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Solving Quadratic Equations by Completing the Square Worksheet Answers can be a fun and easy way to learn about different methods for solving quadratic equations. I learned about this method as a way to learn about different method for solving quadratic equations in an introductory class. It was designed for students who are new to solving equations.
One of the most confusing aspects of solving quadratic equations is when the method becomes difficult because the student has to interpret what the student already knows into some other form. This method was designed to help students get past this problem.
In order to solve a quadratic equation by completing the square, one first needs to identify a right triangle. The right triangle can be derived by creating a 90-degree right triangle by joining the left side of the two-sided triangle with the side of the triangle opposite it. This creates a quadratic that has both a right triangle at the beginning and a right triangle at the end. The square of a quadratic has two sides and a hypotenuse, which are the sides that do not have a corresponding side.
Now that the right triangle is known, we can create a quadratic equation by completing the square. There are four factors involved when solving quadratic equations by completing the square, and we can use them to find the x and y coefficients. Using these results, we can either use the relationship between the quadratic and its coefficients, or we can use the square root of the term to find the term of the quadratic.
Solving quadratic equations by completing the square will usually involve a very close shave. There are many opportunities for the student to make mistakes, and many times it will take more than one attempt to get it right. I felt that the time my professor spent on solving quadratic equations by completing the square seemed like overkill to me, so I tried out the same method myself.
In my first attempt at solving a quadratic equation by completing the square, I found that the method was very easy to do, but that I wasted a lot of time trying to figure out how to complete the quadratic equation. My second attempt involved much more time, but I did find that the method was not too hard to figure out on my own.
This method of solving a quadratic equation by completing the square worksheet answers is very much like the method I learned in school. I also learned about how to calculate the x and y coefficients using the method and how to find the equation of a quadratic using this method.
Using this method, I was able to solve a quadratic equation by completing the square in only 5 minutes! My math teacher told me that I should try to solve quadratic equations by completing the square as much as possible. I am glad he did because now I am using this method to solve my own quadratic equations. |
Logarithms give us another method to write exponents. We can convert back and forth between exponential form and logarithmic form. When we want the exponential form, we are solving for the power. In the case of logarithmic form, we are solving for the exponent. In order to solve a logarithmic equation, we convert the equation into exponential form, then solve.
Test Objectives
• Demonstrate the ability to convert from exponential form to logarithmic form
• Demonstrate the ability to convert from logarithmic form to exponential form
• Demonstrate the ability to solve logarithmic equations
Logarithmic Functions Practice Test:
#1:
Instructions: Write each in logarithmic form.
a) $$2^6=64$$
b) $$81^{\frac{1}{2}}=9$$
c) $$\sqrt[3]{125}=5$$
d) $$10^{-3}=\frac{1}{1000}$$
#2:
Instructions: Write each in exponential form.
a) $$\log_{10}(10,000)=4$$
b) $$\log_{11}\left(\frac{1}{121}\right)=-2$$
c) $$\log_{216}(6)=\frac{1}{3}$$
d) $$\log_{4}(256)=4$$
e) $$\log_{1327}(1)=0$$
#3:
Instructions: Solve each equation.
a) $$\log_{125}(25)=x$$
#4:
Instructions: Solve each equation.
a) $$\log_{x}(625)=4$$
#5:
Instructions: Solve each equation.
a) $$\log_{x}(1024)=\frac{5}{2}$$
Written Solutions:
#1:
Solutions:
a) $$\log_{2}(64)=6$$
b) $$\log_{81}(9)=\frac{1}{2}$$
c) $$\log_{125}(5)=\frac{1}{3}$$
d) $$\log_{10}\left(\frac{1}{1000}\right)=-3$$
#2:
Solutions:
a) $$10^4=10,000$$
b) $$11^{-2}=\frac{1}{121}$$
c) $$216^{\frac{1}{3}}=6$$
d) $$4^4=256$$
e) $$1327^0=1$$
#3:
Solutions:
a) $$x=\frac{2}{3}$$
#4:
Solutions:
a) $$x=5$$
#5:
Solutions:
a) $$x=16$$ |
Home >> CBSE XII >> Math >> Matrices
# A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: $$\text{(a) Rs 1800} \qquad \qquad \text{(b) Rs 2000}$$
Toolbox:
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
Let Rs30,000 be divided into two parts ,Assume one bond as x,then second one is Rs(30,000-x).Let us represent this by $1\times 2$ matrix.[x(30000-x)]
Given Rate of interest as 0.05&0.07 per rupee.Let us denote with matrix R of order $2\times 1$
$R=\begin{bmatrix}0.05\\0.07\end{bmatrix}$
$(i)\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}0.05\\0.07\end{bmatrix}$
$\Rightarrow \begin{bmatrix}x\times (0.05)+(30000-x)\times 0.07\end{bmatrix}=1800$
$\frac{5x}{100}+\frac{7(30000-x)}{100}=1800$
Multiplying by 100
5x+210000-7x=180000
2x=210000-180000=30000
x=15000
$\Rightarrow 30-x=30000-15000=15000.$
Thus the two part are 15000 each.
(ii)Given total interest 2000
$\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}0.05\\0.07\end{bmatrix}=2000$
$\frac{5x}{100}+\frac{7(30000-x)}{100}=2000$
Multiplying by 100 on both sides
5x+210000-7x=200000
2x=210000-200000
x=5000
30000-x=30000-5000=25000
Thus the two parts are 25000&5000. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Heights of Cylinders Given Volume
## h = V/πr^2
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Heights of Cylinders Given Volume
License: CC BY-NC 3.0
Gregory's family just bought a hot tub for their lake house. The hot tub company told his family that the tub holds 125 cubic feet of water. Gregory is interested to know how deep the hot tub is. He measures the diameter of the top and finds that the hot tub is 6 feet across. What is the height, or depth, of the hot tub?
In this concept, you will learn how to calculate the height of a cylinder when given the volume and radius or diameter.
### Finding the Height of a Cylinder Given Volume
Sometimes you will know the volume and radius of a cylinder and you won’t know the height of it. Think about a water tower that is cylindrical in shape. You might know how much volume the tank will hold and the radius of the tank, but not the height of it. When this happens, you can use the formula for the volume of a cylinder to find the missing height:
\begin{align*}V & = \pi r^2h\\ V & = \pi (2)^2 (10)\\ V & = \pi (4) (10)\\ V & = 40 \pi\\ V & = 125.6 \ in.^3\end{align*}
License: CC BY-NC 3.0
Let's look at an example.
A cylinder with a radius of 2 inches has a volume of 125.6 cubic inches. What is the height of the cylinder?
The volume and the radius are given, so substitute these into the formula and then solve for \begin{align*}h\end{align*}, the height.
\begin{align*}V& = \pi{r}^{2}h\\ 125.6&= (3.14)({2}^{2})h\\ 125.6&= (3.14)(4)h\\ 125.6&= 12.56h\\ 125.6 \div\ 12.56&=12.56h \div\ 12.56\\ 10\ in&= h\end{align*}
The height of the cylinder is 10 inches.
Check your work by substituting the answer in for the height. You should get a volume of 125.6 cubic inches.
\begin{align*}V & = \pi r^2h\\ V & = \pi (2)^2 (10)\\ V & = \pi (4) (10)\\ V & = 40 \pi\\ V & = 125.6 \ in.^3\end{align*}
What is the height of a cylinder that has a radius of 6 cm and a volume of 904.32 cubic cm?
Again, you have been given the volume and the radius. Put this information into the formula along with the value of pi and solve for \begin{align*}h\end{align*}, the height.
\begin{align*}V & = \pi r^2h\\ 904.32 & = (3.14) ({6}^{2}) h\\ 904.32 & = (3.14)(36) h\\ 904.32 & = 113.04 \ h\\ 904.32 \div 113.04 & = 113.04h \div\ 113.04\\ 8 \ cm & = h\end{align*}
The height of this cylinder is 8 centimeters.
### Examples
#### Example 1
Earlier, you were given a problem about Gregory and his family's hot tub.
To figure this out, use the formula for volume of a cylinder. He already knows the volume of the tub is 125 cubic feet and the diameter is 6 feet.
First, divide the diameter by 2 and plug the values for volume, pi, and radius into the formula for volume of a cylinder.\begin{align*}r&\ = 6 \div\ 2\\ r&\ = 3\end{align*} \begin{align*}V & = \pi r^2h\\ 125 & = (3.14) ({3}^{2}) h\\ \end{align*}
Next, square the radius and multiply the values together.\begin{align*}125& = (3.14) ({3}^{2}) h\\ 125 & = (3.14)(9) h\\\end{align*}
\begin{align*}125 &= 28.26h\end{align*}
Last, divide both sides by 200.96 for the answer, remembering to include the appropriate unit of measurement.
\begin{align*}125&= 28.26h\\ 125 \div\ 28.26 &= 28.26h \div\ 28.26\\ 4.42\ ft& = h \end{align*}
The answer is Gregory's hot tub is 4.42 feet deep.
#### Example 2
Javier wants to construct a cylindrical container to hold enough water for his pet fish. He read that the fish needs to live in 2,110.08 cubic inches of water. If he constructs a tank with a diameter of 16 inches, how tall must he make it so that it holds the right amount of water?
First, divide the diameter by 2 and plug the values for volume, pi, and radius into the formula for volume of a cylinder.
\begin{align*}r&\ = 16 \div\ 2\\ r&\ = 8\end{align*}
\begin{align*}V & = \pi r^2h\\ 2,110.08 & = (3.14) ({8}^{2}) h\\ \end{align*}
Next, square the radius and multiply the values together.
\begin{align*}2,110.08 & = (3.14) ({8}^{2}) h\\ 2,110.08 & = (3.14)(64) h\\\end{align*}
\begin{align*}2,110.08&= 200.96h\end{align*}
Then, divide both sides by 200.96 for the answer, remembering to include the appropriate unit of measurement.
\begin{align*}2,110.08&= 200.96h\\ 2,110.08 \div\ 200.96 &= 200.96h \div\ 200.96\\ 10.5\ in& = h \end{align*}
The answer is Javier must make his tank 10.5 inches tall for his tank to hold 2,110.08 cubic inches of water.
#### Example 3
Find the height of a cylinder with radius = 6 inches and volume = 904.32 cubic inches.
First, plug the values of the volume, pi, and radius into the formula for volume of a cylinder.
\begin{align*}V&= \pi\ {r}^{2}h\\ 904.32&= (3.14)({6}^{2})h\\\end{align*} Next, square the radius and multiply the values together.
\begin{align*}904.32&= (3.14)({6}^{2})h\\ 904.32&= (3.14)(36)h\end{align*}
\begin{align*}904.32&= 113.04h\end{align*}
Last, divide each side by 113.04 for the answer, remembering to include the appropriate unit of measurement.
\begin{align*}904.32&= 113.04h\\ 904.32 \div\ 113.04&= 113.04h \div\ 113.04\\ 8\ in &=h \end{align*}
The answer is the height of the cylinder is 8 inches.
#### Example 4
Find the height of a cylinder with radius = 3 meters and volume = 354.34 cubic meters.
First, plug the values of the volume, pi, and radius into the formula for volume of a cylinder.
\begin{align*}V&= \pi\ {r}^{2}h\\ 354.34&= (3.14)({3}^{2})h\\\end{align*} Next, square the radius and multiply the values together.
\begin{align*}354.34&= (3.14)({3}^{2})h\\ 904.32&= (3.14)(9)h\end{align*}
\begin{align*}354.34&= 28.26h\end{align*}
Last, divide each side by 28.26 for the answer, remembering to include the appropriate unit of measurement.
\begin{align*}354.34&= 28.26h\\ 354.34 \div\ 28.26&= 28.26h \div\ 28.26\\ 9\ m &=h \end{align*}
The answer is the height of the cylinder is 9 meters.
#### Example 5
Find the height of a cylinder with radius = 5 feet and volume = 785 cubic feet.
First, plug the values of the volume, pi, and radius into the formula for volume of a cylinder.
\begin{align*}V&= \pi\ {r}^{2}h\\ 785&= (3.14)({5}^{2})h\\\end{align*}
Next, square the radius and multiply the values together.\begin{align*}785&= (3.14)({5}^{2})h\\ 785&= (3.14)(25)h\end{align*}
\begin{align*}785&= 78.5h\end{align*}
Last, divide each side by 78.5 for the answer, remembering to include the appropriate unit of measurement.
\begin{align*}785&= 78.5h\\ 785 \div\ 78.5&= 78.5 \div\ 78.5\\ 10\ ft &=h \end{align*}
The answer is the height of the cylinder is 10 feet.
### Review
Given the volume and the radius, find the height of each cylinder.
1. \begin{align*}r = 6 \ in, \ V = 904.32 \ in^3\end{align*}
2. \begin{align*}r = 5 \ in, \ V = 706.5 \ in^3\end{align*}
3. \begin{align*}r = 7 \ ft, \ V = 2307.9 \ ft^3\end{align*}
4. \begin{align*}r = 8 \ ft, \ V = 4019.2 \ ft^3\end{align*}
5. \begin{align*}r = 7 \ ft, \ V = 1538.6 \ ft^3\end{align*}
6. \begin{align*}r = 12 \ m, \ V = 6330.24 \ m^3\end{align*}
7. \begin{align*}r = 9 \ m, \ V = 4069.49 \ m^3\end{align*}
8. \begin{align*}r = 10 \ m, \ V = 5652 \ m^3\end{align*}
9. \begin{align*}r = 12 \ in, \ V = 11304 \ in^3\end{align*}
10. \begin{align*}r = 11 \ ft, \ V = 3039.52 \ ft^3\end{align*}
11. \begin{align*}r = 10 \ in, \ V = 1570 \ in^3\end{align*}
12. \begin{align*}r = 9.5 \ in, \ V = 1700.31 \ in^3\end{align*}
13. \begin{align*}r = 8 \ m, \ V = 1808.64 \ m^3\end{align*}
14. \begin{align*}r = 14 \ ft, \ V = 5538.96 \ ft^3\end{align*}
15. \begin{align*}r = 13.5 \ in, \ V = 4005.85 \ in^3\end{align*}
To see the Review answers, open this PDF file and look for section 10.13.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English
Cubic Units
Cubic units are three-dimensional units of measure, as in the volume of a solid figure.
Volume
Volume is the amount of space inside the bounds of a three-dimensional object.
1. [1]^ License: CC BY-NC 3.0
2. [2]^ License: CC BY-NC 3.0
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Fraction Worksheets for Year 3 (age 7-8)
Finding tenths of numbers, equivalence, adding and subtracting fractions.
Children coming into Year 3 should have a sound understanding of simple fractions, especially halves, quarters and thirds. Â They should have had plenty of practical experience dividing shapes and sets of objects into quarters and thirds.
There are plenty of new concepts on fractions to be understood in Year 3,but it is still important to use practical work, using shapes and objects, to help with understanding. Â A key step is to begin working with tenths and recognising that tenths arise from dividing an object into ten equal parts. Later this can be developed to dividing one-digit numbers or quantities by 10. This is crucial for understanding our decimal system and understanding that a fraction such as three tenths can also be represented as a decimal (0.3) and connects to dividing 3 by 10. Counting in tenths can also help enormously with this. We have a great selection of worksheets which deal with writing tenths, counting in tenths and converting tenths to decimals.
Other fractions are also introduced, such as fifths, and equivalence between fractions is developed with the help of pictures and diagrams. This leads on to ordering fractions with the same denominator. One of the quirks of fractions, which some children find hard to understand, is that as the denominator gets larger the size of the number gets smaller (e.g. one tenth is smaller than one fifth; one twentieth is smaller than one tenth and so on). We have some excellent ordering fractions pages for Year 3 to help consolidate this concept. As well as these, some of our most popular worksheets are the sets of finding fractions of numbers, with questions such as ‘What is one sixth of 30?’ Again, the link between fractions and division needs to be constantly reinforced.
As the year progresses children will be introduced to adding and subtracting fractions, but only with the same denominator and with totals up to one whole one. Simplifying fractions is important at this stage and children should be encouraged to write fractions in their simplest form. (e.g. two sixths can be simplified to one third).
All this will lead onto much more on decimals and fractions in Year 4, including hundredths and further work on equivalence. If you child is finding the Year 3 work difficult then it is important to go back to Year 2, or even Year 1, and see what is understood and what concepts have not been fully understood. |
## Order of Operations and Integers
In calculations involving several operations, the word BEDMAS shows the order in which each operation should be carried out.
### BEDMAS
BEDMAS means:
Brackets: Work out the value of any expressions inside brackets. Exponents: Work out the value of any numbers or expressions with exponents or indices. Sometimes shown as 'of ' meaning multiply. Division Multiplication If these two operations occur together, work from left to right. Addition Subtraction If these two operations occur together, work from left to right.
Examples Answers Calculate: a. 3 + 4 × 5 a. 3 + 4 × 5 = 3 + 20 = 23 b. 4 + (6 − 2) b. 4 + (6 − 2) = 4 + 4 = 8 c. 13 − 2 × 5 + 4 c. 13 − 2 × 5 + 4 = 13 − 10 + 4 = 7 d. 3 × 6 2 d. 3 × 6 2 = 3 × 36 = 108
Care should be taken when using a calculator to do calculations of this type.
### Integers
Integers are positive and negative whole numbers and include zero.
Use a number line when adding and subtracting integers.
The first integer is the starting point on the number line.
• When adding a positive integer, move to the right.
e.g. 4 + 2 = 6
• When adding a negative integer, move to the left.
e.g. 4 + − 2 = 2
Subtraction
• When subtracting a positive integer, move to the left.
e.g. 4 − 2 = 2
• When subtracting a negative integer, move to the right
e.g. 4 − − 2 = 6
Multiplication and division
The rule is the same for both multiplying and dividing two integers.
• If the signs of the two integers are the same, the answer is positive.
• If the signs of the two integers are different, the answer is negative.
The table illustrates these rules:
×/÷ +ve -ve +ve + − -ve − +
Examples:
For multiplication For division 4 × 2 = 8 4 ÷ 2 = 2 - 4 × − 2 = 8 - 4 ÷ − 2 = 2 4 × − 2 = − 8 4 ÷ − 2 = − 2 - 4 × 2 = − 8 - 4 ÷ 2 = − 2
For practice with integer calculations − |
# Evaluating Expressions with Exponents and Roots
Exponents and Roots
In this video, I’m going to talk about exponents and roots. An exponent is a number that tells you how many times to multiply the base number by itself. For instance, if we have 2 raised to the second power, $$2^2$$, what this is telling us is that we need to multiply 2 times 2.
Now, if we had 2 raised to the third power,$$2^3$$, this would be the same as 2 times 2 times 2. The exponent is saying how many times to multiply this number by itself. A root is kind of the opposite of this. Let’s take, for instance, 16.
The radical symbol tells us that we’re taking a root.
If there is no number in the index spot, the number is assumed to be 2. Without any numbers there, the radical sign designates a square root, or a root to the power of 2. The square root of 16 is a number that has to be multiplied by itself twice in order to equal 16.
The square root of 16, if it is squared or raised to the second power, it equals 16. The square root of 16 is 4. 4 raised to the second power is equal to 16.
Roots and exponents are kind of inverse operations, but you can actually express roots as exponent.
For instance, if we wanted to go back to the square root of 16, we can write the square root of 16 as 16 to the $$\frac{1}{2}$$: $$16^{\frac{1}{2}}$$
Essentially, if you have 1 over a number as an exponent, you’re taking that number’s root of the number in question. Like I was saying, the 2 that’s omitted for a square root, that’s the same as $$\frac{1}{2}$$ (the power of $$\frac{1}{2}$$).
You can also have other fractions or other powers of roots. For instance, you might take 8 to the $$\frac{1}{3}$$, or third root of 8. This would be a number that you have to multiply by itself three times in order to achieve this. The third power 8 raised to the third power is equal to 8.
This number would be the same as 2, because 2 times 2 times 2, or 2 to the third power, is equal to 8.
A couple other things about exponents. Any number raised to the 1 is just the number itself. 2 raised to the 1 is just 2. Any real number raised to the 0 is equal to 1. It doesn’t matter if it’s a positive number or a negative number. Any real non-zero number raised to the power of 0 is equal to 1.
You may notice that all of the exponents and roots we’ve looked at so far here are whole numbers or fractions of whole numbers.
You can, in fact, have roots that are, for instance, say this is 2 raised to the 2.3. You can have this sort of quantity, but it’s not going to be easy to calculate by hand, and really you’re going to use a calculator if you encounter anything that has a decimal in the exponent.
That’s a good overview of exponents and roots.
## Exponent of a Square Root Review PDF
114162
by Mometrix Test Preparation | Last Updated: August 30, 2024 |
## How do you tell if a graph has a positive discriminant?
A positive discriminant indicates that the quadratic has two distinct real number solutions. A discriminant of zero indicates that the quadratic has a repeated real number solution. A negative discriminant indicates that neither of the solutions are real numbers.
## How can you use the discriminant when you are graphing a quadratic function?
You can use the discriminant, b2 4ac, to find the number of solutions of a quadratic equation in the standard form ax2 bx c 0. A positive value indicates two solutions, zero indicates one solution, and a negative value indicates no real solution.
## Which graph shows a quadratic function with a discriminant value of 0?
Which graph shows a quadratic function with a discriminant value of 0? On a coordinate plane, a parabola opens up.
## Is the discriminant negative or positive?
A quadratic expression which always takes positive values is called positive definite, while one which always takes negative values is called negative definite. Quadratics of either type never take the value 0, and so their discriminant is negative.
## What can the discriminant tell you?
The discriminant tells you how many solutions there are to quadratic equation or how many x intercepts there are for a parabola. If the discriminant is less than zero, there are no solutions and if the discriminant is equal to zero, there is one solution.
## How many solutions does a quadratic equation have if it’s discriminant is zero?
The discriminant is the part under the square root in the quadratic formula, b²-4ac. If it is more than 0, the equation has two real solutions. If it’s less than 0, there are no solutions. If it’s equal to 0, there is one solution.
## How do you use the discriminant?
NotesThe quadratic equation we’re working with is x2-2x+3=0. The discriminant comes from the Quadratic Formula. Remember, the discriminant tells us how many and what type of solutions a quadratic equation has. Remember, the discriminant is equal to b2-4ac. The standard form of a quadratic equation is ax2+bx+c=0.
You might be interested: Conditional probability equation
## What is a repeated real number solution?
When the left side factors into two linear equations with the same solution, the quadratic equation is said to have a repeated solution. We also call this solution a root of multiplicity 2, or a double root. SOLVING BY THE SQUARE.
## How do you find the discriminant on a calculator?
ax2 + bx + c = 0 The discriminant calculator is an online calculator tool, which calculates the discriminant of a given quadratic equation. For a quadratic equation ax2 + bx + c = 0, where a ≠ 0, the formula of discriminant is b2 – 4ac. ◾ Number of roots – whether the quadratic equation has two roots, one root or none.
So, given a quadratic function, y = ax2 + bx + c, when “a” is positive, the parabola opens upward and the vertex is the minimum value. On the other hand, if “a” is negative, the graph opens downward and the vertex is the maximum value.
## How do I find the discriminant of a parabola?
Quadratic EquationsThe number D = b2 – 4ac is called “discriminant”. If D < 0, then the quadratic equation has no real solutions(it has 2 complex solutions). The vertex of the parabola is x = − b 2 a displaystyle x = -frac{b}{2a} x=−2ab.Problem 3. Solve the equation: Problem 4. Solve the equation:
## When solving a quadratic equation What is the difference between a root and a solution?
Zeros (roots) of a function are the values of x for which f(x)=0 while solutions are the values of the x which make the equation f(x) = c true where c is any constant . for example if f(a) = 0 then a is called root and (a,0) called solution of f(x) while if f(b) = c then (b,c) is called solution.
## What happens if the discriminant is less than 0?
If the discriminant of a quadratic function is less than zero, that function has no real roots, and the parabola it represents does not intersect the x-axis.
## How many solutions if the discriminant is negative?
If the value of the discriminant is zero, the quadratic equation has one real solution. If the value of the discriminant is negative, the quadratic equation has no real solutions.
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#### Bernoulli’s equation example
What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […] |
# Distance & Displacement
This post will take a look at distance and displacement—two core ideas in classical physics.
To understand concepts such as distance and displacement, you first need to understand position. Position is the location of an object. When we are speaking of distance and displacement, this involves an object changing position. However, an object can only change position if it is relative to something else. For example, if a person walks 5 meters from their bedroom to the kitchen, this distance can only exist because there are two positions
• The bedroom
• The kitchen
The position of the person is defined by what we call a reference frame. The reference frame is a stationary object from which position is determined. In the example above, there are three positions
• The bedroom
• The kitchen
• The walking person
The bedroom is the reference frame. In other words, wherever the walking person goes, their position is measured in relation to the bedroom. The kitchen is simply the goal destination or where the walking person is going.
Distance
Distance is a measure of the length of the path between a person’s initial and final position. In our walking example, the person traveled 5 meters from the bedroom to the kitchen. This 5 meters is the distance.
bedroom walk 5 meters to the kitchen
Displacement is a little more complicated. Displacement is the net change in position. In other words, displacement compares your final position to your initial position and determines if there is a difference in these values. For example, if we use the walking example again, the displacement is the same as the distance. This is because the person walked 5 meters and did not move anymore. Doing some simple math, we can calculate the displacement as shown below
Final position – original position = displacement
5 – 0 = 5
The person traveled 5 meters to the final position of the kitchen. The original position is 0 meters because the bedroom is where the person started at and is the reference frame.
The value of the distance and the displacement are not always the same. For example, if the walking person goes to the kitchen and then returns to the bedroom, we get the following numbers
bedroom walk 5 meters to the kitchen
Kitchen walk 5 meters to the bedroom
The distance travel here is 10 meters as we went 5 meters to the kitchen and 5 meters back to the bedroom. However, the displacement is zero because our final position and our initial position are the same in that we left the bedroom and came back to the bedroom. There was a change in the distance but not in the person’s position when the walking was over.
Scalar and Vector
Distance is a measure of magnitude (amount of) length. In this situation, only the magnitude is being measured, so this is a scalar quantity. In the example above, the person walked 5 meters. We know the direction, but this is not needed to understand that the person walked 5 meters
Displacement is a measure of magnitude and also direction. When magnitude and direction are considered, it is called a vector quantity. In our example above, the person walked to the kitchen and then walked backed to the bedroom. Walking to the kitchen could be considered a positive distance while walking back to the bedroom would be regarded as a negative distance. This is why the displacement is zero in the second example. Walking back and forth essentially cancels the values out.
This examination of motion is called kinematics, which is the study of motion without being concerned with what causes this motion.
Conclusion
Distance and displacement are foundational concepts in physics on which many other complex ideas are built upon. Therefore, understanding how things move in relation to space is critical to appreciate for students studying this subject. |
# What does volume of a circle mean?
## What does volume of a circle mean?
Like all geometric prisms, a cylinder’s volume is the product of its cross-sectional area and its length. The cross-sectional area is the area of its circle, and you can calculate it from the circle’s radius. Multiply the circle’s radius by 2 to calculate its diameter.
## How do you determine the volume of a circle?
For example, to calculate the volume of a cone with a radius of 5cm and a height of 10cm: The area within a circle = πr2 (where π (pi) is approximately 3.14 and r is the radius of the circle). In this example, area of base (circle) = πr2 = 3.14 × 5 × 5 = 78.5cm2.
Does a circle have a volume?
No, a circle doesn’t have a volume. A circle is a two-dimensional shape, it does not have volume. A circle only has an area and perimeter/circumference.
### What is the definition of volume in math?
In math, volume can be defined as the 3-dimensional space enclosed by a boundary or occupied by an object. Here, for example, the volume of the cuboid or rectangular prism, with unit cubes has been determined in cubic units.
### What is volume of a shape?
Volume is the amount of space a 3D shape takes up. You can work out the volume of a shape by multiplying height × width × depth. If the shape is made of cubic cm blocks, you can count the cubes to find the shape’s volume.
Where is the radius of a circle?
The radius of a circle is the length of the line segment from the center of a circle to a point on the circumference of the circle. It is generally abbreviated as ‘r’.
#### What is an example of volume?
Volume is the measure of the capacity that an object holds. For example, if a cup can hold 100 ml of water up to the brim, its volume is said to be 100 ml. Volume can also be defined as the amount of space occupied by a 3-dimensional object. |
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## AP®︎/College Calculus AB
### Course: AP®︎/College Calculus AB>Unit 1
Lesson 16: Working with the intermediate value theorem
# Intermediate value theorem review
Review the intermediate value theorem and use it to solve problems.
## What is the intermediate value theorem?
The intermediate value theorem describes a key property of continuous functions: for any function $f$ that's continuous over the interval $\left[a,b\right]$, the function will take any value between $f\left(a\right)$ and $f\left(b\right)$ over the interval.
More formally, it means that for any value $L$ between $f\left(a\right)$ and $f\left(b\right)$, there's a value $c$ in $\left[a,b\right]$ for which $f\left(c\right)=L$.
This theorem makes a lot of sense when considering the fact that the graphs of continuous functions are drawn without lifting the pencil. If we know the graph passes through $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right)$...
... then it must pass through any $y$-value between $f\left(a\right)$ and $f\left(b\right)$.
## What problems can I solve with the intermediate value theorem?
Consider the continuous function $f$ with the following table of values. Let's find out where must there be a solution to the equation $f\left(x\right)=2$.
$x$$-2$$-1$$0$$1$
$f\left(x\right)$$4$$3$$-1$$1$
Note that $f\left(-1\right)=3$ and $f\left(0\right)=-1$. The function must take any value between $-1$ and $3$ over the interval $\left[-1,0\right]$.
$2$ is between $-1$ and $3$, so there must be a value $c$ in $\left[-1,0\right]$ for which $f\left(c\right)=2$.
Problem 1
$f$ is a continuous function.
$f\left(-2\right)=3$ and $f\left(1\right)=6$.
Which of the following is guaranteed by the Intermediate Value Theorem?
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• If an equation of a cube root function is given and you are asked to find an interval that has least one solution, how would you go about that. I understand the Intermediate Value Theorem, but I'm not sure how to find the the interval given an equation.
• You find an x-coordinate where you know the function is negative and another where you know the function is positive. Then the interval with those endpoints must contain a solution, by Intermediate Value Theorem
• What if you're not given an interval to test within?
For example: Use the Intermediate Value Theorem to show that the equation x^3 + x + 1 = 0 has at least 1 solution.
• Then find an interval. By just picking x-values, we can see that your polynomial is positive at x=1 and negative at x= -1. So it must have a solution in (-1, 1). If you want a smaller interval than that, you can check the value at the halfway point. This will either force your zero into one half of (-1, 1) or the other, or it will actually find your zero. Once you have your new, smaller interval, you can continue to shrink it as much as you like.
• How does the intermediate value theorem work?
• The theorem basically says "If I pick an X value that is included on a continuous function, I will get a Y value, within a certain range, to go with it." We know this will work because a continuous function has a predictable Y value for every X value. By "predictable" I mean that the limits for a point from the right and left side are the same as the point's value in the function.
A simple way to imagine this is to pretend the continuous function occupies a box. We don't necessarily know what the function looks like, but we know where it can and can't go. Let's say our interval is [0, 10] and the end points are (0, 1) and (10, 5). In that case we know the function can't go any further left than 0, or any further right than 10. On the y-axis, our function could go lower than 1 or higher than 5 - we don't know. But the important thing is, it definitely has to go between one and five, otherwise it wouldn't connect our end points!
In answering a multiple-choice question about values given by the Intermediate Value Theorem, imagine this box on your graph. If the any option makes assumptions about what happens outside that box, don't select it, because there's no way for you to know those things from the information you've been given.
• what if you're asked to use IVT to show that x^1/2 + (x+1)^1/2 =4 has a root
• We have f(x) = x^(1/2) + (x+1)^(1/2) - 4
We need to show some closed interval [a, b] where f(x) is continuous.
We also need to show that 0 is between f(a) and f(b)
First lets establish a closed interval where the function is continuous.
f(x) is continuous for x >= 0 since the function is made by adding multiple square root functions which are also continuous for x>= 0.
Second, lets find a, and b by experimenting with different x-values.
f(0) = 0^(1/2) + (0+1)^(1/2) - 4
f(0) = 0 + 1 - 4
f(0) = -3
f(5) = 5^(1/2) + (5+1)^(1/2) - 4
f(5) is approximately 0.6856
Notice that 0 is between f(0) and f(5).
Finally, lets summarize the information we found.
We know that f(x) is continuous for x >= 0.
This means that f(x) is also continuous across [0, 5].
We found that f(0) = -3 and f(5) is approximately 0.6856
We now know that 0 is between f(0) and f(5).
Thus, f(x) has a root somewhere in the interval 0 <= x <= 5.
You could make a similar argument with other intervals that you find.
• One question asks "Why doesn't the IVT apply to g(x) on the interval [0, 4]. g(x)=2x^2-8x-5.
It's continuous, so why doesn't it apply?
• I think the point of the question is to notice that g(0)=g(4), so there are no values between them.
• what the IVT,IRCand ARC stands for
(1 vote)
• IVT= Intermediate Value Theorem
IRC= Instantaneous Rate of Change
ARC= Average Rate of Change
• How to show the existence of a real number whose square is two using intermediate value theorem?
(1 vote)
• The first question to ask is: What is the function? i.e. what is f(x)
The second question to ask is: What is a value for x that will make the function output a value less than 2? i.e. find a value for a so that f(a) < 2
The third question to ask is: What is a value for x that will make the function output a value greater than 2? i.e. find a value for b so that f(b) > 2
You then apply the intermediate value theorem to find that there must be a number between a and b that must give you 2 when squared.
Does that help?
• How would you solve this with a step by step answer:
Suppose that f is a continuous function on the interval [0,1] such that 0 smaller than or equal to f(x) is greater than or equal to 1 for each x in [0,1]. Show that there is a number c in [0,1] such f(c)=c
• I assume you mean 0 smaller than or equal to f(x) is smaller than or equal to 1 for each x in [0,1].
Define the function g(x) = f(x) - x. Because x is a continuous function, f(x) is a continuous function, and the difference of two continuous functions is continuous, g(x) is continuous.
Note that g(0) = f(0) and g(1) = f(1) - 1.
Since 0 <= f(x) <= 1 for each x in [0, 1], g(0) >= 0 and g(1) <= 0. Therefore, because g(x) is continuous, it follows from the intermediate value theorem that g(c) = 0 for some number c in [0,1]. Because g(c) = f(c) - c, f(c) = c for this number c. |
Concept: Quadratic and Higher Degree Equations
INTRODUCTION
A polynomial in single variable can be written as: anxn + an-1an-1 + an-2xn-2 + … + a1x + a0
A second-degree polynomial is called a quadratic polynomial. An equation of the form ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0, is called a quadratic equation in variable x. The values of x for which the equation holds true are called the roots or zeros of the equation. Since a quadratic equation is of degree 2, it can have at most two roots.
Here are some examples of quadratic equations.
4x2 = 0
4x2 + 5x = 0
x2 – 5x + 6 = 0
Solving a quadratic equation is finding all its roots. This can be done in the following ways.
Splitting the middle term / Factorisation
Consider the quadratic equation: ax2 + bx + c = 0
In this method we need to think of two numbers such that the sum of these two numbers is equal to ‘b’ (i.e., the coefficient of x) and their product is equal to ‘a × c’ (i.e., the product of coefficient of x2 and the constant term).
Consider the quadratic equation 2x2 + 7x + 6 = 0.
Step 1: Here we need to think of two numbers ‘p’ and ‘q’ such that
p + q = 7 and
p × q = 2 × 6 = 12
The two such number are 3 and 4.
Step 2: Now, split the coefficient of x as (p + q).
⇒ 2x2 + (3 + 4)x + 6 = 0
⇒ 2x2 + 3x + 4x + 6 = 0
Step 3: Now we have four terms. We combine the first two terms and the last two terms.
⇒ x(2x + 3) + 2(2x + 3) = 0
Step 4: Now we have two terms. We again take the common term out.
⇒ (2x + 3)(x + 2) = 0
Step 5: (2x + 3)(x + 2) = 0
⇒ Either (2x + 3) = 0 or (x + 2) = 0
⇒ Either x = -3/2 or x = -2
∴ The two solutions/roots/zeros of quadratic equation 2x2 + 7x + 6 = 0 are -3/2 and -2
i.e., x = -3/2 or x = -2 will satisfy the given equation.
Note: This method may be useful when it is possible to find value of ‘p’ and ‘q’. Otherwise, Formula method should be used to solve quadratic equations.
Remember:
Step 5 can be re-written as
(2x + 3)(x + 2) = 0
$\left(x-\left(\frac{-3}{2}\right)\right)$$\left(x-\left(-2\right)\right)$ = 0
Here, -3/2 and -2 are the roots of the equation
∴ A quadratic equation whose roots are α and β can be written as (x - α)(x - β) = 0
Example: Find the roots of the equation, x2 + 12x + 20 = 0.
Solution:
Given, x2 + 12x + 20 = 0
⇒ x2 + 10x + 2x + 20 = 0
⇒ x(x + 10) + 2(x + 10) = 0
⇒ (x + 10)(x + 2) = 0
⇒ Either (x + 10) = 0 or (x + 2) = 0
⇒ Either x = -10 or x = -2
Hence, the roots of the equation are -10 and -2.
Example: Find the roots of the equation, 3x2 – 17x + 24 = 0.
Solution:
Given, 3x2 – 17x + 24 = 0
⇒ 3x2 – 8x – 9x + 24 = 0
⇒ x(3x – 8) – 3(3x + 8) = 0
⇒ (3x – 8)(x – 3) = 0
⇒ x = 8/3 or 3.
Hence, the roots of the equation are 8/3 and 3.
Example: Find the roots of the equation, 2x2 – 3x - 9 = 0.
Solution:
Given, 2x2 – 3x - 9 = 0
⇒ 2x2 – 6x + 3x - 9 = 0
⇒ 2x(x – 3) + 3(x – 3) = 0
⇒ (x – 3)(2x + 3) = 0
⇒ x = 3 or x = -3/2
Hence, the roots of the equation are -3/2 and 3.
Example: Find the integral values of x if, 3x + $\frac{5}{x}$ = 8.
Solution:
3x + $\frac{5}{x}$ = 8
∴ 3x2 + 5 = 8x
∴3x2 – 8x + 5 = 0
∴3x2 - 3x – 5x + 5 = 0
∴ 3x(x – 1) – 5(x – 1) = 0
∴ (x – 1)(3x – 5) = 0
∴ x – 1 = 0 or 3x – 5 = 0
∴ x = 1 or $\frac{5}{3}$
∴ Integral value of x = 1.
The method of completing squares can be used to find a general formula for finding the roots of a quadratic equation.
The two roots of the quadratic equation, ax2 + bx + c = 0 are given by:
x = $\frac{-\mathrm{b}±\sqrt{{\mathrm{b}}^{2}-4ac}}{2\mathrm{a}}$
The smaller root x1 = $\frac{-\mathrm{b}+\sqrt{{\mathrm{b}}^{2}-4ac}}{2\mathrm{a}}$
The larger root x2 = $\frac{-\mathrm{b}-\sqrt{{\mathrm{b}}^{2}-4ac}}{2\mathrm{a}}$
Explanation
ax2 + bx + c = 0
Dividing both the sides by a we get
x2 + $\frac{b}{a}$x + $\frac{c}{a}$ = 0
∴ x2 + 2 × $\frac{b}{2a}$ × x + $\frac{c}{a}$ = 0
Adding $\frac{{b}^{2}}{4{a}^{2}}$ both sides
∴ x2 + 2 × $\frac{b}{2a}$ × x + $\frac{{b}^{2}}{4{a}^{2}}$ + $\frac{c}{a}$ = $\frac{{b}^{2}}{4{a}^{2}}$
${\left(x+\frac{b}{2a}\right)}^{2}$ = $\frac{{b}^{2}}{4{a}^{2}}$ - $\frac{c}{a}$
∴ x + $\frac{b}{2a}$ = $\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}$ = $\frac{±\sqrt{{b}^{2}-4ac}}{2a}$
∴ x = $\frac{-b}{2a}$ ± $\frac{\sqrt{{b}^{2}-4ac}}{2a}$
Hence, x = $\frac{-\mathrm{b}±\sqrt{{\mathrm{b}}^{2}-4ac}}{2\mathrm{a}}$
Example: Find the roots of the equation, x2 – 8x + 11 = 0.
Solution:
Here, splitting the middle term would not be possible, hence it is best to use the Formula method.
In the equation a = 1, b = –8, c = 11
Hence, the roots of the equation are:
x = $\frac{-\left(-8\right)±\sqrt{{\left(-8\right)}^{2}-4×1×11}}{2×1}$
x = $\frac{8±\sqrt{20}}{2×1}$
x = $\frac{8±2\sqrt{5}}{2×1}$
x = 4 ± √5
Hence, the two roots of the equation are (4 + √5) and (4 - √5)
Example: Find the roots of the equation, 7x2 + 12x – 2 = 0.
Solution:
In the quadratic equation, 7x2 + 12x – 2 = 0, we have a = 7, b = 12, c = -2.
Hence, the roots of the equation are:
x = $\frac{-12±\sqrt{{12}^{2}-4×7×-2}}{2×7}$
x = $\frac{-12±\sqrt{200}}{14}$
x = $\frac{-12±10\sqrt{2}}{14}$
x = $\frac{-6±5\sqrt{2}}{7}$
Hence, the two roots of the equation are $\frac{-6-5\sqrt{2}}{7}$ and $\frac{-6+5\sqrt{2}}{7}$
Example:
Find the value of x if, $\frac{\left(x+1\right)}{\left(x+2\right)}$ + $\frac{\left(2x+1\right)}{\left(x-1\right)}$ = 0
Solution:
$\frac{\left(x+1\right)}{\left(x+2\right)}$ + $\frac{\left(2x+1\right)}{\left(x-1\right)}$ = 0
Multiplying both the sides by (x + 2)(x – 1),
(x + 1)(x – 1) + (2x + 1)(x + 2) = 0
∴ x2 – 1 + 2x2 + 4x + x + 2 = 0
∴ 3x2 + 5x + 1 = 0
∴x = $\frac{-5±\sqrt{{5}^{2}-4×3×1}}{2×3}$
⇒x = $\frac{-5±\sqrt{13}}{6}$
Thus, the roots are $\frac{-5-\sqrt{13}}{6}$ and $\frac{-5+\sqrt{13}}{6}$
Example: There are as many oranges in a crate as there are crates in a fruit shop. Prakhar takes away one crate and then the shop has only 72 oranges left. How many oranges does each crate have?
Solution:
Let there be x oranges in each crate. Then there are x crates in the shop.
Total number of oranges in the shop = x2
Prakhar takes away one crate i.e. x oranges.
∴ x2 – x = 72
∴ x2 – x – 72 = 0
∴ x2 – 9x + 8x – 90 = 0
∴ (x – 9)(x + 8) = 0
∴ x = 9 or x = –8
As the number of oranges cannot be negative,
∴ x = 9.
⇒ Each crate has 9 oranges.
As we saw earlier in the factorisation method of solving quadratic equation, a quadratic equation can be factorized as (x - α)(x - β) = 0
Let us now expand these terms.
⇒ x2 - αx - βx + α × β = 0
⇒ x2 - (α + β)x + αβ = 0
∴ Any quadratic equation can be written as x2 – (sum of the roots)x + product of the roots = 0
Example: Form a quadratic equation whose roots are 4 and 5.
Solution:
A quadratic equation with roots α and β can be formed as (x - α)(x - β) = 0
∴ The given quadratic equation can be formed as (x - 4)(x - 5) = 0
⇒ x2 – 4x – 5x + 20 = 0
⇒ x2 – 9x + 20 = 0
Alternately,
We know a quadratic equation with roots α and β can be formed as x2 – (sum of the roots)x + (product of the roots) = 0
∴ The given quadratic equation can be formed as x2 – (4 + 5)x + (4 × 5) = 0
⇒ x2 – 9x + 20 = 0
Example: Form a quadratic equation whose roots are 2 and -3.
Solution:
A quadratic equation with roots α and β can be formed as (x - α)(x - β) = 0
∴ The given quadratic equation can be formed as (x - 2)(x – (-3)) = 0
⇒ (x - 2)(x + 3)) = 0
⇒ x2 – 2x + 3x - 6 = 0
⇒ x2 + x + 6 = 0
Alternately,
We know a quadratic equation with roots α and β can be formed as x2 – (sum of the roots)x + (product of the roots) = 0
∴ The given quadratic equation can be formed as x2 – (2 + (-3))x + (2 × -3) = 0
⇒ x2 + x - 6 = 0
Example: Form a quadratic equation whose roots are -2/3 and -1.
Solution:
The given quadratic equation is (x –(-2/3))(x-(-1)) = 0
⇒ (x+2/3)(x+1)= 0
⇒ (3x + 2)(x + 1)= 0
⇒ 3x2 + 2x + 3x + 2 = 0
⇒ 3x2 + 5x + 2 = 0
Alternately,
We know a quadratic equation with roots α and β can be formed as x2 – (sum of the roots)x + (product of the roots) = 0
∴ The given quadratic equation can be formed as x2 – (-2/3 + (-1))x + (-2/3 × -1) = 0
⇒ x2 – (-5/3)x + 2/3 = 0
⇒ 3x2 + 5x + 2 = 0
Sum and Product of Roots of a Quadratic Equation
Consider the equation, ax2 + bx + c = 0.
This equation can be written as x2 + $\frac{b}{a}$x + $\frac{c}{a}$ = 0 …(1)
If α and β are the roots of this equation, the quadratic equation can also be written as x2 – (α + β)x + αβ = 0 …(2)
Comparing equation (1) and (2),
Since both are same equations, corresponding coefficients should be same.
Coefficient of x2: Is already same i.e., 1 in both equations.
Coefficient of x: $\frac{b}{a}$ = -(α + β)
⇒ (α + β) = -$\frac{b}{a}$
⇒ Sum of the roots of a quadratic equation = -$\frac{b}{a}$ = -
Constant term: $\frac{c}{a}$ = α × β
⇒ α × β = $\frac{c}{a}$
⇒ Product of the roots of a quadratic equation = $\frac{c}{a}$ =
Example: If α and β are the roots of the quadratic equation 2x2 + 3x + 1 = 0, the find the value of $\frac{1}{\alpha }$ + $\frac{1}{\beta }$.
Solution:
The given quadratic equation is 2x2 + 3x + 1 = 0.
We know, sum of the roots is -b/a and product of the roots is c/a
∴ α + β = -3/2 and αβ = ½
Now,
$\frac{1}{\alpha }$ + $\frac{1}{\beta }$ = $\frac{\alpha +\beta }{\alpha × \beta }$ = $\frac{-3/2}{1/2}$ = -3
Alternately,
Here we can also find the roots of the equation.
Given, 2x2 + 3x + 1 = 0.
⇒ 2x2 + 2x + x + 1 = 0.
⇒ (2x + 1)(x + 1) = 0
⇒ x = -1/2 or -1
Let α = -1/2 and β = -1
$\frac{1}{\alpha }$ + $\frac{1}{\beta }$ = -2 – 1 = -3
Example: If α and β are the roots of the quadratic equation x2 + 3x - 1 = 0, the find the value of $\frac{1}{{\alpha }^{2}}$ + $\frac{1}{{\beta }^{2}}$.
Solution:
The given quadratic equation is x2 + 3x - 1 = 0.
We know, sum of the roots is -b/a and product of the roots is c/a
∴ α + β = -3 and αβ = -1
Now,
$\frac{1}{{\alpha }^{2}}$ + $\frac{1}{{\beta }^{2}}$ = $\frac{{\mathrm{\alpha }}^{2}+{\mathrm{\beta }}^{2}}{{\left(\alpha \beta \right)}^{2}}$ = $\frac{{\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^{2}-2\mathrm{\alpha \beta }}{{\left(\alpha \beta \right)}^{2}}$ = $\frac{9-2×-1}{{\left(-1\right)}^{2}}$ = 11
Example:
If one root of the equation ax2 + bx + c = 0 is triple of the other, then 3b2 =
(1) 16ac (2) c√3 a (3) 6ac (4) None of the above
Solution:
Given, ax2 + bx + c = 0
Let the roots of the given equation are α and 3α.
∴ Sum of roots 4α = -$\frac{b}{a}$
⇒ α = -$\frac{b}{4a}$ …(1) and
Product of roots = 3α2 = $\frac{c}{a}$
⇒ 3${\left(-\frac{\mathrm{b}}{4\mathrm{a}}\right)}^{2}$ = $\frac{\mathrm{c}}{\mathrm{a}}$
$\frac{3{\mathrm{b}}^{2}}{16{\mathrm{a}}^{2}}$ = $\frac{\mathrm{c}}{\mathrm{a}}$
⇒ 3b2 = 16ac
Hence, option 1.
DISCRIMINANT AND NATURE OF ROOTS
In the Formula method of solving a quadratic equation we saw that x = $\frac{-\mathrm{b}±\sqrt{{\mathrm{b}}^{2}-4\mathrm{ac}}}{2\mathrm{a}}$
Here, the term (b2 – 4ac) is very important since it lies under the square root. It is called the discriminant (D) of the quadratic equation.
∴ D = b2 – 4ac
If D ≥ 0, the roots will be real whereas if D < 0, the roots of the quadratic equation will be complex.
∴ In a quadratic equation ax2 + bx + c = 0, if a, b and c are rational numbers, then:
Remember:
• If D < 0 ⇒ Roots will be Complex, Unequal and exist in Conjugate pair
• If D = 0 ⇒ Roots will be Real, Equal and Rational
• If D > 0 ⇒ Roots will be Real and Unequal
• If D is a perfect square, roots will be Rational
• If D is a not perfect square, roots will be Irrational
Remember:
• If c = a, then roots are reciprocal of each other.
• If b = 0, then roots are equal in magnitude but opposite in sign.
• When roots are complex, if one root is p + iq, then the other root will be p - iq (provided a, b and c are rational numbers)
(These are called conjugate pairs, i = iota = √(-1) )
• When roots are real, if one root is m + √n, then the other root will be m - √n (provided a, b and c are rational numbers)
(These are called conjugate pairs)
Example: Find the nature of the roots of the following quadratic equations.
(1) x2 + 2x - 6 = 0
(2) 3x2 - 7x + 2 = 0
(3) 4x2 - 12x + 9 = 0
(4) x2 - 3x + 12 = 0
Solution:
The coefficients of each of these given equations are rational
(1) D = b2 - 4ac = 4 – 4 × 1 × -6 = 4 + 24 = 28
Since D > 0 the roots are real and unequal.
Further, 28 is not a perfect square and hence the roots are irrational and are conjugates of each other.
Real, Distinct, and Irrational (Conjugate pairs)
(2) D = b2 - 4ac = 49 – 4 × 3 × 2 = 64 – 24 = 25
Since 25 > 0, the roots are real
Further 25 is a perfect square and hence the roots are rational.
Real, Distinct, and Rational
(3) D = b2 - 4ac = 144 – 4 × 4 × 9 = 144 - 144 = 0
Since D = 0, the roots are real and equal.
Real, Equal, and Rational.
(4) D = b2 - 4ac = = 9 – 4 × 1 × 12 = 9 - 48 = -37
Since D < 0, the roots are complex and are conjugates of each other.
Complex (conjugate pair)
Example: Find the nature of the roots of the quadratic equation x2 - 4√3x + 12 = 0.
Solution:
The coefficients of the given equation are not rational
The roots are x = $\frac{-\left(-4\sqrt{3}\right)±\sqrt{{\left(-4\sqrt{3}\right)}^{2}-4×1×12}}{2×1}$ = $\frac{4\sqrt{3}±\sqrt{48-48}}{2}$ = 2√3
Here, D = 0, hence the roots are real and equal.
But b is irrational hence, the roots are also irrational.
Roots are Real, Equal and Irrational.
GRAPHICAL REPRESENTATION OF A QUADRATIC EQUATION
Let f(x) = ax2 + bx + c (Quadratic function)
The shape of a quadratic function is a U-shaped curve called a parabola. The sign on the coefficient a of the quadratic function affects whether the graph opens up or down. If a > 0 then the graph makes a smile (opens up) and if a < 0 , the graph makes a frown (opens down).
The graph of a quadratic function (i.e., parabola) is also symmetric.
The value of discriminant (D) determined whether the curve cuts the x-axis or not.
Case 1 If D < 0, then there is no real value of x for which f(x) = 0.
Hence, the graph does not touch the x-axis and lies completely above or below x-axis depending on the value of a.
Case 2: If D = 0, then both the real roots are same.
Hence, the graph touches the x-axis at exactly one point.
Case 3: If D > 0, then both the real roots are distinct.
Hence, the graph cuts the x-axis at exactly two different points.
Example: f(x) is a quadratic expression such that the coefficient of x2 is positive. If the roots of f(x) = 0 lie in the interval (–1, 1), then which of the following is necessarily true?
(a) f(1) > 0 & f(-1) > 0 (b) f(1) > 0 & f(-1) < 0 (c) f(1) < 0 & f(-1) > 0 (d) f(1) < 0 & f(-1) < 0
Solution:
Since the coefficient of x2 is positive, the graph of the quadratic expression will be upward facing.
Since the roots lie between -1 and 1, the graph will cut the x-axis at two points between x = -1 and x = 1.
The graph would look something like this.
We can see that f(1) and f(-1), both are greater than ‘0’.
Hence, option 1.
MAXIMUM / MINIMUM VALUE OF A QUADRATIC FUNCTION
Let f(x) = ax2 + bx + c (Quadratic function)
Case 1: a > 0
If a > 0 , the graph for the quadratic expression will be upward facing.
∴ The maximum value of the quadratic function will be ∞.
To calculate the minimum value of the function we need to first calculate the value of x for which f(x) will be minimum.
Since the curve is symmetric, the two roots lie at the same distance on either side from the point of symmetry.
Therefore, the value of x when f(x) is minimum is exactly in the middle of the two roots.
∴ x = $\frac{{\mathrm{x}}_{1}+{\mathrm{x}}_{2}}{2}$ = $\frac{-\mathrm{b}/\mathrm{a}}{2}$ = $-\frac{\mathrm{b}}{2\mathrm{a}}$
Therefore, at x = -b/2a, f(x) is least possible.
The least value of f(x) = $f\left(\frac{-\mathrm{b}}{2\mathrm{a}}\right)$ = $\mathrm{a}×{\left(\frac{-\mathrm{b}}{2\mathrm{a}}\right)}^{2}$ + $\mathrm{b}×\left(\frac{-\mathrm{b}}{2\mathrm{a}}\right)$ + c = $\frac{4\mathrm{ac}-{\mathrm{b}}^{2}}{4\mathrm{a}}$
Case 2: a < 0
If a < 0 , the graph for the quadratic expression will be downward facing.
∴ The minimum value of the quadratic function will be -∞.
To calculate the maximum value of the function we need to first calculate the value of x for which f(x) will be maximum.
Since the curve is symmetric, the two roots lie at the same distance on either side from the point of symmetry.
Therefore, the value of x when f(x) is maximum is exactly in the middle of the two roots.
∴ x = $\frac{{\mathrm{x}}_{1}+{\mathrm{x}}_{2}}{2}$ = $\frac{-\mathrm{b}/\mathrm{a}}{2}$ = $-\frac{\mathrm{b}}{2\mathrm{a}}$
Therefore, at x = -b/2a, f(x) is highest possible.
The highest value of f(x) = $f\left(\frac{-\mathrm{b}}{2\mathrm{a}}\right)$ = $\mathrm{a}×{\left(\frac{-\mathrm{b}}{2\mathrm{a}}\right)}^{2}$ + $\mathrm{b}×\left(\frac{-\mathrm{b}}{2\mathrm{a}}\right)$ + c = $\frac{4\mathrm{ac}-{\mathrm{b}}^{2}}{4\mathrm{a}}$
Note: You don’t need to remember the formula for highest value of f(x). Just remember that f(x) is highest at x = -b/2a and substitute the value of x in f(x) to find its highest value.
If the roots of a quadratic equation ax2 + bx + c = 0 are α and β, then:
Form a Quadratic Equation whose roots are k more than original equation
Quadratic equation whose roots are α + k and β + k will be obtained by replacing x by x – k in the original equation
i.e., a(x - k)2 + b(x – k) + c = 0
Example: Form a quadratic equation whose roots are 2 more than the roots of 3x2 - 2x - 5 = 0.
Solution:
Here we will substitute x with x – 2.
∴ The new quadratic equation is 3(x - 2)2 - 2(x - 2) - 5 = 0
⇒ 3(x2 – 4x + 4) – 2x + 4 – 5 = 0
⇒ 3x2 – 14x + 11 = 0
Alternately,
Given, 3x2 - 2x - 5 = 0
⇒ 3x2 – 5x + 3x – 5 = 0
⇒ (x + 1)(3x - 5) = 0
∴ The roots of the given equation are -1 and 5/3
⇒ Roots of the new quadratic equation are -1 + 2 = 1 and 5/3 + 2 = 11/3
∴ The new equation is (x - 1)(x – 11/3) = 0
⇒ (x - 1)(3x - 11) = 0
⇒ 3x2 – 14x + 11 = 0
Form a Quadratic Equation whose roots are k less than original equation
Quadratic equation whose roots are α - k and β - k will be obtained by replacing x by x + k in the original equation
i.e., a(x + k)2 + b(x + k) + c = 0.
Example: Form a quadratic equation whose roots are 2 less than the roots of 3x2 - 2x - 5 = 0.
Solution:
Here we will substitute x with x + 2.
∴ The new quadratic equation is 3(x + 2)2 - 2(x + 2) - 5 = 0
⇒ 3(x2 + 4x + 4) – 2x - 4 – 5 = 0
⇒ 3x2 + 10x + 3 = 0
Alternately,
Given, 3x2-2x-5=0
⇒ 3x2 – 5x + 3x – 5 = 0
⇒ (x + 1)(3x - 5) = 0
∴ The roots of the given equation are -1 and 5/3
⇒ Roots of the new quadratic equation are -1 - 2 = -3 and 5/3 - 2 = -1/3
∴ The new equation is (x – (-3))(x – (-1/3)) = 0
⇒ (x + 3)(3x + 1) = 0
⇒ 3x2 + 10x + 3 = 0
Form a Quadratic Equation whose roots are k times that of the original equation
Quadratic equation whose roots are kα and kβ will be obtained by replacing x by x/k in the original equation
i.e., a(x/k)2 + b(x/k) + c = 0
Example: Form a quadratic equation whose roots are three times the roots of 3x2-2x-5=0.
Solution:
Here we will substitute x with x/3.
∴ The new quadratic equation is 3(x/3)2 - 2(x/3) - 5 = 0
⇒ x2/3 - 2x/3 - 5 = 0
⇒ x2 - 2x - 15 = 0
Alternately,
Given, 3x2-2x-5=0
⇒ 3x2 – 5x + 3x – 5 = 0
⇒ (x + 1)(3x - 5) = 0
∴ The roots of the given equation are -1 and 5/3
⇒ Roots of the new quadratic equation are -1 × 3 = -3 and 5/3 × 3 = 5
∴ The new equation is (x – (-3))(x – 5) = 0
⇒ (x + 3)(x – 5) = 0
⇒ x2 - 2x - 15 = 0
Form a Quadratic Equation whose roots are 1/k times that of the original equation
Quadratic equation whose roots are α/k and β/k will be obtained by replacing x by kx in the original equation
i.e., a(kx)2 + b(kx) + c = 0.
Example: Form a quadratic equation whose roots are half the roots of 3x2-2x-5=0.
Solution:
Here we will substitute x with 2x.
∴ The new quadratic equation is 3(2x)2 - 2(2x) - 5 = 0
⇒ 12x2 - 4x - 5 = 0
Alternately,
Given, 3x2-2x-5=0
⇒ 3x2 – 5x + 3x – 5 = 0
⇒ (x + 1)(3x - 5) = 0
∴ The roots of the given equation are -1 and 5/3
⇒ Roots of the new quadratic equation are -1 ÷ 2 = -1/2 and 5/3 ÷ 2 = 5/6
∴ The new equation is (x – (-1/2))(x – 5/6) = 0
⇒ (2x + 1)(6x - 5) = 0
⇒ 12x2 - 4x - 5 = 0
• Form a Quadratic Equation whose roots are reciprocal of the roots of original equation
Quadratic equation whose roots are 1/α and 1/β will be obtained by replacing x by 1/x in the original equation
i.e., a(1/x)2 + b(1/x) + c = 0.
• Form a Quadratic Equation whose roots are equal in value but opposite in sign of the roots of original equation
Quadratic equation whose roots are -α and -β will be obtained by replacing x by -x in the original equation
i.e., a(-x)2 + b(-x) + c = 0.
Common roots of two Quadratic Equations
Let f(x) = 0 and g(x) = 0 be two quadratic equations in x such that they have a common root(s). You would be asked to find the common root(s).
Let’s say α is the common root of the two equations.
Instead of finding roots of f(x) and g(x), we can simply find the roots of f(x) = g(x), i.e., f(x) – g(x) = 0
Note: If f(α) = 0 and g(α) = 0, the f(α) - g(α) will also be equal to 0 but the converse is not true.
i.e., solving for f(x) – g(x) = 0, will give us two roots. Now we need to check for which of these two roots both f(x) and g(x) are equal to ‘0’.
Step 1: Find f(x) - g(x)
Step 2: Find the root(s) of f(x) - g(x) = 0
Step 3: Substitute the roots of f(x) - g(x) = 0 in f(x) and g(x). If the roots satisfy both the equations, then they are the common roots.
