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# Important Mensuration (2D) Formulas
By Parnab Mallick|Updated : January 27th, 2021
It is very important to have an understanding of different formulas of quadrilaterals and circle to comfortably attempt Maths questions which covers a good portion of the Quant Section of Competitive Exams. Here we are providing you formulas and shortcuts on how to solve mensuration questions.
It is very important to have an understanding of different formulas of quadrilaterals and circle to comfortably attempt Maths questions which covers a good portion of the Quant Section of Competitive Exams. Here we are providing you formulas and shortcuts on how to solve mensuration questions.
## Important Formulas on Quadrilateral and Circle
Rectangle
A four-sided shape that is made up of two pairs of parallel lines and that has four right angles; especially: a shape in which one pair of lines is longer than the other pair.
The diagonals of a rectangle bisect each other and are equal.
Area of rectangle = length x breadth = l x b
OR Area of rectangle = if one sides (l) and diagonal (d) are given.
OR Area of rectangle = if perimeter (P) and diagonal (d) are given.
Perimeter (P) of rectangle = 2 (length + breadth) = 2 (l + b).
OR Perimeter of rectangle = if one side (l) and diagonal (d) are given.
Square
A four-sided shape that is made up of four straight sides that are the same length and that has four right angles.
The diagonals of a square are equal and bisect each other at 900.
(a) Area (a) of a square
Perimeter (P) of a square
= 4a, i.e. 4 x side
p= 2√2d
Length (d) of the diagonal of a square
Circle
A circle is a path travelled by a point which moves in such a way that its distance from a fixed point remains constant.
The fixed point is known as center and the fixed distance is called the radius.
(a) Circumference or perimeter of circle =
where r is radius and d is diameter of circle
(b) Area of circle
is circumference
Sector :
A sector is a figure enclosed by two radii and an arc lying between them.
here AOB is a sector
length of arc AB= 2πrΘ/360°
Ring or Circular Path:
area=π(R2-r2)
Perimeter=2π(R+r)
Rhombus
Rhombus is a quadrilateral whose all sides are equal.
The diagonals of a rhombus bisect each other at 900
Area (a) of a rhombus
= a * h, i.e. base * height
Product of its diagonals
since d2
since d2
Perimeter (P) of a rhombus
= 4a, i.e. 4 x side
Where d1 and d2 are two-diagonals.
Side (a) of a rhombus
Parallelogram
A quadrilateral in which opposite sides are equal and parallel is called a parallelogram. The diagonals of a parallelogram bisect each other.
Area (a) of a parallelogram = base × altitude corresponding to the base = b × h
Area (a) of a parallelogram
where a and b are adjacent sides, d is the length of the diagonal connecting the ends of the two sides and
In a parallelogram, the sum of the squares of the diagonals = 2
(the sum of the squares of the two adjacent sides).
i.e.,
Perimeter (P) of a parallelogram
= 2 (a+b),
Where a and b are adjacent sides of the parallelogram.
Trapezium (Trapezoid)
A trapezoid is a 2-dimensional geometric figure with four sides, at least one set of which are parallel. The parallel sides are called the bases, while the other sides are called the legs. The term ‘trapezium,’ from which we got our word trapezoid has been in use in the English language since the 1500s and is from the Latin meaning ‘little table.’
Area (a) of a trapezium
1/2 x (sum of parallel sides) x perpendicular
Distance between the parallel sides
i.e.,
Where, l = b – a if b > a = a – b if a > b
And
Height (h) of the trapezium
Pathways Running across the middle of a rectangle:
X is the width of the path
Area of path= (l+b-x)x
perimeter= 2(l+b-2x)
Outer Pathways:
Area=(l+b+2x)2x
Perimeter=4(l+b+2x)
Inner Pathways:
Area=(l+b-2x)2x
Perimeter=4(l+b-2x)
Some useful Short trick:
• If there is a change of X% in defining dimensions of the 2-d figure then its perimeter will also change by X%
• If all the sides of a quadrilateral are changed by X% then its diagonal will also change by X%.
• The area of the largest triangle that can be inscribed in a semicircle of radius r is r2.
• The number of revolution made by a circular wheel of radius r in travelling distance d is given by
number of revolution =d/2πr
• If the length and breadth of the rectangle are increased by x% and y% then the area of the rectangle will be increased by.
(x+y+xy/100)%
• If the length and breadth of a rectangle are decreased by x% and y% respectively then the area of the rectangle will decrease by:
(x+y-xy/100)%
• If the length of a rectangle is increased by x%, then its breadth will have to be decreased by (100x/100+x)% in order to maintain the same area of the rectangle.
• If each of the defining dimensions or sides of any 2-D figure is changed by x% its area changes by
x(2+x/100)%
where x=positive if increase and negative if decreases.
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# Experimental Probability Formula
## Sample Space
A Sample space is the list of all outcomes in a random experiment.
### Simulation
When it is impossible to list all
the outcomes, we create a model of the experiment. This is called simulation.
### Experimental definition of Probability
By simulating the experiment, we
will find the number of times a particular event or outcomes occur out of the
total number of trials. Based on this
we have a new definition of probability called Experimental definition of
Probability
## Experimental Probability Formula
Let us consider a random experiment.
Let A be an outcome of the random experiment . Then A is called an event.
The Experimental probability of the event A is given by
P(A) = Number of times the event occurs/ Total number of trials
Experimental probability is also known as Relative frequency
definition of Probability.
Lets consider a few examples
Example 1: -
A die is thrown 100 times out of which 5 appears 28 times. Find the experimental probability of getting the number 5?
Solution: -
The Experimental probability of the event A is given by
P(A) = Number of times the event occurs/ Total number of trials
Here die is thrown 100 times. So total number of trails =100
The number 5 occurs 28 times. So the number of times the required event
occurs = 28
Therefore probability of getting the number 5 = 28/100 = 0.28
Example 2: -
A Box contains 15 red balls, 12 blue balls and 13 green
marbles. Find the experimental probability of getting a green ball.
Solution: -
The Experimental probability of the event A is given by
P(A) = Number of times the event occurs/ Total number of trials
Take a ball from the box. Note the color and return the ball.
Repeat a few times (maybe 200 times). Note the number of times a green ball
was picked (Suppose it is 120).
The experimental probability of getting a green ball from
the box is 120/200 = 60/100 = 0.6
Example 3: -
The following are the marks obtained by 1200 students in
a particular examination.
Marks: 100-120 120-140 140-160 160-180 180-200
No of 63 142 500 320 175
Students
Find the probability that a student selected has marks
(i) under 140
(ii) above 180
(iii) between 140 and 200
Solution: -
The Experimental probability of the event A is given by
P(A) = Number of times the event occurs/ Total number of trials
We can see that the total of the marks is 63 + 142 + 500 + 320 + 175 = 1200
(i) We have to find the probability that a selected student get marks under 40.
There are 63 + 142 = 205 students getting marks under 140.
Therefore P(student getting marks under 140)= 205/1200 = 0.17
(ii) We have to find the probability that a selected student get marks above 180.
There are 175 students getting marks above180.
Therefore P(student getting marks above 180) = 175/1200 = 0.15
(iii) We have to find the probability that a selected student get marks between 140 and 200.
There are 500 + 320 + 175 = 995 students getting marks between 140 and 200. Therefore P(student getting marks between 140 and 200) = 995/1200 = 0.83
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Save 10% on your first order with code EXTRA10
# Multiplication Facts x 11 FUN Digital Activities
\$3.00
SKU: 5376047 Categories: ,
## Description
The FUN and ENGAGING Multiplication Practice Activities are Perfect for Math Stations, Homework, Tutoring or Distance Learning. Students always need to practice their math facts, and they will enjoy these Simple and Easy-to-Use Google Slides.
⭐ Watch the Video Preview
This set includes all of the x11 math facts. All Facts 11 x 1 through 11 x 12 and their reverse facts (12 x 11) are included in this resource.
Topics Include:
• Math Facts practice with a mix of movable pieces and places to type in answers to show true memorization
• Missing Factor Equations such as 11 x ? = 33 to help students build the skills needed for division
• Input-Output Tables and Number Machines
• Multiplication Number Bonds
What’s Included?
• 2 Sets of Digital Flashcards for students to use again and again
• 9 Interactive Activities
• Teacher Key
• Directions on how to assign the Google Slides activity to students in Google Classroom™
Standards Alignment
CCSS.3.OA.C.7 Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g. knowing that 8 x 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
CCSS.3.OA.A.4 Determine the unknown whole number in a multiplication or division equation relating three whole numbers. For example, determine the unknown number that makes the equation true in each of the equations 8 x ? = 48, 5 = ? ÷ 3, 6 x 6 = ?
TEKS 3.4F Recall facts to multiply up to 10 by 10 with automaticity and recall the corresponding division facts. (While x 11 facts are beyond this standard, this resource supports TEKS 3.4K, where students need to solve multiplication and division problems within 100, including those with a factor of 11.)
TEKS 3.5D Determine the unknown whole number in a multiplication or division equation relating three whole numbers when the unknown is either a missing factor or product.
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# How many multiples of 3 are there?
## How many multiples of 3 are there?
How to list multiples of a number?
Multiples of 1 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Multiples of 3 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
Multiples of 4 4, 8, 12, 16, 20, 24, 28, 32, 36, 40.
Multiples of 5 5, 10, 15, 20, 25, 30, 35, 40, 45, 50.
Multiples of 6 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.
Is 3 a factor or a multiple?
What are the differences between a factor and a multiple?
Factor Multiple
3 Factors are a number are limited Multiples are endless
4 Factors are generally less or equal to the number Multiples are greater or equal to the given number.
5 Factors are obtained by division Multiples s are obtained by multiplication
What are the multiples of 3 from 1 to 100?
Playing With Numbers | Exercise 3.4 SOLUTION: multiple of 3: 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99. multiple of 4: 4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96. therefore, first three common multiples between 3 and 4 12,24,36.
### What are factors of 3?
Factors of 3 are 1 and 3 only. Note that -1 × -3 = 3.
How do you find the multiples of 3?
For getting the multiples of 3, we need to multiply 3 by 1, 2, 3, 4, and so on.
1. 3 × 1 = 3.
2. 3 × 2 = 6.
3. 3 × 3 = 9.
4. 3 × 4 = 12.
5. 3 × 5 = 15.
What is the factors of 3?
## What are multiples example?
A multiple is the product that we get when one number is multiplied by another number. For example, if we say 4 × 5 = 20, here 20 is a multiple of 4 and 5. The other multiples of 4 can be listed as 4 (4 ×1 = 4), 8 (4 × 2 = 8), 12 (4 × 3 = 12), and so on.
How do you find common multiples?
We can find the common multiples of two or more numbers by listing the multiples of each number and then finding their common multiples. For example, to find the common multiples of 3 and 4, we list their multiples and then find their common multiples. Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36.
Where are the prime numbers?
Prime Numbers : Example Question #1 Explanation: The first seven primes are 2, 3, 5, 7, 11, 13, and 17. Don’t forget about 2, the smallest prime number, and also the only even prime!
### What is the prime factorization of 90?
Therefore the prime factorization of 90 is 90 =3 × 3 × 5 × 2. Taking another prime number which is a factor of 90.
What are the first three multiples of 90?
What are the first 3 common multiples of 90 and 60 Multiples of 90 are 90, 180, 270, 360, 450, 540….. Multiples of 60 are 60, 120, 180, 240, 300, 360, 420, 480, 540…….
What are the answers to the multiples of 3?
Answer Multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147 Back to What are the multiples of 2?
## Are there any numbers that can be divided by 90?
Multiples of 90 are the numbers that can be divided by 90 without leaving any remainder. Interestingly, the multiples of 90 can be derived from the multiples of 9 by multiplying them by 10. The first five multiples of 9 are 9, 18, 27, 36, and 45.
Is the number 270 a multiple of 3?
The statement ‘270 is a multiple of 3’ is equivalent ‘270 is divisible by 3’, or that 3 is a divider of 270. So to find the multiples of 90, simply multiply this number by a number of the set of natural numbers as many times as we want.
Begin typing your search term above and press enter to search. Press ESC to cancel.
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What is 10 of 15?
Multiplication is an essential arithmetic operation; using this multiplication, we can combine quantities, determine the total of repeated additions, and calculate the product of two or more numbers. In this article, we will learn about the product of 10 and 15 in a detailed manner by breaking down the process, examining different methods, and exploring real-world examples; after that, we will understand how these numbers interact and how their product is determined.
Understanding Multiplication
Multiplication is a process that combines two or more numbers to obtain a result which we call a product. The symbol is "*," and "x" represents the multiplication. Multiplication is usually defined as a shortcut to repeated additions; for example, consider the multiplication of 2 and 4, i.e., 8, and we can also write this as adding 4 two times or adding 2 four times; both are identical, i.e., 4 + 4 = 8 or 2 + 2 + 2 + 2 = 8.
Multiplying 10 and 15
We multiply the two numbers to determine the product of 10 and 15. Let's explore the methods to calculate the product:
• Method 1: The Traditional multiplication: This traditional method requires multiplying each value and adding the resulting products; using this traditional method, we get the product of 10, and 15 is 150.
• Method 2: Expanded Multiplication: Another method to calculate the product is by expanding the numbers into their place value form.
10 × 15 = 10 x (10 + 5) = (10 x 10) + (10 x 5) = 100 +50 = 150
Using this, we get the product of 10, and 15 is 150.
• Method 3: Mental Calculation: For smaller numbers like 10 and 15, mental calculation can be a quick and efficient method. This can be done by recognizing patterns and applying basic arithmetic operations, we can find the product without relying on written methods. In this case, we know that multiplying any number by 10 involves adding a zero at the end. Therefore, multiplying 15 by 10 would result in 150. Using this method, we get the same number as the product, 150.
Real-World Applications
Understanding multiplication and the product of numbers has many practical applications in everyday life. Let's explore some real-world scenarios where the product of 10 and 15 is relevant.
1. Unit Conversion: If we convert 10 meters into centimeters, we need to know that 1 meter equals 100 centimeters. By multiplying 10 meters by 100 centimeters per meter, you find that 10 meters is equivalent to 1000 centimeters.
2. Area Calculation: Suppose we have a rectangular garden with a length of 10 meters and a width of 15 meters. To find the area of the garden, we multiply the length by the width; in this case, the area of that rectangle would be 10 meters × 15 meters, which equals 150 square meters.
3. Cost Calculation: Imagine we want to buy 10 books, each costing \$15. By multiplying the number of books (10) by the price per book (\$15), we can calculate the total cost of the books; in this case, the total cost would be \$150.
4. Time and Speed: Suppose we are traveling at a speed of 10 kilometers per hour, and you need to cover a distance of 150 kilometers. We can calculate the total distance covered by multiplying the speed (10 kilometers per hour) by the time taken (15 hours). In this example, the distance covered would be 10 kilometers per hour × 15 hours, resulting in 150 kilometers.
Conclusion
Multiplication is an essential mathematical operation that can allow us to combine quantities, determine repeated addition, and find the product of the numbers. In our case, the multiplication of 10 and 15, the product is 150. Understanding this concept can help us solve real-life problems like unit conversion, calculating cost prizes, and calculating areas and distances. After getting knowledge of this powerful mathematical tool, we have gained the power to solve practical problems.
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# If the remainder on division of ${x^3} + 2{x^2} + kx + 3$ by $x - 3$ is 21, find the quotient and value of k. Hence, find the zeros of the cubic polynomial ${x^3} + 2{x^2} + kx - 18$.
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Hint: We will use the remainder formula to find k from the given equation and long division process to find the quotient. We will also use the division formula of polynomials to get the zeros of the polynomial.
Given that the remainder on division of ${x^3} + 2{x^2} + kx + 3$ by $x - 3$ is 21
We have the following terms:
Dividend: $f(x) = {x^3} + 2{x^2} + kx + 3$
Divisor: ${\text{ }}g\left( x \right){\text{ }} = {\text{ }}x{\text{ }} - {\text{ }}3$ and remainder, $r{\text{ }}\left( x \right){\text{ }} = {\text{ }}21$
Using the remainder formula, we have the following expression:
$f\left( 3 \right) = 21$
$\Rightarrow {(3)^3} + 2.{(3)^2} + k.(3) + 3 = 21 \\ \Rightarrow 27 + 18 + 3k + 3 = 21 \\ \Rightarrow 3k = - 27 \\ \Rightarrow k = - 9 \\$
So, the polynomial is, $p(x) = {x^3} + 2{x^2} - 9x + 3$
Now, from the long division, we get,
${x^3}\; + {\text{ }}2{x^2}\; - {\text{ }}9x{\text{ }} + {\text{ }}3{\text{ }} = {\text{ }}\left( {x{\text{ }} - {\text{ }}3{\text{ }}} \right){\text{ }}({x^2}\; + {\text{ }}5x{\text{ }} + {\text{ }}6){\text{ }} + {\text{ }}21$
∴ The quotient $= {\text{ }}{x^2}\; + {\text{ }}5x{\text{ }} + {\text{ }}6$
Clearly, ${x^3}\; + {\text{ }}2{x^2}\; - {\text{ }}9x{\text{ }}-21 + 3 =$ ${x^3}\; + {\text{ }}2{x^2}\; - {\text{ }}9x{\text{ }}-{\text{ }}18$ is divisible by, $x - 3$
$= {x^3} - 3{x^2} + 5{x^2} - 15x + 6x - 18 \\ = {x^2}(x - 3) + 5x(x - 3) + 6(x - 3) \\$
$= {\text{ }}\left( {x{\text{ }} - {\text{ }}3{\text{ }}} \right){\text{ }}\left( {{x^2}\; + {\text{ }}5x{\text{ }} + {\text{ }}6} \right)\;$
On further splitting of middle terms we get,
$= {\text{ }}\left( {x{\text{ }} - {\text{ }}3{\text{ }}} \right){\text{ }}\left( {{x^2} + 3x + 2x + {\text{ }}6} \right)$
On further simplification we get,
$= {\text{ }}\left( {x{\text{ }} - {\text{ }}3{\text{ }}} \right){\text{ }}\left( {x{\text{ }} + {\text{ }}2} \right)\left( {x{\text{ }} + {\text{ }}3} \right)$
For, now, $(x - 3)$we have, $x = 3$
Then, for, ($x + 2$) we have, $x = - 2$
And also, for, ($x + 3$) we have, $x = - 3$
Therefore, the zeroes of ${x^3}\; + {\text{ }}2{x^2}\; - {\text{ }}9x\; - {\text{ }}18$ are 3, -2 and -3.
Note: We have the remainder theorem as , $f(x) = g(x).h(x) + r(x)$. Where $f(x)$is the dividend and $g(x)$is the divisor. We also have $r\left( x \right)$as the reminder. This type of problems are built with the concept of long division altogether.
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# Lesson 15: More Nets, More Surface Area
Let’s draw nets and find the surface area of polyhedra.
## 15.1: Notice and Wonder: Wrapping Paper
Kiran is wrapping this box of sports cards as a present for a friend.
What do you notice? What do you wonder?
## 15.2: Building Prisms and Pyramids
Your teacher will give you a drawing of a polyhedron. You will draw its net and calculate its surface area.
1. What polyhedron do you have?
2. Study your polyhedron. Then, draw its net on graph paper. Use the side length of a grid square as the unit.
3. Label each polygon on the net with a name or number.
4. Find the surface area of your polyhedron. Show your thinking in an organized manner so that it can be followed by others.
## 15.3: Comparing Boxes
Here are the nets of three cardboard boxes that are all rectangular prisms. The boxes will be packed with 1-centimeter cubes. All lengths are in centimeters.
1. Compare the surface areas of the boxes. Which box will use the least cardboard? Show your reasoning.
2. Now compare the volumes of these boxes in cubic centimeters. Which box will hold the most 1-centimeter cubes? Show your reasoning.
## Summary
The surface area of a polyhedron is the sum of the areas of all of the faces.
Because a net shows us all faces of a polyhedron at once, it can help us find the surface area. We can find the areas of all polygons in the net and add them.
A square pyramid has a square and four triangles for its faces. Its surface area is the sum of the areas of the square base and the four triangular faces: $$(6\boldcdot 6) + 4\boldcdot \left(\frac12 \boldcdot 5 \boldcdot 6\right) = 96$$ The surface area of this square pyramid is 96 square units.
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3-D Solids: Faces, Edges and Vertices
by on February 28, 2012
A 3-D solid (sometimes called a 3-D shape) is a figure that is not flat, it is three-dimensional. Some examples of 3-D solids include a cube, rectangular prism, cone, cylinder, pyramid, sphere and so on. Once your child has had an opportunity to explore various 3-D solids, she will be ready to begin looking at the main components of 3-D solids: faces, edges and vertices.
Faces
A face is a flat surface on a 3-D solid. Students are often asked to identify the number of faces on 3-D solids. It is important to note that a face must be flat, so a sphere technically does not have a face, it has a curved surface. Another noteworthy point: Your child may also learn about a base. A base is a special kind of face. Cylinders and prisms actually have two bases that are both parallel and congruent.
Edges
An edge is a line segment where two faces meet. To help my students better understand edges, I ask them to run their finger along the edge of a tissue box. Try challenging your child to try to find all the edges on a box. Identifying the edges of 3-D solids is typically more difficult for students.
Vertices
“Vertices” is the plural of one vertex. Vertices are corner points. Vertices are found where edges meet.
Here is a chart with the numbers of faces, edges and vertices of some common 3-D solids. Please note that your child may not learn all of these solids at one time.
3-D Solid FACES EDGES VERTICES CUBE 6 12 8 RECTANGULAR PRISM 6 12 8 CYLINDER 2 0 0 CONE 1 0 0 SQUARE PYRAMID 5 8 5 TRIANGULAR PRISM 5 9 6 SPHERE 0 0 0 TRIANGULAR PYRAMID (TETRAHEDRON) 4 6 4 HEXAGONAL PRISM 8 18 12
What to Watch For:
• Children often confuse 3-D solids (sometimes called 3-D shapes) with 2-D shapes. They often learn about shapes (square, rectangle, triangle, pentagon, hexagon, octagon, rhombus, trapezoid, etc…) at the same time as 3-D solids, which can also be confusing. Be sure your child understand the concept that 3-D solids have 3 dimensions. You will likely need to explain what 3-dimensional means (length, width, and height).
WANT MORE?
• As always, playing math games at home is a great way to reinforce math skills learned in school.
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# How do you solve 2x^3 + x^2 - 5x + 2 < 0?
Oct 1, 2015
$2 {x}^{3} + {x}^{2} - 5 x + 2 = \left(x + 2\right) \left(2 x - 1\right) \left(x - 1\right)$
Hence $2 {x}^{3} + {x}^{2} - 5 x + 2 < 0$ when $x \in \left(- \infty , - 2\right) \cup \left(\frac{1}{2} , 1\right)$
#### Explanation:
Let $f \left(x\right) = 2 {x}^{3} + {x}^{2} - 5 x + 2$.
By the rational root theorem, any rational roots of $f \left(x\right) = 0$ must be of the form $\frac{p}{q}$ in lowest terms, where $p , q \in \mathbb{Z}$, $q \ne 0$, $p$ a divisor of the constant term $2$ and $q$ a divisor of the coefficient $2$ of the leading term.
That means that the only possible rational roots are $\pm \frac{1}{2}$, $\pm 1$ and $\pm 2$.
We find $f \left(- 2\right) = f \left(\frac{1}{2}\right) = f \left(1\right) = 0$, so $- 2$, $\frac{1}{2}$ and $1$ are the three roots.
$f \left(x\right)$ can potentially change sign at each of these roots, and will do since none of them is a repeated root.
$f \left(x\right) = \left(x + 2\right) \left(2 x - 1\right) \left(x - 1\right)$
So when $x < - 2$, the signs of the three factors are $-$, $-$ and $-$, so their product $f \left(x\right) < 0$.
When $- 2 < x < \frac{1}{2}$, the signs of the three factors are $+$, $-$ and $-$, so $f \left(x\right) > 0$.
When $\frac{1}{2} < x < 1$, the signs of the factors are $+$, $+$ and $-$, so $f \left(x\right) < 0$.
When $1 < x$, the signs of the factors are $+$, $+$ and $+$, so $f \left(x\right) > 0$.
So we find $f \left(x\right) < 0$ when $x \in \left(- \infty , - 2\right) \cup \left(\frac{1}{2} , 1\right)$
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# Ratio of the areas | PRMO-2019 | Problem 19
Try this beautiful problem from PRMO, 2019 based on Ratio of the areas.
## Ratio of the areas | PRMO | Problem-19
Let $\mathrm{AB}$ be a diameter of a circle and let $\mathrm{C}$ be a point on the segment $\mathrm{AB}$ such that $\mathrm{AC}: \mathrm{CB}=6: 7 .$ Let $\mathrm{D}$ be a point on the circle such that $\mathrm{DC}$ is perpendicular to $\mathrm{AB}$. Let DE be the diameter through $\mathrm{D}$. If $[\mathrm{XYZ}]$ denotes the area of the triangle XYZ. Find [ABD] / $[\mathrm{CDE}]$ to the nearest integer.
• $20$
• $91$
• $13$
• $$23$$
### Key Concepts
Geometry
Triangle
Area
Answer:$$13$$
PRMO-2019, Problem 19
Pre College Mathematics
## Try with Hints
$$\angle \mathrm{AOC} \quad=\frac{6 \pi}{13}, \angle \mathrm{BOC}=\frac{7 \pi}{13}$$
$\mathrm{Ar} \Delta \mathrm{ABD}=\mathrm{Ar} \Delta \mathrm{ABC}=\frac{1}{2} \mathrm{AB} \times \mathrm{OC} \sin \frac{6 \pi}{13}$
$\mathrm{Ar} \Delta \mathrm{CDE}=\frac{1}{2} \mathrm{DE} \times \mathrm{OC} \sin \left(\frac{7 \pi}{13}-\frac{6 \pi}{13}\right)$
$\frac{[\mathrm{ABD}]}{[\mathrm{CDE}]}=\frac{\sin \frac{6 \pi}{13}}{\sin \frac{\pi}{13}}=\frac{1}{2 \sin \frac{\pi}{26}}=\mathrm{p}$
because $\sin \theta \cong \theta$ if $\theta$ is small
$\Rightarrow \sin \frac{\pi}{26} \cong \frac{\pi}{26}$
$\mathrm{p}=\frac{13}{\pi} \Rightarrow$ Nearest integer to $\mathrm{p}$ is 4
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# Lesson 20
Writing and Solving Inequalities in One Variable
### Problem 1
Solve $$2x < 10$$. Explain how to find the solution set.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 2
Lin is solving the inequality $$15 - x < 14$$. She knows the solution to the equation $$15 - x = 14$$ is $$x = 1$$
How can Lin determine whether $$x > 1$$ or $$x < 1$$ is the solution to the inequality?
### Solution
For access, consult one of our IM Certified Partners.
### Problem 3
A cell phone company offers two texting plans. People who use plan A pay 10 cents for each text sent or received. People who use plan B pay 12 dollars per month, and then pay an additional 2 cents for each text sent or received.
1. Write an inequality to represent the fact that it is cheaper for someone to use plan A than plan B. Use $$x$$ to represent the number of texts they send.
2. Solve the inequality.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 4
Clare made an error when solving $$\text-4x+3<23$$.
Describe the error that she made.
\displaystyle \begin{align} \text-4x+3<23 \\ \text-4x<20 \\ x< \text-5 \end{align}
### Solution
For access, consult one of our IM Certified Partners.
### Problem 5
Diego’s goal is to walk more than 70,000 steps this week. The mean number of steps that Diego walked during the first 4 days of this week is 8,019.
1. Write an inequality that expresses the mean number of steps that Diego needs to walk during the last 3 days of this week to walk more than 70,000 steps. Remember to define any variables that you use.
2. If the mean number of steps Diego walks during the last 3 days of the week is 12,642, will Diego reach his goal of walking more that 70,000 steps this week?
### Solution
For access, consult one of our IM Certified Partners.
### Problem 6
Here are statistics for the length of some frog jumps in inches:
• the mean is 41 inches
• the median is 39 inches
• the standard deviation is about 9.6 inches
• the IQR is 5.5 inches
How does each statistic change if the length of the jumps are measured in feet instead of inches?
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 1, Lesson 15.)
### Problem 7
Solve this system of linear equations without graphing: $$\begin{cases} 3y+7=5x \\ 7x-3y=1 \\ \end{cases}$$
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 2, Lesson 15.)
### Problem 8
Solve each system of equations without graphing.
1. $$\begin{cases} 5x+14y=\text-5 \\ \text-3x+10y=72 \\ \end{cases}$$
2. $$\begin{cases}20x-5y=289 \\ 22x + 9y=257 \\ \end{cases}$$
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 2, Lesson 16.)
### Problem 9
Noah and Lin are solving this system: $$\begin{cases} 8x+15y=58 \\ 12x-9y=150 \end{cases}$$
Noah multiplies the first equation by 12 and the second equation by 8, which gives:
$$\displaystyle \begin{cases} 96x+180y=696 \\ 96x-72y=1,\!200 \\ \end{cases}$$
Lin says, “I know you can eliminate $$x$$ by doing that and then subtracting the second equation from the first, but I can use smaller numbers. Instead of what you did, try multiplying the first equation by 6 and the second equation by 4."
1. Do you agree with Lin that her approach also works? Explain your reasoning.
2. What are the smallest whole-number factors by which you can multiply the equations in order to eliminate $$x$$?
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 2, Lesson 16.)
### Problem 10
What is the solution set of the inequality $$\dfrac{x+2}{2}\geq \text-7-\dfrac {x}{2}$$ ?
A:
$$x\leq \text-8$$
B:
$$x\geq \text-8$$
C:
$$x \geq - \frac92$$
D:
$$x\geq 8$$
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 2, Lesson 19.)
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# Solving for x for equation $y=x(8-x)$
I am trying to find the equation of $y=x(8-x)$
So what I did so far ...
$y=8x-x^2$
But no matter what I did after, I still will have $x$ of different degrees? eg. 1 squared, 1 "normal"? else it will be 1 square root, 1 "normal"?
How do I solve this?
Given answer: $f^{-1}=4-\sqrt{16-x}$ (I'm solving for x to find the inverse of the function actually)
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You could use the quadratic formula on $x^2-8x+y=0$ – J. M. Sep 4 '11 at 2:32
$y$ is a variable, can I just treat it as $c$? If so I will get $x=\frac{-(-8) \pm \sqrt{(-8)^2 - 4y}}{2}$ then $x=\frac{8 \pm \sqrt{64-4y}}{2}$, then how do I simplify further? – Jiew Meng Sep 4 '11 at 2:51
you factor a $\sqrt 4 = 2$ out from under the radical – The Chaz 2.0 Sep 4 '11 at 3:04
To find the inverse of a function like this, you can switch the roles of $x$ and $y$ and solve the new equation for $y$. So, you would want to solve $x = -y^2 + 8y$ for $y$.
Rewriting that a little, we get $y^2 - 8y + x = 0$, which is a quadratic in $y$. That is, you would be taking $a = 1$, $b = -8$, and $c = x$ in the quadratic formula. Don't let the fact that $x$ is a variable bother you: the function $y^2 - 8y + x$ is quadratic in $y$ and that's all that matters here.
Applying the formula, you would get
\begin{align*} y &= \frac{8 \pm \sqrt{64 - 4x}}{2}\\ &= \frac{8 \pm \sqrt{4(16 - x)}}{2}\\ &= \frac{8 \pm 2\sqrt{16 - x}}{2}\\ &= 4 \pm \sqrt{16 - x} \end{align*}
The fact that you get $\pm$ in your solution means that the original function doesn't have an inverse on the entire real line. Depending on how you restrict the domain of the original function, your inverse can be either $4 - \sqrt{16 - x}$ or $4 + \sqrt{16 - x}$.
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Could someone please edit my question or explain why the align environment is behaving strangely? – Austin Mohr Sep 4 '11 at 3:08
You either need to double up the backslashes, i.e., use \\\\ instead of \\ or enclose the align environment in double dollar signs. That's because backslashes have act as escape characters in markdown, so in order for the MathJaX engine to see a backslash you need to escape it first, using a backslash. By using dollar signs you deactivate the markdown engine. – t.b. Sep 4 '11 at 3:13
Good answer. Better for worrying about the $\pm$ sign. – Ross Millikan Sep 4 '11 at 3:45
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# Converting 2.5 to a Percentage – Simple Guide!
Have you ever wondered what is 2.5 as a percentage? Maybe you need to convert 2.5 to a percentage for a calculation or analysis. Well, look no further! In this simple guide, we’ll show you how to calculate 2.5 as a percentage quickly and easily.
To convert 2.5 to a percentage, you simply multiply it by 100. This is because a percentage is a way of expressing a part of a whole as a fraction of 100. So, when we say 2.5 as a percent, we mean 2.5 out of 100. Therefore, 2.5 as a percent is 250%.
Converting 2.5 to a percentage is that straightforward! It’s a simple multiplication that allows you to express a value as a portion of 100. Understanding percentages is essential for many calculations, comparisons, and practical applications.
### Key Takeaways:
• Converting 2.5 to a percentage is done by multiplying it by 100.
• A percentage represents a part of a whole out of 100.
• 2.5 as a percent is equivalent to 250%.
• Understanding percentages is important for various calculations and comparisons.
• Knowing how to convert values to percentages is a useful skill in everyday life and various fields.
## Understanding Percentages: A Practical Guide
Percentages are a common way of expressing fractions or ratios as a part of a whole out of 100. Whether you’re analyzing data, comparing prices, or understanding probabilities, percentages play a crucial role in various fields, including finance, retail, and everyday life.
So why use percentages? Well, they make calculations and comparisons quick and easy. Instead of dealing with complex fractions, percentages simplify the process by scaling everything down to a base of 100. This allows for straightforward comparisons between different values and proportions.
At its core, a percentage is a way to express a fraction or ratio as a part of 100. For example, 50% represents 50 out of 100 units or 50/100. By using percentages, it becomes much simpler to understand the relationship between different quantities and make informed decisions.
### The Fundamentals of Percentages
To better understand percentages, let’s consider a practical example. Suppose you’re comparing the prices of two items, A and B. Item A costs \$40, while item B costs \$60. To determine the price difference as a percentage, you can use the following formula:
Percentage difference = (Change / Original value) * 100
Using our example, the change would be \$60 – \$40 = \$20. The original value is \$40. Plugging these values into the formula, we get:
Percentage difference = (\$20 / \$40) * 100 = 50%
Therefore, item B is 50% more expensive than item A.
Understanding percentages allows you to quickly analyze data, compare prices, and interpret information. It simplifies complex calculations and enables you to make informed decisions in various situations.
### Visual Representation of Percentages
Visualizing percentages can be helpful in grasping their practical applications. Consider the following table:
ItemPriceDiscount
Item A\$8020%
Item B\$12030%
In the table above, you can see how percentages are used to express discounts. Item A is priced at \$80 with a 20% discount, resulting in a final price of \$64. Item B, priced at \$120, has a 30% discount, bringing the total down to \$84.
This visual representation helps you quickly compare the original prices, discounts, and final prices, allowing you to make informed purchasing decisions.
Now that we’ve covered the fundamentals of percentages, let’s explore how to convert fractions and decimals into percentages in the next section.
## Converting Fractions to Decimals and Percents
Converting fractions to decimals and percents involves different methods. Visually, fractions can be represented as parts of a whole. To convert a fraction to a decimal, divide the numerator by the denominator. To convert a fraction to a percent, divide the numerator by the denominator and multiply by 100.
When converting a fraction to a decimal, divide the numerator by the denominator. For example, to convert 3/4 to a decimal, divide 3 by 4 to get 0.75. This means that 3/4 is equivalent to 0.75.
When converting a fraction to a percent, divide the numerator by the denominator and multiply by 100. For example, to convert 3/4 to a percent, divide 3 by 4 and multiply by 100 to get 75%. This means that 3/4 is equivalent to 75%.
### Example: Converting 3/4
FractionDecimalPercent
3/40.7575%
Converting fractions to decimals and percents is a useful skill in various fields, such as mathematics, finance, and science. It allows for easy comparison and calculation of values. By mastering this conversion process, you can confidently work with fractions, decimals, and percents in your everyday life and academic or professional endeavors.
### Quote:
“Converting fractions to decimals and percents enables us to express fractions as parts of a whole in different forms, providing versatility and ease of understanding.”
## Decimal to Percent Conversion Made Easy
Converting a decimal to a percent is a straightforward process that involves multiplying the decimal by 100. This is because a percent is a ratio out of 100, representing a fraction or portion of a whole. By understanding this concept, you can easily convert a decimal to a percentage without any hassle.
For example, let’s convert the decimal 0.25 to a percent. Multiply 0.25 by 100:
0.25 * 100 = 25%
So, the decimal 0.25 is equivalent to 25% in percentage form. This means that 25% of a whole is represented by the decimal 0.25.
If you would like to convert other decimals to percentages, you can use a decimal to percent calculator, which simplifies the conversion process and saves you time. Simply input the decimal value, and the calculator will instantly provide you with the equivalent percentage.
Here is an example of a straightforward decimal to percent calculator:
“`html
“`
With the help of a decimal to percent calculator, you can effortlessly convert any decimal to its corresponding percentage, making complex calculations much easier and more efficient.
DecimalPercent
0.110%
0.220%
0.330%
0.440%
0.550%
This table showcases the decimal-to-percent conversion for some common decimal values. It serves as a quick reference for understanding the relationship between decimals and percentages.
Converting decimals to percents is a fundamental mathematical concept that has various practical applications, such as calculating discounts, understanding financial data, or interpreting statistics. By mastering this skill, you’ll be able to navigate these scenarios with ease and confidence.
## Fraction to Percent Conversion: Step-by-Step Guide
Converting fractions to percentages is a straightforward process that allows you to express fractions as parts of a whole out of 100. By following a few simple steps, you can convert any fraction to a percent quickly and easily.
To convert a fraction to a percent:
1. Step 1: Divide the numerator by the denominator.
2. Step 2: Multiply the result by 100.
For example, let’s convert the fraction 3/4 to a percent:
Step 1: Divide 3 by 4. The result is 0.75.
Step 2: Multiply 0.75 by 100. The result is 75%.
So, 3/4 as a percent is 75%.
It’s important to note that percentages can also be expressed as fractions. To convert a percent to a fraction, divide the percent by 100 and simplify the fraction if possible. For example, the percent 75% can be expressed as the fraction 3/4.
Here’s a table summarizing some common fraction to percent conversions:
FractionPercent
1/250%
1/425%
3/475%
2/366.67%
Remember, the process of converting fractions to percents is as simple as dividing the numerator by the denominator and multiplying by 100. Knowing how to convert between fractions and percents is essential for various calculations and comparisons, making it a valuable skill to have.
## Why Percentages Are Useful in Everyday Life
Percentages play a crucial role in our daily lives, providing a practical way to understand and represent various concepts. From calculating shopping discounts to analyzing probabilities, percentages simplify complex calculations and allow for easy comparisons. Let’s explore the importance of percentages and their practical applications in different aspects of life.
### Practical Uses of Percentages
One of the most common uses of percentages is in financial contexts. For example, when we see a sale advertisement with a 50% discount, we instantly understand that we will pay only half the original price. Percentages help us make informed decisions while shopping and budgeting our expenses.
Percentages are also valuable when it comes to personal finance and investments. Whether calculating interest rates, returns on investments, or planning a budget, percentages allow us to analyze and comprehend financial data easily. They help us gauge the growth or decline of our assets and make informed decisions accordingly.
In addition to finances, percentages are crucial in understanding proportions and ratios in daily life. For instance, when following a recipe, we rely on percentages to determine the right proportions of ingredients. Similarly, percentages are used in sports statistics, such as shooting percentages in basketball or batting averages in baseball, giving us insights into performance levels.
### Importance of Percentages
Percentages bring clarity and simplicity to complex situations. They allow us to compare values and make logical decisions based on the given percentages. Whether it’s evaluating market share, calculating test scores, or understanding population growth, percentages provide a common language for expressing proportions, making it easier to analyze and interpret data.
Moreover, percentages help us understand probabilities. In fields like statistics, economics, and medicine, percentages are used to assess risks, predict outcomes, and make informed choices. They allow us to evaluate the likelihood of an event occurring and understand the implications based on the given probability.
“The beauty of percentages lies in their ability to simplify complex data, making it accessible and relatable to individuals across various fields of study and everyday life.”
### A Practical Perspective
To further emphasize the practicality of percentages, let’s consider a real-world example. Imagine you’re comparing two job offers: Offer A has a base salary of \$60,000 per year, while Offer B offers a base salary of \$70,000 per year with a 10% commission on sales. By comparing the percentage-based commission, you can estimate your potential earnings based on your sales performance. This percentage-based comparison enables you to make an informed decision about which offer aligns better with your financial goals and aspirations.
### The Power of Percentages in a Nutshell
Percentages play an integral role in our daily lives, simplifying calculations, aiding decision-making, and providing valuable insights across a multitude of contexts. Their practical uses in personal finance, budgeting, proportion analysis, and understanding probabilities make percentages an essential tool for everyone.
Practical Uses of PercentagesImportance of PercentagesA Practical Perspective
Financial calculations (discounts, investments)Simplifies complex data analysisComparing job offers based on percentages
Recipe proportions and ingredient measurementsAids in decision-making based on dataEstimating potential earnings with commission percentages
Sports statistics and performance measurementEnhances understanding of probabilities
## The History and Origin of Percentages
The use of percentages dates back to Ancient Rome, where they were used for taxation and calculating auction fees. In fact, the origin of the word “percent” can be traced back to the Latin phrase “per centum,” which means “by the hundred.” This reflects the fundamental concept of expressing a fraction or ratio as a part of 100.
Percentages played a crucial role in Ancient Rome’s financial and economic systems. They were used to determine the amount of taxes that individuals and businesses had to pay. Additionally, percentages were employed in calculating auction fees, ensuring fair transactions.
Over time, the concept of using fractions based on 100 spread throughout Europe and became widely used in various fields. Today, percentages are an integral part of our everyday lives, serving as a universal language for expressing proportions, ratios, and comparisons.
Understanding the history and origin of percentages gives us a deeper appreciation for their significance and utility in modern society. It highlights how a mathematical concept developed centuries ago continues to shape our understanding and calculations.
### The Significance of Percentages in Ancient Rome
“Percentages played a vital role in the financial and economic systems of Ancient Rome, serving as a means of taxation and facilitating fair transactions through auction fees.”
Ancient Rome’s Use of PercentagesExamples
TaxationPercentage-based calculations were employed to determine the amount of taxes paid by individuals and businesses.
Auction FeesPercentages were used to calculate the fees associated with auctions, ensuring fairness and transparency in transactions.
## The Limitations of Base 100 in Percentages
While percentages serve as a convenient way to express ratios and facilitate comparisons, they do have some limitations when it comes to using a base of 100. Base 100 is not divisible by all numbers, which can result in repeating decimals when dividing certain numbers by 100.
When dealing with fractions or decimals that cannot be evenly divided by 100, it is important to be aware of these limitations. Repeating decimals can sometimes pose challenges in practical calculations and comparisons.
However, it is worth noting that in most practical cases, rounding off decimals is sufficient and allows for practical use of base 100. Rounding off decimals helps avoid complicated calculations and provides a close approximation for everyday situations.
Let’s take a look at an example to illustrate the limitations of base 100 in percentages:
Example:
Consider the fraction 1/3. When converted to a decimal, it becomes approximately 0.33333. Multiplying this decimal by 100 results in 33.3333%, which is a repeating decimal. Rounding off to the nearest whole number, we get 33% as the more practical representation.
Despite the limitations of base 100, percentages remain an essential tool for expressing ratios and proportions. They allow for easy comparison between different values and provide a standardized way to communicate proportions in various contexts.
Base 100 LimitationsFactors to Consider
Repeating decimalsSome fractions or decimals result in repeating decimals when divided by 100.
Rounding offRounding off decimals to the nearest whole number can help overcome limitations and maintain practicality.
Understanding these limitations and how to work around them is important for accurately interpreting and utilizing percentages in various calculations and scenarios.
### Key Takeaways:
• Base 100 has limitations when dividing certain numbers by 100, resulting in repeating decimals.
• Rounding off decimals is often sufficient for practical calculations and comparisons.
• Despite limitations, percentages remain a useful tool for expressing proportions and facilitating comparisons.
## The Practicality of Percentage Calculations
Percentage calculations are a practical and useful tool in various situations. Whether you are calculating discounts, determining proportions, or understanding percent change, working with percentages simplifies complex calculations and provides valuable insights.
To perform practical percentage calculations, you can rely on the basic percent formula:
(x/y) * 100 = z%
This formula allows you to easily compute the percentage of a whole, find the exact number associated with a percentage, and calculate percent change. Let’s explore the practicality of these calculations in more detail:
### Calculating Discounts
Percentage calculations are frequently used in determining discounts during sales and promotions. For example, if an item is discounted by 25%, you can easily find the discounted price by subtracting 25% of the original price from the original price. This helps consumers make informed decisions while shopping and allows businesses to manage their pricing strategies effectively.
### Determining Proportions
Working with percentages is also essential when dealing with proportions. For instance, if you have a recipe that calls for 2 cups of flour and you only have 1 cup, you can determine the proportion of flour you have by dividing 1 cup by 2 cups and multiplying the result by 100%. This allows you to adjust the recipe accordingly and achieve the desired outcome.
### Understanding Percent Change
Percent change is frequently used in analyzing trends and measuring growth or decline. By comparing the difference between two values using percentages, you gain valuable insights into the magnitude of the change. For example, if a company’s revenue increased from \$100,000 to \$150,000, you can calculate the percent change by dividing the difference (\$50,000) by the original value of \$100,000 and multiplying it by 100%. This provides a clear indication of the percentage increase.
Working with percentages not only simplifies calculations but also facilitates data interpretation and decision-making. Whether you are a student, professional, or business owner, mastering percentage calculations will enhance your numerical skills and give you a competitive edge.
## Conclusion
In conclusion, converting 2.5 to a percentage is a straightforward process – simply multiply it by 100. Percentages are a valuable tool for expressing fractions or ratios as parts of a whole out of 100. Understanding percentages and their practical applications can greatly benefit us in everyday life and various fields.
By following the steps outlined in this guide and utilizing tools such as a decimal to percent calculator, you can easily convert fractions, decimals, and ratios to percentages. This knowledge enables you to make quick calculations, compare values, and simplify complex fractions, making it an invaluable skill to have.
Whether you’re dealing with financial calculations, analyzing retail discounts, or interpreting data, percentages provide a clear and simplified way to express numbers and understand proportions. So, embrace the power of percentages, and unlock a world of convenience in your calculations and comparisons!
## FAQ
### What is 2.5 as a percentage?
To convert 2.5 to a percentage, you multiply it by 100. So, 2.5 as a percent is 250%.
### How do you convert fractions to decimals and percents?
To convert a fraction to a decimal, divide the numerator by the denominator. To convert a fraction to a percent, divide the numerator by the denominator and multiply by 100.
### How do you convert decimals to percents?
To convert a decimal to a percent, multiply the decimal by 100. For example, to convert 0.25 to a percent, multiply it by 100 to get 25%.
### How do you convert fractions to percents?
To convert a fraction to a percent, divide the numerator by the denominator and multiply by 100. For example, to convert 3/4 to a percent, divide 3 by 4 to get 0.75, then multiply by 100 to get 75%.
### Why are percentages useful in everyday life?
Percentages are used in various practical situations, such as understanding discounts, probabilities, and financial calculations. They simplify complex fractions and allow for easy comparisons between different values.
### What is the history and origin of percentages?
Percentages were used in Ancient Rome for taxation and calculating auction fees. The word “percent” comes from the Latin phrase “per centum,” meaning “by the hundred.” The concept of using fractions based on 100 spread throughout Europe and became widely used in various fields.
### What are the limitations of base 100 in percentages?
Base 100 is not divisible by all numbers, which can result in repeating decimals when dividing certain numbers by 100. However, in most practical cases, rounding off decimals is sufficient.
### How are percentage calculations practical?
Percentage calculations are practical for various situations, such as calculating discounts, proportions, and percent change. By using the basic percent formula (x/y) * 100 = z%, you can easily compute percentages and compare values.
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If you find any mistakes, please make a comment! Thank you.
## Solution to Linear Algebra Hoffman & Kunze Chapter 9.2
#### Exercise 9.2.1
Solution:
(a) No. Since $f(0,\beta)\ne 0$.
(b) No. Since $f((0,0),(1,0))\ne 0$.
(c) Yes. Since $f(\alpha,\beta)=4x_1\bar y_1$.
(d) No. Because of $\bar x_2$ there, it is not linear on $\alpha$.
#### Exercise 9.2.2
Solution: It is not a form! I think it should be $$f((x_1,y_1),(x_2,y_2))=x_1x_2+y_1y_2.$$The matrices are$\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix},\begin{pmatrix}2 & 0\\ 0& 2\end{pmatrix},\begin{pmatrix}5 & 11\\ 11 & 25\end{pmatrix},$respectively.
#### Exercise 9.2.3
Solution: Because $A=A^*$, we have$\overline{g(X,Y)}=(Y^*AX)^*=X^*AY=g(Y,X).$It is also clear that $g$ defines a form. Hence we only need to check that $g(X,X)>0$ if $X\ne 0$.
