1},cu_{2},cu_{3}\rangle +\langle cv_{1},cv_{2},cv_{3}\rangle }$ ${\displaystyle =c\langle u_{1},u_{2},u_{3}\rangle +c\langle v_{1},v_{2},v_{3}\rangle }$
${\displaystyle =c\mathbf {u} +c\mathbf {v} }$.
${\displaystyle c(\mathbf {u} +\mathbf {v} )=c(\langle u_{1},u_{2},u_{3}\rangle +\langle v_{1},v_{2},v_{3}\rangle )}$ ${\displaystyle =c\langle u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}\rangle }$ ${\displaystyle =\langle c(u_{1}+v_{1}),c(u_{2}+v_{2}),c(u_{3}+v_{3})\rangle }$ ${\displaystyle =\langle cu_{1}+cv_{1},cu_{2}+cv_{2},cu_{3}+cv_{3}\rangle }$ ${\displaystyle =\langle cu_{1},cu_{2},cu_{3}\rangle +\langle cv_{1},cv_{2},cv_{3}\rangle }$ ${\displaystyle =c\langle u_{1},u_{2},u_{3}\rangle +c\langle v_{1},v_{2},v_{3}\rangle }$
${\displaystyle =c\mathbf {u} +c\mathbf {v} }$.
65. Find all unit vectors orthogonal to ${\displaystyle 3\mathbf {i} +4\mathbf {j} }$ in ${\displaystyle \mathbb {R} ^{2}}$
Rotating ${\displaystyle \mathbf {u} =\langle 3,4\rangle }$ ${\displaystyle 90^{\circ }}$ counterclockwise gives ${\displaystyle \mathbf {u} ^{\perp }=\langle -4,3\rangle }$. ${\displaystyle \mathbf {u} ^{\perp }}$ is orthogonal to ${\displaystyle \mathbf {u} }$, and the normalization of ${\displaystyle \mathbf {u} ^{\perp }}$ and its negative are the only unit vectors that are orthogonal to ${\displaystyle \mathbf {u} }$.
The magnitude of ${\displaystyle \mathbf {u} ^{\perp }}$ is ${\displaystyle {\sqrt {(-4)^{2}+3^{2}}}={\sqrt {16+9}}={\sqrt {25}}=5}$ so the only unit vectors that are orthogonal to ${\displaystyle \mathbf {u} }$ are ${\displaystyle \pm \left\langle {\frac {-4}{5}},{\frac {3}{5}}\right\rangle }$.
Rotating ${\displaystyle \mathbf {u} =\langle 3,4\rangle }$ ${\displaystyle 90^{\circ }}$ counterclockwise gives ${\displaystyle \mathbf {u} ^{\perp }=\langle -4,3\rangle }$. ${\displaystyle \mathbf {u} ^{\perp }}$ is orthogonal to ${\displaystyle \mathbf {u} }$, and the normalization of ${\displaystyle \mathbf {u} ^{\perp }}$ and its negative are the only unit vectors that are orthogonal to ${\displaystyle \mathbf {u} }$.
The magnitude of ${\displaystyle \mathbf {u} ^{\perp }}$ is ${\displaystyle {\sqrt {(-4)^{2}+3^{2}}}={\sqrt {16+9}}={\sqrt {25}}=5}$ so the only unit vectors that are orthogonal to ${\displaystyle \mathbf {u} }$ are ${\displaystyle \pm \left\langle {\frac {-4}{5}},{\frac {3}{5}}\right\rangle }$.
66. Find all unit vectors orthogonal to ${\displaystyle 3\mathbf {i} +4\mathbf {j} }$ in ${\displaystyle \mathbb {R} ^{3}}$
All vectors ${\displaystyle \mathbf {u} =\langle u_{1},u_{2},u_{3}\rangle }$ that are orthogonal to ${\displaystyle \langle 3,4,0\rangle }$ must satisfy ${\displaystyle \langle 3,4,0\rangle \cdot \mathbf {u} =0}$ ${\displaystyle \iff 3u_{1}+4u_{2}=0}$ ${\displaystyle \iff u_{2}=-{\frac {3}{4}}u_{1}}$.
The set of possible values of ${\displaystyle \mathbf {u} }$ is ${\displaystyle \left\{\left\langle x,-{\frac {3}{4}}x,z\right\rangle {\bigg |}x,z\in \mathbb {R} \right\}}$. The restriction that ${\displaystyle |\mathbf {u} |=1}$ becomes ${\displaystyle {\sqrt {u_{1}^{2}+u_{2}^{2}+u_{3}^{2}}}=1}$ ${\displaystyle \iff x^{2}+{\frac {9}{16}}x^{2}+z^{2}=1}$ ${\displaystyle \iff {\frac {25}{16}}x^{2}+z^{2}=1}$ ${\displaystyle \iff \left({\frac {x}{4/5}}\right)^{2}+z^{2}=1}$.
The set of possible ${\displaystyle x}$ and ${\displaystyle z}$ is an ellipse with radii ${\displaystyle 4/5}$ and ${\displaystyle 1}$. One possible parameterization of ${\displaystyle x}$ and ${\displaystyle z}$ is ${\displaystyle x={\frac {4}{5}}\cos(\theta )}$ and ${\displaystyle z=\sin(\theta )}$ where ${\displaystyle \theta \in \mathbb {R} }$. This parameterization yields ${\displaystyle \mathbf {u} =\left\langle {\frac {4}{5}}\cos(\theta ),-{\frac {3}{5}}\cos(\theta ),\sin(\theta )\right\rangle }$ where ${\displaystyle \theta \in \mathbb {R} }$ as the complete set of unit vectors that are orthogonal to ${\displaystyle \langle 3,4,0\rangle }$.
Re-parameterizing by letting ${\displaystyle c=\cos(\theta )}$ gives the set ${\displaystyle \left\langle {\frac {4}{5}}c,-{\frac {3}{5}}c,\pm {\sqrt {1-c^{2}}}\right\rangle ,\ c\in [-1,1]}$
All vectors ${\displaystyle \mathbf {u} =\langle u_{1},u_{2},u_{3}\rangle }$ that are orthogonal to ${\displaystyle \langle 3,4,0\rangle }$ must satisfy ${\displaystyle \langle 3,4,0\rangle \cdot \mathbf {u} =0}$ ${\displaystyle \iff 3u_{1}+4u_{2}=0}$ ${\displaystyle \iff u_{2}=-{\frac {3}{4}}u_{1}}$.