Example: The number of roots common between the two equations x3 - 6x2 + 11x - 6 = 0 and x3 + 2x2 - x - 2 = 0 is:
(1) 0 (2) 1 (3) 2 (4) 3
Solution:
f(x) = x3 - 6x2 + 11x - 6 = 0 ...(1)
g(x) = x3 + 2x2 - x - 2 = 0 ...(2)
Equating equations f(x) and g(x), we get,
⇒ x3 - 6x2 + 11x - 6 = x3 + 2x2 - x - 2
⇒ 8x2 – 12x + 4 = 0
⇒ 2x2 – 3x + 1 = 0
⇒ (2x – 1)(x – 1) = 0
∴ x = 1/2 or x = 1
Only for x = 1, both f(x) = 0 and g(x) = 0
Hence, option 2.
Example: The number of roots common between the two equations x3 + 3x2 + 4x + 5 = 0 and x3 + 2x2 + 7x + 3 = 0 is:
(1) 0 (2) 1 (3) 2 (4) 3
Solution:
f(x) = x3 + 3x2 + 4x + 5 = 0 ...(1)
g(x) = x3 + 2x2 + 7x + 3 = 0 ...(2)
Step 1: Equating equations f(x) and g(x), we get,
⇒ x3 + 3x2 + 4x + 5 = x3 + 2x2 + 7x + 3
⇒ x2 – 3x + 2 = 0
⇒ (x – 1)(x – 2) = 0
∴ x = 1 or x = 2
Since both x = 1 and x = 2, do not satisfy either (1) or (2), there exists no common roots for equations (1) and (2).
Hence, option 1.
Example: Find the value(s) of ‘m’, so that the equations x2 – x – 12 = 0 & mx2 + 10x + 3 = 0 have exactly one root in common.
(a) 3 (b) -43/16 (c) 4 (d) Either (a) or (b)
Solution:
f(x) = x2 – x – 12
g(x) = mx2 + 10x + 3
Solving for f(x)
⇒ (x + 3)(x - 4) = 0
⇒ x = -3 or 4
The common root will be either -3 or 4
Case 1: -3 is the common root
⇒ g(x) = mx2 + 10x + 3 = 0
⇒ 9m – 30 + 3 = 0
⇒ m = 3
Case 2: 4 is the common root
⇒ g(x) = mx2 + 10x + 3 = 0
⇒ 16m + 40 + 3 = 0
⇒ m = -43/16
Hence, option 4.
HIGHER DEGREE POLYNOMIALS
A Polynomial Expression of degree n can be written as:
anxn + an-1xn-1 + an-2xn-2 + ... + a2x2 + a1x + a0
A Polynomial Equation of degree n can be written as:
anxn + an-1xn-1 + an-2xn-2 + ... + a2x2 + a1x + a0 = 0
The above polynomial equation has 'n' roots / solutions / zeroes.
Sum of all the roots of the equation =
Product of all the roots of the equation =
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Food for thought
Solution
Take any two points $A$ and $B$ on the parabola $y = x^2$.
Draw the line $OC$ through the origin, parallel to $AB$, cutting the parabola again at $C$.
Let $A$ have coordinates $(a,a^2)$, let $B$ have coordinates $(b,b^2)$ and let $C$ have coordinates $(c,c^2)$.
Prove that $a+b = c$.
We know that
$A=(a,a^2)$,
$B=(b,b^2)$,
$C=(c,c^2)$.
We can calculate the gradient of a straight line between two points $(x_1,y_1)$ and $(x_2,y_2)$ as $\frac{y_1-y_2}{x_1-x_2}$.
So the gradient of line $AB$ is equal to $\frac{a^2-b^2}{a-b}=a+b$.
Also, the gradient of line $OC$ is equal to $\frac{0-c^2}{0-c}=c$.
However, we know from the question that these two lines are parallel. Two lines are parallel if and only if they have the same gradient, which means that we must have $a+b=c$.
Imagine drawing another parallel line $DE$, where $D$ and $E$ are two other points on the parabola. Extend the ideas of the previous result to prove that the midpoints of each of the three parallel lines lie on a straight line.
Now, suppose that we have two points $D=(d,d^2)$ and $E=(e,e^2)$ on the parabola making another line parallel to $OC$.
The gradient of line $DE$ is equal to $\frac{d^2-e^2}{d-e}=d+e$. As $DE$ is parallel to $OC$, we also have $d+e=c$.
We can calculate the midpoint of a straight line between two points $(x_1,y_1)$ and $(x_2,y_2)$ as $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.
So the midpoint of line $OC$ is $\left(\frac{0+c}{2},\frac{0+c^2}{2}\right) = \left(\frac{c}{2},\frac{c^2}{2}\right).$
The midpoint of line $AB$ is $\left(\frac{a+b}{2},\frac{a^2+b^2}{2}\right) = \left(\frac{c}{2},\frac{a^2+b^2}{2}\right).$
The midpoint of line $DE$ is $\left(\frac{d+e}{2},\frac{d^2+e^2}{2}\right) = \left(\frac{c}{2},\frac{d^2+e^2}{2}\right).$
All three of these points have an $x$-coordinate of $\frac{c}{2}$. This means that the line $x=\frac{c}{2}$ passes through the midpoints of the lines $OC$, $AB$ and $DE$. |
# Dr. Fowler CCM Solving Systems of Equations By Elimination – Harder.
## Presentation on theme: "Dr. Fowler CCM Solving Systems of Equations By Elimination – Harder."— Presentation transcript:
Dr. Fowler CCM Solving Systems of Equations By Elimination – Harder
Solving a system of equations by elimination using addition and subtraction. Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Add or subtract the equations. Step 4: Plug back in to find the other variable. Step 5: Check your solution. Standard Form: Ax + By = C Look for variables that have the same coefficient. Solve for the variable. Substitute the value of the variable into the equation. Substitute your ordered pair into BOTH equations. ALREADY IN NOTES – Read Only for Review
Elimination using Multiplication 1) Solve the system. Adding or subtracting will not eliminate x + 2y = 6 3x + 3y = -6 But, we can multiply the first equation by -3 to eliminate the x term
Elimination using Multiplication x + 2y = 6 3x + 3y = -6 -3 ( ) 1) Solve the system.
Elimination using Multiplication -3x + -6y = -18 3x + 3y = -6 + -3y = -24 y = 8 ANS: (x, 8) Be sure to distribute the -3 to ALL in the equation. 1) Solve the system.
Elimination using Multiplication x + 2y = 6 3x + 3y = -6 ANS: (x, 8) Substitute y = 8 into equation y =8 x + 2(8) = 6 x + 16 = 6 x = -10 1) Solve the system.
Elimination using Multiplication x + 2y = 6 3x + 3y = -6 Answer: ( -10, 8) Substitute y = 8 into equation y =8 x + 2(8) = 6 x + 16 = 6 x = -10 1) Solve the system.
2) Solve the system: 2(2) + 2y = 6 4 + 2y = 6 2y = 2 y = 1 2x + 2y = 6 3x – y = 5 If we multiply the bottom equation by 2 we can eliminate y: 2x + 2y = 6 (2)(3x – y = 5) 2x + 2y = 6 (+) 6x – 2y = 10 8x = 16 x = 2 (2, 1) Substitute x = 2 into either original equation:
More complex Problems 3x + 4y = -25 2x - 3y = 6 Multiply by 2 Multiply by -3. This will get X’s to MATCH 3) Solve the system
More complex Problems 3x + 4y = -25 2x - 3y = 6 2( ) -3( ) 3) Solve the system
More complex Problems 6x + 8y = -50 -6x + 9y = -18 + 17y = -68 y = -4 Answer: (x, -4) 3) Solve the system
More complex Problems 3x + 4y = -25 2x - 3y = 6Substitute y = -4 2x - 3(-4) = 6 2x - -12 = 6 2x + 12 = 6 2x = -6 x = -3 Answer: (-3, -4) 3) Solve the system
3x + 4y = -1 4x – 3y = 7 4) Solve the system using elimination. 3(1) + 4y = -1 3 + 4y = -1 4y = -4 y = -1 Multiply both equations (3)(3x + 4y = -1) (4)(4x – 3y = 7) 9x + 12y = -3 (+) 16x – 12y = 28 25x = 25 x = 1 (1, -1)
Excellent Job !!! Well Done
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# Use of Integers
The use of integers is used to express our day-to-day situations in Mathematical terms.
Examples:
(i) If profit are represented by positive integer then losses by negative integers.
(ii) If heights above sea level by positive integers then depths below sea level by negative integers.
(iii) If rise in price is represented by positive integers, then fall in price by negative integers and so on.
Thus, if +256 represent a profit of $256; then a loss of$ 256 is represented by -256.
Similarly, if a depth of 37 below sea level is represented by -37; then +37 represents a height of 37 above sea level and so on.
Use of integers as directed numbers:
When numbers represent direction, then numbers are called directed numbers.
For example:
(i) If moving 10 m towards North is represented by +10;
-10 represents moving 10 m towards South, opposite direction of North.
If a positive (+ve) integer indicates a particular direction; then the negative (-ve) integer indicates the opposite direction. Conversely, if a negative (-ve) integer indicates any particular direction; then the positive (+ve) integer indicates the opposite direction.
For example:
If +5 represents 5 m towards East; then -4 represents 4 m towards its opposite direction i.e., towards West.
Similarly, if +9 represents 9 m due South, -6 represents 6 m due North. Again if -4 represents 4 km due East; +2 represents 2 km due West
(ii) If 12 m above the earth’s surface is represented by +12; then 18 m below the earth’s surface is represented by -18 and so on.
(iii) If +23 represents a profit of $23; then -19 represents a loss of$19.
(iv) If -63 indicate giving of $63; then taking of$91 is denoted by +91.
(v) If the rise in temperature by 63° C is denoted by +63; then -45 indicates the fall in temperature by 45° C
Write the opposite of the following expressions:
(i) An expenditure of $15. (ii) A profit of$115.
(iii) A loss of $55. (iv) Descending -15 m. (v) Increase in weight by 17 kg. Solution: (i) Expenditure of$15 = Income of -15 dollars.
(ii) Profit of $115 = Loss of 115 dollars. (iii) Loss of$55 = Profit of 55 dollars.
(iv) Descending -15 m or -15 m downwards movement = 15 m upward movement.
(v) Increase in weight by 17 kgs. = Decrease in weight by -17 kgs.
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# Why You Never Need Cylindrical Shells
Don’t get me wrong; finding volumes of solids of rotation by the method of cylindrical shells is a great method. It’s just that you can always work around it; you don’t ever need to use it. The work-around is often longer and involves more work, but it is interesting mathematically. So here’s an example of how to do it.
The region bounded by the graph of $y={{x}^{3}}+x+1$, and the lines y = 1 and x = 2, is revolved around the y-axis. What is the volume of the resulting solid?
Supposedly this volume must be found by the shell method. Using the washer method the volume is set up as the integral of the area of the outside circle with constant radius of 2, minus the area of the inside circle of radius x, times the “thickness” of a slice. This “thickness” is in the y-direction and so is dy. The dy also determines the limits of integration.
$\displaystyle V=\int_{1}^{11}{\left( \pi {{2}^{2}}-\pi {{x}^{2}} \right)\,dy}$
Since the integral contains a dy the usual way is to change the x to a function of y, which in this case involves solving a cubic polynomial. In this case that is very difficult to find x as a function of y and with a different function that may even be impossible. But do we really have to do that? In fact, what is required is to have only one variable and the variable may be x! So we find dy in terms of x and dx and substitute into the expression above. We also change the limits of integration to the corresponding x-values.
$dy=\left( 3{{x}^{2}}+1 \right)dx$
$\displaystyle V=\int_{0}^{2}{\left( \pi {{2}^{2}}-\pi {{x}^{2}} \right)\,\left( 3{{x}^{2}}+1 \right)dx}$
$\displaystyle V=\pi \int_{0}^{2}{\left( 4+11{{x}^{2}}-3{{x}^{4}}\, \right)dx}$
This integral is easy to evaluate and will give the same value, $\tfrac{272}{15}\pi$, as the shell integral.
In order to use this idea, the function must either be one-to-one on the interval or the solid must be broken into sections that are one-to-one. This may make the problem longer. Most problems that you want to do by shells are easier by shells. The point is that washers (or disks) may always be used; not that the washer approach is the easiest way.
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# Solve my math equation
In this blog post, we will show you how to Solve my math equation. Our website will give you answers to homework.
## Solving my math equation
As a student, there are times when you need to Solve my math equation. Absolute value is a concept in mathematics that refers to the distance of a number from zero on a number line. The absolute value of a number can be thought of as its magnitude, or how far it is from zero. For example, the absolute value of 5 is 5, because it is five units away from zero on the number line. The absolute value of -5 is also 5, because it is also five units away from zero, but in the opposite direction. Absolute value can be represented using the symbol "| |", as in "|5| = 5". There are a number of ways to solve problems involving absolute value. One common method is to split the problem into two cases, one for when the number is positive and one for when the number is negative. For example, consider the problem "find the absolute value of -3". This can be split into two cases: when -3 is positive, and when -3 is negative. In the first case, we have "|-3| = 3" (because 3 is three units away from zero on the number line). In the second case, we have "|-3| = -3" (because -3 is three units away from zero in the opposite direction). Thus, the solution to this problem is "|-3| = 3 or |-3| = -3". Another way to solve problems involving absolute value is to use what is known as the "distance formula". This formula allows us to calculate the distance between any two points on a number line. For our purposes, we can think of the two points as being 0 and the number whose absolute value we are trying to find. Using this formula, we can say that "the absolute value of a number x is equal to the distance between 0 and x on a number line". For example, if we want to find the absolute value of 4, we would take 4 units away from 0 on a number line (4 - 0 = 4), which tells us that "the absolute value of 4 is equal to 4". Similarly, if we want to find the absolute value of -5, we would take 5 units away from 0 in the opposite direction (-5 - 0 = -5), which tells us that "the absolute value of -5 is equal to 5". Thus, using the distance formula provides another way to solve problems involving absolute value.
If you’re good at math, you can be a better engineer, accountant and financial analyst. You can also be a better manager and decision maker. However, if you’re bad at math, it can hold you back from doing so many things in life. Plus, it can also lead to anxiety and depression. So, if you want to succeed in life, you need to work hard at math. One of the best ways to do this is by using a math equation solver app. These apps are designed to make it easier for people who are struggling with math to solve equations quickly and accurately. By using them, you can learn how to solve math equations more efficiently and effectively.
Solving math equations is a fundamental skill in mathematics. It’s also a great way to practice your critical thinking skills and develop your problem solving abilities. You can use a variety of resources to help you learn how to solve math equations. For example, you can read books, watch educational videos, take online courses, or enroll in math tutoring programs. And you can get extra practice whenever you have free time at school or at home. So don’t let your math skills slip! If you have any questions about solving math equations, please reach out to a teacher or tutor for help. They’re more than happy to lend a hand!
Solving trinomials is a process that can be broken down into a few simple steps. First, identify the coefficients of the terms. Next, use the quadratic formula to find the roots of the equation. Finally, plug the roots back into the original equation to verify your results. While this process may seem daunting at first, with a little practice it will become second nature. With so many trinomials to solve, there's no time to waste - get started today!
## We solve all types of math problems
Absolute necessity for us parents that have not done algebra, geometry, etc. since we were in school! This app has been such a great help, it makes teaching your child math as easy as taking a picture, literally! I had my doubts about how well the camera would read the problem but after weeks of homework every day, it has not failed even one time. My only possible suggestion would be to have an option to hide the answer to the problem for teaching purposes. Can't recommend it enough! Excellent!
Audrey Robinson
It's an amazing app. It's a blessing for students. Loved it. I scanned some intermediate integration problems and magic it shows every step wise step detailed solution. Must have it in phones of students or anyone who deals with daily mathematics. A great thanks to the app inc. for making such a wonderful application. Again, thanks a lot
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### LOGO Challenge - Following On
Remember that you want someone following behind you to see where you went. Can yo work out how these patterns were created and recreate them?
### Pick's Theorem
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Pentagonal
Can you prove that the sum of the distances of any point inside a square from its sides is always equal (half the perimeter)? Can you prove it to be true for a rectangle or a hexagon?
# Star Gazing
##### Stage: 4 Challenge Level:
Ned, who has just left Christ Church Cathedral School in Oxford, sent a nice solution saying that: The Star Gazing problem is easy. The 6-point star is made up of 6 equilateral triangles, and so is the hexagon inside it, so the ratio of the points' to the inside' is 1:1''.
Ned then found the area of the octagon by dividing it up (a square, 4 rectangles and 4 triangles) and used Pythagoras theorem to find the lengths needed and finally found the ratio of the areas of the points' to the octagon inside' to be 1: (1 + $\sqrt 2$).
The second part is most easily done by taking the shorter sides of the triangles which make up the star to have length 1 unit, so that the area of each triangle is 1/2 and the total area of the `points' is 4 square units . The hypotenuse of each triangle has length$\sqrt 2$ (by Pythagoras Theorem) which gives the lengths of the edges of the octagon. To find the area of the octagon take the square and chop off four triangles at the corners.
The length of the side of the square is (2 + $\sqrt 2$)
the area of the octagon = (2 + $\sqrt 2)^2$ - 2
which simplifies to 4 + 4$\sqrt 2$.
Finally the ratio of areas is 4 : (4 + 4$\sqrt 2$) which simplifies to 1 : (1 + $\sqrt 2$ ). |
Tree Diagrams
Tree Diagrams, like Venn Diagrams, Â provide a simple visual way of representing events and their associated probabilities. They can be used to calculate more complicated probabilities. Tree Diagrams display the probabilities different event outcomes on the ‘branches’ of the diagram. Typically, the events are considered sequentially (one after the other) moving along the branches from left to right.
Independent Events in Tree Diagrams
Consider the very simple case of flipping a coin twice – each toss is a separate event. If the coin is fair, the probability of tossing heads and tails is clearly 0.5. In the second toss, these probabilities still hold.
In this example, all sets of branches have identical probabilities. This is typical of a tree diagram where the events are independent (learn about Independent Events).
We can see from the tree diagram that the probability of tossing two tails in a row, i.e. P(TT), is 1/4. Note that to specify that two event outcomes occur, we move along the branches and so we must multiply the probabilities on the branches together (one outcome AND another outcome implies MULTIPLY). One outcome OR another implies addition. For example, suppose we want to know the probability that at least one head was thrown in two coin tosses. In this case, we can have HH, HT or TH and so we add their associated probabilities together. It follows that the probability of tossing at least one head is 3/4. This makes sense since the only way in which we don’t toss at least one head is if we toss no heads, i.e. two tails.
Conditional Probability on Tree Diagrams
If the probabilities on the second set of branches were different, there is dependence on the outcome of the first event. This is known as conditional probability. Consider the slightly more complicated example of drawing counters from a bag without replacement. The bag starts off with 4 red counters and 6 blue counters. Two counters are drawn from the bag one after the other.
We can see from this diagram, for example, that the probability of drawing one of each is:
Examples
Three fair dice are rolled simultaneously. Find the probability that at least one dice has a number less than or equal to 2.
Solution:
This seems like a very confusing question at first glance. However, although the three dice are rolled simultaneously, we can treat the outcome on each dice as separate events and hence branches on the tree diagram. Think of the outcome for each set of branches as a number less than or equal to 2′ or a number more than 2′. It follows that to find the probability that AT LEAST ONE dice has a number less than or equal to 2, we must first find the probability that NO dice have a number that is less than or equal to 2. In other words, we must find the probability that ALL dice have a number greater than 2. Fill in the probabilities on a tree diagram:
From the tree diagram we can see that the probability that all dice have a number greater than 2 is 8/27. It follows that the probability that at least one dice has a number less than or equal to 2 is given by
.
A false positive diagnosis in medical testing is a positive test result when the disease IS NOT present. Similarly, a false negative is one in which a medical test gives a negative result when the disease IS present. Click here for more details.
2% of people are living with a given condition. The chances of a screening procedure for the condition giving a positive result when the condition is present is 95%. There is a 0.5% chance that the screen diagnoses the condition when it is NOT present. Calculate the probability that the screening procedure gives a correct diagnosis.
Solution:
Start by creating a tree diagram and filling in the branch probabilities:
The diagnosis is correct if the disease is present (D) with a positive diagnosis (P) OR the disease is not present (ND) with a negative diagnosis (N). Hence the probability of a correct result is:
. |
New Zealand
Level 7 - NCEA Level 2
# Period changes for tangents
Lesson
We saw, in another chapter, how the period of the sine and cosine functions is affected by the coefficient $k$k that multiplies $x$x in $\sin kx$sinkx and $\cos kx$coskx.
The fact that the sine and cosine functions both have period $360^\circ$360° is expressed by the statements
$\sin(x+360^\circ n)=\sin x$sin(x+360°n)=sinx and
$\cos(x+360^\circ n)=\cos x$cos(x+360°n)=cosx
for all integers $n$n. So it must be true that since $\tan x=\frac{\sin x}{\cos x}$tanx=sinxcosx, then $\tan(x+360^\circ n)=\tan x$tan(x+360°n)=tanx. However, $360^\circ$360° is not the smallest interval at which the tangent function repeats.
A glance at the graph of the tangent function, shown below, should convince you that this function has a period of $180^\circ$180°. That is, $\tan(x+180^\circ n)=\tan x$tan(x+180°n)=tanx for all integers $n$n.
The fact that the tangent function repeats at intervals of $180^\circ$180° can be verified by considering the unit circle diagram. If $180^\circ$180° is added to an angle $\alpha$α, then the diagram shows that $\sin(\alpha+180^\circ)$sin(α+180°) has the same magnitude as $\sin\alpha$sinα but opposite sign. The same relation holds between $\cos(\alpha+180^\circ)$cos(α+180°) and $\cos\alpha$cosα.
We make use of the definition: $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$tanα=sinαcosα
$\tan(\alpha+180^\circ)$tan(α+180°) $=$= $\frac{\sin(\alpha+180^\circ)}{\cos(\alpha+180^\circ)}$sin(α+180°)cos(α+180°) $=$= $\frac{-\sin\alpha}{-\cos\alpha}$−sinα−cosα $=$= $\tan\alpha$tanα
We are now in a position to determine the period of the function $\tan kx$tankx where $k$k is any number.
We can define a new variable $x'=kx$x=kx so that $\tan kx=\tan x'$tankx=tanx. But, we have seen that $\tan x'=\tan(x'+180^\circ)$tanx=tan(x+180°). So,
$\tan x'$tanx′ $=$= $\tan(kx+180^\circ)$tan(kx+180°) $=$= $\tan k\left(x+\frac{180^\circ}{k}\right)$tank(x+180°k)
This says, we need to advance $x$x by an amount $\frac{180^\circ}{k}$180°k in order to reach the same function value as $\tan kx$tankx. We conclude that the coefficient $k$k in $\tan kx$tankx changes the period by the factor $\frac{1}{k}$1k compared with the period of $\tan x$tanx.
#### Example 1
What is the period of the function $\tan\frac{3x}{2}$tan3x2?
In this example, $k=\frac{3}{2}$k=32. So the period of the function is $\frac{1}{k}$1k times the period of $\tan x$tanx. Thus, $\tan\frac{3x}{2}$tan3x2 has period $\frac{2}{3}\times180^\circ$23×180°, or$120^\circ$120°.
#### Example 2
You are given the graph of a function that looks like the graph of $\tan x$tanx except that it repeats at intervals of $540^\circ$540° rather than at intervals of $180^\circ$180°. Assuming the function has the form $\tan kx$tankx, what is $k$k?
The period of $\tan kx$tankx is $\frac{180^\circ}{k}$180°k and we know this to be $540^\circ$540°.
So, $\frac{180^\circ}{k}=540^\circ$180°k=540° and hence, $k=\frac{1}{3}$k=13.
#### Further Examples
##### QUESTION 1
Consider the equation $y=\tan9x$y=tan9x.
1. Complete the table of values for $y=\tan9x$y=tan9x.
$x$x $y$y $\left(-5\right)^\circ$(−5)° $0^\circ$0° $5^\circ$5° $15^\circ$15° $20^\circ$20° $25^\circ$25° $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
2. Graph the equation.
##### QUESTION 2
The function $f\left(x\right)$f(x) on a restricted domain has the form $f\left(x\right)=\tan bx$f(x)=tanbx. Two neighbouring asymptotes of this function are known to be at $x=30^\circ$x=30° and $x=90^\circ$x=90°.
1. Find the $x$x-intercept on the same domain. Give your answer in exact form.
##### QUESTION 3
The graph of a function in the form $y=\tan bx$y=tanbx is plotted below.
1. State the period of the function.
### Outcomes
#### M7-2
Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs
#### 91257
Apply graphical methods in solving problems |
# Converting Infinite Periodic Decimal into Proper Fraction
If we need to multiply infinite decimal fraction by 10, 100, 1000 and so on we will shift a dot at one, two or three and so on digits to the right such as it do in the finite decimal fraction.
For example, 0.1(23)*100=0.1232323...*100=12.323232...=12.(32) .
Example 1. Converting into proper fraction number: 0.(13).
Suppose x=0.(13)=0.131313... . Let′s multiply this purely periodic fraction x by such number that dot will shift exactly at the period to the right. As the period has two digits, we need to shift the dot to the right at the two periods and for this we multiply x by 100, then 100x=0.131313...*100=13.1313...=13.(13). Then we subtract x from 100x and we will obtain 100x-x=13.(13)-0.(13). So, 99x=13 , whence we obtain x=13/99.
Example 2. Converting into proper fraction number: 0.2(54).
Suppose x=0.2(54).Let′s shift the dot in such mixed periodic fraction to the right so as to obtain purely periodic fraction. For this we multiply x by 10 and we will obtain 10x=2.(54).
Suppose y=2.(54).Let′s convert such purely periodic fraction into proper fraction, as we made earlier. We have: 100y=254.(54); 100y-y=254.(54)-2.54 ;
99y=252; y=252/99=28/11 .
So, 10x=28/11 , from which x=28/(11*10)=14/55. |
Beating older or younger players in badminton tournament
$n\geq 5$ people of different ages play badminton. Each player plays four times, each time against a different player. Prove that there exists a player who beats two people older than her, or who beats two people younger than her.
Since each player plays four times, the average number of wins is two. So there exists a player winning at least two games. But how can we include the age condition?
Suppose there is not.
• Then nobody can win more than two games.
• So everybody wins exactly two games.
• So the youngest player wins two games.
So there is at least one example.
If some person $p$ beats $n\ge3$ people, then this is trivial, as there must be two people out of the $n$ people younger or older than $p,$ at the same time (pigeonhole principle).
If everyone beats exactly $2$ people, then consider the youngest person $p_0:$ this person beats two people, and these two people are older than $p_0$ by assumption, so this player $p_0$ is as required.
And we generalise this problem to the following
Proposition
Assume given $n\gt2$ points $p_1,\cdots,p_n,$ and, for each pair $(p_i,p_j),$ an integer $n(i,j)\in\{0,1\}$ such that $n(i,j)+n(j,i)=1, \forall i\not=j$ and $n(i,i)=0, \forall i.$ Then $\exists i=1,\cdots,n$ such that either the set $\{j\gt i\mid n(i,j)=1\}$ or the set $\{j\lt i\mid n(i,j)=1\}$ is of cardinality $\gt\frac{n-1}{4}.$
Proof.
Define $W_i=|\{j\not=i\mid n(i,j)=1\}|, W^+_i=|\{j\gt i\mid n(i,j)=1\}|,$ and $W_i^-=\{j\lt i\mid n(i,j)=1\},$ so $W_i=W^+_i+W^-_i.$
It is evident that $\sum_iW_i/n=(n-1)/2.$ And we consider two cases:
I. If there is $i_0$ such that $W_{i_0}\gt(n-1)/2,$ then one of $W_{i_0}^+$ and $W_{i_0}^-$ must be $\gt(n-1)/4$ II. If $W_{i}=(n-1)/2, \forall i,$ then $W_1=W^+_1\gt(n-1)/4.$
$\square$
Hope this helps. |
Chapter 8 Electromagnetic Induction, AC Circuits, and Electrical Technologies
# 8.5 Electric Generators
### Summary
• Calculate the emf induced in a generator.
• Calculate the peak emf which can be induced in a particular generator system.
Electric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in Chapter 23.1 Induced Emf and Magnetic Flux. We will now explore generators in more detail. Consider the following example.
### Example 1: Calculating the Emf Induced in a Generator Coil
The generator coil shown in Figure 1 is rotated through one-fourth of a revolution (from θ = 0o to θ = 90o) in 15.0 ms. The 200-turn circular coil has a 5.00 cm radius and is in a uniform 1.25 T magnetic field. What is the average emf induced?
Strategy
We use Faraday’s law of induction to find the average emf induced over a time Δt.
We know that N = 200 and Δt = 15.0 ms, and so we must determine the change in flux Δφ to find emf.
Solution
Since the area of the loop and the magnetic field strength are constant, we see that
Δφ =Δ (BA) cos θ = AB Δ (cosθ )
Now, Δ (cos θ ) = -1.0, since it was given that θ goes from 0o to 90o . Thus Δφ = -AB, and
. The area of the loop is A = π r2 = (3.14 … )(0.0500 m2) = 7.85 x 10-3 m2. Entering this value gives
.
Discussion
This is a practical average value, similar to the 120 V used in household power.
The emf calculated in Example 1 is the average over one-fourth of a revolution. What is the emf at any given instant? It varies with the angle between the magnetic field and a perpendicular to the coil. We can get an expression for emf as a function of time by considering the motional emf on a rotating rectangular coil of width w and height in a uniform magnetic field, as illustrated in Figure 2.
Charges in the wires of the loop experience the magnetic force, because they are moving in a magnetic field. Charges in the vertical wires experience forces parallel to the wire, causing currents. But those in the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. We can thus find the induced emf by considering only the side wires. Motional emf is given to be , where the velocity v is perpendicular to the magnetic field B. Here the velocity is at an angle θ with B, so that its component perpendicular to B is vsinθ (see Figure 2). Thus in this case the emf induced on each side is , and they are in the same direction. The total emf around the loop is then
.
This expression is valid, but it does not give emf as a function of time. To find the time dependence of emf, we assume the coil rotates at a constant angular velocity ω. The angle θ is related to angular velocity by θ = ωt, so that
Now, linear velocity v is related to angular velocity ω by v = rω. Here r = w/2 , so that v = (w/2) ω, and
.
Noting that the area of the loop is , and allowing for loops, we find that
is the emf induced in a generator coil of turns and area A rotating at a constant angular velocity ω in a uniform magnetic field B. This can also be expressed as
emf = emfo sinωt
where
emfo=NAB ω
is the maximum (peak) emf. Note that the frequency of the oscillation is f = ω / 2 π , and the period is . Figure 3 shows a graph of emf as a function of time, and it now seems reasonable that AC voltage is sinusoidal.
The fact that the peak emf, emfo = NAB ω, makes good sense. The greater the number of coils, the larger their area, and the stronger the field, the greater the output voltage. It is interesting that the faster the generator is spun (greater ω), the greater the emf. This is noticeable on bicycle generators—at least the cheaper varieties. One of the authors as a juvenile found it amusing to ride his bicycle fast enough to burn out his lights, until he had to ride home lightless one dark night.
Figure 4 shows a scheme by which a generator can be made to produce pulsed DC. More elaborate arrangements of multiple coils and split rings can produce smoother DC, although electronic rather than mechanical means are usually used to make ripple-free DC.
### Example 2: Calculating the Maximum Emf of a Generator
Calculate the maximum emf, emf0, of the generator that was the subject of Example 1.
Strategy
Once ω, the angular velocity, is determined, emf0 = NABω can be used to find emf0. All other quantities are known.
Solution
Angular velocity is defined to be the change in angle per unit time:
.
One-fourth of a revolution is π / 2 radians, and the time is 0.0150 s; thus,
104.7 rad/s is exactly 1000 rpm. We substitute this value for ω and the information from the previous example emf0 = NAB ω, yielding
Discussion
The maximum emf is greater than the average emf of 131 V found in the previous example, as it should be.
In real life, electric generators look a lot different than the figures in this section, but the principles are the same. The source of mechanical energy that turns the coil can be falling water (hydropower), steam produced by the burning of fossil fuels, or the kinetic energy of wind. Figure 5 shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator.
Generators illustrated in this section look very much like the motors illustrated previously. This is not coincidental. In fact, a motor becomes a generator when its shaft rotates. Certain early automobiles used their starter motor as a generator. In Chapter 23.6 Back Emf, we shall further explore the action of a motor as a generator.
# Section Summary
• An electric generator rotates a coil in a magnetic field, inducing an emf given as a function of time by
,
where is the area of an N-turn coil rotated at a constant angular velocity ω in a uniform magnetic field B.
• The peak emf emf0 of a generator is
emf0 = NAB ω
### Conceptual Questions
1: Using RHR-1, show that the emfs in the sides of the generator loop in Figure 4 are in the same sense and thus add.
2: The source of a generator’s electrical energy output is the work done to turn its coils. How is the work needed to turn the generator related to Lenz’s law?
### Problems & Exercises
1: Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field.
2: At what angular velocity in rpm will the peak voltage of a generator be 480 V, if its 500-turn, 8.00 cm diameter coil rotates in a 0.250 T field?
3: What is the peak emf generated by rotating a 1000-turn, 20.0 cm diameter coil in the Earth’s 5.00 x 10-5 T. magnetic field, given the plane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in 10.0 ms?
4: What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.)
5: (a) A bicycle generator rotates at 1875 rad/s, producing an 18.0 V peak emf. It has a 1.00 by 3.00 cm rectangular coil in a 0.640 T field. How many turns are in the coil? (b) Is this number of turns of wire practical for a 1.00 by 3.00 cm coil?
6: Integrated Concepts
This problem refers to the bicycle generator considered in the previous problem. It is driven by a 1.60 cm diameter wheel that rolls on the outside rim of the bicycle tire. (a) What is the velocity of the bicycle if the generator’s angular velocity is 1875 rad/s? (b) What is the maximum emf of the generator when the bicycle moves at 10.0 m/s, noting that it was 18.0 V under the original conditions? (c) If the sophisticated generator can vary its own magnetic field, what field strength will it need at 5.00 m/s to produce a 9.00 V maximum emf?
7: (a) A car generator turns at 400 rpm when the engine is idling. Its 300-turn, 5.00 by 8.00 cm rectangular coil rotates in an adjustable magnetic field so that it can produce sufficient voltage even at low rpms. What is the field strength needed to produce a 24.0 V peak emf? (b) Discuss how this required field strength compares to those available in permanent and electromagnets.
8: Show that if a coil rotates at an angular velocity ω, the period of its AC output is 2 π / ω.
9: A 75-turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.25 T field, starting with the plane of the coil parallel to the field. (a) What is the peak emf? (b) At what time is the peak emf first reached? (c) At what time is the emf first at its most negative? (d) What is the period of the AC voltage output?
10: (a) If the emf of a coil rotating in a magnetic field is zero at t=0, and increases to its first peak at t = 0.100 ms, what is the angular velocity of the coil? (b) At what time will its next maximum occur? (c) What is the period of the output? (d) When is the output first one-fourth of its maximum? (e) When is it next one-fourth of its maximum?
11: Unreasonable Results
A 500-turn coil with a 0.250 m2 area is spun in the Earth’s 5.00 x 10-5 T field, producing a 12.0 kV maximum emf. (a) At what angular velocity must the coil be spun? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible?
## Glossary
electric generator
a device for converting mechanical work into electric energy; it induces an emf by rotating a coil in a magnetic field
emf induced in a generator coil
emf = NAB ω sinωt, where A is the area of an N-turn coil rotated at a constant angular velocity ω in a uniform magnetic field B, over a period of time t.
peak emf
emf0 = NABω
### Solutions
Problems & Exercises
1: 474 V
3: 0.247 V
5: (a) 50 (b) yes
7: (a) 0.477 T (b) This field strength is small enough that it can be obtained using either a permanent magnet or an electromagnet.
9: (a) 5.89 V (b) At t=0 (c) 0.393 s (d) 0.785 s
11: (a) 1.92 x 106 radians/second . (b) This angular velocity is unreasonably high, higher than can be obtained for any mechanical system. (c) The assumption that a voltage as great as 12.0 kV could be obtained is unreasonable. |
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# Proving the chain rule
Proving the chain rule for derivatives.
The chain rule tells us how to find the derivative of a composite function:
$\frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]={f}^{\prime }\left(g\left(x\right)\right){g}^{\prime }\left(x\right)$
The AP Calculus course doesn't require knowing the proof of this rule, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.
## First, we would like to prove two smaller claims that we are going to use in our proof of the chain rule.
(Claims that are used within a proof are often called lemmas.)
### 1. If a function is differentiable, then it is also continuous.
Proof: Differentiability implies continuitySee video transcript
### 2. If function $u$ is continuous at $x$ , then $\mathrm{\Delta }u\to 0$ as $\mathrm{\Delta }x\to 0$ .
If function u is continuous at x, then Δu→0 as Δx→0 See video transcript
## Now we are ready to prove the chain rule!
Chain rule proofSee video transcript
## Bonus: We can use the chain rule and the product rule to prove the quotient rule.
The quotient rule tells us how to find the derivative of a quotient:
$\begin{array}{rl}\frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]& =\frac{\frac{d}{dx}\left[f\left(x\right)\right]\cdot g\left(x\right)-f\left(x\right)\cdot \frac{d}{dx}\left[g\left(x\right)\right]}{\left[g\left(x\right){\right]}^{2}}\\ \\ \\ & =\frac{{f}^{\prime }\left(x\right)g\left(x\right)-f\left(x\right){g}^{\prime }\left(x\right)}{\left[g\left(x\right){\right]}^{2}}\end{array}$
Quotient rule from product & chain rulesSee video transcript
## Want to join the conversation?
• In the proof of differentiability implies continuity, you separate the limits saying that the limit of the products is the same as the product of the limits. But the limit of x*1/x at zero cannot be divided as the limit of x times the limit of 1/x as the latter one does not exist.
I do understand that this works in this case because both the limits exist.
• Good! You have found where the theorem fails: that is the limit of the product is the product of the limit remains true only when both limits actually exist! Clearly if one of the limits fails to exist then any equality would be meaningless as you have argued above!!
• In the chain rule proof, at what point did you use the lemma 'differentiability implies continuity' (that you proved in a previous video)?
• He uses it when using the second lemma because the second lemma assumes the continuity.
(1 vote)
• In video 1, why is it okay to divide by (x-c)? Isn't that 0 when x -> c?
• Only if x = c. If it's approaching, then it's not quite equal yet.
• I didn't understand why if function u is continuous at x, then Δu→0 as Δx→0.
Can't a continuous function u(x) have a vertical line somewhere in its graph? In the graph of a vertical line, as the change in x approaches 0, the change in u(x) approaches infinity.
(1 vote)
• A function cannot contain a vertical line in its graph because a vertical line would imply that one input value is being mapped to multiple output values (which would violate the definition of a function).
• In the proof of the chain rule by multiplying delta u by delta y over delta x it assumes that delta u is nonzero when it is possible for delta u to be 0 (if for example u(x) =2 then the derivative of u at x would be 0) and then delta y over delta u would be undefined? This would be multiplying by 0 as we never proved that delta u was nonzero and only that it was approaching 0 in the proof above the chain ule proof. The limit of delta u as delta x approaches 0 is still 0 is delta u =0 but we then couldn’t multiply by delta u. How is it allowed to multiply delta y over delta x by delta u over delta u if we do not know that delta u is nonzero?
• In the second video at , shouldn't it be:
[lim x->c (u(x))] - u(c) ?
• Actually, with limit rules if you had lim(x-3) you could turn it into lim(x)-3, so Sal is using that property in reverse. He also says around that you can treat u(c) as a constant. Though I think it works even if you do it the way you explain.
• In the second video, we suppose that function is differentiable when rewriting the limit? Because continuity does not imply differentiability.
• There is no need to worry about differentiability in the second video because Sal never differentiates anything during the proof.
• It seems that using the same "reasoning" ("trick") used in the first video to demonstrate that differentiability implies continuity, we can prove statements which aren't (generally) true. Consider the following application to f(x) and an arbitrary number "a":
limit[x->c](f(x)-a)=
limit[x->c]((x-c)(f(x)-a)/(x-c))=
limit[x->c](x-c)limit[x->c](f(x)-a)=
0limit[x->c](f(x)-a)=
0
Is this a proof that limit of f(x) at an arbitrary "c" equals any arbitrary "a"?
(1 vote)
• The third step is wrong.
lim(𝑥 → 𝑐) [(𝑓(𝑥) − 𝑎)∕(𝑥 − 𝑐)] = ±∞ ∙ lim(𝑥 → 𝑐) [𝑓(𝑥) − 𝑎]
• Video 2 is supposed to prove that as the change in x APPROACHES 0, the change in u APPROACHES 0.
But at the end, Sal writes that the limit of the change in u is EQUAL to 0 as the change in x APPROACHES 0.
EQUAL and APPROACHES are not the same thing, are they? How can we conclude EQUALS from APPROACHING?
(1 vote)
• The sentences 'The limit as x goes c of Δu is 0' means the same thing as 'Δu approaches 0 as x approaches c'.
The quantity u is approaching 0, because the limit of u equals 0.
• in the chain rule video he says we need to assume y and u are differentiable at x, but doesn't y need to be differentiable at u(x) instead of at x? |
### 2D Rotational Symmetry
#### Lesson Objective
This lesson shows you the ideas behind 2D rotational symmetry and you will get to see a quick video demonstration on it.
When we rotate a shape about its center point, we may notice that at a certain angle, the rotated shape coincides with its 'not rotated' self (see picture).
When this happens, the shape is said to have rotational symmetry. This lesson will show you the ideas behind it.
You can proceed by reading the study tips first or watch the math video. You can try out the practice questions after that.
### Study Tips
#### Tip #1
When we rotate the triangle about its center point for 360o, we will notice that it fits onto itself for 3 times for every 120o rotation.
By definition, the number of times a shape fits onto itself when rotated is called the order of symmetry.
Hence, we can see that the order of symmetry for this triangle is 3.
The math video below will show you more on this visually.
#### Tip #2
Now, this shape will only fits onto itself for 1 time after it is been rotated for 360o. Hence, the order of symmetry is 1.
However, for any shape that has rotation symmetry of order 1, that shape is considered as not having any rotational symmetry.
Hence, this shape has no rotational symmetry.
### Math Video
#### Math Video Transcript
00:00:03.100
This is a quick demonstration on two-dimensional rotational symmetry.
00:00:08.150
Now, let's consider this card. when I rotate this card about this point, for 360 degrees, observe how many times this card fits onto itself.
00:00:20.060
Let's start. One. Two.
00:00:29.180
So, we can see that, the card fits onto itself for 2 times.
00:00:35.080
Hence, with this observation, we can say that this card has rotation symmetry of order, 2.
00:00:43.020
Alright, let's take another example. Consider this flower.
00:00:48.100
When I rotate this flower about this point, for 360 degrees, observe how many times this flower fits onto itself.
00:00:57.020
Let's start. one. two. three. four. five.
00:01:07.140
So, we can see that, the card fits onto itself for 5 times.
00:01:13.050
Hence, we can say that this flower has rotation symmetry of order, 5.
00:01:20.040
That is all for this lesson. Try out the practice question to further your understanding.
### Practice Questions & More
#### Multiple Choice Questions (MCQ)
Now, let's try some MCQ questions to understand this lesson better.
You can start by going through the series of questions on 2d rotational symmetry or pick your choice of question below.
#### Site-Search and Q&A Library
Please feel free to visit the Q&A Library. You can read the Q&As listed in any of the available categories such as Algebra, Graphs, Exponents and more. Also, you can submit math question, share or give comments there. |
What Are The 3 Consecutive Odd Integers?
Why can’t we write the form as 2n 1?
Multiplication of two odd numbers gives an odd number.
(2n-1)*(2n+1)= 4n²-1, when u divide this by 2 u get remainder as 1 so it’s odd.
So both represent an odd number, depending on the limits for n in an equation ( where many other things depend on n)you can write 2n-1 or 2n+1 to represent an odd number..
What are 2 consecutive integers?
Even and Odd Consecutive Integers Even consecutive integers are even numbers that follow each other in order. The easiest example would be 2, 4, 6, 8 and 10. 246, 248, 250 and 252 is another example of even consecutive integers.
What is an even integer?
An integer is even if it is divisible by two and odd if it is not even. For example, 6 is even because there is no remainder when dividing it by 2. By contrast, 3, 5, 7, 21 leave a remainder of 1 when divided by 2. Examples of even numbers include −4, 0, 82 and 178.
What 3 numbers add up to 51?
Three consecutive integers add up to 51. The three consecutive integers are 16,17 and 18.
What is the formula for sum of odd numbers?
If the number is starting from 1 then you can do simply n^2. for example: if n=6 like 1,3,5,7,9,11 then it’s sum will be 6^2=36. But if odd number series not start from 1 then formula will be sn=n/2(2a+(n-1)d).
How do you find the sum of even integers?
The sum of even natural numbers between 1 and 2n can be written as the sum of 2*i from i=1 to i=n. Factoring out the 2, it is equal to 2 times the sum of i from i=1 to n. Using the formula for the sum of consecutive numbers, we get that the sum of all even numbers from 1 to 2n equals n*(n+1).
What is the odd number?
more … Any integer (not a fraction) that cannot be divided exactly by 2. The last digit is 1, 3, 5, 7 or 9. Example: −3, 1, 7 and 35 are all odd numbers.
What is the sum of 3 consecutive odd numbers is 69?
Let x + 4 = the third number. can you finish it? The numbers are 21, 23, 25.
What are the consecutive odd integers?
Consecutive odd integers are odd integers that follow each other in sequence. You may find it hard to believe, but just like even integers, a pair of any consecutive odd integers are also 2 units apart.
How do you represent an odd integer?
You can represent an odd integer with the expression 2n+1, where “n” is any integer. What are three consecutive odd integers that have a sum of 63.
What do you notice about the sum of 3 consecutive numbers?
If you add any 3 consecutive numbers together it will always equal a multiple of 3, e.g.
How do you find consecutive numbers?
To represent consecutive numbers algebraically, let one of the numbers be x. Then the next consecutive numbers would be x + 1, x + 2, and x + 3. If the question calls for consecutive even numbers, you would have to ensure that the first number you choose is even.
What are four consecutive odd integers?
Correct answer: Explanation: Consecutive odd integers can be represented as x, x+2, x+4, and x+6. Our integers are 5, 7, 9, and 11.
How do you find odd consecutive integers?
If we start with an even number and each number in the sequence is 2 more than the previous number then we will get consecutive even integers. For example: 16,18, 20, … If we start with an odd number and each number in the sequence is 2 more than the previous number then we will get consecutive odd integers.
What are two consecutive odd integers?
If x is any odd number, then x and x + 2 are consecutive odd numbers. E.g. 7 and 9 are consecutive odd numbers, as are 31 and 33.
What is the difference between odd and even consecutive integers?
Odd consecutive numbers Odd consecutive integers are odd integers that follow each other. They have a difference of 2 between every two numbers. If n is an odd integer, then n, n+2, n+4 and n+6 will be odd consecutive integers. … In even consecutive numbers, odd numbers ( like 1,3,5,7,9 etc.) will not be included.
Can a sum of three odd numbers be even?
Without cheating, it’s not possible because sum of three odd numbers can’t be even. Let k be any real number. … 2k+1 will be an odd number & 2k will be an even number.
What are three consecutive odd integers?
Explanation: Three odd consecutive numbers can be written as x, x+2, and x+4. Now we find the next two odd numbers by adding 2 and 4 to this number. So your answer is 37, 39 and 41.
What are 3 consecutive numbers?
Numbers that follow each other continuously in the order from smallest to largest are called consecutive numbers. For example: 1, 2, 3, 4, 5, 6, and so on are consecutive numbers.
What are 4 consecutive even numbers?
That means the numbers will be 24, 26, 28, and 30. The average of four consecutive even numbers is 27. Let us denote the first number with x. |
# How to Change Any Number to a Percent, With Examples
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Understanding and calculating percentages can help you in many ways: working out the correct tip at a restaurant, knowing how much you are saving on that mega sale or allowing you to interpret data from mathematical and scientific research. In short, learning more about percentages is important for chemistry and all other areas.
A percentage is a way of expressing one number as a portion or share of a whole number, and percentages are always based on their relation to 100, which represents the whole number or object. For example, 75% is the same as 75 out of 100. Any percentage lower than 100 is just part of the whole or total.
Percentages are ratios, they can, therefore, be written as fractions and then decimals. Converting answers from percentages to fractions to decimals can be a good exercise to check the accuracy of your work.You can convert any number to a percent.
## Whole Numbers to Percents
To find a percentage of a whole, you first need to know exactly what you're measuring. For instance, if 38 people in your community have library cards, and you wish to work out the percentage of library card holders in the population, you'll first need to know the total size of your community.
A percentage of a total is equal to the subset of a total divided by the total, then multiplied by 100. If your community has 230 people in it, you can calculate the percentage of people with a library card. First, you divide 38 by 230, then multiply that total by 100.
38/230 = .165 .165 x 100 = 16.5
Therefore 16.5 percent of the people in your community have a library card.
## Fractions Into Percents
A fraction has two parts, a numerator on the top and a denominator on the bottom. Any percentage can be expressed as a fraction with the percentage value as the numerator and 100 as the denominator. So 80 percent is also 80/100. You can use this to convert a fraction into a percentage as well.
For example, to turn the fraction 4/25 into a percent, you'll need a fraction with 100 as the denominator. So you'll need to multiply the denominator by 4 since this will produce 100. You then need to also multiply the numerator by the same amount.
4/25 = (4 x 4)/(25 x 4) = 16/100
You now have a fraction of 100, so you can convert it to a percentage quite easily: 16/100 is equal to 16 percent. Here's another example:
3/5 = (3 x 20)/(5 x 20) = 60/100 = 60%
If you have access to a calculator, you could also divide the numerator by the denominator and turn the fraction into a decimal. Decimals are simple to convert into percent notation.
## Decimals Into Percents
Decimals are even easier to convert into percentages. To convert a decimal into a percentage, multiply the decimal by 100 and add the percent sign. To multiply a decimal by 100, move the decimal point two digits to the right:
0.8= 0.8 x 100 = 80% 0.53: 0.53 x 100 = 53% 0.173: 0.173 x 100 = 17.3%
You can also use this method to convert decimal numbers that are larger than 1. Say you know there are 1.34 times as many books in your library this year than last year, and you want to know what percentage the number of books you have this year is of the books you had last year. As before, you would multiply 1.34 by 100 and add the percent sign:
1.34 x 100 = 134%
Your number exceeds 100 percent because you're comparing this year's book total to the total books from last year, a number represented by 100 percent. Since there are more books this year, that percentage is larger than 100.
## Bonus: Percents Into Decimals
You can convert percentages to decimals using the opposite operation. Instead of multiplying the decimal by 100, divide the percent by 100. If 56 percent of people eat roast beef sandwiches you could also say that for every 100 people, 56 eat roast beef. Divide 56 by 100. Because you're dividing by 100, you'll move the decimal point two places to the left:
56/100 = 0.56
Other conversion examples are: 5%: 5/100 = 0.05, 79%: 79/100 = 0.79, 295%: 295/100 = 2.95 |
# Wikijunior:Introduction to Mathematics/Numbers
• Place Amount
• Numeral
## Lesson
### Normal Numbers
Normal numbers are the numbers you use most often. They show an amount of separate things. The picture under shows how normal numbers show how many of a thing. 0 (Zero) is a special number for when there is nothing.
$0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $20$ $100$
### Numerals and Place Amounts
A number is one or more signs that are called numerals (like a word is one or more signs that are called letters). There are ten numerals (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) that can make most numbers. Each numeral has a place in a number and shows a different amount when it is in a different place. The place amount is the amount a numeral shows in a place.
The first place (going from right to left) has a place amount of one, and it is the Ones Place. A numeral in the Ones Place shows the normal amount. The next place has a place amount of ten, and it is the Tens Place. Numerals in the Tens Place show how many groups of ten. The next place is the Hundred Place because one hundred is ten groups of ten. The next place is the Thousand Place because one thousand is ten groups of one hundred. A place amount is a group of ten of the place amount one place to the right.
$346289$7 7 ones $34628$9$7$ 9 tens $3462$8$97$ 8 hundred $346$2$897$ 2 thousand $34$6$2897$ 6 ten thousand $3$4$62897$ 4 hundred thousand 3$462897$ 3 million
If every number had a special sign, then you would need a big book of signs to know all the numbers. This system of numerals and place amounts is useful because ten signs can make a very many numbers.
### More, less, and equal
Often it is useful to look at two amounts and show which one is more, less or equal to the other. Math has special signs to show this.
The equals sign is two lines like this: " = ". Use it between two amounts that are the same. These are some examples:
• $5=5$
• $=$
• $3462897 = 3462897$
The more than sign is a bent line like this: " > ". Use it to show that an amount on the left of the sign is more than an amount on the right. These are some examples:
• $10>5$
• $>$
• $485>394$
The less than sign is a bent line like this: " < ". Use it to show that an amount on the left of the sign is less than an amount on the right. These are some examples:
• $2<3$
• $<$
• $3887 < 3902$
These are some signs people use less often:
• The not equal sign is this: " ". Use it between two amounts that are not the same. An alternative is " != ".
• The more or equal sign is this: " ". Use it to show that an amount on the left of the sign is more than or the same as an amount on the right. An alternative is " >= ".
• The less or equal sign is this: " ". Use it to show that an amount on the left of the sign is less than or the same as an amount on the right. An alternative is " =< ". |
# Types of Vectors: Definition & Properties of Different types of vectors
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### What is a Vector?
The vectors are specified as an object including magnitude plus direction. It depicts the movement of the object from one location to another. If we consider vectors as a straight line, then the extent of the line denotes the magnitude, and the pointer on this line is the direction in which the vector is traveling as shown below:
The vectors are named individually based on their features such as magnitude, direction, and their association with the different types of vectors. These different types of vectors help implement various arithmetic operations and calculations concerning vectors. Let us understand them one by one.
### Types of Vectors
There are 12 types of vectors that are as follows:
### 1. What is a Zero Vector?
A zero vector is a vector with zero magnitude. This implies that the starting point of the vector matches the final point. Such a vector has zero (0) magnitude and is denoted by 0. Consider an example of zero vector to understand the same.
If for a vector say $$\vec{PQ}$$, the coordinates of the point P lie at the same position as that of the point Q then the vector is declared to be a zero vector.
This reflects that the magnitude of the zero vector is always zero plus the direction for such a vector is indeterminate. Also, the vector does not aim in any direction. The zero vectors are known as null vectors.
### 2. What is a Unit Vector?
A unit vector is a vector whose magnitude is of unit length. If $$\vec{x}$$ is a vector whose magnitude is x, then unit vector of $$\vec{x}$$ in the direction of x is denoted by $$\widehat{x}$$ and is defined as$$\widehat{x}=\dfrac{\overrightarrow{x}}{\left| x\right| }$$.
Where $${\left| x\right| }$$ denotes the magnitude of vector x.
The measure of unit vectors is one(1). We should be careful that if two vectors are said to be unit vectors, then they don’t need to be equal. They might have an equal magnitude but can vary in their direction.
Learn about Sets and Complex Numbers here.
### 3. What is a Position Vector?
A position vector is a vector that symbolizes either the position or the location of any given point with respect to any arbitrary reference point like the origin. The direction of the position vector always points from the origin of that vector toward a given point.
Position vectors help us to determine the position along with the direction of movement of the vectors in a 3D dimension Cartesian system.
Consider O as the reference/ origin point and Y be an arbitrary point in space then the vector $$\vec{OY}$$ is known as the position vector of the point Y.
### 4. What is an Equal Vector?
Two or more vectors are said to be equal vectors when their magnitude is equal and also their direction is the same.
An equal vector is defined when two vectors or more than two vectors possess the same magnitude, as well as the same direction.
In the above diagram, the vectors $$\vec{XY}$$ and vector $$\vec{MN}$$ are equal as they both have the same magnitude as well as direction.
These vectors can have different initial and terminal points but their length and directions should be equal. You can also check MCQs on equal vectors.
### 5. What is a Negative Vector?
A negative vector can be defined when one vector is supposed to be the negative of another vector if they have equal magnitudes with opposite directions. Consider there are two vectors P and Q, such that these vectors have the same magnitude but are opposite in direction then these vectors can be presented by:
P = – Q
Also, P and Q are said to be the negative vector to one another.
### 6. What is a Collinear Vector?
A collinear vector can be defined when two or more than two vectors are parallel to one another irrespective of the magnitude or the direction. The parallel nature of vectors indicates that they never meet or intersect with each other. Consider the below image to understand the same.
Thus, we can estimate any two vectors as collinear vectors if and only if these two vectors are either along the identical line or the vectors are parallel to one another in the same direction/opposite direction. Therefore collinear vectors are also known as parallel vectors.
Also, learn about Vector Algebra here.
### 7. What is a Co-initial Vector?
Co-initial vectors come under the type of vectors wherein two or more than two separate vectors have alike/same initial points. This states that in this type of vector, all vectors begin from the same initial position i.e. the origin spot is identical for the vectors.
For example, if we consider two vectors namely $$\vec{AX}$$ and $$\vec{AY}$$ as shown below then these vectors are termed co-initial vectors as they both possess a similar initial point that is A.
### 8. What are Like and Unlike Vectors?
Like vectors are vectors that have the same direction. On the other hand, if the vectors possess opposite directions w.r.t to one another then they are said to be unlike vectors.
### 9. What is a Co-Planar Vector?
Co-planar vectors are the type of vectors where three or more than three vectors rest in the same plane or lie in the parallel plane.
### 10. What is a Displacement Vector?
A displacement vector can be defined if a given point is displaced from position say P to Q then the displacement PQ draws a vector $$\vec{PQ}$$.
### 11. What is an Orthogonal Vector?
Two or more than two vectors in space are considered to be orthogonal if the angle between them is 90 degrees.
### 12. What are Concurrent vectors?
Concurrent vectors are those types of vectors that pass through the same point. This implies that a concurrent vector arrangement is a set of two or more vectors whose lines of action meet at a point.
Now, you have learned about the types of vectors. Try to attempt Types of Vector MCQs.
### Properties of Different Types of Vectors
We can apply different mathematical operations to vectors such as addition, subtraction, and multiplication. The different properties of vectors are listed below:
• Addition of vectors is commutative and associative
• Dot Product of two vectors is a scalar and lies in the plane of the two vectors.
• Cross product of two vectors is a vector, which is perpendicular to the plane containing these two vectors.
• $$\vec{A}$$.$$\vec{B}$$ = $$\vec{B}$$.$$\vec{A}$$
• $$\vec{A}$$ × $$\vec{B}$$ ≠ $$\vec{B}$$ × $$\vec{A}$$
• $$\widehat{i}$$.$$\widehat{i}$$ = $$\widehat{j}$$.$$\widehat{j}$$ = $$\widehat{k}$$.$$\widehat{k}$$ = 1
• $$\widehat{i}$$.$$\widehat{j}$$ = $$\widehat{j}$$.$$\widehat{k}$$ = $$\widehat{k}$$.$$\widehat{i}$$ = 0
• $$\widehat{i}$$ × $$\widehat{i}$$ = $$\widehat{j}$$ × $$\widehat{j}$$ = $$\widehat{k}$$ × $$\widehat{k}$$ = 0
• $$\widehat{i}$$ × $$\widehat{j}$$ = $$\widehat{k}$$; $$\widehat{j}$$ × $$\widehat{k}$$ = $$\widehat{i}$$; $$\widehat{k}$$ × $$\widehat{i}$$ = $$\widehat{j}$$
• $$\widehat{j}$$ × $$\widehat{i}$$ = -$$\widehat{k}$$; $$\widehat{k}$$ × $$\widehat{j}$$ = -$$\widehat{i}$$; $$\widehat{i}$$ × $$\widehat{k}$$ = -$$\widehat{j}$$
Summing up: A vector in science, math or physics points towards the carrier and is applied to represent physical quantities such as displacement, velocity, acceleration, force etc. Moreover, vectors have various applications in physics as well as engineering. These are used to depict physical objects as they hold both magnitudes and directions.
Stay tuned to the Testbook App or visit the testbook website for more updates on similar topics from mathematics, science, and numerous such subjects, and can even check the test series available to test your knowledge regarding various exams.
If you are checking Types of Vector article, also check the related maths articles in the table below: Lagrang’s mean value theorem Domain of a function Transitive relations Modulus function Types of functions Exponential functions
### Types of Vectors FAQs
Q.1 What is a vector?
Ans.1 A vector can be interpreted by a line with a pointer tending towards its direction, and its length outlines the magnitude of the vector.
Q.2 What are Types of Vectors?
Ans.2 The various types of vectors are zero vector, unit vector, co-initial vector, position vector, like and unlike vector, collinear vector, equal vector, coplanar vector, displacement vector, negative vector, etc.
Q.3 How do we define a collinear vector?
Ans.3 When two or more than two vectors are parallel to one another irrespective of the magnitude or the direction are said to be collinear vectors.
Q.4 For two vectors to be equal they should have the?
Ans.4 For two vectors to be equal, they should have the same magnitude and direction. |
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# Combined Operations Problems
Have you tried to do our combined operations problems?
When we do a problem, there are times when there is no need to do any operation and other times when you have to do one or many.
