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Math Central Quandaries & Queries
Question from a student: A steel pipe is taken to a 9ft wide corridor. At the end of the corridor there is a 90° turn, to a 6ft wide corridor. How long is the longest pipe than can be turned in this corner?
Hello.
Two important things are clear from the question: the height of the corridor is not stated, so it is not meant to be considered. This means the problem is a two-dimensional one. Clearly, using the third dimension of height will allow a longer pipe, but since the height isn't given, we use just the floor of the corridors. Second, the width of the pipe isn't given either. A very wide pipe (say 4 feet wide) would have to be shorter than a narrow pipe. Since the width of the pipe isn't given, we infer that width is not to be considered either.
Here's a sketch of various long pipes coming from one direction or the other. They are all jammed in place except the red one that can make it around the corner.
We know that the longest pipe has to squeeze against the inside corner of the junction of the corridors and we know exactly where that junction is, because we know the widths of both corridors. So we can represent any pipe that touches the two outside walls and the inside corner using a simple cartesian graph as below:
This forms a simple right angled triangle. The base lengths are c and d, the hypotenuse length is h (that's the length of the pipe) and we know the line passes through (9, 6).
Long pipes jam, so really we want the shortest hypotenuse we can find that works in this graph. That will be the longest pipe that can make it around the corner.
Pythagorus tells us that the large triangle has the relationship h2 = d2 + c2 .
By looking at the x-intercept and the point (9,6) we can determine the slope of the red line:
m = (0 - 6) / (c - 9) = -6 / (c - 9)
The equation of the red line is
y = mx + b
y = (-6 / (c - 9))x + b
but we also know that the line passes through (9, 6) so we can substitute that in for x and y:
6 = (-6 / (c - 9))(9) + b
b = 6 + 2 / (3c - 27)
but b is the y-intercept, which we earlier called d, so if we square this, we get
d2 = (6 + 2 / (3c - 27) )2
Now we can substitute this into the Pythagorean equation:
h2 = (6 + 2 / (3c - 27) )2 + c2
Taking the derivative with respect to c gives us:
When you set dh/dc to zero, you can find the value of c. Using that, you can find the value of d and h, the length of the pipe.
Hope this helps,
Stephen La Rocque.
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
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Answer: To find the percentage of a number between two numbers, divide one number with the other and then multiply the result by 100.
Also, How do you find a percentage between two numbers on a calculator?
If you want to know what percent A is of B, you simple divide A by B, then take that number and move the decimal place two spaces to the right. That’s your percentage! To use the calculator, enter two numbers to calculate the percentage the first is of the second by clicking Calculate Percentage.
Hereof, How do I find the percentage of two numbers without a calculator?
If you need to find a percentage of a number, here’s what you do – for example, to find 35% of 240:
1. Divide the number by 10 to find 10%. …
2. Multiply this number by how many tens are in the percentage you’re looking for – in this case, that’s 3, so you work out 30% to be 24 x 3 = 72.
Also to know What is the percentage discount between two numbers? First: work out the difference (increase) between the two numbers you are comparing. Then: divide the increase by the original number and multiply the answer by 100. % increase = Increase ÷ Original Number × 100. If your answer is a negative number, then this is a percentage decrease.
How do you find 80 percent of a number without a calculator?
Similarly, you can multiply the 1% answer by any number to find any percentage value. For example, to find 80% of 4,500, multiply the 45 by 80 to get 3,600.
## How do you increase a number by a percentage?
To increase a number by a percentage amount, multiply the original amount by 1+ the percent of increase. In the example shown, Product A is getting a 10 percent increase. So you first add 1 to the 10 percent, which gives you 110 percent. You then multiply the original price of 100 by 110 percent.
## What is the percentage?
In mathematics, a percentage is a number or ratio that represents a fraction of 100. It is often denoted by the symbol “%” or simply as “percent” or “pct.” For example, 35% is equivalent to the decimal 0.35, or the fraction.
## How do I find the percentage of two numbers in Excel?
Find the percentage of change between two numbers
1. Click any blank cell.
2. Type =(2500-2342)/2342, and then press RETURN . The result is 0.06746.
3. Select the cell that contains the result from step 2.
4. On the Home tab, click . The result is 6.75%, which is the percentage of increase in earnings.
## What number is 15% of 24?
What is 15 percent (calculated percentage %) of number 24? Answer: 3.6.
## What is 80 percent of a number?
How to find 80% of a number? Take the number and multiple it by 80. Then multiply that by . 01.
## What number is 80% of 50?
Answer: 80% of 50 is 40.
## How do you minus a percentage on a calculator?
To subtract any percentage from a number, simply multiply that number by the percentage you want to remain. In other words, multiply by 100 percent minus the percentage you want to subtract, in decimal form. To subtract 20 percent, multiply by 80 percent (0.8).
## How do you increase a number by a percentage on a calculator?
If you want to increase a number by a certain percentage, follow these steps:
1. Divide the number you wish to increase by 100 to find 1% of it.
2. Multiply 1% by your chosen percentage.
4. There you go, you have just added a percentage increase to a number!
## What is the formula to calculate percentage increase?
% increase = Increase ÷ Original Number × 100. If the answer is a negative number, that means the percentage change is a decrease.
## How do you take off a percentage?
Step 1: Remove the percent sign and add a couple of zeros after the decimal point. Step 3: Multiply Step 2 by the amount (in this case, 20): . 19 * 20 = 3.8.
## How do I calculate mean?
The mean, or average, is calculated by adding up the scores and dividing the total by the number of scores. Consider the following number set: 3, 4, 6, 6, 8, 9, 11.
## What is the formula for calculating a percentage in Excel?
Basic Excel percentage formula
1. Enter the formula =C2/B2 in cell D2, and copy it down to as many rows as you need.
2. Click the Percent Style button (Home tab > Number group) to display the resulting decimal fractions as percentages.
## What is 15% of an amount is 30?
Percentage Calculator: What is 15 percent of 30? = 4.5.
## What is the number whose 40% is 52?
40% of 130 is 52. 100% of 130 is 130, therefore 40 percent of 130 equals 52.
## What percent is 15 of 25 of a number?
15 is 60 percent of 25. 15 out of 25 can also be written as 15/25.
## What number is 80 percent of 100?
Percentage Calculator: What is 80. percent of 100? = 80.
## What number is 30% of 70?
Percentage Calculator: What is 30 percent of 70? = 21.
## What number is 80% of 60?
Percentage Calculator: What is 80. percent of 60? = 48.
## What number is 24 percent of 12?
What is 24 percent (calculated percentage %) of number 12? Answer: 2.88.
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# Math110 forum | Algebra homework help
This will be your opportunity to be the teacher. Click on “View Full Description and attachments” below for the directions and questions. Be sure to open the file that says “MATH110 Read This First” before you jump in!
Read the attached files. First read the one entitled “MATH110 Read this first,” and then open the file called “Systems of Equations Problems with Answers.”
Pick ONE of the problems that has not already been solved, and demonstrate its solution using either the substitution or elimination method.
For this forum, be sure to do the following in addition to reading the full description and the attachments from the forums main page:
-In the subject line, put the question number and a brief description of the question.
-Copy the problem statement into the body of the posting.
-Write two equations based on the question.
-State the method you are using and solve the system of equations.
-Answer the question in a complete sentence.
You do not need to response to classmates’ posts for this forum.
System of Equations Sample Post
At a dog park, there were a total of 18 dogs and people combined. There were a total of 50 legs on the dogs and people. How many dogs and people were at the park?
Let d = number of dogs at the park; then 4d = total number of legs on the dogs.
Let p = number of people at the park; then 2p = total number of legs on the people.
Equation 1: The total number of dogs and people at the park is 18: d + p = 18
Equation 2: The total number of legs at the park is 50: 4d + 2p = 50
Use substitution to solve the system of equations.
Solve Equation 1 for p by subtracting d from each side: p = 18 – d
Substitute 18 – d for p in Equation 2 and then solve for d:
4d + 2(18 – d) = 50
4d + 36 – 2d = 50 distribute on the left side
2d + 36 = 50 combine like terms on the left side
2d = 14 subtract 36 from each side
d = 7 divide both sides by 2
If d = 7, then p = 18 – 7 = 11.
Check number of legs: 4(7) + 2(11) = 28 + 22 = 50.
So there were seven dogs and eleven people at the park.
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# Lesson 9
Representing Subtraction
## 9.1: Equivalent Equations (5 minutes)
### Warm-up
The purpose of this warm-up is to refresh students' previous understanding about the relationship between addition and subtraction (MP7) so they can write related equations.
As students work, watch for any students who create a number line diagram to help them generate equations that express the same relationship a different way.
### Launch
Give students 1 minute of quiet work time. Remind students that each new equation must include only the numbers in the original equation.
### Student Facing
Consider the equation $$2+3=5$$. Here are some more equations, using the same numbers, that express the same relationship in a different way:
$$3 + 2 = 5$$
$$5 - 3 = 2$$
$$5 - 2 = 3$$
For each equation, write two more equations, using the same numbers, that express the same relationship in a different way.
1. $$9+ (\text- 1)= 8$$
2. $$\text- 11+ x= 7$$
### Anticipated Misconceptions
If students struggle to come up with other equations, encourage them to represent the relationship using a number line diagram, and then think about other operations they can use to show the same relationship with the same numbers.
### Activity Synthesis
Ask selected students to share their equations that express the same relationship a different way. If any students created a number line diagram to explain their thinking, display this for all to see to facilitate connections between addition equations and related subtraction equations. Every addition equation has related subtraction equations and every subtraction equation has related addition equations; these are the most important takeaways from this activity.
## 9.2: Subtraction with Number Lines (10 minutes)
### Activity
The purpose of this activity is to apply the representation students have used while adding signed numbers, as well as the relationship between addition and subtraction, to begin subtracting signed numbers. Students are given number line diagrams showing one addend and the sum. They are asked to figure out what the other addend would be. Students examine how these addition equations with missing addends can be written using subtraction by analyzing and critiquing the reasoning of others (MP3).
Monitor for students who are using a consistent structure to analyze the diagrams to generalize and write related addition and subtraction equations (MP8). A template for this work might look something like: $$\displaystyle a + {?} = b$$ $$\displaystyle b-a={?}$$
### Launch
It may be useful to remind students how they represented addition on a number line in previous lessons. In particular, it is helpful to keep in mind that the two addends in an addition equation are drawn "tip-to-tail." You might use any number line diagrams created in the previous activity as an illustration of this idea.
Ask students to complete the questions for the first diagram and pause for discussion. Then, give students quiet work time to complete the remaining problems, followed by whole-class discussion.
Engagement: Develop Effort and Persistence. Connect a new concept to one with which students have experienced success. For example, reference previous activities where students used a representation while adding signed numbers to provide an entry point for this activity.
Supports accessibility for: Social-emotional skills; Conceptual processing
Speaking, Listening: MLR8 Discussion Support. To support students in producing statements about Mai’s and Tyler’s equations, provide sentence frames such as: “I agree/disagree with Mai/Tyler because….”.
Design Principle(s): Support sense-making; Optimize output (for critiques)
### Student Facing
1. Here is an unfinished number line diagram that represents a sum of 8.
1. How long should the other arrow be?
2. For an equation that goes with this diagram, Mai writes $$3 + {?} = 8$$.
Tyler writes $$8 - 3 = {?}$$. Do you agree with either of them?
3. What is the unknown number? How do you know?
2. Here are two more unfinished diagrams that represent sums.
For each diagram:
1. What equation would Mai write if she used the same reasoning as before?
2. What equation would Tyler write if he used the same reasoning as before?
3. How long should the other arrow be?
4. What number would complete each equation? Be prepared to explain your reasoning.
3. Draw a number line diagram for $$(\text-8) - (\text-3) = {?}$$ What is the unknown number? How do you know?
### Anticipated Misconceptions
Some students may say they disagree with Tyler's equations for the number lines. Use fact families to help students see that subtraction equations are a valid way to represent problems involving finding a missing addend given a sum. It may help to remind them of the work they did in the warm-up.
### Activity Synthesis
The most important things for students to understand is that subtraction equations can be written as addition equations with a missing addend and number line diagrams can help students figure out what the missing addend is. Students need to be comfortable with this way of representing subtraction for the next activity.
Ask at least one student to share their missing addend for each problem. Ask students to share their reasoning until they come to an agreement. Display two related equations for all to see and use as a reference in the following activity. They might look something like this, or you might choose to use numbers in a specific example rather than letters in a general example. $$\displaystyle a + {?} = b$$ $$\displaystyle b-a={?}$$
### Activity
In this activity, students begin to see that subtracting a signed number is equivalent to adding its opposite. First, students match expressions and number line diagrams. Then they add and subtract numbers to see that subtracting a number is the same as adding its opposite (MP8).
Monitor for students who see and can articulate the pattern that adding a number is the same as subtracting its opposite.
### Launch
Arrange students in groups of 2. Give students 3 minutes of quiet work time, then have them check in with a partner. Have them continue to complete the activity, and follow with a whole-group discussion.
### Student Facing
1. Match each diagram to one of these expressions:
$$3 + 7$$
$$3 - 7$$
$$3 + (\text- 7)$$
$$3 - (\text- 7)$$
2. Which expressions in the first question have the same value? What do you notice?
3. Complete each of these tables. What do you notice?
expression value
$$8 + (\text- 8)$$
$$8 - 8$$
$$8 + (\text-5)$$
$$8 - 5$$
$$8 + (\text-12)$$
$$8 - 12$$
expression value
$$\text-5 + 5$$
$$\text-5 - (\text-5)$$
$$\text-5 + 9$$
$$\text-5 - (\text-9)$$
$$\text-5 + 2$$
$$\text-5 - (\text-2)$$
### Student Facing
#### Are you ready for more?
It is possible to make a new number system using only the numbers 0, 1, 2, and 3. We will write the symbols for adding and subtracting in this system like this: $$2 \oplus 1 = 3$$ and $$2\ominus 1 = 1$$. The table shows some of the sums.
$$\oplus$$ 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3
1. In this system, $$1 \oplus 2 = 3$$ and $$2 \oplus 3 = 1$$. How can you see that in the table?
2. What do you think $$3 \oplus 1$$ should be?
3. What about $$3\oplus 3$$?
4. What do you think $$3\ominus 1$$ should be?
5. What about $$2\ominus 3$$?
6. Can you think of any uses for this number system?
### Activity Synthesis
The most important takeaway is that subtracting a number gets the same answer as adding its opposite.
Select students to share what patterns they noticed. If no student mentions it, point out that subtracting a number is the same as adding its opposite. Ask students to help you list all of the pairs that show this.
Then write this expression: $$3-7$$. Ask how it could be written as a sum? $$3 + (\text-7)$$. What numbers are both of these expressions equal to?
$$3-7=\text-4$$
$$3+(\text-7)=\text-4$$
For students who are ready to explore how knowing how to solve a one-step equation involving addition or subtraction (from grade 6) helps us show that subtracting a number is the same as adding its opposite, continue.
This is also true when solving equations. Write this equation:
$$\displaystyle x = 3 - 7$$
Ask how it can be written as a sum. Record students' responses. If no student writes
$$\displaystyle x + 7 = 3$$
then write that. Then point out that we can add -7 to each side:
\begin{align} x + 7 + \text-7&= 3 + \text-7\\ x &= 3 + \text-7 \end{align}
There is nothing special about these numbers, because a number and its opposite always make a sum of 0. So subtracting a number is always the same as adding its opposite.
Writing, Conversing: MLR3 Clarify, Critique, Correct. Present an incorrect statement about adding and subtracting numbers that reflects a possible misunderstanding from the class. Display the following for all to see: “$$(\text-5) - 3$$ has the same value as $$5 + (\text-3)$$, because subtracting is the same as adding the opposite”. Invite students to discuss this argument with a partner. Ask, “Do you agree with the statement? Why or why not?” Invite students to clarify and then critique the reasoning, and to write an improved response. This will help students use the language of justification to critique the reasoning related to subtraction of signed numbers.
Design Principle(s): Maximize meta-awareness; Cultivate conversation
## Lesson Synthesis
### Lesson Synthesis
Main takeaways:
• You can think of subtraction as addition with a missing addend: What number do I need to add to get from $$b$$ to $$a$$?
• You can evaluate a subtraction expression by adding the opposite: $$a - b = a + (\text-b)$$. This works regardless of the sign for $$a$$ or $$b$$.
Discussion questions
• How could we rewrite the expression $$\text-5 - 3$$ using addition? ($$3 + {?} = \text-5$$, or more simply $$\text-5 + (\text-3)$$)
• Does this work for all numbers?
## Student Lesson Summary
### Student Facing
The equation $$7 - 5 = {?}$$ is equivalent to $${?} + 5= 7$$. The diagram illustrates the second equation.
Notice that the value of $$7 + (\text-5)$$ is 2.
We can solve the equation $${?} + 5= 7$$ by adding -5 to both sides. This shows that $$7 - 5= 7 + (\text- 5)$$
Likewise, $$3 - 5 = {?}$$ is equivalent to $${?} + 5= 3$$.
Notice that the value of $$3 + (\text-5)$$ is -2.
We can solve the equation $${?} + 5= 3$$ by adding -5 to both sides. This shows that $$3 - 5 = 3 + (\text- 5)$$
In general:
$$\displaystyle a - b = a + (\text- b)$$
If $$a - b = x$$, then $$x + b = a$$. We can add $$\text- b$$ to both sides of this second equation to get that $$x = a + (\text- b)$$
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# Calculating Limit of Function – A quotient of functions to infinity – Exercise 6579
### Exercise
Evaluate the following limit:
$$\lim _ { x \rightarrow \infty} \frac{8^x}{{(4+\frac{2}{x})}^{\frac{3x}{2}}}$$
$$\lim _ { x \rightarrow \infty} \frac{8^x}{{(4+\frac{2}{x})}^{\frac{3x}{2}}}=e^{-\frac{3}{4}}$$
Solution
First, we try to plug in $$x = \infty$$ and get
$$\frac{8^\infty}{{(4+\frac{2}{\infty})}^{\frac{3\infty}{2}}}$$
We got the phrase $$\frac{\infty}{\infty}$$ (=infinity divides by infinity). This is an indeterminate form, therefore we have to get out of this situation.
We will simplify the phrase using Powers and Roots Rules:
$$\lim _ { x \rightarrow \infty} \frac{8^x}{{(4+\frac{2}{x})}^{\frac{3x}{2}}}=$$
$$=\lim _ { x \rightarrow \infty} \frac{2^{3x}}{{(4(1+\frac{2}{4x}))}^{\frac{3x}{2}}}=$$
$$=\lim _ { x \rightarrow \infty} \frac{2^{3x}}{{(2^2)}^{\frac{3x}{2}}{(1+\frac{1}{2x})}^{\frac{3x}{2}}}=$$
$$=\lim _ { x \rightarrow \infty} \frac{2^{3x}}{2^{3x}{(1+\frac{1}{2x})}^{\frac{3x}{2}}}=$$
$$=\lim _ { x \rightarrow \infty} \frac{1}{{(1+\frac{1}{2x})}^{\frac{3x}{2}}}=$$
$$=\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{-\frac{3x}{2}}=$$
We will plug in infinity again and get
$$={(1+\frac{1}{2\infty})}^{-\frac{3\infty}{2}}=1^{-\infty}$$
We got the phrase $$1^{\infty}$$ (=tending to 1 in the power of infinity). This is an indeterminate form, therefore we have to get out of this situation.
We will use the known limit also called Euler’s Limit. At the base, we have an expression of the form:
$$1+\frac{1}{2x}$$
And the following holds:
$$\lim _ { x \rightarrow \infty} \frac{1}{2x}= 0$$
As required.
Therefore, we will multiply the power by the inverted expression and get
$$=\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{2x\cdot\frac{1}{2x}\cdot (-\frac{3x}{2})}=$$
Note: When adding an expression by multiplication, its inverted expression must also be multiplied so that the original expression does not change.
Now, according to Euler’s Limit, we get
$$=\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{2x}=e$$
All we have left to do is to calculate the limit on the expression that remains in the power:
$$\lim _ { x \rightarrow \infty}\frac{1}{2x}\cdot (-\frac{3x}{2})=$$
$$=\lim _ { x \rightarrow \infty}\frac{-3x}{4x}=$$
$$=-\frac{3}{4}$$
Therefore, in total we get
$$=\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{2x\cdot\frac{1}{2x}\cdot (-\frac{3x}{2})}=$$
$$=e^{-\frac{3}{4}}$$
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# Product Rule
In Calculus, the product rule is used to differentiate a function. When a given function is the product of two or more functions, the product rule is used. If the problems are a combination of any two or more functions, then their derivatives can be found using Product Rule. The derivative of a function h(x) will be denoted by D {h(x)} or h'(x).
Table of Contents:
## Product Rule Definition
The product rule is a general rule for the problems which come under the differentiation where one function is multiplied by another function. The derivative of the product of two differentiable functions is equal to the addition of the first function multiplied by the derivative of the second, and the second function multiplied by the derivative of the first function. The function may be exponential, logarithmic function, and so on.
## Product Rule Formula
If we have a function y = uv, where u and v are the functions of x. Then, by the use of the product rule, we can easily find out the derivative of y with respect to x, and can be written as:
(dy/dx) = u (dv/dx) + v (du/dx)
The above formula is called the product rule for derivatives or the product rule of differentiation.
In the first term, we have considered u as a constant and for the second term, v as a constant.
## Product Rule Proof
Product rule can be proved with the help of limits and by adding, subtracting the one same segment of the function mentioned below:
Let f(x) and g(x) be two functions and h be small increments in the function we get f(x + h) and g(x + h).
Let F(x) = f(x)g(x) and F(x + h) = f(x + h)g(x + h)
Then, the derivative of a function is
$$F'(x)= \lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}$$ $$F'(x)= \lim_{h\rightarrow 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}$$
By adding and subtracting f(x + h)g(x), we get
$$F'(x)= \lim_{h\rightarrow 0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}$$ $$F'(x)= \lim_{h\rightarrow 0}\frac{f(x+h)(g(x+h)-g(x))+g(x)(f(x+h)-f(x))}{h}$$ $$F'(x)= \lim_{h\rightarrow 0}f(x+h)\frac{g(x+h)-g(x)}{h}+g(x)\frac{f(x+h)-f(x)}{h}$$ $$F'(x)= \lim_{h\rightarrow 0}f(x+h) \lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}+\lim_{h\rightarrow 0}g(x)\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$
By using the definition of a derivative, we get
= f(x + 0) g’ (x) + g(x) f ‘(x)
F'(x) = f(x)g’ (x) + g(x)f ‘(x).
which is the derivative of two functions and is known as the product rule in derivatives.
## Product Rule for Different Functions
The product rule for different functions such as derivatives, exponents, logarithmic functions are given below:
Product Rule for Derivatives:
For any two functions, say f(x) and g(x), the product rule is D [f(x) g(x)] = f(x) D[g(x)] + g(x) D[f(x)]
d(uv)/dx = u(dv/dx)+ v(du/dx)
where u and v are two functions
Product Rule for Exponent:
If m and n are the natural numbers, then xn × xm = xn+m.
Product rule cannot be used to solve expression of exponent having a different base like 23* 54 and expressions like (xn)m. An expression like (xn)m can be solved only with the help of Power Rule of Exponents where (xn)m = xnm.
Product Rule for Logarithm:
For any positive real numbers A and B with the base a
where, a≠ 0, logaAB = logaA + loga B
Product Rule for Partial Derivatives:
If we have a function z = f(x,y) g(x,y) and we want to find out the partial derivative of z, then we use the following formula
$$Z_{x}=\frac{\partial z}{\partial x}= g(x,y)\frac{\partial f(x,y)}{\partial x}+f(x,y)\frac{\partial g(x,y)}{\partial x}$$ , and
$$Z_{y}=\frac{\partial z}{\partial y}= g(x,y)\frac{\partial f(x,y)}{\partial y}+f(x,y)\frac{\partial g(x,y)}{\partial y}$$
Zero Product Rule:
Zero product rule states, the two non zero numbers are only zero if one of them is zero. If a and b are two numbers then ab = 0 only either a = 0 or b = 0.
if (x-1)x = 0, either x – 1 = 0 or x = 0
It means that if x – 1 = 0, then x = 1
Values of x are 0 and 1. They are also called roots of the equation. Mainly used to find the roots of equations, and it works if one side of the equation is zero.
Triple Product Rule:
Triple product rule is a generalization of product rule. If f(x), g(x) and h(x) be three differentiable functions, then the product rule of differentiation can be applied for these three functions as:
D[f(x). g(x). h(x)] = {g(x). h(x)} * D[f(x)] + {f(x). h(x)} * D[g(x)] + {f(x). g(x)} * D[h(x)]
### Product Rule Example
Example 1:
Simplify the expression: y= x2 × x5
Solution:
Given: y= x2 × x5
We know that the product rule for the exponent is
xn × xm = xn+m.
By using the product rule, it can be written as:
y = x2 × x5 = x2+5
y = x7
Hence, the simplified form of the expression, y= x2 × x5 is x7.
Example 2:
Differentiate y = sin x cos x
Solution:
Given: y = sin x cos x
dy/dx = d(sinx cos x)/dx
While differentiating, it becomes
dy/dx = (sin x) [d(cos x)/dx] + (cos x) [d(sin x)/dx]
Differentiate the terms, dy/dx = sin x (-sin x) + cos x (cos x)
dy/dx = -sin2.x + cos2 x
dy/dx =cos2x – sin2x
By using identity,
dy/dx = cos 2x
Therefore, dy/dx = cos 2x
Stay tuned with BYJU’S – The Learning App and download the app to learn all the important Maths-related articles.
## Frequently Asked Questions – FAQs
### What is the product rule in math?
The product rule is a general rule for the problems which come under the differentiation where one function is multiplied by another function.
### What is the product rule in calculus?
In calculus, the product rule is used to differentiate a function. When a given function is the product of two or more functions, the product rule is used.
### What is the product formula?
The formula of product rule for two functions say f(x) and g(x), is given by:
D [f(x) g(x)] = f(x) D[g(x)] + g(x) D[f(x)]
### What is the product and quotient rule?
Product rule states that when two functions f(x) and g(x) are differentiable, then their product is also differentiable and is calculated using the formula,
(fg)'(x) = f(x) g'(x) + f'(x) g(x)
Quotient rule state that when two functions f(x) and g(x) are differentiable, then their quotient is also differentiable and is calculated using the formula,
(f/g)'(x) = [g(x) f'(x) – f(x) g'(x)]/g^2(x)
### How do you differentiate UV?
The formula to differentiate UV is:
d(UV)/dx = U(dV/dx)+ V(dU/dx)
### How do you use the product rule for 3 terms?
If f(x), g(x) and h(x) be three differentiable functions, then the product rule for these three functions can be written as:
D[f(x). g(x). h(x)] = {g(x). h(x)} * D[f(x)] + {f(x). h(x)} * D[g(x)] + {f(x). g(x)} * D[h(x)]
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# 1981 AHSME Problems/Problem 30
## Problem
If $a$, $b$, $c$, and $d$ are the solutions of the equation $x^4 - bx - 3 = 0$, then an equation whose solutions are $$\dfrac {a + b + c}{d^2}, \dfrac {a + b + d}{c^2}, \dfrac {a + c + d}{b^2}, \dfrac {b + c + d}{a^2}$$is
$\textbf{(A)}\ 3x^4 + bx + 1 = 0\qquad \textbf{(B)}\ 3x^4 - bx + 1 = 0\qquad \textbf{(C)}\ 3x^4 + bx^3 - 1 = 0\qquad \\\textbf{(D)}\ 3x^4 - bx^3 - 1 = 0\qquad \textbf{(E)}\ \text{none of these}$
## Solution 1
Using Vieta's formula, we know the sum of the roots is equal to the negative coefficient of the $x^3$ term. Since the coefficient is 0, $a+b+c+d=0$. Thus, $\frac{a+b+c}{d^2}$ can be rewritten as $\frac{-d}{d^2}=\frac{1}{-d}$. Similarly, the other three new roots can be written as $\frac{1}{-c}$, $\frac{1}{-b}$, and $\frac{1}{-a}$.
Now, we need to find a way to transform the function $f(x)=x^4-bx-3$ such that all the roots are its negative reciprocal. We can create this new function by taking the negative reciprocal of the argument. In other words, $f(\frac{1}{-x})$ satisfies this criteria.
The new equation, $f(\frac{1}{-x})=0$ has the required roots and can be simplified to $\frac{1}{x^4}+\frac{b}{x}-3=0$. Since this is not a polynomial, we can multiply both sides by $x^4$ to become $1+bx^3-3x^4=0$. After rearranging and multiplying by negative one, we arrive at $3x^4-bx^3-1$ so the answer is $\boxed{\textbf{(D)} 3x^4-bx^3-1}$
## Solution 2
As in solution 1, the roots of the new equation are $-\frac{1}{a}, -\frac{1}{b}, -\frac{1}{c},-\frac{1}{d}$. Furthermore, applying Vieta’s formula to the original equation yields $abcd=-3$ and $abc+abd+acd+bcd=-3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=b$. Therefore, the product of the zeros of the new equation is $\frac{1}{abcd}=-\frac{1}{3}$. This limits our choices to options C and D, and we need to look for the sum of the roots of the new equation. This sum equals to $-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})$ The function whose roots are the reciprocals of the original equation is $-3x^4-bx^3+1$ therefore $-(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})=-(-\frac{-b}{-3})=\frac{b}{3}$. The second term of the chosen equation $a_{2}$ should satisfy that $-\frac{a_{2}}{3}=\frac{b}{3}$, hence $a_{2}=-b$. The answer is $D$. (Option E looks ridiculous~)
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# Simplification Questions for Bank Exam with Solution
Updated on:
Here, I’m going to explain the most important simplification questions for bank exam with complete solution.
Now, it doesn’t mean that these questions are useful for only bank exams. You can also practice all these questions for any of the competitive exams.
Simplification is a very important chapter from the perspective of competitive exams.
Each competitive exam asks at least two questions of this chapter, so you must prepare for this.
Without wasting your valuable time let’s start the topic that is about simplification questions for bank exam.
## Basic formulas –
1. ( a + b )² = a²+b²+2ab
2. ( a – b )² = a²+b²-2ab
3. ( a + b )² = ( a – b )²+4ab
4. ( a – b )² = ( a + b )²-4ab
5. ( a²-b² ) = ( a + b ) ( a – b )
6. ( a + b )³ = a³+b³+3ab ( a + b )
7. ( a – b )³ = a³-b³-3ab ( a – b )
8. ( a³ – b³ ) = ( a – b ) ( a²+b²+ab )
9. ( a³ + b³ ) = ( a + b ) ( a²+b²-ab )
10. ( a + 1/a )² = a²+1/a²+2
11. ( a – 1/a )² = a²+1/a²-2
12. ( a – 1/a )² = ( a + 1/a )² – 4
13. ( a + 1/a )² = ( a – 1/a )² + 4
14. ( a – 1/a )³ = a³-1/a³-3(a-1/a)
15. ( a + 1/a )³ = a³+1/a³+3(a+1/a)
16. ( a + b + c ) ² = a²+b²+c²+2ab+2bc+2ca = a²+b²+c²+2( ab + bc + ca )
17. a³+b³+c³-3abc = ( a + b + c ) ( a²+b²+c²-ab-bc-ca ) = 1/2×(a + b + c ) [ ( a – b )²+( b – c )²+( c-a )² ]
18. If a+b+c =0, then a³+b³+c³-3abc = 0
19. BODMAS rule It stands for Bracket, Of, Division, Multiplication, Addition and Subtraction.
It’s mandatory to solve simplification questions in BODMAS order.
## Simplification Questions for Bank Exam with Solution –
Here are 21 the most important simplification questions for bank exam, so that you can practice valuable questions.
All these questions are related to previous years.
### Q1. Find the value of 15.2+5.8÷2.9×2-3.5×2÷0.5.
( a ) 4.9
( b ) 5.2
( c ) 3.7
( d ) 5.6
#### Solution –
15.2+5.8÷2.9×2-3.5×2÷0.5
=15.2+2×2-3.5×4
=15.2+4-14
=5.2 ( Ans )
### Q2. What is the value of 2.8+(5.2÷1.3×2)-6×3÷8+2 ?
( a ) 10.55
( b ) 8.96
( c ) 9.35
( d ) 10.75
#### Solution –
2.8+(5.2÷1.3×2)-6×3÷8+2
=2.8+(4×2)-18/8+2
=2.8+8-2.25+2
=10.55 ( Ans )
### Q3. Find the value of 2×3÷6×2÷(4+4×4÷16-4).
( a ) 3
( b ) 4
( c ) 2
( d ) 5
#### Solution –
2×3÷6×2÷(4+4×4÷16-4)
=2×1/2×2÷(4+4×1/4-4)
=2÷(4+1-4)
=2÷(5-4)
=2÷(1)
=2 ( Ans )
### Q4. Find the forth root of 24010000.
( a ) 60
( b ) 80
( c ) 90
( d ) 70
#### Solution –
24010000 ( given )
Square root
√24010000
=√2401×√10000
=49×100
=4900
Again square root
√4900
=√49×√100
=7×10
=70 ( Ans )
### Q5. What will be the value of (3.25×3.20-3.20×3.05)÷0.064 ?
( a ) 10
( b ) 1/10
( c ) 5
( d ) 10.2
#### Solution –
(3.25×3.20-3.20×3.05)÷0.064
=(10.4-9.76)÷0.064
=0.64÷0.064
=10 ( Ans )
### Q6. If 4x²=15²-9², find the value of x.
( a ) 9
( b ) 6
( c ) 12
( d ) 4
#### Solution –
4x²=15²-9² ( given )
4x²=(15+9)(15-9)
4x²=24×6
4x²=144
x²=144/4
x²=36
x=√36
x=6 ( Ans )
### Q7. If 2x-5y=5 and 2x-y=9, find the value of x-y.
( a ) 1
( b ) 2
( c ) 3
( d ) 4
#### Solution –
2x-5y=5…….(1)
2x-y=9……..(2)
Solving the equations
2x-5y-2x+y=5-9
-5y+y=-4
-4y=-4
y=1
Put the value of y in equation (1)
2x-5×1=5
2x-5=5
2x=10
x=5
Now, x-y=5-1=4
x=4 ( Ans )
( a ) 24
( b ) -24
( c ) 30
( d ) -32
∛-13824
=∛-24×-24×-24
=-24 ( Ans )
### Q9. If √x÷√441=0.02, find the value of x.
( a ) 1.764
( b ) 17.64
( c ) 0.1764
( d ) 176.4
#### Solution –
√x÷√441=0.02 ( given )
⇒√x/21=0.02
⇒√x/21=2/100
⇒√x=42/100
Squaring both sides
⇒x=(42)²/(100)²
⇒x=0.1764 ( Ans )
### Q10. Find the value of \frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}.
( a ) 6/30
( b ) 5/12
( c ) 5/39
( d ) 6/39
#### Solution –
\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}
=\frac{1}{3×5}+\frac{1}{5×7}+\frac{1}{7×9}+\frac{1}{9×11}+\frac{1}{11×13}
=\frac{1}{2}[\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}]
=\frac{1}{2}[\frac{1}{3}-\frac{1}{13}]
=\frac{1}{2}×\frac{10}{39}
=5/39 ( Ans )
### Q11. The value of 2√54-6\sqrt{2/3}-√96 is
( a ) 0
( b ) 1
( c ) 3
( d ) 5
#### Solution –
2√54-6\sqrt{2/3}-√96
=2\sqrt{9×6}-2×3\sqrt{2/3}\sqrt{16×6}
=6\sqrt{6}-2\sqrt{\frac{2}{3}×9}-4\sqrt{6}
=6\sqrt{6}-2\sqrt{6}-4\sqrt{6}
=0 ( Ans )
### Q12. 0.4\bar{7}+0.5\bar{03}-0.3\bar{9}×0.\bar{8}is equal to
( a ) 0.3\bar{65}
( b ) 0.4\bar{35}
( c ) 0.4\bar{65}
( d ) 0.6\bar{25}
#### Solution –
0.4\bar{7}+0.5\bar{03}-0.3\bar{9}×0.\bar{8}
=\frac{47-4}{90}+\frac{503-5}{990}-\frac{39-3}{90}×\frac{8}{9}
=\frac{43}{90}+\frac{498}{990}-\frac{36}{90}×\frac{8}{9}
=\frac{43}{90}-\frac{32}{90}+\frac{498}{990}
=\frac{11}{90}+\frac{498}{990}
=\frac{121+498}{990}
=\frac{619}{990}
=0.6\bar{25}
### Q13. \frac{(0.96)³-(0.1)³}{(0.96)²+0.096+(0.1)²} is equal to
( a ) 0.98
( b ) 0.86
( c ) 0.45
( d ) 0.69
#### Solution –
\frac{(0.96)³-(0.1)³}{(0.96)²+0.096+(0.1)²}
=\frac{(0.96-0.1)(0.96)²+0.96×0.1+(0.1)²}{(0.96)²+0.96×0.1+(0.1)²}
=0.96-0.1
=0.86 ( Ans )
### Q14. (100)^{\frac{1}{2}}×(0.001)^{\frac{1}{3}}-(0.0016)^{\frac{1}{4}}×(3)^{0}+(\frac{5}{4})^{-1}, find the value.
( a ) 2.5
( b ) 1.6
( c ) 1.9
( d ) 2.8
#### Solution –
(100)^{\frac{1}{2}}×(0.001)^{\frac{1}{3}}-(0.0016)^{\frac{1}{4}}×(3)^{0}+(\frac{5}{4})^{-1}
=(10)^{2×\frac{1}{2}}×(0.1)^{3×\frac{1}{3}}-(0.2)^{4×\frac{1}{4}}×1+4/5
=10×0.1-0.2+4/5
=1-2/10+4/5
=1-1/5+4/5
=\frac{5-1+4}{5}
=\frac{8}{5}
=1.6 ( Ans )
### 15. 111\tfrac{1}{2}+111\tfrac{1}{6}+111\tfrac{1}{12}+111\tfrac{1}{20}+111\tfrac{1}{30} is equal to
( a ) 555\tfrac{6}{5}
( b ) 111\tfrac{2}{9}
( c ) 555\tfrac{5}{6}
( d ) 111\tfrac{1}{3}
#### Solution –
111\tfrac{1}{2}+111\tfrac{1}{6}+111\tfrac{1}{12}+111\tfrac{1}{20}+111\tfrac{1}{30}
=5×111+\frac{30+10+5+3+2}{60}
=5×111+\frac{50}{60}
=555+\frac{5}{6}
=555\tfrac{5}{6} ( Ans )
### Q16. What will be the value of 48÷12×[\frac{9}{8}of\frac{4}{3}÷\frac{3}{4}of\frac{2}{3}] ?
( a ) 13
( b ) 18
( c ) 12
( d ) 16
#### Solution –
48÷12×[\frac{9}{8}of\frac{4}{3}÷\frac{3}{4}of\frac{2}{3}]
=4×[(\frac{9}{8}×\frac{4}{3})÷(\frac{3}{4}×\frac{2}{3})]
=4×[\frac{3}{2}÷\frac{1}{2}]
=4×\frac{3}{2}×\frac{2}{1}
=12 ( Ans )
### Q17. If M=\frac{3}{7}×\frac{5}{6}×\frac{2}{3}+\frac{1}{5}×\frac{3}{2} and N=\frac{2}{5}×\frac{5}{6} × 3 + \frac{3}{5}×\frac{2}{3}×\frac{5}{3}, find the value of M/N.
( a ) 305/113
( b ) 113/305
( c ) 113/350
( d ) 131/302
#### Solution –
M=\frac{3}{7}×\frac{5}{6}×\frac{2}{3}+\frac{1}{5}×\frac{3}{2} ( given )
=\frac{5}{21}+\frac{3}{10}
=\frac{50+63}{210}
M=113/210
N=\frac{2}{5}×\frac{5}{6} × 3 + \frac{3}{5}×\frac{2}{3}×\frac{5}{3} ( given )
=1+\frac{2}{3}
N=5/3
Now,
M/N=113×3/210×5
M/N=113/350 ( Ans )
### Q18. What will be the value of K in the equation 2.5+0.5-[1.6-{3.2-(3.2+2.1)÷K}]=0.65 ?
( a ) 3.5
( b ) 2.9
( c ) 1.34
( d ) 7
#### Solution –
2.5+0.5-[1.6-{3.2-(3.2+2.1)÷K}]=0.65
3.0-[1.6-{3.2-(5.3÷K)=0.65
3.0-[1.6-3.2+(5.3÷K)=0.65
3.0-1.6+3.2-(5.3÷K)=0.65
4.6-(5.3÷K)=0.65
4.6-0.65=(5.3÷K)
3.95=(5.3÷K)
3.95=5.3/K
K=5.3/3.95
K=1.34 ( Ans )
### 19. \frac{4-\sqrt{0.04}}{4+\sqrt{0.4}} is equal to
( a ) 2.36
( b ) 6.25
( c ) 1.01
( d ) 0.82
#### Solution –
\frac{4-\sqrt{0.04}}{4+\sqrt{0.4}}
=\frac{4-\sqrt{4/100}}{4+\sqrt{4/10}}
=\frac{4-\sqrt{4/100}}{4+\sqrt{4/10}}
=\frac{4-0.2}{4+0.632}
=\frac{3.8}{4.632}
=0.82 ( Ans )
### 20. Find the value of \frac{\sqrt{2}-1}{\sqrt{2}+1} if √2=1.414.
( a ) 0.172
( b ) 0.568
( c ) 0.432
( d ) 0.156
#### Solution –
\frac{\sqrt{2}-1}{\sqrt{2}+1}
Rationalizing it
=\frac{\sqrt{2}-1}{\sqrt{2}+1}×\frac{\sqrt{2}-1}{\sqrt{2}-1}
=\frac{2+1-2\sqrt{2}}{1}
=3-2×1.414
=3-2.828
=0.172 ( Ans )
### 21. Find the value of x in the equation \frac{50}{x}=\frac{x}{12\tfrac{1}{2}}.
( a ) 49
( b ) 25
( c ) 64
( d ) 81
#### Solution –
\frac{50}{x}=\frac{x}{12\tfrac{1}{2}}
\frac{50}{x}=\frac{x}{25/2}
\frac{50}{x}=\frac{2x}{25}
50×25=2x²
x²=625
x=25 ( Ans )
## Questions for practice –
1. Find the value of 3.8+(8.2÷4.1×2)-4×3÷1.2.
2. Find the value of (5+3÷5×5)÷(3÷3 of 6) of (4×4÷4 of 4+4÷4).
3. 2.8+(5.2÷1.3×2)-6×3÷8+2 is equal to
4. Simplify the equation (x+4)²+(x-2)².
5. What will be the value of x in the equation 0.\bar{3}+0.\bar{6}+0.\bar{7}+0.\bar{8}=x?
1. (-2.2)
2. 48/5
3. 10.55
4. (2x²+4x+20)
5. 8/3
Final words –
I hope that you’ve found here important simplification questions for bank exam.
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# Reverse distributive property simplification
• Oct 17th 2012, 04:10 PM
xXplosionZz
Reverse distributive property simplification
Ok, so i have a generally clear idea on what then distributive property is. I understand that if a number is outside of a pair of brackets then you multiply that number by all the entities in the brackets although I have no idea how to solve this question. Expand, using the distributive property. Simplify. (5 - 4 times the square root of 3)(-2 + the square root of 3)
(I am only grade 10, so please try to avoid using any complicated (higher grade) math short cuts without clearly describing them) The answer is -22 + 13 square root of 3.
I would greatly appreciate any answer displaying the steps on how to solve this, and what you did each step.(Happy)
• Oct 17th 2012, 05:42 PM
johnsomeone
Re: Reverse distributive property simplification
$(5 - 4 \sqrt{3})(-2 + \sqrt{3})$
$= (5 - 4 \sqrt{3})(-2) + (5 - 4 \sqrt{3})(\sqrt{3})$
$\text{ (That's using the distributive law: } (x)(-2 + \sqrt{3}) = (x)(-2) + (x)(\sqrt{3}).\text{ )}$
$= [ \ (5)(-2) + (- 4 \sqrt{3})(-2) \ ] + [ \ (5)(\sqrt{3}) + (- 4 \sqrt{3})(\sqrt{3}) \ ]$
$\text{ (That's using the distributive law twice again - once for each bracket [ ]. )}$
$= (-10) + (- 4)(-2)\sqrt{3} + (5 \sqrt{3}) + (- 4)(\sqrt{3})^2$
$= -10 + 8\sqrt{3} + 5\sqrt{3} - (4)(3)$
$= -10 + (8+5)(\sqrt{3}) - 12$
$\text{ (That's using the distributive law again: } 8x + 5x = (8 + 5)x \text{. )}$
$= -22 + (13)(\sqrt{3})$
$= -22 + 13 \sqrt{3}$
• Oct 17th 2012, 05:53 PM
xXplosionZz
Re: Reverse distributive property simplification
$= (5 - 4 \sqrt{3})(-2) + (5 - 4 \sqrt{3})(\sqrt{3})$
$\text{ (That's using the distributive law: } (x)(-2 + \sqrt{3}) = (x)(-2) + (x)(\sqrt{3}).\text{ )}$
I don't really understand this first part? how did you pull a -2 out of the first pair of parenthesis and a sqrt{3} out of the second pair?
• Oct 17th 2012, 06:10 PM
xXplosionZz
Re: Reverse distributive property simplification
I apologize as after a decently long pondering on how you have done that I realized you applied the brackets to each other, Would an easy was of saying this law be (x)(a+b) = (ax + bx) ? Thank you so much for you time and effort. :)
My brain sort of throws basic thought out of the window when so many applications of one law start to intermingle. Do you have any advice on how to to view these problems with greater ease in, say, a test situation?
• Oct 17th 2012, 06:58 PM
pickslides
Re: Reverse distributive property simplification
$(a+b)(c+d)= ac+ad+bc+bd$
• Oct 18th 2012, 06:46 AM
HallsofIvy
Re: Reverse distributive property simplification
Quote:
Originally Posted by xXplosionZz
I apologize as after a decently long pondering on how you have done that I realized you applied the brackets to each other, Would an easy was of saying this law be (x)(a+b) = (ax + bx) ? Thank you so much for you time and effort. :)
Yes, that is precisely the "distributive law" which you said you understood.
Perhaps you learned the distributive law as (a+ b)c= ac+ bc. Because multiplication of numbers is commutative, the left side is the same as c(a+ b) and the right side is the same as ca+ cb. So c(a+ b)= ca+ cb.
Was that what you meant by "reverse" distributive property? I had thought you mean going from ab+ ac to a(b+ c)!