The set of possible values of ${\displaystyle \mathbf {u} }$ is ${\displaystyle \left\{\left\langle x,-{\frac {3}{4}}x,z\right\rangle {\bigg |}x,z\in \mathbb {R} \right\}}$. The restriction that ${\displaystyle |\mathbf {u} |=1}$ becomes ${\displaystyle {\sqrt {u_{1}^{2}+u_{2}^{2}+u_{3}^{2}}}=1}$ ${\displaystyle \iff x^{2}+{\frac {9}{16}}x^{2}+z^{2}=1}$ ${\displaystyle \iff {\frac {25}{16}}x^{2}+z^{2}=1}$ ${\displaystyle \iff \left({\frac {x}{4/5}}\right)^{2}+z^{2}=1}$.
The set of possible ${\displaystyle x}$ and ${\displaystyle z}$ is an ellipse with radii ${\displaystyle 4/5}$ and ${\displaystyle 1}$. One possible parameterization of ${\displaystyle x}$ and ${\displaystyle z}$ is ${\displaystyle x={\frac {4}{5}}\cos(\theta )}$ and ${\displaystyle z=\sin(\theta )}$ where ${\displaystyle \theta \in \mathbb {R} }$. This parameterization yields ${\displaystyle \mathbf {u} =\left\langle {\frac {4}{5}}\cos(\theta ),-{\frac {3}{5}}\cos(\theta ),\sin(\theta )\right\rangle }$ where ${\displaystyle \theta \in \mathbb {R} }$ as the complete set of unit vectors that are orthogonal to ${\displaystyle \langle 3,4,0\rangle }$.
Re-parameterizing by letting ${\displaystyle c=\cos(\theta )}$ gives the set ${\displaystyle \left\langle {\frac {4}{5}}c,-{\frac {3}{5}}c,\pm {\sqrt {1-c^{2}}}\right\rangle ,\ c\in [-1,1]}$
67. Find all unit vectors that make an angle of ${\displaystyle \pi /3}$ with the vector ${\displaystyle \langle 1,2\rangle }$
The angle that ${\displaystyle \langle 1,2\rangle }$ makes with the x-axis is ${\displaystyle \arctan(2)}$ counterclockwise.
Making a both a clockwise and a counterclockwise rotation of ${\displaystyle \pi /3}$ gives ${\displaystyle \langle \cos(\arctan(2)\mp \pi /3),\sin(\arctan(2)\mp \pi /3)\rangle }$ ${\displaystyle =\langle \cos(\arctan(2))\cos(\mp \pi /3)-\sin(\arctan(2))\sin(\mp \pi /3),\sin(\arctan(2))\cos(\mp \pi /3)+\cos(\arctan(2))\sin(\mp \pi /3)\rangle }$ ${\displaystyle =\left\langle {\frac {1}{\sqrt {1+2^{2}}}}\cdot {\frac {1}{2}}-{\frac {2}{\sqrt {1+2^{2}}}}\cdot {\frac {\mp {\sqrt {3}}}{2}},{\frac {2}{\sqrt {1+2^{2}}}}\cdot {\frac {1}{2}}+{\frac {1}{\sqrt {1+2^{2}}}}\cdot {\frac {\mp {\sqrt {3}}}{2}}\right\rangle }$ ${\displaystyle =\left\langle {\frac {1}{2{\sqrt {5}}}}\pm {\frac {\sqrt {3}}{\sqrt {5}}},{\frac {1}{\sqrt {5}}}\mp {\frac {\sqrt {3}}{2{\sqrt {5}}}}\right\rangle }$ ${\displaystyle ={\frac {1}{2{\sqrt {5}}}}\left\langle 1\pm 2{\sqrt {3}},2\mp {\sqrt {3}}\right\rangle }$
${\displaystyle ={\frac {\sqrt {5}}{10}}\left\langle 1\pm 2{\sqrt {3}},\ 2\mp {\sqrt {3}}\right\rangle }$
The angle that ${\displaystyle \langle 1,2\rangle }$ makes with the x-axis is ${\displaystyle \arctan(2)}$ counterclockwise.
Making a both a clockwise and a counterclockwise rotation of ${\displaystyle \pi /3}$ gives ${\displaystyle \langle \cos(\arctan(2)\mp \pi /3),\sin(\arctan(2)\mp \pi /3)\rangle }$ ${\displaystyle =\langle \cos(\arctan(2))\cos(\mp \pi /3)-\sin(\arctan(2))\sin(\mp \pi /3),\sin(\arctan(2))\cos(\mp \pi /3)+\cos(\arctan(2))\sin(\mp \pi /3)\rangle }$ ${\displaystyle =\left\langle {\frac {1}{\sqrt {1+2^{2}}}}\cdot {\frac {1}{2}}-{\frac {2}{\sqrt {1+2^{2}}}}\cdot {\frac {\mp {\sqrt {3}}}{2}},{\frac {2}{\sqrt {1+2^{2}}}}\cdot {\frac {1}{2}}+{\frac {1}{\sqrt {1+2^{2}}}}\cdot {\frac {\mp {\sqrt {3}}}{2}}\right\rangle }$ ${\displaystyle =\left\langle {\frac {1}{2{\sqrt {5}}}}\pm {\frac {\sqrt {3}}{\sqrt {5}}},{\frac {1}{\sqrt {5}}}\mp {\frac {\sqrt {3}}{2{\sqrt {5}}}}\right\rangle }$ ${\displaystyle ={\frac {1}{2{\sqrt {5}}}}\left\langle 1\pm 2{\sqrt {3}},2\mp {\sqrt {3}}\right\rangle }$
${\displaystyle ={\frac {\sqrt {5}}{10}}\left\langle 1\pm 2{\sqrt {3}},\ 2\mp {\sqrt {3}}\right\rangle }$
## Cross Product
Find ${\displaystyle \mathbf {u} \times \mathbf {v} }$ and ${\displaystyle \mathbf {v} \times \mathbf {u} }$
80. ${\displaystyle \mathbf {u} =\langle -4,1,1\rangle }$ and ${\displaystyle \mathbf {v} =\langle 0,1,-1\rangle }$
${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (1)(-1)-(1)(1),(1)(0)-(-4)(-1),(-4)(1)-(1)(0)\rangle =\langle -2,-4,-4\rangle }$ ${\displaystyle \mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle 2,4,4\rangle }$
${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (1)(-1)-(1)(1),(1)(0)-(-4)(-1),(-4)(1)-(1)(0)\rangle =\langle -2,-4,-4\rangle }$ ${\displaystyle \mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle 2,4,4\rangle }$
81. ${\displaystyle \mathbf {u} =\langle 1,2,-1\rangle }$ and ${\displaystyle \mathbf {v} =\langle 3,-4,6\rangle }$
${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (2)(6)-(-1)(-4),(-1)(3)-(1)(6),(1)(-4)-(2)(3)\rangle =\langle 8,-9,-10\rangle }$ ${\displaystyle \mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle -8,9,10\rangle }$
${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (2)(6)-(-1)(-4),(-1)(3)-(1)(6),(1)(-4)-(2)(3)\rangle =\langle 8,-9,-10\rangle }$ ${\displaystyle \mathbf {v} \times \mathbf {u} =-(\mathbf {u} \times \mathbf {v} )=\langle -8,9,10\rangle }$
Find the area of the parallelogram with sides ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$.