Today we are going to see these problems where various operations are necessary in order to arrive at a solution. We call them combined operations problems, and we are going to order them according to the operations that need to be done:
### 1. Additive Structure Problems
In this type, we include all the problems that involve doing addition, subtraction, or addition and subtraction. For example:
Yesterday, Thomas bought a t-shirt for $15 and a backpack for$23, but they gave him a discount and, in total, he only spent $35. How much of a discount did they give him? We can solve the problem directly, using the following operation: They gave him a discount of$3.
Or we can form an intermediary question and answer it first to make it easier for ourselves:
How much would Thomas have paid if he did not get the discount?
Without the discount, he would have paid $38. They gave him a discount of$3.
### 2. Multiplicative Structure Problems
In these types of problems, we include all of the problems that involve doing multiplication, division, or multiplication and division. For example:
At the amusement park, we rode ”The Crazy Wheel,” which is very fun. The security guard told us that the wheel has run 40 times today and it has always been full, carrying 5 children each time. Another attraction, “The Purple Dragon”, has carried 3 times as many children as “The Crazy Wheel.” How many children have ridden on “ The Purple Dragon?”
Let’s do the same as in the previous one.
If we know how to respond to the question directly, we can do the next operation:
In total, 600 children have ridden.
If not, we came up with an intermediary question to make it easier:
How may children have ridden on “The Crazy Wheel”?
And solve operation by operation:
200 children have ridden “The Crazy Wheel.”
600 children have ridden “The Purple Dragon.”
### 3. Mixed Structure Problems
In these problems, there is a mix of additive structure operations (addition and/or subtraction) and multiplicative structure operations (multiplication and/or division).
For example, the following problem:
The pirate Silver Beard told me that he has found a treasure on a deserted island that had a total of 3000 gold coins spread equally across three chests. Additionally, in every chest, there were also 200 silver coins and twice as many bronze coins than silver. How many coins were there in total in each chest?
Now it is much more difficult to calculate the solution all at once, right?
For that we are going to come up with and answer the necessary intermediary questions:
How many gold coins were in each chest?
In each chest, there were 1000 gold coins.
How many bronze coins were in each chest?
In each chest, there were 400 bronze coins.
Now we are able to answer the last question:
How many coins were there in total in each chest?
In total, each chest had 1600 coins.
Remember that in order to learn more about combined operations problems and the rest of the primary math problems, you can register now for Smartick free trial.
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# Lesson 18
Using Long Division
Let’s divide whole numbers.
### Problem 1
Andre and Jada both found $$657 \div 3$$ using the partial quotients method, but they did the calculations differently, as shown here.
1. How is Jada's work the same as Andre’s work? How is it different?
2. Explain why they have the same answer.
### Problem 2
Here is a long-division calculation of $$917 \div 7$$.
1. There is a 7 under the 9 of 917. What does this 7 represent?
2. What does the subtraction of 7 from 9 mean?
3. Why is a 1 written next to the 2 from $$9-7$$?
### Problem 3
Han's calculation of $$972 \div 9$$ is shown here.
1. Find $$180 \boldcdot 9$$.
2. Use your calculation of $$180 \boldcdot 9$$ to explain how you know Han has made a mistake.
3. Identify and correct Han’s mistake.
### Problem 4
Find each quotient.
### Problem 5
The mass of one coin is 16.718 grams. The mass of a second coin is 27.22 grams. How much greater is the mass of the second coin than the first? Show your reasoning.
(From Unit 3, Lesson 15.)
### Problem 6
One micrometer is a millionth of a meter. A certain spider web is 4 micrometers thick. A fiber in a shirt is 1 hundred-thousandth of a meter thick.
1. Which is wider, the spider web or the fiber? Explain your reasoning.
2. How many meters wider?
(From Unit 3, Lesson 15.) |
# How do you write the partial fraction decomposition of the rational expression (3x + 2) / [(x - 1)(x + 4)]?
Jan 2, 2016
$\frac{3 x + 2}{\left(x - 1\right) \left(x + 4\right)} = \frac{1}{x - 1} + \frac{2}{x + 4}$
#### Explanation:
Since the denominator already in the factor form we can rewrite it as
$\frac{3 x + 2}{\left(x - 1\right) \left(x + 4\right)} = \frac{A}{x - 1} + \frac{B}{x + 4}$
Multiply both sides by the LCD to get
$3 x + 2 = A \left(x + 4\right) + B \left(x - 1\right)$
Multiple the expression to get
$3 x + 2 = A x + 4 A + B x - B$
Set up the system of equation like this
$x : \text{ " " } A + B = 3$
${x}^{0} \text{ " " } 4 A - B = 2$
Solve system by elimination method
$A + B = 3$
$4 A - B = 2$
$5 A \text{ " = 5}$
$\implies A = 1$
Solve for B
$\left(1\right) + B = 3 \implies B = 2$
Partial fraction of $\frac{3 x + 2}{\left(x - 1\right) \left(x + 4\right)} = \frac{1}{x - 1} + \frac{2}{x + 4}$ |
How will Tom Brady fare on 02/15/2020 and the days ahead? Let’s use astrology to perform a simple analysis. Note this is just for fun – don’t get too worked up about the result. I will first find the destiny number for Tom Brady, and then something similar to the life path number, which we will calculate for today (02/15/2020). By comparing the difference of these two numbers, we may have an indication of how well their day will go, at least according to some astrology practitioners.
PATH NUMBER FOR 02/15/2020: We will analyze the month (02), the day (15) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. We’ll show you how it works now. First, for the month, we take the current month of 02 and add the digits together: 0 + 2 = 2 (super simple). Then do the day: from 15 we do 1 + 5 = 6. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 2 + 6 + 4 = 12. This still isn’t a single-digit number, so we will add its digits together again: 1 + 2 = 3. Now we have a single-digit number: 3 is the path number for 02/15/2020.
DESTINY NUMBER FOR Tom Brady: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Tom Brady we have the letters T (2), o (6), m (4), B (2), r (9), a (1), d (4) and y (7). Adding all of that up (yes, this can get tiring) gives 35. This still isn’t a single-digit number, so we will add its digits together again: 3 + 5 = 8. Now we have a single-digit number: 8 is the destiny number for Tom Brady.
CONCLUSION: The difference between the path number for today (3) and destiny number for Tom Brady (8) is 5. That is larger than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t get too worked up about that! As mentioned earlier, this is just for fun. If you want to see something that we really strongly recommend, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
Reciprocal and division of fractions space two various methods. When the numerator and also denominator that a portion are interchanged, then it is stated to it is in it’s reciprocal. Suppose a fraction is a/b, climate it’s reciprocal will certainly be b/a. A fraction is a numerical amount that is not a entirety number. Rather it to represent a component of the whole. Because that example, it speak how numerous slices of a pizza are continuing to be or eat of the entirety pizza, such together one-half (½), three-quarters (¾) etc. Division of fountain is an procedure performed top top fractions v multiple steps. Also, learn separating fractions here.
You are watching: What is the reciprocal of 3/4
Parts of FractionThe fraction has two parts:
NumeratorDenominator.
Types the Fraction: Fractions room basically of three types, proper, improper and also mixed. Discover the interpretations below.
Proper Fraction: If both the numerator and denominator room positive, and also the molecule is less than the denominator, then it is a suitable fraction.
Example: 2/5, 1/3, 3/6, 7/8. 9/11, etc.
Improper Fractions: Fractions having actually numerator greater than the denominator are dubbed Improper fractions.
Example: 8/3, 3/2, 6/3, 11/9, etc
Mixed Fraction: When a totality number and a proper fraction are combined, that is well-known as a mixed fraction.
All these details to be the basics that fractions. Now let us discover reciprocal that fractions together with its division.
## Reciprocal of Fractions
The portion obtained by swapping or interchanging Numerator and also Denominator v each various other is known as mutual of the offered fraction.
For example, a reciprocal of 5 is 1/5, a reciprocal of 8/3 is 3/8.
The mutual of a mixed portion can be acquired by convert it into an improper fraction and climate swap the numerator and also denominator.
For example, to find the mutual of (small 2frac13);
Convert the mixed portion into wrong fraction:(small 2frac13=frac73)Now invert the fraction: 7/3 and 3/7, whereby 3/7 is referred to as reciprocal the 7/3 or (small 2frac13).
Note: The product that a fraction and it’s mutual is always 1.
Also, read:
## Division the Fractions
Division entailing a portion follows specific rules. To carry out any division involving fraction just main point the very first number through the reciprocal of the second number. Actions are together follows:
Step 1: an initial change the department sign (÷) to multiplication sign (×)
Step 2: If we change the authorize of division to multiplication, in ~ the exact same time we have to write the mutual of the second term or fraction.
Step 3: Now, main point the numbers and also simplify the result.
These rule are typical for:
Division the the entirety number through a fraction.Division the a portion by a whole numberDivision that a fraction by one more fraction.
Note: that is to be provided that division of fountain is usually the multiplication of portion obtained by reciprocal of the denominator (i.e. Divisor).
### Examples of divisions of Fractions
Examples for each the the condition as mentioned previously are defined below.
Division the the entirety Number by a Fraction
Example 1: 16 ÷ 4/3
Solution: 16 ÷ 4/3 = 16/1 × 3/4
3/4 is the reciprocal of 4/3.
Hence, (16 × 3)/(1×4)
4 × 3 = 12
Therefore,
16 ÷ 4/3 = 12
Division of a portion by a totality Number
Example 2: Divide 8/3 by 3
Solution: We need to simplify, 8/3 ÷ 3
The reciprocal of 3 is 1/3.
Now creating the provided expression right into multiplication form,
8/3 × 1/3 = 8 /9
Therefore,
8/3 ÷ 3 = 8/9
Division of a fraction by another Fraction
Example 3: 8/3 ÷ 4/3
Solution: 8/3 ÷ 4/3
Reciprocal of second term 4/3 is 3/4.
Now main point the very first term with the reciprocal of the second term.
8/3 × 3/4 = 8/4 = 2
Hence,
8/3 ÷ 4/3 = 2
To perform department involving blended fraction, transform the mixed portion into one improper portion and follow the over steps.
Study more on the connected topics such together representing fractions on a number line at BYJU’S today!
The reciprocal of a portion will acquire by interchanging the numerator and denominator. For example, y/x is the mutual of the fraction x/y, i.e. Y/x = 1/(x/y).
When separating fractions by entirety numbers, we should transform the department into multiplication by composing the reciprocal of the divisor, i.e. A totality number. For example, separating 2/3 through 2 deserve to be performed by converting together (2/3) × (1/2). Hence, the simplification becomes basic now.
To simplify the department process when dividing fractions, reciprocals are provided so that division will be converted to multiplication. For example, (4/5) ÷ (8/7) can be composed as (4/5) × (7/8).
See more: Be-19 What Must A Type Iii Marine Sanitation Device Have When Boating On Inland Waters?
The reciprocal rule of division method is “Multiply the dividend by the reciprocal of the divisor”. In simple words, invert the divisor and also multiply through the dividend. |
## Wednesday, November 11, 2009
### Vedic Mathematics Lesson 26: Solving Equations 1
Solving equations is an essential skill that is an integral part of Algebra. In this lesson, we will learn about some patterns of equations and formulae for solving them, so that instead of deriving the solution from first principles, we can apply these simple formulae to get the answer right away.
You can find all the previous posts about Vedic Mathematics below:
Multiplication Special Case 3
Vertically And Crosswise I
Vertically And Crosswise II
Squaring, Cubing, Etc.
Subtraction
Division By The Nikhilam Method I
Division By The Nikhilam Method II
Division By The Nikhilam Method III
Division By The Paravartya Method
Digital Roots
Straight Division I
Straight Division II
Vinculums
Divisibility Rules
Simple Osculation
Multiplex Osculation
Why is the ability to identify patterns of equations, and solve them by applying formulae important? After all, algebra is about deriving the answer by isolating the unknown variable and equating it to the solution. Consider the case of a generic quadratic equations of the form ax^2 + bx + c = 0. Students are initially taught how to solve such equations by separating b into two parts such that the parts b1 and b2 add up to b, and the ratios a/b1 and b2/c are equal.
That procedure will then lead to the formation of an equation of the form below:
dx(ex + f) + g(ex + f) = 0
which in turn is then simplified to the form:
(dx + g)(ex + f) = 0.
This then gives us the solutions -g/d and -f/e.
Afterwards, we find that this process only works when it is easy to split b up according to the rules. When the required split of b involves fractions and decimals, it is not easy to derive b1 and b2 by trial and error without a lot of effort.
At that point, one may have been taught the derivation of the standard quadratic formula which involves the process of "completing the square". The end result of this derivation is a formula that is etched deep in most high-schoolers' brains.
Going forward, when encountering a quadratic equation that needs to be solved, students are expected to use the formula directly rather than going through the process of justifying the use of it by deriving the formula from first principles. In fact, the formula takes the place of going through the process of trying to isolate the unknown variable in order to solve the equation. The formula is a time-saving device that tells us that if and when we go through the process of isolating x (if we actually want to do it that way), we will end up with solutions that are predicted by this formula. The use of the formula is now so widespread that most people who are perfectly capable of solving quadratic equations can not derive the quadratic formula from scratch!
In this lesson, we will deal with four simple types of equations for which one can derive formulae similar to the quadratic formula. Going forward, one can then use these formulae directly instead of either deriving them or going through the steps involved in isolating the unknown variable following first principles. This will not only save us lots of time and effort, but also reduce the probability of errors during the process of isolation itself. Just as the quadratic formula is used widely to solve quadratic equations, the formulae in this lesson should be taught and used widely to solve equations instead of expecting students and others to go through the process of deriving them from first principles.
Note that all the simple forms of equations are provided in their most general form. Many equations may be missing some elements, which makes their identification and classification into one of these general forms a little tricky, but, with practice, it should become easier. As always, the application of the formulae itself will become much easier (and can easily be done mentally) with practice.
All these formulae are derived primarily by using a Vedic Sutra that reads "Paravartya Yojayet". The literal translation of this sutra is "transpose and adjust". This is a fairly general phrase that covers pretty much all that we do to solve simple equations.
For instance, given an equation such as 2x - 4 = 0, we "transpose" the -4 to the other side and get 2x = 4. We then "adjust" the coefficient of x to isolate it and get x = 4/2, which is the same as x = 2.
In fact, the example above is a simple form of a more general type of simple linear equation for which a formula is readily available to derive the solution. The general form is: ax + b = cx + d. This is the first type of simple equations we will solve using the Paravartya Yojayet sutra. The solution after the necessary transpositions and adjustments is x = (d-b)/(a-c).
So, for instance, if one were to encounter an equation such as 13x + 11 = 3x - 6, one should be able to solve it instantly using the above formula: x = (-6 - 11)/(13 - 3) = -1.7. No actual transposing or adjusting needs to be done from first principles to derive the solution. That is the power of using a formula, just the same way as use of the quadratic formula allows us to solve a quadratic equation without completing the square, transposing or adjusting anything.
Similarly, the solution of the equation 4x - 1 = 3x + 5 is simply x = (5 + 1)/(4 -3) = 6. What about 2x - 4 = 0? In this case, we simply have to recognize that c and d are both zero. We are left with a = 2 and b = -4. Substituting these values in the formula readily gives us the solution x = -(-4)/2 = 2.
How about an equation such as x + 2 = x + 5? Obviously, it is possible to identify on sight that this is not a valid equation with a valid solution since 2 is not equal to 5. Applying the formula to this equation validates this by giving us a solution of x = (5 - 2)/(1 - 1). Since (1 - 1) = 0, the solution involves a division by zero, which means that the solution is undefined, as it should be!
The second type of simple equation is of the form (ax + b)/(cx + d) = p/q. After one goes through the first principles process of cross-multiplying, transposing and isolating x, we would get the solution as x = (pd - bq)/(aq - cp). It is now possible to use this formula directly without having to go through the labor involved in deriving it whenever we encounter an equation of the above variety.
Thus, for instance, if we need to solve (3x + 2)/(x - 5) = 1/3, we would simply use the formula and say that x = (1*(-5) - 2*3)/(3*3 - 1*1) = -11/8.
Similarly, if (3x + 5)/(2x + 1) = 2, notice that this is actually an equation of the second type, even though the right hand side is not in the form of a fraction. In this case, p/q = 2, which can be rewritten as 2/1. This simply means that p = 2 and q = 1. Using these values, we should be able to solve the equation mentally using our formula, and get x = (2*1 - 5*1)/(3*1 - 2*2) = 3.
What about an equation such as (2x + 1)/x = 3/4? We have to recognize, once again, that this is an equation of the second type which simply has d = 0. Thus we have a = 2, b = 1, c = 1, d = 0, p = 3 and q = 4. Plugging these values into the formula x = (pd - bq)/(aq - cp), we readily get x = (3*0 - 1*4)/(2*4 - 1*3) = -4/5.
Now consider the equation 2(4x + 2) = 4(x - 5). One can either convert it into an equation of the first form by expanding out the multiplications to get 8x + 4 = 4x - 20. We can then apply the formula for the first type and get x = -6. One can also manipulate the equation into one of the second form and get (4x + 2)/(x - 5) = 4/2. Now applying the formula for the second form of equations gives us the same solution, x = -6. In fact, equations of the first and second forms can easily be transformed back and forth between the two types, and depending on the state they are in at any given moment, one can use the appropriate formula to solve them.
What happens when we try to solve (3x + 2)/(x + 3) = 3? Once again, if we cross-multiply the terms, we get a simpler-to-analyze linear equation that reads 3x + 2 = 3x + 9. This is obviously a meaningless equation with no well-defined solution. Applying the formula to this situation (after setting q = 1), gives us x = (3*3 - 2*1)/(3*1 - 1*3). The denominator becomes zero, alerting us immediately that the equation is not a valid, meaningful one with any valid solutions.
The third type of equations looks a lot scarier, but is equally easy to deal with once the formula has been derived and is ready for use. These are equations of the form (x + a)(x + b) = (x + c)(x + d). The reason this may look scary at first glance is because it looks as if this should lead to a quadratic equation once the terms have been expanded out. But in reality, the x^2 terms on either side of the equation cancel out, leaving us with a simple linear equation.
After expanding out the terms and transposing and adjusting, we can derive the solution to this equation as x = (cd - ab)/(a + b - c - d). And, in the future, we can use this formula to solve all such equations instead of going through the laborious process of expanding out the terms and solving the equation from first principles.
Thus, if (x + 2)(x + 6) = (x + 1)(x+3), then x = (1*3 - 2*6)/(2 + 6 - 1 - 3) = -9/4. Isn't it amazing that this is all there is to solving an equation which would take at least a few minutes just to expand out to its linear form?
What about (x + 1)(x + 4) = (x + 2)^2? Here, we need to understand that (x + 2)^2 is just a shorthand for writing (x + 2)(x + 2), which should then enable us to correctly identify and classify this equation as belonging to the third type. After that, it is easy to solve it and get the solution as x = (2*2 - 1*4) / (1 + 4 - 2 - 2) = 0.
Now, let us see what happens when we try to solve an equation such as (x + 1)(x + 3) = (x + 2)(x + 2). It is tricky to identify at a glance that this equation is actually one of those meaningless, invalid equations with no solution. But our formula can tell us this right away because (a + b - c - d), which is the denominator in our formula, becomes zero. At this point, out of curiosity, we may choose to expand out the equation to its linear form. We then find that it actually reads 4x + 3 = 4x + 4. This is obviously meaningless, but the fact could be well-concealed by the way the original equation is written. The formula, though, alerts us to the problem right away even if we don't choose to expand out the equation to its linear form.
This third form can be generalized to an equation of the form (ax + b)(cx + d) = (ex + f)(gx + h). This equation simplifies to a linear form ONLY WHEN a*c = e*g. When that condition is true, we can then do the necessary expansion of the terms, transpositions and adjustments to derive the formula for the solution as: x = (fh - bd)/(ad + bc - eh - fg).
For instance, (2x + 3)(3x + 2) = (6x + 1)(x + 8) can be solved easily using this formula to give us x = -1/18. However, the formula is not applicable to (x + 5)(2x + 3) = (3x + 5)(2x -3) because a*c = 2, and e*g = 6, and therefore, a*c is not equal to e*g.
This generalized form can be used to solve certain equations that may not look like they fall into the pattern covered by this type of equation. Consider the equation (3x + 4)/(x + 5) = (6x + 1)/(2x + 2). This looks complicated and does not seem to fall under this category of equations, but once you cross-multiply the terms, you get (3x + 4)(2x + 2) = (6x + 1)(x + 5). Since a*c = e*g, we can actually solve this equation using the general formula to get x = 3/17.
The fourth type of equation we will deal with in this lesson is of the form (p)/(x + a) + (q)/(x + b) = 0. This also looks like a pretty messy type of equation that would take some time to unwind to a simple form that is readily amenable to solution.
If we do go through the process though, we will eventually find that the solution to this equation is simply x = (-pb - qa)/(p + q). Obviously, this is much easier to deal with than going through the laborious, time-consuming and error-prone process of deriving the solution from first principles!
Applying this method to 3/(x + 3) + 2/(x + 2) = 0, for instance, we instantly get the answer as (-3*2 - 2*3)/(3 + 2) = -12/5. Taking another example, we can solve 1/(x + 5) - 2/(x + 1) = 0 and get x = (-1*1 + 2*5)/(1 - 2) = -9.
Let us take the example now of 3/x - 2/(x - 2) = 0. This is actually an equation of the fourth type with a = 0 and b = -2. Substituting these values into the formula gives us the solution x = (-3*(-2) - 0*(-2))/(3 - 2) = 6.
With equations of this type, it can become harder to identify meaningless equations at first glance, but the formula can provide us instant confirmation as to whether a particular equation is valid or not. For instance take the case of 3/(x + 9) - 3/(x - 5) = 0. The formula instantly tells us that since p + q = 3 - 3 = 0, this equation has no valid solutions.
Similarly, take the equation 2/x - 3/x = 0. Here the solution actually works out to zero because both a and b are zero, and therefore -pb - qa = 0. However, at first glance this may seem counter-intuitive since 2/0 - 3/0 is actually undefined. However, if we expand out the equation to linear form first, we actually see that the equation becomes 2x = 3x. At this point, we can tell that zero is indeed the correct solution to the problem.
The fourth type of equation can also be expanded to a more general form. In this general form, p/(ax + b) + q/(cx + d) = 0. The solution to this general form of the equation can be derived as x = (-pd - qb)/(pc + aq). Using this more general form of the formula, one can solve the equation 1/(3x + 4) = 3/(5x + 1) to get x = -11/4 (make sure to note that the first step in applying the formula is to bring the right hand term of the equation to the left hand side to get the standard form of the equation on which the formula is based. This makes q = -3, not q = 3. The astute reader may have also noticed that it is easy to transform equations of the fourth form into ones of the first form and vice versa also).
Thus, not only do these formulae allow us to solve complex-looking equations on sight, mentally, they also allow us to identify when equations have no valid solutions, and sometimes they give us the right solution even when the solution is not intuitive. Just as the quadratic formula is used as soon as quadratic equation presents itself, without regard for first principles, these four types of linear equations should be identified on sight and the appropriate formula should be applied to solve them, without regard for first principles. We are not really ignoring first principles when we do this, it is just that we have taken the labor out of applying the first principles by deriving the end result directly.
The table below summarizes the different types of equations and their solutions into a handy cheat-sheet:
Equation Form Solution ax + b = cx + d x = (d – b)/(a – c) (ax + b)/(cx + d) = p/q x = (pd – bq)/(aq – cp) (x + a)(x + b) = (x + c)(x + d) x = (cd – ab)/(a + b – c – d) (ax + b)(cx + d) = (ex + f)(gx + h) Valid only if a*c = e*g x = (fh – bd)/(ad + bc – eh – fg) p/(x + a) + q/(x + b) = 0 x = (–pb – qa)/(p + q) p/(ax + b) + q/(cx + d) = 0 x = (–pd – qb)/(pc + aq)
Obviously, identifying the equations by type, and learning to apply the appropriate formula mentally and quickly, comes only with practice. Hope you will spend some time practicing with these different types of equations. As the examples illustrate, one sometimes has to manipulate the given equation in some simple fashion (cross-multiply, move a term from the left hand side to the right hand side or vice versa, etc.) to get it into a form covered by the types above, before the appropriate formula can be applied to the equation. It takes some practice to identify these manipulations also. Good luck, and happy computing!
Anonymous said...
Yes, why try to actually understand the process of solving linear equations when we can just memorize a huge list of forms? If the problem gets more complicated, not to worry, I'll just have to make an even bigger list of forms and memorize those.
What if I accidentally forget the exact form of one of the solutions? Well, I can't reconstruct it the solution for myself, since I have wisely devoted all my time to memorizing a list rather than understanding the process. No problem, I'll just have to memorize them better next time!
Math is easy when it's reduced to memorizing a massive (infinitely long) list! Why didn't anybody ever think of this before?
Now, if only I could find a list to help me solve equations of the type:
y''(t) + f(t)*y'(t) + g(t)*y(t) = h(t)
with appropriate boundary/initial conditions of course. I could waste time trying to understand what this equation means and therefore develop powerful methods for solving these equations, but it would be much better to just have a list and practice rote memorization until I go blind.
Blogannath said...
I am sure you have never used the quadratic formula, ever in your life (and would not consider using it in the future either), but instead go through the process of completing the squares or deriving the quadratic formula every time you need to solve a quadratic equation. Good for you!
I also hope that your "thorough understanding" of mathematics because of your unwillingness to use any formulae has lead to several great and exciting discoveries in mathematics, and has put you in line for a Fields Medal. Congratulations! Thanks for your gracious visit!!
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Bonds to 100 – Tens Homework Extension Year 2 Addition and Subtraction | Classroom Secrets
Maths Resources & WorksheetsYear 2 Maths LessonsAutumn Block 2 (Addition and Subtraction)05 Bonds To 100 – Tens › Bonds to 100 – Tens Homework Extension Year 2 Addition and Subtraction
# Bonds to 100 – Tens Homework Extension Year 2 Addition and Subtraction
## Step 5: Bonds to 100 - tens Homework Extension Year 2 Autumn Block 1
Bonds to 100 - tens Homework Extension provides additional questions which can be used as homework or an in-class extension for the Year 2 Bonds to 100 - tens Resource Pack. These are differentiated for Developing, Expected and Greater Depth.
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### What's included in the pack?
This pack includes:
• Bonds to 100 - tens Homework Extension with answers for Year 2 Autumn Block 2
#### National Curriculum Objectives
Differentiation:
Questions 1, 4 and 7 (Varied Fluency)
Developing Draw the missing piece of equipment and fill in the missing numbers so that the pictorial and number sentences match. One missing number. Using known facts for 10 and applying to numbers within 100.
Expected Draw the missing piece of equipment and fill in the missing numbers so that the pictorial and number sentences match. Some missing numbers. Using known facts for 10 and applying to numbers within 100.
Greater Depth Draw the missing piece of equipment and fill in the missing numbers so that the pictorial and number sentences match. Mostly missing numbers. Using known facts for 10 and applying to numbers within 100.
Questions 2, 5 and 8 (Varied Fluency)
Developing Shade the number frame to match the numbers given in digits. One part of the bond already shaded on the frame. Multiples of 10 within 100, increasing by 10.
Expected Shade the number frame to match the numbers given in words. Multiples of 10 within 100, increasing by 10.
Greater Depth Shade the number frame to match the numbers given in words. Decreasing in multiples of 20 from a multiple of 5.
Questions 3, 6 and 9 (Reasoning and Problem Solving)
Developing Calculate the value of 4 symbols using number bonds for multiples of 10. Clues all given as addition statements. Numbers represented pictorially.
Expected Calculate the value of 4 symbols using number bonds for multiples of 10. Two addition and two subtraction statements given as clues. Numbers given in digits.
Greater Depth Calculate the value of 4 symbols using number bonds for multiples of 5. Clues involve addition or subtraction statements with 2 or more symbols involved. Numbers given in digits.
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Trust your intuition. Did you know that even 4- and 5-year-olds can develop math intuition? One way is by building a sense of equality (the same) and inequality (more). And you don’t have to wait until they know how to count to teach these comparisons!
This post will give you some strategies and ideas for teaching students visual measurement, so they can compare without counting. Stick around until the end of the post for FREE printables you can use in your classroom!
Corresponding Common Core Standards: K.CC.C.6, K.CC.B.4.A
## Step 1: Understanding equality and one-to-one matching
How can you teach students to compare when they can’t yet count the objects? Begin with a concrete approach using manipulatives. Build a tower using 2-5 uniform cubes (same size, shape, and color) and ask students to build a tower with “the same number”:
They can easily compare height to arrive at equality without having to count.
After students respond, reinforce the concept of “the same number” and the one-to-one match of cubes by comparing each cube of the two towers. Point with two fingers and say, “Same, same, same….”
If they respond incorrectly, you’ll get to a single cube and say, “…different.” For example, at Happy Numbers, we respond to incorrect answers by illustrating this difference:
Soon, students can take over the role of comparing aloud. This approach helps them move toward a counting comparison. They learn that the towers are the same height because they are made up of the same number of cubes. It also reinforces the vocabulary of “the same” and “different.”
## Step 2: Understanding inequality
Next, introduce the vocabulary “more.” By using familiar, uniform cube towers, you’ll build upon existing knowledge through scaffolding. Build two towers of differing height and ask which has “more” cubes:
Students understand the new term visually by comparing the height of the towers. They see that the two towers are different and learn that the term for the taller one is “more.”
Again, evaluate each response by pointing to cubes and saying, “Same, same, same, different” to reinforce visual one-to-one matching and to build the concepts of equality and inequality:
## Step 3: Using non-uniform objects
The next scaffolded step is to maintain the vertical alignment, familiar task, and familiar vocabulary while switching to non-uniform objects. Combine different sets of cubes (math counters, stringing beads, etc.) as long as they are roughly the same size:
Students learn that in determining “more,” it is the height that matters and not the appearance of the cubes. This is one of the first steps in transitioning to abstract thinking.
One way to add complexity to the task is by combining the concepts of “the same” and “more.” Build two towers of identical cubes and increase the number of answer options. Ask students whether one tower has more, the other tower has more, or they are the same:
Another way to increase complexity is by presenting two sets of scattered cubes. Give students the option to align them in towers if they want to. Those who are beginning to rely on counting or have a strong ability to match one-to-one can skip that step. Either way, using manipulatives during instruction helps students to adapt and strengthen their thinking.
If students choose not to align the cubes by building towers and answer incorrectly, prompt them to build the towers and answer again.
## Step 5: Applying the skill in a new environment
By now, students should have a good feel for “the same” and “more” and can apply this understanding in a new environment. Present familiar tasks (align objects and compare) using horizontal alignment and varied objects. This requires students to transfer skills and helps them avoid situated cognition.
Begin with uniform objects in a horizontal alignment — for example, at Happy Numbers we use train cars as kindergarteners like to build trains. After identifying “the same” and “more,” present scattered sets and have students align them horizontally to compare:
Next, try varying the objects by presenting different objects for the same task. Matchbox cars, Shopkins, or mini erasers all work well (as long as they are roughly the same size)! Again, give students the option of whether or not to align the objects before comparing:
If students answer incorrectly without first aligning the objects, prompt them to do so and then answer again. If the response is still incorrect, demonstrate one-to-one matching by pointing and saying, “Same, same, ….”:
### 1. Get online: HappyNumbers.com ←
HappyNumbers.com provides interactive, scaffolded lessons for use on PCs or tablets. The exercises discussed in this post are part of our kindergarten curriculum.
### 2. Use our FREE printables: your copy is here ←your copy is here ←
We’re sharing a set of printables based on strategies from this lesson. Use them for independent practice, homework, or reteaching. Be sure to share them with colleagues!
Yours,
Evgeny & Happy Numbers Team |
# Analyzing Exponential Functions
Now we’ll take a look at an example where we analyze an exponential function. The method is as follows:
Rule
### AnalyzingExponentialFunctions
1.
Find the zeros.
2.
Find the stationary points.
3.
Find the inflection points.
Example 1
Analyze the function $f\left(x\right)=2{x}^{2}\cdot {e}^{x}$
1.
Find the zeros by setting $f\left(x\right)=0$:
$2{x}^{2}\cdot {e}^{x}=0$
The zero product property gives that $2{x}^{2}=0$ or ${e}^{x}=0$. However, ${e}^{x}$ is always positive, so $\begin{array}{llll}\hfill 2{x}^{2}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
This gives a zero at the origin $\left(0,0\right)$.
2.
Find the maxima and minima by setting ${f}^{\prime }\left(x\right)=0$.
First, find the derivative of $f\left(x\right)=2{x}^{2}\cdot {e}^{x}$: $\begin{array}{llll}\hfill {f}^{\prime }\left(x\right)& =4x\cdot {e}^{x}+2{x}^{2}\cdot {e}^{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(4x+2{x}^{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2x\cdot {e}^{x}\left(2+x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Then, find where ${f}^{\prime }\left(x\right)$ is equal to 0:
$2x\cdot {e}^{x}\left(2+x\right)=0$
Again, ${e}^{x}$ is always positive, so $\begin{array}{llllllllllll}\hfill 2x& =0\phantom{\rule{2em}{0ex}}& \hfill ⇒& \phantom{\rule{2em}{0ex}}& \hfill x& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2+x& =0\phantom{\rule{2em}{0ex}}& \hfill ⇒& \phantom{\rule{2em}{0ex}}& \hfill x& =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
You then need the corresponding $y$-values to find the point. You do this by inputting your $x$-values back into the main function $f\left(x\right)$:
$\begin{array}{llll}\hfill y& =f\left(0\right)=2\cdot {0}^{2}\cdot {e}^{0}=0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =f\left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{\left(-2\right)}^{2}\cdot {e}^{-2}=8{e}^{-2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{8}{{e}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
$\begin{array}{llll}\hfill y& =f\left(0\right)=2\cdot {0}^{2}\cdot {e}^{0}=0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =f\left(-2\right)=2{\left(-2\right)}^{2}\cdot {e}^{-2}=8{e}^{-2}=\frac{8}{{e}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
You now need to determine which point is a maximum and which is a minimum. You do that by drawing a sign chart.
From this, you see that the maximum is $\phantom{\rule{-0.17em}{0ex}}\left(-2,\frac{8}{{e}^{2}}\right)$ and the minimum is $\left(0,0\right)$.
3.
Find the inflection points by setting ${f}^{″}\left(x\right)=0$.
First, you find the second derivative by differentiating ${f}^{\prime }\left(x\right)={e}^{x}\left(4x+2{x}^{2}\right)$:
$\begin{array}{llll}\hfill {f}^{″}\left(x\right)& ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(4x+2{x}^{2}\right)+{e}^{x}\left(4+4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(4x+2{x}^{2}+4+4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{2}+8x+4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
$\begin{array}{llll}\hfill {f}^{″}\left(x\right)& ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(4x+2{x}^{2}\right)+{e}^{x}\left(4+4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(4x+2{x}^{2}+4+4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{2}+8x+4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
Then, let ${f}^{″}\left(x\right)=0$ and solve the equation:
${e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{2}+8x+4\right)=0$
As ${e}^{x}$ is always positive, you get
$2{x}^{2}+8x+4=0$
You solve this using the quadratic formula and get the solutions $x\approx -0.6$ and $x\approx -3.4$. You find the corresponding $y$-values by putting your new $x$-values back into the main function $f\left(x\right)$. You then get: $\begin{array}{llll}\hfill y& =f\left(-3.4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\cdot {\left(-3.4\right)}^{2}\cdot {e}^{-3.4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 0.772\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =f\left(-0.6\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{\left(-0.6\right)}^{2}\cdot {e}^{-0.6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =8{e}^{-0.6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 0.395\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
which means that you have inflection points at $\left(-3.4,\phantom{\rule{0.17em}{0ex}}0.772\right)$ and $\left(-0.6,\phantom{\rule{0.17em}{0ex}}0.395\right)$.
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Home Middle-School Fractions: Tips And Tricks
Middle-School Fractions: Tips And Tricks
When it comes to middle school math, quite often, students are continuing their math learning journeys with gaps in their knowledge. Not only do middle school learners need to master algebra and geometry, but they also need to hone other skills that will let them compete in math competitions.
As you’re preparing for an upcoming math competition, there are lots of different skills you’ll need to be able to perform at your best. For middle-school learners, in particular, a math competition offers an early chance to put recently learned skills to the test.
As these children progress through their math education, they are expected to use not just their newly learned skills but also become faster and more confident in applying their previously learned skills to more complex problems. Luckily, there are some tips and tricks that can be learned to help these young learners
Percentages Trick
Fractions and percentages: two cousins in the math world. Finding the percentage of some numbers can be quite easy. On the other hand, finding the percentage of other numbers can be much, much more difficult. Did you know that there was a simple trick that can help middle-school learners calculate percentages quickly and excel in math competitions?
All you need to know is the following:
X% of A is the same as A% of X
Let’s try applying this with an example. Let’s say I wanted to find out what 6% of 80 was. Not an easy calculation to do in your head. Now, let’s swap the numbers and see if the problem becomes more simple.
Using the shortcut above, we look for 80% of 6. We know 80% is the same as ⅘, and it is much easier to calculate ⅘ of 6 (simply multiply 6 by 0.8) than it is to calculate 6% of 80 (dividing 80 by 100 and multiplying by 6).
As you can see, this trick can help you reduce the necessary calculations and give you an edge over your competitors during math competitions. Without the tip, we would have to make two calculations, whereas, with the tip in mind, we can do one simple calculation.
The Multiples of 9 Tip
Let’s take another quick look at another of my favorite math tricks and tips: the multiples of 9 tip. This one is even easier than the percentages trick.
The two individual numbers that make up the first 10 multiples of 9 {9, 18, 27, 36, 45, 54, 63, 72, 81, 90} always add up to 9. Here are some examples:
9 x 8 = 72 → 7 + 2 = 9 9 x 4 = 36 → 3 + 6 = 9 9 x 7 = 63 → 6 + 3 = 9
This tip can help you make sure your multiples of 9 are correct, faster than the speed of light!
Of course, percentages, fractions, and multiples are not the only skills that middle-school students will need to excel in math competitions. They will need to cover a range of topics from the basics of algebra, through logical thinking, and all the way to geometry
On top of this, middle school students often have to contend with oversized classrooms, where the needs of the individual students are often lumped together, with some students’ additional needs sadly falling through the cracks.
That’s where the Online Math Center comes in
Not only do we offer courses for young learners of all levels, but we have specific courses for middle school students, as well as courses tailored to prepare students for math competitions.
With classes no bigger than 5 students and led by our fully-qualified and experienced tutors, there is no danger of any topics falling through the cracks, as they might in a class of 15, 20, or even 30 students. Students can also avail of one-to-one tutoring, to reinforce some of the topics that may be tougher to grasp. We want every student’s concerns to be addressed and every student’s voice to be heard. We believe in a holistic approach to math tutoring and we’ve seen our students benefit greatly from this.
Our tutors have years of experience working with thousands of students. So far, the students have found the programs to be as fun as they are educational!
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# Lesson 11
Using an Algorithm to Divide Fractions
### Problem 1
Select all the statements that show correct reasoning for finding $$\frac{14}{15}\div \frac{7}{5}$$.
A:
Multiplying $$\frac{14}{15}$$ by 5 and then by $$\frac{1}{7}$$.
B:
Dividing $$\frac{14}{15}$$ by 5, and then multiplying by $$\frac{1}{7}$$.
C:
Multiplying $$\frac{14}{15}$$ by 7, and then multiplying by $$\frac{1}{5}$$.
D:
Multiplying $$\frac{14}{15}$$ by 5 and then dividing by 7.
E:
Multiplying $$\frac{15}{14}$$ by 7 and then dividing by 5.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 2
Clare said that $$\frac{4}{3}\div\frac52$$ is $$\frac{10}{3}$$. She reasoned: $$\frac{4}{3} \boldcdot 5=\frac{20}{3}$$ and $$\frac{20}{3}\div 2=\frac{10}{3}$$
Explain why Clare’s answer and reasoning are incorrect. Find the correct quotient.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 3
Find the value of $$\frac{15}{4}\div \frac{5}{8}$$. Show your reasoning.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 4
Consider the problem: Kiran has $$2\frac34$$ pounds of flour. When he divides the flour into equal-sized bags, he fills $$4\frac18$$ bags. How many pounds fit in each bag?
Write a multiplication equation and a division equation to represent the question. Then, find the answer and show your reasoning.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 5
Divide $$4\frac12$$ by each of these unit fractions.
1. $$\frac18$$
2. $$\frac14$$
3. $$\frac16$$
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 4, Lesson 10.)
### Problem 6
Consider the problem: After charging for $$\frac13$$ of an hour, a phone is at $$\frac25$$ of its full power. How long will it take the phone to charge completely?
Decide whether each equation can represent the situation.
1. $$\frac13\boldcdot {?}=\frac25$$
2. $$\frac13\div \frac25={?}$$
3. $$\frac25 \div \frac13 ={?}$$
4. $$\frac25 \boldcdot {?}=\frac13$$
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 4, Lesson 9.)
### Problem 7
Elena and Noah are each filling a bucket with water. Noah’s bucket is $$\frac25$$ full and the water weighs $$2\frac12$$ pounds. How much does Elena’s water weigh if her bucket is full and her bucket is identical to Noah’s?
1. Write multiplication and division equations to represent the question.
2. Draw a diagram to show the relationship between the quantities and to find the answer.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 4, Lesson 8.) |
# Convert exbi to gibi
Learn how to convert 1 exbi to gibi step by step.
## Calculation Breakdown
Set up the equation
$$1.0\left(exbi\right)={\color{rgb(20,165,174)} x}\left(gibi\right)$$
Define the prefix value(s)
$$The \text{ } value \text{ } of \text{ } exbi \text{ } is \text{ } 2.0^{60}$$
$$The \text{ } value \text{ } of \text{ } gibi \text{ } is \text{ } 2.0^{30}$$
Insert known values into the conversion equation to determine $${\color{rgb(20,165,174)} x}$$
$$1.0\left(exbi\right)={\color{rgb(20,165,174)} x}\left(gibi\right)$$
$$\text{Insert known values } =>$$
$$1.0 \times {\color{rgb(89,182,91)} 2.0^{60}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 2.0^{30}}}$$
$$\text{Or}$$
$$1.0 \cdot {\color{rgb(89,182,91)} 2.0^{60}} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 2.0^{30}}$$
$$\text{Conversion Equation}$$
$$2.0^{60} = {\color{rgb(20,165,174)} x} \times 2.0^{30}$$
Cancel factors on both sides
$$\text{Cancel factors}$$
$${\color{rgb(255,204,153)} \cancelto{2^{30}}{2.0^{60}}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancel{2.0^{30}}}$$
$$\text{Simplify}$$
$$2^{30} = {\color{rgb(20,165,174)} x}$$
Switch sides
$${\color{rgb(20,165,174)} x} = 2^{30}$$
Solve $${\color{rgb(20,165,174)} x}$$
$${\color{rgb(20,165,174)} x} = 1073741824\approx1.0737 \times 10^{9}$$
$$\text{Conversion Equation}$$
$$1.0\left(exbi\right)\approx{\color{rgb(20,165,174)} 1.0737 \times 10^{9}}\left(gibi\right)$$ |
## Intermediate Algebra (12th Edition)
$-m^2p\sqrt[4]{mp^2}$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $2\sqrt[4]{m^9p^6}-3m^2p\sqrt[4]{mp^2} ,$ simplify first each term by expressing the radicand as a factor that is a perfect power of the index. Then, extract the root. Finally, combine the like radicals. $\bf{\text{Solution Details:}}$ Expressing the radicand as an expression that contains a factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} 2\sqrt[4]{m^8p^4\cdot mp^2}-3m^2p\sqrt[4]{mp^2} \\\\= 2\sqrt[4]{(m^2p)^4\cdot mp^2}-3m^2p\sqrt[4]{mp^2} .\end{array} Extracting the roots of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} 2m^2p\sqrt[4]{mp^2}-3m^2p\sqrt[4]{mp^2} .\end{array} By combining like radicals, the expression above is equivalent to \begin{array}{l}\require{cancel} (2m^2p-3m^2p)\sqrt[4]{mp^2} \\\\= -m^2p\sqrt[4]{mp^2} .\end{array} |
# Rescaling the Pythagorean Theorem
The Pythagorean theorem can apply to any shape, not just triangles. It can measure nearly any type of distance. And yet this 2000-year-old formula is still showing us new tricks.
Re-arranging the formula from this:
$\displaystyle{c = \sqrt{a^2 + b^2}}$
to this:
$\displaystyle{c= a \cdot \sqrt{1 + (b/a)^2}}$
helps us understand the relationship between slope (steepness) and distance. Let’s take a look.
Scaling leads to new insights. Yes, \$500k/year is a lot; but it really comes alive when you imagine things costing 10x less (A new laptop? \$150. A new porsche? \$6000). Rescaling formulas can be eye-opening as well. Let’s start with our favorite 3-4-5 triangle and divide every side by 3: What happened? Well, we have a smaller red triangle with sides 3/3 (aka 1), 4/3 and 5/3. We’ve got a mini version of the large triangle, and the Pythagorean Theorem still holds: $\displaystyle{1^2 + (4/3)^2 = (5/3)^2}$ ## So Why’s This Special? It doesn’t seem like much, but there’s some surprising insights: First, we can rescale any triangle to have 1 as the smallest side (divide by “a”). All similar triangles (i.e. those with the same ratios, like 3-4-5 and 6-8-10) will shrink into the same mini triangle. This mini triangle has an interesting property: it only cares about the ratio b/a. The only “meaningful” numbers are 1 and (b/a), giving: $\displaystyle{\text{mini hypotenuse} = \sqrt{1 + (b/a)^2}}$ And what’s special about b/a? It’s the slope of the hypotenuse line! It’s called the slope, the gradient, the derivative, rise over run — whatever the label, b/a is the rate at which the hypotenuse changes! This makes sense. For every unit traveled along the short leg, we gain “slope units (b/a)” on the other leg. In a 3-4-5 triangle, we go 4/3 units “North” for every 1 unit “East”. And the length of our hypotenuse increases 5/3 (1.66) for every 1 unit East. The result is pretty cool: we used the steepness of the hypotenuse (b/a) to find the distance traveled per unit East,$\sqrt{1 + (b/a)^2}$. ## An Example, Please This is a bit weird, so let’s do an example. Suppose we’ve gone 5 units East and 12 units North. What’s our distance from the starting point? The traditional approach plugs in the Pythagorean Theorem to get$c = \sqrt{5^2 + 12^2} = 13$. It works, but let’s try our mini-triangle method: Instead of a large triangle with sides 5 and 12, scale down by 5: we get a mini triangle with sides 5/5 (or 1) and 12/5. The “mini hypotenuse” is then$\sqrt{1 + (12/5)^2} = 2.6$. This means we travel 2.6 units along the hypotenuse for every 1 unit East. Going the full 5 units East (our original triangle) is 5 * 2.6 = 13 units. Neato — we got the same answer both ways. But silly me, I made a mistake. Instead of 5 units on that trajectory, I meant 6. No 7. No wait, 8. 9, for sure. Normally, we’d be furiously hammering that square root button to find the new distance. Maybe even using trigonometry to “make it easier”. But not today — since we’re on the same trajectory, we can re-use our scaling constant of 2.6: We can find the new distance traveled with regular multiplication, with nary a square root in sight. Cool! This approach is faster for humans and computers alike — you wouldn’t believe the crazy approaches programmers take to avoid a square root. ## Static and Dynamic Formulas I’ve realized that our venerable Pythagorean Theorem focuses on a and b separately: $\displaystyle{c = \sqrt{a^2 + b^2}}$ We consider a and b as separate elements, to be squared and summed. This approach is straightforward, and helps when designing bridges or making pictures of triangles. The traditional formula focuses on final values. But the rescaled version has a new twist: $\displaystyle{c = a \cdot \sqrt{ \left(1 + (b/a)^2 \right)} }$ We’re not that interested in the separate quantities — we want the ratio b/a, or the slope of the hypotenuse. This slope creates a scaling constant,$\sqrt{1 + (b/a)^2}\$, that tells us how our “Eastward” motion translates to distance along our path. The dynamic formula focuses on rates of change.
If we have a hypothetical function f(x), we might write the dynamic Pythagorean Theorem this way:
$\displaystyle{\text{distance along path} = x \cdot \sqrt{1 + (slope)^2}}$
This concept is used in calculus to find the length of any line or curve — but we’ll save that for another day.
The key is to realize a single formula can be re-arranged and lead to new insights. Stay curious — we stop learning when we think we’ve “got it all figured out”.
## Appendix 1: Slope vs. Distance
One point that confused me was separating the idea of slope (b/a) from distance traveled (the hypotenuse, c).
Slope is b/a, rise over run — how much height you get when you increase width. How “steep” the hill is, so to speak. Unfortunately, the word “slope” makes us think of the side of the hill — but slope is really about height.
Distance (the hypotenuse) is about the side of the hill — how far you’ve walked. The “steepness” isn’t that important — you’re laying a measuring tape on the ground, which could be flat, vertical or upside-down. Does the length of a board depend on how you hold it?
But, in our man-made world, slope and distance are related because we often express locations in terms of “units East (x coordinate)” and not “units along a path”. So when a map says “go 1 mile due East” and you’re in front a mountain (large slope), you end up traveling a large distance (more than 1 mile). When on a flat road (zero slope), 1 mile East is simply 1 mile East. The bigger the slope, the more distance you must travel to “go 1 mile East”.
Again, we see that the Pythagorean Theorem is not just about triangles — it can convert slope (steepness) into distance traveled. Happy math. |
## Equation of a line: The derivation of y = mx + b
We have discussed in context the origin (click here and here) of the linear equation $y = ax + b$, where $a$ and $b$ are real numbers. We have also talked about the slope of a line and many of its properties. In this post, we will discuss the generalization of the equation of a line in the coordinate plane based on its slope and y-intercept.
We have learned that to get a slope of a line, we only need two points. We have also learned that given two points on a line, its slope is described as the rise (difference in the y-coordinates) over the run (difference in x-coordinates). Therefore, if we have two points with coordinates $(x_1,y_1)$ and $(x_2,y_2)$, the slope $m$ is defined the formula
$m = \displaystyle\frac{y_2 - y_1}{x_2 - x_1}$.
All the points on a vertical line have similar x-coordinates; therefore, the run ${x_2 - x_1}$ is equal to $0$ making $m$ undefined. From here, we can conclude a vertical line has no slope. » Read more
## Matchsticks, Linear Relations, and Multiple Representations
Introduction
We have mentioned the different types of functions in the Introductions to Functions post. In this post, we are going to learn about linear function and its characteristics.
To start, let us examine the problem below taken from the TIMSS 2003 released items given to Grade 10 students in more than 40 countries all over the world.
Matchsticks are arranged as shown in the figures.
If the pattern is continued, how many matchsticks would be used to make figure 10.
A. 30 B. 33 C. 36 D. 39 E. 42
The problem is too easy that even a first grade pupil would be able to answer it given enough time. Smart students would be able to easily see patterns. For example, they can relate the number of squares to the number of matchsticks. If they cannot find a pattern, the last resort would be by brute force; that is, by manually drawing the tenth figure. » Read more |
# File
```Mathematics is often described as the study of patterns, so it is not surprising
that area models have many patterns. You saw one important pattern in
Lesson 4.1.1 (Casey’s pattern from problem 4-4). Today you will continue to
use patterns while you develop a method to factor trinomial expressions.
4-12. Examine the area model shown at
right.
a.
Review what you learned in Lesson 4.1.1 by
writing the area of the rectangle at right as a
sum and as a product.
(5x – 2)(2x – 7) = 10x2 − 39x + 14
b.
Does this area model fit Casey’s pattern for
diagonals? State the pattern and justify your
The products of the diagonals are equal;
−35x · −4x = 10x2 · 14 = 140x2
4.1.2 Factoring with Area Models
November 16, 2015
Objectives
CO: SWBAT
develop an algorithm to
using algebra tiles.
LO: SWBAT
explain to a partner
4-13. FACTORING
EXPRESSIONS
a.
A polynomial in the form ax2 + bx + c, with a ≠ 0, is called a quadratic
expression in standard form. To develop a method for factoring
quadratic expressions without using algebra tiles, you will first model
how to factor with algebra tiles, and then look for connections within
an area model.
Using algebra tiles, factor 2x2 + 5x + 3; that is,
use the tiles to build a rectangle, and then
write its area as a product.
(2x + 3)(x + 1)
b.
When you factor using tiles, you need to
determine how to arrange the tiles to form a
rectangle. Using an area model to factor
requires a slightly different process.
Miguel wants to use an area model to factor
3x2 + 10x + 8. He knows that 3x2 and 8 go
into the rectangle along one of the diagonals,
as shown at right. Finish the rectangle by
deciding how to split and place the remaining
term. Then write the area as a product.
(3x + 4)(x + 2)
Algebra tiles
EXPRESSIONS
c.
Kelly wants to determine a method for factoring
2x2 + 7x + 6. She remembers Casey’s pattern for
diagonals. She places 2x2 and 6 into the rectangle
in the locations shown at right. Without actually
factoring yet, what do you know about the missing
two parts of the area model?
Their sum is 7x, and their product is 12x2
d.
To complete Kelly’s area model, create and solve a
Diamond Problem using the information from part
(c).
The product 12x2 should be placed at the top of the
diamond problem, 7x at the bottom, and the terms 3x and
4x should be in the middle.
e.
Use your results from the Diamond Problem to
complete the area model for 2x2 + 7x + 6, and then
write the area as a product of factors.
(2x + 3)(x + 2)
So, then…?
1.
2.
3.
4.
5.
6.
7.
Place the x2-term and the constant term of the quadratic
expression in opposite corners of the area model.
Determine the product of the x2-term and the constant
term. Place this product in the top of a Diamond Problem.
Place the x-term from the expression in the bottom (sum) of
a Diamond Problem.
Solve the Diamond Problem by identifying two terms whose
product and sum match those in the diamond.
Place the solutions from the Diamond Problem into the two
remaining corners of the area model.
Determine the dimensions of the area model.
Write the area as the product of the dimensions.
4-14. Factoring with an area model is especially convenient when algebra tiles are not
available or when the number of necessary tiles becomes too large to manage. Using a
Diamond Problem helps avoid guessing and checking, which can at times be
challenging. Use the process from problem 4-13 to factor 6x2 + 17x + 12.
1.
2.
3.
4.
5.
(2x + 3)(3x + 4)
Place x2 & # in box
Set up diamond
Solve diamond
Find dimensions
Write product
4-16. Use the process you developed in problems 4-13 and 4-14 to
cannot be factored, justify your conclusion.
a.
b.
c.
d.
x2 + 9x + 18
4x2 + 17x – 15
4x2 – 8x + 3
3x2 + 5x – 3
a.
b.
c.
d.
(x + 3)(x + 6)
(4x − 3)(x + 5)
(2x − 3)(2x − 1)
not factorable
``` |
# janmr blog
## Fractions and Circles February 06, 2010
Fractions produced by mediants have some very interesting properties. We saw some of them in connection with the Stern-Brocot tree. This articles explores a more curious property, relating fractions to circles in the plane. It was discovered in 1938 by Lester R. Ford and is also mentioned in Conway and Guy's The Book of Numbers.
Let us consider a way to construct fractions. We always start out with the sequence
(1)
$\frac{0}{1}, \frac{1}{0}$
and then repeatedly obtain a new sequence from an existing one. Given a sequence, we pick two consequtive fractions and insert their mediant between them. Recall that the mediant of $\frac{m_1}{n_1}$ and $\frac{m_2}{n_2}$ is $\frac{m_1+m_2}{n_1+n_2}$. We will call sequences created in this manner mediant sequences.
So the first (and only) mediant sequence possible from the initial sequence (1) is
$\frac{0}{1}, \left<\frac{1}{1}\right>, \frac{1}{0}$
where the new fraction is surrounded by angle brackets. From this point on, we have multiple choices each time we produce a new mediant sequence. An example of the following sequences could be
$\frac{0}{1}, \left<\frac{1}{2}\right>, \frac{1}{1}, \frac{1}{0}$ $\frac{0}{1}, \frac{1}{2}, \frac{1}{1}, \left<\frac{2}{1}\right>, \frac{1}{0}$ $\frac{0}{1}, \left<\frac{1}{3}\right>, \frac{1}{2}, \frac{1}{1}, \frac{2}{1}, \frac{1}{0}$
An essential property, in regard to this article, is this: For all mediant sequences, any fraction $\frac{m_1}{n_1}$ followed by $\frac{m_2}{n_2}$ fulfills $n_1 m_2 - m_1 n_2 = 1$. This is easily shown by induction by considering the initial mediant sequence (1) and that if $n_1 m_2 - m_1 n_2 = 1$ for a subsequence
$\ldots, \frac{m_1}{n_1}, \frac{m_2}{n_2}, \ldots$
then it also holds after inserting the mediant
$\ldots, \frac{m_1}{n_1}, \frac{m_1+m_2}{n_1+n_2}, \frac{m_2}{n_2}, \ldots$
(two cases to check).
Note how this property implies that the fractions in any mediant sequence are ordered by size. But the most important use of this property will appear shortly.
We now do the following in a Cartesian coordinate system: For each fraction $\frac{m}{n}$ in a given mediant sequence, draw a circle centered at $\left(\frac{m}{n}, \frac{1}{2 n^2}\right)$ with radius $\frac{1}{2 n^2}$. An example can be seen in Figure 1.
By construction, the x-axis is obviously a tangent to all such circles. Furthermore, two circles, corresponding to two fractions $\frac{m_1}{n_1}$ and $\frac{m_2}{n_2}$, touch at exactly one point if and only if $| n_1 m_2 - m_1 n_2 | = 1$. This is shown by an application of the Pythagorean Theorem:
\begin{aligned} \left( \frac{m_2}{n_2} - \frac{m_1}{n_1} \right)^2 + \left( \frac{1}{2 n_2^2} - \frac{1}{2 n_1^2} \right)^2 &= \left( \frac{1}{2 n_2^2} + \frac{1}{2 n_1^2} \right)^2 \quad \Leftrightarrow \\ \left( \frac{n_1 m_2 - m_1 n_2}{n_1 n_2} \right)^2 &= \frac{1}{n_1^2 n_2^2} \quad \Leftrightarrow \\ | n_1 m_2 - m_1 n_2 | &= 1. \end{aligned}
This is exactly the property that consequtive fractions in any mediant sequence fulfills.
Zooming into the dashed region in Figure 1 leads to Figure 2.
So given a mediant sequence, we can draw such circles for each fraction in the sequence, except for the final $\frac{1}{0}$ (which is just an auxillary element). The circles with diameter $1$ correspond to the seqeuence $\frac{0}{1}, \frac{1}{1}, \frac{2}{1}, \ldots$ and will all touch each other. See Figure 1. Any circle with diameter less than $1$ will touch exactly two larger circles, namely the two from which it was created when considering their mediant sequence representatives. To be more specific, if we went from a mediant sequence
$\ldots, \frac{m_1}{n_1}, \frac{m_2}{n_2}, \ldots$
with $n_2 \neq 0$ to
$\ldots, \frac{m_1}{n_1}, \frac{m_1+m_2}{n_1+n_2}, \frac{m_2}{n_2}, \ldots,$
then the three circles corresponding to the fractions of this second subsequence will all touch, and the middle one will obviously be the smallest (has the largest denominator). See, e.g., the fractions $\frac{1}{2}$, $\frac{3}{5}$, $\frac{2}{3}$ in Figure 2.
Zooming in once again we get Figure 3.
Note the many self-similarities.
Commenting is not possible for this post, but feel free to leave a question, correction or any comment by using the contact page |
# Exercise 14.3
QUESTION 1
In the given figure, compute the area of quadrilateral ABCD.
Sol :
Given : Here from the given figure we gwt
(1) ABCD is a quadrilateral with base AB .
(2) ΔABD is a right angled triangle.
(3) ΔBCD is a right angled triangle with base BC right angled at B
To find : Area of quadrilateral ABCD
Calculation :
In right triangle ΔBCD , by using Pythagoreans theorem
CD2 = BD2 + BD2
⇒ 172 = BD2 + 82
⇒ BD2 = 172 – 82
⇒ BD2 = 289 – 64
⇒ BD2 = 225
⇒ BD = 15 cm
since area of triangle
So , Area of triangle ΔBCD
= 60 cm2
In right angled triangle ABD
BD2 = AB2 + AD2
152 = AB2 + 92
= 12 cm
Area of right triangle
= 54 cm2
Area of parallelogram ABCD = area ( ΔABD) + area ( ΔBCD)
= 54 + 60
= 114 cm2
Hence we get Area of quadrilateral ABCD = 114 cm2
QUESTION 2
In the given figure , PQRS is a square and T and U are respectively , the mid – points of PS and QR . Find the area of ΔOTS if PQ = 8 cm
Sol :
Given : Here from the given figure we get
(1) PQRS is a square .
(2) T is a mid-point of PS which means
(3) U is the mid-point of PS which means
(4) QU = 8 cm
To find : Area of ΔOTS
Calculation:
Since it is given that PQ = 8 cm . So
PS = SR = QR = 8 cm (side of square are equal) |
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# Given that ${m_1} = \,10\,kg$ and ${m_2} = 12\,kg$ connected as shown in the figure given below. Find tension in the string and acceleration of the two blocks.
Last updated date: 18th Jun 2024
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Hint: In this solution, we will find the components of the blocks that will exert tension on the string i.e. parallel to the slope of the inclined plane. Both the blocks will have the same acceleration since they are connected with a string which will be under constant tension.
Looking at the figure, the block of mass 12 kg will be heavier and as a result, it will move downwards and at the same time, pull the block of mass 10 kg upwards. Both the block will have the same acceleration since they are tied by the same string under acceleration and hence the same tension in the string as well.