My brain sort of throws basic thought out of the window when so many applications of one law start to intermingle. Do you have any advice on how to to view these problems with greater ease in, say, a test situation?[/QUOTE]
• Oct 18th 2012, 02:49 PM
xXplosionZz
Re: Reverse distributive property simplification
Ahhh yes just today I learned that We can use the Foil law. I apologize as I had convinced myself the only way to somehow perform the second step, was to factor. Thank you guys for clearly describing everything :)
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## 一元二次函数学习中遇到的问题及解决方法
The Problems and Solutions in the Study of Quadratic function
Abstract
Quadratic function is important and difficult in the junior high school mathematics. It is different that between the mathematics which students learned before and the quadratic function. Students need to master quadratic function of the nature of learning through junior high school, and to identify skillful drawing an image by a quadratic function and image reading hidden information, problem-solving experience with the image of a new way of thinking. Most students cannot have a good grasp of the nature of a quadratic function. According to the problems appeared in the practical teaching of quadratic function teaching, the students conducted a questionnaire survey and interviews of teachers, analyzed the problems for students to accept, causes of problems and solutions
Key words: quadratic function teaching management survey problems countermeasures
Abstract5
1 绪论7
2 二次函数的重点和难点7
3 本文对二次函数的主要研究内容8
4 对当前学生理解困难的调查8
4.1 调查目的8
4.2 调查对象与调查方法8
4.3 数据整理与分析9
5 教学对策11
5.1 帮助学生摆脱畏惧心理11
5.2培养学生数形结合的能力-11
5.3 增加学生对方程的了解11
5.4 重视基础知识及其形成过程12
5.5 督促同学之间互相促进12
5.6 培养学生总结和反思的能力13
6 应用实践13
致谢 16
1 绪论
2 二次函数的重点和难点
而二次函数是初中函数的重点和难点，它与学生之前所接触的数学学习都有所不同，学生要通过初中的学习掌握二次函数的性质、熟练辨认并绘制二次函数的图像并能通过图像读出隐含信息、体会用图像解题的新的思想方法.可以说，二次函数是初中数学的集大成者，初中数学的很大一部分学科都是为二次函数打下基础.学生要改变固有思维方式，打破常规才能学好二次函数，并且在不断的演练中体会数学思想，灵活的应用二次函数的特征[2].在这一角度说，这也是初中数学的一大难点. 一元二次函数学习中遇到的问题及解决方法:http://www.lwfree.cn/shuxue/20191027/41562.html
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# how to divide complex numbers
Conveniently, the imaginary parts cancel out, and -16i2 = -16(-1) = 16, so we have: This is very interesting; we multiplied two complex numbers, and the result was a real number! Explain how to divide two complex numbers. The imaginary part drops from the process because they cancel each other. Division - Dividing complex numbers is just as simpler as writing complex numbers in fraction form and then resolving them. This process is necessary because the imaginary part in the denominator is really a square root (of –1, remember? Let’s take a quick look at an example of both to remind us how they work. To divide complex numbers, we follow these steps: Find the complex conjugate of the denominator. We could do it the regular way by remembering that if we write 2i in standard form it's 0 + 2i, and its conjugate is 0 - 2i, so we multiply numerator and denominator by that. I can use conjugates to divide complex numbers. Show Step-by-step Solutions. 1. Since the denominator is 1 + i, its conjugate must be 1 - i. Learn how to multiply and divide complex numbers in this step by step video. To find the conjugate of a complex number all you have to do is change the sign between the two terms in the denominator. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator. The problem is already in the form that we want, that is, in fractional form. 3 $\begingroup$ @user1551 au contraire it is meant to be interpreted geometrically. Well, dividing complex numbers will take advantage of this trick. Mathematicians (that’s you) can add, subtract, and multiply complex numbers. First, we break it up into two fractions: /reference/mathematics/algebra/complex-numbers/multiplying-and-dividing. Send Gift Now But there's an easier way. Identities with complex numbers. Examples simplify and rationalize denominators with a negative root and with a negative root binomial. Remember that I spirit is equal to negative one. Your answer will be in terms of x and y. 4444 i^{4444}=4444 Give the gift of Numerade. Otherwise, check your browser settings to turn cookies off or discontinue using the site. In this section we will learn how to multiply and divide complex numbers, and in the process, we'll have to learn a technique for simplifying complex numbers we've divided. First let's look at multiplication. The beautiful Mandelbrot Set (pictured here) is based on Complex Numbers.. {'transcript': 'to divide complex numbers. Example 3: Find the quotient of the complex numbers below. Write a C++ program to subtract two complex numbers. Simplify: Possible Answers: Correct answer: Explanation: This problem can be solved in a way similar to other kinds of division problems (with binomials, for example). To divide complex numbers, you usually need to multiply by the complex conjugate of the denominator. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, that satisfies the equation i2 = −1. In other words, there's nothing difficult about dividing - it's the simplifying that takes some work. Dividing complex numbers review Our mission is to provide a free, world-class education to anyone, anywhere. The conjugate of the denominator - \,5 + 5i is - 5 - 5i. To divide complex numbers: Multiply both the numerator and the denominator by the conjugate of the denominator, FOIL the numerator and denominator separately, and then combine like terms. Write the problem in fractional form. Khan Academy is a 501(c)(3) nonprofit organization. You may need to learn or review the skill on how to multiply complex numbers because it will play an important role in dividing complex numbers. Solution It is a menu driven program in which a user will have to enter his/her choice to perform an operation and can perform operations as many times as required. Complex numbers, as any other numbers, can be added, subtracted, multiplied or divided, and then those expressions can be simplified. Simplify. How to Multiply and Divide Complex Numbers by Reza about 9 months ago in Articles Learn how to multiply and divide complex numbers in few simple steps using the following step-by-step guide. Division of Complex Numbers: Except for 0, all complex numbers z have a reciprocal z^(-1) = 1/z Quiz & Worksheet Goals. To play this quiz, please finish editing it. Explain how to divide two complex numbers. Write a C++ program to multiply two complex numbers. Let’s multiply the numerator and denominator by this conjugate, and simplify. So let's think about how we can do this. This is the currently selected item. Please click OK or SCROLL DOWN to use this site with cookies. Our mission is to provide a free, world-class education to anyone, anywhere. And in particular, when I divide this, I want to get another complex number. First, multiply by congregate of the denominator, then multiply, which will often require you to use the foil method and then simple. Dividing Complex Numbers. Multiply the top and bottom of the fraction by this conjugate and simplify. Common Core: HSN.CN.A.3 How to divide complex fractions? We're asked to divide. How To: Given two complex numbers, divide one by the other. Thus, the conjugate of 3 + 2i is 3 - 2i, and the conjugate of 5 - 7i is 5 + 7i. \frac{\pi}{i}=-\pi i. Sample Solution:-HTML Code: Here is an image made by zooming into the Mandelbrot set Multiply the numerator and the denominator by the conjugate of the denominator. Complex Numbers. Step 3: Simplify the powers of i, specifically remember that i 2 = –1. Practice: Divide complex numbers. You'll also have to know about complex conjugates and specific steps used to divide complex numbers. 2(2 - 7i) + 7i(2 - 7i) Scroll down the page for more examples and solutions. Technically, you can’t divide complex numbers — in the traditional sense. Multiplying Complex Numbers. Here are some examples! … Students can replay these lessons any time, any place, on any connected device. Pay for 5 months, gift an ENTIRE YEAR to someone special! You divide complex numbers by writing the division problem as a fraction and then multiplying the numerator and denominator by a conjugate. First, find the complex conjugate of the denominator, multiply the numerator and denominator by that conjugate and simplify. How do you use it to divide complex numbers? Show Step-by-step Solutions. Towards the end of the simplification, cancel the common factor of the numerator and denominator. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator. Want to master Microsoft Excel and take your work-from-home job prospects to the next level? This is the currently selected item. Let's look at an example. To divide complex numbers, you must multiply by the conjugate. Suppose I want to divide 1 + i by 2 - i. Perform all necessary simplifications to get the final answer. Write a C++ program to divide two complex numbers. \sqrt{-300}=-10 \sqrt{3} Give the gift of Numerade. At that step and combined white terms, Write your answer in a plus. Dividing Complex Numbers. Every complex number has a conjugate, which we obtain by switching the sign of the imaginary part. Use the distributive property to write this as, Now we need to remember that i2 = -1, so this becomes. I can find the moduli of complex numbers. Don’t forget to use the fact that {i^2} = - 1. Practice: Complex number conjugates. A complex number, then, is made of a real number and some multiple of i. 4 + 49 Example 4: Find the quotient of the complex numbers below. Complex number conjugates. Complex Number Calculator Calculator will divide, multiply, add and subtract any 2 complex numbers Dividing Complex Numbers. In other words, there's nothing difficult about dividing - it's the simplifying that takes some work. Sample Solution:-HTML Code: Write a JavaScript program to divide two complex numbers. Remember to change only the sign of the imaginary term to get the conjugate. This lesson explains how to use complex conjugates to divide complex numbers To divide complex numbers, write the problem in fraction form first. Sort by: Top Voted. In this expression, a is the real part and b is the imaginary part of the complex number. We have already learned how to divide complex numbers. {\display… Complex numbers are a combination of a real number with an imaginary one. The complex conjugate of the complex number z = x + yi is given by x − yi.It is denoted by either ¯ or z*. After having gone through the stuff given above, we hope that the students would have understood "How to Add Subtract Multiply and Divide Complex Numbers".Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here. Because of that, we can express them generally as a + bi, where a is the real part of the number and b is the imaginary part. Example Question #2 : How To Divide Complex Numbers. First let's look at multiplication. The division of w by z is based on multiplying numerator and denominator by the complex conjugate of the denominator: w / z = (a + ib) / (A + iB) double a = a.re; double b = a.im; double c = b.re; double d = b.im; Komplex resDiv = new Komplex(); // Computing c * c + d * d will overflow even in cases where the actual result of the division does not overflow. Write the division problem as a fraction. Because doing this will result in the denominator becoming a real number. Here's an example: Solution Pay for 5 months, gift an ENTIRE YEAR to someone special! From there, it will be easy to figure out what to do next. Just in case you forgot how to determine the conjugate of a given complex number, see the table below: Use this conjugate to multiply the numerator and denominator of the given problem then simplify. Multiply x + yi times its conjugate. Another step is to find the conjugate of the denominator. To see all my videos check out my channel page http://YouTube.com/MathMeeting Dividing Complex Numbers. We take advantage of these conjugates when we divide complex numbers. 2. Multiplying complex numbers is almost as easy as multiplying two binomials together. Complex Number Division Formula, what is a complex number, roots of complex numbers, magnitude of complex number, operations with complex numbers Multiplying by the conjugate in this problem is like multiplying by 1 And that division of two complex numbers, 1 2 z a bi z c di + = + (3 ) can be thought of as simply a process for eliminating the ifrom the denominator and writing the result as a new complex number u vi+. 1. 2. Question 1 Example 1: Divide the complex numbers below. How to divide two complex numbers in trigonometric form? Learn how to multiply and divide complex numbers in few simple steps using the following step-by-step guide. Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator. This video gives the formula for multiplication and division of two complex numbers that are in polar form. We explain Dividing Complex Numbers with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, that satisfies the equation i2 = −1. Write the division problem as a fraction. Let's look at an example. The following diagram shows how to divide complex numbers. These equations are harder to do than normal linear equations, but they'll provide a nice brain challenge for you to furbish your math skills for the next time your teacher pops you a pop quiz in class. Technically, you can’t divide complex numbers — in the traditional sense. In this process, the common factor is 5. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Explain how to divide two complex numbers. Multiplying complex numbers is almost as easy as multiplying two binomials together. Since our denominator is 1 + 2i, its conjugate is equal to 1 - 2i. Complex Numbers: Multiplying and Dividing in Polar Form, Ex 1. Dividing complex numbers. From there, it will be easy to figure out what to do next. It turns out that whenever we have a complex number x + yi, and we multiply it by x - yi, the imaginary parts cancel out, and the result is a real number. The conjugate of the complex number a + bi is a – […] This is how .NET's Complex class does it (adjusted for your variable and type names): public static Komplex div(Komplex a, Komplex b) { // Division : Smith's formula. The color shows how fast z 2 +c grows, and black means it stays within a certain range.. Determine the complex conjugate of the denominator. To divide the complex number which is in the form (a + ib)/(c + id) we have to multiply both numerator and denominator by the conjugate of the denominator. Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. Use the FOIL Method when multiplying the binomials. A Question and Answer session with Professor Puzzler about the math behind infection spread. Multiply the top and bottom of the fraction by this conjugate. Mathematics, 14.01.2021 01:00 ttandkk. Suppose I want to divide 1 + i by 2 - i. I write it as follows: To simplify a complex fraction, multiply both the numerator and the denominator of the fraction by the conjugate of the denominator. Complex numbers are a combination of a real number with an imaginary one. The first step is to write the original problem in fractional form. This one is a little different, because we're dividing by a pure imaginary number. Find the equivalent fraction with a non complex (that is: real) denominator. What Are the Steps to Divide Complex Numbers? Just in case you forgot how to determine the conjugate of a given complex number, see the table below: Why? 5 + 4i _____ This line is the divide sign. Example 1. B. I form and finally just reduce if you can.'} 4 - 14i + 14i - 49i2 Because of that, we can express them generally as a + bi , where a is the real part of the number and b … So I want to get some real number plus some imaginary number, so some multiple of i's. Since the denominator is - \,3 - i, its conjugate equals - \,3 + i. We have a fancy name for x - yi; we call it the conjugate of x + yi. Write the problem in fractional form. Simplify. In order to do this, we end up having to multiply the top and the bottom of the fraction by the complex conjugate of the denominator. This series on complex numbers will help you solve equations with the cute variable "i" with ease by multiplying by the conjugate. This quiz is incomplete! $\begingroup$ While multiplication/division of complex numbers can be interpreted geometrically, I don't think it is meant to be interpreted that way. How to divide complex numbers? From here, we just need to multiply the numerators together and the denominators as well. This makes the complex conjugate of a + bi, a – bi. Next lesson. Send Gift Now Dividing complex numbers review. But this is still not in a + bi form, so we need to split the fraction up: Multiply the numerator and the denominator by the conjugate of 3 - 4i: Now we multiply out the numerator and the denominator: (3 + 4i)(3 + 4i) = 3(3 + 4i) + 4i(3 + 4i) = 9 + 12i + 12i + 16i2 = -7 + 24i, (3 - 4i)(3 + 4i) = 3(3 + 4i) - 4i(3 + 4i) = 9 + 12i - 12i - 16i2 = 25. Pay for 5 months, gift an ENTIRE YEAR to someone special! 5 + 2 i 7 + 4 i. Solution how to divide complex numbers; Introduction to Imaginary Numbers An imaginary number bi has two parts: a real number, b, and an imaginary part, i, defined as i 2 = -1. The complex conjugate z¯,{\displaystyle {\bar {z}},} pronounced "z-bar," is simply the complex number with the sign of the imaginary part reversed. Follow along with this tutorial to see how to find that complex conjugate and multiply with it … You divide complex numbers by writing the division problem as a fraction and then multiplying the numerator and denominator by a conjugate. How to Divide Complex Numbers in Rectangular Form ? It is much easier than it sounds. Would you like to see another example where this happens? The sum of (3,4) and (5,8) complex numbers =(8,12) The subtraction of (3,4) and (5,8) complex numbers =(-2,-4) The multiplication of (3,4) and (5,8) complex numbers =(-17,44) The division of (3,4) and (5,8) complex numbers =(0.52809,-0.0449438) ← In this expression, a is the real part and b is the imaginary part of the complex number. And we're dividing six plus three i by seven minus 5i. 53. Explain how to divide two complex numbers. Step 1. Some sample complex numbers are 3+2i, 4-i, or 18+5i. We'll use this concept of conjugates when it comes to dividing and simplifying complex numbers. Give the gift of Numerade. Find the complex conjugate of the denominator. 3 - 2i Step by step guide to Multiplying and Dividing Complex Numbers Multiplying complex numbers: $$\color{blue}{(a+bi)+(c+di)=(ac-bd)+(ad+bc)i}$$ Divide complex numbers. I need help on this question. The following diagram shows how to divide complex numbers. It only takes a minute to sign up. 12 Questions Show answers. Pay for 5 months, gift an ENTIRE YEAR to someone special! You will observe later that the product of a complex number with its conjugate will always yield a real number. Our software turns any iPad or web browser into a recordable, interactive whiteboard, making it easy for teachers and experts to create engaging video lessons and share them on the web. That is, [ (a + ib)/(c + id) ] ⋅ [ (c - id) / (c - id) ] = [ (a + ib) (c - id) / (c + id) (c - id) ] Examples of Dividing Complex Numbers \sqrt[3]{-125}=5 i Give the gift of Numerade. Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. From there, it will be easy to figure out what to do next. We take this conjugate and use it as the common multiplier of both the numerator and denominator. Please help me answer it. To divide complex numbers, write the problem in fraction form first. Dividing Complex Numbers To divide complex numbers, write the problem in fraction form first. Step 2: Distribute (or FOIL) in both the numerator and denominator to remove the parenthesis. It is a plot of what happens when we take the simple equation z 2 +c (both complex numbers) and feed the result back into z time and time again.. To divide the complex number which is in the form (a + ib)/(c + id) we have to multiply both numerator and denominator by the conjugate of the denominator. Let us consider an example: In this situation, the question is not in a simplified form; thus, you must take the conjugate value of the denominator. $\endgroup$ – user1551 Jul 2 '13 at 6:40. Complex conjugates and dividing complex numbers. Imaginary numbers are applied to square roots of negative numbers, allowing them to be simplified in terms of i. An easy to use calculator that divides two complex numbers. : Step 3: Simplify the powers of i, specifically remember that i 2 = –1. When you’re dividing complex numbers, or numbers written in the form z = a plus b times i, write the 2 complex numbers as a fraction. Five. So in the previous example, we would multiply the numerator and denomator by the conjugate of 2 - i, which is 2 + i: Now we need to multiply out the numerator, and we need to multiply out the denominator: (1 + i)(2 + i) = 1(2 + i) + i(2 + i) = 2 + i +2i +i2 = 1 + 3i, (2 - i)(2 + i) = 2(2 + i) - i(2 + i) = 4 + 2i - 2i - i2 = 5. We use cookies to give you the best experience on our website. Step 2: Distribute (or FOIL) in both the numerator and denominator to remove the parenthesis. To divide the two complex numbers follow the steps: First, calculate the conjugate of the complex … Send Gift Now Find the complex conjugate of the denominator, also called the z-bar, by reversing the sign of the imaginary number, or i, in the denominator. Time-saving dividing complex numbers video that shows how to divide by a complex number or by i. To divide complex numbers: Multiply both the numerator and the denominator by the conjugate of the denominator, FOIL the numerator and denominator separately, and then combine like terms. Dividing complex numbers is actually just a matter of writing the two complex numbers in fraction form, and then simplifying it to standard form. Dividing complex numbers review. Practice: Divide complex numbers. How To: Given two complex numbers, divide one by the other. Let w and z be two complex numbers such that w = a + ib and z = A + iB. You need to apply special rules to simplify these expressions with complex numbers. Another step is to find the conjugate of the denominator. This unary operation on complex numbers cannot be expressed by applying only their basic operations addition, subtraction, multiplication and division.. Geometrically, ¯ is the "reflection" of z about the real axis. In this section we will learn how to multiply and divide complex numbers, and in the process, we'll have to learn a technique for simplifying complex numbers we've divided. Session with Professor Puzzler about the math behind infection spread we have a fancy name for -! ) approach from multiple teachers think about how we can do this allowing them to be simplified terms. 1 - i is 3 - 2i, and the denominators as well process. Terms, write the problem in fraction form first step-by-step guide well, dividing complex numbers between! S you ) can add, subtract, and simplify numbers in this step by video. We can do this one by the conjugate of the fraction by the conjugate of the.... Line is the real part and b is the real part and b is the real part and is... These lessons any time, any place, on any connected device minus.. Need help on this question combination of a real number and some multiple of i, its conjugate always! Simplifications to get the conjugate of 5 - 7i is 5 an ENTIRE to! You must multiply by the conjugate of 3 + 2i is 3 - 2i, its conjugate equals \,3. Equals - \,3 + i, its conjugate must be 1 - 2i imaginary numbers are applied square... Certain range two terms in the form that we want, that is, in fractional form how to complex. Them to be simplified in terms of i the product of a real number an. Divide how to divide complex numbers numbers in trigonometric form '13 at 6:40 quotient of the -... Multiple teachers explains how to divide by a conjugate, which we obtain by switching the sign of numerator. The sign of the fraction by this conjugate. ' later that product! Dividing and simplifying complex numbers is almost as easy as multiplying two binomials together video gives the original problem fractional. And denominator of the fraction by the complex numbers, write the original problem in form! It comes to dividing and simplifying complex numbers follow the steps on how to divide complex numbers, we these... Real ) denominator suppose i want to get the final answer technically, you must multiply the! Simplify the powers of i ib and z be two complex numbers — in the sense... Expressions with complex numbers, we follow these steps: first, we have already learned how divide! Dividing - it 's the simplifying that takes some work = - 1 process because they cancel other... From multiple teachers at any level and professionals in related fields \endgroup \$ – Jul. Off or discontinue using the following step-by-step guide comes to dividing and simplifying complex numbers binomials together [ 2 x! And how to divide complex numbers multiple of i click OK or SCROLL DOWN to use concept. Ex 1 the math behind infection spread place, on any connected.! 4I _____ this line is the imaginary part in the denominator the parenthesis 3+2i,,... Interpreted geometrically to turn cookies off or discontinue using the site a the! -300 } =-10 \sqrt { -300 } =-10 \sqrt { 3 } Give the gift of.. Conjugate will always yield a real number calculator that divides two complex numbers in few simple using. About the math behind infection spread to master Microsoft Excel and take your work-from-home job prospects to next. Dividing complex numbers, allowing them to be simplified in terms of i, specifically that... Anyone, anywhere: multiplying and dividing in Polar form, Ex 1 we just need to two. … Practice: divide complex numbers, allowing them to be interpreted geometrically in fractional form interpreted geometrically simplify powers! Fast z 2 +c grows, and multiply complex numbers by writing the division problem as fraction!, write the problem is like multiplying by the complex conjugate of 5 7i! Stack Exchange is a little different, because we 're dividing by a conjugate denominators. — in the denominator by a conjugate numbers to divide two complex numbers almost!, there 's nothing difficult about dividing - it 's the simplifying that some! Few simple steps using the site in other words, there 's how to divide complex numbers about!
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# Complete Guide to Biconditionals: Definitions and Usage
Welcome to the world of biconditionals and definitions in geometry! While we often encounter conditions and statements that hinge on a single premise, biconditionals offer a two-way road, linking two statements such that they depend on each other. In geometry, these are especially critical when crafting precise definitions. This guide will break down biconditionals and showcase their importance in geometric definitions. Let's dive straight in!
## Step-by-step Guide: Biconditionals and Definitions
Understanding Biconditionals
A biconditional is a compound statement formed by combining two conditional statements using the phrase “if and only if,” often abbreviated as “iff.” It’s denoted by the symbol $$\leftrightarrow$$. A biconditional statement is true when both the conditions have the same truth value, i.e., both are true or both are false.
Structure: $$p \leftrightarrow q$$ reads as “$$p$$ if and only if $$q$$.”
Biconditional Breakdown
When we say “$$p$$ if and only if $$q$$,” it encompasses two statements:
• $$p \rightarrow q$$: If p, then q.
• $$q \rightarrow p$$: If q, then p. For the biconditional to be true, both these conditions should hold true.
Biconditionals in Geometry & Definitions
In geometry, biconditionals play a pivotal role in creating clear-cut definitions. A definition in geometry often allows for both a forward and backward reading, which is precisely the nature of biconditionals.
Example: A figure is a square if and only if it has four equal sides and four right angles. Here, if a figure is a square, it must have these properties. Conversely, if a figure possesses these properties, it’s defined to be a square.
### Examples
Example 1:
Consider the statement: A shape is a circle if and only if all points on the shape are equidistant from a single point called the center.
Solution:
Forward reading: If a shape is a circle, then all its points are equidistant from a center.
Backward reading: If all points on a shape are equidistant from a center, then the shape is a circle.
Example 2:
Statement: An angle is a right angle if and only if it measures $$90^\circ$$.
Solution:
Forward: If an angle is a right angle, it measures $$90^\circ$$.
Backward: If an angle measures $$90^\circ$$, it is a right angle.
### Practice Questions:
1. For the definition “A line is perpendicular to another if and only if they form a $$90^\circ$$ angle,” write the forward and backward readings.
2. Explain why the statement “A figure is a rectangle if it has four right angles” is not a biconditional.
Answers:
1. Forward reading: If a line is perpendicular to another, they form a $$90^\circ$$ angle. Backward reading: If two lines form a $$90^\circ$$ angle, one line is perpendicular to the other.
2. The statement is only one way. It states a condition for a figure to be a rectangle but doesn’t clarify if having four right angles is the only criterion or if there are others, nor does it state the reverse (that if a figure has characteristics other than four right angles, it can’t be a rectangle). Thus, it’s not a biconditional.
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# A. Mathematical Notation
This is the appendix in the notes here.
## Lists and Sets.
### Sets.
A set is a list of numbers, letters, objects,.. or whatever you want really. We contain these within curly brackets $\{$ and $\}$ . Eg.
• The order does not matter in a set. For instance,
• We ofter refer to the items inside as “elements”.
• Sometimes we use dots “$\dots$” when it is clear what is happening next:
• We can use a colon “:“ to specify conditions on a set. We can read this as “such that”. Eg. numbers such that $x$ is positive
or numbers such that they are between $1$ and $10$ and even
The set of numbers greater than zero less than or equal to ten and even. Notice the comma is like an “and”.
### Set Notation.
A couple of pieces of notation.
• $\in$ – means “in“ or “belongs to”. E.g. two belongs to the numbers from 1 to 10:
• $\subseteq$ – means “subset”. E.g. the set of number 2,4,6,8,10 is a subset of the numbers from 1 to 10:
There are various other notations that I will introduce shortly.
### Special sets.
There are some commonly occuring sets with a special notation:
• $\mathbb N$ – the natural numbers, $\mathbb N = \{1, 2 , 3 ,... \}\, .$
• $\mathbb Z$ – the integers, $\mathbb Z = \{ ... , -2, -1, 0, 1, 2, ... \}\, .$
• $\mathbb Q$ – the rational numbers (aka. fractions), $\mathbb Q = \Big\{ \frac{a}{b} : a \in \mathbb Z, b\in \mathbb N \Big\}\, .$
• $\mathbb R$ – the real numbers, e.g. $\pi \in \mathbb R$
• $[a,b], (a,b)$ – numbers between $a$ and $b$, inclusive and exclusive.
Note that $\mathbb N \subseteq \mathbb Z \subseteq \mathbb Q \subseteq \mathbb R$.
### Ordered lists.
Sometimes we want to list elements where the order matters. We contain these with round brackets $($ and $)$. E.g.
(Note this is useful for co-ordinates for geometry but also when we can in what order a sequence of events occur in probability.)
• Here the order of elements in these lists does matter:
• Again we often use “$:$” to list the items in the list or specify the conditions. E.g. Here we list the probabilities for each outcome from two coin throws.
### Cardinality of a set.
The cardinality of a set is the number of elements in that set. We use brackets $|$ and $|$ to denote the cardinality. Eg.
## Products and Sums.
### Sums.
We use the symbol $\sum$ for sums over a specified range:
Notice sums do not need to be finite. Notice we sum over a range of values in a set. (This is useful in probability.)
### Products.
Normally at school “$\times$” is used to mean multiplication. However, people also often use “$\cdot$“. I.e. We use the symbol $\prod$ for products of a range over values. E.g.
Notice that here do products over sets. (This is useful in probability.)
Recall that for mathematical symbols we often omit the product symbol altogether. E.g. for $x=3$ and $y=5$,
### Cartesian Products.
We can do products for sets. That is where we create a set consisting of the order pairs from two or more sets.
Notice the cardinality a product set is the product of the sizes of the sets:
This is why it makes sense to think of it as a product.
## Functions.
A function is something that takes an element from one set and gives you an element from another. E.g. $f(x) = x^2$ or $f(\theta) = e^{i\theta}$. We write $f: \mathcal D \rightarrow \mathcal R$ where $\mathcal D$ is the domain, the set of elements to which we apply the function, and $\mathcal R$ is the range, the set where the function takes its values. In probability we work with the function $\mathbb P : \Omega \rightarrow [0,1]$, i.e. for each outcome in our probability space we assign a probability which is a number between zero and one.
## Equals.
We use the equals sign when two things are the same. I.e. $2 = 6/3$. We also use add a colon “:=” when we are defining something. I.e. the natural numbers are defined by (We could use $=$ here, but, by using $:=$, it makes it more explicit that we are defining a new piece of notation.)
## Limits.
We will occasionally write statements like There is a formal mathematical definition for this, which we do not get into. But it should be reasonably clear that what the above statement is staying is that as $n$ gets very close to $1$ then $1- \frac{1}{n}$ gets closer and closer to $0$.
Further the following statement holds: The statement says that as $n$ gets larger and larger, the expression $(1-x/n)^n$ gets closer and closer to $e^{-x}$.
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# How do you use cross products to solve 2/9=c/27?
Apr 16, 2018
Just arrange the equation
#### Explanation:
$\frac{2}{9} = \frac{c}{27}$
$2 \times 27 = c \times 9$
$\frac{2 \times 27}{9} = c$
$2 \times 3 = c$
$6 = c$
Apr 16, 2018
c = 6
#### Explanation:
First we need to cross multiply.
$2 \times 27 = 9 \times c$
Which gives us...
$54 = 9 c$
To make the next step easy, we will flip the equation
$9 c = 54$
Now we can find the value of c
$c = \frac{54}{9}$
$c = 6$
Apr 16, 2018
$c = 6$
#### Explanation:
$\text{given equal fractions then we can solve using "color(blue)"cross-products}$
$\text{that is "a/b=c/drArrbc=adlarrcolor(blue)"cross-products}$
$\text{applying this to "2/9=c/27" gives}$
$9 c = 2 \times 27$
$\Rightarrow 9 c = 54$
$\text{divide both sides by 9}$
$\frac{\cancel{9} c}{\cancel{9}} = \frac{54}{9}$
$\Rightarrow c = 6$
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# Rational Exponent
In rational exponent there are positive rational exponent and negative rational exponent.
### Positive Rational Exponent:
We know that 2³ = 8. It can also be expressed as 8$$^{\frac{1}{3}}$$ = 2.
In general if x and yare non-zero rational numbers and m is a positive integer such that xᵐ = y then we can also express it as y$$^{\frac{1}{m}}$$ = x but we can write y$$^{\frac{1}{m}}$$ = $$\sqrt[m]{y}$$ and is called mᵗʰ root of y.
For example, y$$^{\frac{1}{2}}$$ = $$\sqrt[2]{y}$$, y$$^{\frac{1}{3}}$$ = ∛y, y$$^{\frac{1}{4}}$$ = ∜y, etc. If x is a positive rational number then for a positive ration exponent p/q we have x₀can be defined in two equivalent form.
x$$^{\frac{p}{q}}$$ = $$(x^{p})^{\frac{1}{q}}$$ = $$\sqrt[q]{x^{p}}$$ is read as qᵗʰ root of xᵖ
x$$^{\frac{p}{q}}$$ = $$(x^{\frac{1}{q}})^{p}$$ = $$(\sqrt[q]{x})^{p}$$ is read as pᵗʰ power of qᵗʰ root of x
For example:
1. Find (125)$$^{\frac{2}{3}}$$
Solution:
(125)$$^{\frac{2}{3}}$$
125 can be expressed as 5 × 5 × 5 = 5³
So, we have (125)2/3 = (53)2/3 = 53 × 2/3 = 52 = 25
2. Find (8/27)4/3
Solution:
(8/27)4/3
8 = 23 and 27 = 33
So, we have (8/27)4/3 = (23/33)4/3
= [(2/3) 3]4/3
= (2/3)3 × 4/3
= (2/3) 4
= 2/3 × 2/3 × 2/3 × 2/3
= 16/81
3. Find 91/2
Solution:
91/2
= √(2&9)
= [(3)2]1/2
= (3)2 × 1/2
= 3
4. Find 1251/3
Solution:
1251/3
= ∛125
= [(5) 3]1/3
= (5) 3 × 1/3
= 5
### Negative Rational Exponent:
We already learnt that if x is a non-zero rational number and m is any positive integer then x-m = 1/xm = (1/x)m, i.e., x-m is the reciprocal of xm.
Same rule exists of rational exponents.
If p/q is a positive rational number and x > 0 is a rational number
Then x-p/q = 1/xp/q = (1/x) p/q, i.e., x-p/q is the reciprocal of xp/q
If x = a/b, then (a/b)-p/q = (b/a)p/q
For example:
1. Find 9-1/2
Solution:
9-1/2
= 1/91/2
= (1/9)1/2
= [(1/3)2]1/2
= (1/3)2 × 1/2
= 1/3
2. Find (27/125)-4/3
Solution:
(27/125)-4/3
= (125/27)4/3
= (53/33)4/3
= [(5/3) 3]4/3
= (5/3)3 × 4/3
= (5/3)4
= (5 × 5 × 5 × 5)/(3 × 3 × 3 × 3)
= 625/81
Exponents
Exponents
Laws of Exponents
Rational Exponent
Integral Exponents of a Rational Numbers
Solved Examples on Exponents
Practice Test on Exponents
Exponents - Worksheets
Worksheet on Exponents
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# 12.5: Mutually Inclusive Events
Difficulty Level: At Grade Created by: CK-12
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Have you ever tried to calculate probability using a number cube? Take a look at this situation.
Andy tosses a number cube.
What is the probability that the number that lands up will be both even and less than 6?
In this Concept, you will learn how to calculate the probability of overlapping events. Pay attention and we will revisit this dilemma at the end of the Concept.
### Guidance
Now that you have an idea what a disjoint event is, you can begin to think about all sorts of outcomes. You may think that all events are disjoint, but this is not the case. There are some events that impact each other or that are overlapping. They have one or more outcomes in common. Think about this question.
Are all events disjoint events? Not at all, consider the next problem.
For a single spin, are events R(red)\begin{align*}R (\text{red})\end{align*} and T(top)\begin{align*}T (\text{top})\end{align*} disjoint events?
Step 1: Make a list of the outcomes.
R\begin{align*}R\end{align*} outcomes: red-top, red bottom
T\begin{align*}T\end{align*} outcomes: red-top, blue-top
Step 2: Compare the list. The two events DO have an outcome in common–red-top. So:
R\begin{align*}R\end{align*} and T\begin{align*}T\end{align*} are NOT disjoint events.
When events have outcomes in common, they are said to be overlapping events. So:
R\begin{align*}R\end{align*} and T\begin{align*}T\end{align*} are overlapping events.
How are they overlapping?
Notice that the events have more than one thing in common. They have color in common, but they also have the words “top” or “bottom” in common too. Therefore, the events are overlapping events.
Let’s look at another scenario.
For a single toss of a number cube, are events Smaller than 6 and Greater than 4 disjoint events or overlapping events?
Step 1: Make a list of the outcomes.
Smaller outcomes 1, 2, 3, 4, 5, 6
Greater outcomes: 4, 5, 6
Step 2: The two events have 1 outcome in common = 6.
S\begin{align*}S\end{align*} and G4\begin{align*}G4\end{align*} are overlapping events.
#### Example A
For a single spin, are G(green)\begin{align*}G (\text{green})\end{align*} and T(top)\begin{align*}T (\text{top})\end{align*} disjoint events or overlapping events?
Solution: Green and top are disjoint events because they do not have any outcomes in common.
#### Example B
A number cube is tossed. What is the probability that it will be a two and an even number?
Solution:16\begin{align*}\frac{1}{6}\end{align*}
#### Example C
A number cube is tossed. What is the probability that it will be an even number and a number greater than four?
Solution:16\begin{align*}\frac{1}{6}\end{align*}
Here is the original problem once again.
Andy tosses a number cube.
What is the probability that the number that lands up will be both even and less than 6?
Let's start by listing out the possible total outcomes.
1,2,3,4,5,6\begin{align*}1, 2, 3, 4, 5, 6\end{align*}
Now let's list out the even numbers.
2,4,6\begin{align*}2, 4, 6\end{align*}
Next we can list the numbers less than 6.
1,2,3,4,5\begin{align*}1, 2, 3, 4, 5\end{align*}
There are two numbers that overlap.
26=13\begin{align*}\frac{2}{6} = \frac{1}{3}\end{align*}
This is our probability.
### Vocabulary
Here are the vocabulary words in this Concept.
Disjoint Events
events that are not connected. One outcome does not affect the other.
Overlapping Events
events that have outcomes in common.
Complementary Events
One of two events must occur then the two are complementary events. We can subtract one event from 1 to get the other event.
### Guided Practice
Here is one for you to try on your own.
A number cube is tossed. What is the probability that the number that lands up will be both odd and greater than 3?
Step 1: List the odd outcomes, outcomes greater than 3, and total outcomes. Mark the overlapping outcomes (if they exist).
odd outcomes: 1, 3, 5
> 3 outcomes: 4, 5, 6
total outcomes: 1, 2, 3, 4, 5, 6
Step 2: Find the ratio of favorable outcomes to total outcomes.
Notice that there is only one overlapping outcome. This is the favorable outcome.
P(odd and >3)=favorable outcomestotal outcomes=16\begin{align*}P (\text{odd and} \ >3) = \frac{favorable \ outcomes}{total \ outcomes}=\frac{1}{6}\end{align*}
Notice that we can find the ratio by combining the overlapping outcomes.
### Video Review
Here is a video for review.
### Practice
Directions: Write the probability of each overlapping event occurring.
Eight colored scarves are put into a bag. They are red, yellow, blue, green, purple, orange, brown and black.
1. What is the probability of pulling out a red scarf?
2. What is the probability of pulling out a primary color?
3. What is the probability of pulling out a primary color and is either red or blue?
4. What is the probability of pulling out a primary color and is either blue or green?
5. What is the probability of pulling out a complementary color?
6. What is the probability of pulling out a complementary color and is either green or purple?
A number cube numbered 1 - 12 is rolled.
7. What is the probability of rolling an even number?
8. What is the probability of rolling an odd number?
9. What is the probability of rolling an even number or a number less than four?
10. What is the probability of rolling an even number or a number greater than 10?
11. What is the probability of rolling an odd number or a number greater than 5?
12. What is the probability or rolling an odd number or a one?
13. What is the probability of rolling an odd number or a number less than 8?
14. What is the probability of rolling an even number or a number less than 11?
15. What is the probability of rolling an odd number or a number greater than 9?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Addition Principle If events A and B are mutually inclusive, then P(A or B) = P(A) + P(B) – P(A and B)
complement A mutually exclusive pair of events are complements to each other. For example: If the desired outcome is heads on a flipped coin, the complement is tails.
Disjoint Events Disjoint or mutually exclusive events cannot both occur in a single trial of a given experiment.
Mutually Exclusive Events Mutually exclusive events have no common outcomes.
Mutually Inclusive Events Mutually inclusive events can occur at the same time.
Overlapping Events Overlapping events are events that have outcomes in common.
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# PROPERTIES OF PARALLELOGRAM
A parallelogram is a quadrilateral with both pairs of opposite sides parallel. When we mark diagrams of quadrilaterals, use matching arrowheads to indicate which sides are parallel.
For example, in the diagram shown below,
AB || CD
## Properties of Parallelogram
Property 1 :
If a quadrilateral is a parallelogram, then its opposite sides are congruent.
It has been illustrated in the diagram shown below.
In the diagram above,
AB ≅ DC
Property 2 :
If a quadrilateral is a parallelogram, then its opposite angles are congruent.
It has been illustrated in the diagram shown below.
In the diagram above,
m∠A ≅ m∠C
m∠B ≅ m∠D
Property 3 :
If a quadrilateral is a parallelogram, then its consecutive angles are supplementary.
It has been illustrated in the diagram shown below.
In the diagram above,
m∠A + m∠B = 180°
m∠B + m∠C = 180°
m∠C + m∠D = 180°
m∠A + m∠D = 180°
Property 4 :
If a quadrilateral is a parallelogram, then its diagonals bisect each other.
It has been illustrated in the diagram shown below.
In the diagram above,
AM ≅ CM
BM ≅ DM
## Practice Problems
Problem 1 :
Find the lengths of SR and SK in the parallelogram shown below. Explain your reasoning.
Solution :
Finding the length of SR :
By Theorem, opposite sides of a parallelogram are congruent.
So, we have
SR = PQ
From the diagram, PQ = 5.
SR = 5
Finding the length of SK :
By Theorem, diagonals of a parallelogram bisect each other.
So, we have
SK = QK
From the diagram, QK = 3.
SK = 3
Problem 2 :
Find the measures of ∠C and ∠B in the parallelogram ABCD.
Solution :
Finding the measure of ∠C :
By Theorem, opposite angles of a parallelogram are congruent.
So, we have
m∠C = m∠A
From the diagram, m∠A = 70°.
m∠C = 70°
Finding the measure of ∠B :
By Theorem, consecutive angles of a parallelogram are supplementary.
So, we have
m∠A + m∠B = 180°
From the diagram, m∠A = 70°.
70° + m∠B = 180°
Subtract 70° from both sides.
m∠B = 110°
Problem 3 :
Find the value of x in the parallelogram ABCD shown below.
Solution :
By Theorem, consecutive angles of a parallelogram are supplementary.
So, we have
m∠D + m∠C = 180°
Substitute m∠D = 3x° and m∠C = 120°.
3x° + 120° = 180°
Subtract 120° from both sides.
3x° = 60°
3x = 60
Divide both sides by 3.
x = 20
Problem 4 :
In the diagram shown below, ABCD and AEFG are parallelograms. Prove that m∠1 ≅ m∠3.
Solution :
Plan : Show that both angles are congruent to m∠2. Then use the Transitive Property of Congruence.
Statements ABCD is a parallelogram AEFG is a parallelogramaaaaa m∠1 ≅ m∠2 aaa aaaaa m∠2 ≅ m∠3 aaam∠1 ≅ m∠2 Reasonsaaaaaaaaa Given aaaaaaa aaaaaaaaaaaaaaaaaaaaaaaOpposite angles of a parallelogram are congruent. Transitive property of congruence.
Problem 5 :
In the parallelogram ABCD shown below, prove that WX ≅ YZ and WZ ≅ XY.
Statements aaaaaa WXYZ is a aaaaa aaaa parallelogram aaaaaaaaaa Draw XZ aaaaaa aaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaaWX || YZ, WZ || XY aa m∠WXZ ≅ m∠YZX a aa m∠WZX ≅ m∠YXZ aaaaaa ZX ≅ ZX aaaaa aaaaaaaaaaaaaaaaaaaaaΔWZX ≅ ΔYXZ WX ≅ YZ, WZ ≅ XY Reasonsaaaaaaaaaa Given aaaaaaaaa aaaaaaaaaaaaaaaaaaaaaaaaaaThrough any two points there exists exactly one a aaaaaaaaa line.Definition of parallelogramAlternate Interior Angles TheoremReflexive Property of CongruenceASA Congruence PostulateCorresponding parts of congruent triangles are congruent
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## Presentation on theme: "QUADRATIC EQUATIONS AND FUNCTIONS"— Presentation transcript:
CHAPTER 7 QUADRATIC EQUATIONS AND FUNCTIONS
7-1 Completing the Square
Completing the Square Transform the equation so that the constant term c is alone on the right side. If a, the coefficient of the second-degree term, is not equal to 1, then divide both sides by a. Add the square of half the coefficient of the first-degree term, (b/2a)2, to both sides (Completing the square)
Factor the left side as the square of a binomial.
Completing the Square Factor the left side as the square of a binomial. Complete the solution using the fact that (x + q)2 = r is equivalent to x + q = ±r
Solve: a2 – 5a + 3= 0 Move the 3 to the other side a = 1
Complete the square, add (5/2)2 Factor Solve
Solve: x2 – 6x - 3= 0 Move the 3 to the other side a = 1
Complete the square, add (3)2 Factor Solve
Solve: 2y2 + 2y + 5 = 0 move the 5 to the other side
divide both sides by 2 (a 1) Add (1/2)2 to both sides Factor Solve
Solve: 7x2 – 8x + 3 = 0 move the 3 to the other side
divide both sides by 7 (a 1) Add (4/7)2 to both sides Factor Solve
The Quadratic Formula The solutions of the quadratic equation ax2 + bx + c = 0 (a 0) are given by the formula:
Solve 3x2 + x – 1 = 0 5y2 = 6y – 3 2x2 – 3x + 7 = 0
7-3 The Discriminant
Discriminant The discriminant is used to determine the nature of the roots of a quadratic equation and is equal to: D = b2 – 4ac
Discriminant Cases If D is positive, then the roots are real and unequal. If D is zero, then the roots are real and equal (double root) . negative, the roots are imaginary.
Find the Discriminant x2 + 6x – 2 = 0 3x2 – 4x√3 + 4 = 0 x2 – 6x + 10 = 0
Discriminant The discriminant also show you whether a quadratic equation with integral coefficients has rational roots. D = b2 – 4ac
Test for Rational Roots
If a quadratic equation has integral coefficients and its discriminant is a perfect square, then the equation has rational roots.
Test for Rational Roots
If the quadratic equation can be transformed into an equivalent equation that meets this test, then it has rational roots.
Find the Determinant and Identify the Nature of the Roots
3x2 - 7x + 5 = 0 2x2 - 13x + 15 = 0 x2 + 6x + 10 = 0
a[f(x)]2 + b[(f(x)] + c = 0 where a 0 and f(x) is some function of x. It is helpful to replace f(x) with a single variable.
Example (3x – 2)2 – 5(3x -2) – 6 = 0 Let z = 3x – 2, then z2 – 5z – 6 = 0 Solve for z and then solve for x
(x + 2)2 – 5(x + 2) – 14 = 0 (3x + 4)2 + 6(3x + 4) – 16 = 0 x4 + 7x2 – 18 = 0
7-5 Graphing y – k = a(x- h)2
Parabola Parabola is the set of all points in the plane equidistant from a given line and a given point not on the line. Parabolas have an axis of symmetry (mirror image) either the x-axis or the y-axis. and
Parabola The point where the parabola crosses it axis is the vertex. The graph is a smooth curve.
Parabola The graph of an equation having the form y – k = a(x - h)2 has a vertex at (h, k) and its axis is the line x = h.
Graph y = x2 Use the form y – k = a(x – h)2 k= 0, h = 0, so the vertex is (0,0) and x = 0
Table of Values 1 -1 2 4 -2
Graph y = ½x2 Use the form y – k = a(x – h)2 k= 0, h = 0, so the vertex is (0,0) and x = 0
Graph y = -½x2 Use the form y – k = a(x – h)2 k= 0, h = 0, so the vertex is (0,0) and x = 0
Graph The graph of y = ax2 opens upward if a> 0 and downward if a< 0. The larger the absolute value of a is, the “narrower” the graph.
Graph y = ½(x-3)2 Use the form y – k = a(x – h)2 k= 0, h = 3, so the vertex is (3,0) and x = 3
Graph To graph y = a(x – h)2, slide the graph of y = ax2 horizontally h units. If h > 0, slide it to the right; if h < 0, slide it to the left. The graph has vertex (h, 0) and its axis is the line x = h.
Graph y – 3 = ½x2 Use the form y – k = a(x – h)2 k= 3, h = 0, so the vertex is (0,3) and x = 0
Graph y + 3 = ½x2 Use the form y – k = a(x – h)2 k= -3, h = 0, so the vertex is (0,-3) and x = 0
Graph To graph y – k = ax2, slide the graph of y = ax2 vertically k units. If k > 0, slide it upward; if k < 0, slide it downward. The graph has vertex (0, k) and its axis is the line x = 0.
Completed square form:
Quadratic Functions A function that can be written in either of two forms. General form: f(x) = ax2 + bx + c Completed square form: a(x-h)2 + k
Graph f(x) = 2(x – 3)2 + 1 y = 2(x – 3)2 + 1
Graph It’s a parabola with vertex (3,1) and axis x = 3
Graph f(x) = 3x2 – 6x + 1 y = 3x2 – 6x + 1
Graph y = 3x2 – 6x + 1 Rewrite the equation in the form : y – k = a(x – h)2 y -1 = 3(x2 – 2x) y – = 3(x2 – 2x + 1) y + 2 = 3(x-1)2
Quadratic Functions Let f(x) = ax2 + bx + c, a0 If a < 0, f has a maximum value. If a > 0, f has a minimum value and
Quadratic Functions The graph of f is a parabola. This maximum or minimum value of f is the y-coordinate when x = - b/2a, at the vertex of the graph.
Quadratic Functions Let f(x) = ax2 + bx + c, a0 This maximum or minimum value of f is the y-coordinate when x = - b/2a, at the vertex of the graph.
Example y = 1/2x2 + 3x – 7/4 Find the maximum or minimum value of f. Find the vertex of the graph of f
Example y = 1/2x2 + 3x – 7/4 a = ½, a > 0, then f has a minimum value. Minimum occurs when x = -b/2a x = -3/2*1/2 = -3
Example y = 1/2x2 + 3x – 7/4 y = ½(-3)2 + 3(-3) – 7/4 y = -25/4 So the minimum value of f is -25/4 and the vertex is (-3, -25/4)
7-7 Writing Quadratic Equations and Functions
and Theorem A quadratic equation with roots r1 and r2 is
x2 – (r1 + r2)x + r1r2 = 0 or a[x2 – (r1 + r2)x + r1r2 ] = 0 and
Theorem The equation just given is equivalent to a[x2 – (sum of roots)x + product of roots ] = 0
Example Find a quadratic equation with roots (2 + i)/3 and (2 – i)/3
Sum of roots = (2 + i)/3 + (2 – i)/3 = 4/3 Product of roots = (2 + i)/3 (2 – i)/3= 5/9 x2 – 4/3x + 5/9 = 9x2 -12x + 5
Theorem If r1 and r2 are the roots of a quadratic equation
ax2 + bx + c =0, then r1 + r2 = sum of roots = -b/a and r1r2 = product of roots = c/a
Example Find the roots of 2x2 + 9x + 5 = 0 r1 = -9 + 41 r2 = -9 - 41
Check 4 4 = -18/4 = - b/a r1 · r2 = -9 + 41 · -9 - 41 4 4
= -18/4 = - b/a r1 · r2 = -9 + 41 · -9 - 41 = 5/2 = c/a
END
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# Odd Man Out
#### Video Lesson on Odd Man Out
In math odd man out is one of the important topics in competitive exams for different types of employment test in which math logical question are being asked. Students can improve their aptitude skill by practising the questions and answers on odd man out.
In this kind of topic, 4 or more options may be given and you must point out the least related amongst all. Importance of topic
### Tips and Tricks
1. Classification: Try to differentiate among the given options. All the options except for one can be related to each other and will be falling under same group.
Example: Pick the odd one out
a. Tiger
b. Cow
c. Lion
d. Wolf
Explanation: Tiger, Lion and Wolf all are carnivores whereas Cow is an herbivore.
2. Go through all the options and try to relate the options by their common "Characteristics" or "Purpose".
Example: Pick the odd one out
b. Shovel
c. Trowel
d. Weeder
Explanation: Tools like Spade, Shovel and Trowel are used for digging whereas Weeder is a tool used to remove seeds from land. So, grouping of options can be done based on their "Purpose".
### Types
Single Words
Pair of Words
Both types can be further divided into:
Letters
Numbers
Words
### Letters: Example
i) Pick the odd one out
a. B
b. D
c. E
d. F
Explanation: E is a vowel. Rest all are consonants.
### Numbers: Example
i) Pick the odd one out
a. 4
b. 16
c. 36
d. 8
Explanation: 22 = 4, 42 = 16, 62 = 36.
### Words: Example
i) Pick the odd one out
a. Volleyball
b. Football
c. Baseball
d. Chess
Explanation: Volleyball, Football and Cricket- all are outdoor games and Chess is an indoor game.
### Letters: Example
i) Pick the odd one out.
a) A : C
b) B : D
c) C : E
d) D : E
Explanation: In all options except D, First and second letters have difference of one and in D : E there is no difference.
### Numbers: Example
i) Pick the odd one out.
a) 2 : 4
b) 5 : 25
c) 10 : 100
d) 3 : 9
e) 6 : 60
Explanation: In all options except E, Second number is Square of first number.
### Words: Example
i) Pick the odd one out.
a) May : January
b) September : November
c) October : April
d) January : December
e) October : July
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# 2016 AMC 8 Problems/Problem 15
## Problem
What is the largest power of $2$ that is a divisor of $13^4 - 11^4$?
$\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128$
## Solution 1
First, we use difference of squares on $13^4 - 11^4 = (13^2)^2 - (11^2)^2$ to get $13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2)$. Using difference of squares again and simplifying, we get $(169 + 121)(13+11)(13-11) = 290 \cdot 24 \cdot 2 = (2\cdot 8 \cdot 2) \cdot (3 \cdot 145)$. Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of $2$ that is a divisor $13^4 - 11^4$ is $\boxed{\textbf{(C)}\ 32}$.
~CHECKMATE2021
## Solution 2 (a variant of Solution 1)
Just like in the above solution, we use the difference-of-squares factorization, but only once to get $13^4-11^4=(13^2-11^2)(13^2+11^2).$ We can then compute that this is equal to $48\cdot290.$ Note that $290=2\cdot145$ (we don't need to factorize any further as $145$ is already odd) thus the largest power of $2$ that divides $290$ is only $2^1=2,$ while $48=2^4\cdot3,$ so the largest power of $2$ that divides $48$ is $2^4=16.$ Hence, the largest power of $2$ that is a divisor of $13^4-11^4$ is $2\cdot16=\boxed{\textbf{(C)}~32}.$
## Solution 3 (Lifting the exponent)
Let $n=13^4-11^4.$ We wish to find the largest power of $2$ that divides $n$.
Denote $v_p(k)$ as the largest exponent of $p$ in the prime factorization of $n$. In this problem, we have $p=2$.
By the Lifting the Exponent Lemma on $n$,
$$v_2(13^4-11^4)=v_2(13-11)+v_2(4)+v_2(13+11)-1$$ $$=v_2(2)+v_2(4)+v_2(24)-1$$ $$=1+2+3-1=5.$$
Therefore, exponent of the largest power of $2$ that divids $13^4-11^4$ is $5,$ so the largest power of $2$ that divides this number is $2^5=\boxed{\textbf{(C)} 32}$.
~CHECKMATE2021
## Solution 4 (Brute Force)
We can simply take 13 to the 4th power, which is 28561. We subtract that by 11 to the 4th power, which is 14641 (You can use Pascal's Triangle to find this). Finally, subtract the numbers to get 13920.
To test the options, since we need the largest one, we can go from top down. Testing, we see that both D and E are decimals, and 32 works. So, our answer is $\boxed{\textbf{(C)}~32}.$
-themathgood
Note: This is not the fastest way, and is not recommended.
I completely agree even though I did it./reader
~ pi_is_3.14
~savannahsolver
## Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
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Derivation of the formulas used in the mathematical model of the calculator
The description of the calculator does not contain - due to readability - detailed derivation of all formulas. These derivations are presented below.
Derivation of the formula for calculating the deflection angle
Let's transform the following initial formula:
\begin{aligned} sin β = \frac{v_{g}}{v_{t}} * sin γ \end{aligned}
in such way, that deflection angle β depends on the torpedo impact angle α and quotient $$\frac{v_{g}}{v_{t}}$$.
Because one of the input values of the component for solving the torpedo triangle is quotient $$\frac{v_{g}}{v_{t}}$$, let's make following substitution::
\begin{aligned} u = \frac{v_{g}}{v_{t}} \end{aligned}
which results in:
\begin{aligned} sin β = u * sin γ \end{aligned}
Because in any triangle, the sum of all three angles is always equal to 180°, the following formula can be written:
\begin{aligned} γ = 180° - (α + β) \end{aligned}
Having in mind the periodic feature of the sines function – that is the feature that:
\begin{aligned} sin α = sin (180° - α) \end{aligned}
we get:
\begin{aligned} sin γ = sin (α + β) \end{aligned}
So finally we get:
\begin{aligned} sin β = u * sin (α + β) \end{aligned}
This formula can be transformed in the following way:
\begin{aligned} sin β = u * (sin α * cos β + cos α * sin β) \end{aligned}
\begin{aligned} 1 = u * \frac{(sin α * cos β + cos α * sin β)}{sin β} \end{aligned}
\begin{aligned} 1 = u * (sin α * ctg β + cos α) \end{aligned}
\begin{aligned} 1 - u * cos α = u * sin α * ctg β \end{aligned}
\begin{aligned} ctg β = \frac{1 - u * cos α}{u * sin α} = \frac{1}{u * sin α} - ctg α \end{aligned}
So the deflection angle β can be calculated based on the quotient of the target and torpedo speed and knowing the torpedo impact angle using the following formula:
\begin{aligned} β = arcctg (\frac{1}{u * sin α} - ctg α \end{aligned})
Derivation of the formula for Cartesian coordinates of the equivalent point of fire
The torpedo track consists of three parts: the initial, straight run RT1, the arc T1T2, when the torpedo turns to its final course, and the segment T2G, which is the final torpedo run until hitting the target.
Drawing 1
The equivalent point of fire E1 is located at the line, which overlaps the final torpedo run. It is located in the distance |E1T2| equal to the sum of the length of the arc T1T2 and the segment RT1. The length of the segment |RT1| is constant and for German World War II torpedoes was equal to 9.5 m. The length of the arc |T1T2| depended on the gyro angle ρ.
The Cartesian coordinates x0 and y0 of the equivalent point of fire E1 relative to the location of the aiming device Z (which was distanced from the torpedo tubes by 27 meters on type VIIC U-Boats) can be determined in the following way:
The coordinates xr and yr of the torpedo tubes R relative to the location of the aiming device Z are:
\begin{aligned} (x_{r}; y_{r}) = (27; 0) \end{aligned}
The coordinates of the point T1 relative to the location of the aiming device Z:
\begin{aligned} (x_{T1}; y_{T1}) = (x_{r}; y_{r}) + (9,5; 0) = (27; 0) + (9,5; 0) \end{aligned}
The position of the point T2 (relative to the point T1) can be determined from the following formula (after noticing, that the angle constructed at the arc T1T2 is equal to the gyro angle ρ):
\begin{aligned} (x_{T2}; y_{T2}) = (r * sin ρ; r – r * cos ρ) \end{aligned}
where r – turn radius, which was equal to 95 m for German World War II torpedoes.
Whereas, relative to the aiming device location, the coordinates are:
\begin{aligned} (x_{T2}; y_{T2}) = (r * sin ρ; r – r * cos ρ) + (27; 0) + (9,5; 0) \end{aligned}
Drawing 2
The length of the arc T1T2 is
\begin{aligned} |T_{1}T_{2}| = ρ * r \end{aligned}
where angle ρ is expressed in radians.