Let $X=(\alpha,\beta)$, then\begin{align*}g(X,X)&=(\bar\alpha,\bar \beta)\begin{pmatrix}1 & i\\ -i & 2\end{pmatrix}\begin{pmatrix}\alpha\\ \beta\end{pmatrix}\\&=\alpha\bar{\alpha}+2\beta\bar\beta-i(\alpha\bar\beta-\bar\alpha\beta)\\&=(\alpha-i\beta)\overline{(\alpha-i\beta)}+\beta\bar\beta\\&=|\alpha-i\beta|^2+|\beta|^2 > 0.\end{align*}Suppose $g(X,X)=0$, then $\alpha-i\beta=0$ and $\beta=0$. Hence $X=0$. Therefore, we are done.
#### Exercise 9.2.4
Solution: Since $f$ is a form, we have $f$ is linear on $\alpha$. Since $f(\alpha,\beta)=f(\beta,\alpha)$, we also have $f$ is linear on $\beta$. Therefore, $f$ is a form which is also a bilinear form. Then we have$-if(\alpha,\beta)=f(\alpha,i\beta)=f(i\beta,\alpha)=if(\beta,\alpha)=if(\alpha,\beta).$Thus $f(\alpha,\beta)=0$. That is $f=0$.
#### Exercise 9.2.5
Solution: The matrix of $f$ in the standard basis is given by $\begin{pmatrix}1 & 2\\2& 4\end{pmatrix}$. Its eigenvalues are zero and 5. The corresponding eigenvectors are $(2,-1)$ and $(1,2)$. Now it is easy to check that the matrix of $f$ in the basis $(2,-1)$ and $(1,2)$ is diagonal $\mathrm{diag}(0,25)$.
#### Exercise 9.2.6
Solution: If $f$ is non-degenerate, we show that $T_f$ is non-singular. Suppose $T_f\alpha=0$ for some $\alpha\in V$, then we have $$f(\alpha,\beta)=(T\alpha|\beta)=0$$ for all $\beta$. Since $f$ is non-degenerate, we must have $\alpha=0$. Hence $T_f$ is non-singular.
Conversely, suppose $f(\alpha,\beta)=0$ for all $\beta$, then we have $$f(\alpha,\beta)=(T_f\alpha|\beta)=0,$$for all $\beta$. In particular, we have $0=(T_f\alpha|T_f\alpha)$. Hence $T_f\alpha=0$. But $T_f$ is non-singular, therefore $\alpha=0$. Thus $f$ is non-degenerate.
#### Exercise 9.2.7
Solution:
Call the form $f$ right non-degenerate if $0$ is the only vector $\alpha$ such that $f(\beta,\alpha)=0$ for all $\beta$.
Suppose $f$ is left non-degenerate, we show that $f$ is right non-degenerate. Suppose $f(\beta,\alpha)=0$ for all $\beta$. Hence $(T_f\beta|\alpha)=0$ for all $\beta$. Because $f$ is left non-degenerate, $T_f$ is non-singular by Exercise 9.2.6. Since $V$ is finite-dimensional, we have $T_f$ is invertible. Hence there exists $\beta$ such that $T_f\beta=\alpha$. Therefore, we conclude that $$0=(T_f\beta|\alpha)=(\alpha|\alpha).$$Thus $\alpha=0$, namely $f$ is right non-degenerate.
Similar to Exercise 9.2.6, one can show that $f$ is right non-generate if and only if $T_f^*$ is non-singular. Then repeat our argument, we can show the other direction.
#### Exercise 9.2.8
Solution: By Theorem 6 of page 291, there exists a vector $\beta_0$ such that $L(\alpha)=(\alpha|\beta_0)$. On the other hand, we also have $$f(\alpha,\beta)=(T_f\alpha|\beta)=(\alpha|T_f^*\beta).$$Since $f$ is non-degenerate, $T_f$ is non-singular. Because $V$ is finite-dimensional, we conclude that $T_f$ is invertible. So is $T_f^*$. Therefore, there exists a unique $\beta\in V$ such that $T_f^*\beta=\beta_0$. We have$L(\alpha)=(\alpha|\beta_0)=(\alpha|T_f^*\beta)=f(\alpha,\beta).$We showed the existence.
Suppose $\beta_1$ and $\beta_2$ are vectors such that $$L(\alpha)=f(\alpha,\beta_1)=f(\alpha,\beta_2).$$Then $f(\alpha,\beta_1-\beta_2)=0$ for all $\alpha$. Since $f$ is non-degenerate, we have $\beta_1-\beta_2=0$, which shows the uniqueness.
#### Exercise 9.2.9
Solution: As we have seen in previous exercises, $T_f^*$ is invertible. We have$f(S\alpha,\beta)=(T_fS\alpha|\beta)=(\alpha|S^*T_f^*\beta),$$f(\alpha,S’\beta)=(T_f\alpha|S’\beta)=(\alpha|T_f^*S’\beta).$Hence it suffices to choose some $S’$ such that $S^*T_f^*=T_f^*S’$, that is $S’=(T_f^*)^{-1}S^*T_f^*$.
Indeed, such $S’$ is unique! Try to prove it.
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Which Expression Represents The Volume, In Cubic Units, Of The Composite Figure?
WILL GIVE BRAINLIEST Which expression represents the volume, in cubic units, of the composite figure? A. (8)(6)(6) + (8)(6)(4)B. (8)(6)(10) + (8)(6)(4)C. (8)(6)(4) – (8)(6)(6)D. (8)(6)(4) – (8)(6)(10)
Solution:
The figure is splitted into two shapes.
One is rectangular pyramid and the other is rectangular prism.
Rectangular pyramid:
Length = 8
Width = 6
Height = 10 – 4 = 6
Volume of rectangular pyramid =
Volume of rectangular pyramid
Rectangular prism:
Length = 8
Width = 6
Height = 4
Volume of rectangular prism = length × width × height
= 8 × 6 × 4
Volume of rectangular prism = (8)(6)(4)
Volume of figure = Volume of pyramid + volume of prism
The volume of the composite figure is cubic units.
“Let”s see if it works with the two consecutive integers 5 and 6:
let x = 5
so x + 1 = 6
5 + 6 = 11, so:
2x + 1 = 11
2x = 10 (subtracted 1 from both sides)
x = 5 (divided both sides by 2)
it works!”
represent the sequence 16,24,32,40…
Step-by-step explanation:
The sequence given is
16,24,32,40.
And the options given are
an=4n+1
an=8n+8
an=8n+16
We will check for all so as to find the correct expression.
In the given sequence the first term is 16,
i.e a1 = 16
1). We will check for an = 4n+1 whether the first term is 16 or not.
We will put the value of n as 1;
With this option we get the first term as 5 but in the given sequence first term is 16.
Hence this option is Incorrect.
2). We will check for an = 8n+8 whether the first term is 16 or not.
We will put the value of n as 1;
With this option we get the first term as 16 and in the given sequence also first term is 16.
Hence this option is correct.
3). We will check for an = 8n+16 whether the first term is 16 or not.
We will put the value of n as 1;
With this option we get the first term as 24 but in the given sequence first term is 16.
Hence this option is Incorrect.
Hence we can say that an = 8n+8 represent the sequence 16,24,32,40…
D. 6x³ + 5x² – 3x – 2
Step-by-step explanation:
First assign each variable with its number.
l = 2x + 1
w = x + 1
h = 3x – 2
Now, plug in.
v = (2x + 1)(x +1)(3x – 2)
Now combine the first two pairs following the FOIL method. (First Outer Inner Last)
(2x² + 3x + 1)(3x – 2)
Now you would multiply the 3x on to each term and the -2 on to each term.
6x³ + 9x² + 3x – 4x² – 6x – 2
Now combine like terms.
6x³ + 5x² – 3x – 2
p:20
p:25
Step-by-step explanation:
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Miscellaneous
Chapter 1 Class 11 Sets
Serial order wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Misc 6 Introduction Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B) Let U = {1, 2, 3, 4, 5} A = {1, 2} B = {2, 3, 4} A – B = A – (A ∩ B) = {1, 2} – {2} = {1} We use the result A – B = A ∩ B’ in this question Also, B’ = U – B = {1, 2, 3, 4, 5} – {2, 3, 4} = {1, 5} A – B = A ∩ B’ = {1, 2} ∩ {1, 5} = {1} Misc 6 Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B) To prove : A = (A ∩ B) ∪ (A – B) Solving R.H.S (A ∩ B) ∪ (A – B) Using A – B = A – (A ∩ B) = A ∩ B’ = (A ∩ B) ∪ (A ∩ B’) = A ∩ (B ∪ B’) ∪ Union - Combination of two sets ∩ Intersection - Common of two sets (Distributive law: A ∩ (B ∪ C)= (A ∩ B) ∪ (A ∩ C)) = A ∩ (U) = A = L.H.S Hence proved (As B ∪ B’ = U) (As A ∩ U = A ) To prove : A ∪ (B – A) = (A ∪ B) Taking L.H.S A ∪ (B – A) Using B – A = B – (A ∩ B) = B ∩ A’ = A ∪ (B ∩ A’) Using distributive law :A ∪ (B ∩ C)= (A ∪ B) ∩ (A ∪ C) = (A ∪ B) ∩ (A ∪ A’) = (A ∪ B) ∩ (U) = (A ∪ B) = R.H.S Hence proved (As A ∪ A’ = U )
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# UNIT RATE WORD PROBLEMS WORKSHEET
## About "Unit rate word problems worksheet"
Unit rate word problems worksheet is much useful to the students who would like to practice problems on "Unit rate"
First let us understand what is unit rate.
Unit rate definition (or) Definition of unit rate :
Unit rate compares the given amount to one unit of another measure
(or)
The ratio between the given value and 1
(or)
Comparing the given amount or value to 1
Examples :
1. If 8 dolls are made in 4 days,
then number of dolls made in 1 day = 2
2. If David earns \$180 in 9 hours,
then number of dollars earned by him in 1 hour = \$ 20
3. If there are 16 cups in 4 quarts,
then the number of cups in 1 quart = 4
## Liquid Measurements - Unit rates
From the above picture, we can get the following unit rates related liquid measurements.
1 gallon = 16 cups
1 gallon = 8 pints
1 gallon = 4 quarts
1 quart = 2 pints
1 quart = 4 cups
1 pint = 2 cups
## Unit rate word problems worksheet - Problems
Here we are going to look at unit rate word problems.
1. In a business, if A can earn \$ 7500 in 2.5 years, find the unit rate of his earning per month.
2. If David can prepare 2 gallons of juice in 4 days, how many cups of juice can he prepare per day ?
3. If John can cover 360 miles in 3 hours, find the number of miles covered by John in 1 minute.
4. 75 basketballs cost \$1,143.75. Find the unit rate in price per basketball.
5. In 36.5 weeks, Miguel raised \$2,372.50 for cancer research. How was his unit rate in price per week?
6. Shanel walks 2/ 5 of a mile every 1/7 hour. Express her speed as a unit rate in miles per hour.
7. Declan use 2 /35 of a gallon of gas for every 4 /5 of a mile that he drives. At this rate, how many miles can he drive on one gallon of gas?
8. A person can cover a distance of 84 miles in 4 gallons of fuel. If he has 2.5 gallons, how many miles can he cover ?
9. If a person drinks 8 cups of apple juice per month, how many gallons will he drink in one year?
10. My David earns \$416 in 8 hours. How much does earn in 2.8 hours ?
Here they are .
## Step by step solution
Problem 1 :
In a business, if A can earn \$ 7500 in 2.5 years, find the unit rate of his earning per month.
Solution :
Given : Earning in 2.5 years = \$ 7500
1 year = 12 months
2.5 years = 2.5 x 12 = 30 months
Then, earning in 30 months = \$ 7500
Therefore, earning in 1 month = 7500 / 30 = \$ 250
Hence, the unit rate of his earning per month is \$ 250
Let us look at the next problem on "Unit rate word problems worksheet"
Problem 2 :
If David can prepare 2 gallons of juice in 4 days, how many cups of juice can he prepare per day ?
Solution :
No of gallons of juice prepared in 4 days = 2 gallons
1 gallon = 16 cups
So, no. of cups of juice prepared in 4 days = 2 x 16 = 32 cups
Therefore, David can prepare 32 cups of juice in 4 days.
Then, no. of cups of juice prepared in 1 day = 32 / 4 = 8
Hence, David can prepare 8 cups of juice in 1 day.
Let us look at the next problem on "Unit rate word problems worksheet"
Problem 3 :
If John can cover 360 miles in 3 hours, find the number of miles covered by John in 1 minute.
Solution :
No of miles covered in 3 hours = 360
Then, no. of miles covered in 1 hour = 360 / 3 = 180
1 hour = 60 minutes
So, no. of miles covered in 60 minutes = 180
Then, no. of miles covered 1 minute = 180 / 60 = 3
Hence, John can cover 3 miles in 1 minute.
Let us look at the next problem on "Unit rate word problems worksheet"
Problem 4 :
75 basketballs cost \$1,143.75. Find the unit rate in price per basketball.
Solution :
Given : 75 basketballs cost \$1,143.75
Then, price pf one basket ball = 1143.75 / 75 = 15.25
Hence, the unit rate in price per basket ball is \$ 15.25
Let us look at the next problem on "Unit rate word problems worksheet"
Problem 5 :
In 36.5 weeks, Miguel raised \$2,372.50 for cancer research. How was his unit rate in price per week?
Solution :
Given : Miguel raised \$2, 372.50 in 36.5 weeks
Then, amount raised in one week = 2372.5 / 36.5 = 65
Hence, the unit rate in price per week was \$ 65
Let us look at the next problem on "Unit rate word problems worksheet"
Problem 6 :
Shanel walks 2/ 5 of a mile every 1/7 hour. Express her speed as a unit rate in miles per hour.
Solution :
Given : Shanel walks 2/ 5 of a mile every 1/7 hour
We know the formula for speed.
That is, Speed = Distance / time
Speed = (2/5) / (1/7)
Speed = (2/5) x (7/1)
Speed = 14 / 5
Speed = 2.8 miles per hour.
Hence, the speed of Shanel is 2.8 miles per hour
Let us look at the next problem on "Unit rate word problems worksheet"
Problem 7 :
Declan use 2 /35 of a gallon of gas for every 4 /5 of a mile that he drives. At this rate, how many miles can he drive on one gallon of gas?
Solution :
Given : In 2 /35 of a gallon of gas, 4 /5 of a mile is traveled
Then, in 1 gallon of gas = (4/5) x (35/2) miles traveled.
in 1 gallon of gas = 14 miles traveled.
Hence, Declan can drive 14 miles in 1 gallon of gas
Let us look at the next problem "Unit rate word problems worksheet"
Problem 8 :
A person can cover a distance of 84 miles in 4 gallons of fuel. If he has 2.5 gallons, how many miles can he cover ?
Solution :
Given : 84 miles can be traveled in 4 gallons.
Then, no. of miles traveled in 1 gallon = 84 / 4 = 21
Therefore, no. of miles traveled in 2.5 gallons = 21 x 2.5 = 52.5
Hence, 52.5 miles traveled in 2.5 gallons of fuel
Let us look at the next problem on "Unit rate word problems worksheet"
Problem 9 :
If a person drinks 8 cups of apple juice per month, how many gallons will he drink in one year?
Solution :
Given : 8 cups in one month
1 year = 12 months
So, no. of cups in 1 year = 8 x 12 = 96 cups
1 gallon = 16 cups
Therefore, no. of gallons in 1 year = 96 / 16 = 6
Hence, he will drink 6 gallons of apple juice in 1 year
Let us look at the next problem on "Unit rate word problems worksheet"
Problem 10 :
My David earns \$416 in 8 hours. How much does earn in 2.8 hours ?
Solution :
Given : Earning in 8 hours = \$ 416
Earning in 1 hour = \$ 52
Earning in 2.8 hours = 52 x 2.8 = 145.6
Hence, Mr. David will earn \$145.6 in 2.8 hours
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How do you solve and graph x^3<=4x^2+3x?
Jun 25, 2017
The solution is $x \in \left(- \infty , 2 - \sqrt{7}\right] \cup \left[0 , 2 + \sqrt{7}\right]$
Explanation:
We solve this inequality with a sign chart.
${x}^{3} \le 4 {x}^{2} + 3 x$
${x}^{3} - 4 {x}^{2} - 3 x \le 0$
$x \left({x}^{2} - 4 x - 3\right) \le 0$
We need the roots of the quadratic equation
${x}^{2} - 4 x - 3 = 0$
The discrimenant is
$\Delta = {b}^{2} - 4 a c = {\left(- 4\right)}^{2} - 4 \cdot \left(1\right) \cdot \left(- 3\right) = 16 + 12 = 28$
As, $\Delta > 0$, there are 2 real roots
${x}_{2} = \frac{- b + \sqrt{\Delta}}{2} = \frac{1}{2} \left(4 + \sqrt{28}\right) = 2 + \sqrt{7} = 4.646$
${x}_{1} = \frac{- b - \sqrt{\Delta}}{2} = \frac{1}{2} \left(4 + \sqrt{28}\right) = 2 - \sqrt{7} = - 0.646$
Let $f \left(x\right) = x \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$
We can build the sign chart
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$${x}_{1}$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a a a}$${x}_{2}$$\textcolor{w h i t e}{a a a a}$$+ \infty$
$\textcolor{w h i t e}{a a a a}$${x}_{1}$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$${x}_{2}$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$
Therefore,
$f \left(x\right) \le 0$ when $x \in \left(- \infty , 2 - \sqrt{7}\right] \cup \left[0 , 2 + \sqrt{7}\right]$
graph{x^3-4x^2-3x [-12.34, 12.97, -9.01, 3.65]}
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# Tangents and Limits
A tangent to a curve is a straight line that touches the curve at a single point but does not intersect it at that point. For example, in the figure to the right, the y-axis would not be considered a tangent line because it intersects the curve at the origin. A secant to a curve is a straight line that intersects the curve at two or more points.
In the figure to the right, the tangent line intersects the curve at a single point P but does not intersect the curve at P. The secant line intersects the curve at points P and Q.
The concept of limits begins with the tangent line problem. We want to find the equation of the tangent line to the curve at the point P. To find this equation, we will need the slope of the tangent line. But how can we find the slope when we only know one point on the line? The answer is to look at the slope of the secant line. It's slope can be determined quite easily since there are two known points P and Q. As you slide the point Q along the curve, towards the point P, the slope of the secant line will become closer to the slope of the tangent line. Eventually, the point Q will be so close to P, that the slopes of the tangent and secant lines will be approximately equal.
A limit of a function is written as :
We want to find the limit of f(x) as x approaches a. To do this, we try to make the values of f(x) close to the limit L, by taking x values that are close to, but not equal to, a. In short, f(x) approaches L as x approaches a.
# Examples
1 | Find the equation of the tangent line to the curve at point P
# Left and Right Hand Limits
The previous example shows that the value a can be approached from both the left and right sides. Each side has its own limit. For example, as x approaches a from the right side we have and as x approaches a from the left we have .
The graph to the right shows an example of a function with different right and left hand limits at the point x = 1. As x approaches 1 from the left side, the limit of f(x) approaches 1. As x approaches 1 from the right side however, the limit of f(x) approaches 4. In this case, the limit of f(x) as x approaches 1 does not exist, because the left and right hand limits do not approach the same value. This idea leads to the following theorem:
For the limit to exist, the left and right hand limits must approach the same value. In our example, as x approaches 3, the left and right hand limits both approach a value of 4. Since the left and right hand limits are the same, the limit of f(x) as x approaches 3 exists and is equal to 4. Even though the actual value of f(3) is equal to 2, the limit is equal to 4. This gives the following theorem:
# Examples
2 | Evaluate the limit, if it exists
Note: The example above uses basic concepts that are covered in the sections below.
# Infinite Limits
If a function is defined on either side of a, but the limit as x approaches a is infinity or negative infinity, then the function has an infinite limit. The graph of the function will have a vertical asymptote at a. A curve y=f(x) will have a vertical asymptote at x = a if any of the following conditions hold:
Note: When dealing with rational functions, often there is a vertical asymptote at values of x that make the denominator equal to 0. However, you must still show that one of the conditions above holds true to prove that there is a vertical asymptote at that point. For example, function f(x)=(x2-1)/(x-1) is undefined at x=1 but does not have a vertical asymptote at x=1. In fact, f(x) = x+1 for all x1.
# Examples
3 | Find the vertical asymptotes of the function
# Laws of Limits
Calculating limits using graphs and tables takes a lot of unnecessary time and work. Using the limit laws listed below, limits can be calculated much more quickly and easily.
From the limit laws above, comes the property of direct substitution. This property makes it possible to solve most rational and polynomial functions. The property of direct substitution states: For any rational or polynomial function f, if a is in the domain of f then
Often, the method of direct substitution cannot be used because a is not in the domain of f. In these cases, it is sometimes possible to factor the function and eliminate terms so that the function is defined at the point a. For an example of factoring, see example 6 below.
# Examples
4 | Evaluate the limit using limit laws
5 | Evaluate the limit using the property of direct substitution
6 | Evaluate the limit by factoring f(x) and eliminating terms
# The Squeeze Theorem
The squeeze theorem is an important concept that will be very helpful in upper year calculus courses. The squeeze theorem states:
In simpler terms, the squeeze theorem states that if the graph of g is squeezed between the graphs of f and h when x is near a, and if f and h have the same limit L as x approaches a, then the limit of g as x approaches a is also L. The graph to the right illustrates the squeeze theorem.
Note: The function g does not have to be completely contained between f and h. It must only be contained between f and h while x is near a. The graph illustrates where the functions f and g cross each other.
# Examples
7 | Evaluate the limit using the squeeze theorem
# Continuity and Discontinuity
A function f is continuous at a number a if:
Note: When proving that a function is continuous, you may only show that the limit of f(x) as x approaches a is equal to f(a). This property implies that f(a) is defined and that the limit exists.
A function is continuous on an interval if it is continuous at every number that falls within that interval. Continuous functions have the following properties for simple operations. If functions f and g are continuous at a, and c is a constant, then the following functions will also be continuous at a:
A function is discontinuous at a if it is defined near a but not continuous at a. To prove that a function is discontinous, we must show which of the requirements of continuous functions that it fails to hold for. There are several different types of discontinuity, which are listed below.
Removable discontinuity: A function has a removable discontinuity at a if the limit as x approaches a exists, but either f(a) is different from the limit or f(a) does not exist. It is called removable discontuniuity because the discontinuity can be removed by redefining the function so that it is continuous at a. In example #6 above, the function has a removable discontinuity at x = 3 because if the function is redefined so that f(3) = -4/7, it will be continuous at x = 3.
Infinite discontinuity: A function has an infinite discontinuity at a if the limit as x approaches a is infinite. In example #3 above, the function has an infinite discontinuity at every point a = k*pi, since each point has an infinite limit.
Jump discontinuity: A function has a jump discontinuity at a if the left- and right-hand limits as x approaches a exist, but are different. It is called jump discontinuity because the function jumps from the left-hand limit to the right-hand limit at each point. In example #2 above, the function has a jump discontinuity at x = 0, since the right and left hand limits approach different values.
Note: Polynomial functions are continuous everywhere. Rational, root and trigonometric functions are continuous at every number in their domain. In other words, they are continuous wherever they are defined.
# Examples
8 | Determine where the function is continuous
9 | Explain why the function is discontinuous at each point
# The Intermediate Value Theorem
The intermediate value theorem states: If f is a continuous function on the closed interval [a, b] and N is a number between f(a) and f(b), where f(a) does not equal f(b), then there exists a number c in interval (a, b) such that f(c) = N.
The intermediate value theorem simply says that if a function is continuous on a closed interval [a, b], and f(a) is different from f(b), then every value on the y-axis between f(a) and f(b) has a corresponding value on the x-axis between a and b.
Note: As the graph to the right illustrates, there can be more than one value c in (a, b) for a number N betwen f(a) and f(b). The graph shows that there exist 3 numbers, c0, c1 and c2, such that f(c0)=f(c1)=f(c2)=N.
The intermediate value theorem is helpful when proving that a root of a function exists in a certain interval.
# Examples
10 | Show that the function has a root between a and b
# Tangents and Limits (Revisited)
As explained at the beginning of this tutorial, a tangent to a curve is a line that touches the curve at a single point, P(a,f(a)). The tangent line T is the line through the point P with the slope :
given that this limit exists. The graph to the right illustrates how the slope of the tangent line is derived. The slope of the secant line PQ is given by f(x)-f(a)/x-a. As x approaches a, the slope of PQ becomes closer to the slope of the tangent line T. If we take the limit of the slope of the secant line as x approaches a, it will be equal to the slope of the tangent line T.
The slope of the tangent line becomes much easier to calculate if we consider the following conditions. If we let the distance between x and a be h, so that x=a+h, and substitute that equality for x in the slope formula, we get:
Note: Either of the limit formulas above can be used to find the slope. You will obtain the same answer using either formula.
These formulas have many practical applications. They can be used to find the instantaneous rates of change of variables. For example, if we use the formula above, the instantaneous velocity at time t=a is equal to the limit of f(a+h)-f(a)/h as h approaches 0.
# Examples
11 | Find the equation of the tangent line to the curve at the point P
12 | Find the instantaneous rate of change
For more practice with the concepts covered in the limits tutorial, visit the Limit Problems page at the link below. The solutions to the problems will be posted after the limits chapter is covered in your calculus course.
To test your knowledge of limits, try taking the general limits test on the iLrn website or the advanced limits test at the link below.
Limit Problems
General Limit Test on iLrn
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# Derivation of the Proof of the Area of a Rectangle
In getting the areas of figures, calculating the area of a square is probably the easiest. We can easily calculate the area of a square by squaring its side length . For example, a square with side length 5 units has area 25 square units. What about the area of a rectangle?
Based on the finding the area of a square, our intuition tells us that it is the product of the length and the width. However, how do we prove that it is indeed true? In this post, we are going to discuss the proof of the area of a rectangle. We show that it is the product of its length and its width.
Theorem: The area of a rectangle is the product of its length and width.
Consider the square below with side length $x + y$ units. The square is divided into four parts: two squares and two rectangles. We already know that the area of the two squares are $x^2$ and $y^2$. We do not know the area of the rectangle yet because that is what we are trying to prove.
Now, let $A$ be the area of each rectangle shown above. Clearly, the area of the largest square is the sum of the areas of the two smaller squares and the two rectangles. In equation form, we have
$(x + y)(x + y) = x^2 + y^2 + 2A$.
Expanding the left hand side, we have
$x^2 + 2xy + y^2 = x^2 + y^2 + 2A$.
Subtracting x^2 + y^2 from both sides results to
$2xy = 2A$.
Solving for $A$ gives us
$A = xy$.
But $x$ and $y$ are the length and width of the rectangle, therefore, the area of any rectangle is the product of its length and its width.
Reference: Geometry by Edwin E Moise and Floyd L. Downs, Jr.
## 7 thoughts on “Derivation of the Proof of the Area of a Rectangle”
1. Nice! Its helpful for some of us (maths teacher). We really have to show students where the area came from.
2. Thanks for your proof. But one thing pupil who aren’t aware (x+y)^2 at all how can we prove it?
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# Linear independence and dependence of vectors
I am really stuck in this problem, I have only 2 days to learn matrix's base, and its generator. My problem is that I know definitions but I don't understand intuitively what they mean.
What I know: base = vectors which generate the matrix. generator = vectors which generate matrix, but there are more than enough (unlike in base)
But what is linear independence and dependence? Can you please give me trivial examples where I distinctly see the difference between them. Why dependence? What does it depend on? Why do I call (in)dependent?
Thanks in tons..
• You use the set of generators to obtains any other vector. But sometimes you can obtain one vector from the generator set using the other vectors from the same set. this implies that the set is not composed only by linearly independent vectors. Feb 14, 2013 at 18:58
• @Sigur, thanks. but what is dependence here? how can i directly see whether it is dependent or independent? Feb 14, 2013 at 19:02
• Here is a good basic text: en.wikipedia.org/wiki/Linear_independence Feb 14, 2013 at 19:03
• @Sigur, you helped me!. now i know what to do! Feb 15, 2013 at 4:45
Do you remember the stories of treasure maps? We were told that to get to the treasure we needed to take five steps North and three steps East. We can make this mathematical. Let $\vec{n}$ be a step North and $\vec{e}$ a vector East. Then taking five steps North and then three steps East can be written $5\vec{n}+3\vec{e}$. Going South is negative North and going West is negative East: three steps South and then one step West can be written $-3\vec{n}-\vec{e}$.
The vectors $\vec{n}$ and $\vec{e}$ span the treasure map because you can get to any point on the map by going some steps North and some steps East. (Including "negative steps".) In linear algebra speak, the vectors $\vec{n}$ and $\vec{e}$ span the treasure map. Moreover, they are linearly independent: there is no combination of North/South steps that can make up for an East/West step. You need both choices of direction to be able to reach all of the points on the map.
Let me introduce a new movement: a diagonal step $\vec{d}$. (Imagine this as a movement to the North-East direction.) Sure, $\vec{n}$, $\vec{e}$ and $\vec{d}$ still span the treasure map; $\vec{n}$ and $\vec{e}$ did that without $\vec{d}$'s help. However, these are not linearly independent. Adding $\vec{d}$ does not help me to get to points on the map that I couldn't have otherwise got to. For example, a diagonal step can be done by making a partial step to the North and a partial step to the East. Mathematically:
$$\vec{d} = \frac{1}{\sqrt{2}}\vec{n} + \frac{1}{\sqrt{2}}\vec{e} \, .$$
Generally: a set of vectors span a space if a mixture of them allows you to "get to" each point of the space. They are linearly independent if you need all of them to do that, i.e. taking one away (e.g. not being able to move North) will stop you doing this. They are linearly dependent if you have more possibilities than you need, e.g. having a diagonal step when a small North and small East step will do the job.
• very nice example. now i got it. but i still dont know how to see this (in)dependence directly before calculating the vectors. Feb 15, 2013 at 4:35
• @doniyor Being able to "see" if some vectors are linearly independent or linearly dependent is very difficult. There are some big give-aways. If you have three vectors for a two dimensional space then clearly one of them is redundant (this is the definition of dimension). If one of the vectors is a multiple of one of the others then clearly the vectors are linearly dependent (adding 2$\vec{n}$ on top of $\vec{n}$ gives nothing new). In general, with lots of vectors, it's very hard. You have to solve a set of simultaneous equations. Feb 15, 2013 at 17:58
• @doniyor If you have, say, five vectors and each vector has, say, eight, components, then put them in to a five-by-eight matrix and find its rank. If the rank is five then they are linearly independent. I can't do this by eye. I have to do some maths. Have a good read of en.wikipedia.org/wiki/Rank_%28linear_algebra%29 Feb 15, 2013 at 18:05
Example of linear dependence: If $\overrightarrow{a}=\left(1,1,1\right)$, $\overrightarrow{b}=\left(2,3,4\right)$ and $\overrightarrow{c}=\left(4,5,6\right)$, then $2\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}$. One vector can be expressed as linear combination of others, so these three vectors are linearly dependent ("value of one vector depends on values of others").
Example of linear independence: $\overrightarrow{a}=\left(1,1,2\right)$, $\overrightarrow{b}=\left(2,3,4\right)$ and $\overrightarrow{c}=\left(4,5,6\right)$ - there no vector can be expressed as linear combination of other vectors.
Also... I suppose you meant basis matrix which is formed from linearly independent vectors, using them as columns. (See more in Wiki link, given by Sigur.)
My way to think about independence is as follows:
Given the set of vectors $S = \{a,b,c\}$ Then $S$ is linearly independent when: $ma + nb + qc = 0$ if and only if the coefficients $m, n, q$ are all zero.
In other words, one of vectors $a, b$, or $c$ can't be expressed as a linear combination of the remaining vectors.
Or if we want to say this in the matrix interpretation, independent means we only have trivial solution for the coefficients.
** A simple example is the standard basis of R3: Let $S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}$
(1) Then, if I use the "number" interpretation, I should set $k1(1,0,0) + k2(0,1,0) + k3(0,0,1) = (0,0,0)$ where k1,k2,k3 are coefficients. Then by playing around with the system of equations, we see that $k1=k2=k3=0$ is the only way to make the above equation work
(2) In matrix notation, my matrix representation for the coefficients is a $3\times 3$ matrix, where the $1$'s indicate how many copies of coefficient $k1, k2$, or k3 is needed. The last column contains all 0's. This is exactly similar to what I said in (1), but it's only that we have a column vector of all zeros instead of a row vector of all zeros
$$\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right]$$
Clearly by reading off the rows, we see $(0,0,0)$ is the only valid solution, and this solution is trivial.
So if you want to know where does the "independence" really happen, I guess the the word implies that the coefficients have no relation with the entries of the vectors. I don't care what numbers are in the vectors $a,b,c$. As long as I can set them into a linear combination, set it equal to the zero vector, and get zero coefficients, then even if my vectors $a,b,c$ are something crazy like (chair, cat, house), I still can exact information.
I hope this helps.
• great, but at the end, you are calling $a,b,c$ as vectors in contrast to your 1). this confused me a bit Feb 15, 2013 at 4:30
• @doniyor I change the letters to make things clear. Thanks Feb 16, 2013 at 5:32
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# geometric linear transformation find the angle
T is a reflection on the $x$ axis and after a reflection on the line $y = x$
Show that T is a rotation and give the angle.
So my matrix of transformation would be $$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$
and I know that a matrix of rotation is $$\begin{bmatrix} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{bmatrix}$$
By logic I know that the angle would be $90^\circ$, but there is an approach of how can I demonstrate it? And how to show that T is a rotation?
Thank you
When linear transformations are expressed through the same matrix, they are the same. In your case, you showed that the matrix $$A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$ corresponds to
1. reflection around the $x$-axis followed by reflection around $y=x$;
2. rotation of $90^\circ$.
If you are asking about why these come out the same, think about what happens to the point $(x,y)$ under each one geometrically. The first way flips it around $x$ axis to map $(x,y) \mapsto (x,-y)$. Then you flip around $x=y$ which basically exchanges the coordinates, you your final map is $$(x,y) \mapsto (x, -y) \mapsto (-y, x).$$ Meanwhile, directly rotating $(x,y)$ by $90^\circ$ clockwise does exactly the same thing. To convince yourself of that, note that you map $$(x,0) \mapsto (0,x) \text{ and } (0,y) \mapsto (-y, 0)$$ and since the transformation is linear, $$T(x,y) = T(x,0) + T(0,y) = (0,x) + (-y,0) = (-y,x)$$ as desired.
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# How do you find a power series representation for $\ln \left( 1-{{x}^{2}} \right)$ and what is the radius of convergence?
Last updated date: 20th Jun 2024
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Hint: In mathematics, a power series (in one variable) is an infinite series. Power series are useful in mathematical analysis, where they arise as Taylor series of infinitely differentiable functions. To solve this question, we should know how to write the power series/ MacLaurin series of a function of form $\ln \left( 1+x \right)$. The power series representation for $\ln \left( 1+x \right)$ is $\sum\limits_{n=0}^{\infty }{\dfrac{{{(-1)}^{n}}{{x}^{n+1}}}{n+1}}$. Also to find the radius of convergence, use the ratio test which states that if $\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|<1$, then $\sum\limits_{n=0}^{\infty }{{{a}_{n}}}$ converges.
Complete step by step solution:
We are asked to express the power series expansion of $\ln \left( 1-{{x}^{2}} \right)$, and to find the radius of convergence. We know that power series representation for $\ln \left( 1+x \right)$ is $\sum\limits_{n=0}^{\infty }{\dfrac{{{(-1)}^{n}}{{x}^{n+1}}}{n+1}}$.
To find the representation for $\ln \left( 1-{{x}^{2}} \right)$, we substitute $-{{x}^{2}}$ at the place of x in the summation formula, by doing this we get
$\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{\left( -{{x}^{2}} \right)}^{n+1}}}{n+1}}$
Using the exponential property ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$, the above expression can also be written as
$\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}{{\left( -1 \right)}^{n+1}}{{\left( {{x}^{2}} \right)}^{n+1}}}{n+1}}$
Again using the property ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$ in the reverse direction this time, we can simplify the above expression as
$\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{2n+1}}{{\left( x \right)}^{2n+2}}}{n+1}}$
As 2n+1 is an odd number, ${{\left( -1 \right)}^{2n+1}}$ will always be $-1$. Thus, we get
$\Rightarrow \sum\limits_{n=0}^{\infty }{\dfrac{-{{\left( x \right)}^{2n+2}}}{n+1}}$
To find the radius of convergence, use the ratio test which states that if $\displaystyle \lim_{x \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|<1$, then $\sum\limits_{n=0}^{\infty }{{{a}_{n}}}$ converges
$\Rightarrow \displaystyle \lim_{x \to \infty }\left| \dfrac{\dfrac{-{{\left( x \right)}^{2\left( n+1 \right)+2}}}{\left( n+1 \right)+1}}{\dfrac{-{{\left( x \right)}^{2n+2}}}{n+1}} \right|<1$
Simplifying the above expression, we get
$\Rightarrow \displaystyle \lim_{x \to \infty }\left| \dfrac{-{{\left( x \right)}^{2n+4}}}{n+2}\times \dfrac{n+1}{-{{\left( x \right)}^{2n+2}}} \right|<1$
$\Rightarrow \left| {{x}^{2}} \right|<1$
Solving the above inequality, we get
$\Rightarrow -1< x <1$
Thus, the convergence radius of the given expression is 1.
Note:
To solve these types of problems, one should know the expansions of different types of series, for example, the Taylor series. The Maclaurin series is a special case of the Taylor series. Calculation mistakes should be avoided.
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# How do you use the product to sum formulas to write cos2thetacos4theta as a sum or difference?
Feb 17, 2017
$= \frac{\cos 6 t + \cos 2 t}{2}$
#### Explanation:
Use trig identity:
$\cos a + \cos b = 2 \cos \left(\frac{a - b}{2}\right) . \cos \left(\frac{a + b}{2}\right)$
In this case, we have:
$\frac{a - b}{2} = 2$ --> a - b = 4 (1)
$\frac{a + b}{2} = 4$ --> a + b = 8 (2)
Solving the system of equation (1) and (2), we get:
2a = 4 + 8 = 12 --> $a = 6$
$b = a - 4 = 2$
There for:
$\frac{\cos \left(6 t\right) + \cos \left(2 t\right)}{2} = \cos \left(2 t\right) . \cos \left(4 t\right)$
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1. Chapter 4 Class 12 Determinants
2. Serial order wise
3. Examples
Transcript
Example 24 If A = 1 3 3 1 4 3 1 3 4 , then verify that A adj A = |A| I. Also find A 1. Taking L.H.S A (adj A) First Calculating adj A adj A = A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 Now, A = 1 3 3 1 4 3 1 3 4 M11 = 4 3 3 4 = 4(4) 3(3) = 7 M12 = 1 3 1 4 = 1(4) 1(3) = 1 M13 = 1 4 1 3 = 1(3) 1(4) = 1 M21 = 3 3 3 4 = 3(4) 3(3) = 3 M22 = 1 3 1 4 = 1(4) 1(3) = 1 Thus, adj (A) = A 11 A 21 A 31 A 12 A 22 A 32 A 13 A 23 A 33 = 7 3 3 1 1 0 1 0 1 Finding A (adj A) A adj (A) = 1 3 3 1 4 3 1 3 4 7 3 3 1 1 0 1 0 1 = 1 7 +3 1 +3( 1) 1 3 +3 1 +3(0) 1 3 +3 0 +3(1) 1 7 +4 1 +3( 1) 1 3 +4 1 +3(0) 1 3 +4 0 +3(1) 1 7 +3 1 +4( 1) 1 3 +3 1 +4(0) 1 3 +3 0 +4(1) = 7 3 3 3+3+0 3+0+3 7 4 3 3+4+0 3+0+3 7 3 4 3+3+0 3+0+4 = 1 0 0 0 1 0 0 0 1 Taking R.H.S |A| I Calculating |A| |A| = 1 3 3 1 4 3 1 3 4 = 1 (4(4) 3(3)) 3(1(4) 1(3)) + 3(1(4) 1(3)) = 1(7) 3(1) +3( 1) = 7 3 3 = 1 Now, |A| I = 1 1 0 0 0 1 0 0 0 1 = 1 0 0 0 1 0 0 0 1 = L.H.S Thus, A(adj A) = |A| I Hence proved Finding A-1 We know that A-1 = 1 |A| (adj A) exists if |A| 0 Here, |A| = 1 0 Thus A-1 exists So, A-1 = 1 |A| (adj A) = 1 1 7 3 3 1 1 0 1 0 1 =
Examples
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# How to Find Domain and Range of Relation
Understanding the domain and range of a relation is crucial in mathematics, particularly when delving into functions. The domain consists of all permissible input values (typically the $$x$$-values), while the range encompasses all possible output values (usually the $$y$$-values). In relation to higher complexity and variation, these concepts may seem daunting, but with a systematic approach, it becomes more manageable.
## Step-by-Step Guide to Find Domain and Range of a Relation
Here is a step-by-step guide to finding the domain and range of a relation:
### Step 1: Distinguish the Relation Type
• Determine whether your relation is described by a table, a graph, an equation, or a written description. The method for finding the domain and range may vary based on this.
### Step 2: For Tabulated Relations (Ordered Pairs)
• Domain: List out all the unique $$x$$-values.
• Range: List out all the unique $$y$$-values.
### Step 3: For Graphical Representations
• Domain: Identify the leftmost and rightmost points of the graph. This gives the range of $$x$$-values the relation spans.
• Range: Identify the highest and lowest points of the graph to determine the $$y$$-values the relation encompasses.
### Step 4: For Equations
• Rational Functions: For fractions, the denominator should never be zero. Solve for values of $$x$$ that would make the denominator zero and exclude them from the domain.
• Radical Functions: For even roots, the radicand (value inside the root) should be non-negative. Determine the $$x$$-values for which the radicand is non-negative.
• Logarithmic Functions: The argument of the log must be greater than zero.
• Domain: Solve for $$x$$-values taking into account the restrictions from the above function types.
• Range: Depending on the function type, deducing the range might be straightforward or require deeper analysis.
### Step 5: Consider Restrictions
• Mathematical restrictions (like those mentioned in the equations section).
• Physical or real-world constraints (e.g., a relation describing time and growth might have a domain limited to positive values).
### Step 6: State the Domain and Range in Proper Notation
• Interval Notation: Uses brackets and parentheses to describe intervals of values. For example, $$[a,b]$$ denotes all values between a and b, inclusive.
• Set-builder Notation: Defines a set by specifying its properties. For instance, {$$x∣x>2$$} represents all $$x$$-values greater than $$2$$.
### Step 7: Examine for Implicit Restrictions
• Sometimes, the domain and range might be restricted by the context or the nature of the relation, even if not explicitly mentioned.
### Step 8: Cross-verify Using Multiple Methods:
• Whenever possible, approach the problem using different techniques. For instance, if you’ve derived the domain from an equation, try sketching a graph to visually verify the domain.
### Step 9: Complex Relations and Composite Functions
• For relations involving multiple functions or operations (like the composition of functions), determine the domain and range of each part individually before combining them.
### Step 10: Iterative Reflection and Refinement
• Reevaluate your conclusions. Sometimes, upon deeper reflection or by considering another perspective, you might spot oversights or nuances you missed in your initial analysis.
By meticulously following these steps and maintaining a comprehensive approach, the intricate task of determining the domain and range of a relation, even one of high complexity, can be systematically and effectively tackled.
### Examples:
Example 1:
Consider the relation $$S=$${$$(4,6),(5,7),(6,8),(7,6)$$}.
Determine the domain and range of the relation.
Solution:
To determine the domain, list all the $$x$$-values from the ordered pairs: $$4,5,6,7$$
So, the domain of relation $$S$$ is {$$4, 5, 6, 7$$}.
Next, to determine the range, list all the $$y$$-values from the ordered pairs: $$6,7,8,6$$
After removing the duplicates, the range of relation $$S$$ is {$$6,7,8$$}.
Example 2:
Consider the relation $$T=$${$$(−2,0),(0,−1),(2,1),(3,0)$$}.
Determine the domain and range of the relation.
Solution:
To determine the domain, list all the $$x$$-values from the ordered pairs: $$−2,0,2,3$$
So, the domain of relation $$T$$ is {$$-2, 0, 2, 3$$}.
Next, to determine the range, list all the $$y$$-values from the ordered pairs: $$0,−1,1,0$$
After removing the duplicates, the range of relation $$T$$ is {$$-1, 0, 1$$}.
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# How to write quadratic function given the following information?
## An apartment complex has 1600 units available of which 800 are currently rented for 300 dollars per month. A market survey indicated that a each five dollar decrease in monthly rent will result in 20 new renters. Write a function that models the monthly income, where x is the number of five dollar decreases in monthly rent. Then find the rent which yields the maximum monthly income, and what that income is.
Feb 20, 2018
$y = - 100 {x}^{2} - 4000 x + 246000$
#### Explanation:
Well first off, you have to figure out how to write all of the terms in terms of $x$. So here, there are two terms to figure out: the amount of money your current renters will pay you, plus the amount of rent your new renters will pay you. However, before we do this, we have to figure out a function for the rent, because it is going to change.
This part is pretty straightforward. The rent is going to start at $300 and continue to go down by $5 each time there is a decrease. So we get
$300 - 5 x$
as the rent with each decrease. This is because you start at $300 and take away $5 each time there is a decrease. This makes the next part fairly simple.
So now, we have to find out how many renters we have at a given time. So the function is going to be your current renters, plus the amount of new renters you will get. Since the original $800$ renters is always constant, it will not have a variable in our function.
However, you will get new $20$ new renters with each decrease, so this will be $20 x$ in the function. So our function for how many renters we have at any given time is
$800 + 20 x$
Now, finally, from logic, we know that the amount of renters at a given time $\left(800 + 20 x\right)$ times the rent at a given time $\left(300 - 5 x\right)$, because each person pays the rent, is equal to the total profit.
$y = \left(800 + 20 x\right) \left(300 - 5 x\right)$
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# 1033 and Level 3
To solve a level 3 puzzle begin with 80, the clue at the very top of the puzzle. Clue 48 goes with it. What are the factor pairs of those numbers in which both factors are between 1 and 12 inclusive? 80 can be 8×10, and 48 can be 4×12 or 6×8. What is the only number that listed for both 80 and 48? Put that number in the top row over the 80. Put the corresponding factors where they go starting at the top of the first column.
Work down that first column cell by cell finding factors and writing them as you go. Three of the factors have been highlighted because you have to at least look at the 55 and the 5 to deal with the 20 in the puzzle. Have fun!
Print the puzzles or type the solution in this excel file: 12 factors 1028-1034
Here are a few facts about the number 1033:
It is a twin prime with 1031.
32² + 3² = 1033, so it is the hypotenuse of a Pythagorean triple:
192-1015-1033 calculated from 2(32)(3), 32² – 3², 32² + 3²
1033 is a palindrome in two other bases:
It’s 616 in BASE 13 because 6(13²) + 1(13) + 6(1) = 1033
1J1 in BASE 24 (J is 19 base 10) because 24² + 19(24) + 1 = 1033
8¹ + 8º + 8³ + 8³ = 1033 Thanks to Stetson.edu for that fun fact!
• 1033 is a prime number.
• Prime factorization: 1033 is prime.
• The exponent of prime number 1033 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 1033 has exactly 2 factors.
• Factors of 1033: 1, 1033
• Factor pairs: 1033 = 1 × 1033
• 1033 has no square factors that allow its square root to be simplified. √1033 ≈ 32.1403
How do we know that 1033 is a prime number? If 1033 were not a prime number, then it would be divisible by at least one prime number less than or equal to √1033 ≈ 32.1. Since 1033 cannot be divided evenly by 2, 3, 5, 7, 3, 13, 17, 19, 23, 29 or 31, we know that 1033 is a prime number.
Here’s another way we know that 1033 is a prime number: Since its last two digits divided by 4 leave a remainder of 1, and 32² + 3² = 1033 with 32 and 3 having no common prime factors, 1033 will be prime unless it is divisible by a prime number Pythagorean triple hypotenuse less than or equal to √1033 ≈ 32.1. Since 1033 is not divisible by 5, 13, 17, or 29, we know that 1033 is a prime number.
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# A balloon with a volume of 5.3 L is taken from an indoor temperature of 24 C to the outdoors. The volume of the balloon outside is 4.9 L. How do you determine the C temperature outside?
Aug 31, 2016
The outside temperature is 0 °C.
#### Explanation:
Given
The volume ${V}_{1}$ of a gas at a temperature ${T}_{1}$.
A second volume ${V}_{2}$
Find
The second temperature ${T}_{2}$
Strategy
A problem involving two gas volumes and two temperatures must be a Charles' Law problem.
The formula for Charles' Law is
$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {V}_{1} / {T}_{1} = {V}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$
Solution
We can rearrange Charles' Law to get
T_2 = T_1 × V_2/V_1
${V}_{1} = \text{5.3 L}$; ${T}_{1} = \left(24 + 273.15\right) K = \text{297.15 K}$
${V}_{2} = \text{4.9 L}$; ${T}_{2} = \text{?}$
${T}_{2} = \text{297.15 K" × (4.9 cancel("L"))/(5.3 cancel("L")) = "274 K}$
$\text{200 K" = "(274 - 273.15) °C" = "0 °C}$
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# Lesson 5
Triangles in Circles
• Let’s see how perpendicular bisectors relate to circumscribed circles.
### 5.1: One Perpendicular Bisector
The image shows a triangle.
1. Construct the perpendicular bisector of segment $$AB$$.
2. Imagine a point $$D$$ placed anywhere on the perpendicular bisector you constructed. How would the distance from $$D$$ to $$A$$ compare to the distance from $$D$$ to $$B$$? Explain your reasoning.