Since the block of mass 10 kg will be moving upwards, the pseudo acceleration force and the tension in the string will both be towards the pulley, parallel to the surface of the power and the component of the weight for the block will be downwards. So, we can write
${m_1}a = T - {m_1}g\sin {\theta _1}$
$\Rightarrow T = {m_1}g\sin {\theta _1} + {m_1}a$
Similarly, for the second block, we can write, the weight and the acceleration will be downwards
${m_2}a = {m_2}g\sin {\theta _2} - T$
$\Rightarrow T = {m_2}g\sin {\theta _2} - {m_2}a$
So comparing tension in both these equations, we can write
$\dfrac{1}{2}{m_1}g + {m_1}a = \dfrac{{\sqrt 3 }}{2}{m_2}g - {m_2}a$
Solving for $a$, we get
$22a = \dfrac{{\sqrt 3 }}{2}(12)g - \dfrac{1}{2}(10)g$
On substituting $g = 10\,m/{s^2}$, we get
$22a = 54$
Hence $a = 2.45\,m/{s^2}$
Then the tension in the string can be calculated as
$T = {m_1}g\sin {\theta _1} + {m_1}a$
$\Rightarrow T = 50 + 24.5$
Hence the tension in the string will be
$T = 74.5\,N$
Note: Here we have assumed that there is no friction acting on the blocks and the ramps which can hinder the motion of the blocks. Since the second block is heavier and also at a more inclined plane, we can assume that it will be pulling the first block. |
# The Fade Mathematics: Kite Shawl Calculations
Yesterday I wrote about the construction of kite shawls and I promise you two things: The formula for kite shawl calculations, and a pattern template for them. Here we go!
This article contains a bit of math. If you feel intimidated no worries, you’ll find a simple pattern template using a percentage system at the end of this article!
For the impatient, here are the key facts about this shawl shape (an extensive overview of kite shawls can be found in yesterday’s article The Secrets of Kite Shawls):
• Percentages of yardage for each section vary.
• Percentages depend on the ratio of rows in the increase vs. straight section.
• For regular kites this ratio is 2:1.
• You can knit kites without a straight section.
If you work the regular kite start decreasing before you have only 46% of your yardage available, otherwise you will run out of yarn before finishing. In other words: the decrease section uses almost half of your yardage!
Details can be found below, as well as a pattern template at the very end.
## Kite Shawl Calculations
### Increase Section
Let m be the number of rows in the increase section and assume starting with one stitch only. Our total increase rate for each half is one stitch every other row, so after m rows we have m/2 stitches on each side.
The area covered by the increase section therefor is m*m/2.
### Straight Section
Let n be the number of rows in the straight section. Regular kite shawls have n = m/2 rows in their straight section. The area covered by the straight section is m/2*n (left side) and m/2*n + n*n/2 (right side).
If you omit the straight section, which is totally possible, this area obviously equals zero.
### Decrease Section
The number of rows in the decrease section is determined by m as we decrease all stitches we created during the increase section. We decrease one stitch every other row on the left half thus the number of rows we need equals m.
On the left half we have an area of (m*m/2)/2, on the right side the area equals (n+m)/2*(m/2) + (m/2 * m/4)/2.
### Section Percentages
The percentage of each section is calculated by summing up all section areas to get the total area. Let the total area be A and the areas for each section be a1, a2 and a3 (a1 = increase, a2 = straight, a3 = decrease).
The increase section uses a1/A * 100% of the total yardage, the straight section a2/A *100% and the decrease section a3/A *100%.
These percentages strongly depend of the ratio m/n (for n not zero).
For n=m/2 the percentages are
• 24% (increase section)
• 30% (straight section)
• 46% (decrease section)
You see that if you’re starting to decrease after you have used up a bit more than a half of your yarn available you will run out of yarn before finishing.
## Kite Shawl Pattern Template
Setup
• CO 5 sts and knit one row.
• Next Row: K1, YO, k to last st, YO, k1.
• Next Row: Knit.
• Next Row: K1, YO, k1, cdd, k1, YO, k1.
• Next Row: K1, YO, k to last st, YO, k1.
• Next Row: K1, YO, k2, cdd, k2, YO, k1.
• Next Row: K1, YO, k to last st, YO, k1.
You can see the center spine stitch now (the position of the cdd on each RS row). This stitch is called CSS from now on, and is used as your main orientation landmark.
### Increase Section
• Next Row: K1, YO, k to 1 st before CSS, cdd, k to last st, YO, k1.
• Next Row: K1, YO, k to CSS, purl CSS, k to last st, YO, k1.
• Repeat the last two rows until you have worked m rows.*
*Please see above for an explanation what m and n stand for. The abbreviation “st” means “stitch”.
### Straight Section
• Next Row: K1, YO, k to 1 st before CSS, cdd, k to last st, YO, k1.
• Next Row: K1, k to CSS, purl CSS, k to last st, YO, k1.
• Repeat the last two rows until you have worked n rows.
### Decrease Section
• Next Row: K1, YO, k to 1 st before CSS, cdd, k to last st, k1.
• Next Row: K1, k to CSS, purl CSS, k to last st, YO, k1.
• Repeat the last two rows until you have only one stitch between CSS and the left edge of the knitting.
• Bind off all stitches
I really hope this article helps you understanding the principles of kite shawl calculations a bit better, and hopefully it prevents you from running out of yarn knitting kite shawls in the future (like I did two – yes, two! times).
Let me know if you still have any questions about kite shawls by leaving a comment below!
Save
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### 28 thoughts on “The Fade Mathematics: Kite Shawl Calculations”
• Hi Julia
On the left half we have an area of (m*m/2)/2, on the right side the area equals (n+m)/2*(m/2) + (m/2 * m/4)/2.
Left half area is ok for me. I think there are the mistake in the right half area. What do you think about this formula (n+m)/2*m + (m * m/2)/2
• Hello and thank you for that post. I have a question concerning yarn percentage for shawls in which the straight section is omitted: Can you explain after what percentage of yarn available the decrease section should start. Thank you for your answer in advance.
• Ritika Chandak
Thankyou so much for this lovely maths 🙂 !
Started the shawal !
• Nichole Lubcke
Great formula. The math has me stumped (I’m better with words than numbers!) How do I turn m into a number? You state to knit m rows in the increase section and I get that each side of the centre is half the number of stitches at there are rows (I think) I but how do I know how many rows m should be?
Thank you!
• “Let m be the number of rows in the increase section” means “m” is the number of rows you decide to knit during your increase section. You decide what “m” will be.
• Thank you for posting this explanation. I had to put all distractions aside, and it took me a while, but I got it. I stand in awe of you.
I do have a question regarding your sample “Kite Shawl Pattern Template” though.
Directions read as follows for the “Setup” (with my st count in parentheses) :
CO 5 sts and knit one row. (5 sts)
Next Row: K1, YO, k to last st, YO, k1. (7 sts)
Next Row: Knit. (7 sts)
Repeat the last two rows until you have 7 sts total. (already have 7 sts)
Do I just do the obvious and simply skip the last line of directions? Thank you!
• Julia
Thanks for your comment! The post has been updated according to your suggestions.
• barbara
If CSS and CDD are the same thing why are both used in the same line. that is confusing to me. CDD is center double decrease, what is CSS?
• Julia
It’s not: “css” means center spine stitch, “cdd” means central double decrease.
• Sherry McCary
Awesome article. Thank you so much for the formula and template pattern!
• Thank you for this article! I’ve been looking at kite shawls and trying to figure out how they are constructed without having one in hand or having made one. You have saved me a bunch of work.
• sidney
Hi,
I’m trying to do a kite shawl but I don’t get the math in the decrease section…
The right row is: K1, YO, k to 1 st before CSS, cdd, k to last st, k1.
There is 1 st increase and 2 stitches decrease, we finish the right side row with 1 stitch decrease.
Then the wrong side row is: K1, k to CSS, purl CSS, k to last st, YO, k1.
The YO counts as an increase.
At the end of both rows the stitch count is back to normal.
There is no decrease…
Is there something I’m missing ??
Thanks for the help 🙂
• You are in fact not really decreasing stitches, but the CSS shifts so it ‘looks’ like a decrease. It’s a bit hard to explain. It seems like you’re knitting sideways. Hope this helps!
• Linda Shelhamer
I am so glad you explained this because I couldn’t figure out the decrease that is not decreasing. Why do you make the stitch in the center a purl stitch?
• First of all, thank you so much for all the information you share on your blog, especially the series on shawl design, because that is what I needed just now. I am now in my shawl stage because they are so easy and satisfying to knit up!
I just wanted to let you know that the ratio 40-60% for the kite shawl did not work for me. I carefully weighed my yarn, stopped increasing on the left side a little before reaching 40%, and when I was attaching the last ball of yarn I realized I would not be able to finish it, I would need at least another 82 grams of yarn, almost a whole extra ball. So taking this amount into account, the ratio turns out to be 33%. If I want to finish the shawl – meaning to knit until there is no stitch left to the left of the center spine – I have to stop increasing on the left side when I reach 30% of my proposed yardage.
• Thanks for your comment! I’ll have another look (the ratio in question worked well for me when I wrote the article, but I haven’t knitted or designed another kite shawl since) and get back to you.
• Sonya Johnson
Brilliant article – thank you so much! I work at a LYS, and have designed a multi-color boomerang bias shawl pattern for our shop, after figuring out the simple maths for it. Given the popularity of these “kite” shawls (I didn’t even know this type of asymmetrical triangle had a name!), I’ve been wanting to make another for the shop, and while I’d figured out the maths for the symmetrical part of it, the yarn usage is something I’d have not figured out. So, your article is very timely, as I’ve been sitting down drawing up schematics for various ways to incorporate my own design ideas into this shawl style, based around yarns we carry at our shop.
I do great with the simple math and geometry when it comes to knitting, but the trig part that you described above I would not have been able to ever figure out (barely got through trig in college), since I’ve forgotten the little bit about it I did manage to learn.
I found your site via a Pinterest link, and am so glad I did. Will look forward to subscribing!
Happy knitting,
-Sonya
• Glad the article is of help! 🙂
• It was only with the help of a tutor that I passed geometry and had to take basic algebra 3 times! However, I love reading articles by those who live and breathe the math and can apply it to their knitting. Since I never made it to advanced math, I appreciate the written instructions! I love the diagrams. I appreciate how you have dissected shawl shapes and explained their basic patterns.
• Michelle
hi, I love the shawl pictured. Is there a designer credit or pattern link?
• Julia
Michelle, this shawl is one of the Plant Anatomy patterns. It’s my design and not yet published but will go live within the next days. I’ll post an article and notify my mailing list as soon as it’s available!
• Cora
Sorry, but in german: ich finde deine Design- und Constructionsposts unglaublich großartig! Hilft mir so sehr dabei, eigene Idee umzusetzen. Ganz herzlichen Dank für all deine wunderbaren Informationen!
Liebe Grüße, Cora
• Julia
Gern geschehen – my pleasure! 🙂 |
# How Many Containers in One Cup / Cups in One Container?
Alignments to Content Standards: 6.NS.A.1
1. If $\frac12$ cup of water fills $\frac23$ of a plastic container, how many containers will 1 cup fill?
• Solve the problem by drawing a picture.
• Which of the following multiplication or divisions problems represents this situation? Explain your reasoning.
$$\frac12 \times \frac23=? \qquad \frac12 \div \frac23=? \qquad \frac23 \div \frac12=?$$
• Solve the arithmetic problem you chose in part (3) and verify that you get the same answer as you did with your picture.
2. If $\frac12$ cup of water fills $\frac23$ of a plastic container, how many cups of water will the full container hold?
• Solve the problem by drawing a picture.
• Which of the following multiplication or divisions equations represents this situation? Explain your reasoning.
$$\frac12 \times \frac23=? \qquad \frac12 \div \frac23=? \qquad \frac23 \div \frac12=?$$
• Solve the arithmetic problem you chose in part (3) and verify that you get the same answer as you did with your picture.
## IM Commentary
These two fraction division tasks use the same context and ask “How much in one group?” but require students to divide the fractions in the opposite order. Students struggle to understand which order one should divide in a fraction division context, and these two tasks give them an opportunity to think carefully about the meaning of fraction division.
The purpose of this problem is to help students deepen their understanding of the meaning of fractions and fraction division and to see that they get the same answer using standard algorithm as they do just reasoning through the problem. Later they can build on this to explain why dividing by, for example, $\frac23$ is the same as multiplying by $\frac32$ based on the meaning of multiplication and division by fractions.
## Solution
• Below is a series of pictures that can be used to solve the problem as well as an explanation of what the picture represents at each step.
$\frac12$cup of water fills $\frac23$ of a plastic container, as shown below.
We know that $\frac12$ cup of water fills $\frac23$ of a plastic container. The question is asking, "How many containers can be filled with 1 cup of water." Since 1 cup of water is twice as much as $\frac12$ cup of water, we know that 1 cup of water will fill twice as much of a container.
The answer to the question can easily be found by relabeling the above picture.
Twice as much water will fill twice as many containers. Thus, 1 cup of water will fill $\frac43$ of a container.
• We are looking for the number of containers filled by 1 cup, and we know $\frac12$ of that unknown amount fills $\frac23$ of a container. We can write this symbolically as $$\frac12 \times ? = \frac23$$ which is equivalent to $\frac23 \div \frac12$ Computing this by multiplying by the reciprocal, we find that \begin{align} \frac23 \div \frac12 &=\\ &=\frac23 \times \frac21 \\ &= \frac43 \end{align}
• When we solved the problem with a picture, we took the fraction of the container filled by $\frac12$ cup and doubled it to find the fraction of a container filled by one cup. This is exactly what we did when we did it through computations: to divide $\frac23$ by $\frac12$ we multiplied it by 2. The two ways of solving it do result in the same answer and even follow the same process.
• Below is a series of pictures that can be used to solve the problem as well as an explanation of what the picture represents at each step.
$\frac12$ cup fills $\frac23$ of the container, as shown in the picture below.
Since the picture represents $\frac23$ of the container, $\frac12$ of $\frac23$ will be $\frac13$ of the container. In the picture below, each piece represents $\frac12$ of $\frac23$, and also represents $\frac12$ of $\frac12$. Since $\frac12 \times \frac12 = \frac14$, each piece also represents $\frac14$ cup.
We have $\frac23$ of the container, but we need to know how many cups it takes to fill $\frac33$ of the container, so if we take 3 of the thirds, we will have the amount of water needed for the full container. This is the same thing as multiplying $3 \times \frac14$.
We can see that $\frac34$ cup is needed to fill the whole container.
• We are looking for the number of cups needed to fill one container, and we know $\frac23$ of that unknown amount requires $\frac12$ a cup. We can write this symbolically as $$\frac23 \times ? = \frac12$$ which is equivalent to $\frac12 \div \frac23$ Computing this by multiplying by the reciprocal, we find that \begin{align} \frac12 \div \frac23 &=\\ &=\frac12 \times \frac32 \\ &= \frac34 \end{align}
• When we solved the problem using a picture, we found $\frac12$ of the amount (because we knew it was 2 thirds of the container, and so $\frac12$ of that would be 1 third of the container) and then we multiplied that amount by 3 to get the amount in the whole container. This is exactly what we did when we multiplied by the reciprocal of $\frac23$: multiplying by $\frac12$ and then by $3$ is the same as multiplying by $\frac32$. So whether we solve it with a picture or via computation, we get the same answer (in fact, the process itself is the same). |
0
What is 4k plus 1 equals 3.25?
Updated: 9/21/2023
Wiki User
12y ago
Let's start by writing the problem.
4K + 1 = 3.25
Now we begin by removing the single term (1) from the left side. We must perform the opposite or inverse operation to remove it. In our case, the opposite or inverse of addition is subtraction.
4K + 1 - 1 = 3.25 - 1
We're left with the following problem: 4K = 2.25. Our next step is to disconnect the variable K from the 4. Remember in Algebra, we do the opposite or inverse operation and it needs to be done to both sides.
4K ÷ 4 = 2.25 ÷ 4
Lastly, we're left with our solution which is K = 0.5625
If we plug in our number we come to:
4(0.5625) + 1 = 3.25
2.25 + 1 = 3.25
3.25 = 3.25
So we'd rewrite the problem as: 4K + 1 = 3.25 when K = 0.5625.
Wiki User
12y ago
Earn +20 pts
Q: What is 4k plus 1 equals 3.25? |
# Understanding Imaginary Numbers
By Kathleen Knowles, 03 Apr 2021
The term "imaginary number" describes any number that, when squared, gives a negative result. When you consider that man invented all numbers, you can also consider working with imaginary numbers. It's acceptable to invent new numbers as long as it works within the bounds of the rules that are already in place.
Simply put, an imaginary number is the square root of a negative number and does not have a tangible value. Although imaginary numbers are not real numbers and you cannot quantify them on a number line, these numbers are "real." We use them all the time in advanced Mathematics classes.
## Solving Problems Involving Imaginary Numbers
For a time, the belief held that you can't get the square root of a negative number. This resulted from the "non-existence" of numbers that were negative after you squared them. It was impossible to work backward by taking the square root since every number was positive after you squared them.
So you couldn't square root a negative number and expect to come up with anything practical. But you don't have to worry about working out the square roots of negative numbers. To solve this problem, you can use a new number.
This new number was invented during the Reformation period. During this time, nobody believed that you could use this number for any "real world" use. It was strictly for making it easy to work out the computations in solving certain equations. As such, the new number was generally viewed as "a pretend number" invented only for convenience.
The new number we are talking about is called "i," denoting "imaginary." People believed that this number—the square root of a negative number—wasn't real. You write this imaginary number as:
• i = √-1
• Thus: i² = (√-1)² = -1
• And: i² = (√-1)² = √(-1)² = √1=1
To determine the square root of a negative number in terms of the imaginary unit "i," use the following property where represents any non-negative real number:
√-a = √-1.a =√-1.√a= I√a
With this information, we can write:
√-9 = √-1.9 = √-1.√9= i.3 =3i
We would expect that 3i squared equals -9. So (3i)² =9i² =9(-1) = -9. You can write the square root of any negative number in terms of the imaginary unit. Such numbers are called imaginary numbers.
## Solving Imaginary Numbers Involving Radicals
Since multiplication is commutative, the imaginary numbers are equivalent and are often misinterpreted as part of the radicand. To deal with this confusion, place the imaginary number in front of the radical, then solve the problem. Let's consider the complex number 21-20i.
### Example 1
Solve the equation 21-20i.
#### Solution
• From the definition of a square root, this number satisfies the equation: 21-20i=x2
• Now express x as where a and b are real numbers: 21-20i=(a+bi)²
• Then multiple out the term on the right-hand side: 21-20i=a²+(2ab)i+(b2)i²
• As i²=-1 by definition of "i", rearrange the equation to give: 21-20i=(a²-b²)+(2ab)i.
Now that both sides of the equation are in the same form, compare the coefficients to obtain two equations in a and b. You have a²-b²=21 (call this equation 1). Next, compare the imaginary parts of the equation (the coefficients of i). You have 2ab=-20 (call this equation 2). You'll now have two equations with two unknowns. You can solve the simultaneous equations for a and b.
• First, can make b the subject of equation 2 by dividing both sides by 2a like this: b=-10/a.
• You can then substitute this expression for b into equation 1 like this: a²-(-10/a)²=21.
• Simplify and factorize this equation to get the following: (a²+4)(a²-25)=0
• The resultant equation is quadratic in disguise. Therefore: a²=-4 or a²=25
Remember the assumption that a and b are real numbers. a²=-4 has no solutions of interest to you. This means your solutions are a=5 and a=-5. Now substitute each a value into your earlier expression for b. This means that when a=5, b=-2 and when a=-5, b=2. Lastly, put a and b into the context of the question and get the solution as 5-2i and -5+2i.
## Solving Imaginary Numbers With A Single Radical
If you have an equation with a single radical, follow the procedure below:
1. Isolate the radical to the left side of the equation and leave everything else on the equation's left.
2. Square both sides of the equation.
3. Get the value of the unknown.
4. Substitute the value of the unknown in the original equation to verify.
### Example
Solve the equation √7x -2 = 5
#### Solution
• √7x + 4 = 7
• 7x +4 = 7²
• 7x + 4 = 49
• 7x = 45
• x = 45/7
You can now substitute x = 45/7 into the original equation √7x -2 =5. Thus: 7-2 = 5
## Solving Equations of Imaginary Numbers Involving Division
To divide imaginary numbers, you multiply the numerator and denominator by the complex conjugate a - bi. In this case, assuming a - bi is a complex number, then you will have:
(a + bi) (a - bi) = a²+ b²
### Example
Divide
x = (5 - 3i) / 4 + 2i
#### Solution
Multiply top and bottom by 4 - 2i
• (5 - 3i)(4 - 2i) / (4 + 2i)(4 - 2i)
• (20-10i-12i+6i²) / (16 - 8i + 8i - 4i2)
• (20 - 22i + 6i²) / 16 - 4i2
• (14 - 22i) / 20
## Understanding the Practical Application of the Concept of Imaginary Numbers
Also called complex numbers, imaginary numbers are applicable in real life. For instance, in quadratic planes, these numbers show up in equations that do not touch the x-axis. Imaginary Numbers are especially very useful in advanced calculus.
Imaginary Numbers are also very essential in electricity, especially in alternating current (AC) electronic devices. Here, the AC electricity alternates between positive and negative in a sine wave. As such, combining AC currents can be extremely challenging. So using imaginary currents and real numbers has helped solve this problem by making it possible to do the calculations to avoid electrocution.
Lastly, imaginary numbers are essential in signal processing. This is especially so if what you're measuring relies on cosine or sine wave. Signal processing is vital in cellular and wireless technologies as well as radar and brain waves.
### Imaginary Doesn't Mean Impossible
Initially, imaginary numbers were considered impossible to solve. From this discussion, however, it's evident that they're not as complex as they seem. You can actually solve problems involving these types of numbers. Knowledge of imaginary numbers has deep significance and profound importance to the understanding of Physics and Mathematics.
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# 2.7: Mixed Numbers in Applications
Difficulty Level: Basic Created by: CK-12
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Practice Mixed Numbers in Applications
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Suppose you're skateboarding at an average rate of \begin{align*}15 \frac{1}{2}\end{align*} miles per hour, and you skateboard for \begin{align*}\frac{3}{5}\end{align*} of an hour. How far did you travel? In this Concept, you'll learn how to multiply mixed numbers and fractions so that you can answer real-world questions such as this.
### Guidance
You’ve decided to make cookies for a party. The recipe you’ve chosen makes 6 dozen cookies, but you only need 2 dozen. How do you reduce the recipe?
In this case, you should not use subtraction to find the new values. Subtraction means to make less by taking away. You haven’t made any cookies; therefore, you cannot take any away. Instead, you need to make \begin{align*}\frac{2}{6}\end{align*} or \begin{align*}\frac{1}{3}\end{align*} of the original recipe. This process involves multiplying fractions.
For any real numbers \begin{align*}a, b, c,\end{align*} and \begin{align*}d\end{align*}, where \begin{align*}\ b \neq 0\end{align*} and \begin{align*}\ d \neq 0\end{align*}:
\begin{align*}\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}\end{align*}
#### Example A
The original cookie recipe calls for 8 cups flour. How much is needed for the reduced recipe?
Solution: Begin by writing the multiplication situation. \begin{align*}8 \cdot \frac{1}{3}\end{align*}. You need to rewrite this product in the form of the property above. In order to perform this multiplication, you need to rewrite 8 as the fraction \begin{align*}\frac{8}{1}\end{align*}.
\begin{align*}8 \times \frac{1}{3} = \frac{8}{1} \times \frac{1}{3} = \frac{8 \cdot 1}{1 \cdot 3} = \frac{8}{3} = 2 \frac{2}{3}\end{align*}
You will need \begin{align*}2 \ \frac{2}{3}\end{align*} cups of flour.
Multiplication Properties
Properties that hold true for addition such as the Associative Property and Commutative Property also hold true for multiplication. They are summarized below.
The Associative Property of Multiplication: For any real numbers \begin{align*}a, \ b,\end{align*} and \begin{align*}c,\end{align*}
\begin{align*}(a \cdot b)\cdot c = a \cdot (b \cdot c)\end{align*}
The Commutative Property of Multiplication: For any real numbers \begin{align*}a\end{align*} and \begin{align*}b,\end{align*}
\begin{align*}a(b) = b(a)\end{align*}
The Same Sign Multiplication Rule: The product of two positive or two negative numbers is positive.
The Different Sign Multiplication Rule: The product of a positive number and a negative number is a negative number.
#### Example B
Ayinde is making a dog house that is \begin{align*}3\frac{1}{2}\end{align*} feet long and \begin{align*}2\frac{2}{3}\end{align*} feet wide. How many square feet of area will the dog house be?
Solution:
Since the formula for area is
\begin{align*} area=length \times width\end{align*},
we plug in the values for length and width:
\begin{align*} area=3\frac{1}{2} \times 2\frac{2}{3} .\end{align*}
We first need to turn the mixed fractions into improper fractions:
\begin{align*} area=3\frac{1}{2} \times 2\frac{2}{3} =\frac{7}{2} \times \frac{8}{3}= \frac{7\times 8}{2\times 3} = \frac{56}{6}. \end{align*}
Now we turn the improper fraction back into a mixed fraction. Since 56 divided by 6 is 9 with a remainder of 2, we get:
\begin{align*} \frac{56}{6}=9\frac{2}{6}=9\frac{1}{3}\end{align*}
The dog house will have an area of \begin{align*}9\frac{1}{3}\end{align*} square feet.
Note: The units of the area are square feet, or feet squared, because we multiplied two numbers each with units of feet:
\begin{align*}feet \times feet=feet^2\end{align*},
which we call square feet.
#### Example C
Doris’s truck gets \begin{align*}10 \frac{2}{3}\end{align*} miles per gallon. Her tank is empty so she puts in \begin{align*}5 \frac{1}{2}\end{align*} gallons of gas.
How far can she travel?
Solution: Begin by writing each mixed number as an improper fraction.
\begin{align*}10 \frac{2}{3} = \frac{32}{3} && 5 \frac{1}{2} = \frac{11}{2}\end{align*}
Now multiply the two values together.
\begin{align*}\frac{32}{3} \cdot \frac{11}{2} = \frac{352}{6} = 58\frac{4}{6} = 58\frac{2}{3}\end{align*}
Doris can travel \begin{align*}58 \ \frac{2}{3}\end{align*} miles on 5.5 gallons of gas.
### Guided Practice
Anne has a bar of chocolate and she offers Bill a piece. Bill quickly breaks off \begin{align*}\frac{1}{4}\end{align*} of the bar and eats it. Another friend, Cindy, takes \begin{align*}\frac{1}{3}\end{align*} of what was left. Anne splits the remaining candy bar into two equal pieces, which she shares with a third friend, Dora. How much of the candy bar does each person get?
Solution: Think of the bar as one whole.
\begin{align*}1- \frac{1}{4} = \frac{3}{4}\end{align*}. This is the amount remaining after Bill takes his piece.
\begin{align*}\frac{1}{3} \times \frac{3}{4} = \frac{1}{4}\end{align*}. This is the fraction Cindy receives.
\begin{align*}\frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\end{align*}. This is the amount remaining after Cindy takes her piece.
Anne divides the remaining bar into two equal pieces. Every person receives \begin{align*}\frac{1}{4}\end{align*} of the bar.
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Multiplication of Rational Numbers (8:56)
Multiply the following rational numbers.
1. \begin{align*}\frac{3}{4} \times \frac{1}{3}\end{align*}
2. \begin{align*}\frac{15}{11} \times \frac{9}{7}\end{align*}
3. \begin{align*}\frac{2}{7} \cdot -3.5\end{align*}
4. \begin{align*}\frac{1}{13} \times \frac{1}{11}\end{align*}
5. \begin{align*}\frac{7}{27} \times \frac{9}{14}\end{align*}
6. \begin{align*}\left (\frac{3}{5} \right )^2\end{align*}
7. \begin{align*}\frac{1}{11} \times \frac{22}{21} \times \frac{7}{10}\end{align*}
8. \begin{align*}5.75 \cdot 0\end{align*}
In 9 – 11, state the property that applies to each of the following situations.
1. A gardener is planting vegetables for the coming growing season. He wishes to plant potatoes and has a choice of a single 8 by 7 meter plot, or two smaller plots of 3 by 7 meters and 5 by 7 meters. Which option gives him the largest area for his potatoes?
2. Andrew is counting his money. He puts all his money into $10 piles. He has one pile. How much money does Andrew have? 3. Nadia and Peter are raising money by washing cars. Nadia is charging$3 per car, and she washes five cars in the first morning. Peter charges \$5 per car (including a wax). In the first morning, he washes and waxes three cars. Who has raised the most money?
1. Teo is making a flower box that is 5-and-a-half inches by 15-and-a-half inches. How many square inches will he have in which to plant flowers?
Mixed Review
1. Compare these rational numbers: \begin{align*}\frac{16}{27}\end{align*} and \begin{align*}\frac{2}{3}\end{align*}.
2. Define rational numbers.
3. Give an example of a proper fraction. How is this different from an improper fraction?
4. Which property is being applied? \begin{align*}16 - (-14) = 16 + 14 = 30\end{align*}
5. Simplify \begin{align*}11 \frac{1}{2} + \frac{2}{9}\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
TermDefinition
Associative Property of Multiplication For any real numbers $a, \ b,$ and $c,$ $(a \cdot b)\cdot c = a \cdot (b \cdot c)$.
Commutative Property of Multiplication For any real numbers $a$ and $b,$ $a(b) = b(a)$.
Different Sign Multiplication Rule The product of a positive number and a negative number is a negative number.
Same Sign Multiplication Rule The product of two positive or two negative numbers is positive.
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# Lesson 26
Modeling with Systems of Inequalities in Two Variables
The organizers of a conference needs to prepare at least 200 notepads for the event and have a budget of \$160 for the notepads. A store sells notepads in packages of 24 and packages of 6. This system of inequalities represent these constraints: $$\begin{cases} 24x+6y\geq200\\16x + 5.40y\leq160 \end{cases}$$ 1. Explain what the second inequality in the system tells us about the situation. 2. Here are incomplete graphs of the inequalities in the system, showing only the boundary lines of the solution regions. Which graph represents the boundary line of the second inequality? 3. Complete the graphs to show the solution set to the system of inequalities. 4. Find a possible combination of large and small packages of notepads the organizer could order. ### Solution For access, consult one of our IM Certified Partners. (From Unit 2, Lesson 25.) ### Problem 2 A certain stylist charges \$15 for a haircut and \$30 for hair coloring. A haircut takes on average 30 minutes, while coloring takes 2 hours. The stylist works up to 8 hours in a day, and she needs to make a minimum of \$150 a day to pay for her expenses.
1. Create a system of inequalities that describes the constraints in this situation. Be sure to specify what each variable represents.
2. Graph the inequalities and show the solution set.
3. Identify a point that represents a combination of haircuts and and hair-coloring jobs that meets the stylist’s requirements.
4. Identify a point that is a solution to the system of inequalities but is not possible or not likely in the situation. Explain why this solution is impossible or unlikely.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 2, Lesson 25.)
### Problem 3
Choose the graph that shows the solution to this system: $$\begin{cases} y > 3x + 2 \\ \text-4x+3y \leq12 \end{cases}$$
A:
B:
C:
D:
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 2, Lesson 24.)
### Problem 4
Match each inequality to the graph of its solution.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 2, Lesson 23.) |
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# Unit 3
Inferential Statistics
Chapter 6: Confidence Intervals
Chapter 7: Hypothesis Testing
6-1
Chapter 6
Confidence Intervals
6.1: Estimating ( known)
6.2: Estimating ( unknown)
6.3: Estimating p
6-2
## STA 2023: Elementary Statistics
Section 6.1: Estimating a Population Mean ( known)
Chapters 6 and 7 start inferential statistics: drawing conclusions about population using sample data.
There are two major types of inferential statistics. Use sample data to:
1. Estimate the value of a population parameter (Confidence Intervals)
2. Test some claim about a population parameter (Hypothesis Testing).
## Parameter and Statistic Notation
Parameter
Idea:
Statistic
x
N
Quantitative Data
(Means)
Population Mean: =
Qualitative Data
(Proportions)
Population Proportion: p =
Sample Mean: x =
x
N
x
n
Sample Proportion: p =
x
n
## 1. Estimate the mean weight of an adult male ().
2. Estimate the proportion of adults who currently approve of the president (p).
Point Estimates
A point estimate for a parameter is a single value estimate.
1. For , the best point estimate is x :
p p
## Example 1: Find the point estimate.
a. A sample of 100 adult men has a mean weight of x = 170 lbs.
Find the point estimate for , the mean weight of all adult males.
## b. A sample of 500 adults showed 240 approved of the president.
Find the point estimate for p, the proportion of all adults who approve of the president.
6-3
## Interval Estimates (Confidence Intervals)
1. An interval estimate is a range of values that is likely to contain the parameter.
2. Format:
= x E
or
p= p E
## 3. A confidence interval is an interval estimate that comes with a level of confidence, c,
where c is the probability the interval actually contains the parameter.
## Common Confidence Levels:
Interpreting confidence levels:
Confidence Level, c
c
1-c
90%
0.90
0.10
95%
0.95
0.05
99%
0.99
0.01
## Example 2: Write the confidence interval.
a. A sample of 100 adult men has a mean weight of x = 170 lbs. A 95% confidence interval for
the mean has a margin of error of 3 lbs.
b. A sample of 500 adults gave a 48% approval rating for the president. (90% CI, MOE 3%).
.
Confidence Intervals for Population Mean ( known)
Estimating a population mean, = x E :
1. Get a random sample of size n and compute x .
2. If population is normal or n 30 and is known:
x N ,
z distribution
n
## 3. To compute the E, first need to find critical values:
Critical Values, zc
1. Confidence levels, c, are associated with special z-scores, called critical values, zc
2. To find zc, let the confidence level c represent the middle area of the standard normal curve.
Then find the z-scores than mark the boundaries.
6-4
Example 3.
Find the critical values associated with the following confidence levels.
a. 95%
b. 90%
c. 99%
## Confidence Interval for ( known)
Given: is known, the sample is random, and either the population is normal or n 30 :
1. Find the sample statistics n and x
2. Compute the ( c 100% ) confidence interval for a population mean, using:
=
x E , where E =
zc
n
Interval Notation:
= x E
or
x E < < xE
or
( x E, x + E )
Rounding: Round confidence intervals the same as the statistics: one more decimal than data.
6-5
Example 4: A study by researchers at the University of Maryland found the body temperatures of 106
healthy adults had a mean temperature of 98.20F. Assume the population standard deviation is
known to be 0.62F.
a.
Construct a 90%, 95%, and 99% confidence interval for the mean body temperature of all
## c. As confidence increases, what happens to the margin of error?
6-6
Example 5:
The Center for Education Reform conducted a study to find the mean salary of public school
teachers. A random sample of 300 teachers had a mean salary of \$41,820. From past studies, the
population standard deviation is known to be \$2700.
a. Find the 95% confidence interval for the mean salary of all public elementary teachers.
b. If 2000 teachers had been sampled with the same results, what happens to the margin of
error?
6-7
## Determining the Sample Size Required to Estimate
To determine what sample size is needed to estimate , first choose:
1. Level of confidence
2. desired Margin of Error
z
Then, minimum sample size required to estimate a population mean is n = c ,
E
where n is always rounded up* to the nearest whole number.
2
## If is unknown (common), there are two common ways to estimate it:
1. Get a preliminary sample with n 30 and then use s to approximate , or,
range
2. If you can estimate the range, then approximate using
4
Example 6.
a. A realtor agency wants to estimate the mean number of days a home is on the market before it
sells. How many homes must be sampled if they want to be 95% confident the sample mean is
within 10 days of the true population mean? Assume a preliminary study suggests = 30 days.
b. Repeat the above example with different margin of errors (5 days, 3 days, 2 days, 1 day, etc.)
What level of confidence and margin of error do you think leads to an acceptable sample size?
6-8
Example 7.
Suppose you want estimate the mean weight of an adult male, to be in error by no
more than 3 pounds, and with 95% confidence.
a. Use your knowledge of mens weights to estimate the range and then estimate .
b. How many men would you have to weigh if you want to estimate the mean weight within 3
pounds, and with 95% confidence?
c. The CDC conducts annual health surveys to guide research and policies to for preventing
premature mortality (BRFSS System). If they wanted to estimate the mean weight of an
adult male within 3 pounds with 99% confidence, what sample size would they need?
6-9
## STA 2023: Elementary Statistics
Section 6.2: Estimating a Population Mean ( unknown)
Choosing Probability Distributions for Confidence Intervals
known:
unknown:
If is known and either the population is normal or n 30, then x N ,
:
n
## use the standard normal distribution and critical z-scores, zc find E.
If is unknown and either the population is normal or n 30,
the x ' s will have a t-distribution and use critical t-scores, tc, to find E.
## Students t-distribution (with n 1 degrees of freedom)
The t-distribution is a family of curves, each determined by a parameter called the degrees of freedom.
The degrees of freedom are equal to 1 less than the sample size: d.f. = n -1:
Sample Size, n
n = 25
n = 15
n=9
Degrees of Freedom, n 1
d.f = 24
d.f. = 14
d.f = 8
## Properties of the t distribution:
1. The t-distribution is bell-shaped and symmetric about its mean.
2. The mean = 0, but varies more than the normal distribution.
3. As the degrees of freedom increase, the t-distribution approaches the normal distribution.
For n > 30, the t-distribution the z-distribution.
## C100% Confidence Interval for ( unknown)
Given: is unknown, the sample is random, and either the population is normal or n 30 :
1. Find the sample statistics n, x and s
2. Compute the ( c 100% ) confidence interval for a population mean, using
s
n
x E , where E =
tc
=
, and t has n 1 degrees of freedom
6-10
## Critical Values, tc , for Confidence Intervals
Example 1: Use the t-distribution table to find the critical values. Make a sketch for each.
a. 90% confidence interval, n = 10
## c. 99% confidence interval, n = 25
d. Compare tc to zc .
6-11
Example 2: A city health department wishes to determine whether the mean bacteria count per unit
volume of water at a lake beach is within the safety level of 200. A researcher collected 32 water
samples and found the mean bacteria count to be 204.2 bacteria/100 ml and the standard deviation
of 14.6 bacterial/100 ml.
a. Construct a 95% confidence interval for the mean bacteria count per unit volume in the
entire lake.
b. Is it possible the bacteria is within the prescribed limits?
Example 3: How much do mountain lions weigh? The 77th Annual Report of the New Mexico
Department of Game and Fish gave the following information. Adult mountain lions (18 months and
older) captured and released for the first time in the San Andres Mountains gave the following weights
(in pounds). Assume the weights of mountain lions are approximately normally distributed.
68
104
128
122
60
64
90
80
112
82
a. Find a 90% confidence interval for the mean weight of all mountain lions in the specified region.
6-12
## Estimating : Choosing the correct distribution
When finding confidence intervals, it is very important to use the correct distribution.
Besides using whether is known or unknown, the distribution of the population and n is also needed:
**Note:
If the population is not normally distributed and n < 30, neither the z or t distributions can be used!!
Example 4: State which distribution (z or t) would be used to create a 95% confidence interval for
the population mean, , in the following examples. If neither distribution could be used, explain why.
a. The gestation period of humans is normally distributed with = 16 days. A sample of size
n = 12 has a mean of 267 days.
b. For estimating the mean amount of rainfall during the month of April in Chicago, a simple
random sample of 36 years has a mean of 3.63 inches and a standard deviation of 1.63 inches.
c. The starting salaries of law school graduates are skewed right with a population standard
deviation of \$24,000. A random sample of 15 law school graduates has a mean starting salary
of \$63,000.
6-13
## STA 2023: Elementary Statistics
Section 6.3 Estimating a Population Proportion, p
## Qualitative data and Proportions
Population proportion: p =
Sample proportion: p =
X
; the proportion (or %) of the population that has the attribute.
N
x
; the proportion (or %) of a sample that has the attribute.
n
Other notation: q = 1 p ; the proportion (%) of the population that does not have the attribute.
q = 1 p ; the proportion (%) of the sample that does not have the attribute.
## Estimating a population proportion, p
1. Point estimate: p p
2. Confidence Interval estimate: p= p E
3. Notation:
p= p E or p E < p < p + E
or ( p E , p + E )
pq
p=
p E ; where E =
zc
n
## 1. Use the z-distribution provided: np 5 and nq 5
2. Round proportions and confidence intervals to 3 significant digits.
6-14
Example 1:
In a study conducted by the Centers for Disease Control in 2007, 490 out of 1,700 randomly
selected Americans with a high school diploma were found to be obese.
a. Find a 95% confidence interval for the proportion of high school graduates who are obese.
b. When the same study was performed on Americans with 4 or more years of college education,
if was reported that 20.2% of college graduates are obese (based on 95% confidence, and a
margin of error of 2% ). Based on these results, does the obesity rate for high school
Example 2:
An internet company claims that 95% of the products ordered will be mailed within 48 hours of an
order being placed. A random sample of 200 orders showed 174 of the orders were mailed on time.
a. Using a 99% level of confidence, construct a confidence interval for the percentage of all
orders that were shipped within 48 hours.
b. Based on your result, does it appear the companys claim is correct? Explain.
6-15
## Sample Size for Estimating Population Proportion p :
To determine what sample size is needed to estimate p, first choose:
1. Level of confidence
2. desired Margin of Error
Then, minimum sample size required to estimate a population proportion is:
1. When an estimate of p is known:
or
z 2 pq
n= c 2
E
zc 2 ( 0.25 )
n=
E2
## Again, n is always rounded up to the nearest whole number.
Example 3: For the internet company in the previous example,
a. What sample size should have been used if they wanted a 3% MOE?
b. What sample size would be needed if they wanted to be 95% confident with a 3% MOE?
Example 4: The National Institute of Drug Abuse wants to estimate the percentage of U.S. teenagers
aged 12-18 who have used any illicit drug other than marijuana in the past year. How many
teenagers must be surveyed if they want to be 95% confident that the sample has a margin of error
of no more than 2% ?
a. Assume that there is no available information that could be used as an estimate of p .
b. Assume they use a 2014 estimate that 11% of teenagers had used an illicit drug in the past
year.
6-16
Chapter 7
Hypothesis Testing
7.1: Basic of Hypothesis Testing
7.2: Tests for ( known)
7.3: Tests for ( unknown)
7.4: Tests for p
7-1
## STA 2023: Elementary Statistics
Section 7.1/7.2: Fundamentals of Hypothesis Testing
Recall: Inferential Statistics draw conclusions about a population based on sample data:
Confidence Intervals: Estimate the value of a population parameter (using sample statistics).
Hypothesis Tests: Tests a claim (hypothesis) about a population parameter (using sample statistics).
Hypothesis test: Tests a claim (hypothesis) about a population parameter, using sample statistics.
The claim can be a historical value, a business claim, a product specification, etc.
Components of a hypothesis test:
A.
B.
C.
D.
## The Null and Alternative Hypotheses
The Test Statistic and making decisions
Types of Errors in hypothesis testing
A. The Hypotheses
Null hypothesis, H0 :
## - statement about the parameter (, p, )
- assumed true until proven otherwise
- must contain equality sign: , , or =
Alternative Hypothesis, Ha :
## - statement about the parameter that must be true if Ho is false.
- must contain: >, <, or
H o : = 0
H a : 0
or
H o : 0
H a : < 0
or
H o : 0
H a : > 0
Example 1:
## Write the claim, the null and alternative hypotheses symbolically.
Identify which hypothesis is the original claim.
Make sure you use the correct symbol for the parameter ( , p,or )
a. Claim: A science group claims normal body temperatures is less than 98.6F.
b. A safety agency claims that at least 70% of cell phone users text while driving.
c. A quality control engineer says the diameters of bike tires have a standard deviation of 0.05
inches.
7-2
B. Making Decisions
1. State hypotheses. **Assume H 0 is true.
2. Collect sample data (test statistic).
3. Make a decision about H 0 :
a. Reject H0 ( accept H a ) ,or,
b. Fail to Reject H0 ( cant accept Ha).
Example 2. Consider the scenario of a trial, where the defendant is presumed innocent until proven
otherwise. Set up a hypothesis test and explain the two decisions that can be made.
C. Types of Errors
1. Type I error: Reject H 0 H 0 is actually true
2. Type II error: FTR H 0 H 0 is actually false
The symbol is used to represent the probability of a Type I error: P ( Type I error ) =
The symbol is used to represent the probability of a Type II error: P ( Type II error ) =
Note: is set in advance of starting the test, and is also known as the level of significance of the test.
Example 3: Identify the Type I and Type II errors associated with the following claims or
hypotheses, and the possible consequences of those errors.
a. From the trial in the previous example, the claim of innocent until proven guilty.
b. A bottling company claims the mean amount of coke in a coke can is 12 oz. This is
periodically checked by quality control personnel.
7-3
## D. Writing conclusions (in terms of the original claim)
After making a decision about the null, a final conclusion is written about the original claim. How the
conclusion is worded depends on the decision, and whether the claim was in Ho or Ha.
Ho contains claim:
1. Reject Ho: There is enough evidence to reject claim that Ho
2. FTR Ho: There is not enough evidence to reject the claim that Ho
Ha contains claim:
1. Reject Ho: There is enough evidence to support the claim that Ha
2. FTR Ho: There is not enough evidence to support the claim that Ha
Example 4: Write the conclusion for the hypothesis test based on the given decision.
a. Researchers at the University of Maryland claims that the mean body temperature of healthy
adults is less than 98.6F. The decision was Reject H 0 .
b. A safety agency claims that at least 70% of cell phone users text while driving.
The decision was FTR H 0 .
7-4
## E. How to make a decision about the Null Hypothesis
1. Write claim. State H o , H a . Assume H 0 is true. ( = 0 ). Determine the tails of the test.
2. Get test statistic: value from sample data; used to make a decision about H 0 .
For tests about , use x . Convert x to a standard score ( z or t )
## a. Critical Region Method, or
b. P-value method
## Tails of the Test and the Critical Region Method
Tails of the test: Assuming the null hypothesis H 0 is true, where would your sample mean
x have to fall to convince you to reject
H 0 (and accept H a )?
Critical Region: Region under the curve where values of test statistic that would be unlikely,
assuming H 0 is true.
Area of critical region = ( is the significance level of the test)
Critical Values: Values (zc or tc) that bound the critical region
Decision:
## If test statistic x falls in the critical region, reject Ho.
If test statistic x falls in the non-critical region, FTR Ho.
Critical Region
H a : < 0
H a : > 0
H a : 0
7-5
## The P-value Method of Making Decisions
Instead of determining whether your sample mean is unusual by whether or not it falls in a
critical region, be more precise and find the p-value:
P-value: Probability of getting a test statistic as extreme or more than the one from the sample data,
assuming Ho is true.
1.
2.
2.
3.
4.
Select level of significance, . (How unusual would your sample have to be to reject Ho?)
Gather a sample. Compute x and covert to a standard test statistic, z *.
Determine the tails of the test.
Compute the P-value.
Make a decision: a. If P-value (unusual), Reject Ho.
b. If P-value > (not unusual), FTR Ho.
## 1. Left Tailed Test
H 0 : 0 (Assume true)
H a : < 0
H 0 : 0
H a : > 0
## 3. Two Tailed Test
H 0 : = 0
H a : 0
7-6
Example 5:
Make a decision about the Ho, under the given conditions with the indicated method.
## Critical Region Method
Sketch, shade the critical region (area = ),
find the critical values, make a decision
left-tailed test, = 0.05, z = 1.73
P-value Method
find the P-value, make a decision
left-tailed test, = 0.05, z = 1.73
right-tailed test,
=
0.01,
=
z 2.07
right-tailed test,
=
0.01,
=
z 2.07
two-tailed test,
z 1.81
=
0.05,
=
two-tailed test,
=
0.05,
=
z 1.81
7-7
## STA 2023: Section 7.2
Hypothesis Test about Population Mean ( known)
Method: Given is known, the sample is random, and either the population is normal or n 30,
use the z-test for the population mean :
1. Write the claim in terms of .
Use the claim to write H o , H a . Assume H 0 is true. (i.e., = 0 )
Determine whether it is a left tailed test, right-tailed test, or 2-tailed test.
Note the level of significance, .
2. Get sample test statistic, x .
x
n
## 3. Sketch the sampling distribution for x .
Indicate on the sketch whether it is a left tailed test, right-tailed test, or 2-tailed test.
Make a decision about H 0 based on the test statistic and given significance level , using:
a. Critical Region method: Shade the critical region, find the critical values, zc
If z is in the critical region, reject H 0
If z is not in the critical region, fail to reject H 0
b. P-value method: Find the probability of the test statistic being as extreme or more as the
sample:
If P < , reject H 0
If P , fail to reject H 0
4. Interpret your decision by writing a conclusion in terms of the original claim.
7-8
For each of the following: State the claim, write the hypotheses, find the test statistic, sketch the sampling
distribution, use both the critical region method and the p-value method to make a decision about Ho, and
interpret your decision in terms of the original claim.
Example 1:
Fire insurance rates for a certain suburb are based on past data that states mean distance from a home
in the community to the nearest fire department is 4.7 miles. Residents in the community claim the
mean distance is less, and want their rates lowered. From a random sample of 64 homes in the
suburb, the mean distance was 4.3 miles. Assume the population standard deviation was 2.4 miles.
Use the data to test the claim that the mean distance to the nearest fire station is less than 4.7 miles.
Use a 1% level of significance.
7-9
Example 2:
Golf course designers are concerned that old courses are becoming obsolete because new equipment
enables golfers to hit the ball so far: In effect, golf courses are shrinking. One designer claims that
new courses need to be built with the expectation that players will be able to hit the ball more than 250
yards (with their drivers), on average. Suppose a sample of 135 golfers is tested, and their mean
driving distance is 256.3 yards. Assume the population standard deviation is 43.4 yards. At = 0.05 ,
can you support the designers claim?
7-10
Example 3:
Many colleges across a certain state have long used the CPT placement test for placing students into
the appropriate math courses. Historically, the mean score on the CPT has been 82. The school is
considering switching to a new placement test called COMPASS. The COMPASS test is easier to
administer and less expensive, but the college is unwilling to switch if the COMPASS test results in a
different mean than the CPT. An independent testing agency tested 36 students, which gave a mean
of 79. Past studies show the population is normally distributed with a standard deviation of 8. Use
this information to test the COMPASS claim that their test has the same mean as the CPT. Use
= 0.05.
7-11
## STA 2023: Elementary Statistics
Section 7.3: Hypothesis Test for Population Mean ( unknown)
Method: Given is unknown, the sample is random, and either the population is normal or n 30,
use the t-test for the population mean :
## 1. Write the claim in terms of .
Use the claim to write H o , H a . Assume H 0 is true. (i.e., = 0 )
Determine whether it is a left tailed test, right-tailed test, or 2-tailed test.
Note the level of significance, .
2. Get the sample test statistic, x , and s.
For unknown and either x N or n 30, use the t-distribution with n 1 degrees of freedom.
x
Convert x to a standard test statistic: t =
s n
## 3. Sketch the sampling distribution for x .
Indicate on the sketch whether it is a left tailed test, right-tailed test, or 2-tailed test.
Make a decision about H 0 based on the test statistic and given significance level , using:
a. Critical Region method: Shade the critical region, find the critical values, zc
If z is in the critical region, reject H 0
If z is not in the critical region, fail to reject H 0
b. P-value method: Find the probability of the test statistic being as extreme or more as the
sample:
If P < , reject H 0
If P , fail to reject H 0
## 4. Interpret your decision by writing a conclusion in terms of the original claim.
7-12
Example 1: Use the t-distribution to make a decision about the Ho under the given conditions.
Critical Region Method
Sketch, shade the critical region (area = ),
find the critical values, make a decision
## P-value Method (software)
find the P-value, make a decision
## a. right tailed test,
=
0.01,
=
=
n 25,
t 2.516
=
=
=
0.01,
right tailed test,
n 25,
t 2.516
## left tailed test, = 0.05, n = 10, t = 1.63
c. 2 tailed test,
=
0.05,
=
n 35,
=
t 2.103
=
0.05,
=
=
2 tailed test,
n 35,
t 2.103
7-13
For each of the following: State the claim, write the hypotheses, find the test statistic, sketch the sampling
distribution, use either the critical region method or the p-value method to make a decision about Ho, and
write your conclusion in terms of the original claim.
Example 2:
A group of researchers conducted a study of the birth weights of babies born to women who smoked
regularly while they were pregnant. For a sample of 32 such women, the mean birth weight of the
babies was 3,160 grams and the standard deviation of 440 grams. Test the hypothesis that the mean
birth weight of babies born to mothers who smoke is less than 3370 g (the mean birth weight for all
babies). Use = 0.01.
7-14
Example 3.
a. A car company claims that the mean gas mileage for its luxury sedan is more than 28 mpg in
the city. To test the claim, the company randomly selects 10 sedans, giving the mpg shown below.
Use a level of significance of = 0.01. Assume the mpg for all such sedans has an approximately
normal distribution.
29.1
30.6
27.8
32.4
28.0
27.9
29.2
30.1
28.0
28.4
b. Notice the claim was more than 28 mpg. Why not say at least 28 mpg?
7-15
## STA 2023: Elementary Statistics
Section 7.4: Testing a claim about a Population Proportion
## Recall: Qualitative data and Proportions
Population proportion: p =
X
; the proportion of the population that has the attribute.
N
q = 1 p ; the proportion of the population that does not have the attribute.
Sample proportion: p =
x
; the proportion of a sample that has the attribute.
n
q = 1 p ; the proportion of the sample that does not have the attribute. .
H o : p = p0
H a : p p0
H o : p p0
H o : p p0
or
( Two tailed )
H a : p < p0
or
( Left tailed )
H a : p > p0
( Right tailed )
## Hypothesis Test for Population Proportion
1. Write the claim in terms of p.
Use the claim to write H o , H a . Assume H 0 is true. (i.e. p = p0 )
Determine whether you are doing a left tailed, right tailed, or 2-tailed test.
pq
For large samples, np 5 and nq 5 : p N p,
.
n
p p
**
Convert p to a standard test statistic, z: z =
pq
n
3. Sketch the sampling distribution for p .
Indicate on the sketch whether youre doing a left-tailed test, a right-tailed test, or a 2-tailed test.
Based on the test statistics, make a decision about H 0 using:
a. Critical Region Method (shade critical region, find critical values, zc).
b. P-value method: shade and find the P-value area.
## 4. Interpret your decision in terms of the original claim.
Note**: When the calculation of p results in a decimal with many places, store the number on your
calculator and use all the decimals when evaluating the z test statistic.
Large errors in z and the P-value can result from rounding p too much.
7-16
For each of the following: State the claim, write the hypotheses, find the test statistic, sketch the sampling
distribution, use the critical region or P-value method to make a decision about Ho, and interpret your
decision in terms of the original claim.
Example 1:
In a survey conducted by the Gallop Organization in 2011, 456 of 1012 adults aged 18 years or older
said they had a gun in the house. In 2010, 41% of households had a gun. Is there sufficient evidence
to support the claim that the proportion of households that have a gun has changed from 2010 to
2011? Use a 5% level of significance.
7-17
Example 2:
Pepcid is a drug that can be used to heal duodenal ulcers. Suppose the manufacturer of Pepcid claims
that more than 80% of patients are healed after taking 40 mg of Pepcid every night for 8 weeks. In
clinical trials, 148 of 178 patients suffering from duodenal ulcers were healed after 8 weeks. Test the
manufacturers claim at = 0.01 level of significance.
7-18
## Exam 3 Review: Chapters 6 & 7 (Larson, 6th ed)
Below is a list of topics for the exam. Review class notes, homework, and quizzes.
Complete the review problems in MML.
Additional review can be found in the electronic textbook (see next page).
Chapter 6: Confidence Intervals
1. Construct and interpret confidence intervals
2. Estimating :
= x E
a. x N or n 30 and known:
b. x N or n 30 and unknown:
3. Estimating p :
p=
p E ;
n
s
use t distribution: E = tc
n
use z distribution: E = zc
pq
E=
zc
n
(provided np and nq 5 )
4. Sample sizes
a. for :
z
n= c
E
b. for p:
n=
zc 2 ( 0.25 )
zc 2 pq
Proportion
,
unknown
p
q
n
=
Proportion ( p , q known )
(
)
2
E
E2
## Chapter 7: Hypothesis Testing
1. Understand: claim, null and alternative hypothesis, decisions, Type I and Type II errors,
significance level
2. Conduct hypothesis tests for population mean or a population proportion p by:
State the claim, write the hypotheses, find the standard test statistic, sketch the sampling
distribution, use either the critical region method or p-value method to make a decision about Ho,
and interpret your decision in terms of the original claim.
a. Critical region method: Shade the critical region (area = ), find the critical values
b. P-value method: Find probability of getting a test statistic as extreme or more as the
sample; compare to
3. Test statistics and distributions:
x N or n 30, known:
## b. For tests about :
x N or n 30, unknown:
## c. For tests about p :
( np > 5, nq > 5 ):
use z distribution: z =
n
x
use t distribution: t =
s n
use z distribution, z = p p
pq
n
7-19
Review Problems
*1. MyMathLab: Test 3 Review (Chapters 6 and 7)
2. Optional Review: Chapter Quizzes with Videos
Chapter Quizzes: MML, Electronic Textbook, Chapter #, Chapter Quiz and Videos
Student Solutions Manual: MML, Student Solutions Manual, Chapter #
Chapter 6 Quiz: All 1-5;
Chapter 7 Quiz: All 1-5
7-20 |
# Techniques Of Evaluating Limits
## EVALUATION OF LIMITS:
Now we discuss the various methods used in obtaining limits. Each method will be accompanied by some examples illustrating that method.
(A) DIRECT SUBSTITUTION
This already finds mention at the start of the current section, where we saw that for a continuous function, the limit can be obtained by direct substitution.
This is because, by definition of a continuous function (at x = a):
$${\text{LHL}}\;\left( {{\text{at}}\;\,x = a} \right) = {\text{RHL}}\;\left( {{\text{at}}\;\,x = a} \right)\; = f\left( a \right)$$
Hence, for example, all polynomial limits can be evaluated by direct substitution.
Some examples make all this clear:
(i)$$\mathop {\lim }\limits_{x \to 1} \,{x^3} + 1 = 2$$ (ii) $$\mathop {\lim }\limits_{x \to 2} \,\,5{x^2} + 3x + 1 = 27$$ (iii) $$\mathop {\lim }\limits_{x \to - 1} \,\,4{x^3} + 4 = 0$$ (iv) $$\mathop {\lim }\limits_{y \to 1} \,\,|y| + 1 = 2$$ (v) $$\mathop {\lim }\limits_{x \to 5} \,\,\frac{{5{x^2} + 4}}{{2x + 7}} = \frac{{129}}{{17}}$$ (vi) $$\mathop {\lim }\limits_{x \to 8} \,\,\frac{{{x^3} + 1}}{{x + 1}} = \frac{{513}}{9} = 57$$
and so on
(B) FACTORIZATION
We saw an example of this method in evaluating \begin{align}\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x - 1}}\end{align} .
In such forms, the limit is indeterminate due to a certain factor occuring in the expression (For example, in the limit above, (x – 1) occurs in both the numerator and denominator and makes the limit indeterminate, of the form $$\frac{0}{0}$$ ) . Factorization leads to cancellation of that common factor and reduction of the limit to a determinate form.
(i) \begin{align}\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{(x - 1)({x^2} + x + 1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + x + 1} \right) = 3\end{align}
Note that this limit is also of the form \begin{align}\mathop {\lim }\limits_{x \to 1} \frac{{{x^m} - 1}}{{x - 1}}\end{align} (whose limit is m)
(ii) \begin{align}\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 3x + 2}}{{{x^2} - 4}} = \mathop {\lim }\limits_{x \to 2} \frac{{(x - 1)(x - 2)}}{{(x - 1)(x + 2)}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 1}}{{x + 2}} = \frac{1}{4}\end{align}
(iii) \begin{align}\mathop {\lim }\limits_{x \to 0} \frac{{(1 + x)(1 + 2x)(1 + 3x) - 1}}{x}\end{align}
$$= \mathop {\lim }\limits_{x \to 0} \frac{{1 + 6x + 11{x^2} + 6{x^3} - 1}}{x}$$ $$=\mathop {\lim }\limits_{x \to 0} \frac{{6x + 11{x^2} + 6{x^3}}}{x}$$ $$= \mathop {\lim }\limits_{x \to 0} \,\,(6 + 11x + 6{x^2}) = 6$$
(iv) \begin{align}\mathop {\lim }\limits_{x \to 1} \,\,\frac{{{x^4} - 3x + 2}}{{{x^5} - 4x + 3}}\end{align}
We see that upon substitution of x = 1, both the numerator and denominator become 0.
Hence, (x – 1) is a factor of both the numerator and denominator (Factor theorem)
(C) RATIONALIZATION
In this method, the rationalization of an indeterminate expression leads to determinate one. The following examples elaborate this method.