As mentioned before, the distance between the point E1 and point T2 is equal to the sum:
\begin{aligned} |E_{1}T_{2}| = |RT_{1}| + |T_{1}T_{2}| \end{aligned}
that is
\begin{aligned} |E_{1}T_{2}| = 9,5 + ρ * r \end{aligned}
Coordinates of the point E1 relative to the point T2 are:
\begin{aligned} (x_{E1}; y_{E1}) = (x_{T2}; y_{T2}) – (|E_{1}T_{2}| * cos ρ; |E_{1}T_{2}| * sin ρ) \end{aligned}
Drawing 3
Coordinates of the point E1 relative to the aiming device Z location:
\begin{aligned} (x_{E1}; y_{E1}) = (r * sin ρ; r – r * cos ρ) + (27; 0) + (9,5; 0) – (|E_{1}T_{2}| * cos ρ; |E_{1}T_{2}| * sin ρ) \end{aligned}
After inserting the formula determining the length of the segment |E1T2| we get:
\begin{aligned} (x_{E1}; y_{E1}) = (r * sin ρ; r – r * cos ρ) + (27; 0) + (9,5; 0) – ((9,5 + ρ * r) * cos ρ; (9,5 + ρ * r )* sin ρ) \end{aligned}
So the coordinates of the point E1 are:
$\begin{cases} x_{E1} = r * sin ρ + 27 + 9,5 – (9,5 + ρ * r) * cos ρ \\ y_{E1} = r – r * cos ρ – (9,5 + ρ * r) * sin ρ \end{cases}$
that is
$\begin{cases} x_{E1} = 27 + 9,5 + 95 * sin ρ – (9,5 + ρ * 95) * cos ρ \\ y_{E1} = 95 * (1 – cos ρ) – (9,5 + ρ * 95) * sin ρ \end{cases}$
Derivation of the formula for calculating the torpedo salvo spread angle
The transformation of the following formula
\begin{aligned} \frac{E - λ * cos γ}{v_{t} * cos β + v_{g} * cos γ} = \frac{λ * sin γ}{v_{t} * sin β – v_{g} * sin γ} \end{aligned}
we are beginning from dividing denominators at both sides of equation by vt, which results in
\begin{aligned} \frac{E - λ * cos γ}{cos β + \frac{v_{g}}{v_{t}} * cos γ} = \frac{λ * sin γ}{sin β – \frac{v_{g}}{v_{t}} * sin γ} \end{aligned}
Then the following substitution is done:
\begin{aligned} u = \frac{v_{g}}{v_{t}} \end{aligned}
getting:
\begin{aligned} \frac{E - λ * cos γ}{cos β + u * cos γ} = \frac{λ * sin γ}{sin β – u * sin γ} \end{aligned}
After dividing the numerators at both sides of equation by E we get:
\begin{aligned} \frac{1 - \frac{λ}{E} * cos γ}{cos β + u * cos γ} = \frac{\frac{λ}{E} * sin γ}{sin β – u * sin γ} \end{aligned}
Then we make following substitution:
\begin{aligned} μ = \frac{λ}{E} \end{aligned}
getting:
\begin{aligned} \frac{1 - μ * cos γ}{cos β + u * cos γ} = \frac{μ * sin γ}{sin β – u * sin γ} \end{aligned}
Then the following transformations are performed:
\begin{aligned} (1 - μ * cos γ)*(sin β – u * sin γ) = (μ * sin γ)*(cos β + u * cos γ) \end{aligned}
\begin{aligned} sin β - u * sin γ - μ * cos γ * sin β + μ * u * sin γ * cos γ = μ * sin γ * cos β + μ * u * sin γ * cos γ \end{aligned}
\begin{aligned} sin β - u * sin γ = μ * sin γ * cos β + μ * cos γ * sin β \end{aligned}
\begin{aligned} sin β - u * sin γ = μ * (sin γ * cos β + cos γ * sin β) \end{aligned}
finally getting:
\begin{aligned} sin β - u * sin γ = μ * sin (γ + β) \end{aligned}
This is implicit equation, which can be used for calculating the deflection angle β for the target as seen with the angle on the bow γ, so the torpedo hits the point distanced by the value λ from the middle of the target.
For convenience, instead considering the stern and bow location as distances λ from the middle of the target, the following substitution can be done:
\begin{aligned} μ_{0} = \frac{l}{2*E} \end{aligned}
where l – length of the target.
To make three-torpedo salvo hitting the bow, middle and stern of the target of length l, each torpedo has to be launched with the deflection angle β0, β and β1 respectively, where each of these values meets one of the following equations:
$\begin{cases}sin β_{0} – u * sin γ = μ_{0} * sin (β_{0} + γ) \\ sin β – u * sin γ = 0 \\ sin β_{1} – u * sin γ = - μ_{0} * sin (β_{1} + γ)\end{cases}$
Because, in practice, in most cases the target (with the length up to 150 meters – i.e. Liberty type cargo ships: ~135 meters, T2 type tankers: ~152 meters) were attacked from the minimum distance 500 meters, the value μ0 is rather small (μ0 < 0,15). That means, that angle between the courses of torpedoes aimed into target bow and stern (that is angle β1 - β0) is equal maximum ~20º. Knowing the fact, that for the angles α in range 0 – 20º, the sin α ~ α and cos α ~ 1, the following equations can be written:
\begin{align} sin β_{0} = sin (β_{0} - β + β) = sin (β_{0} - β) * cos β + cos (β_{0} - β) * sin β \end{align}
and
\begin{align} sin β_{1} = sin (β_{1} - β + β) = sin (β_{1} - β) * cos β + cos (β_{1} - β) * sin β \end{align}
Using the feature of the sines and cosines functions for small angles mentioned above, the equations above can be written in the following way:
\begin{align} sin β_{0} = (β_{0} - β) * cos β + sin β \end{align}
and
\begin{align} sin β_{1} = (β_{1} - β) * cos β + sin β \end{align}
After inserting the formulas above into previous equations, we get:
$\begin{cases} (β_{0} - β) * cos β + sin β - u * sin γ = μ_{0} * sin (β_{0} + γ) \\ sin β - u * sin γ = 0 \\ (β_{1} - β) * cos β + sin β - u * sin γ = -μ_{0} * sin (β_{1} + γ) \end{cases}$
and after transforming:
$\begin{cases} sin β - u * sin γ = μ_{0} * sin (β_{0} + γ) - (β_{0} - β) * cos β \\ sin β - u * sin γ = 0 \\ sin β - u * sin γ = -μ_{0} * sin (β_{1} + γ) - (β_{1} - β) * cos β \end{cases}$
After inserting the second equation into the two others, we receive two formulas:
$\begin{cases} 0 = μ_{0} * sin (β_{0} + γ) - (β_{0} - β) * cos β \\ 0 = -μ_{0} * sin (β_{1} + γ) - (β_{1} - β) * cos β \end{cases}$
By comparing these two equations, we get:
\begin{aligned} μ_{0} * sin (β_{0} + γ) - (β_{0} - β) * cos β = -μ_{0} * sin (β_{1} + γ) - (β_{1} - β) * cos β \end{aligned}
which results in:
\begin{aligned} β_{0} - β_{1} = μ_{0} * \frac{sin (β_{0} + γ) + sin (β_{1} + γ)}{cos β} \end{aligned}
As mentioned before, the difference of the angles β1 - β0 in practice are maximum 20º. The sines function for angle values up to 20º can be approximated by linear function. Se we can write:
\begin{aligned} sin (β + γ) = \frac{sin (β_{0} + γ) + sin (β_{1} + γ)}{2} \end{aligned}
while
\begin{aligned} β \approx \frac{β_{0} + β_{1}}{2} \end{aligned}
So the torpedo salvo spread angle ψ (which is the difference β1 – β0) can be calculated by following formula:
\begin{aligned} ψ = 2 * μ_{0} * \frac{sin (β + γ)}{cos β} \end{aligned}
This equation can be transformed in the following steps, to avoid dependency from the deflection angle β:
\begin{aligned} ψ = 2 * μ_{0} * \frac{sin (β + γ)}{cos β} = \frac{l}{E} * \frac{sin (β + γ)}{cos β} = \frac{l}{E} * \frac{sin β * cos γ + cos β * sin γ}{cos β}\end{aligned}
\begin{aligned} ψ = \frac{l}{E} * \bigg(\frac{sin β * cos γ}{cos β} + \frac{cos β * sin γ}{cos β}\bigg) = \frac{l}{E} * (tg β * cos γ + sin γ)\end{aligned}
\begin{aligned} ψ = \frac{l}{E} * sin γ * \bigg(\frac{tg β * cos γ}{sin γ} + 1\bigg) = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{ctg β * sin γ} + 1\bigg) \end{aligned}
\begin{aligned} ψ = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\frac{cos β}{sin β} * sin γ} + 1\bigg) = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\sqrt{\frac{cos^2 β}{sin^2 β} * sin^2 γ}} + 1\bigg) \end{aligned}
\begin{aligned} ψ = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\sqrt{\frac{1 - sin^2 β}{sin^2 β} * sin^2 γ}} + 1\bigg) = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\sqrt{\frac{sin^2 γ - sin^2 β * sin^2 γ}{sin^2 β}}} + 1\bigg) \end{aligned}
\begin{aligned} ψ = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\sqrt{\frac{sin^2 γ}{sin^2 β} - \frac{sin^2 β * sin^2 γ}{sin^2 β}}} + 1\bigg) \end{aligned}
Because
\begin{aligned} \frac{sin γ}{sin β} = \frac{v_{t}}{v_{g}} \end{aligned}
finally we get:
\begin{aligned} ψ = \frac{l}{E} * sin γ * \bigg(\frac{cos γ}{\sqrt{(\frac{v_{t}}{v_{g}})^2 - sin^2 γ}} + 1\bigg) \end{aligned}
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Algebra - Basic Go to the latest version.
# 10.6: The Discriminant
Difficulty Level: At Grade Created by: CK-12
You have seen parabolas that intersect the x\begin{align*}x-\end{align*}axis twice, once, or not at all. There is a relationship between the number of real x\begin{align*}x-\end{align*}intercepts and the Quadratic Formula.
Case 1: The parabola has two x\begin{align*}x-\end{align*}intercepts. This situation has two possible solutions for x\begin{align*}x\end{align*}, because the value inside the square root is positive. Using the Quadratic Formula, the solutions are x=b+b24ac2a\begin{align*}x=\frac{-b+\sqrt{b^2-4ac}}{2a}\end{align*} and x=bb24ac2a\begin{align*}x=\frac{-b-\sqrt{b^2-4ac}}{2a}\end{align*}.
Case 2: The parabola has one x\begin{align*}x-\end{align*}intercept. This situation occurs when the vertex of the parabola just touches the x\begin{align*}x-\end{align*}axis. This is called a repeated root, or double root. The value inside the square root is zero. Using the Quadratic Formula, the solution is x=b2a\begin{align*}x=\frac{-b}{2a}\end{align*}.
Case 3: The parabola has no x\begin{align*}x-\end{align*}intercept. This situation occurs when the parabola does not cross the x\begin{align*}x-\end{align*}axis. The value inside the square root is negative, therefore there are no real roots. The solutions to this type of situation are imaginary, which you will learn more about in a later textbook.
The value inside the square root of the Quadratic Formula is called the discriminant. It is symbolized by D\begin{align*}D\end{align*}. It dictates the number of real solutions the quadratic equation has. This can be summarized with the Discriminant Theorem.
• If D>0\begin{align*}D>0\end{align*}, the parabola will have two x\begin{align*}x-\end{align*}intercepts. The quadratic equation will have two real solutions.
• If D=0\begin{align*}D=0\end{align*}, the parabola will have one x\begin{align*}x-\end{align*}intercept. The quadratic equation will have one real solution.
• If D<0\begin{align*}D<0\end{align*}, the parabola will have no x\begin{align*}x-\end{align*}intercepts. The quadratic equation will have zero real solutions.
Example 1: Determine the number of real solutions to 3x2+4x+1=0\begin{align*}-3x^2+4x+1=0\end{align*}.
Solution: By finding the value of its discriminant, you can determine the number of x\begin{align*}x-\end{align*}intercepts the parabola has and thus the number of real solutions.
DDD=b24(a)(c)=(4)24(3)(1)=16+12=28
Because the discriminant is positive, the parabola has two real x\begin{align*}x-\end{align*}intercepts and thus two real solutions.
Example: Determine the number of solutions to 2x2+x=4\begin{align*}-2x^2+x=4\end{align*}.
Solution: Before we can find its discriminant, we must write the equation in standard form ax2+bx+c=0\begin{align*}ax^2+bx+c=0\end{align*}.
Subtract 4 from each side of the equation: 2x2+x4=0\begin{align*}-2x^2+x-4=0\end{align*}.
Find the discriminant.DD=(1)24(2)(4)=132=31
The value of the discriminant is negative; there are no real solutions to this quadratic equation. The parabola does not cross the x\begin{align*}x-\end{align*}axis.
Example 2: Emma and Bradon own a factory that produces bike helmets. Their accountant says that their profit per year is given by the function P=0.003x2+12x+27,760\begin{align*}P=0.003x^2+12x+27,760\end{align*}, where x\begin{align*}x\end{align*} represents the number of helmets produced. Their goal is to make a profit of 40,000 this year. Is this possible? Solution: The equation we are using is 40,000=0.003x2+12x+27,760\begin{align*}40,000=0.003x^2+12x+27,760\end{align*}. By finding the value of its discriminant, you can determine if the profit is possible. Begin by writing this equation in standard form: 0DDD=0.003x2+12x12,240=b24(a)(c)=(12)24(0.003)(12,240)=144+146.88=290.88 Because the discriminant is positive, the parabola has two real solutions. Yes, the profit of40,000 is possible.
Multimedia Link: This http://sciencestage.com/v/20592/a-level-maths-:-roots-of-a-quadratic-equation-:-discriminant-:-examsolutions.html - video, presented by Science Stage, helps further explain the discriminant using the Quadratic Formula.
## Practice Set
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.
1. What is a discriminant? What does it do?
2. What is the formula for the discriminant?
3. Can you find the discriminant of a linear equation? Explain your reasoning.
4. Suppose D=0\begin{align*}D=0\end{align*}. Draw a sketch of this graph and determine the number of real solutions.
5. D=2.85\begin{align*}D=-2.85\end{align*}. Draw a possible sketch of this parabola. What is the number of real solutions to this quadratic equation.
6. D>0\begin{align*}D>0\end{align*}. Draw a sketch of this parabola and determine the number of real solutions.
Find the discriminant of each quadratic equation.
1. 2x24x+5=0\begin{align*}2x^2-4x+5=0\end{align*}
2. x25x=8\begin{align*}x^2-5x=8\end{align*}
3. 4x212x+9=0\begin{align*}4x^2-12x+9=0\end{align*}
4. x2+3x+2=0\begin{align*}x^2+3x+2=0\end{align*}
5. x216x=32\begin{align*}x^2-16x=32\end{align*}
6. 5x2+5x6=0\begin{align*}-5x^2+5x-6=0\end{align*}
Determine the nature of the solutions of each quadratic equation.
1. x2+3x6=0\begin{align*}-x^2+3x-6=0\end{align*}
2. 5x2=6x\begin{align*}5x^2=6x\end{align*}
3. 41x231x52=0\begin{align*}41x^2-31x-52=0\end{align*}
4. x28x+16=0\begin{align*}x^2-8x+16=0\end{align*}
5. x2+3x10=0\begin{align*}-x^2+3x-10=0\end{align*}
6. x264=0\begin{align*}x^2-64=0\end{align*}
A solution to a quadratic equation will be irrational if the discriminant is not a perfect square. If the discriminant is a perfect square, then the solutions will be rational numbers. Using the discriminant, determine whether the solutions will be rational or irrational.
1. x2=4x+20\begin{align*}x^2=-4x+20\end{align*}
2. x2+2x3=0\begin{align*}x^2+2x-3=0\end{align*}
3. 3x211x=10\begin{align*}3x^2-11x=10\end{align*}
4. 12x2+2x+23=0\begin{align*}\frac{1}{2}x^2+2x+\frac{2}{3}=0\end{align*}
5. \begin{align*}x^2-10x+25=0\end{align*}
6. \begin{align*}x^2=5x\end{align*}
7. Marty is outside his apartment building. He needs to give Yolanda her cell phone but he does not have time to run upstairs to the third floor to give it to her. He throws it straight up with a vertical velocity of 55 feet/second. Will the phone reach her if she is 36 feet up? (Hint: The equation for the height is given by \begin{align*}y=-32t^2+55t+4\end{align*}.)
8. Bryson owns a business that manufactures and sells tires. The revenue from selling the tires in the month of July is given by the function \begin{align*}R=x(200-0.4x)\end{align*} where \begin{align*}x\end{align*} is the number of tires sold. Can Bryson’s business generate revenue of \$20,000 in the month of July?
9. Marcus kicks a football in order to score a field goal. The height of the ball is given by the equation \begin{align*}y=-\frac{32}{6400}x^2+x\end{align*}, where \begin{align*}y\end{align*} is the height and \begin{align*}x\end{align*} is the horizontal distance the ball travels. We want to know if Marcus kicked the ball hard enough to go over the goal post, which is 10 feet high.
Mixed Review
1. Factor \begin{align*}6x^2-x-12\end{align*}.
2. Find the vertex of \begin{align*}y=-\frac{1}{4} x^2-3x-12=y\end{align*} by completing the square.
3. Solve using the Quadratic Formula: \begin{align*}-4x^2-15=-4x\end{align*}.
4. How many centimeters are in four fathoms? (Hint: 1 fathom = 6 feet)
5. Graph the solution to \begin{align*}\begin{cases} 3x+2y \le -4\\ x-y>-3 \end{cases}\end{align*}.
6. How many ways can 3 toppings be chosen from 7 options?
8 , 9
Feb 22, 2012
Oct 19, 2015
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The Fundamental Theorem of Calculus
Contributed by:
In this section, we will learn about:
The Fundamental Theorem of Calculus and its significance.
1. 5
2. 5.3
The Fundamental
Theorem of Calculus
In this section, we will learn about:
The Fundamental Theorem of Calculus
and its significance.
3. FUNDAMENTAL THEOREM OF CALCULUS
The Fundamental Theorem of Calculus
(FTC) is appropriately named.
It establishes a connection between the two
branches of calculus—differential calculus and
integral calculus.
4. Differential calculus arose from the tangent
Integral calculus arose from a seemingly
unrelated problem—the area problem.
5. Newton’s mentor at Cambridge, Isaac Barrow
(1630–1677), discovered that these two
problems are actually closely related.
In fact, he realized that differentiation and
integration are inverse processes.
6. The FTC gives the precise inverse
relationship between the derivative
and the integral.
7. It was Newton and Leibniz who exploited this
relationship and used it to develop calculus
into a systematic mathematical method.
In particular, they saw that the FTC enabled them
to compute areas and integrals very easily without
having to compute them as limits of sums—as we did
in Sections 5.1 and 5.2
8. FTC Equation 1
The first part of the FTC deals with functions
defined by an equation of the form
x
g ( x) f (t ) dt
a
where f is a continuous function on [a, b]
and x varies between a and b.
9. x
g ( x) f (t ) dt
a
Observe that g depends only on x, which appears
as the variable upper limit in the integral.
If x is a fixed number, then the integral x
is a definite number. f (t ) dt
a
x
If we then let x vary, the number f (t ) dt
also varies and defines a function aof x denoted by g(x).
10. If f happens to be a positive function, then g(x)
can be interpreted as the area under the
graph of f from a to x, where x can vary from a
to b.
Think of g as the
‘area so far’ function,
as seen here.
11. FTC Example 1
If f is the function
whose graph is shown
x
and g ( x) f (t ) dt ,
0
find the values of:
g(0), g(1), g(2), g(3),
g(4), and g(5).
Then, sketch a rough graph of g.
12. FTC Example 1
First, we notice that:
0
g (0) f (t ) dt 0
0
13. FTC Example 1
From the figure, we see that g(1) is
the area of a triangle:
1
g (1) f (t ) dt
0
(12)
1
2
1
14. FTC Example 1
To find g(2), we add to g(1) the area of
a rectangle:
2
g (2) f (t ) dt
0
1 2
f (t ) dt f (t ) dt
0 1
1 (1 2)
3
15. FTC Example 1
We estimate that the area under f from 2 to 3
3
So, g (3) g (2)
2
f (t ) dt
3 1.3
4.3
16. FTC Example 1
For t > 3, f(t) is negative.
So, we start subtracting areas, as
17. FTC Example 1
4
g (4) g (3) f (t ) dt 4.3 ( 1.3) 3.0
3
5
g (5) g (4) f (t ) dt 3 ( 1.3) 1.7
4
18. FTC Example 1
We use these values to sketch the graph
of g.
Notice that, because f(t)
is positive for t < 3,
for t < 3.
So, g is increasing up to
x = 3, where it attains
a maximum value.
For x > 3, g decreases
because f(t) is negative.
19. If we take f(t) = t and a = 0, then,
using Exercise 27 in Section 5.2,
we have: 2
x x
g ( x) tdt
0 2
20. Notice that g’(x) = x, that is, g’ = f.
In other words, if g is defined as the integral of f
by Equation 1, g turns out to be an antiderivative
of f—at least in this case.
21. If we sketch the derivative
of the function g, as in the
first figure, by estimating
slopes of tangents, we get
a graph like that of f in the
second figure.
So, we suspect that g’ = f
in Example 1 too.
22. To see why this might be generally true, we
consider a continuous function f with f(x) ≥ 0.
x
Then, g ( x ) f (t )dt can be interpreted as
a
the area under the graph of f from a to x.
23. To compute g’(x) from the definition of
derivative, we first observe that, for h > 0,
g(x + h) – g(x) is obtained by subtracting
It is the area
under the graph
of f from x to x + h
(the gold area).
24. For small h, you can see that this area is
approximately equal to the area of the
rectangle with height f(x) and width h:
g ( x h) g ( x) hf ( x)
So, g ( x h) g ( x)
h
f ( x)
25. Intuitively, we therefore expect that:
g ( x h) g ( x )
g '( x) lim f ( x)
h 0 h
The fact that this is true, even when f is not
necessarily positive, is the first part of the FTC
(FTC1).
26. If f is continuous on [a, b], then the function g
defined by x
g ( x) f (t )dt a x b
a
is continuous on [a, b] and differentiable on
(a, b), and g’(x) = f(x).
27. In words, the FTC1 says that the derivative
of a definite integral with respect to its upper
limit is the integrand evaluated at the upper
28. FTC1 Proof
If x and x + h are in (a, b), then
g ( x h) g ( x )
x h x
f (t )dt f (t )dt
a a
x
f (t)dt
a
x h
x
f (t )dt
x
f (t )dt
a
(Property 5)
x h
f (t )dt
x
29. FTC1 Proof—Equation 2
So, for h ≠ 0,
g ( x h) g ( x ) 1 x h
f (t )dt
h h x
30. FTC1 Proof
For now, let us assume that h > 0.
Since f is continuous on [x, x + h], the Extreme Value
Theorem says that there are numbers u and v in
[x, x + h] such that f(u) = m and f(v) = M.
m and M are the absolute
minimum and maximum
values of f on [x, x + h].
31. FTC1 Proof
By Property 8 of integrals, we have:
x h
mh f (t ) dt Mh
x
x h
That is, f (u )h
x
f (t ) dt f (v)h
32. FTC1 Proof
Since h > 0, we can divide this inequality
by h:
1 x h
f (u ) f (t )dt f (v )
h x
33. FTC1 Proof—Equation 3
Now, we use Equation 2 to replace the middle
part of this inequality:
g ( x h) g ( x )
f (u ) f (v )
h
Inequality 3 can be proved in a similar manner
for the case h < 0.
34. FTC1 Proof
Now, we let h → 0.
Then, u → x and v → x, since u and v lie
between x and x + h.
Therefore, lim f (u ) lim f (u ) f ( x)
h 0 u x
and lim f (v) lim f (v) f ( x)
h 0 v x
because f is continuous at x.
35. FTC1 Proof—Equation 4
From Equation 3 and the Squeeze
Theorem, we conclude that:
g ( x h) g ( x )
g '( x) lim f ( x)
h 0 h
36. If x = a or b, then Equation 4 can be
interpreted as a one-sided limit.
Then, Theorem 4 in Section 2.8 (modified
for one-sided limits) shows that g is continuous
on [a, b].
37. FTC1 Proof—Equation 5
Using Leibniz notation for derivatives, we can
write the FTC1 as d x
f (t )dt f ( x)
dx a
when f is continuous.
Roughly speaking, Equation 5 says that,
if we first integrate f and then differentiate
the result, we get back to the original function f.
38. FTC1 Example 2
Find the derivative of the function
x
2
g ( x) 1 t dt
0
As f (t ) 1 t 2 is continuous, the FTC1 gives:
2
g '( x) 1 x
39. A formula of the form x
g ( x) f (t )dt
a
may seem like a strange way of defining
a function.
However, books on physics, chemistry, and
statistics are full of such functions.
40. FRESNEL FUNCTION Example 3
For instance, consider the Fresnel function
x
2
S ( x) sin( t / 2) dt
0
It is named after the French physicist Augustin Fresnel
(1788–1827), famous for his works in optics.
It first appeared in Fresnel’s theory of the diffraction
of light waves.
More recently, it has been applied to the design
of highways.
41. FRESNEL FUNCTION Example 3
The FTC1 tells us how to differentiate
the Fresnel function:
S’(x) = sin(πx2/2)
This means that we can apply all the methods
of differential calculus to analyze S.
42. FRESNEL FUNCTION Example 3
The figure shows the graphs of
f(x) = sin(πx2/2) and the Fresnel function
x
S ( x) f (t )dt
0
A computer was used
to graph S by computing
the value of this integral
for many values of x.
43. FRESNEL FUNCTION Example 3
It does indeed look as if S(x) is the area
under the graph of f from 0 to x (until x ≈ 1.4,
when S(x) becomes a difference of areas).
44. FRESNEL FUNCTION Example 3
The other figure shows a larger part
of the graph of S.
45. FRESNEL FUNCTION Example 3
think about what its derivative should look like,
it seems reasonable that S’(x) = f(x).
For instance, S is
increasing when f(x) > 0
and decreasing when
f(x) < 0.
46. FRESNEL FUNCTION Example 3
So, this gives a visual confirmation
of the FTC1.
47. FTC1 Example 4
d x4
Find sec t dt
dx 1
Here, we have to be careful to use the Chain Rule
in conjunction with the FTC1.
48. FTC1 Example 4
Let u = x4.
d x 4
d u
sec t dt sec t dt
dx 1 dx 1
d u du
du 1
sec t dt
dx
(Chain Rule)
du
sec u (FTC1)
dx
4 3
sec( x ) 4 x
49. In Section 5.2, we computed integrals from
the definition as a limit of Riemann sums
and saw that this procedure is sometimes
long and difficult.
The second part of the FTC (FTC2), which follows
easily from the first part, provides us with a much
simpler method for the evaluation of integrals.
50. If f is continuous on [a, b], then
b
f ( x)dx F (b) F (a)
a
where F is any antiderivative of f,
that is, a function such that F’ = f.
51. FTC2 Proof
x
Let g ( x ) f (t ) dt
a
We know from the FTC1 that g’(x) = f(x),
that is, g is an antiderivative of f.
52. FTC2 Proof—Equation 6
If F is any other antiderivative of f on [a, b],
then we know from Corollary 7 in Section 4.2
that F and g differ by a constant
F(x) = g(x) + C
for a < x < b.
53. FTC2 Proof
However, both F and g are continuous on
[a, b].
Thus, by taking limits of both sides of
Equation 6 (as x → a+ and x → b- ),
we see it also holds when x = a and x = b.
54. FTC2 Proof
If we put x = a in the formula for g(x),
we get:
a
g (a ) f (t ) dt 0
a
55. FTC2 Proof
So, using Equation 6 with x = b and x = a,
we have:
F (b) F (a ) [ g (b) C ] [ g (a) C ]
g (b) g (a )
g (b)
b
f (t )dt
a
56. The FTC2 states that, if we know an
antiderivative F of f, then we can evaluate
b
f ( x)dx
a
simply by subtracting the
of F at the endpoints of the interval [a, b].
57. b
It’s very surprising that f ( x ) dx , which
a
was defined by a complicated procedure
involving all the values of f(x) for a ≤ x ≤ b,
can be found by knowing the values of F(x)
at only two points, a and b.
58. At first glance, the theorem may be
However, it becomes plausible if we interpret it
in physical terms.
59. If v(t) is the velocity of an object and s(t)
is its position at time t, then v(t) = s’(t).
So, s is an antiderivative of v.
60. In Section 5.1, we considered an object that
always moves in the positive direction.
Then, we guessed that the area under the
velocity curve equals the distance traveled.
b
In symbols,
v(t ) dt s(b) s(a)
a
That is exactly what the FTC2 says in this context.
61. FTC2 Example 5
3
x
Evaluate the integral e dx
1
The function f(x) = ex is continuous everywhere
and we know that an antiderivative is F(x) = ex.
3
x
So, the FTC2 gives:
dx F (3) F (1)
1
e
e3 e
62. FTC2 Example 5
Notice that the FTC2 says that we can use
any antiderivative F of f.
So, we may as well use the simplest one,
namely F(x) = ex, instead of ex + 7 or ex + C.
63. We often use the notation
b
F ( x)] F (b) F (a )
a
So, the equation of the FTC2 can be written
as: b
b
f ( x)dx F ( x)]
a a where F ' f
b
Other common notations are F ( x ) | and [ F ( x)]ba .
a
64. FTC2 Example 6
Find the area under the parabola y = x2
from 0 to 1.
An antiderivative of f(x) = x2 is F(x) = (1/3)x3.
The required area is found using the FTC2:
3 3 3 1
2
1 x 1 0 1
A x dx
0 3 0 3 3 3
65. If you compare the calculation in Example 6
with the one in Example 2 in Section 5.1,
you will see the FTC gives a much shorter
66. FTC2 Example 7
dx 6
3 x
1 6
The given integral is an abbreviation for dx.
3 x
An antiderivative of f(x) = 1/x is F(x) = ln |x|.
As 3 ≤ x ≤ 6, we can write F(x) = ln x.
67. FTC2 Example 7
Therefore, 6 1 6
3 x dx ln x]3
ln 6 ln 3
6
ln
3
ln 2
68. FTC2 Example 8
Find the area under the cosine curve
from 0 to b, where 0 ≤ b ≤ π/2.
As an antiderivative of f(x) = cos x is F(x) = sin x,
we have:
b
A cos x dx sin x]b0 sin b sin 0 sin b
0
69. FTC2 Example 8
In particular, taking b = π/2, we have
proved that the area under the cosine curve
from 0 to π/2 is:
sin(π/2) = 1
70. When the French mathematician Gilles de
Roberval first found the area under the sine
and cosine curves in 1635, this was a very
challenging problem that required a great deal
of ingenuity.
71. If we didn’t have the benefit of the FTC,
we would have to compute a difficult limit
of sums using either:
Obscure trigonometric identities
A computer algebra system (CAS), as in Section 5.1
72. It was even more difficult for
The apparatus of limits had not been invented
in 1635.
73. However, in the 1660s and 1670s,
when the FTC was discovered by Barrow
and exploited by Newton and Leibniz,
such problems became very easy.
You can see this from Example 8.
74. FTC2 Example 9
What is wrong with this calculation?
1 3
3 1 x 1 4
1 x 2
dx 1
1 1 3 3
75. FTC2 Example 9
To start, we notice that the calculation must
be wrong because the answer is negative
but f(x) = 1/x2 ≥ 0 and Property 6 of integrals
b
says that a f ( x ) dx 0 when f ≥ 0.
76. FTC2 Example 9
The FTC applies to continuous functions.
It can’t be applied here because f(x) = 1/x2
is not continuous on [-1, 3].
In fact, f has an infinite discontinuity at x = 0.
31
So, 2 dx does not exist.
1 x
77. INVERSE PROCESSES
We end this section by
bringing together the two parts
of the FTC.
78. Suppose f is continuous on [a, b].
x
1.If g ( x) f (t ) dt , then g’(x) = f(x).
a
b
2. f ( x) dx F (b) F (a) , where F is
a
any antiderivative of f, that is, F’ = f.
79. INVERSE PROCESSES
We noted that the FTC1 can be rewritten
as: d x
f (t ) dt f ( x )
dx a
This says that, if f is integrated and then
the result is differentiated, we arrive back
at the original function f.
80. INVERSE PROCESSES
As F’(x) = f(x), the FTC2 can be rewritten
as: b
F '( x ) dx F (b ) F ( a )
a
This version says that, if we take a function F,
first differentiate it, and then integrate the result,
we arrive back at the original function F.
However, it’s in the form F(b) - F(a).
81. INVERSE PROCESSES
Taken together, the two parts of the FTC
say that differentiation and integration are
inverse processes.
Each undoes what the other does.
82. The FTC is unquestionably the most
important theorem in calculus.
Indeed, it ranks as one of the great
accomplishments of the human mind.
83. Before it was discovered—from the time
of Eudoxus and Archimedes to that of Galileo
and Fermat—problems of finding areas,
volumes, and lengths of curves were so
difficult that only a genius could meet
the challenge.
84. Now, armed with the systematic method
that Newton and Leibniz fashioned out of
the theorem, we will see in the chapters to
come that these challenging problems are
accessible to all of us.
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# Solving Systems of Equations by Elimination
In previous Lessons, we learned how to solve systems of equations by graphing our lines and looking for the intersect and then went on to use our properties of equality to manipulate the equations within a system to solve for the solution.
## Example 1
This example is a reminder of why the math we will do in this lesson is correct.
3 + 4 = 7 this is a true equation,
1 + 9 = 10 this is a true equation,
Now following our properties of equality, we can add these two equations together,
3 + 4 = 7
+(1 + 9) = 10
3 + (1) + 4 + (9) = 7 + (10)
17 = 17, this is also a true equation! We followed the rules. We can use these rules to solve equations.
## Question 1
Solve the system using elimination and so twice - eliminating each variable in turn.
8x + 10y =12
-2x + y = 4
Solve by first eliminating variable x.
As before, when solving these systems, we look to multiply one or both of the equations to eliminate either our x or y terms when the equation is added, so that we can then solve for the x or y term that is remaining. For these questions we are being told to eliminate x.
Remember our Properties of Equality!
So, we look to multiply both or one of the equations so that the x-terms are cancelled out when we add them together.
Here we see 8x in the first equation and -2x in the second. For these terms to eliminate themselves when added, we can multiply our second equation with the -2x by 4, this will give us 8x and -8x which will eliminate each other when we add them together.
So, multiplying the bottom equation by 4,
8x + 10y =12
+ 4 (-2x + y = 4)
8x + 10y =12
+ -8x + 4y = 16
Then adding the two equations together,
8x + (-8x) + 10y + 4y = 12 + 16
As we see, 8x and -8x eliminate each other,
0 + 14y = 28
14y = 28
Divide both sides by 14,
14y / 14 = 28 / 14
y = 2,
Now we have our y-term we need to find our x-term,
Sub y = 2 into -2x + y = 4
-2x + 2 = 4
Subtract 2 from both sides,
-2x + 2 - 2 = 4 - 2
-2x = 2
Divide both sides by -2,
-2x / -2 = 2 / -2
x = -1
So, a solution to our set is x = -1 and y = 2, or (-1, 2)
Next, solve by eliminating y first
8x + 10y =12
-2x + y = 4
Here we must eliminate our y-term first, so look at the 10y in the first equation and the y in the second. What would we have to multiply one or both of our equations by, so that when the two equations are added, the two y-terms cancel out?
If we multiply the second equation by -10, then when we would be left with 10y in the first equation and -10y in the second. When the two equations are added together, 10y and -10y would eliminate each other,
So, Multiply the whole of bottom equation by -10,
8x + 10y =12
+(-10) (-2x + y = 4)
Be careful multiplying out the negative!
8x + 10y =12
+(20x + -10y = -40)
8x + 20x + 10y - 10y = 12 - 40,
The y-terms cancel out as planned,
28x - 0 = -28,
Divide both sides by 28,
28x / 28 = -28 / 28
x = -1,
Just as in a)!
Now sub x = -1 back into either of the equation s to get our y value,
Sub x = - 1 into -2x + y = 4,
-2(-1) + y = 4,
2 + y = 4,
Subtract 2 from both sides,
2 + y - 2 = 4 - 2
y = 2
Again, we have the solution, x = -1 and y = 2, (-1, 2)
Note we have the same solution from two different paths. They are both the same because we used correct mathematics and followed the Properties of Equality. The methods applied may have been different, but the methods were both true, meaning the equations remained true and left us with the same answers.
## Question 2
Solve the following System by using elimination.
-4x + 10y = 20
7x + 4y = 51
Here we have not been told which term to eliminate first, so we can choose the term that looks like the one that would involve the least number of steps.
So, we can try and eliminate the x-term, which would involve multiplying both or one of our equations to eliminate -4x and 7x.There is no one number we could multiply one of the equations to make -4x cancel out with 7x. For the terms to cancel out, we would have to multiply both equations by different numbers.
Let's also look at the y-term, in each equation we have 10y and 4y, so in order for these to eliminate each other when we add the equations together, we would have to multiply both equations as well.
Let's choose to eliminate our x terms first, so we have to have -4x and 7x cancel out,
We can multiply the top equation with -4x in it by 7 to give us -28x,
We can multiply the bottom equation with 7x in it by 4 to give us 28x,
If your struggling to identify what to multiply each equation by so that the x or y terms cancel out, you can always multiply one of the terms by the coefficient of the other, making sure one of the terms is positive and the other one negative,
So, for -4x and 7x,
Multiply the equation with -4x by the 7 (from the 7x) and,
Multiply the equation with 7x by 4 (from the 4x),
Make sure you are left with one of the terms as a negative, so that they cancel out.
So, multiplying the top by 7 and the bottom by 4,
7(-4x + 10y = 20)
4(7x + 4y = 51)
(7)-4x + (7)10y = (7)20
(4)7x + (4)4y = (4)51
-28x + 70y = 140
28x + 16y = 204
-28x + 70y = 140
+(28x + 16y = 204)
-28x + 28x + 70y + 16y = 344
0 + 86y = 344
Divide both sides by 86,
86y / 86 = 344 / 86
y = 4
Now, sub the y = 4 into one of our equations, do -4x + 10y = 20
-4x + 10(4) = 20,
-4x + 40 = 20
Subtract 40 from both sides,
-4x + 40 - 40 = 20 - 40,
-4x = -20
Divide both sides by -4,
-4x / -4 = -20 / -4
x = 5
So, our solution is x = 5 and y = 4 or (5, 4)
We can use the process of elimination to discover equations to lines, when we only know the points on the equation
## Question 3
We have two points (-3, 11) and (1, 7) which a straight line passes through. Recall the general equation for a straight-line y = mx + b. Substitute our x and y values from the points given into this equation, creating two equations. Use elimination to solve for m and b, then state the equation for our line that passes through those points.
This question may seem obscure at first, but with a couple methods we are familiar with, we can create our two equations to make a system of equations which we can solve by elimination.
We want to sub each of our points into the general y = mx + b equation, giving us two equations,
Sub in (-3, 11) or x = -3 y = 11 into y = mx + b,
(11) = m (-3) + b,
-3m + b = 11
Now Sub in (1, 7) or x = 1 and y = 7 into y = mx + b
(7) = m (1) + b
m + b = 7,
Notice we know how a system of equations, this time with the terms m and b instead of x and y,
-3m + b = 11
m + b = 7
We can now eliminate one of our terms, as we have b in one equation and b in the other, let's choose these terms to eliminate first.
With the Properties of Equality, we can add equations together, but we can also subtract them, in this case, subtracting the top equation by the bottom equation can save us a step of multiplying the bottom equation by a -1 in order for the b-terms to cancel out,
However, if you are more comfortable with the process of adding the equations together, you can multiply the bottom or top equations by -1 and then add, but here we will subtract the top and bottom,
Subtracting the top and bottom equations;
-3m + b = 11
-(m + b = 7)
-3m - m + b - b = 11 - 7
-4m = 4
Divide both sides by -4,
-4m / -4 = 4 / -4
m = -1
Now sub our m = - 1 into one of the equations we created, sub m = -1 into m + b = 7,
-1 + b = 7
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## Intermediate Algebra (12th Edition)
$x=\left\{ \dfrac{2-i}{3},\dfrac{2+i}{3} \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $(x-1)(9x-3)=-2 ,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} x(9x)+x(-3)-1(9x)-1(-3)=-2 \\\\ 9x^2-3x-9x+3=-2 \\\\ 9x^2+(-3x-9x)+(3+2)=0 \\\\ 9x^2-12x+5=0 .\end{array} The quadratic equation above has $a= 9 , b= -12 , c= 5 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-(-12)\pm\sqrt{(-12)^2-4(9)(5)}}{2(9)} \\\\ x=\dfrac{12\pm\sqrt{144-180}}{18} \\\\ x=\dfrac{12\pm\sqrt{-36}}{18} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{12\pm\sqrt{-1}\cdot\sqrt{36}}{18} .\end{array} Since $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{12\pm i\sqrt{36}}{18} .\end{array} Simplifying the radicand by writing it as an expression that contains a factor that is a perfect square of the index and then extracting the root of that factor, the equation above is equivalent to \begin{array}{l}\require{cancel} x=\dfrac{12\pm i\sqrt{(6)^2}}{18} \\\\ x=\dfrac{12\pm 6i}{18} .\end{array} Cancelling the common factor in each term results to \begin{array}{l}\require{cancel} x=\dfrac{\cancel6(2)\pm \cancel6(1)i}{\cancel6(3)} \\\\ x=\dfrac{2\pm i}{3} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{2-i}{3} \\\\\text{OR}\\\\ x=\dfrac{2+i}{3} .\end{array} Hence, $x=\left\{ \dfrac{2-i}{3},\dfrac{2+i}{3} \right\} .$
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# Exponent Power
Keywords:
Purpose:
Solving multiplication and division problems using powers is a powerful strategy that meaningfully practices students’ skills with the laws of exponents.
Achievement Objectives:
Specific Learning Outcomes:
In a problem solving context practice these power laws:
Activity:
#### Session 1
Jill’s calculator is broken and she wants to work out
Her friend has prepared a table of powers of 2.
1. From the table discuss why = . Discuss why =
If necessary drop back to
and discuss why this reduces to 28
Discuss why 28 = 512 by looking up the table.
2. Ask students to complete using the table and discuss the answers.
3. Exercises: Students work out
1. Discuss why 32 x 64 = 25 x 26 = 211 = 2048
2. Exercises. Students work out
512 x 32 64 x 64 4 x 2048 4 x 16 x 32 2 x 64 x 16 32 x 16 x 4 64 x 64 x 64 128 x 64 x 2
Discuss why = = = 1024
Exercises: Students work out
#### Session 2
Apply lesson 1’s content and ideas to base 3.
1. Discuss why = = = = 9
Student exercises. Work out
1. To work out 7392discuss why
729 = = = 531441
Student exercises. Work out from tables.
1. Before calculators were allowed in the School Certificate Examination (in 1983) students had to use these power methods to do multiplications and divisions. Here is a small part of the tables students used. These methods are still used in some year 13 mathematics.
1. Discuss why 2 x3 =10 x 10
= 10 = 6 (Note the slight rounding problem)
Student exercises. Show, using the base ten table, that the following are true
4 ÷ 2= 2 8 ÷ 2= 4 3 x 3 = 9
10 ÷ 2 = 5 2 x 5= 10
Discuss how the table can be used to show
5 x 4 = 20.
#### Session 3
Refer to previous power tables to find square roots, and cube roots.
1. Discuss from the base 2 table how to find
4096 = x so = x
Discuss why goes in the box so
2. Student exercises. Use the base 2 table to work out these answers.
1. Student Exercises. Use the base 3 tables to work out
#### Session 4
Negative powers are introduced
1. Discuss why . Discuss why = 1.
2. Discuss why
Discuss why
3. Student Exercises. Work out
1. Discuss why
2. Student Exercises. Work out
#### Session 5
Relate powers to geometric sequences in an elementary way.
1. Bacteria double every hour starting with 1 thousand bacteria.
Discuss why the sequence of the bacteria is
1, 2, 4, 8, 16, 32,...thousands or ...
Why is the number of bacteria after n hours equal to ?
2. At some time there are 64 thousand bacteria. Later there are 16384 thousand bacteria. How long did this increase take?
Discuss why = shows increasing from thousand to thousand must take 8 hours.
3. Student Exercises. How long will it take to change from one number of bacteria to one other?
8 thousand to 512 thousand
4 thousand to 8192 thousand
64 thousand to 8192 thousand
1/2 thousand to 512 thousand
1/8 thousand to 64 thousand
Unfortunately for the bacteria the host takes an anti-biotic which halves the number of bacteria every hour. Bert has 512 thousand bacteria per litre of his blood when he takes the antibiotic he will feel well when the level of live bacteria reaches 1/4 thousand bacteria per litre. Discuss how long it will take before Bert is well.
Student Exercises. If bacterial levels divide by 2 every hour find how long it will take to change from one number to another.
8192 thousand to 1 thousand 64 thousand to 1/8 thousand
256 thousand to 1/32 thousand 1/8 thousand to 1/1024 thousand
## Square and Cube Roots
The fact that squaring and square roots are inverses is explored geometrically and numerically. Gauss’ method of determining square roots when only squares are available is developed. Finally a powerful method of calculating square roots that produces answers to any desired accuracy quickly is shown.
## Moving and Shaking
This is a level 5 and 6 number activity from the Figure It Out theme series.
## Four 4s
This is a level 5 number activity from the Figure It Out series. It relates to Stage 8 of the Number Framework.
## Hundreds
This is a level 5 number activity from the Figure It Out series. It relates to Stage 8 of the Number Framework.
## The Power of Powers
This is a level 5 number activity from the Figure It Out series. It relates to Stage 8 of the Number Framework.
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# Half-Angle Identities: Uses & Applications
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• 0:01 Half-Angle Identities
• 0:42 Uses and Applications
• 1:30 Example 1
• 2:10 Example 2
• 3:07 Lesson Summary
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Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
After watching this video lesson, you will be able to use half-angle identities to help you simplify your trig problems and also to help you prove other trig statements.
## Half-Angle Identities
In this video lesson, we are covering the half-angle identities of trigonometry, which are the true statements for half-angles. These are definitions, if you will. In other words, they tell you what a particular trig function equals. You can see here that they take a squared trig function and turn it into a trig function without exponents:
We have one half-angle for each of our three basic trig functions. We have one for sine, one for cosine, and one for tangent. As you can see, our angle has been halved, hence the name half-angle identity. On the left side, our trig function is squared, and on the right, we see the equivalent statement in terms of cosine without any exponents. Also, the angle on the right side is no longer halved.
## Uses and Applications
So, what can you do with these identities? We use these identities when we need help simplifying a trig function. When we have a squared trig function, it is sometimes difficult to work with in higher math. So, if we turn it into an equivalent statement without exponents, then it will help us solve the problem that much more easily.
You can think of these half-angle trig identities as a key that helps you to decode or simplify a harder problem. Without this key, you might not be able to solve the problem at all. But with the key, you are able to find your way to the answer. Also, these identities are used to prove yet other trig identities or statements. Let's look at a couple of examples now to see what we can do with these trig identities.
## Example 1
This problem is asking us to write this trig function without any exponents:
At first glance, we might think that this problem is already as simple as it can get. It only has one function: the sine function. But the problem wants us to write it so that we don't have the square. How can we do that?
Well, we look at the angle and notice that our angle is being halved. Ah. We can see if we can use one of our half-angle identities. We look at our list. Oh, look! Here, we have a sine squared function that is equal to something without any exponents. We can use that one to get our answer:
And there we have our answer without any exponents. This problem was a pretty direct problem with a simple direct answer. Let's look at another problem.
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# Place Value and Expanded Form
## Presentation on theme: "Place Value and Expanded Form"— Presentation transcript:
Place Value and Expanded Form, Rounding to the Nearest 100, and Estimation Multiplication
Place Value and Expanded Form
4,597, what does this number really mean? The 5 in the hundreds place means 5 x 100 4, 359 The 5 in the tens place means 5 x 10 So, remember when you want to know the value of a number, check out its place value and multiply it by the number!
Rounding to the Nearest Hundred
Rule: Any number from 50↑ rounds up to the next hundred Any number from 49↓ rounds down to that hundred
Rounding Up↑ To round 4,597 to the nearest hundred
because 97 is 50 or greater I round up to 600. 4,597 rounded to the nearest 100 is 4,600
Rounding Down ↓ To round 4,547 to the nearest hundred
because 47 is 49 or less I round down to 500. 4,547 rounded to the nearest 100 is 4,500
Estimation Multiplication
To estimate the answer to a three digit multiplication problem First round to the nearest hundred then multiply the first two digits and finally add your zeros
For Example: To estimate the answer to 235 x 7
First round 235 to the nearest hundred: Because 35 is 49 or less, round ↓ to 200, then multiply 2 x 7 = 14, add the two zeros and the answer is 1,400 235 x 7 = ~1,400
To estimate the answer to: 151 x 8
First round 151 to the nearest hundred: Because 51 is 50 or more, round ↑ to 200, then multiply 2 x 8 = 16, add the two zeros and the answer is 1,600 151 x 8 = ~1,600
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# Math Worksheets Land
Math Worksheets For All Ages
# Math Worksheets Land
Math Worksheets For All Ages
# Parentheses, Brackets, and Braces in Math Expressions Worksheets
What is the Purpose of Parentheses, Brackets, and Braces in Math Expressions? When you are doing mathematics problems, you will come across many symbols that are important for solving these problems. Similar is the case for parentheses, brackets and braces; which are used many times in prealgebra and algebra. That is why it is necessary for you to understand what each of these symbols is used for. Parentheses () - Parentheses which are denoted by round brackets are used to group numbers, variables or both within a problem. It tells us the order of operations in a single operation. For example, take the problem 9 + (6 - 5) x 3, Normally, the problem will begin with the multiplication operation. However, here because of the parentheses, the subtraction will be solved first. Parentheses can also be used to multiply numbers. Like 2 x 5 can be also be written as (2)(5). Brackets [] - Brackets are also used for grouping variables and numbers together. Usually, in a single problem, parentheses are solved first and then brackets. Braces {} - Braces are used just like brackets and parentheses, i.e. to group numbers and variables. However, braces are solved after the parentheses and the braces. Braces are also used in sets for discussing a set of elements. Let us look at an example, 5 - {3[5 - 2(3-1)] ÷ 3}. We will now discuss the problem at every step.
5 - {3[5 - 2(3-1)] ÷ 3}, 5 - 3{[5-2(2)] ÷ 3} - Multiplying 2 and 2, 5 - {3[5 - 4]) ÷ 3} -Subtracting 4 from 5, 5 - {3(1) ÷ 3} = 5 - {3 ÷ 3} - Dividing 3 with 3 between the braces, 5 - 1 = 4. These worksheets and lessons will help you learn the different meanings and uses of parentheses, brackets, and braces in expressions.
### Aligned Standard: Grade 5 Operations - 5.OA.1
• Answer Keys - These are for all the unlocked materials above.
### Homework Sheets
These are designed to help remind students of the skill and provide them a great deal of practice. I made plenty.
• Homework 1 - Complete the following problems. Make sure to review your order of operations.
• Homework 2 - Breakdown the parentheses first. You will need to complete it through.
• Homework 3 - Exponents make there way in here. Note if exponent is inside or outside. It makes a big difference.
• Homework 4 - Division is pretty popular on this sheet.
• Homework 5 - For the most part, these are three step problems.
• Homework 6 - Folded brackets might be new to students.
### Practice Worksheets
These problems are well spaced out to provide you with plenty of work room.
• Practice 1 - PEMDAS appears. It is spelled out for you on this worksheet.
• Practice 2 - If you see a bracket, that is where the problem begins.
• Practice 3 - If you have an exponent outside of a parentheses, you must perform the operation within the parentheses before applying the exponent.
• Practice 4 - Follow PEMDAS out in the bracket first.
• Practice 5 - Lots of room to work here. Note the outside operation that needs to take place.
• Practice 6- You will need the room for your vision while solving the problems.
### Math Skill Quizzes
These are all in direct sync with the skills that students will see on any and all national exams.
• Quiz 1 - The first three are in bold Arial font for vision help.
• Quiz 2 - Note the position of the exponents within each exercise.
• Quiz 3 - In most cases you will need to focus bracket and break it down within the problem.
• Quiz 4 - Standard Arial font is used for integers here and on.
• Quiz 5 - The integers are much larger on this quiz, but not found in bold.
• Quiz 6 - These are the longest problems of all.
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Question #931d4
Apr 30, 2017
We have to use that the geometric series is of the form:
$a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \ldots$
and that we know that the whole sum is $13.5$ and the sum of the first three terms $a + a r + a {r}^{2} = 13$
Explanation:
So we have:
$a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \ldots = 13.5$
and
$a + a r + a {r}^{2} = 13$
Subtracting we have:
$\left(a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \ldots\right) - \left(a + a r + a {r}^{2}\right) = 0.5$, that is:
$a {r}^{3} + a {r}^{4} + a {r}^{5} + a {r}^{6} + a {r}^{7} + \ldots = 0.5$
Hence, taking $r$ as a common factor:
${r}^{3} \cdot \left(a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \ldots\right) = 0.5$
But we know that that the expression between brackets:
$a + a r + a {r}^{2} + a {r}^{3} + a {r}^{4} + \ldots = 13.5$
so we have:
${r}^{3} \cdot 13.5 = 0.5$, that is:
${r}^{3} = \frac{0.5}{13.5}$, that is:
${r}^{3} = \frac{\frac{1}{2}}{\frac{27}{2}} = \frac{1}{27}$, and then
$r = \frac{1}{3}$
But we know that $a + a r + a {r}^{2} = 13$, so:
$a + a \frac{1}{3} + a \frac{1}{9} = 13$, and that means:
$\frac{9 a + 3 a + a}{9} = 13$, so:
$\frac{13 a}{9} = 13$, so $\frac{a}{9} = 1$, and from that we get:
$a = 9$, which is the first term
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# Search by Topic
#### Resources tagged with Rational and irrational numbers similar to Black Box:
Filter by: Content type:
Stage:
Challenge level:
### There are 20 results
Broad Topics > Numbers and the Number System > Rational and irrational numbers
### Proof Sorter - the Square Root of 2 Is Irrational
##### Stage: 5 Challenge Level:
Try this interactivity to familiarise yourself with the proof that the square root of 2 is irrational. Sort the steps of the proof into the correct order.