### 5.2: Three Perpendicular Bisectors
1. Construct the perpendicular bisector of segment $$BC$$ from the earlier activity. Label the point where the 2 perpendicular bisectors intersect as $$P$$.
2. Use a colored pencil to draw segments $$PA,PB,$$ and $$PC$$. How do the lengths of these segments compare? Explain your reasoning.
3. Imagine the perpendicular bisector of segment $$AC$$. Will it pass through point $$P$$? Explain your reasoning.
4. Construct the perpendicular bisector of segment $$AC$$.
5. Construct a circle centered at $$P$$ with radius $$PA$$.
6. Why does the circle also pass through points $$B$$ and $$C$$?
Points $$A,B,$$ and $$C$$ are graphed. Find the coordinates of the circumcenter and the radius of the circumscribed circle for triangle $$ABC$$.
### 5.3: Wandering Centers
Move the vertices of triangle $$ABC$$ and observe the resulting location of the triangle’s circumcenter, point $$D$$. Determine what seems to be true when the circumcenter is in each of these locations:
1. outside the triangle
2. on one of the triangle’s sides
3. inside the triangle
### Summary
We saw that some quadrilaterals have circumscribed circles. Is the same true for triangles? In fact, all triangles have circumscribed circles. The key fact is that all points on the perpendicular bisector of a segment are equidistant from the endpoints of the segment.
Suppose we have triangle $$ABC$$ and we construct the perpendicular bisectors of all 3 sides. These perpendicular bisectors will all meet at a single point called the circumcenter of the triangle (label it $$D$$). This point is on the perpendicular bisector of $$AB$$, so it’s equidistant from $$A$$ and $$B$$. It’s also on the perpendicular bisector of $$BC$$, so it’s equidistant from $$B$$ and $$C$$. So, it is actually the same distance from $$A,B,$$ and $$C$$. We can draw a circle centered at $$D$$ with radius $$AD$$. The circle will pass through $$B$$ and $$C$$ too because the distances $$BD$$ and $$CD$$ are the same as the radius of the circle.
In this case, the circumcenter happened to fall inside triangle $$ABC$$, but that does not need to happen. The images show cases where the circumcenter is inside a triangle, outside a triangle, and on one of the sides of a triangle.
### Glossary Entries
• circumcenter
The circumcenter of a triangle is the intersection of all three perpendicular bisectors of the triangle’s sides. It is the center of the triangle’s circumscribed circle.
• circumscribed
We say a polygon is circumscribed by a circle if it fits inside the circle and every vertex of the polygon is on the circle.
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Dnt know how to work these out:
$\displaystyle \sqrt{\frac{2}{5}x + 3} - \frac{4}{5} = \frac{2}{5}$
$\displaystyle 6 + \sqrt{7x+3} = 6$
2. $\displaystyle \sqrt{\frac{2}{5}x + 3} - \frac{4}{5} = \frac{2}{5}$
Add $\displaystyle \frac{4}{5}$ to both sides of the equation:
$\displaystyle \sqrt{\frac{2}{5}x + 3} = \frac{6}{5}$
Square both sides of the equation:
$\displaystyle \frac{2}{5}x + 3 = \frac{36}{25}$
Subtract 3 from both sides:
$\displaystyle \frac{2}{5}x = -\frac{39}{25}$
Multiply both sides by 5:
$\displaystyle 2x = -\frac{195}{25}$
Divide both sides by 2:
$\displaystyle x = -\frac{195}{50} = -\frac{39}{10}$
3. $\displaystyle 6 + \sqrt{7x+3} = 6$
Subtract 6 from both sides:
$\displaystyle \sqrt{7x+3} = 0$
Square both sides:
$\displaystyle 7x+3 = 0$
Subtract 3 from both sides:
$\displaystyle 7x = -3$
Divide both sides by 7:
$\displaystyle x = -\frac {3}{7}$
4. So on a problem like this: $\displaystyle 4- \sqrt{2x+3} = 5$ do I add 4 to both sides?
And on this one, I add 2 to both sides then subtract 5? Not sure since there isn't an addition sign after it like the problem before. $\displaystyle 5 \sqrt{\frac{4x}{3}} - 2 = 0$
5. So on a problem like this: $\displaystyle 4- \sqrt{2x+3} = 5$ do I add 4 to both sides?
No. If you added 4 to both sides you would not simplify anything at all. Try rearranging the equation first like this:
$\displaystyle - \sqrt{2x+3} + 4 = 5$
Now do you see that you have to subtract 4 from both sides?
6. Originally Posted by Belanova
So on a problem like this: $\displaystyle 4- \sqrt{2x+3} = 5$ do I add 4 to both sides? Mr F says: Noooo. You >subtract< 4 from both sides.
And on this do I subtract? Not sure since there isn't an addition sign after it like the problem before. $\displaystyle 5 \sqrt{\frac{4x}{3}} - 2 = 0$ Mr F says: You >add< 2 to both sides.
..
7. And on this one, I add 2 to both sides then subtract 5? Not sure since there isn't an addition sign after it like the problem before. $\displaystyle 5 \sqrt{\frac{4x}{3}} - 2 = 0$
You would add 2 to both sides, but instead of subtracting 5 you would divide both sides by 5.
8. You try to isolate the square root expression all by itself as seen in your previous two questions. That way, you can simply square both sides without having to go through too much algebra.
For the first one:
$\displaystyle 4 - \sqrt{2x + 3} = 5$
$\displaystyle \sqrt{2x + 3} = -1$
You can square both sides much more easily than if you did it to the first line.
For the second one, again isolate the square root expression to one side so you can square it.
9. $\displaystyle 4- \sqrt{2x+3} = 5$
$\displaystyle = -1 = no solution$
$\displaystyle 5 \sqrt{\frac{4x}{3}} - 2 = 0$
$\displaystyle = \frac{3}{25}$
10. $\displaystyle 4- \sqrt{2x+3} = 5 = -1 = no solution$
No not quite. To get rid of a square root, you do not take the square root of it, you square it. So, instead of taking the square root of -1, square -1.
Then you have that:
$\displaystyle 2x+3 = 1$
Subtract 3 from both sides:
$\displaystyle 2x = -2$
Divide both sides by 2:
$\displaystyle x = -1$
11. $\displaystyle 5 \sqrt{\frac{4x}{3}} - 2 = 0 = \frac{3}{25}$
Yes, that is correct.
12. Originally Posted by topher0805
No not quite. To get rid of a square root, you do not take the square root of it, you square it. So, instead of taking the square root of -1, square -1.
Then you have that:
$\displaystyle 2x+3 = 1$
Subtract 3 from both sides:
$\displaystyle 2x = -2$
Divide both sides by 2:
$\displaystyle x = -1$
You made a mortal mistake there. After each step, check the domain and the range! Especially after taking the square of both sides!
You say that,
$\displaystyle 4 -\sqrt{2x+3} = 5$
$\displaystyle \sqrt{2x+3} = -1$
We could just take the square of both sides and go on and find x=-1.. If $\displaystyle \sqrt{2x+3}$ hadn't had to be equal to or greater than zero..
You probably made this mistake because you didn't notice that $\displaystyle \sqrt{x}$ is not the inverse function of $\displaystyle x^2$.
$\displaystyle \sqrt{x}$ is never negative.
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# Sam ran 63,756 feet in 70 minutes. What is Sam's rate in miles per hour?
##### 2 Answers
Aug 6, 2016
Sam's rate is $\left(10.35 \text{miles")/(hour}\right)$
#### Explanation:
Let's break the answer down into three parts:
color (magenta) ("First," we're going to convert feet to miles using this conversion factor:
color(white)(aaaaaaaaaaaaaaaaa 1 mile = 5,280 feet
color (blue) ("Then," we're going to convert minutes to hours using the relationship below:
color(white)(aaaaaaaaaaaaaaaaa 1 hour = 60 minutes
color (red) ("Finally," we are going to divide the value that we get in miles by the value that we get in hours, since the word "per" means to divide.
color (magenta) ("Step 1:"
63,756 cancel"feet"xx("1mile"/("5,280"cancel"feet")) = $12.075 \text{miles}$
color (blue) ("Step 2:"
70 cancel"minutes"xx((1"hour")/(60cancel"minutes")) $= 1.1667 \text{hours}$
color (red) ("Step 3:"
$\left(12.075 \text{miles")/(1.167 "hours}\right)$ $= \left(10.35 \text{miles")/(hour}\right)$
Aug 6, 2016
Sam's speed is about 10.3 mi/h.
#### Explanation:
Use dimensional analysis.
Determine the equality between miles and feet, and hours and minutes.
$\text{1 mi=5280 ft}$
$\text{1 h=60 m}$
Each equality can make two conversion factors, which are equal to one.
$\text{1 mi"/"5280 ft"="1"="5280 ft"/"1mi}$
$\text{1 h"/"60 min"="1"="60 min"/"1 h}$
Multiply the given value by the conversion factor that has the desired unit in the numerator. This will leave you with the desired unit, and the undesired unit in the denominator will cancel.
Convert feet to miles.
$63756 \cancel{\text{ft"xx(1"mi")/(5280cancel"ft")="12.075 mi}}$
Convert minutes to hours.
$70 \cancel{\text{min"xx(1"h")/(60cancel"min")="1.167 hour}}$
Divide miles by hours.
(12.075"mi")/(1.167"h")="10.3 mi/h"
Sam's speed is about 10.3 mi/h.
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# Government Standards
## Common Core
Assessment Exam - Common Core Math - High School: Number and Quantity
The Real Number System eTAP Lesson
Extend the properties of exponents to rational exponents.
Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 51/3 to be the cube root of 5 because we want (51/3)3 = 5(1/3)3 to hold, so (51/3)3 must equal 5.
CCSS.Math.Content.HSN.RN.A.1
Real Numbers
Square Roots
Rewrite expressions involving radicals and rational exponents using the properties of exponents.
CCSS.Math.Content.HSN.RN.A.2
Simplifying Square Roots
Use properties of rational and irrational numbers.
Explain why the sum or product of two rational numbers is rational; that the sum of a rational number and an irrational number is irrational; and that the product of a nonzero rational number and an irrational number is irrational.
CCSS.Math.Content.HSN.RN.B.3
Quantities eTAP Lesson
Reason quantitatively and use units to solve problems.
Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.
CCSS.Math.Content.HSN.Q.A.1
Word Problems of Measurements & Four Operations
Define appropriate quantities for the purpose of descriptive modeling.
CCSS.Math.Content.HSN.Q.A.2
Cubic Units for Volume
Interpreting Categorical and Quantitative Data eTAP Lesson
Summarize, represent, and interpret data on two categorical and quantitative variables
Summarize categorical data for two categories in two-way frequency tables. Interpret relative frequencies in the context of the data (including joint, marginal, and conditional relative frequencies). Recognize possible associations and trends in the data.
CCSS.Math.Content.HSS.ID.B.5
Data Classification, Range, and Midrange
Quantities eTAP Lesson
Reason quantitatively and use units to solve problems.
Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.
CCSS.Math.Content.HSN.Q.A.3
The Complex Number System eTAP Lesson
Perform arithmetic operations with complex numbers.
Know there is a complex number i such that i2 = -1, and every complex number has the form a + bi with a and b real.
CCSS.Math.Content.HSN.CN.A.1
Complex Numbers
Use the relation i2 = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.
CCSS.Math.Content.HSN.CN.A.2
Commutative and Associative Property
Find the conjugate of a complex number; use conjugates to find moduli and quotients of complex numbers.
CCSS.Math.Content.HSN.CN.A.3
Represent complex numbers and their operations on the complex plane.
Represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number.
CCSS.Math.Content.HSN.CN.B.4
Complex Plane
Represent addition, subtraction, multiplication, and conjugation of complex numbers geometrically on the complex plane; use properties of this representation for computation. For example, (-1 + √3 i)3 = 8 because (-1 + √3 i) has modulus 2 and argument 120°.
CCSS.Math.Content.HSN.CN.B.5
Calculate the distance between numbers in the complex plane as the modulus of the difference, and the midpoint of a segment as the average of the numbers at its endpoints.
CCSS.Math.Content.HSN.CN.B.6
Complex Numbers
Use complex numbers in polynomial identities and equations.
Solve quadratic equations with real coefficients that have complex solutions.
CCSS.Math.Content.HSN.CN.C.7
Extend polynomial identities to the complex numbers. For example, rewrite x2 + 4 as (x + 2i)(x - 2i).
CCSS.Math.Content.HSN.CN.C.8
Multiplying and Dividing Polynomials
Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials.
CCSS.Math.Content.HSN.CN.C.9
De Moivre’s Theorem
Vector and Matrix Quantities eTAP Lesson
Represent and model with vector quantities.
Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g., v, |v|, ||v||, v).
CCSS.Math.Content.HSN.VM.A.1
Two Dimensional Vectors
Find the components of a vector by subtracting the coordinates of an initial point from the coordinates of a terminal point.
CCSS.Math.Content.HSN.VM.A.2
Solve problems involving velocity and other quantities that can be represented by vectors.
CCSS.Math.Content.HSN.VM.A.3
One Dimensional Motion Problems (Newton's Second Law)
Perform operations on vectors.
Add vectors end-to-end, component-wise, and by the parallelogram rule. Understand that the magnitude of a sum of two vectors is typically not the sum of the magnitudes.
CCSS.Math.Content.HSN.VM.B.4.a
Given two vectors in magnitude and direction form, determine the magnitude and direction of their sum.
CCSS.Math.Content.HSN.VM.B.4.b
Two Dimensional Trajectory Problems
Understand vector subtraction v - w as v + (-w), where -w is the additive inverse of w, with the same magnitude as w and pointing in the opposite direction. Represent vector subtraction graphically by connecting the tips in the appropriate order, and perform vector subtraction component-wise.
CCSS.Math.Content.HSN.VM.B.4.c
Represent scalar multiplication graphically by scaling vectors and possibly reversing their direction; perform scalar multiplication component-wise, e.g., as c(vx, vy) = (cvx, cvy).
CCSS.Math.Content.HSN.VM.B.5.a
Two Dimensional Problems Involving Balanced Forces
Compute the magnitude of a scalar multiple cv using ||cv|| = |c|v. Compute the direction of cv knowing that when |c|v ≠ 0, the direction of cv is either along v (for c < 0) or against v (for c < 0).
CCSS.Math.Content.HSN.VM.B.5.b
Perform operations on matrices and use matrices in applications.
Use matrices to represent and manipulate data, e.g., to represent payoffs or incidence relationships in a network.
CCSS.Math.Content.HSN.VM.C.6
What are Matrices
Add, subtract, and multiply matrices of appropriate dimensions.
CCSS.Math.Content.HSN.VM.C.7
Subtracting Matrices
Understand that, unlike multiplication of numbers, matrix multiplication for square matrices is not a commutative operation, but still satisfies the associative and distributive properties.
CCSS.Math.Content.HSN.VM.C.9
Understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1 in the real numbers. The determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse.
CCSS.Math.Content.HSN.VM.C.10
Solving Linear Systems with Matrices
Multiply a vector (regarded as a matrix with one column) by a matrix of suitable dimensions to produce another vector. Work with matrices as transformations of vectors.
CCSS.Math.Content.HSN.VM.C.11
Work with 2 × 2 matrices as a transformations of the plane, and interpret the absolute value of the determinant in terms of area.
CCSS.Math.Content.HSN.VM.C.12
Determinant
|
Decimals - Easy Mathematics Step-by-Step
## Easy Mathematics Step-by-Step (2012)
### Chapter 6. Decimals
In this chapter, you learn how to work with decimals.
Decimal Concepts
Decimal fractions are fractions with a denominator that is some positive power of 10, such as , , , and so forth. To represent these numbers, you extend the place-value system of numbers and use a decimal point to separate whole numbers from decimal fractions. The number 428.36 is a mixed decimal and means . You read 428.36 as “Four hundred twenty-eight and thirty-six hundredths” or as “Four hundred twenty-eight point thirty-six.”
Don’t say “and” when reading whole numbers. For instance, 203 is “two hundred three,” not “two hundred and three.”
Hereinafter, mixed decimals and decimal fractions will be called simply decimals.
A place-value diagram for some of the positional values of the decimal system is shown in Figure 6.1.
Figure 6.1 Place values in the decimal system
The use of the decimal point is a very convenient way to represent decimal fractions. For instance, using the decimal point, you write as 0.365.
The 0 in front of the decimal point is used as a way to make the decimal point noticeable for decimal fractions that are less than 1.
You deal with the decimal point in arithmetic calculations by following some simple “rules” that are just mathematical shortcuts to ensure accuracy of the calculations.
You read a decimal point as either “and” or “point.” For example, 35.6 is read “Thirty-five and six-tenths” or as “Thirty-five point six.”
Adding and subtracting are only done for like amounts. A statement such as “4 inches plus 7 inches” is meaningful, while “7 apples plus 5 inches” is meaningless. Similarly, with decimals, you combine tenths with tenths, hundredths with hundredths, and so on. Thus, when you add or subtract decimals, keep the decimal points lined up in the computation so that you are adding digits of like place values.
Problem Add 25.78, 241.342, and 12.5.
Solution
Step 1. Write the numbers in an addition column, being sure to line up the decimal point.
Zeros are inserted where needed to make sure that all place values are in all the numbers.
Step 2. Add as with whole numbers.
Problem Subtract 168.274 from 6547.34.
Solution
Step 1. Write the numbers in a subtraction column, being sure to line up the decimal point.
When adding or subtracting decimals, you should insert zeros for missing place values.
Step 2. Subtract as you would with whole numbers.
The carrying and borrowing procedure is the same as with whole numbers.
Multiplying Decimals
A simple rule for multiplying two decimals is to sum the number of decimal places in the multiplicand (first factor) and the multiplier (second factor) and put this number of places in the product.
Problem Multiply 47.63 by 32.57.
Solution
Step 1. Write the numbers in a multiplication column.
You do not have to line up the decimal points for a multiplication problem.
Step 2. Perform the multiplication as you would with whole numbers.
Step 3. Sum the number of decimal places in both the multiplicand and the multiplier, which is four in this problem, and put that number of decimal places in the final product.
Dividing Decimals
To devise a rule for dividing decimals, you employ the use of the cancellation law of fractions. For example, . The technique, then, in dividing decimals, is to multiply both dividend (numerator) and divisor (denominator) by the appropriate power of 10 that will make the divisor a whole number. In practice, this strategy amounts to moving the decimal point to the right the appropriate number of places to make the divisor a whole number. Of course, the decimal point in the dividend must be moved in the same manner. Once this is done, the decimal point in the dividend and the quotient (answer) must be aligned.
Problem Divide 6.25 by 2.5.
Solution
Step 1. Write as a long division problem.
Step 2. Move the decimal point one place to the right in both numbers to make the divisor a whole number.
Step 3. Divide as you would whole numbers, keeping the decimal points aligned but ignoring the decimal point in the intermediate multiplications.
Step 4. State the main result.
Calculators will automatically place the decimal point, but you should know these rules to check for errors that may occur in entering data into your calculator.
Rounding Decimals
Working with decimals can sometimes result in lengthy decimal expressions. In application, you may be interested in the decimal expression to only a few places. In this case, you use the technique of rounding to determine the final approximation. For example, if you want the decimal to only two places, you look at the digit in the third place to the right of the decimal point. If the third digit is 5 or greater, increase the digit in the second place by 1 and drop all digits past the second digit to the right of the decimal point. If the third digit is less than 5, leave the digit in the second place as is and drop all digits past the second digit to the right of the decimal point. For example, 45.57689 rounded to two places is 45.58. The process for all places is the same.
Problem Use a calculator to compute and then round the answer to two decimal places.
. Recall from previous chapters that the fraction bar indicates division. It is important to recognize this use of the fraction bar.
Solution
Step 1. Do the division by calculator.
Step 2. Check whether the digit in the third place to the right of the decimal point is 5 or greater or less than 5.
The digit in the third place is 4, which is less than 5.
Step 3. Leave the digit in the second place as is and drop all digits past the second digit to the right of the decimal point.
Handheld calculators have taken the tedium out of arithmetic calculations and for that all are glad. Do the following problems by hand or with a calculator, but before you do them, predict the number of decimal places in the answer or, in the case of division, where the decimal point will be located.
Exercise 6
2. Subtract 1.8264 from 23.3728.
3. Multiply 0.214 by 1.93.
4. Multiply 1.21 by 0.0056.
5. Divide 0.1547 by 0.014.
6. Divide 2.916 by 0.36.
7. Divide 2.917 by 0.37 and round to three places.
8. Multiply 6.678 by 0.37 and round to two places.
9. Divide 3.977 by 0.0372 and round to three places.
10. Multiply 45.67892 by 0.0374583 and round to four places.
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# What is factoring?
What is factoring in algebra? Factoring is the process by which one tries to make a mathematical expression look like a multiplication problem by looking for factors. Basically, factoring reverses the multiplication process.
• Factoring can be as easy as looking for 2 numbers to multiply to get another number. For example, It is not hard to see that 32 = 4 × 8 once you know your multiplication table.
• When factoring, you could also be looking for the prime factorization of a number. For example 81 = 3 × 3 × 3 × 3.
• Or you may try to factor out the greatest common factor. For example, 2x + 10 = 2(x + 5) and 2 is the greatest common factor.
• Finally, you may try to factor expressions as complicated as x2 - 14x - 32, 15x2 - 26x + 11, or 150x3 + 350x2 + 180x + 420.
Both numerical and algebraic expressions can be factored using some specific method(s). A list of the different types of factoring are given in this lesson. Check them out so you can learn the specific method of factoring.
The simplest expressions to factor are of course numerical expressions. However, looking for the prime factorization of a big number like 240 may require a lot more work.
The lesson below about factoring integers will show you how to factor 240 and other big numbers. Then, you will be ready to factor algebraic expressions such as x2 + 5x + 6 and more complicated expressions using a variety of methods.
There are three concepts you will need to understand very well before you attempt to factor an expression. Start by studying the topics in the introduction.
Prerequisites
Factoring integers
Important to understand in order to grasp the meaning of factor.
Finding the greatest common factor
Important to understand before factoring polynomials or expressions with at least two terms.
Multiplying binomials
Important to understand before factoring trinomials
## What is factoring in algebra? Here are the different types of factoring
Factoring algebraic expressions
Learn how to factor a polynomial or an algebraic expression with two or more terms.
Factoring trinomials
Learn to factor a trinomial that has the form x2 + bx + c
Factoring by grouping
Learn how to factor a trinomial of the form ax2 + bx + c by grouping terms.
How to factor a trinomial by getting rid of the impostor.
Learn how to factor a trinomial of the form ax2 + bx + c by getting rid of the impostor.
Factor using the box method
Probably the most straightforward way to factor a trinomial.
Factoring perfect square trinomials
Learn to factor perfect square trinomials.
Learn to factor using the quadratic formula x2 + bx + c.
Learn how to factor and simplify radicals.
## Have A Great Problem About Factoring?
Do you have a problem about factoring ? Share it here with the solution!
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Solution: Problem Challenge 3: Minimum Window Sort
Go Back
## Problem Statement
Given an array, find the length of the smallest subarray in it which when sorted will sort the whole array.
Example 1:
``````Input: [1, 2, 5, 3, 7, 10, 9, 12]
Output: 5
Explanation: We need to sort only the subarray [5, 3, 7, 10, 9] to make the whole array sorted
``````
Example 2:
``````Input: [1, 3, 2, 0, -1, 7, 10]
Output: 5
Explanation: We need to sort only the subarray [1, 3, 2, 0, -1] to make the whole array sorted
``````
Example 3:
``````Input: [1, 2, 3]
Output: 0
Explanation: The array is already sorted
``````
Example 4:
``````Input: [3, 2, 1]
Output: 3
Explanation: The whole array needs to be sorted.
``````
Constraints:
• 1 <= arr.length <= 10<sup>4</sup>
• -10<sup>5</sup> <= arr[i] <= 10<sup>5</sup>
## Solution
As we know, once an array is sorted (in ascending order), the smallest number is at the beginning and the largest number is at the end of the array. So if we start from the beginning of the array to find the first element which is out of sorting order i.e., which is smaller than its previous element, and similarly from the end of array to find the first element which is bigger than its previous element, will sorting the subarray between these two numbers result in the whole array being sorted?
Let’s try to understand this with Example-2 mentioned above. In the following array, what are the first numbers out of sorting order from the beginning and the end of the array:
`````` [1, 3, 2, 0, -1, 7, 10]
``````
Starting from the beginning of the array the first number out of the sorting order is ‘2’ as it is smaller than its previous element which is ‘3’. Starting from the end of the array the first number out of the sorting order is ‘0’ as it is bigger than its previous element which is ‘-1’ As you can see, sorting the numbers between ‘3’ and ‘-1’ will not sort the whole array. To see this, the following will be our original array after the sorted subarray:
`````` [1, -1, 0, 2, 3, 7, 10]
``````
The problem here is that the smallest number of our subarray is ‘-1’ which dictates that we need to include more numbers from the beginning of the array to make the whole array sorted. We will have a similar problem if the maximum of the subarray is bigger than some elements at the end of the array. To sort the whole array we need to include all such elements that are smaller than the biggest element of the subarray.
### Step-by-step Algorithm
1. Initialize Pointers: Set `low` to 0 and `high` to the last index of the array.
2. Find Left Boundary:
• Move `low` to the right while the current element is less than or equal to the next element.
3. Check If Sorted:
• If `low` reaches the end, the array is already sorted. Return 0.
4. Find Right Boundary:
• Move `high` to the left while the current element is greater than or equal to the previous element.
5. Find Min and Max:
• Iterate from `low` to `high` to find the minimum and maximum values in this subarray.
6. Extend Left Boundary:
• Move `low` to the left while the previous element is greater than the subarray's minimum.
7. Extend Right Boundary:
• Move `high` to the right while the next element is less than the subarray's maximum.
8. Calculate Length:
• The length of the subarray to be sorted is `high - low + 1`.
### Algorithm Walkthrough
Using the input `[1, 3, 2, 0, -1, 7, 10]`:
• Initialize Pointers: `low = 0`, `high = 6`.
• Find Left Boundary:
• Compare `1` and `3`, move `low` to `1`.
• Compare `3` and `2`, stop. `low = 1`.
• Find Right Boundary:
• Compare `10` and `7`, move `high` to `5`.
• Compare `7` and `-1`, move `high` to `4`.
• Compare `-1` and `0`, stop at `4`.
• Find Min and Max:
• Subarray is `[3, 2, 0, -1]`.
• Minimum is `-1`, Maximum is `3`.
• Extend Left Boundary:
• `1` is greater than `-1`, `low` stays `1`.
• Extend Right Boundary:
• `7` is not less than `3`, `high` stays `4`.
• Calculate Length:
• Length is `high - low + 1` = `4 - 0 + 1 = 5`.
Here is the visual representation of this algorithm for Example 1:
## Code
Here is what our algorithm will look like:
Python3
Python3
Time Complexity
The time complexity of the above algorithm will be O(N).
Space Complexity
The algorithm runs in constant space O(1).
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# Identify and write the like terms in the following group of terms.$7p,8pq, - 5pq, - 2p,3p$
Last updated date: 10th Aug 2024
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Hint:The terms having the same variable with same exponents are called like terms. Check the variables and exponents of all the variables. Then the terms with the similar exponents and variables will come under like terms.
Here, the given terms are $7p,8pq, - 5pq, - 2p,3p$
Among the given terms we should find the like terms. Initially for that we should consider the terms having same variables as a group
In the given group we have two different variables they are $p{\text{ & }}pq$
Let us group the terms that have $p$ as its variable.
$7p, - 2p$ and $3p$ are the terms which have variable $p$.
These three terms are known as terms of variable $p$, since they contain the same variable p to the same power. (The power of p is 1 in all the three terms)
Now let us group all the terms that contain the variable $pq$.
In the given group of variables there are two terms with variable $pq$.
$8pq$ and $- 5pq$ are the two terms which have the variable $pq$
These two terms are known as terms of variable $pq$, since they contain the same variable p and q to the same power. (The powers of p and q is 1 in both terms)
Hence,
We have found that the like terms in the given group are $\left( {7p, - 2p,3p} \right)$ and $\left( {8pq, - 5pq} \right)$
Note:We know that like terms are terms that contain the same variables raised to the same exponent (power). Only the numerical coefficients are different.
Constants are always said to be like terms because in every constant term there may be any number of variables which have the exponent zero. Unlike terms are the terms which have different variables and exponents.
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# if A gives B $3, B will have twice as much as A. If B gives A$5, A will have twice as much as...
## Question:
If A gives B {eq}$3, {/eq} B will have twice as much as A. If B gives A {eq}$5, {/eq} A will have twice as much as B.How much does each have?
## Algebraic equation word problems
An algebraic equation is formed when a relation between any variable is equated to a constant.
In the word problems of algebraic equation, the relationship is given in statement and we need to form an algebraic equation out of it. Assuming a base variable and determining other unknowns in its term also helps in simplifying the equation.
For example: It costs $400 for 4 adult tickets and 3 children ticket. It can be expressed as 4x +3y = 400 as an algebraic equation where x and y are variables representing the cost of one adult and children ticket each and 400 is the equivalent constant of the relation shown in the equation. ## Answer and Explanation: Let the amount of money with A and B be {eq}A and B {/eq} respectively. According to the problem: A gives B$3, B will have twice as much as A
$$2*(A-3) = B+3$$
$$2A-6 = B+3$$
$$2A-B=9$$
According to the problem: If B gives A $5, will have twice as much as B. $$A+5 = 2(B-5)$$ $$A+5 = 2B-10$$ $$2B -A=15$$ Solving equations simultaneously: $$(2B -A=15) + 2(2A-B=9)$$ $$3A = 33$$ $$A = 11$$ Using the value of A in the equation {eq}2B -A=15 {/eq} $$2B -A=15$$ $$2B -11=15$$ $$2B =26$$ $$B =13$$ The amount of money with A and B is {eq}$11\ and\ \$13 {/eq} respectively.
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# If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is
Question:
If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is
(a) $\frac{p-q}{q-r}$
(b) $\frac{q-r}{p-q}$
(c) pqr
(d) none of these
Solution:
(b) $\frac{q-r}{p-q}$
Let a be the first term and d be the common difference of the given A.P.
Then, we have:
$\mathrm{p}^{\text {th }}$ term, $a_{p}=a+(p-1) d$
$q^{\text {th }}$ term, $a_{q}=a+(q-1) d$
$r^{\text {th }}$ term, $a_{r}=a+(r-1) d$
Now, according to the question the $p^{\text {th }}$, the $q^{\text {th }}$ and the $r^{\text {th }}$ terms are in G.P.
$\therefore(a+(q-1) d)^{2}=(a+(p-1) d) \times(a+(r-1) d)$
$\Rightarrow a^{2}+2 a(q-1) d+((q-1) d)^{2}=a^{2}+a d(r-1+p-1)+(p-1)(r-1) d^{2}$
$\Rightarrow a d(2 q-2-r-p+2)+d^{2}\left(q^{2}-2 q+1-p r+p+r-1\right)=0$
$\Rightarrow a(2 q-r-p)+d\left(q^{2}-2 q-p r+p+r\right)=0 \quad(\because d$ cannot be 0$)$
$\Rightarrow a=-\frac{\left(q^{2}-2 q-p r+p+r\right) d}{(2 q-r-p)}$
$\therefore$ Common ratio, $r=\frac{a_{q}}{a_{p}}$
$=\frac{a+(q-1) d}{a+(p-1) d}$
$=\frac{\frac{\left(q^{2}-2 q-p r+p+r\right) d}{(p+r-2 q)}+(q-1) d}{\frac{\left(q^{2}-2 q-p r+p+r\right) d}{(p+r-2 q)}+(p-1) d}$
$=\frac{q^{2}-2 q-p r+p+r+p q+r q-2 q^{2}-p-r+2 q}{q^{2}-2 q-p r+p+r+p^{2}+p r-2 p q-p-r+2 q}$
$=\frac{p q-p r-q^{2}+q r}{p^{2}+q^{2}-2 p q}$
$=\frac{p(q-r)-q(q-r)}{(p-q)^{2}}$
$=\frac{(p-q)(q-r)}{(p-q)^{2}}$
$=\frac{(q-r)}{(p-q)}$
|
Simplifying Exponential Expressions
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1 Simplifying Eponential Epressions
2 Eponential Notation Base Eponent Base raised to an eponent Eample: What is the base and eponent of the following epression? 7 is the base 7 is the eponent
3 Goal To write simplified statements that contain distinct bases, one whole number in the numerator and one in the denominator, and no negative eponents. E: ab 9bc 6a b c 4a
4 Multiplying Terms When we are multiplying terms, it is easiest to break the problem down into steps. First multiply the number parts of all the terms together. Then multiply the variable parts together. Eamples: a. ( 4 )( -5 ) = ( )(. ) = -0 Only the z is squared b. (5z)(z)(4y) = (5... 4)(y. z. z) = 0yz
5 Eploration Evaluate the following without a calculator: 4 = = = = Describe a pattern and find the answer for: 0 = 8 7 9
6 Zero Power a 0 = Anything to the zero power is one Can a equal zero? No. You can t divide by 0.
7 Eploration Simplify: 4 Use the definition of eponents to epand 4 There are 7 variables Notice (from the initial epression) +4 is 7! 7
8 Product of a Power If you multiply powers having the same base, add the eponents. a m n
9 Simplify: Add the eponents since the bases are the same Eample 9 y 0 9 Anything raised to the 0 power is
10 Practice Simplify the following epressions: ) ) z y 5 y 4 ) y
11 Eploration Simplify: 5 The Product of a Power Rule says to add all the s Adding five times is equivalent to multiplying by 5. The same eponents from the initial epression! 5 5 Use the definition of eponents to epand
12 Power of a Power a mn To find a power of a power, multiply the eponents.
13 Eample 6 Simplify: Multiply the powers of a eponent raised to another power Any base without a power, is assumed to have an eponent of s s t 4t s s 6 t 9 ss t 4t 4 8s t 4t s t 9 Multiply numbers without eponents and add the eponents when the bases are the same
14 Practice Simplify the following epressions: ) y 4 8 y ) 5 a 4 a a ) y y 7 y
15 Eploration Simplify: z 5 z z z z z Adding five times is equivalent to multiplying by 5 Notice: Both the z and were raised to the 5 th power! z z The Product of a Power Rule says to add the eponents with the same bases Use the definition of eponents to epand
16 Power of a Product a m b m If a base has a product, raise each factor to the power
17 Eample Simplify: 4 5 Everything inside the parentheses is raised to the eponent outside the parentheses y y 5 y 0 0 y Multiply the powers of a eponent raised to another power y 45 Multiply numbers without eponents and add the eponents when the bases are the same
18 Practice Simplify the following epressions: ) pqr ab a pqr 4 5 ) 4 ) - yz 54 y z 9 8 5a b 7
19 First Four a 5 b d 6 0. r 8 s a 7 b 7 c. 6z 8. 9a m 6 n z y r 5 s 5 t 5 4. a 4 b 4 c 6. 08a 7. 7b y
20 Eploration Complete the tables (with fractions) by finding the pattern /5 /5 /5 / / /6 /8 ¼ ½ 4 8 6
21 Negative Powers A simplified epression has no negative eponents. a m a m Negative Eponents flip and become positive a m
22 Eample Simplify: 0 4 All of the old rules still apply for negative eponents Flip ONLY the thing with the negative eponent to the bottom and the eponent becomes positive 4a b 5a 45 a 0 4 0a b 0b a 6 b 6 This is not simplified since there is a negative eponent
23 Simplify: Everything with a positive eponent stays where it is. Eample 8y 8 y y y Everything with a negative eponent is flipped and eponent becomes positive. Since all of the negative eponents are gone, apply all of the old rules to simplify.
24 Practice Simplify the following epressions: ) 8 ) ) y y 8 4 4) a b 5 7 y y 7 8 a 4b 6
25 Eploration Simplify: 0 Use the definition of eponents to epand The 6 s in the denominator cancel 6 out of the 0 s in the numerator. This is the same as subtracting the eponents from the initial epression! Since everything is multiplied, you can cancel common factors Only 4 s remain in the numerator
26 Quotient of a Power a m n To find a quotient of a power, subtract the denominator s eponent from the numerator s eponent if the bases are the same. a 0
27 Eample Simplify: Divide the base numbers first 6 6 y 6y 6 y Subtract the eponents of the similar bases since there is division Not simplified since there is a negative eponents 4 y 4 y Flip any negative eponents
28 Practice Simplify the following epressions: 6 0 ab ) 5 a a 5 ) 4 y 6 y 5 ) 9 4y 4y 7 6 y
29 Eploration Simplify: Use the definition of eponents to epand Use the definition of eponents to rewrite. Notice: Both the numerator and denominator were raised to the 6 th power! a b a a a a a a b b b b b b aaaaaa bbbbbb 6 a 6 b 6 Multiply the fractions
30 Power of a Quotient a m b To find a power of a quotient, raise the denominator and numerator to the same power. m
31 Eample Simplify: Everything in the fraction is raised to the power out side the parentheses. Subtract the eponents when there is division, and add when there is multiplication 7 y y 5 y 6 8 y 5 y 7 y 5 y 8 y y y 9 9 Multiply the fractions
32 Practice Simplify the following epressions: ) ) ) a bc y s f zr 4 8 a b y f r s z
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# A direct current of $5{\text{A}}$ is superimposed on an alternating current $I = 10\sin \omega t$ flowing through a wire, The effective value of the resulting current will be(A) $\left( {\dfrac{{15}}{2}} \right){\text{A}}$(B) $5\sqrt 3 {\text{A}}$(C) $5\sqrt 5 {\text{A}}$(D) $15{\text{A}}$
Last updated date: 20th Jun 2024
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Hint We can obtain the equation of the resulting current through the wire by adding the direct current to the alternating current. The effective value of current is equal to the root mean square value of the current. So we need to take the square, calculate its mean by integrating it over a time period, and take the square root to obtain the RMS value which will be the required effective current.
Complete step-by-step solution:
Since the direct current is superimposed over the alternating current, the resulting current will be equal to the sum of the direct and the alternating current. So we have
${I_R} = {I_{dc}} + {I_{ac}}$
According to the question, ${I_{dc}} = 5{\text{A}}$ and ${I_{ac}} = 10\sin \omega t$. Substituting these in the above equation, we get the resultant current as
${I_R} = 5 + 10\sin \omega t$
Now, we know that the effective value of the current is nothing but the RMS, or the root mean square value of the current. By the definition of RMS, we have to take the square, then take the mean, and finally take the square root of the current. So on squaring both sides of the above equation, we get
${I_R}^2 = {\left( {5 + 10\sin \omega t} \right)^2}$
We know that ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$. So we write the above equation as
${I_R}^2 = {5^2} + 2\left( 5 \right)\left( {10\sin \omega t} \right) + {\left( {10\sin \omega t} \right)^2}$
$\Rightarrow {I_R}^2 = 25 + 100\sin \omega t + 100{\sin ^2}\omega t$ (1)
Now, we know that the mean of a function is given by
$M = \dfrac{1}{T}\int_0^T {f\left( t \right)dt}$
The period of the function ${I_R}^2$ is clearly equal to $\dfrac{{2\pi }}{\omega }$. So we integrate it from $0$ to $\dfrac{{2\pi }}{\omega }$ to get its mean as
$M = \dfrac{1}{{2\pi /\omega }}\int_0^{\dfrac{{2\pi }}{\omega }} {{I_R}^2dt}$
$\Rightarrow M = \dfrac{\omega }{{2\pi }}\int_0^{\dfrac{{2\pi }}{\omega }} {{I_R}^2dt}$
Putting (1) above, we get
$\Rightarrow M = \dfrac{\omega }{{2\pi }}\int_0^{\dfrac{{2\pi }}{\omega }} {\left( {25 + 100\sin \omega t + 100{{\sin }^2}\omega t} \right)dt}$
$\Rightarrow M = \dfrac{\omega }{{2\pi }}\left( {\int_0^{\dfrac{{2\pi }}{\omega }} {25dt} + \int_0^{\dfrac{{2\pi }}{\omega }} {100\sin \omega tdt} + \int_0^{\dfrac{{2\pi }}{\omega }} {100{{\sin }^2}\omega tdt} } \right)$
We know that the average value of the sinusoidal functions is equal to zero. So we can put $\int_0^{\dfrac{{2\pi }}{\omega }} {100\sin \omega tdt} = 0$ in the above equation to get
$M = \dfrac{\omega }{{2\pi }}\left( {\int_0^{\dfrac{{2\pi }}{\omega }} {25dt} + \int_0^{\dfrac{{2\pi }}{\omega }} {100{{\sin }^2}\omega tdt} } \right)$
We know that ${\sin ^2}\theta = \dfrac{{1 - \cos 2\theta }}{2}$. So we write the above equation as$M = \dfrac{\omega }{{2\pi }}\left( {\int_0^{\dfrac{{2\pi }}{\omega }} {25dt} + \int_0^{\dfrac{{2\pi }}{\omega }} {100\left( {\dfrac{{1 - \cos 2\omega t}}{2}} \right)dt} } \right)$
$\Rightarrow M = \dfrac{\omega }{{2\pi }}\left( {25\int_0^{\dfrac{{2\pi }}{\omega }} {dt} + 50\int_0^{\dfrac{{2\pi }}{\omega }} {\left( {1 - \cos 2\omega t} \right)dt} } \right)$
$\Rightarrow M = \dfrac{\omega }{{2\pi }}\left( {25\int_0^{\dfrac{{2\pi }}{\omega }} {dt} + 50\int_0^{\dfrac{{2\pi }}{\omega }} {dt} - \int_0^{\dfrac{{2\pi }}{\omega }} {\cos 2\omega tdt} } \right)$
We know that $\int {dt} = t$. So we get
$\Rightarrow M = \dfrac{\omega }{{2\pi }}\left( {25\left[ t \right]_0^{\dfrac{{2\pi }}{\omega }} + 5025\left[ t \right]_0^{\dfrac{{2\pi }}{\omega }} - 50\int_0^{\dfrac{{2\pi }}{\omega }} {\cos 2\omega tdt} } \right)$
$\Rightarrow M = \dfrac{\omega }{{2\pi }}\left( {25\left( {\dfrac{{2\pi }}{\omega }} \right) + 50\left( {\dfrac{{2\pi }}{\omega }} \right) - \int_0^{\dfrac{{2\pi }}{\omega }} {\cos 2\omega tdt} } \right)$
Since $\cos 2\omega t$ is also sinusoidal, its average over a cycle will also be zero, that is, $\int_0^{\dfrac{{2\pi }}{\omega }} {\cos 2\omega tdt} = 0$.
$\Rightarrow M = \dfrac{\omega }{{2\pi }}\left( {25\left( {\dfrac{{2\pi }}{\omega }} \right) + 50\left( {\dfrac{{2\pi }}{\omega }} \right)} \right)$
$\Rightarrow M = \left( {25 + 50} \right) = 75{{\text{A}}^2}$
Now, we take the square root of this mean value to get the final RMS value of the resulting current as
$RMS = \sqrt M$
$\Rightarrow RMS = \sqrt {75} {\text{A}} = 5\sqrt 3 {\text{A}}$
Thus the effective value of the resulting current is equal to $5\sqrt 3 {\text{A}}$.
Hence, the correct answer is option B.
Note: We can integrate the square of the current between any time interval which makes a complete time period. But it will be convenient to set the lower limit of the integral equal to zero. So we chose the period from $0$ to $2\pi$.
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# Linear Relationships – first differences
### QUESTION
What does it mean to use “first differences” to determine which of the 4 tables shown at right represent non-linear relationships?
To find first differences, look at column 2: subtract the 1st number from the 2nd, the 2nd from the 3rd, etc.
• If these differences are all the same, then you have a linear relationship.
• If not, then the relationship is non-linear.
Why?
• Well, linear relationships have graphs that are straight lines, so their slopes are constant.
• A linear relationship is like a typical staircase, where all steps are the same height.
• When you find first differences, it’s like finding the “height” of each of the steps.
We’re making some assumptions here about the numbers in column 1:
• they are in numerical order
• their first differences are all the same: for example, 1,2,3,4,5 or 1,3,5,7,9 would be fine, but 1,2,4,5,7 or 1,3,2,5,4 would not
### BIGGER PICTURE
First differences (and second and third differences) help determine whether there is a pattern in a set of data, as well as the nature of the pattern.
For example, let’s say you invest \$1,000. Below are 3 growth options, tracked for 5 years. Which would you prefer?
Which option would you prefer?
Let’s look at how your investment grows in each option:
1. In the first option, your investment grows each year by \$0. (first difference is always 0)
2. In the second option, your investment grows each year by \$1000. (first difference is always 1,000)
3. In the third option, your investment grows by \$1,000 in the 1st year, \$1,100 in the 2nd, \$1,210 in the 3rd, and so on. The annual growth increases as the years pass. (first difference keeps changing/growing)
Options 1 and 2 have constant growth (\$0 and \$1,000 annually, respectively). Option 3 does not.
Data that has constant growth represent linear relations.
Linear relations, as the name suggests, have graphs whose points are in a straight line.
If you plotted the 3 options as a scatter plot (and extend for 20 years), it would look like this:
Surprised by this graph?
Why does the 3rd option so much better than the 2nd option, when at first it seemed so similar to the 2nd option?
• Option 1 gives you an additional \$1,000 every year, which is constant growth.
• Option 2 gives you an additional 10% every year, which is exponential growth.
Here are the equations for the 3 options.
(x represents the year)
### SECOND DIFFERENCES
What about second differences? Let’s look at the table on the right.
• The first differences are in red.
• The second differences are in blue.
Here’s a context for this:
• column 1 could represent time
• column 2 could represent distance travelled by a falling object
• the first differences relate to the speed of the object (notice the speed keeps increasing, which makes sense for falling objects)
• the second differences would relate to the speed of the speed, or the acceleration of the object, like the acceleration due to gravity, which is constant.
If you studied Calculus:
• the first differences relate to the first derivative (y’)
• the second differences relate to the second derivative (y”)
##### Linear and non-linear relationships across the grades
What are the first differences in the Grade 1 growing pattern shown below?
What linear relationship does the above pattern represent?
See http://learnx.ca/growing-patterns/ for more like this, such as:
• linear and other relationships across grades 1, 4 and 9
• an interview with applied mathematician Lindi Wahl, and
• a music video from Grades 1/2 classrooms, with lyrics based on parent comments
### COMPUTATIONAL MODELLING
Click on the link below to use code to investigate first differences as well as graphs of linear relations.
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# 022 Sample Final A, Problem 3
Find the antiderivative: ${\displaystyle \int {\frac {6}{x^{2}-x-12}}\,dx.}$
Foundations:
1) What does the denominator factor into? What will be the form of the decomposition?
2) How do you solve for the numerators?
3) What special integral do we have to use?
1) Since ${\displaystyle x^{2}-x-12=(x-4)(x+3)}$ , and each term has multiplicity one, the decomposition will be of the form: ${\displaystyle {\frac {A}{x-4}}+{\frac {B}{x+3}}}$.
2) After writing the equality, ${\displaystyle {\frac {6}{x^{2}-x-12}}\,=\,{\frac {A}{x-4}}+{\frac {B}{x+3}}}$, clear the denominators, and evaluate both sides at ${\displaystyle x=4,-3}$. Each evaluation will yield the value of one of the unknowns.
3) We have to remember that ${\displaystyle \int {\frac {c}{x-a}}dx=c\ln(x-a)}$ , for any numbers ${\displaystyle c,a}$.
Solution:
Step 1:
First, we factor: ${\displaystyle x^{2}-x-12=(x-4)(x+3)}$ .
Step 2:
Now we want to find the partial fraction expansion for ${\displaystyle {\frac {6}{(x-4)(x+3)}}}$ , which will have the form ${\displaystyle {\frac {A}{x-4}}+{\frac {B}{x+3}}}$.
To do this, we need to solve the equation ${\displaystyle 6\,=\,A(x+3)+B(x-4)}$.
Plugging in ${\displaystyle -3}$ for ${\displaystyle x}$, we find that ${\displaystyle 6\,=\,-7B}$ , and thus ${\displaystyle B=-{\frac {6}{7}}}$ .
Similarly, we can find ${\displaystyle A}$ by plugging in ${\displaystyle 4}$ for ${\displaystyle x}$. This yields ${\displaystyle 6=7A}$ , so ${\displaystyle A={\frac {6}{7}}}$ .
This completes the partial fraction expansion:
${\displaystyle {\frac {6}{x^{2}-x-12}}\,=\,{\frac {6}{7(x-4)}}-{\frac {6}{7(x+3)}}.}$
Step 3:
By the previous step, we have
${\displaystyle \int {\frac {6}{x^{2}-x-12}}dx\,=\,\int {\frac {6}{7(x-4)}}dx-\int {\frac {6}{7(x+3)}}dx.}$
Integrating by the rule in 'Foundations',
${\displaystyle \int {\frac {6}{x^{2}-x-12}}dx\,=\,{\frac {6}{7}}\ln(x-4)-{\frac {6}{7}}\ln(x+3).}$
Step 4:
Now, make sure you remember to add the ${\displaystyle +C}$ to the integral at the end.
${\displaystyle {\frac {6}{7}}\ln(x-4)-{\frac {6}{7}}\ln(x+3)+C.}$
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Triangle ABC is an isosceles triangle with AB...
### Related Test
Triangle ABC is an isosceles triangle with AB = AC and AD is the perpendicular dropped from vertex A on the side BC. What is the perimeter of triangle ABC?