(i) $$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {{x^2} + 1} - 1}}{{\sqrt {{x^2} + 16} - 4}}\,\,\left( {{\rm{of\; the\;indeterminate\;form }}\frac{{\rm{0}}}{{\rm{0}}}} \right)$$
\begin{align}{\rm\bf{(ii)}} \quad \mathop {\lim }\limits_{x \to \infty } &\left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} + 1} } \right) \left( {{\rm{\;of\;the\;indeterminate\;form\;}}\infty - \infty } \right)\\&= \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + x + 1} - \sqrt {{x^2} + 1} } \right) \times \frac{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} }}\\&= \mathop {\lim }\limits_{x \to \infty } \frac{{({x^2} + x + 1) - ({x^2} + 1)}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} }}\\&= \mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} }}\end{align}
Rationalization has led us to another indeterminate form of $$\frac{\infty }{\infty }$$ . However, it can easily be made determinate in the following manner:
$\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} + 1} }}$
Now, as $$x \to \infty ,\frac{1}{x} \to 0$$ and $$\frac{1}{{{x^2}}} \to 0$$
Hence, the limit above reduces to
$\frac{1}{{\sqrt {1 + 0 + 0} + \sqrt {1 + 0} }} = \frac{1}{2}$
(D) REDUCTION TO STANDARD FORMS
In this method, we try to reduce the given limit to one of the standard forms we studied earlier.
(i) $$\mathop {\lim }\limits_{x \to 0} {\left( {1 + \sin x} \right)^{2\cot x}}$$
This limit is of the indeterminate form $${1^\infty }$$ . $$\left( {{\rm{as}}\,x \to 0,\,\,\sin x \to 0\,\,{\rm{and}}\,\,\cot x \to \infty } \right)$$
We proceed as follows:
$$= {e^{\mathop {\lim }\limits_{x \to 0} 2\cos x}} = {e^2}(\cos x \to 1\,\,\,{\rm{as}}\,\,\,x \to 0)$$
(ii) \begin{align}\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{(\pi - 2x)}^3}}}\end{align}
This limit is of the indeterminate form $$\frac{0}{0}$$
Let $$x = \frac{\pi }{2} + h$$ so that as $$x \to \frac{\pi }{2},\,\,h \to 0.$$
\begin{align}\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{(\pi - 2x)}^3}}} &= \mathop {\lim }\limits_{h \to 0} \frac{{\cot \left( {\frac{\pi }{2} + h} \right) - \cos \left( {\frac{\pi }{2} + h} \right)}}{{{{( - 2h)}^3}}}\\ & = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{ - \tan \,h + \sin \,h}}{{ - 8{h^3}}}} \right)\\ &= \frac{1}{8}\,\,\,\mathop {\lim }\limits_{h \to 0} \,\,\frac{{ - \sin \,h + \frac{{\sin \,h}}{{\cos \,h}}}}{{{h^3}}}\ \frac{1}{8}\,\,\,\mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sin \,h}}{h}.\frac{{1 - \cos \,h}}{{{h^2}}}.\frac{1}{{\cos \,h}}\\ &= \frac{1}{8}\,\,\,\mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sin \,h}}{h}\,\,.\,\,\frac{{2{{\sin }^2}h/2}}{{4{{\left( {h/2} \right)}^2}}}\,\,.\,\,\frac{1}{{\cos \,h}}\\ &= \frac{1}{{16}}\,\,\,\mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sin \,h}}{h}\,\,.\,\,{\left( {\frac{{\sin\;h /2}}{{\left( {h/2} \right)}}} \right)^2}\,.\,\,\frac{1}{{\cos \,h}} \end{align}
This expression now only contains the limits $$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$$ and $$\mathop {\lim }\limits_{x \to 0} \cos x = 1$$
Hence, the final result is $$\frac{1}{{16}}$$
We will now see examples based on the methods discussed above. We urge you to first try out all these examples on your own before viewing the solutions. |
# function definition algebra
1
Dec
## function definition algebra
Evaluation is really quite simple. {\displaystyle y={\sqrt[{n}]{p(x)}}} {\displaystyle y=\pm {\sqrt {1-x^{2}}}.\,}. An algebraic function in m variables is similarly defined as a function We’ve now reached the difference. However, since functions are also equations we can use the definitions for functions as well. One more evaluation and this time we’ll use the other function. x Okay we’ve got two function evaluations to do here and we’ve also got two functions so we’re going to need to decide which function to use for the evaluations. {\displaystyle f(x)=\cos(\arcsin(x))={\sqrt {1-x^{2}}}} So, for each of these values of $$x$$ we got a single value of $$y$$ out of the equation. So, in the absolute value example we will use the top piece if $$x$$ is positive or zero and we will use the bottom piece if $$x$$ is negative. ) We know that we evaluate functions/equations by plugging in the number for the variable. . 3 Next we need to talk about evaluating functions. We do have a square root in the problem and so we’ll need to worry about taking the square root of a negative numbers. The fact that we found even a single value in the set of first components with more than one second component associated with it is enough to say that this relation is not a function. Indeed, interchanging the roles of x and y and gathering terms. Definition: A function is a relation from a domain set to a range set, where each element of the domain set is related to exactly one element of the range set. Which half of the function you use depends on what the value of x is. Now I know what you're asking. A letter such as f, g or h is often used to stand for a function.The Function which squares a number and adds on a 3, can be written as f(x) = x 2 + 5.The same notion may also be used to show how a function affects particular values. y Be careful with parenthesis in these kinds of evaluations. To avoid square roots of negative numbers all that we need to do is require that. where the coefficients ai(x) are polynomial functions of x, with integer coefficients. We will have some simplification to do as well after the substitution. = The list of second components associated with 6 is then : 10, -4. $$y$$ out of the equation. , {\displaystyle y=p(x)} 3 Thus, a function f should be distinguished from its value f(x0) at the value x0 in its domain. It is important to note that not all relations come from equations! A function may be thought of as a rule which takes each member x of a set and assigns, or maps it to the same value y known at its image.. x → Function → y. Bet I fooled some of you on this one! Note that we can have values of $$x$$ that will yield a single $$y$$ as we’ve seen above, but that doesn’t matter. = 1. + In that part we determined the value(s) of $$x$$ to avoid. We’ll evaluate $$f\left( {t + 1} \right)$$ first. This means that it is okay to plug $$x = 4$$ into the square root, however, since it would give division by zero we will need to avoid it. . ( While we are on the subject of function evaluation we should now talk about piecewise functions. Consider for example the equation of the unit circle: In this final part we’ve got both a square root and division by zero to worry about. y Recall the mathematical definition of absolute value. Functions are ubiquitous in mathematics and are essential for formulating physical relationships in the sciences. This is also an example of a piecewise function. y This is read as “f of $$x$$”. Note that it is okay to get the same $$y$$ value for different $$x$$’s. 4 Make sure that you deal with the negative signs properly here. In this section we will formally define relations and functions. ( We introduce function notation and work several examples illustrating how it works. p Let’s take a look at the following example that will hopefully help us figure all this out. Furthermore, even if one is ultimately interested in real algebraic functions, there may be no means to express the function in terms of addition, multiplication, division and taking nth roots without resorting to complex numbers (see casus irreducibilis). Recall, that from the previous section this is the equation of a circle. However, evaluation works in exactly the same way. Hopefully these examples have given you a better feel for what a function actually is. Circles are never functions. So, we will get division by zero if we plug in $$x = - 5$$ or $$x = 2$$. Now, let’s see if we have any division by zero problems. x For some reason students like to think of this one as multiplication and get an answer of zero. For the function f + g, f - g, f.g, the domains are defined as the inrersection of the domains of f and g For f/g , the domains is the intersection of the domains of f and g except for the points where g(x) = 0 ( There is however a possibility that we’ll have a division by zero error. Here are the evaluations. ) Note as well that the value of $$y$$ will probably be different for each value of $$x$$, although it doesn’t have to be. Quite often algebraic functions are algebraic expressions using a finite number of terms, involving only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power. There are various standard ways for denoting functions. If you put in … ) ) y There is only one arrow coming from each x; there is only one y for each x.It just so happens that it's always the same y for each x, but it is only that one y. The number under a square root sign must be positive in this section p Likewise, we will only get a single value if we add 1 onto a number. Let’s take the function we were looking at above. On the other hand, it’s often quite easy to show that an equation isn’t a function. G {\displaystyle G} is a subset of A × B {\displaystyle A\times B} , called the graph of f {\displaystyle \operatorname {f} } In addition the following two properties hold: 1. For the final evaluation in this example the number satisfies the bottom inequality and so we’ll use the bottom equation for the evaluation. , Therefore, let’s write down a definition of a function that acknowledges this fact. x From these ordered pairs we have the following sets of first components (i.e. , , q the list of values from the set of second components) associated with 2 is exactly one number, -3. , ln This can also be true with relations that are functions. On the other hand, relation #2 has TWO distinct y values 'a' and 'c' for the same x value of '5' . ) So the output for this function with an input of 7 is 13. Therefore, it seems plausible that based on the operations involved with plugging $$x$$ into the equation that we will only get a single value of We are much more interested here in determining the domains of functions. the first number from each ordered pair) and second components (i.e. Evaluating a function is really nothing more than asking what its value is for specific values of $$x$$. That means that we’ll need to avoid those two numbers. Now, at this point you are probably asking just why we care about relations and that is a good question. Now, go back up to the relation and find every ordered pair in which this number is the first component and list all the second components from those ordered pairs. For each $$x$$, upon plugging in, we first multiplied the $$x$$ by 5 and then added 1 onto it. As we’ve done with the previous two equations let’s plug in a couple of value of $$x$$, solve for $$y$$ and see what we get. i Do not get so locked into seeing $$f$$ for the function and $$x$$ for the variable that you can’t do any problem that doesn’t have those letters. Be careful. To some extent, even working mathematicians will conflate the two in informal settings for convenience, and to avoid appearing pedantic. x arcsin Hence there are only finitely many such points c1, ..., cm. It is easy to mess up with them. Formally, let p(x, y) be a complex polynomial in the complex variables x and y. Let’s do a couple of quick examples of finding domains. In mathematics, an algebraic function is a function that can be defined A function is said to be a One-to-One Function, if for each element of range, there is a unique domain. This is just a notation used to denote functions. The domain of an equation is the set of all $$x$$’s that we can plug into the equation and get back a real number for $$y$$. For example. Definition of Limit of a Function Cauchy and Heine Definitions of Limit Let $$f\left( x \right)$$ be a function that is defined on an open interval $$X$$ containing $$x = a$$. ± Think of an algebraic function as a machine, where real numbers go in, mathematical operations occur, and other numbers come out. that are polynomial over a ring R are considered, and one then talks about "functions algebraic over R". So, to keep the square root happy (i.e. An equivalent definition: A function (f) is a relation from a set A to a set B (denoted f: A ® B), such that for each element in the domain of A (Dom(A)), the f-relative set of A (f(A)) contains exactly one element. {\displaystyle \exp(x),\tan(x),\ln(x),\Gamma (x)} ( x With this case we’ll use the lesson learned in the previous part and see if we can find a value of $$x$$ that will give more than one value of $$y$$ upon solving. {\displaystyle a_{i}(x)} . If even one value of $$x$$ yields more than one value of $$y$$ upon solving the equation will not be a function. However, strictly speaking, it is an abuse of notation to write "let $${\displaystyle f\colon \mathbb {R} \to \mathbb {R} }$$ be the function f(x) = x ", since f(x) and x should both be understood as the value of f at x, rather than the function itself. > Sometimes, coefficients Therefore, the list of second components (i.e. The input of 2 goes into the g function. Let’s take a look at evaluating a more complicated piecewise function. An algebraic functionis a function that involves only algebraic operations, like, addition, subtraction, multiplication, and division, as well as fractional or rational exponents. Before starting the evaluations here let’s notice that we’re using different letters for the function and variable than the ones that we’ve used to this point. x ( In this case we won’t have division by zero problems since we don’t have any fractions. ( We also give a “working definition” of a function to help understand just what a function is. With the exception of the $$x$$ this is identical to $$f\left( {t + 1} \right)$$ and so it works exactly the same way. To see why this relation is a function simply pick any value from the set of first components. So, with these two examples it is clear that we will not always be able to plug in every $$x$$ into any equation. We also define the domain and range of a function. That isn’t a problem. In that example we constructed a set of ordered pairs we used to sketch the graph of $$y = {\left( {x - 1} \right)^2} - 4$$. The key here is to notice the letter that is in front of the parenthesis. To gain an intuitive understanding, it may be helpful to regard algebraic functions as functions which can be formed by the usual algebraic operations: addition, multiplication, division, and taking an nth root. As a final topic we need to come back and touch on the fact that we can’t always plug every $$x$$ into every function. The following definition tells us just which relations are these special relations. At this stage of the game it can be pretty difficult to actually show that an equation is a function so we’ll mostly talk our way through it. n , We looked at a single value from the set of first components for our quick example here but the result will be the same for all the other choices. |
# All the Trigonometric identities for the class 10 explained
Learn and know what are the important trigonometric identities for the class 10 students. In trigonometry chapter, after trigonometric ratios, trigonometric identities plays a crucial role.
For the students who are in class 10, trigonometric identities are useful in understanding further trigonometry concepts that will come in higher grade. At present, we will know what is trigonometric identity and how many trigonometric identities are there those comes in class 10.
## Meaning of identity in trigonometry:
We know that, in algebra we have identities. For example, a plus b whole square, a square minus b square and so on. Like in algebra, in trigonometry also we have identities. Identity meaning is same in both the cases. For any value of θ, the equation in which trigonometric ratios are involved will be satisfied, which means L.H.S and R.H.S equal we get.
### How many trigonometric identities are there in class 10 syllabus?
Especially, for the class 10 there are 3 main trigonometric identities are there. All these three trigonometric identities are very important to class 10 students.
#### List of trigonometric identities for the class 10:
Important trigonometric identities for class 10 are as follows
The first trigonometric identity is
$\sin ^{ 2 }{ \theta }$ + $\cos ^{ 2 }{ \theta }$ = 1
From the above trigonometric identity we also get,
$\sin ^{ 2 }{ \theta }$ = 1 – $\cos ^{ 2 }{ \theta }$
$\cos ^{ 2 }{ \theta }$ = 1- $\sin ^{ 2 }{ \theta }$
The second trigonometric identity is
$\sec ^{ 2 }{ \theta }$$\tan ^{ 2 }{ \theta }$ = 1
Note:
Here we should not consider value as odd multiple of 90 degrees.
From the above trigonometric identity we also get,
$\sec ^{ 2 }{ \theta }$ = 1 + $\tan ^{ 2 }{ \theta }$
$\tan ^{ 2 }{ \theta }$ = $\sec ^{ 2 }{ \theta }$ – 1
The third trigonometric identity is
$\cosec ^{ 2 }{ \theta }$$\cot ^{ 2 }{ \theta }$ = 1
Note:
Value we should not take integral multiple of 180 degrees.
From the above trigonometric identity we also get,
$\cosec ^{ 2 }{ \theta }$ = 1 + $\cot ^{ 2 }{ \theta }$
$\cot ^{ 2 }{ \theta }$ = $\cosec ^{ 2 }{ \theta }$ – 1 |
# Writing absolute value equations from a graph
Write a function that models this situation. That costs more than a human haircut at least my haircuts! This is the solution for equation 2. Plug these values into both equations.
And, even better, a site that covers math topics from before kindergarten through high school. This means we can write this absolute value function as a piecewise function. If you already know the solution, you can tell immediately whether the number inside the absolute value brackets is positive or negative, and you can drop the absolute value brackets.
To solve this, you have to set up two equalities and solve each separately. This is solution for equation 1. You may also be asked to take an absolute value graph and write it as a piecewise function: We learned how about Parent Functions and their Transformations here in the Parent Graphs and Transformations section.
Here are the graphs, with explanations on how to derive their piecewise equations: Writing an Equation with a Known Solution If you have values for x and y for the above example, you can determine which of the two possible relationships between x and y is true, and this tells you whether the expression in the absolute value brackets is positive or negative.
Sciencing Video Vault 1. You plan to sell She Love Math t-shirts as a fundraiser. You might want to review Solving Absolute Value Equations and Inequalities before continuing on to this topic. If you plot the above two equations on a graph, they will both be straight lines that intersect the origin.
Equation 2 is the correct one.
Learn these rules, and practice, practice, practice! Therefore, the piecewise function is: We have to start at 0, since dogs have to weigh over 0 pounds: To review how to obtain equations from linear graphs, see Obtaining the Equations of a Line, and from quadratics, see Finding a Quadratic Equation from Points or a Graph.
Set Up Two Equations Set up two separate and unrelated equations for x in terms of y, being careful not to treat them as two equations in two variables: Put in numbers and try it!
Plug in known values to determine which solution is correct, then rewrite the equation without absolute value brackets. Piecewise Function Word Problems Problem: So, the piecewise function is: Obtaining Equations from Piecewise Function Graphs You may be asked to write a piecewise function, given a graph.
You can now drop the absolute value brackets from the original equation and write instead: When you take the absolute value of a number, the result is always positive, even if the number itself is negative. Note that this piecewise equation is non-continuous.
For a random number x, both the following equations are true: This means that any equation that has an absolute value in it has two possible solutions. The piecewise function is: Here are more examples, with explanations. You might want to review Quadratic Inequalities for the second example below: Welcome to She Loves Math!
So the whole piecewise function is:The lesson Solving and Graphing Absolute Value Inequalities: Practice Problems covers the following objectives: Understanding the difference between absolute value equations and inequalities.
Video: Graphing Absolute Value Equations: Dilations & Reflections Although a basic absolute value graph isn't complicated, transformations. How To: Solve an absolute value equation and graph the answer on a number line By getexcellent; 8/11/10 AM How To: Solve linear absolute value equations & inequalities How To: Write and graph an equation in slope intercept form.
Here are the graphs, with explanations on how to derive their piecewise equations: Absolute Value as a Piecewise Function. We can write absolute value functions as piecewise functions You may also be asked to take an absolute value graph and write it as a piecewise function. Chapter 2 Linear Equations and Functions Absolute Value Functions REPRESENTING ABSOLUTE VALUE FUNCTIONS In Lesson you learned that the absolute value of x is defined by: |x| = The graph of this piecewisefunction consists of two rays, is V-shaped, and opens up.
The corner point of the graph, called the occurs at the origin. Transcript of Graphing Absolute Value Equations. What is Absolute Value? Absolute value is how far a number is from zero A V-Shape graph that points upward or downward is the graph of an absolute value equation.
In this lesson you will graph by translating the graph of y = | x | For writing absolute value equations, you write an.
Writing absolute value equations from a graph
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# A pharmaceutical company received \$3 million in royalties…
## Solution:
This is a percent decrease problem. We will use the formula: percent change = (new – old)/old x 100 to calculate the final answer.
We first set up the ratios of royalties to sales. The first ratio will be for the first 20 million in sales, and the second ratio will be for the next 108 million in sales. Because all of the sales are in millions, we do not have to express all the trailing zeros in our ratios.
First 20 Million
royalties/sales = 3/20
Next 108 Million
royalties/sales = 9/108 = 1/12
Because each ratio is not an easy number to use, we can simplify each one by multiplying each by the LCM of the two denominators, which is 60. Keep in mind that we are able to do this only because our answer choices are expressed in percents.
First 20 Million
royalties/sales = (3/20) x 60 = 9
Next 108 Million
royalties/sales = 9/108 = (1/12) x 60 = 5
We can plug 9 and 5 into our percent change formula:
(new – old)/old x 100
[(5 – 9)/9] x 100
-4/9 x 100
At this point we can stop and consider the answer choices. Since we know that 4/9 is just a bit less than ½, we know that -4/9 x 100 is about a 45% decrease. |
Early Years
Age 3-6
# Curriculum
Demonstrating Literacy and Mathematics Behaviour
• Collect, organize, display, and interpret data to solve problems and to communicate information, and explore the concept of probability in everyday contexts (#19).
# Context
• Students and teacher are in a seated circle on the carpet.
# Materials
• 3 different coloured hula hoops to correspond with the colours of sorting materials
• Different coloured leaves or any other material that can be sorted by colour
# Lesson
• Introduce the lesson by laying the three hula hoops flat on the ground (not overlapping).
• Ask students, How many hoops do I have? What colour are they?
• Without actually showing students the leaves, ask students, Which leaves would be placed in the yellow, green, and red hoop?
• Keeping the hoops apart (not overlapping), one at a time distribute one solid-coloured leaf to each child.
• Have the child place the leaf in the appropriate hoop (e.g. red leaf would go into the red hoop etc.) Ask students to justify their decisions.
• Then, begin to hand out leaves consisting of two colours.
• Ask, Which hoop does this leaf belong in if it is both yellow and green?
• Give students the opportunity to ponder which hoop the leaf should be placed in.
• Some students may express that the leaf should be placed in between two hoops.
• Ask students, How could we arrange the hoops so that the leaf could be sorted as both yellow and green?
• Students may suggest that the hoops be overlapped. If not, hint at the idea.
• Modify the three hoops so that they are now overlapping (see image above).
• Continue to hand out multi-coloured leaves and have children sort them. If children do not put their leaf in the correct place, leave it.
• Once all the leaves are sorted, ask children, Are there any leaves in wrong place?
• If so, discuss why a particular leaf may be placed in one section over another.
# Look Fors
• Do children understand the idea of sorting by different attributes?
• Can children explain their sorting decisions?
• What techniques do children use when sorting? Can they easily sort by more than one colour? |
# Video: GCSE Mathematics Foundation Tier Pack 2 • Paper 3 • Question 9
GCSE Mathematics Foundation Tier Pack 2 • Paper 3 • Question 9
02:19
### Video Transcript
A bag contains blue, red, green, and yellow counters only. In total, there are 528 counters in the bag. One-third of the counters are green. 67 of the counters are red. 115 of the counters are blue. Work out how many of the counters are yellow.
So we’re told in the question that the bag only contains blue, red, green, and yellow counters. And we’re also told that the total number of counters in the bag is 528. This means that the numbers of blue, red, green, and yellow counters must sum to 528.
We’re also told in the question that 67 of the counters are red and 115 of the counters are blue. So we can substitute these values into the equation that we’ve written. We have 115 plus 67 plus the number of green counters plus the number of yellow counters is equal to 528.
The other piece of information we’re given in the question is that one-third of the counters are green. To find one-third of a number, we just divide it by three. So we divide the total number of counters, which is 528, by three, giving 176. We can perform this division on a calculator if it’s available or we can use a short division method.
Now we’ve worked out that the number of green counters in the bag is 176, so we can substitute this value into our equation. We don’t know what the number of yellow counters in the bag is, but this is what we’re looking to work out. And we now have all the information to enable us to do so. We just need to solve the equation that we’ve got.
115, 67, and 176 sum to 358. So our equation becomes 358 plus 𝑦, the number of yellow counters, is equal to 528. To solve this equation, we just need to subtract 358 from each side. And it gives 𝑦 is equal to 170. This is the number of yellow counters in the bag. |
# If R (x, y) is a point on the line segment joining the points
Question:
If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that y = a + b.
Solution:
The formula for the area ' $A$ ' encompassed by three points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is given by the formula,
$\Delta=\frac{1}{2}\left|\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right|$
If three points are collinear the area encompassed by them is equal to 0.
It is said that the point R(x, y) lies on the line segment joining the points P(a, b) and Q(b, a). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.
$\Delta=\frac{1}{2}\left|\left(a y+x a+b^{2}\right)-\left(x b+b y+a^{2}\right)\right|$
$0=\frac{1}{9}\left|a y+x a+b^{2}-x b-b y-a^{2}\right|$
$0=a y+x a+b^{2}-x b-b y-a^{2}$
$a^{2}-b^{2}=a x+a y-b x-b y$
$(a+b)(a-b)=(a-b)(x+y)$
$(a+b)=(x+y)$
Hence under the given conditions we have proved that $x+y=a+b$. |
# Finding Equivalent Ratios Given the Total Quantity
Video solutions to help Grade 7 students learn how to to find an equivalent ratio of two partial quantities given a part-to-part ratio and the total of those quantities.
## New York State Common Core Math Module 2, Grade 7, Lesson 13
### Lesson 13 Outcome
• Students use tables to find an equivalent ratio of two partial quantities given a part-to-part ratio and the total of those quantities, in the third column, including problems with ratios of fractions.
### Lesson 13 Summary
• To find missing quantities in a ratio table where a total is given, determine the unit rate from the ratio of two given quantities and use it to find the missing quantities in each equivalent ratio.
### NYS Math Module 2 Grade 7 Lesson 13 Examples
Example 1
A group of 6 hikers are preparing for a one-week trip. All of the group's supplies will be carried by the hikers in backpacks. The leader decided that it would be fair for each hiker to carry a backpack that is the same fraction of his weight as all of the other hikers'. In this set-up, the heaviest hiker would carry the heaviest load. The table below shows the weight of each hiker and the weight of his/her backpack.
Complete the table. Find the missing amounts of weight by applying the same ratio as the first 2 rows.
Example 2
When a business buys a fast food franchise, it is buying the recipes used at every restaurant with the same name. For example, all Pizzeria Specialty House Restaurants have different owners, but they must all use the same recipes for their pizza, sauce, bread, etc. You are now working at your local Pizzeria Specialty House restaurant and listed below are the amounts of meat used on one meat-lovers pizza.
1/4 cup of sausage
1/3 cup of pepperoni
1/6 cup of bacon
1/8 cup of ham
1/8 cup of beef
What is the total amount of toppings used on a meat-lovers pizza?
The meat must be mixed using this ratio to ensure that customers will receive the same great tasting meat-lovers pizza from every Pizzeria Specialty House Restaurant nationwide. The table below shows 3 different orders for meat-lovers pizza on Superbowl Sunday. Using the amounts and total for one pizza given above, fill in every row and column of the table so the mixture tastes the same.
Exercise 1
1. The table below shows 6 different-sized pans of the same recipe for macaroni and cheese. If the recipe relating the ratio of ingredients stays the same, how might it be altered to account for the different sized pans? |
Paul's Online Math Notes
[Notes]
Calculus III - Notes
Line Integrals Previous Chapter Surface Integrals of Vector Fields Previous Section Next Section Divergence Theorem
## Stokes’ Theorem
In this section we are going to take a look at a theorem that is a higher dimensional version of Green’s Theorem. In Green’s Theorem we related a line integral to a double integral over some region. In this section we are going to relate a line integral to a surface integral. However, before we give the theorem we first need to define the curve that we’re going to use in the line integral.
Let’s start off with the following surface with the indicated orientation.
Around the edge of this surface we have a curve C. This curve is called the boundary curve. The orientation of the surface S will induce the positive orientation of C. To get the positive orientation of C think of yourself as walking along the curve. While you are walking along the curve if your head is pointing in the same direction as the unit normal vectors while the surface is on the left then you are walking in the positive direction on C.
Now that we have this curve definition out of the way we can give Stokes’ Theorem.
Stokes’ Theorem
Let S be an oriented smooth surface that is bounded by a simple, closed, smooth boundary curve C with positive orientation. Also let be a vector field then,
In this theorem note that the surface S can actually be any surface so long as its boundary curve is given by C. This is something that can be used to our advantage to simplify the surface integral on occasion.
Let’s take a look at a couple of examples.
Example 1 Use Stokes’ Theorem to evaluate where and S is the part of above the plane . Assume that S is oriented upwards. Solution Let’s start this off with a sketch of the surface. In this case the boundary curve C will be where the surface intersects the plane and so will be the curve So, the boundary curve will be the circle of radius 2 that is in the plane . The parameterization of this curve is, The first two components give the circle and the third component makes sure that it is in the plane . Using Stokes’ Theorem we can write the surface integral as the following line integral. So, it looks like we need a couple of quantities before we do this integral. Let’s first get the vector field evaluated on the curve. Remember that this is simply plugging the components of the parameterization into the vector field. Next, we need the derivative of the parameterization and the dot product of this and the vector field. We can now do the integral.
Example 2 Use Stokes’ Theorem to evaluate where and C is the triangle with vertices , and with counter-clockwise rotation. Solution We are going to need the curl of the vector field eventually so let’s get that out of the way first. Now, all we have is the boundary curve for the surface that we’ll need to use in the surface integral. However, as noted above all we need is any surface that has this as its boundary curve. So, let’s use the following plane with upwards orientation for the surface. Since the plane is oriented upwards this induces the positive direction on C as shown. The equation of this plane is, Now, let’s use Stokes’ Theorem and get the surface integral set up. Okay, we now need to find a couple of quantities. First let’s get the gradient. Recall that this comes from the function of the surface. Note as well that this also points upwards and so we have the correct direction. Now, D is the region in the xy-plane shown below, We get the equation of the line by plugging in into the equation of the plane. So based on this the ranges that define D are, The integral is then, Don’t forget to plug in for z since we are doing the surface integral on the plane. Finishing this out gives,
In both of these examples we were able to take an integral that would have been somewhat unpleasant to deal with and by the use of Stokes’ Theorem we were able to convert it into an integral that wasn’t too bad.
Surface Integrals of Vector Fields Previous Section Next Section Divergence Theorem Line Integrals Previous Chapter
[Notes]
© 2003 - 2018 Paul Dawkins |
# How to Reflect About a Line
Video Crash Courses
Want to watch animated videos and solve interactive exercises about mirror symmetry?
When you reflect a figure, you draw its mirror image. A figure has a mirror image just like you do—you can imagine looking at yourself in the mirror in the morning. Now, you’ll learn how to draw these kinds of mirror images of geometric figures, whether the mirror line is located outside of, or on top of, the figures.
When reflecting a figure, it’s smart to use a draft compass. Put the point of the draft compass where the mirror line intersects the auxiliary line. Mark the distance between this point and the point on the figure you want to mirror with your draft compass. Make a small arc where that point needs to be on the other side of the mirror line. The small arc will then intersect the auxiliary line at the point you need to draw the reflected figure.
## Reflecting a Circle About a Line
A circle is a smooth figure without sides or vertices. Below, on the left, there’s a picture showing a circle reflected about a line outside of it, to its right. The reflection across this line is in a lighter shade of blue.
On the right, there’s a circle that has been reflected about a line on top of the symmetry line of the circle. You can also see that the reflected portion is a lighter blue.
To measure the radius of the circle, put the point of your draft compass in the center and put the other leg on the boundary of the circle.
## Reflecting a Triangle About a Line
A triangle has three vertices and three sides. When reflecting a triangle about a line, you need to draw lines from the vertices straight towards the mirror line—and further past it. Make the lines extra long, just to be sure.
Below, on the left, you can see a triangle reflected about a line outside the triangle. The reflected figure is a lighter shade of blue.
Then, to the right, you can see a triangle reflected about a line that is on top of the symmetry line of the triangle. The reflection is also a lighter shade of blue.
A quadrilateral has four vertices and four sides. When you need to mirror a quadrilateral about a line, draw lines from each of the vertices straight towards the mirror line—and further past it. Make the lines extra long just in case.
Below, on the left, you can see a quadrilateral that has been reflected about a line outside itself. The reflected figure is light blue.
And to the right, you can see a quadrilateral that has been reflected about a line that is on top of its line of symmetry. Then nothing changes! The reflected portion is now a lighter shade of blue for the sake of demonstration. |
# For the 31 used cars sold last month at Car Dealership X , which of
USED CARS SOLD LAST MONTH AT CAR DEALERSHIP \$X\$
For the \$31\$ used cars sold last month at Car Dealership \$X\$ , which of the following could be the median price?
Indicate all such prices.
1. \$\\$5{,}500\$
2. \$\\$6{,}500\$
3. \$\\$7{,}000\$
4. \$\\$8{,}500\$
5. \$\\$9{,}500\$
6. \$\\$10{,}500\$
So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. Buuuut then you had some questions about the quant section—specifically question 8 of the second Quantitative section of Practice Test 1. Those questions testing our knowledge of Graphical Methods for Describing Data can be kind of tricky, but never fear, PrepScholar has got your back!
## Survey the Question
Let’s search the problem for clues as to what it will be testing, as this will help shift our minds to think about what type of math knowledge we’ll use to solve this question. Pay attention to any words that sound math-specific and anything special about what the numbers look like, and mark them on our paper.
The question gives us a table displaying numerical information, so we will likely draw upon what we’ve learned about Graphical Methods for Describing Data. Let’s keep what we’ve learned about this skill at the tip of our minds as we approach this question.
## What Do We Know?
Let’s carefully read through the question and make a list of the things that we know.
1. We have a table containing price ranges and the number of cars sold in each price range
2. A total of \$31\$ cars were sold
3. We want to know which answer choices could possibly be the median price of the cars sold
## Develop a Plan
We want to find which answer choices could be the median price of \$31\$ cars sold. The median of a list of numbers is the value right in the middle IF we place the values in order from least to greatest. So half of the values should be less than the median, and half should be greater than the median. If we divide \$31\$ by \$2\$, we get \$15.5\$. Hmm, we can’t have \$15.5\$ prices less than the median price. Ah! That extra half of a car must be the car with the median price, since it’ll be split between the lower half and upper half of the set of values. So we should use \$15\$ instead of \$15.5\$. That makes a bit more sense. Okay, so with \$31\$ different prices, we’ll have \$15\$ prices below the median, then the median price, and then \$15\$ prices above the median. So the median price would be the \$16\t\h\$ price if we were counting from the least price toward the greatest price. Let’s plan to use our table to find out which car is the \$16\t
h\$ if we count from the lowest price and up.
## Solve the Question
From the table, we can see that, going from least to greatest in terms of car price, the first \$7\$ cars are under \$\\$5{,}000\$.
Then the next \$10\$ prices are between \$\\$5{,}000\$ and \$\\$7{,}499\$. So we have the first \$7\$ cars, then cars \$8\$ through \$17\$ fall in the \$\\$5{,}000\$ to \$\\$7{,}499\$ price range. Ah ha! The \$16\t\h\$ car must be in the \$\\$5{,}000\$ to \$\\$7{,}499\$ price range!
Choosing any answers between \$\\$5{,}000\$ and \$\\$7{,}499\$, we find that the correct answers are A, B, and C.
## What Did We Learn
Well, this was an interesting way to test our knowledge about medians. Looks like a good way to find the position of the median term is to just divide the number of values by \$2\$. We’ll just need to keep in mind that if we have an odd number of values, then the median value will be split between the upper and lower halves of our set of values. But we can fix that by just subtracting \$0.5\$ from our result, just like how we found that if we have \$31\$ cars, then \$31/2-0.5\$, or \$15\$ cars will be below the median price.
Want more expert GRE prep? Sign up for the five-day free trial of our PrepScholar GRE Online Prep Program to access your personalized study plan with 90 interactive lessons and over 1600 GRE questions.
Have questions? Leave a comment or send us an email at [email protected]. |
# How do you find the slope of 3y=x+3?
##### 1 Answer
Jul 13, 2016
Slope: $\frac{1}{3}$
#### Explanation:
Slope-intercept form: $y = m x + b$; where $m$ is the slope.
If we put the given equation into slope-intercept form, we can identify the slope.
First get $y$ all by itself.
$3 y = x + 3$
$y = \frac{x}{3} + 1$
This equation is now in slope intercept form. We must find the slope. We saw at the top that the slope, $m$, comes right before the variable. Here is our variable term:
$\frac{x}{3}$
It is in the form of a fraction. What do we do to clearly identify the slope? Rewrite the variable term, since $\frac{x}{3}$ is the same as $\frac{1}{3} \times x$.
$\frac{x}{3} = \frac{1}{3} \times x = \frac{1}{3} x$
Now let's rewrite the whole equation.
$y = \frac{1}{3} x + 1$
$\textcolor{g r e y}{y = m x + b}$
Look for the number before $x$, or the number combined with it. In this case, the slope appears to be $\frac{1}{3}$.
If we were to graph the equation and look at the slope that way, it would be the same.
graph{1/3x+1 [-10, 10, -5, 5]} |
# Modular square root
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
A modular square root $r$ of an integer number $a$ modulo an integer $m$ greater than 1 is an integer such that:
$r^2 \equiv a\ \pmod m$
In this article we will consider the case when the modulus is prime. Otherwise we can compute the square roots modulo the prime factors of $m$ and then generate a solution using the Chinese Remainder Theorem.
When the argument is congruent to zero, there is only one modular square root, namely zero. Otherwise, the number of square roots can be two or zero depending on whether the argument is a quadratic residue modulo $m$ or not. The sum of both square roots is congruent to zero.
In order to compute the square root, we will consider different cases, depending on the modulus. When this modulus is odd, we assume that the quantity $a^{(m-1)/2} \bmod m$ equals 1 (otherwise there is no square root if $a \not\equiv 0\ \pmod m$).
### Modulus equal to 2
In this case the square root is congruent to the argument $r$.
### Modulus congruent to 3 modulo 4
$r\equiv \pm a^{(m+1)/4}\ \pmod m$
### Modulus congruent to 5 modulo 8
First compute the square root of -1 ($i$) as follows:
$v\equiv (2a)^{(p-5)/8}\ \pmod m$
$i\equiv 2av^2\ \pmod m$
Then compute the square root as follows:
$r\equiv \pm av(i-1)\ \pmod m$
### Modulus congruent to 1 modulo 8
In this case we can use the Shanks method.
1. Set $e$ and an odd $q$ such that $m = 2^e q+1$.
2. Choose $x$ at random in the range $1 \lt x \lt m$ and set $z \leftarrow x^q\,\bmod m$. If $z^{2^{e-1}}\equiv 1\,\pmod m$, repeat this step. (This generates a quadratic non-residue of $m$.)
3. Set $y \leftarrow z$, $r \leftarrow e$, $x \leftarrow a^{(q-1)/2} \bmod m$, $v \leftarrow ax\bmod m$, $w \leftarrow vx \bmod m$.
4. If $w=1$, the computation ends with $\pm v$ as the square root.
5. Find the smallest value of $k$ such that $w^{2^k}\equiv 1\,\pmod m$.
6. Set $d \leftarrow y^{2^{r-k-1}} \bmod m$, $y \leftarrow d^2\,\bmod m$, $r \leftarrow k$, $v \leftarrow dv\,\bmod m$, $w \leftarrow wy\,\bmod m$.
7. Go to step 4.
### Example 1
Find the square root of 58 modulo 101.
First of all we check that the modulus 101 is prime. Then we find that it is congruent to 5 mod 8.
Now we compute $58^{(101-1)/2} \bmod 101 = 1$ so there are two square roots to be computed.
$v = (2*58)^{(101-5)/8} \bmod 101 = 15^{12} \bmod 101 = 88$
$i = 2*58*88^2 \bmod 101 = 10$
Notice that $i*i \equiv -1 \,\pmod {101}$
$r = 58*88*(10-1) \bmod 101 = 82$
So the modular square roots are 82 and its negative 19, which can be easily verified: $82*82 \equiv 58 \,\pmod {101}$, $19*19 \equiv 58 \,\pmod {101}$.
### Example 2
Find the square root of 111 modulo 113.
First of all we check that the modulus 113 is prime. Then we find that it is congruent to 1 mod 8.
Now we compute $111^{(113-1)/2} \bmod 113 = 1$ so there are two square roots to be computed.
• Step 1: $e = 4$, $q = 7$.
• Step 2: $x = 2$, $z = 2^7 \bmod 113 = 15$, $z^{2^3} \bmod 113 = 1$, so we have to repeat step 2.
• Step 2: $x = 3$, $z = 3^7 \bmod 113 = 40$, $z^{2^3} \bmod 113 = 112$.
• Step 3: $y = 40$, $r = 4$, $x = 111^{(7-1)/2} \bmod 113 = 105$, $v = 111*105 \bmod 113 = 16$, $w = 16*105 \bmod 113 = 98$.
• Step 4: $w$ is not equal to 1 so the computation continues.
• Step 5: $k = 2$
• Step 6: $d = 40^{2^{4-2-1}} \bmod 113 = 18$, $y = 18^2 \bmod 113 = 98$, $r = 2$, $v = 18*16 \bmod 113 = 62$, $w = 98*98 \bmod 113 = 112$.
• Step 4: $w$ is not equal to 1 so the computation continues.
• Step 5: $k = 1$
• Step 6: $d = 98^{2^{2-1-1}} \bmod 113 = 98$, $y = 98^2 \bmod 113 = 112$, $r = 1$, $v = 98*62 \bmod 113 = 87$, $w = 112*112 \bmod 113 = 1$.
• Step 4: $w = 1$, so the square root is $\pm v = \pm 87$.
So the modular square roots are 87 and its negative 26, which can be easily verified: $87*87 \equiv 111 \,\pmod {113}$, $26*26 \equiv 111 \,\pmod {113}$. |
# ML Aggarwal Solutions for Chapter 1 Rational and Irrational Numbers Class 9 Maths ICSE
Here, we are providing the solutions for Chapter 1 Rational and Irrational Numbers from ML Aggarwal Textbook for Class 9 ICSE Mathematics. Solutions of the first chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 9 Chapter 1 Rational and Irrational Numbers ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on the equations of algebra, word problems, representation on number line and solutions of set.
Exercise 1.1
1. Insert a rational number between and 2/9 and 3/8 arrange in descending order.
Solution
Given,
Rational numbers: 2/9 and 3/8
Let us rationalize the numbers,
By taking LCM for denominators 9 and 8 which is 72.
2/9 = (2×8)/(9×8) = 16/72
3/8 = (3×9)/(8×9) = 27/72
Since,
16/72 < 27/72
So, 2/9 < 3/8
The rational number between 2/9 and 3/8 is
Hence, 3/8 > 43/144 > 2/9
The descending order of the numbers is 3/8, 43/144, 2/9
2. Insert two rational numbers between 1/3 and 1/4 and arrange in ascending order.
Solution
Given,
The rational numbers 1/3 and ¼
By taking LCM and rationalizing, we get
= 7/24
Now let us find the rational number between ¼ and 7/24
By taking LCM and rationalizing, we get
= 13/48
So,
The two rational numbers between 1/3 and ¼ are
7/24 and 13/48
Hence, we know that, 1/3 > 7/24 > 13/48 > ¼
The ascending order is as follows: ¼, 13/48, 7/24, 1/3
3. Insert two rational numbers between – 1/3 and – 1/2 and arrange in ascending order.
Solution
Given:
The rational numbers -1/3 and -1/2
By taking LCM and rationalizing, we get
= -5/12
So, the rational number between -1/3 and -1/2 is -5/12
-1/3 > -5/12 > -1/2
Now, let us find the rational number between -1/3 and -5/12
By taking LCM and rationalizing, we get
= -9/24
= -3/8
So, the rational number between -1/3 and -5/12 is -3/8
-1/3 > -3/8 > -5/12
Hence, the two rational numbers between -1/3 and -1/2 are
-1/3 > -3/8 > -5/12 > -1/2
The ascending is as follows: -1/2, -5/12, -3/8, -1/3
4. Insert three rational numbers between 1/3 and 4/5, and arrange in descending order.
Solution
Given,
The rational numbers 1/3 and 4/5
By taking LCM and rationalizing, we get
= 17/30
So, the rational number between 1/3 and 4/5 is 17/30
1/3 < 17/30 < 4/5
Now, let us find the rational numbers between 1/3 and 17/30
By taking LCM and rationalizing, we get
= 27/60
So, the rational number between 1/3 and 17/30 is 27/60
1/3 < 27/60 < 17/30
Now, let us find the rational numbers between 17/30 and 4/5
By taking LCM and rationalizing, we get
= 41/60
So, the rational number between 17/30 and 4/5 is 41/60
17/30 < 41/60 < 4/5
Hence, the three rational numbers between 1/3 and 4/5 are
1/3 < 27/60 < 17/30 < 41/60 < 4/5
The descending order is as follows: 4/5, 41/60, 17/30, 27/60, 1/3
5. Insert three rational numbers between 4 and 4.5.
Solution
Given:
The rational numbers 4 and 4.5
By rationalizing, we get
= (4 + 4.5)/2
= 8.5 / 2
= 4.25
So, the rational number between 4 and 4.5 is 4.25
4 < 4.25 < 4.5
Now, let us find the rational number between 4 and 4.25
By rationalizing, we get
= (4 + 4.25)/2
= 8.25 / 2
= 4.125
So, the rational number between 4 and 4.25 is 4.125
4 < 4.125 < 4.25
Now, let us find the rational number between 4 and 4.125
By rationalizing, we get
= (4 + 4.125)/2
= 8.125 / 2
= 4.0625
So, the rational number between 4 and 4.125 is 4.0625
4 < 4.0625 < 4.125
Hence, the rational numbers between 4 and 4.5 are
4 < 4.0625 < 4.125 < 4.25 < 4.5
The three rational numbers between 4 and 4.5
4.0625, 4.125, 4.25
6. Find six rational numbers between 3 and 4.
Solution
Given:
The rational number 3 and 4
So let us find the six rational numbers between 3 and 4,
First rational number between 3 and 4 is
= (3 + 4) / 2
= 7/2
Second rational number between 3 and 7/2 is
= (3 + 7/2) / 2
= (6+7) / (2 × 2) [By taking 2 as LCM]
= 13/4
Third rational number between 7/2 and 4 is
= (7/2 + 4) / 2
= (7+8) / (2 × 2) [By taking 2 as LCM]
= 15/4
Fourth rational number between 3 and 13/4 is
= (3 + 13/4) / 2
= (12+13) / (4 × 2) [By taking 4 as LCM]
= 25/8
Fifth rational number between 13/4 and 7/2 is
= [(13/4) + (7/2)] / 2
= [(13+14)/4] / 2 [By taking 4 as LCM]
= (13 + 14) / (4 × 2)
= 27/8
Sixth rational number between 7/2 and 15/4 is
= [(7/2) + (15/4)] / 2
= [(14 + 15)/4] / 2 [By taking 4 as LCM]
= (14 + 15) / (4 × 2)
= 29/8
Hence, the six rational numbers between 3 and 4 are
25/8, 13/4, 27/8, 7/2, 29/8, 15/4
7. Find five rational numbers between 3/5 and 4/5.
Solution
Given:
The rational numbers 3/5 and 4/5
Now, let us find the five rational numbers between 3/5 and 4/5
So we need to multiply both numerator and denominator with 5 + 1 = 6
We get,
3/5 = (3 × 6) / (5 × 6) = 18/30
4/5 = (4 × 6) / (5 × 6) = 24/30
Now, we have 18/30 < 19/30 < 20/30 < 21/30 < 22/30 < 23/30 < 24/30
Hence, the five rational numbers between 3/5 and 4/5 are
19/30, 20/30, 21/30, 22/30, 23/30
8. Find ten rational numbers between -2/5 and 1/7.
Solution
Given:
The rational numbers -2/5 and 1/7
By taking LCM for 5 and 7 which is 35
So, -2/5 = (-2 × 7) / (5 × 7) = -14/35
1/7 = (1 × 5) / (7 × 5) = 5/35
Now, we can insert any10 numbers between -14/35 and 5/35
i.e., -13/35, -12/35, -11/35, -10/35, -9/35, -8/35, -7/35, -6/35, -5/35, -4/35, -3/35, -2/35, -1/35, 1/35, 2/35, 3/35, 4/35
Hence, the ten rational numbers between -2/5 and 1/7 are
-6/35, -5/35, -4/35, -3/35, -2/35, -1/35, 1/35, 2/35, 3/35, 4/35
9. Find six rational numbers between 1/2 and 2/3.
Solution
Given:
The rational number ½ and 2/3
To make the denominators similar let us take LCM for 2 and 3 which is 6
½ = (1 × 3) / (2 × 3) = 3/6
2/3 = (2 × 2) / (3 × 2) = 4/6
Now, we need to insert six rational numbers, so multiply both numerator and denominator by 6 + 1 = 7
3/6 = (3 × 7) / (6 × 7) = 21/42
4/6 = (4 × 7) / (6 × 7) = 28/42
We know that, 21/42 < 22/42 < 23/42 < 24/42 < 25/42 < 26/42 < 27/42 < 28/42
Hence, the six rational numbers between ½ and 2/3 are
22/42, 23/42, 24/42, 25/42, 26/42, 27/42
Exercise 1.2
1. Prove that, √5 is an irrational number.
Solution
Let us consider √5 be a rational number, then
√5 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).
So,
5 = p2 / q2
p2 = 5q2 …. (1)
As we know, ‘5’ divides 5q2, so ‘5’ divides p2 as well. Hence, ‘5’ is prime.
So 5 divides p
Now, let p = 5k, where ‘k’ is an integer
Square on both sides, we get
p2 = 25k2
5q2 = 25k2 [Since, p2 = 5q2, from equation (1)]
q2 = 5k2
As we know, ‘5’ divides 5k2, so ‘5’ divides q2 as well. But ‘5’ is prime.
So 5 divides q
Thus, p and q have a common factor 5. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √5 is not a rational number.
√5 is an irrational number.
Hence proved.
2. Prove that, √7 is an irrational number.
Solution
Let us consider √7 be a rational number, then
√7 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).
So,
7 = p2 / q2
p2 = 7q2 …. (1)
As we know, ‘7’ divides 7q2, so ‘7’ divides p2 as well. Hence, ‘7’ is prime.
So 7 divides p
Now, let p = 7k, where ‘k’ is an integer
Square on both sides, we get
p2 = 49k2
7q2 = 49k2 [Since, p2 = 7q2, from equation (1)]
q2 = 7k2
As we know, ‘7’ divides 7k2, so ‘7’ divides q2 as well. But ‘7’ is prime.
So 7 divides q
Thus, p and q have a common factor 7. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √7 is not a rational number.
√7 is an irrational number.
Hence proved.
3. Prove that √6 is an irrational number.
Solution
Let us consider √6 be a rational number, then
√6 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).
So,
6 = p2 / q2
p2 = 6q2 …. (1)
As we know, ‘2’ divides 6q2, so ‘2’ divides p2 as well. Hence, ‘2’ is prime.
So 2 divides p
Now, let p = 2k, where ‘k’ is an integer
Square on both sides, we get
p2 = 4k2
6q2 = 4k2 [Since, p2 = 6q2, from equation (1)]
3q2 = 2k2
As we know, ‘2’ divides 2k2, so ‘2’ divides 3q2 as well.
‘2’ should either divide 3 or divide q2.
But ‘2’ does not divide 3. ‘2’ divides q2 so ‘2’ is prime.
So 2 divides q
Thus, p and q have a common factor 2. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √6 is not a rational number.
√6 is an irrational number.
Hence proved.
4. Prove that 1/√11 is an irrational number.
Solution
Let us consider 1/√11 be a rational number, then
1/√11 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).
So,
1/11 = p2 / q2
q2 = 11p2 …. (1)
As we know, ‘11’ divides 11p2, so ‘11’ divides q2 as well. Hence, ‘11’ is prime.
So 11 divides q
Now, let q = 11k, where ‘k’ is an integer
Square on both sides, we get
q2 = 121k2
11p2 = 121k2 [Since, q2 = 11p2, from equation (1)]
p2 = 11k2
As we know, ‘11’ divides 11k2, so ‘11’ divides p2 as well. But ‘11’ is prime.
So 11 divides p
Thus, p and q have a common factor 11. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, 1/√11 is not a rational number.
1/√11 is an irrational number.
Hence proved.
5. Prove that √2 is an irrational number. Hence show that 3 — √2 is an irrational.
Solution
Let us consider √2 be a rational number, then
√2 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).
So,
2 = p2 / q2
p2 = 2q2 …. (1)
As we know, ‘2’ divides 2q2, so ‘2’ divides p2 as well. Hence, ‘2’ is prime.
So 2 divides p
Now, let p = 2k, where ‘k’ is an integer
Square on both sides, we get
p2 = 4k2
2q2 = 4k2 [Since, p2 = 2q2, from equation (1)]
q2 = 2k2
As we know, ‘2’ divides 2k2, so ‘2’ divides q2 as well. But ‘2’ is prime.
So 2 divides q
Thus, p and q have a common factor 2. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √2 is not a rational number.
√2 is an irrational number.
Now, let us assume 3 – √2 be a rational number, ‘r’
So, 3 – √2 = r
3 – r = √2
We know that, ‘r’ is rational, ‘3- r’ is rational, so ‘√2’ is also rational.
This contradicts the statement that √2 is irrational.
So, 3- √2 is irrational number.
Hence proved.
6. Prove that, √3 is an irrational number. Hence, show that 2/5×√3 is an irrational number.
Solution
Let us consider √3 be a rational number, then
√3 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).
So,
3 = p2 / q2
p2 = 3q2 …. (1)
As we know, ‘3’ divides 3q2, so ‘3’ divides p2 as well. Hence, ‘3’ is prime.
So 3 divides p
Now, let p = 3k, where ‘k’ is an integer
Square on both sides, we get
p2 = 9k2
3q2 = 9k2 [Since, p2 = 3q2, from equation (1)]
q2 = 3k2
As we know, ‘3’ divides 3k2, so ‘3’ divides q2 as well. But ‘3’ is prime.
So 3 divides q
Thus, p and q have a common factor 3. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √3 is not a rational number.
√3 is an irrational number.
Now, let us assume (2/5)√3 be a rational number, ‘r’
So, (2/5)√3 = r
5r/2 = √3
We know that, ‘r’ is rational, ‘5r/2’ is rational, so ‘√3’ is also rational.
This contradicts the statement that √3 is irrational.
So, (2/5)√3 is irrational number.
Hence proved.
7. Prove that √5 is an irrational number. Hence, show that -3 + 2√5 is an irrational number.
Solution
Let us consider √5 be a rational number, then
√5 = p/q, where ‘p’ and ‘q’ are integers, q 0 and p, q have no common factors (except 1).
So,
5 = p2 / q2
p2 = 5q2 …. (1)
As we know, ‘5’ divides 5q2, so ‘5’ divides p2 as well. Hence, ‘5’ is prime.
So 5 divides p
Now, let p = 5k, where ‘k’ is an integer
Square on both sides, we get
p2 = 25k2
5q2 = 25k2 [Since, p2 = 5q2, from equation (1)]
q2 = 5k2
As we know, ‘5’ divides 5k2, so ‘5’ divides q2 as well. But ‘5’ is prime.
So 5 divides q
Thus, p and q have a common factor 5. This statement contradicts that ‘p’ and ‘q’ has no common factors (except 1).
We can say that, √5 is not a rational number.
√5 is an irrational number.
Now, let us assume -3 + 2√5 be a rational number, ‘r’
So, -3 + 2√5 = r
-3 – r = 2√5
(-3 – r)/2 = √5
We know that, ‘r’ is rational, ‘(-3 – r)/2’ is rational, so ‘√5’ is also rational.
This contradicts the statement that √5 is irrational.
So, -3 + 2√5 is irrational number.
Hence proved.
8. Prove that the following numbers are irrational:
(i) 5 +√2
(ii) 3 – 5√3
(iii) 2√3 – 7
(iv) √2 +√5
Solution
(i) 5 +√2
Now, let us assume 5 + √2 be a rational number, ‘r’
So, 5 + √2 = r
r – 5 = √2
We know that, ‘r’ is rational, ‘r – 5’ is rational, so ‘√2’ is also rational.
This contradicts the statement that √2 is irrational.
So, 5 + √2 is irrational number.
(ii) 3 – 5√3
Now, let us assume 3 – 5√3 be a rational number, ‘r’
So, 3 – 5√3 = r
3 – r = 5√3
(3 – r)/5 = √3
We know that, ‘r’ is rational, ‘(3 – r)/5’ is rational, so ‘√3’ is also rational.
This contradicts the statement that √3 is irrational.
So, 3 – 5√3 is irrational number.
(iii) 2√3 – 7
Now, let us assume 2√3 – 7 be a rational number, ‘r’
So, 2√3 – 7 = r
2√3 = r + 7
√3 = (r + 7)/2
We know that, ‘r’ is rational, ‘(r + 7)/2’ is rational, so ‘√3’ is also rational.
This contradicts the statement that √3 is irrational.
So, 2√3 – 7 is irrational number.
(iv) √2 +√5
Now, let us assume √2 +√5 be a rational number, ‘r’
So, √2 +√5 = r
√5 = r – √2
Square on both sides,
(√5)2 = (r – √2)2
5 = r2 + (√2)2 – 2r√2
5 = r2 + 2 – 2√2r
5 – 2 = r2 – 2√2r
r2 – 3 = 2√2r
(r2 – 3)/2r = √2
We know that, ‘r’ is rational, ‘(r2 – 3)/2r’ is rational, so ‘√2’ is also rational.
This contradicts the statement that √2 is irrational.
So, √2 +√5 is irrational number.
Exercise 1.3
1. Locate √10 and √17 on the amber line.
Solution
√10
√10 = √(9 + 1) = √((3)2 + 12)
Now let us construct:
• Draw a line segment AB = 3cm.
• At point A, draw a perpendicular AX and cut off AC = 1cm.
• Join BC.
BC = √10cm
√17 = √(16 + 1) = √((4)2 + 12)
Now let us construct:
• Draw a line segment AB = 4cm.
• At point A, draw a perpendicular AX and cut off AC = 1cm.
• Join BC.
BC = √17cm
2. Write the decimal expansion of each of the following numbers and say what kind of decimal expansion each has:
(i) 36/100
(ii) 4 1/8
(iii) 2/9
(iv) 2/11
(v) 3/13
(vi) 329/400
Solution
(i) 36/100
36/100 = 0.36
It is a terminating decimal.
(ii) 4 1/8
4 1/8 = (4×8 + 1)/8 = 33/8
33/8 = 4.125
It is a terminating decimal.
(iii) 2/9
2/9 = 0.222
It is a non-terminating recurring decimal.
(iv) 2/11
2/11 = 0.181
It is a non-terminating recurring decimal.
(v) 3/13
3/13 = 0.2317692307
It is a non-terminating recurring decimal.
(vi) 329/400
329/400 = 0.8225
It is a terminating decimal.
3. Without actually performing the king division, State whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:
(i) 13/3125
(ii) 17/8
(iii) 23/75
(iv) 6/15
(v) 1258/625
(vi) 77/210
Solution
We know that, if the denominator of a fraction has only 2 or 5 or both factors, it is a terminating decimal otherwise it is non-terminating repeating decimals.
(i) 13/3125
3125 = 5 × 5 × 5 × 5 × 5
Prime factor of 3125 = 5, 5, 5, 5, 5 [i.e., in the form of 2n, 5n]
It is a terminating decimal.
(ii) 17/8
8 = 2 × 2 × 2
Prime factor of 8 = 2, 2, 2 [i.e., in the form of 2n, 5n]
It is a terminating decimal.
(iii) 23/75
75 = 3 × 5 × 5
Prime factor of 75 = 3, 5, 5
It is a non-terminating repeating decimal.
(iv) 6/15
Let us divide both numerator and denominator by 3
6/15 = (6 ÷ 3) / (15 ÷ 3)
= 2/5
Since the denominator is 5.
It is a terminating decimal.
(v) 1258/625
625 = 5 × 5 × 5 × 5
Prime factor of 625 = 5, 5, 5, 5 [i.e., in the form of 2n, 5n]
It is a terminating decimal.
(vi) 77/210
Let us divide both numerator and denominator by 7
77/210 = (77 ÷ 7) / (210 ÷ 7)
= 11/30
30 = 2 × 3 × 5
Prime factor of 30 = 2, 3, 5
It is a non-terminating repeating decimal.
4. Without actually performing the long division, find if 987/10500 will have terminating or non-terminating repeating decimal expansion. Give reasons for your answer.
Solution
Given:
The fraction 987/10500
Let us divide numerator and denominator by 21, we get
987/10500 = (987 ÷ 21) / (10500 ÷ 21)
= 47/500
So,
The prime factors for denominator 500 = 2 × 2 × 5 × 5 × 5
Since it is of the form: 2n, 5n
Hence it is a terminating decimal.
5. Write the decimal expansions of the following numbers which have terminating decimal expansions:
(i) 17/8
(ii) 13/3125
(iii) 7/80
(iv) 6/15
(v) 2²×7/54
(vi) 237/1500
Solution
(i) 17/8
Denominator, 8 = 2 × 2 × 2 = 23
It is a terminating decimal.
When we divide 17/8, we get
17/8 = 2.125
(ii) 13/3125
3125 = 5 × 5 × 5 × 5 × 5
Prime factor of 3125 = 5, 5, 5, 5, 5 [i.e., in the form of 2n, 5n]
It is a terminating decimal.
When we divide 13/3125, we get
13/3125 = 0.00416
(iii) 7/80
80 = 2 × 2 × 2 × 2 × 5
Prime factor of 80 = 24, 51 [i.e., in the form of 2n, 5n]
It is a terminating decimal.
When we divide 7/80, we get
7/80 = 0.0875
(iv) 6/15
Let us divide both numerator and denominator by 3, we get
6/15 = (6 ÷ 3) / (15 ÷ 3)
= 2/5
Since the denominator is 5,
It is terminating decimal.
6/15 = 0.4
(v) (2²×7)/54
We know that the denominator is 54
It is a terminating decimal.
(2²×7)/54 = (2 × 2 × 7) / (5 × 5 × 5 × 5)
= 28/625
28/625 = 0.0448
It is a terminating decimal.
(vi) 237/1500
Let us divide both numerator and denominator by 3, we get
237/1500 = (237 ÷ 3) / (1500 ÷ 3)
= 79/500
Since the denominator is 500,
Its factors are, 500 = 2 × 2 × 5 × 5 × 5
= 22 × 53
It is terminating decimal.
237/1500= 79/500 = 0.1518
6. Write the denominator of the rational number 257/5000 in the form 2× 5where m, n is non-negative integers. Hence, write its decimal expansion on without actual division.
Solution
Given:
The fraction 257/5000
Since the denominator is 5000,
The factors for 5000 are:
5000 = 2 × 2 × 2 × 5 × 5 × 5 × 5
= 23 × 54
257/5000 = 257/(23 × 54)
It is a terminating decimal.