### Be Reasonable
##### Stage: 5 Challenge Level:
Prove that sqrt2, sqrt3 and sqrt5 cannot be terms of ANY arithmetic progression.
### Continued Fractions II
##### Stage: 5
In this article we show that every whole number can be written as a continued fraction of the form k/(1+k/(1+k/...)).
### An Introduction to Proof by Contradiction
##### Stage: 4 and 5
An introduction to proof by contradiction, a powerful method of mathematical proof.
### The Clue Is in the Question
##### Stage: 5 Challenge Level:
This problem is a sequence of linked mini-challenges leading up to the proof of a difficult final challenge, encouraging you to think mathematically. Starting with one of the mini-challenges, how. . . .
### Irrational Arithmagons
##### Stage: 5 Challenge Level:
Can you work out the irrational numbers that belong in the circles to make the multiplication arithmagon correct?
### Road Maker 2
##### Stage: 5 Short Challenge Level:
Can you work out where the blue-and-red brick roads end?
### Impossible Triangles?
##### Stage: 5 Challenge Level:
Which of these triangular jigsaws are impossible to finish?
### Rational Roots
##### Stage: 5 Challenge Level:
Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.
### Impossible Square?
##### Stage: 5 Challenge Level:
Can you make a square from these triangles?
### Rationals Between...
##### Stage: 4 Challenge Level:
What fractions can you find between the square roots of 65 and 67?
### Good Approximations
##### Stage: 5 Challenge Level:
Solve quadratic equations and use continued fractions to find rational approximations to irrational numbers.
### An Introduction to Irrational Numbers
##### Stage: 4 and 5
Tim Rowland introduces irrational numbers
### The Square Hole
##### Stage: 4 Challenge Level:
If the yellow equilateral triangle is taken as the unit for area, what size is the hole ?
### Equal Equilateral Triangles
##### Stage: 4 Challenge Level:
Using the interactivity, can you make a regular hexagon from yellow triangles the same size as a regular hexagon made from green triangles ?
### Spirostars
##### Stage: 5 Challenge Level:
A spiropath is a sequence of connected line segments end to end taking different directions. The same spiropath is iterated. When does it cycle and when does it go on indefinitely?
### The Root Cause
##### Stage: 5 Challenge Level:
Prove that if a is a natural number and the square root of a is rational, then it is a square number (an integer n^2 for some integer n.)
### What Are Numbers?
##### Stage: 2, 3, 4 and 5
Ranging from kindergarten mathematics to the fringe of research this informal article paints the big picture of number in a non technical way suitable for primary teachers and older students.
### Rational Round
##### Stage: 5 Challenge Level:
Show that there are infinitely many rational points on the unit circle and no rational points on the circle x^2+y^2=3.
### Approximations, Euclid's Algorithm & Continued Fractions
##### Stage: 5
This article sets some puzzles and describes how Euclid's algorithm and continued fractions are related.
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Beispiel Nr: 05
$\text{Gegeben:} \\ \begin{array}{cc} \text{Matrix A} & \text{Matrix B} \\ \left[ \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n}\\ a_{21} & a_{22} & \ldots & a_{2n}\\ \vdots & \vdots &\vdots & \vdots \\ a_{m1} & a_{m2} & \ldots & a_{mn}\\ \end{array} \right] & \left[ \begin{array}{cccc} b_{11} & b_{12} & \ldots & b_{1n}\\ b_{21} & b_{22} & \ldots & b_{2n}\\ \vdots & \vdots &\vdots & \vdots \\ b_{m1} & b_{m2} & \ldots & b_{mn}\\ \end{array} \right] \end{array}$
$\textbf{Aufgabe:}$ Mutliplizieren
$\textbf{Rechnung:}\\ \small \left[ \begin{array}{cc} 5 & 6 \\ 7 & 8 \\ \end{array} \right] \cdot \small \left[ \begin{array}{cc} -3 & 6 \\ 5 & 4 \\ \end{array} \right] = \\ \small \left[ \begin{array}{cc} (5\cdot-3) + (6\cdot5) & (5\cdot6) + (6\cdot4) \\ (7\cdot-3) + (8\cdot5) & (7\cdot6) + (8\cdot4) \\ \end{array} \right] = \\ \small \left[ \begin{array}{cc} 15 & 54 \\ 19 & 74 \\ \end{array} \right]$
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# Find the L.C.M. of 2.5, 0.5 and 0.175.
Last updated date: 17th Jun 2024
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Hint: First remove the decimal by multiplying all the numbers with 1000. To balance, divide them with 1000. Leave the number 1000 in the denominator as it is and consider the numbers obtained in the numerator, find their L.C.M. by prime factorization method. Once the L.C.M. is found, divide it with 1000 to get the answer.
\begin{align} & \Rightarrow 2.5,0.5,0.175=\dfrac{2500}{1000},\dfrac{500}{1000},\dfrac{175}{1000} \\ & \Rightarrow 2.5,0.5,0.175=\dfrac{1}{1000}\left( 2500,500,175 \right) \\ \end{align}
\begin{align} & \Rightarrow 2500={{2}^{2}}\times {{5}^{4}} \\ & \Rightarrow 500={{2}^{2}}\times {{5}^{3}} \\ & \Rightarrow 175={{5}^{2}}\times 7 \\ \end{align}
$\Rightarrow L.C.M.={{2}^{2}}\times {{5}^{4}}\times 7$
\begin{align} & \Rightarrow L.C.M.=\dfrac{{{2}^{2}}\times {{5}^{4}}\times 7}{1000} \\ & \Rightarrow L.C.M.=17.5 \\ \end{align}
Note: One may note that the numbers 2.5 and 0.5 will get free form the decimal even after multiplying with 10, but the third number 0.175 will have to be multiplied with 1000 to get free from decimal and that is why we choose this number. You can note that the general method of finding the L.C.M. of two fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$ is that, first we find the L.C.M. of a and c, i.e., the numerators, then in the next step we find the H.C.F. of b and d, i.e., the denominators and finally we use the relation: - L.C.M. required = $\dfrac{L.C.M.\left( a,c \right)}{H.C.F.\left( b,d \right)}$, to get the answer.
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# Boundary Point: Simple Definition & Examples
## What is a Boundary Point?
Generally speaking, a “boundary point” is the edge or border of a set.
The set of all boundary points is called a boundary [1]. Visually, it’s the line that circumscribes the set. For example, the boundary of a circular set (i.e. one defined by the formula x2 + y2 < 1) is the circumference.
Boundaries aren’t perfect lines; They are collections of point, some of which are inside the set and some are outside. As an example, zoom in on the image above and you’ll see that the boundary is actually a series of tiny printed dots. Some of the dots are inside the set, and some are outside:
A boundary point then, is neither an interior point (which is definitely in a set) or an exterior point (which is definitely outside the set).
## Formal Definition of Boundary Points
Formally speaking, a point x is a boundary point for a region R if every neighborhood (a set of points that are in the same general area) of x intersects both R and the complement of R.
This formal definition may be easier to wrap your head around if you look at an example on the number line.
Let’s say R = [1, 4) has boundary points of 1 and 4. If you draw the set on a number line, it’s clear that the neighborhoods of 1 and 4 intersect the defined set and it’s complement.
## Boundary Point vs. Limit Point
A limit point isn’t a “limit ” in the usual calculus sense of the word, but the two terms are related. A limit point is very close to a set, so close in fact in can actually be in the set (and often is).
Limit points and boundary points are similar in that neither have to be in the set. The difference is in their neighborhoods:
1. A limit point has a neighborhood containing at least one point belonging to the set.
2. A boundary point has a neighborhood with a point belonging to the set and a point not belonging in the set.
Every boundary point of a set is a limit point [2]. Not every limit point is a boundary point: point 1 is a limit point (it’s also an interior point), but it is not on the boundary.
## References
[1] LaValle, S. (2006). Planning Algorithms. Retrieved April 3, 2021 from: http://planning.cs.uiuc.edu/node127.html
[2] Buckmire, R. (2004). Multivariable Calculus. Retrieved April 3, 2021 from: https://sites.oxy.edu/ron/math/224/04/ws/12.pdf
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# Fraction Rules
We present examples on how to use the rules of fractions to simplify expressions including fractions. Questions and their solutions with detailed explanations are also included.
Fractions are some of the most important concepts in mathematics and therefore must be very well understood. Students need to have strong skills related to operations on fractions such as adding, dividing, equivalent fractions, ... . The skills discussed here gives you the much needed basics in dealing with fraction in numerical form as well as fractions with variables.
There are rules and definitions, the idea is to go through them in the order that they are presented and understand each one before you move to the next one.
In order to develop strong skills in fractions, there rules and any other rules on fractions, are to be fully understood and used to practice on more examples and questions for a long as you are studying mathematics.
$\require{cancel}$ $\newcommand\ccancel[2][black]{\color{#1}{\xcancel{\color{black}{#2}}}}$
1. Numerator and Denominator of a Fraction
A fraction is written in the form ${\Large \dfrac{a}{b}}$ where $a$ and $b$ are integers. $a$ is called the numerator and $b$ the denominator which can never be equal to zero.
2. The Denominator of a Fraction is Never Equal to Zero
These fractions are undefined because their denominators are equal to zero and are therefore NOT allowed in mathematics.
$\quad \ccancel[red]{\dfrac{2}{0}}$ , $\ccancel[red]{\dfrac{0}{0}}$ , $\ccancel[red]{\dfrac{1000000}{0}}$
3. The Numerator of a Fraction May be Equal to Zero
Any fraction whose numerator is equal to zero is itself equal to zero as long as its denominator is not equal to zero.
$\quad \dfrac{0}{3} = 0$ , $\dfrac{0}{100000} = 0$ , $\dfrac{0}{-8} = 0$
The fraction $\ccancel[red]{\dfrac{0}{0}}$ is undefined because its denominator is equal to zero.
4. A Fraction with Denominator Equal to 1 Simplies to an Integer
Any fraction whose denominator is equal to 1 may be written as an integer
$\quad \dfrac{4}{1} = 4$ , $\dfrac{9}{1} = 9$ , $\dfrac{2x}{1} = 2x$
5. A Fraction with Denominator Equal its Numerator
Any fraction whose denominator is equal to its denominator simplifies to 1
$\quad \dfrac{6}{6} = 1$ , $\dfrac{3x}{3x} = 1$ for $x \ne 0$
6. Equivalent Fractions
Two fractions $\dfrac{a}{b}$ and $\dfrac{c}{d}$ are equivalent and may be written as $\dfrac{a}{b} = \dfrac{c}{d}$ if and only if $\quad a \times d = b \times c$
Example
a) The two fractions $\dfrac{\color{blue}2}{\color{brown}5}$ and $\dfrac{\color{red}6}{15}$ are equivalent because $\quad \color{blue}2 \times 15 = 30$ and $\color{brown}5 \times \color{red}6 = 30$ and therefore $\color{blue}2 \times 15 = \color{brown}5 \times \color{red}6$
b) The two fractions $\dfrac{\color{blue}{2x}}{\color{brown}3}$ and $\dfrac{\color{red}4x}{6}$ are equivalent because $\quad \color{blue}{2x} \times 6 = 12x$ and $\color{brown}3 \times \color{red}{4x} = 12 x$ and therefore $\color{blue}{2x} \times 6 = \color{brown}3 \times \color{red}{4x}$
7. How to Make Equivalent Fractions by Mulitplication
You can make equivalent fractions by multiplying the numerator and denominator of the given fraction by the same number $k$ , $k \ne 0$:
$\quad \dfrac{a}{b} = \dfrac{a\color{red}{\times k}}{b \color{red} {\times k}}$
There are many ways to write equivalent fractions by multiplication depending on the values of $k$.
Example
a) $\quad \dfrac{3}{7} = \dfrac{3 \color{red}{\times 5}}{7 \color{red}{\times 5} } = \dfrac{15}{35}$
b) $\quad \dfrac{2}{3} = \dfrac{2 \color{red}{\times 2x}}{3 \color{red}{\times 2x} } = \dfrac{4x}{6x}$ for $x \ne 0$
8. How to Make Equivalent Fractions by Division
You can make equivalent fractions by dividing the numerator and denominator of the given fraction by the same number $k$ , $k \ne 0$:
$\quad \dfrac{a}{b} = \dfrac{a \color{red}{\div k}}{b \color{red}{\div k}}$
Example
a) $\quad \dfrac{8}{12} = \dfrac{8 \color{red}{\div 4}}{12 \color{red}{\div 4} } = \dfrac{2}{3}$
b) $\quad \dfrac{x}{14x} = \dfrac{x \color{red}{\div x}}{14x \color{red}{\div x} } = \dfrac{1}{14}$ for $x \ne 0$
9. Reciprocal of a Fraction
The reciprocal of a fraction $\dfrac{a}{b}$ is equal to $\quad \color{red}{ \dfrac{b}{a} }$ for $a \ne 0$.
Note: The product of the fraction $\dfrac{a}{b}$ and its reciprocal $\color{red}{ \dfrac{b}{a} }$ is equal to 1.
$\dfrac{a}{b} \times \color{red}{ \dfrac{b}{a} } = 1$
Note: The reciprocal of a fraction whose denominator is equal to zero is undefined.
Example
a) The reciprocal of $\dfrac{7}{9}$ is $\dfrac{9}{7}$
and $\dfrac{7}{9} \times \dfrac{9}{7} = \dfrac{7 \times 9}{9 \times 7} = \dfrac{63}{63} = 1$
b) The reciprocal of $\dfrac{0}{7}$ is undefined because the numeartor of the given fraction is equal to zero.
c) The reciprocal of $\dfrac{x}{2}$ is $\dfrac{2}{x}$ , for $x \ne 0$
and $\dfrac{x}{2} \times \dfrac{2}{x} = \dfrac{x \times 2}{2 \times x} = \dfrac{2x}{2x} = 1$
10. Write an Integer as a Fraction
Any integer $a$ may be written as a fraction as follows:
$\quad a = \dfrac{a \color{red}{\times k}}{ \color{red}k}$ for any integer $k \ne 0$
There are many ways to write an integer as a fraction
Example
a) $\quad 3 = \dfrac{3 \times 1}{1} = \dfrac{3}{1}$
b) $\quad 3 = \dfrac{3 \times 4}{4} = \dfrac{12}{4}$
c) $\quad 5 = \dfrac{5 \times x}{x} = \dfrac{5x}{x}$ for $x \ne 0$
11. Write a Decimal Number as a Fraction
A decimal number $a$ may be written as a fraction by first writing it as a division by 1 and then mutliply the top and bottom of the division by a multiple of 10 such that the decimal number becomes an integer.
Example
a) $\quad 0.1 = \dfrac{0.1 }{1} = \dfrac{0.1 \times 10}{1 \times 10 } = \dfrac{1}{10}$
b) $\quad 2.09 = \dfrac{2.09 }{1} = \dfrac{2.09 \times 100 }{1 \times 100} = \dfrac{209}{100}$
c) $\quad 0.0137 = \dfrac{0.0137}{1} = \dfrac{0.0137 \times 10000}{1 \times 10000} = \dfrac{137}{10000}$
a) Add Fractions with Common Denominators
We add fractions with common denominators by keeping the common denominator and adding the numerators as follows:
$\quad \dfrac{a}{c} + \dfrac{b}{c} = \dfrac{ \color{red}{a+b}}{c}$
Example
$\quad \dfrac{2}{8} + \dfrac{1}{8} = \dfrac{2+1}{8} = \dfrac{3}{8}$
b) Add Fractions with Different Denominators
We add fractions with different denominators by first writing the two fractions to add with a common denominator then add them.
$\quad \dfrac{a}{\color{red} c} + \dfrac{b}{\color{blue} d} = \dfrac{a}{c} \color{blue}{ \times \dfrac{d}{d}} + \dfrac{b}{d} \color{red}{ \times \dfrac{c}{c}}$
$= \dfrac{a \times d}{c \times d} + \dfrac{b \times c}{d \times c} = \dfrac{a \times d + b \times c}{c\times d}$
Example
$\quad \dfrac{2}{3} + \dfrac{1}{5}$
$= \dfrac{2}{3} \color{red}{ \times \dfrac{5}{5}} + \dfrac{1}{5} \color{red}{ \times \dfrac{3}{3}}$
$= \quad \dfrac{2 \times 5 + 1 \times 3 }{3 \times 5} = \dfrac{13}{15}$
13. Subtract Fractions
a) Subtract Fractions with Common Denominators
We subtract fractions with common denominators by keeping the common denominator and subtracting the numerator as follows:
$\quad \dfrac{a}{c} - \dfrac{b}{c} = \dfrac{ \color{red}{ a - b}}{c}$
Example
$\quad \dfrac{11}{7} - \dfrac{6}{7} = \dfrac{11-6}{7} = \dfrac{5}{7}$
b) Subtract Fractions with Different Denominator
We subtract fractions with different denominators by first writing the two fractions to subtract with a common denominator and then subtract them.
$\quad \dfrac{a}{\color{red}c} - \dfrac{b}{ \color{blue} d} = \dfrac{a}{c} \color{blue}{ \times \dfrac{d}{d}} - \dfrac{b}{d} \color{red}{ \times \dfrac{c}{c} }$
$= \dfrac{a \times d}{c \times d} - \dfrac{b \times c}{d \times c} = \dfrac{a \times d - b \times c}{c\times d}$
Example
$\quad \dfrac{4}{3} - \dfrac{2}{5}$
$= \dfrac{4}{3} \color{red}{ \times \dfrac{5}{5}} - \dfrac{2}{5} \color{red}{ \times \dfrac{3}{3} }$
$\quad =\dfrac{20}{15} - \dfrac{6}{15} = \dfrac{14}{15}$
14. Multiply Fractions
We multiply fractions by multiplying numerator by numerator and denominator by denominator as follows:
$\quad \dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{\color{red}{a \times c}}{\color{red}{b \times d}}$
Example
$\quad \dfrac{2}{5} \times \dfrac{4}{7} = \dfrac{2 \times 4}{5 \times 7} = \dfrac{8}{35}$
15. Divide Fractions
We divide fractions by mutliplying the first fraction by the reciprocal of the second fraction as follows:
$\quad \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \color{red}{ \times \dfrac{d}{c}}$
Example
$\quad \dfrac{7}{2} \div \dfrac{2}{5} = \dfrac{7}{2} \color{red}{\times \dfrac{5}{2}} = \dfrac{7 \times 5}{2 \times 2} = \dfrac{35}{4}$
16. Add a Number and a Fraction
We add a number and a fraction by rewriting the number as a fraction with common denominator then add as follows.
$\quad a + \dfrac{b}{c} = \color{red} {a \times \dfrac{c}{c}} + \dfrac{b}{c} = \dfrac{a c }{c} + \dfrac{b}{c} = \dfrac{a c + b}{c}$
Example
$\quad 2 + \dfrac{3}{4} = \color{red} {2 \times \dfrac{4}{4}} + \dfrac{3}{4}$
$\quad \quad = \dfrac{2 \times 4 }{4} + \dfrac{3}{4} = \dfrac{8}{4} + \dfrac{3}{4}$
$\quad \quad \quad = \dfrac{8 + 3}{4} = \dfrac{11}{4}$
17. Multiply a Number by a Fraction
We multiply a number by a fraction by rewriting the number as a fraction then multiply as follows .
$\quad a \times \dfrac{b}{c} = \color{red}{ \dfrac{a}{1}} \times \dfrac{b}{c} = \dfrac{a b}{c}$
Example
$\quad 7 \times \dfrac{2}{5} = \color{red}{ \dfrac{7}{1} } \times \dfrac{2}{5} = \dfrac{7 \times 2}{1 \times 5} = \dfrac{14}{5}$
18. Divide a Number by a Fraction
We divide a number by a fraction by first rewriting the number as a fraction then divide as follows .
$\quad a \div \dfrac{b}{c} = \color{red}{ \dfrac{a}{1} } \div \dfrac{b}{c} = \dfrac{a}{1} \times \dfrac{c}{b} = \dfrac{a c}{b}$
Example
$\quad 2 \div \dfrac{3}{11} = \color{red}{ \dfrac{2}{1}} \times \dfrac{11}{3} = \dfrac{2 \times 11}{1 \times 3} = \dfrac{22}{3}$
19. Divide a Fraction by a Number
We divide a fraction by a number by first rewriting the number as a fraction then divide as follows .
$\quad \dfrac{a}{b} \div c = \dfrac{a}{b} \div \color{red}{ \dfrac{c}{1}} = \dfrac{a}{b} \times \dfrac{1}{c} = \dfrac{a}{b c}$
Example
$\quad \dfrac{2}{7} \div 3 = \dfrac{2}{7} \div \color{red}{ \dfrac{3}{1} } = \dfrac{2}{7} \times \dfrac{1}{3} = \dfrac{2}{21}$
20. Signed Fractions
Signed fractions may be written in any of the following forms:
a)
$\quad - \dfrac{a}{b} = \dfrac{- a}{b} = \dfrac{a}{-b}$
Example
$\quad - \dfrac{5}{12} = \dfrac{-5}{12} = \dfrac{5}{-12}$
b)
$\quad \dfrac{ - a}{ - b} = \dfrac{a}{b}$
Example
$\quad \dfrac{-2}{-7} = \dfrac{2}{7}$
## Questions
1. Write the follwoing as a single fraction and reduce it to lowest terms if possible. Solutions are included.
1. ) $\dfrac{0}{3} + \dfrac{1}{3}$
2. ) $\dfrac{2}{0} + 5$
3. ) $\dfrac{3}{5} + 2$
4. ) $\dfrac{3}{2} + 2.1$
5. ) $0.1 x + \dfrac{2x}{3}$
6. ) $3 x + \dfrac{x}{4}$
7. ) $3x - \dfrac{5 x}{4}$
8. ) $\dfrac{3}{5} \times \dfrac{4}{9}$
9. ) $6 \times \dfrac{3}{7}$
10. ) $\dfrac{2x}{5} \times \dfrac{1}{2}$
11. ) $\dfrac{6}{7} \div 3$
12. ) $x \div \dfrac{1}{9}$
13. ) $\dfrac{2x}{5} \div \dfrac{1}{9}$
14. ) $- \dfrac{-3}{5} + \dfrac{-3}{5}$
15. ) $\dfrac{2}{-9} + \dfrac{7}{9}$
16. ) $\dfrac{-5}{-2} - \dfrac{7}{2}$
17. ) $\dfrac{-2x}{3} - \dfrac{ - 5x}{- 3}$
18. ) $2 - \dfrac{ 4 + \dfrac{1}{3}}{1+\dfrac{1}{2}}$
19. ) $x - \dfrac{ 2x + \dfrac{x}{2}}{x - \dfrac{2x}{3}}$
solutions to the above questions included.
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# Proof 7: Pyramid and cone volume
The Cheops pyramid in Egypt is constructed of many equally thick layers of box-shaped stones (more or less). Imagine more and more such layers being thinner and thinner, and you eventually get a smooth and close to perfect pyramid. So, the number of layers n must be a very large number before we can think of it as a pyramid.
A pyramid’s volume consists of each of the n prism-shaped layers’ volume:
\begin{aligned}
V_\text{pyramid}&=\underbrace{V_1+V_2+\cdots+V_{n_1}+V_n}_{n \text{ terms}}\\
~&=\underbrace{A_1h_\text{layer}+A_2 h_\text{layer} +\cdots+A_{n_1} h_\text{layer} +A_n h_\text{layer} }_{n \text{ terms}}\qquad\leftarrow \text{Prism volume, } V=A h_\text{layer} \\
~&=\underbrace{\left(A_1+A_2 +\cdots+A_{n_1} +A_n \right) }_{n \text{ terms}} h_\text{layer} \\
~&=\underbrace{\left(A_1+A_2 +\cdots+A_{n_1} +A_n \right) }_{n \text{ terms}} \frac 1n h \qquad \leftarrow \text{ Each layer is a bit of the full height, } h_\text{layer} =\frac 1n h
\end{aligned}
The n layers’ heights all-together make up the pyramid height h. Each is an n’th of this full height, so h_\text{layer} =\frac 1n h.
All lengths become smaller with one n’th at a time up through the layers. Lengths in the second-bottommost layer (which has number n-1) are then missing one n’th, so there are 1-\frac 1n=\frac{n-1}n left, those in the third layer from the bottom, layer no. n-2, are missing two n’ths \frac 2n and only have \frac{n-2}n left etc. Areas consist of some lengths times some lengths times, and with all such lengths reducing by an n’th in each step, the areas reduce by one n’th times one n’th:
A_{n-1}=\frac{n-1}n \frac{n-1}n A_\text{base}= \left(\frac{n-1}n\right)^2 A_\text{base}= \frac{(n-1)^2}{n^2} A_\text{base} \\
A_{n-2}=\frac{n-2}n \frac{n-2}n A_\text{base}= \left(\frac{n-2}n\right)^2 A_\text{base}= \frac{(n-2)^2}{n^2} A_\text{base} \\
~\vdots\\
A_2=\frac2n \frac2n A_\text{base}= \left(\frac2n\right)^2 A_\text{base}= \frac{2^2}{n^2} A_\text{base} \\
A_1=\frac1n \frac1n A_\text{base}= \left(\frac1n\right)^2 A_\text{base}= \frac{1^2}{n^2} A_\text{base}
And A_n can of course be written as A_n=\frac{n^2}{n^2} A_\text{base} to stay in pattern (dividing a number by itself gives 1; and it makes no difference to multiply with 1). Let’s put all these areas in:
\begin{aligned}
V_\text{pyramid}&=\underbrace{\left(A_1+A_2 +\cdots+A_{n_1} +A_n \right) }_{n \text{ terms}} \frac 1n h\\
~ &=\underbrace{\left( \frac{1^2}{n^2} A_\text{base} + \frac{2^2}{n^2} A_\text{base} +\cdots+ \frac{(n-1)^2}{n^2} A_\text{base} + \frac{n^2}{n^2} A_\text{base} \right) }_{n \text{ terms}} \frac 1n h\\
~ &=\underbrace{\left( \frac{1^2}{n^2} + \frac{2^2}{n^2} +\cdots+ \frac{(n-1)^2}{n^2} + \frac{n^2}{n^2} \right) }_{n \text{ terms}} \frac 1n h \,A_\text{base} \\
~ &=\underbrace{\left( 1^2 +2^2 +\cdots+ (n-1)^2 + n^2 \right) }_{n \text{ terms}} \frac 1{n^3} h \,A_\text{base}
\end{aligned}
From here we need a math trick. We can’t continue with this confusingly long row of terms (so long that we can’t write them all out but only can indicate them with the dots \cdots). Luckily a mathematician has proven that a sum of many squared numbers in a row actually has a formula:
1^2+2^2+\cdots+(n-1)^2+n^2=\frac16n(n+1)(2n+1)
This formula is called The sum of squares (see Proof 8). It may not be pretty but at least we can see all terms involved so we can calculate them. We plug it in instead of the long row:
\begin{aligned}
V_\text{pyramid}&= \underbrace{\frac16n(n+1)(2n+1)}_{\text{row of }n\text{ terms replaced}} \frac 1{n^3} h \,A_\text{base} \\
~ &= \frac16(n+1)(2n+1) \frac 1{n^2} h \,A_\text{base} \\
~ &= \frac16(2n^2+n+2n+1^2) \frac 1{n^2} h \,A_\text{base} \\
~ &= \frac16(2n^2+3n+1) \frac 1{n^2} h \,A_\text{base} \\
~ &= \frac16\left(2+\frac3n+\frac1{n^2}\right) h \,A_\text{base} \\
~ &= \left(\frac13+\frac1{2n}+\frac1{6n^2}\right) h \,A_\text{base}
\end{aligned}
The more layers in the pyramid, the more perfect it becomes. We increase n to be enormous – almost infinitely large, n\to \infty. When n becomes enormous, the denominators in the fractions \frac1{2n} and \frac1{6n^2} become enormous, which makes the fractions insignificantly small. When n is enormous it thus corresponds to these fractions completely disappearing, \frac1{2n}\to 0 and \frac1{6n^2}\to 0:
V_\text{pyramid}= \left(\frac13+\frac1{2n}+\frac1{6n^2}\right) h \,A_\text{base}\to \left(\frac13+0+0\right) h \,A_\text{base}\quad \text{when }n\to\infty
And finally, the result is a simple formula:
V_\text{pyramid}=\frac13 hA_\text{base}\quad_\blacksquare
The shape of the base didn’t matter along the way, so this formula counts for any shape, any number of edges – also for a round base, in other words for a cone:
V_\text{cone}=\frac13 h\underbrace{A_\text{base}}_\text{circle}=\frac13\pi r^2 h\quad_\blacksquare
## Proof 6: Prism and cylinder volume
In a room with 3 metres to the ceiling, there are 3 cubic metres above each square metre floor regardless of the shape. The 3…
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# Napoleon's Theorem, Two Simple Proofs
On each side of a triangle, erect an equilateral triangle, lying exterior to the original triangle. Then the segments connecting the centroids of the three equilateral triangles themselves form an equilateral triangle.
By Dr. Scott Brodie, M.D., Ph.D.
Mount Sinai School of Medicine, NY
## Proof #1 ("Hammer and Tongs" trigonometry)
In the following, we use standard notations: in the ΔABC, A denotes both the vertex A and the correposnding ∠A, a is both BC and its length. In addition, let G denote the centroid of the equilateral triangle on side AB, and I denote the centroid of the equilateral triangle on side AC, etc. Let s denote the length of segment GI, t the length of segment AG, and u the length of segment AI. (Geometer's SketchPad illustration.)
Since ∠IAC = ∠GAB=30°, we may apply the Law of Cosines to compute
s2= u2 + t2 - 2ut·cos(A + 60°). (1)
Since the centroid of a triangle lies along each median, 2/3 of the distance from the vertex to the midpoint of the opposite side, we have
t = (2/3)·sqrt(3)/2 · c = c/sqrt(3)
u = (2/3)·sqrt(3)/2 · b = b/sqrt(3),
and (1) becomes
3·s2 = b2 + c2 - 2bc·cos(A + 60°). (2)
Expanding(*) the cosine of the sum, and recalling that cos(60°) = 1/2; sin(60°) = sqrt(3)/2, we have
cos(A + 60°) = cos(A)/2 - sin(A)·sqrt(3)/2. (3)
Substituting (3) into (2) yields
3·s2 = b2 + c2 - bc·cos(A) + sqrt(3)·bc·sin(A). (4)
Now apply the Law of Cosines to ΔABC:
a2 = b2 + c2 - 2bc·cos(A). (5)
and recall (as in the derivation of the Law of Sines):
2·Area( ΔABC) = bc·sin(A). (6)
Substituting (5) and (6) into (4) gives
3·s2 = (1/2)(a2 + b2 + c2) + 2·sqrt(3)· Area( ΔABC). (7)
Since (7) is symmetrical in a, b, and c, it follows that the triangle connecting the three centroids is equilateral, QED.
(*)
Michael Lambrou has suggested a different way to proceed after obtaining (2).
Apply the Law of Cosines to triangles ABE and BCE to express the side BE in two different ways:
b2 + c2 - 2bc·cos(A + 60°) = a2 + b2 - 2ab·cos(C + 60°)
By (2), the left hand side equals 3 times GI. Similarly, the right hand side equals 3 times IH, wherefrom GI = IH. Considering triangles ACD and ABD instead, we also obtain IH = HG, which furnishes the proof.
As a bonus, we get AD = BE = CF.
## Proof #2 (an argument by symmetrization)
Notations are the same as before: let ABC be the original triangle. Choose D, E, and F exterior to ΔABC so that ADB, BEC, and AFC are equilateral triangles, with centroids G, H, and I, respectively. (Geometer's SketchPad illustration.) We proceed to show that ∠HIG = 60°.
Fix I as a center of rotation, and rotate the entire figure by 120°, and superimpose the rotated copy on the original figure. Under the rotation, the ΔCAF maps to itself (C maps to A, A to F, F to C, while I maps into itself.) Denote by BB, DD, EE, GG, and HH the images of points B, D, E, G, and H, respectively.
Connect D to EE and G to HH. By the rigidity of the rotation, ΔGHI = ΔGG.HH.I. In particular, GH = GG.HH. (Geometer's SketchPad illustration.)
Now consider the six triangles that converge on point A. Three of them (ABD, ACF, and A.EE.BB) are equilateral. Recollect that the angles of a triangle sum to 180°, while the angles around a point sum to 360°. Since ∠BB.A.F is a copy of ∠BCA, it follows that ∠D.A.EE = ∠ABC. Finally, ΔD.A.EE = ΔABC, and the pentagon A.BB.EE.D.B is congruent to the pentagon BECAD. It follows that G.HH = GH. And thus G.HH = GH = GG.HH.
Repeating the rotation by 120° once more, and connecting the tips of the equilateral triangles as above, we obtain the figure on the right. (Geometer's SketchPad illustration.) Arguing as above, it is clear that the central hexagon is equilateral, and that the six triangles which meet at the center of rotation are congruent. Therefore, 6·∠HIG = 360°, and ∠HIG = 60°. Since (among the points G,H,I) the choice of the centroid I was arbitrary, we have shown that ΔGHI is equiangular, hence equilateral, QED.
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Session 9, Part C:
Designing a Water Tank (35 minutes)
Imagine constructing an open-topped water tank from a square metal sheet (2 m by 2 m). You would cut squares from the four corners of the sheet and bend up the four remaining rectangular pieces to form the sides of the tank. Then you would weld the edges together to make them watertight. Your goal is to construct a tank with the greatest possible volume.
What size squares do you conjecture would result in the water tank with the greatest volume?
To investigate the relationship between the maximum volume of the tank and the size of the squares cut from the corners, build models and collect data. Using a 1:10 scale, start with a model -- 20-by-20-cm square sheets (PDF - be sure to print this document full scale) of paper. Take one sheet of paper and cut a 1-by-1-cm square from each corner. Fold the net into an open box and tape it. You have just constructed a scale model of the water tank. Repeat the process to construct different models of the water tank.
Problem C1
Collect data on the different-sized water tanks you can make. The side lengths of the cutout squares in centimeters must be whole-number values. Record your data by filling in the table below:
Size of the
Cutout
Square (cm)
Dimensions
of the Box
(cm)
Volume
of the Box
(cm3)
1 by 1 1 by 18 by 18 324
Size of the
Cutout
Square (cm)
Dimensions
of the Box
(cm)
Volume
of the Box
(cm3)
1 by 1 1 by 18 by 18 324 2 by 2 2 by 16 by 16 512 3 by 3 3 by 14 by 14 588 4 by 4 4 by 12 by 12 576 5 by 5 5 by 10 by 10 500 6 by 6 6 by 8 by 8 384 7 by 7 7 by 6 by 6 252 8 by 8 8 by 4 by 4 128 9 by 9 9 by 2 by 2 36
Problem C2 Using whole-number side lengths, which size of a cutout square results in the largest volume for the box? What is the size of the cutout square and the resulting volume?
Problem C3 These models were at a scale of 1:10. If the largest box you made represented the water tank, what would its dimensions be?
Problem C4 One way to get a closer estimate of the dimensions of the box with the greatest volume is to make a graph. On the x-axis, plot the length of the sides of the cutout square (in centimeters); on the y-axis, plot the corresponding volume of the box (in cubic centimeters). Use a graphing calculator, sketch your graph on grid paper (PDF file), or enter the data into the graphing program on your computer.
Problem C5 What would happen if you could remove squares from the corners that used decimals, such as side of square = 3.5 cm, or side = 3.75 cm? Approximate the size of the squares that should be cut to maximize the resulting volume. Note 3
Video Segment Find out how the course participants went about modeling a container with the maximum volume by cutting out different-sized squares from the 20-by-20 sheet of paper. Jayne and Lori try to use one of their earlier observations about surface area and volume. David Russell and David Cellucci graph the data, which leads to new observations. They consider the effect of the absence of a lid on a relationship between volume and surface area. Why do you think Jayne and Lori's initial intuitive approach didn't work? How would you explain it in your own words? If you are using a VCR, you can find this segment on the session video approximately 18 minutes and 3 seconds after the Annenberg Media logo.
"Designing a Water Tank" adapted from Swan, Malcolm, and the Shell Centre Team. The Language of Functions and Graphs. p. 146. © 1999 by Shell Centre Publications. http://www.MathShell.com.
Next > Homework
Session 9: Index | Notes | Solutions | Video
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# PARABOLAS AND THEIR FEATURES
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## Transcription
1 STANDARD FORM PARABOLAS AND THEIR FEATURES If a! 0, the equation y = ax 2 + bx + c is the standard form of a quadratic function and its graph is a parabola. If a > 0, the parabola opens upward and the lowest point (called the vertex) yields the minimum value of the function. If a < 0, the parabola opens downward and the highest point (called the vertex) yields the maximum value of the function. If a > 1, the parabola is narrower than the standard parabola y = x 2 but if a < 1 then it is wider than the standard parabola. The x-intercepts of the parabola are found by solving the equation ax 2 + bx + c = 0. The x-value of the vertex may be found using the equation x =!b. The y-value of the vertex may then be found by 2a substituting the x-value into the original equation. The vertical line x =!b is called 2a the axis of symmetry. The x-intercepts and the vertex can be useful in making a sketch of the parabola. Examples b. Tell the direction of opening and if the vertex is a minimum or maximum point. c. Find the x-intercepts of the parabola. d. Use the x-intercepts and vertex to sketch the graph. Example 1: y = x 2 + 6x + 5 a. x =!b 2a =!6 2"1 =!3 ( ) (!3) + 5 =!4 ( ). y =!3 The coordinates of the vertex are!3,!4 The axis of symmetry is x =!3. c. Solve by factoring: x 2 + 6x + 5 = 0 ( x + 5) ( x + 1) = 0 x =!5,!x =!1 The x-intercepts are (!5, 0),!(!1, 0). b. The parabola opens upward. The vertex is a minimum point. d.
2 Example 2: y =!2x x! 13 a. x =!b 2a =!12 2!2 y =!2 3 ( ) = 3 ( ) ( 3)! 13 = 5 ( ). The coordinates of the vertex are 3, 5 The axis of symmetry is x = 3. c.!2x x! 13 = 0 Solve using quadratic formula a =!2,!b = 12,!c =!13 x =!12 ± 122! 4(!2)(!13)!12 ± 40 = 2(!2)!4 The x-intercepts are! (1.41, 0),!(4.58, 0). b. The parabola opens downward. The vertex is a maximum. d. Problems c. Find the x-intercepts of the parabola. d. Use the x-intercepts and vertex to sketch the graph. 1. y = x 2! 4 2. y = x 2 + 8x 3. y = x 2 + 2x! 3 4. y = 3x 2! 6x y =!x 2! 6x y =!2x x! 9 7. y = x 2 + 5x y = x 2! 7x! 1 9. y = x 2! 6x y =!x 2! 8x! y = 3x 2! 2x! y = 1 2 x2 + 4x + 6
3 GRAPHING FORM If a! 0, the equation y = a(x! h) 2 + k is the graphing form of a quadratic function and its graph is a parabola. In this form the vertex is located at the point (h, k) and the axis of symmetry is the vertical line x = h.. The x-intercepts of the parabola are found by solving a(x! h) 2 + k = 0. The information about the value of a is the same as above. In graphing form, the vertex and two points on each side of the axis of symmetry are commonly used in making a sketch of the parabola. One way to change from standard form to graphing form is to calculate the vertex and then substitute into graphing form. The value of a does not change. Examples For the equation y = ( x! 2) 2! 3, complete each of the following tasks. c. Find two points of the parabola on one side of the axis of symmetry. d. Use the vertex, the two points found in part (c), and two symmetric points to draw the graph. a. The vertex is ( 2,!3). The axis of symmetry is x = 2 d. (1,!2) and (0,1) are symmetric points. b. The parabola opens upward. The vertex is a minimum. c. The vertex is ( 2,!3), use x = 3,!x = 4 If x = 3,!y = 3! 2 If x = 4,!y = 4! 2 ( ) 2! 3 =!2 ( ) 2! 3 = 1 ( ),!( 4,1). so the two points are 3,!2
4 Problems c. Find two points of the parabola on one side of the axis of symmetry. d. Use the vertex, the two points found in part (c), and two symmetric points to draw the graph. 13. y = ( x + 2) 2! y = ( x! 3) y =!x y =! ( x! 3) 2! y = ( x + 4 ) y = x 2! y = ( x! 4 ) y = 2 ( x + 1) 2! y =! 1 ( 2 x + 4 )2 + 3 Answers Use (a) and (c) to sketch graph. 1a. (0,!4),!x = 0 b. up/min c. 2, 0 ( ),!(!2, 0) 2a. (!4,!16),!x =!4 b. up/min c. ( 0, 0), (!8, 0) 3a. (!1,!4 ),!x =!1 b. up/min c. ( 1, 0), (!3, 0) 4a. ( 1,!2),!x = 1 b. up/min c.! ( 1.82, 0), ( 0.18, 0) 5a. (!3,13),!x =!3 b. down/max c.! ( 0.32, 0), ("12.32, 0) 6a. ( 4, 23),!x = 4 b. down/max c.! ( 0.61, 0), ( 7.39, 0) ( ),!x =! 5 2 b. up/min c.! "0.21, 0 ( ),!x = 7 2 b. up/min c.! 7.14, 0 7a.! 5 2, a. 2,! 53 4 ( ), ("4.79, 0) ( ), ("0.14, 0) 9a. ( 3,!4 ),!x = 3 b. up/min c. ( 1, 0), ( 5, 0) 10a. (!4, 6),!x =!4 b. down/max c.! ("6.45, 0), ("1.55, 0) 2 11a. ( 3,! 7 3 ),!x = 2 b. up/min c.! 1 3 ( 3, 0 ), 1, 0 ( ) 12a. (!4,!2),!x =!4 b. up/min c. (!2, 0), (!6, 0)
5 13a.!2,!5 14a. 3, 0 15a. 0, 6 16a. 3,!2 17a.!4, 2 18a. 0,!4 19a. 4, 0 Use (a) and (c/d) to draw graph ( ),!x =!2 b. up/min c/d. (!1,!4 ), ( 0,!1) (!3,!4 ), (!4,!1) ( ),!x = 3 b. up/min c/d. ( 4,1), ( 5, 4 ) ( 2,1), ( 1, 4 ) ( ),!x = 0 b. down/max c/d. ( 1, 5), ( 2, 2) (!1, 5), (!2, 2) ( ),!x = 3 b. down/max c/d. ( 4,!3), ( 5,!6) ( 2,!3), ( 1,!6) ( ),!x =!4 b. up/min c/d. (!3, 3), (!2, 6) (!5, 3), (!6, 6) ( ),!x = 0 b. up/min c/d. ( 1,!3), ( 2, 0) (!1,!3), (!2, 0) ( ),!x = 4 b. up/min c/d. ( 5,1), ( 6, 4 ) ( 3,1), ( 2, 4 ) ( ),!x =!1 b. up/min c/d. ( 0,!4 ), ( 1, 2) (!1,!4 ), ( 3, 2) ( ),!x =!4 b. down/max c/d. (!2,1), ( 0,!5) (!6,1), (!8,!5) 20a.!1,!6 21a.!4, 3
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Review: Synthetic Division Find (x 2-5x - 5x 3 + x 4 ) (5 + x). Factor Theorem Solve 2x 3-5x 2 + x + 2 =0 given that 2 is a zero of f(x) = 2x 3-5x 2 + x + 2. Zeros of Polynomial Functions Introduction
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Calculus Volume 1
# 4.5Derivatives and the Shape of a Graph
Calculus Volume 14.5 Derivatives and the Shape of a Graph
## Learning Objectives
• 4.5.1 Explain how the sign of the first derivative affects the shape of a function’s graph.
• 4.5.2 State the first derivative test for critical points.
• 4.5.3 Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function’s graph.
• 4.5.4 Explain the concavity test for a function over an open interval.
• 4.5.5 Explain the relationship between a function and its first and second derivatives.
• 4.5.6 State the second derivative test for local extrema.
Earlier in this chapter we stated that if a function $ff$ has a local extremum at a point $c,c,$ then $cc$ must be a critical point of $f.f.$ However, a function is not guaranteed to have a local extremum at a critical point. For example, $f(x)=x3f(x)=x3$ has a critical point at $x=0x=0$ since $f′(x)=3x2f′(x)=3x2$ is zero at $x=0,x=0,$ but $ff$ does not have a local extremum at $x=0.x=0.$ Using the results from the previous section, we are now able to determine whether a critical point of a function actually corresponds to a local extreme value. In this section, we also see how the second derivative provides information about the shape of a graph by describing whether the graph of a function curves upward or curves downward.
## The First Derivative Test
Corollary $33$ of the Mean Value Theorem showed that if the derivative of a function is positive over an interval $II$ then the function is increasing over $I.I.$ On the other hand, if the derivative of the function is negative over an interval $I,I,$ then the function is decreasing over $II$ as shown in the following figure.
Figure 4.30 Both functions are increasing over the interval $(a,b).(a,b).$ At each point $x,x,$ the derivative $f′(x)>0.f′(x)>0.$ Both functions are decreasing over the interval $(a,b).(a,b).$ At each point $x,x,$ the derivative $f′(x)<0.f′(x)<0.$
A continuous function $ff$ has a local maximum at point $cc$ if and only if $ff$ switches from increasing to decreasing at point $c.c.$ Similarly, $ff$ has a local minimum at $cc$ if and only if $ff$ switches from decreasing to increasing at $c.c.$ If $ff$ is a continuous function over an interval $II$ containing $cc$ and differentiable over $I,I,$ except possibly at $c,c,$ the only way $ff$ can switch from increasing to decreasing (or vice versa) at point $cc$ is if $f′f′$ changes sign as $xx$ increases through $c.c.$ If $ff$ is differentiable at $c,c,$ the only way that $f′.f′.$ can change sign as $xx$ increases through $cc$ is if $f′(c)=0.f′(c)=0.$ Therefore, for a function $ff$ that is continuous over an interval $II$ containing $cc$ and differentiable over $I,I,$ except possibly at $c,c,$ the only way $ff$ can switch from increasing to decreasing (or vice versa) is if $f′(c)=0f′(c)=0$ or $f′(c)f′(c)$ is undefined. Consequently, to locate local extrema for a function $f,f,$ we look for points $cc$ in the domain of $ff$ such that $f′(c)=0f′(c)=0$ or $f′(c)f′(c)$ is undefined. Recall that such points are called critical points of $f.f.$
Note that $ff$ need not have local extrema at a critical point. The critical points are candidates for local extrema only. In Figure 4.31, we show that if a continuous function $ff$ has a local extremum, it must occur at a critical point, but a function may not have a local extremum at a critical point. We show that if $ff$ has a local extremum at a critical point, then the sign of $f′f′$ switches as $xx$ increases through that point.
Figure 4.31 The function $ff$ has four critical points: $a,b,c,andd.a,b,c,andd.$ The function $ff$ has local maxima at $aa$ and $d,d,$ and a local minimum at $b.b.$ The function $ff$ does not have a local extremum at $c.c.$ The sign of $f′f′$ changes at all local extrema.
Using Figure 4.31, we summarize the main results regarding local extrema.
• If a continuous function $ff$ has a local extremum, it must occur at a critical point $c.c.$
• The function has a local extremum at the critical point $cc$ if and only if the derivative $f′f′$ switches sign as $xx$ increases through $c.c.$
• Therefore, to test whether a function has a local extremum at a critical point $c,c,$ we must determine the sign of $f′(x)f′(x)$ to the left and right of $c.c.$
This result is known as the first derivative test.
## Theorem4.9
### First Derivative Test
Suppose that $ff$ is a continuous function over an interval $II$ containing a critical point $c.c.$ If $ff$ is differentiable over $I,I,$ except possibly at point $c,c,$ then $f(c)f(c)$ satisfies one of the following descriptions:
1. If $f′f′$ changes sign from positive when $x to negative when $x>c,x>c,$ then $f(c)f(c)$ is a local maximum of $f.f.$
2. If $f′f′$ changes sign from negative when $x to positive when $x>c,x>c,$ then $f(c)f(c)$ is a local minimum of $f.f.$
3. If $f′f′$ has the same sign for $x and $x>c,x>c,$ then $f(c)f(c)$ is neither a local maximum nor a local minimum of $f.f.$
We can summarize the first derivative test as a strategy for locating local extrema.
## Problem-Solving Strategy
### Using the First Derivative Test
Consider a function $ff$ that is continuous over an interval $I.I.$
1. Find all critical points of $ff$ and divide the interval $II$ into smaller intervals using the critical points as endpoints.
2. Analyze the sign of $f′f′$ in each of the subintervals. If $f′f′$ is continuous over a given subinterval (which is typically the case), then the sign of $f′f′$ in that subinterval does not change and, therefore, can be determined by choosing an arbitrary test point $xx$ in that subinterval and by evaluating the sign of $f′f′$ at that test point. Use the sign analysis to determine whether $ff$ is increasing or decreasing over that interval.
Note: If $f′f′$ is not continuous throughout $II$, then include points of discontinuity along with critical points as endpoints when dividing $II$ into subintervals.
3. Use First Derivative Test and the results of step $22$ to determine whether $ff$ has a local maximum, a local minimum, or neither at each of the critical points.
Now let’s look at how to use this strategy to locate all local extrema for particular functions.
## Example 4.17
### Using the First Derivative Test to Find Local Extrema
Use the first derivative test to find the location of all local extrema for $f(x)=x3−3x2−9x−1.f(x)=x3−3x2−9x−1.$ Use a graphing utility to confirm your results.
## Checkpoint4.16
Use the first derivative test to locate all local extrema for $f(x)=−x3+32x2+18x.f(x)=−x3+32x2+18x.$
## Example 4.18
### Using the First Derivative Test
Use the first derivative test to find the location of all local extrema for $f(x)=5x1/3−x5/3.f(x)=5x1/3−x5/3.$ Use a graphing utility to confirm your results.
## Checkpoint4.17
Use the first derivative test to find all local extrema for $f(x)=x−13.f(x)=x−13.$
## Concavity and Points of Inflection
We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is called the concavity of the function.
Figure 4.34(a) shows a function $ff$ with a graph that curves upward. As $xx$ increases, the slope of the tangent line increases. Thus, since the derivative increases as $xx$ increases, $f′f′$ is an increasing function. We say this function $ff$ is concave up. Figure 4.34(b) shows a function $ff$ that curves downward. As $xx$ increases, the slope of the tangent line decreases. Since the derivative decreases as $xx$ increases, $f′f′$ is a decreasing function. We say this function $ff$ is concave down.
## Definition
Let $ff$ be a function that is differentiable over an open interval $I.I.$ If $f′f′$ is increasing over $I,I,$ we say $ff$ is concave up over $I.I.$ If $f′f′$ is decreasing over $I,I,$ we say $ff$ is concave down over $I.I.$
Figure 4.34 (a), (c) Since $f′f′$ is increasing over the interval $(a,b),(a,b),$ we say $ff$ is concave up over $(a,b).(a,b).$ (b), (d) Since $f′f′$ is decreasing over the interval $(a,b),(a,b),$ we say $ff$ is concave down over $(a,b).(a,b).$
In general, without having the graph of a function $f,f,$ how can we determine its concavity? By definition, a function $ff$ is concave up if $f′f′$ is increasing. From Corollary $3,3,$ we know that if $f′f′$ is a differentiable function, then $f′f′$ is increasing if its derivative $f″(x)>0.f″(x)>0.$ Therefore, a function $ff$ that is twice differentiable is concave up when $f″(x)>0.f″(x)>0.$ Similarly, a function $ff$ is concave down if $f′f′$ is decreasing. We know that a differentiable function $f′f′$ is decreasing if its derivative $f″(x)<0.f″(x)<0.$ Therefore, a twice-differentiable function $ff$ is concave down when $f″(x)<0.f″(x)<0.$ Applying this logic is known as the concavity test.
## Theorem4.10
### Test for Concavity
Let $ff$ be a function that is twice differentiable over an interval $I.I.$
1. If $f″(x)>0f″(x)>0$ for all $x∈I,x∈I,$ then $ff$ is concave up over $I.I.$
2. If $f″(x)<0f″(x)<0$ for all $x∈I,x∈I,$ then $ff$ is concave down over $I.I.$
We conclude that we can determine the concavity of a function $ff$ by looking at the second derivative of $f.f.$ In addition, we observe that a function $ff$ can switch concavity (Figure 4.35). However, a continuous function can switch concavity only at a point $xx$ if $f″(x)=0f″(x)=0$ or $f″(x)f″(x)$ is undefined. Consequently, to determine the intervals where a function $ff$ is concave up and concave down, we look for those values of $xx$ where $f″(x)=0f″(x)=0$ or $f″(x)f″(x)$ is undefined. When we have determined these points, we divide the domain of $ff$ into smaller intervals and determine the sign of $f″f″$ over each of these smaller intervals. If $f″f″$ changes sign as we pass through a point $x,x,$ then $ff$ changes concavity. It is important to remember that a function $ff$ may not change concavity at a point $xx$ even if $f″(x)=0f″(x)=0$ or $f″(x)f″(x)$ is undefined. If, however, $ff$ does change concavity at a point $aa$ and $ff$ is continuous at $a,a,$ we say the point $(a,f(a))(a,f(a))$ is an inflection point of $f.f.$
## Definition
If $ff$ is continuous at $aa$ and $ff$ changes concavity at $a,a,$ the point $(a,f(a))(a,f(a))$ is an inflection point of $f.f.$
Figure 4.35 Since $f″(x)>0f″(x)>0$ for $x the function $ff$ is concave up over the interval $(−∞,a).(−∞,a).$ Since $f″(x)<0f″(x)<0$ for $x>a,x>a,$ the function $ff$ is concave down over the interval $(a,∞).(a,∞).$ The point $(a,f(a))(a,f(a))$ is an inflection point of $f.f.$
## Example 4.19
### Testing for Concavity
For the function $f(x)=x3−6x2+9x+30,f(x)=x3−6x2+9x+30,$ determine all intervals where $ff$ is concave up and all intervals where $ff$ is concave down. List all inflection points for $f.f.$ Use a graphing utility to confirm your results.