(2) The perimeter of triangle ADB is 15 + 5√3
• a)
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
• b)
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
• c)
BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
• d)
• e)
Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
Steps 1 & 2: Understand Question and Draw Inferences
Given:
• Isosceles ΔABC
• AB = AC
• So, perpendicular AD will bisect side BC.
• If we assume ∠ ABC to be xo, the different ∠s in the figure can be represented as below:
To find: Perimeter of ΔABC
• =AB + BC + CA
• = 2AB + BC
So, in order to find the perimeter, we need to know AC and BC
Knowing the ∠s may help us (because we may then employ trigonometric ratios)
Step 3: Analyze Statement 1 independently
90°- x° = 2x°
⇒ 3x° = 90°
⇒ x° = 30°
But we don’t know the magnitude of any side of the triangle. So, we cannot yet use trigonometric ratios to find the sides of the triangle.
Not sufficient.
Step 4: Analyze Statement 2 independently
(2) The perimeter of triangle ADB is 15 + 5√3
The value of 2AB + BC depends on the value of AD.
Since we do not know the value of AD, we cannot find the required value
Not sufficient.
• From Statement 2
• Using Equation II, a unique value of AD will be obtained
• Using Equation I, the value of 2AB + BC will be obtained
Sufficient.
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# Maths - Matrix Arithmetic
To add matrices just add the corresponding elements, the matrices being added must have the same dimensions, example are shown on the following pages:
### Matrix Subtraction
To subtract matrices just subtract the corresponding elements, the matrices being subtracted must have the same dimensions, example are shown on the following pages:
### Matrix Multiplication
To multiply matrices,
[M] = [A][B]
mik = sums=1p(aisbsk)
In other words, to work out each entry in the matrix, we take the row from the first operand and the column from the second operand:
a00 a01 a10 a11
b00 b01 b10 b11
=
a00 a01
b00 b10
a00 a01
b01 b11
a10 a11
b00 b10
a10 a11
b01 b11
This single row times a single column is equivalent to the dot product:
a00 a01 a10 a11
b00 b01 b10 b11
=
a00*b00 + a01*b10 a00*b01 + a01*b11 a10*b00 + a11*b10 a10*b01 + a11*b11
Example are shown on the following pages:
It is important to realise that the order of the multiplicands is significant, in other words [A][B] is not necessarily equal to [B][A]. In mathematical terminology matrix multiplication is not commutative.
It we need to change the order of the terms being multiplied then we can use the following:
([A] * [B])T = [B]T * [A]T
### Identity Matrix
The identity matrix is the do nothing operand for matrix multiplication, so if the identity matrix is denoted by [I] then,
[I][a] = [a]
The identity matrix is a square matrix with the leading diagonal terms set to 1 and the other terms set to 0, for example:
1 0 0 0 1 0 0 0 1
Since matrix multiplication is not commutative if [I][a] = [a] then is it necessarily true that [a][I] = [a] ?
Also if [b][b]-1=[I] then does [b]-1[b] =[I] ?
try calculating the following:
a00 a01 a02 a10 a11 a12 a20 a21 a22
1 0 0 0 1 0 0 0 1
and
1 0 0 0 1 0 0 0 1
a00 a01 a02 a10 a11 a12 a20 a21 a22
### Division and Inverse matrix
We don't tend to use the notation for division, since matrix multiplication is not commutative we need to be able to distinguish between [a][b]-1 and [b]-1[a].
So instead of a divide operation we tend to multiply by the inverse, for instance if,
[m] = [a][b]
then,
[m][b]-1 = [a][b][b]-1
because [b][b]-1=[I] we can remove [b][b]-1 -- is this true???
[m][b]-1 = [a]
Where I can, I have put links to Amazon for books that are relevant to the subject, click on the appropriate country flag to get more details of the book or to buy it from them.
Geometric Algebra for Physicists - This is intended for physicists so it soon gets onto relativity, spacetime, electrodynamcs, quantum theory, etc. However the introduction to Geometric Algebra and classical mechanics is useful.
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# How To Solve Ratio And Proportion Problems – Tips And Tricks
0
3900
Ratio and Proportion are commonly asked questions in aptitude test and competitive exams. These problems require special arithmetic calculation ability to solve it quickly. Questions from these category could be so tricky, so you must equipped yourself with all the shortcut methods and fast calculation is one of the plus for ratio and proportion problems. Let’s start with some formula and definitions :
Ratio : The ratio of two quantities a and b in the same units , is the fraction a/b and we write it as a : b.
Proportion : The equality of ratios is called proportion.
=> a : b : : c : d <=> (b x c) = (a x d)
Product of means (b,c) = Product of extremes (c,d)
Componendo and Dividendo : If a / b = c / d , then (a + b) / (a – b) = (c + d) / (c – d)
You would have studied these definitions and formulas many times, let’s start with solving some problems :
Problem 1 : If a : b = 5 : 9 and b : c = 4 : 7, find a : b : c
Solution 1 : Actually its very tricky question if you do not know how to solve it. In this kind of problems , first see the common term and b is the common term in here. So the trick in here is to make the value of b common in both the ratio and that could be done by multiplying the values with the corresponding ratio values i.e.
a : b = 5 : 9 ( Multiply it by 4 which is the ratio value of b in b : c)
a : b = 5 *4 / 9*4 => 20 : 36
Similarly, b : c = 4 : 7 (Multiply it by 9 which the ratio value of b in a : b)
b : c = 4*9 / 7*9 => 36 : 63
So, now the value of b has been equated in both the ratio , so we can finally conclude
Therefore , a : b : c = 20 : 36 : 63
Problem 2 : If A : B = (1/2) : (3/8) , B : C = (1/3) : (5/9) and C : D = (5/6) : (3/4), then find A : B : C : D
Solution 2 : So now here 4 terms are given instead of 3 like the previous problem ,
Now to solve this problem we need to first find out A : B : C using the method we used in the previous problem
Therefore , A : B = (1/2 * 1/3 ) : (3/8 * 1/3) = (1/6) : (1/8) and B : C = (1/3 * 3/8) : (5/9 * 3/8) = (1/8) : (5/24)
Hence , A : B : C = (1/6) : (1/8) : (5/24)
Since C : D = (5/6) : (3/4) , So C is the common term in here ,
Therefore, A : B : C = (1/6 * 5/6) : (1/8 * 5/6) : (5/24 * 5/6) = 5/36 : 5/48 : 25/144
Hence C : D = (5/6 * 5/24) : (3/4 * 5/24) = 25/144 : 15/96
Therefore, A : B : C : D = 5/36 : 5/48 : 25/144 : 15/96 = 1/6 : 1/8 : 5/24 : 3/16 (Take 5/6 as common)
Multiply the numerators with the LCM of the denominators (48)
Thus A : B : C : D = 8 : 6 : 10 : 9
Problem 3 : A bag contains 50 paise, 25 paise and 10 paise coins in the ratio 5 : 9 : 4, amounting to Rs 206. Find the number of coins of each type.
Solution 3 : Let the number of coins of 50p , 25p and 10p be 5x, 9x and 4x
Therefore, (5x/2) + (9x/4) + (4x/10) = 206 => x = 40
Hence, No of 50 p coins = 5x = 5 * 40 = 200; 25 p coins = 9x = 9*40= 360 and 10 p coins = 4x = 4*40=160
Problem 4 : In a mixture of is 60 liters , the ratio of milk and water is 2 : 1. If this ratio is to be 1 : 2, then the quantity of water to be further added is ?
Solution 4 : Quantity of milk = 60 * (2/3) = 40 liters,
Thus quantity of water = 60 – 40 = 20 liters
New ratio of milk and water is 1 : 2
Let quantity of water to be added further is x liters.
There ratio of milk to water = 40 / (20+x)
Hence, 40 / (20+x) = 1/2 => x = 60
There quantity of water to be added is 60 liters.
Problem 5 : Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should these be mixed to get an alloy 15 times as heavy as water ?
Solution 5 : Given Gold (G) = 19 times Water (W) = 19W
Therefore Copper (C) = 9W
New alloy = 15W
Lets mix 1 gm of Gold with x gm of Copper , which would result in (x + 1)gm of alloy
1 gm of Gold + x gm of Copper = (x+1) gm of alloy
19W + (9W)* x = (x+1)* 15W
=> x = 2/3
Therefore ratio Gold to Copper in the alloy = 1 : (2/3) => 3 : 2
Problem 6 : Based on Componendo and Dividendo
If p = 4xy / (x + y), find the value of [ ( (p + 2x) / (p – 2x) ) + ( (p + 2y) / (p – 2y) ) ]
Solution 6 : Given p = 4xy / (x + y) ,
Therefore p / 2x = 2y / (x+y)
Apply Componendo and Dividendo,
p+2x / p-2x = 2y+x+y / 2y-x-y => p+2x / p-2x = 3y + x / y-x
Similarly, p+2y / p-2y = 3x + y / x-y
Problem 7 : The ratio of incomes of A and B is 5 : 4 and the ratio of their expenditures is 3 : 2. If at the end of the year, each saves Rs 1600, then what is the income of A ?
Solution 7 : Suppose the income of A and B is 5x and 4x respectively and their expenditures is 3y and 2y respectively.
Therefore , 5x – 3y = 1600 and 4x – 2y = 1600 , On solving both the equations ,
x = 800 , y = 800
Thus income of A = 5x = 5*800 = Rs 4000
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# How do you write the quadratic function in standard form y=4(x-1)^2+5?
Jun 25, 2017
$y = 4 {x}^{2} - 8 x + 6$
#### Explanation:
$y = 4 {\left(x - 1\right)}^{2} + 5$ This is Vertex form.
Expand & simplify
$y = 4 \setminus \cdot \left(x - 1\right) \left(x - 1\right) + 5$
$y = 4 \setminus \cdot \left({x}^{2} - 2 x + 1\right) + 5$
$y = 4 {x}^{2} - 8 x + 4 + 5$
$y = 4 {x}^{2} - 8 x + 9$ This is Standard form.
Jun 25, 2017
$y = 4 {x}^{2} - 8 x + 9$
#### Explanation:
$\text{the equation of a parabola in "color(blue)"standard form}$ is.
color(red)(bar(ul(|color(white)(2/2)color(black)(y=ax^2+bx+c ; a!=0)color(white)(2/2)|)))
$\text{expand " (x-1)^2" using FOIL}$
$y = 4 \left({x}^{2} - 2 x + 1\right) + 5$
$\textcolor{w h i t e}{y} = 4 {x}^{2} - 8 x + 4 + 5$
$\textcolor{w h i t e}{y} = 4 {x}^{2} - 8 x + 9 \leftarrow \textcolor{red}{\text{ in standard form}}$
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# 1979 AHSME Problems/Problem 24
## Problem 24
Sides $AB,~ BC$, and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5$, and $20$, respectively. If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5}$, then side $AD$ has length
• A polygon is called “simple” if it is not self intersecting.
$\textbf{(A) }24\qquad \textbf{(B) }24.5\qquad \textbf{(C) }24.6\qquad \textbf{(D) }24.8\qquad \textbf{(E) }25$
## Solution
We know that $\sin(C)=-\cos(B)=\frac{3}{5}$. Since $B$ and $C$ are obtuse, we have $\sin(180-C)=\cos(180-B)=\frac{3}{5}$. It is known that $\sin(x)=\cos(90-x)$, so $180-C=90-(180-C)=180-B$. We simplify this as follows:
$$-90+C=180-B$$
$$B+C=270^{\circ}$$
Since $B+C=270^{\circ}$, we know that $A+D=360-(B+C)=90^{\circ}$. Now extend $AB$ and $CD$ as follows:
$[asy] size(10cm); label("A",(-1,0)); dot((0,0)); label("B",(-1,4)); dot((0,4)); label("E",(-1,7)); dot((0,7)); label("C",(4,8)); dot((4,7)); label("D",(24,8)); dot((24,7)); draw((0,0)--(0,4)); draw((0,4)--(4,7)); draw((4,7)--(24,7)); draw((24,7)--(0,0)); draw((0,4)--(0,7), dashed); draw((0,7)--(4,7), dashed); //diagram by WannabeCharmander [/asy]$
Let $AB$ and $CD$ intersect at $E$. We know that $\angle AED=90^{\circ}$ because $\angle E = 180 - (A+D)=180-90 = 90^{\circ}$.
Since $\sin BCD = \frac{3}{5}$, we get $\sin ECB=\sin(180-BCD)=\sin BCD = \frac{3}{5}$. Thus, $EB=3$ and $EC=4$ from simple sin application.
$AD$ is the hypotenuse of right $\triangle AED$, with leg lengths $AB+BE=7$ and $EC+CD=24$. Thus, $AD=\boxed{\textbf{(E)}25}$
-WannabeCharmander
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# How do you solve abs(5n-4)=16?
Jul 20, 2017
See a solution process below:
#### Explanation:
The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
Solution 1
$5 n - 4 = - 16$
$5 n - 4 + \textcolor{red}{4} = - 16 + \textcolor{red}{4}$
$5 n - 0 = - 12$
$5 n = - 12$
$\frac{5 n}{\textcolor{red}{5}} = - \frac{12}{\textcolor{red}{5}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} n}{\cancel{\textcolor{red}{5}}} = - \frac{12}{5}$
$n = - \frac{12}{5}$
Solution 2
$5 n - 4 = 16$
$5 n - 4 + \textcolor{red}{4} = 16 + \textcolor{red}{4}$
$5 n - 0 = 20$
$5 n = 20$
$\frac{5 n}{\textcolor{red}{5}} = \frac{20}{\textcolor{red}{5}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} n}{\cancel{\textcolor{red}{5}}} = 4$
$n = 4$
The Solutions Are: $n = - \frac{12}{5}$ and $n = 4$
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# 6.15: Percent of Decrease
Difficulty Level: At Grade Created by: CK-12
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At the end of the summer season, the local farmer’s market marked down its swan gourds to $1.00 apiece. Throughout the summer, the swan gourds had sold for$3.50 each. What was the percent of decrease in cost?
In this concept, you will learn how to calculate the percent of decrease.
### Finding Percent of Decrease
Prices can increase. Costs can increase. Numbers can increase. You can find the percent of increase when dealing with an increase. All of these things can also decrease. When there has been a decrease from an original amount to a new lower amount, you can find the percent of decrease.
The percent of decrease from one amount to another is the ratio of the amount of decrease to the original amount.
To find the percent of decrease, follow these steps:
First, find the amount of decrease by subtracting the lower amount from the higher amount.
Next, write a fraction in which the numerator is the amount of decrease and the denominator is the original amount.
Percent of Decrease =Amount of DecreaseOriginal Amount\begin{align*}\text{Percent of Decrease }=\frac{\text{Amount of Decrease}}{\text{Original Amount}}\end{align*}
Then, write the fraction as a percent.
Now let’s look at how to apply these steps.
Find the percent of decrease from 50 to 40.
First, subtract 40 from 50.
5040=10\begin{align*}50 - 40 = 10\end{align*}
Next, write the fraction:
Percent of Decrease =Amount of DecreaseOriginal Amount=1050\begin{align*}\text{Percent of Decrease }=\frac{\text{Amount of Decrease}}{\text{Original Amount}}=\frac{10}{50}\end{align*}
Then, write the fraction as a percent.
One Way
105050x50x50x====x1001,0001,0005020\begin{align*}\begin{array}{rcl} \frac{10}{50}&=&\frac{x}{100} \\ 50x &=& 1,000 \\ \frac{\cancel{50}x}{\cancel{50}} &=& \frac{1,000}{50} \\ x &=& 20 \end{array}\end{align*}
Another Way
1050=10÷1050÷10=155)1.00¯¯¯¯¯¯¯¯¯¯¯¯ 0.20Divide to 2 decimal places.0.20=20%\begin{align*}&\frac{10}{50}=\frac{10 \div 10}{50 \div 10}=\frac{1}{5} \\ & \overset{ \ \ 0.20}{5 \overline{ ) {1.00 \;}}} \qquad \leftarrow \text{Divide to} \ 2 \ \text{decimal places.} \\ &0.20 = 20 \% \end{align*}The answer is the percent of decrease from 50 to 40 is 20%.
Let’s look at another example.
Find the percent of decrease from 200 to 170.
First, subtract 170 from 200.
200170=30\begin{align*}200 - 170 = 30\end{align*}
Next, write the fraction: Percent of Decrease =Amount of DecreaseOriginal Amount=30200\begin{align*}\text{Percent of Decrease }=\frac{\text{Amount of Decrease}}{\text{Original Amount}}=\frac{30}{200}\end{align*}
Then, write the fraction as a percent.
One Way
30200200x200x200x====x1003,0003,00020015\begin{align*}\begin{array}{rcl} \frac{30}{200}&=&\frac{x}{100} \\ 200x &=& 3,000 \\ \frac{\cancel {200}x}{\cancel{200}} &=& \frac{3,000}{200} \\ x &=& 15 \end{array}\end{align*}
Another Way
30200=30÷10200÷10=32020)3.00¯¯¯¯¯¯¯¯¯¯¯¯ 0.15Divide to 2 decimal places.0.15=15%\begin{align*}&\frac{30}{200}=\frac{30 \div 10}{200 \div 10}=\frac{3}{20} \\ & \overset{ \ \ 0.15}{20 \overline{ ) {3.00 \;}}} \quad \leftarrow \text{Divide to} \ 2 \ \text{decimal places.} \\ &0.15 = 15 \%\end{align*}The answer is the percent of decrease from 200 to 170 is 15%.
### Examples
#### Example 1
Earlier, you were given a problem about the end of summer sale at the farmer’s market.
The swan gourds were marked down from $3.50 to$1.00. What was the percent of decrease?
First, subtract to find the difference.
3.501.00=2.50\begin{align*}3.50 - 1.00 = 2.50\end{align*}
Next, divide that difference by the original amount.
2.50÷3.50=0.714\begin{align*}2.50 \div 3.50 = 0.714\end{align*}
Finally, convert this decimal to a percent.
0.714=71.4%\begin{align*}0.714 = 71.4 \%\end{align*}
The answer is the percent of decrease from $3.50 to$1.00 is 71.4%.
#### Example 2
Jessie’s work schedule went from 20 hours to 18 hours. What was the percent of the decrease?
First, subtract to find the difference.
2018=2\begin{align*}20 - 18 = 2\end{align*}
Next, divide that difference by the original amount.
2÷20=0.1\begin{align*}2 ÷ 20 = 0.1\end{align*}
Finally, convert this decimal to a percent.
0.1=10%\begin{align*}0.1 = 10 \%\end{align*}
The answer is the percent of decrease from 20 hours to 18 hours was 10%.
#### Example 3
Find the percent of decrease from 10 to 5.
First, subtract 5 from 10.
105=5\begin{align*}10 - 5 = 5\end{align*}
Next, write the fraction: Percent of Decrease =Amount of DecreaseOriginal Amount=510\begin{align*}\text{Percent of Decrease }=\frac{\text{Amount of Decrease}}{\text{Original Amount}}=\frac{5}{10}\end{align*}
Then, write the fraction as a percent:
51010x10x10x====x1005005001050\begin{align*}\begin{array}{rcl} \frac{5}{10}&=&\frac{x}{100} \\ 10x &=& 500 \\ \frac{10x}{10} &=& \frac{500}{10} \\ x &=& 50 \end{array}\end{align*}
The answer is the percent decrease from 10 to 5 is 50%.
#### Example 4
Find the percent of decrease from 25 to 15.
First, subtract to find the difference.
2515=10\begin{align*}25 - 15 = 10\end{align*}
Next, divide that difference by the original amount.
10÷25=0.4\begin{align*}10 \div 25 = 0.4\end{align*}
Finally, convert this decimal to a percent.
0.4=40%\begin{align*}0.4 = 40 \%\end{align*}
The answer is the percent of decrease from 25 to 15 is 40%.
#### Example 5
Find the percent of decrease from 125 to 70.
First, subtract 70 from 125.
12570=55\begin{align*}125 -70 = 55\end{align*}
Next, write the fraction: Percent of Decrease =Amount of DecreaseOriginal Amount=55125\begin{align*}\text{Percent of Decrease }=\frac{\text{Amount of Decrease}}{\text{Original Amount}}=\frac{55}{125}\end{align*}
Then, write the fraction as a percent:
55125125x10x125x====x1005,5005,50012544\begin{align*}\begin{array}{rcl} \frac{55}{125}&=&\frac{x}{100} \\ 125x &=& 5,500 \\ \frac{10x}{125} &=& \frac{5,500}{125} \\ x &=& 44 \end{array} \end{align*}
The answer is the percent decrease from 125 to 70 is 44%.
### Review
Find the percent of decrease given the original amount. You may round to the nearest whole percent when necessary or leave your answer as a decimal.
1. From 25 to 10
2. From 30 to 11
3. From 18 to 8
4. From 30 to 28
5. From 12 to 8
6. From 90 to 85
7. From 200 to 150
8. From 97 to 90
9. From 56 to 45
10. From 15 to 2
11. From 220 to 110
12. From 75 to 66
13. From 180 to 121
14. From 1500 to 1275
15. From 18,000 to 900
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Percent of Decrease The percent of decrease is the percent that a value has decreased by.
Percent of Increase The percent of increase is the percent that a value has increased by.
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Next: Quadratic Coupled Map Up: Synchronization of Chaotic Maps Previous: Introduction
# Background
We will start by introducing the necessary background ideas to understand what is happening. First we will introduce the notion of a map, by giving an analogy to differential equations (DE). A system of DE's is a set of equations which tells how some things change with respect to some other variable. For example, the equation:
tells us how the variable x, which represents the position of a particle, changes with respect to time. This is a continuous relationship, so for any time t, there is a value for . The idea of maps is very similar, but now the relationship is discrete. Things no longer change continuously, but in steps. A differential equation can be regarded as a difference equation in a limit as the space in between the steps goes to zero. A map is like this, but there the space in between the steps is fixed and finite. Let's look at a simple example.
So now is not a function of a continuous variable, but we get the value for x at different steps n. Just like a differential equation, we start off with an initial condition, say =1 in our example. Now plug this in and we get =2. Now plug this back in again and we get =3 and so on. This process is called iterating the map. This is the basic idea of a map and while it might seem redundant now, keeping this idea in mind will help make things clearer later.
Now we introduce some other notions which will also be necessary. One key idea is the notion of a fixed point. A fixed point is a point which does not change under iteration of the map; specifically, . This is what is called a period one fixed point. Fixed points of higher period occur as the map is iterated. For example, suppose , but . In this case, we say is a point of period k. It is easy to see from this that a set of period n points will contain all the points of period k, when k|n. Again for example, the set of all period six points for some map will contain all the period one, two and three points, plus some extra points which are specific to the period six.
We ask, what good is a fixed point? Linearization about a fixed point tells us the behavior in the neighborhood of the point. Points near the fixed point act in one of four ways, depending on the nature of the fixed point. If you start at a point in the neighborhood of the fixed point and iterate the map at that initial point, the behavior of the solution will either fall into the fixed point (in which case it is called a sink), move away from it (a source), experience a combination of those two (a saddle) or revolve around it in an orbit (a center). This is illustrated in Fig.1, where the lines represent the behavior of a solution as time goes on.
Figure 1: Behavior in the Neighborhood of Fixed Points
A useful technique to determine the behavior near a fixed point is to linearize the equations of the map about that fixed point. We can then write these equations in matrix form, which we call the Jacobian (we will call it the matrix A later). You can then find values called the eigenvalues of this matrix. We do this by finding the determinant of the difference between the Jacobian and a matrix which is all zero except for 's along the diagonal. Specifically, we have an equation which looks like where I is the identity matrix. It is these values which are our eigenvalues. Depending on how many dimensions your system is, you might get a complicated expression for . For example, if you have a three dimensional system, will be the root of a cubic equation, which might be easily solvable or it might not.
These eigenvalues tell you the stability of the fixed point as prescribed above. If the absolute value of the eigenvalues is less than 1, then the fixed point is a sink. If they are greater than 1, the fixed point is a source. If they are a combination of these, the fixed point is a saddle. If they are exactly equal to 1, then the fixed point of the linearized system is a center, and the quadratic terms control the behavior. The behavior of the map in neighborhoods of periodic points can be characterized in the same way as fixed points
This might be confusing so let's look at an example and see how we can use the machinery defined above. Let's look at the system . This is a two dimensional map and can be thought of as the system:
Now as specified above, we will have a period one fixed point when and . Using a substitution and a little help from the quadratic formula, we see that this occurs when (x,y) = (0,0) or (3,9). So we have two period one fixed points for our system. As noted above, these will always be fixed points, so we are interested in their stability. Let's look at the point (x,y)=(3,9). If we can find the eigenvalues for this point, then we will have a sense for the behavior of points nearby (3,9). So let's look at an arbitrary point close to the fixed point. Let and where and are numbers very close to zero. We want to plug these into the equations and eliminate x and y. So for example, we would write . Working out the algebra, we get a system which looks like:
Now we have a nice system which tells us the behavior close to the fixed point. Since we are close by, we can neglect all higher order terms and only look at the linear terms. So we write out this system in matrix form:
This is our linearized system in matrix form, where the matrix A is the Jacobian for that point. So now we want to solve the equation to get the eigenvalues. So we have:
This gives values for of 3 and -2. The absolute value of these numbers is greater than one, so our fixed point is unstable. This tells us that points near (3,9) will diverge away from that point as we iterate the map. This example may seem a bit long winded, but it serves to show how the basic machinery works. The same technique can be used to find the stability for any fixed point of any period.
Another useful technique to look at is the idea of a bifurcation. When we have a system, there are certain parameters which can be varied. For example, let's take a look at the quadratic map, which will one of the main focuses later on. The map is given by the equation:
Where a is a constant. Since a can be any constant number, changing its value will have the effect of changing the fixed points and their associated eigenvalues. So the inherent behavior of the system depends on the value for this parameter. We call this a bifurcation parameter. We can write a program which iterates the map for a certain value of a, assuming we give an acceptable initial condition for . If we start off at too large a value, the iterates will diverge off to infinity and we get nothing interesting. But if we stay in an acceptable range where is small enough and make a plot of the iterates as a function of our bifurcation parameter a, we get an interesting pattern. This is called the period doubling cascade, as shown in Figure 1. This shows us that the number of fixed points in the system grows faster and the complexity of the system increases.
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# Vector projection
## Why vector projection
A projection of a vector onto another vector has many applications.
Imagine for example a cart on a slope. Given a vector made from two points on the slope and the gravity vector, we can use the projection of the gravitational acceleration vector onto the slope vector to calculate the direction and magnitude of the acceleration of the cart in a simple game physics setup.
Or think of the situation where a character jumps along an irregular polygon wall. When the collision circle of the character collides with the wall, we can calculate the closest point to the wall by projecting the circle’s center onto the wall.
If we want more precise collision detection and response, we can do this as well using vector projection, though we’ll need to learn a bit more about lines and polygons before we can get into that.
But before we can project points onto lines or start colliding polygons, we first need to get a basic understanding of how a vector can be projected onto another vector.
## Calculating the projection of a vector onto another vector
If we look at the picture below, we see $$\vec{a_1}$$ which is the projection of the vector $$\vec{a}$$ onto $$\vec{b}$$. The goal is to calculate $$\vec{a_1}$$ in terms of $$\vec{a}$$ and $$\vec{b}$$.
To make this task easier, or at least more intuitive, we are going to rotate this setup. Let’s arrange the vector $$\vec{b}$$ so that it is parallel with the x-axis.
If we look at the vector $$\vec{a}$$ now, we see that $$\vec{a_1}=\langle x, 0\rangle$$ if $$\vec{a}=\langle x,y\rangle$$. Now let’s scale our setup, so that $$\vec{a}$$ is lying on the unit circle, a circle with as radius 1. In this case we see that $$x$$ is equal to the cosine of $$\alpha$$, which is the angle between $$\vec{a}$$ and $$\vec{b}$$.
$x=cos(\alpha)$
Of course in our second rotated but unscaled scenario, the vector $$\vec{a}$$ doesn’t necessarily have a length equal to 1. Both $$\vec{a}$$ and $$\vec{a_1}$$ are scaled by the length of $$\vec{a}$$. This means that to go back to the second scenario, have to multiply by the length of $$\vec{a}$$ to correct the scale.
$x=\lvert\vec{a}\lvert cos(\alpha)\quad(1)$
As we stated before, we don’t want the angle $$\alpha$$ in our equation, as we want a solution with just the two vectors $$\vec{a}$$ and $$\vec{b}$$. We can use the dot product, since the cosine of an angle is equal to the dot product of the vectors that form the angle if the vectors are normalized
$cos(\alpha)=\hat{a}.\hat{b}$
A normalized vector is one whose length is 1, thus it is obtained by dividing a vector by its length. Doing this with vectors $$\vec{a}$$ and $$\vec{b}$$ gives
$cos(\alpha)=\frac{\vec{a}.\vec{b}}{\lvert \vec{a}\lvert\lvert \vec{b}\lvert}\quad(2)$
Substituting the cosine in (1) by (2) gives
$x=\lvert \vec{a}\lvert\frac{\vec{a}.\vec{b}}{\lvert \vec{a}\lvert\lvert \vec{b}\lvert}$
Or simplified, since we can ignore multiplying and dividing by the same factor $$\lvert \vec{a}\lvert$$
$x=\frac{\vec{a}.\vec{b}}{\lvert \vec{b}\lvert}$
Our projection of vector $$\vec{a}$$ in scenario 2 is was the vector $$\vec{a_1}=\langle x,0\rangle$$ or $$\vec{a_1}=x\langle 1,0\rangle$$, substituting $$x$$ gives
$\vec{a_1}=\frac{\vec{a}.\vec{b}}{\lvert \vec{b}\lvert}\langle1, 0\rangle$
We chose $$\vec{b}$$ to be parallel with the x-axis to clearly see that $$x$$ was equal to the scaled cosine. But in our original scenario $$\vec{b}$$ is not parallel to the x-axis. the only difference with scenario one and two is that we rotated everything. This rotation kept the sizes of the vectors as well as the angle between them intact. The only difference that affects the outcome is that $$\vec{b}$$ is rotated. As we can see, the projected point $$\vec{a_1}$$ always lies somewhere along $$\vec{b}$$. In scenario two, $$\vec{a_1}$$ only lies along the x-axis because $$\vec{b}$$ is parallel with it. The vector $$\langle 1,0\rangle$$ is nothing more than the normalized vector $$\hat{b}$$. Thus our projection of $$\vec{a}$$ on $$\vec{b}$$ in scenario one is
$proj(\vec{a},\vec{b})=\frac{\vec{a}.\vec{b}}{\lvert \vec{b}\lvert}\hat{b}$
Since $$\hat{b}$$ is equal to $$\vec{b}$$ divided by it’s length $$\lvert \vec{b}\lvert$$ we can write
$proj(\vec{a},\vec{b})=\frac{\vec{a}.\vec{b}}{ {\lvert \vec{b}\lvert}^2}\vec{b}$
And since the length $$\lvert \vec{b}\lvert$$ is the square root of the dot product of $$\vec{b}$$ with itself or $$\lvert \vec{b}\lvert=\sqrt{\vec{b}.\vec{b}}$$ we get
$proj(\vec{a},\vec{b})=\frac{\vec{a}.\vec{b}}{\vec{b}.\vec{b}}\vec{b}$
So the projection of $$\vec{a}$$ on $$\vec{b}$$ is the dot product of $$\vec{a}$$ and $$\vec{b}$$ divided by the dot product of $$\vec{b}$$ with itself multiplied with $$\vec{b}$$.
### Projecting a point on a line
Now that we can project a vector onto another vector, it is easy to do things like projecting a point onto a line.
If we have a line AB defined by two points A and B on the line, and C the point we want to project, we build the vector $$\vec{AC}$$, and project it onto the vector $$\vec{AB}$$. From the resulting vector, $$\vec{AC'}$$ we can then make the projected point C’ by adding A again.
Basically what happens is that first we translate everything so that A becomes the origin.
Then we project AC onto AB.
And finally we translate everything back where it was.
### Distance of a point to a line
We can now calculate the distance from a point to a line, as the projected point is the point on the line which lies closest to our point.
### Distance of a circle to a line
Similarly we can calculate the distance from a circle to a line. We just measure the distance between the circle center and the line, and subtract the radius.
While you might be tempted to use this for collision detection, by checking whether the distance is negative, don’t forget we’re using a square root in our distance calculation, which we don’t need. A more optimal way is using the squared distance, and looking whether that is smaller than the square of the radius.
Remember, only use distances when you actually need the distance, for comparing distances, use squared distances instead.
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# Number of permutations of matrices with unique rows and columns
Consider an $m$ by $n$ matrix filled with integers in $\left[0, b\right[$.
There would be $b^{mn}$ possible matrices.
Two matrices would be considered equivalent (in this system) iff you can switch some rows and columns in the matrix and they are the same (So the order doesn't matter). For example:
$$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}$$
How can you get the number of unique matrices?
For $n=m=b=2$, the number of unique matrices is $7$:
$$\text{(The above example)}\tag1$$$$\begin{bmatrix}0&0\\0&0\end{bmatrix}\tag2$$$$\begin{bmatrix}1&0\\0&0\end{bmatrix}=\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&1\end{bmatrix}=\begin{bmatrix}0&0\\1&0\end{bmatrix}\tag3$$$$\begin{bmatrix}1&1\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\1&1\end{bmatrix}\tag4$$$$\begin{bmatrix}1&0\\1&0\end{bmatrix}=\begin{bmatrix}0&1\\0&1\end{bmatrix}\tag5$$$$\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\tag6$$$$\begin{bmatrix}1&1\\1&1\end{bmatrix}\tag7$$
For $n=2; m=3; b=4$, the answer is $430$.
How would I go about calculating this?
### Progress
(Using $n=2; m=3; b=4$ as an example)
You can create a set of possible rows $N$. This would have $b^m$ items and start:
{(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 1, 0), (0, 1, 1), ...}
Which will be written:
{N_0, N_1, N_2, N_3, N_4, ...}
Then, matrices can be defined as of $n$ items in $N$, being rows. Since the order of the rows do not matter, it can be a multiset. So, the first few matrices would be:
{{N_0, N_0}, {N_0, N_1}, {N_0, N_2}, ...}
Before removing duplicates, there would be $\left(b^m\right)^n=b^{mn}$ of these. After removing duplicates there would be $\begin{pmatrix}n+b^m-1\\n\end{pmatrix}$ (Choose $n$ rows from $b^m$ possible with repeats) matrices with unique rows.
Some of these would have more than one unique matrix defined by the same rows but with different columns. So, for each one, I would have to look at the number of unique ways to define them as columns. I don't know how I would go about this in a timely manner.
I then found out about Burnside's Lemma, which I got that that
$$|X/G|=\frac1{|G|}\sum_{g\in G}|X^g|$$
Where $|X/G|$ is the number of oribits (Or unique matrices), $X$ is the set of all possible matrices (So $|X|=b^{mn}$), $G$ is a group of "the actions", like swapping some rows and / or columns, and $X^g$ is the set of the matrices in $X$ where applying the action $g$ leaves the matrix unchanged, but I could not figure out how to calculate the group of possible actions.
• This appeared at the following MSE link if I am interpreting the question correctly. There is a worked example at this MSE link II. – Marko Riedel Mar 12 '17 at 21:32
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You are here Home » Innov8tiv U » Exponent Rules: 5 Key Strategies to Solve Tough Equations
# Exponent Rules: 5 Key Strategies to Solve Tough Equations
Exponent rules explain the ways to simplify various equations with different types of exponents. Equations that contain exponents with the same base can be solved quickly, whereas, in the case of other equations, it is required to use logs to solve them. There are different kinds of exponent equations based on positive exponent, negative exponents, and rational exponent, which may look intimidating at first. However, with various learning strategies, you can make exponent rules easy to follow.
Before going to the strategies, let’s have a brief idea about what are exponents and their rules:
## What are Exponents?
Exponents are also known as powers. It is a value that shows how many times a base number will multiply by itself. The number raised by power is known as the base, while the superscript number above it is the exponent or power. Types of exponents are
positive exponents, negative exponents, zero exponents, and rational exponents.
## Exponent Rules
There are seven exponent rules, or laws of exponents, that apply to these exponents equations, and every student needs to learn them. Each rule shows how to solve different types of math equations and add, subtract, multiply and divide exponents. These seven exponent rules are:
• Product of powers rule: If multiplying two bases with the same value, write the base same and then add the exponents to get the solution.
• Quotient of powers rule: When dividing two bases of the same value, keep the base the same and subtract the exponent values.
• When dividing exponent expressions with the same bases, just subtract the exponent’s power, keeping the same base.
• Power of a power rule: This rule is used to solve exponent equations where one exponent power is raised by another power. In such equations, multiply the exponents together by keeping the same base.
• Power of a product rule: When any base is being multiplied by an exponent, distribute the exponent to each part of the base.
• Power of a quotient rule: This rule is about raising a quotient by a power. In this rule, the exponent needs to be distributed to all values of a quotient within the brackets.
• Zero power rule: Any base raised to the power of zero is equal to one.
• Negative exponent rule: When a number has a negative exponent, flip it into a reciprocal to turn the exponent into a positive exponent.
## 5 Key Strategies to Solve Tough Equations
### 1. Isolate the Exponential Expression by Keeping them on One Side of the Equation:
Make sure that there is an exponential expression on one side of the equation and a whole number on the other side. You need to rearrange the equation if it is not to keep the exponent on one side of the equation.
### 2. Check if any Two Exponents have the Same Base
If the base number is the same in an exponential expression, you can easily solve that equation by placing the exponent on either side.
### 3. Determine Whether any Two Exponents can be Canceled
While simplifying the exponent equations, check if there are terms that can be reduced or canceled. It will help in reducing the complexity of the equation.
### 4. Apply the Same Rule on Both Sides
While solving any exponent equation, remember to apply the same rules on both sides of the equation. If you perform some operation on one side of an equation, you must do it to the other side of the equation. For example, get the logarithms of both sides of the equation. You can use any bases for logs.
### 5. Factor the Terms to Yield Common Terms
Look for terms that can be factored in to yield a common term or solution.
### Conclusion
Rote learning of various mathematical concepts leads to poor subject knowledge and lack of fundamental understanding to learn advanced concepts. This is a major cause of kids lacking interest in math. When kids learn math through active learning, reasoning, and logic, they gain the conceptual fluency required to form a solid math foundation. Cuemath encourages kids to learn math logically by strengthening their reasoning skills. Deep knowledge of each topic enables kids to form links and understanding through various math topics.
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# Horizontal Asymptote Rules, Task Solving Tips, And Examples
By: Angelina Grin
8 min
0
10.04.2022
In the modern world, mathematics is used everywhere, even despite technological progress. That is why educational institutions are so insistently forcing students to study this science. High school students experience the most difficulties because they have to build graphing, find a degree of the numerator or a degree of the denominator, and many other complex tasks. Today we want to pay attention to the rules of horizontal asymptote - with these leading terms, students have the most questions.
## Asymptote - What Mathematical Concept Is It?
If you first plot the asymptotes of the curve, then in many cases, the construction of the graph of the function is more straightforward. The fate of the asymptote is full of tragedy. Imagine what it is like all your life to move in a straight line towards the cherished goal, to come as close to it as possible, but never reach it. So it is with the asymptote: it constantly seeks to reach the curve graph of a function, approaches it at the minimum possible distance, but never touches it.
Scientific definition: Asymptotes are straight lines of unlimited length that approach the function you created as close as possible but do not touch it. At the same time, the variable tends to plus infinity or negative infinity. There are such varieties of asymptotes: vertical, oblique, and horizontal.
### First Type: Vertical Asymptote
The vertical line x = a is the vertical asymptote of the graph of a function if the point x = a is a breakpoint of the second kind. It follows from the definition that the straight line x = a is the vertical asymptote of the graph of the function f (x) if at least one of the conditions is satisfied:
• the limit of the function with the value of the argument tending to some value "a" on the far left is equal to plus or minus infinity;
• the limit of the function with the value of the argument tending to some value of "a" from the right is equal to plus or minus infinity.
You can search for the vertical asymptotes of the function's graph at the points of discontinuity and the boundaries of the domain of definition. The graph of a function that is continuous on the whole number line has no vertical asymptotes.
### Second Type: Oblique Asymptote
The vertical asymptote, which we described above, is always parallel to the coordinate axes. Therefore, to construct it, we only need a certain number - a point on the x-axis or ordinates through which the asymptote passes. For the slant asymptote, more is required—the angular leading coefficient k, which shows the slope of the line. The x-intercept shows how much the line is above or below the origin.
Thus, the graph of a function cannot contain three or more oblique asymptotes in any way - only two. Otherwise, you performed incorrect calculations. When the graph approaches a single oblique asymptote, it is customary to combine infinities under a single record. It would help if you had the following equation of the oblique asymptote:
### Third Type: Horizontal Asymptote
The first thing to know about the horizontal asymptotes is that they are parallel to the Ox axis. If we have a function like this:
Then y = b is the horizontal line of the curve line y = f (x). The horizontal asymptote is a particular case of an oblique one at k = 0. We will go over horizontal asymptote in more detail since it is most often found in geometry homework.
## Asymptote Rules
Finding the asymptotes of the graph of a function is based on the following rules:
• Theorem 1. Let the function y = x be defined at least in some semi-neighborhood of the point x = a, and at least one of its one-sided limits at this point is infinite, that is, equal to + ∞ or - ∞. Then the straight line is the vertical asymptote of the graph of the function. Thus, the vertical asymptotes of the function's graph should be sought at the points of discontinuity of the function. Or the ends of its domain of definition.
• Theorem 2. Let the function y = f (x) is defined for the argument values large enough in absolute value, and there is a finite limit of the function. The straight line y = b is the horizontal asymptote of the graph of the function y = f (x). It may happen that: and . In given functions, b1 and b2 are finite numbers, and then the graph has two different horizontal asymptotes: left-sided and right-sided. If there are only one of the finite limits, b1 or b2, then the graph has either one left-sided or one right-sided horizontal asymptote.
• Theorem 3. Let the function y = f (x) be defined for the values of the argument large enough in absolute value, and there are finite limits: and . Then the straight line is the oblique asymptote of the graph of the function. Note that if at least one of the indicated limits is infinite, there is no oblique asymptote. The oblique asymptote, like the horizontal one, can be one-sided.
## How to Find Horizontal Asymptotes?
As we noted above, the horizontal asymptote of a function f (x) is a straight line parallel to the x-axis to which the function f (x) approaches as it tends to infinity. Then we can write the equation of the horizontal asymptote in this form: y = y0, where y0 is some constant (finite number).
To find the horizontal asymptote of the function f (x), you necessary to find y0. You can get the value of y0 by calculating the limits:
If the value of at least one limit is equal to a finite number y0, then y = y0 is the horizontal asymptote of the function f (x).
Are you having a hard time understanding geometry and formulas? Our paper writing service will help you understand the calculation of degrees of the numerator, degrees of the polynomials, construction of rational expressions for any horizontal asymptote.
### Example How to Find Horizontal Asymptotes
There are many geometric assignments associated with finding horizontal asymptote y and x values. We will consider one of these tasks and show the solution algorithm.
Task: Find all the asymptotes of the graph of a function:
Decision: The function is defined at x є (-∞, -1), U (1, + ∞). Let us find its one-sided limits at the points x = ± 1.
Based on the task, we get the following calculations:
As you can see, we can no longer find two other one-sided limits. The straight lines x = -1 and x = 1 are the vertical asymptotes of the function's graph. Let's calculate:
Hence, the straight line is the horizontal asymptote. Since the horizontal asymptote exists, we no longer look for oblique ones (they are not).
Answer: the graph has two vertical asymptotes x = ± 1 and one horizontal asymptote y = 2.
## Glossary of Terms You Need When Finding a Horizontal Asymptote
For you to better understand how to solve tasks with the calculation of horizontal asymptote, you need to learn the meanings of such words:
• The number above the fractional bar is called the numerator. The numerator shows how many parts were taken from the whole.
• The number below the fractional bar is called the denominator. The denominator shows how many equal parts the whole is divided into.
• A rational function (also polynomial function) is a function resulting from a finite number of arithmetic operations (addition, multiplication, and division) over the variable x and arbitrary polynomial.
• Exponential growth is multiple multiplications. The more value the function takes, the faster it grows further.
• The graph of a function is a set of points (x; y), where x is the argument, and y is the value of the function corresponding to the given argument.
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Creative Writer and Blog Editor
Despite my relatively young age, I am a professional writer with more than 14 years of experience. I studied journalism at the university, worked for media and digital agencies, and organized several events for ed-tech companies. Yet for the last 6 years, I've worked mainly in marketing. Here, at Studybay, my objective is to make sure all our texts are clear, informative, and engaging.
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# Bitwise multiplication
Mr. EG has come up with a great coding scheme for (non-negative) integers in binary. To show off his creation, he demonstrates the following multiplications:
010 x 011 = 010011
011 x 010010 = 010010011
010011 x 010010011 = 010010010011011
01000101 x 01001000111 = 0100100100010100111
What is the result of the following multiplication?
0100010100101 x 01001001001000101
Can you translate all those confusing 1s and 0s back to good old decimal integers?
Note: x represents standard integer multiplication.
The result of the multiplication:
0100010100101 x 01001001001000101 = 010010010010010001010010100101
Description of Mr. EG's coding scheme:
To code a positive integer, factor it as a product of primes, and write those primes in increasing order of size. Write each prime factor in binary, and prepend a number of 0's which is one less than the number of digits. Then concatenate the resulting strings.
As an example, to code $20$, we factor $20=2\cdot 2\cdot 5$. The binary representations of $2$ and $5$ are 10 and 101. We prepend one and two 0's, respectively, to get 010 and 00101. Finally, we concatenate two copies of 010 and one copy of 00101 to get 01001000101.
The purpose of the leading 0's is to indicate the number of digits in each prime, so that a positive integer can be unambiguously recovered from its representation. For instance, if we did not prepend 0's, the string 11111 could either come from $21=3\cdot 7$, or from $31$.
Translating back into decimal, the given multiplications are:
$2\times 3 = 6$,
$3\times 4=12$,
$6\times 12=72$,
$10\times 28=280$.
The missing multiplication becomes $(2\cdot 5\cdot 5)\times(2\cdot2\cdot 2\cdot 2\cdot 5)=4000=(2\cdot2\cdot2\cdot2\cdot2\cdot5\cdot 5\cdot 5)$, which is coded as 010 010 010 010 010 00101 00101 00101.
• Almost perfect (and surprisingly quick!) - you have cracked the encoding and the puzzle. However, your interpretation of 01001000111 is slightly off - it would actually yield 01001000101 when encoded. Fix it and I'll accept your answer. (On a minor note, you write "to encode 30", and then proceed to encode 20...) Apr 17, 2015 at 13:24
2 x 3 = 23 (2,3) -- [Decimal: 19]
3 x 18 = 183 (18,3) -- [Decimal: 187]
19 x 147 = 14719 (147,19) -- [Decimal: 9371]
69 x 583 = 69583 (69,583) -- [Decimal: 149671]
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# How do you simplify root3(1/7)?
Aug 15, 2017
$\frac{\sqrt[3]{49}}{7}$
#### Explanation:
Since ${x}^{\frac{m}{n}} = \sqrt[n]{{x}^{m}}$, we can write $\sqrt[3]{\frac{1}{7}}$ as ${\left(\frac{1}{7}\right)}^{\frac{1}{3}}$.
According to the power of a quotient rule, ${\left(\frac{a}{b}\right)}^{m} = {a}^{m} / {b}^{m}$. Thus, we can say ${\left(\frac{1}{7}\right)}^{\frac{1}{3}} = {1}^{\frac{1}{3}} / {7}^{\frac{1}{3}}$.
From here, we can say ${1}^{\frac{1}{3}} / {7}^{\frac{1}{3}} = \frac{1}{7} ^ \left(\frac{1}{3}\right)$, since $1$ raised to any power is $1$.
We are left with $\frac{1}{7} ^ \left(\frac{1}{3}\right)$ or $\frac{1}{\sqrt[3]{7}}$.
However, we cannot have a radical in the denominator. To rationalize this expression, we must try to make the denominator $7$. To do this, multiply both the numerator and denominator by ${7}^{\frac{2}{3}} / {7}^{\frac{2}{3}}$ or $\frac{\sqrt[3]{{7}^{2}}}{\sqrt[3]{{7}^{2}}}$, respectively.
$\frac{1}{7} ^ \left(\frac{1}{3}\right) \cdot \textcolor{b l u e}{{7}^{\frac{2}{3}} / {7}^{\frac{2}{3}}} = {7}^{\frac{2}{3}} / {7}^{\frac{3}{3}} = {7}^{\frac{2}{3}} / 7 = \frac{\sqrt[3]{{7}^{2}}}{7} = \frac{\sqrt[3]{49}}{7}$
$\frac{1}{\sqrt[3]{7}} \cdot \textcolor{b l u e}{\frac{\sqrt[3]{{7}^{2}}}{\sqrt[3]{{7}^{2}}}} = \frac{\sqrt[3]{49}}{\sqrt[3]{343}} = \frac{\sqrt[3]{49}}{7}$
So, ${\left(\frac{1}{7}\right)}^{\frac{1}{3}}$ simplifies to $\frac{\sqrt[3]{49}}{7}$.