82. ${\displaystyle \mathbf {u} =\langle -3,0,2\rangle }$ and ${\displaystyle \mathbf {v} =\langle 1,1,1\rangle }$
The cross product of vectors ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$ is a vector with length ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ where ${\displaystyle \theta }$ is the angle between ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$. ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ is the area of the parallelogram with sides ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$, so this area is found by computing ${\displaystyle |\mathbf {u} \times \mathbf {v} |}$.
${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (0)(1)-(2)(1),(2)(1)-(-3)(1),(-3)(1)-(0)(1)\rangle =\langle -2,5,-3\rangle }$
${\displaystyle A=|\mathbf {u} \times \mathbf {v} |={\sqrt {(-2)^{2}+5^{2}+(-3)^{2}}}={\sqrt {4+25+9}}={\sqrt {38}}}$
The cross product of vectors ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$ is a vector with length ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ where ${\displaystyle \theta }$ is the angle between ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$. ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ is the area of the parallelogram with sides ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$, so this area is found by computing ${\displaystyle |\mathbf {u} \times \mathbf {v} |}$.
${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (0)(1)-(2)(1),(2)(1)-(-3)(1),(-3)(1)-(0)(1)\rangle =\langle -2,5,-3\rangle }$
${\displaystyle A=|\mathbf {u} \times \mathbf {v} |={\sqrt {(-2)^{2}+5^{2}+(-3)^{2}}}={\sqrt {4+25+9}}={\sqrt {38}}}$
83. ${\displaystyle \mathbf {u} =\langle 8,2,-3\rangle }$ and ${\displaystyle \mathbf {v} =\langle 2,4,-4\rangle }$
The cross product of vectors ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$ is a vector with length ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ where ${\displaystyle \theta }$ is the angle between ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$. ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ is the area of the parallelogram with sides ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$, so this area is found by computing ${\displaystyle |\mathbf {u} \times \mathbf {v} |}$.
${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (2)(-4)-(-3)(4),(-3)(2)-(8)(-4),(8)(4)-(2)(2)\rangle =\langle 4,26,28\rangle }$
${\displaystyle A=|\mathbf {u} \times \mathbf {v} |={\sqrt {4^{2}+26^{2}+28^{2}}}={\sqrt {2^{2}(2^{2}+13^{2}+14^{2})}}=2{\sqrt {369}}=6{\sqrt {41}}}$
The cross product of vectors ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$ is a vector with length ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ where ${\displaystyle \theta }$ is the angle between ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$. ${\displaystyle |\mathbf {u} ||\mathbf {v} |\sin(\theta )}$ is the area of the parallelogram with sides ${\displaystyle \mathbf {u} }$ and ${\displaystyle \mathbf {v} }$, so this area is found by computing ${\displaystyle |\mathbf {u} \times \mathbf {v} |}$.
${\displaystyle \mathbf {u} \times \mathbf {v} =\langle (2)(-4)-(-3)(4),(-3)(2)-(8)(-4),(8)(4)-(2)(2)\rangle =\langle 4,26,28\rangle }$
${\displaystyle A=|\mathbf {u} \times \mathbf {v} |={\sqrt {4^{2}+26^{2}+28^{2}}}={\sqrt {2^{2}(2^{2}+13^{2}+14^{2})}}=2{\sqrt {369}}=6{\sqrt {41}}}$
84. Find all vectors that satisfy the equation ${\displaystyle \langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle }$
The cross product is orthogonal to both multiplicand vectors. ${\displaystyle \langle 0,1,1\rangle }$ should be orthogonal to both ${\displaystyle \langle 1,1,1\rangle }$ and ${\displaystyle \mathbf {u} }$. However, ${\displaystyle \langle 1,1,1\rangle \cdot \langle 0,1,1\rangle =2\neq 0}$ so ${\displaystyle \langle 1,1,1\rangle }$ and ${\displaystyle \langle 0,1,1\rangle }$ are not orthogonal. The equation ${\displaystyle \langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle }$ is never true, and therefore the set of vectors ${\displaystyle \mathbf {u} }$ that satisfy the equation is ${\displaystyle \emptyset =}$"None".
The cross product is orthogonal to both multiplicand vectors. ${\displaystyle \langle 0,1,1\rangle }$ should be orthogonal to both ${\displaystyle \langle 1,1,1\rangle }$ and ${\displaystyle \mathbf {u} }$. However, ${\displaystyle \langle 1,1,1\rangle \cdot \langle 0,1,1\rangle =2\neq 0}$ so ${\displaystyle \langle 1,1,1\rangle }$ and ${\displaystyle \langle 0,1,1\rangle }$ are not orthogonal. The equation ${\displaystyle \langle 1,1,1\rangle \times \mathbf {u} =\langle 0,1,1\rangle }$ is never true, and therefore the set of vectors ${\displaystyle \mathbf {u} }$ that satisfy the equation is ${\displaystyle \emptyset =}$"None".