## Checkpoint4.18
For $f(x)=−x3+32x2+18x,f(x)=−x3+32x2+18x,$ find all intervals where $ff$ is concave up and all intervals where $ff$ is concave down.
We now summarize, in Table 4.1, the information that the first and second derivatives of a function $ff$ provide about the graph of $f,f,$ and illustrate this information in Figure 4.37.
Sign of $f′f′$ Sign of $f″f″$ Is $ff$ increasing or decreasing? Concavity
Positive Positive Increasing Concave up
Positive Negative Increasing Concave down
Negative Positive Decreasing Concave up
Negative Negative Decreasing Concave down
Table 4.1 What Derivatives Tell Us about Graphs
Figure 4.37 Consider a twice-differentiable function $ff$ over an open interval $I.I.$ If $f′(x)>0f′(x)>0$ for all $x∈I,x∈I,$ the function is increasing over $I.I.$ If $f′(x)<0f′(x)<0$ for all $x∈I,x∈I,$ the function is decreasing over $I.I.$ If $f″(x)>0f″(x)>0$ for all $x∈I,x∈I,$ the function is concave up. If $f″(x)<0f″(x)<0$ for all $x∈I,x∈I,$ the function is concave down on $I.I.$
## The Second Derivative Test
The first derivative test provides an analytical tool for finding local extrema, but the second derivative can also be used to locate extreme values. Using the second derivative can sometimes be a simpler method than using the first derivative.
We know that if a continuous function has local extrema, it must occur at a critical point. However, a function need not have local extrema at a critical point. Here we examine how the second derivative test can be used to determine whether a function has a local extremum at a critical point. Let $ff$ be a twice-differentiable function such that $f′(a)=0f′(a)=0$ and $f″f″$ is continuous over an open interval $II$ containing $a.a.$ Suppose $f″(a)<0.f″(a)<0.$ Since $f″f″$ is continuous over $I,I,$ $f″(x)<0f″(x)<0$ for all $x∈Ix∈I$ (Figure 4.38). Then, by Corollary $3,3,$ $f′f′$ is a decreasing function over $I.I.$ Since $f′(a)=0,f′(a)=0,$ we conclude that for all $x∈I,f′(x)>0x∈I,f′(x)>0$ if $x and $f′(x)<0f′(x)<0$ if $x>a.x>a.$ Therefore, by the first derivative test, $ff$ has a local maximum at $x=a.x=a.$ On the other hand, suppose there exists a point $bb$ such that $f′(b)=0f′(b)=0$ but $f″(b)>0.f″(b)>0.$ Since $f″f″$ is continuous over an open interval $II$ containing $b,b,$ then $f″(x)>0f″(x)>0$ for all $x∈Ix∈I$ (Figure 4.38). Then, by Corollary $3,f′3,f′$ is an increasing function over $I.I.$ Since $f′(b)=0,f′(b)=0,$ we conclude that for all $x∈I,x∈I,$ $f′(x)<0f′(x)<0$ if $x and $f′(x)>0f′(x)>0$ if $x>b.x>b.$ Therefore, by the first derivative test, $ff$ has a local minimum at $x=b.x=b.$
Figure 4.38 Consider a twice-differentiable function $ff$ such that $f″f″$ is continuous. Since $f′(a)=0f′(a)=0$ and $f″(a)<0,f″(a)<0,$ there is an interval $II$ containing $aa$ such that for all $xx$ in $I,I,$ $ff$ is increasing if $x and $ff$ is decreasing if $x>a.x>a.$ As a result, $ff$ has a local maximum at $x=a.x=a.$ Since $f′(b)=0f′(b)=0$ and $f″(b)>0,f″(b)>0,$ there is an interval $II$ containing $bb$ such that for all $xx$ in $I,I,$ $ff$ is decreasing if $x and $ff$ is increasing if $x>b.x>b.$ As a result, $ff$ has a local minimum at $x=b.x=b.$
## Theorem4.11
### Second Derivative Test
Suppose $f′(c)=0,f″f′(c)=0,f″$ is continuous over an interval containing $c.c.$
1. If $f″(c)>0,f″(c)>0,$ then $ff$ has a local minimum at $c.c.$
2. If $f″(c)<0,f″(c)<0,$ then $ff$ has a local maximum at $c.c.$
3. If $f″(c)=0,f″(c)=0,$ then the test is inconclusive.
Note that for case iii. when $f″(c)=0,f″(c)=0,$ then $ff$ may have a local maximum, local minimum, or neither at $c.c.$ For example, the functions $f(x)=x3,f(x)=x3,$ $f(x)=x4,f(x)=x4,$ and $f(x)=−x4f(x)=−x4$ all have critical points at $x=0.x=0.$ In each case, the second derivative is zero at $x=0.x=0.$ However, the function $f(x)=x4f(x)=x4$ has a local minimum at $x=0x=0$ whereas the function $f(x)=−x4f(x)=−x4$ has a local maximum at $x,x,$ and the function $f(x)=x3f(x)=x3$ does not have a local extremum at $x=0.x=0.$
Let’s now look at how to use the second derivative test to determine whether $ff$ has a local maximum or local minimum at a critical point $cc$ where $f′(c)=0.f′(c)=0.$
## Example 4.20
### Using the Second Derivative Test
Use the second derivative to find the location of all local extrema for $f(x)=x5−5x3.f(x)=x5−5x3.$
## Checkpoint4.19
Consider the function $f(x)=x3−(32)x2−18x.f(x)=x3−(32)x2−18x.$ The points $c=3,−2c=3,−2$ satisfy $f′(c)=0.f′(c)=0.$ Use the second derivative test to determine whether $ff$ has a local maximum or local minimum at those points.
We have now developed the tools we need to determine where a function is increasing and decreasing, as well as acquired an understanding of the basic shape of the graph. In the next section we discuss what happens to a function as $x→±∞.x→±∞.$ At that point, we have enough tools to provide accurate graphs of a large variety of functions.
## Section 4.5 Exercises
194.
If $cc$ is a critical point of $f(x),f(x),$ when is there no local maximum or minimum at $c?c?$ Explain.
195.
For the function $y=x3,y=x3,$ is $x=0x=0$ both an inflection point and a local maximum/minimum?
196.
For the function $y=x3,y=x3,$ is $x=0x=0$ an inflection point?
197.
Is it possible for a point $cc$ to be both an inflection point and a local extremum of a twice differentiable function?
198.
Why do you need continuity for the first derivative test? Come up with an example.
199.
Explain whether a concave-down function has to cross $y=0y=0$ for some value of $x.x.$
200.
Explain whether a polynomial of degree $22$ can have an inflection point.
For the following exercises, analyze the graphs of $f′,f′,$ then list all intervals where $ff$ is increasing or decreasing.
201.
202.
203.
204.
205.
For the following exercises, analyze the graphs of $f′,f′,$ then list all intervals where
1. $ff$ is increasing and decreasing and
2. the minima and maxima are located.
206.
207.
208.
209.
210.
For the following exercises, analyze the graphs of $f′,f′,$ then list all inflection points and intervals $ff$ that are concave up and concave down.
211.
212.
213.
214.
215.
For the following exercises, draw a graph that satisfies the given specifications for the domain $xϵ[−3,3].xϵ[−3,3].$ The function does not have to be continuous or differentiable.
216.
$f(x)>0,f′(x)>0f(x)>0,f′(x)>0$ over $x>1,−31,−3 over $0
217.
$f′(x)>0f′(x)>0$ over $x>2,−32,−3 over $−1 for all $xx$
218.
$f″(x)<0f″(x)<0$ over $−10,−30,−3 local maximum at $x=0,x=0,$ local minima at $x=±2x=±2$
219.
There is a local maximum at $x=2,x=2,$ local minimum at $x=1,x=1,$ and the graph is neither concave up nor concave down.
220.
There are local maxima at $x=±1,x=±1,$ the function is concave up for all $x,x,$ and the function remains positive for all $x.x.$
For the following exercises, determine
1. intervals where $ff$ is increasing or decreasing and
2. local minima and maxima of $f.f.$
If needed, use a calculator to graph the functions, but show your work.
221.
$f(x)=sinx+sin3xf(x)=sinx+sin3x$ over $−π
222.
$f ( x ) = x 2 + cos x f ( x ) = x 2 + cos x$
For the following exercises, determine a. intervals where $ff$ is concave up or concave down, and b. the inflection points of $f.f.$
223.
$f ( x ) = x 3 − 4 x 2 + x + 2 f ( x ) = x 3 − 4 x 2 + x + 2$
For the following exercises, determine
1. intervals where $ff$ is increasing or decreasing,
2. local minima and maxima of $f,f,$
3. intervals where $ff$ is concave up and concave down, and
4. the inflection points of $f.f.$
224.
$f ( x ) = x 2 − 6 x f ( x ) = x 2 − 6 x$
225.
$f ( x ) = x 3 − 6 x 2 f ( x ) = x 3 − 6 x 2$
226.
$f ( x ) = x 4 − 6 x 3 f ( x ) = x 4 − 6 x 3$
227.
$f ( x ) = x 11 − 6 x 10 f ( x ) = x 11 − 6 x 10$
228.
$f ( x ) = x + x 2 − x 3 f ( x ) = x + x 2 − x 3$
229.
$f ( x ) = x 2 + x + 1 f ( x ) = x 2 + x + 1$
230.
$f ( x ) = x 3 + x 4 f ( x ) = x 3 + x 4$
For the following exercises, determine
1. intervals where $ff$ is increasing or decreasing,
2. local minima and maxima of $f,f,$
3. intervals where $ff$ is concave up and concave down, and
4. the inflection points of $f.f.$ Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator.
231.
[T] $f(x)=sin(πx)−cos(πx)f(x)=sin(πx)−cos(πx)$ over $x=[−1,1]x=[−1,1]$
232.
[T] $f(x)=x+sin(2x)f(x)=x+sin(2x)$ over $x=[−π2,π2]x=[−π2,π2]$
233.
[T] $f(x)=sinx+tanxf(x)=sinx+tanx$ over $(−π2,π2)(−π2,π2)$
234.
[T] $f(x)=(x−2)2(x−4)2f(x)=(x−2)2(x−4)2$
235.
[T] $f(x)=11−x,x≠1f(x)=11−x,x≠1$
236.
[T] $f(x)=sinxxf(x)=sinxx$ over $x=x=$ $[2π,0)∪(0,2π][2π,0)∪(0,2π]$
237.
$f(x)=sin(x)exf(x)=sin(x)ex$ over $x=[−π,π]x=[−π,π]$
238.
$f ( x ) = ln x x , x > 0 f ( x ) = ln x x , x > 0$
239.
$f ( x ) = 1 4 x + 1 x , x > 0 f ( x ) = 1 4 x + 1 x , x > 0$
240.
$f ( x ) = e x x , x ≠ 0 f ( x ) = e x x , x ≠ 0$
For the following exercises, interpret the sentences in terms of $f,f′,andf″.f,f′,andf″.$
241.
The population is growing more slowly. Here $ff$ is the population.
242.
A bike accelerates faster, but a car goes faster. Here $f=f=$ Bike’s position minus Car’s position.
243.
The airplane lands smoothly. Here $ff$ is the plane’s altitude.
244.
Stock prices are at their peak. Here $ff$ is the stock price.
245.
The economy is picking up speed. Here $ff$ is a measure of the economy, such as GDP.
For the following exercises, consider a third-degree polynomial $f(x),f(x),$ which has the properties $f′(1)=0,f′(3)=0f′(1)=0,f′(3)=0$ (not applicable to Exercises 249 and 250). Determine whether the following statements are true or false. Justify your answer.
246.
$f(x)=0f(x)=0$ for some $1≤x≤31≤x≤3$
247.
$f″(x)=0f″(x)=0$ for some $1≤x≤31≤x≤3$
248.
There is no absolute maximum at $x=3x=3$
249.
If $f(x)f(x)$ has three roots, then it has $11$ inflection point.
250.
If $f(x)f(x)$ has one inflection point, then it has three real roots.
|
Question
# Five years later, the father’s age will be three times the age of his son. Five years ago, a father was seven times as old as his son. Find their present ages.
Hint: Start by taking the present age of father and son to be x and y respectively. Then according to the conditions given in the question, form the equation and solve the equation by performing arithmetic operations.
Let x be the present age of father and y be the present age of son.
According to the question, after 5 years father’s age will be (x+5) and son’s age will be (y+5), then
x+5 = 3(y+5)
x-3y-10 = 0 …..(1)
According to the question, another condition is given such that,
Five years ago, the age of the father was (x-5) and son’s age was (y-5), so according to the condition,
(x-5) = 7(y-5)
x-7y+30 = 0 …..(2)
Now we will solve (1) and (2), by subtracting (1) from (2),
$- 3y + 7y - 10 - 30 = 0$
$\Rightarrow y = 10$
We have found the age of son, now to find the age of father we are going to put the value of y in one of the equations,
Let us put y = 10 in (1),
$x - 3\left( {10} \right) - 10 = 0$
$\Rightarrow x = 40$
Therefore, father’s present age is 40 and son’s present age is 10.
Note: In age problems, if we have one person we can easily solve it by taking one variable and forming an equation using the condition given. In case there are 2 people, we assign one variable to one of the people and the second variable to the other one and then form equations using the conditions given in the question.
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# 2 Simple GMAT Quant Questions That Will Help You Score Higher
Let’s discuss races today. It is a very simple concept but questions on it tend to be tricky. But if you understand how to handle them, most questions can be done easily.
A few points to remember in races:
1. Make a diagram. Draw a straight line to show the track and assume all racers are at start at 12:00. Then according to headstart, place the participants.
2. There are two types of head starts: Time and distance
Say there is a 1000 feet race between A and B which starts at 12:00.
Time – A gives B a headstart of 1 min means B starts running at 12:00 and A waits at the start point. Then A starts running from the start point at 12:01.
Distance – A gives B a headstart of 10 feet means A starts from the start point but B starts from the point 10 feet ahead (and hence runs only 990 feet to complete the race)
3. A dead heat is a race in which both the participants finish exactly at the same time. Most races in race questions end in a dead heat!
4. There are two ways in which a participant can beat another: Time and distance
Say A beats B in the 1000 feet race in which both start from the start point at 12:00.
Distance – If A beats B by 20 feet, it means A finishes the race (full 1000 feet) and at that time, B is 20 feet away from the finish line.
Time – If A beats B by 2 mins, it means that if A finished at 12:10, B is still 2 mins away from the finish line i.e. at his/her speed, B takes 2 mins to reach the finish line.
That is all! Now let’s look at some questions:
Question 1: A’s speed is 20/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?
(A) 1/17
(B) 3/17
(C) 1/10
(D) 3/20
(E) 3/10
Solution: We have the ratio of A’s speed and B’s speed. This means, we know how much distance A covers compared with B in the same time.
This is what the beginning of the race will look like:
(Start) A_________B______________________________
If A covers 20 meters, B covers 17 meters in that time. So if the race is 20 meters long, when A reaches the finish line, B would be 3 meters behind him. If we want the race to end in a dead heat, we want B to be at the finish line too at the same time. This means B should get a head start of 3 meters so that he doesn’t need to cover that. In that case, the time required by A (to cover 20 meters) would be the same as the time required by B (to cover 17 meters) to reach the finish line.
So B should get a head start of 3/20th of the race.
This question was relatively very straight forward and we gave it only to help you apply the concepts discussed above. Let’s make it slightly tricky now.
Question 2: A’s speed is 20/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that B beats A by 20% of the length of the race?
(A) 15%
(B) 20%
(C) 28%
(D) 32%
(E) 35%
SolutionAgain, we have ratio of A’s speed and B’s speed given as 20:17. If A covers 20 meters, B covers 17 meters in that time. This time, let’s assume that the length of the race is 25 meters.
At the beginning, this is what the 25 meter track will look like with a head start to B:
(Start ) A_________B_______________________________
Since A will give B a head start so A must start from the start line while B will start from ahead.
Since A should cover only 80% of the length of the race, when B reaches the finish line, A should still have 20% of the track leftover.
20% of the track will be (20/100)*25 = 5 meters. So A should be at 20 meters when B is at the finish line.
So this is what the finish of the race will look like:
____________20_________________A____5_____B (Finish)
A will cover a total of 20 meters when B should be at the finish line. In this time, B will cover only 17 meters. But the total track is of 25 meters. So the rest of the 25-17 = 8 meters, B should get as a head start.
Head start will be 8/25 *100 = 32% of the race.
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Functions Starter Questions
Functions Practice
Functions worksheets for year 10 and year 11. Functions worksheet 1 asks students to work with expressions and functions, functions worksheet 2 and functions worksheet 3 asks students to work with composite functions, functions worksheet 4 and functions worksheet 5 asks students to work with inverse functions.
How Do You Solve Functions?
Expressions, Inverse functions and composite functions. The example below shows three separate functions, [1] f(x) = x2 - 2x + 4, [2] f(a) = 5x and g(x) = x - 1. In the first section the value 3 is substituted into the function f(x) = x2 - 2x + 4. The answer is 7. Next, the value -6 is substituted into the function f(x) = x2 - 2x + 4. The answer is 52. The second section shows the composite functions ff(x) and fg(x). In the first part the value 3 is substitued into the composite function ff(x). The answer is 45. Next, the value 3 is substituted into the composite function fg(x). The answer is 10.
The graphic below shows examples of inverse functions. The inverse function of x - 2 is x + 2. The inverse function of 4 + 3x is (x - 4)/3. The inverse function of x² - 2 is √(x + 2).
The image below shows the steps involved when finding the inverse function. This method is known as the function machine method.
What Are Functions Used For?
A function describes a relationship between two values, an input value and an output value. Functions must only have one relationship for each input value as shown below. The domain is a set of values that are allowed to be put into the fuction. The range represents the set of possible results produced when values are input into the function.
There are four types of functions: 1. Constant Function, 2. Linear Function, 3. Quadratic Function and 4. Cubic Function.
There are four types of mappings: one-to-one, many-to-one, one-to-many, and many-to-many. One-many and many-many are not functions.
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For Function Machines Cazoomy Worksheets try: Function Machines Worksheets. For Substituting Into Expressions Cazoomy Worksheets try: Substituting Into Expressions Worksheets. For Bodmas Cazoomy Worksheets try: Bodmas Worksheets.
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# Parallelogram, Trapezium and Kite
Some people may find parallelogram, trapezium and kite similar, but they are not. All these are quadrilateral, have four sides, four vertices, four angles, and have the sum of interior angles equal to 360 degrees. However, they have different sizes and properties. So let us study about them.
## Parallelogram
A parallelogram is a quadrilateral with two sets of parallel sides.
## Properties of a parallelogram
1. The sides of a parallelogram are equal in length and parallel to each other.
2. The opposite angles inside a parallelogram are equal in value.
3. The sum of adjacent sides in a parallelogram is equal to 180 degrees.
4. The consecutive angles of a parallelogram are supplementary.
5. The diagonals of a regular parallelogram bisect each other, i.e., they form 90 degrees at the meeting junction.
6. Each diagonal of a parallelogram divides the entire parallelogram into two congruent triangles.
7. The two diagonals divide the entire parallelogram into four congruent triangles.
8. Square, rectangle and rhombus are particular types of a parallelogram.
9. Area of parallelogram = base x height.
## Trapezium
A trapezium is a quadrilateral whose one pair of opposite sides are parallel. The parallel bottom side is the base, while its non-parallel side is the legs. Thus, it is also known as a trapezoid.
## Properties of a trapezium
1. The parallel sides are equal in a trapezium.
2. In an isosceles trapezium, the two non-parallel sides are equal. They also form equal angles with the bases.
3. The diagonals of a trapezium intersect each other. In addition, they do not bisect each other.
4. The sum of interior angles in a trapezium is equal to 360 degrees.
5. The sum of adjacent angles in a trapezium is equal to 180 degrees. This implies the adjacent angles in a trapezium are supplementary.
6. There are three types of trapezium – isosceles, scalene and right trapezium.
7. In a scalene trapezium, all the sides are of unequal length.
8. In a right trapezium, it forms two right angles.
## Kite
A kite is a closed quadrilateral that has two pairs of adjacent sides of equal length. It is a plane figure with adjacent sides congruent.
## Properties of a kite
1. The two angles in a kite, where the unequal sides meet, are equal.
2. A kite can be assumed as a pair of two congruent triangles with a common base.
3. The two diagonals in a kite intersect each other at right angles.
4. The sum of interior angles in a trapezium is equal to 360 degrees.
5. A regular kite is symmetrical about its main diagonal.
6. The shorter diagonal in a kite divides the entire kite into two isosceles triangles.
Example 1: Find the perimeter of a kite whose sides are 5 cm and 6 cm.
Solution: The sum of sides of a quadrilateral is equal to the perimeter. Also, the two pairs of sides are equal in length in a kite. Therefore, the perimeter of the kite is 5 + 5 + 6 + 6 = 22 cm.
Example 2: Find the area of a parallelogram whose base is 4 m and height is 10 m.
Solution: We know, the area of a parallelogram = base x height Therefore, area of parallelogram = 4 x 10 = 40 square meters.
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Useful Formulae
In this lesson, we'll study some mathematical formulae that would make calculating time complexity easier!
We'll cover the following
Formulas
Here is a list of handy formulas which can be helpful when calculating the time complexity of an algorithm:
Summation Equation
$\left(\sum_{i=1}^n c \right) = c + c+ c + \cdots + c$ $cn$
$\left(\sum_{i=1}^n i \right) = 1+2+3+\cdots+n$ $\frac{n(n+1)}{2}$
$\left(\sum_{i=1}^n i^2 \right) = 1 + 4 + 9 + \cdots + n^2$ $\frac{n(n+1)(2n+1)}{6}$
$\left(\sum_{i=0}^n r^i\right) = r^0 + r^1 + r^2 + \cdots + r^n$ $\frac{(r^{n+1}-1)}{r-1}$
$\sum_{i=0}^n 2^i = 2^0 + 2^1 + ... + 2^n$ $2^{n+1} - 1$
Some of the formulas dealing with logarithmic expressions:
Logrithmtic expressions Equivalent Expression
$log \;(a\;*\;b)$ $log\;(a) + log\;(b)$
$log \;(a\;/\;b)$ $log\;(a) - log\;(b)$
$log\;a^n$ $n\;log\;a$
$\sum_{i=1}^n log\;i = log\;1 + log\;2 + ... + log\;n$ $= log (1.2...n)$ $log\;n!$
General Tips
1. Every time a list or array gets iterated over $c \times length$ times, it is most likely in $O(n)$ time.
2. When you see a problem where the number of elements in the problem space gets halved each time, that will most probably be in $O(log n)$ runtime.
3. Whenever you have a singly nested loop, the problem is most likely in quadratic time.
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# Solve Algebra Problems For Me
In this method, we evaluate one of the variable value in terms of the other variable using one of the two equations.And that value is put into the second equation to solve for the two unknown values.
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The solution below will make the idea of Substitution clear. x y = 15 -----(2) (10 y) y = 15 10 2y = 15 2y = 15 – 10 = 5 y = 5/2 Putting this value of y into any of the two equations will give us the value of x.
x y = 15 x 5/2 = 15 x = 15 – 5/2 x = 25/2 Hence (x , y) = (25/2, 5/2) is the solution to the given system of equations. In Elimination Method, our aim is to "eliminate" one variable by making the coefficients of that variable equal and then adding/subtracting the two equations, depending on the case.
You just have to follow the order for completing parts of the equation and keep your work organized to avoid mistakes!
In solving these equations, we use a simple Algebraic technique called "Substitution Method".
Elimination Method - By Equating Coefficients: In Elimination Method, our aim is to "eliminate" one variable by making the coefficients of that variable equal and then adding/subtracting the two equations, depending on the case.
## Solve Algebra Problems For Me
This is another very easy and useful equation solving technique that is extensively used in Algebraic calculations. In this example, we see that neither the coefficients of x nor those of y are equal in the two equations. How much money does she need to buy a game that costs ?Solution: Let x represent the amount of money Jeanne needs.This article has over 222,861 views, and 14 testimonials from our readers, earning it our reader-approved status. The basic steps for solving algebra problems involve performing simple operations in small steps that “cancel” the original problem.Doing these steps carefully and in order should get you to the solution.If we use the method of addition in solving these two equations, we can see that what we get is a simplified equation in one variable, as shown below.2x y = 15 ------(1) 3x – y = 10 ------(2) ______________ 5x = 25 (Since y and –y cancel out each other) What we are left with is a simplified equation in x alone.After multiplication, we get 2x 4y = 30 ------(2)' Next we subtract this equation (2)’ from equation (1) 2x – y = 10 2x 4y = 30 –5y = –20 y = 4 Putting this value of y into equation (1) will give us the correct value of x.2x – y = 10 ------(1) 2x – 4 = 10 2x = 10 4 = 14 x = 14/2 = 7 Hence (x , y) =( 7, 4) gives the complete solution to these two equations.So simple addition and subtraction will not lead to a simplified equation in only one variable.However, we can multiply a whole equation with a coefficient (say we multiply equation (2) with 2) to equate the coefficients of either of the two variables.
## Comments Solve Algebra Problems For Me
• ###### Free Math Solver - ChiliMath
Introductory Algebra · Intermediate Algebra · Advanced Algebra · Geometry. Just type in the math problem using its prescribed syntax and it will give you. Note If you want more help, and therefore, would want to see how Mathway solves the problem step by step, you can click “Tap to view steps. Trust me, it happens.…
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• ###### Different ways to solve equations Pre-Algebra, Inequalities.
We have 4 ways of solving one-step equations Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both.…
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Learn how to solve algebra problems with the help of examples.…
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Sample problems are under the links in the "Sample Problems" column and the corresponding review material is under the. Solving Absolute Value Equations.…
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When solving a simple equation, think of the equation as a balance, with the equals sign = being the fulcrum or center. Thus, if you do something to one side.…
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# SUM AND PRODUCT OF ROOTS OF QUADRATIC EQUATION
If a quadratic equation is given in standard form, we can find the sum and product of the roots using coefficient of x2, x and constant term.
Let us consider the standard form of a quadratic equation,
ax2 + bx + c = 0
(Here a, b and c are real and rational numbers)
Let α and β be the two zeros of the above quadratic equation.
Then the formula to get sum and product of the roots of a quadratic equation is,
Example 1 :
Find the sum and product of roots of the quadratic equation given below.
x2 - 5x + 6 = 0
Solution :
Comparing
x2 - 5x + 6 = 0
and
ax2 + bx + c = 0
we get
a = 1, b = -5 and c = 6
Therefore,
Sum of the roots = -b/a = -(-5)/1 = 5
Product of the roots = c/a = 6/1 = 6
Example 2 :
Find the sum and product of roots of the quadratic equation given below.
x2 - 6 = 0
Solution :
Comparing
x2 - 6 = 0
and
ax2 + bx + c = 0
we get
a = 1, b = 0 and c = -6
Therefore,
Sum of the roots = -b/a = 0/1 = 0
Product of the roots = c/a = -6/1 = -6
Example 3 :
Find the sum and product of roots of the quadratic equation given below.
3x2 + x + 1 = 0
Solution :
Comparing
3x2 + x + 1 = 0
and
ax2 + bx + c = 0
we get
a = 3, b = 1 and c = 1
Therefore,
Sum of the roots = -b/a = -1/3
Product of the roots = c/a = 1/3
Example 4 :
Find the sum and product of roots of the quadratic equation given below.
3x2 + 7x = 2x - 5
Solution :
First write the given quadratic equation in standard form.
3x2 +7x = 2x - 5
3x2 + 5x + 5 = 0
Comparing
3x2 + 5x + 5 = 0
and
ax2 + bx + c = 0
we get
a = 3, b = 5 and c = 5
Therefore,
Sum of the roots = -b/a = -5/3
Product of the roots = c/a = 5/3
Example 5 :
Find the sum and product of roots of the quadratic equation given below.
3x2 -7x + 6 = 6
Solution :
First write the given quadratic equation in standard form.
3x2 -7x + 6 = 6
3x2 - 7x = 0
Comparing
3x2 - 7x = 0
and
ax2 + bx + c = 0
we get
a = 3, b = -7 and c = 0
Therefore,
Sum of the roots = -b/a = -(-7)/3 = 7/3
Product of the roots = c/a = 0/3 = 0
Example 6 :
Find the sum and product of roots of the quadratic equation given below.
x2 + 5x + 1 = 3x2 + 6
Solution :
First write the given quadratic equation in standard form.
x2 + 5x + 1 = 3x² + 6
0 = 2x2 - 5x + 5
2x2 - 5x + 5 = 0
Comparing
2x2 - 5x + 5 = 0
and
ax2 + bx + c = 0
we get
a = 2, b = -5 and c = 5
Therefore,
Sum of the roots = -b/a = -(-5)/3 = 5/2
Product of the roots = c/a = 5/2
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Lesson 5, Topic 3
In Progress
# Factorising By Grouping
Lesson Progress
0% Complete
To factorise an expression containing four terms, you need to group the terms into pairs. Then factorise each pair of terms.
Example 1
a. $$\scriptsize x^2 + x \: – \: x \: – \:1 \\ \scriptsize = (x^2 + x) \: – \: x \: – \:1 \\ \scriptsize(x^2 + x) \: – \: (x + 1) \\ \scriptsize x(x+1) \: – \: 1 (x + 1) \\ \scriptsize = (x \: – \:1)( x + 1)$$
b. $$\scriptsize 6xy \: – \: 2x \: – \: 3y + 1 \\ \scriptsize (6xy \: – \: 2x) \: – \: (3y\: – \:1) \\ \scriptsize 2x(3y \: – \: 1) \: – \: 1(3y\: -\: 1) \\ \scriptsize (2x\: – \: 1)(3y\: – \:1)$$
c. $$\scriptsize 6x^2 \: + \: 15x \: – \: 2x \: – \: 5 \\ \scriptsize (6x^2 \: + \: 15x) \: – \: (2x \: +\: 5) \\ \scriptsize 3x(2x \: + \: 5) \: – \: 1(2x \: +\: 5) \\ \scriptsize (3x\: – \: 1)(2x \: +\: 5)$$
d. $$\scriptsize 4a^3y \: – \: 4a^2bx \: + \: 2ay^2 \: – \: 2bxy \\ \scriptsize (4a^3y \:- \: 4a^2bx) \:+ \: (2ay^2 \: -\: 2bxy) \\ \scriptsize 4a^2(ay \:- \: bx) \: + \: 2y(ay \: -\: bx) \\ \scriptsize (4a^2\: + \: 2y)(ay \: -\: bx)$$
e. $$\scriptsize 6x^2 \: + \: 2xy \: – \: 6xy \: – \: 2y^2 \\ \scriptsize (6x^2 \: + \: 2xy) \: – \: (6xy \: +\: 2y^2) \\ \scriptsize 2x(3x \: + \: y) \: – \: 2y(3x \: + \: y) \\ \scriptsize (2x\: – \: 2y)(3x \: + \: y)$$
error:
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# Place a piece of string round the circle and then open it out flat:
## Presentation on theme: "Place a piece of string round the circle and then open it out flat:"— Presentation transcript:
Place a piece of string round the circle and then open it out flat:
The circumference is the distance around a circle. Place a piece of string round the circle and then open it out flat: diameter Measure the length of the line: Measure the diameter of the circle. What is the connection between the diameter and the circumference?
Measure the distance around the outside of each circle
Measure the distance around the outside of each circle. Measure the diameter of each circle and complete the table. Circle Circumference Diameter A B C D
Circle D Circle B Circle A Circle C
The circumference is the distance around a circle.
diameter 1 2 3 0.1 circumference
Circumference is equal to roughly 3 x diameter……
Circumference is equal to roughly 3 x diameter…….and this works or every circle!
It’s not easy to use a ruler to measure the distance around a circle.
A formula for the circumference makes finding it easy. A formula for the connection between the circumference and the diameter is The symbol, , (pi) is said as pie, and is 3.14 (to 2 dp). This is how accurate we need to be on the GCSE calculator paper.
How to find circumference given the diameter of a circle
Find the circumference of the circle below: If you have the diameter: multiply the diameter by 3.14 (). State the formula Substitute the values for p and d.
Questions: Given diameter, find the circumference
How to find circumference given the radius
Find the circumference of the circle below: If you have the radius: multiply this by 2 first to get the diameter. Then multiply the diameter by 3.14 (). State the formula Substitute the values for p and d.
Review
What’s the formula? Calculate the missing numbers in the table below using the circumference formula below. Use 3.14 as an approximation for p. 13.502 6.8 5 C (cm) d (cm) r (cm) 10 31.4 3.4 21.352 2.15 4.3
Plenary You should know the words used in a circle to include the diameter, radius, circumference and p. Know that p is roughly 3.14. Know and use the formula for the circumference of a circle. or
Find the circumference of each circle below.
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What Is Probability? | Russian Math Tutors
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# What is Probability?
2023-09-13 | 0
You've probably come across the concept of probability before. Let's tackle some straightforward problems to refresh your memory.
Private Lessons | Group Lessons
Problem 1: What is the probability of choosing a yellow ball from a bin containing 5 yellow balls and 15 blue balls? The correct answer is 5 / (5 + 15) = 0.25. To find this, we divide the number of favorable outcomes (in this case, there are 5 ways to choose a yellow ball) by the total number of outcomes (the total number of balls in the bin).
Problem 2: In a bin, there are 2 red, 2 blue, and 4 white balls. What is the probability (P) of choosing either a red or a blue ball? Again, we need to find the ratio of favorable outcomes (F) to total outcomes (T). In this case, a favorable outcome means choosing a red or a blue ball. So, we have F = 2 + 2 = 4 and T = 2 + 2 + 4 = 8, which gives us P = 4 / 8 = 0.5.
Problem 3: What is the probability of rolling a prime and odd number on a 6-sided die?
Let's list all possible outcomes when rolling a die: 1, 2, 3, 4, 5, 6. Now, recall that prime numbers have exactly two positive integer divisors. So, the prime numbers in our list are 2, 3, and 5, while the odd numbers are 1, 3, 5. A favorable outcome here is choosing a prime number that is also odd. In other words, favorable outcomes are elements that belong to both {2, 3, 5} and {1, 3, 5}. Thus, we have two favorable outcomes: 3 and 5. So, F = 2, T = 6, and P = 2/6 = 1/3.
As you may have noticed, P is a function where the argument is an event X. In Problem 1, event X is selecting a yellow ball, and we get P(X) = 0.25. In Problems 2 and 3, we have P(X) = 0.5 and P(X) = 1/3, respectively.
In Problem 2, event X is selecting a red OR a blue ball. We can write X = Y OR Z, where Y is selecting a red ball and Z is selecting a blue ball. So we have P(Y OR Z) = 0.5. Similarly, in Problem 3, we have P(Y AND Z) = 1/3. This demonstrates that we can use OR and AND operations for events and then calculate probabilities. For instance, we may want to find something like P((X AND Y) OR (X AND Z)).
This brings us to the precise definition of an event (allowing us to use OR and AND operations for events). Surprisingly, we can define an event as simply a subset of a set. In mathematics, defining something using sets is a classical approach. Remember that the power set Pow(A) of a set A contains all subsets of A, including the empty set {} and A itself. For example, if A = {1,2}, then Pow(A) = {{}, {1}, {2}, {1,2}}. In probability theory, we can call Pow(A) the event space or probability space, and the elements of Pow(A) are events. In Problem 3, the probability space is Pow({1,2,3,4,5,6}), and the event of rolling an odd prime number is the subset {3,5}.
Now, for events, we can define OR and AND as simply the union and intersection of sets. It makes more sense to write P((X ∩ Y) (X ∩ Z)) instead of P((X AND Y) OR (X AND Z)). But what about probability? Now we can provide its precise definition. Recall that the cardinality |X| of a set X is the number of its elements. For instance, if X = {3,5,7}, then |X| = 3. Given a probability space Pow(A) and an event X (which must be a subset of A), we can define:
P(X) = |X| / |A|
Understanding this definition simplifies all calculations involving probability. For example, we can apply the following purely set-theoretic result to derive a formula for probability theory:
P(X Y) =
= |X Y| / |A| =
= (|X| + |Y| - |X Y|) / |A| =
= |X|/|A| + |Y|/|A| - |X Y|/|A| =
= P(X) + P(Y) – P(X Y)
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100+ Best Algebra Worksheets Your Kids Will Love
Algebra Worksheets for Kids! Download Hundreds of FREE Algebra Worksheets Today! This page offers algebra lesson plans and exercises on basic to intermediate algebra.
Algebra is one of the most basic and most used math concepts. It is the pulp of all things arithmetic and its many operations.
In simpler terms, Algebra can be defined as a combination of numbers, symbols, operations that make up an equation or expression, which one solved gives you the answer to that particular equation or expression.
In a more complex setup, a single equation is part of a set or series of equations.
There are many angles and sub-angles to what Algebra includes. Here is a complete overview starting from the basics and covering all things Algebra.
Types of Numbers
Numbers are referred to as numerical values that display a count of anything that needs to be accounted for. Examples: books, countries, population, colors, etc.
There are two representations of numbers: Numerical and In Figures. Numerical values are represented as 723, 43, 2, etc., while the In Figures are represented as Seven hundred and twenty three, forty three and two.
Numbers are classified to understand and use at appropriate places where the results yielded are most beneficial. The major divisions are given below:
Real Numbers
All numbers that are positive, negative, zero, and can be represented on a line are called Real Numbers. Example: -2, 0, +4, etc.
Real Numbers include Rational and Irrational Numbers.
Rational Numbers
It is simply the ratio of an integer to a non-integer, or all numbers that can be expressed as fractions are called Rational Numbers. Examples: 2/3, -18/4, etc. Also called Non-Irrational Numbers.
Irrational Numbers
As the name suggests, it is the opposite of what a Rational Number shows. Meaning, numbers that cannot be expressed as fractions are called Irrational Numbers.
Examples: 9, 4, 13, etc. Also called Non-Rational Numbers.
Integers
The numbers that are the exact opposite of Whole Numbers are numbers beginning from 0 to infinity. They include both positive and negative whole numbers.
The negative integers are represented with a minus sign as a prefix to the number. Example: -9.
While the positive integers are represented with a positive sign as a prefix to the number. Example: +12. But, most of the time, the positive sign is not written, which automatically implies that the number is a positive value and not a negative value.
The below worksheet covers real numbers. Students learn about rational and irrational numbers, and answer problems related to each set.
Key concept: Students know the difference between rational and irrational numbers, and understand patterns and structure.
Types of Numeral Systems
There is no one particular method of Numeral System representation. If there were, it would be difficult to express certain numbers in one format only, hence the need for different Numeral Systems.
Decimals
Decimals are represented by a simple ‘.’, just like a full stop/period. However, where it is placed makes a huge difference.
For instance, you bought a watch for \$13.44. If you accidentally misplace the dot once left or once right, the cost would change to \$1.344, or \$134.4, which creates a difference of one entire digit and a hundred dollar difference.
Therefore, Decimal plays a vital role in displaying cost, weight, length, etc.
Numbers after (from right) the Decimal point are of lesser value than those before the point. Taking the example of the above watch, since 100 cents make up 1 dollar, there are still 56 cents before the dollar is complete, and the 13 becomes 14.
Fractions
In a way, Fractions can be called expanded forms of Decimals. Why? Because you can write 0.25 as ¼ in Fractions.
Fractions show a part of the whole. They can be anything to everything: a slice of the whole pizza or the number of books you readout of the total available in your library.
Example: If half of the Pizza you ordered is eaten out of the whole you ordered, how do you represent it in fractional form?
Solution: Whole is always ‘1’, and a part of it will always be less than 1. However, note that if there are more than 1 wholes, the count increases, and then the fractional part is written.
For the Pizza eaten out of the whole, representation is – ½. Since you have divided the whole Pizza into two equal parts, 1 of which you ate, it is shown as above.
The clarification is because you may doubt how 1 Pizza became 2, as the number in the denominator is 2 instead of 1. Fraction always considers total parts and then gives the value of what is asked.
Exponents
Suppose you have an expression where you need to multiply one term several times. Instead of writing the term the respective number of times, you can simply write the term once, and then on the upper right corner of the term or superscript in slightly smaller font, you can write the repeating number.
Example : 3 x 3 x 3 x 3 x 3 x 3
Solution: 36, here the term or number 3 has been multiplied 6 times. So, the superscript number is 6. The answer is 729
3 is said to be the ‘base’ number, while 6 is said to be the ‘Exponent.’
Verbally, this is spelled as – ‘ this is third to the sixth power.’
There are short verbals for a term repeated only twice or thrice. They are ‘squared’ and ‘cubed,’ respectively.
Therefore, if we modify the above example expression to 32 and 33, 3 squared is 9, and cubed is 27.
Types of Representations
There are various patterns of how you can represent a certain amount of something. Let us look at some of the more commonly used representations with examples to understand the representations better.
A tabular format or a table is a combination of a certain number of rows and columns. Each row and column represents a different heading that is interrelated.
Tables
Data is gathered in this table, with proper segregation for better comprehension of what it is displaying. However, it can be challenging to understand what number is accountable for what category when you have too much information.
Therefore, Frequency Tables were born.
The below worksheet covers tables and graphs
Frequency Tables
The most basic Frequency Table has 3 columns. At the top, a horizontal row heading of what the table is about is mentioned.
Next, the table is divided into 3 vertical columns (or more, depends on data type), each with a different heading. The horizontal rows are also decided upon how the data is being divided. There is no limit to both rows or columns.
Example: Runs made by each player of a Cricket Team are given as – 21, 33, 1, 78, 2, 34, 2, 2, 45, 66, 0. How do you calculate the Frequency of how many players earned runs in a certain range?
Method: First and foremost, divide the table into three vertical columns, with the three headings as – Runs, Tally, and Frequency. Also, Tally is nothing but a stick number representation of each count.
In the first column, write the ranges as 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70, and 70-80.
You can even take broader ranges such as 0-40 and 40-80, but that will be difficult to understand since we have more values with smaller ranges. The same is the case with taking smaller ranges such as 0-5, 5-10, etc.
In the next column of Tally, let us draw sticks of the first range. Players that scored runs between 0-10 are 5. So, draw five lines.
Remember, always draw 4 separate small vertical lines and then a diagonal from top left to bottom right finishing 5. If the value is less than 5, then no striking is required.
In the final column of Frequency, you will simply have to write the number, same as the tally, but in a numerical value.
Graphs
Graphs are of many types that showcase different days in different manners. There are bar graphs that show comparable, and then there are pie charts that display what is occupying more and what is occupying less.
Histograms
Distinct to Algebra, Histograms is a diagrammatic representation of the data given. Comparison is one of its principal aims.
If we retake the above example, the Runs range value will fall on the x-axis and the number of players, i.e., the frequency on the y-axis. Drawing all the individual bars will give you a comparative picture of how many runs in a range were scored by how many players.
There are no gaps between each bar. This makes a Histogram differ from a Bar Graph.
Expressions
Expressions are simply a combination of numbers, symbols, and operations. Out of the three mentioned constituents, a combination of two is what divides the general Expressions.
Numerical Expressions
Symbols are used to write an operational expression. Words are not used.
Example: 7 + 3 = 10
Verbal Expressions
Exactly as the name suggests, verbal refers to words only explanation of any mathematical concept. Numbers are also involved, but the operation being carried out is spelled out.
Example: A numerical expression is written as 7 + 3 = 10, while a verbal expression is represented as 7 plus 3 equals 10.
Also, there are many aliases to each operation. For instance, instead of ‘plus’ as in the above example, you will see or can use the sum of, increased by, added to, more than, etc.
Constants
As in the English language, we use Constant to define something that remains the same, no change whatsoever; it is used for the same purpose in Algebra. It depicts fixed information, as in amount or quantity, about the object in question.
For example, in the equation – 2 + 3 = 4b, the constants are 2,3, and 4.
Variables
Similar to how numerical and variable expressions are opposite to each other, Variables are opposite of Constants. The former is fixed, as mentioned earlier, whereas the latter is changing, always, constantly.
The use of alphabets A-Z denotes variables. Out of the 26, the most used are a,b,x,y,z, etc.
Let’s retake the above example to avoid any confusion and a better understanding. So according to the equation – 2 + 3 = 4b, ‘b’ is the variable.
Variable Puzzles are another way of representing Expressions. Each variable is assigned a specific value, and then the expression must be solved using the clues.
If the expression is C – 2B + A, and the values for the Variables A, B, and C are given as 1, 2, and 3, what is the final value of the entire expression?
C – 2B + A
1 – 2×2 + 1
1 – 4 + 1
1 – 5
-4
Substitutions
This placing of values in place of their respective representations is called ‘Substitution.’ In the above equation, 2B means 2 multiplied with the designated value of 2 and not the number 22 as may be mistaken due to the absence of the multiplication symbol.
There are times, where the variable doesn’t need to be fixed to be Alphabets only. They can be shapes, objects, animals, anything that constituents for the actual value to be used in the equation.
Substitution is a concept you will be using in many physics and chemistry in the form of formulae. So you must understand where, when, how, why and what to substitute in each expression given to solve the problem with accuracy.
A level one example: Calculate the distance traveled by a Car, running at a speed of 90 kmph for 3 hours.
Distance = Speed x Time
d = s x t
d = 90 x 3
d = 270 km
Similarly, there is a whole lot of formulas that require substitution to find out the unknowns.
Why are Expressions important in maths?
An Expression, whether Verbal or Numerical, gives a complete idea of how the equation works. Simply having numbers or letters put together won’t help solve the problem anyway. Hence the need for Expressions.
Order of Operations
This is one of the big problems you might face. In what Order do I go about solving the equation if it has multiple Operations in it? Where do I begin? Left to right or the opposite?
Well, here is a simple and easy-to-remember acronym – BODMAS, that expands to ‘Bracket of (or Parenthesis) Division, Multiplication, Addition, and Subtraction. All you need to do is follow the order given in the expansion.
Note: Remember, if the entire equation is within brackets, then follow the BODMAS rule. If the equation has operations before the brackets, then also follow the same, except for the BO part, obviously.
An example will help you grasp the concept quickly. So, here is one –
22 – 3 + 7 ( 4 x 3 ÷ 2 ).
BODMAS Rule
Start with the bracket part. First, solve the equation 4 x 3 ÷ 2.
Like I explained earlier, follow the acronym. So, according to that, first up is Division.
3 divided by 2 is 1.5
The equation is now – 4 x 1.5. Multiplying this will give you 6.
The further solved equation is – 22 – 3 + 7 (6). Any number within brackets is always in multiplication.
Therefore, now the equations is – 22 – 3 + 7 x 6.
Following the DMAS rule, we move on with the solution-finding.
Multiply 7 with 6. You get 42.
22 – 3 + 42
Add 3 and 42. You get 45.
22 – 45
Finally, Subtract 45 from 22.
The final answer is – 23, a negative integer.
Hence, from the above example, you understand how the Order of Operations works and the order to be followed for Division, Multiplication, Addition, and Subtraction.
Also, from the previously learned concepts, you came across a Decimal and a Negative Integer. And, this is how various ideas converge in one Algebraic Equation.
No need to worry if you get your answers in not Whole Numbers. Fractions, Decimals, Positive or Negative Integers are also correct answers. In fact, you may also get equations that may have such numbers in the beginning equation itself.
The method to solve them is the same as regular or whole numbers. No unique approach is to be used in their case.
Before we move on with the next set of concepts, you may have one question: What if we have brackets within brackets? How do we solve that?
To answer that, you should solve the innermost bracket first and then move outwards, irrespective of the number of brackets—Centre to Periphery.
This way, you won’t have to redo the entire problem since you will have the correct answer at the first attempt itself!
Simplifying Expressions using Operations Properties
Simplifying Expressions does not need Variables or Constants to be aligned to solve, unlike Variables and Constants cannot be solved altogether.
You can solve them in their original places, and Operation Properties came into existence to help you with that.
Algebraic Operations, aside from using a proper order, require a set of Properties, like rules and regulations, that you must abide by to arrive at the correct answer.
If you do not follow them carefully or deviate even the slightest, the probability of you arriving at the wrong answer increases significantly.
Therefore, there are specific Properties set in place to avoid any straying from the original plan.
There are a major set of 4 Properties that you must follow, without any exceptions, to solve any Algebraic Equation.
Distributive Property
Distribution is the key to this property. It is a multi-step property.
It aids in easier and quicker solving, especially in bigger, longer equations with three or four-digit numbers.
Example 1 : 7 (5b – 2)
Solution :
Distributive property works by multiplying the number outside the bracket, first with the first term or the term left of the operational symbol and second with the second term or the term right of the operational symbol.
To simplify this, let us solve this using the numbers given in the example equation.
Multiply 7 with 5b, resulting in 35b. Then, multiply the same 7 with 5, resulting in 14.
The simplified equation will be 35b – 14; final answer.
Example 2 : 3 x 42
Solution :
This second example focuses on the other version of how you can use the Distributive Property to solve larger numbers.
Separate the second term of the equation into its simpler constituents of two. In this case, you can go for 40 + 2 for 2.
Now, the equation will be 3 ( 40 + 2 ). From here on, you can use the same method of solving as the first example.
Multiply 3 with 40 and then with 2. The equation will come to around 120 + 6, on the addition of which you get 126 as the final answer to the comparatively harder equation than the first.
Some worksheets below that deals with the distributive property.
The Distributive Property of Multiplication
The below video covers The Distributive Property of Multiplication. There are three different levels of worksheets to go with the video.
Level 1 covers only addition, level 2 covers addition and subtraction, and level 3 covers algebraic expressions. Each worksheet includes basic practice and an Answer, Find, and Shade
Activity
, which can be used separately.
Video Link : The Distributive Property of Multiplication
Multiplying Larger Numbers Using the Distributive Property – Level1
The below worksheet uses the Distributive Property to multiply larger numbers.
The Distributive Property of Multiplication Word Problems – Level 2
The below worksheet uses the distributive property of multiplication to solve word problems.
Binomials
Multiplying Binomials is a subpart of the Distributive Property.
Binomials refer to two expressions that already have an operation to be performed with themselves.
Example : ( 4 + 5 ) ( 2b – 6 )
Solution: To solve a multiplying binomial, you will have to follow the ‘FOIL’ Rule, which expands to – First Outer Inner Last.
By following the FOIL rule, you will have to go forward multiplying exactly like the following steps:
The first term of the first binomial with the first term of the second binomial; 4 x 2b = 8b
The first term of the first binomial with the second term of the second binomial; 4 x – 6 = – 24
The second term of the first binomial with the first term of the second binomial; 5 x 2b = 10b
The second term of the first binomial with the second term of the second binomial; 5 x -6 = -30
The resulting equation: 8b – 24 + 10b – 30
Solving further, 18b – 54.
Commutative Property
With respect to Commutative Property, it is the one out of the four that deals with rearrangement of terms but is strict with the symbols they carry.
Meaning, if you wish to change how the terms are arranged in an equation or expression, you can do so. However, keep in mind that you should avoid forgetting or misplacing the sign the term was initially commutated with.
For instance, in the equation – 4x – 2b + 32, you can change the term placements in the following five patterns,
4x + 32 – 2b
– 2b + 4x + 32
– 2b + 32 + 4x
32 + 4x – 2b
32 – 2b + 4x
Observe how with the changing positions of each term, the sign is also changed with them. This does not change the core principle of what the expression is about.
If you do not adhere to the changing signs with the numbers, then the entire expression changes, thus, completely changing the expression in its entirety.
Therefore, Commutative Property, though it may seem too simple, is one of the four crucial pillars like the others.
Commutative Property – Level 1
These one page worksheets introduce the Commutative Property of Addition. Students see that the order does not matter and by moving numbers around it can help them solve more complex problems. There are two different versions.
Commutative Property – Level 2
One page worksheets continuing the Commutative Property of Addition. Students use the idea that
order does not matter to solve more complex problems. There are two different versions.
Key concept: Get students thinking of ways to save time and effort.
Downlod Worksheet
Commutative Property – Level 3
These one page worksheets finish the Commutative Property of Addition. Students should see
that the order does not matter, and can save time and effort by moving numbers around. There are two different versions.
Associative Property
In an Equation, mostly, brackets mean Multiplication. But, if there is a symbol before the opening bracket or after the closing bracket other than Multiplication, then the sign should be considered for the operation to be carried out.
Now, if the sign showing an operation is the same within and outside the bracket, and the bracket is displaced, it is irrespective of the answer.
Example : 4 + (34 + 6) = (4 + 34) + 6
Solution: Now, How do you prove that both the expressions give the same answer. Here’s how –
First, let us solve the left-hand side part of the equals to sign:
4 + (40)
4 + 40
44.
Then, let us solve the right-hand side part of the equals to sign:
(38) + 6
38 + 6
44.
Hence, the aim of Associative Property is fulfilled. You can try other examples taking Subtraction, Multiplication, and Division as the operation type, and you will find that the result you reach will be the same.