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Paul's Online Notes
Home / Algebra / Solving Equations and Inequalities / Quadratic Equations - Part II
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### Section 2-6 : Quadratic Equations - Part II
2. Complete the square on the following expression.
${u^2} - 11u$
Show All Steps Hide All Steps
Start Solution
First, we need to identify the number we need to add to this. Recall that we will need the coefficient of the $$u$$ to do this. The number we need is,
${\left( {\frac{{ - 11}}{2}} \right)^2} = \frac{{{{\left( { - 11} \right)}^2}}}{{{{\left( 2 \right)}^2}}} = \frac{{121}}{4}$ Show Step 2
To complete the square all we need to do then is add this to the expression and factor the result. Doing this gives,
$\require{color}\require{bbox} \bbox[2pt,border:1px solid black]{{{u^2} - 11u \,{\color{Red} + \frac{{121}}{4}} = {{\left( {u - \frac{{11}}{2}} \right)}^2}}}$
Recall that this will always factor as $$u$$ plus the number inside the parenthesis in the first step, $$- \frac{{11}}{2}$$ in this case.
Do not get too excited about the fractions that can show up in these problems. They will be there occasionally and so we need to be able to deal with them. Luckily, if you can recall the “trick” to the factoring they aren’t all that bad.
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# Writing Equations In Slope-Intercept Form Worksheet Answers
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Writing Equations In Slope-Intercept Form Worksheet Answers – There are many forms that are used to represent a linear equation, one that is commonly found is the slope intercept form. It is possible to use the formula for the slope-intercept to solve a line equation as long as that you have the straight line’s slope , and the y-intercept. It is the y-coordinate of the point at the y-axis meets the line. Learn more about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three basic forms of linear equations: the standard slope-intercept, the point-slope, and the standard. Even though they can provide similar results when used in conjunction, you can obtain the information line produced more quickly using an equation that uses the slope-intercept form. It is a form that, as the name suggests, this form utilizes a sloped line in which you can determine the “steepness” of the line indicates its value.
This formula is able to calculate the slope of a straight line. It is also known as the y-intercept or x-intercept in which case you can use a variety of available formulas. The line equation in this particular formula is y = mx + b. The slope of the straight line is signified in the form of “m”, while its intersection with the y is symbolized with “b”. Each point of the straight line is represented as an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” must remain as variables.
## An Example of Applied Slope Intercept Form in Problems
The real-world, the slope intercept form is commonly used to depict how an object or issue changes over its course. The value that is provided by the vertical axis represents how the equation tackles the magnitude of changes in what is represented via the horizontal axis (typically times).
An easy example of using this formula is to determine how the population grows in a certain area in the course of time. Based on the assumption that the area’s population grows annually by a certain amount, the value of the horizontal axis will grow one point at a time with each passing year and the point worth of the vertical scale will grow to show the rising population by the amount fixed.
You can also note the starting value of a problem. The beginning value is at the y-value of the y-intercept. The Y-intercept is the place where x is zero. Based on the example of a problem above the starting point would be at the time the population reading begins or when time tracking begins , along with the associated changes.
Thus, the y-intercept represents the point in the population at which the population begins to be tracked in the research. Let’s say that the researcher begins to do the calculation or measure in 1995. This year will represent considered to be the “base” year, and the x 0 points would be in 1995. So, it is possible to say that the population in 1995 represents the “y”-intercept.
Linear equations that use straight-line formulas are nearly always solved in this manner. The beginning value is expressed by the y-intercept and the change rate is represented by the slope. The principal issue with the slope-intercept form generally lies in the interpretation of horizontal variables, particularly if the variable is attributed to the specific year (or any kind or unit). The first step to solve them is to make sure you understand the variables’ definitions clearly.
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# Elementary Algebra Review Part 7
Taught by YourMathGal
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Part of video series
Lesson Summary:
In this Elementary Algebra Review lesson, students will learn how to simplify numerical expressions using order of operations, and evaluate variable expressions by plugging in given values. The lesson includes six problems that cover a range of concepts, such as subtraction, division, and multiplication. Each problem is explained step-by-step, and students are encouraged to write each number as an integer or reduced fraction, rather than a decimal. By the end of the lesson, students will have strengthened their algebra skills in a practical, real-world way.
Lesson Description:
Simplify numerical expressions using order of operations, evaluating variable expressions by plugging in given values
More free YouTube videos by Julie Harland are organized at http://yourmathgal.com
• How do you simplify numerical expressions using order of operations?
• How do you simplify (-3 - 4^2)/(-2^5) using order of operations?
• How do you simplify (5 - 4*-3)/(5^2) using order of operations?
• How do you simplify (7 - 4(1 - 3))/(6^2) using order of operations?
• How do you evaluate variable expressions with given values of variables?
• How do you evaluate 5x^3 when x = 2
• How do you evaluate 7yz - 8x when x = 2, y = -3, and z = -4?
• How do you evaluate (4xy - z)/(3x + 1) when x = 2, y = -3, and z = -4?
• How do you evaluate (x + 2y)^3 when x = 2 and y = -3?
• How do you evaluate (3y + x)/(z + 5) when x = 2, y = -3, and z = -4?
• How do you evaluate y^2/(x - yz) when x = 2, y = -3, and z = -4?
• #### Staff Review
• Currently 4.0/5 Stars.
This lesson includes more simplifying of numerical expressions and evaluating variable expressions by using values for each variable. Again, all steps are shown and explained.
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# How to find out the formula for acceleration?
## How to find out the formula for acceleration?
Formula of Acceleration 1 Final Velocity is v 2 Initial velocity is u 3 Acceleration is a 4 Time taken is t 5 Distance traveled is s
How to calculate the acceleration of a meter per second?
It is denoted by symbol a and is articulated as- meter per second squared or m/s 2 is the S.I unit for Acceleration, If t (time taken), v (final velocity) and u (initial velocity) are provided. Then the acceleration is given by formula
### Do you have to subtract initial velocity from final velocity to calculate acceleration?
You need to subtract the initial velocity from the final velocity. If you reverse them, you will get the direction of your acceleration wrong. If you don’t have a starting time, you can use “0”. If the final velocity is less than the initial velocity, the acceleration will be negative, meaning that the object slowed down.
Which is the S.I unit for acceleration?
The S.I unit for acceleration is meter per second square or m/s 2. If t (time taken), v (final velocity) and u (initial velocity) are provided. Then the acceleration is given by the formula Underneath we have provided some sample numerical based on acceleration which might aid you to get an idea of how the formula is made use of:
## Which is the equation for the acceleration vector?
The acceleration vector is →a =a0x^i +a0y^j. a → = a 0 x i ^ + a 0 y j ^. Each component of the motion has a separate set of equations similar to (Figure) – (Figure) of the previous chapter on one-dimensional motion. We show only the equations for position and velocity in the x – and y -directions.
What is the acceleration of a skier at 10.0?
The magnitude of the velocity of the skier at 10.0 s is 25 m/s, which is 60 mi/h. It is useful to know that, given the initial conditions of position, velocity, and acceleration of an object, we can find the position, velocity, and acceleration at any later time.
### What are the equations for one dimensional acceleration?
Allowing the acceleration to have terms up to the second power of time leads to the following motion equations for one dimensional motion. Variable acceleration Polynomial integrals
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Spectrum Math Grade 6 Chapter 1 Lesson 11 Answer Key Problem Solving
Go through the Spectrum Math Grade 6 Answer Key Chapter 1 Lesson 1.11 Problem Solving and get the proper assistance needed during your homework.
Spectrum Math Grade 6 Chapter 1 Lesson 1.11 Problem Solving Answers Key
Solve each problem.
Question 1.
A package weighs 2.6 pounds. How much do 8 of the same-sized packages weigh?
The packages weigh ___ pounds.
Total number of pounds all packages weighs = 20.8.
=> The packages weigh 20.8 pounds.
Explanation:
Number of pounds a package weighs = 2.6.
Number of same-sized packages weigh = 8.
Total number of pounds all packages weighs = Number of pounds a package weighs × Number of same-sized packages weigh
= 2.6 × 8
= 20.8.
Question 2.
It takes Maxine 0.3 hours to make a potholder. How many potholders can she make in 4.5 hours?
She can make ___________ potholders.
Number of potholders can she make in 4.5 hours = 15.
=> She can make 15 potholders.
Explanation:
Number of hours to make a potholder it takes Maxine = 0.3.
Total number of hours = 4.5.
Number of potholders can she make in 4.5 hours = Total number of hours ÷ Number of hours to make a potholder it takes Maxine
= 4.5 ÷ 0.3
= 15.
Question 3.
A box of grass seed weighs 0.62 pounds. How much does a box containing 0.75 times as much grass seed weigh?
The box weighs ____ pounds.
Number of pounds a box containing 0.75 times as much grass seed weigh = 0.465.
=> The box weighs 0.465 pounds.
Explanation:
Number of pounds a box of grass seed weighs = 0.62.
Number of times much grass seed weigh = 0.75.
Number of pounds a box containing 0.75 times as much grass seed weigh = Number of pounds a box of grass seed weighs × Number of times much grass seed weigh
= 0.62 × 0.75
= 0.465.
Question 4.
A collection of nickels is worth $18.60. How many nickels are in the collection? There are ____ nickels in the collection. Answer: Number of nickels are in the collection = 372. => There are 372 nickels in the collection. Explanation: Number of dollars a collection of nickels = 18.60. Conversion: 1 nickles = 0.05 dollars. ?? nickles = 18.60 => 1 × 18.60 = 0.05 × ?? => 18.60 ÷ 0.05 = ?? => 372 nickles = ?? Question 5. Mrs. Anderson bought party favors for the 24 students in her class. Each favor costs$2.27. How much did all the party favors cost?
The favors cost ____.
Total cost of all the party favors = $54.48. => The favors cost$54.48.
Explanation:
Number of students in her class Mrs. Anderson bought party favors = 24.
Cost of each favor = $2.27. Total cost of all the party favors = Number of students in her class Mrs. Anderson bought party favors × Cost of each favor = 24 ×$2.27
= $54.48. Question 6. Each prize for a carnival booth costs$0.32. How many prizes can you buy with $96? You can buy ____ prizes. Answer: Number of prizes can be bought with$96 = 300.
=> You can buy 300 prizes.
Explanation:
Cost of each prize for a carnival booth = $0.32. Total cost of prizes bought =$96.
Number of prizes can be bought with $96 = Total cost of prizes bought ÷ Cost of each prize for a carnival booth =$96 ÷ $0.32 = 300. Question 7. Brittany has a pack of 24 pencils. Each pencil weighs 0.9 grams. How much does the pack of pencils weigh? The pack of pencils weighs ____ grams. Answer: Number of grams the pack of pencils weigh = 21.6. => The pack of pencils weighs 21.6 grams. Explanation: Number of pencils a pack of Brittany has = 24. Number of grams each pencil weighs = 0.9. Number of grams the pack of pencils weigh = Number of pencils a pack of Brittany has × Number of grams each pencil weighs = 24 × 0.9 = 21.6. Solve each problem. Question 1. Workers are using a piece of iron that is 0.324 millimeters thick and a piece of copper that is 0.671 millimeters thick. How much thicker is the copper? The copper is ____ millimeters thicker. Answer: Number of millimeters thicker is the copper = 0.347. => The copper is 0.347 millimeters thicker. Explanation: Number of millimeters thick workers are using a piece of iron = 0.324. Number of millimeters thick a piece of copper = 0.671. Number of millimeters thicker is the copper = Number of millimeters thick a piece of copper – Number of millimeters thick workers are using a piece of iron = 0.671 – 0.324 = 0.347. Question 2. Lenora bought a book for$ 12.36 and some school supplies for $7.29 and$5.47. How much did she spend?
She spent ____.
Total amount she spent = $25.12. => She spent$25.12.
Explanation:
Cost of the book Lenora bought = $12.36. Cost of the school supplies she bought =$7.29 and $5.47. Total amount she spent = Cost of the book Lenora bought + Cost of the school supplies she bought =$12.36 + $7.29 +$5.47
= $19.65 +$5.47
= $25.12. Question 3. Joe’s bill at the grocery store came to$6.08. He paid with a ten-dollar bill and a dime. How much change did he get?
Amount of change she gets = 40.2 Dimes.
=> He received 40.2 Dimes in change.
Explanation:
Cost of the Joe’s bill at the grocery store = $6.08. Amount she paid = ten dollar and a Dime. Conversion: 1 Dollar = 10 Dimes. => 10 Dollars = ?? => 1 × ?? = 10 × 10 => ?? = 100 Dimes. Amount she paid = 100 + 1 = 101 dimes. Conversion: 1 Dollar = 10 Dimes. 6.08 Dollars = ?? Dimes. => 1 × ?? = 10 × 6.08 => ?? = 60.8Dimes. => Amount of change she gets = Cost of the Joe’s bill at the grocery store – Amount she paid = 101 – 60.8 = 40.2 Dimes. Question 4. One bottle holds 67.34 ounces and another bottle holds 48.5 ounces. Combined, how much do they hold? The bottles hold ___ ounces combined. Answer: Total number of ounces two bottles holds = 115.84. => The bottles hold 115.84 ounces combined. Explanation: Number of ounces one bottle holds = 67.34. Number of ounces another bottle holds = 48.5. Total number of ounces two bottles holds = Number of ounces one bottle holds + Number of ounces another bottle holds = 67.34 + 48.5 = 115.84. Question 5. A basic stereo system costs$ 189.67. An upgraded model costs $212.09. How much more does the upgraded model cost? The upgraded model costs ___ more. Answer:$22.42 more the upgraded model costs.
=> The upgraded model costs $22.42 more. Explanation: Cost of a basic stereo system =$189.67.
Cost of an upgraded model = $212.09. Difference: Cost of an upgraded model – Cost of a basic stereo system =$212.09 – $189.67 =$22.42.
Question 6.
Lin ran 0.683 kilometers on Wednesday and 0.75 kilometers on Thursday. How far did he run on the two days combined?
He ran ____ kilometers over both days.
Total number of kilometers he ran on the two days combined = 1.433.
=> He ran 1.433 kilometers over both days.
Explanation:
Number of kilometers on Wednesday Lin ran = 0.683.
Number of kilometers on Thursday Lin ran = 0.75.
Total number of kilometers he ran on the two days combined = Number of kilometers on Wednesday Lin ran + Number of kilometers on Thursday Lin ran
= 0.683 + 0.75
= 1.433.
Question 7.
A certain cabinet door is actually made of three thin boards that are pressed together. The boards are 0.371 inches, 0.13 inches, and 0.204 inches thick. How thick is the cabinet door?
The door is ___ inches thick.
|
Finding the average distance in an equilateral triangle
In the previous post I looked at the average distance between 2 points in a rectangle. In this post I will investigate the average distance between 2 randomly chosen points in an equilateral triangle.
Drawing a sketch.
The first step is to start with an equilateral triangle with sides 1. This is shown above. I sketched this using Geogebra – and used some basic Pythagoras to work out the coordinates of point C.
I can then draw a square of sides 1 around this triangle as shown above. I’m then going to run a Python program to randomly select points and then work out the distance between them – but I need to make sure that the 2 points chosen are both inside this triangle. For this I need to work out the equation of the line AC and CB.
Using basic coordinate geometry we can see that the line AC has equation y = √3x. We want the inequality y < √3x so that we are on the correct side of this line.
The line BC has equation y = -√3x + √3. Therefore the triangle must also satisfy the inequality y < -√3x + √3.
I can then run the following code on Python, with finds the average distance between points (a,c) and (b,d) both within the unit square but also subject to the 2 inequality constraints above.
When this is run it performs 999,999 trials and then finds the average distance. This returns the following value:
So we can see that the average distance is just over a third of a unit.
Finding the average distance of an equilateral triangle of length n.
We can then draw the sketch above to find the equation of lines AC and CB for an equilateral triangle with lengths n. This leads to the following inequalities:
y < √3x
y < -√3x + √3n
So we can then modify the code as follows:
This then returns the average distances for equilateral triangles of sizes 1 to 10.
And when we plot this on Desmos we can see that there is a linear relationship:
The regression line has gradient 0.36 (2sf) so we can hypothesise that for an equilateral triangle of size n, the average distance between 2 points is approximately 0.36n.
Checking the maths
I then checked the actual equation for the average distance between 2 points in an equilateral triangle of sides n:
This gives us:
So we can see that we were accurate to 2 significant figures. So this is a nice mixture of geometry, graphing and computational power to provide a result which would be otherwise extremely difficult to calculate.
|
## Book: Mathematics Part-II
### Chapter: 13. Probability
#### Subject: Maths - Class 12th
##### Q. No. 5 of Exercise 13.5
Listen NCERT Audio Books - Kitabein Ab Bolengi
5
##### The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs(i) none(ii) not more than one(iii) more than one(iv) at least one will fuse after 150 days of use.
Let us assume that the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials be x.
As we can see that the trial is made with replacement, thus, the trials will be Bernoulli trials.
It is already mentioned in the question that, p = 0.05
Thus, q = 1 – p = 1 – 0.05 = 0.95
Here, we can clearly observe that x has a binomial representation with n = 5 and p = 0.05
Thus, P(X = x) = nCxqn-xpx , where x = 0, 1, 2, …n
= 5Cx(0.95)5-x(0.05)x
(i) Probability of no such bulb in a random drawing of 5 bulbs = P(X = 0)
= 5C0(0.95)5-0(0.05)0
= 1× 0.955
= (0.95)5
(ii) Probability of not more than one such bulb in a random drawing of 5 bulbs = P(X≤ 1)
= P(X = 0) + P(X = 1)
= 5C0(0.95)5-0(0.05)0+ 5C1(0.95)5-1(0.05)1
= 1× 0.955 + 5 × (0.95)4 × 0.05
= (0.95)4 (0.95 +0.25)
= (0.95)4 × 1.2
(iii) Probability of more than one such bulb in a random drawing of 5 bulbs = P(X>1)
= 1 – P(X ≤ 1)
= 1 – [(0.95)4 × 1.2]
(iv) Probability of at least one such bulb in a random drawing of 5 bulbs = P(X ≥ 1)
= 1 – P(X < 1)
= 1 – P(X = 0)
= 1 –(0.95)5
5
1
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# Iterative Methods
GCSELevel 6-7AQAEdexcelOCR
## Iterative Methods
Iterative methods or iterations is the idea of repeating a process over and over with the purpose of getting closer to an answer. In maths, iterative methods are often used when finding an exact answer is not so simple. There are 3 key skills involved with iterative method questions at GCSE level, which are shown below.
Make sure you are happy with the following topics before moving onto Iterative methods
Level 6-7GCSEAQAEdexcelOCR
## Skill 1: Trial and Improvement
Trial and improvement is an iterative process whereby you try different solutions for an equation until you get the degree of accuracy that you want. This is easiest to see with an example.
Example: Use trial and improvement show that $\textcolor{blue}{x}^3+6\textcolor{blue}{x}=4$ has a solution between $0$ and $1$.
Give your answer to $1$ dp.
First pick the largest number in the range ($\textcolor{blue}{x=1}$). Then, substitute this value into the equation.
$\textcolor{blue}{x}^3 +6\textcolor{blue}{x} = (\textcolor{blue}{1})^3 +6(\textcolor{blue}{1}) = 1 + 6 = 7$
We know $7>4$ (it’s too big), so we must try a smaller number. $\textcolor{blue}{x=0}$
$\textcolor{blue}{x}^3+6\textcolor{blue}{x}=(\textcolor{blue}{0})^3 + 6(\textcolor{blue}{0})=0$
$0$ is too small, so we know the solution must be somewhere in between $0$ and $1$.
We now repeat this process,
\begin{aligned}(\textcolor{blue}{0.5})^3 + 6(\textcolor{blue}{0.5})&=3.125.. \, \text{Too small}& \\ (\textcolor{blue}{0.7})^3 + 6(\textcolor{blue}{0.7})&=4.543..\, \text{Too big}& \\ (\textcolor{blue}{0.6})^3 + 6(\textcolor{blue}{0.6})&=3.816..\, \text{Too small}& \\ (\textcolor{blue}{0.65})^3 + 6(\textcolor{blue}{0.65})&=4.17..\,\,\,\, \text{Too big}&\end{aligned}
As we can now see, $0.6 =$ too small, but $0.65=$ too big. This means the actual solution must be between these two values.
We know that any number between $0.6$ and $0.65$ must round to $0.6$ to $1$ dp, so the solution must be $0.6$ to $1$dp.
Level 6-7GCSEAQAEdexcelOCR
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Level 6-7GCSEAQAEdexcelOCR
## Skill 2: Using Iteration Machines
An iteration machine allows us to find an approximate solution to an equation we may not be able to solve any other way.
Example: Use the iterative formula shown to find the value of $x$ to $1$ dp
$\textcolor{red}{x_{n+1}} = \dfrac{4-\textcolor{limegreen}{x_{n}}^3}{6}$
Use $x_1 = 1$.
Step 1: Use $x_1$ as $\textcolor{limegreen}{x_n}$ in the equation to give $\textcolor{red}{x_{x+1}}$
$x_1 = 1 \,\,\ \text{gives } \,\,\, \textcolor{red}{x_2} = \dfrac{4-{\textcolor{limegreen}{x_{1}}}^3}{6}=\dfrac{4-(\textcolor{limegreen}{1})^3}{6}=\textcolor{red}{0.5}$
Step 2: Repeat the process using $x_2$ as $\textcolor{limegreen}{x_n}$ to give $\textcolor{red}{x_3}$
$x_2 = 0.5 \,\,\, \text{ gives } \,\,\, \textcolor{red}{x_3} = \dfrac{4-{\textcolor{limegreen}{x_{2}}}^3}{6}=\dfrac{4-(\textcolor{limegreen}{0.5})^3}{6}=\textcolor{red}{0.6548333...}$
Step 3: Repeat the process to give $x_4$, $x_5$, ….
$x_3=0.6548333... \,\,\, \text{ gives } \,\,\, \textcolor{red}{x_4}=\dfrac{4-{\textcolor{limegreen}{x_{3}}}^2}{6}=\dfrac{4-(\textcolor{limegreen}{0.6548333...})^3}{6}=\textcolor{red}{0.6217...}$
$x_4=0.6217... \,\,\, \text{ gives } \,\,\, \textcolor{red}{x_5}=\dfrac{4-{\textcolor{limegreen}{x_{4}}}^2}{6}=\dfrac{4-(\textcolor{limegreen}{0.6217...})^3}{6}=\textcolor{red}{0.6266...}$
Step 4: Once two consecutive answers round to the same $1$ dp answer, we have our final answer.
$\xcancel{0.6548333... \,\,\, \text{rounds to} \,\,\, 0.7}$
$0.6217... \,\,\, \text{rounds to} \,\,\, 0.6$
$0.6266... \,\,\, \text{rounds to} \,\,\, 0.6$
Final answer $x = 0.6$ ($1$dp)
Note: When doing a question like this, it makes your life a lot easier if you use the ANS key on your calculator.
Level 6-7GCSEAQAEdexcelOCR
## Skill 3: Creating the iteration formula
It is often necessary to form the iterative methods formula (e.g. $x_{n+1}=f(x_n))$ by rearranging an equation.
Example: Show that $x^3+x^2 =7$ can be rearranged to give to give $x_{n+1} = \sqrt[3]{7-{x_{n}}^2}$
In order to do this we need to rearrange the equation to be in the form stated, then add in the notation required.
\begin{aligned} (-x^2) \,\,\,\,\,\,\,\,\, x^3+x^2 &=7 \\ (\sqrt[3]{}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x^3 &= 7-x^2 \\ x &= \sqrt[3]{7-x^2}\end{aligned}
Finally we add in the required notation to complete our answer.
\begin{aligned}x &= \sqrt[3]{7-x^2} \\ x_{n+1} &= \sqrt[3]{7-{x_{n}}^2} \end{aligned}
Level 6-7GCSEAQAEdexcelOCR
## Iterative Methods Example Questions
We will form a table with one column of $x$ values, one column on the results of calculating $2x^3-6x$, and one column stating if the answer is bigger or smaller than the desired $1$.
So, if $1.8$ gives a result that is too small and $1.85$ gives a result that is too big, then the actual solution must be somewhere between these two values.
Given that any number between $1.8$ and $1.85$ must round to $1.8$, the solution must be $1.8$ to $1$ decimal place.
Save your answers with
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To find a solution we will use the recursive formula, until we get two consecutive terms which round to the same number to $2$ decimal places.
\begin{aligned} x_1&=2 \\ x_2&=\sqrt[3]{3(2)+9}=2.4662 \\ x_3&=\sqrt[3]{3(2.4662)+9}=2.54060 \\ x_4&=\sqrt[3]{3(2.54060 )+9}=2.55207 \\ x_5&=\sqrt[3]{3(2.55207 )+9}=2.55383\end{aligned}
These last two results both round to $2.55$ to $2$dp, so the solution must be $2.55$ to $2$ decimal places.
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To find a solution we will use the recursive formula, until we get two consecutive terms which round to the same number to $2$ decimal places.
\begin{aligned}x_1&=1 \\ x_2&=\dfrac{-3}{(1)^2+5}=-0.5 \\ x_3&=\dfrac{-3}{(-0.5)^2+5}=-0.5714 \\ x_4&=\dfrac{-3}{(-0.5714 )^2+5}=-0.5632 \\ x_5&= \dfrac{-3}{(-0.5632 )^2+5}= -0.564\end{aligned}
These last two results both round to $-0.56$ to $2$dp, so the solution must be $-0.56$ to $2$ decimal places.
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To find a solution we will use the recursive formula, until we get two consecutive terms which round to the same number to $3$ decimal places.
\begin{aligned}x_1&=4 \\ x_2&=\dfrac{3}{(4)^2}+3=3.1875 \\ x_3&=\dfrac{3}{(3.1875)^2}+3=3.29527 \\ x_4&=\dfrac{3}{(3.29527)^2}+3=3.27627 \\ x_5&=\dfrac{3}{(3.27627)^2}+3=3.27949 \\ x_6&=\dfrac{3}{(3.27949)^2}+3=3.27894 \\ x_7&=\dfrac{3}{(3.27894)^2}+3=3.27903\end{aligned}
These last two results both round to $3.279$ to $3$dp, so the solution must be $3.279$ to $3$ decimal places.
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To find a solution we will use the recursive formula, until we get two consecutive terms which round to the same number to $2$ decimal places.
\begin{aligned} x_1&=2 \\ x_2&=\sqrt[3]{6(2)+5}= 2.57128 \\ x_3&=\sqrt[3]{6(2.57128)+5}= 2.73363 \\ x_4&=\sqrt[3]{6(2.73363)+5}= 2.77641 \\ x_5&=\sqrt[3]{6(2.77641)+5}= 2.78746 \\ x_6&=\sqrt[3]{6(2.78746)+5}= 2.79030 \\ x_7&=\sqrt[3]{6(2.79030)+5}= 2.791036\end{aligned}
These last two results both round to $2.79$ to $2$dp, so the solution must be $2.79$ to $2$ decimal places.
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## Iterative Methods Worksheet and Example Questions
### (NEW) Iterative Methods (Higher) Exam Style Questions - MME
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## Iterative Methods Drill Questions
### Iterative Methods
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#### Transcript Sample
```Parameters VS. Statistics
and
Samples & Census
How do we compare
statistics and parameters?
M2 Unit 4: Day 3
Statistics VS Parameters
Statistics (for samples) :
x and S
Parameters (for populations) :
and
Today…
You will be given the population
parameters and
You will be given a data set for a sample
and you will find the statistics x and S
Then you will compare the statistics to
the parameters
Example1
For a large population, the mean =
25.1 and the s.d =5.3.
A random sample produced the following data values:
25, 21, 32, 14, 17, 22, 29.
Compare the mean and s.d of the random sample to
the population parameters.
Sample
x 22.86
S x 6.36
Population
25.1
x 5.3
The mean of the sample is
less than population mean
while the s.d. of the sample
is greater than the
population s.d.
Example 2
For a large population, the mean =
4.8 and the s.d =3.6.
One random sample produced the following data
values: 5, 1, 3, 4, 7, 6, 8, 2, 1, 3.
Compare the means and s.d’s of the random samples
to the population parameters.
Sample 1
x4
S x 2.45
Population
4.8
x 3.6
The mean and standard
deviation of the sample are
less than population mean
and the population s.d.
Example 3
For a large population, the mean =
12.3 and the s.d =4.24.
One random sample produced data values of
21, 13, 15, 12, 20, 22, 16, 18.
Another random sample produced data values of
14, 15, 12, 20, 29, 18.
Compare the means and s.d’s of the random samples
to the population parameters. The mean of the both
Sample 1
Sample 2
x 17.13
S x 3.72
x 18
S x 6.10
Population
12.3
x 4.24
samples are greater than
population mean. The s.d. of
the 1st sample is less than
that of the population s.d
while the s.d. of the 2nd
sample is greater than the
population s.d.
Example 4: Compare Statistics and Parameters
Population
18.4
x 15.6
Gillian
Ted
x 17.8
S x 12.35
x 21.1
S x 14.52
less than
greater than
less than
Example 5: Compare Statistics and Parameters
A teacher wants to know how often her students study at
night. Random samples are collected from John and
Sally, two students. Their results are given below. The
population mean is 1.24 and the population standard
deviation is about 2.32. Compare the means and
standard deviations of the random samples to the
populations parameters. John: .35, 1.23, .55, 2, 3.1
Sally: 1.1, .46, 2.3, .25, 2.2
John
Sally
x 1.45
S x 1.13
x 1.26
S x 0.96 x 2.32
Population The mean of the both samples are
1.24 greater than population mean.
The standard deviations of both
samples are less than that of the
population standard deviation.
Comparing
A Population and A Sample
A population is a group of people or objects
• “ the whole”, everything or everyone which is
relevant
A sample is a subset of the population
• “the part”, just a piece of the population
Example: ECHS student survey
• sample = 1 student from each class
• population = all the students in the school
Example:
Each week, the Gallup Poll questions about
1500 adult US residents to determine
national opinion on a wide variety of issues.
What is the population?
Population = all US residents that week
Who is the sample?
Sample = 1500 US residents questioned
Random Sample
A sample in which each member of the
population has an equal chance of being
selected.
Examples:
Putting everyone’s name in a hat and pulling one out.
Having a computer randomly generate a list of items.
Representative samples
A)
B)
C)
A principal wants to know if teachers would
be willing to give \$1.00 each week to help
provide new text books for the students.
Which would give a representative sample?
Ask teachers who have first period planning
Ask the teachers that want new text books in
their classrooms
Representative random samples
The following list provides the average number of
cookies baked for a bake sale. A food inspector
wants to choose samples of cookies to inspect.
Choose a method of random selection that is
Joe
20
Sally
12
Jane
28
A) choose the first cookie baked by each person
B) choose at random, 2 cookies from each person
from Sally and 7 cookies from Jane
Representative random samples
The following list provides the average number of tshirts made by a t-shirt company. The manager
wants to choose samples of t-shirts to inspect.
Choose a method of random selection that is
Black
1500
representative of all t-shirts.
White
1250
Red
1000
A) choose 5 black t-shirts, 10 white and 50 red
B) choose at random, 30 Black shirts, 25 White shirts
and 20 Red shirts
C) choose at random, one of each shirt
Samples
Unbiased vs. Biased
Unbiased (FAIR) Sample:
A representation of the population you
Biased Sample:
A sample that over represents the
population or under represents part of
the population.
Examples:
1. A pharmacy wants to find out if its customers
would be interested in ordering their
prescription through their website.
Asking every third prescription customer if they
would be interested would be an ______
unbiased sample.
2. A movie theater wants to know which day of
the week its customers prefer to see movies.
A biased sample would be to ask the question to
all of its customers on Thursday.
Examples:
3. A school is trying to find out what is the
favorite sport among the students.
Asking every other football player, “What is your
biased sample.
favorite sport?” would be a
4. A store wants to know which style of jeans
customers prefer.
An unbiased sample would be to ask every
other customer what their favorite style of jeans
is.
Census
Every individual in the population is
contacted
• not really practical because they are very time
consuming and extremely costly.
Example:
• the US does a census every 10 years.
It tries
to contact everyone in the US and asks them
questions about race, ethnicity, # of people in
their house hold, etc.
The next census will occur in 2020.
Assignment:
Pg. 275 # 6 – 8, 11
Pg. 276 # 1, 2, 4 – 6
```
|
### Home > CCAA > Chapter 8 Unit 7 > Lesson CC2: 8.3.1 > Problem8-62
8-62.
Simplify each expression.
1. $\frac { 9 } { 15 } \div \frac { 4 } { 3 }$
When you divide by a fraction, you are multiplying by its reciprocal.
$\frac{9}{20}$
1. $- \frac { 19 } { 20 } + \frac { 4 } { 5 }$
Find a common denominator.
$-\frac{3}{20}$
1. $-\frac{8}{9}\div\left(-\frac{2}{5}\right)$
See part (a).
1. $3 \frac { 1 } { 2 } \div 1 \frac { 1 } { 7 }$
Convert these mixed numbers into fractions greater than one.
$\frac{49}{16}$
1. $-\frac{3}{4}-\left(-\frac{11}{16}\right)$
This is the same as $-\frac{3}{4}+\frac{11}{16}$.
$-\frac{1}{16}$
1. $\frac{2}{9}\cdot\frac{14}{15}\cdot\left(-\frac{9}{10}\right)$
$\frac{2}{9}\Big(-\frac{9}{10}\Big)\cdot\frac{14}{15}$
1. $-10\frac{4}{5}+\left(-\frac{3}{8}\right)$
You may add whole numbers and fractions separately.
$(-10)+\left(-\frac{4}{5}+-\frac{3}{8}\right)$
$-11\frac{7}{40}$
1. $\frac{12}{5}\div\left(-\frac{1}{10}\right)$
See part (a).
|
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## SAT (Fall 2023)
### Course: SAT (Fall 2023)>Unit 6
Lesson 5: Problem Solving and Data Analysis: lessons by skill
# Percents | Lesson
## What are percentages, and how frequently do they appear on the test?
A percentage is a ratio out of 100 that represents a part-to-whole relationship. Percent (percent) means parts per hundred.
In this lesson, we'll learn to:
1. Calculate percentages using part and whole values
2. Switch between equivalent forms of percentages
3. Calculate percent change
On your official SAT, you'll likely see 1 to 3 questions about percentages.
You can learn anything. Let's do this!.
## How do we calculate percentages?
### Finding a percentage
Finding a percentSee video transcript
### Calculating a percent value
There are two values that are important for finding a percentage: a part and a whole. To calculate a percentage, use the following formula:
percent, equals, start fraction, start text, p, a, r, t, end text, divided by, start text, w, h, o, l, e, end text, end fraction, dot, 100
For example, say you took a quiz in math class and got 21 out of the 24 questions correct. We could calculate the percentage of questions you got correct as follows:
• The part is 21.
• The whole is 24.
start fraction, 21, divided by, 24, end fraction, dot, 100, equals, 87, point, 5, percent
If we have any two of the part, the whole, and the percentage, we can solve for the missing value!
Note: be careful when identifying the part and the whole; the part won't necessarily be the smaller number!
Example: What is 150, percent of 8 ?
### Finding complementary percentages
Since all parts of a whole should add up to 100, percent, we can also use percentages to determine the value of any missing parts.
Example: A bag is filled with red and blue marbles. If there are 25 marbles in the bag, and 56, percent of the marbles are blue, how many red marbles are there in the bag?
### Try it!
Try: Identify parts and wholes
Of the juniors at South High School, 45, percent played for a school sports team. If 117 juniors played for a school sports team, how many students are in the junior class?
• 45 is the
.
• 117 is the
.
• We need to solve for the
.
Try: Find complementary percentages
A group of 1, comma, 300 people took a survey in which they had to select their favorite of the four seasons (winter, spring, summer, and fall). The incomplete table below shows the results.
Seasonpercent of respondents
Winter16
Spring22
Summer
Fall28
What percentage of respondents said summer was their favorite season? (Ignore the percent symbol when entering your answer. For example, if the answer is 12, percent, enter 12.)
What number of respondents said summer was their favorite season?
## What forms can percentages take?
### Converting percentages to decimals and fractions
Converting percents to decimals & fractions exampleSee video transcript
### Switching between forms of percentages
We can use equivalent forms of percentages interchangeably and choose the one(s) that best suit our purpose.
For example, 50, percent is equivalent to the following:
• The ratio 50, colon, 100, which reduces to 1, colon, 2.
• The fraction start fraction, 50, divided by, 100, end fraction, which reduces to start fraction, 1, divided by, 2, end fraction.
• The decimal value 0, point, 5.
Note: a useful shortcut for converting percentages to decimals is to remove the percent symbol and move the decimal point 2 places to the left.
Decimal equivalents for percentages are highly useful when making calculations. For example, if we wanted to find 112, percent of value x, we could simply multiply x by the decimal equivalent, 1, point, 12.
Example: What is 25, percent of 364 ?
### Translating percentage word problems
You'll frequently see percentages referenced in word problems. Luckily, there's an easy way to translate these word problems into arithmetic:
• "what" means x
• "is" means equals
• "of" means multiplied by
• "percent" means divided by 100
So:
36 is what percent of 60? → 36, equals, start fraction, x, divided by, 100, end fraction, dot, 60
### In what form should I enter my answer?
Questions on the SAT may ask for "what percent" and require you to enter that value into the answer field.
In these instances, you should not enter decimal or fractional equivalents, but instead enter the percent value as an integer (without a percent sign). So, if the answer is 50, percent, you should simply enter 50.
### Try it!
Try: switch between forms of percentages
In 2019, 29, percent of Major League Baseball starting pitchers were left-handed.
The equivalent fraction is
.
The equivalent decimal is
.
Try: use decimal equivalents in calculations
In testing two engines, a mechanic finds that Engine A puts out 125, percent of the horsepower put out by Engine B. If Engine B puts out 120 horsepower, how many horsepower does Engine A put out?
First, we can convert 125, percent to the decimal
.
We can then multiply that decimal by
to find the number of horsepowers put out by Engine A.
Engine A puts out
horsepower.
## How do we calculate percent changes?
### Percentage word problems
Percent word problem: guavasSee video transcript
### Calculating percent change
We're often asked to calculate by what percent a quantity changes relative to an initial value: the percent discount on jeans, the percent increase in population, etc. When calculating a percent change from an initial value to a final value:
1. Find the difference between the initial and final values.
2. Divide the difference by the initial value.
3. Convert the decimal to a percentage by multiplying the quotient by 100.
percent, start text, c, h, a, n, g, e, end text, equals, start fraction, start text, d, i, f, f, e, r, e, n, c, e, end text, divided by, start text, i, n, i, t, i, a, l, end text, end fraction, dot, 100
Example: The price of a vacuum was reduced from dollar sign, 200 to dollar sign, 170. What was the percent reduction in price?
If we have any two of the percent change, the initial value, and the final value, we can solve for the missing value! And remember: decimal equivalents for percentages are highly useful when making calculations.
Example: The price of a pair of shoes is dollar sign, 40 after a 20, percent discount. What is the price of the shoes before discount?
### Try it!
Try: Identify Initial value, final value, and percent change
A customer’s monthly internet bill was dollar sign, 63, point, 89. Due to a rate increase, her monthly bill is now dollar sign, 68, point, 86. To the nearest tenth of a percent, by what percent did the amount of the customer’s internet bill increase?
• 63, point, 89 is the
.
• 68, point, 86 is the
.
• We need to solve for the
.
The difference between initial and final values is
.
The percent change (to the nearest tenth of a percent) is
. (Ignore the percent symbol when entering your answer. For example, if the answer is 12, point, 4, percent, enter 12, point, 4.)
Practice: Calculate a percentage
Where Do People Get Most of Their News?
SourcePercent of those surveyed
Internet/social media43, percent
Television23, percent
Newspapers15, percent
Others/none of the above8, percent
The table above shows a summary of 1, comma, 800 responses to a survey question. Based on the table, how many of those surveyed get most of their news from either newspapers or the radio?
Practice: USE COMPLEMENTARY PERCENTAGES
Tara has read 95, percent of the books she owns. If Tara owns 160 books, how many of her books has she NOT read?
Practice: CALCULATE PERCENT INCREASE
The price of a particular brand of headphones was dollar sign, 7 in 2016. In 2018, the price of the same brand of headphones was dollar sign, 10. What is the approximate percent increase in the price of the headphones?
Practice: Calculate percent change
YearSubscriptions sold
201810, comma, 500
201912, comma, 390
The manager of a streaming video service received the report above on the number of subscriptions sold by the service. The manager estimated that the percent increase from 2019 to 2020 would be double the percent increase from 2018 to 2019. To the nearest integer, how many subscriptions did the manager expect would be sold in 2020 ?
## Things to remember
Percent means parts per hundred.
p, percent, equals, start fraction, p, divided by, 100, end fraction
A shortcut for converting percentages to decimals is to remove the percent symbol and move the decimal point left 2 places.
When translating word problems:
• "what" means x
• "is" means equals
• "of" means multiplied by
• "percent" means divided by 100
The sum of all parts of a whole is 100, percent.
When calculating a percent change from an initial value to a final value:
1. Find the difference between the initial and final values.
2. Divide the difference by the initial value.
3. Convert the resulting decimal to a percentage.
percent, start text, c, h, a, n, g, e, end text, equals, start fraction, start text, d, i, f, f, e, r, e, n, c, e, end text, divided by, start text, i, n, i, t, i, a, l, end text, end fraction, dot, 100
## Want to join the conversation?
• 44% times 25 should be 11 not 12 as stated in the "Finding complementary percentages."
• You're correct. If you want to make sure nobody else has the same confusion, you can always tell Khan Academy about the mistake by clicking the "Report a Mistake" link when you go to ask a question.
• I have my SAT on 4th December. Is Khan Academy enough for practise? The books are a bit expensive. Are there other online resources?
• It is! Very effective and good to teach you all you need to know on what's going to be on the test and what to look forward to when taking it. Khan Academy has helped me know what to focus on and to sharpen the skills I don't do very well; skills I would've probably failed on the SAT.
I wish you luck on your SAT!
• is doing this for sat prep helpful?
• The answer to the one on subscriptions is 16,850.4. why do we round down and not up?
(1 vote)
• The question has "To the nearest integer" in it, which more often than not signifies that we round as normal to the nearest whole subscription. The answer we get from doing the math is 16,850.4. To round to a place, we look at the place behind it. If it's 0-4, we round down. Else, we round up. Here, it's a 4 in the tenths place, so we round the answer down to 16, 850.
|
# Into Math Grade 4 Module 7 Lesson 1 Answer Key Represent Division with Regrouping
We included HMH Into Math Grade 4 Answer Key PDF Module 7 Lesson 1 Represent Division with Regrouping to make students experts in learning maths.
## HMH Into Math Grade 4 Module 7 Lesson 1 Answer Key Represent Division with Regrouping
I Can represent and record division problems with a 1-digit divisor and regrouping.
Dani manages a charter boat business. She has these fishing lures for 3 boats. She wants to put the same number of lures on each boat. How many lures should Dani put on each boat?
Dani manages a charter boat business.
She has these fishing lures for 3 boats.
She wants to put the same number of lures on each boat.
Total lures = 4 x 10 + 8 = 48
So, 48 ÷ 3 =
Dani put 16 lures equally on each boat.
Turn and Talk Why does Dani need to open one package of lures in order to divide all the lures equally?
In order to get equal lures for each boat, she has to open one package of lures. Because 4 packets i.e., 4 x 10 = 40 can not be shared equally, one lure will be remained after sharing 39, 39 = 3 x 13. So, she has to open one package of lures in order to divide all the lures equally.
Build Understanding
1. Frank owns a charter boat business. He buys 56 fishing rods. Frank divides the rods equally among his 4 boats. How many fishing rods does each boat get?
Find 56 ÷ 4.
Use base-ten blocks to show 56 as 5 tens 6 ones.
Try to put all the blocks into 4 equal groups.
Draw a quick picture to show your work.
A. Which blocks are not in a group? Why?
___________________________
B. How can you regroup the leftover ten with the leftover ones?
___________________________
C. Use your blocks to show the regrouping. Then finish putting all the blocks in 4 equal groups.
• How many tens and ones are in each group? ___________
• How many fishing rods does each boat get? _________
Turn and Talk How can regrouping 1 ten as 10 ones help when you divide?
Step It Out
2. Use quick pictures to represent division,
Find 85 ÷ 3.
The picture shows 85 as 8 tens 5 ones. The 3 circles represent the equal groups. Cross off tens and ones as you put them into the groups.
A. Share the 8 tens equally among the 3 groups. How many tens are in each group? ___
How many tens are left over? _______
B. Regroup the leftover tens as ones.
How many ones are there now? ________
C. Share the ones equally among the 3 groups.
How many ones are in each group? ________
How many ones are left over? ________
D. Write the whole-number quotient and remainder.
85 ÷ 3 is ____ r ___.
Turn and Talk Why do you need to regroup in Step B?
Check Understanding Math Board
Question 1.
A Lana has 76 sand dollars. She puts the same number in 4 boxes. How many sand dollars are in each box? Use base-ten blocks to represent the division. Record the results.
_____ sand dollars are in each box.
Question 2.
Use Tools Marco puts all of his scallop shells in 6 jars. He puts the same number in each jar. How many scallop shells are in each jar? Use base- ten blocks to represent the division. Then draw a quick picture to show your work.
Use Tools Use base-ten blocks or quick pictures to represent the division. Record the results.
Question 3.
56 ÷ 4 ___
Question 4.
85 ÷ 7 ___
Question 5.
38 ÷ 6 ___
|
# Geometry
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### Using the Angle-Side-Angle Method to Prove Triangles Congruent
The ASA (Angle-Side-Angle) postulate states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent
### Using the Isosceles Triangle Theorems to Solve Proofs
The following two theorems — If sides, then angles and If angles, then sides — are based on a simple idea about isosceles triangles that happens to work in both directions:
### Properties of Rhombuses, Rectangles, and Squares
The three special parallelograms — rhombus, rectangle, and square — are so-called because they’re special cases of the parallelogram. (In addition, the square is a special case or type of both the rectangle
### The Properties of a Kite
A kite is a quadrilateral in which two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means that one side can’t be used in both pairs). Check out the kite in the below figure.
### How to Prove that a Quadrilateral Is a Parallelogram
There are five ways in which you can prove that a quadrilateral is a parallelogram. The first four are the converses of parallelogram properties (including the definition of a parallelogram). Make sure
### How to Prove that a Quadrilateral Is a Rectangle
There are three ways to prove that a quadrilateral is a rectangle. Note that the second and third methods require that you first show (or be given) that the quadrilateral in question is a parallelogram
### How to Prove that a Quadrilateral Is a Rhombus
You can use the following six methods to prove that a quadrilateral is a rhombus. The last three methods in this list require that you first show (or be given) that the quadrilateral in question is a parallelogram
### How to Prove that a Quadrilateral Is a Square
There are four methods that you can use to prove that a quadrilateral is a square. In the last three of these methods, you first have to prove (or be given) that the quadrilateral is a rectangle, rhombus
### How to Calculate the Area of a Parallelogram, Kite, or Trapezoid
The area formulas for the parallelogram, kite, and trapezoid are based on the area of a rectangle. The following figures show you how each of these three quadrilaterals relates to a rectangle, and the
### A Rhombus Area Problem
Here’s a rhombus area problem involving triangles and ratios: Find the area of rhombus RHOM given that MBis 6 and that the ratio of RB to BH is 4 : 1, as shown in the following figure.
### How to Calculate the Area of a Kite
You calculate the area of a kite by using the lengths of its diagonals. Here’s an example: what’s the area of kite KITE in the following figure?
### How to Calculate the Area of a Trapezoid
You can use the right-triangle trick to find the area of a trapezoid. The following trapezoid TRAP looks like an isosceles trapezoid, doesn’t it? Don’t forget — looks can be deceiving.
### The Properties of Trapezoids and Isosceles Trapezoids
A trapezoid is a quadrilateral with exactly one pair of parallel sides (the parallel sides are called bases). The following figure shows a trapezoid to the left, and an isosceles trapezoid on the right
### How to Calculate the Area of a Regular Octagon
You can calculate the area of a regular octagon with the standard regular polygon method, but there’s a nifty alternative method based on the fact that a regular octagon is a square with its four corners
### Interior and Exterior Angles of a Polygon
Everything you need to know about a polygon doesn’t necessarily fall within its sides. You may need to find exterior angles as well as interior angles when working with polygons:
### How to Find the Number of Diagonals in a Polygon
To find the number of diagonals in a polygon with nsides, use the following formula:
### How to Identify and Name Similar Polygons
You can identify similar polygons by comparing their corresponding angles and sides. As you see in the following figure, quadrilateral WXYZ is the same shape as quadrilateral
### How to Align Similar Polygons
If you get a problem with a diagram of similar polygons that aren’t lined up in the same orientation, consider redrawing one of them so that they’re both aligned in the same way. This may make the problem
### Determining a Triangle's Area from Its Base and Height
If you know the base and height of a triangle, you can use a tried-and-true formula to find its area. You likely first ran into the basic triangle area formula in about sixth or seventh grade. If you’ve
### Determining a Triangle's Area from Its Three Sides
When you know the length of a triangle’s three sides and you don’t know an altitude, you can use Hero’s formula to find the area. Check it out:
### Determining the Area of an Equilateral Triangle
To calculate a triangle’s area — for most types of triangles — you need to know the triangle’s base and height. However, with an equilateral triangle, all you need to know is the length of one of its sides
### How to Find the Centroid of a Triangle
The three medians of a triangle intersect at its centroid. The centroid is the triangle’s balance point, or center of gravity. (In other words, if you made the triangle out of cardboard, and put its centroid
### How to Do a Parallelogram Proof
A good way to begin a proof is to think through a game plan that summarizes your basic argument or chain of logic. The following examples of parallelogram proofs show game plans followed by the resulting
### How to Prove that a Quadrilateral Is a Kite
Proving that a quadrilateral is a kite is a piece of cake. Usually, all you have to do is use congruent triangles or isosceles triangles. Here are the two methods:
### How to Calculate the Area of a Quadrilateral
There are five formulas that you can use to calculate the area of the seven special quadrilaterals. There are only five formulas because some of them do double duty — for example, you can calculate the
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The word problems on the proportion aid you in a mathematical comparison between two numbers. The proportion gets majorly based on ratio and fractions. This article on Proportion Word Problems Worksheet with Answers PDF gives you a various number of proportion problems. At the end of this page, you can get clear knowledge of the concept of proportion. It says when two ratios are equivalent they are in proportion.