85. Find the volume of the parallelepiped with edges given by position vectors ${\displaystyle \langle 5,0,0\rangle }$, ${\displaystyle \langle 1,4,0\rangle }$, and ${\displaystyle \langle 2,2,7\rangle }$
The volume of a parallelepiped with edges defined by the vectors ${\displaystyle \mathbf {u} }$, ${\displaystyle \mathbf {v} }$, and ${\displaystyle \mathbf {w} }$ is the absolute value of the scalar triple product: ${\displaystyle |(\mathbf {u} \times \mathbf {v} )\cdot \mathbf {w} |}$.
${\displaystyle |(\langle 5,0,0\rangle \times \langle 1,4,0\rangle )\cdot \langle 2,2,7\rangle |}$ ${\displaystyle =|\langle (0)(0)-(0)(4),(0)(1)-(5)(0),(5)(4)-(0)(1)\rangle \cdot \langle 2,2,7\rangle |}$ ${\displaystyle =|\langle 0,0,20\rangle \cdot \langle 2,2,7\rangle |}$ ${\displaystyle =140}$
The volume of a parallelepiped with edges defined by the vectors ${\displaystyle \mathbf {u} }$, ${\displaystyle \mathbf {v} }$, and ${\displaystyle \mathbf {w} }$ is the absolute value of the scalar triple product: ${\displaystyle |(\mathbf {u} \times \mathbf {v} )\cdot \mathbf {w} |}$.
${\displaystyle |(\langle 5,0,0\rangle \times \langle 1,4,0\rangle )\cdot \langle 2,2,7\rangle |}$ ${\displaystyle =|\langle (0)(0)-(0)(4),(0)(1)-(5)(0),(5)(4)-(0)(1)\rangle \cdot \langle 2,2,7\rangle |}$ ${\displaystyle =|\langle 0,0,20\rangle \cdot \langle 2,2,7\rangle |}$ ${\displaystyle =140}$
86. A wrench has a pivot at the origin and extends along the x-axis. Find the magnitude and the direction of the torque at the pivot when the force ${\displaystyle \mathbf {F} =\langle 1,2,3\rangle }$ is applied to the wrench n units away from the origin.
The moment arm is ${\displaystyle \mathbf {r} =\langle n,0,0\rangle }$, so the torque applied is ${\displaystyle \tau =\mathbf {r} \times \mathbf {F} =\langle n,0,0\rangle \times \langle 1,2,3\rangle =\langle (0)(3)-(0)(2),(0)(1)-(n)(3),(n)(2)-(0)(1)\rangle =\langle 0,-3n,2n\rangle }$ The magnitude of the torque is ${\displaystyle |\tau |={\sqrt {0^{2}+(-3n)^{2}+(2n)^{2}}}=n{\sqrt {13}}}$. The torque's direction is ${\displaystyle {\hat {\tau }}={\frac {\tau }{|\tau |}}=\langle 0,-3/{\sqrt {13}},2/{\sqrt {13}}\rangle }$.
The moment arm is ${\displaystyle \mathbf {r} =\langle n,0,0\rangle }$, so the torque applied is ${\displaystyle \tau =\mathbf {r} \times \mathbf {F} =\langle n,0,0\rangle \times \langle 1,2,3\rangle =\langle (0)(3)-(0)(2),(0)(1)-(n)(3),(n)(2)-(0)(1)\rangle =\langle 0,-3n,2n\rangle }$ The magnitude of the torque is ${\displaystyle |\tau |={\sqrt {0^{2}+(-3n)^{2}+(2n)^{2}}}=n{\sqrt {13}}}$. The torque's direction is ${\displaystyle {\hat {\tau }}={\frac {\tau }{|\tau |}}=\langle 0,-3/{\sqrt {13}},2/{\sqrt {13}}\rangle }$.
Prove the following identities or show them false by giving a counterexample.
87. ${\displaystyle \mathbf {u} \times (\mathbf {u} \times \mathbf {v} )=\mathbf {0} }$
False: ${\displaystyle \mathbf {i} \times (\mathbf {i} \times \mathbf {j} )=-\mathbf {j} }$
False: ${\displaystyle \mathbf {i} \times (\mathbf {i} \times \mathbf {j} )=-\mathbf {j} }$
88. ${\displaystyle \mathbf {u} \cdot (\mathbf {v} \times \mathbf {w} )=\mathbf {w} \cdot (\mathbf {u} \times \mathbf {v} )}$
True: Once expressed in component form, both sides evaluate to ${\displaystyle u_{1}v_{2}w_{3}-u_{1}v_{3}w_{2}+u_{2}v_{3}w_{1}-u_{2}v_{1}w_{3}+u_{3}v_{1}w_{2}-u_{3}v_{2}w_{1}}$
True: Once expressed in component form, both sides evaluate to ${\displaystyle u_{1}v_{2}w_{3}-u_{1}v_{3}w_{2}+u_{2}v_{3}w_{1}-u_{2}v_{1}w_{3}+u_{3}v_{1}w_{2}-u_{3}v_{2}w_{1}}$
89. ${\displaystyle (\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=2(\mathbf {u} \times \mathbf {v} )}$
True: ${\displaystyle (\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=(\mathbf {u} -\mathbf {v} )\times \mathbf {u} +(\mathbf {u} -\mathbf {v} )\times \mathbf {v} }$${\displaystyle =\mathbf {u} \times \mathbf {u} -\mathbf {v} \times \mathbf {u} +\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {v} }$${\displaystyle =\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {u} }$${\displaystyle =\mathbf {u} \times \mathbf {v} +\mathbf {u} \times \mathbf {v} }$${\displaystyle =2(\mathbf {u} \times \mathbf {v} )}$
True: ${\displaystyle (\mathbf {u} -\mathbf {v} )\times (\mathbf {u} +\mathbf {v} )=(\mathbf {u} -\mathbf {v} )\times \mathbf {u} +(\mathbf {u} -\mathbf {v} )\times \mathbf {v} }$${\displaystyle =\mathbf {u} \times \mathbf {u} -\mathbf {v} \times \mathbf {u} +\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {v} }$${\displaystyle =\mathbf {u} \times \mathbf {v} -\mathbf {v} \times \mathbf {u} }$${\displaystyle =\mathbf {u} \times \mathbf {v} +\mathbf {u} \times \mathbf {v} }$${\displaystyle =2(\mathbf {u} \times \mathbf {v} )}$
## Calculus of Vector-Valued Functions
100. Differentiate ${\displaystyle \mathbf {r} (t)=\langle te^{-t},t\ln t,t\cos(t)\rangle }$.
${\displaystyle \langle e^{-t}-te^{-t},\ln(t)+1,\cos(t)-t\sin(t)\rangle }$
${\displaystyle \langle e^{-t}-te^{-t},\ln(t)+1,\cos(t)-t\sin(t)\rangle }$
101. Find a tangent vector for the curve ${\displaystyle \mathbf {r} (t)=\langle 2t^{4},6t^{3/2},10/t\rangle }$ at the point ${\displaystyle t=1}$.