So,
Let us multiply both numerator and denominator by 2, we get
257/5000 = (257 ×2) / (5000 ×2)
= 514/10000
= 0.0514
7. Write the decimal expansion of 1/7. Hence, write the decimal expression of? 2/7, 3/7, 4/7, 5/7 and 6/7.
Solution
Given:
The fraction: 1/7
1/7 = 0.142857142857
Since it is recurring,
8. Express the following numbers in the form p/q’. Where p and q are both integers and q≠0;
Solution
Let x == 0.3333…
Since there is one repeating digit after the decimal point,
Multiplying by 10 on both sides, we get
10x = 3.3333…
Now, subtract both the values,
9x = 3
x = 3/9
= 1/3
= 1/3
Let x == 5.2222…
Since there is one repeating digit after the decimal point,
Multiplying by 10 on both sides, we get
10x = 52.2222…
Now, subtract both the values,
9x = 52 – 5
9x = 47
x = 47/9
= 47/9
Let x = 0.404040
Since there is two repeating digit after the decimal point,
Multiplying by 100 on both sides, we get
100x = 40.404040…
Now, subtract both the values,
99x = 40
x = 40/99
0.404040… = 40/99
Let x == 0.47777…
Since there is one non-repeating digit after the decimal point,
Multiplying by 10 on both sides, we get
10x = 4.7777
Since there is one repeating digit after the decimal point,
Multiplying by 10 on both sides, we get
100x = 47.7777
Now, subtract both the values,
90x = 47 – 4
90x = 43
x = 43/90
= 43/90
Let x == 0.13434343…
Since there is one non-repeating digit after the decimal point,
Multiplying by 10 on both sides, we get
10x = 1.343434
Since there is two repeating digit after the decimal point,
Multiplying by 100 on both sides, we get
1000x = 134.343434
Now, subtract both the values,
990x = 133
x = 133/990
= 133/990
Let x == 0.001001001…
Since there is three repeating digit after the decimal point,
Multiplying by 1000 on both sides, we get
1000x = 1.001001
Now, subtract both the values,
999x = 1
x = 1/999
= 1/999
9. Classify the following numbers as rational or irrational:
(i) √23
(ii) √225
(iii) 0.3796
(iv) 7.478478
(v) 1.101001000100001…
Solution
(i) √23
Since, 23 is not a perfect square,
√23 is an irrational number.
(ii) √225
√225 = √(15)2 = 15
Since, 225 is a perfect square,
√225 is a rational number.
(iii) 0.3796
0.3796 = 3796/1000
Since, the decimal expansion is terminating decimal.
0.3796 is a rational number.
(iv) 7.478478
Let x = 7.478478
Since there is three repeating digit after the decimal point,
Multiplying by 1000 on both sides, we get
1000x = 7478.478478…
Now, subtract both the values,
999x = 7478 – 7
999x = 7471
x = 7471/999
7.478478 = 7471/999
Hence, it is neither terminating nor non-terminating or non-repeating decimal.
7.478478 is an irrational number.
(v) 1.101001000100001…
Since number of zero’s between two consecutive ones are increasing. So it is non-terminating or non-repeating decimal.
1.101001000100001… is an irrational number.
Let x = 345.0456456
Multiplying by 10 on both sides, we get
10x = 3450.456456
Since there is three repeating digit after the decimal point,
Multiplying by 1000 on both sides, we get
1000x = 3450456.456456…
Now, subtract both the values,
10000x – 10x = 3450456 – 345
9990x = 3450111
x = 3450111/9990
Since, it is non-terminating repeating decimal.
is a rational number.
10. The following real numbers have decimal expansions as given below. In each case, state whether they are rational or not. If they are rational and expressed in the form p/q, where p, q are integers, q≠ 0 and p, q are co-prime, then what can you say about the prime factors of q?
(i) 37.09158
(ii)
(iii) 8.9010010001…
(iv) 2.3476817681…
Solution
(i) 37.09158
We know that
It has terminating decimal
Here
It is a rational number and factors of q will be 2 or 5 or both.
(ii)
We know that
It has non-terminating recurring decimals
Here
It is a rational number.
(iii) 8.9010010001…
We know that
It has non-terminating, non-recurring decimal.
Here
It is not a rational number.
(iv) 2.3476817681…
We know that
It has non-terminating, recurring decimal.
Here
It is a rational number and the factors of q are prime factors other than 2 and 5.
11. Insert an irrational number between the following.
(i) 1/3 and ½
(ii) -2/5 and ½
(iii) 0 and 0.1
Solution
(i) One irrational number between 1/3 and ½
1/3 = 0.333…
½ = 0.5
So, there are infinite irrational numbers between 1/3 and ½.
One irrational number among them can be 0.4040040004…
(ii) One irrational number between -2/5 and ½
-2/5 = -0.4
½ = 0.5
So there are infinite irrational numbers between -2/5 and ½.
One irrational number among them can be 0.1010010001…
(iii) One irrational number between 0 and 0.1
There are infinite irrational numbers between 0 and 1.
One irrational number among them can be 0.06006000600006…
12. Insert two irrational numbers between 2 and 3.
Solution
2 is expressed as √4
And 3 is expressed as √9
So, two irrational numbers between 2 and 3 or √4 and √9 are √5, √6
13. Write two irrational numbers between 4/9 and 7/11.
Solution
4/9 is expressed as 0.4444…
7/11 is expressed as 0.636363…
So, two irrational numbers between 4/9 and 7/11 are 0.4040040004… and 0.6060060006…
14. Find one rational number between √2 and √3.
Solution
√2 is expressed as 1.4142…
√3 is expressed as 1.7320…
So, one rational number between √2 and √3 is 1.5.
15. Find two rational numbers between 2√3 and √15.
Solution
√12 = √(4×3) = 2√3
Since, 12 < 12.25 < 12.96 < 15
So, √12 < √12.25 < √12.96 < √15
Hence, two rational numbers between √12 and √15 are [√12.25, √12.96] or [√3.5, √3.6].
16. Insert irrational numbers between √5 and √7.
Solution
Since, 5 < 6 < 7
So, irrational number between √5 and √7 is √6.
17. Insert two irrational numbers between √3 and √7.
Solution
Since, 3 < 4 < 5 < 6 < 7
So,
√3 < √4 < √5 < √6 < √7
But √4 = 2, which is a rational number.
So,
Two irrational numbers between √3 and √7 are √5 and √6.
Exercise 1.4
1. Simplify the following:
(i) √45 – 3√20 + 4√5
(ii) 3√3 + 2√27 + 7/√3
(iii) 6√5 × 2√5
(iv) 8√15 ÷ 2√3
(v) √24/8 + √54/9
(vi) 3/√8 + 1/√2
Solution
(i) √45 – 3√20 + 4√5
Let us simplify the expression,
√45 – 3√20 + 4√5
= √(9×5) – 3√(4×5) + 4√5
= 3√5 – 3×2√5 + 4√5
= 3√5 – 6√5 + 4√5
= √5
(ii) 3√3 + 2√27 + 7/√3
Let us simplify the expression,
3√3 + 2√27 + 7/√3
= 3√3 + 2√(9×3) + 7√3/(√3×√3) (by rationalizing)
= 3√3 + (2×3)√3 + 7√3/3
= 3√3 + 6√3 + (7/3) √3
= √3 (3 + 6 + 7/3)
= √3 (9 + 7/3)
= √3 (27+7)/3
= 34/3 √3
(iii) 6√5 × 2√5
Let us simplify the expression,
6√5 × 2√5
= 12 × 5
= 60
(iv) 8√15 ÷ 2√3
Let us simplify the expression,
8√15 ÷ 2√3
= (8 √5 √3) / 2√3
= 4√5
(v) √24/8 + √54/9
Let us simplify the expression,
√24/8 + √54/9
= √(4×6)/8 + √(9×6)/9
= 2√6/8 + 3√6/9
= √6/4 + √6/3
By taking LCM
= (3√6 + 4√6)/12
= 7√6/12
(vi) 3/√8 + 1/√2
Let us simplify the expression,
3/√8 + 1/√2
= 3/2√2 + 1/√2
By taking LCM
= (3 + 2)/(2√2)
= 5/(2√2)
By rationalizing,
= 5√2/(2√2 × 2√2)
= 5√2/(2×2)
= 5√2/4
2. Simplify the following:
(i) (5 + √7) (2 + √5)
(ii) (5 + √5) (5 – √5)
(iii) (√5 + √2)2
(iv) (√3 – √7)2
(v) (√2 + √3) (√5 + √7)
(vi) (4 + √5) (√3 – √7)
Solution
(i) (5 + √7) (2 + √5)
Let us simplify the expression,
= 5(2 + √5) + √7(2 + √5)
= 10 + 5√5 + 2√7 + √35
(ii) (5 + √5) (5 – √5)
Let us simplify the expression,
By using the formula,
(a)2 – (b)2 = (a + b) (a – b)
So,
= (5)2 – (√5)2
= 25 – 5
= 20
(iii) (√5 + √2)2
Let us simplify the expression,
By using the formula,
(a + b)2 = a2 + b2 + 2ab
(√5 + √2)2 = (√5)2 + (√2)2 + 2√5√2
= 5 + 2 + 2√10
= 7 + 2√10
(iv) (√3 – √7)2
Let us simplify the expression,
By using the formula,
(a – b)2 = a2 + b2 – 2ab
(√3 – √7)2 = (√3)2 + (√7)2 – 2√3√7
= 3 + 7 – 2√21
= 10 – 2√21
(v) (√2 + √3) (√5 + √7)
Let us simplify the expression,
= √2(√5 + √7) + √3(√5 + √7)
= √2×√5 + √2×√7 + √3×√5 + √3×√7
√10 + √14 + √15 + √21
(vi) (4 + √5) (√3 – √7)
Let us simplify the expression,
= 4(√3 – √7) + √5(√3 – √7)
= 4√3 – 4√7 + √15 – √35
3. If √2 = 1.414, then find the value of
(i) √8 + √50 + √72 + √98
(ii) 3√32 – 2√50 + 4√128 – 20√18
Solution
(i) √8 + √50 + √72 + √98
Let us simplify the expression,
√8 + √50 + √72 + √98
= √(2×4) + √(2×25) + √(2×36) + √(2×49)
= √2 √4 + √2 √25 + √2 √36 + √2 √49
= 2√2 + 5√2 + 6√2 + 7√2
= 20√2
= 20 × 1.414
= 28.28
(ii) 3√32 – 2√50 + 4√128 – 20√18
Let us simplify the expression,
3√32 – 2√50 + 4√128 – 20√18
= 3√(16×2) – 2√(25×2) + 4√(64×2) – 20√(9×2)
= 3√16 √2 – 2√25 √2 + 4√64 √2 – 20√9 √2
= 3.4√2 – 2.5√2 + 4.8√2 – 20.3√2
= 12√2 – 10√2 + 32√2 – 60√2
= (12 – 10 + 32 – 60) √2
= -26√2
= -26 × 1.414
= -36.764
4. If √3 = 1.732, then find the value of
(i) √27 + √75 + √108 – √243
(ii) 5√12 – 3√48 + 6√75 + 7√108
Solution
(i) √27 + √75 + √108 – √243
Let us simplify the expression,
√27 + √75 + √108 – √243
= √(9×3) + √(25×3) + √(36×3) – √(81×3)
= √9 √3 + √25 √3 + √36 √3 – √81 √3
= 3√3 + 5√3 + 6√3 – 9√3
= (3 + 5 + 6 – 9) √3
= 5√3
= 5 × 1.732
= 8.660
(ii) 5√12 – 3√48 + 6√75 + 7√108
Let us simplify the expression,
5√12 – 3√48 + 6√75 + 7√108
= 5√(4×3) – 3√(16×3) + 6√(25×3) + 7√(36×3)
= 5√4 √3 – 3√16 √3 + 6√25 √3 + 7√36 √3
= 5.2√3 – 3.4√3 + 6.5√3 + 7.6√3
= 10√3 – 12√3 + 30√3 + 42√3
= (10 – 12 + 30 + 42) √3
= 70√3
= 70 × 1.732
= 121.24
5. State which of the following are rational or irrational decimals.
(i) √(4/9), -3/70, √(7/25), √(16/5)
(ii) -√(2/49), 3/200, √(25/3), -√(49/16)
Solution
(i) √(4/9), -3/70, √(7/25), √(16/5)
√(4/9) = 2/3
-3/70 = -3/70
√(7/25) = √7/5
√(16/5) = 4/√5
So,
√7/5 and 4/√5 are irrational decimals.
2/3 and -3/70 are rational decimals.
(ii) -√(2/49), 3/200, √(25/3), -√(49/16)
-√(2/49) = -√2/7
3/200 = 3/200
√(25/3) = 5/√3
-√(49/16) = -7/4
So,
-√2/7 and 5/√3 are irrational decimals.
3/200 and -7/4 are rational decimals.
6. State which of the following are rational or irrational decimals.
(i) -3√2
(ii) √(256/81)
(iii) √(27×16)
(iv) √(5/36)
Solution
(i) -3√2
We know that √2 is an irrational number.
So, -3√2 will also be irrational number.
(ii) √(256/81)
√(256/81) = 16/9 = 4/3
It is a rational number.
(iii) √(27×16)
√(27×16) = √(9×3×16) = 3×4√3 = 12√3
It is an irrational number.
(iv) √(5/36)
√(5/36) = √5/6
It is an irrational number.
7. State which of the following are irrational numbers.
(i) 3 – √(7/25)
(ii) -2/3 + ∛2
(iii) 3/√3
(iv) -2/7 ∛5
(v) (2 – √3) (2 + √3)
(vi) (3 + √5)2
(vii) (2/5 √7)2
(viii) (3 – √6)2
Solution
(i) 3 – √(7/25)
Let us simplify,
3 – √(7/25) = 3 – √7/√25
= 3 – √7/5
Hence, 3 – √7/5 is an irrational number.
(ii) -2/3 + ∛2
Let us simplify,
-2/3 + ∛2 = -2/3 + 21/3
Since, 2 is not a perfect cube.
Hence it is an irrational number.
(iii) 3/√3
Let us simplify,
By rationalizing, we get
3/√3 = 3√3/(√3×√3)
= 3√3/3
= √3
Hence, 3/√3 is an irrational number.
(iv) -2/7 ∛5
Let us simplify,
-2/7 ∛5 = -2/7 (5)1/3
Since, 5 is not a perfect cube.
Hence it is an irrational number.
(v) (2 – √3) (2 + √3)
Let us simplify,
By using the formula,
(a + b) (a – b) = (a)2 (b)2
(2 – √3) (2 + √3) = (2)2 – (√3)2
= 4 – 3
= 1
Hence, it is a rational number.
(vi) (3 + √5)2
Let us simplify,
By using (a + b)2 = a2 + b2 + 2ab
(3 + √5)2 = 32 + (√5)2 + 2.3.√5
= 9 + 5 + 6√5
= 14 + 6√5
Hence, it is an irrational number.
(vii) (2/5 √7)2
Let us simplify,
(2/5 √7)2 = (2/5 √7) × (2/5 √7)
= 4/ 25 × 7
= 28/25
Hence it is a rational number.
(viii) (3 – √6)2
Let us simplify,
By using (a – b)2 = a2 + b2 – 2ab
(3 – √6)2 = 32 + (√6)2 – 2.3.√6
= 9 + 6 – 6√6
= 15 – 6√6
Hence it is an irrational number.
8. Prove the following are irrational numbers.
(i) ∛2
(ii) ∛3
(iii) ∜5
Solution
(i) ∛2
We know that ∛2 = 21/3
Let us consider 21/3 = p/q, where p, q are integers, q>0.
p and q have no common factors (except 1).
So,
21/3 = p/q
2 = p3/q3
p3 = 2q3 … (1)
We know that, 2 divides 2q3 then 2 divides p3
So, 2 divides p
Now, let us consider p = 2k, where k is an integer
Substitute the value of p in (1), we get
p3 = 2q3
(2k)3 = 2q3
8k3 = 2q3
4k3 = q3
We know that, 2 divides 4k3 then 2 divides q3
So, 2 divides q
Thus p and q have a common factor ‘2’.
This contradicts the statement, p and q have no common factor (except 1).
Hence, ∛2 is an irrational number.
(ii) ∛3
We know that ∛3 = 31/3
Let us consider 31/3 = p/q, where p, q are integers, q>0.
p and q have no common factors (except 1).
So,
31/3 = p/q
3 = p3/q3
p3 = 3q3 ….. (1)
We know that, 3 divides 3q3 then 3 divides p3
So, 3 divides p
Now, let us consider p = 3k, where k is an integer
Substitute the value of p in (1), we get
p3 = 3q3
(3k)3 = 3q3
9k3 = 3q3
3k3 = q3
We know that, 3 divides 9k3 then 3 divides q3
So, 3 divides q
Thus p and q have a common factor ‘3’.
This contradicts the statement, p and q have no common factor (except 1).
Hence, ∛3 is an irrational number.
(iii) ∜5
We know that ∜5 = 51/4
Let us consider 51/4 = p/q, where p, q are integers, q>0.
p and q have no common factors (except 1).
So,
51/4 = p/q
5 = p4/q4
P4 = 5q4 ... (1)
We know that, 5 divides 5q4 then 5 divides p4
So, 5 divides p
Now, let us consider p = 5k, where k is an integer
Substitute the value of p in (1), we get
P4 = 5q4
(5k)4 = 5q4
625k4 = 5q4
125k4 = q4
We know that, 5 divides 125k4 then 5 divides q4
So, 5 divides q
Thus p and q have a common factor ‘5’.
This contradicts the statement, p and q have no common factor (except 1).
Hence, ∜5 is an irrational number.
9. Find the greatest and the smallest real numbers.
(i) 2√3, 3/√2, -√7, √15
(ii) -3√2, 9/√5, -4, 4/3 √5, 3/2√3
Solution
(i) 2√3, 3/√2, -√7, √15
Let us simplify each fraction
2√3 = √(4×3) = √12
3/√2 = (3×√2)/(√2×√2) = 3√2/2 = √((9/4)×2) = √(9/2) = √4.5
-√7 = -√7
√15 = √15
So,
The greatest real number = √15
Smallest real number = -√7
(ii) -3√2, 9/√5, -4, 4/3 √5, 3/2√3
Let us simplify each fraction
-3√2 = -√(9×2) = -√18
9/√5 = (9×√5)/(√5×√5) = 9√5/5 = √((81/25)×5) = √(81/5) = √16.2
-4 = -√16
4/3 √5 = √((16/9)×5) = √(80/9) = √8.88 = √8.8
3/2√3 = √((9/4)×3) = √(27/4) = √6.25
So,
The greatest real number = 9√5
Smallest real number = -3√2
10. Write in ascending order.
(i) 3√2, 2√3, √15, 4
(ii) 3√2, 2√8, 4, √50, 4√3
Solution
(i) 3√2, 2√3, √15, 4
3√2 = √(9×2) =√18
2√3 = √(4×3) =√12
√15 = √15
4 = √16
Now, let us arrange in ascending order
√12 < √15 < √16 < √18
So,
2√3 < √15 < 4 < 3√2
(ii) 3√2, 2√8, 4, √50, 4√3
3√2 = √(9×2) =√18
2√8 = √(4×8) =√32
4 = √16
√50 = √50
4√3 =√(16×3) = √48
Now, let us arrange in ascending order
√16 < √18 < √32 < √48 < √50
So,
4 < 3√2 < 2√8 < 4√3 < √50
11. Write in descending order.
(i) 9/√2, 3/2 √5, 4√3, 3√(6/5)
(ii) 5/√3, 7/3 √2, -√3, 3√5, 2√7
Solution
(i) 9/√2, 3/2 √5, 4√3, 3√(6/5)
9/√2 = (9×√2)/(√2×√2) = 9√2/2 = √((81/4)×2) = √(81/2) = √40.5
3/2 √5 = √((9/4)×5) = √(45/4) = √11.25
4√3 = √(16×3) = √48
3√(6/5) = √((9×6)/5) = √(54/5) = √10.8
Now, let us arrange in descending order
√48 > √40.5 > √11.25 > √10.8
So,
4√3 > 9/√2 > 3/2√5 > 3√(6/5)
(ii) 5/√3, 7/3 √2, -√3, 3√5, 2√7
5/√3 = √(25/3) = √8.33
7/3 √2 = √((49/9) ×2) = √98/9 = √10.88
-√3 = -√3
3√5 = √(9×5) =√45
2√7 = √(4×7) = √28
Now, let us arrange in descending order
√45 > √28 > √10.88.. > √8.33.. > -√3
So,
3√5 > 2√7 > 7/3√2 > 5/√3 > -√3
12. Arrange in ascending order.
∛2, √3, 6√5
Solution
Here we can express the given expressions as:
∛2 = 21/3
√3 = 31/2
6√5 = 51/6
Let us make the roots common so,
21/3= 2(2× 1/2 × 1/3) = 41/6
31/2 = 3(3× 1/3 × 1/2) = 271/6
51/6 = 51/6
Now, let us arrange in ascending order,
41/6, 51/6, 271/6
So,
21/3 <51/6< 31/2
So,
∛2< 6√5< √3
Exercise 1.5
1. Rationalize the denominator of the following:
(i) 3/4√5
(ii) 5√7 / √3
(iii) 3/(4 – √7)
(iv) 17/(3√2 + 1)
(v) 16/ (√41 – 5)
(vi) 1/ (√7 – √6)
(vii) 1/ (√5 + √2)
(viii) (√2 + √3) / (√2 – √3)
Solution
(i) 3/4√5
Let us rationalize,
3/4√5 = (3×√5) /(4√5×√5)
= (3√5) / (4×5)
= (3√5) / 20
(ii) 5√7 / √3
Let us rationalize,
5√7 / √3= (5√7×√3) / (√3×√3)
= 5√21/3
(iii) 3/(4 – √7)
Let us rationalize,
3/(4 – √7) = [3×(4 + √7)] / [(4 – √7) × (4 + √7)]
= 3(4 + √7) / [42 – (√7)2]
= 3(4 + √7) / [16 – 7]
= 3(4 + √7) / 9
= (4 + √7) / 3
(iv) 17/(3√2 + 1)
Let us rationalize,
17/(3√2 + 1) = 17(3√2 – 1) / [(3√2 + 1) (3√2 – 1)]
= 17(3√2 – 1) / [(3√2)2 – 12]
= 17(3√2 – 1) / [9.2 – 1]
= 17(3√2 – 1) / [18 – 1]
= 17(3√2 – 1) / 17
= (3√2 – 1)
(v) 16/ (√41 – 5)
Let us rationalize,
16/ (√41 – 5) = 16(√41 + 5) / [(√41 – 5) (√41 + 5)]
= 16(√41 + 5) / [(√41)2 – 52]
= 16(√41 + 5) / [41 – 25]
= 16(√41 + 5) / [16]
= (√41 + 5)
(vi) 1/ (√7 – √6)
Let us rationalize,
1/ (√7 – √6) = 1(√7 + √6) / [(√7 – √6) (√7 + √6)]
= (√7 + √6) / [(√7)2 – (√6)2]
= (√7 + √6) / [7 – 6]
= (√7 + √6) / 1
= (√7 + √6)
(vii) 1/ (√5 + √2)
Let us rationalize,
1/ (√5 + √2) = 1(√5 – √2) / [(√5 + √2) (√5 – √2)]
= (√5 – √2) / [(√5)2 – (√2)2]
= (√5 – √2) / [5 – 2]
= (√5 – √2) / [3]
= (√5 – √2) /3
(viii) (√2 + √3) / (√2 – √3)
Let us rationalize,
(√2 + √3) / (√2 – √3) = [(√2 + √3) (√2 + √3)] / [(√2 – √3) (√2 + √3)]
= [(√2 + √3)2] / [(√2)2 – (√3)2]
= [2 + 3 + 2√2√3] / [2 – 3]
= [5 + 2√6] / -1
= – (5 + 2√6)
2. Simplify each of the following by rationalizing the denominator:
(i) (7 + 3√5) / (7 – 3√5)
(ii) (3 – 2√2) / (3 + 2√2)
(iii) (5 – 3√14) / (7 + 2√14)
Solution
(i) (7 + 3√5) / (7 – 3√5)
Let us rationalize the denominator, we get
(7 + 3√5) / (7 – 3√5) = [(7 + 3√5) (7 + 3√5)] / [(7 – 3√5) (7 + 3√5)]
= [(7 + 3√5)2] / [72 – (3√5)2]
= [72 + (3√5)2 + 2.7. 3√5] / [49 – 9.5]
= [49 + 9.5 + 42√5] / [49 – 45]
= [49 + 45 + 42√5] / [4]
= [94 + 42√5] / 4
= 2[47 + 21√5]/4
= [47 + 21√5]/2
(ii) (3 – 2√2) / (3 + 2√2)
Let us rationalize the denominator, we get
(3 – 2√2) / (3 + 2√2) = [(3 – 2√2) (3 – 2√2)] / [(3 + 2√2) (3 – 2√2)]
= [(3 – 2√2)2] / [32 – (2√2)2]
= [32 + (2√2)2 – 2.3.2√2] / [9 – 4.2]
= [9 + 4.2 – 12√2] / [9 – 8]
= [9 + 8 – 12√2] / 1
= 17 – 12√2
(iii) (5 – 3√14) / (7 + 2√14)
Let us rationalize the denominator, we get
(5 – 3√14) / (7 + 2√14) = [(5 – 3√14) (7 – 2√14)] / [(7 + 2√14) (7 – 2√14)]
= [5(7 – 2√14) – 3√14 (7 – 2√14)] / [72 – (2√14)2]
= [35 – 10√14 – 21√14 + 6.14] / [49 – 4.14]
= [35 – 31√14 + 84] / [49 – 56]
= [119 – 31√14] / [-7]
= -[119 – 31√14] / 7
= [31√14 – 119] / 7
3. Simplify:
[7√3 / (√10 + √3)] – [2√5 / (√6 + √5)] – [3√2 / (√15 + 3√2)]
Solution
Let us simplify individually,
[7√3 / (√10 + √3)]
Let us rationalize the denominator,
7√3 / (√10 + √3) = [7√3(√10 – √3)] / [(√10 + √3) (√10 – √3)]
= [7√3.√10 – 7√3.√3] / [(√10)2 – (√3)2]
= [7√30 – 7.3] / [10 – 3]
= 7[√30 – 3] / 7
= √30 – 3
Now,
[2√5 / (√6 + √5)]
Let us rationalize the denominator, we get
2√5 / (√6 + √5) = [2√5 (√6 – √5)] / [(√6 + √5) (√6 – √5)]
= [2√5.√6 – 2√5.√5] / [(√6)2 – (√5)2]
= [2√30 – 2.5] / [6 – 5]
= [2√30 – 10] / 1
= 2√30 – 10
Now,
[3√2 / (√15 + 3√2)]
Let us rationalize the denominator, we get
3√2 / (√15 + 3√2) = [3√2 (√15 – 3√2)] / [(√15 + 3√2) (√15 – 3√2)]
= [3√2.√15 – 3√2.3√2] / [(√15)2 – (3√2)2]
= [3√30 – 9.2] / [15 – 9.2]
= [3√30 – 18] / [15 – 18]
= 3[√30 – 6] / [-3]
= [√30 – 6] / -1
= 6 – √30
So, according to the question let us substitute the obtained values,
[7√3 / (√10 + √3)] – [2√5 / (√6 + √5)] – [3√2 / (√15 + 3√2)]
= (√30 – 3) – (2√30 – 10) – (6 – √30)
= √30 – 3 – 2√30 + 10 – 6 + √30
= 2√30 – 2√30 – 3 + 10 – 6
= 1
4. Simplify:
[1/(√4 + √5)] + [1/(√5 + √6)] + [1/(√6 + √7)] + [1/(√7 + √8)] + [1/(√8 + √9)]
Solution
Let us simplify individually,
[1/(√4 + √5)]
Rationalize the denominator, we get
[1/(√4 + √5)] = [1(√4 – √5)] / [(√4 + √5) (√4 – √5)]
= [(√4 – √5)] / [(√4)2 – (√5)2]
= [(√4 – √5)] / [4 – 5]
= [(√4 – √5)] / -1
= -(√4 – √5)
Now,
[1/(√5 + √6)]
Rationalize the denominator, we get
[1/(√5 + √6)] = [1(√5 – √6)] / [(√5 + √6) (√5 – √6)]
= [(√5 – √6)] / [(√5)2 – (√6)2]
= [(√5 – √6)] / [5 – 6]
= [(√5 – √6)] / -1
= -(√5 – √6)
Now,
[1/(√6 + √7)]
Rationalize the denominator, we get
[1/(√6 + √7)] = [1(√6 – √7)] / [(√6 + √7) (√6 – √7)]
= [(√6 – √7)] / [(√6)2 – (√7)2]
= [(√6 – √7)] / [6 – 7]
= [(√6 – √7)] / -1
= -(√6 – √7)
Now,
[1/(√7 + √8)]
Rationalize the denominator, we get
[1/(√7 + √8)] = [1(√7 – √8)] / [(√7 + √8) (√7 – √8)]
= [(√7 – √8)] / [(√7)2 – (√8)2]
= [(√7 – √8)] / [7 – 8]
= [(√7 – √8)] / -1
= -(√7 – √8)
Now,
[1/(√8 + √9)]
Rationalize the denominator, we get
[1/(√8 + √9)] = [1(√8 – √9)] / [(√8 + √9) (√8 – √9)]
= [(√8 – √9)] / [(√8)2 – (√9)2]
= [(√8 – √9)] / [8 – 9]
= [(√8 – √9)] / -1
= -(√8 – √9)
So, according to the question let us substitute the obtained values,
[1/(√4 + √5)] + [1/(√5 + √6)] + [1/(√6 + √7)] + [1/(√7 + √8)] + [1/(√8 + √9)]
= -(√4 – √5) + -(√5 – √6) + -(√6 – √7) + -(√7 – √8) + -(√8 – √9)
= -√4 + √5 – √5 + √6 – √6 + √7 – √7 + √8 – √8 + √9
= -√4 + √9
= -2 + 3
= 1
5. Give a and b are rational numbers. Find a and b if:
(i) [3 – √5] / [3 + 2√5] = -19/11 + a√5
(ii) [√2 + √3] / [3√2 – 2√3] = a – b√6
(iii) {[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]} = a + 7/11 b√5
Solution
(i) [3 – √5] / [3 + 2√5] = -19/11 + a√5
Let us consider LHS
[3 – √5] / [3 + 2√5]
Rationalize the denominator,
[3 – √5] / [3 + 2√5] = [(3 – √5) (3 – 2√5)] / [(3 + 2√5) (3 – 2√5)]
= [3(3 – 2√5) – √5(3 – 2√5)] / [32 – (2√5)2]
= [9 – 6√5 – 3√5 + 2.5] / [9 – 4.5]
= [9 – 6√5 – 3√5 + 10] / [9 – 20]
= [19 – 9√5] / -11
= -19/11 + 9√5/11
So when comparing with RHS
-19/11 + 9√5/11 = -19/11 + a√5
Hence, value of a = 9/11
(ii) [√2 + √3] / [3√2 – 2√3] = a – b√6
Let us consider LHS
[√2 + √3] / [3√2 – 2√3]
Rationalize the denominator,
[√2 + √3] / [3√2 – 2√3] = [(√2 + √3) (3√2 + 2√3)] / [(3√2 – 2√3) (3√2 + 2√3)]
= [√2(3√2 + 2√3) + √3(3√2 + 2√3)] / [(3√2)2 – (2√3)2]
= [3.2 + 2√2√3 + 3√2√3 + 2.3] / [9.2 – 4.3]
= [6 + 2√6 + 3√6 + 6] / [18 – 12]
= [12 + 5√6] / 6
= 12/6 + 5√6/6
= 2 + 5√6/6
= 2 – (-5√6/6)
So when comparing with RHS
2 – (-5√6/6) = a – b√6
Hence, value of a = 2 and b = -5/6
(iii) {[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]} = a + 7/11 b√5
Let us consider LHS
Since there are two terms, let us solve individually
{[7 + √5]/[7 – √5]}
Rationalize the denominator,
[7 + √5]/[7 – √5] = [(7 + √5) (7 + √5)] / [(7 – √5) (7 + √5)]
= [(7 + √5)2] / [72 – (√5)2]
= [72 + (√5)2 + 2.7.√5] / [49 – 5]
= [49 + 5 + 14√5] / [44]
= [54 + 14√5] / 44
Now,
{[7 – √5]/[7 + √5]}
Rationalize the denominator,
[7 – √5]/[7 + √5] = (7 – √5) (7 – √5)] / [(7 + √5) (7 – √5)]
= [(7 – √5)2] / [72 – (√5)2]
= [72 + (√5)2 – 2.7.√5] / [49 – 5]
= [49 + 5 – 14√5] / [44]
= [54 – 14√5] / 44
So, according to the question
{[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]}
By substituting the obtained values,
= {[54 + 14√5] / 44} – {[54 – 14√5] / 44}
= [54 + 14√5 – 54 + 14√5]/44
= 28√5/44
= 7√5/11
So when comparing with RHS
7√5/11 = a + 7/11 b√5
Hence, value of a = 0 and b = 1
6. If {[7 + 3√5] / [3 + √5]} – {[7 – 3√5] / [3 – √5]} = p + q√5, find the value of p and q where p and q are rational numbers.
Solution
Let us consider LHS
Since there are two terms, let us solve individually
{[7 + 3√5] / [3 + √5]}
Rationalize the denominator,
[7 + 3√5] / [3 + √5] = [(7 + 3√5) (3 – √5)] / [(3 + √5) (3 – √5)]
= [7(3 – √5) + 3√5(3 – √5)] / [32 – (√5)2]
= [21 – 7√5 + 9√5 – 3.5] / [9 – 5]
= [21 + 2√5 – 15] / [4]
= [6 + 2√5] / 4
= 2[3 + √5]/4
= [3 + √5] /2
Now,
{[7 – 3√5] / [3 – √5]}
Rationalize the denominator,
[7 – 3√5] / [3 – √5] = [(7 – 3√5) (3 + √5)] / [(3 – √5) (3 + √5)]
= [7(3 + √5) – 3√5(3 + √5)] / [32 – (√5)2]
= [21 + 7√5 – 9√5 – 3.5] / [9 – 5]
= [21 – 2√5 – 15] / 4
= [6 – 2√5]/4
= 2[3 – √5]/4
= [3 – √5]/2
So, according to the question
{[7 + 3√5] / [3 + √5]} – {[7 – 3√5] / [3 – √5]}
By substituting the obtained values,
= {[3 + √5] /2} – {[3 – √5] /2}
= [3 + √5 – 3 + √5]/2
= [2√5]/2
= √5
So when comparing with RHS
√5 = p + q√5
Hence, value of p = 0 and q = 1
7. Rationalise the denominator of the following and hence evaluate by taking √2 = 1.414, √3 = 1.732, upto three places of decimal:
(i) √2/(2 + √2)
(ii) 1/(√3 + √2)
Solution
(i) √2/(2 + √2)
By rationalizing the denominator,
√2/(2 + √2)= [√2(2 – √2)] / [(2 + √2) (2 – √2)]
= [2√2 – 2] / [22 – (√2)2]
= [2√2 – 2] / [4 – 2]
= 2[√2 – 1] / 2
= √2 – 1
= 1.414 – 1
= 0.414
(ii) 1/(√3 + √2)
By rationalizing the denominator,
1/(√3 + √2) = [1(√3 – √2)] / [(√3 + √2) (√3 – √2)]
= [(√3 – √2)] / [(√3)2 – (√2)2]
= [(√3 – √2)] / [3 – 2]
= [(√3 – √2)] / 1
= (√3 – √2)
= 1.732 – 1.414
= 0.318
8. If a = 2 + √3, find 1/a, (a – 1/a)
Solution
Given:
a = 2 + √3
So,
1/a = 1/ (2 + √3)
By rationalizing the denominator,
1/ (2 + √3) = [1(2 – √3)] / [(2 + √3) (2 – √3)]
= [(2 – √3)] / [22 – (√3)2]
= [(2 – √3)] / [4 – 3]
= (2 – √3)
Then,
a – 1/a = 2 + √3 – (2 – √3)
= 2 + √3 – 2 + √3
= 2√3
9. Solve:
If x = 1 – √2, find 1/x, (x – 1/x)4
Solution
Given:
x = 1 – √2
so,
1/x = 1/(1 – √2)
By rationalizing the denominator,
1/ (1 – √2) = [1(1 + √2)] / [(1 – √2) (1 + √2)]
= [(1 + √2)] / [12 – (√2)2]
= [(1 + √2)] / [1 – 2]
= (1 + √2) / -1
= -(1 + √2 )
Then,
(x – 1/x)4 = [1 – √2 – (-1 – √2)]4
= [1 – √2 + 1 + √2]4
= 24
= 16
10. Solve:
If x = 5 – 2√6, find 1/x, (x2 – 1/x2)
Solution
Given:
x = 5 – 2√6
so,
1/x = 1/(5 – 2√6)
By rationalizing the denominator,
1/(5 – 2√6) = [1(5 + 2√6)] / [(5 – 2√6) (5 + 2√6)]
= [(5 + 2√6)] / [52 – (2√6)2]
= [(5 + 2√6)] / [25 – 4.6]
= [(5 + 2√6)] / [25 – 24]
= (5 + 2√6)
Then,
x + 1/x = 5 – 2√6 + (5 + 2√6)
= 10
Square on both sides we get
(x + 1/x)2 = 102
x2 + 1/x2 + 2x.1/x = 100
x2 + 1/x2 + 2 = 100
x2 + 1/x2 = 100 – 2
= 98
11. If p = (2-√5)/(2+√5) and q = (2+√5)/(2-√5), find the values of
(i) p + q
(ii) p – q
(iii) p2 + q2
(iv) p2 – q2
Solution
Given:
p = (2-√5)/(2+√5) and q = (2+√5)/(2-√5)
(i) p + q
[(2-√5)/(2+√5)] + [(2+√5)/(2-√5)]
So by rationalizing the denominator, we get
= [(2 – √5)2 + (2 + √5)2] / [22 – (√5)2]
= [4 + 5 – 4√5 + 4 + 5 + 4√5] / [4 – 5]
= [18]/-1
= -18
(ii) p – q
[(2-√5)/(2+√5)] – [(2+√5)/(2-√5)]
So by rationalizing the denominator, we get
= [(2 – √5)2 – (2 + √5)2] / [22 – (√5)2]
= [4 + 5 – 4√5 – (4 + 5 + 4√5)] / [4 – 5]
= [9 – 4√5 – 9 – 4√5] / -1
= [-8√5]/-1
= 8√5
(iii) p2 + q2
We know that (p + q)2 = p2 + q2 + 2pq
So,
p2 + q2 = (p + q)2 – 2pq
pq = [(2-√5)/(2+√5)] × [(2+√5)/(2-√5)]
= 1
p + q = -18
so,
p2 + q2 = (p + q)2 – 2pq
= (-18)2 – 2(1)
= 324 – 2
= 322
(iv) p2 – q2
We know that, p2 – q2 = (p + q) (p – q)
So, by substituting the values
p2 – q2 = (p + q) (p – q)
= (-18) (8√5)
= -144√5
12. If x = (√2 – 1)/( √2 + 1) and y = (√2 + 1)/( √2 – 1), find the value of x2 + 5xy + y2.
Solution
Given:
x = (√2 – 1)/( √2 + 1) and y = (√2 + 1)/( √2 – 1)
x + y= [(√2 – 1)/( √2 + 1)] + [(√2 + 1)/( √2 – 1)]
By rationalizing the denominator,
= [(√2 – 1)2 + (√2 + 1)2] / [(√2)2 – 12]
= [2 + 1 – 2√2 + 2 + 1 + 2√2] / [2 – 1]
= [6] / 1
= 6
xy = [(√2 – 1)/( √2 + 1)] × [(√2 + 1)/( √2 – 1)]
= 1
We know that
x2 + 5xy + y2 = x+ y2 + 2xy + 3xy
It can be written as
= (x + y)2 + 3xy
Substituting the values
= 62 + 3 × 1
So we get
= 36 + 3
= 39
Chapter test
1. Without actual division, find whether the following rational numbers are terminating decimals or recurring decimals:
(i) 13/45
(ii) -5/56
(iii) 7/125
(iv) -23/80
(v) – 15/66
In case of terminating decimals, write their decimal expansions.
Solution
(i) We know that
The fraction whose denominator is the multiple of 2 or 5 or both is a terminating decimal
In 13/45
45 = 3 × 3 × 5
Hence, it is not a terminating decimal.
(ii) In -5/56
56 = 2 × 2 × 2 × 7
Hence, it is not a terminating decimal.
(iii) In 7/125
125 = 5 × 5 × 5
We know that
Hence, it is a terminating decimal.
(iv) In -23/80
80 = 2 × 2 × 2 × 2 × 5
We know that
Hence, it is a terminating decimal.
(v) In – 15/66
66 = 2 × 3 × 11
Hence, it is not a terminating decimal.
2. Express the following recurring decimals as vulgar fractions:
Solution
(i) We know that
Now multiply both sides of equation (1) by 10
10x = 13.4545 ... (2)
Again multiply both sides of equation (2) by 100
1000x = 1345.4545 … (3)
By subtracting equation (2) from (3)
990x = 1332
By further calculation
x = 1332/990 = 74/55
(ii) We know that
Now multiply both sides of equation (1) by 1000
1000x = 2357.357357 ... (2)
By subtracting equation (1) from (2)
999x = 2355
By further calculation
x = 2355/999
3. Insert a rational number between 5/9 and 7/13, and arrange in ascending order.
Solution
We know that
A rational number between 5/9 and 7/13
= 64/117
Here,
7/13< 64/117< 5/9
Therefore, in ascending order – 7/13, 64/117, 5/9.
4. Insert four rational numbers between 4/5 and 5/6.
Solution
We know that
Rational numbers between 4/5 and 5/6
Here LCM of 5, 6 = 30
So the four rational numbers are
121/150, 122/150, 123/150, 124/150
By further simplification
121/150, 61/75, 41/50, 62/75
5. Prove that the reciprocal of an irrational number is irrational.
Solution
Consider x as an irrational number
Reciprocal of x is 1/x
If 1/x is a non-zero rational number
Then x × 1/x will also be an irrational number.
We know that the product of a non-zero rational number and irrational number is also irrational.
If x × 1/x = 1 is rational number
Our assumption is wrong
So 1/x is also an irrational number.
Therefore, the reciprocal of an irrational number is also an irrational number.
6. Prove that the following numbers are irrational:
Solution
(i) √8
If √8 is a rational number
Consider √8 = p/q where p and q are integers
q ˃ 0 and p and q have no common factor
By squaring on both sides
8 = p2/q2
So we get
p2 = 8q2
We know that
8p2 is divisible by 8
p2 is also divisible by 8
p is divisible by 8
Consider p = 8k where k is an integer
By squaring on both sides
p2 = (8k)2
p2 = 64k2
We know that
64k2 is divisible by 8
p2 is divisible by 8
p is divisible by 8
Here p and q both are divisible by 8
So our supposition is wrong
Therefore, √8 is an irrational number.
(ii) √14
If √14 is a rational number
Consider √14 = p/q where p and q are integers
q ≠ 0 and p and q have no common factor
By squaring on both sides
14 = p2/q2
So we get
p2 = 14q2 ... (1)
We know that
p2 is also divisible by 2
p is divisible by 2
Consider p = 2m
Substitute the value of p in equation (1)
(2m)2 = 13q2
So we get
4m2 = 14q2
2m2 = 7q2
We know that
q2 is divisible by 2
q is divisible by 2
Here p and q have 2 as the common factor which is not possible
Therefore, √14 is an irrational number.
If is a rational number
Consider = p/q where p and q are integers
q ˃ 0 and p and q have no common factor
By cubing on both sides
2 = p3/q3
So we get
p3 = 2q3 ….. (1)
We know that
2q3 is also divisible by 2
p3 is divisible by 2
p is divisible by 2
Consider p = 2k where k is an integer
By cubing both sides
p3 = (2k)3
p3 = 8k3
So we get
2q3 = 8k3
q3 = 4k3
We know that
4k3 is divisible by 2
q3 is divisible by 2
q is divisible by 2
Here p and q are divisible by 2
So our supposition is wrong
Therefore, is an irrational number.
7. Prove that √3 is a rational number. Hence show that 5 – √3 is an irrational number.
Solution
If √3 is a rational number
Consider √3 = p/q where p and q are integers
q ˃ 0 and p and q have no common factor
By squaring both sides
3 = p2/q2
So we get
P2 = 3q2
We know that
3q2 is divisible by 3
p2 is divisible by 3
p is divisible by 3
Consider p = 3 where k is an integer
By squaring on both sides
P2 = 9k2
9k2 is divisible by 3
p2 is divisible by 3
3q2 is divisible by 3
q2 is divisible by 3
q is divisible by 3
Here p and q are divisible by 3
So our supposition is wrong
Therefore, √3 is an irrational number.
In 5 – √3
5 is a rational number
√3 is an irrational number (proved)
We know that
Difference of a rational number and irrational number is also an irrational number
So 5 – √3 is an irrational number.
Therefore, it is proved.
8. Prove that the following numbers are irrational:
(i) 3 + √5
(ii) 15 – 2√7
Solution
(i) If 3 + √5 is a rational number say x
Consider 3 + √5 = x
It can be written as
√5 = x – 3
Here x – 3 is a rational number
√5 is also a rational number.
Consider √5 = p/q where p and q are integers
q ˃ 0 and p and q have no common factor
By squaring both sides
5 = p2/q2
p2 = 5q2
We know that
5q2 is divisible by 5
p2 is divisible by 5
p is divisible 5
Consider p = 5k where k is an integer
By squaring on both sides
p2 = 25k2
So we get
5q2 = 25k2
q2 = 5k2
Here
5k2 is divisible by 5
q2 is divisible by 5
q is divisible by 5
Here p and q are divisible by 5
So our supposition is wrong
√5 is an irrational number
3 + √5 is also an irrational number.
Therefore, it is proved.
(ii) If 15 – 2√7 is a rational number say x
Consider 15 – 2√7 = x
It can be written as
2√7 = 15 – x
So we get
√7 = (15 – x)/ 2
Here
(15 – x)/ 2 is a rational number
√7 is a rational number
Consider √7 = p/q where p and q are integers
q ˃ 0 and p and q have no common factor
By squaring on both sides
7 = p2/q2
p2 = 7q2
Here
7q2 is divisible by 7
p2 is divisible by 7
p is divisible by 7
Consider p = 7k where k is an integer
By squaring on both sides
p2 = 49k2
It can be written as
7q2 = 49k2
q2 = 7k2
Here
7k2 is divisible by 7
q2 is divisible by 7
q is divisible by 7
Here p and q are divisible by 7
So our supposition is wrong
√7 is an irrational number
15 – 2√7 is also an irrational number.
Therefore, it is proved.
By rationalizing the denominator
Here
¾ is a rational number and √5/4 is an irrational number
We know that
Sum of a rational and an irrational number is an irrational number.
Therefore, it is proved.
9. Rationalise the denominator of the following:
Solution
10. If p, q are rational numbers and p – √15q = 2√3 – √5/4√3 – 3√5, find the values of p and q.
Solution
It is given that
By comparing both sides
p = 3 and q = -2/3
11. If x = 1/3 + 2√2, then find the value of x – 1/x.
Solution
Here
1/x = 3 + 2√2/1 = √3 + 2√2
We know that
x – 1/x = (3 – 2√2) – (3 + 2√2)
By further calculation
= 3 – 2√2 – 3 – 2√2
So we get
= -4√2
13. (i) If x = 7+3√5/7-3√5, find the value of x2 + 1/x2
(ii) If x = √5-√3/√5+√2 and y = √5+√2/√5-√2, then find the value of x2 + xy + y2
(iii) If x = √3-√2/√3+√2 and y = √x2 +
Solution
Dividing by 2
=1000 – 30
= 970
13. Write the following real numbers in descending order:
Solution
We know that
√2 = √2
3.5 = √12.25
√10 = √10
Writing the above numbers in descending order
√18.75, √12.25, √10, √2, – √12.5
So we get
5/2 √3, 3.5, √10, √2, -5/√2
14. Find a rational number and an irrational number between √3 and √5.
Solution
Let (√3)2 = 3 and (√5)2 = 5
(i) There exists a rational number 4 which is the perfect square of a rational number 2.
(ii) There can be much more rational numbers which are perfect squares.
We know that, one irrational number between √3 and √5 = ½ (√3 + √5) = (√3 + √5)/2
15. Insert three irrational numbers between 2√3 and 2√5, and arrange in descending order.
Solution
Take the square
(2√3)2 = 12 and (2√5)2 = 20
So the number 13, 15, 18 lie between 12 and 20 between (√12)2 and (√20)2
√13, √15, √18 lie between 2√3 and 2√5
Therefore, three irrational numbers between
2√3 and 2√5 are √13, √15, √18 or √13, √15 and 3√2.
Here
√20 ˃ √18 ˃ √15 ˃ √13 ˃ √12 or,
2√5 ˃ 3√2 ˃ √15 ˃ √13 ˃ 2√3
Therefore, the descending order: 2√5, 3√2, √15, √13 and 2√3.
16. Give an example each of two different irrational numbers, whose
(i) sum is an irrational number.
(ii) product is an irrational number.
Solution
(i) Consider a = √2 and b = √3 as two irrational numbers
Here,
a + b = √2 + √3 is also an irrational number.
(ii) Consider a = √2 and b = √3 as two irrational numbers
Here,
ab = √2 √3 = √6 is also an irrational number.
17. Give an example of two different irrational numbers, a and b, where a/b is a rational number.
Solution
Consider a = 3√2 and b = 5√2 as two different irrational numbers
Here
a/b = 3√2/5√2 = 3/2 is a rational number.
18. If 34.0356 is expressed in the form p/q, where p and q are coprime integers, then what can you say about the factorization of q?
Solution
We know that,
34.0356 = 340356/10000 (in p/q form)
= 85089/2500
Here,
85089 and 2500 are coprime integers
So the factorization of q = 2500 = 22× 54
Is of the form (2m × 5n)
where m and n are positive or non-negative integers.
19. In each case, state whether the following numbers are rational or irrational. If they are rational and expressed in the form p/q, where p and q are coprime integers, then what can you say about the prime factors of q?
(i) 279.034
(iii) 3.010010001…
(iv) 39.546782
(v) 2.3476817681…
(vi) 59.120120012000…
Solution
(i) 279.034 is a rational number because it has terminating decimals
279.034 = 279034/1000 (in p/q form)
= 139517/500 (Dividing by 2)
We know that
Factors of 500 = 2 × 2 × 5 × 5 × 5 = 22 × 53
Which is of the form 2m × 5n where m and n are positive integers.
It is a rational number as it has recurring or repeating decimals
Consider x =
= 76.17893 17893 17893 …..
100000x = 7617893.178931789317893…..
By subtraction
99999x = 7617817
x = 7617817/99999 which is of p/q form
We know that
Prime factor of 99999 = 3 × 3 × 11111
q has factors other than 2 or 5 i.e. 32 × 11111
(iii) 3.010010001….
It is neither terminating decimal nor repeating
Therefore, it is an irrational number.
(iv) 39.546782
It is terminating decimal and is a rational number
39.546782 = 39546782/1000000 (in p/q form)
= 19773391/500000
We know that p and q are coprime
Prime factors of q = 25 × 56
Is of the form 2m × 5n where m and n are positive integers
(v) 2.3476817681…
Is neither terminating nor repeated decimal
Therefore, it is an irrational number.
(vi) 59.120120012000….
It is neither terminating decimal nor repeated
Therefore, it is an irrational number. |
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# Problem Set 7
Thursday, November 10, or Friday, November 11. For example, if your section
section classroom at 10:00am on Friday, November 11. Late problems earn zero
points.
Note: you can work on these problems or your own, or in a small group with
other current Econ 1 students. If you choose to work in a group, each student
needs to hand in a separate, individual copy to his/her TA.
## 1. Taylor, chapter 18, problem 7
2. Taylor, chapter 18, problem 10
## Note: Taylor, fourth edition
Question 1 (Chapter 18, Problem 7)
(a) We calculate nominal GDP for 2001 by multiplying the value and the quantity
for each good and then summing over the goods: (\$2.50 x 1,000) + (\$1.25 x
500) + (\$100.00 x 10) = \$2,500 + \$625 + \$1,000 = \$4,125.
We calculate nominal GDP for 2002 in the same way: (\$3.50 x 800) + (\$2.25
x 400) + (\$100.00 x 14) = \$2,800 + \$900 + \$1,400 = \$5,100.
## (b) Using 2001 prices:
2001 GDP is \$4,125 as shown in part (a). 2002 GDP is (\$2.50 x 800) +
(\$1.25 x 400) + (\$100.00 x 14) = \$2,000 + \$500 + \$1,400 = \$3,900. The
difference is \$3,900 – \$4,125 = – \$225. The percentage change in GDP is –
\$225 divided by \$4,125 which is equal to –.0545 or – 5.45%.
## Using 2002 prices:
2001 GDP is (\$3.50 x 1,000) + (\$2.25 x 500) + (\$100.00 x 10) = \$3,500 +
\$1,125 + \$1,000 = \$5,625. 2002 GDP is \$5,100 as shown in part (a). The
difference is \$5,100 – \$5,625 = – \$525. The percentage change in GDP is –
\$525 divided by \$5,625 which is equal to –.0933 or – 9.33%.
## (c) The percentage change in GDP is slightly different depending on whether
2001 or 2002 prices are used. “Such differences are inevitable, because there
is no reason to prefer the prices in one year to those of another year when
controlling for inflation. Economists arrive at a single percentage by simply
averaging the two percentages” (Taylor, p. 469). Here we have a 5.45%
decrease in GDP if we use 2001 prices and a 9.33 % decrease if we use 2002
prices. To average, we add 5.45 and 9.33 and then divide by 2. This gives us
that the percent change in real GDP is equal to – 7.39%.
A geometric average can be used. The geometric average is the square root of
the product of the two numbers: (0.0545 × 0.0933) = 7.13 . Therefore, using
a geometric average, the percent change in real GDP is equal to – 7.13%.
(d) In part (a) we found that 2002 nominal GDP is \$5,100. We are given that
2001 is the base year, so we can find 2002 real GDP using 2001 prices and
2002 quantities. From part (b) we found 2002 real GDP using 2001 prices to
be \$3,900. The 2002 deflator is found by dividing 2002 nominal GDP by
\$5,100
2002 real GDP: = 1.31
\$3,900
Alternate Answer 1: We are given that the GDP deflator for 2001 is equal to
1.0. This means that 2001 nominal GDP is equal to 2001 real GDP. In (c) we
found that real GDP declined by 7.39 % (using a simple average). Therefore,
2002 real GDP is equal to (1-.0739) x \$4,125 = \$3,820. The 2002 deflator is
\$5,100
found by dividing 2002 nominal GDP by 2002 real GDP: = 1.34
\$3,820
Alternate Answer 2: We are given that the GDP deflator for 2001 is equal to
1.0. This means that 2001 nominal GDP is equal to 2001 real GDP. In (c) we
found that real GDP declined by 7.13 % (using a geometric average).
Therefore, 2002 real GDP is equal to (1-.0713) x \$4,125 = \$3,820. The 2002
deflator is found by dividing 2002 nominal GDP by 2002 real GDP:
\$5,100
= 1.33
\$3,830
## a) –\$5 billion: Inventory investment = change in inventory stock = 5 – 10 = –5
b) \$4 billion: Net exports (X) = exports – imports = 21 – 17 = 4
c) \$231 billion: GDP = Consumption (C) + Investment (I) + Government expenditures
(G) + Net exports (X) = 140 + (27 – 5) + 65 + 4 = 231
d) –\$10 billion: Statistical discrepancy = (C+I+G+X) – (labor income + capital income +
depreciation + indirect business taxes + net income of foreigners) = 231 – (126 + 70 + 12
+ 28 + 5) = 231 – 241 = –10
e) \$26 billion: National savings = GDP – C – G = 231 – 140 – 65
National Saving = 26
Investment + Net Exports = I + X = (27– 5) + 4 = 26 |
# Introduction
An ANOVA is a statistical test used to compare a quantitative variable between groups, to determine if there is a statistically significant difference between several population means. In practice, it is usually used to compare three or more groups. However, in theory, it can also be done with only two groups.1
In a previous post, we showed how to perform a one-way ANOVA in R. In this post, we illustrate how to conduct a one-way ANOVA by hand, via what is usually called an “ANOVA table”.
# Data and hypotheses
To illustrate the method, suppose we take a sample of 12 students, divided equally into three classes (A, B and C) and we observe their age. Here is the sample:
We are interested in comparing the population means between classes.
Remember that the null hypothesis of the ANOVA is that all means are equal (i.e., age is not significantly different between classes), whereas the alternative hypothesis is that at least one mean is different from the other two (i.e., age is significantly different in at least one class compared to the other two). Formally, we have:
• $$\mu_A = \mu_B = \mu_C$$
• at least one mean is different
# ANOVA by hand
As mentioned above, we are going to do an ANOVA table to conclude the test.
Note that the ANOVA requires some assumptions (i.e., independence, equality of variances and normality). The aim of this post is to illustrate how to do an ANOVA by hand and not how to verify these assumptions, so we suppose they are met without any verification. See how to test these assumptions in R if you are interested.
## Overall and group means
We first need to compute the mean age by class (referred as the group means):
• class A: $$\frac{24 + 31 + 26 + 23}{4} = 26$$
• class B: $$\frac{24 + 21 + 19 + 24}{4} = 22$$
• class C: $$\frac{15 + 21 + 18 + 18}{4} = 18$$
and the mean age for the whole sample (referred as the overall mean):
$$$\begin{split} & \frac{24 + 31 + 26 + 23 + 24 + 21 + 19 }{12} \\ &\frac{+ 24 + 15 + 21 + 18 + 18}{12} = 22 \end{split}$$$
## SSR and SSE
We then need to compute the sum of squares regression (SSR), and the sum of squares error (SSE).
The SSR is computed by taking the square of the difference between the mean group and the overall mean, multiplied by the number of observations in the group:
and then taking the sum of all cells:
$64+0+64 = 128 = SSR$
The SSE is computed by taking the square of the difference between each observation and its group mean:
and then taking the sum of all cells:
$$$\begin{split} & 4+25+0+9+4+1+9+4 \\ & +9+9+0+0 = 74 = SSE \end{split}$$$
For those interested in computing the sum of square total (SST), it is simply the sum of SSR and SSE, that is,
$$$\begin{split} SST &= SSR + SSE\\ &= 128 + 74 \\ & =202 \end{split}$$$
## ANOVA table
The ANOVA table looks as follows (we leave it empty and we are going to fill it in step by step):
We start to build the ANOVA table by plugging the SSR and SSE values found above into the table (in the “Sum.of.Sq.” column):
The “Df” column corresponds to the degrees of freedom, and is computed as follows:
• for the line regression: number of groups - 1 = 3 - 1 = 2
• for the line error: number of observations - number of groups = 12 - 3 = 9
With this information, the ANOVA table becomes:
The “Mean.Sq.” column corresponds to the Mean Square, and is equal to the sum of square divided by the degrees of freedom, so the “Sum.of.Sq.” column divided by the “Df” column:
Finally, the F-value corresponds to the ratio between the two mean squares, so $$\frac{64}{8.222} = 7.78$$:
This F-value gives the test statistic (also referred as $$F_{obs}$$), which needs to be compared with the critical value found in the Fisher table to conclude the test.
We find the critical value in the Fisher table based on the degrees of freedom (those used in the ANOVA table) and based on the significance level. Suppose we take a significance level $$\alpha = 0.05$$, the critical value can be found in the Fisher table as follows:
So we have
$F_{2; 9; 0.05} = 4.26$
If you are interested to find this value with R, it can be found with the qf() function, where 0.95 corresponds to $$1 - \alpha$$:
qf(0.95, 2, 9)
## [1] 4.256495
## Conclusion of the test
The rejection rule says that, if:
• $$F_{obs} > F_{2; 9; 0.05} \Rightarrow$$ we reject the null hypothesis
• $$F_{obs} \le F_{2; 9; 0.05} \Rightarrow$$ we do not reject the null hypothesis
In our case,
$F_{obs} = 7.78 > F_{2; 9; 0.05} = 4.26$
$$\Rightarrow$$ We reject the null hypothesis that all means are equal. In other words, it means that at least one class is different than the other two in terms of age.2
To verify our results, here is the ANOVA table using R:
## Df Sum Sq Mean Sq F value Pr(>F)
## class 2 128 64.00 7.784 0.0109 *
## Residuals 9 74 8.22
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
We found the same results by hand, but note that in R, the $$p$$-value is computed instead of comparing the $$F_{obs}$$ with the critical value. The $$p$$-value can easily be found in R based on the $$F_{obs}$$ and the degrees of freedom:
pf(7.78, 2, 9,
lower.tail = FALSE
)
## [1] 0.010916
# Conclusion
I hope this article helped you to conduct a one-way ANOVA by hand. See this tutorial if you want to learn how to do it in R.
As always, if you have a question or a suggestion related to the topic covered in this article, please add it as a comment so other readers can benefit from the discussion.
1. In that case, a Student’s t-test is usually preferred over an ANOVA, although both tests will lead to the exact same conclusions.↩︎
2. Remember that an ANOVA cannot tell you which group is different than the other in terms of the quantitative dependent variable, nor whether they are all different or if only one is different. To answer this question, post-hoc tests are required. This is beyond the scope of the present post, but it can easily be done in R (see this tutorial).↩︎
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# Linear Transformation $T:\R^2 \to \R^2$ Given in Figure
## Problem 610
Let $T:\R^2\to \R^2$ be a linear transformation such that it maps the vectors $\mathbf{v}_1, \mathbf{v}_2$ as indicated in the figure below.
Find the matrix representation $A$ of the linear transformation $T$.