Proportion encourages solving many real-life problems. The ratio and proportion are key foundations to grasp the various concepts in mathematics. Proportions are denoted by the symbol “::”, “=”. This free printable Worksheet on Word Problems on Proportion provides various types of questions with answers to make you understand clearly how to solve the proportions problems in exams. By solving proportion examples with answers pdf, you can also improve your problem-solving skills & math skills.
Do Refer:
## Proportion Word Problems with Answers
Example 1:
If the numbers are 4, 15, 7, and 20. What number should be added to make the numbers proportional?
Solution:
Given numbers are 4, 15, 7, and 20.
Let the needed number be k.
Now, we write the numbers according to the problem
4+k, 15+k, 7+k, and 20+k are proportional numbers.
Here,
$$\frac{4+k}{15+k}$$ = $$\frac{7+k}{20+k}$$
⇒ (4+k) × (20+k) = (7+k) × (15+k)
⇒ 80+ 4k +20k +k² = 105+ 7k+ 15k+ k²
⇒ 24k+80 = 22k+42
⇒ 24k- 22k = 105-80
⇒ 2k = 25
⇒ k = $$\frac{25}{2}$$
⇒ k = 12.5
Thus, the required number is 12.5.
Example 2:
Priya enlarged the size of a photo to a height of 20 inches. What is the new width if it was originally 4 inches tall and 2 inches in width?
Solution:
The height of a photo is 20 inches.
Now, to find the new width.
The ratio to calculate is 4: 2 = 20: w
⇒ $$\frac{4}{2}$$ = $$\frac{20}{w}$$
⇒ 4w = 20×2
⇒ 4w = 40
⇒ w = $$\frac{40}{4}$$
⇒ w = 10.
Thus, the new width of the photo is 10inches.
Example 3:
If P is directly proportional to n when P is 4 and n is 6. Find the value of P when n is 8.
Solution:
Given P is proportional to n ( P ∝ n ).
Now, convert to an equation multiplied by k the constant of variation.
⇒ P = nk
To find k use the given condition
P = 4 when n = 6
P = nk
⇒ k = $$\frac{n}{P}$$ = $$\frac{6}{4}$$
Now, the equation is P = $$\frac{6}{4}$$ × n = $$\frac{6n}{4}$$
When n = 8, then
P= $$\frac{6n}{4}$$ = $$\frac{6×8}{4}$$ = $$\frac{48}{4}$$ = 12.
Hence, the value of P is 12 when n is 8.
Example 4:
Find the third proportion of 14 and 22?
Solution:
The proportions given are 14 and 22.
Let the third proportion be x.
According to the problem, 14 and 22 are in continued proportion.
Now,
$$\frac{14}{22}$$ = $$\frac{22}{x}$$
⇒ 14× x = 22 × 22
⇒ 14x = 484
⇒ x = 484/14 = 34.57
Therefore, the third proportion is 34.57.
Example 5:
Ram, Prudhvi, and Mahesh have $15,$22, and $25 respectively with them. Father asks them to give him an equal amount so that the money held by them now is in continued proportion. Find the amount taken from each of them? Solution: Let the amount taken from each of them be$m.
Now,
15-m, 22-m, and 25-m are in continued proportion.
Thus, $$\frac{15-m}{22-m}$$ = $$\frac{22-m}{25-m}$$
⇒ (15-m)(25-m) = (22-m)(22-m)
⇒ 375 – 15m -25m +m² = 484 – 44m +m²
⇒ 375 – 40m = 484 -44m
⇒ 375 – 484 = -44m +40m
⇒ -109 = -4m
⇒ m = 109/4
⇒ m = 27.25
Therefore, the required amount is \$27.25.
Example 6:
Keerti ran 150meters in 25seconds. How long did she take to run 3meters?
Solution:
Given that
keerti ran 150metres in 25seconds.
Now, to find the time she takes to run 3mts.
Let k be the time required.
$$\frac{25}{150}$$ = $$\frac{k}{3}$$
⇒ k = $$\frac{25}{150}$$ × 3
⇒ k = $$\frac{75}{150}$$
⇒ k = 0.5
Thus, keerti took 0.5seconds to complete 3meters.
Example 7:
Check whether the following numbers form a proportion or not.
(i) 4.5, 3.5, 6.6, and 8.8
(ii) 2$$\frac{2}{4}$$, 1$$\frac{3}{2}$$, 2.2, and 5.5
Solution:
(i) Given 4.5: 3.5 = $$\frac{4.5}{3.5}$$ = $$\frac{45}{35}$$ = $$\frac{9}{7}$$
6.6: 8.8 = $$\frac{6.6}{8.8}$$ = $$\frac{66}{88}$$ = $$\frac{6}{8}$$
Therefore, $$\frac{4.5}{3.5}$$ ≠ $$\frac{6.6}{8.8}$$
Thus, 4.5, 3.5, 6.6, and 8.8 are not in proportion.
(ii) Given 2$$\frac{2}{4}$$, 1$$\frac{3}{2}$$, 2.2, and 5.5
2$$\frac{2}{4}$$: 1$$\frac{3}{2}$$
= $$\frac{10}{4}$$: $$\frac{5}{2}$$
= $$\frac{10}{4}$$ × 4: $$\frac{5}{2}$$ × 4
= 10: 10 = 1: 1
2.2: 5.5 = $$\frac{2.2}{5.5}$$ = $$\frac{22}{55}$$ = $$\frac{2}{5}$$ = 2: 5
Thus, the given numbers 2$$\frac{2}{4}$$, 1$$\frac{3}{2}$$, 2.2, and 5.5 are not in proportion.
Example 8:
Find the fourth proportional of numbers 3, 6, and 12?
Solution:
Given three proportional numbers are 5, 6, and 12.
Let the fourth proportional be x.
Now, 3, 6,12, and x be the proportionality numbers.
Thus,
$$\frac{3}{6}$$ = $$\frac{12}{x}$$
⇒ 3x = 12×6
⇒ 3x = 72
⇒ x = $$\frac{72}{3}$$
⇒ x = 24.
Hence, the fourth proportional number is 24.
Example 9:
If the mean proportion is 16 and the third proportional is 64 then find the two numbers?
Solution:
Given mean proportion is 16 and the third proportion is 64.
Let the required numbers be a and b.
According to the given problem,
√ab = 16
⇒ ab = 16²
⇒ ab = 256
Now, $$\frac{b²}{a}$$ = 64
⇒ b² = 64a
⇒ a = $$\frac{b²}{64}$$
Substitute, a = $$\frac{b²}{64}$$ in ab = 256
⇒ $$\frac{b²}{64}$$ × b = 256
⇒ $$\frac{b³}{64}$$ = 256
⇒ b³ = 256 × 64
⇒ b³ = 28 × 26
⇒ b³ = 212 × 2²
⇒ b = 24 ×2²
⇒ b = 26
⇒ b = 64
So, from the equation a = $$\frac{b²}{64}$$, we get
a = $$\frac{64²}{64}$$
⇒ a = $$\frac{4096}{64}$$
⇒ a = 64
Therefore, the required two numbers are 64 and 64.
|
Illinois State University Mathematics Department
MAT 312: Probability and Statistics for Middle School Teachers Dr. Roger Day (day@ilstu.edu)
### Assignment #3: Possible Solutions
Assignment due on Thursday, 6/24/04
Use the Couples' Ages data set, handed out in class, to complete the following tasks.
1. Create the equation for the median-median line of this data set, using husband's ages as the first element (x) of each ordered pair and wife's ages as the second element (y) in each ordered pair. Show all steps in the process as we illustrated in class and express your equation in slope-intercept form. You can check the equation of your median-median line using your calculator.
Here is the scatter plot of the data. Below it, I show each of the three clusters of data points from which we generate the summary points.
Summary Point (x1,y1) = (25,24) Summary Point (x2,y2) = (31,28) Summary Point (x3,y3) = (54,46)
Using the outer summary points, the equation of the line containing them is
y=(22/29)x + 146/29
or, approximately,
y=0.7586x + 5.0345.
When x = 31, this equation gives us y = 828/29, or approximately 28.5517.
Our second summary point, however, has a y-value of 28, so we need to move our line one-third of the distance between y = 28 and y = 828/29. This distance is one third of 16/29, or 16/87, which is approximately 0.1839.
So we decrease the y-intercept of our first line by 16/87, giving us the equation
y=(22/29)x + 422/87
or, approximately,
y=0.7586x + 4.8506.
Your calculator can confirm for you the ordered pairs comprising the summary points and the equation of this median-median line. Here is the scatter plot of the data with the median-median line on it.
2. Use your calculator to generate the least-squares linear regression equation for this data set, using husband's ages as the first element (x) of each ordered pair and wife's ages as the second element (y) in each ordered pair. Copy all the information your calculator shows you when you generate the least-squares equation. Express your equation in slope-intercept form.
After entering the data into a TI-83 calculator and selecting the command for Linear Regression, the calculator shows the following:
LinReg y=ax+b a=.8789973259 b=2.445803822 r^2=.8909477545 r=.9439002884
So the least-squares linear regression equation is
y=0.8789973259x+2.445803822
Here is a scatter plot of the data with both regression lines on it. The red line is the median-median line and the green line is the least-squares line.
3. Calculate the sum of the squared error terms (SSE) for each of the models you just created (i.e., for the median-median line and for the least-squares line). You are encouraged to use your calculator's "lists" features to carry out your calculations. Express each sum rounded to the nearest hundredth of a unit.
The SSE for the median-median line is 617.95 and the SSE for the least-squares line is 461.22. Look at the last plot above and convince yourself, graphically, why the SSE is smaller for the least-squares line.
|
# A circle has a chord that goes from ( pi)/2 to (11 pi) / 6 radians on the circle. If the area of the circle is 5 pi , what is the length of the chord?
##### 1 Answer
Aug 8, 2016
$= 9.4$
#### Explanation:
A chord that goes from $\frac{\pi}{2}$to $11 \frac{\pi}{6}$
so it travels the distance $11 \frac{\pi}{6} - \frac{\pi}{2} = 4 \frac{\pi}{3}$;
or
$4 \frac{\pi}{3} \div 2 \pi = \frac{2}{3}$ of the Circumference of the Circle
Area of the Circle$= \pi {r}^{2} = 5 \pi$
or
${r}^{2} = 5$
or
$r = \sqrt{5}$
or
$r = 2.24$
Circumference of the circle$= 2 \pi r = 2 \left(\pi\right) \left(2.24\right) = 14.07$
Therefore Length of the chord$= 14.07 \left(\frac{2}{3}\right) = 9.4$
|
# 34 6 req 4 for example 29 2 req 6 4 resistor in
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Unformatted text preview: 6×3 =2 6+3 (The symbol is used to indicate a parallel combination.) Also, the 1and 5- resistors are in series; hence their equivalent resistance is 6 3 2Ω Req +2 Figure 2.34 =6 Req = 4 For Example 2.9. 2Ω Req 6Ω =4 resistor in Fig. 2.35(a); (a) 4Ω = + 2.4 +8 Req 2.4 Ω 8Ω = 14.4 (b) Figure 2.35 PRACTICE PROBLEM 2.9 2Ω By combining the resistors in Fig. 2.36, find Req . Answer: 6 . Req v v Figure 2.36 | e-Text Main Menu | Textbook Table of Contents | Equivalent circuits for Example 2.9. 3Ω 6Ω 1Ω | 2Ω 8Ω 4×6 = 2.4 4+6 The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In Fig. 2.35(b), the three resistors are in series. Hence, the equivalent resistance for the circuit is 6 3Ω 4Ω This 4- resistor is now in parallel with the 6their equivalent resistance is 4 6Ω 8Ω Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In Fig. 2.35(a), we notice that the two 2- resistors are in series, so the equivalent resistance is 2 5Ω = +5 1 1Ω 4Ω 4Ω 3Ω For Practice Prob. 2.9. Problem Solving Workbook Contents 5Ω 46 PART 1 DC Circuits EXAMPLE 2.10 Calculate the equivalent resistance Rab in the circuit in Fig. 2.37. 10 Ω c a 1Ω 1Ω d 6Ω Rab 4Ω 3Ω 5Ω 12 Ω b b Figure 2.37 10 Ω a c 1Ω d 2Ω b b 3Ω 6Ω b b (a) 10 Ω c a 3Ω 2Ω b b b (b) Figure 2.38 Equivalent circuits for Example 2.10. b For Example 2.10. Solution: The 3- and 6- resistors are in parallel because they are connected to the same two nodes c and b. Their combined resistance is 3×6 3 6= (2.10.1) =2 3+6 Similarly, the 12- and 4- resistors are in parallel since they are connected to the same two nodes d and b. Hence 12 × 4 12 4= (2.10.2) =3 12 + 4 Also the 1- and 5- resistors are in series; hence, their equivalent resistance is 1 +5 =6 (2.10.3) With these three combinations, we can replace the circuit in Fig. 2.37 with that in Fig. 2.38(a). In Fig. 2.38(a), 3- in parallel with 6- gives 2- , as calculated in Eq. (2.10.1). This 2- equivalent resistance is now in series with the 1-...
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# Equivalent fractions and comparing fractions: FAQ
## What are equivalent fractions?
Equivalent fractions are fractions that represent the same part of a whole, even though they look different. For example, $\frac{1}{2}$ and $\frac{2}{4}$ are equivalent fractions because they both represent half of a whole.
Try it yourself with this exercise:
## How can I use fraction models to find equivalent fractions?
You can use fraction models, like pie charts or bar models, to visually see how two fractions might be equivalent. For example, if you draw two pie charts and divide one in half and the other into four equal parts, you can see that $\frac{2}{8}$ and $\frac{1}{4}$ are equivalent.
$\frac{2}{8}=\frac{1}{4}$
Try it yourself with this exercise:
## How can I use number lines to find equivalent fractions?
You can use number lines to show how two fractions might be equivalent. For example, if you mark $\frac{1}{2}$ and $\frac{4}{8}$ on a number line, you'll see that they both land on the same spot.
Try it yourself with this exercise:
## How can I compare fractions with unlike denominators?
To compare fractions with unlike denominators, you can either use a common denominator or use benchmarks.
Common denominators are when two or more fractions have the same denominator (the bottom number in a fraction). To use a common denominator, you can find equivalent fractions for both fractions so that they both have the same denominator.
Benchmarks are fractions that are commonly used to compare other fractions to. The most common benchmark fractions are $\frac{1}{4}$, $\frac{1}{2}$, and $\frac{3}{4}$. To use benchmarks, you can compare the fractions to a benchmark fraction, like $\frac{1}{2}$, to see which fraction is larger or smaller.
Try it yourself with these exercises:
## Why is it important to learn about comparing fractions and equivalent fractions?
Comparing fractions and equivalent fractions are important skills for a few reasons. For one, they help you build a deeper understanding of what fractions represent and how they work. Additionally, knowing how to compare fractions and find equivalent fractions can be helpful in real life situations, such as when you're cooking or baking and need to measure out ingredients.
## Want to join the conversation?
• Do I have to do all of the exercises?
• Yeah what the heck?
• when are we gona do 100 fractions
• in 3 years
• what even r faqs
• plz tell me
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# How do you teach repeated addition?
## How do you teach repeated addition?
A good way of teaching repeated addition is to help your child visualise the question. They could use a number line to track each step of ‘4+4+4+4’. They could also group images into sets of 4 to help them realise that ‘4×4’ = 4 lots of 4.
### What is a repeated addition model?
Repeated addition is adding groups of numbers together multiple times. It is a type of multiplication that is used to teach children how to multiply.
Why do we teach repeated addition?
Why Teaching Repeated Addition is Important. Repeated addition is the easiest way to progress from additive to multiplicative understanding. It helps kids build rock solid mathematical understandings, compared to solving pages of multiplication problems.
What is the repeated addition of 3×5?
A teacher may ask a child to work out what 3 lots of 5 are. They may be asked to make 3 groups of 5 counters. Then they may be asked to write the number sentence 5 + 5 + 5 = and then work out the answer, which is 15. They need to repeatedly be made aware that 3 x 5 is ‘3 lots of 5’.
The addition is taking two or more numbers and adding them together, that is, it is the total sum of 2 or more numbers. Example: How many apples are there in all? There are 7 apples in one basket and 4 apples in the other.
### Why repeated addition is such an important mathematical concept to develop?
Repeated addition is the easiest way to progress from additive to multiplicative understanding. It helps kids build rock solid mathematical understandings, compared to solving pages of multiplication problems. If this step is missed students will struggle with multiplication down the line.
What is the short way to repeated addition of the same number?
Repeated addition of the same number is called multiplication.
What is wrong with repeated addition?
If we need a simple, elementary-level catchphrase to replace “multiplication is repeated addition,” how about this? Multiplication is counting by this-per-that groups. As with any such phrase, this statement fails to capture all that multiplication entails.
## How do you write an addition lesson plan?
Introduction
1. Call the students together as a group.
3. Take responses by raised hands.
4. Show the students two manipulatives.
6. Take responses by raised hands.
7. Reiterate to the students one plus one equals two.
### How do you introduce an addition lesson?
Here’s a 7-step process for teaching addition that creates easier lesson plans for you, and better understanding for your students.
1. Introduce the concept using countable manipulatives.
2. Transition to visuals.
3. Use a number line.
4. Counting Up.
5. Finding the ten.
6. Word problems.
7. Memorize the math facts.
What is the objective of teaching addition?
The objective of this lesson is for students to represent addition and subtraction with objects and actions to understand the concepts of adding to and taking from. The key vocabulary words in this lesson are addition, subtraction, together and apart.
What is the easiest way to teach a child addition?
How to Teach Addition | 7 Simple Steps
1. Introduce the concept using countable manipulatives. Using countable manipulatives (physical objects) will make addition concrete and much easier to understand.
2. Transition to visuals.
3. Use a number line.
4. Counting Up.
5. Finding the ten.
6. Word problems.
7. Memorize the math facts.
## How do you explain addition?
### What is a repeated addition problem?
Ask students what addition problem you’ve drawn ( 5 + 5 ). Write the addition problem on the board. Review the term repeated addition and explain: Each of these circles have five dots, so we are adding the same number twice. This is called repeated addition because we are adding the same number, or equal groups, repeatedly.
Teach your students how repeated addition relates to multiplication. This lesson builds number sense and supports a conceptual understanding of multiplication. Students will be able to use repeated addition as a strategy to multiply two single-digit factors. Draw two circles on the board with five dots in each.
What is 6 times 2 as repeated addition?
So we will use this X-looking thing. And so, six times two can be viewed as repeated addition in exactly this same way. So six times two would be equal to 12. And we could go the other way around. If someone were to ask you, what is four times three?
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Algebra I Recipe: The Slope of a Line
A. Slope Facts
1. Positive slope – when the line slants upward from left to right.
2. Negative slope – when the line slants downward from left to right.
3. There are two directions or changes with slope.
• The up and down change or vertical change is the change in the y-values.
• The left and right change or horizontal change is the change in the x-values.
4. The slope formula is m = (y2 - y1) / (x2 - x1) and used when you know two points on the line.
• Label the points (x1, y1) & (x2, y2).
• Substitute the numbers into the formula.
• Perform the operation in the numerator and denominator.
• Reduce the fraction completely.
• DO NOT write slope as a mixed number.
5. A horizontal line with equation y = # has a slope of zero.
• The y-values would be the same therefore zero would be obtained in the numerator of the formula.
• Zero divided by any number equals zero.
6. A vertical line with equation x = # has no slope or undefined slope.
• The x-values would be the same therefore zero would be obtained in the denominator of the formula.
• Any number divided by zero is undefined.
Examples:
(7, 2) & (1, 1)
(3, -2) & (-1, -2)
(6, -3) & (3, -1)
(3, 0) & (3, -2)
B. How to Determine the Slope of a Graphed Line Using (rise)/(run)
1. Pick any two points on the line.
2. Determine the rise by counting the spaces you move up or down.
• Move up – positive number
• Move down – negative number
3. Determine the run by counting the spaces you move right or left.
• Move right – positive number
• Move left – negative number
C. How to Graph a Line with a Given Slope and a Given Point the Line Goes Through
1. Graph the given point.
2. Use the movements of slope (or rise/run) from the graphed point.
3. Make a point after making the two movements and repeat to graph more points.
Examples:
(1, 2) and m = -3/2
(-4, 3) and m = 5
(6, -2) and m = ¼
(-3, -5) and m = -2
G Redden
Show Related AlgebraLab Documents
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# 1.9 Proficiency exam
Page 1 / 1
This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module is a proficiency exam to the chapter Addition and Subtraction of Whole Numbers. Each problem is accompanied with a reference link pointing back to the module that discusses the type of problem demonstrated in the question. The problems in this exam are accompanied by solutions.
## Proficiency exam
( [link] ) What is the largest digit?
9
( [link] ) In the Hindu-Arabic number system, each period has three values assigned to it. These values are the same for each period. From right to left, what are they?
ones, tens, hundreds
( [link] ) In the number 42,826, how many hundreds are there?
8
( [link] ) Is there a largest whole number? If so, what is it?
no
( [link] ) Graph the following whole numbers on the number line: 2, 3, 5.
( [link] ) Write the number 63,425 as you would read it aloud.
Sixty-three thousand, four hundred twenty-five
( [link] ) Write the number eighteen million, three hundred fifty-nine thousand, seventy-two.
18,359,072
( [link] ) Round 427 to the nearest hundred.
400
( [link] ) Round 18,995 to the nearest ten.
19,000
( [link] ) Round to the most reasonable digit: During a semester, a mathematics instructor uses 487 pieces of chalk.
500
For problems 11-17, find the sums and differences.
675
( [link] ) $\text{3106}+\text{921}$
4,027
188
( [link] ) $\begin{array}{c}\hfill 5,189\\ \hfill 6,189\\ \hfill 4,122\\ \hfill \underline{+8,001}\end{array}$
23,501
( [link] ) $\text{21}+\text{16}+\text{42}+\text{11}$
90
( [link] ) $\text{520}-\text{216}$
304
70,123
( [link] ) Subtract 425 from 816.
391
( [link] ) Subtract 712 from the sum of 507 and 387.
182
( [link] ) Is the sum of 219 and 412 the same as the sum of 412 and 219? If so, what makes it so?
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
In the number 779,844,205 how many ten millions are there?
From 1973 to 1979, in the United States, there was an increase of 166.6% of Ph.D. social scientists to 52,000. How many were there in 1973?
7hours 36 min - 4hours 50 min
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# 2023 AMC 10A Problems/Problem 18
## Problem
A rhombic dodecahedron is a solid with $12$ congruent rhombus faces. At every vertex, $3$ or $4$ edges meet, depending on the vertex. How many vertices have exactly $3$ edges meet?
$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$
## Solution 1
Note Euler's formula where $\text{Vertices}+\text{Faces}-\text{Edges}=2$. There are $12$ faces and the number of edges is $24$ because there are 12 faces each with four edges and each edge is shared by two faces. Now we know that there are $14$ vertices on the figure. Now note that the sum of the degrees of all the points is twice the number of edges. Let $x=$ the amount of vertices with $3$ edges. Now we know $\frac{3x+4(14-x)}{2}=24$. Solving this system of equations gives $x = 8$ so the answer is $\boxed{\textbf{(D) }8}$. ~aiden22gao ~zgahzlkw (LaTeX) ~ESAOPS (Simplified)
## Solution 2 (Quick solution)
Let $x$ be the number of vertices with three edges, and $y$ be the number of vertices with four edges. Since there are $\frac{4*12}{2}=24$ edges on the polyhedron, we can see that $\frac{3x+4y}{2}=24$. Then, $3x+4y=48$. Notice that by testing the answer choices, (D) is the only one that yields an integer solution for $y$. Thus, the answer is $\boxed{\textbf{(D) }8}$.
~Mathkiddie
## Solution 3
With $12$ rhombi, there are $4\cdot12=48$ total boundaries. Each edge is used as a boundary twice, once for each face on either side. Thus we have $\dfrac{48}2=24$ total edges.
Let $A$ be the number of vertices with $3$ edges (this is what the problem asks for) and $B$ be the number of vertices with $4$ edges. We have $3A + 4B = 48$.
Euler's formula states that, for all convex polyhedra, $V-E+F=2$. In our case, $V-24+12=2\implies V=14.$ We know that $A+B$ is the total number of vertices as we are given that all vertices are connected to either $3$ or $4$ edges. Therefore, $A+B=14.$
We now have a system of two equations. There are many ways to solve for $A$; choosing one yields $A=\boxed{\textbf{(D) }8}$.
Even without Euler's formula, we can do a bit of answer guessing. From $3A+4B=48$, we take mod $4$ on both sides.
$$3A+4B\equiv48~(\mod{4})$$ $$3A\equiv0~(\mod{4})$$
We know that $3A$ must be divisible by $4$. We know that the factor of $3$ will not affect the divisibility by $4$ of $3A$, so we remove the $3$. We know that $A$ is divisible by $4$. Checking answer choices, the only one divisible by $4$ is indeed $A=\boxed{\textbf{(D) }8}$.
~Technodoggo ~zgahzlkw (small edits) ~ESAOPS (LaTeX)
## Solution 4
Note that Euler's formula is $V+F-E=2$. We know $F=12$ from the question. We also know $E = \frac{12 \cdot 4}{2} = 24$ because every face has $4$ edges and every edge is shared by $2$ faces. We can solve for the vertices based on this information.
Using the formula we can find: $$V + 12 - 24 = 2$$ $$V = 14$$ Let $t$ be the number of vertices with $3$ edges and $f$ be the number of vertices with $4$ edges. We know $t+f = 14$ from the question and $3t + 4f = 48$. The second equation is because the total number of points is $48$ because there are 12 rhombuses of $4$ vertices. Now, we just have to solve a system of equations. $$3t + 4f = 48$$ $$3t + 3f = 42$$ $$f = 6$$ $$t = 8$$ Our answer is simply just $t$, which is $\boxed{\textbf{(D) }8}$ ~musicalpenguin
## Solution 5
Each of the twelve rhombi has two pairs of angles across from each other that must be congruent. If both pairs of angles occur at $4$-point intersections, we have a grid of squares. If both occur at $3$-point intersections, we would have a cube with six square faces. Therefore, two of the points must occur at a $3$-point intersection and two at a $4$-point intersection.
Since each $3$-point intersection has $3$ adjacent rhombuses, we know the number of $3$-point intersections must equal the number of $3$-point intersections per rhombus times the number of rhombuses over $3$. Since there are $12$ rhombuses and two $3$-point intersections per rhombus, this works out to be:
$\frac{2\cdot12}{3}$
Hence: $\boxed{\textbf{(D) }8}$ ~hollph27 ~Minor edits by FutureSphinx
## Solution 6 (Based on previous knowledge)
Note that a rhombic dodecahedron is formed when a cube is turned inside out (as seen here), thus there are 6 4-vertices (corresponding to each face of the cube) and 8 3-vertices (corresponding to each corner of the cube). Thus the answer is $\boxed{\textbf{(D) }8}$
## Solution 7 (Using Answer Choices)
Let $m$ be the number of $4$-edge vertices, and $n$ be the number of $3$-edge vertices. The total number of vertices is $m+n$. Now, we know that there are $4 \cdot 12 = 48$ vertices, but we have overcounted. We have overcounted $m$ vertices $3$ times and overcounted $n$ vertices $2$ times. Therefore, we subtract $3m$ and $2n$ from $48$ and set it equal to our original number of vertices. $$48 - 3m - 2n = m+n$$ $$4m + 3n = 48$$ From here, we reduce both sides modulo $4$. The $4m$ disappears, and the left hand side becomes $3n$. The right hand side is $0$, meaning that $3n$ must be divisible by $4$. Looking at the answer choices, this is only possible for $n = \boxed{8}$.
-DEVSAXENA
(Isn't this the same as the last half of Solution 2?)
## Solution 8 (Dual)
Note that a rhombic dodecahedron is the dual of a cuboctahedron. A cuboctahedron has $8$ triangular faces, which correspond to $\boxed{\textbf{(D) }8}$ vertices on a rhombic dodecahedron that have $3$ edges.
## Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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Week 5 – Precalc 11
This week in Precalculus 11, we started the Solving Quadratic Equations Unit. We first reviewed factoring polynomial expressions.
Factoring: separating an expression into its components
Polynomial expression: an expression of numbers and variables being added, subtracted or multiplied (example: 2x)
Greatest common factor: the greatest number that can divide into all the terms in the expression (example: 2, 4, 6, 8, 10 GCF = 2)
Term: a number or a group of numbers being multiplied (example: 2$x^2$)
Binomial: a polynomial expression with 2 terms (example: x + 1)
Difference of squares: a polynomial expression in which subtraction takes place between 2 perfect square terms (example: 4x – 1)
Trinomial: a polynomial expression with 3 terms (example: $x^2$ + x + 1)
Conjugates: 2 terms with opposite addition/subtraction signs (example: 1 + 1 & 1 – 1)
To factor a polynomial expression, we first look for the greatest common factor between the terms and divide each term by that number.
Example:
2x + 2
= 2(x + 1)
If the expression is a binomial, we check if it is a difference of squares. When factored, a difference of squares results in conjugates.
Example:
4x – 1
= (2x + 1)(2x – 1)
If the expression is a trinomial, we check if it is in the form a$x^2$ + bx + c. If a = 1, we separate bx into 2 terms that multiply to c and add to bx.
Example:
$x^2$ + 2x + 1
= (x + 1)(x + 1)
= $(x + 1)^2$
If a ≠ 1, we separate bx into 2 terms that multiply to ac and add to bx, then find the greatest common factor of each side, and divide each term by that number.
Example:
2$x^2$ + 4x + 2
= 2$x^2$ + 2x + 2x + 2
= 2x(x + 1) + 2(x + 1)
= (2x + 2)(x + 1)
If the expression is in a different form than a$x^2$ + bx + c, the expression can sometimes be changed to this form.
Example:
$(x + 1)^2$ + 2(x + 1) + 1
= $a^2$ 2a + 1
= (a + 1)(a + 1)
= $(a + 1)^2$
= $[(x + 1) + 1]^2$
= $(x + 2)^2$
Week 4 – Precalc 11
Radical expression: the term a$\sqrt[b]{c}$
Radical: a root sign (represented by √)
Coefficient: the number outside the radical, it is being multiplied by the radical
When adding or subtracting radical expressions, the radicals (the radicand and the root) must be like terms. Some radicals can be simplified to achieve like terms. The coefficients are then added or subtracted, and the radical remains the same.
Example:
$\sqrt{2}$ + $\sqrt{8}$
= $\sqrt{2}$ + $\sqrt{4}$$\sqrt{2}$
= $\sqrt{2}$ + 2$\sqrt{2}$
= 3$\sqrt{2}$
Week 3 – Precalc 11
This week in Precalculus 11, we started the Absolute Value and Radicals unit. We first learned about the absolute value of a real number.
Absolute value of a real number: its distance from zero on a number line/the principal square root of its square
Principal square root: the positive (+) square root
Example: $\sqrt{1}$ = ±1 +1 = principal square root -1 = negative (-) square root
The symbols “||” represent absolute value.
Example: <-0-1-2-3-4-5-> |0| = 0
|0|
= $\sqrt{0^2}$
= 0
This is a new concept I had not yet learned about and can add to my knowledge of Pre-Calculus.
Week 2 – Precalc 11
This week in Precalculus 11, we learned about geometric sequences.
Geometric sequence: a list of terms in which each term is multiplied by a common ratio to equal the next term
Term: a number (represented by “$t_n$“, “n” being the term’s place in the geometric sequence)
Common ratio: a common number multiplied by each term to equal the next term (represented by “r”)
Example: 1, 2, 4, 8, 16 r = 2
To find the value of a term in a geometric sequence, we use the formula $t_n$ = $t_1$ · $r^{n - 1}$.
Example: Find $t_{10}$: 1, 2, 4, 8, 16
$t_{10}$ = ? $t_1$ = 1 r = 2 n = 10
$t_n$ = $t_1$ · $r^{n - 1}$
($t_{10}$) = (1)$(2)^{(10) - 1}$
$t_{10}$ = 1 · $2^9$
$t_{10}$ = 1 · 512
$t_{10}$ = 512
This method of finding the value of terms in an geometric sequence is faster than extending the list.
Week 1 – My Arithmetic Sequence
13, 26, 39, 52, 65…
$t_n$ = $t_1$ + d(n – 1)
($t_{50}$) = (13) + (13)[(50) – 1]
$t_{50}$ = 13 + 13 · 49
$t_{50}$ = 13 + 637
$t_{50}$ = 650
$S_n$ = $\frac{n}{2}$($t_1$ + $t_n$)
($S_{50}$) = $\frac{(50)}{2}$[(13) + (650)]
$S_{50}$ = 25 · 663
$S_{50}$ = 16575
Week 1 – Precalc 11
This week in Pre-Calculus 11, we started the Sequences and Series unit. We first learned about arithmetic sequences.
Arithmetic sequence: a list of terms in which a common difference is added to each term to equal the next term
Term: a number (represented by “$t_n$“, “n” being the terms place in the arithmetic sequence)
Common difference: a common number added to each term to equal the next term (represented by “d”)
Example: 0, 1, 2, 3, 4 d = +1
t₁ t₂ t₃ t₄ t₅
To find the value of a term in an arithmetic sequence, we use the formula $t_n$ = t₁ + d(n – 1).
Example: Find t₁₀: 0, 1, 2, 3, 4
$t_{10}$ = ? $t_1$ = 0 d = +1 n = 10
$t_n$ = t₁ + d(n – 1)
(t₁₀) = (0) + (1)[(10) – 1]
(t₁₀) = 0 + 1 · 9
(t₁₀) = 0 + 9
(t₁₀) = 9
This method of finding the values of terms in an arithmetic sequence is faster than extending the list.
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# Disjoint vs. Independent Events: What’s the Difference?
Two terms that students often confuse are disjoint and independent.
Here’s the difference in a nutshell:
We say that two events are disjoint if they cannot occur at the same time.
We say that two events are independent if the occurrence of one event has no effect on the probability of the other event occurring.
The following examples illustrate the difference between these two terms in various scenarios.
### Example 1: Flipping a Coin
Scenario 1: Suppose we flip a coin once. If we define event A as the coin landing on heads and we define event B as the coin landing on tails, then event A and event B are disjoint because the coin can’t possibly land on heads and tails.
Scenario 2: Suppose we flip a coin twice. If we define event A as the coin landing on heads on the first flip and we define event B as the coin landing on heads on the second flip, then event A and event B are independent because the outcome of one coin flip doesn’t affect the outcome of the other.
### Example 2: Rolling a Dice
Scenario 1: Suppose we roll a dice once. If we let event A be the event that the dice lands on an even number and we let event B be the event that the dice lands on an odd number, then event A and event B are disjoint because the dice can’t possibly land on an even number and an odd number at the same time.
Scenario 2: Suppose we roll a dice twice. If we define event A as the dice landing on a “5” on the first roll and we define event B as the dice landing on a “5” on the second roll, then event A and event B are independent because the outcome of one dice roll doesn’t affect the outcome of the other.
### Example 3: Selecting a Card
Scenario 1: Suppose we select a card from a standard 52-card deck. If we let event A be the event that the card is a Spade and we let event B be the event that the card is a Diamond, then event A and event B are disjoint because the card can’t possibly be a Spade and a Diamond at the same time.
Scenario 2: Suppose we select a card from a standard 52-card deck twice in a row with replacement. If we define event A as the card being a Spade on the first draw and we define event B as the card being a Spade on the second draw, then event A and event B are independent because the outcome of one draw doesn’t affect the outcome of the other.
### Probability Notation: Disjoint vs. Independent Events
Written in probability notation, we say that events A and B are disjoint if their intersection is zero. This can be written as:
• P(A∩B) = 0
For example, suppose we roll a dice once. Let event A be the event that the dice lands on an even number and let event B be the event that the dice lands on an odd number.
We would define the sample space for the events as follows:
• A = {2, 4, 6}
• B = {1, 3, 5}
Notice that there is no overlap between the two sample spaces. Thus, events A and B are disjoint events because they both cannot occur at the same time.
Thus, we could write:
• P(A∩B) = 0
Similarly, written in probability notation, we say that events A and B are independent if the following is true:
• P(A∩B) = P(A) * P(B)
For example, suppose we roll a dice twice. Let event A be the event that the dice lands on a “5” on the first roll and let event B be the event that the dice lands on a “5” on the second roll.
If we write out all of the 36 possible ways for the dice to land, we would find that in only 1 out of the 36 scenarios the dice lands on a “5” both times. Thus, we would say P(A∩B) = 1/36.
We also know that the probability of the dice landing on a “5” during the first roll is P(A) = 1/6.
We also know that the probability of the dice landing on a “5” during the second roll is P(B) = 1/6.
Thus, we could write:
• P(A∩B) = P(A) * P(B)
• 1/36 = 1/6 * 1/6
• 1/36 = 1/36
Since this equation holds true, we could indeed say that event A and event B are independent in this scenario.
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# 8th Grade FSA Math FREE Sample Practice Questions
Preparing your student for the 8th Grade FSA Math test? To succeed on the FSA Math test, students need to practice as many real FSA Math questions as possible. There’s nothing like working on FSA Math sample questions to measure your student’s exam readiness and put him/her more at ease when taking the FSA Math test. The sample math questions you’ll find here are brief samples designed to give students the insights they need to be as prepared as possible for their FSA Math test.
Check out our sample FSA Math practice questions to find out what areas your student needs to practice more before taking the FSA Math test!
Start preparing your student for the 2021 FSA Math test with our free sample practice questions. Also, make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice.
## 10 Sample 8th Grade FSA Math Practice Questions
1- Five years ago, Amy was three times as old as Mike was. If Mike is 10 years old now, how old is Amy?
A. 4
B. 8
C. 12
D. 20
2- What is the length of AB in the following figure if AE$$=4$$, CD$$=6$$ and AC$$=12$$?
$$\img {https://appmanager.effortlessmath.com/public/images/questions/test2121212121212.JPG }$$
A. 3.8
B. 4.8
C. 7.2
D. 24
3- If a gas tank can hold 25 gallons, how many gallons does it contain when it is $$\frac{2}{5}$$ full?
A. 50
B. 125
C. 62.5
D. 10
4- In the xy-plane, the points (4,3) and (3,2) are on line A. Which of the following equations of lines is parallel to line A?
A. $$y=3x$$
B. $$y=\frac{x}{2}$$
C. $$y=2x$$
D. $$y=x$$
5- If $$x$$ is directly proportional to the square of $$y$$, and $$y=2$$ when $$x=12$$, then when $$x=75 y=$$ ?
A. $$\frac{1}{5}$$
B. 1
C. 5
D. 12
6- Jack earns $616 for his first 44 hours of work in a week and is then paid 1.5 times his regular hourly rate for any additional hours. This week, Jack needs$826 to pay his rent, bills and other expenses. How many hours must he work to make enough money in this week?
A. 40
B. 48
C. 53
D. 54
7-
$$\img {https://appmanager.effortlessmath.com/public/images/questions/test252525252525252525252525.JPG }$$
If a is the mean (average) of the number of cities in each pollution type category, b is the mode, and c is the median of the number of cities in each pollution type category, then which of the following must be true?
A. $$a<b<c$$
B. $$b<a<c$$
C. $$a=c$$
D. $$b<c=a$$
8-
$$\img {https://appmanager.effortlessmath.com/public/images/questions/test252525252525252525252525.JPG }$$
What percent of cities are in the type of pollution A, C, and E respectively?
A. $$60\%, 40\%, 90\%$$
B. $$30\%, 40\%, 90\%$$
C. $$30\%, 40\%, 60\%$$
D. $$40\%, 60\%, 90\%$$
9-
$$\img {https://appmanager.effortlessmath.com/public/images/questions/test252525252525252525252525.JPG }$$
How many cities should be added to type of pollutions B until the ratio of cities in type of pollution B to cities in type of pollution E will be 0.625?
A. 2
B. 3
C. 4
D. 5
10- In the following right triangle, if the sides AB and AC become twice longer, what will be the ratio of the perimeter of the triangle to its area?
$$\img {https://appmanager.effortlessmath.com/public/images/questions/test3030303030303030.JPG }$$
A. $$\frac{1}{2}$$
B. 2
C. $$\frac{1}{3}$$
D. 3
## Best 8th Grade FSA Math Practice Resource for 2021
1- D
Five years ago, Amy was three times as old as Mike. Mike is 10 years now. Therefore, 5 years ago Mike was 5 years.
Five years ago, Amy was:
$$A=3×5=15$$
Now Amy is 20 years old:
$$15 + 5 = 20$$
2- B
Two triangles $$∆$$BAE and $$∆$$BCD are similar. Then:
$$\frac{AE}{CD}=\frac{AB}{BC}=\frac{4}{6}=\frac{x}{12}$$
$$→48-4x=6x→10x=48→x=4.8$$
3- D
$$\frac{2}{5}×25=\frac{50}{5}=10$$
4- D
The slop of line A is:
m $$= \frac{y_2-y_1}{x_2-x_1}=\frac{3-2}{4-3}=1$$
Parallel lines have the same slope and only choice D $$(y=x)$$ has slope of 1.
5- C
$$x$$ is directly proportional to the square of $$y$$. Then:
$$x=cy^2$$
$$12=c(2)^2→12=4c→c=\frac{12}{4}=3$$
The relationship between $$x$$ and $$y$$ is:
$$x=3y^2$$
$$x=75$$
$$75=3y^2→y^2=\frac{75}{3}=25→y=5$$
6- D
he amount of money that jack earns for one hour: $$\frac{616}{44}=14$$
Number of additional hours that he work to make enough money is: $$\frac{826-616}{1.5×14}=10$$
Number of total hours is: $$44+10=54$$
7- C
Let’s find the mean (average), mode and median of the number of cities for each type of pollution.
Number of cities for each type of pollution: $$6, 3, 4, 9, 8$$
Mean $$= \frac{sum \space of \space terms}{number \space of \space terms}=\frac{6+3+4+9+8}{5}=6$$
Median is the number in the middle. To find median, first list numbers in order from smallest to largest.
$$3, 4, 6, 8, 9$$
Median of the data is 6.
Mode is the number which appears most often in a set of numbers. Therefore, there is no mode in the set of numbers.
Median $$=$$ Mean, then, $$a=c$$
8- A
Percent of cities in the type of pollution A:
$$\frac{6}{10} × 100=60\%$$
Percent of cities in the type of pollution C:
$$\frac{4}{10} × 100 = 40\%$$
Percent of cities in the type of pollution E:
$$\frac{9}{10}× 100 = 90\%$$
9- A
Let the number of cities should be added to type of pollutions B be $$x$$. Then:
$$\frac{x + 3}{8}=0.625→x+3=8×0.625→x+3=5→x=2$$
10- A
AB$$=12$$ And AC$$=5$$
BC$$=\sqrt{(12^2+5^2 )} = \sqrt{(144+25)} = \sqrt{169}=13$$
Perimeter $$=5+12+13=30$$
Area $$=\frac{5×12}{2}=5×6=30$$
In this case, the ratio of the perimeter of the triangle to its area is:
$$\frac{30}{30}= 1$$
If the sides AB and AC become twice longer, then:
AB$$=24$$ And AC$$=10$$
BC$$=\sqrt{(24^2+10^2 )} = \sqrt{(576+100)} = \sqrt{676} = 26$$
Perimeter $$=26+24+10=60$$
Area $$=\frac{10×24}{2}=10×12=120$$
In this case the ratio of the perimeter of the triangle to its area is:
$$\frac{60}{120}=\frac{1}{2}$$
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# Mathematics - Discrete Mathematics - Implication and Equivalence Statements
[Image 1]
## Introduction
Hey it's a me again @drifter1!
Today we continue with Mathematics, and more specifically the branch of "Discrete Mathematics", in order to get into more on Implication and Equivalence Statements.
Of course, you should check out the previous part before getting into this one...
So, without further ado, let's dive straight into it!
## Truth Tables
In the previous article, a proposition was defined as a declarative sentence that is either true or false. So, even in compound propositions (which are the combination of multiple propositions) the only thing that matters is the final truth value.
The truth value depends on the chosen truth value of each propositional variable that makes up the statement. There are only two possibilities for each such variable. This allows for a visual representation of all possible combinations via something known as a truth table.
### Logical Connectives
The truth tables of the main logical connectives are the following:
where T means true and F means false.
In the context of computer science and mainly logic gates, truth tables are mostly in 0-1 form, where 0 means false and 1 means true.
It's easy to generalize conjunction and disjunction for any number of propositions (inputs):
• the conjunction will have a T only in the row where all propositions are T
• the disjunction will have a F only in the row where all propositions are F
### Exercise
What's the truth table of an Exclusive-OR (XOR) with 3 inputs?
Hint: In the case of two inputs the output is true only when the propositions are not equal. For any number of inputs it is generalized into being true only if the number of true inputs is odd.
The inverse of the XOR is known as a XNOR. It is true when the two inputs are equal, and generally the number of true inputs is even.
## Implication (or Conditional)
An implication is a statement of the form p → q, where p is the hypothesis (or antecedent) and q the conclusion (or consequent). Such mathematical statements can be proven directly by assuming p (p is true) and deducing q from it.
Implications basically specify the relationship between two propositions. If p → q is a proposition then:
• q → p is the converse, which is NOT logically equivalent and a completely independent implication
• ¬q → ¬p is the contrapositive, which IS logically equivalent to the original implication
• ¬p → ¬q is the inverse
## Equivalence (or Bi-Conditional)
If both p → q and q → p are true then p ↔ q. Such a statement is called a bi-conditional statement or an equivalence.
Proving logical equivalence is as simple as proving that two statements have the same truth table or that each can be deduced from the other (proving p → q and q → p respectively).
Logical equivalence is commonly denoted using the symbol, which basically means equivalence.
### Propositional Algebra
Based on logical equivalence it's possible to simplify statements using various laws of algebra. These are:
### Example
Let's proof the following equivalence:
For the left-hand side, replacing the conditionals using the respective laws, and simplifying, yields:
For the right-hand side now, replacing the conditional and then applying De Morgan's law yields:
So, both result in the same statement, and are thus logically equivalent!
## RESOURCES:
### Images
Block diagrams and other visualizations were made using draw.io.
## Previous articles of the series
• Introduction → Discrete Mathematics, Why Discrete Math, Series Outline
• Sets → Set Theory, Sets (Representation, Common Notations, Cardinality, Types)
• Set Operations → Venn Diagrams, Set Operations, Properties and Laws
• Sets and Relations → Cartesian Product of Sets, Relation and Function Terminology (Domain, Co-Domain and Range, Types and Properties)
• Relation Closures → Relation Closures (Reflexive, Symmetric, Transitive), Full-On Example
• Equivalence Relations → Equivalence Relations (Properties, Equivalent Elements, Equivalence Classes, Partitions)
• Partial Order Relations and Sets → Partial Order Relations, POSET (Elements, Max-Min, Upper-Lower Bounds), Hasse Diagrams, Total Order Relations, Lattices
• Combinatorial Principles → Combinatorics, Basic Counting Principles (Additive, Multiplicative), Inclusion-Exclusion Principle (PIE)
• Combinations and Permutations → Factorial, Binomial Coefficient, Combination and Permutation (with / out repetition)
• Combinatorics Topics → Pigeonhole Principle, Pascal's Triangle and Binomial Theorem, Counting Derangements
• Propositions and Connectives → Propositional Logic, Propositions, Connectives (∧, ∨, →, ↔ and ¬)
## Final words | Next up
And this is actually it for today's post!
Next time we will continue on with strategies for proving mathematical statements...
See ya!
Keep on drifting!
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Solution to Exercise:
A 3-input XOR has the following truth table:
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• Numbers are four types: Natural Number, Whole Number, Integer, and Rational Numbers.
# Numbers in general form
• Any two digit number xy made of digits x and y can be written as = 10 X x + 1X y
• Any three digit number xyz made of digits x, y, and z can be written as = 100 X x + 10Xy + 10Xz
EXAMPLE 1: Write the number 12 in generalized form.
SOLUTION: 12 = 10X1 + 1X2 = 10+2
EXAMPLE 2: Write the number 10X3 + 4 in usual form.
SOLUTION: 10X3 + 4 = 30 + 4 = 34
# Games with numbers
• Reversing a two-digit number:
STEP 1: Suppose we choose a number xy, which is a short form for the 2-digit number 10x + y.
STEP 2: On reversing the digits, we get the number yx = 10y + x.
STEP 3: If we add the two numbers we get: (10x + y) + (10y + x) = 11x + 11y = 11 (x + y).
So, the sum is always a multiple of 11.
Observe here that if we divide the sum by 11, the quotient is x + y, which is exactly the sum of the digits of chosen number xy.