${\displaystyle {\dot {\mathbf {r} }}(t)=\langle 8t^{3},9t^{1/2},-10/t^{2}\rangle }$ so a possible a tangent vector at ${\displaystyle t=1}$ is ${\displaystyle {\dot {\mathbf {r} }}(1)=\langle 8,9,-10\rangle }$
${\displaystyle {\dot {\mathbf {r} }}(t)=\langle 8t^{3},9t^{1/2},-10/t^{2}\rangle }$ so a possible a tangent vector at ${\displaystyle t=1}$ is ${\displaystyle {\dot {\mathbf {r} }}(1)=\langle 8,9,-10\rangle }$
102. Find the unit tangent vector for the curve ${\displaystyle \mathbf {r} (t)=\langle t,2,2/t\rangle ,\ t\neq 0}$.
${\displaystyle {\dot {\mathbf {r} }}(t)=\langle 1,0,-2/t^{2}\rangle }$ ${\displaystyle |{\dot {\mathbf {r} }}(t)|={\sqrt {1^{2}+0^{2}+(-2/t^{2})^{2}}}={\sqrt {1+4/t^{4}}}={\frac {\sqrt {t^{4}+4}}{t^{2}}}}$ so the unit tangent vector is ${\displaystyle \mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle t^{2},0,-2\rangle }{\sqrt {t^{4}+4}}}}$
${\displaystyle {\dot {\mathbf {r} }}(t)=\langle 1,0,-2/t^{2}\rangle }$ ${\displaystyle |{\dot {\mathbf {r} }}(t)|={\sqrt {1^{2}+0^{2}+(-2/t^{2})^{2}}}={\sqrt {1+4/t^{4}}}={\frac {\sqrt {t^{4}+4}}{t^{2}}}}$ so the unit tangent vector is ${\displaystyle \mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle t^{2},0,-2\rangle }{\sqrt {t^{4}+4}}}}$
103. Find the unit tangent vector for the curve ${\displaystyle \mathbf {r} (t)=\langle \sin(t),\cos(t),e^{-t}\rangle ,\ t\in [0,\pi ]}$ at the point ${\displaystyle t=0}$.
${\displaystyle {\dot {\mathbf {r} }}(t)=\langle \cos(t),-\sin(t),-e^{-t}\rangle }$
${\displaystyle |{\dot {\mathbf {r} }}(t)|={\sqrt {(\cos(t))^{2}+(-\sin(t))^{2}+(-e^{-t})^{2}}}={\sqrt {1+e^{-2t}}}}$ so the unit tangent vector is ${\displaystyle \mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle \cos(t),-\sin(t),-e^{-t}\rangle }{\sqrt {1+e^{-2t}}}}}$
At ${\displaystyle t=0}$: ${\displaystyle \mathbf {T} (0)={\frac {\langle 1,0,-1\rangle }{\sqrt {2}}}}$
${\displaystyle {\dot {\mathbf {r} }}(t)=\langle \cos(t),-\sin(t),-e^{-t}\rangle }$
${\displaystyle |{\dot {\mathbf {r} }}(t)|={\sqrt {(\cos(t))^{2}+(-\sin(t))^{2}+(-e^{-t})^{2}}}={\sqrt {1+e^{-2t}}}}$ so the unit tangent vector is ${\displaystyle \mathbf {T} (t)={\frac {{\dot {\mathbf {r} }}(t)}{|{\dot {\mathbf {r} }}(t)|}}={\frac {\langle \cos(t),-\sin(t),-e^{-t}\rangle }{\sqrt {1+e^{-2t}}}}}$
At ${\displaystyle t=0}$: ${\displaystyle \mathbf {T} (0)={\frac {\langle 1,0,-1\rangle }{\sqrt {2}}}}$
104. Find ${\displaystyle \mathbf {r} }$ if ${\displaystyle \mathbf {r} '(t)=\langle {\sqrt {t}},\cos(\pi t),4/t\rangle }$ and ${\displaystyle \mathbf {r} (1)=\langle 2,3,4\rangle }$.
For an arbitrary ${\displaystyle t>0}$ the position ${\displaystyle \mathbf {r} (t)}$ can be computed by the integral ${\displaystyle \mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du}$.
${\displaystyle \mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du}$ ${\displaystyle =\langle 2,3,4\rangle +\int _{u=1}^{t}\langle {\sqrt {u}},\cos(\pi u),4/u\rangle du}$ ${\displaystyle =\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}u^{3/2},{\frac {\sin(\pi u)}{\pi }},4\ln(u)\right\rangle {\Bigg |}_{u=1}^{t}}$ ${\displaystyle =\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}(t^{3/2}-1),{\frac {\sin(\pi t)}{\pi }},4\ln(t)\right\rangle }$
${\displaystyle =\left\langle {\frac {2t^{3/2}+4}{3}},{\frac {\sin(\pi t)}{\pi }}+3,4\ln(t)+4\right\rangle }$
For an arbitrary ${\displaystyle t>0}$ the position ${\displaystyle \mathbf {r} (t)}$ can be computed by the integral ${\displaystyle \mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du}$.