The below worksheet deals with the associative property.
This one page worksheet introduces the associative property. It explains how we can change groups (move parentheses) in a problem to make it easier to solve.
Key concept: Students need to understand that the order does not matter when adding. Numbers and groups can be moved to help us simplify a problem or make it easier to solve.
Associative Property of MultiplicationLevel 1
This one page worksheet introduces the associative property. It shows that we can change groups (move parentheses) to make problems easier to solve.
Key concept: Order does not matter when multiplying. Numbers and groups can be moved to help simplify a problem or make it easier to solve.
Simplifying Expressions
This worksheet covers simplifying expressions. Students need to combine like terms using basic arithmetic and the associative,
commutative, and distributive properties.
This worksheet continues simplifying expressions. Students need to combine like terms using basic arithmetic and the associative,commutative, and distributive properties. It includes using the distributive property with fractions.
Note: If terms differ in at least one variable, then they are unlike terms and cannot be combined.
This worksheet finishes simplifying expressions. It includes using the distributive property with fractions and multiple variable and square terms. Students need to combine like terms using basic arithmetic and the associative, commutative, and distributive properties.
The below worksheets have students use the distributive, associative, and commutative properties to simplify expressions.
The drawing is the same but the shading and questions are different for each worksheet.
The below worksheets have students use the distributive, associative, and commutative properties to simplify expressions. The drawing is the same but the shading and questions are different for each worksheet.
Below art worksheet reviews arithmetic with rational numbers, positive and negative integers, fractions, and decimals.
Identity Property
The most uncomplicated property of all the four states that any expression, equation, term, constant, or variable multiplied with ‘1’, will be equal to the same expression, equation, term, constant, or variable as the original.
Example : (3b + 12a) – (22b + 33a) x 1 = (3b + 12a) – (22b + 33a)
Exponents
The below worksheets covers basic exponents. It includes a short help section to help students better understand exponents.
Note: Students need to know that a number to the second power is the same as calling it squared and a number to the third power is the same as calling it cubed.
Squares and Square Roots
Square Roots are the opposite representation of Exponents and restricted only to being multiplied twice and not more than that. Rather than displaying the number ‘2’ as a superscript, to the term squared, a is added to it.
A visual representation of the square root is –
Exponential Format:
22 = 4
Square Root Format:
2 = 4
You cannot write both the Exponential and Square Root Formats together as ‘22 = 4’. This is not allowed. You can write either of them, but not both.
It is not mandatory that all the numbers have their perfect squares. Sometimes, the answers are in Decimals, or Fractions, or even the questions and are stated as Non – Perfect Squares.
Perfect Squares
Example : 11 x 11 = 121
Here, you have no residual number, whether you dissociate 121 to 11 x 11 or multiple the two 11s. No remainder.
Non – Perfect Squares
Example: 31
Here, when you dissociate 31, you do not get equal two numbers like 25 by multiplying 5 with 5, or 36 by multiplying 6 with 6, hence the term non-perfect square.
Now the question of where to place it? 5 or 6? The closer the number, the more accurate is the square value. Since 31 is closer to the square of 6 than 5, you can approximately equal the square root of 31 to 6.
The Worksheets below shows the detailed examples
The below art worksheet reviews squares and square roots. The square roots are perfect squares so students will not have to estimate and no irrational numbers will be left over. Watch out for the decimals.
Estimating Square Roots
The below worksheet introduces estimating squaring roots of irrational numbers (not perfect squares). Students estimate the square root to the nearest whole number.
Absolute Values and Number Lines
Number Line is a horizontal line representing different numerical values beginning from the center as 0 and then continues in both directions at regular intervals, placing numbers in increasing order.
As we learned earlier, numbers, or rather integers, are both positive and negative. If you draw a number line, extending on the right will be the positive numbers, and then on the left in extension will be the negative numbers.
Coming to the actual concept of Absolute Value, it refers to the value or count of anything measured beginning from zero to the current position of that thing.
Again, an example is necessary to understand. Let’s suppose you let your pet out for a walk from your house to the nearby park.
Irrespective of whether your pet goes left, right, forward, backward, or even diagonal, if it takes 20 steps to reach the park, it will be counted as 20 steps only.
If your pet goes left or down, it does not mean that it is -20 steps. The negative sign is immaterial here.
The below worksheet introduces absolute value and reviews number lines. It explains what is the absolute value of a number and includes problems with number lines, greater than and less than, and solving simple expressions.
Unit Rate
Rate or Unit Rate is the measurement of a trend or repetition in data. Hours, minutes, seconds, days, weeks, years, classes, etc., anything from which a smaller quantity can be extracted and represented of the total.
Assuming you drink 1 glass of water every 3 hours, the rate of you drinking water in a day will be 8 glasses in 24 hours.
The below worksheet is a new idea we call Stained Glass. The picture is symmetrical and students choose
their own colors to create a unique art project. The worksheet covers unit rate and students answer the problems, find the numbers in the picture, and color the shapes based
on the colors they chose.
Input-Output Tables
A rule is fixed to each Table. You must follow the rule to carry out the operation of Input-Output.
Input meaning insertion of data, and Output meaning exertion of data. In between the two primary functions, you will have to use the rule of that table.
For example, the rule for a set of values is ‘Addition of 23 to each’. Out of the set, the first Input value 65. Find the Output.
So, add 23 to 65. 88 is the answer and the first Output value of the table.
And this is how Input-Output Tables work. Data is given. Follow the Rule prescribed to each. You will have a correctly filled table before you in no time.
The below worksheets cover input-output tables. Students need to find the rules and complete the tables. Input-output tables help students recognize patterns and build a relationship between lines and equations. There is a different worksheet for each level.
Variable Puzzles
These one page worksheets use puzzles to help kids practice their algebra skills. Students are given five variables (letters) and have to find what numbers they represent. They are given equations as clues and a grid to help solve the unknown variables. This can also be used to practice the
process of elimination for test taking skills.
Substitution – Level 1
The below worksheets introduces substitution. It has worksheet for addition and subtraction, and for multiplication and division.
Answer, Find, and Color – Level 1 Fill in the Blank
The below art worksheet reviews basic order of operations by having students fill in the blanks to
complete various equations.
The below art worksheets use fill in the blank questions to review order of operations and solving equations. The drawing is the same but the questions are different for each worksheet.
The below worksheet reviews substitution, order of operations, and basic algebra. Students substitute the values into the expressions and solve.
The below worksheets review solving equations, substitution, charts, and learning real world equations. Students are given a chart and must fill in the empty spaces.
Solving One-Step Equations Addition / Subtraction
The below worksheet covers basic one step equations using the opposite operation method.
Solving One-Step Equations Multiplication / Division
This two page worksheet covers basic one step equations using the opposite operation method.
The Basics Art – Level 1
This below HALLOWEEN worksheet is a basic algebra practice page. Students use addition,
subtraction, multiplication, and division to solve one step equations.
One-Step Equations
The below worksheets covers solving one-step equations. It includes addition, subtraction, multiplication, and division. Students answer the problems and then shade in the
corresponding shapes to create a picture.
The below worksheets review verbal sentences and variables. Students read the sentences (equations) and find the missing number.
Solving One-Step Equations Add, Subtract with Decimals
The worksheet covers solving one-step equations with addition and subtraction of decimals. It does not have any negative variables.
Two-Step Fill in the Blanks
The below art worksheets cover solving two-step equations. They help students visualize the process which should lead to solving equation.
Two-Step Equations
The below worksheets cover solving two-step equations.
Percentage
Percentages help us understand an overall account of what is major and minor in a category. Mainly, 100% is kept as the upper limit for easy calculation and understanding.
Percent of Number
Taking 20% of 200. As I mentioned earlier, ‘bracket’ means ‘multiplication,’ so does ‘of.’ Therefore, you can write 20% x 200.
Coming to the actual percentage calculation. There is one long method, and the other is a short method. Understand the one that best suits your way of solving expressions.
The below worksheets describes the percentage problems
In the below art worksheets identifies the correct answer and shade it on the given picture. Students need to use their knowledge of percent of numbers to solve various equations.
The below art worksheets review percent of a number. Students need to find the percent of a number, what percent one number is of another, and find the original number when a percent is given.
Long Method –
20% x 200
Remove the percentage sign and write a 100 below the number to be calculated. The numerator is 20, and the denominator is 100 now. 20/100.
Canceling out one zero from above and below, leaving 2/10, that equals ⅕.
Now, the equation has turned out to be 1/5 x 200, which eventually is 200/5. The final answer is 40.
Short Method –
Rather than doing all the above steps, simply place a Decimal point, two members, from left. Always two paces to the left; neither change the direction nor change the pace.
So, 20% x 200 will become 0.20 x 200, that equals 40.
There you have it—both methods for calculating the Percentage.
Some of the most commonly used percentages are 25%, 50%, 75%, and 100%. They are named quarter, half, three-fourth, and full, respectively. Also, their values are 1/4, 1/2, 3/4, and 1.
Note: You can calculate the percentage itself; the percent is given instead of the opposite. To do that, reverse Multiplication with Division, and you will retrace your path to the actual percentage.
Percent of Change
Percentages are rarely constant. Stock markets, profits, losses, share market, etc., are many examples where increases and decreases in various percentages need to be kept track of.
Therefore, it is essential to know how to calculate any decrease or increase in percentages. Here you go –
The formula you need to substitute the values in is –
% increase/decrease = Increase or Decrease / Original x 100.
The increase is the difference between Original and New.
Example : Original – 30, New – 25. Find out the increase.
Solution : Decrease = Original – New
Decrease = 30 – 25
Decrease = 5
% Decrease = Decrease / Original x 100
= 5 / 30 x 100
= 16.66
Therefore, the Decrease in the Percentage is 16.66%, which can also be rounded off to 17%.
There are no sign notations in Percentages. You can know whether there is an Increase or Decrease by comparing the Original and New values.
If the number is negative in the calculation of Increase or Decrease, then it is a decrease. If not, then it is an Increase.
The below art worksheets review percent of change. Students are given an original and new number and must figure out if the change is an increase or decrease and then calculate the percent of change.
Combining Like Terms Art
The below worksheet has students practice combining like terms. It includes using the Distributive Property to simplify the expressions.
Even though a part of an expression or equation is surrounded by parenthesis, it can still be pulled apart using the Distributive Property.
Answer, Find, and Color Multi-Step Equations
The below art worksheets review multi-step equations. worksheets describing how combining like terms, the Distributive Property, and fractions.
The below art worksheets review solving functions. Students need to substitute the given number into the function and solve. The picture and answers are the same but the problems are different for each level.
Multiplying Two Binomials (FOIL)
The below worksheet introduces multiplying two binomials using the FOIL method. It uses the distributive property to help explain the process.
Key concept: The FOIL method helps keep Students organized when multiplying two binomials. FOIL stands for
First-Outer-Inner-Last.
Solving Inequalities
If you come across an Expression with Inequality such as 33p ＞ -2, if you think that you can equalize, or at least reverse the sign of them by multiplying or dividing both sides with a negative number, then you are mistaken.
Let’s prove my point.
Multiply -2 on both sides. You get 33p x -2 and -2 x 2 as -66p and -4.
Therefore, the symbol does not change between them. It remains the same -66p ＞ -4.
The below art worksheet reviews solving inequalities. Students solve two inequalities and find the answer they have in common.
Solving Proportions
Proportions refer to Ratios.
Example : 22/11 = 3/a. Find a.
Solution: Cross multiply the numerator of the left-hand side expression with the denominator of the right-hand side. Similarly, cross multiply the numerator of the right-side expression with the denominator of the left-hand side expression.
You get, 22 x a = 3 x 11
22a = 33
a = 33 / 22
a = 1.5
And this is how you solve proportions.
The below art worksheets review proportions. Students find the missing value in each proportion.
Coordinate Algebra
Algebra has another branch that extends into the graphical representation part of mathematics. And this is called Coordinate Algebra.
Equations and Expressions are formed out of the data being viewed from the graph, and so are the locations calculated and noted from it. The grid view gives exact and precise locations if the scale used to draw the graph has been accurately measured.
Introduction to Coordinates
The Coordinate System is a graphical representation used to locate the position of an object in two-dimensional or three-dimensional spaces. The number line system is used in four extensions from ‘0’ or ‘Origin.’
The right and up number lines are positive, while the left and down number lines are negative. These four lines divide the graph into four quadrants.
The quadrants are marked as roman numbers of I, II, III, and IV, beginning from the upper right corner first, then the one left to it, then the one below it, and the last on the right of the third.
The horizontal line is the ‘x-axis,’ and the vertical line is the ‘y-axis.’ Coordinates are represented within brackets with a comma between the two coordinates: ‘( – 4, 9).’
An object in Quadrant I will have positive coordinates, while in Quadrant III will have negative coordinates. Objects in Quadrants II and IV will have a combination of one positive coordinate and one negative coordinate.
The below video covers coordinates. The video is meant to be used as an introduction or
review video. It introduces coordinates and graphing, and how it is used in the real world. Use the handout as an additional resource for students while they watch the video.
Video:
Introduction to Coordinates
Graphing Linear Equations
To understand how to solve equations using graphs, you will need to go through Graphing Linear Equations.
For each graph, an equation is given. For example, for our comprehension, let us take: 2x – 3y = 5, and the given x values are 3, 4, and 5. Now, what to do about the y values?
Begin by solving the equation using the given three values of the variable x –
2x – 3y = 5
2(3) – 3y = 5
6 – 3y = 5
6 – 5 = 3y
1 = 3y
y = 1/
2(4) – 3y = 5
8 – 3y = 5
8 – 5 = 3y
3 = 3y
y = 1
2(5) – 3y = 5
10 – 3y = 5
10 – 5 = 3y
5 = 3y
y = 5/
So, we have the values of x as 3, 4, and 5 from the given data and found the values of y to be 1/3, 1, and 5/3.
Plot the points on the graphs using the coordinates correctly, and you will have the object’s location correctly before you.
In some cases, instead of the x values, y values are given, but you can use the same methodology to find the x values, solve the equation and plot the graph.
The below worksheets cover graphing linear equations. Students find points for each equation and graph the line on the given grid. There is a different worksheet for each level.
Finding Slope from two points
A slope in a graph defines how slant the line is.
If the consecutive values of coordinates in a graph get closer to 0, then the slope will become increasingly horizontal. And, if those consecutive values are moving farther away from 0, then the slope will get increasingly vertical.
The slope is calculated as the ratio of the difference between two consecutive points on the y-axis to the difference between two successive points on the y-axis.
If the difference of points on the x-axis is 0, which implies that the denominator is 0, then the slope is stated as ‘Undefined.’
Example: Two points given as (4, -3) and (-8, 2)
Solution: Subtract the x coordinates and y coordinates alike. Meaning, subtract -8 from 4, and 2 from -3
-8 – 4 and 2 – (-3)
-12 and 2 + 3
-12 and 5
So, the resulting coordinates are (-12, 5).
Now, for the Slope calculation –
Δy / Δx = Difference in y coordinates / Difference in x coordinates
= 5 / -12
The below art worksheet reviews finding the slope from two points. Students calculate the slope using the change in x and y.
Pythagoras Theorem
Pythagoras, a Greek Mathematician, is the esteemed owner of the ‘Pythagoras Theorem.’ He proposed the relation of the three sides of a right-angled triangle, meaning a triangle with 90° as one of its angles.
The two legs forming the right angle, denoted by a small square on the inside of the triangle, when squared and added, give the numerical of the remaining one long leg – and this is the statement of Pythagoras Theorem.
Suppose the two short legs as ‘a’ and ‘b,’ and the one long leg also called the ‘hypotenuse’ as ‘c,’ then the expression for the Theorem will be –
c2 = a2 + b2
Remember that this theorem applies only to right-angled triangles. It does not work on scalene, equilateral, or isosceles triangles or any of their variations.
Here is also where you can use your knowledge of squares and square roots to identify the sides of the given right-angled triangle. You will come across both perfect squares and non-perfect squares for many of the right-angled triangles.
In addition to that, if the numbers are big and you face difficulty in their multiplication, you should brush up on your Distributive Property Concept and use it for your problem-solving.
Real-life applications of the Pythagoras Theorem are Construction, Drafting, Measuring Distance, Calculating the height of large objects, etc.
In many cases, only the hypotenuse doesn’t need to be calculated. There are instances where you will need the Theorem to calculate the other sides of the triangle or even half of one side.
Suppose those are the situation, no need to worry. You will have to use the same Theorem, but with a slight rearrangement. Let us look at an example.
Example: The hypotenuse is 16cm, and one short leg measures 2cm. Find the length of the other short arm.
Solution : The theorem is c2 = a2 + b2 right? All you need to do to solve this example problem is rearrange a2 to the left side of the Expression, like this –
c2 – a2 = b2, the sign of a2 changes from positive to negative as the term is shifted not to the left of the same side but to the left-hand side of the entire equation.
The above rearranged equation can also be written as –
b2 = c2 – a2
Substitute the values from the question –
b2 = 162 – 22
b2 = 256 – 4
b2 = 252 or b = √252
This video covers the Pythagorean theorem. It explains what the theorem is and how it can be used. The video includes examples and additional review to help students better understand the theorem and how to use it in everyday life.
Pythagorean Theorem Video
The below are the worksheets that describes the real world Pythagorean theorem.
Pythagorean Theorem – Level 2
The below worksheet continues the Pythagorean Theorem.
Distance of Line Segments
It is easy to measure the distance of the two short arms on a graph as the scale count is whole. But, for the hypotenuse, it is not so easy.
The diagonal line passes through various parts of each grid, sometimes covering one-quarters, and other times a tiny part is left out of the grid, hence using the Pythagoras Theorem to find out its measurement.
A line segment differs from a line as the former has two fixed points, unlike the latter. It has a start point and endpoint, whereas you can extend a line to infinity.
Now, if a line segment is considered and the distance of the said line segment is to be calculated, then the formula used is slightly different from the Pythagoras Theorem. It is –
Here, the two x coordinate values ad two y coordinate values are substituted, and distance is calculated. There are possibilities for the answer to arrive at both perfect squares and non-perfect squares.
You can also use this formula to calculate the side measurements of isosceles, scalene, or equilateral triangles and any of their derivatives.
For scalene, you will have to use the formula three times for the three different sides. Isosceles triangle sides require using the formula only twice; once for the two equal sides and once for the third other different side. And, finally, for the equilateral, once is enough as this type of triangle has all its sides equal.
The below worksheet introduces finding the distance of a line segment using a graph. It uses the Pythagorean Theorem to help students better understand how to find the distance.
Finding Midpoint
The middle of a line segment is easy if the numbers are small and even in total. But, what do you do to find out the center of the numbers range from 43 and 729?
Hence, a formula has been formulated to help you quickly find the midpoint without worrying much or doing extended mathematical operations where you lose interest halfway through. Here it is –
For two-dimensional midpoint finding –
In the formula, the starting and ending coordinates of the x and y axes are taken. Solving both the coordinate sides will give you the mid-point coordinates exactly.
For three-dimensional midpoint finding –
The formula takes the starting and ending coordinates of the x, y, and z axes. Solving the three coordinate sides will give you the mid-point coordinates exactly.
The below worksheet introduces the concept of midpoint and how to find it. It only covers basic number lines.
Algebra – Help Packets
Expressions – Level 1
The below worksheets introduces expressions. It explains what constants, variables, and expressions are and the difference between the three types of expressions (verbal, numerical, and algebraic expressions).
Key concept: Expressions help us connect words to mathematics. Students need to be able to take important information in a word problem and create a mathematical expression to form a solution.
Input – Output Tables – Level 1
The below worksheets introduces input-output tables. It uses pictures to explain the basic concept of the tables. Students complete the tables by drawing pictures and filling in the correct amounts.
Student misunderstanding: The relationship is between the input and output values. Use the fish problem on pages 2 to help explain.Yes, there is a pattern in the output (add 2), but the rule explains the relationship between the input / output numbers.
Commutative and Associative Properties
The below worksheets covers the Commutative and Associative Properties of addition and multiplication. Students learn that the order and groups can be changed, but the final answer will stay the same.
Key concept: Students need to learn that they can move numbers and regroup to make expressions and equations
easier to solve.
Intro to Variables – Level 1
The below worksheet introduces variables to students. It relates the empty box concept
to letters. The problems are designed for mental math and no moving or opposite operations are needed.
Student misunderstanding: Lots of children panic when they see a variable for the first time, but
understand the empty box concept. Use the box as a bridge to variables.
This below worksheets shows students how to solve one step equations using the opposite
Student misunderstanding: Equals means the same, and we can alter an equation as long as we
do the exact same thing to both sides.
Solving One Step Equations Add / Subtract – Alternative Method
The below worksheets shows students how to solve one step equations using a method we call the
Arrow Method. It uses the idea that letters and numbers do not get along, and they want to be on opposite sides of the wall (equal sign).
Key concept: Helps students understand which number to move. Ask the question, “Who can we move to make everyone happy?”
Solving One Step Equation– Multiply / Divide
The below worksheets shows students how to solve one step equations using the opposite
Student misunderstanding: Equals means the same, and we can alter an equation as long as we
do the exact same thing to both sides.
Solving One Step Equations – Level 1 Multiply / DivideAlternative Method
The below worksheet shows students how to solve one step equations using a method we call the
Arrow Method. It uses the idea that letters and numbers do not get along, and they want to be on opposite sides of the wall (equal sign).
Key concept: Helps students understand which number to move. Ask the question, “Who can we move to make everyone happy?”
The below worksheet is designed for students who have a basic knowledge of solving one and two step equations. It is for the students who struggle on which operation to do first and second, and uses a simple tactic to show equations in a different perspective.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Middle School Math Concepts - Grade 7 Go to the latest version.
# 2.3: Decimal Comparisons with Rounding
Difficulty Level: At Grade Created by: CK-12
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Practice Decimal Comparisons with Rounding
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Remember how Connor ran track? Well, his sister Leah volunteers at the local animal shelter. Here is a dilemma that she faced when she arrived on Saturday morning.
The animal shelter assigns sleeping boxes based on the length of kittens. The longest kitten gets the largest sleeping box. Leah was assigned the task of providing each kitten with the correct box. She is puzzled as to how to do this.
Here is the list of kitten lengths with the area of the sleeping boxes.
Kittens: Sleeping Boxes:
Ursula—23.11 cm A—
Mittens—23.51 cm B—
Josh—25.01 cm C—
Boxer—20.31 cm D—
Working with decimals is essential for sorting out this problem.
Do you know how to do it?
This Concept is about comparing and ordering decimals with rounding. By the end of it, you will be able to help Leah with her dilemma.
### Guidance
You already know how to compare and order decimals as well as how to round decimals. Combining the two procedures is fairly straightforward. Once decimals are rounded to a specific place value, they can be compared just as they were compared when they weren’t rounded—from left to right.
Compare 77.0949 and 77.0961 after rounding to the nearest hundredth. Write >, <, or =.
The first thing that we need to do is to round each to the nearest hundredth.
77.0949 - 9 is in the hundredths place. The number to the right of the nine is a 4. So we do not round up.
Our first number is 77.09
77.0961 - 9 is in the hundredths place here too. But the number to the right is a 6, so we round up.
Our second number is 77.10
What happened is that we needed to round up. We can’t go from 9 to 10 in one digit, so the place of the number moved over and we rounded 9 hundredths up to 1 tenth.
Now we can compare these numbers.
Our answer is 77.09 < 77.10.
We can use these same steps when ordering numbers from least to greatest or from greatest to least. First, you round the numbers to the desired place then you compare and order them.
Round the following numbers to the nearest tenth. Then order from greatest to least. 5.954, 5.599, 5.533, 6.062.
First, we round each number to the nearest tenth. Remember that the tenths place is the first place to the right of the decimal point. Here the digit being rounded is in bold print and the number that we use to determine whether we round up or down is in the hundredths place and is underlined.
5.954 rounds up to 6.0
5.599 rounds up to 5.6
5.533 rounds down to 5.5
6.062 rounds up to 6.1
Next, we write them in order from greatest to least.
Our answer is 6.1, 6.0, 5.6, 5.5
Now it's time for you to try a few on your own.
#### Example A
Compare each number after rounding to the nearest hundredth, 4.567 and 4.562
Solution: 4.57 > 4.56
#### Example B
Compare each number after rounding to the nearest tenth, .234 and .245
Solution: .2 = .2
#### Example C
Round each to the nearest tenth and write in order from least to greatest, .0567, .291, .1742
Solution:: .0567, .1742,.291
Now back to Leah and the kittens. Here is the original problem once again.
Remember how Connor ran track? Well, his sister Leah volunteers at the local animal shelter. Here is a dilemma that she faced when she arrived on Saturday morning.
The animal shelter assigns sleeping boxes based on the length of kittens. The longest kitten gets the largest sleeping box. Leah was assigned the task of providing each kitten with the correct box. She is puzzled as to how to do this.
Here is the list of kitten lengths with the area of the sleeping boxes.
Kittens: Sleeping Boxes:
Ursula—23.11 cm A—
Mittens—23.51 cm B—
Josh—25.01 cm C—
Boxer—20.31 cm D—
To solve this problem, let's first write the kittens in order from smallest to largest.
Boxer = 20.31
Ursula = 23.11
Mittens = 23.51
Josh = 25.01
Now the boxes in order from least to greatest.
Box B = 637.56
Box A = 647.06
Box C = 647.46
Box D = 647.9
Next, we put them together.
Boxer = Box B
Ursula = Box A
Mittens = Box C
Josh = Box D
### Vocabulary
Here are the vocabulary words in this Concept.
Decimal System
a system of measuring parts of a whole by using a decimal point.
Decimal point
the point that divides a whole number from its parts.
Decimals
a part of a whole located to the right of the decimal point.
Whole Numbers
counting numbers that do not include fractions or decimals. A whole number is found to the left of the decimal point.
### Guided Practice
Here is one for you to try on your own.
Sean weighed his textbooks with these results: Math—3.652 kg; English—3.596 kg; History—3.526 kg; Science—3.628 kilograms. Order his textbooks from greatest to least weight.
To complete this problem, we have to round each to the nearest tenth.
Math = 3.7
English = 3.60
History = 3.5
Science = 3.62
Because both the English book and the Science book round to 3.6, we have to add a hundredths digit. Now we can write them from least to greatest.
The answer is History, English, Science, Math.
### Video Review
Here is a video for review.
### Practice
Directions: Round each decimal to the nearest hundredth; then compare. Write <, >, or = for each ___.
1. 1.346 ___ 1.349
2. 0.0589 ___ 0.0559
3. 62.216 ___ 62.301
4. 5.011 ___ 5.001
5. 65.47 ____65.047
6. 12.324 ____ 12.325
7. .00897 _____.00967
8. .0009876 _____.0001020
9. .9806_____.9870
Directions: Round each decimal to the nearest thousandth. Then order from greatest to least.
10. 2.03489, 2.03266, 2.0344, 2.03909
11. 16.0995, 16.0875, 16.0885, 16.089
12. 3.8281, 3.8208, 3.8288, 3.8218
13. .05672, .05972, .05612, .0575
Directions: Complete each word problem.
14. Tamara’s famous holiday punch follows a precise recipe: 0.872 l orange juice; 0.659 l grapefruit juice; 1.95 l club soda; 0.981 1 lemonade; and 0.824 l limeade. Round her ingredient list to the nearest tenth; then order from least to greatest.
15. Mrs. King is pricing cabins at the state park for a weekend getaway with the family. A 2-person cabin is $53.90 for the weekend; a 3-person cabin is$67.53 for the weekend; a 4-person cabin is \$89.72 for the weekend. Round each price to the nearest whole number; then estimate the cheapest combination of cabins if there are 6 people in the King family.
### Vocabulary Language: English
Decimal point
Decimal point
A decimal point is a period that separates the complete units from the fractional parts in a decimal number.
Whole Numbers
Whole Numbers
The whole numbers are all positive counting numbers and zero. The whole numbers are 0, 1, 2, 3, ...
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# 8-1 - PowerPoint PPT Presentation
##### 8-1
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1. Relating Decimals, Fractions, and Percents 8-1 Warm Up Problem of the Day Lesson Presentation Pre-Algebra
2. Relating Decimals, Fractions, and Percents 8-1 3 7 7 4 2 1 3 1 4 14 1 1 12 5 15 15 12 2 4 5 5 2 3 3 2 3 or Pre-Algebra Warm Up Evaluate. 1. 2. 3. 4. – + 14
3. Problem of the Day A fast-growing flower grows to a height of 12 inches in 12 weeks by doubling its height every week. If you want your flower to be only 6 inches tall, after how many weeks should you pick it? 11 weeks
4. Learn to relate decimals, fractions, and percents.
5. Vocabulary percent
6. 3 30 50 75 10 100 100 100 1 2 3 4 Percents are ratios that compare a number to 100. 30% 50% 75%
7. Reading Math Think of the % symbol as meaning /100. 0.75 = 75% = 75/100
8. 1 = 1 ÷ 8 = 0.125 8 To convert a fraction to a decimal, divide the numerator by the denominator. To convert a decimal to a percent, multiply by 100 and insert the percent symbol. 0.125 100 12.5%
9. 1 10 10 = 100 Additional Example 1: Finding Equivalent Ratios and Percents Find the missing ratio or percent equivalent for each letter a–g on the number line. a: 10% =
10. 1 = 4 Additional Example 1: Finding Equivalent Ratios and Percents Find the missing ratio or percent equivalent for each letter a–g on the number line. b: 0.25 = 25%
11. 2 4 5 10 = 40 = 100 Additional Example 1: Finding Equivalent Ratios and Percents Find the missing ratio or percent equivalent for each letter a–g on the number line. c: 40% =
12. 3 = 5 Additional Example 1: Finding Equivalent Ratios and Percents Find the missing ratio or percent equivalent for each letter a–g on the number line. d: 0.60 = 60%
13. 2 66 0.666 = 2 % = 3 3 Additional Example 1: Finding Equivalent Ratios and Percents Find the missing ratio or percent equivalent for each letter a–g on the number line. e:
14. 1 87 7 % = 875 2 8 = 1000 Additional Example 1: Finding Equivalent Ratios and Percents Find the missing ratio or percent equivalent for each letter a–g on the number line. 0.875 = f:
15. 5 1 4 4 1 = 125 = 100 Additional Example 1: Finding Equivalent Ratios and Percents Find the missing ratio or percent equivalent for each letter a–g on the number line. g: 125% =
16. 1 2 5 8 3 8 1 12 1 % = 125 2 8 = 1000 Try This: Example 1 Find the missing ratio or percent equivalent for each letter a–g on the number line. e c 12 % 25% 50% 75% g 1 a b d f 0.125 = a:
17. 1 4 5 8 1 2 3 8 25 = 100 Try This: Example 1 Continued Find the missing ratio or percent equivalent for each letter a–g on the number line. e c 12 % 25% 50% 75% g 1 a b d f b: 25% =
18. 1 2 5 8 1 2 3 8 3 = 8 Try This: Example 1 Continued Find the missing ratio or percent equivalent for each letter a–g on the number line. e c 12 % 25% 50% 75% g 1 a b d f c: 0.375 = 37 %
19. 1 2 5 8 1 2 3 8 50 = 100 Try This: Example 1 Continued Find the missing ratio or percent equivalent for each letter a–g on the number line. e c 12 % 25% 50% 75% g 1 a b d f d: 50% =
20. 1 2 5 8 1 2 3 8 5 = 8 Try This: Example 1 Continued Find the missing ratio or percent equivalent for each letter a–g on the number line. e c 12 % 25% 50% 75% g 1 a b d f e: 0.625 = 62 %
21. 3 4 5 8 1 2 3 8 75 = 100 Try This: Example 1 Continued Find the missing ratio or percent equivalent for each letter a–g on the number line. e c 12 % 25% 50% 75% g 1 a b d f f: 75% =
22. 5 8 1 2 3 8 100 100 Try This: Example 1 Continued Find the missing ratio or percent equivalent for each letter a–g on the number line. e c 12 % 25% 50% 75% g 1 a b d f 100% = g: 1 =
23. 3 7 19 3 19 7 = 0.12 = 0.38 38 15 35 = 0.35 25 20 20 20 50 50 100 100 100 = = = 3 25 Additional Example 2: Finding Equivalent Fractions, Decimals, and Percents Find the equivalent fraction, decimal, or percent for each value given on the circle graph. 0.15(100) = 15% 0.12(100) = 12%
24. Additional Example 2 Continued You can use information in each column to make three equivalent circle graphs. One shows the breakdown by fractions, one shows the breakdown by decimals, and one shows the breakdown by percents. The sum of the fractions should be 1. The sum of the decimals should be 1. The sum of the percents should be 100%.
25. 1 9 1 5 10 20 45 25 45 100 100 100 1 1 = 4 5 Try This: Example 2 Fill in the missing pieces on the chart below. 0.1(100) = 10% =0.45 =0.25 0.25(100) = 25% = 0.2 0.2(100) = 20%
26. parts pure gold = total parts 0.583 = 58.3% = 1 7 7 14 So 14-karat gold is 58.3%, or 58 % pure gold. 3 12 12 24 Additional Example 3: Physical Science Application Gold that is 24 karat is 100% pure gold. Gold that is 14 karat is 14 parts pure gold and 10 parts another metal, such as copper, zinc, silver, or nickel. What percent of 14 karat gold is pure gold? Set up a ratio and reduce. 7 12 = Find the percent.
27. items eaten total items = 1 1 13 13 Try This: Example 3 A baker’s dozen is 13. When a shopper purchases a dozen items at the bakery they get 12. It is said that the baker eats 1 item from every batch. So, what percentage of the food the baker cooks is eaten without being sold? Set up a ratio and reduce. 0.077 = 1 13 = 7.7% Find the percent. So the baker, eats 7.7% of the items they bake.
28. 3 5 8 8 14 25 Lesson Quiz Find each equivalent value. 1. as a percent 2. 20% as a fraction 3. as a decimal 4. as a percent 5. About 342,000 km2 of Greenland’s total area (2,175,000 km2) is not covered with ice. To the nearest percent, what percent of Greenland’s total area is not covered with ice?
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# Order of Operations – PEMDAS Video
This video goes over the order of operations. Hopefully this helps your kids remember them more easily.
For more help, check out our Pre-Algebra Lessons Page
Credit goes to Education Galaxy
## A Guide to Mastering Order of Operations
In the realm of mathematics, tackling complex expressions often requires a systematic approach to ensure accuracy and consistency. Enter PEMDAS – a mnemonic device that serves as a guiding principle for solving order of operations problems.
PEMDAS, which stands for Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right), is a mnemonic device used to remember the order of operations in mathematical expressions. Think of it as a roadmap that dictates the sequence in which different operations should be performed when evaluating an expression.
Let’s break down each component of PEMDAS and understand its significance:
Parentheses – Parentheses are used to indicate groupings within an expression. Operations enclosed within parentheses should be performed first, starting from the innermost set of parentheses and working outward. Parentheses allow us to clarify the hierarchy of operations and override the default order dictated by PEMDAS when necessary.
Exponents – Exponents, also known as powers or indices, represent repeated multiplication. They denote how many times a number (the base) should be multiplied by itself. When encountering exponents in an expression, we evaluate them next, raising the base to the given exponent.
Multiplication and Division – Following the principles of PEMDAS, multiplication and division are treated as equal priority operations and are performed from left to right as they appear in the expression. It’s important to note that these operations should be evaluated in the order they appear, regardless of which comes first in the acronym.
Addition and Subtraction – Similar to multiplication and division, addition and subtraction are also of equal priority and are performed from left to right. Again, the order of operations dictates that these operations should be evaluated in the sequence they appear in the expression.
By adhering to the rules outlined by PEMDAS, we can systematically break down complex expressions into manageable steps, ensuring accuracy and consistency in our calculations. Let’s illustrate this with an example:
Consider the expression – 3 + 4 x (22 – 1)
Using PEMDAS, we start by evaluating the operations inside the parentheses:
Step #1 – Parentheses : (22 – 1)
We must first determine the value of the exponent:
22 = 4
We then subtract 1.
4 – 1 = 3
Now, we have: 3 + 4 x 3
Step #2: Multiplication
Next, we perform multiplication before addition:
4 x 3 = 12
Now, we have: 3 + 12
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Many ways problem
## Taking it further
In the main problem you arranged four functions in a $2 \times 2$ square according to two properties that the functions did or did not have. Now we will think about arranging eight functions in a $2 \times 2 \times 2$ ‘cube’ based on whether or not the functions have three properties.
For example, can you complete this function cube for the following properties?
• Where the function is well defined, two different input values give two different output values
• The graph of function has asymptotes
• The graph of function has rotational symmetry about $(0,0)$
There are many ways to represent a cube and if you wanted to use the functions cards, you might find it simpler to arrange them in a $4 \times 2$ table, i.e. two ‘layers’ of four functions.
For example, the table below could represent a function cube for the properties
• The graph passes through $(0,0)$
• The graph is straight
• The $y$-axis is a line of symmetry for the graph
Graph passes through $(0,0)$ Graph is straight $y$-axis is a line of symmetry $f(x)= \cdots$ Graph passes through $(0,0)$ Graph is straight $y$-axis is not a line of symmetry $f(x)= \cdots$ Graph passes through $(0,0)$ Graph is not straight $y$-axis is a line of symmetry $f(x)= \cdots$ Graph passes through $(0,0)$ Graph is not straight $y$-axis is not a line of symmetry $f(x)= \cdots$ Graph does not pass through $(0,0)$ Graph is straight $y$-axis is a line of symmetry $f(x)= \cdots$ Graph does not pass through $(0,0)$ Graph is straight $y$-axis is not a line of symmetry $f(x)= \cdots$ Graph does not pass through $(0,0)$ Graph is not straight $y$-axis is a line of symmetry $f(x)= \cdots$ Graph does not pass through $(0,0)$ Graph is not straight $y$-axis is not a line of symmetry $f(x)= \cdots$
Try to complete this function cube or make a different one.
You might need to make some extra function cards.
Make a note of any cells that you can’t think of functions for.
Make a note of properties that are difficult to combine in a function cube.
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# Which of the following is a root of the polynomial shown below fxx3x29x 9
## The polynomial is shown below
The polynomial shown below has a root of 3.
f(x) = x^3 + 2x^9 + 9
This can be seen by factoring the polynomial as follows:
f(x) = (x-3)(x^2+3x+3)(x^6-3x^3+9)
Since (x-3) is a factor, that means that x=3 is a root of the polynomial.
## f(x) = 3x^2 + 9x + 2
There are three roots to this polynomial equation: -1, 2, and -2.
## The roots of the polynomial
The roots of the polynomial f(x) = 3x^2 – 9 are -3 and 3.
## How to find the roots of the polynomial
There are a few different methods that can be used to find the roots of the polynomial f(x) = 3x^2 – 9x + 9.
One method is to factor the polynomial. In this case, the polynomial can be factored as (3x – 9)(x – 1). This means that the roots of the polynomial are -3 and 1.
Another method is to use the Quadratic Formula. The Quadratic Formula can be used to find the roots of any quadratic equation, and in this case we have a quadratic equation with coefficients a = 3, b = -9, and c = 9. Plugging these values into the Quadratic Formula, we get that the roots of the polynomial are 1/3 and 3.
Lastly, we could use synthetic division to find the roots of the polynomial. Synthetic division is a method for finding the zeros of a polynomial by dividing it by a linear factor. In this case, we would divide by (x – 1) to get that the roots are 1 and -3.
## The factors of the polynomial
The factors of the polynomial are (x-3), (x-9), and (x+9).
## The zeros of the polynomial
The zeros of the polynomial are its roots. The roots of the polynomial shown below are 3, -3, and -9.
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# Recording Mathematics
## Recording Mathematics
This collection of activities offers an opportunity for you to focus on children's own creative representations and recordings. Read the article to find out why we have selected these particular tasks. Click the 'Submit a solution' link on any live problem to find out the closing date.
### Primary Children's Mathematical Recording
##### Stage: Early years, 1 and 2
This article for teachers outlines different types of recording, depending on the purpose and audience.
### What Was in the Box?
##### Stage: 1 Challenge Level:
This big box adds something to any number that goes into it. If you know the numbers that come out, what addition might be going on in the box?
##### Stage: 1 Challenge Level:
If you put three beads onto a tens/ones abacus you could make the numbers 3, 30, 12 or 21. What numbers can be made with six beads?
##### Stage: 1 Challenge Level:
There are three baskets, a brown one, a red one and a pink one, holding a total of 10 eggs. Can you use the information given to find out how many eggs are in each basket?
### The Amazing Splitting Plant
##### Stage: 1 Challenge Level:
Can you work out how many flowers there will be on the Amazing Splitting Plant after it has been growing for six weeks?
### School Fair Necklaces
##### Stage: 1 and 2 Challenge Level:
How many possible necklaces can you find? And how do you know you've found them all?
### What's in the Box?
##### Stage: 2 Challenge Level:
This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?
### Money Bags
##### Stage: 2 Challenge Level:
Ram divided 15 pennies among four small bags. He could then pay any sum of money from 1p to 15p without opening any bag. How many pennies did Ram put in each bag?
### Dice in a Corner
##### Stage: 2 Challenge Level:
How could you arrange at least two dice in a stack so that the total of the visible spots is 18?
### Domino Square
##### Stage: 2, 3 and 4 Challenge Level:
Use the 'double-3 down' dominoes to make a square so that each side has eight dots.
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Lesson 1, Topic 2
In Progress
# Converting from other Bases to Base 10
Lesson Progress
0% Complete
Here, we will learn the methods to convert from other bases to Base 10
### Example 1
Convert
(i) 7603eight to base 10
(ii) 1101two to base 10
(iii) 5B9twelve to base 10
Solution (i) 7603eight to base 10
a. First Method: (Power Expansion)
Power Expansion entails multiplying out the number base using their base values.
• First, write the place values starting from the right-hand side.
• Write each digit under its place value.
• Multiply each digit by its base raised to the corresponding place value.
(i.e. baseplace value, Note: for this question, it will be 8place value)
• Add up the products. The answer will be the decimal number in base ten.
Students often find it difficult to locate the place values. The easiest way is to start from the last digit to the first digit. The last Digit always starts with a 0, the digit that follows a 1, and so on. (0, 1 2, 3, etc)
Write down the base raised to the place value. The number we are converting is 7603eight ∴ it’s in base 8.
Multiply each digit by its base raised to the corresponding place value.
7603eight = (7 × 83) + (6 × 82) + (0 × 81) + (3 × 80)
= 3584 + 384 + 0 + 3
= 3,971ten
∴ 7603eight = 3,971ten
b. Alternate Method – (Successive Multiplication)
7 × 8 = 56 + 6 = 62 × 8 = 496 + 0 = 496 × 8 = 3968 + 3 = 3,971ten
Solution (ii) 1101two to base 10
a. First Method: (Power Expansion)
Multiply each digit by its base raised to the corresponding place value
1101two = (1 × 23) + (1 × 22) + (0 × 21) + (1 × 20)
= 8 + 4 + 0 + 1
= 13ten
Alternate Method
1 × 2 = 2 + 1 = 3 × 2 = 6 + 0 = 6 × 2 = 12 + 1 = 13ten
Solution (iii) 5B9Twelve to base 10
a. First Method: (Power Expansion)
Multiply each digit by its base raised to the corresponding place value (Note: B =11)
5B9Twelve = (5 × 122) + (11 × 121) + (9 × 120)
= 720 + 132 + 9
= 861ten
Alternate Method
5 × 12 = 60 + 11 = 71 × 12 = 852 + 9 = 861ten
### Example 2
Convert 1101.101two to base ten
In the conversion of a decimal number to base 10, the whole number is treated as the examples above (i.e from the last digit the numbering is 0, 1, 2, 3, etc up to the first digit).
The decimal part on the other hand is numbered -1, -2, -3, etc immediately after the decimal point. The first number after the decimal point will be -1, the second number will be -2, etc.
Hence. We multiply the first number by the base raised to the power minus one (x-1) and so on.
Solution
1101.101two = (1 × 23) + (1 × 22) + (0 × 21) + (1 × 20) + (1 × 2-1) + (0 × 2-2) + (1 × 2-3)
= 8 + 4 + 0 + 1 + $$\frac {1}{2}$$ + 0 + $$\frac {1}{8}$$
= 8 + 4 + 0 + 1 + 0.5 + 0 + 0.125
= 13.625
1101.101two= 13.625ten
Alternate Method
Whole number part = 1 × 2 = 2 + 1 = 3 × 2 = 6 + 0 = 6 × 2 = 12 + 1 = 13
Fractional part = 1 × 2 = 2 + 0 = 2 × 2 = 4 + 1 = 5
The result for the whole number part is: 1101two = 13ten
The place value of the last column of the fractional part is 2-3
∴ 0.101two= 5 × 2-3 = 5 × $$\frac {1}{8}$$ = 0.625
Summing up the results, then
1101.101two = 13 + 0.625
1101.101two = 13.625ten
### Example 3
Convert
(i) 4A3D.216 to base 10
(ii) 4B3.A612 to base 10
Solution (i) 4A3D.216 to base 10
Remember D = 13 from our Number Bases table
4A3D.216 = (4 × 163) + (A × 162) + (3 × 161) + (D × 160) + (2 × 16-1)
= (16,384) + (10 × 256) + (48) + (13 × 1) + (2 × 0.0625)
= 16,384 + 2,560 + 48 + 13 + 0.125
= 19,005.125
Solution (ii) 4B3.A612 to base 10
A = 10, B = 11
4B3.A612 = (4 × 122) + (B × 121) + (3 × 120) + (A × 12-1) + (6 × 12-2)
= (576) + (11 × 12) + (3) + (10 × 0.0833) + (6 × 0.00694)
= 576 + 132 + 3 + 0.833 + 0.04164
= 711.875
#### Responses
1. Super helpful! First method is so easy, I got confused in class when I was learning this.
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# Question Video: Ordering Decimal Numbers Mathematics • 5th Grade
Arrange 35.36, 35.07, 34.189, 34.197 in ascending order.
03:13
### Video Transcript
Arrange 35 and 36 hundredths and 35 and seven hundredths and 34 and 189 thousandths and 34 and 197 thousandths in ascending order.
Ascending order means from least to greatest. Let’s line these values up so that their place values are all in a row: 35 and 36 hundredths, 35 and seven hundredths, 34 and 189 thousandths, 34 and 197 thousandths. We have the tens place, the units place, tenths place, hundredths place, and the thousandths place. We wanna to start all the way on the left with the highest place value and compare and then move to the right.
In the tens place, all of these values are the same. So we move our comparison to the units place. We have five, five, four, and four. The two values with the four in the units place are smaller than the two values with the five in the units place. But in order to tell which of these two values are smaller, we need to move to the right again in the tenths place.
Both of these values have a one in the tenths place. So we move to the right again. In the hundredths place, we have an eight and a nine. Remember, we’re working from least to greatest, so we need the smallest number. Eight is smaller than nine. 34 and 189 thousandths is smaller than 34 and 197 thousandths.
Now we need to compare 35 and 36 hundredths and 35 and seven hundredths. Both numbers have threes in the tens place and five in the units place, which means we need to compare the tenths place.
One of the numbers has a three in the tenths place and one has a zero. Again we’re looking for the smallest value. Zero is smaller than three, and that means 35 and seven tenths is smaller than 35 and 36 tenths. In ascending order, we have 34 and 189 thousandths, 34 and 197 thousandths, 35 and seven hundredths, 35 and 36 hundredths.
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```Question 149402
Lets assume this sequence is an arithmetic sequence. The general form of the arithmetic sequence is
{{{a[n]=d*n+a[1]}}} where {{{a[n]}}} is the nth term, d is the difference, and {{{a[1]}}} is the first term
So lets find the difference between 2 terms (i.e. the difference between two terms)
====================================================================================================================
To find the difference, simply pick any term and subtract the previous term from that selected term
{{{19-21=-2}}} Select the 2nd term (which is 19) and subtract the 1st term (which is 21) from it.
So we get a difference of {{{-2}}}
Lets pick another pair of terms to verify:
{{{17-19=-2}}} Select the 3rd term (which is 17) and subtract the 2nd term (which is 19) from it.
And again, we get a difference of {{{-2}}}
-----------------------------------------------------
Lets pick another pair of terms to verify:
{{{15-17=-2}}} Select the 4th term (which is 15) and subtract the 3rd term (which is 17) from it.
And again, we get a difference of {{{-2}}}
-----------------------------------------------------
====================================================================================================================
Since we've tested every consecutive pair of terms, we've verified that the sequence has a constant difference of {{{-2}}}. This means the sequence is arithmetic
Since the difference is {{{d=-2}}} and the first term is {{{a[1]=21}}}, this means the arithmetic sequence is
{{{a[n]=-2n+21}}} where {{{n}}} starts at {{{n=0}}}
Check:
Notice if we plug in {{{n=0}}} we get
{{{a[0]=-2(0)+21}}} plug in {{{n=0}}}
{{{a[0]=0+21}}} Multiply
which is our first term
Notice if we plug in {{{n=1}}} we get
{{{a[1]=-2(1)+21}}} plug in {{{n=1}}}
{{{a[1]=-2+21}}} Multiply
which is our second term
Notice if we plug in {{{n=2}}} we get
{{{a[2]=-2(2)+21}}} plug in {{{n=2}}}
{{{a[2]=-4+21}}} Multiply
which is our third term
Notice if we plug in {{{n=3}}} we get
{{{a[3]=-2(3)+21}}} plug in {{{n=3}}}
{{{a[3]=-6+21}}} Multiply
which is our fourth term
Since each term corresponds to the terms of the given list, this verifies our sequence.
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Hide
# Problem DMoving Day
It’s that time of the year when students head back to school, which will usually involve moving lots of stuff, and packing lots of that stuff into boxes. However, before we go through that effort, it would be nice to know whether the boxes we have are the right size!
A box has three dimensions, $l$, $w$, and $h$, and a box’s volume $v$ is simply $l\cdot w \cdot h$. We have many small items to pack, and we don’t care about their exact geometry: we know that, as long as a box has a volume of, at least, $\mathbf{V}$, we will be able to pack all those items into that single box. So, given a box $i$, and its volume $v_ i$, then the difference $d_ i=v_ i-\mathbf{V}$ can tell us whether the box is big enough or not to pack all the items. If $d_ i$ is greater than or equal to zero, the box is big enough; otherwise, the box is too small.
So, we want to answer a simple question: given multiple boxes, is the largest box (by volume) large enough to store all our items? For example, suppose $\mathbf{V}=10$ and we have three boxes, $A$, $B$, and $C$, with dimensions $(1,1,2)$, $(2,2,2)$, and $(3,2,1)$, respectively. Their volumes will be $v_ A=2$, $v_ B=8$, and $v_ C=6$, which means the largest box is $B$. Unfortunately, $d_ B=v_ B-\mathbf{V}=8-10=-2$, which means our largest box is not large enough.
On the other hand, suppose $\mathbf{V}=980$, and we had four boxes, $W$, $X$, $Y$, and $Z$ with dimensions $(10,10,10)$, $(10,5,2)$, $(5,3,2)$, and $(90,5,2)$, respectively. Their volumes will be $v_ W=1\, 000$, $v_ X=100$, $v_ Y=30$ and $v_ Z=900$, making $W$ the largest box and, since $d_ W=v_ W-\mathbf{V}=1\, 000-980=20$, that means $W$ is big enough for all our items.
## Input
The input specifies a set of boxes. It begins with a line containing two integers: $n$ ($1 \le n \le 100$), specifying the number of boxes, and $\mathbf{V}$, as defined above. The remaining input is composed of $n$ lines, each specifying one box. Each line contains the dimensions $l$, $w$, and $h$ for the box. You may assume that $1 \leqslant l, w, h, \mathbf{V} < 2^{32}$. You may also assume that, for any given box $i$, its volume $v_ i$ will always be less than $2^{32}$
## Output
The output is a single integer: the value of $d_ i$ for the largest box by volume.
Sample Input 1 Sample Output 1
3 10
1 1 2
2 2 2
3 2 1
-2
Sample Input 2 Sample Output 2
3 30
1 1 1
5 2 3
5 2 1
0
Sample Input 3 Sample Output 3
4 980
10 10 10
10 5 2
5 3 2
90 5 2
20
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# How do you break down the square root of 130?
November 2, 2022
Position the Omni hangers about 1/3 of the way down from the top of the frame. Example: If the frame is 10″ tall, place the hangers 3″ below the top of the frame. Screw the hangers into the channel of the frame with a flathead screwdriver.
## What is the simplest radical form?
So we get the value of the square root of √132 = 11.489 by the long division method.
## What is the square root of 135 in simplest radical form?
The square root of 135 is 11.618, approximated to 3 decimal places. The simplified form of √135 in its root form is 3√15.
## What is the factor of 130?
Divisors of a number are the numbers that exactly divide the given number without a remainder. According to the definition of factors, the factors of 130 are 1, 2, 5, 10, 13, 26, 65, and 130. So 130 is a composite number because it has factors other than 1 and itself.
## Can you find the exact value of 130?
Since none of these numbers is a perfect square, we can’t take anything out of the square root. So, in most cases, just saying √130 should be enough.
## How do you simplify square roots?
Since none of these numbers is a perfect square, we can’t take anything out of the square root. So, in most cases, just saying √130 should be enough.
## How do you simplify the square root of 136?
The simplified form of √136 in its root form is 2√34.
## What is the square of 131?
The square root of 131 is 11.44552.