• Reversing a 3-digit number:
STEP 1: Let the 3-digit number be xyz = 100x + 10y + z.
STEP 2: After reversing the order of the digits, we get the number zyx= 100z + 10y + x.
STEP 3: On subtraction:
If x > z, then the difference between the numbers is (100x + 10y + z) – (100z + 10y + x) = 100x + 10y + z – 100z – 10y – x = 99x – 99z = 99(x – z).
• If z>a, then the difference between the numbers is (100z + 10y + x) – (100x + 10y + z) = 99z – 99x = 99(z – x).
• And, of course, if x = z, the difference is 0. In each case, the resulting number is divisible by 99. So the remainder is 0.
Observe that quotient is x – z or z – x.
• Forming three digit numbers with given 3-digits:
xyz = 100x + 10y + z
zxy = 100z + 10x + y
yzx = 100y + 10z + x
xyz + zxy + yzx = 111(x + y + z)
= 37 × 3(x + y + z), which is divisible by 37
EXAMPLE 1: Prove that the addition of a 2-digit no. and its reversed number is equal to the product of 11 and sum of the digits.
SOLUTION: 12= 10X1 + 2
21= 10X2 + 1
12+21= 11X3= 11X(1+2) [PROVED]
Hence, the addition of a 2-digit no. and its reversed number is equal to the product of 11 and sum of the digits.
EXAMPLE 2: Form 3-digit numbers with given 3 digit no. 123
SOLUTION: Combinations of the number 123 are 231,312 [Keeping in mind the no.s are in order]
123+231+312= 666= (1+2+3)X111= (1+2+3)X3X37
Hence, a 3 digit number formed by another 3-digit number can always be divided by 37.
# Letter for digits
• Here letters take the place of digits in an arithmetic ‘addition’, and the we have to find out which letter represents which digit.
• Each letter must stand for just one digit. Each digit must be represented by just one letter.
• The first digit of a number can’t be zero. Hence, we write the no. ‘ninety-five’ as 95, and not as 095, or 0095.
EXAMPLE 1: Find X and Y in the addition. X + X + X
Y X
SOLUTION: This has two letters X and Y whose values are to be found.
Study the addition in the one's column: the sum of three X’s is a number whose ones digit is X.
Therefore, the sum of two X’s must be a number whose ones digit is 0.
This happens only for X = 0 and X = 5.
If X = 0, then the sum is 0 + 0 + 0 = 0, which makes Y = 0 too.
We do not want this (as it makes X = Y, and then the tens digit of YX too becomes 0), so we reject this possibility.
So, X = 5.
Therefore, the puzzle is solved as shown below.
5
+ 5
+ 5
15 That is, X = 5 and Y = 1.
EXAMPLE 2: Find the digits X and Y.
Y X
× Y 3
5 7 X
SOLUTION: This also has two letters X and Y whose values are to be found.
Since the ones digit of 3 × X is X, it must be that X = 0 or X = 5.
Now, look at Y.
If Y = 1, then YX × Y3 would at most be equal to 19 × 19; that is, it would at most be equal to 361.
But the product here is 57X, which is more than 500.
So we cannot have Y = 1.
If Y = 3, then YX × Y3 would be more than 30 × 30; that is, more than 900.
But 57X is less than 600.
So, Y cannot be equal to 3.
Hence, Y = 2 .
Therefore, the multiplication is either 20 × 23 or 25 × 23.
But, 20 × 23 = 460. Thus, the second one is the correct multiplication.
The second one works out correctly, 25 × 23 = 575. The answer is X = 5, Y = 2.
2 5
×2 3
575
# Tests of Divisibility
• Divisibility by 10: If the ones digit of a number is 0, then the number is a multiple of 10; and if the ones digit is not 0, then the number is not a multiple of 10. So, we get a test of divisibility by 10.
• Divisibility by 5: If the ones digit of a number is 0 or 5, then it is divisible by 5.
• Divisibility by 2: If the one’s digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.
• Divisibility by 9 and 3:
(a) A number is divisible by 9 if the sum of its digits is divisible by 9.
(b) A number X is divisible by 3 if the sum of its digits is divisible by 3. Otherwise, it is not divisible by 3.
EXAMPLE 1: Check the divisibility of 15284 by 3.
SOLUTION: The sum of the digits of 15284 is 1 + 5 + 2 + 8 + 4 = 20. This number is not divisible by 3. We conclude that 15284 too is not divisible by 3.
EXAMPLE 2: If the three-digit number 12a is divisible by 9, what is the value of a?
SOLUTION: Since 12a is divisible by 9, sum of its digits, i.e., 1 + 2 + a should be divisible by 9, i.e., 3 + a should be divisible by 9.
This is possible when 3 + a = 9 or 18, .... But, since a is a digit, therefore, 3 + a = 9, i.e., x = 6.
# Practice these questions
Q1) If 12a3 is a multiple of 3, where a is a digit, what might be the values of a?
Q2) If 34a is a multiple of 3, where a is a digit, what is the value of a?
Q3) If 56a7 is a multiple of 9, where a is a digit, what is the value of a?
Q4) Find the value of the letters 1 2 X
+ 6 A Y
X 0 9
Q5) Find the value of the letters 2 A B
+A B 1
B 1 8
Q6) If 417 is added with its other combinations then check if the sum is divided by 37 or not.
Q7) Write the number 100 × x + 10 × y + z in its usual form.
Q8) Write the number 302 in its generalised form.
# Recap
• Numbers are four types: Natural Number, Whole Number, Integer, and Rational Numbers.
• Any three digit number xyz made of digits x, y, and z can be written as = 100 X x + 10Xy + 10Xz
• xyz = 100x + 10y + z
zxy = 100z + 10x + y
yzx = 100y + 10z + x
xyz + zxy + yzx = 111(x + y + z) = 37 × 3(x + y + z), which is divisible by 37
• If the ones digit of a number is 0, then the number is a multiple of 10
• If the digit at one’s place of a number is 0 or 5, then the number is divisible by 5.
• If the ones digit of a number is 0, 2, 4, 6 or 8 then the number is divisible by 2.
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# Subsets
A set is a collection of objects. A subset is a set that consists of members of another set. So, a subset is a set that is contained inside another set.
then some of its subsets are
However, it has other subsets as well.
• There's the subset containing no elements. We call this the empty set and write it as $\{\}$ or $\emptyset$.
• Then, there are subsets containing one element. There are four of these: $\{\text{frog}\}, \{\text{duck}\}, \{\text{elephant}\},\{\text{cow}\}$
• Next come the subsets with two elements. There are six:$\{\text{frog},\text{duck}\}, \{\text{frog},\text{elephant}\}, \{\text{frog},\text{cow}\}, \{\text{duck},\text{elephant}\}, \{\text{duck},\text{cow}\}, \{\text{elephant},\text{cow}\}$.
• There are four three element subsets: $\{\text{frog},\text{duck}, \text{elephant}\}, \{\text{frog},\text{elephant},\text{cow}\}, \{\text{frog},\text{cow},\text{duck}\}, \{\text{elephant},\text{cow},\text{duck}\}$
• Finally, the whole set is a subset of itself: $\{\text{frog},\text{duck},\text{elephant},\text{cow}\}$
So, this four element set has $1 + 4 + 6 + 4 + 1 = 16 = 2^4$ subsets.
Let's look at some sets of other sizes and see how many subsets they have.
## The Empty Set
The empty set is the only set that has only one subset. Every set has the set itself and the empty set as subsets, but in this case, they are the same thing.
The only subset of the empty set is the empty set.
The empty set has $1 = 2^0$ subsets.
## A Set with One Element
How many subsets does a set with one element have?
Let's say our one element set is
$\{ \text{frog}\}$
We need to find all the sets that are contained in it.
There are only two choices:
• The emptyset $\emptyset$
• The whole set: $\{\text{frog}\}$
So the set with $1$ element in it has $2$ subsets.
Just, by the way, $2 = 2^1$.
## A Set with Two Elements
How many subsets does a set with two elements have?
Let's say our two element set is
$\{ \text{frog}, \text{duck}\}$
We need to find all the sets that are contained in it.
There are a couple of choices this time, but of course, we have:
• The emptyset $\emptyset$
• The whole set: $\{\text{frog}, \text{duck}\}$
We can also form subsets with one element. There are two possibilities:
• $\{\text{frog}\}$
• $\{\text{duck}\}$
So the set with $2$ elements has $4$ subsets.
Did you notice that $2 = 2^2$?
## A Set with Three Elements
How many subsets does a set with three elements have?
Let's say our three element set is
$\{ \text{frog}, \text{duck}, \text{elephant} \}$
We need to find all the sets that are contained in it.
There are a couple of choices this time, but of course, we have:
• The emptyset $\emptyset$
• The whole set: $\{\text{frog}, \text{duck}, \text{elephant}\}$
We can also form subsets with one element. There are three possibilities:
• $\{\text{frog}\}$
• $\{\text{duck}\}$
• $\{\text{elephant}\}$
Finally, we can form subsets with two elements. There are three of these as well:
• $\{\text{frog}, \text{duck}\}$
• $\{\text{duck}, \text{elephant}\}$
• $\{\text{frog},\text{elephant}\}$
So the set with $3$ elements has $8$ subsets.
Did you notice that $2 = 2^3$?
Now I'm going to leave the work to you.
## Number of Subsets of a Set with Four Elements
How many subsets does a set with four elements have?
Let's say our four element set is
$\{ \text{frog}, \text{duck}, \text{elephant}, \text{cow} \}$
We need to find all the sets that are contained in it. Let's work systematically, and list them by their sizes. I'll name the sizes, you'll do the listing. Be careful, it's really important to get all of them.
• Subsets with 0 elements:
• Subsets with 1 element:
• Subsets with 2 elements:
• Subsets with 3 elements:
• Subsets with 4 elements:
Note: If you got them all, there should be $16$ of them.
## Number of Subsets of a Set with Five Elements
How many subsets does a set with five elements have?
Let's say our five element set is
$\{ \text{frog}, \text{duck}, \text{elephant}, \text{cow}, \text{snail} \}$
We need to find all the sets that are contained in it. Let's work systematically, and list them by their sizes. I'll name the sizes, you'll do the listing. Be careful, it's really important to get all of them.
• Subsets with 0 elements:
• Subsets with 1 element:
• Subsets with 2 elements:
• Subsets with 3 elements:
• Subsets with 4 elements:
• Subsets with 5 elements:
Note: If you got them all, there should be $32$ of them.
## Number of Subsets of a Set with Six Elements
How many subsets does a set with six elements have?
Let's say our six element set is
$\{ \text{frog}, \text{duck}, \text{elephant}, \text{cow}, \text{snail}, \text{echidna} \}$
Hold your horses! There's no need to write them all down. You should have worked out the pattern by now.
How many do you think there should be?
## The Number of Subsets of a Set with $n$ Elements
Did you notice the pattern? Each time we increased the number of elements by $1$, we doubled the number of subsets.
So, the number of subsets of a set with $n$ elements is
$2^n$
A set with $6$ elements has ________________ subsets.
A set with $7$ elements has ________________ subsets.
A set with $8$ elements has ________________ subsets.
## The Numbers of Subsets of Each Size
There's another pattern to spot when we're counting subsets:
• A set with 0 elements has 1 subset.
• A set with 1 element has 1 subset with 0 elements and 1 subset with 1 element.
• A set with 2 elements has 1 subset with 0 elements, 2 subsets with 1 element and 1 subset with 2 elements
• A set with 3 elements has 1 subset with 0 elements, 3 subsets with 1 element, 3 subsets with 2 elements, and 1 subset with 3 elements
• etc.
To me, the green numbers look a lot like the rows of Pascal's triangle:
In fact, that's absolutely correct: the rows of Pascal's triangle tell you how many subsets of each size a given set has. For example, the row starting with $1,6$ tells you that a set with 6 elements has
• 1 subset with no elements
• 6 subsets with one element
• 15 subsets with two elements
• 20 subsets with three elements
• 15 subsets with four elements
• 6 subsets with five elements
• 1 subset with six elements
So, the rows of Pascal's triangle let you check that you've written down all the subsets of a set of a particular size, and adding up the numbers in the row lets you know how many subsets a set of that size has.
Since the rth entry in row n of Pascal's triangle is equal to $\begin{pmatrix}n\\r\end{pmatrix}$, we can use this formula to quickly calculate how many subsets of a given size a set has.
For example, a set with $123$ elements has $\begin{pmatrix}123\\ 24\end{pmatrix}$ subsets with 24 elements. We say that this is the number of ways of choosing 24 elements from a set of 123 elements, where the order doesn't matter and repetition is not allowed.
### Description
This chapter series is for Year 10 or higher students, topics include
• Arranging Objects in Lines
• Factorial
• Subsets
• Four colour theorem
and more
### Audience
Year 10 or higher students
### Learning Objectives
These chapters are related to data and in particular "Counting" topics such as Binomial Theorem, Subsets etc
Author: Subject Coach
You must be logged in as Student to ask a Question.
|
Equations & Brackets.. You are now going to solve more complex equations by combining together two ideas that you have seen already. Try the following.
Presentation on theme: "Equations & Brackets.. You are now going to solve more complex equations by combining together two ideas that you have seen already. Try the following."— Presentation transcript:
Equations & Brackets.
You are now going to solve more complex equations by combining together two ideas that you have seen already. Try the following questions to make sure you are ready: Idea 1: Solve the equation 3x – 6 = 18 Solution : x = 8 Idea 2 : Multiply out the bracket 3 ( 2x – 4 ) Solution : 6x – 12 We are now going to combine these two ideas together to solve the equations on the following pages.
Solving Equations. Example 1. Solve the equation : 2 ( x + 4 ) = 14 Solution. 2( x + 4 ) = 14Multiply out the bracket first. 2x + 8 = 14 Subtract 8 from both sides. 2x = 6 Divide both sides by 2. x = 3Hint: check the solution by substituting x = 3 into the original equation.
Example 2. Solve Solve the equation : 2 ( 3x + 2 ) = 28 Solution: 2( 3x + 2 ) = 28 What should you do first ? Multiply out the bracket. 6x + 4 = 28 What comes next ? Subtract 4 from each side. 6x = 24 What comes next ? Divide each side by 6. x = 4 Check your answer by putting x = 4 back into the bracket.
Example 3. Solve the equation : 5( x – 3 ) = 20 Solution. 5 ( x – 3 ) = 20 What does 5 times –3 equal. -155x – 15 = 20 What must you do to both sides? Add 15 5x = 35 x = 7 Check the answer.
What Goes In The Box 1 ? Solve the following equations: (1) 2 ( x + 4 ) = 14 (2) 3 ( 2x + 5 ) = 27 (3) 5 ( 2x + 6 ) = 100 (4) 4 ( 3x – 5 ) = 40 (5) 7 ( 2 x – 6 ) = 28 (6) 3 ( 4 + 2x ) = 30 x = 3 x = 2 x = 7 x = 5 x = 3
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Algebra II Chapter 5
```Algebra II
Chapter 5
The graph of a quadratic function is a parabola, as shown at rig.
Standard Form:
f ( x) = ax 2 + bx + c
⎛ b ⎛ −b ⎞ ⎞
vertex: (x, y) = ⎜ − , f ⎜ ⎟ ⎟
⎝ 2a ⎝ 2a ⎠ ⎠
a<0
graph opens down
a>0
graph opens up
b
axis of symmetry: x = −
2a
Larson
Text, Ch
5, p. 249
Graph each function given in Standard Form:
1. y = x 2 + 4x + 3
2. y = x 2 + 6x + 11
a=
a=
b=
b=
x=−
b
=
2a
⎛ b⎞
y = f ⎜− ⎟ =
⎝ 2a ⎠
b
=
2a
⎛ b⎞
y = f ⎜− ⎟ =
⎝ 2a ⎠
(x, y) =
(x, y) =
x=−
Vertex form for a quadratic function (Let’s think about absolute value functions again first to
help us understand this.)
Recall: Vertex form for the equation of an absolute value function is f (x) = a x − h + k
What do the constants in the equation represent?
Graph the following absolute value equations using the vertex form.
1.
f (x) = −2 x − 1 + 3
2.
f (x) =
1
x +1 − 5
3
vertex =
vertex =
slope of the right branch of the graph =
slope of the right branch of the graph =
Vertex Form:
f ( x) = a( x − h) 2 + k
vertex: (h, k)
axis of symmetry: x = h
a<0
graph opens down
a>0
graph opens up
Graph each function given in Vertex Form.
1.
y = −(x − 1)2 + 4
2.
(h, k) =
(h, k) =
(x, y) =
(x, y) =
a=
a=
1
y = (x + 3)2 − 2
2
Intercept Form:
f (x) = a(x − r1 )(x − r2 )
x-intercepts: (r1 , 0) and (r2 , 0)
⎛r +r
⎛ r + r ⎞⎞
vertex: ⎜ 1 2 , f ⎜ 1 2 ⎟ ⎟
⎝ 2 ⎠⎠
⎝ 2
a<0
a>0
graph opens down
graph opens up
Graph each function in Intercept Form.
1.
f (x) = (x − 2)(x + 4)
2.
f (x) = −2(x + 1)(x − 5)
r1 =
r1 =
r2 =
r2 =
vertex =
vertex =
a=
a=
5.2 Solving Quadratic Equations by Factoring
Zero Product Property
Let A and B be real numbers or algebraic expressions.
If AB = 0, then A = 0 or B = 0.
Solve each equation by using the Zero Product Property:
1.
0 = (x − 1)(x − 3)
2.
0 = x(x + 4)
3.
0 = x2 − 9
4.
0 = x 2 + 3x − 18
5.
2x 2 − 17x + 45 = 3x − 5
6.
4x 2 + 12x − 7 = 0
7.
0 = 6x 2 − 16x + 8
8.
125x 2 − 5 = 0
5.3 Solving Quadratic Equations by Finding Square Roots
Simplify expressions using properties of square roots:
1)
48
2)
90
3)
3
16
4)
25
3
5)
6 i 10
6)
5
3
125
Solve the quadratic equation by finding square roots.
7)
2x 2 + 1 = 17
8)
x 2 − 9 = 16
9)
4x 2 + 7 = 23
10)
5(x − 1)2 = 50
11)
1
(x + 8)2 = 14
2
5.5 Completing the Square
Factor the Perfect Square Trinomials.
A.) x 2 − 8 x + 16 =
B.) x 2 + 5 x +
25
=
4
C.) x 2 − 7 x +
49
=
4
Can you see any pattern on how the second term in factored form is related to the middle term of
Find the value of c that makes the quadratic equation a perfect square
trinomial. Then write the quadratic in vertex form.
1.) y = x 2 − 14 x + c
2.) y = x 2 +
10
x+c
3
Completing the Square to Graph a Quadratic Function
Rewrite the equation in vertex form by completing the square. Find the vertex. Then solve for the
x intercepts. Verify your x intercepts on the calculator. Graph the parabola.
1.) y = x 2 + 10 x − 3
Vertex Form :______________
(h, k):
a=
(x, y):
2.) y = x 2 + 6 x − 8
Vertex Form :______________
(h, k):
(x, y):
a=
3.) y = −x 2 + 4x − 1
Vertex Form :______________
(h, k):
a=
(x, y):
4.) y = 2 x 2 − 12 x + 14
Vertex Form :______________
(h, k):
(x, y):
a=
5.) y = 4x 2 − 6x + 1
Vertex Form :______________
(h, k):
a=
(x, y):
6.) y = −3x 2 − 6x − 8
Vertex Form :______________
(h, k):
a=
(x, y):
Completing the Square to solve a Quadratic Equation
Solve the following equations by completing the square.
1.)
x 2 − 12x = −28
2.)
x 2 + 3x − 1 = 0
3.)
−3x 2 + 24x = 27
4.)
4x 2 − 40x − 8 = 0
5.)
3x 2 − 26x + 2 = 5x 2 + 1
6.)
2x 2 + 3x + 1 = 0
7.)
−4x 2 − 2x = −5
8.)
−3x 2 + 5x = −7
5.4 Complex Numbers (part 1)
What happens when you try to solve:
x 2 = −1
???
By definition:
i = −1
i =
2
(
)
2
−1 = −1
−25
−12
−48
From Larson Textbook, page 272
Solving Equations over the Complex Numbers
Solve the equation for x, giving your answer in terms of i.
1.) x 2 + 64 = 0
Check:
2.) 2x 2 + 26 = −10
Check:
3.) 3x 2 + 10x = −26
4.) 6x 2 − 2x + 2 = 4x 2 + x
1
5.) − (x + 1)2 = 5
2
6.) −6(x + 5)2 = 120
Plot the complex numbers in the complex plane:
imaginary axis
1.
2 – 3i
2.
-3 + 4i
3.
2i
4.
-i + 7
real axis
Adding and subtracting complex numbers: Combine the real parts and
combine the imaginary parts.
1.
(3 + 4i) + (6 + i)
2.
(1− i) − (7 + 3i)
3.
(6 − 2i) − (5 + i) − (10 + 5i)
5.4 Complex Numbers (Part 2)
Multiplying Complex Numbers
Simplify each expression as a complex number in standard form: a + bi
1.
3i(5 − i)
2.
−7i(3 − 2i)
3.
(2 + 3i)(5 − 6i)
4.
(2 − 5i)(2 + 5i)
5.
(1+ i)(1− i)
Write each expression as a complex number in standard form: a + bi
1.
3
1+ i
2.
6+i
1− 2i
3.
5+i
−3 − i
4.
8−i
8+i
5.
6+i
2i
6.
6
i
Why do we need complex numbers?
Complex numbers are at the heart of understanding Fractal Geometry. See pages 275 and 276 of
your textbook. Fractal Geometry is used to model a variety of natural phenomena. Check out
Complex numbers are also used in electronics to describe electrical circuits.
Complex numbers are used in a variety of higher level mathematics.
Find the x-intercepts for the following equations by completing the square :
y = −2 x 2 − 2 x + 7
y = ax 2 + bx + c
y = ax 2 + bx + c
−b ± b 2 − 4ac
x=
2a
Use the quadratic formula to solve for x in each equation.
1.
x 2 + 3x − 2 = 0
2.
−7x 2 + 2x + 9 = 0
a = _____ b = _____ c = _____
a = _____ b = _____ c = _____
x=
x=
3.
5x 2 + 9x = −x 2 + 5x + 1
a = _____ b = _____ c = _____
x=
Use the quadratic formula, factoring or taking square roots to solve for x in each equation. Use
your graphing calculator to check your solutions. (Graph the equation using y = , then use 2nd
Calc, 2: Zero to find where the graph crosses the x-axis.)
1.
5(x − 2)2 + 1 = 0
3.
2x 2 + 5x + 1 = 0
2.
2x 2 + 5x − 3 = 0
5.6 The Discriminant
For each equation, find the value of the discriminant, determine how many solutions, and then
1.
x 2 − 6x + 10 = 0
2.
x 2 − 6x + 9 = 0
3.
x 2 − 6x + 8 = 0
5.6 Motion Problems
Warm up:
Graph the following parabola by finding the vertex and the x-intercepts:
1.) y = x 2 − 2 x − 3
Vertex: _______________
x-intercepts: ____________
Now what if the original equation was written in the form:
y ≥ x 2 − 2x − 3
How would that change the graph?
Choose 2 points inside the parabola:
Choose 2 points outside the parabola:
1.)
1.)
Solution? _________
Solution? _________
2.)
2.)
Solution? _________
Solution? _________
1.) y < 2 x 2 − 5 x − 3
Vertex: __________
X intercepts: _______
2.)
A.) y ≥ x 2 − 4
B.) y < − x 2 − x + 2
Equation A
Equation B
Vertex: _____
Vertex: _____
x ints: _____
x ints: ______
3.)
A.) y ≤ − x 2 + 9
B.) y > x 2 + 5 x − 6
Equation A
Equation B
Vertex: _____
Vertex: _____
x ints: _____
x ints: ______
You can easily solve quadratic inequalities by using the following algebraic method.
Step 1:
Step 2:
Step 3:
Step 4.
Step 5:
Find the zeros of the quadratic and use them as boundaries for your solutions.
Plot the boundaries on a number line.
Use a test point to test the intervals between the boundaries.
Summarize your solutions as a compound inequality or using set notation.
FYI: What is set notation??
3≤ x ≤ 5
−1 ≤ x ≤ 10
4<x<7
−1 ≤ x < 8
5<x≤6
can be written as [3,5]
can be written as [-1, 10]
can be written as (4, 7)
can be written as [-1, 8)
can be written as (5, 6]
Use a bracket [ ] for less than or equal to, use a parenthesis ( ) for less than. Easy!
1.
2x 2 − 7x + 3 ≥ 0
2.
3x 2 − 16x + 5 ≤ 0
3.
−x 2 − 12x < 32
−x 2 + x + 5 < 0
4.
(If you can’t easily factor, use the quadratic formula to find the zeros.)
J
1.) Write an equation in vertex form for the parabola shown.
Recall vertex form: y = a(x − h)2 + k
2.) Write the equation of the parabola in vertex form with vertex (-2,3) and
passes through the point (2,-5).
3.) Write an equation in standard form for the parabola shown.
Recall standard form: ax 2 + bx + c = 0
4.) Write the standard form of the equation of the parabola with xintercepts of (-3,0) and (2,0) and passes through the point (-1,4).
5.) Write the intercept form of the equation of the parabola with x1
1
intercepts of ( ,0) and ( − ,0) and passes through the point (-1,2).
3
5
Recall intercept form: y = a(x − r1 )(x − r2 )
6.) A study compared the speed x (in miles per hour) and the average fuel economy y ( in miles
per gallon) for cars. The results are shown in the table.
Speed,x
Fuel
economy,y
15
22.3
20
25.5
25
27.5
30
29
35
28.8
40
30
45
29.9
50
30.2
55
30.4
60
28.8
65
27.4
70
25.3
a.) Graph a scatter plot of Fuel Economy vs. Speed on your graphing
calculator. Do you think that a linear or quadratic regression model fits
the data better?
b.) Find a linear regression model for the data using your graphing calculator. Report the
equation of the linear model below. Report the r2 value for this model.
c.) Find a quadratic regression model for the data using your graphing calculator. Report the
equation of the quadratic model below. Report the r2 value for this model.
d.) Using the model that fits the data best, predict the fuel economy for a car travelling at a speed
of 80 mph.
e.) Find the speed that maximizes fuel economy.
```
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# How To Calculate Tangent Of An Angle
Friday, January 13th 2023. | Sample Templates
How To Calculate Tangent Of An Angle – Any high school student who has encountered trigonometry in their math curriculum has likely encountered the laws of sine, cosine, and area, which are commonly used to solve triangles, where some information is given and the rest must be calculated. Somewhat surprisingly (since it’s relatively simple to obtain), the “tan rule” is usually not included in this particular trigonometric toolkit. However, as we hope to show in this article, this principle can be very useful under certain circumstances. Especially when two sides and an angle included in a triangle are given. In this situation, it is not possible to immediately apply the law of sines to determine the other angles, since none of the given sides is opposite to the given angle. To determine the third side, first apply the law of cosines, then apply the law of sines to one (or both) of the remaining angles. However, the tan rule allows us to determine one of the two missing angles without reference to the law of sines or cosines.
This derivation follows in the footsteps of Appendix A. The latter is a translation (from Stephen Hawking’s book: On the Shoulders of Giants) of an extract from “Liber Primus” from “De Revolutionibus Orbium Coelestium” written by a certain Nicolaus Copernicus and published after his death in 1543. It should be noted that the “theorem” is older because Copernicus derived his geometry from Greek sources, especially Euclid and Ptolemy. He (Copernicus), however, notes the later introduction of Arabic numerals:
## How To Calculate Tangent Of An Angle
“This notation surpasses any other – Greek or Latin – in its exceptional ease of use, and is easily adapted to any computing class.”
#### Secants, Chords, And Tangents (video)
The diagram of triangle ABC below shows the sides AB (i.e. c) and BC (i.e. a) and the included angle ABC ##(hat)##. From point A drop a perpendicular to point D to the side BC, extended if necessary according to whether D is inside the triangle ##;(hat## acute) or outside the triangle ##;( hat## open). The triangle is now divided into two right triangles ABD and ADC. Since the angle at D is a right angle and given ##hat##, we can calculate AD as ##csinhat## and BD as ##ccoshat##. CD can then be defined as the difference between BC and BD: ##a-ccoshat##.
(Note that this difference will be negative if ##hat## is blunt, in which case ##tanhat## will be negative as expected.)
Therefore, in the right-angled triangle ADC, given the sides AD and CD, we can find AC and the angle ACD.
### Solved Show Your Matlab Commands, M Files, And Results. 1.
And it will not be different if the angle B is obtuse as shown in the diagram above. The perpendicular from point A to line BC is extended to form a right-angled triangle ABD, where AD=c sin(180-##hat##)=c sin##hat## and BD=c cos(180-## hat##)=-c cos##hat##.
Since BA and BC are in a certain ratio (BA:BC = c:a = k:1), AB is given by (c=ka) in the same parts as BD (##=-kacoshat ##) and all CBD (##=a-kacoshat##) are given.
Accordingly, in the right-angled triangle ADC, since two sides AD and CD are given, then AC (i.e. b) is also given the side and angle ##hat##.
### Question Video: Finding The Measure Of An Angle Given The Measure Of An Arc By Using The Properties Of Tangents To The Circle
An important point to note in the second half of the above derivation (##hat##) is that Copernicus moves from reference to specific parties to a relationship of specific parties. When solving for angle C using the “tan law”, the absolute values of a given pair of sides are unnecessary as long as the ratio between them is known. This reflects the last formula ##tanhat## and is illustrated in Practical Examples 1 and 4.
PF user kuruman notes that given the formula (say) ##tanhat=frac}}}##, the other two tan coefficients can be found using cyclic permutation, ##Arightarrow B rightarrow C rightarrow A## lowercase and uppercase. Or if we start with ##tanhat=frac}}}}## with a similar circular permutation in the opposite direction ##Arightarrow C rightarrow B rightarrow A##.
We will explore three physical applications where the “Tan Law” can be usefully applied. The first is Compton scattering, in which a photon strikes a stationary electron and a momentum transfer occurs, resulting in the photon being “scattered” at an angle that depends on the change in frequency (energy) of the incident and scattered photon. Generally, two angles need to be specified, namely the angle of the scattered photon and the angle of the electron. Both are measured from the original direction of motion of the incident photon.
## Solved] Figure With Each Point Is Must.. Question # 1: Two Tangents…
The second objective is to determine the projectile’s launch angle(s) if it hits a target that has known horizontal and vertical coordinates (measured from the point of launch). In this application we will use the ##vectimesvec=vectimesvec## vector equation described in the previous article titled “Quarter Projectile in Motion”.
The third is two-dimensional elastic collisions between two masses – one moving and the other stationary. As with Compton scattering, we find that the angles of the post-collision trajectory of the two masses (relative to the direction of the moving mass before the collision) depend on the energy transferred from one to the other, as well as (depending on the case). ) mass ratio. And after determining one angle based on the energy relationship, we can determine the other using the arson rule.
You can see that there are some interesting similarities between these seemingly different physical applications. Impulse and/or momentum are physical vector quantities in any situation, and the angle of deflection is related to the change in energy. In each case, the 3 vector quantities involved form a triangle in which 2 sides and the included angle are known or can be calculated. The second angle can then be determined using the “tan law”.
#### What Is The Inverse Of Cosine?
Although we focus on momentum and energy applications in this article, the “tan law” is obviously as versatile in use as its “siblings”, the law of sine and cosine. In the first of the “worked examples” provided, the “tanning rule” is illustrated in a problem involving 3 forces in equilibrium. Of course, the vector sum of such forces will be zero, and the vector associated with them is a closed triangle in which the ratio of the two sides is given along with the included angle, and we need to find a pair of unknowns. angle. The other 3 practical examples refer to the above physical applications.
Below is a vector diagram representing Compton scattering. The momentum of the incident photon is equal to the sum of the momentum of the scattered photon and the incident electron. The electron is assumed to be initially at rest.
In order to use the tan law to determine the trajectory angle ##phi## of the electron, we first need the deflection angle ##theta_d## between the incident and scattered photons. We could just use the known result, but for completeness we will include a short derivation of it.
#### Law Of Tangents
As mentioned above, we will use the vector equation ##vectimesvec=vectimesvec## to determine the two possible projectile projectile angles aimed at a target with known coordinates (measured from the starting point). ##vec a## is the acceleration, ##vec## is the displacement from the launch point, and ##vec u## and ##vec v## are the initial and final velocities of the projectile, respectively. For projectile motion ###vec a=0;hat} -g;hat}## and ###vec s=R;hat} + h;hat}##, where R is the target horizontal range and h is its vertical height above the starting point. Hence ##vectimesvec=gR ;hat}##. We can write ##vectimesvec## as ##|v| |u| sintheta_d; hat}## where ##theta_d## is the angle between the initial and final velocity vectors. From this we get the equation: \$\$sintheta_d=frac.\$\$ Here we follow the method described by J. Gibson Winans in an article titled “Convenient Equations for Shotile Motion” from the American Journal of Physics. This paper uses a structural technique to determine the elevation angles, while here we have used an analytical approach using the “tan law” described above. Also in this paper, the above vector equation was derived from the mathematics of quaternion products, while here we present a derivation based on Newton’s second law applied to angular momentum.
We start by “transforming” the vector equations into an angular momentum equation by simply multiplying both sides by the ball’s mass m: \$\$mvectimesvec=mvectimesvec.\$\$ Then, in terms of the ball’s motion # #m vectimesvec## represents the gravitational moment of the projectile relative to its launch point. Taking an example from PF user @TSny (who originally identified the physical meaning of the vector equation as described here), we can show that the right-hand side of the equation ##(mvec v times vec u)## represents the rate of change of angular momentum according to with Newton’s second law applied to angular motion: \$\$L=mvectimesvec=mvectimesfrac+vec)t}=fractimesvec) t }\$\$
\$\$implies dot L=mleft(fractimesvec}+fractimesvec)t}right).\$\$ Write ##vec a## as ##frac##, the above will simplify to \$\$dot L =mleft(fractimesvec}+fractimesvec}right)=mvec v times vec u\$\$
### If The Angle Between Two Tangents Drawn From An External Point ‘p’ To
Using the “tan law” we can now “solve” the following velocity vector plot to obtain an expression for the launch angle(s) versus the initial horizontal velocity of the projectile.
How to find tangent of an angle, how to calculate an angle, how to calculate roof angle, how to calculate angle, tangent of an acute angle, calculate angle from tangent, tangent of an angle, tangent of an angle formula, how to calculate a angle, calculate angle using tangent, how to find an angle using tangent, how to calculate tangent
|
# Deriving the Sum of the Arithmetic Sequence
We have discussed arithmetic progression or arithmetic sequence and you have learned how to find its nth term. In the previous post, you have also learned how to find the sum of the first n positive integers. Notice that first n positive integers 1, 2, 3, 4, 5, all the way up to n is also an arithmetic sequence with first term 1 and constant difference 1.
Now, the question that comes to mind is, how do we find the sum of an arithmetic sequence?
Recall the method that we used in “Finding the sum of the first n positive integers.” We added the integers twice with the order of the terms reversed as shown above. Clearly, we can use this method to find the sum of the arithmetic sequence 3, 7, 11, 15, 19, 23, 27. Continue reading
# Sum of the First n Positive Integers: Another Solution
Late last month, we have seen how Gauss had possibly calculated the sum of the first n positive integers. In this post, we explore another solution. As we have done before, let us first start with a specific example, and then generalize later. What is the sum of the first 6 positive integers?
If we let S be the sum of the first 6 positive integers, then S = 1 + 2 + 3 + 4 + 5 + 6. Since the order of the addends does not change the sum, S = 6 + 5 + 4 + 3 + 2 + 1. Adding the two equation, we have
so, 2S = 6(7) = 42 and S = 21.
We can generalize the addition above as shown in the following image.
As we can see, each sum is equal to n + 1 and there are n of them. Therefore,
$latex 2S = n(n+1)$
which implies that
$latex S = \displaystyle\frac{n(n+1)}{2}$
|
### How to Analyze a GMAT Integrated Reasoning Two-Part Question
This is the latest in a series of How To Analyze articles that began with the general How To Analyze A Practice Problem article (click on the link to read the original article). This week, we’re going to analyze a specific IR question from the Two-Part prompt category. First, give yourself up to 2.5 minutes to try the below GMATPrep problem.
An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and workers who will be applying a finish to the exterior of the spheres need to know the surface area of each sphere. The finishing process costs $92 per square meter. The surface area of a sphere is equal to 4Ï€r2, where r is the radius of the sphere. In the table, select the value that is closest to the cost of finishing a sphere with a 5.50-meter circumference as well as the cost of finishing a sphere with a 7.85-meter circumference. Make only two selections, one in each column. Circumference 5.50 m Circumference 7.85 m Finishing cost$900 $1,200$1,800 $2,800$3,200 $4,500 After trying the problem, checking the answer, and reading the given solution (if any), I then try to answer the questions listed below. First, I’ll give you what I’ll call the standard solution (that is, one we might see in an Official Guide book if this were an official guide problem “ a correct solution but not necessarily one that shows us the easiest way to do the problem). Then we’ll get into the analysis. Standard solution: The formula for circumference is C = 2Ï€r. We can use this to calculate the radii of the two spheres (note that the problem asks us to find the closest values, so we can estimate): 5.50 m circumference sphere: 5.5 = 2Ï€r. (5.5/2Ï€) = r. (Use calculator) r = 0.876 (approx.). 7.85 m circumference sphere: 7.85 = 2Ï€r. (7.85/2Ï€) = r. (Use calculator) r = 1.25 (approx.). Next, the problem tells us that the formula for surface area is SA = 4Ï€r2. We can plug in to calculate the surface area of each sphere: 5.50 m circumference sphere: r = 0.876. SA = 4 Ï€(0.876)2. (Use calculator) SA = 9.63 7.85 m circumference sphere: r = 1.25. SA = 4 Ï€(1.25)2. (Use calculator) SA = 19.625 Finally, we can multiply each surface area by$92 per square meter to find the cost required to finish each sphere:
5.50 m circumference sphere: cost = (SA)(cost per square meter) = (9.63)(92) = $886 7.85 m circumference sphere = (19.625)(92) =$1,805
The correct answers are $900 for the first column and$1,800 for the second column.
1. Did I know WHAT they were trying to test?
– Was I able to CATEGORIZE this question by topic and subtopic? By process / technique? If I had to look something up in my books, would I know exactly where to go?
→ The question is an IR Two-Part prompt. The question prompt is pretty distinctive: the answer choices are presented in table form and I have to select two of them. This problem is wordy, but it’s not a verbal problem “ it definitely falls on the quant side of the fence.
– Did I COMPREHEND the symbols, text, questions, statements, and answer choices? Can I comprehend it all now, when I have lots of time to think about it? What do I need to do to make sure that I do comprehend everything here? How am I going to remember whatever I’ve just learned for future?
→ I got the question right but I spent way too much time doing it “ nearly 4 minutes. The math was tedious and I had to use the calculator repeatedly. Reading the standard solution really didn’t help much “ that is pretty much what I did, but it’s just so much work! I can already tell that I’m going to have to try to figure out a more efficient way to do this one (see section 2, below).
– Did I understand the actual CONTENT (facts, knowledge) being tested?
→ This part was fine I think. I knew the formula for circumference. They gave me the surface area formula, and I did put everything together okay. It just took me too long.
2. How well did I HANDLE what they were trying to test?
– Did I choose the best APPROACH? Or is there a better way to do the problem? (There’s almost always a better way!) What is that better way? How am I going to remember this better approach the next time I see a similar problem?
→ When I was solving, I remember thinking one thing was annoying: at one point, I divided by Ï€ and then a few steps later I had to multiply by Ï€ again. Maybe there’s a way I could’ve cut out that step altogether. I’m going to go play around with just the formulas and see what happens.
→ Let’s see, first I had C = 2Ï€r and the point here was that I had to solve for the radius: C/2Ï€ = r. My next step was to plug that r into the surface area formula:
SA = 4Ï€r2
SA = 4Ï€(C/2Ï€)2
→ Oh, wow, check it out! I’m going to be able to cancel out some stuff and make this way easier!
SA = C2/Ï€
→ That’s amazing. All I have to do is square the circumference figure and divide by Ï€. Let’s see, first I’m going to square 5.5. That’s about 30 (since 52 = 25 and 62 = 36). Also, I can estimate not only because the problem told me I could (by asking for the closest number) but also because of those answer choices. Look how far apart they are! Okay, now divide by 3 (close enough “ we’re estimating!) and the surface area is 10. The cost per square meter is $92, so the approximate cost is$920. Only one answer is close. Wow “ I just did that whole thing without having to use my calculator once!
→ Let’s do the other one. I need to square 7.85 which is almost 8. 82 = 64, so let’s say 7.852 = 60. Divide by 3 to get 20. Then, 20 ´ $92 is about$1,800 (because 2 ´ 9 = 18).
→ So a little work up front made my life immensely easier in the end. How would I know to do this next time? Well certainly, first, I’m going to write down the two formulas (circumference and surface area) side-by-side. That will let me notice that they both have r variables and pi symbols. Next, the problem literally tells me that I can approximate (closest to), plus the answers are really far apart. But when I look at the formulas separately, there doesn’t seem to be any reasonable way to approximate “ for instance, to find the radius for the first sphere, I’d have to divide 5.5 by 2 and by 3, or 5.5 / 6 “ which is smaller than 1 “ should I estimate to 1 or would I have to use 0.9? Hmm, this seems messy “ maybe there’s some algebra manipulation I can do first yes, yes, there is.
→ That’s how I’d get myself to realize that I should combine those two equations up front. That might take about 45 to 60 seconds, then I’d use another 45 to 60 seconds to do the algebra, and now I’ve got 30 to 60 seconds left to do the last bit of math, which has now become so easy that I don’t even need the calculator.
– Did I have the SKILLS to follow through? Or did I fall short on anything?
→ I had the math skills, yes, but not the test taker savvy skills to realize that I should’ve done some algebra simplification before I started to plug and chug. I need to retrain myself to look before I leap “ in other words, to see how I can simplify things before I just dive right in.
– Did I make any careless mistakes? If so, WHY did I make each mistake? What habits could I make or break to minimize the chances of repeating that careless mistake in future?
→ I didn’t on this one, although I could certainly imagine that it would have been easy to do so!
– Am I comfortable with OTHER STRATEGIES that would have worked, at least partially? How should I have made an educated guess?
→ Certainly, the smaller sphere should cost less, so I would’ve kept that in mind if I’d had to guess. Alternatively, I could’ve worked backwards starting from the answers “ but I’m thinking that would be too time consuming on this one, because I have 6 answers and have to find values for 2 spheres.
– Do I understand every TRAP & TRICK that the writer built into the question, including wrong answers?
→ Actually, from the way that they wrote this, I do think that they were trying to get me to just start plugging and chugging (and wasting a bunch of time). Next time I have the urge to hop on the calculator immediately, I’m going to make sure to write stuff out first and see whether that’s really my best course of action.
3. How well did I or could I RECOGNIZE what was going on?
– Did I make a CONNECTION to previous experience? If so, what problem(s) did this remind me of and what, precisely, was similar? Or did I have to do it all from scratch? If so, see the next bullet.
– Can I make any CONNECTIONS now, while I’m analyzing the problem? What have I done in the past that is similar to this one? How are they similar? How could that recognition have helped me to do this problem more efficiently or effectively? (This may involve looking up some past problem and making comparisons between the two!)
→ They got me to default to pulling up the calculator and chugging away. Now that I’ve seen this, though, I’m going to be prepared for next time (see below).
– HOW will I recognize similar problems in the future? What can I do now to maximize the chances that I will remember and be able to use lessons learned from this problem the next time I see a new problem that tests something similar?
→ The key thing here, I think, was the two formulas that shared a variable “ that, coupled with the fact that they told me I could estimate, yet it didn’t seem reasonable at all (at first) to estimate. Those two things together are a really good clue that perhaps the two formulas can be combined and simplified, and then I will actually be able to estimate (and possibly entirely avoid using the calculator in the first place!).
And that’s it! Note that, of course, the details above are specific to each individual person “ such a write-up would be different for every single one of you, depending upon your particular strengths, weaknesses, and mistakes. Hopefully, though, this gives you a better idea of the way to analyze an IR problem. This framework also gives you a valuable way to discuss problems with fellow online students or in study groups “ this is the kind of discussion that really helps to maximize scores.
* GMATPrep question courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.
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6. Stacey Koprince January 7, 2013 at 6:14 pm
Nope, that’s typical. They expect us to know certain things, including a rough approximation of the value of pi. If they want to ask us about something that they do *not* expect us to know (such as the surface area of a sphere), then they’ll tell it to us.
It’s important, of course, to know which things they already expect us to know!
7. Aditya December 27, 2012 at 3:52 am
The best way to solve such problems, especially when we are not given value of certain variables by GMAT is to first understand that we are not supposed to know the value of those variables at all. Now coming back to the problem – easiest way to solve is as below:-
1) given 2Ï€r = 5.5 and 7.85 and 4Ï€r2 = surface area.
2) we need to find out possible values of 4*Ï€*5.5*5.5*92 and 4*Ï€*7.85*7.85*92
3) without thinking too much, just divide 4*Ï€*7.85*7.85*92 by 4*Ï€*5.5*5.5*92, so you will get an approximate factor of around 1.95 to 2.05 or say ~ 2 (this is very easy if you are good at basic maths)
4) so now we know that the 2nd value is around 2ice the first value, which means first value can only be the first 3 choices.
5) just by looking at the 3 choices we can easily make out that 900 and 1800 are the right answers for this problem.
Good luck !!
8. Udni December 27, 2012 at 3:34 am
I am really surprised that GMAT is providing formula for surface area of the sphere in the problem but did not bother to provide the value of “pi”. ..Strange ??
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10. PraPon December 22, 2012 at 11:40 pm
Great explanation Stacy. It was immensely helpful!!! Thank you.
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# How do you complete a square in algebra?
## How do you complete a square in algebra?
Steps
1. Step 1 Divide all terms by a (the coefficient of x2).
2. Step 2 Move the number term (c/a) to the right side of the equation.
3. Step 3 Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.
## What is an example of coefficient in math?
What is a Coefficient in Math? A coefficient is a number or an alphabet that is multiplied by a variable of a single term or the terms of a polynomial. For example, in the term 7x, 7 is the coefficient. The coefficient of x in the term 3xy is 3y.
What are the steps to completing the square?
The completing the square method involves the following steps:
1. Step 1) Divide all terms by the coefficient of .
2. Step 2) Find.
3. Step 3) Find.
4. Step 4) Add to both sides of the equation.
5. Step 5) Complete the square on the left-hand-side of the equation.
6. Step 7) Take the square root of both sides and solve for the variable.
Does the coefficient have to be 1 to complete the square?
Even if \begin{align*}b\end{align*} is not divisible by \begin{align*}a\end{align*}, the coefficient of \begin{align*}x^2\end{align*} needs to be 1 in order to complete the square. Step 3: Now, complete the square. Determine what number would make a perfect square trinomial.
### What are coefficients in algebra?
Coefficient is a number that is being multiplied by the variable. 2x+6x+14. The 2x, 6x, and 14 are terms because they are being added together. 2 and x; 6 and x are factors because they are being multiplied together. 2 from 2x and 6 from 6x are the coefficients because they are being multiplied by the variable.
### What is a coefficient in an algebraic expression?
coefficients are the number when you multiply a number and a variable. For example 5a the coefficient in that term is 5 if you have 48e the coefficient is 48. So the coefficient is the number when you multiply a number times a variable.
How do you complete the square with 2 terms?
1. Step 1: Rearrange–Divide (as needed)
3. Step 3: Factor Left–Simplify Right.
4. Step 4: Solve!
5. Step 1: Divide & Group, Move Constant Rt.
6. Step 1: Group & Factor.
7. Step 2: Complete the Square Twice, (Add)
8. Step 2: Complete the Square, (Add-Mult.-Subtract)
How to calculate completing the square?
Enter the expression in the input box
• To get the result,click “Solve by Completing the Square”
• In the new window,the variable value will be displayed for the given expression
• ## How to factor by completing the square?
– Determine the values of 𝑎 and 𝑏. – Calculate 2 𝑎 𝑏. – Add 2 𝑎 𝑏 and − 2 𝑎 𝑏 to the expression. – Factor 𝑎 + 2 𝑎 𝑏 + 𝑏 as ( 𝑎 + 𝑏) , or factor 𝑎 − 2 𝑎 𝑏 + 𝑏 as ( 𝑎
## How do you solve the equation by completing the square?
Transform the equation so that the constant term,c,is alone on the right side.
• If a,the leading coefficient (the coefficient of the x 2 term),is not equal to 1,divide both sides by a .
• Add the square of half the coefficient of the x -term,( b 2 a) 2 to both sides of the equation.
• Factor the left side as the square of a binomial.
• What are the steps to complete the square?
Step 1 Divide all terms by a (the coefficient of x 2).; Step 2 Move the number term (c/a) to the right side of the equation.; Step 3 Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.; We now have something that looks like (x + p) 2 = q, which can be solved rather easily: Step 4 Take the square root on both sides of
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We've updated our
TEXT
# Summary: Using Intercepts to Graph Lines
## Key Concepts
• Intercepts
• The $x$-intercept is the point, $\left(a,0\right)$ , where the graph crosses the $x$-axis. The $x$-intercept occurs when $y$ is zero.