${\displaystyle \mathbf {r} (t)=\mathbf {r} (1)+\int _{u=1}^{t}\mathbf {r} '(u)du}$ ${\displaystyle =\langle 2,3,4\rangle +\int _{u=1}^{t}\langle {\sqrt {u}},\cos(\pi u),4/u\rangle du}$ ${\displaystyle =\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}u^{3/2},{\frac {\sin(\pi u)}{\pi }},4\ln(u)\right\rangle {\Bigg |}_{u=1}^{t}}$ ${\displaystyle =\langle 2,3,4\rangle +\left\langle {\frac {2}{3}}(t^{3/2}-1),{\frac {\sin(\pi t)}{\pi }},4\ln(t)\right\rangle }$
${\displaystyle =\left\langle {\frac {2t^{3/2}+4}{3}},{\frac {\sin(\pi t)}{\pi }}+3,4\ln(t)+4\right\rangle }$
105. Evaluate ${\displaystyle \displaystyle \int _{0}^{\ln 2}(e^{-t}\mathbf {i} +2e^{2t}\mathbf {j} -4e^{t}\mathbf {k} )dt}$
${\displaystyle \int _{t=0}^{\ln 2}\langle e^{-t},2e^{2t},-4e^{t}\rangle dt}$
${\displaystyle =\langle -e^{-t},e^{2t},-4e^{t}\rangle {\bigg |}_{t=0}^{\ln 2}}$ ${\displaystyle =\langle -1/2-(-1),4-1,-8-(-4)\rangle }$
${\displaystyle =\langle 1/2,3,-4\rangle }$
${\displaystyle \int _{t=0}^{\ln 2}\langle e^{-t},2e^{2t},-4e^{t}\rangle dt}$
${\displaystyle =\langle -e^{-t},e^{2t},-4e^{t}\rangle {\bigg |}_{t=0}^{\ln 2}}$ ${\displaystyle =\langle -1/2-(-1),4-1,-8-(-4)\rangle }$
${\displaystyle =\langle 1/2,3,-4\rangle }$
## Motion in Space
120. Find velocity, speed, and acceleration of an object if the position is given by ${\displaystyle \mathbf {r} (t)=\langle 3\sin(t),5\cos(t),4\sin(t)\rangle }$.
${\displaystyle \mathbf {v} =\langle 3\cos(t),-5\sin(t),4\cos(t)\rangle }$, ${\displaystyle |\mathbf {v} |=5}$, ${\displaystyle \mathbf {a} =\langle -3\sin(t),-5\cos(t),-4\sin(t)\rangle }$
${\displaystyle \mathbf {v} =\langle 3\cos(t),-5\sin(t),4\cos(t)\rangle }$, ${\displaystyle |\mathbf {v} |=5}$, ${\displaystyle \mathbf {a} =\langle -3\sin(t),-5\cos(t),-4\sin(t)\rangle }$
121. Find the velocity and the position vectors for ${\displaystyle t\geq 0}$ if the acceleration is given by ${\displaystyle \mathbf {a} (t)=\langle e^{-t},1\rangle ,\ \mathbf {v} (0)=\langle 1,0\rangle ,\ \mathbf {r} (0)=\langle 0,0\rangle }$.
${\displaystyle \mathbf {v} (t)=\mathbf {v} (0)+\int _{u=0}^{t}\mathbf {a} (u)du}$
${\displaystyle =\langle 1,0\rangle +\int _{u=0}^{t}\langle e^{-u},1\rangle du}$ ${\displaystyle =\langle 1,0\rangle +\langle -e^{-u},u\rangle {\bigg |}_{u=0}^{t}}$ ${\displaystyle =\langle 1,0\rangle +\langle -e^{-t}-(-1),t-0\rangle }$ ${\displaystyle =\langle 2-e^{-t},t\rangle }$
${\displaystyle \mathbf {r} (t)=\mathbf {r} (0)+\int _{u=0}^{t}\mathbf {v} (u)du}$ ${\displaystyle =\langle 0,0\rangle +\int _{u=0}^{t}\langle 2-e^{-u},u\rangle du}$ ${\displaystyle =\langle 2u+e^{-u},{\frac {1}{2}}u^{2}\rangle {\bigg |}_{u=0}^{t}}$ ${\displaystyle =\langle (2t+e^{-t})-1,{\frac {1}{2}}t^{2}-0\rangle }$
${\displaystyle =\langle e^{-t}+2t-1,t^{2}/2\rangle }$
${\displaystyle \mathbf {v} (t)=\mathbf {v} (0)+\int _{u=0}^{t}\mathbf {a} (u)du}$
${\displaystyle =\langle 1,0\rangle +\int _{u=0}^{t}\langle e^{-u},1\rangle du}$ ${\displaystyle =\langle 1,0\rangle +\langle -e^{-u},u\rangle {\bigg |}_{u=0}^{t}}$ ${\displaystyle =\langle 1,0\rangle +\langle -e^{-t}-(-1),t-0\rangle }$ ${\displaystyle =\langle 2-e^{-t},t\rangle }$
${\displaystyle \mathbf {r} (t)=\mathbf {r} (0)+\int _{u=0}^{t}\mathbf {v} (u)du}$ ${\displaystyle =\langle 0,0\rangle +\int _{u=0}^{t}\langle 2-e^{-u},u\rangle du}$ ${\displaystyle =\langle 2u+e^{-u},{\frac {1}{2}}u^{2}\rangle {\bigg |}_{u=0}^{t}}$ ${\displaystyle =\langle (2t+e^{-t})-1,{\frac {1}{2}}t^{2}-0\rangle }$
${\displaystyle =\langle e^{-t}+2t-1,t^{2}/2\rangle }$
## Length of Curves
Find the length of the following curves.
140. ${\displaystyle \mathbf {r} (t)=\langle 4\cos(3t),4\sin(3t)\rangle ,\ t\in [0,2\pi /3].}$
For an infinitesimal step ${\displaystyle dt}$, the length traversed is approximately
${\displaystyle ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle -12\sin(3t),12\cos(3t)\rangle |dt=12dt}$.
The total length is therefore:
${\displaystyle L=\int _{t=0}^{2\pi /3}ds=\int _{t=0}^{2\pi /3}12dt=12t{\Bigg |}_{t=0}^{2\pi /3}=8\pi }$
For an infinitesimal step ${\displaystyle dt}$, the length traversed is approximately
${\displaystyle ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle -12\sin(3t),12\cos(3t)\rangle |dt=12dt}$.
The total length is therefore:
${\displaystyle L=\int _{t=0}^{2\pi /3}ds=\int _{t=0}^{2\pi /3}12dt=12t{\Bigg |}_{t=0}^{2\pi /3}=8\pi }$
141. ${\displaystyle \mathbf {r} (t)=\langle 2+3t,1-4t,3t-4\rangle ,\ t\in [1,6].}$
For an infinitesimal step ${\displaystyle dt}$, the length traversed is approximately
${\displaystyle ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 3,-4,3\rangle |dt={\sqrt {9+16+9}}\cdot dt={\sqrt {34}}\cdot dt}$.