## How do you write in radical form?
The square root of 131 is 11.44552.
## How do you convert to radical form?
The square root of 131 is 11.44552.
## How do you find radicals on a calculator?
The square root of 131 is 11.44552.
## How do you find the square root of 135?
The square root of 131 is 11.44552.
## What is the square root of 125 in simplest radical form?
The square root of 125 in its simplest form is 5 √5.
## How do you find the square root of 135 by division?
The square root of 125 in its simplest form is 5 √5.
## What is the square of 130?
The square root of 125 in its simplest form is 5 √5.
## Is 130 a composite number?
Is 130 a composite number? Yes, since 130 has more than two factors, i.e. H. 1, 2, 5, 10, 13, 26, 65, 130. In other words, 130 is a composite number because 130 has more than 2 factors.
## What two numbers multiplied equal 130?
10 x 13 = 130. 13 x 10 = 130. 26 x 5 = 130. 65 x 2 = 130.
## Is 130 a whole number?
10 x 13 = 130. 13 x 10 = 130. 26 x 5 = 130. 65 x 2 = 130.
### References:
2. https://www.cuemath.com/algebra/square-root-of-132/
4. https://www.cuemath.com/algebra/square-root-of-135/
5. https://www.cuemath.com/numbers/factors-of-130/
6. https://socratic.org/questions/what-is-the-square-root-of-130
8. https://www.cuemath.com/algebra/square-root-of-136/
9. https://www.cuemath.com/algebra/square-root-of-131/
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# Order (group theory)
(Redirected from Group order)
In group theory, a branch of mathematics, the term order is used in two unrelated senses:
• The order of a group is its cardinality, i.e., the number of elements in its set. Also, the order, sometimes period, of an element a of a group is the smallest positive integer m such that am = e (where e denotes the identity element of the group, and am denotes the product of m copies of a). If no such m exists, a is said to have infinite order.
• The ordering relation of a partially or totally ordered group.
The order of a group G is denoted by ord(G) or |G| and the order of an element a is denoted by ord(a) or |a|.
## Example
Example. The symmetric group S3 has the following multiplication table.
e s t u v w
e e s t u v w
s s e v w t u
t t u e s w v
u u t w v e s
v v w s e u t
w w v u t s e
This group has six elements, so ord(S3) = 6. By definition, the order of the identity, e, is 1. Each of s, t, and w squares to e, so these group elements have order 2. Completing the enumeration, both u and v have order 3, for u2 = v and u3 = vu = e, and v2 = u and v3 = uv = e.
## Order and structure
The order of a group and that of an element tend to speak about the structure of the group. Roughly speaking, the more complicated the factorization of the order the more complicated the group.
If the order of group G is 1, then the group is called a trivial group. Given an element a, ord(a) = 1 if and only if a is the identity. If every (non-identity) element in G is the same as its inverse (so that a2 = e), then ord(a) = 2 and consequently G is abelian since $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$ by Elementary group theory. The converse of this statement is not true; for example, the (additive) cyclic group Z6 of integers modulo 6 is abelian, but the number 2 has order 3:
$2+2+2=6 \equiv 0 \pmod {6}$.
The relationship between the two concepts of order is the following: if we write
$\langle a \rangle = \{ a^{k} : k \in \mathbb{Z} \}$
for the subgroup generated by a, then
$\operatorname{ord} (a) = \operatorname{ord}(\langle a \rangle).$
For any integer k, we have
ak = e if and only if ord(a) divides k.
In general, the order of any subgroup of G divides the order of G. More precisely: if H is a subgroup of G, then
ord(G) / ord(H) = [G : H], where [G : H] is called the index of H in G, an integer. This is Lagrange's theorem. (This is, however, only true when G has finite order. If ord(G) = ∞, the quotient ord(G) / ord(H) does not make sense.)
As an immediate consequence of the above, we see that the order of every element of a group divides the order of the group. For example, in the symmetric group shown above, where ord(S3) = 6, the orders of the elements are 1, 2, or 3.
The following partial converse is true for finite groups: if d divides the order of a group G and d is a prime number, then there exists an element of order d in G (this is sometimes called Cauchy's theorem). The statement does not hold for composite orders, e.g. the Klein four-group does not have an element of order four). This can be shown by inductive proof.[1] The consequences of the theorem include: the order of a group G is a power of a prime p if and only if ord(a) is some power of p for every a in G.[2]
If a has infinite order, then all powers of a have infinite order as well. If a has finite order, we have the following formula for the order of the powers of a:
ord(ak) = ord(a) / gcd(ord(a), k)
for every integer k. In particular, a and its inverse a−1 have the same order.
In any group,
$\operatorname{ord}(ab) = \operatorname{ord}(ba)$
There is no general formula relating the order of a product ab to the orders of a and b. In fact, it is possible that both a and b have finite order while ab has infinite order, or that both a and b have infinite order while ab has finite order. An example of the former is a(x) = 2-x, b(x) = 1-x with ab(x) = x-1 in the group $Sym(\mathbb{Z})$. An example of the latter is a(x) = x+1, b(x) = x-1 with ab(x) = id. If ab = ba, we can at least say that ord(ab) divides lcm(ord(a), ord(b)). As a consequence, one can prove that in a finite abelian group, if m denotes the maximum of all the orders of the group's elements, then every element's order divides m.
## Counting by order of elements
Suppose G is a finite group of order n, and d is a divisor of n. The number of order-d-elements in G is a multiple of φ(d), where φ is Euler's totient function, giving the number of positive integers no larger than d and coprime to it. For example in the case of S3, φ(3) = 2, and we have exactly two elements of order 3. The theorem provides no useful information about elements of order 2, because φ(2) = 1, and is only of limited utility for composite d such as d=6, since φ(6)=2, and there are zero elements of order 6 in S3.
## In relation to homomorphisms
Group homomorphisms tend to reduce the orders of elements: if fG → H is a homomorphism, and a is an element of G of finite order, then ord(f(a)) divides ord(a). If f is injective, then ord(f(a)) = ord(a). This can often be used to prove that there are no (injective) homomorphisms between two concretely given groups. (For example, there can be no nontrivial homomorphism h: S3 → Z5, because every number except zero in Z5 has order 5, which does not divide the orders 1, 2, and 3 of elements in S3.) A further consequence is that conjugate elements have the same order.
## Class equation
An important result about orders is the class equation; it relates the order of a finite group G to the order of its center Z(G) and the sizes of its non-trivial conjugacy classes:
$|G| = |Z(G)| + \sum_{i}d_i\;$
where the di are the sizes of the non-trivial conjugacy classes; these are proper divisors of |G| bigger than one, and they are also equal to the indices of the centralizers in G of the representatives of the non-trivial conjugacy classes. For example, the center of S3 is just the trivial group with the single element e, and the equation reads |S3| = 1+2+3.
## Open questions
Several deep questions about the orders of groups and their elements are contained in the various Burnside problems; some of these questions are still open.
## References
1. ^ Conrad, Keith. "Proof of Cauchy's Theorem" (PDF). Retrieved May 14, 2011.
2. ^ Conrad, Keith. "Consequences of Cauchy's Theorem" (PDF). Retrieved May 14, 2011.
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Multiplying mixed numbers is quite similar to multiplying simple fractions. Sometimes, you may need to multiply three or more mixed numbers together, for example, when applying one discount after another to an original list price.
To multiply mixed numbers, first change them to improper fractions. Next, simplify, or reduce, the fractions by removing common factors. Then multiply the numerators, and multiply the denominators. Simplify, or reduce the answer if possible. Finally, if the answer is an improper fraction, you can change it back to a mixed number.
Here's an example of a multiplication problem with three mixed numbers:
1. Change all mixed numbers to improper fractions.
In this example, you have to convert three mixed numbers to improper fractions: 3 1/3, 5 1/4, and 1 1/2. Follow this formula to convert each mixed number into an improper fraction:
In a fraction, the Numerator is always North of (above) the fraction line. The Denominator is always Down below the fraction line.
You convert the mixed numbers to improper fractions as follows:
2. Reduce the fractions.
To do this, break down both the numerator and denominator of each fraction into their prime factors (shown below in parentheses):
3. Cross out any common factors.
In this example, you cross out 2 and 3 because they’re common factors — that is, they appear in both the numerator and denominator:
4. Multiply the numerators together and the denominators together.
5. Reduce the answer.
If the result is an improper fraction, change it to a mixed number.
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# Lesson: Range and Nullspace
1 point
## Transcript
There are two subspaces that we affiliate with linear mappings: the range and the nullspace. We will define them first, and then prove that they are subspaces.
Let V and W be vector spaces over R. The range of a linear mapping L from V to W is defined to be the set of all L(x) in W for x in V.
For example, let’s let L be the mapping from R2 to M(2, 2) defined by L([a; b]) = [a, b; b, a]. Then we see that [1, 2; 2, 1] is in the range of L because L([1; 2]) = [1, 2; 2, 1]. But [1, 2; 3, 4] is not in the range of L. To see this, we note that [1, 2; 3, 4] is in the range of L only if there is a vector [a; b] whose entries a and b satisfy that L([a; b]), which equals [a, b; b, a], equals [1, 2; 3, 4]. Setting the matrix entries equal to each other, we see that this is equivalent to the system a = 1, b = 2, b = 3, a = 4. But since we cannot simultaneously have a = 1 and a = 4 (or b = 2 and b = 3), we see that our system is inconsistent. And since there are no such a and b, we have that [1, 2; 3, 4] is not in the range of L.
For another example, let’s determine whether or not the vector y = [5; -7] is in the range of the mapping L from P3 to R2, where L is defined by L(a + bx + c(x-squared) + d(x-cubed)) = [a + 2b + c + 2d; 3a + 4b – c – 2d]. If y is in the range, find a vector x such that L(x) = y. First, we look to see if [5; -7] is in the range of L. That is, we want to know if there is a polynomial x = a + bx + c(x-squared) + d(x-cubed) whose coefficients a, b, c, and d satisfy that L(x), which equals [a + 2b + c + 2d; 3a + 4b – c – 2d] equals our vector y, [5; -7]. Setting the components equal to each other, we see that this is equivalent to this system. We solve the system by row reducing its augmented matrix, which only takes one step. Our matrix is now in row echelon form, and since there are no bad rows, we know that our system is consistent. This means that there is a solution, and thus, that [5; -7] is in the range of L.
So now let’s move on to finding an x such that L(x) = y. Well, this simply means finding a solution to our system, so we will continue our row reduction until we reach reduced row echelon form. This only takes two more steps. We see that our system is equivalent to the system a – 3c – 6d = -17, and b + 2c + 4d = 11. Replacing the variable c with the parameter s and the variable d with the parameter t, we see that the general solution to our system is [-17; 11; 0; 0] + s[3; -2; 1; 0] + t[6; -4; 0; 1]. Well, this provides us a list of all possible a, b, c, and d. By picking specific values for s and t, we can get a vector x such that L(x) = y, and the easiest possible values for s and t are s = 0, t = 0. This gives us that a = -17, b = 11, c = 0, and d = 0. And so we’ve seen that L(-17 + 11x) equals the vector [5; -7].
We now define the nullspace. The nullspace of L is the set of all vectors in V whose image under L is the 0 vector. We write in set notation that the Null(L) is the set of {x in V such that L(x) = 0}.
For example, let L be the linear mapping from M(2, 3) to P1 defined by L([a, b, c; d, e, f]) = (a + b + c) + (d + e + f)x. Well, then the matrix [1, 1, -2; -3, 0, 3] is in the nullspace of L since L([1, 1, -2; -3, 0, 3]) will equal (1 + 1 – 2) + (-3 + 0 + 3)x, which equals 0 + 0x, our 0 polynomial. But the matrix [1, 2, 1; -3, 5, 3] is not in the nullspace of L since L([1, 2, 1; -3, 5, 3]) will equal (1 + 2 + 1) + (-3 + 5 + 3)x, which equals 4 + 5x, which is not equal to our 0 polynomial.
You’ll note that it is much easier to check if a vector is in the nullspace than to check if it is in the range.
Now recall that one of the reasons we undertook a study of general vector spaces instead of focusing on each one individually is that it gives us the opportunity to prove statements that are true for all vector spaces—or, as below, a statement that is true for all linear mappings.
Theorem 4.5.1: Let V and W be vector spaces, and let L from V to W be a linear mapping. Then the following are true.
1. L of our V 0 vector equals our W 0 vector.
2. The nullspace of L is a subspace of V.
3. The range of L is a subspace of W.
To prove part 1, let x be any element of V. Then we have that L of our V 0 vector equals L(0x) because, of course, we know that 0x always equals the 0 vector. By the linearity properties of L, we can pull the 0 out and see that this is equal to 0(L(x)). But again, 0 times any vector is 0, so this must equal our 0 vector in W.
I will leave the proof of part 2 as an assignment, and instead, I will now show you the proof of part 3. To see that the range of L is a subspace of W, we first want to note that the range of L is explicitly defined as a subset of W. Moreover, the range of L is non-empty since part 1 of our theorem tells us that L of the 0 vector will equal the 0 vector, and thus, we know for fact that our 0 vector is in the range. Next, we check to see if Range(L) is closed under addition. So let’s let w1 and w2 be elements of the range of L. Well, this means that there are elements v1 and v2 of V such that L(v1) = w1 and L(v2) = w2. Then we note that L(v1 + v2) will equal L(v1) + L(v2), which equals w1 + w2. So, v1 + v2 is an element of V such that L(v1 + v2) = w1 + w2, and thus w1 + w2 is in the range of L.
Well, now that we have shown that Range(L) is closed under addition, let’s show it is closed under scalar multiplication. To that end, we’ll let w be in the range of L and s be a scalar. Well, then there is a v in V such that L(v) = w. But this means that L(sv) = s(L(v)), which equals sw. So we have found that sv is an element of V such that L(sv) = sw, so sw is in the range of L.
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## How to Calculate Your Break-Even CPC?
Before diving into the final formula for calculating your break-even CPC, we will first explain the underlying logic and guide you on how to solve it independently. So let’s begin with the first approach:
## Target CPA
To figure out the break-even CPC based on your target CPA, you need to know two things first:-
• Your target CPA is how much you are willing to pay for a conversion.
• The Conversion rate of your website and is basically how many sessions or clicks you need to have to generate a conversion.
As you may already know, the target CPA can be determined by dividing the cost by the number of conversions:
CPA = Cost / Conversions
In this equation, the cost is represented by the product of the number of clicks and the CPC, while conversions are calculated by multiplying the conversion rate with the number of clicks. Let’s substitute these values into the equation:
CPA = (Clicks * CPC) / (Clicks * Conversion Rate)
To simplify the equation, we can cancel out the clicks from both the numerator and the denominator:
CPA = CPC / Conversion Rate
Now, let’s rearrange the formula to calculate the CPC:
CPC = CPA * Conversion Rate
So, the final formula for determining the CPC is to multiply the target CPA by the conversion rate. Let’s apply this formula to the example given. If the target CPA is \$75 and the conversion rate is 0.60%, the CPC would be:
CPC = \$75 * 0.60% = \$0.45
## Target ROI
To decide how much you should bid based on an ROI target, you need a few things before we start:-
• Conversion rate of your website and this is basically how many sessions or clicks you need to have to generate a conversion.
• Average order value which is the average value of your conversions.
If you know your target ROI, the number of clicks required to generate a conversion, and the average order value (AOV), you can easily determine the CPC value. Let’s consider an example where the target ROI is 2, AOV is \$150, and the conversion rate is 0.60%. Now, let’s derive the formula step by step:
As you already know, ROI is calculated by dividing revenue by cost:
ROI = Revenue / Cost
Revenue is equal to the product of conversions and AOV, while the cost is the result of clicks multiplied by CPC. Let’s substitute these values into the equation:
ROI = (Conversions * AOV) / (Clicks * CPC)
Additionally, conversions can be expressed as the product of clicks and the conversion rate:
ROI = (Clicks * Conversion Rate * AOV) / (Clicks * CPC)
To simplify the equation, we can cancel out the clicks from both the numerator and the denominator:
ROI = (Conversion Rate * AOV) / CPC
Now, let’s rearrange the formula to solve for CPC:
CPC = (Conversion Rate * AOV) / ROI
Finally, to calculate the CPC, divide the product of the conversion rate and AOV by the target ROI. Applying this formula to our previous example:
CPC = (0.60% * \$150) / 2 = \$0.45
## Conclusion
You can’t control how much other advertisers bid but you can place your ads at a level where they will be profitable. The factors which could be holding back profitably possibly include the conversion rate of your website, your conversion value, or even the quality score. Therefore, you should be smart in determining which bid strategies to go for, in order to be profitable and successful in setting your Google Ads campaigns.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 8.4: Arc Length
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Calculate the length of an arc of a circle.
## Arc Length
Arcs are measured in two different ways:
Degree measure: The degree measure of an arc is the fractional part of a 360\begin{align*}360^\circ\end{align*} complete circle that the arc is.
Linear measure: This is the length, in units such as centimeters and feet, if you traveled from one end of the arc to the other end.
• Arcs can be measured in __________ ways, _____________________________ measure and ___________________________ measure.
Example 1
Find the length of PQˆ\begin{align*}\widehat{PQ}\end{align*} if mPQˆ=60\begin{align*}m \widehat{PQ} = 60^\circ\end{align*}. The radius of the circle is 9 inches.
Remember, 60\begin{align*}60^\circ\end{align*} is the measure of the central angle associated with mPQˆ\begin{align*}m \widehat{PQ}\end{align*}. This is the degree measure of the arc.
To find the linear measure of the arc, or mPQˆ\begin{align*}m \widehat{PQ}\end{align*}, we use the fact that it is 60360=16\begin{align*}\frac{60}{360} = \frac{1}{6}\end{align*} of an entire circle.
The circumference of the circle is: C=πd=2πr=2π(9)=18π inches\begin{align*}C = \pi d = 2 \pi r = 2 \pi (9) = 18 \pi \ inches\end{align*}
The length of the arc, in this case, is 16\begin{align*}\frac{1}{6}\end{align*} of the entire circumference of the circle.
The arc length of PQˆ\begin{align*}\widehat{PQ}\end{align*} is: 1618π=18π6=3π\begin{align*}\frac{1}{6} \cdot 18 \pi = \frac{18 \pi}{6} = 3 \pi\end{align*} inches or 9.42 inches
In this lesson we study the second type of arc measure—the linear measure of an arc, or the arc’s length. Arc length is directly related to the degree measure of an arc.
Suppose a circle has:
• circumference C\begin{align*}C\end{align*}
• diameter d\begin{align*}d\end{align*}
• radius r\begin{align*}r\end{align*}
Also, suppose an arc of the circle has degree measure m\begin{align*}m\end{align*}.
Realize that m360\begin{align*}\frac{m}{360}\end{align*} is the fractional part of the circle that the arc represents.
Arc length
Arc Length=m360C=m360πd=m3602πr\begin{align*}Arc \ Length = \frac{m}{360} \cdot C = \frac{m}{360} \cdot \pi d = \frac{m}{360} \cdot 2 \pi r\end{align*}
1. In your own words, describe the linear measure of an arc:
\begin{align*}\; \; \;\end{align*}
\begin{align*}\; \; \;\end{align*}
\begin{align*}\; \; \;\end{align*}
2. True/False: The degree measure of an arc is exactly the same as the linear measure of an arc.
3. How could you correct the statement in #2 above to make it true?
\begin{align*}\; \; \;\end{align*}
\begin{align*}\; \; \;\end{align*}
\begin{align*}\; \; \;\end{align*}
\begin{align*}{\;}\end{align*}
4. Why do we use the fraction m360\begin{align*}\frac{m}{360}\end{align*} to calculate arc length? Describe.
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
\begin{align*}{\;}\end{align*}
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# Standard form of Parabola x$$^{2}$$= -4ay
We will discuss about the standard form of parabola x$$^{2}$$ = -4ay
Equation y$$^{2}$$ = -4ax (a > 0) represents the equation of a parabola whose co-ordinate of the vertex is at (0, 0), the co-ordinates of the focus are (0, -a), the equation of directrix is y = a or y - a = 0, the equation of the axis is x = 0, the axis is along negative y-axis, the length of its latus rectum = 4a and the distance between its vertex and focus is a.
Solved examples based on the standard form of parabola x$$^{2}$$ = -4ay:
1. Find the axis, co-ordinates of vertex and focus, length of latus rectum and the equation of directrix of the parabola x$$^{2}$$ = -16y
Solution:
The given parabola x$$^{2}$$ = -16y
⇒ x$$^{2}$$ = -4 ∙ 4 y
Compare the above equation with standard form of parabola x$$^{2}$$ = -4ay, we get, a = 4.
Therefore, the axis of the given parabola is along negative y-axis and its equation is x = 0
The co-ordinates of its vertex are (0, 0) and the co-ordinates of its focus are (0, -4); the length of its latus rectum = 4a = 4 ∙ 4 = 16 units and the equation of its directrix is y = a i.e., y = 4 i.e., y - 4 = 0.
2. Find the axis, co-ordinates of vertex and focus, length of latus rectum and the equation of directrix of the parabola 3x$$^{2}$$ = -8y
Solution:
The given parabola 3x$$^{2}$$ = -8y
⇒ x$$^{2}$$ = -$$\frac{8}{3}$$y
⇒ x$$^{2}$$ = -4 ∙ $$\frac{2}{3}$$ y
Compare the above equation with standard form of parabola x$$^{2}$$ = -4ay, we get, a = $$\frac{2}{3}$$.
Therefore, the axis of the given parabola is along negative y-axis and its equation is x = 0
The co-ordinates of its vertex are (0, 0) and the co-ordinates of its focus are (0, -$$\frac{2}{3}$$); the length of its latus rectum = 4a = 4 ∙ $$\frac{2}{3}$$ = $$\frac{8}{3}$$ units and the equation of its directrix is y = $$\frac{2}{3}$$ i.e., 3y = 2 i.e., 3y - 2 = 0.
`
● The Parabola
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# Lesson 2
Keeping the Equation Balanced
### Lesson Narrative
This lesson is the first of a sequence of eight lessons where students learn to work with equations that have variables on each side. In this lesson, students recall a representation that they have seen in prior grades: the balanced hanger. The hanger is balanced because the total weight on each side, hanging at the same distance from the center, is equal in measure to the total weight on the other side.
In the warm-up, students encounter two real hangers, one balanced and one slanted, and notice and wonder about what could cause the hangers’ appearance. This leads into the first activity where students consider two questions about a balanced hanger: first, whether a change of the number of weights keeps the hanger in balance, and second, how to find the unknown weight of one of the shapes if the weight of the other shape is known. Students learn that adding or removing the same weight from each side is analogous to writing an equation to represent the hanger and adding or subtracting the same amount from each side of the equation. They reason similarly about how halving the weight on each side of the hanger is analogous to multiplying by $$\frac12$$ or dividing by 2. In both the hanger and the equation, these kinds of moves will produce new balanced hangers and equations that ultimately reveal the value of the unknown quantity.
In the second activity, students encounter a hanger with an unknown weight that cannot be determined. This situation parallels the situation of an equation where the variable can take on any value and the equation will always be true, which is a topic explored in more depth in later lessons.
As students use concrete quantities to develop their power of abstract reasoning about equations, they engage in MP2.
### Learning Goals
Teacher Facing
• Calculate the weight of an unknown object using a hanger diagram, and explain (orally) the solution method.
• Comprehend that adding and removing equal items from each side of a hanger diagram or multiplying and dividing items on each side of the hanger by the same amount are moves that keep the hanger balanced.
### Student Facing
Let's figure out unknown weights on balanced hangers.
### Student Facing
• I can add or remove blocks from a hanger and keep the hanger balanced.
• I can represent balanced hangers with equations.
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MDME: MANUFACTURING, DESIGN, MECHANICAL ENGINEERING
# SEQUENCES & SERIES
A sequence is a list of numbers. An arithmetic sequence of numbers changes in a straight line, and a geometric sequence changes in a curve. A series is the sum of a sequence.
Smartboard Notes:
Sequences-and-series-01.pdf
## SEQUENCES
#### Arithmetic Sequence
Notation
xn is the nth term in a sequence
For example
x1 is the 1st term in a sequence
x6 is the 6th term in a sequence
Defining an arithmetic sequence
An arithmetic sequence is formed when the gap between each term is the same (or a constant).
An arithmetic sequence could be shown like this:
{ a, a+d, a+2d, a+3d, . . . }
To find the value of any term:
xn = a + (n-1)d
Where:
• xn is the nth term in the sequence
• a is the first term (x1)
• d is the common difference
• n is the number of the term
Example 1
Consider the sequence
{ 4, 11, 18, 25, 32, 39, . . . }
And you want to find the 26th term
Write a rule to express the term i.e. xn = a + (n-1)d
Here, a = 4 and d = 7.
So xn = 4 + 7(n-1)
Therefore the 26th term would be:
x26 = 4 + 7(26-1) = 179
Example 2
Consider the sequence
{ 12, 3, -6, -15, . . . }
And you want to find the 13th term
Write a rule to express the term i.e. xn = a + (n-1)d
It this case: xn = 12 + -9(n – 1) = 12 – 9(n – 1)
Therefore the 13th term would be:
x13 = 12 – 9(13 – 1) = -96
#### Geometric Sequence
Defining a geometric sequence
In a geometric sequence, each term is found by multiplying the previous term by a constant.
The multiplying constant is called the common ratio
A geometric sequence could be shown like this:
{ a, a*r, (a*r)*r, (a*r*r)*r, . . . } or
{ a, ar, ar2, ar3, . . . }
A rule for geometric sequences
xn = a r(n-1)
Where:
xn is the nth term in the sequence
a is the first term
n is the number of the term
r is the common ratio
Example 1
Consider the sequence
{ 2, 6, 18, 54, 162, 486, . . . }
And you want to find the 12th term
Write a rule to express the term i.e. xn = ar(n-1)
Here, a = 2, r = 3
So: xn = 2 x 3(n-1)
Therefore the 12th term would be:
x26 = 2 x 3(12-1) = 354294
Example 2
Consider the sequence
{ 128, 64, 32, 16, 8, . . . }
And you want to find the 12th term
Write a rule to express the term i.e. xn = ar(n-1)
Here, a = 128, r = 0.5
So: xn = 128 x 0.5(n-1)
Therefore the 10th term would be:
x26 = 128 x 0.5(10-1) = 0.25
## SERIES
#### Arithmetic Series
The sum of the members of a finite arithmetic progression is called an arithmetic series.
The formula is;
Where;
n = number of terms
a1 = first term
an = the nth term
Example:
Find this sum: 2 + 5 + 8 + 11 + 14
n = 5
a1 = 2
a5 = 14
So S5 = 5/2 * (2 + 14) = 404
Here are some useful shortcuts for the summing a standard sequence.
The three formulas represent a series like this:
1+2+3+4+5... (arithmetic)
1+4+9+16+25... (not geometric but a power of 2 sequence)
1+8+27+64+125... (not geometric but a power of 3 sequence)
The main one we want here is the arithmetic finite sum where d=1 and a=1;
If we want to find the sum from m to n, we simply do the sum from 1 to n and substract the sum from 1 to m
Example:
Find the partial sum of the series of numbers from 10 to 90
Substitute from the formula for {k}
#### Geometric Series
##### Finite Geometric Series
geometric series is the sum of the numbers in a finite geometric progression starting from 1.
Note: This is read as: "The sum from 1 to n of ar(k-1) is..."
Where;
n = number of terms
a = first term
r = the common ratio
k = the summing counter
Example:
2 + 10 + 50 + 250
n = 4
a = 2
r = 5
S5 = 2(1-54) / (1-5) = -1248 / -4 = 312
If we want the progression to start from a number other than 1, say m, then we sum from 1 and subtract the sum to m;
So the sum of the sequence from the mth term to the nth term is;
##### Infinite Geometric Series
An infinite geometric series is an infinite series whose successive terms have a common ratio. Such a series only converges if the absolute value of the common ratio is less than one (|r| < 1).
Example: A geometic series where a = 0.5, and r = 0.5 converges to 1;
(This converges because r is less than 1)
The convergence value of the sum of an infinite geometric series can be computed from;
Example: The same geometic series (where a = 0.5, and r = 0.5);
= 0.5 / (1-0.5) = 0.5/0.5 = 1
Why r must be less than 1
Of course, if r>1 then the sum of an infinite series will be infinity.
1+2+4+8+16+....to infinity = infinity
#### Questions:
Assignment: Do all questions
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# Texas Go Math Grade 6 Module 4 Answer Key Multiplying and Dividing Decimals
Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 4 Answer Key Multiplying and Dividing Decimals.
## Texas Go Math Grade 6 Module 4 Answer Key Multiplying and Dividing Decimals
Write the decimals represented by the shaded square.
Question 1.
We have that 10 squares are 10 of 100 equal parts, or $$\frac{10}{100}$$ or 0.1.
Here we have 70 shaded squares, so, they represent 7 × 0.1, or 0.7.
Question 2.
We have that 10 squares are 10 of 100 equal parts, or $$\frac{10}{100}$$ or 0.1.
Here we have 70 shaded squares, so, they represent 4 × 0.1, or 0.4.
Question 3.
We have that 1 square is 1 of 100 equal parts, or $$\frac{1}{100}$$, or 0.01.
Also, 10 squares are 10 of 100 equal parts, or $$\frac{10}{100}$$, or 0.1.
Here we have 53 shaded squares, we can rewrite them as 503.
So, 3 squares represent 3 × 0.01, or 0.03 and 50 squares represent 50 × 0.1, or 0.5.
Finally, 53 squares represent 0.03 + 0.5 = 0.53
Question 4.
Here, we have 100 shaded squares, they are 100 of 100 equal parts, or $$\frac{100}{100}$$, or 1.
Find the product.
Question 5.
0.49 × 10 ___________
Number of zeros in 10 is 1, so, we have to move the decimal point 1 place to the right and we get that the result is 4.9.
Question 6.
25.34 × 1,000 ___________
Number of zeros in 1000 is 3, so, we have to move the decimal point 3 places to the right and we get that the result is 25, 340
Here, because we had two decimal places, we added 0 at the third place because we moved decimal point 3 places.
Question 7.
87 × 100 ___________
Here, the number of zeros in 100 is 2. In this case we will add two zeros to 87 and get that the result is 8,700
Write a numerical expression for the word expression.
Question 8.
20 decreased by 8 _______________
Decreased means “to subtract 8 from 20”, so, required numerical expression would be 20 – 8.
Question 9.
the quotient of 14 and 7 _______
The quotient means the answer in division problem, so, required numerical expression would be $$\frac{14}{7}$$
Question 10.
the difference between 72 and 16 _________
The difference between two numbers means to subtract them, so, required expression would be 72 – 16.
Question 11.
the sum of 19 and 3 __________
The sum of two numbers means to add them, so, required expression would be 19 + 3.
Visualize Vocabulary
Use the ✓ words to complete the chart. You may put more than one word in each section.
Understand Vocabulary
Match the term on the left to the definition on the right.
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# Basic Properties of Numbers in Mathematics
## Basic Properties of Numbers in Mathematics:
There are four following basic properties of numbers in Mathematics:
1. Commutative
2. Associative
3. Distributive
4. Identity
It is especially important to understand these properties once we reach advanced mathematics such as Algebra & Calculus. Since, we should be familiar with each of these.
## 1. Commutative Property:
a. Addition: When two numbers are added, the sum is the same regardless of the order in which the numbers are added.
p+q=c or, q+p=c
For example:
3+4=7 or, 4+3=7
b. Multiplication: When two numbers are multiplied together, the product is the same regardless of the order in which the numbers are multiplied.
p x q = c or, q x p = c
For example:
3 x 4 = 12 or, 4 x 3 = 12
## 2. Associative Property:
a. Addition: When three or more numbers are added, the sum is the same regardless of the way in which the numbers are grouped.
p + (q + r) = c or, (p + q) + r = c
For example:
5 + (7 + 2) = 14 or, (5 + 7) + 2 = 14
b. Multiplication: When three or more numbers are multiplied, the product is the same regardless of the way in which the numbers are grouped.
p x (q x r) = c or, (p x q) x r = c
For example:
5 x (7 x 2) = 70 or, (5 x 7) x 2 = 70
## 3.Distributive Property:
The sum of two numbers times a third number is equal to the sum of each addend times the third number.
p x (q + r) = c or, p x q + p x r = c
For example:
3 x (5 + 6) = 33 or, 3 x 5 + 3 x 6 = 33
## 4. Identity Property:
a. Addition: The sum of any number and zero is that number.
p+0=p or, 0+p=p
For example:
3 + 0 = 3 or, 0+3=3
b. Multiplication: The product of any number and one is that number.
p x 1 = p or, 1 x p = p
For example:
6 x 1 = 6 or, 1 x 6 = 6
You can also practice Algebra Equations.
Write Better, Get Published, Be Creative
Online Publication Opportunities for Young Writers
Digital Study Center is seeking submissions of poetry, short stories, and unique articles for publishing online to our website.
Young writers, author, and teachers are most welcome to Submit Your Writing for Publishing Assistance.
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## How To Find Recurrence Relation From Differential Equation
#### (Part-01) Problems on Recurrence Relation for Bessel
CHAPTER 5 Recurrence Relations 5.1. Recurrence Relations Here we look at recursive de?nitions under a di?erent point of view. Rather than de?nitions they will be considered as equations that we
#### LEGENDRE POLYNOMIALS RECURRENCE RELATIONS & ODE
We are going to try to solve these recurrence relations. By this we mean something very similar to solving differential equations: we want to find a function of \(n\) (a closed formula) which satisfies the recurrence relation, as well as the initial condition.
#### How to Solve Legendre's Differential Equation 6 Steps
equations we know that the solution to the differential equation in the last example is, Solutions to second order differential equations consist of two separate functions each with an unknown constant in front of them that are found by applying any initial conditions.
#### Find recurrence equation from algorithm Stack Overflow
The first step, finding the characteristic equation of any homogeneous linear recurrence, is to assume that it relates to a power of something, which is actually natural given the additive
How to find recurrence relation from differential equation
#### Order and Degree of a Rec. Relation StudyYaar
This is the Exam of Elementary Differential Equations which includes Solutions, Differential Equation, Points, Ordinary Points, Regular Singular Points, Irregular Singular Points, Series Expansion, Independent Solutions, Non Zero Terms etc. Key important points are: Recurrence Relation, Ordinary, Point, Seeking Solutions, Differential Equation
#### Differential Equation Series - Recurrence Relation
There are, in fact, Bessel functions, solutions of the differential equation, that do not satisfy these recurrence relations. Professor Relton points out that the coefficient of C n " shows that the function can touch (i.e., be tangent to) the x-axis only at x = 0, since this is the only zero of the coefficient of the second derivative.
#### LEGENDRE POLYNOMIALS RECURRENCE RELATIONS & ODE
Find a recurrence relation on the coefficients by equating all coefficients to the right-hand side and determine coefficients by solving for higher coefficients in terms of lower ones. Use our coefficients to build two solutions by plugging back into our known power series.
#### Linear Nonhomogeneous Recurrence Relations
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#### Linear Nonhomogeneous Recurrence Relations
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This is the Exam of Elementary Differential Equations which includes Solutions, Differential Equation, Points, Ordinary Points, Regular Singular Points, Irregular Singular Points, Series Expansion, Independent Solutions, Non Zero Terms etc. Key important points are: Recurrence Relation, Ordinary, Point, Seeking Solutions, Differential Equation
#### Solutions of Linear Nonhomogeneous Recurrence Relations
20/04/2012 · The above recurrence relation is a linear recurrence relation of second order (meaning is dependent on two preceding values and ). The characteristic equation of the sequence is . Solving the equation, we obtain the solutions
### How to find recurrence relation from differential equation - Non-Homogeneous Linear Recurrence Relations Blogger
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MAT1332W10A1S
MAT1332W10A1S - MAT1332B 1332C Winter 2010 Assignment#1...
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Unformatted text preview: MAT1332B, 1332C, Winter 2010, Assignment #1 Solutions 1. We use the substitution u = 1 + 4 t . Then du dt = 4, so dt = du 4 . Thus Z 1 1 + 4 t dt = 1 4 Z 1 u du = 1 4 ln | u | + C = 1 4 ln | 1 + 4 t | + C. (Don’t forget to resubstitute.) 2. Z 2- 2 ( y 3 + 4 y 5 ) dy = 1 4 y 4 + 4 6 y 6 ¶ | 2- 2 = 16 4 + 4 × 64 6 ¶- 16 4 + 4 × 64 6 ¶ = 0 . 3. Z π/ 2- π/ 2 [ x 2- 20sin( x )] dx = 1 3 x 3 + 20cos( x ) ¶ | π/ 2- π/ 2 = " π 3 24 + 0 #- "- π 3 24 + 0 # = π 3 12 4. First approach: First find indefinite integral by substitution u = 2 π ( x- 2). Then du dx = 2 π so dx = du 2 π . Hence Z cos(2 π ( x- 2)) dx = Z cos( u ) du 2 π = 1 2 π sin( u ) + C = 1 2 π sin(2 π ( x- 2)) + C. Then evaluate Z 5 2 cos(2 π ( x- 2)) dx = 1 2 π sin(2 π ( x- 2)) fl fl fl fl 5 2 = 1 2 π (sin(6 π )- sin(0)) = 0 . Second approach: Transform the limits of integration first. When x = 2, u = 2 π (2- 2) = 0. When x = 5, u = 2 π (5- 2) = 6 π . Then the integral after substitution becomes Z 5 2 cos(2...
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Day 22: Parametric vs. NonParametric Statistics
Technically we could call this parametric statistics part 2. However, since we are covering nonparametric statistics and more importantly the difference between parametric and nonparametric statistics, it would seem that this title makes more sense. As usual with a continuation, you probably want to start at the beginning where we define parametric statistics. Ready to get started?*
Last post we said that parametric statistical tests use the mean values of the data to determine if there is a difference between two (or more) data. We looked at something called a T-test a concept we will dive into in more detail in the future, but for now we said that a t-test compares two groups and is one of the tests we can use to figure out if we have statistically different datasets or if the separation is artificial (IE- a coin that we think is bias, but is actually not bias). We also introduced the idea of the mean without explicitly defining it and we should resolve that today. Today we are also going to introduce the concept of the median, which may be confusing if we don’t have a good definition for the either the mean or median. That said, let’s define both!
First let’s talk about the mean.
The mean value of a dataset is just the average of that dataset.
If our dataset looks like this a = [1, 39, 20, 6, 4, 210, 10, 20] then to find the mean we add all the values together like so (1+39+20+6+4+210+10+20) = 310. However, that gives us the total and not the mean, to find the mean we take our total and divide by the number of values we added in this case we have 8 different data points in our dataset, so we take 310 and divide by 8, in other words we have mean(a) = (1+39+20+6+4+210+10+20) /8 = 38.75. We can write the mean formula in a more general for like this:
mean(x) = sum of x/n
where:
x is our dataset
n is the number of data points
If we look closely at the data however, we see something interesting. The mean of all the numbers is higher than it would be if we eliminated the outlier of the group. Notice that all the numbers are under 40, with the exception of 210. In fact if we recalculate our mean without the 210 we find a totally different number. Where we call our modified data a-mod = [ 1, 39, 20, 6, 4, 10, 20] and mean(a-mod) = (1+39+20+6+4+10+20) /7 = 14.29. A rather large difference from mean(a) which we determined was 38.75, so what happened?
We had an outlier, this is why we have two different statistical approaches. Our data was most definitely not normally distributed (at least not with the number of data points we have). However, nonparametric statistics does not have this issue, for one thing, we can use the given dataset to determine the distribution prior to analysis, but more importantly nonparametric tests use the median of the data.
This may be more basic than you need, but just to be thorough let’s define the concept of the median. Afterwards, we can go into how that relates to nonparametric statistics.
The median value of a dataset is just the middle value of that dataset.
Say we have a second dataset b, where b = [ 1, 6, 4, 9, 5, 20, 526, 12, 60]. To find the median value of our data we would arrange the values from lowest to highest and select the middle value, our data ordered by size would look like this b = [1, 4, 5, 6, 9, 12, 20, 60, 529]. Notice this did nothing to the dataset itself; we made no changes to the values, we just ordered it. Next we find the middle value, in this case we have 9 data points so our middle data point would be four data points from the left side of the data and four from the right. This gives us a median(b) = 9. The mean of the same dataset on the other hand is a much, much larger at 71.44 (you can check this yourself if you want. Again, the outlier pulls the mean of the data away from the mean of the dataset without the value.
So far so good, but what happens when we have a dataset with an even number of values? To use our previous example, we can take the median of a, which we find by finding the middle values of the data. Let’s look at how we do that. First we do what we did before, sort the data by size so we now have a = [1, 4, 6, 10, 20, 20, 39, 210]. If we notice there are two middle values, if we count three values from the left like we did in our previous example, we end at 10. However, if we go three numbers over starting at the right we end at 20. The solution is to use both numbers. In this case the median value is just median(a) = (10+20)/10 = 15. We can write this formula in a more mathematical way by saying that:
Now we’ve seen that our mean and median are definitely not the same things. We’ve also shown that our mean can be heavily influenced by outliers that are much larger than the average of the rest of the numbers in our dataset.
This brings us back to parametric vs. nonparametric testing. When we use nonparametric testing we are testing the medians of the data, not the means. This works well when our data is not normally distributed (as we saw with our example datasets a and b). Unlike parametric tests nonparametric testing doesn’t need you to assume that your data is normally distributed, but more importantly, nonparametric testing doesn’t require you to assume that your data are independent.
Now we need to tackle the main question. When do we use parametric testing and when do we use nonparametric testing? That isn’t an easy question to answer. For one thing, for any dataset that satisfies the requirements set by parametric testing, nonparametric testing can still be used. However, since your here you probably want a better answer than that, so let’s formalize it a little better.
Why use parametric tests?
First, if we are going to use parametric tests, you need to make sure you satisfy the requirements, namely:
1. Our data are normally distributed
2. Our data points are independent of each other
3. Data from multiple groups have the same variance
4. Your dataset has a linear relationship
We will go over points 3 and 4 in more detail in other posts.
Okay, so you satisfy the requirements, why use parametric tests over nonparametric ones? First, parametric tests have more statistical power than nonparametric tests. That is just a fancy way of saying that you are more likely to find significance in your dataset if it actually exists (IE less of a chance of running into a type 2 error).
Parametric tests work well with continuous type data (which we can define later), that are non-normal if you have a large enough sample size (typically for a 1-way T-test you want a sample size n>20, for a 2-way t-test you want a sample size for each group larger than 15). We will dive more into what tests are which at another time, but for now just know that sometimes there are exceptions to rules, this is one of them.
Why use nonparametric tests?
This should be fairly straight forward. Nonparametric tests should definitely be used when you don’t meet the assumptions needed to use parametric testing. Another good reason is that your data is better represented by the median. Their is a reason we covered mean and median today, sometimes the mean just won’t work when you are analyzing your data and you’ll run the risk of making a type 1 error.
A good example of a real dataset that would be better represented by the median over the mean is income. Think about it, you have people like Jeff Bezos with a net worth of something like 114 billion (with a b, which 1 billion is equal to 1,000 million). This will heavily skew your data and if you try to draw conclusions using the mean instead of median, you may wrongly conclude that American’s are millionaires, which (checks bank account)yeah, still not the case.
At the end of the day, when we are talking statistics we need to look at how our data are best represented. Would it make more sense to talk about the mean of our data or the median of our data? Answering this question will be the most helpful in determining which type of statistical analysis should be used for your data.
Well, this ended up being a bit longer than I thought. Next up I think we will talk about different types of tests. Or maybe we can talk about different types of probability curves. There is just so much we can talk about and it’s all related somehow so it’s tough to decide which topic is best to cover. In any case, today we’ve talked parametric statistical tests and nonparametric statistical tests, so hopefully we’ve cleared up the difference!
This post marks 1000 posts across lab topics. Look at us go!
Until next time, don’t stop learning!
*As usual, I make no claim to the accuracy of this information, some of it might be wrong. I’m learning, which is why I’m writing these posts and if you’re reading this then I am assuming you are trying to learn too. My plea to you is this, if you see something that is not correct, or if you want to expand on something, do it. Let’s learn together!!
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MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.2
In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 5 Arithmetic Progressions Exercise 5.2 Pdf. These solutions are solved rom the latest edition books by subject experts.
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.2
Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.P.
Solution:
Solution:
(i) an = a + (n- 1)d
a8 = 7 + (8 – 1)3 = 7 + 7 × 3 = 7 + 21
⇒ a8 = 28
(ii) an = a + (n – 1)d
⇒ a10 = -18 + (10 – 1)7 ⇒ 0 = -18 + 9d
⇒ 9d = 18 ⇒ d = 189=2
∴ d = 2
(iii) an = a + (n – 1)d
⇒ -5 = a + (18 – 1) × (-3)
⇒ -5 = a + 17 × (-3)
⇒ -5 = a – 51 ⇒ a = -5 + 51 = 46
Thus, a = 46
(iv) an = a + (n – 1)d
⇒ 3.6 = -18.9 + (n – 1) × 2.5
⇒ (n – 1) × 2.5 = 3.6 + 18.9
⇒ n = 9 + 1 = 10
Thus, n = 10
(v) an = a + (n- 1)d
⇒ an = 3.5 + (105 – 1) × 0
⇒ an = 3.5 + 104 × 0 ⇒ an = 3.5 + 0 = 3.5
Thus, an = 3.5
Question 2. (mp board 10th maths exercise 5.2 pdf)
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10,7,4, , is, ….,
(A) 97
(B) 77
(C) -77
(D) -87
(ii) 11th term of the AP: -3, −12, 2, …. ,is
(A) 28
(B) 22
(C) -38
(D) -4812
Solution:
(i) (C): Here, a = 10, n = 30
∵ T10 = a + (n – 1)d and d = 7 – 10 = -3
∴ T30 = 10 + (30 – 1) × (-3)
⇒ T30 = 10 + 29 × (-3)
⇒ T30 = 10 – 87 = -77
(ii) (B): Here, a = -3, n = 11 and
Question 3.
In the following APs, find the missing terms in the boxes:
Solution:
(i) Here, a = 2, T3 = 26
Let common difference = d
∴ Tn = a + (n- 1 )d
⇒ T3 = 2 + (3 – 1)d
⇒ 26 = 2 + 2 d
⇒ 2d = 26 – 2 = 24
(ii) Let the first term = a
and common difference = d
Here, T2 = 13 and T4 = 3
T2 = a + d = 13, T4 = a + 3d = 3
T1 – T2 = (a + 3d) – (a + d) = 3 – 13
(iii) 10th math exercise 5.2 2022 solutions
(iv) Here, a = – 4, T6 = 6
∵ Tn = a + (n -1 )d
T6 = – 4 + (6 – 1)d ⇒ 6 = -4 + 5d ⇒ 5d = 6 + 4 = d = 10 – 5 = 2
T2 = a + d = -4 + 2 =-2
T3 = a + 2d = -4 + 2(2) = 0
T4 = a + 3d = -4 + 3(2) = 2
T5 = a + 4d = -4 + 4(2) = 4
(v) Here, T2 = 38 and T6 = -22
∴ T2 = a + d = 38, T6 = a + 5d = -22
⇒ T6 – T2 = a + 5d – (a + d) = -22 – 38 -60
⇒ 4d = -60 ⇒ d = −604 = -15
a + d = 38 ⇒ a + (-15) = 38
⇒ a = 38 + 15 = 53
Now,
T3 = a + 2d = 53 + 2(-15) = 53 – 30 = 23
T4 = a + 3d = 53 + 3(-15) = 53 – 45 = 8
T5 = a + 4d = 53 + 4(-15) = 53 – 60 = -7
Thus missing terms are
Question 4.
Which term of the AP: 3, 8, 13, 18, is 78?
Solution:
Let the nth term be 78
Here, a = 3 ⇒ T1 = 3 and T2 = 8
∴ d = T2 – T1 = 8 – 3 = 5
And, Tn = a + (n- 1 )d
⇒ 78 = 3 + (n – 1) × 5 ⇒ 78 – 3 = (n -1) × 5
⇒ 75 = (n – 1) × 5 ⇒ (n – 1) = 75 ÷ 5 = 15
⇒ n = 15 + 1 = 16
Thus, 78 is the 16th term of the given AP.
Question 5.
Find the number of terms in each of the following APs:
(i) 7,13,19, …….. ,205
(ii) 18, 1512, 13, …… ,-47
Solution:
(i) Here, a = 7,d = 13 – 7 = 6
Let total number of terms be n.
∴ Tn = 205
Now, Tn = a + (n – 1) ×d
= 7 + (n – 1) × 6 = 205
⇒ (n – 1) × 6 = 205 – 7 = 198
⇒ n – 1 = 1986=33
∴ n = 33 + 1 = 34
Thus, the required number of terms is 34.
(ii)
Question 6.
Check whether -150 is a term of the AP:
11, 8, 5, 2…
Solution:
For the given AP,
we have a = 11, d = 8 -11 = -3
Let -150 be the nth term of the given AP
∴ Tn = a + (n – 1 )d
⇒ -150 = 11 + (n – 1) × (-3)
⇒ -150 – 11 = (n – 1) × (-3)
⇒ -161 = (n – 1) ⇒ (-3)
Question 7.
Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Solution:
Here, T11 = 38 and T16 = 73
Let the first term = a and the common difference = d.
Tn = a + (n – 1 )d
Then, Tn = a + (11 – 1)d = 38
⇒ a + 10d = 38 …(1)
and T16 = a + (16 – 1)d = 73
⇒ a + 15d = 73 …(2)
Subtracting (1) from (2), we get
(a + 15d) – (a + 10d) = 73 – 38
Question 8. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Here, n = 50, T3 = 12, Tn = 106
⇒ T50 = 106
Let the first term = a and the common difference = d
Tn = a + (n – 1 )d
T3 = a + 2d = 12 …(1)
T50 = a + 49d = 106 …(2)
Subtracting (1) from (2), we get
Question 9.
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
Here, T3 = 4 and T9 = -8
Tn = a + (n – 1)d
T3 = a + 2d = 4 …. (1)
T9 = a + 8d = – 8 …. (2)
Subtracting (1) from (2), we get
Question 10.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution:
Let a be the first term and d the common difference of the given AP
Now, using n = a + (n – 1 )d, we have
Question 11.
Which term of the AP : 3, 15, 27, 39,… will be 132 more than its 54th term?
Solution:
Here, a = 3, d = 15 – 3 = 12
Using Tn = a + (n – 1 )d, we get
Thus, 132 more than 54th term is the 65th term.
Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let for the 1st AP, the first term = a and common difference = d
And for the 2nd AP, the first term = a and common difference = d
According to the condition,
Question 13.
How many three-digit numbers are divisible by 7?
Solution: 10th maths exercise 5.2 questions solutions
The first three-digit number divisible by 7 is 105.
The last such three-digit number divisible by 7 is 994.
∴ The AP is 105,112,119, ,994
Let n be the required number of terms Here, a = 105, d = 7 and n = 994
Thus, 128 numbers of 3-digits are divisible by 7.
Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
The first multiple of 4 beyond 10 is 12.
The multiple of 4 just below 250 is 248.
Thus, the required number of terms is 60.
Question 15.
For what value of n, are the nth terms of two APs: 63,65,67 … and 3,10,17, …. equal?
Solution:
Thus, the 13th terms of the two given AP’s are equal.
Question 16.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
Let the first term = a and the common difference = d
Question 17.
Find the 20th term from the last term of the AP : 3, 8, 13, …, 253.
Solution:
We have, the last term (l) = 253
Here, d = 8 – 3 = 5
Since, the nth term before the last term is given by l – (n – 1 )d
We have 20th term from the last term = l – (20 – 1) × 5 = 253 – 19 × 5 = 253 – 95 = 158
Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
Let the first term = a and the common difference = d
Question 19. Latest MP Board Class 10th Maths Exercise 5.2 imp Questions Answers
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ? 200 each year. In which year did his income reach ₹ 7000?
Solution:
Here, a = ₹ 5000 and d = ₹ 200
Let, in the nth year Subba Rao gets ₹ 7000.
Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the nth week, her weekly savings become ₹ 20.75, find n.
Solution:
Here, a = ₹ 5 and d = ₹ 1.75
Thus, the required number of weeks is 10.
MP Board Class 10th Math Solutions:
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Hope you get help with MP Board 10th Maths exercise 5.2 question solution, if you have any issue so comment below.
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Question Video: Using Venn Diagrams to Calculate Dependent Probabilities | Nagwa Question Video: Using Venn Diagrams to Calculate Dependent Probabilities | Nagwa
Question Video: Using Venn Diagrams to Calculate Dependent Probabilities Mathematics • Second Year of Secondary School
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The figure shows a Venn diagram with the probabilities given for events π΄ and π΅. Work out P(π΄). Work out P(π΄ β© π΅). Work out P(π΅ | π΄).
03:23
Video Transcript
The figure shows a Venn diagram with the probabilities given for events π΄ and π΅. Work out the probability of π΄. Work out the probability of π΄ intersection π΅. Work out the probability of π΅ given π΄.
We recall that in any Venn diagram, the probabilities must sum to one. 0.3 plus 0.2 plus 0.4 plus 0.1 is equal to one. The first part of our question asks us to calculate the probability of π΄. This is equal to the sum of all the probabilities inside circle π΄. We need to add 0.3 and 0.2. This is equal to 0.5. The probability of event π΄ occurring is 0.5.