• The $y$-intercept is the point, $\left(0,b\right)$ , where the graph crosses the $y$-axis. The $y$-intercept occurs when $y$ is zero.
• Find the x and y intercepts from the equation of a line
• To find the $x$-intercept of the line, let $y=0$ and solve for $x$.
• To find the $y$-intercept of the line, let $x=0$ and solve for $y$.
• Graph a line using the intercepts
1. Find the x- and y- intercepts of the line.
• Let $y=0$ and solve for $x$.
• Let $x=0$ and solve for $y$.
2. Find a third solution to the equation.
3. Plot the three points and then check that they line up.
4. Draw the line.
• Choose the most convenient method to graph a line
1. Determine if the equation has only one variable. Then it is a vertical or horizontal line.
• $x=a$ is a vertical line passing through the $x$-axis at $a$.
• $y=b$ is a vertical line passing through the $y$-axis at $b$.
2. Determine if y is isolated on one side of the equation. Then graph by plotting points. Choose any three values for x and then solve for the corresponding y- values.
3. Determine if the equation is of the form $Ax+By=C$ , find the intercepts. Find the x- and y- intercepts and then a third point.
## Glossary
intercepts of a line
Each of the points at which a line crosses the $x$-axis or the $y$-axis is called an intercept of the line.
## Contribute!
Did you have an idea for improving this content? We’d love your input.
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Class 12 Maths NCERT Solutions for Chapter 9 Differential Equations Exercise 9.4
Differential Equations Exercise 9.4 Solutions
1. For the differential equations find the general solution :
dy/dx = (1 - cos x)/(1 + cos x)
Solution
The given differential equation is :
Separating the variables, we get :
Now, integrating both sides of this equation, we get :
⇒ y = 2 tan(x/2) - x + C
This is the required general solution of the given differential equation.
2. For the differential equation find the general solution :
dy/dx = √(4 – y2) (-2 < y < 2)
Solution
The given differential equation is :
dy/dx = √(4 – y2
Separating the variables, we get :
⇒ dy/√(4 – y2) = dx
Now, integrating both sides of this equation, we get :
⇒ y/2 = sin(x + C)
⇒ y = 2 sin (x + C)
This is the required general solution of the given differential equation.
3. For the differential equations find the general solution :
dy/dx + y = 1(y ≠ 1)
Solution
The given differential equation is :
dy/dx + y = 1
⇒ dy + y dx = dx
⇒ dy = (1 - y) dx
Separating the variables, we get :
⇒ dy/(1 - y) = dx
Now, integrating both sides, we get :
⇒ -log(1 - y) = x + log C
⇒ - log C - log (1 - y) = x
⇒ log C(1 - y) = -x
⇒ C(1 - y) = e-x
⇒ y = 1 + Ae-x (where A = -1/C)
This is the required general solution of the given differential equation.
4. For the differential equations find the general solution:
sec2 x tan y dx + sec2 y tan x dy = 0
Solution
The given differential equation is :
sec2 x tan y dx + sec2 y tan x dy = 0
⇒ sec2 x = dt/dx
⇒ sec2 x dx = dt
= log t
= log (tan x)
Similarly,
Substituting these values in equation (1), we get :
log (tan x) = - log (tan y) + log C
⇒ log (tan x) = log (C/tan y)
⇒ tan x = C/tan y
⇒ tan x tan y = C
This is the required general solution of the given differential equation.
5. For the differential equations find the general solution:
(ex + e–x) dy – (ex – e–x) dx = 0
Solution
The given differential equation is :
(ex + e–x) dy – (ex – e-x ) dx = 0
⇒ (ex + e-x ) dy = (ex - e-x )dx
Let (ex + e–x) = t
Differentiating both sides with respect to x, we get :
⇒ (ex + e–x) dx = dt
Substituting his value in equation (1), we get :
y = ∫(1 + t) dt + C
⇒ y = log(t) + C
⇒ y = log (ex + e–x) + C
This is the required general solution of the given differential equation.
6. For the differential equations find the general solution:
dy/dx = (1 + x2 )(1+ y2
Solution
The given differential equation is :
dy/dx = (1 + x2 )(1+ y2)
⇒ dy/(1 + y2 ) = (1 + x2)dx
Integrating both sides of this equation, we get :
⇒ tan-1 y = ∫dx + ∫x2 dx
⇒ tan-1 y = x + x3/3 + C
This is the required general solution of the given differential equation.
7. For the differential equations find the general solution :
y log y dx - x dy = 0
Solution
The given differential equation is :
y log y dx - x dy = 0
⇒ y log y dx = x dy
⇒ dy/(y log y) = dx/ x
Integrating both sides, we get :
∫dy/(y log y) = ∫ dx/x ...(1)
Let log y = t
Substituting this value in equation (1), we get :
∫dt/t = ∫dx/x
⇒ log t = log x + log C
⇒ log (log y) = log Cx
⇒ log y = Cx
⇒ y = ecx
This is the required general solution of the given differential equation.
8. For the differential equations find the general solution :
x5 (dy/dx) = -y5
Solution
The given differential equation is :
Integrating both sides, we get :
⇒ x-4 + y-4 = -4k
⇒ x-4 + y-4 = C (C = -4k)
This is the required general solution of the given differential equation.
9. For the differential equation find the general solution :
ex tan y dx + (1 – ex) sec2 y dy = 0
Solution
The given differential equation is :
dy/dx = sin-1 x
⇒ dy = sin-1 x dx
Integrating both sides, we get :
Substituting this value in equation (1), we get :
⇒ y = x sin-1 x + √t + C
⇒ y = x sin-1 x + √(1 - x2) + C
This is the required general solution of the given differential equation.
10. For the differential equations find the general solution:
ex tan y dx + (1 – ex) sec2 y dy = 0
Solution
The given differential equation is :
ex tan y dx + (1 – ex) sec2 y dy = 0
(1 - ex )sec2 y dy = -ex tan y dx
Separating the variables, we get :
Now, let 1 - ex = t.
∴ d/dx (1 - ex) = dt/dx
-ex = dt/dx
⇒ -ex dx = dt
⇒ log (tan y) = log (1 - ex ) + log C
⇒ log (tan y) = log [C(1 - ex )]
⇒ tan y = C(1 - ex )
This is the required general solution of the given differential equation.
11. For each of the differential equations find a particular solution satisfying the given condition:
(x3 + x2 + x + 1) dy/dx = 2x2 + x; y = 1 when x = 0
Solution
The given differential equation is :
⇒ 2x2 + x = Ax2 + A + Bx2 + Bx + Cx + C
⇒ 2x2 + x = (A + B)x2 + (B + C)x + (A + C)
Comparing the coefficients of x2 and x, we get :
A + B = 2
B + C = 1
A + C = 0
Solving these equations, we get :
A = 1/2, B = 3/2 and C = -1/2
Substituting the values of A, B and C in equation (2), we get :
Now, y = 1 when x = 0.
⇒ 1 = (1/4) log (1) - (1/2) tan-1 0 + C
⇒ 1 = (1/4) × 0 - (1/2) × 0 + C
⇒ C = 1
Substituting C = 1 in equation (3), we get :
y = [(1/4) log (x + 1)2 (x2 + 1)3] - (1/2) tan-1 x + 1
12. For each of the differential equations find a particular solution satisfying the given condition :
x(x2 - 1) dy/dx = 1, y = 0 when x = 2
Solution
x(x2 - 1) dy/dx = 1
Comparing the coefficients of x2, x and constant, we get :
A = -1
B - C = 0
A + B + C = 0
Solving these equations, we get B = 1/2 and C = 1/2 .
Substituting the values of A, B, and C in equation (2), we get :
⇒ 3k2/4 = 1
⇒ 3k2 = 4
⇒ k2 = 4/3
Substituting the value of k2 in equation (3), we get :
13. For each of the differential equations find a particular solution satisfying the given condition:
cos (dx/dy) = a (a ∈ R); y = 1 when x = 0
Solution
cos (dy/dx) = a
⇒ dy/dx = cos-1 a
⇒ dy = cos-1 a dx
Integrating both sides, we get :
∫dy = cos-1 a ∫dx
⇒ y = cos-1 a . x + C
⇒ y = x cos-1 a + C ...(1)
Now, y = 1 when x = 0
⇒ 1 = 0. cos-1 a + C
⇒ C = 1
Substituting C = 1 in equation (1), we get :
y = x cos-1 a + 1
⇒ (y -1)/x = cos-1 a
⇒ cos[(y -1)/x] = a
14. For each of the differential equations find a particular solution satisfying the given condition:
dy/dx = y tan x; y = 1 when x = 0
Solution
dy/dx = y tan x
⇒ dy/y = tan x dx
Integrating both sides, we get :
∫dy/y = ∫tan x dx
⇒ log y = log(sec x) + log C
⇒ log y = log (C sec x)
⇒ y = C sec x ...(1)
Now, y = 1 when x = 0.
⇒1 = C × sec 0
⇒1 = C × 1
⇒ C = 1
Substituting C = 1 in equation (1), we get :
y = sec x
15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = ex sin x.
Solution
The differential equation of the curve is :
y' = ex sin x
⇒ dy/dx = ex sin x
⇒ dy = ex sin x dx
Integrating both sides, we get :
∫dy = ∫ex sin x dx ...(1)
Let I = ∫ex sin x dx.
⇒ I = ex sin x - ex sin x - I
⇒ 2I = ex (sin x - cos x)
⇒ I = [ex (sin x - cos x)]/2
Substituting this value in equation (1), we get :
y = [ex (sin x - cos x)]/2 + C ...(2)
Now, the curve passes through point (0, 0)
⇒ 2y = ex (sin x - cos x) + 1
⇒ 2y - 1 = ex (sin x - cos x)
Hence, the required equation of the curve is 2y - 1 = ex (sin x - cos x).
16. For the differential equation xy (dy/dx) = (x + 2)(y + 2) find the solution curve passing through the point (1, -1).
Solution
The differential equation of the given curve is :
⇒ y - 2log(y + 2) = x + 2log x + C
⇒ y - x - C = log x2 + log (y + 2)2
⇒ y - x - C = log [x2 (y + 2)2 ] ...(1)
Now, the curve passes through point (1, -1).
⇒ -1 - 1 - C = log[(1)2 (-1 + 2)2 ]
⇒ -2 - C = log 1 = 0
⇒ C = -2
Substituting C = -2 in equation (1), we get :
y - x + 2 = log[x2 (y + 2)2 ]
This is the required solution of the given curve.
17. Find the equation of a curve passing through the point (0, –2) given that at any point (x ,y) on the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.
Solution
Let x and y be the x - coordinate and y - coordinate of the point on the curve respectively.
We know that the slope of a tangent to the curve in the coordinate axes is given by the relation.
dy/dx
According to the given information, we get :
y (dy/dx) = x
⇒y dy = x dx
Integrating both sides, we get:
∫y dy = ∫x dx
⇒ y2/2 = x2 /2 + C
⇒ y2 - x2 = 2C ...(1)
Now, the curve passes through point (0, -2).
∴ (-2)2 - 22 = 2C
⇒ 2C = 4
Substituting 2C = 4 in equation (1), we get :
y2 - x2 = 4
This is the required equation of the curve.
18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).
Solution
It is given that (x, y) is the point of contact of the curve and its tangent.
The slope (m1 ) of the line segment joining (x, y) and (-4, -3) is (y +3)/(x + 4)
We know that the slope of the tangent to the curve is given by the relation,
dy/dx
∴ Slope (m2 ) of the tangent = dy/dx
According to the given information :
m2 = 2m1
⇒ log (y + 3) = 2 log (x + 4) + log C
⇒ log (y + 3) = log C (x + 4)2
⇒ y + 3 = C(x + 4)2 ...(1)
This is the general equation of the curve.
It is given that it passes through point (-2, 1).
⇒ 1 + 3 = C(-2 + 4)2
⇒ 4 = 4C
⇒ C = 1
Substituting C = 1 in equation (1), we get :
y + 3 = (x + 4)2
This is the required equation of the curve.
19. The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Solution
Let the rate of change of the volume of the balloon be k (where k is a constant )
⇒ 4Ï€r3 = 3(kt + C) ...(1)
Now, at t = 0, r = 3,
⇒ 4Ï€ × 3= 3 (k × 0 + C)
⇒ 108Ï€ = 3C
⇒ C = 36Ï€
At = 3, r = 6:
⇒ 4Ï€ × 63 = 3 (k × 3 + C)
⇒ 864Ï€ = 3 (3k + 36Ï€)
⇒ 3k = –288Ï€ – 36Ï€ = 252Ï€
⇒ k = 84Ï€
Substituting the values of k and C in equation (1), we get:
4Ï€r3 = 3(84Ï€t + 36Ï€)
⇒ 4Ï€r3 = 4Ï€(63t + 27)
⇒ r3 = 63t + 27
⇒ r = (63t + 27)1/3
Thus, the radius of the balloon after t seconds is (63t + 27)1/3 .
20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (loge2 = 0.6931).
Solution
Let p, t, and r represent the principal, time, and rate of interest respectively.
It is given that the principal increases continuously at the rate of r% per year.
It is given that when t = 0, p = 100.
⇒ 100 = ek ...(2)
Now, if t = 10, then p = 2× 100 = 200.
⇒ r/10 = 0.6931
⇒ r = 6.931
Hence, the value of r is 6.93 %.
21. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).
Solution
Let p and t be the principal and time respectively.
It is given that the principal increases continuously at the rate of 5 % per year.
Now, when t = 0, p = 1000.
1000 = ec ...(2)
At t = 10, years the amount will worth Rs 1648.
22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Solution
Let y be the number of bacteria at any instant t.
It is given that the rate of growth of the bacteria is proportional to the number peresent.
∴ dy/dt ∝ y
⇒ dy/dt = ky (where k is a constant)
⇒ dy/y = kdt
Integrating both sides, we get:
∫dy/y = k ∫dt
⇒ log y = kt + C ...(1)
Let y0 be the number of bacteria at t = 0 .
log y0 be the number of bacteria at t = 0.
log y0 = C
Substituting the value of C in equation (1), we get :
log y = kt + log y0
⇒ log y - log y0 = kt
⇒ log (y/y0) = kt
⇒ kt = log (y/y0) ...(2)
Also, it is given that the number of bacteria increases by 10% in 2 hours.
⇒ y = (110/100)y0
⇒ y/y0 = 11/10 ...(3)
Substituting this value in equation (2), we get :
k⋅2 = log (11/10)
⇒ k = (1/2) log (11/10)
Therefore, equation (2) becomes :
Now, let the time when the number of bacteria increases from 100000 to 200000 be t1 .
y = 2y0 at t = t1
From equation (4), we get:
Hence, in (2 log 2)/[log (11/10)] hours the number of bacteria increases from 100000 to 200000.
23. The general solution of the differential equation dy/dx = ex+y is
(A) ex + e–y = C
(B) ex + ey = C
(C) e–x + ey = C
(D) e–x + e–y = C
Solution
Integrating both sides, we get :
∫e–y dy = ∫ex dx
⇒ -e–y = ex + k
⇒ ex + e–y = -k
⇒ ex + e–y = c (c = -k )
Hence, the correct answer is A.
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# 6.3: Differentiation and Integration of Logarithmic and Exponential Functions
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
A student will be able to:
• Understand and use the rules of differentiation of logarithmic and exponential functions.
• Understand and use the rules of integration of logarithmic and exponential functions.
In this section we will explore the derivatives of logarithmic and exponential functions. We will also see how the derivative of a one-to-one function is related to its inverse.
## The Derivative of a Logarithmic Function
Our goal at this point to find an expression for the derivative of the logarithmic function Recall that the exponential number is defined as
(where we have substituted for for convenience). From the definition of the derivative of that you already studied in Chapter 2,
We want to apply this definition to get the derivative to our logarithmic function Using the definition of the derivative and the rules of logarithms from the Lesson on Exponential and Logarithmic Functions,
At this stage, let the limit of then becomes Substituting, we get
Inserting the limit,
But by the definition
From the box above, we can express in terms of natural logarithm by the using the formula Then
Thus we conclude
and in the special case where
To generalize, if is a differentiable function of and if then the above two equations, after the Chain Rule is applied, will produce the generalized derivative rule for logarithmic functions.
Derivatives of Logarithmic Functions
Remark: Students often wonder why the constant is defined the way it is. The answer is in the derivative of With any other base the derivative of would be equal a more complicated expression than Thinking back to another unexpected unit, radians, the derivative of is the simple expression only if is in radians. In degrees, , which is more cumbersome and harder to remember.
Example 1:
Find the derivative of
Solution:
Since , for
Example 2:
Find .
Solution:
Example 3:
Find
Solution:
Here we use the Chain Rule:
Example 4:
Find the derivative of
Solution:
Here we use the Product Rule along with
Example 5:
Find the derivative of
Solution:
We use the Quotient Rule and the natural logarithm rule:
## Integrals Involving Natural Logarithmic Function
In the last section, we have learned that the derivative of is . The antiderivative is
If the argument of the natural logarithm is then thus
Example 6:
Evaluate
Solution:
In general, whenever you encounter an integral with an integrand as a rational function, it might be possible that it can be integrated with the rule of natural logarithm. To do so, determine the derivative of the denominator. If it is the numerator itself, then the integration is simply the of the absolute value of the denominator. Let’s test this technique.
Notice that the derivative of the denominator is , which is equal to the numerator. Thus the solution is simply the natural logarithm of the absolute value of the denominator:
The formal way of solving such integrals is to use substitution by letting equal the denominator. Here, let and Substituting,
Remark: The integral must use the absolute value symbol because although may have negative values, the domain of is restricted to
Example 7:
Evaluate
Solution:
As you can see here, the derivative of the denominator is Our numerator is However, when we multiply the numerator by we get the derivative of the denominator. Hence
Again, we could have used substitution.
Example 8:
Evaluate .
Solution:
To solve, we rewrite the integrand as
Looking at the denominator, its derivative is . So we need to insert a minus sign in the numerator:
## Derivatives of Exponential Functions
We have discussed above that the exponential function is simply the inverse function of the logarithmic function. To obtain a derivative formula for the exponential function with base we rewrite as
Differentiating implicitly,
Solving for and replacing with
Thus the derivative of an exponential function is
In the special case where the base is since the derivative rule becomes
To generalize, if is a differentiable function of with the use of the Chain Rule the above derivatives take the general form
And if
Derivatives of Exponential Functions
Example 9:
Find the derivative of .
Solution:
Applying the rule for differentiating an exponential function,
Example 10:
Find the derivative of .
Solution:
Since
Example 11:
Find if
where and are constants and
Solution:
We apply the exponential derivative and the Chain Rule:
## Integrals Involving Exponential Functions
Associated with the exponential derivatives in the box above are the two corresponding integration formulas:
The following examples illustrate how they can be used.
Example 12:
Evaluate .
Solution:
Example 13:
Solution:
In the next chapter, we will learn how to integrate more complicated integrals, such as , with the use of substitution and integration by parts along with the logarithmic and exponential integration formulas.
For a video presentation of the derivatives of exponential and logarithmic functions (4.4), see Math Video Tutorials by James Sousa, The Derivatives of Exponential and Logarithmic Functions (8:26).
## Review Questions
1. Find of
2. Find of
3. Find of
4. Find of
5. Find of
6. Find of
7. Evaluate
8. Evaluate
9. Evaluate
10. Evaluate
11. Evaluate
12. Evaluate
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# 3.7: Solving Multi-Step Inequalities
Difficulty Level: At Grade Created by: CK-12
## Introduction
Marching Times
“We have a lot to do to get ready for the big parade,” Mrs. Kline announced on Monday.
And Mrs. Kline was definitely correct. The students in the band had learned three new routines that they didn’t quite have down yet. It was going to take extra marching practice to put all of the pieces together. To do this, Mrs. Kline had the students outside marching for extra time.
On Tuesday, the band spent thirty fewer minutes marching than they did on Thursday. All in all, everyone was tired.
“We marched for more than three hours between Tuesday and today,” Juan said on the ride home on Thursday. His feet could definitely tell how long they had been marching, but all in all the hard work was paying off because the pieces were looking much better.
Given this information, what are three possible times that the band marched on Thursday?
Once again, you will need to know about inequalities to solve this problem. This problem will require you to write and solve an inequality that will have more than one step. This lesson will give you all the information that you need to accomplish this task.
What You Will Learn
In this lesson, you will learn how to complete the following skills.
• Solve inequalities involving combinations of like terms.
• Solve inequalities involving parentheses and the distributive property.
• Solve inequalities involving both distributing and combining.
• Model and solve real-world problems using multi-step inequalities.
Teaching Time
I. Solve Inequalities Involving Combinations of Like Terms
As you already know, some inequalities can be solved in a single step. For example, we could solve \begin{align*}b+4<10\end{align*} in one step—by subtracting 4 from each side.
However, two or more steps may be required to solve some inequalities. Inequalities that need more than one inverse operation to solve them can be called multi-step inequalities.
Let’s start by looking at combining like terms when we solve an inequality.
Example
\begin{align*}4x+3x<21\end{align*}
First, you can see that we have two terms that have the same variable. These are like terms. To solve an inequality with like terms, we will need to combine the like terms and then we can solve the inequality using the methods that we have already learned.
\begin{align*}7x<21\end{align*}
Here we divide both sides of the inequality by 7. Multiplication is the inverse operation of division.
\begin{align*}x<3\end{align*}
Now let’s look at one that is a bit more complicated.
Example
Solve for \begin{align*}n\end{align*}: \begin{align*}7n-8n-3>23\end{align*}.
First, subtract \begin{align*}7n-8n\end{align*} because \begin{align*}7n\end{align*} and \begin{align*}8n\end{align*} are like terms. Remember, you will need the rules for working with positive and negative integers when solving these inequalities.
\begin{align*}7n-8n-3 &> 23\\ 7n+(-8n)-3 &> 23\\ [7n+(-8n)]-3 &> 23\\ -1n-3 &> 23\\ -n-3 &> 23.\end{align*}
The expression on the left side of the inequality, \begin{align*}-n-3\end{align*}, is now in simplest form. The 3 cannot be subtracted from the \begin{align*}-n\end{align*} because they are not like terms.
The next step is to isolate the term with the variable, \begin{align*}-n\end{align*}, on one side of the inequality. Since 3 is subtracted from \begin{align*}-n\end{align*}, we should add 3 to both sides of the inequality to do this.
\begin{align*}-n-3 &> 23\\ -n-3+3 &> 23+3\\ -n+(-3+3) &> 26\\ -n+0 &> 26\\ -n &> 26\end{align*}
Since \begin{align*}-n\end{align*} means \begin{align*}-1n\end{align*} or \begin{align*}-1 \times n\end{align*}, we can divide each side of the inequality by -1 to get a positive \begin{align*}n\end{align*} by itself on one side of the equation. Since that involves dividing both sides of the inequality by a negative number, we must reverse the inequality symbol.
\begin{align*}-n &> 26\\ -1n &> 26\\ \frac{-1n}{-1} &< \frac{26}{-1}\\ 1n &< -26\\ n &< -26\end{align*}
The solution for this inequality is \begin{align*}n < -26\end{align*}.
Notice that you performed several operations when solving this inequality. Just as we could solve two step equations, we can also solve two step inequalities.
Let’s look at an example of a two – step inequality.
Example
Solve for \begin{align*}b\end{align*}: \begin{align*}3b+4<10\end{align*}.
Notice that there are two terms on the left side of the inequality, \begin{align*}3b\end{align*} and 4. Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}3b\end{align*}, by itself on one side of the inequality.
In the inequality, 4 is added to \begin{align*}3b\end{align*}. So, we can use the inverse of addition—subtraction. We can subtract 4 from both sides of the inequality. We do not need to change the inequality symbol during this step because we are subtracting a number, not multiplying or dividing by a negative number.
\begin{align*}3b+4 &< 10\\ 3b+4-4 &< 10-4\\ 3b+0 &< 6\\ 3b &< 6\end{align*}
Now, the term that includes a variable, \begin{align*}3b\end{align*}, is by itself on one side of the equation.
We can now use inverse operations to get the \begin{align*}b\end{align*} by itself. Since \begin{align*}3b\end{align*} means \begin{align*}3 \times b\end{align*}, we can divide both sides of the inequality by 3 to isolate the variable. Since we are dividing by a positive number, not a negative number, the inequality symbol should not change.
\begin{align*}3b &< 6\\ \frac{3b}{3} &< \frac{6}{3}\\ 1b &< 2\\ b &< 2\end{align*}
The solution is \begin{align*}b<2\end{align*}.
Now let’s look at solving another multi-step equation.
Example
Solve for \begin{align*}k\end{align*}: \begin{align*}3k+20+2k+10 \le 5\end{align*}.
First, combine the like terms on the left side of the inequality. \begin{align*}3k\end{align*} and \begin{align*}2k\end{align*} are like terms since each has the same variable, \begin{align*}k\end{align*}. The numbers 20 and 10 are also like terms, so they can be combined as well.
Use the commutative property of addition to help you reorder the terms being added. This property states that terms can be added in any order. Then use the associative property to group the terms so like terms are being added. The associative property of addition states that the grouping of terms being added does not matter. Once you reordered the terms, combine the like terms.
\begin{align*}3k+20+2k+10 & \le 5\\ 3k+(20+2k)+10 & \le 5\\ 3k+(2k+20)+10 & \le 5\\ (3k+2k)+(20+10) & \le 5\\ 5k + 30 & \le 5\end{align*}
The next step is to get the term with the variable, \begin{align*}5k\end{align*}, on one side of the inequality. Since 30 is added to \begin{align*}5k\end{align*}, we should subtract 30 from both sides of the inequality to do this.
\begin{align*}5k+30 & \le 5\\ 5k+30-30 & \le 5-30\\ 5k+0 & \le 5+(-30)\\ 5k & \le -25\end{align*}
Since \begin{align*}5k\end{align*} means \begin{align*}5 \times k\end{align*}, we should divide each side of the inequality by 5—a positive number—to get the \begin{align*}k\end{align*} by itself on one side of the inequality. Be careful! It is true that you will need to divide -25 by 5 to find the solution. However, since you are not dividing both sides of the inequality by a negative number, you do not reverse the inequality symbol.
\begin{align*}5k & \le -25\\ \frac{5k}{5} & \le \frac{-25}{5}\\ 1k & \le -5\\ k & \le -5\end{align*}
The solution is \begin{align*}k \le -5\end{align*} .
Now let’s look at some other types of inequalities.
II. Solve Inequalities Involving Parentheses and the Distributive Property
The inequalities that you will see in this section involve parentheses. We know from our earlier work that we can simplify an equation with parentheses by using the distributive property. We can do this with inequalities as well. Using the distributive property can help you to simplify an inequality so that it is easier to solve.
Let’s look at an example.
Example
Solve for \begin{align*}q\end{align*}: \begin{align*}-9(q+3)<45\end{align*}
Apply the distributive property to the left side of the inequality. Multiply each of the two numbers inside the parentheses by -9 and then add those products.
\begin{align*}-9(q+3) &< 45\\ (-9 \times q)+(-9 \times 3) &< 45\\ -9q+(-27) &< 45\end{align*}
Now, solve as you would solve any two-step inequality. Since -27 is added to \begin{align*}-9q\end{align*}, we can get \begin{align*}-9q\end{align*} by itself on one side of the inequality by subtracting -27 from both sides. Remember, subtracting -27 from a number is the same as adding its opposite, 27, to that number.
\begin{align*}-9q+(-27) &< 45\\ -9q+(-27)-(-27) &< 45-(-27)\\ -9q+(-27+27) &< 45+27\\ -9q+0 &< 72\\ -9q &< 72\end{align*}
To get \begin{align*}q\end{align*} by itself on one side of the inequality, we need to divide both sides by -9. Since we are dividing both sides by a negative number, you need to reverse the inequality symbol.
\begin{align*}-9q &< 72\\ \frac{-9q}{-9} &> \frac{72}{-9}\\ 1q &> -8\\ q &> -8\end{align*}
The solution is \begin{align*}q > -8\end{align*}.
Yes, there are a lot of steps to keep track of, but if you follow each part and don’t skip steps you will get there. Remember, you are trying to solve the inequality. Keeping your focus on this will help you keep from getting confused.
Example
\begin{align*}\frac{1}{2} (x+4) \le 10\end{align*}
First, we use the distributive property to multiply one-half with both of the terms inside the parentheses.
\begin{align*}\frac{1}{2} x+2 \le 10\end{align*}
Next, we subtract two from both sides of the inequality.
\begin{align*}\frac{1}{2} x \le 8\end{align*}
Now we can multiply both sides by the reciprocal of one-half which will cancel out the one-half leaving our variable alone. This is an example of the multiplicative inverse property.
\begin{align*}\frac{2}{1} \left(\frac{1}{2}\right) x \le 8 \left(\frac{2}{1}\right)\end{align*}
The answer is that \begin{align*}x \le 16\end{align*}.
Some inequalities will require that you use combining like terms and the distributive property. Let’s take a look at these problems now.
III. Solve Inequalities Involving Both Distributing and Combining
Now let’s apply both combining like terms and the distributive property when working with the same inequality.
Example
Solve for \begin{align*}w\end{align*}: \begin{align*}-2(8+w)+18<28\end{align*}.
First, we should apply the distributive property to the left side of the inequality. We can multiply each of the two numbers inside the parentheses by -2 and then add those products.
\begin{align*}-2(8+w)+18 &< 28\\ (-2 \times 8)+(-2 \times w)+18 &< 28\\ -16+(-2w)+18 &< 28\end{align*}
Next, we can add the like terms (-16 and 18) on the left side of the inequality. Using the commutative and associative properties to reorder the terms on the left side of the equation can make it easier to see how to do this.
\begin{align*}-16+(-2w)+18 &< 28\\ -16+[(-2w)+18] &< 28\\ -16+[18+(-2w)] &< 28\\ (-16+18)+(-2w) &< 28\\ 2+(-2w) &< 28\end{align*}
Finally, we solve as you would solve any two-step inequality. Since 2 is added to \begin{align*}-2w\end{align*}, our first step should be to subtract 2 from both sides of the inequality.
\begin{align*}2+(-2w) &< 28\\ 2-2+(-2w) &< 28-2\\ 0+(-2w) &< 26\\ -2w &< 26\end{align*}
Now, we can isolate the variable, \begin{align*}w\end{align*}, by dividing both sides of the inequality by -2. Since we are dividing both sides by a negative number, we need to reverse the inequality symbol.
\begin{align*}-2w &< 26\\ \frac{-2w}{-2} &> \frac{26}{-2}\\ 1w &> -13\\ w &> -13\end{align*}
The solution is \begin{align*}w > -13\end{align*}.
All of these skills can help us when we work with real – world problems. Let’s look at solving real – world problems with multi – step inequalities.
IV. Model and Solve Real – World Problems Using Multi-Step Inequalities
Sometimes, we can write a multi-step inequality to represent a real-world problem situation. Once again, you will be using key words to work on these. Take a minute and refer back to your notebook. Then continue.
Now let’s look at an example.
Example
Ms. Layne wants to build a rectangular deck in her back yard. She wants the length of the deck to be exactly 9 feet. She wants the perimeter of her deck to be, at most, 28 feet. The perimeter of any rectangle can be found by using the expression \begin{align*}P=2(l+w)\end{align*}, where \begin{align*}l\end{align*} represents the length and \begin{align*}w\end{align*} represents the width.
a. Write an inequality that could be used to represent \begin{align*}w\end{align*}, the possible widths, in feet, she could use for her deck.
b. Would a deck with a width of 6 feet result in a deck with the perimeter she wants?
Consider part a first.
You know that the length is 9 feet, so substitute 9 for \begin{align*}l\end{align*} into the expression \begin{align*}2(l+w)\end{align*}. This expression represents the actual perimeter of the deck.
\begin{align*}\text{actual perimeter}=2(l+w)=2(9+w)\end{align*}
Since she wants the perimeter to be “at most” 28 feet, you should use the “less than or equal to” \begin{align*}(\le)\end{align*} symbol. Translate this problem into an inequality.
\begin{align*}& She \ wants \ the \ \underline{perimeter} \ of \ her \ deck \ to \ be, \ \underline{at \ most}, \ \underline{28 \ feet}.\\ & \qquad \qquad \qquad \quad \ \ \downarrow \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \quad \ \downarrow\\ & \qquad \qquad \qquad \ 2(9+w) \qquad \qquad \qquad \qquad \quad \ \ \le \qquad \quad 28\end{align*}
So, this problem can be represented by the inequality \begin{align*}2(9+w) \le 28\end{align*}.
Next, consider part b.
To find all the possible values of \begin{align*}w\end{align*}, solve the inequality. First, apply the distributive property to the right side.
\begin{align*}2(9+w) & \le 28\\ (2 \times 9)+(2 \times w) & \le 28\\ 18+2w & \le 28\end{align*}
Now, solve as you would solve any two-step inequality. First, subtract 18 from both sides of the inequality.
\begin{align*}18+2w & \le 28\\ 18-18+2w & \le 28-18\\ 0+2w & \le 10\\ 2w & \le 10\end{align*}
Next, divide both sides of the inequality by 2. Since you are dividing by a positive number, the inequality symbol should stay the same.
\begin{align*}2w & \le 10\\ \frac{2w}{2} & \le \frac{10}{2}\\ 1w & \le 5\\ w & \le 5\end{align*}
The value of \begin{align*}w\end{align*} must be less than or equal to 5. Since 6 is greater than, not less than, 5, it is not a possible value of \begin{align*}w\end{align*}. So, if she built her deck so it was 6 feet wide, it would have a larger perimeter than she wants.
Now let’s look at the problem from the introduction. We can use what we have learned to help us in writing and solving an inequality.
## Real-Life Example Completed
Marching Times
Here is the original problem once again. First, reread it. Then write an inequality and solve it for three possible times for Thursday’s practice. There are four parts to your answer.
“We have a lot to do to get ready for the big parade,” Mrs. Kline announced on Monday.
And Mrs. Kline was definitely correct. The students in the band had learned three new routines that they didn’t quite have down yet. It was going to take extra marching practice to put all of the pieces together. To do this, Mrs. Kline had the students outside marching for extra time.
On Tuesday, the band spent thirty fewer minutes marching than they did on Thursday. All in all, everyone was tired.
“We marched for more than three hours between Tuesday and today,” Juan said on the ride home on Thursday. His feet could definitely tell how long they had been marching, but all in all the hard work was paying off because the pieces were looking much better.
Given this information, what are three possible times that the band marched on Thursday?
Solution to Real – Life Example
First, write an inequality using the given information.
\begin{align*}m - 30\end{align*} is the time that the students practiced on Tuesday
\begin{align*}m\end{align*} is the time that they practiced on Thursday
\begin{align*} >3 \end{align*} hours is the time that they practiced in all.
Here is the inequality.
\begin{align*}m + m - 30 > 3\end{align*} hours or 180 minutes
It makes sense to work with minutes because the given times are in minutes.
Now we solve the inequality.
\begin{align*}2m - 30 &> 180\\ 2m &> 210\\ m &> 105 \ minutes\end{align*}
The band marched for more than 105 minutes on Thursday. You could assume that they marched for 110 minutes, 115 minutes or 120 minutes. There are many possible options.
## Vocabulary
Here are the vocabulary words that are found in this lesson.
Inequality
a mathematical statement where one quantity can be less than, greater than or equal to another quantity.
Inverse Operation
the opposite operation. Inverse operations are used to solve equations.
Like Terms
terms in an equation or inequality that have common variables or no variables.
Distributive Property
a term outside of a set of parentheses can be multiplied by each of the terms inside the parentheses. This simplifies the parentheses.
## Time to Practice
Directions: Solve each inequality.
1. \begin{align*}2x+5>13\end{align*}
2. \begin{align*}4x-2<10\end{align*}
3. \begin{align*}6y+9>69\end{align*}
4. \begin{align*}2x-3 \le -4\end{align*}
5. \begin{align*}5x+2 \ge -8\end{align*}
6. \begin{align*}2x-9 \le -5\end{align*}
7. \begin{align*}\frac{x}{3}+1>5\end{align*}
8. \begin{align*}\frac{x}{2}-1<-3\end{align*}
9. \begin{align*}\frac{x}{5}+3>-9\end{align*}
10. \begin{align*}\frac{x}{2}-5>-10\end{align*}
11. \begin{align*}6k-3 > 15\end{align*}
12. \begin{align*}11 - \frac{x}{4} \le 12\end{align*}
13. \begin{align*}12+9j+j<72\end{align*}
14. \begin{align*}12b-3b+5 \ge -31\end{align*}
15. \begin{align*}18+7n+3+6n \le 86\end{align*}
16. \begin{align*}3z-15z-30>54\end{align*}
17. \begin{align*}-2(10+h)<-40\end{align*}
18. \begin{align*}\frac{1}{3}(15+t) \le 26\end{align*}
19. \begin{align*}8(p-3)-2 \ge 14\end{align*}
20. \begin{align*}-3(2+4y)+y>16\end{align*}
Directions: Use inequalities to solve each problem.
Lori went hiking last weekend. On Sunday, she hiked 8 more kilometers than she hiked on Saturday. The total number of kilometers she hiked on both days was less than 40 kilometers.
1. Let \begin{align*}s\end{align*} represent the number of kilometers she hiked on Saturday. Write an inequality to represent this problem.
2. Find three possible values of \begin{align*}s\end{align*}.
Mr. Chang wants to use a rectangular section of his backyard for a vegetable garden. He wants the width of the garden to be exactly 7 feet. He wants the perimeter of the rectangular garden to be no less than 36 feet. The perimeter of any rectangle can be found by using the formula \begin{align*}P=2(l+w)\end{align*}, where \begin{align*}l\end{align*} represents the length and \begin{align*}w\end{align*} represents the width.
1. Write an inequality that could be used to represent \begin{align*}l\end{align*}, the possible lengths, in feet, Mr. Chang could use for his rectangular garden.
2. Would a rectangular garden with a length of 12 feet result in a garden with the perimeter he wants?
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# Base of an Isosceles Triangle
Last updated date: 17th Apr 2024
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## What is an Isosceles Triangle?
Triangles classified as isosceles have at least two sides with equal sides or angles. As three-sided unconstrained polygons, triangles are known to exist. Based on the length of their sides, triangles are called equilateral or isosceles.
No matter which way the triangle's peak (or apex) points, an isosceles triangle always has two equal sides. The following are some tips on isosceles triangles:
• There are equal sides to it. The basis angles are two equal angles that make up the object.
• A right isosceles triangle is one where the third angle is 90 degrees.
As we can see, the given figure $OD = OC$ and $\angle D$ are equal to $\angle C$, so we can say that this given triangle is an isosceles triangle.
Isosceles Triangle
## Properties of an Isosceles Triangle
The properties of the isosceles triangle are given below:
• Two equal sides and two equal angles.
• The two equal sides of an isosceles triangle are called the legs and the angle between them is called the vertex or apex angle.
• The side opposite the vertex angle is called the base and base angles are equal.
• The perpendicular from the apex angle bisects the base and the apex angle.
• The perpendicular drawn from the apex angle divides the isosceles triangle into two congruent triangles and is also known as its line of symmetry.
## Isosceles Triangle Diagram
Three Types of Triangles
In the above figure, we can find the visual difference between the three different types of triangles to compare with an isosceles triangle. The first figure is of an equilateral triangle and second is an isosceles triangle and the last one is Scalene.
Different Triangles
In the above-given figure, we can see many types of triangles. All triangles given above state their features on their own. First three have their properties by sides, and the remaining three by their angles. Each triangle has its features and properties. As we can see in the diagram that equilateral triangle has three equal angles, the isosceles have two equal angles, and the scalene has no equal angles. Similarly, the acute triangle has three angles less than 90, and the right has only one angle equal to 90, and the obtuse has an angle of more than 90.
## The Base of an Isosceles Triangle
A triangle with two equal-length sides is termed an isosceles triangle. The base refers to the triangle's third side. The angle between the legs is known as the vertex angle. The base angles are those angles that have the base as one of their sides.
Q. Given an isosceles triangle measure of the unequal angle is 70° and the other two equal angles measure x; then what is the value of x?
Ans: The given angle is 70.
${\bf{70}}^\circ {\rm{ }} + {\rm{ }}{\bf{x}}{\rm{ }} + {\rm{ }}{\bf{x}}{\rm{ }} = {\rm{ }}{\bf{180}}^\circ$
${\bf{70}}^\circ {\rm{ }} + {\rm{ }}{\bf{2x}}{\rm{ }} = {\rm{ }}{\bf{180}}^\circ$
${\bf{2x}}{\rm{ }} = {\rm{ }}{\bf{180}}{\rm{ }}-{\rm{ }}{\bf{70}}{\rm{ }} = {\rm{ }}{\bf{110}}^\circ$
${\bf{x}}{\rm{ }} = {\rm{ }}\frac{{110}}{2}{\rm{ }} = {\rm{ }}{\bf{55}}^\circ$
Hence, The value of x is 55.
## Area of an Isosceles Triangle
The region that an isosceles triangle occupies in two dimensions is referred to as its area. Typically, the base and height of an isosceles triangle are divided by two to create the triangle. The following equation can be used to determine an isosceles triangle's area:
An isosceles triangle's area is A = ½* b* h square units.
## Examples
Example 1: Calculate the area of an isosceles triangle given $b{\rm{ }} = {\rm{ }}12{\rm{ }}cm$ and $h{\rm{ }} = {\rm{ }}17{\rm{ }}cm$?
Ans:
Given , $b{\rm{ }} = {\rm{ }}12{\rm{ }}cm$
$h{\rm{ }} = {\rm{ }}17{\rm{ }}cm$
Area of Isosceles Triangle $= {\rm{ }}\left( {1/2} \right){\rm{ }} \times {\rm{ }}b{\rm{ }} \times {\rm{ }}h$
$= {\rm{ }}\left( {1/2} \right){\rm{ }} \times {\rm{ }}12{\rm{ }} \times {\rm{ }}17$
$= {\rm{ }}6{\rm{ }} \times {\rm{ }}17$
$= {\rm{ }}102{\rm{ }}c{m^2}$
Example 2: Find the length of the base of an isosceles triangle whose area is$243{\rm{ }}c{m^2}$, and the altitude of the triangle is $27{\rm{ }}cm.$
Ans: Area of the triangle is $243{\rm{ }}c{m^2}$
Height of the triangle is $27{\rm{ }}cm$
The base of the triangle = b =?
Area of Isosceles Triangle is $\left( {1/2} \right){\rm{ }} \times {\rm{ }}b{\rm{ }} \times {\rm{ }}h$
$243{\rm{ }} = {\rm{ }}\left( {1/2} \right){\rm{ }} \times {\rm{ }}b{\rm{ }} \times {\rm{ }}27$
$243{\rm{ }} = {\rm{ }}\left( {b \times 27} \right)/2$
$b{\rm{ }} = {\rm{ }}\left( {243 \times 2} \right)/27$
$b{\rm{ }} = {\rm{ }}18{\rm{ }}cm$
Hence, the base of the triangle is 18 cm.
## Summary
In this article, we have covered various topics regarding isosceles triangles. We understood that the basic properties of the triangle are that both the sides and angles are equal. We have also solved the examples and solved questions to understand them better.
## FAQs on Base of an Isosceles Triangle
1. What is the difference between the isosceles triangle and the isosceles right angle triangle?
A triangle with at least two equal sides is said to be isosceles. Due to the equality of the triangle's two legs, the corresponding angles are also equal. The isosceles right triangle has two equal sides and one angle that is exactly 90 degrees.
2. What do you mean by the area of an isosceles triangle?
The area of an isosceles triangle is nothing but the area occupied by the triangle.
3. What is the angle distribution of the right isosceles triangle?
The right isosceles triangle has one 90-degree and two 45-degree angles.
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## College Algebra (10th Edition)
$f(x)=(x-1)(2x+1)(x+\sqrt{2})(x-\sqrt{2})$ Zeros: $-\displaystyle \sqrt{2},\ \ -\frac{1}{2},\ \ 1,\ \ \sqrt{2},$ all with multiplicity 1.
$f(x)$ has at most $4$ real zeros, as the degree is $4$. Rational zeros: $\displaystyle \frac{p}{q},$ in lowest terms, $p$ must be a factor of $2,\qquad \pm 1,\pm 2,$ and $q$ must be a factor of $2,\qquad \pm 1,\pm 2$ $\displaystyle \frac{p}{q}=\pm 1,\pm\frac{1}{2},\pm 2$ Testing with synthetic division, try $x-1$ $\left.\begin{array}{l} 1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 2 &-1 &-5 &2 &2 \\\hline &2 &1 &-4 & -2 \\\hline 2& 1 &-4 &-2& |\ \ 0 \end{array}$ $f(x)=(x-1)(2x^{3}+x^{2}-4x-2)$ Try synthetic division again, $x+\displaystyle \frac{1}{2}$ $\left.\begin{array}{l} -1/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 2 &1 &-4 &-2 \\\hline &-1 &0 & +2 \\\hline 2 & 0 &-4 & |\ \ 0 \end{array}$ $f(x)=(x-1)(x+\displaystyle \frac{1}{2})(2x^{2}-4)=(x-1)(x+\frac{1}{2})\cdot 2(x^{2}-2)$ $=(x-1)(2x+1)(x^{2}-2)$ Solving $x^{2}-2=0,$ $x=\pm\sqrt{2}$, we have the other two zeros. $f(x)=(x-1)(2x+1)(x+\sqrt{2})(x-\sqrt{2})$ Zeros: $-\displaystyle \sqrt{2},\ \ -\frac{1}{2},\ \ 1,\ \ \sqrt{2},$ all with multiplicity 1.
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Computation Operations For Class 3 | Practice Questions
# Computation Operations
## Computation Operations - Sub Topics
• Computation Operations
• Subtraction
• Multiplication
• Division
• Solved Questions on Computation Operation
• ## Computation Operations
The fundamental operations used to calculate values are computations. The computation operations include addition, subtraction, multiplication and division.
The process of addition involves combining two or more numbers to find their sum. The plus sign (+) is used to represent addition.
For example,
The addition of 529 and 213 is show as:
Thus, 529 + 213 = 742
### Subtraction
Subtraction is the act of calculating the difference between two quantities, or integers. The subtraction symbol is "" often known as the minus sign.
For example,
The subtraction of 529 and 213 is shown as:
Thus, 529 − 213 = 316
### Multiplication
Multiplication is used to find the product of two or more numbers. The symbol used for multiplication is ‘×’.
For example, one car has 4 wheels.
So, 3 cars have 4 + 4 + 4 = 12
= 4 × 3 = 12 wheels.
The multiplication of 45 and 23 is shown as:
Thus, 45 × 23 = 1035
### Division
Division is the process of dividing a number into equal parts and determining how many of those parts there are in total. The symbol used for division is '÷'.
For example, There are 20 balls to be put equally in 4 bags.
The arrangements can be made like this:
This is expressed as 20 ÷ 4 = 5.
Here, 5 balls can be put in each bag
Here, 20 is the Dividend, 4 is the Divisor, 5 is the quotient and 0 is the remainder.
Divide 930 by 15 is shown as:
Thus, 930 ÷ 15 = 62
Example:Which mathematical operator in the blank is used to make the given expression true?
982 _______ 492 = 490
a) +
b) −
c) ÷
d) ×
Explanation: Themathematical operator is used:
982492 = 490
Example: 25 buses are carrying 600 passengers. Each bus contains an equal number of passengers. How many people does a bus carry?
a) 22
b) 24
c) 32
d) 34
Explanation: Number of buses = 25
Total number of passengers = 600
Number of passengers in a bus = 600 ÷ 25 = 24
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# boolean algebra, why is this try to simplify wrong?
so I have been trying to learn simplifying boolean Algebra, lets look at this term : $$(a*b)+(a*\neg b)+(\neg a * \neg b)$$
I have several questions, I know how the right solution looks like but why couldn't I collect $a$ and $\neg a$?
so :
1. $a* \neg a *(b+\neg b+\neg b)$
2. $0*(1)$
is the way I simplifyied it right, is it just non-sense doing it like this or what is I wrong with this?
• Because you have to collect $a$ from the first two terms... Apr 28 '17 at 12:15
• I did collect $a$ and $\neg a$, can you explain it a little bit to me? Apr 28 '17 at 12:17
• It works exactly as with numbers; you can collect $2$ from $(2 \times a) + (2 \times b)$ but you cannot colelct $2$ and $-2$ from $(2 \times a) + (-2 \times b)$ Apr 28 '17 at 12:17
• They are called Laws of boolean algebra; very similar to arithmetical laws. Apr 28 '17 at 12:19
• In the first equation if you set $a = 1 = b$ it evaluates to $1$, but equation $(1)$ evaluates to $0$. Here is a derivation: $(a\cdot b) + (a \cdot \neg b) + (\neg a)\cdot (\neg b) = (a \cdot b) + (\neg b) = (a+\neg b) \cdot (b + \neg b) = a + \neg b$
– Mike
Apr 28 '17 at 12:28
For the derivation, notice that there are 4 possible states of $a$ and $b$:
$$a = 0, b = 0 \\ a=0, b=1 \\ a=1,b=0 \\ a=1, b=1$$
Now, the expression $(a*b)$ is true for the fourth state, the expression $(a*\neg b)$ is true for the third state and the expression $(\neg a * \neg b)$ is true for the first state. So your expression is equivalent to: $$(a*b)+(a*\neg b)+(\neg a * \neg b) \equiv \neg(\neg a * b)$$ ("Just not the second option!"). By De-Morgan's rule: $$\neg(\neg a * b) \equiv a + \neg b$$
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