## Solution 1.
From the figure, we see that
$\mathbf{v}_1=\begin{bmatrix} -3\\ 1 \end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix} 5\\ 2 \end{bmatrix},$ and
$T(\mathbf{v}_1)=\begin{bmatrix} 2\\ 2 \end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix} 1\\ 3 \end{bmatrix}.$
Let $A$ be the matrix representation of the linear transformation $T$. By definition, we have $T(\mathbf{x})=A\mathbf{x}$ for any $\mathbf{x}\in \R^2$.
We determine $A$ as follows.
We have
\begin{align*}
\begin{bmatrix}
2& 1 \\
2& 3
\end{bmatrix}
&=[T(\mathbf{v}_1), T(\mathbf{v}_2)]\\
&=[A\mathbf{v}_1, A\mathbf{v}_2]\\
&=A[\mathbf{v}_1, \mathbf{v}_2]\\
&=A\begin{bmatrix}
-3& 5 \\
1& 2
\end{bmatrix}.
\end{align*}
Note that the determinant of the leftmost matrix is $-11$, hence it is invertible and the inverse is given by
$\begin{bmatrix} -3& 5 \\ 1& 2 \end{bmatrix}^{-1}=\frac{1}{11}\begin{bmatrix} -2& 5 \\ 1& 3 \end{bmatrix}.$ Hence
\begin{align*}
A&= \begin{bmatrix}
2& 1 \\
2& 3
\end{bmatrix} \begin{bmatrix}
-3& 5 \\
1& 2
\end{bmatrix}^{-1}\6pt] &=\frac{1}{11}\begin{bmatrix} 2& 1 \\ 2& 3 \end{bmatrix}\begin{bmatrix} -2& 5 \\ 1& 3 \end{bmatrix}. \\[6pt] &=\frac{1}{11}\begin{bmatrix} -3& 13 \\ -1& 19 \end{bmatrix}. \end{align*} Therefore, the matrix representation of T is \[A=\frac{1}{11}\begin{bmatrix} -3& 13 \\ -1& 19 \end{bmatrix}.
## Solution 2.
Let $\{\mathbf{e}_1, \mathbf{e}_2\}$ be the standard basis for $\R^2$.
Then the matrix representation $A$ of the linear transformation $T$ is given by
$A=[T(\mathbf{e}_1), T(\mathbf{e}_2)].$ From the figure, we see that
$\mathbf{v}_1=\begin{bmatrix} -3\\ 1 \end{bmatrix} \text{ and } \mathbf{v}_2=\begin{bmatrix} 5\\ 2 \end{bmatrix},$ and
$T(\mathbf{v}_1)=\begin{bmatrix} 2\\ 2 \end{bmatrix} \text{ and } T(\mathbf{v}_2)=\begin{bmatrix} 1\\ 3 \end{bmatrix}.$
The values of $T(\mathbf{e}_1)$ and $T(\mathbf{e}_2)$ are not indicated in the figure but we can determine these values as follows.
Note that $\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$ (as they are linearly independent from the figure).
Let us express $\mathbf{e}_1$ as a linear combination of $\mathbf{v}-1$ and $\mathbf{v}_2$.
Let
$\mathbf{e}_1=c_1\mathbf{v}_1+c_2\mathbf{v}_2,$ where $c_1, c_2$ are scalars to be determined.
This is equivalent to
$\begin{bmatrix} 1 \\ 0 \end{bmatrix} =[\mathbf{v}_1, \mathbf{v}_2]\begin{bmatrix} c_1\\ c_2 \end{bmatrix} =\begin{bmatrix} -3 & 5 \\ 1& 2 \end{bmatrix} \begin{bmatrix} c_1\\ c_2 \end{bmatrix}.$
As we have
$\begin{bmatrix} -3& 5 \\ 1& 2 \end{bmatrix}^{-1}=\frac{1}{11}\begin{bmatrix} -2& 5 \\ 1& 3 \end{bmatrix},$ it follows that
\begin{align*}
\begin{bmatrix}
c_1\\ c_2
\end{bmatrix}
=\begin{bmatrix}
-3& 5 \\
1& 2
\end{bmatrix}^{-1}\begin{bmatrix}
1\\ 0
\end{bmatrix}=
\frac{1}{11}\begin{bmatrix}
-2& 5 \\
1& 3
\end{bmatrix}
\begin{bmatrix}
1\\ 0
\end{bmatrix}
= \frac{1}{11}\begin{bmatrix}
-2
\\ 1
\end{bmatrix}.
\end{align*}
Thus, we obtain the linear combination
$\mathbf{e}_1=-\frac{2}{11}\mathbf{v}_1+\frac{1}{11}\mathbf{v}_2,$
By the linearity of $T$, we have
\begin{align*}
T(\mathbf{e}_1)&=-\frac{2}{11}T(\mathbf{v}_1)+\frac{1}{11}T(\mathbf{v}_2)\\
&=-\frac{2}{11}\begin{bmatrix}
2
\\ 2
\end{bmatrix}
+\frac{1}{11}\begin{bmatrix}
1\\ 3
\end{bmatrix}
\\
&=\frac{1}{11}\begin{bmatrix}
-3\\ -1
\end{bmatrix}.
\end{align*}
Similarly, let
$\mathbf{e}_2=d_1\mathbf{v}_1+d_2\mathbf{v}_2,$ where $d_1, d_2$ are scalars to be determined.
Then we have
$\begin{bmatrix} 0\\ 1 \end{bmatrix} =\begin{bmatrix} -3 & 5 \\ 1& 2 \end{bmatrix} \begin{bmatrix} d_1\\ d_2 \end{bmatrix}.$ It follows that
\begin{align*}
\begin{bmatrix}
d_1\\ d_2
\end{bmatrix}
=\begin{bmatrix}
-3& 5 \\
1& 2
\end{bmatrix}^{-1}\begin{bmatrix}
0\\ 1
\end{bmatrix}=
\frac{1}{11}\begin{bmatrix}
-2& 5 \\
1& 3
\end{bmatrix}
\begin{bmatrix}
0\\ 1
\end{bmatrix}
= \frac{1}{11}\begin{bmatrix}
5
\\ 3
\end{bmatrix}.
\end{align*}
Thus, we have the linear combination
$\mathbf{e}_2=\frac{5}{11}\mathbf{v}_1+\frac{3}{11}\mathbf{v}_2,$ and by linearity of $T$ we obtain
\begin{align*}
T(\mathbf{e}_2)&=\frac{5}{11}T(\mathbf{v}_1)+\frac{3}{11}T(\mathbf{v}_2)\\
&=\frac{5}{11}T(\mathbf{v}_1)+\frac{3}{11}T(\mathbf{v}_2)\\
&=\frac{5}{11}\begin{bmatrix}
2
\\ 2
\end{bmatrix}
+\frac{3}{11}\begin{bmatrix}
1\\ 3
\end{bmatrix}
\\
&=\frac{1}{11}\begin{bmatrix}
13\\ 19
\end{bmatrix}.
\end{align*}
In conclusion, the matrix representation for $T$ is
$A=[T(\mathbf{e}_1), T(\mathbf{e}_2)]=\frac{1}{11}\begin{bmatrix} -3& 13 \\ -1& 19 \end{bmatrix}.$
##### Eigenvalues of $2\times 2$ Symmetric Matrices are Real by Considering Characteristic Polynomials
Let $A$ be a $2\times 2$ real symmetric matrix. Prove that all the eigenvalues of $A$ are real numbers by... |
# Some students planned a picnic. The total budget for food was ₹2000.
Question:
Some students planned a picnic. The total budget for food was ₹2000. But, 5 students failed to attend the picnic and thus the cost for food for each member increased by ₹20. How many students attended the picnic and how much did each student pay for the food?
Solution:
Let x be the number of students who planned a picnic.
$\therefore$ Original cost of food for each member $=₹ \frac{2000}{x}$
Five students failed to attend the picnic. So, (− 5) students attended the picnic.
$\therefore$ New cost of food for each member $=₹ \frac{2000}{(x-5)}$
According to the given condition,
$₹ \frac{2000}{x-5}-₹ \frac{2000}{x}=₹ 20$
$\Rightarrow \frac{2000 x-2000 x+10000}{x(x-5)}=20$
$\Rightarrow \frac{10000}{x^{2}-5 x}=20$
$\Rightarrow x^{2}-5 x=500$
$\Rightarrow x^{2}-5 x-500=0$
$\Rightarrow x^{2}-25 x+20 x-500=0$
$\Rightarrow x(x-25)+20(x-25)=0$
$\Rightarrow(x-25)(x+20)=0$
$\Rightarrow x-25=0$ or $x+20=0$
$\Rightarrow x=25$ or $x=-20$
∴ x = 25 (Number of students cannot be negative)
Number of students who attended the picnic = − 5 = 25 − 5 = 20
Amount paid by each student for the food $=₹ \frac{2000}{(25-5)}=₹ \frac{2000}{20}=₹ 100$ |
# Transformations of the Sine Function
y = a sin(bx + c)
Examine the graphs of y = a sin (bx + c) for different values of a, b, and c.
First, let's consider y = sin x, where a = 1, b = 1 and c = 0:
This graph represents the parent function whose maximum value is 1, minimum value is -1 and the period is 2p.
Consider the graph of y = a sin x for different values of a.
What is the effect of the value of a on the graph? Consider the following:
g y = sin x
g y = 2 sin x
g y = 3 sin x
g y = 4.5 sin x
Notice that each of the graphs have the same x-intercepts and the same period; however, their respective distances from the x-axis (i.e., their maximums and minimums) are different. The absolute value of the a value represents the amplitude of the graph, which in turn describes the vertical stretch (a > 1) or compression (0 < a < 1). If the value of a is negative, the absolute value of a is the amplitude, but the graph is reflected over the x-axis:
g y = sin x
g y = -2 sin x
Here the amplitude of the second graph is 2, but notice that instead the first peak is not at 2 as we would expect, but at -2.
Consider the graph of y = sin (bx) for different values of b.
As stated earlier, the period for the sine function is 2p. As we consider the following graphs, note what effect b has on the regular period.
g y = sin x
g y = sin 2x
g y = sin .5x
The period for the new graphs is 2p/b. For 0 < b < 1, the graph is horizontally stretched, therefore the period (the time it takes for the function values to begin repeating) is increased: 2p/.5 = 4p. For b > 1, the graph is horizontally compressed, therefore the period is decreased: 2p/ 2 = p.
Now consider the graph of y = sin (x + c) for different values of c.
g y = sin x
g y = sin (x + p)
g y = sin (x + p/2)
The value of c represents a horizontal translation of the graph, also called a phase shift. To determine the phase shift, consider the following: the function value is 0 at all x- intercepts of the graph, i.e. at all points x + c = 0. The phase shift is represented by x = -c. When c > 0, the graph shifts to the left c spaces, and if c < 0, the graph shifts to the right c spaces. For y = sin (x + p), the phase shift is -p. Notice that all points on the original graph are shifted to the left p spaces. The phase shift for y = sin (x - p/2) is p/2. All points on the original graph are shifted to the right p/2 spaces.
Now, let's determine the amplitude, period, and phase shift of the following graph:
The amplitude of the graph is 2.5. The period of the new function is 2p/3.
For the phase shift: 3x - p/2 = 0, x = p/6.
So where did the sine function and the other five trig functions originate? Click here to find out. |
# How do you solve 25-4x+6x+15?
Aug 5, 2018
Use the Order of Operation
#### Explanation:
The order of operations is PEMDAS Think about it this way
If you get hurt in PE call an MD ASap
There are no parenthesis or exponents so start with o and Division going from left to right ( Remember that MD is one person so do Multiplication and Division at the same time)
there is no Multiplication and Division so go on to Addition and Subtraction.
Start by adding ( - 4x + 6x)
$25 \left(- 4 x + 6 x\right) + 15 = 25 + 2 x + 15$
Use the communicative property to group the like terms.
$25 + 2 x + 15 = \left(25 + 15\right) + 2 x$
( 25 + 15 + 2x = 40 + 2x
This can be reduced to
$2 \left(20 + x\right)$
It is not really possible to solve this because there is no equal sign
Aug 5, 2018
$2 x + 40$
#### Explanation:
We have the following:
$\textcolor{s t e e l b l u e}{25} \textcolor{p u r p \le}{- 4 x + 6 x} + \textcolor{s t e e l b l u e}{15}$
We can combine our variables terms to get $\textcolor{p u r p \le}{2 x}$, and our constants to get $\textcolor{s t e e l b l u e}{40}$. Putting it all together, we get
$2 x + 40$
Hope this helps! |
# An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
By BYJU'S Exam Prep
Updated on: September 13th, 2023
Given
Height of object = 5 cm
Position of object u = – 25 cm
Focal length of lens f = 10 cm
We have to find the position of image v, size, and nature of the image.
Let us use the formula
1/v – 1/u = 1/f
Now substitute the values
1/v + 1/25 = 1/10
1/v = 1/10 – 1/25
1/v = (5 – 2)/ 50
1/v = 3/50
v = 50/3 = 16.66 cm
The distance of the image on the opposite side of the lens is 16.66 cm.
We know that
Magnification = v/u
m = 16.66/-25
m = -0.66
m = height of the image/ height of the object
-0.66 = height of the image/ 5 cm
Height of the image = -3.3 cm
A negative sign shows that an inverted image is formed
The position of the image is at 16.66 cm on the opposite side of the lens
The size of the image is -3.33 cm on the opposite side of the lens
The nature of the image will be real and inverted
Therefore, the position is at 16.66 cm on the opposite side of the lens, the size is -3.33 cm on the opposite side of the lens and the nature of the image formed will be real and inverted.
Summary:
## An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size, and nature of the image formed.
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. The position is at 16.66 cm on the opposite side of the lens, the size is -3.33 cm on the opposite side of the lens and the nature of the image formed will be real and inverted.
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Ratio Of Two Triangles | AMC-10A, 2004 | Problem 20
Try this beautiful problem from AMC 10A, 2004 based on Geometry: Ratio Of Two Triangles
Ratio Of Two Triangles – AMC-10A, 2004- Problem 20
Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$
• $\frac{4}{3}$
• $\frac{3}{2}$
• $\sqrt 3$
• $2$
• $1+\sqrt 3$
Key Concepts
Square
Triangle
Geometry
Answer: $2$
AMC-10A (2002) Problem 20
Pre College Mathematics
Try with Hints
We have to find out the ratio of the areas of two Triangles $\triangle DEF$ and $\triangle ABE$.Let us take the side length of $AD$=$1$ & $DE=x$,therefore $AE=1-x$
Now in the $\triangle ABE$ & $\triangle BCF$ ,
$AB=BC$ and $BE=BF$.using Pythagoras theorm we may say that $AE=FC$.Therefore $\triangle ABE \cong \triangle CEF$.So $AE=FC$ $\Rightarrow DE=DF$.Therefore the $\triangle DEF$ is an isosceles right triangle. Can you find out the area of isosceles right triangle $\triangle DEF$
Can you now finish the problem ……….
Length of $DE=DF=x$.Then the the side length of $EF=X \sqrt 2$
Therefore the area of $\triangle DEF= \frac{1}{2} \times x \times x=\frac{x^2}{2}$ and area of $\triangle ABE$=$\frac{1}{2} \times 1 \times (1-x) = \frac{1-x}{2}$.Now from the Pythagoras theorm $(1-x)^2 +1 =2x^2 \Rightarrow x^2=2-2x=2(1-x)$
can you finish the problem……..
The ratio of the area of $\triangle DEF$ to that of $\triangle ABE$ is $\frac{\frac{x^2}{2}}{\frac{(1-x)}{2}}$=$\frac{x^2}{1-x}$=$\frac {2(1-x)}{(1-x)}=2$
Categories
Length of a Tangent | AMC-10A, 2004 | Problem 22
Try this beautiful problem from American Mathematics Competitions, AMC 10A, 2004 based on Geometry: Length of a Tangent.
Length of a Tangent – AMC-10A, 2004- Problem 22
Square $ABCD$ has a side length $2$. A Semicircle with diameter $AB$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $AD$at $E$. What is the length of $CE$?
• $\frac{4}{3}$
• $\frac{3}{2}$
• $\sqrt 3$
• $\frac{5}{2}$
• $1+\sqrt 3$
Key Concepts
Square
Semi-circle
Geometry
Answer: $\frac{5}{2}$
AMC-10A (2004) Problem 22
Pre College Mathematics
Try with Hints
We have to find out length of $CE$.Now $CE$ is a tangent of inscribed the semi circle .Given that length of the side is $2$.Let $AE=x$.Therefore $DE=2-x$. Now $CE$ is the tangent of the semi-circle.Can you find out the length of $CE$?
Can you now finish the problem ……….
Since $EC$ is tangent,$\triangle COF$ $\cong$ $\triangle BOC$ and $\triangle EOF$ $\cong$ $\triangle AOE$ (By R-H-S law).Therefore $FC=2$ & $EC=x$.Can you find out the length of $EC$?
can you finish the problem……..
Now the $\triangle EDC$ is a Right-angle triangle……..
Therefore $ED^2+ DC^2=EC^2$ $\Rightarrow (2-x)^2 + 2^2=(2+x)^2$ $\Rightarrow x=\frac{1}{2}$
Hence $EC=EF+FC=2+\frac{1}{2}=\frac{5}{2}$
Categories
Diamond Pattern | AMC-10A, 2009 | Problem 15
Try this beautiful problem from AMC 10A, 2009 based on Diamond Pattern.
Diamond Pattern – AMC-10A, 2009- Problem 15
The figures $F_1$, $F_2$, $F_3$, and $F_4$ shown are the first in a sequence of figures. For $n\ge3$, $F_n$ is constructed from $F_{n – 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n – 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$?
• $756$
• $761$
• $786$
Key Concepts
Pattern
Sequence
Symmetry
Answer: $761$
AMC-10A (2009) Problem 15
Pre College Mathematics
Try with Hints
From the above diagram we observe that in $F_1$ the number of diamond is $1$.in $F_2$ the number of diamonds are $5$.in $F_3$ the number of diamonds are $13$. in $F_4$ the numbers of diamonds are $25$.Therefore from $F_1$ to $F_2$ ,$(5-1)$=$4$ new diamonds added.from $F_2$ to $F_3$,$(13-5=8$ new diamonds added.from $F_3$ to $F_4$,$(25-13)=12$ new diamonds added.we may say that When constructing $F_n$ from $F_{n-1}$, we add $4(n-1)$ new diamonds.
Can you now finish the problem ……….
so we may construct that Let $S_n$ be the number of diamonds in $F_n$. We already know that $P_1$=1 and for all $n >1$ ,$P_n=P_{n-1}+4(n-1)$.now can you find out $P_{20}$
can you finish the problem……..
Now $P_{20}$=$P_{19} + 4(20-1)$
$\Rightarrow P_{20}$=$P_{19} + 4.19)$
$\Rightarrow P_{20}$=$P_{18}+(4 \times 18) +( 4 \times 19)$
$\Rightarrow P_{20}$= …………………………………..
$\Rightarrow P_{20}$=$1+4(1+2+3+……..+18+19)$
$\Rightarrow P_{20}$=$1+ \frac{ 4 \times 19 \times 20}{2}$
$\Rightarrow P_{20}$=$761$
Categories
Problem on Semicircle | AMC 8, 2013 | Problem 20
Try this beautiful problem from Geometry based on Semicircle.
Area of the Semicircle – AMC 8, 2013 – Problem 20
A $1\times 2$ rectangle is inscribed in a semicircle with the longer side on the diameter. What is the area of the semicircle?
• $\frac{\pi}{2}$
• $\pi$
• $\frac{\pi}{3}$
Key Concepts
Geometry
Square
semi circle
Answer:$\pi$
AMC-8 (2013) Problem 20
Pre College Mathematics
Try with Hints
At first we have to find out the radius of the semicircle for the area of the semicircle.Now in the diagram,AC is the radius of the semicircle and also AC is the hypotenuse of the right Triangle ABC.
Can you now finish the problem ……….
Now AB=1 AND AC=1 (As ABDE $1\times 2$ rectangle .So using pythagorean theorm we can eassily get the value of AC .and area of semicircle =$\frac{\pi r^2}{2}$
can you finish the problem……..
Given that ABDE is a square whose AB=1 and BD=2
Therefore BC=1
Clearly AC be the radius of the given semi circle
From the $\triangle ABC$,$(AB)^2+(BC)^2=(AC)^2\Rightarrow AC=\sqrt{(1^2+1^2)}=\sqrt2$
Therefore the area of the semicircle=$\frac{1}{2}\times \pi(\sqrt 2)^2$=$\pi$
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Radius of semicircle | AMC-8, 2013 | Problem 23
Try this beautiful problem from Geometry: Radius of the semicircle
Radius of the semicircle- AMC-8, 2013- Problem 23
Angle $ABC$ of $\triangle ABC$ is a right angle. The sides of $\triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $\overline{AB}$ equals $8\pi$, and the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$. What is the radius of the semicircle on $\overline{BC}$?
• $9$
• $7.5$
• $6$
Key Concepts
Geometry
Triangle
Semi-circle
Answer: $7.5$
AMC-8 (2013) Problem 23
Pre College Mathematics
Try with Hints
We have to find out the radius of the semi-circle on ${BC}$? Now ABC is a Right angle Triangle.so if you find out AB and AC then BC will be easily calculated…..
Can you now finish the problem ……….
To find the value of AB and AC, notice that area of the semi-circle on AB is given and length of the arc of AC is given….
can you finish the problem……..
Let the length of AB=$2x$,So the radius of semi-circle on AB=$x$.Therefore the area =$\frac{1}{2}\times \pi (\frac{x}{2})^2=8\pi$$\Rightarrow x=4$
Theregfore length of AB=$2x$=8
Given that the arc of the semi-circle on $\overline{AC}$ has length $8.5\pi$,let us take the radius of the semicircle on AB =$r$.now length of the arc of the semicircle on $\overline{AC}$ has length $8.5\pi$ i.e half perimeter=$8.5$
so $\frac{1}{2}\times {2\pi r}=8.5\pi$$\Rightarrow r=8.5$$\Rightarrow 2r=17$.so AC=17
The triangle ABC is a Right Triangle,using pythagorean theorm……
$(AB)^2 + (BC)^2=(AC)^2$$\Rightarrow BC=\sqrt{(AC)^2 -(AB)^2}$$\Rightarrow BC =\sqrt{(17)^2 -(8)^2}$$\Rightarrow BC= \sqrt{289 -64 }$$\Rightarrow BC =\sqrt {225}=15$
Therefore the radius of semicircle on BC =$\frac{15}{2}=7.5$
Categories
Area of a Regular Hexagon | AMC-8, 2012 | Problem 23
Try this beautiful problem from Geometry: Area of the Regular Hexagon – AMC-8, 2012 – Problem 23.
Area of the Regular Hexagon – AMC-8, 2012- Problem 23
An equilateral triangle and a regular hexagon have equal perimeters. If the triangle’s area is 4, what is the area of the hexagon?
• $8$
• $6$
• $10$
Key Concepts
Geometry
Triangle
Hexagon
Answer: $6$
AMC-8 (2012) Problem 23
Pre College Mathematics
Try with Hints
To find out the area of the Regular hexagon,we have to find out the side length of it.Now the perimeter of the triangle and Regular Hexagon are same….from this condition you can easily find out the side length of the regular Hexagon
Can you now finish the problem ……….
Let the side length of an equilateral triangle is$x$.so the perimeter will be $3x$ .Now according to the problem the perimeter of the equiliteral triangle and regular hexagon are same,i.e the perimeter of regular hexagon=$3x$
So the side length of be $\frac{3x}{6}=\frac{x}{2}$
can you finish the problem……..
Now area of the triangle $\frac{\sqrt 3}{4}x^2=4$
Now the area of the Regular Hexagon=$\frac{3\sqrt3}{2} (\frac{x}{2})^2=\frac{3}{2} \times \frac{\sqrt{3}}{4}x^2=\frac{3}{2} \times 4$=6
Categories
Area of a Triangle | AMC-8, 2000 | Problem 25
Try this beautiful problem from Geometry: Area of a Triangle
Area of the Triangle- AMC-8, 2000- Problem 25
The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $BC$ and $CD$ are joined to form a triangle, the area of that triangle is
• $25$
• $27$
• $29$
Key Concepts
Geometry
Triangle
square
Answer: $27$
AMC-8 (2000) Problem 25
Pre College Mathematics
Try with Hints
Area of the triangle =$\frac{1}{2} \times base \times height$
Can you now finish the problem ……….
Therefore area of the shaded region i.e area of the $\triangle AEF$=area of the square- area of $(\triangle ADE +\triangle EFC +\triangle ABF)$
can you finish the problem……..
Given that area of the rectangle ABCD=72
Let length AB=$x$ and length of CD=$y$
Therefore DE=EC=$\frac{y}{2}$ and BF=FC=$\frac{x}{2}$
Area of ABCD=$xy$=72
Area of the $\triangle ADE=\frac{1}{2}\times DE \times AD= \frac{1}{2}\times \frac{x}{2} \times y =\frac{xy}{4}=\frac{72}{4}=18$
Area of the $\triangle EFC=\frac{1}{2}\times EC \times FC= \frac{1}{2}\times \frac{x}{2} \times \frac{y}{2} =\frac{xy}{8}=\frac{72}{8}=9$
Area of the $\triangle ABF=\frac{1}{2}\times AB \times BF= \frac{1}{2}\times y\times \frac{x}{2}=\frac{xy}{4}=\frac{72}{4}=18$
Therefore area of the shaded region i.e area of the $\triangle AEF$=area of the square- area of $(\triangle ADE +\triangle EFC +\triangle ABF)=72-(18+8+18)=27$
Categories
Area of the Trapezoid | AMC 8, 2002 | Problem 20
Try this beautiful problem from Geometry based on Area of Trapezoid.
Area of the Trapezoid – AMC- 8, 2002 – Problem 20
The area of triangle XYZ is 8 square inches. Points A and B are midpoints of congruent segments XY and XZ . Altitude XC bisects YZ.What is the area (in square inches) of the shaded region?
• $6$
• $4$
• $3$
Key Concepts
Geometry
Triangle
Trapezoid
Answer:$3$
AMC-8 (2002) Problem 20
Pre College Mathematics
Try with Hints
Given that Points A and B are midpoints of congruent segments XY and XZ and Altitude XC bisects YZ
Let us assume that the length of YZ=$x$ and length of $XC$= $y$
Can you now finish the problem ……….
Therefore area of the trapezoid= $\frac{1}{2} \times (YC+AO) \times OC$
can you finish the problem……..
Let us assume that the length of YZ=$x$ and length of $XC$= $y$
Given that area of $\triangle xyz$=8
Therefore $\frac{1}{2} \times YZ \times XC$=8
$\Rightarrow \frac{1}{2} \times x \times y$ =8
$\Rightarrow xy=16$
Given that Points A and B are midpoints of congruent segments XY and XZ and Altitude XC bisects YZ
Then by the mid point theorm we can say that $AO=\frac{1}{2} YC =\frac{1}{4} YZ =\frac{x}{4}$ and $OC=\frac{1}{2} XC=\frac{y}{2}$
Therefore area of the trapezoid shaded area = $\frac{1}{2} \times (YC+AO) \times OC$= $\frac{1}{2} \times ( \frac{x}{2} + \frac{x}{4} ) \times \frac{y}{2}$ =$\frac{3xy}{16}=3$ (as $xy$=16)
Categories
Intersection of two Squares | AMC 8, 2004 | Problem 25
Try this beautiful problem from Geometry based on Intersection of two Squares.
When 2 Squares intersect | AMC-8, 2004 | Problem 25
Two $4\times 4$ squares intersect at right angles, bisecting their intersecting sides, as shown. The circle’s diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the squares?
• $28-2\pi$
• $25-2\pi$
• $30-2\pi$
Key Concepts
Geometry
square
Circle
Answer: $28-2\pi$
AMC-8, 2004 problem 25
Pre College Mathematics
Try with Hints
Area of the square is $\pi (r)^2$,where $r$=radius of the circle
Can you now finish the problem ……….
Clearly, if 2 squares intersect, it would be a smaller square with half the side length, 2.
can you finish the problem……..
Clearly the intersection of 2 squares would be a smaller square with half the side length, 2.
The area of this region =Total area of larger two squares – the area of the intersection, the smaller square i.e $4^2+4^2 -2^2=28$
Now The diagonal of this smaller square created by connecting the two points of intersection of the squares is the diameter of the circle
Using the Pythagorean th. diameter of the circle be $\sqrt{2^2 +2^2}=2\sqrt 2$
Radius=$\sqrt 2$
area of the square=$\pi (\sqrt2)^2$=$2\pi$
Area of the shaded region= 28-2$\pi$
Categories
Radius of the Circle | AMC-8, 2005 | Problem 25
Try this beautiful problem from Geometry: Radius of a circle
Radius of a circle – AMC-8, 2005- Problem 25
A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
• $\frac{5}{\sqrt \pi}$
• $\frac{2}{\sqrt \pi}$
• $\sqrt \pi$
Key Concepts
Geometry
Cube
square
Answer: $\frac{2}{\sqrt \pi}$
AMC-8 (2005) Problem 25
Pre College Mathematics
Try with Hints
The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square
Can you now finish the problem ……….
Region within the circle and square be $x$ i.e In other words, it is the area inside the circle and the square
The area of the circle -x=Area of the square – x
can you finish the problem……..
Given that The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square
Let the region within the circle and square be $x$ i.e In other words, it is the area inside the circle and the square .
Let r be the radius of the circle
Therefore, The area of the circle -x=Area of the square – x
so, $\pi r^2 – x=4-x$
$\Rightarrow \pi r^2=4$
$\Rightarrow r^2 = \frac{4}{\pi}$
$\Rightarrow r=\frac{2}{\sqrt \pi}$ |
### Scientific Notation
```Scientific Notation
Scientific Notation
Consists of two parts
1. mantissa
The full size number that comes before x 10
2. characteristic
The exponent that comes after x 10 and is
superscript
Scientific Notation
Example:
15
3.00 x 10
Mantissa
Characteristic
Scientific Notation
To convert a normal number to scientific
notation, move the decimal either to the
left or right until you have a mantissa which
is greater than -10 or less than +10
Scientific Notation
If the absolute value of the number is
smaller than 1, the characteristic will be
negative
If the absolute value of the number is larger
than 1, the characteristic will be positive
Scientific Notation
Example:
542
The absolute value of the number is greater than
one so the characteristic is positive
542
Moved to the left two spaces
5.42 x 102
Scientific Notation
Example:
-45000000
The absolute value of the number is greater than
one so the characteristic is positive
-45000000
Moved to the left seven spaces
-4.5 x 107
Example:
0.0040
The absolute value of the number is less than one so
the characteristic is negative
0.0040
Moved to the right three spaces
4.0
-3
x 10
Example:
-0.209
The absolute value of the number is less than one so
the characteristic is negative
-0.209
Moved to the right one space
-2.09 x
-1
10
Example:
9.245
The absolute value of the number is between -10 and
+10.
9.245
The decimal is not moved. In this case, the
characteristic is zero.
9.245 x100
Expanding Scientific Notation to
Regular notation
If the characteristic is positive,
move the decimal to the right.
If the characteristic is negative,
move the decimal to the left.
Expanding Scientific Notation to
Regular notation
Example
6.50 x 102
The characteristic tells us to move two
places to the right.
650.
Expanding Scientific Notation to
Regular notation
Example
6.50 x 10-2
The characteristic tells us to move two
places to the left.
0.0650
Fixing Incorrect Scientific Notation
Fix the mantissa by moving the decimal
Add the “fixed” characteristic to the
original characteristic
Example
20.4 x 106 (incorrect)
Moved over one space to the left (+) to
correct the mantissa
2.04 x
(6+1)
10
2.04 x 107 (correct)
Example
200.4 x 10-6 (incorrect)
Moved over two spaces to the left (+) to
correct the mantissa
2.004
(-6+2)
x 10
2.004 x 10-4 (correct)
Example
0.00340 x 108 (incorrect)
Moved over three spaces to the right (-) to
correct the mantissa
3.40
(8
+
-3)
x 10
3.40 x 105 (correct)
Example
0.00340 x 10-8 (incorrect)
Moved over three spaces to the right (-) to
correct the mantissa
3.40
(-8
+
-3)
x 10
3.40 x 10-11 (correct)
``` |
# Diff eq solver with steps
Diff eq solver with steps is a mathematical instrument that assists to solve math equations. Math can be difficult for some students, but with the right tools, it can be conquered.
## The Best Diff eq solver with steps
A statistics math solver is a tool that can be used to solve problems related to statistics. It can be used to calculate things like mean, median, mode, and standard deviation. It can also be used to create graphs and charts.
Solve slope intercept form is an algebraic equation that can be used to find the y-intercept of a line. It uses the slope of two points on a graph and the y-intercet to find the y-intercept. It is used in algebra classes and in statistics. To solve it, first find the equation of the line: b>y = mx + c/b> where b>m/b> is the slope and b>c/b> is the y-intercept. Add them up for both sides: b>y + mx = c/b>. Solve for b>c/b>: b>c = (y + mx) / (m + x)/b>. Substitute into your original equation: b>y = mx + c/b>. Finally, take your original data points and plug them into this new equation to find the y-intercept: b>y = mx + c/b>. In words, solve "for c" by plugging your data into both sides of your equation as you would solve any algebraic equation. Then solve for "y" by adjusting one side until you get "c" back on top. Example 1: Find the y-intercept if this line is graphed below.
There are a few different ways to solve for slope, but the most common is to use the slope formula. To use this formula, you'll need to know the coordinates of two points on the line. Once you have those, simply plug them into the formula and solve.
A system of equations is a set of two or more equations that share the same variables. In order to solve a system of equations, all of the equations must be satisfied simultaneously. This can be done by using a variety of methods, including substitution, elimination, and graphing. One useful tool for solving systems of equations is a compound inequality solver. A compound inequality solver is a tool that allows
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# Problem of the Week
## Updated at Aug 3, 2020 5:16 PM
For this week we've brought you this equation problem.
How would you solve the equation $$3-4\times \frac{5}{4q}=\frac{4}{3}$$?
Here are the steps:
$3-4\times \frac{5}{4q}=\frac{4}{3}$
1 Cancel $$4$$.$3-\frac{5}{q}=\frac{4}{3}$2 Subtract $$3$$ from both sides.$-\frac{5}{q}=\frac{4}{3}-3$3 Simplify $$\frac{4}{3}-3$$ to $$-\frac{5}{3}$$.$-\frac{5}{q}=-\frac{5}{3}$4 Multiply both sides by $$q$$.$-5=-\frac{5}{3}q$5 Simplify $$\frac{5}{3}q$$ to $$\frac{5q}{3}$$.$-5=-\frac{5q}{3}$6 Multiply both sides by $$3$$.$-5\times 3=-5q$7 Simplify $$-5\times 3$$ to $$-15$$.$-15=-5q$8 Divide both sides by $$-5$$.$\frac{-15}{-5}=q$9 Two negatives make a positive.$\frac{15}{5}=q$10 Simplify $$\frac{15}{5}$$ to $$3$$.$3=q$11 Switch sides.$q=3$Doneq=3 |
# To make 10 servings of soup you need 4 cups of broth. You want to know how many servings you can make with 8 pints of broth. What proportion should you use?
Jun 27, 2018
see below
#### Explanation:
Remember $1$ pint $= 2$ cups
$\Rightarrow$
$\textcolor{w h i t e}{\text{XXX}} 8$ pints $= 16$ cups
or (equivalently)
$\textcolor{w h i t e}{\text{XXX}} 2$ pints $= 4$ cups
So the given ratio $10 \text{ servings" : 4" cups}$ could be expressed as $10 \text{ servings" : 2" pints}$;
and the desired $8$ pints could be expressed as $16$ cups
If the number of servings per cups is denoted as ${x}_{c}$
and the number of servings per pint is denoted as ${x}_{p}$
The ratio could be expressed
{: ("either as ","servings : cups"," or ", "servings : pints"), (,10 : 4,,10 : 2), (,x_c : 16,,x_p : 8), ("so",," or ",), (,(x_c)/16=10/4,,(x_p)/8=10/2), (,x_c=40,,x_p=40) :}
Jun 27, 2018
$40 \text{ servings for 8 pints}$
#### Explanation:
We are mixing units of measurement. To make the calculation more straight forward lets use pints through out. This means we have to determine how many pints are in 4 cups.
$\textcolor{b l u e}{\text{Converting cups to pints}}$
Assumption: 2 cups have the same volume as 1 pint.
So by ratio we have: $\left(\text{pints")/("cups}\right) \to \frac{1}{2}$
So for 4 cups: $\left(\text{pints")/("cups}\right) \to \frac{1}{2} \times 1 \to \frac{1}{2} \times \frac{2}{2} = \frac{2}{4}$
Thus 4 cups is the same volume as 2 pints:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$
Let the unknown count of servings be $x$
$\left(\text{servings")/("pints}\right) \to \frac{10}{2} = \frac{x}{8}$
Multiply both sides by 8
$x = \frac{10 \times 8}{2} = 40 \text{ servings for 8 pints}$ |
Chapters
## Integer
In order to understand radicals, you should first know what an integer is. An integer is part of a group of numbers known as real numbers. There are two main types of numbers that are not real numbers:
• Imaginary numbers
• Infinity
Remember, these are not real numbers. You are probably more familiar with real numbers, which include:
• Whole numbers
• Rational numbers
• Irrational numbers
Take a look at the image and table below to better understand the difference between these two types of numbers.
Description More examples Whole Positive real number without decimal/fraction 188, 7 000 Rational Positive or negative with decimal/fraction 1.1, -888.99 Irrational Positive or negative, can’t be written as decimal/fraction ,
You may be wondering, what happens when you have a negative whole number? That is actually an integer.
## Real Number
Real numbers are the basis of most of the maths you will learn or use. Whenever you’re in doubt, it can be helpful to look at a number line, also called a “real number line.”
As you can see, the real number line includes both positive and negative numbers.
## Exponents
We’re almost to radicals. One last concept to understand before diving into radicals is that of an exponent. An exponent can be thought of as the opposite of a radical. Another name for exponents is ‘powers.’ The table below describes some exponents.
Meaning Common Name Written 2 x 2 Square Two to the power of 2; Two squared 3 x 3 x 3 Cube 3 to the power of 3; 3 cubed 4 x 4 Square 4 to the power of 2; 4 squared
Recall that a radical, also called a ‘root,’ is the opposite of an exponent. Because of this, you can think of each operation as ‘undoing’ the other. Let’s first go through the most common radical: the square root.
Square Root Result Written = 4 , 2 Square root of 2 = 49 , 7 Square root of 3 = 16 , 4 Square root of 2
Notice what the operations in the table have in common? The square roots are all ‘undoing’ powers of 2. In other words, the square root takes any number and tries to find what number ‘fits’ exactly twice when multiplied by itself.
When you want to take the nth root of a number x, it means you want to find the number k that, multiplied by itself n times, will give you the original number x.
Take a look at some examples below.
Square Root Result Result = k * k * k 3 = k * k * k * k 4 = k * k * k * k * k 5
## Notation of Radicals
Radicals have some common notations. The first thing you should note is that a radical can actually be written as an exponent. Take a look below.
While this may seem confusing, we break down each below.
Explanation Example = A square root is simply a number to the 2nd root, because they are the most common the 2 is usually not written = A number to the nth root = A number to the nth root can also be written as an exponent, which is 1 over the nth root =
Let’s take a look at some examples of radicals. Let's complete the following task: convert the sentence into the common notation for the radical.
• The 4th root of 16
• The square root of 225
• 4 to the power of
After this, solve the problem. These kinds of problems are often solved by calculators, which usually have the square root symbol and a separate symbol: .
There are a couple of rules you should keep in mind when you’re dealing with radicals. These rules are written in the table below, along with their descriptions.
Rule Example = x = 15 = * = * = 2*3 = 6 = = \frac{}{} = * = x * = 400
There are many ways that you can simplify a radical. The most common way you can try to simplify a radical is to try and divide the number up inside of the radical so that you can get a perfect square.
A perfect square is a radical that contains a number that can be simplified into a square. Take four as an example.
As you can see, the number 4 can be broken down into 2 squared. We look for these types of numbers inside of a radical because they are easy to take out.
To simplify larger radicals, you can apply the same approach. Take the number 20 as an example.
As you can see, we first:
• Break down the original number
• See if there are any perfect squares
• We are left with 2 *
In order to simplify radical division, we need to use a combination of the rules we learned above. First, we have the following.
We use the rules above to solve by first rewriting it, then splitting it into smaller numbers.
Simplifying the nominator and denominator, we simply get:
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# Sample Exercise 5.1 Describing and Calculating Energy Changes
## Presentation on theme: "Sample Exercise 5.1 Describing and Calculating Energy Changes"— Presentation transcript:
Sample Exercise 5.1 Describing and Calculating Energy Changes
A bowler lifts a 5.4-kg (12-lb) bowling ball from ground level to a height of 1.6 m (5.2 ft) and then drops it. (a) What happens to the potential energy of the ball as it is raised? (b) What quantity of work, in J, is used to raise the ball? (c) After the ball is dropped, it gains kinetic energy. If all the work done in part (b) has been converted to kinetic energy by the time the ball strikes the ground, what is the ball’s speed just before it hits the ground? (Note: The force due to gravity is F = m × g, where m is the mass of the object and g is the gravitational constant; g = 9.8 m ⁄ s2.) Solution Analyze We need to relate the potential energy of the bowling ball to its position relative to the ground. We then need to establish the relationship between work and the change in the ball’s potential energy. Finally, we need to connect the change in potential energy when the ball is dropped with the kinetic energy attained by the ball. Plan We can calculate the work done in lifting the ball by using Equation 5.3: w = F × d. The kinetic energy of the ball just before it hits the ground equals its initial potential energy. We can use the kinetic energy and Equation 5.1 to calculate the speed, v, just before impact. Solve (a) Because the ball is raised above the ground, its potential energy relative to the ground increases. (b) The ball has a mass of 5.4 kg and is lifted 1.6 m. To calculate the work performed to raise the ball, we use Equation 5.3 and F = m × g for the force that is due to gravity: W = F × d = m × g × d = (5.4 kg)(9.8 m ⁄ s2)(1.6 m) = 85 kg-m2 ⁄ s2 = 85 J Thus, the bowler has done 85 J of work to lift the ball to a height of 1.6 m.
Sample Exercise 5.1 Describing and Calculating Energy Changes
Continued (c) When the ball is dropped, its potential energy is converted to kinetic energy. We assume that the kinetic energy just before the ball hits the ground is equal to the work done in part (b), 85 J: Ek = mv2 = 85 J = 85 kg-m2 ⁄ s2 We can now solve this equation for v: Check Work must be done in (b) to increase the potential energy of the ball, which is in accord with our experience. The units are appropriate in (b) and (c). The work is in units of J and the speed in units of m ⁄ s. In (c) we carry an additional digit in the intermediate calculation involving the square root, but we report the final value to only two significant figures, as appropriate. Comment A speed of 1 m ⁄ s is roughly 2 mph, so the bowling ball has a speed greater than 10 mph just before impact.
Sample Exercise 5.1 Describing and Calculating Energy Changes
Continued Practice Exercise 1 Which of the following objects has the greatest kinetic energy? (a) a 500-kg motorcycle moving at 100 km ⁄ h, (b) a 1,000-kg car moving at 50 km ⁄ h, (c) a 1,500-kg car moving at 30 km ⁄ h, (d) a 5,000-kg truck moving at 10 km ⁄ h, (e) a 10,000-kg truck moving at 5 km ⁄ h. Practice Exercise 2 What is the kinetic energy, in J, of (a) an Ar atom moving at a speed of 650 m ⁄ s, (b) a mole of Ar atoms moving at 650 m ⁄ s? (Hint: 1 amu = 1.66 ×10−27 kg.)
Sample Exercise 5.2 Relating Heat and Work to Changes of Internal Energy
Gases A(g) and B(g) are confined in a cylinder-and-piston arrangement like that in Figure 5.4 and react to form a solid product C(s): A(g) + B(g) → C(s). As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system? Solution Analyze The question asks us to determine ΔE, given information about q and w. Plan We first determine the signs of q and w (Table 5.1) and then use Equation 5.5, ΔE = q + w, to calculate ΔE. Solve Heat is transferred from the system to the surroundings, and work is done on the system by the surroundings, so q is negative and w is positive: q = −1150 J and w = 480 kJ. Thus, ΔE = q + w = (−1150 J) + (480 J) = −670 J The negative value of ΔE tells us that a net quantity of 670 J of energy has been transferred from the system to the surroundings.
Sample Exercise 5.2 Relating Heat and Work to Changes of Internal Energy
Continued Comment You can think of this change as a decrease of 670 J in the net value of the system’s energy bank account (hence, the negative sign); 1150 J is withdrawn in the form of heat while 480 J is deposited in the form of work. Notice that as the volume of the gases decreases, work is being done on the system by the surroundings, resulting in a deposit of energy. Practice Exercise 1 Consider the following four cases: (i) A chemical process in which heat is absorbed, (ii) A change in which q = 30 J, w = 44 J, (iii) A process in which a system does work on its surroundings with no change in q, (iv) A process in which work is done on a system and an equal amount of heat is withdrawn. In how many of these cases does the internal energy of the system decrease? (a) 0, (b) 1, (c) 2, (d) 3, (e) 4. Practice Exercise 2 Calculate the change in the internal energy for a process in which a system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings.
Sample Exercise 5.3 Solution
A fuel is burned in a cylinder equipped with a piston. The initial volume of the cylinder is L, and the final volume is L. If the piston expands against a constant pressure of 1.35 atm, how much work (in J) is done? (1 L-atm = J) Solution Analyze We are given an initial volume and a final volume from which we can calculate ΔV. We are also given the pressure, P. We are asked to calculate work, w. Plan The equation w = −PΔV allows us to calculate the work done by the system from the given information. Solve The volume change is ΔV = Vfinal − Vinitial = L − L = L Thus, the quantity of work is w = −PΔV = −(1.35 atm)(0.730 L) = − L-atm Converting L-atm to J, we have
Sample Exercise 5.3 Practice Exercise 1 Practice Exercise 2 Continued
Check The significant figures are correct (3), and the units are the requested ones for energy (J). The negative sign is consistent with an expanding gas doing work on its surroundings. Practice Exercise 1 If a balloon is expanded from to L against an external pressure of 1.02 atm, how many L-atm of work is done? (a) −0.056 L-atm, (b) −1.37 L-atm, (c) 1.43 L-atm, (d) 1.49 L-atm, (e) 139 L-atm. Practice Exercise 2 Calculate the work, in J, if the volume of a system contracts from 1.55 to 0.85 L at a constant pressure of atm.
Sample Exercise 5.4 Determining the Sign of ΔH
Indicate the sign of the enthalpy change, ΔH, in the following processes carried out under atmospheric pressure and indicate whether each process is endothermic or exothermic: (a) An ice cube melts; (b) 1 g of butane (C4H10) is combusted in sufficient oxygen to give complete combustion to CO2 and H2O. Solution Analyze Our goal is to determine whether ΔH is positive or negative for each process. Because each process occurs at constant pressure, the enthalpy change equals the quantity of heat absorbed or released, ΔH = qP. Plan We must predict whether heat is absorbed or released by the system in each process. Processes in which heat is absorbed are endothermic and have a positive sign for ΔH; those in which heat is released are exothermic and have a negative sign for ΔH. Solve In (a) the water that makes up the ice cube is the system. The ice cube absorbs heat from the surroundings as it melts, so ΔH is positive and the process is endothermic. In (b) the system is the 1 g of butane and the oxygen required to combust it. The combustion of butane in oxygen gives off heat, so ΔH is negative and the process is exothermic. Practice Exercise 1 A chemical reaction that gives off heat to its surroundings is said to be ____________ and has a ____________ value of ΔH. (a) endothermic, positive (c) exothermic, positive (b) endothermic, negative (d) exothermic, negative
Sample Exercise 5.4 Determining the Sign of ΔH
Continued Practice Exercise 2 Molten gold poured into a mold solidifies at atmospheric pressure. With the gold defined as the system, is the solidification an exothermic or endothermic process?
Sample Exercise 5.5 Relating ΔH to Quantities of Reactants and Products
How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system? (Use the information given in Equation 5.18.) Solution Analyze Our goal is to use a thermochemical equation to calculate the heat produced when a specific amount of methane gas is combusted. According to Equation 5.18, 890 kJ is released by the system when 1 mol CH4 is burned at constant pressure. Plan Equation 5.18 provides us with a stoichiometric conversion factor: (1 mol CH −890 kJ). Thus, we can convert moles of CH4 to kJ of energy. First, however, we must convert grams of CH4 to moles of CH4. Thus, the conversion sequence is Solve By adding the atomic weights of C and 4 H, we have 1 mol CH4 = 16.0 CH4. We can use the appropriate conversion factors to convert grams of CH4 to moles of CH4 to kilojoules: The negative sign indicates that the system released 250 kJ into the surroundings.
Sample Exercise 5.5 Relating ΔH to Quantities of Reactants and Products
Continued Practice Exercise 1 The complete combustion of ethanol, C2H5OH (FW = 46.0 g ⁄ mol), proceeds as follows: C2H5OH(l) + 3→O2(g) 2CO2(g) + 3H2O(l) ΔH = −555 kJ What is the enthalpy change for combustion of 15.0 g of ethanol? (a) −12.1 kJ (b) −181 kJ (c) −422 kJ (d) −555 kJ (e) −1700 kJ Practice Exercise 2 Hydrogen peroxide can decompose to water and oxygen by the reaction 2 H2O2(l) → 2 H2O(l) + O2(g) ΔH = −196 kJ Calculate the quantity of heat released when 5.00 g of H2O2(l) decomposes at constant pressure.
Sample Exercise 5.6 Relating Heat, Temperature Change, and Heat Capacity
(a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 °C (about room temperature) to 98 °C (near its boiling point)? (b) What is the molar heat capacity of water? Solution Analyze In part (a) we must find the quantity of heat (q) needed to warm the water, given the mass of water (m), its temperature change (∆T), and its specific heat (Cs). In part (b) we must calculate the molar heat capacity (heat capacity per mole, Cm) of water from its specific heat (heat capacity per gram). Plan (a) Given Cs, m, and ∆T, we can calculate the quantity of heat, q, using Equation (b) We can use the molar mass of water and dimensional analysis to convert from heat capacity per gram to heat capacity per mole. Solve (a) The water undergoes a temperature change of ΔT = 98 °C − 22 °C = 76 °C = 76 K Using Equation 5.22, we have (b) The molar heat capacity is the heat capacity of one mole of sub- stance. Using the atomic weights of hydrogen and oxygen, we have mol H2O = 18.0 g H2O From the specific heat given in part (a), we have
Sample Exercise 5.6 Relating Heat, Temperature Change, and Heat Capacity
Continued Practice Exercise 1 Suppose you have equal masses of two substances, A and B. When the same amount of heat is added to samples of each, the temperature of A increases by 14 °C whereas that of B increases by 22 °C. Which of the following statements is true? (a) The heat capacity of B is greater than that of A. (b) The specific heat of A is greater than that of B. (c) The molar heat capacity of B is greater than that of A. (d) The volume of A is greater than that of B. (e) The molar mass of A is greater than that of B. Practice Exercise 2 (a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J ⁄ g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 °C. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Sample Exercise 5.7 Measuring ΔH Using a Coffee-Cup Calorimeter When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 to 27.5 °C. Calculate the enthalpy change for the reaction in kJ ⁄ mol HCl, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g ⁄ mL, and that its specific heat is 4.18 J ⁄ g-K. Solution Analyze Mixing solutions of HCl and NaOH results in an acid–base reaction: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) We need to calculate the heat produced per mole of HCl, given the temperature increase of the solution, the number of moles of HCl and NaOH involved, and the density and specific heat of the solution. Plan The total heat produced can be calculated using Equation The number of moles of HCl consumed in the reaction must be calculated from the volume and molarity of this substance, and this amount is then used to determine the heat produced per mol HCl. Solve Because the total volume of the solution is 100 mL, its mass is The temperature change is Using Equation 5.23, we have (100 mL)(1.0 g ⁄ mL) = 100 g ΔT = 27.5 °C − 21.0 °C = 6.5 °C = 6.5 K qrxn = −Cs × m × ∆T = −(4.18 J ⁄ g − K)(100 g)(6.5 K) = −2.7 × 103 J = −2.7 kJ
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)
Sample Exercise 5.7 Measuring ΔH Using a Coffee-Cup Calorimeter Continued Because the process occurs at constant pressure, ΔH = qP = −2.7 kJ To express the enthalpy change on a molar basis, we use the fact that the number of moles of HCl is given by the product of the volume (50 mL = L) and concentration (1.0 M = 1.0 mol/L) of the HCl solution: (0.050 L)(1.0 mol ⁄ L) = mol Thus, the enthalpy change per mole of HCl is ΔH = −2.7 kJ ⁄ mol = −54 kJ ⁄ mol Check ΔH is negative (exothermic), as evidenced by the observed increase in the temperature. The magnitude of the molar enthalpy change seems reasonable. Practice Exercise 1 When g of Mg metal is combined with enough HCl to make 100 mL of solution in a constant-pressure calorimeter, the following reaction occurs: Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) If the temperature of the solution increases from 23.0 to 34.1 ° C as a result of this reaction, calculate ΔH in kJ ⁄ mol Mg. Assume that the solution has a specific heat of 4.18 J ⁄ g- °C. (a) −19.1 kJ ⁄ mol (b) −111 kJ ⁄ mol (c) −191 kJ ⁄ mol (d) −464 kJ ⁄ mol (e) −961 kJ ⁄ mol
AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)
Sample Exercise 5.7 Measuring ΔH Using a Coffee-Cup Calorimeter Continued Practice Exercise 2 When 50.0 mL of M AgNO3 and 50.0 mL of M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from to °C. The temperature increase is caused by the following reaction: AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) Calculate ΔH for this reaction in kJ ⁄ mol AgNO3, assuming that the combined solution has a mass of g and a specific heat of 4.18 J ⁄ g- °C.
2 CH6N2(l) + 5 O2(g) → 2 N2(g) + 2 CO2(g) + 6 H2O(l)
Sample Exercise 5.8 Measuring qrxn Using a Bomb Calorimeter The combustion of methylhydrazine (CH6N2), a liquid rocket fuel, produces N2(g), CO2(g), and H2O(l): 2 CH6N2(l) + 5 O2(g) → 2 N2(g) + 2 CO2(g) + 6 H2O(l) When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from to °C. In a separate experiment the heat capacity of the calorimeter is measured to be kJ ⁄ °C. Calculate the heat of reaction for the combustion of a mole of CH6N2. Solution Analyze We are given a temperature change and the total heat capacity of the calorimeter. We are also given the amount of reactant combusted. Our goal is to calculate the enthalpy change per mole for combustion of the reactant. Plan We will first calculate the heat evolved for the combustion of the 4.00-g sample. We will then convert this heat to a molar quantity. Solve For combustion of the 4.00-g sample of methylhydrazine, the temperature change of the calorimeter is Δ T= (39.50 °C − °C ) = °C We can use ΔT and the value for Ccal to calculate the heat of reaction (Equation 5.24): qrxn = −Ccal × ΔT + −(7.794 kj ⁄ °C)(14.50 °C) = − kj We can readily convert this value to the heat of reaction for a mole of CH6N2:
Sample Exercise 5.8 Measuring qrxn Using a Bomb Calorimeter
Continued Check The units cancel properly, and the sign of the answer is negative as it should be for an Exothermic reaction. The magnitude of the answer seems reasonable. Practice Exercise 1 The combustion of exactly g of benzoic acid in a bomb calorimeter releases kJ of heat. If the combustion of g of benzoic acid causes the temperature of the calorimeter to increase from to °C, calculate the heat capacity of the calorimeter. (a) kJ ⁄ °C (b) 6.42 kJ ⁄ °C (c) 14.5 kJ ⁄ °C (d) 21.2 kJ ⁄ g- °C (e) 32.7 kJ ⁄ °C Practice Exercise 2 A g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is kJ ⁄ °C. The temperature increases from to °C. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole.
Sample Exercise 5.9 Using Hess’s Law to Calculate ΔH
The enthalpy of reaction for the combustion of C to CO2 is −393.5 kJ ⁄ mol C, and the enthalpy for the combustion of CO to CO2 is −283.0 kJ ⁄ mol CO: (1) C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ (2) CO(g) + O2(g) → CO2(g) ΔH = −283.0 kJ Using these data, calculate the enthalpy for the combustion of C to CO: (3) C(s) + O2(g) → CO(g) ΔH = ? Solution Analyze We are given two thermochemical equations, and our goal is to combine them in such a way as to obtain the third equation and its enthalpy change. Plan We will use Hess’s law. In doing so, we first note the numbers of moles of substances among the reactants and products in the target equation (3). We then manipulate equations (1) and (2) to give the same number of moles of these substances, so that when the resulting equations are added, we obtain the target equation. At the same time, we keep track of the enthalpy changes, which we add.
Sample Exercise 5.9 Using Hess’s Law to Calculate ΔH
Continued Solve To use equations (1) and (2), we arrange them so that C(s) is on the reactant side and CO(g) is on the product side of the arrow, as in the target reaction, equation (3). Because equation (1) has C(s) as a reactant, we can use that equation just as it is. We need to turn equation (2) around, however, so that CO(g) is a product. Remember that when reactions are turned around, the sign of ΔH is reversed. We arrange the two equations so that they can be added to give the desired equation: When we add the two equations, CO2(g) appears on both sides of the arrow and therefore cancels out. Likewise O2(g) is eliminated from each side. Practice Exercise 1 Calculate ΔH for 2NO(g) + O2(g) → N2O4(g), using the following information: N2O4(g) → 2NO2(g) ΔH = kJ NO(g) + O2(g) → 2NO2(g) ΔH = −113.1 kJ (a) 2.7 kJ (b) −55.2 kJ (c) −85.5 kJ (d) −171.0 kJ (e) kJ
C(graphite) → C(diamond) ΔH = ?
Sample Exercise 5.9 Using Hess’s Law to Calculate ΔH Continued Practice Exercise 2 Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion of graphite is −393.5 kJ ⁄ mol, and that of diamond is −395.4 kJ ⁄ mol: C(graphite) + O2(g) → CO2(g) ΔH = −393.5 kJ C(diamond) + O2(g) → CO2(g) ΔH = −395.4 kJ Calculate ΔH for the conversion of graphite to diamond: C(graphite) → C(diamond) ΔH = ?
Sample Exercise 5.10 Using Three Equations with Hess’s Law to Calculate ΔH
Calculate ΔH for the reaction 2 C(s) + H2(g) → C2H2(g) Given the following chemical equations and their respective enthalpy changes: C2H2(g) + O2(g) → 2 CO2(g) + H2O(l) ΔH = − kJ C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ H2(g) + O2(g) → H2O(l) ΔH = −285.8 kJ Solution Analyze We are given a chemical equation and asked to calculate its ΔH using three chemical equations and their associated enthalpy changes. Plan We will use Hess’s law, summing the three equations or their reverses and multiplying each by an appropriate coefficient so that they add to give the net equation for the reaction of interest. At the same time, we keep track of the ΔH values, reversing their signs if the reactions are reversed and multiplying them by whatever coefficient is employed in the equation.
Sample Exercise 5.10 Using Three Equations with Hess’s Law to Calculate ΔH
Continued Solve Because the target equation has C2H2 as a product, we turn the first equation around; the sign of ΔH is therefore changed. The desired equation has 2 C(s) as a reactant, so we multiply the second equation and its ΔH by 2. Because the target equation has H2 as a reactant, we keep the third equation as it is. We then add the three equations and their enthalpy changes in accordance with Hess’s law: When the equations are added, there are 2 CO2, O2, and H2O on both sides of the arrow. These are canceled in writing the net equation. Check The procedure must be correct because we obtained the correct net equation. In cases like this you should go back over the numerical manipulations of the ΔH values to ensure that you did not make an inadvertent error with signs.