The total length is therefore:
${\displaystyle L=\int _{t=1}^{6}ds=\int _{t=1}^{6}{\sqrt {34}}\cdot dt={\sqrt {34}}\cdot t{\Bigg |}_{t=1}^{6}=5{\sqrt {34}}}$
For an infinitesimal step ${\displaystyle dt}$, the length traversed is approximately
${\displaystyle ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 3,-4,3\rangle |dt={\sqrt {9+16+9}}\cdot dt={\sqrt {34}}\cdot dt}$.
The total length is therefore:
${\displaystyle L=\int _{t=1}^{6}ds=\int _{t=1}^{6}{\sqrt {34}}\cdot dt={\sqrt {34}}\cdot t{\Bigg |}_{t=1}^{6}=5{\sqrt {34}}}$
## Parametrization and Normal Vectors
142. Find a description of the curve that uses arc length as a parameter: ${\displaystyle \mathbf {r} (t)=\langle t^{2},2t^{2},4t^{2}\rangle \ t\in [1,4].}$
For an infinitesimal step ${\displaystyle dt}$, the length traversed is approximately ${\displaystyle ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 2t,4t,8t\rangle |dt=|t|{\sqrt {4+16+64}}\cdot dt={\sqrt {84}}\cdot tdt=2{\sqrt {21}}\cdot tdt}$
Given an upper bound of ${\displaystyle u\in [1,4]}$, the arc length swept out from ${\displaystyle t=1}$ to ${\displaystyle t=u}$ is: ${\displaystyle s=\int _{t=1}^{u}ds=\int _{t=1}^{u}2{\sqrt {21}}\cdot tdt={\sqrt {21}}\cdot t^{2}{\bigg |}_{t=1}^{u}={\sqrt {21}}(u^{2}-1)}$
The arc length spans a range from ${\displaystyle 0}$ to ${\displaystyle 15{\sqrt {21}}}$. For an arc length of ${\displaystyle s\in [0,15{\sqrt {21}}]}$, the upper bound on ${\displaystyle t}$ that generates an arc length of ${\displaystyle s}$ is ${\displaystyle u={\sqrt {{\frac {s}{\sqrt {21}}}+1}}}$, and the point at which this upper bound occurs is: ${\displaystyle \displaystyle \mathbf {r} (s)=\left({\frac {s}{\sqrt {21}}}+1\right)\langle 1,2,4\rangle }$
For an infinitesimal step ${\displaystyle dt}$, the length traversed is approximately ${\displaystyle ds=\left|{\frac {d\mathbf {r} }{dt}}\right|dt=|\langle 2t,4t,8t\rangle |dt=|t|{\sqrt {4+16+64}}\cdot dt={\sqrt {84}}\cdot tdt=2{\sqrt {21}}\cdot tdt}$
Given an upper bound of ${\displaystyle u\in [1,4]}$, the arc length swept out from ${\displaystyle t=1}$ to ${\displaystyle t=u}$ is: ${\displaystyle s=\int _{t=1}^{u}ds=\int _{t=1}^{u}2{\sqrt {21}}\cdot tdt={\sqrt {21}}\cdot t^{2}{\bigg |}_{t=1}^{u}={\sqrt {21}}(u^{2}-1)}$
The arc length spans a range from ${\displaystyle 0}$ to ${\displaystyle 15{\sqrt {21}}}$. For an arc length of ${\displaystyle s\in [0,15{\sqrt {21}}]}$, the upper bound on ${\displaystyle t}$ that generates an arc length of ${\displaystyle s}$ is ${\displaystyle u={\sqrt {{\frac {s}{\sqrt {21}}}+1}}}$, and the point at which this upper bound occurs is: ${\displaystyle \displaystyle \mathbf {r} (s)=\left({\frac {s}{\sqrt {21}}}+1\right)\langle 1,2,4\rangle }$
143. Find the unit tangent vector T and the principal unit normal vector N for the curve ${\displaystyle \mathbf {r} (t)=\langle t^{2},t\rangle .}$ Check that TN=0.
A tangent vector is ${\displaystyle {\frac {d\mathbf {r} }{dt}}=\langle 2t,1\rangle }$. Normalizing this vector to get the unit tangent vector gives:
${\displaystyle \mathbf {T} (t)={\frac {d\mathbf {r} }{ds}}={\frac {d\mathbf {r} /dt}{|d\mathbf {r} /dt|}}=\displaystyle {\frac {\langle 2t,1\rangle }{\sqrt {4t^{2}+1}}}}$
A vector that has the direction of the principal unit normal vector is ${\displaystyle {\frac {d\mathbf {T} }{dt}}=\left\langle {\frac {2}{\sqrt {4t^{2}+1}}}-{\frac {8t^{2}}{(4t^{2}+1)^{3/2}}},{\frac {-4t}{(4t^{2}+1)^{3/2}}}\right\rangle }$ ${\displaystyle ={\frac {1}{(4t^{2}+1)^{3/2}}}\langle 2(4t^{2}+1)-8t^{2},-4t\rangle }$ ${\displaystyle ={\frac {1}{(4t^{2}+1)^{3/2}}}\langle 2,-4t\rangle }$ ${\displaystyle ={\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle }$
Normalizing ${\displaystyle {\frac {d\mathbf {T} }{dt}}}$ gives the principal unit normal vector: ${\displaystyle \mathbf {N} (t)}$ ${\displaystyle ={\frac {d\mathbf {T} /dt}{|d\mathbf {T} /dt|}}}$ ${\displaystyle ={\frac {{\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle }{{\frac {2}{(4t^{2}+1)^{3/2}}}{\sqrt {4t^{2}+1}}}}}$
${\displaystyle =\displaystyle {\frac {\langle 1,-2t\rangle }{\sqrt {4t^{2}+1}}}}$
A tangent vector is ${\displaystyle {\frac {d\mathbf {r} }{dt}}=\langle 2t,1\rangle }$. Normalizing this vector to get the unit tangent vector gives:
${\displaystyle \mathbf {T} (t)={\frac {d\mathbf {r} }{ds}}={\frac {d\mathbf {r} /dt}{|d\mathbf {r} /dt|}}=\displaystyle {\frac {\langle 2t,1\rangle }{\sqrt {4t^{2}+1}}}}$
A vector that has the direction of the principal unit normal vector is ${\displaystyle {\frac {d\mathbf {T} }{dt}}=\left\langle {\frac {2}{\sqrt {4t^{2}+1}}}-{\frac {8t^{2}}{(4t^{2}+1)^{3/2}}},{\frac {-4t}{(4t^{2}+1)^{3/2}}}\right\rangle }$ ${\displaystyle ={\frac {1}{(4t^{2}+1)^{3/2}}}\langle 2(4t^{2}+1)-8t^{2},-4t\rangle }$ ${\displaystyle ={\frac {1}{(4t^{2}+1)^{3/2}}}\langle 2,-4t\rangle }$ ${\displaystyle ={\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle }$
Normalizing ${\displaystyle {\frac {d\mathbf {T} }{dt}}}$ gives the principal unit normal vector: ${\displaystyle \mathbf {N} (t)}$ ${\displaystyle ={\frac {d\mathbf {T} /dt}{|d\mathbf {T} /dt|}}}$ ${\displaystyle ={\frac {{\frac {2}{(4t^{2}+1)^{3/2}}}\langle 1,-2t\rangle }{{\frac {2}{(4t^{2}+1)^{3/2}}}{\sqrt {4t^{2}+1}}}}}$
${\displaystyle =\displaystyle {\frac {\langle 1,-2t\rangle }{\sqrt {4t^{2}+1}}}}$
## Equations of Lines And Planes
160. Find an equation of a plane passing through points ${\displaystyle (1,1,2),\ (1,2,2),\ (-1,0,1).}$
Let ${\displaystyle P}$ denote a plane that contains points ${\displaystyle (1,1,2)}$, ${\displaystyle (1,2,2)}$, and ${\displaystyle (-1,0,1)}$. Let ${\displaystyle \mathbf {N} }$ denote an arbitrary vector that is orthogonal to ${\displaystyle P}$, and ${\displaystyle \mathbf {r} _{0}}$ denote the position vector of an arbitrary point contained by ${\displaystyle P}$. A point at position vector ${\displaystyle \mathbf {r} }$ is contained by ${\displaystyle P}$ if and only if the displacement from ${\displaystyle \mathbf {r} _{0}}$ |
## Want to keep learning?
This content is taken from the University of Padova's online course, Precalculus: the Mathematics of Numbers, Functions and Equations. Join the course to learn more.
4.2
## Precalculus
Skip to 0 minutes and 11 seconds Hello. Welcome to Week 4. Let us consider the introduction, types of equations step. And let us start with the first exercise. It’s a problem about cherries. Precisely, the question is, how many cherries were on the table if, after we ate 4/5 of those, we remain with three times the initial amount minus 42? This problem can be easily solved with an equation. Let us see. Call x the initial number of cherries on the table.
Skip to 1 minute and 12 seconds Then the exercise gives us some information. What do we know? That after we ate 4/5 of those, we remain with three times the initial amount minus 42. Ok then, this is the initial amount. After we have eaten the 4/5 of those, we remain with the initial amount minus the 4/5 of the initial amount. And we know, from the text of the exercise, that this is equal to three times the initial amount minus 42.
Skip to 2 minutes and 2 seconds This is just a translation of the problem in an equation. And what do you get? This is a linear equation. Very easy to solve, but let us see. We get x minus 4/5 x minus 3 times x equal to minus 42. And then let us collect together the coefficients of x. And we have 5 over 5 which is 1 minus 4 over 5 minus 3 times 5 over 5, x equal to minus 42. And this is exactly, we have 5 as common denominator. And you have 5 minus 4, which is 1, minus 15. 1 minus 15 is minus 14.
Skip to 3 minutes and 18 seconds Then we have that this fraction has to be equal to minus – times x has to be equal to minus 42.
Skip to 3 minutes and 30 seconds Then, now we can multiply on both sides by minus 1. And we get 14 over 5x equal to 42. We can divide on both sides by 14. Indeed, divide by 14 is exactly like to divide by 7, and then divide by 2. If we divide by 7 42, we get 6. And then we divide by 2, we get 3.
Skip to 4 minutes and 14 seconds Therefore, we remain with the equation x equal, you multiply it by 5 on both sides, x equal to 15. Therefore, the initial amount of cherries was 15. Thank you.
# Types of equations in practice - Part 1
The following exercises are solved in this step.
We invite you to try to solve them before watching the video.
In any case, you will find below a PDF file with the solutions.
### Exercise 1.
How many cherries were on the table if, after we ate $$4/5$$ of them, we are left with 3 times the initial amount, minus 42?
### Exercise 2. [Solved only in the PDF file]
The sum of two consecutive integer numbers is equal to 93. What are the two numbers?
### Exercise 3. [Solved only in the PDF file]
The product of two consecutive even integer numbers is $$48$$. What are the two numbers? |
Too Good at Spider Solitaire
Have you ever been punished for being too good at spider solitaire? I mean, have you ever been stuck because you collected too many suits? Many versions of the game don’t allow you to deal from the deck if you have empty columns, nor do they allow you to get back a completed suit. If the number of cards left on the table in the middle of the game is less than ten — the number of columns — you are stuck. I always wondered what the probability is of being stuck. This probability is difficult to calculate because it depends on your strategy. So I invented a boring version of spider solitaire for the sake of creating a math problem. Here it goes:
You start with two full decks of 104 cards. Initially you take 54 cards. At each turn you take all full suits out of your hand. If you have less than ten cards left in your hand, you are stuck. If not, take ten more cards from the leftover deck and continue. What is the probability that you can be stuck during this game?
Let us simplify the game even more by playing the easy level of the boring spider solitaire in which you have only spades. So you have a total of eight full suits of spades. I leave it to my readers to calculate the total probability of being stuck. Here I would like to estimate the easiest case: the probability of being stuck before the last deal.
There are ten cards left in the deck. For you to be stuck, they all should have a different value. The total number of ways to choose ten cards is 104 choose 10. To calculate the number of ways in which these ten cards have different values we need to choose these ten values in 13 choose 10 ways, then multiply by the number of ways each card of a given value can be taken from the deck: 810. The probability is about 0.0117655.
I will leave it to my readers to calculate the probability of being stuck before the last deal at the medium level: when you play two suits, hearts and spades.
No, I will not tell you how many times I played spider solitaire.
Share: |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.