C(s) + H2O(g) → CO(g) + H2(g)
Sample Exercise 5.10 Using Three Equations with Hess’s Law to Calculate ΔH Continued Practice Exercise 1 We can calculate ΔH for the reaction C(s) + H2O(g) → CO(g) + H2(g) using the following thermochemical equations: C(s) + O2(g) → CO2(g) ΔH1 = −393.5 kJ 2 CO(g) + O2(g) → 2 CO2(g) ΔH2 = −566.0 kJ 2 H2(g) + O2(g) → 2 H2O(g) ΔH3 = −483.6 kJ By what coefficient do you need to multiply ΔH2 in determining ΔH for the target equation? (a) −1 ⁄ (b) − (c) (d) 1 ⁄ (e) 2
Sample Exercise 5.10 Using Three Equations with Hess’s Law to Calculate ΔH
Continued Practice Exercise 2 Calculate ΔH for the reaction NO(g) + O(g) → NO2(g) given the following information: NO(g) + O3(g) → NO2(g) + O2(g) ΔH = −198.9 kJ O3(g) → O2(g) ΔH = −142.3 kJ O2(g) → 2 O(g) ΔH = kJ
Sample Exercise 5.11 Equations Associated with Enthalpies of Formation
For which of these reactions at 25 °C does the enthalpy change represent a standard enthalpy of formation? For each that does not, what changes are needed to make it an equation whose ΔH is an enthalpy of formation? (a) 2 Na(s) + O2(g) → Na2O(s) (b) 2 K(l) + Cl2(g) → 2 KCl(s) (c) C6H12O6(s) → 6 C(diamond) + 6 H2(g) + 3 O2(g) Solution Analyze The standard enthalpy of formation is represented by a reaction in which each reactant is an element in its standard state and the product is one mole of the compound. Plan We need to examine each equation to determine (1) whether the reaction is one in which one mole of substance is formed from the elements, and (2) whether the reactant elements are in their standard states. Solve In (a) 1 mol Na2O is formed from the elements sodium and oxygen in their proper states, solid Na and O2 gas, respectively. Therefore, the enthalpy change for reaction (a) corresponds to a standard enthalpy of formation. In (b) potassium is given as a liquid. It must be changed to the solid form, its standard state at room temperature. Furthermore, 2 mol KCI are formed, so the enthalpy change for the reaction as written is twice the standard enthalpy of formation of KCl(s). The equation for the formation reaction of 1 mol of KCl(s) is K(s) + Cl2(g) → KCl(s)
6 C(graphite) + 6 H2(g) + 3 O2(g) → C6H12O6(s)
Sample Exercise 5.11 Equations Associated with Enthalpies of Formation Continued Reaction (c) does not form a substance from its elements. Instead, a substance decomposes to its elements, so this reaction must be reversed. Next, the element carbon is given as diamond, whereas graphite is the standard state of carbon at room temperature and 1 atm pressure. The equation that correctly represents the enthalpy of formation of glucose from its elements is 6 C(graphite) + 6 H2(g) + 3 O2(g) → C6H12O6(s) Practice Exercise 1 If the heat of formation of H2O(l) is −286 kJ ⁄ mol, which of the following thermochemical equations is correct? (a) 2H(g) + O(g) → H2O(l) ΔH = −286 kJ (b) 2H2(g) + O2(g) → 2H2O(l) ΔH = −286 kJ (c) H2(g) + O2(g) → H2O(l) ΔH = −286 kJ (d) H2(g) + O(g) → H2O(g) ΔH = −286 kJ (e) H2O(l) → H2(g) + O2(g) ΔH = −286 kJ
Sample Exercise 5.11 Equations Associated with Enthalpies of Formation
Continued Practice Exercise 2 Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl4) and look up ΔHf ° for this compound in Appendix C.
Sample Exercise 5.12 Calculating an Enthalpy of Reaction from Enthalpies of Formation
(a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g) and H2O(l). (b) Compare the quantity of heat produced by combustion of 1.00 g propane with that produced by 1.00 g benzene. Solution Analyze (a) We are given a reaction [combustion of C6H6(l) to form CO2(g) and H2O(l)] and asked to calculate its standard enthalpy change, ΔH °. (b) We then need to compare the quantity of heat produced by combustion of g C6H6 with that produced by 1.00 g C3H8, whose combustion was treated previously in the text. (See Equations 5.29 and 5.30.) Plan (a) We first write the balanced equation for the combustion of C6H6. We then look up Hf ° values in Appendix C or in Table 5.3 and apply Equation 5.31 to calculate the enthalpy change for the reaction. (b) We use the molar mass of C6H6 to change the enthalpy change per mole to that per gram. We similarly use the molar mass of C3H8 and the enthalpy change per mole calculated in the text previously to calculate the enthalpy change per gram of that substance.
Sample Exercise 5.12 Calculating an Enthalpy of Reaction from Enthalpies of Formation
Continued Solve (a) We know that a combustion reaction involves O2(g) as a reactant. Thus, the balanced equation or the combustion reaction of 1 mol C6H6(l) is C6H6(l) + O2(g) → 6 CO2(g) + 3 H2O(l) We can calculate ΔH ° for this reaction by using Equation 5.31 and data in Table 5.3. Remember to multiply the ΔHf ° value for each substance in the reaction by that substance’s stoichiometric coefficient. Recall also that ΔHf ° = 0 for any element in its most stable form under standard conditions, so ΔHf °[O2(g)] = 0. ∆H °rxn = [(6ΔHf °(CO2) + 3ΔHf °(H2O)] – [ΔHf °(C6H6) + ΔHf °(O2)] = [6(−393.5 kJ) + 3(−285.8 kJ)] – [(49.0 kJ) (0 kJ)] = (−2361 − − 49.0) kJ = −3267 kJ
Sample Exercise 5.12 Calculating an Enthalpy of Reaction from Enthalpies of Formation
Continued (b) From the example worked in the text, ΔH ° = −2220 kJ for the combustion of 1 mol of propane. In part (a) of this exercise we determined that ΔH ° = −3267 kJ for the combustion of 1 mol benzene. To determine the heat of combustion per gram of each substance, we use the molar masses to convert moles to grams: C3H8(g): (−2220 kJ/mol)(1 mol/44.1 g) = −50.3 kJ/g C6H6(l): (−3267 kJ/mol)(1 mol/78.1 g) = −41.8 kJ/g
Sample Exercise 5.12 Calculating an Enthalpy of Reaction from Enthalpies of Formation
Continued Comment Both propane and benzene are hydrocarbons. As a rule, the energy obtained from the combustion of a gram of hydrocarbon is between 40 and 50 kJ. Practice Exercise 1 Calculate the enthalpy change for the reaction 2H2O2(l) → 2H2O(l) + O2(g) using enthalpies of formation: ΔHf °(H2O2) = −187.8 kJ ⁄ mol ΔHf °(H2O) = −285.8 kJ ⁄ mol (a) −88.0 kJ, (b) −196.0 kJ, (c) kJ, (d) kJ, (e) more information needed Practice Exercise 2 Use Table 5.3 to calculate the enthalpy change for the combustion of 1 mol of ethanol: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
Sample Exercise 5.13 Calculating an Enthalpy of Formation Using Enthalpy of Reaction
The standard enthalpy change for the reaction is kJ. Use Table 5.3 to calculate the standard enthalpy of formation of CaCO3(s). Solution Analyze Our goal is to obtain ΔHf °(CaCO3). Plan We begin by writing the expression for the standard enthalpy change for the reaction: ΔH °rxn = ΔHf ° (CaO) + ΔHf °(CO2) – ΔHf °(CaCO3) Solve Inserting the given ΔH °rxn and the known ΔHf ° values from Table 5.3 or Appendix C, we have = −635.5 kJ − kJ − ΔHf °(CaCO3) Solving for ΔHf °(CaCO3) gives ΔHf °(CaCO3) = − kJ ⁄ mol
Sample Exercise 5.13 Calculating an Enthalpy of Formation Using Enthalpy of Reaction
Continued Check We expect the enthalpy of formation of a stable solid such as calcium carbonate to be negative, as obtained. Practice Exercise 1 Given 2 SO2(g) + O2(g) → 2 SO3(g), which of the following equations is correct? (a) ΔHf °(SO3) = ΔH °rxn − ΔHf °(SO2) (b) ΔHf °(SO3) = ΔH °rxn + ΔHf °(SO2) (c) 2ΔHf °(SO3) = ΔH °rxn + 2ΔHf °(SO2) (d) 2ΔHf °(SO3) = ΔH °rxn − 2ΔHf °(SO2) (e) 2ΔHf °(SO3) = 2ΔHf °(SO2) − ΔH °rxn Practice Exercise 2 Given the following standard enthalpy change, use the standard enthalpies of formation in Table 5.3 to calculate the standard enthalpy of formation of CuO(s): CuO(s) + H2(g) → Cu(s) + H2O(l) ΔH ° = −129.7 kJ
Sample Exercise 5.14 Estimating the Fuel Value of a Food from Its Composition
(a) A 28-g (1-oz) serving of a popular breakfast cereal served with 120 mL of skim milk provides 8 g protein, 26 g carbohydrates, and 2 g fat. Using the average fuel values of these substances, estimate the fuel value (caloric content) of this serving. (b) A person of average weight uses about 100 Cal ⁄ mi when running or jogging. How many servings of this cereal provide the fuel value requirements to run 3 mi? Solution Analyze The fuel value of the serving will be the sum of the fuel values of the protein, carbohydrates, and fat. Plan We are given the masses of the protein, carbohydrates, and fat contained in a serving. We can use the data in Table 5.4 to convert these masses to their fuel values, which we can sum to get the total fuel value. Solve
Sample Exercise 5.14 Estimating the Fuel Value of a Food from Its Composition
Continued This corresponds to 160 kcal: Recall that the dietary Calorie is equivalent to 1 kcal. Thus, the serving provides 160 Cal. Analyze Here we are faced with the reverse problem, calculating the quantity of food that provides a specific fuel value. Plan The problem statement provides a conversion factor between Calories and miles. The answer to part (a) provides us with a conversion factor between servings and Calories. Solve We can use these factors in a straightforward dimensional analysis to determine the number of servings needed, rounded to the nearest whole number:
Sample Exercise 5.14 Estimating the Fuel Value of a Food from Its Composition
Continued Practice Exercise 1 A stalk of celery has a caloric content (fuel value) of 9.0 kcal. If 1.0 kcal is provided by fat and there is very little protein, estimate the number of grams of carbohydrate and fat in the celery. (a) 2 g carbohydrate and 0.1 g fat, (b) 2 g carbohydrate and 1 g fat, (c) 1 g carbohydrate and 2 g fat, (d) 2.2 g carbohydrate and 0.1 g fat, (e) 32 g carbohydrate and 10 g fat. Practice Exercise 2 (a) Dry red beans contain 62% carbohydrate, 22% protein, and 1.5% fat. Estimate the fuel value of these beans. (b) During a very light activity, such as reading or watching television, the average adult expends about 7 kJ ⁄ min. How many minutes of such activity can be sustained by the energy provided by a serving of chicken noodle soup containing 13 g protein, 15 g carbohydrate, and 5 g fat?
C3H5N3O9(l) → N2(g) + CO2(g) + H2O(l) + O2(g)
Sample Integrative Exercise Putting Concepts Together Trinitroglycerin, C3H5N3O9 (usually referred to simply as nitroglycerin), has been widely used as an explosive. Alfred Nobel used it to make dynamite in Rather surprisingly, it also is used as a medication, to relieve angina (chest pains resulting from partially blocked arteries to the heart) by dilating the blood vessels. At 1 atm pressure and 25 °C, the enthalpy of decomposition of trinitroglycerin to form nitrogen gas, carbon dioxide gas, liquid water, and oxygen gas is − kJ ⁄ mol. (a) Write a balanced chemical equation for the decomposition of trinitroglycerin. (b) Calculate the standard heat of formation of trinitroglycerin. (c) A standard dose of trinitroglycerin for relief of angina is 0.60 mg. If the sample is eventually oxidized in the body (not explosively, though!) to nitrogen gas, carbon dioxide gas, and liquid water, what number of calories is released? (d) One common form of trinitroglycerin melts at about 3 °C. From this information and the formula for the substance, would you expect it to be a molecular or ionic compound? Explain. (e) Describe the various conversions of forms of energy when trinitroglycerin is used as an explosive to break rockfaces in highway construction. Solution (a) The general form of the equation we must balance is C3H5N3O9(l) → N2(g) + CO2(g) + H2O(l) + O2(g)
Sample Integrative Exercise Putting Concepts Together
Continued We go about balancing in the usual way. To obtain an even number of nitrogen atoms on the left, we multiply the formula for C3H5N3O9 by 2, which gives us 3 mol of N2, 6 mol of CO2 and 5 mol of H2O. Everything is then balanced except for oxygen. We have an odd number of oxygen atoms on the right. We can balance the oxygen by using the coefficient for O2 on the right: 2 C3H5N3O9(l) → 3 N2(g) + 6 CO2(g) + 5 H2O(l) + O2(g) We multiply through by 2 to convert all coefficients to whole numbers: 4 C3H5N3O9(l) → 6 N2(g) + 12 CO2(g) + 10 H2O(l) + O2(g) (At the temperature of the explosion, water is a gas. The rapid expansion of the gaseous products creates the force of an explosion.) (b) We can obtain the standard enthalpy of formation of nitroglycerin by using the heat of decomposition of trinitroglycerin together with the standard enthalpies of formation of the other substances in the decomposition equation: The enthalpy change for this decomposition is 4(− kJ) = − kJ. [We need to multiply by 4 because there are 4 mol of C3H5N3O9(l) in the balanced equation.]
−6165.6 kJ = 12(−393.5 kJ) + 10(−285.8 kJ) − 4ΔHf °[C3H5N3O9(l)]
Sample Integrative Exercise Putting Concepts Together Continued This enthalpy change equals the sum of the heats of formation of the products minus the heats of formation of the reactants, each multiplied by its coefficient in the balanced equation: − kJ = 6ΔHf °[N2(g)] + 12ΔHf °[CO2(g)] + 10ΔHf °[H2O(l)] + ΔHf °[O2(g)] −4ΔHf °[C3H5N3O9(l)] The ΔHf ° values for N2(g) and O2(g) are zero, by definition. Using the values for H2O(l) and CO2(g) from Table 5.3 or Appendix C, we have − kJ = 12(−393.5 kJ) + 10(−285.8 kJ) − 4ΔHf °[C3H5N3O9(l)] ΔHf °[C3H5N3O9(l)] = −353.6 kJ ⁄ mol
Sample Integrative Exercise Putting Concepts Together
Continued (c) Converting 0.60 mg C3H5N3O9(l) to moles and using the fact that the decomposition of 1 mol of C3H5N3O9(l) yields kJ we have:
Sample Integrative Exercise Putting Concepts Together
Continued (d) Because trinitroglycerin melts below room temperature, we expect that it is a molecular compound. With few exceptions, ionic substances are generally hard, crystalline materials that melt at high temperatures (Sections 2.6 and 2.7) Also, the molecular formula suggests that it is a molecular substance because all of its constituent elements are nonmetals. (e) The energy stored in trinitroglycerin is chemical potential energy. When the substance reacts explosively, it forms carbon dioxide, water, and nitrogen gas, which are of lower potential energy. In the course of the chemical transformation, energy is released in the form of heat; the gaseous reaction products are very hot. This high heat energy is transferred to the surroundings. Work is done as the gases expand against the surroundings, moving the solid materials and imparting kinetic energy to them. For example, a chunk of rock might be impelled upward. It has been given kinetic energy by transfer of energy from the hot, expanding gases. As the rock rises, its kinetic energy is transformed into potential energy. Eventually, it again acquires kinetic energy as it falls to Earth. When it strikes Earth, its kinetic energy is converted largely to thermal energy, though some work may be done on the surroundings as well.
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# Difference between revisions of "2013 AMC 10A Problems/Problem 23"
The following problem is from both the 2013 AMC 12A #19 and 2013 AMC 10A #23, so both problems redirect to this page.
## Problem
In $\triangle ABC$, $AB = 86$, and $AC=97$. A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$. Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$?
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$
### Solution 1 (Power of a Point)
Let $BX = q$, $CX = p$, and $AC$ meets the circle at $Y$ and $Z$, with $Y$ on $AC$. Then $AZ = AY = 86$. Using the Power of a Point (Secant-Secant Power Theorem), we get that $p(p+q) = 11(183) = 11 * 3 * 61$. We know that $p+q>p$, so $p$ is either 3,11, or 33. We also know that $p>11$ by the triangle inequality on $\triangle ACX$. $p$ is 33. Thus, we get that $BC = p+q = \boxed{\textbf{(D) }61}$.
### Solution 2 (Stewart's Theorem)
Stewart's Theorem: https://en.wikipedia.org/wiki/Stewart%27s_theorem
Let $x$ represent $CX$, and let $y$ represent $BX$. Since the circle goes through $B$ and $X$, $AB = AX = 86$. Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to $0$, dividing both sides of the equation by $y$ is allowed.)
$x(x+y) = (97+86)(97-86)$
$x(x+y) = 2013$
The prime factors of $2013$ are $3$, $11$, and $61$. Obviously, $x < x+y$. In addition, by the Triangle Inequality, $BC < AB + AC$, so $x+y < 183$. Therefore, $x$ must equal $33$, and $x+y$ must equal $\boxed{\textbf{(D) }61}$.
### Solution 3
Let $CX=x, BX=y$. Let the circle intersect $AC$ at $D$ and the diameter including $AD$ intersect the circle again at $E$. Use power of a point on point C to the circle centered at A.
So $CX*CB=CD*CE=>$ $x(x+y)=(97-86)(97+86)=>$ $x(x+y)=3*11*61$.
Obviously $x+y>x$ so we have three solution pairs for $(x,x+y)=(1,2013),(3,671),(11,183),(33,61)$. By the Triangle Inequality, only$x+y=61$ yields a possible length of $BX+CX=BC$.
Therefore, the answer is $\boxed{\textbf{(D) }61}$. |
# Quick Review: Circles
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In this article, you will get an overview of the concepts and formulae related to Circles.
Consider the given circle:
###### Important Terms
• Circumference: The distance around the circle is called circumference of the circle. It can be determined either by using the diameter d or the radius r.
C= 2πr or 2πd
• Arc: A part of a circle is known as an arc and is named according to its angle. Arcs are divided into 3 parts:
Minor Arcs: 00 < v < 1800
Major Arcs: 1800 < v < 3600
For Semi Circles: v = 1800
The length of the arc l is measured by the degree measure of the arc, v, and the circumference of the whole circle, C, by the following formula:
I = C*(V/360)
• Area: The area of the circle is calculated by the formula: πr2 As we know that radius, r =d/2. Therefore, we can also say that area, A = π(d2/4)
• Inscribed Angle: y = α/2
• Area of Sector: Area of Sector AOC y= πR2α/360
###### Important Points
• Triangle inscribed in a semi-circle = Right Triangle
• A circle is a polygon with infinite number of sides. If the perimeter is same, the polygon with the maximum number of sides has the maximum area. So, the circle has the maximum area for a given perimeter/circumference.
• If the radius of a circle is perpendicular to a chord, then it bisects the chord and vice versa.
• Equidistant chords from the center are equal in length.
• Tangents drawn to a circle from an exterior point are equal in length.
• If a square of maximum dimensions is cut from a circle, then the percentage area wasted is 36.3.
• If a circle of maximum radius is cut from a square, then the percentage area wasted is 21.5%.
• The length of the direct tangent to two circles of radii, r1 and r2 is , where D is the distance between the two centers.
• The length of the transverse tangent to two circles of radii, r1 and r2 is
• A circle is of 3600.
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# Lesson 17: Write Expressions in Which Letters Stand for Numbers
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1 Write Expressions in Which Letters Stand for Numbers Student Outcomes Students write algebraic expressions that record all operations with numbers and/or letters standing for the numbers. Lesson Notes Large paper is needed to complete this lesson. Classwork Fluency Exercise (5 minutes): Addition of Decimals Sprint: Refer to the Sprints and Sprint Delivery Script sections in the Module Overview for directions on how to administer a Sprint. Opening (5 minutes) Discuss the Exit Ticket from Lesson 16. Students continue to work on writing expressions, so discuss any common mistakes from the previous lesson. Exercises (25 minutes) Students work in groups of two or three to complete the stations. At each station, students write down the problem and the expression with variables and/or numbers. Encourage students to underline key words in each problem. Exercises Station One: 1. The sum of a and b a + b 2. Five more than twice a number c Scaffolding: If students struggled during Lesson 16, complete some examples with students before moving into the exercises c or 2c Martha bought d number of apples and then ate 6 of them. d 6 170
2 Station Two: decreased by p 14 p 2. The total of d and f, divided by 8 d+f or (d + f) Rashod scored 6 less than 3 times as many baskets as Mike. Mike scored b baskets. 3b 6 Station Three: 1. The quotient of c and 6 c 6 2. Triple the sum of x and 17 3(x + 17) 3. Gabrielle had b buttons but then lost 6. Gabrielle took the remaining buttons and split them equally among her 5 friends. b 6 or (b 6) 5 5 Station Four: 1. d doubled 2d 2. Three more than 4 times a number x 4x + 3 or 3 + 4x 3. Mali has c pieces of candy. She doubles the amount of candy she has and then gives away 15 pieces. 2c 15 Station Five: 1. f cubed f 3 2. The quantity of 4 increased by a, and then the sum is divided by 9. 4+a or (4 + a) Tai earned 4 points fewer than double Oden s points. Oden earned p points. 2p 4 171
4 Name Date Exit Ticket Write an expression using letters and/or numbers for each problem below. 1. d squared 2. A number x increased by 6, and then the sum is doubled. 3. The total of h and b is split into 5 equal groups. 4. Jazmin has increased her \$45 by m dollars and then spends a third of the entire amount. 5. Bill has d more than 3 times the number of baseball cards as Frank. Frank has f baseball cards. 173
5 Exit Ticket Sample Solutions Write an expression using letters and/or numbers for each problem below. 1. d squared d 2 2. A number x increased by 6, and then the sum is doubled. 2(x + 6) 3. The total of h and b is split into 5 equal groups. h+b or (h + b) Jazmin has increased her \$45 by m dollars and then spends a third of the entire amount m 3 or 1 (45 + m) 3 5. Bill has d more than 3 times the number of baseball cards as Frank. Frank has f baseball cards. 3f + d or d + 3f Problem Set Sample Solutions Write an expression using letters and/or numbers for each problem below less than the quantity of 8 times n 8n times the sum of y and 11 6(y + 11) 3. The square of m reduced by 49 m The quotient when the quantity of 17 plus p is divided by 8 17+p or (17 + p) Jim earned j in tips, and Steve earned s in tips. They combine their tips and then split them equally. j+s or (j + s)
6 6. Owen has c collector cards. He quadruples the number of cards he has and then combines them with Ian, who has i collector cards. 4c + i 7. Rae runs 4 times as many miles as Madison and Aaliyah combined. Madison runs m miles, and Aaliyah runs a miles. 4(m + a) 8. By using coupons, Mary Jo is able to decrease the retail price of her groceries, g, by \$125. g To calculate the area of a triangle, you find the product of the base and height and then divide by 2. bh or bh The temperature today was 10 degrees colder than twice yesterday s temperature, t. 2t
7 Addition of Decimals I Round 1 Directions: Evaluate each expression. Number Correct:
8 Addition of Decimals I Round 1 [KEY] Directions: Evaluate each expression
9 Addition of Decimals I Round 2 Number Correct: Improvement: Directions: Evaluate each expression
10 Addition of Decimals I Round 2 [KEY] Directions: Evaluate each expression
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### ILLINOIS DISTRICT REPORT CARD
-6-525-2- Hazel Crest SD 52-5 Hazel Crest SD 52-5 Hazel Crest, ILLINOIS 2 8 ILLINOIS DISTRICT REPORT CARD and federal laws require public school districts to release report cards to the public each year. |
Anyone who has studied maths at university will have realised quickly that this is not the same subject you learn at school. There are hints of more recognisable subjects, such as statistics and mechanic – however, from the very beginning of the course, you are also introduced to abstract algebra and suddenly fields, rings and groups take on very different meanings. Whether you’re about to start a maths degree and want to give yourself a head start or are just interested in learning something new here’s a short introduction to one of these areas of abstract studies, Group Theory, that we hope will spark your interest!
So to start, what is a group? Slightly unhelpfully a group can be whatever you want it to be... as long as it follows certain rules called axioms.
A group is a set, say G, of elements, combined with some binary operation such that
1) For all elements of the group, e.g a and b, the element produced as a result of applying the binary operation to a and b, ab, is also in the group.
2) For any three elements in the group, say a, b, c, then the element (ab)c must be equal to a(bc) (where ab is the element formed by combining a and b with the binary operation). This property may seem familiar and rightly so: it is called associativity and is something that binary operations we use every day, such as addition and multiplication, also satisfy.
3) A group must have an identity element, some element e such that for any element of the group, say a, ae = a = ea.
4) Every element in the group must have some corresponding ‘inverse element’ that it can be combined under the binary operation to give the identity element.
This set of rules may seem extremely abstract but there are in fact several groups that you have been using since kindergarten. For example, if we take the set of whole numbers or ‘integers’, denoted ℤ, and the binary operation addition, then it’s possible to show that {ℤ,+} is a group. By definition, adding together any two integers will also give an integer, addition is associative as stated above, 0 is an integer and any number add zero will return that same number (so zero is the identity element) and finally for every integer, say a, -a is also an integer and a + -a = 0 and so all the axioms are satisfied.
Groups can be far more abstract than this, however, and are not limited to numbers as their elements. Take the set G = {“cat”, “dog”}. Defining a binary operation ‘*’ as Cat*Cat = Cat, Dog*Dog = Cat and Cat*Dog = Dog*Cat = Dog then {G , * } is a group! To see why again you just have to work through the axioms.
Some of the most important applications of group theory come from these stranger creations. The most common example is the “symmetric groups”. Studying the symmetries of some n-sided regular polygon and defining it as a group has led to applications across other sciences, for example in the structure of a hydrogen atom. If you are fortunate enough to come across group theory then don’t be scared! Abstract algebra can be intimidating at first but once you get your head round it, a symmetry group does become just as clear and natural as the set of real numbers!
- Lucy Chats Maths |
# Disclaimer: The following material is being kept online for archival purposes.
## Although accurate at the time of publication, it is no longer being updated. The page may contain broken links or outdated information, and parts may not function in current web browsers.
Site Map Math Index Glossary Timeline Questions & Answers Lesson Plans #M-6. the Pythagoras Theorem
# (M-5) Deriving Approximate Results
### A Preliminary Derivation
Given a fraction a/b, one may multiply or divide its top and bottom ("numerator and denominator") by the same number c:
(a/b) = (ac)/(bc)
where (remember?) the two letters ac stand for "a times c" and similarly for bc.
That is so because (c/c) = 1, no matter what the value of c is (except of course zero: "Thou Shalt not Divide by Zero") and multiplying anything by 1 does not change its value. In multiplying fractions, the rule is to multiply top with top, bottom with bottom, so we get
(a/b) (c/c) = (ac)/(bc)
As for dividing top and bottom by the same number d
(a/b) = [a/d]/ [b/d]
it follows at once from the preceding, if we choose the number c to equal 1/d.
### Working with Small Quantities
Some equations, identities or formulas contain small quantities, and these can be made much simpler and easier to use by sacrificing a little accuracy. In fact, some equations which have no simple solution at all (like Kepler's equation in section (12a)) can yield in this way an approximate solution, often good enough for most uses, or else open to further improvement.
Many such calculations make use of the following observation. When we derive squares, 3rd powers, 4th powers etc. of numbers larger than 1, the results are always bigger, while for numbers smaller than 1, the results are always smaller. For example:
power More than 1 Less than 1 number 10 0.1 square 100 0.01 3rd power 1000 0.001 4th power 10,000 0.0001
The above also holds for negative numbers, if one understands "bigger" and "smaller" to refer to the absolute value (the value without sign). For instance:
power More than 1 Less than 1 number – 10 – 0.1 square 100 0.01 3rd power – 1000 – 0.001 4th power 10,000 0.0001 5th power – 100,000 – 0.00001
Say z is a number much smaller than 1 (written z << 1, or for absolute values |z| << 1). Then by the identity near the end of of section M-4
(1 – z)(1 + z) = 1 – z2
Since z2 is much smaller than 1 or z, we can write, using the symbol ~ for "approximately equal"
(1 – z)(1 + z) ~ 1
and dividing both sides by (1 – z)
(1+z) ~ 1/(1– z)
(Many texts use the symbol ~ not alone but placed above an equal sign). For example (check with your calculator)
If z = 0.01, (1+z) = 1.01, (1– z) = 0.99,
then 1/(1– z) = 1/0.99 = 1.010101...
which is close enough to (1+z) for many purposes.
The basic rule is: one may neglect small quantities like z, z2, z3 etc. when they are added to (or subtracted from) something much bigger. One may not do so if they are just multiplied or divided, because then, if they are removed, nothing is left of the expression containing them.
Here z can be either positive or negative. If we write z = – y, where y is a small number of opposite sign, we get
(1– y) ~ 1/(1+y)
which is another useful result, valid for any small number. If that small number is again renamed and is now called z (not the same z as before, of course), we get
(1– z) ~ 1/(1+z)
which can also be obtained from the earlier equation
(1 – z)(1 + z) ~ 1
by dividing both sides by (1 + z).
In section (34a) where the distance to the Lagrangian point L1 is derived, it turns out necessary to approximate 1/[1– z]3. You start from (1+z) ~ 1/(1– z) and raise both sides to their 3rd powers:
(1+z)3 ~ 1/(1– z)3
Multiply out the left side:
(1 + z)3 = (1+z)(1+z)(1+z) = (1 + 3z + 3z2 + z3)
However, if z2 and z3 are much smaller than z, then dropping the terms containing them only increases the error slightly, leaving
1/(1– z)3 ~ 1 + 3z
The next section is optional.
### A step beyond: The Binomial Theorem
Formally 1/(1–z)3 is (1– z) to the power –3. It suggests that more generally, for small z and for any value of a
(1–z)a ~ 1 – az
1/(1– z)a ~ 1 + az
and similarly
(1 + z)a ~ 1 + az
1/(1 + z)a ~ 1 – az
(these are the same formula, for positive and negative z and a). That in fact is true, and a may be positive, negative or fractional. It is the consequence of a result first proved by Newton, his so-called binomial theorem. For those interested, that formula states
(1 + z)a = 1 + az + [a(a–1)/2] z2 + [a(a–1)(a–2)/6] z3 + ...
where the denominator of the fraction preceding any power zn is obtained by multiplying together the whole numbers up to n, giving the number generally denoted as n! and called "n factorial." For example
4! = (1)(2)(3)(4) = 24 5! = (1)(2)(3)(4)(5) = 120
If a is a positive whole number, the sequence a, (a-1), (a-2)... ultimately reaches zero, and the term where that first happens itself equals zero, as do all the ones that follow, all of whom contain a multiplier ("factor") zero. The series of powers of z then ends with za and we get formulas like the one derived earlier for a=3:
(1 + z)3 = (1 + 3z + 3z2 + z3)
Those cases of the binomial theorem were known before Newton. In fact Omar Khayyam (1044-1123), according to the site cited here, was considered the first to derive the theorem and its coefficients, for powers which are positive whole numbers (some claim it was known even before that). Khayyam--meaning "tentmaker," his "takhallus" or poetical name--is better known nowadays for his "Rubaiyat" collection of four-line poems, made famous after Edward Fitzgerald in 1839 gave them an inspiring translation into English.
Newton showed that the theorem also held for negative and fractional values of a, where the series on the right side goes on to higher and still higher powers of z, without end. If z is small these powers quickly become negligibly small, and it is no great error to leave them out and write (for z of either sign)
1/(1 + z)a ~ 1/(1 + az) ~ 1 – az
Note: Why not divide by zero? It does not work. There exists no number like 1/0 (except maybe infinity, which is not a regular number), and use of expressions like 0/0 can lead to contradictions such as 2 = 3.
Next Stop: #M-6 The Theorem of Pythagoras
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last edited 27 October 2016 |
# Trig Equations
A LevelAQAEdexcelOCR
## Trig Equations
Solving the basic trig equations is pretty easy, but what happens if we’ve got a function which has been stretched or translated?
Make sure you are happy with the following topics before continuing.
A LevelAQAEdexcelOCR
## Inspection
First, we need to find an initial solution.
So, for example, let’s say we want to find the values of $x$ when $\textcolor{red}{\tan x} = \textcolor{purple}{\dfrac{1}{\sqrt{3}}}$.
We want to draw the graph, and mark on a horizontal line where the condition is met, i.e. where $\textcolor{red}{\tan x} = \textcolor{purple}{\dfrac{1}{\sqrt{3}}}$.
By inspection, we know that the initial solution is at $30°$, and we can see it repeats at every $180°$.
We’ll denote this $30° ± 180°n$.
A LevelAQAEdexcelOCR
@mmerevise
## CAST Diagrams
The CAST diagram is a handy tool to show us where values of the standard trig functions are positive.
So, here’s the breakdown:
• For $0° < x < 90°$ALL of $\textcolor{blue}{\sin x}, \textcolor{limegreen}{\cos x}$ and $\textcolor{red}{\tan x}$ are positive
• For $90° < x < 180°$, ONLY $\textcolor{blue}{\sin x}$ is positive
• For $180° < x < 270°$, ONLY $\textcolor{red}{\tan x}$ is positive
• For $270° < x < 360°$, ONLY $\textcolor{limegreen}{\cos x}$ is positive
Think back to plotting the Unit Circle, in the Trig Basics section.
Let’s say we want to find the values of $x$ such that $\textcolor{red}{\tan x} = \textcolor{purple}{\dfrac{1}{\sqrt{3}}}$, as before.
We know that there is a solution when $x = 30°$.
First, we plot the point on the diagram, then find the corresponding angles:
From here, we can see that $\textcolor{red}{\tan x} = \textcolor{purple}{\dfrac{1}{\sqrt{3}}}$ when $x = 30° ± 360°n$ or $210° ± 360°n$, or, more concisely, $30° ± 180°n$.
We ignored the two solutions where $\textcolor{red}{\tan x}$ is not positive, i.e. $90° < x < 180°$ and $270° < x < 360°$.
A LevelAQAEdexcelOCR
## Dealing With Trig Transformations
Transformations pose a little bit of a problem… See, CAST diagrams become much harder to navigate now. You’re much better off sketching out the function and solving using the horizontal line technique.
So, let’s just begin with an example.
We have $f(x) = \cos 3x$. Find the values of $x \in \lbrack 0°, 360°\rbrack$* such that $f(x) = \textcolor{purple}{\dfrac{1}{2}}$.
* This is just set notation, meaning $0° \leq x \leq 360°$
Well, we have a series of solutions, but they’re not immediately obvious.
What we can do instead is plot the regular $\textcolor{limegreen}{\cos x}$ graph on an interval three times as large as the proposed interval, and divide our solutions there by $3$.
$\textcolor{limegreen}{\cos x} = \textcolor{purple}{\dfrac{1}{2}}$ has solutions $60°, 300°, 420°, 660°, 780°, 1020°$.
Therefore, $\cos 3x = \textcolor{purple}{\dfrac{1}{2}}$ has solutions $20°, 100°, 140°, 220°, 260°, 340°$.
A LevelAQAEdexcelOCR
## Trig Equations Example Questions
$x = \sin ^{-1} \left( \dfrac{\sqrt{3}}{2}\right) = -120°, -60°, 240°, 300°$
$\cos x = 0.2588$ gives $x = 75°$.
By the CAST diagram, we can see that we also have a solution where $x = 270° + 15° = 285°$.
For $f(x) = 1$, we want $\sin \dfrac{3x}{2} = \dfrac{1}{2}$.
If we look for values of $-3\pi \leq x \leq 3\pi$ where $\sin x = \dfrac{1}{2}$, and multiply our values of $x$ by a scale factor of $\dfrac{2}{3}$, we have our new set of solutions.
$\sin x = \dfrac{1}{2}$ occurs at $\dfrac{-11\pi}{6}, \dfrac{-7\pi}{6}, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{13\pi}{6}, \dfrac{17\pi}{6}$
Therefore the solution of $f(x) = 1$ is $x = \dfrac{-11\pi}{9}, \dfrac{-7\pi}{9}, \dfrac{\pi}{9}, \dfrac{5\pi}{9}, \dfrac{13\pi}{9}, \dfrac{17\pi}{9}$.
A Level
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Lines parallel to the same line
If two lines are parallel to the same line, we should prove that the two lines are also parallel to each other.
Let us look at the figure required to prove the statement given above.
Let the line $$AB$$ be parallel to line $$CD$$ and let line $$AB$$ also be parallel to $$EF$$.
We should prove that lines $$CD$$ and $$EF$$ are parallel.
$$\angle 1 = \angle 2 \longrightarrow (1)$$
[By corresponding angles axiom]
Similarly, $$\angle 1 = \angle 3 \longrightarrow (2)$$
On equating $$(1)$$ and $$(2)$$, we get:
$$\angle 2 = \angle 3$$
The angles $$\angle 2 = \angle 3$$ are corresponding angles.
Thus by the converse of corresponding angles axiom, we can prove that $$CD$$ is parallel to $$EF$$.
Theorem $$6$$: Lines which are parallel to the same line are parallel to each other.
Let us imagine that $$AB \parallel CD$$, and $$AB \parallel EF$$. $$PQ$$ intersects $$AB$$, $$CD$$, and $$EF$$.
We should prove that $$CD$$ $$\parallel EF$$.
Let us first consider the condition $$AB \parallel CD$$.
Here, $$\angle 1$$ is the corresponding angle of $$AB$$, and $$\angle 2$$ is the corresponding angle of $$CD$$.
Since $$AB \parallel CD$$, their corresponding angles are equal.
That is, $$\angle 1 = \angle 2 \longrightarrow (1)$$
Let us now consider the condition $$AB \parallel EF$$.
Even here, since $$AB$$ and $$EF$$ are parallel to each other, their corresponding angles are equal.
$$\angle 1$$ is the corresponding angle of $$AB$$, $$\angle 3$$ is the corresponding angle of $$EF$$.
Thus, $$\angle 1 = \angle 3 \longrightarrow (2)$$
On equating $$(1)$$ and $$(2)$$, we get:
$$\angle 2 = \angle 3$$
Here, $$\angle 2$$ is the corresponding angle of $$CD$$, and $$\angle 3$$ is the corresponding angle of $$EF$$.
Since the corresponding angles of $$CD$$ and $$EF$$ are equal, $$CD \parallel EF$$. |
## What is asked problem solving?
Problem – solving interview questions are questions that employers ask related to the candidate’s ability to gather data, analyze a problem, weigh the pros and cons and reach a logical decision.
## What are the steps in problem solving in mathematics?
Four Stages of Problem Solving
1. Understand and explore the problem;
2. Find a strategy;
3. Use the strategy to solve the problem;
4. Look back and reflect on the solution.
## What is problem solving in math?
problem solving includes examining the question to find the key ideas, choosing an appropriate strategy, doing the maths, finding the answer and then re-checking.
## What are the 7 steps in problem solving?
Effective problem solving is one of the key attributes that separate great leaders from average ones.
1. Step 1: Identify the Problem.
2. Step 2: Analyze the Problem.
3. Step 3: Describe the Problem.
4. Step 4: Look for Root Causes.
5. Step 5: Develop Alternate Solutions.
6. Step 6: Implement the Solution.
7. Step 7: Measure the Results.
You might be interested: Quick Answer: How To Use Venn Diagram In Math?
## What are the 10 problem solving strategies?
The 10 problem solving strategies include:
• Guess and check.
• Make a table or chart.
• Draw a picture or diagram.
• Act out the problem.
• Find a pattern or use a rule.
• Check for relevant or irrelevant information.
• Find smaller parts of a large problem.
• Make an organised list.
## What is a good example of problem solving?
Problem – solving starts with identifying the issue. For example, a teacher might need to figure out how to improve student performance on a writing proficiency test. To do that, the teacher will review the writing tests looking for areas of improvement.
## What is problem solving strategy?
A problem – solving strategy is a plan of action used to find a solution. Different strategies have different action plans associated with them ([link]). For example, a well-known strategy is trial and error. When using trial and error, you would continue to try different solutions until you solved your problem.
## How do you solve difficult problems?
How to Solve Tough Problems
1. Recognize the Problem. Jointly agree a problem exists. Describe the problem situation.
2. Determine the Root Cause. Localize where and when the problems occur.
3. Explore Possible Solutions. Discuss and analyze the root causes.
4. Select a Practical Solution. Define success criteria.
5. Implement the Solution. Invest time.
## What are the 4 steps to solving an equation?
We have 4 ways of solving one- step equations: Adding, Substracting, multiplication and division.
## What is problem solving in the classroom?
Problem – solving is the ability to identify and solve problems by applying appropriate skills systematically. Problem – solving is a process—an ongoing activity in which we take what we know to discover what we don’t know.
You might be interested: Question: How Math Is Used In Physics?
## What is the number sentence in problem solving?
Number sentences are simply the numerical expression of a word problem.
## How do you teach math problem solving skills?
The Problem Solving Steps
1. Step 1 – Understand the Problem.
2. Then, we were on to Step 2 – Make a Plan.
3. Step 3 – Solving the problem.
4. Finally, Step 4 – Check It.
5. Stop – Don’t rush with any solution; just take your time and look everything over.
6. Think – Take your time to think about the problem and solution.
## What are 3 key attributes of a good problem solver?
Effective problem solvers share ten common characteristics.
• They have an “ attitude ”!
• They re-define the problem.
• They have a system.
• They avoid the experience trap.
• They consider every position as though it were their own.
• They recognize conflict as often a prerequisite to solution.
• They listen to their intuition.
## What are the 5 steps to problem solving?
5-steps to Problem Solving
1. Define the problem. In understanding and communicating the problem effectively, we have to be clear about what the issue is.
2. Gather information. What were the circumstances?
3. Generate possible solutions. Work together to brainstorm on all possible solutions.
4. Evaluate ideas and then choose one.
5. Evaluate. |
Home | | Surveying II | | Surveying | Law of Weights
# Law of Weights
The weight of the arithmetic mean of the measurements of unit weight is equal to the number of obs
LAW OF WEIGHTS
From the method of least squares the following laws of weights are established:
(i) The weight of the arithmetic mean of the measurements of unit weight is equal to the number of observations.
For example, let an angle A be measured six times, the following being the values:
A Weight A Weight
30 o 20? 8' 1 30 o 20? 10' 1
30 o 20? 10' 1 30 o 20? 9' 1
30 o 20? 7' 1 30 o 20? 10' 1
Arithmetic mean
= 30 o 20? + 1/6 (8' + 10' + 7' + 10' + 9' + 10')
= 30 o 20? 9'.
Weight of arithmetic mean = number of observations = 6.
(2) The weight of the weighted arithmetic mean is equal to the sum of the individual weights.
For example, let an angle A be measured six times, the following being the values :
A Weight A Weight
30 o 20? 8' 2 30 o 20? 10' 3
30 o 20? 10' 3 30 o 20? 9' 4
30 o 20? 6' 2 30 o 20? 10' 2
Sum of weights = 2 + 3 + 2 + 3 + 4 + 2 =16
Arithmetic mean = 30 o 20? + 1/16 (8'X2 + 10' X3+ 7'X2 + 10'X3 + 9' X4+ 10'X2)
= 30 o 20? 9'.
Weight of arithmetic mean = 16.
(3) The weight of algebric sum of two or more quantities is equal to the reciprocals of the individual weights.
For Example angle A = 30 o 20? 8', Weight 2
B = 15 o 20? 8', Weight 3
Weight of A + B =
(4) If a quantity of given weight is multiplied by a factor, the weight of the result is obtained by dividing its given weight by the square of the factor.
(5) If a quantity of given weight is divided by a factor, the weight of the result is obtained by multiplying its given weight by the square of the factor.
(6) If a equation is multiplied by its own weight, the weight of the resulting equation is equal to the reciprocal of the weight of the equation.
(7) The weight of the equation remains unchanged, if all the signs of the equation are changed or if the equation is added or subtracted from a constant.
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
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LCM (739; 15) = ? Calculate the least common multiple, LCM, by two methods: 1) The prime factorization of the numbers and 2) The Euclidean algorithm
The least common multiple, LCM: the latest calculated
The LCM of 739 and 15 = ? May 27 06:54 UTC (GMT) The LCM of 464 and 300 = ? May 27 06:54 UTC (GMT) The LCM of 84 and 3 = ? May 27 06:54 UTC (GMT) The LCM of 33,020 and 231,189 = ? May 27 06:54 UTC (GMT) The LCM of 27 and 77 = ? May 27 06:54 UTC (GMT) The LCM of 252 and 360 = ? May 27 06:54 UTC (GMT) The LCM of 4,652 and 27,912 = ? May 27 06:54 UTC (GMT) The LCM of 6,088 and 228 = ? May 27 06:54 UTC (GMT) The LCM of 124 and 1,248 = ? May 27 06:54 UTC (GMT) The LCM of 45 and 90 = ? May 27 06:54 UTC (GMT) The LCM of 112 and 392 = ? May 27 06:54 UTC (GMT) The LCM of 125 and 9 = ? May 27 06:54 UTC (GMT) The LCM of 900 and 629 = ? May 27 06:54 UTC (GMT) The least common multiple, LCM: the list of all the operations
The least common multiple (lcm). What it is and how to calculate it.
• The number 60 is a common multiple of the numbers 6 and 15 because 60 is a multiple of 6 (60 = 6 × 10) and also a multiple of 15 (60 = 15 × 4).
• There are infinitely many common multiples of 6 and 15.
• If the number "v" is a multiple of the numbers "a" and "b", then all the multiples of "v" are also multiples of "a" and "b".
• The common multiples of 6 and 15 are the numbers 30, 60, 90, 120, and so on.
• Out of these, 30 is the smallest, 30 is the least common multiple (lcm) of 6 and 15.
• Note: The prime factorization of a number: finding the prime numbers that multiply together to give that number.
• If e = lcm (a, b), then the prime factorization of "e" must contain all the prime factors involved in the prime factorization of "a" and "b" taken by the highest power.
• Example:
• 40 = 23 × 5
• 36 = 22 × 32
• 126 = 2 × 32 × 7
• lcm (40, 36, 126) = 23 × 32 × 5 × 7 = 2,520
• Note: 23 = 2 × 2 × 2 = 8. We are saying that 2 was raised to the power of 3. Or, shorter, 2 to the power of 3. In this example 3 is the exponent and 2 is the base. The exponent indicates how many times the base is multiplied by itself. 23 is the power and 8 is the value of the power.
• Another example of calculating the least common multiple, lcm:
• 938 = 2 × 7 × 67
• 982 = 2 × 491
• 743 = is a prime number and cannot be broken down into other prime factors
• lcm (938, 982, 743) = 2 × 7 × 67 × 491 × 743 = 342,194,594
• If two or more numbers have no common factors (they are coprime), then their least common multiple is calculated by simply multiplying the numbers.
• Example:
• 6 = 2 × 3
• 35 = 5 × 7
• lcm (6, 35) = 2 × 3 × 5 × 7 = 6 × 35 = 210 |
# Mathematical writing guidelines
In section 5.3 from the book Book of Proof by Hammack (3rd edition, this link is to the author's website), the author outlines 12 mathematical writing guidelines to help the young mathematician with writing better proofs.
Those guidelines, with their examples are as follows:
1. Begin each sentence with a word, not a mathematical symbol:
Wrong: $$A$$ is a subset of $$B$$.
Correct: The set $$A$$ is a subset of $$B$$.
2. End each sentence with a period, even when the sentence ends with a mathematical symbol or expression:
Wrong: Euler proved that $$\sum_{k=1}^\infty\frac{1}{k^s}=\prod_{p\in P}\frac{1}{1-\frac{1}{p^s}}$$
Correct: Euler proved that $$\sum_{k=1}^\infty\frac{1}{k^s}=\prod_{p\in P}\frac{1}{1-\frac{1}{p^s}}$$.
3. Separate mathematical symbols and expressions with words:
Wrong: Because $$x^2-1=0$$, $$x=1$$ or $$x=-1$$.
Correct: Because $$x^2-1=0$$, it follows that $$x=1$$ or $$x=-1$$.
4. Avoid misuse of symbols:
Wrong: The empty set is a $$\subseteq$$ of every set.
Correct: The empty set is a subset of every set.
5. Avoid using unnecessary symbols:
Wrong: No set $$X$$ has negative cardinality.
Correct: No set has negative cardinality.
6. Use first person plural:
Use the words "we" and "us" rather than "I," "you" or "me."
7. Use the active voice:
Wrong: The value $$x=3$$ is obtained through division of both sides by $$5$$.
Correct: Dividing both sides by $$5$$, we get $$x=3$$.
8. Explain each new symbol:
Wrong: Since $$a\mid b$$, it follows that $$b=ac$$.
Correct: Since $$a\mid b$$, it follows that $$b=ac$$ for some integer $$c$$.
9. Watch out for "it":
Wrong: Since $$X\subseteq Y$$, and $$0<|X|$$, we see that it is not empty.
Correct: Since $$X\subseteq Y$$, and $$0<|X|$$, we see that $$Y$$ is not empty.
10. Since, because, as, for, so:
The following statements all mean that $$P$$ is true (or assumed to be true) and as a consequence $$Q$$ is true also:
• $$Q$$ since $$P$$
• $$Q$$ because $$P$$
• $$Q$$, as $$P$$
• $$Q$$, for $$P$$
• $$P$$, so $$Q$$
• Since $$P$$, $$Q$$
• Because $$P$$, $$Q$$
• As $$P$$, $$Q$$
11. Thus, hence, therefore, consequently:
These adverbs precede a statement that follows logically from previous sentences or clauses:
Wrong: Therefore $$2k+1$$.
Correct: Therefore $$a=2k+1$$.
12. Clarity is the gold standard of mathematical writing:
If you think breaking a rule makes your writing clearer, then break the rule.
Are there any other rules or personal experiences that lead to writing a better proof?
• Your last 3 examples for guideline 10 violate guideline 3, so wouldn't it be better to leave them out? And "$Q$, for $P$" entails using "for" to mean "because", which is a rather poetic usage which the reader won't expect. Jul 16, 2020 at 18:57
• I have voted to close this as being too opinion based. What makes for good mathematical writing is fairly personal, and may depend quite a lot on one's advisor or the editor of the journal where one has submitted work (for example, with respect to (6), I would argue that pronouns should be avoided entirely). Jul 29, 2020 at 14:14 |
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Lesson 4.4 Conditions of Equilibrium
# Lesson 4.4 Conditions of Equilibrium - Equilibrium of Rigid...
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Equilibrium of Rigid Bodies Equilibrium is a state of balance. In order that an object may be in equilibrium, all the forces acting on it must be balanced. F net = 0 This can be summarized as: F y = 0 F x = 0 F z = 0 Consider two equal and opposite forces F acting on the object on the right. These forces are balanced because But the object is not in equilibrium because these forces produce a net torque on the object. Another necessary condition for equilibrium is τ net = 0 F F τ = 0 1
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We call this a counterclockwise torque. We call this a clockwise torque. For convenience we take a counterclockwise torque as positive and a clockwise torque as negative. B C A Axis of rotation. This should be zero for rotational equilibrium. 2
Torque of a couple A couple is formed by two equal and opposite parallel forces on an object so that their points of applications are different We find the torque of these forces about an axis through A Now we take the torque of these two forces about B F 2 B C A F 1 Torque of a couple is the product of one of the forces and the perpendicular distance between the forces . = BA·F1 = AB·F 2 (F 1 = F 2 in magnitude) 3
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Solving problems using equilibrium conditions To solve problems using conditions of equilibrium, we use the following steps: 1. Identify the point or object in equilibrium 2. Draw vectors to represent all the forces acting on the object 3. Resolve the force vectors into x- and y-components 4. Mark all the distances 6. Find the torque of all the forces about any chosen axis 7. Use τ = 0 to form another equation. 8. Solve the equations for the unknown. 4
A uniform meter stick pivoted at its center has a mass 100 g suspended at the 25 cm position. (a) At what position should a 75.0 g mass be suspended to put the system in equilibrium? (b) What mass would have to be suspended at the 90 cm mark for the system to be in equilibrium? (a)
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Lesson 4.4 Conditions of Equilibrium - Equilibrium of Rigid...
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VIEWS: 12 PAGES: 14
• pg 1
``` Homework 3C1
• Define: Tessellation.
• Copy the Yellow box on
book.
• Finish: Worksheet 3-5
Drill 3C1
1. Solve for x
X+2 = 7
10 7
Drill 3C1
2. Copy rectangle RECT and
point P. Rotate the rectangle
Homework 3C1
• Define: Tessellation.
• Copy the Yellow box on
book.
• Finish: Worksheet 3-5
Drill 3C1 X+2 = 7
1. Solve for x 10 7
7(X+2) = 7(10)
7X+14 =70
S14 7X = 56
D7 X=8
Drill 3C1
2. Copy rectangle RECT and
point P. Rotate the rectangle 90°
C’
T’
P
E’
R’
Symmetry
Objective: The student
will find and locate the
line of and point of
symmetry of images.
Vocabulary: Symmetry
Symmetry - a figure has a point or .
line about which it maps onto itself.
Types of Symmetry – Rotational
Symmetry & Line Symmetry
Rotational Symmetry – a figure
rotates 1800 or less about a fixed
point to map onto itself.
Line Symmetry- a figure reflects
over a line to map onto itself.
Example of Symmetry
A’B’C’D’ is the reflected image of
ABCD over the line l .
D’ C’
This is the A B
line of l
reflection. A’ B’
D C
Notice that the A’B’C’D’ is
ABCD is the
orientation has the copy or
reversed original or
preimage. image.
Example of Symmetry
H’G’K’L’ is the image of HGKL rotated
KH’ J
G’
GKJH is the original
or preimage.
P
Notice that the
G J’ H
K’ orientation has
not reversed
G’K’J’H’ is the P is the point
copy or image. of rotation.
problems on
worksheet 3-5 of
Closing Problems
1. Which figure has both line and
rotational symmetry?
F. G.
J. H.
Closing Problems
2. How many lines of symmetry
does the figure below have? |
# Chord properties
Circles are among the most important shapes in Geometry. It is simply an enclosed curve that has various parts. We can compute for its circumference and its area using some formula for circles or we can use the different parts to solve for the circle’s dimension.
In this chapter we will discuss about these parts. In 10.1 we will be discussing the angles found in a circle. First, there’s the Central angle, which is named as such because its vertex is the center of the circle and the sides are both radii of the circle. The intercepted arc that corresponds to the central angle is called the minor arc and has the same measurement as the central angle. The remaining section of the circle is called the major arc which has a measure of 360 degrees minus the measurement of the minor arc. Second, there’s the inscribed angle which has it vertex along the any point of the circle.
In 10.2 we will be looking at another part of the circle which is the chord. Chords are simply lines made from connecting two points on the circle. The diameter is an example of a chord. There are different chord properties that we will be learning like any two congruent chords would mean that there are also two congruent central angles, and two congruent arcs. We are also going to look at the perpendicular bisector. This is a line that bisects a chord and passes through the center of the circle.
We are also going to look into tangents for the last part of the chapter. Tangents are any line that touches one point on the circle. This point on the circle is referred to as the point of tangency. We are also going to look at some properties of the tangent like if a tangent of a circle intersects with a radius of the circle, they make two right angles.
### Chord properties
In a circle, when a line passes through the center of the circle and is perpendicular to a chord, the line will bisect the chord. In other words, the perpendicular bisector of a chord always passes through the center of the circle. |
# What happens when you dot a vector with a unit vector?
## What happens when you dot a vector with a unit vector?
Since the projection of a vector on to itself leaves its magnitude unchanged, the dot product of any vector with itself is the square of that vector’s magnitude. In addition, since a vector has no projection perpendicular to itself, the dot product of any unit vector with any other is zero.
## What is the dot product of two vectors physics?
Algebraically, the dot product is the sum of the products of the corresponding entries of the two sequences of numbers. Geometrically, it is the product of the Euclidean magnitudes of the two vectors and the cosine of the angle between them. These definitions are equivalent when using Cartesian coordinates.
How do you do dot product in physics?
Property 4: The dot product of a vector to itself is the magnitude squared of the vector i.e. a.a = a.a cos 0 = a. Property 5: The dot product follows the distributive law also i.e. a. (b + c) = a.b + a.c.
What is the value of a vector into a vector?
We know that, cross(vector) product of two vectors is a third vector whose magnitude is given by the product of magnitude of given vectors multiplied by sin ratio of the smaller angle between them. In your case, given two vectors are the same, i.e., A and hence, they are equal in magnitude and angle between them is 0°.
### Is the cross product of two unit vectors a unit vector?
The cross product of two unit vectors is always a unit vector. The magnitude of sum of two unit vectors is always greater than the magnitude of their difference.
### What is dot product with example?
we calculate the dot product to be a⋅b=1(4)+2(−5)+3(6)=4−10+18=12. Since a⋅b is positive, we can infer from the geometric definition, that the vectors form an acute angle.
What exactly is the dot product?
The dot product essentially tells us how much of the force vector is applied in the direction of the motion vector. The dot product can also help us measure the angle formed by a pair of vectors and the position of a vector relative to the coordinate axes.
Why dot product is a scalar?
5 Answers. No, it doesn’t give another vector. It gives the product of the length of one vector by the length of the projection of the other. This is a scalar.
## How do you calculate the dot product?
Here are the steps to follow for this matrix dot product calculator: First, input the values for Vector a which are X1, Y1, and Z1. Then input the values for Vector b which are X2, Y2, and Z2. After inputting all of these values, the dot product solver automatically generates the values for the Dot Product and the Angle Between Vectors for you.
## How to compute dot product?
To find the dot product of two vectors in Excel, we can use the followings steps: 1. Enter the data . Enter the data values for each vector in their own columns. For example, enter the data values for… 2. Calculate the dot product. To calculate the dot product, we can use the Excel function
What is the formula for dot product?
Algebraically, the dot product is the sum of products of the vectors’ components. For three-component vectors, the dot product formula looks as follows: a·b = a₁ * b₁ + a₂ * b₂ + a₃ * b₃. In a space that has more than three dimensions, you simply need to add more terms to the summation.
What does the dot product represent?
The dot product is a scalar representation of two vectors, and it is used to find the angle between two vectors in any dimensional space. |
11:02 am Calculus
# Exploring Limit Calculus: A Beginner’s Guide
Limit Calculus – In the world of mathematics especially in calculus, the concept of limits plays an essential role. Limits are fundamental to understanding the behavior of functions and solving complex mathematical problems. In this article, we will delve into the world of limits, exploring what they are, why they matter, and how to work with them.
## Limit Calculus
Whether you’re a student dealing with calculus or just curious about this mathematical concept, we’ll break it down in simple terms.
### Definition of Limit Calculus
In calculus, a limit refers to the value a function approaches as it gets closer and closer to a specific point. This point is typically denoted as ‘c.’ We denote the limit of a function ‘f(x)’ as ‘L’ and write it as:
limx →c f(x) = L
We say that the limit of ‘f(x)’ as ‘x’ approaches ‘c’ is ‘L’ if, for every positive number ‘ε’ (epsilon), there exists a positive number ‘δ’ (delta) such that:
|f(x) – L| < ε whenever 0 < |x – c| < δ
Understanding limits is like having a key to unlock the secrets of calculus. It forms the foundation for various concepts within calculus, such as continuity, derivatives, and integrals. Limits help us analyze the behavior of functions, enabling us to tackle complex mathematical problems with ease.
### Evaluating Limits Algebraically
#### Basic Algebraic Techniques for Evaluating Limits
Algebraic techniques are powerful ways for evaluating limits. Some essential methods include factoring, simplifying, and expanding expressions to make them responsive to limit evaluation.
#### Applying Limit Laws to Simplify Complex Expressions
Limit laws, such as the limit sum rule and limit product rule, allow us to simplify limits by breaking them down into manageable parts. These rules make evaluating the limits of complex functions more straightforward.
#### Solving Indeterminate Forms Using Algebraic Manipulation
Indeterminate forms like 0/0 or ∞/∞ often arise when evaluating limits. Algebraic manipulation techniques can help us resolve these indeterminate forms and find the actual limit.
Techniques for Evaluating Limit Calculus
Direct Substitution Method
The direct substitution method is the simplest way to evaluate limits. It involves substituting the value ‘c’ directly into the function ‘f(x)’ and calculating the limit.
Factoring and Canceling Techniques
Factoring and canceling techniques come in handy when dealing with rational functions. We can simplify the expression by factoring and canceling common factors to make limit evaluation easier.
Rationalizing Techniques
Rationalizing the numerator or denominator can help simplify expressions involving radicals, making them more manageable when evaluating limits.
L’Hôpital’s Rule for Evaluating Limits of Indeterminate Forms
When you encounter indeterminate forms like 0/0 or ∞/∞ that resist simplification, L’Hôpital’s rule provides an elegant solution. It allows us to differentiate the numerator and denominator separately, potentially resolving the limit.
A limit finder is a handy tool for evaluating limits according to the above methods with steps to ease up manual calculations
### Examples For finding Limit Calculus
Example 1
If f(x) = 2x + 1. What happens to ‘f(x)’ as ‘x’ approaches 3?
Solution:
We can calculate the limit as follows:
Step 1: Write the function with the limit notation.
limx →3 (2x + 1)
Step 2: Calculation
Replace x by 3
= 2(3) + 1
= 7
So, the limit of ‘f(x)’ as ‘x’ approaches 3 is 7.
Example 2
g(x) = (x2 – 1) / (x – 1). As ‘x’ approaches to 1.
Solution:
Step 1: Write the function with the limit notation.
limx →1 [(x2 – 1) / (x – 1)]
If we apply a limit on the function; we will get a 0/0 form in the answer.
Step 2: Factorization
limx →1 [(x2 – 1) / (x – 1)]
Making the factors of x2 – 1
x2 – 1 = (x + 1) (x – 1)
Step 3: Calculation
limx →1 {(x + 1) (x – 1) / (x – 1)}
= limx →1 (x + 1)
Plugging in the limit
limx →1 (x + 1) = 1 + 1
limx →1 (x + 1) = 2
### Applications of Limits
#### Determining Continuity and Differentiability of Functions
Limits help us determine whether a function is continuous or differentiable at a specific point. A continuous function has no jumps or holes, while differentiability indicates the smoothness of a function.
#### Finding the Slope of a Curve at a Specific Point
By evaluating limits, we can find the slope of a curve at a particular point. This is crucial for understanding the behavior of functions in calculus.
#### Calculating Instantaneous Rates of Change
Limits enable us to calculate instantaneous rates of change, which are essential for various real-world applications, such as physics and economics.
Understanding the Behavior of Functions Near Certain Points
Limits provide insights into how functions behave as we approach specific points. This information is invaluable for analyzing functions and solving complex problems.
Takeaway
Limits are not just a mathematical concept to memorize; they are the cornerstone of calculus. They open doors to understanding the behavior of functions, solving intricate problems, and making sense of the world through mathematics.
By understanding the basics of limits, their formal definition, and various evaluation techniques, you are well prepared to embark on your journey through the fascinating realm of calculus.
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