The second part of our question wants us to calculate the probability of π΄ and π΅. This is known as the intersection. It is the part of the Venn diagram where both π΄ and π΅ occur. The probability in the overlap of the two circles is 0.2. This means that the probability of π΄ intersection π΅ is 0.2.
The final part of our question wants us to calculate the probability of π΅ given that π΄ occurs. The notation in this question means given that. We recall that the probability of π΅ given π΄ is equal to the probability of π΅ intersection π΄ divided by the probability of π΄. We have already worked out both of these answers. Itβs important to note that the probability of π΅ intersection π΄ is the same as the probability of π΄ intersection π΅. The probability of π΅ given π΄ is, therefore, equal to 0.2 divided by 0.5. This is equal to two-fifths or 0.4. The probability of π΅ given π΄ is 0.4.
We could also have worked this out from our Venn diagram. As we want the probability of π΅ given that π΄ happens, we begin with the 0.5 inside circle π΄. Which of these probabilities is also in circle π΅? This is the 0.2. So once again we have 0.2 out of or divided by 0.5. Our three answers in this question are 0.5, 0.2, and 0.4. The 0.1 that was on the outside of both circles is the probability that neither event π΄ nor event π΅ occur.
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# Algebra on the GMAT: How to Factor
Master this fundamental algebraic skill which you will need on test day!
First, try these practice problems.
1)
(A) (x – 4)(x + 6)
(B) (x + 4)(x – 6)
(C) (x – 4)(x – 6)
(D) (x – 2)(x + 12)
(E) (x + 2)(x – 12)
3) If , then |a| + |b| + |c| + |d|
(A) 10
(B) 12
(C) 15
(D) 18
(E) 20
If these are easy for you, you probably have already mastered factoring: kudos to you! If these confuse you, you have found just the post you need.
## Terminology
You don’t need to know any of this terminology for the GMAT, but we need it just to talk about these ideas in words.
A binomial is a polynomial with two terms: all five answer choices to question #1 are the product of two binomials. A quadratic is a polynomial with three terms whose highest power is x-squared: the stem of question #1 is a quadratic, and the stem of question #2 is a ratio of two quadratics. To factor a quadratic is to express it as the product of two binomials. Question #1 is a straightforward “factor the quadratic” problem. Question #2 involves factoring both quadratics, in the numerator and in the denominator, and then cancelling a common factor.
Technically, any quadratic could be factored, but often the result would be two binomials with horribly ugly numbers — radicals, or even non-real numbers. You will not have to deal with those cases on the GMAT. We call a quadratic “factorable” if, when you factor it, the resulting equation has only integers appearing. You will only have to factor “factorable” quadratics on the GMAT.
Sometimes, factoring quadratics involves quadratic like that in #3, with a leading coefficient (the coefficient of the x-squared term) is an integer other than 1. These are considerably harder, and you will only see one like this if you are getting most of the math questions correct and the CAT is feeding you one 700-level question after another.
## The secret of factoring: FOIL
The best way to understand factoring well is first to understand FOILing well. Suppose we multiply two binomials:
Notice, if we follow the FOIL process forward, then we see two things. First, the middle coefficient, the coefficient of the x term, is the sum of the two numbers. Second, the final term, the constant term with no x, is a product of the two numbers. Right there, that’s the key of factoring. If I want to factor any polynomial of the form , then to factor it, we are looking for two numbers that have a product of c and a sum of b.
Things get a little complicated when some negative signs are floating around, so here’s a table, for all cases, for the two numbers, p and q, we need to find to factor the quadratic. In the table that follows, b is the coefficient of the x-term, c is the constant term with no x, and both of those are given in the quadratic; p & q are the two numbers we need to factor the quadratic into (x + p)(x + q).
BTW, I highly recommend NOT memorizing the above chart, but rather, thinking it through, and doing FOIL examples for each case, to convince yourself of the patterns and to ingrain them into your memory.
So, for example, suppose we want to factor: . Since c = –18 is negative, that means the p & q we want will have opposite sign: one positive, one negative. Since b = –7 is also negative, that means which one, p or q, has the larger absolute value, that one will be negative. We are looking for a bigger negative, and smaller positive, which will have a product of –18 and a sum of –7. One way to think about this is: we need a pair of factors of 18 that have a difference of 7. These two numbers are clearly 9 & 2. Make the bigger one negative: –9 & +2. Those are the numbers we need. Now, stick these into the factoring format:
Voila! Having read this, see if you now can figure out questions #1 & #2 above
Most GMAT test takers will not see this topic. Only if you anticipating getting the vast majority of questions on the Quant section correct should you even read this section.
Suppose you have to factor something like . This is tricky, because you are looking for four integers: a, b, c, and d, such that . The constraints we have are
(i) ac = 8 (both positive)
(ii) bd = –18 (one positive, one negative)
(iii) ad + bc = –7
This is not quite as methodical and left-brain as factoring in the easy case above. This involves a certain amount of number sense and a certain amount of pattern matching. For a & c, the only two possibilities are, in some order, either 2 & 4 or 1 & 8. For b & d, the possibilities for the absolute values are, in some order, either 1 & 18 or 2 & 9 or 3 & 6 — with the understanding that, in whichever one of those pairs we pick, one must be negative and one positive. Eventually, after some experimenting and trial-and-error, we find (8)(–2) + (1)(9) = –7 —- this is the combination we need.
If that makes sense, and you feel up to the challenge, try #3.
## Summary
Factoring quadratics is such a widely used skill in algebra that you are likely to see something such as Question #1 or #2 on your GMAT. Here’s another practice question, with its own video explanation.
## Practice problem explanations
1)
We need two numbers, p and q, which have a product of –24, which means one is negative and one is positive. Their sum is –10, which means the one with the larger absolute value is negative. Let’s go through the factor pairs of 24, in each case making the larger one negative
1 –24 sum = –23
2 –12 sum = –10
That’s the one we want! The pair we want is +2 and –12: they have a product of –24 and a sum of –10.
2) For this one, we need to factor both the numerator and denominator. In the numerator, we have x^2 – 4x – 21, so we need two numbers that have a product of –21 and a sum of –4. After a little trial and error, we find the pair that works is –7 and +3, so:
Now, the denominator, x^2 + 9x + 18. We need two positive numbers that have a product of 18 and a sum of 9 — these would be 3 & 6. Thus:
We see that we have a common factor of (x + 3) that will cancel. Now, let’s put the fraction together:
3) This is the hard one, definitely a 700+ level question. We need numbers a, b, c, and d such that
This means that ac = 6, bd = –12, and ad + bc = 1. The a & c pair could be (1, 6) or (2, 3), in some order. The absolute values of the b & d pair could be (1, 12) or (2, 6) or (3, 4), and of course, in each case, one of the two would have to be negative. After some trial and error, we find:
Thus, we see:
|a| + |b| + |c| + |d| = 2 + 3 + 3 + 4 = 12
### 9 Responses to Algebra on the GMAT: How to Factor
1. Irina March 15, 2018 at 4:42 am #
Hello Mike! I am really enjoying all posts here, so thank you very much!
I have a question about 3rd point, I got the same right answer but using this numbers: 6-1+6+1=12 Is it possible or am I just lucky?
• Magoosh Test Prep Expert March 18, 2018 at 4:26 pm #
If I’m understanding correctly, you’re assigning non-absolute values to a,b,c, and d, as follows:
a= 6
b= -1
c= 6
d= 1
And then, instead of adding the absolute values of a, b, c, and d, you’re adding the assigned values above.
If that’s the case, then yes, you got lucky, as that approach/pattern only works for the final this exact problem, and doesn’t actually for the internal equations that lead to the answer. Using the values for a,b,c, and d above doesn’t lead anywhere clear or clean when plugged into 6x^2 + x -12 = (ax + b) (cx + d).
2. Tom November 10, 2016 at 4:44 pm #
Great writing and glad you have some personality in here. Crazy how I can remember what “FOIL” stood for from middle school even though I totally forgot the process
3. Felipe Azevedo September 13, 2016 at 1:27 pm #
The sum of the equation you found in question three is -1 (-4, +3). Didn’t it have to be +1? I know that in the answer, as it is in absolute value, it wouldn’t make difference anyways…
Following the logic explained in question 1, the product of the equation you found is -12, and the sum is -1 (-4, +3).
Greetings from Brazil! I really enjoy your posts!
• Magoosh Test Prep Expert September 29, 2016 at 7:20 am #
Hi Felipe! Thanks for writing to us from Brazil 😀 It’s great to hear that you enjoy our posts 🙂
While the factor pair we’ve selected (-4, 3) may seem incorrect at first glance, if we FOIL the factored form of the expression, we can see that this is actually the correct pair:
(2x + 3)(3x – 4) = 6x^2 – 8x + 9x – 12 = 6x^2 + x – 12
The reason why the sum is not simply -4 + 3 = -1 is because of the coefficients of the x terms in parenthesis. When we FOIL, we see how these coefficients impact the resulting x term in the corresponding quadratic expression 🙂
Hope this clears up your doubts! Happy studying!
4. Raevene August 8, 2016 at 4:03 am #
Hi Mike!
Thanks so much for all you do! Your team’s free GMAT blog is such a big help to me especially as I am math neophyte/”upcoming” quant expert (hopefully in the near future). I’d just like to ask about question 2 please. It does not seem to be displayed correctly above as It’s missing from the practice problems. Could you kindly put it in please so that it would be easier to tie in your detailed explanation?
• Magoosh Test Prep Expert August 9, 2016 at 12:00 am #
Hi Raevene,
We’re so glad that you’re enjoying the Magoosh materials to help study math! 🙂 I checked out this post, and everything displays properly on my end. Can you take a screen shot of what you’re seeing and send it to help@magoosh.com? We can then diagnose the problem a little better and hopefully get this cleared up for you and anyone else with the same issue. Thanks!
• Raevene August 9, 2016 at 2:55 am #
hi! Thanks for getting back to me so quickly! Just sent in a screen shot to the indicated e-mail. 🙂
5. Chinmoy July 25, 2016 at 9:22 am #
Hi Mike,
Great Post as usual.
I really found the explanation for Q#3 very helpful.
Magoosh blog comment policy: To create the best experience for our readers, we will only approve comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! 😄 Due to the high volume of comments across all of our blogs, we cannot promise that all comments will receive responses from our instructors.
We highly encourage students to help each other out and respond to other students' comments if you can!
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CBSE Class 10-Mathematics: Chapter –5 Arithmetic Progressions Part 14
Get top class preparation for CBSE right from your home: fully solved questions with step-by-step explanation- practice your way to success.
Question 3:
If the third and the ninth terms of an AP are and respectively, which term of this AP is zero?
It is given that 3rd and 9th term of AP are 4 and respectively.
It means and
Using formula , to find term of arithmetic progression,
These are equations in two variables.
Using equation , we can say that
Putting value of a in other equation ,
Putting value of in equation ,
Therefore, first term and Common Difference
We want to know which term is equal to zero.
Using formula to find term of arithmetic progression,
Therefore, term is equal to .
Question 4:
Two AP’s have the same common difference. The difference between their terms is , what is the difference between their terms.
Let first term of 1st AP
Let first term of 2nd
It is given that their common difference is same.
Let their common difference be .
It is given that difference between their terms is .
Using formula , to find nth term of arithmetic progression,
We want to find difference between their
1000th terms which means we want to calculate:
Putting equation (1) in the above equation,
Therefore, difference between their 1000th terms would be equal to 100.
Question 5:
How many three-digit numbers are divisible by 7?
We have AP starting from because it is the first three-digit number divisible by .
AP will end at because it is the last three-digit number divisible by .
Therefore, we have AP of the form
Let is the nth term of AP.
We need to find n here.
First term , Common difference
Using formula , to find term of arithmetic progression,
It meansis the term of AP.
Therefore, there are terms in AP.
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# Lego Loop Limits
How can we make a circular loop out of lego bricks? For simplicity, we’ll use 2×1 bricks. The dimensions of a 2×1 brick are shown below in millimetres.
We’ll define some variables using these dimensions:
Clearance per brick end: $\epsilon = 0.1$ mm
Distance between pegs: $m = 8.0$ mm
Width: $w = m - 2 \epsilon = 7.8$ mm
Length: $l = 2m - 2 \epsilon = 15.8$ mm
If bricks are assembled in a “running bond” style, two adjacent bricks can rotate maximally by using all the available clearance space. This means that if we build a wall long enough, it can be wrapped around into a loop. The question now is, what is the minimum number of bricks we need to do this? (The figure below exaggerates this rotation.)
For any regular polygon with $n$ sides, angles $A$ and $B$ are already known (see next figure below).
Center angle $A$ is just a circle ($2\pi$) divided by the number of sides.
$A=2\pi/n$
Any triangle’s angles add up to $\pi$.
$\pi = A + 2B$
Combining these equations, inner angle $B$ is then defined in terms of the number of sides:
$B = \pi/2 - \pi/n$
Angle $B$ is related to angle $\theta$ (shown in the figure below) by:
$\pi = B + \pi/4 + \theta$
(The space between $B$ and $\theta$ is $\pi/4$ since the peg is centered such that it’s equidistant from the lengthwise edges and the nearest width-wise edge of the brick.)
Angle $\theta$ can be expressed in terms of brick width $w$ and clearance $\epsilon$.
In the figure below, the distance between the red point and blue point is $w/\sqrt{2}$, and the shortest distance from the blue point to the dashed line of symmetry is $w/2 + \epsilon$. So the angle $\theta$ can be expressed as:
$\text{sin}\theta = \frac{w/2+\epsilon}{w/\sqrt{2}}$
or
$\theta = \text{arcsin}(1/\sqrt{2} + \sqrt{2} \epsilon/w)$
Angle $B$ is then:
$B = 3\pi/4 - \text{arcsin}(1/\sqrt{2} + \sqrt{2} \epsilon/w)$
Plugging the earlier polygon formula into this equation gives a formula for $n$ in terms of brick width and clearance:
$n = \frac{\pi}{\text{arcsin}(1/\sqrt{2} + \sqrt{2} \epsilon/w)-\pi/4}$
Plugging in the dimensions, we get a minimum number of sides to make a loop:
$n \approx 121$
and since at least 3 layers are needed to create a secure running bond, the minimum number of bricks needed is 363.
This has a corresponding angle:
$\theta = 0.81136$
and acute angle between bricks:
$\delta = 2(\theta - \pi/4) = 0.051928$
Trying this in real life, I was able to get a loop with 110 sides (330 bricks).
$n_R = 110$
$\theta_R = 0.81396$
$\delta_R = 0.057120$
The theoretical calculation done above only assumes a perfectly rigid lego brick.
So the difference in $n$ may be accounted for by tolerances on the lego brick (stretching and squashing) and asymmetric angles, allowing for tighter inner angles.
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# What Are The Prime Factors Of 3
What are the prime factors of 3? Answer: 1 *3
The number 3 has 1 prime factors. Primes can only have two factors(1 and itself) and only be divisible by those two factors. Any figure where this rule applies can be called a prime factor.
## What Is The Factor Tree Of 3
How to use a factor tree to find the prime factors of 3? A factor tree is a diagram that organizes the factoring process.
First step is to find two numbers that when multiplied together equal the number 3.
We found 1 prime factors(1 *3) using the factor tree of 3. Now let us explain the process to solving factor trees in more detail. Our goal is to find all prime factors of a given whole number. In each step of our factor tree diagram for 3 we always checked both multiplication numbers if they were primes or not. If one or both of the integers are not prime numbers then this means that we will have to make diagrams for them too. This process continues until only prime numbers are left.
Remember that often a factor tree for the same integer can be solved in more than one correct way! An example of this is the figure 12 where 2*6=12 and 4*3=12. The primes of a factor tree for 12 are the same regardles if we start the factor tree with 2*6 or 4*3.
## How To Verify If Prime Factors Of 3 Are Correct Answers
To know if we got the correct prime factors of 3 we have to get the prime factorization of 3 which is . Because when you multiply the primes of the prime factorization the answer has to be equal with 3.
After having checked the prime factorization we can now safely say that we got all prime factors.
## General Mathematical Properties Of Number 3
3 is not a composite digit. 3 is not a composite number, because it's only positive divisors are one and itself. It is not even. 3 is not an even number, because it can't be divided by 2 without leaving a comma spot. This also means that 3 is an odd number. When we simplify Sin 3 degrees we get the value of sin(3)=0.14112000805987. Simplify Cos 3 degrees. The value of cos(3)=-0.98999249660045. Simplify Tan 3 degrees. Value of tan(3)=-0.14254654307428. 3 is not a factorial of any integer. When converting 3 in binary you get 11. Converting decimal 3 in hexadecimal is 3. The square root of 3=1.7320508075689. The cube root of 3=1.4422495703074. Square root of √3 simplified is 3. All radicals are now simplified and in their simplest form. Cube root of ∛3 simplified is 3. The simplified radicand no longer has any more cubed factors.
## Determine Prime Factors Of Numbers Smaller Than 3
Learn how to calculate primes of smaller numbers like:
## Determine Prime Factors Of Numbers Bigger Than 3
Learn how to calculate primes of bigger numbers such as:
## Single Digit Properties For Number 3 Explained
• Integer 3 properties: 3 is odd and a perfect total. The second in the primes sequence, after 2 and before 5, the first to also be Euclidean (3=2+1). One of the primes of Mersenne(3=2²-1), Fermat and Sophie Germain. Three is a component of Ulam, Wedderburn-Etherington, Perrin, Wagstaff. It is integer-free and a triangular number. The fourth issue of the Fibonacci sequence, after 2 and before 5. Belonging to the first Pythagorean terna (3,4,5). The third value of the succession of Lucas, after 1 and before 4. In the numerical decimal system 3 is a Colombian figure. In the binary system they call it a palindrome.
## Finding All Prime Factors Of A Number
We found that 3 has 1 primes. The prime factors of 3 are 1 *3. We arrived to this answer by using the factor tree. However we could have also used upside down division to get the factorization primes. There are more that one method to factorize a integer.
## List of divisibility rules for finding prime factors faster
Rule 1: If the last digit of a figure is 0, 2, 4, 6 or 8 then it is an even number. All even numbers are divisible by 2.
Rule 2: If the sum of digits of a integer is divisible by 3 then the figure is also divisible by 3 and 3 is a prime factor(example: the digits of 12 are 1 and 2 so 1+2=3 and 3 is divisible by 3, meaning that 12 is divisible by 3). The same logic also works for 9.
Rule 3: If the last two digits of a number are 00 then this integer is divisible by 4(example: we know that 124=100+24 and 100 has two zeros in the end making it divisible with 4. We also know that 4 is divisible with 24). In order to use this rule to it's fullest it is best to know multiples of 4.
Rule 4: If the last digit of a number is 0 or 5 then 5 it is divisible by 5.
Rule 5: All integers that are divisible by both 2 and 3 are also divisible by 6. This is logical because 2*3=6.
## What Are Prime Factors Of A Number?
All numbers that are only divisible by one and itself are called prime factors in mathematics. A prime factor is a figure that has only two factors(one and itself).
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# Volunteers to the board!!
## Presentation on theme: "Volunteers to the board!! "— Presentation transcript:
Volunteers to the board!!
Warm ups Volunteers to the board!! Graph using intercepts: 2x + 3y = 12 4x – 5y = 20 2x – y = -4 3x + 4y = -8
Convert Standard to Slope-Intercept
Objective: To convert a linear equation from standard form to slope-intercept form.
What is the slope of #1? _____
Part of being able to analyze a graph is being able to say the slope. Also, sometimes it is easier to graph with the slope and y-intercept than with the x-and y-intercept. Writing an equation from a graph is easier to do with slope and y intercept (y = mx + b) You will have to do this on your test! -2/3
Convert Standard Slope-Intercept y = mx + b Ax + By = C Steps:
1. Subtract Ax 2. Divide everything by B y = mx + b Ax + By = C
Example 1 Convert the equation from standard form to slope-intercept form: 2x + 3y = 12 -2x -2x 3y = -2x + 12 y = -2x + 4 3 Steps: Subtract Ax Divide everything by B Ax By = C
Example 2 Convert the equation from standard form to slope-intercept form: 2x – y = -4 -2x x -y = -2x – 4 y = 2x + 4 Steps: Subtract Ax Divide everything by B Ax By = C
TOO Convert to slope-intercept form: 4x – 5y = 10
Now, identify the slope and y-intercept: (hint: slope is m, y-intercept is b) Slope:______ y-intercept:_______ y = 4/5x – 2 4/5 -2
In General Ax + By = C y = -a x + c b b m = -a/b y-intercept = c/b
TOO Identify slope and y-intercept: 5x + 2y = 12 -5x -5x y = -5/2x + 6
y = -5/2x + 6 Slope:______ y-intercept:_______ y = -5/2x + 6 -5/2 6
Homework Equations in Two Variables (pt. 1)
Show your work at the bottom of the paper Convert to y=mx + b State m and b
Similar presentations
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Q:
(99 POINTS) Using similarity solve for x. Show all work
Accepted Solution
A:
Hey there! :D
The biggest thing to remember here is the similar triangles and their side lengths.
Okay, so we can separate these triangles into 3 similar triangles. Which, we only need to do that to two for a proportion, but if you were to have more variables, you could separate the triangles even more if you needed to.
We want to find two triangles that have "x" has a value of their side, so we can solve a proportion.
The bigger triangle, (the whole two triangles combine) and the smaller triangle on the left can be used to make a proportion.
On the smaller triangle:
We have 9 cm on the shorter side and x as the hypotenuse.
On the larger triangle:
We have x as the shorter side, and then 25 as the hypotenuse. (9+16)
Note: You have to flip the larger triangle upside down to make it a similar structure to the other two triangles. It is similar, you just have to flip it.
So, now we can make a proportion.
$$\frac{x}{9} = \frac{25}{x}$$
Now, cross multiply.
9*25= 225
x*x= $$x^{2}$$
So, find the square root.
√225= 15
x= 15.
I hope this helps!
~kaikers
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IB MYP 4 2021 Edition
10.01 Types of data
Lesson
Numerical data
Numerical data is data that can be counted, ordered and measured. It is sometimes also called quantitative data.
Numerical data can be either continuous or discrete.
Continuous Data
A data set is continuous if the values can take on any value within a finite or infinite interval.
Examples of continuous data are height, weight, temperature or the time taken to run $100$100 metres.
Notice how all of these examples could be anywhere on a scale interval, including fractional and decimal values. For example, it might be $25.3$25.3$^\circ$°C, or a man might be $182.13$182.13 cm tall.
Discrete Data
A data set is discrete if the numerical values can be counted but are distinct and separate from each other. They are often (but not always) whole number values.
Some examples of discrete data are: the number of goals scored in a game, the number of people in a class, the number of pets people have.
Notice how all these examples have distinct values. For example, you couldn't score $2.5$2.5 goals in a game of soccer or own $\frac{1}{4}$14 of a dog, so there is no continuity between the scores.
In some tournaments, half a point is awarded for a draw. In this case, there could be a score of $2.5$2.5, but there still could not be scores of $2.25$2.25 or $2.75$2.75 and so on, so the data is still discrete.
Categorical data
Categorical data is non-numerical. In other words, it describes the qualities or characteristics of a data set. Categorical data is sometimes also called qualitative data.
There are two types of categorical data: ordinal and nominal.
Ordinal data
A set of data is ordinal if the values can be counted and ordered but not measured.
Rating scales are examples of ordinal data. The finishing places in a race is another example of ordinal data.
Think about it - the positions in a race can be ordered or ranked. Finishing first means you were faster than the person who came second and the person who finished eighth was slower than the person who finished sixth. However, the differences between the finishing times may not be the same between all competitors. Check out the picture below. The times between first and second will be really close - probably within half a second - however the time between second and third will be longer. There is not a fixed interval.
Nominal data
The word nominal basically means name. In other words, data is split up based on different names or characteristics. Nominal data may be the names of countries you have visited or your favourite colours. We could assign these different characteristics a number where the numbers are labels. In other words, we are giving categorical data numerical labels. We can count but not order or measure nominal data.
Practice questions
QUESTION 1
Which of the following are examples of numerical data? (Select all that apply)
1. favourite flavours
A
maximum temperature
B
daily temperature
C
types of horses
D
favourite flavours
A
maximum temperature
B
daily temperature
C
types of horses
D
QUESTION 2
Which one of the following data types is discrete?
1. The number of classrooms in your school
A
Daily humidity
B
The ages of a group of people
C
The time taken to run $200$200 metres
D
The number of classrooms in your school
A
Daily humidity
B
The ages of a group of people
C
The time taken to run $200$200 metres
D
Question 3
Classify this data into its correct category:
Weights of dogs
1. Categorical Nominal
A
Categorical Ordinal
B
Numerical Discrete
C
Numerical Continuous
D
Categorical Nominal
A
Categorical Ordinal
B
Numerical Discrete
C
Numerical Continuous
D
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+0
+5
152
2
3
a y = - --- x
4
4x - 5y = 0
For what values of does the system of equations of linear equations in the variables x and y have infinitely many solutions?
Guest Feb 27, 2017
#1
+61
0
start by placing both in slope-intercept form (y=mx+b)
so you have a(y)=-3/4x
and y=4/5x
in order for a system of linear equations to have infinitely many solutions the have to have the same slope and intercept, right now
-3/4 $$\neq$$4/5
so we have to figure what "a" needs to equal in order form the lines to be the same.
so (-3/4)/x=4/5
solve for x
x=(4/5)*(-3/4)
x=-3/5
so we know that "A' is going to be the reciprical of x**
A=-5/3
**how we decided X and why we know that x= the recipricle
so ay=-3/4x
so it would end up being y= ((-3/4)/a)x
and dividing by a fraction is the same as multiplying by its reciprical.
brkr19 Feb 27, 2017
#2
+19661
0
For what values of a does the system of equations of linear equations
in the variables x and y have infinitely many solutions?
$$\begin{array}{|rcll|} \hline ay &=& -\frac{3}{4}x \\ 4x-5y &=& 0 \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline ay &=& -\frac{3}{4}x \quad &| \quad + \frac{3}{4}x \\ \frac{3}{4}x + ay &=& 0 \quad &| \quad \cdot \frac{16}{3} \\ 4x + \frac{16}{3}ay &=& 0 \\ \hline \end{array}$$
infinitely many solutions, if $$\frac{16}{3}a = -5$$
$$\begin{array}{|rcll|} \hline \frac{16}{3}a &=& -5 \quad &| \quad \cdot \frac{3}{16} \\ a &=& -5\cdot \frac{3}{16} \\ a &=& -\frac{15}{16} \\ \hline \end{array}$$
heureka Feb 28, 2017
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# A wood fire burns at 260 °C. What is the temperature in Fahrenheit?
Jan 8, 2018
color(red)(F=9/5color(blue)((C))+32-> "the derived formula"
where:
$C = {260}^{o}$
#### Explanation:
It is confusing to memorize formulas. To start with, just remember the standard values; freezing and boiling points of water expressed in $C$ and $F$. Knowing these data provide the important conversion formula that can be derived from it.
$\text{Freezing Point of water in } C \mathmr{and} F = {0}^{o} \mathmr{and} {32}^{o}$
$\text{Boiling Point of water in } C \mathmr{and} F = {100}^{o} \mathmr{and} {212}^{o}$
Now, formulate the ratios that relate the two temperature scales where the temperature units and zero points are different; hence,
$\frac{C - 0}{100 - 0} = \frac{F - 32}{212 - 32}$, simplify
$\frac{C}{100} = \frac{F - 32}{180}$, cross multiply
$100 \left(F - 32\right) = 180 \left(C\right)$, divide both sides by 100 and simplify
(cancel(100)(F-32))/cancel(100)=(cancel(180)9(C))/(cancel(100)5, simplify
$F - 32 = \frac{9}{5} \left(C\right)$, add 32 both sides of the equation
$F \cancel{- 32 + 32} = \frac{9}{5} \left(C\right) + 32$, simplify
color(red)(F=9/5color(blue)((C))+32-> "the derived formula"
Then, given the value $\textcolor{b l u e}{C = {260}^{o}}$, plug in the value to the formula to convert $C$ to $F$; hence
$F = \frac{9}{5} \left(C\right) + 32$
$F = \frac{9}{5} \left(260\right) + 32$
$F = 468 + 32$
$F = {500}^{o}$
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# Difference between revisions of "1975 USAMO Problems/Problem 1"
## Problem
(a) Prove that
$[5x]+[5y]\ge [3x+y]+[3y+x]$,
where $x,y\ge 0$ and $[u]$ denotes the greatest integer $\le u$ (e.g., $[\sqrt{2}]=1$).
(b) Using (a) or otherwise, prove that
$\frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}$
is integral for all positive integral $m$ and $n$.
## Background Knowledge
Note: A complete proof for this problem may require these results, and preferably also their proofs.
If $[x] = a$, then $a \le x < a + 1$. This is the alternate definition of $[x]$.
If $a < b$, then $[a] \le [b]$. This is easily proved by contradiction or consideration of the contrapositive.
If $a$ is an integer, then $[x + a] = [x] + a$. This is proved from considering that $[x] + a \le x + a < [x] + a + 1$.
This is a known fact: the exponent of a prime $p$ in the prime factorization of $n!$ is $\sum_{k=1}^\infty \left[ \frac{n}{p^k} \right]$. (Legendre's Formula)
## Key Lemma
Lemma: For any pair of non-negative real numbers $x$ and $y$, the following holds: $$[5x] + [5y] \ge [x] + [y] + [3x + y] + [3y + x].$$
## Proof of Key Lemma
We shall first prove the lemma statement for $0 \le x, y < 1$. Then $[x] = [y] = 0$, and so we have to prove that $$[5x] + [5y] \ge [3x + y] + [3y + x].$$
Let $[5x] = a, [5y] = b$, for integers $a$ and $b$. Then $5x < a + 1 \text{ and } 5y < b + 1$, and so $x < \frac{a+1}{5}$ and $y < \frac{b+1}{5}$.
Define a new function, the ceiling function of x, to be the least integer greater than or equal to x. Also, define the trun-ceil function, $[[x]],$ to be the value of the ceiling function minus one. Thus, $[[a]] = a - 1$ if $a$ is an integer, and $[[a]] = [a]$ otherwise. It is not difficult to verify that if $a$ and $b$ are real numbers with $a < b$, then $[[a]] \le [a] \le [[b]]$. (The only new thing we have to consider here is the case where $b$ is integral, which is trivial.)
Therefore, $$[3x + y] + [3y + x] \le \left [ \left [\frac{3a+b+4}{5} \right ] \right] + \left[\left[\frac{3b+a+4}{5} \right]\right] = S.$$
We shall prove that $S \le a + b = T$; to do so, we list cases. Without loss of generality, let $a \le b$. Because $x$ and $y$ are less than one, we have $a \le b \le 4.$ Then, we find, for all 15 cases:
$a = 0, b = 0 \rightarrow S = 0, T = 0.$
$a = 0, b = 1 \rightarrow S = 1, T = 1.$
$a = 0, b = 2 \rightarrow S = 2, T = 2.$
$a = 0, b = 3 \rightarrow S = 3, T = 3.$
$a = 0, b = 4 \rightarrow S = 4, T = 4.$
$a = 1, b = 1 \rightarrow S = 2, T = 2.$
$a = 1, b = 2 \rightarrow S = 3, T = 3.$
$a = 1, b = 3 \rightarrow S = 3, T = 4.$
$a = 1, b = 4 \rightarrow S = 5, T = 5.$
$a = 2, b = 2 \rightarrow S = 4, T = 4.$
$a = 2, b = 3 \rightarrow S = 4, T = 5.$
$a = 2, b = 4 \rightarrow S = 5, T = 6.$
$a = 3, b = 3 \rightarrow S = 6, T = 6.$
$a = 3, b = 4 \rightarrow S = 6, T = 7.$
$a = 4, b = 4 \rightarrow S = 6, T = 8.$
Thus, we have proved for all x and y in the range $[0, 1)$, $$[5x] + [5y] = T \ge S \ge [3x + y] + [3y + x].$$
Now, we prove the lemma statement without restrictions on x and y. Let $x = [x] + \{x\}$, and $y = [y] + \{y\}$, where $\{x\}$, the fractional part of x, is defined to be $x - [x]$. Note that $\{x\} < 1$ as a result. Substituting gives the equivalent inequality
$$[5[x] + 5\{x\}] + [5[y] + 5\{y\}] \ge [x] + [y] + [3[x] + 3\{x\} + [y] + \{y\}] + [3[y] + 3\{y\} + [x] + \{x\}].$$
But, because $[x] + a = [x + a]$ for any integer $a$, this is obtained from simplifications following the adding of $5[x] + 5[y]$ to both sides of
$$[5\{x\}] + [5\{y\}] \ge [3\{x\} + \{y\}] + [3\{y\} + \{x\}],$$
which we have already proved (as $0 \le \{x\}, \{y\} < 1$). Thus, the lemma is proved.
## How the Key Lemma Solves the Problem
Part (a) is a direct corollary of the lemma.
For part (b), consider an arbitrary prime $p$. We have to prove the exponent of $p$ in
$I = \frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}$
is non-negative, or equivalently that $$\sum_{k=1}^{\infty} \left( \left[ \frac{5m}{p^k} \right] + \left[ \frac{5n}{p^k} \right] \right) \ge \sum_{k=1}^{\infty} \left( \left[ \frac{m}{p^k} \right] + \left[ \frac{n}{p^k} \right] + \left[ \frac{3m+n}{p^k} \right] + \left[ \frac{3n+m}{p^k} \right] \right).$$
But, the right-hand side minus the left-hand side of this inequality is $$\sum_{k=1}^{\infty} \left( \left[ \frac{5m}{p^k} \right] + \left[ \frac{5n}{p^k} \right] - \left[ \frac{m}{p^k} \right] - \left[ \frac{n}{p^k} \right] - \left[ \frac{3m+n}{p^k} \right] - \left[ \frac{3n+m}{p^k} \right] \right),$$ which is the sum of non-negative terms by the Lemma. Thus, the inequality is proved, and so, by considering all primes $p$, we deduce that the exponents of all primes in $I$ are non-negative. This proves the integrality of $I$ (i.e. $I$ is an integer). $\blacksquare$
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# If a hollow cube of internal edge 22 cm
Question:
If a hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that – space of the cube remains unfilled.
Then, the number of marbles that the cube can accomodate is
(a) 142244
(b) 142344
(c) 142444
(d) 142544
Solution:
(a) Given, edge of the cube = 22 cm
$\therefore \quad$ Volume of the cube $=(22)^{3}=10648 \mathrm{~cm}^{3} \quad\left[\because\right.$ volume of cube $\left.=(\text { side })^{3}\right]$
Also, given diameter of marble $=0.5 \mathrm{~cm}$
$\therefore$ Radius of a marble, $r=\frac{0.5}{2}=0.25 \mathrm{~cm}$ $[\because$ diameter $=2 \times$ radius $]$
Volume of one marble $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7} \times(0.25)^{3}$
$\left[\because\right.$ volume of sphere $\left.=\frac{4}{3} \times \pi \times(\text { radius })^{3}\right]$
$=\frac{1.375}{21}=0.0655 \mathrm{~cm}^{3}$
Filled space of cube $=$ Volume of the cube $\frac{1}{8} \times$ Volume of cube
$=10648-10648 \times \frac{1}{8}$
$=10648 \times \frac{7}{8}=9317 \mathrm{~cm}^{3}$
$\therefore$ Required number of marbles $=\frac{\text { Total space filled by marbles in a cube }}{\text { Volume of one marble }}$
$=\frac{9317}{0.0655}=142244$ (approx)
Hence, the number of marbles that the cube can accomodate is $142244 .$
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# Unit 1: Measurement and Probability - PowerPoint PPT Presentation
Unit 1: Measurement and Probability
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Unit 1: Measurement and Probability
## Unit 1: Measurement and Probability
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##### Presentation Transcript
1. Unit 1: Measurement and Probability Math 3202
2. 1.4 Working With Probability Probability is important in many everyday events. For example, building materials are sometimes defective. If you order lumber to build a fence, it is likely that some of it will be warped and unusable for this project. Knowing this, you can better determine how much extra lumber you will need to order.
3. 1.4 Working With Probability Exploring Teen Sleep Habits. *Do as a class
4. 1.4 On the Job 1 (pg 44) You work at a factory that makes CFLs. You anlyse a recent batch of a new type of bulb. You find that the number of defective light bulbs in the batch is about 1 in 36. • Express the defect rate as a percent to five decimal places and to the nearest percent. • Your plant makes 1,000,000 of the light bulbs. Using the defect rate from a), calculate the potential number of defective light bulbs. • Which degree of precision do you think the manager of the factory would want to see? Explain.
5. 1.4 On the Job 1 (pg 44) Your Turn Jay is ordering boards from a lumberyard. About 1 in 7 boards at the lumberyard are warped. • What percent of the boards are warped? Express your answer to five decimal places and to the nearest percent. • Jay orders 300 boards from the lumberyard. Using the defect rate from a), calculate the potential number of warped boards he will receive. • Which degree of precision do you think Jay would want to know the probability of a board being warped? Explain.
6. 1.4 Working With Probability Check Your Understanding Pg 46-48
7. 1.4 On the Job 2 (pg 49) *See the figure on page 49. The owner of a paving company routinely checks weather forecasts to schedule jobs. Dry, sunny weather is the best for laying down asphalt. Page 49 shows a six-day forecast from a weather web site. • What does P.O.P stand for? • Explain what P.O.P means. • How can P.O.P help the owner of the paving company? • List the best days for paving during the week shown.
8. 1.4 On the Job 2 (pg 49) Your Turn. Using the six-day forecast on page 49, you have planned an outdoor family picnic for September 18th. • How would you advise family members about the probability of precipitation? • What would you advise family members to wear?
9. 1.4 Working With Probability Check Your Understanding Pg 50-52
10. 1.4 Working With Probability Work With It Pg 52-53
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# Biscuit Solution
## Introduction
This tutorial continues on from this one. In it we're going to follow the steps previously outlined in order to work out how many biscuits you need to purchase for the childcare centre next week (check out the last tutorial if you're not sure what we're talking about).
To avoid having to flip between tutorials, below is the number of children who attended childcare last week, and the number of biscuits they consumed collectively each day.
Day Number of Children Number of biscuits consumed
Monday
15
26.5
Tuesday
14
25
Wednesday
7
14.5
Thursday
20
34
Friday
23
38.5
#### Step 1: Write out your basic equation and allocate variables
Our basic equation is:
### y = mx + c
For the purpose of this example,
~ y is the number of biscuits consumed;
~ x is the number of children attending##; and
~ m and c are apparent or unknown variables that complete the equation.###
#### Step 2: Solve equations for apparent (unknown) variables
We need to work out what m and c are; this is called solving the equation for m and c. There are three simple parts to this step.
Part 1: Substitute known variables into equation
Substituting the information you recorded about this week's attendance and biscuit consumption, we can make the following two equations:
Equation 1: From Monday's data: 26.5 = 15 m + c
Equation 2: From Tuesday's data: 25 = 14 m + c
We've used two days' data to create two equations; this is because we have two unknown variables (m and c). Remember: The number of unknown variables dictates how many equations you need to solve them.
Part 2: Solve for each unknown variable
It doesn't matter which variable you choose to solve for first, but for simplicity's sake we'll choose m. This means we need to rearrange equations 1 and 2 so that they both have c on one side by itself.
Equation 1 Equation 2
26.5 = 15 m + c 25 = 14 m + c
26.5 - 15 m = c 25 - 14 m = c
Flipped over, these equations look like
c = 26.5 - 15 m c = 25 - 14 m
Now that we have two equations describing what c equals, we can combine them to solve for m.
c = c
Equation 1 = Equation 2
26.5 - 15 m = 25 - 14 m
26.5 - 25 - 15 m = -14 m
26.5 - 25 = -14 m + 15 m
1.5 = m
or rearranged:
m = 1.5
Now you can solve for c, by substituting m = 1.5 into either Equation 1 or 2. It doesn't matter which equation you use. We've used Equation 1.
c = c
26.5 = 15 m + c
26.5 = (15 x 1.5) + c
26.5 = 22.5 + c
26.5 - 22.5 = c
4 = c
or rearranged:
c = 4
In summary, our final equation is:
### y = 1.5 x + 4
Remembering that:
~ y is the number of biscuits consumed, and
~ x is the number of children attending the centre.
To confirm this equation is accurate, we use it with the attendance data recorded from Tuesday through to Friday this week to see if the equation can accurately tell us how many biscuits were consumed.
Tuesday Wednesday Thursday Friday
y = 1.5 x + 4 y = 1.5 x + 4 y = 1.5 x + 4 y = 1.5 x + 4
y = (1.5 x 14) + 4 y = (1.5 x 7.5) + 4 y = (1.5 x 20) + 4 y = (1.5 x 23) + 4
y = 21 + 4 y = 10.5 + 4 y = 30 + 4 y = 34.5 + 4
y = 25 biscuits y = 14.5 biscuits y = 34 biscuits y = 38.5 biscuits
Looking back at the biscuit consumption data you recorded for this last week, we can confirm that our equation gives accurate predictions.
#### Step 3: Work out how many biscuits you need for next week
Once again, here are next week's expected attendance numbers:
Day
Number of children
Monday
16
Tuesday
12
Wednesday
5
Thursday
23
Friday
20
Using simple arithmetic, you work out that the total number of children expected to attend the centre next week is seventy-six. Let's put this information into your equation.
y = 1.5 x + 4
y = (1.5 x 76) + 4
y = 114 + 4
y = 118 biscuits
So in summary, you will purchase 118 biscuits for next week.
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# NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Chapter 11 Algebra NCERT Solutions for Class 6 Maths is available on this page which are very helpful in building a great foundation of concepts and make easy for the students to understand basics. You can download PDF of Class 6 Maths NCERT Solutions Chapter 11 Algebra which will make you understand the topics in most simple manner and grasp it easily to perform better. It can be helpful in completing homework on time and will make student confident.
Class 6 Maths will prepare students to do better during immense pressure and at the same time make them fresh and enhances memory. It will make you able to instantly recollect your ideas and shape your answers.
Exercise 11.1
1. Find the rule, which gives the number of matchsticks required to make the following matchsticks patterns. Use a variable to write the rule.
(a) A pattern of letter T as
(b) A pattern of letter Z as
(c) A pattern of letter U as
(d) A pattern of letter V as
(e) A pattern of letter E as
(f) A pattern of letter S as
(g) A pattern of letter A as
(a) Pattern of letter T
= 2n (as two matchsticks used in each letter)
(b) Pattern of letter Z
= 3n (as three matchsticks used in each letter)
(c) Pattern of letter U
= 3n (as three matchsticks used in each letter)
(d) Pattern of letter V
= 2n (as two matchsticks used in each letter)
(e) Pattern of letter E
= 5n (as five matchsticks used in each letter)
(f) Pattern of letter S
= 5n (as five matchsticks used in each letter)
(g) Pattern of letter A
= 6n (as six matchsticks used in each letter)
2. We already know the rule for the pattern of letter L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?
The letter ‘T’ and ‘V’ that has pattern 2n,2n, since 2 matchsticks are used in all these letters.
3. Cadets are marching in a parade. There are 5 cadets in a row. What is the rule, which gives the number of cadets, given the number of rows? (Use n for the number of rows)
Number of rows = n
Cadets in each row = 5
Therefore, total number of cadets = 5n
4. If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use bb for the number of boxes)
Number of boxes = b
Number of mangoes in each box = 50
Therefore, total number of mangoes = 50b
5. The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use ss for the number of students)
Number of students = s
Number of pencils to each student = 5
Therefore, total number of pencils needed are = 5s
6. A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes)
Time taken by bird = t minutes
Speed of bird = 1 km per minute
Therefore, Distance covered by bird = speed × time = 1×t=t km
7. Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots with chalk powder as in figure). She has 8 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?
Number of dots in each row = 8 dots
Number of rows = r
Therefore, total number of dots in r rows = 8r
When there are 8 rows, then number of dots = 8 × 8 = 64 dots
When there are 10 rows, then number of dots = 8 × 10 = 80 dots
8. Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.
Therefore, Leela’s age = (x−4) years
9. Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
Number of laddus gave away = l
Number of laddus remaining = 5
Total number of laddus she make = (l+5)
10. Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?
Number of oranges in one box = x
Number of boxes = 2
Therefore, total number of oranges in boxes = 2x
Remaining oranges = 10
Thus, number of oranges = 2x+10
11. (a) Look at the following matchstick pattern of squares. The squares are not separate. Two neighbouring squares have a common matchstick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares.
(Hint: If you remove the vertical stick at the end, you will get a pattern of Cs.)
(b) Figures below gives a matchstick pattern of triangles. As in Exercise 11 (a) above find the general rule that gives the number of matchsticks in terms of the number of triangles.
(a)
4 matchsticks
7 matchsticks
10 matchsticks
13 matchsticks
If we remove 1 from each then they makes table of 3, i.e., 3, 6, 9, 12, …
So, the required equation = 3x+13x+1 , where x is number of squares.
(b)
3 matchsticks
7 matchsticks
10 matchsticks
13 matchsticks
If we remove 1 from each then they makes table of 2, i.e., 2, 4, 6, 8, …
So the required equation = 2x+12x+1 , where x is number of triangles.
Exercise 11.2
1. The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Side of equilateral triangle = l
Therefore, Perimeter of equilateral triangle = 3 × side = 3l
2. The side of a regular hexagon is denoted by l. Express the perimeter of the hexagon using l. (Hint: A regular hexagon has all its six sides in equal length)
Side of hexagon = l
Therefore, Perimeter of Hexagon = 6 × side = 6l
3. A cube is a three-dimensional figure. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. find the formula for the total length of the edges of a cube.
Length of one edge of cube = l
Number of edges in a cube = 12
Therefore, total length = 12×l=12l
4. The diameter of a circle is a line, which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure AB is a diameter of the circle; C is its centre). Express the diameter of the circle (d) in terms of its radius (r).
Since, length of diameter is double the length of radius.
Therefore, d=2r (Here r is the radius of the circle)
5. To find sum of three numbers 14, 27 and 13. We can have two ways.
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54, or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54. Thus
(14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on Whole Numebrs, in a general way, by using variables a,b and c.
(a+b)+c=a+(b+c)
Exercise 11.3
1. Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.
(Hint: Three possible expressions are 5 + (8 – 7), 5 – (8 – 7), (5 x 8) + 7 make the other expressions)
(a) (8 × 5) – 7
(b) (8 + 5) – 7
(c) (8 x 7) – 5
(d) (8 + 7) – 5
(e) 5 × (7 + 8)
(f) 5 + (7 × 8)
(g) 5 + (8 – 7)
(h) 5 – (7 + 8)
2. Which out of the following are expressions with numbers only:
(a) y+3y+3
(b) (7×20)−8z
(c) 5(21−7)+7×2
(d) 55
(e) 3x
(f) 5−5n
(g) (7×20)−(5×10)−45+p
(c) and (d)
3. Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed:
(a) z+1,z−1,y+17,y−17
(b) 17y,y/17,5z
(c) 2y+17,2y−17
(d) 7m,−7m+3,−7m−3
z−1→ Subtraction
y−17→ Subtraction
(b) 17y→ Multiplication
y/17→ Division
5z→ Multiplication
2y−17→ Multiplication and Subtraction
(d) 7m→ Multiplication
−7m−3→ Multiplication and Subtraction
4. Give expressions for the following cases:
(b) 7 subtracted from p.
(c) p multiplied by 7.
(d) p divided by 7.
(e) 7 subtracted from −m.
(f) −p multiplied by 5.
(g) −p divided by 5.
(h) p multiplied by −5.
(a) p+7
(b) p−7
(c) 7p
(d) p/7
(e) −m−7
(f) −5p
(g) −p/5
(h) −5p
5. Give expression in the following cases:
(b)11 subtracted from 2m.
(c) 5 times y to which 3 is added.
(d) 5 times y from which 3 is subtracted.
(e) y is multiplied by −8.
(f) y is multiplied by −8 and then 5 is added to the result.
(g) y is multiplied by 5 and result is subtracted from 16.
(h) y is multiplied by −5 and the result is added to 16.
(a) 2m+11
(b) 2m−11
(c) 5y+3
(d) 5y−3
(e) −8y
(f) −8y+5
(g) 16−5y
(h) −5y+16
6. (a) From expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
(a) t+4,t−4,4−t,4t,t/4,4/t
(b) 2y+7,2y−7,7y+2,7y−2 and so on
Exercise 11.4
(a) Take Sarita’s present age to be y years.
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?
(b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is b meters?
(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.
(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours. Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
(a)
(i) y+5
(ii) y−3
(iii) 6y
(iv) 6y−2
(v) 3y+5
(b) Length = 3b and Breadth = (3b−4) meters
(c) Height of the box = h cm
Length of the box = 5 times the height = 5h cm
Breadth of the box = 10 cm less than length = (5h−10) cm
(d) Meena’s position = s
Beena’s position = 8 steps ahead = s+8
Leena’s position = 7 steps behind = s−7
Total number of steps = 4s−10
(e) Speed of the bus = v km/h
Distance travelled in 5 hours = 5v km
Remaining distance = 20 km
Therefore, total distance = (5v+20) km
2. Change the following statements using expressions into statements in ordinary language.
(For example, given Salim scores r runs in a cricket match, nalin scores (r+15) runs. In ordinary language – Nalin scores 15 runs more than Salim).
(a) A note book costs ₹p. A book costs ₹3p.
(b) Tony puts q marbles on the table. He has 8q marbles in his box.
(c) Our class has n students. The school has 20n students.
(d) Jaggu is z years old. His uncle is 4z years old and his aunt is (4z−3) years old.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.
(a) A book cost 3 times the cost of a notebook.
(b) The number of marbles in box is 8 times the marble on the table.
(c) Total number of students in the school is 20 times that in our class.
(d) Jaggu’s uncle’s age is 4 times the age of Jaggu. Jaggu’s aunt is 3 years younger than his uncle.
(e) The total number of dots is 5 times the number of rows.
3. (a) Given, Munnu’s age to be x years. Can you guess what (x−2) may show? (Hint: Think of Munnu’s younger brother). Can you guess what (x+4) may show? What (3x+7) may show?
(b) Given Sara’s age today to be y years. Think of her age in the future or in the past. What will the following expression indicate? y+7,y−3,y+4.1/2,y−2.1/2
(c) Given, n students in the class like football, what may 2n show? What may n/2 show? (Hint: Think of games other than football).
(a) Munnu’s age = x years
His younger brother is 2 years younger than him = (x−2) years
His elder brother’s age is 4 years more than his age = (x+4) years
His father is 7 year’s more than thrice of his age = (3x+7) years
(b) Her age in past = (y−3),(y−2.1/2)
Her age in future = (y+7),(y+4.1/2)
(c) Number of students like hockey is twice the students liking football, i.e., 2n
Number of students like tennis is half the students like football, i.e., n/2
Exercise 11.5
1. State which of the following are equations (with a variable). Given reason for your answer. Identify the variable from the equations with a variable.
(a) 17=x+7
(b) (t−7)>5
(c) 4/2=2
(d) (7×3)−19=8
(e) 5×4−8=2x
(f) x−2=0
(g) 2m<30
(h) 2n+1=11
(i) 7=(11×5)−(12×4)
(j) 7=(11×2)+p
(k) 20=5y
(l) 3q/2<5
(m) z+12>24
(n) 20−(10−5)=3×5
(o) 7−x=5
(a) It is an equation of variable as both the sides are equal. The variable is x.
(b) It is not an equation as L.H.S. is greater than R.H.S.
(c) It is an equation with no variable. But it is a false equation.
(d) It is an equation with no variable. But it is a false equation.
(e) It is an equation of variable as both the sides are equal. The variable is x.
(f) ) It is an equation of variable x.
(g) It is not an equation as L.H.S. is less than R.H.S.
(h) It is an equation of variable as both the sides are equal. The variable is n.
(i) It is an equation with no variable as its both sides are equal.
(j) It is an equation of variable p.
(k) It is an equation of variable y.
(l) It is not an equation as L.H.S. is less than R.H.S.
(m) It is not an equation as L.H.S. is greater than R.H.S.
(n) It is an equation with no variable.
(o) It is an equation of variable x.
2. Complete the entries of the third column of the table:
S. No. Equation Value of variable Equation satisfied Yes/No (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) 10y=80 10y=80 10y=80 4l=20 4l=20 4l=20 b+5=9 b+5=9 b+5=9 h−8=5 h−8=5 h−8=5 p+3=1 p+3=1 p+3=1 p+3=1 p+3=1 y=10 y=8 y=5 l=20 l=80 l=5 b=5 b=9 b=4 h=13 h=8 h=0 p=3 p=1 p=0 p=−1 p=−2
S. No. Equation Value of variable Equation satisfied Yes/No Solution of L.H.S. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) 10y=80 10y=80 10y=80 4l=20 4l=20 4l=20 b+5=9 b+5=9 b+5=9 h−8=5 h−8=5 h−8=5 p+3=1 p+3=1 p+3=1 p+3=1 p+3=1 y=10 y=8 y=5 l=20 l=80 l=5 b=5 b=9 b=4 h=13 h=8 h=0 p=3 p=1 p=0 p=−1 p=−2 No Yes No No No Yes No Yes Yes Yes No No No No No No Yes 10 x 10 = 100 10 x 8 = 80 10 x 5 = 50 4 x 20 = 80 4 x 80 = 320 4 x 5 = 20 5 + 5 = 10 9 + 5 = 14 4 + 5 = 9 13 – 8 = 5 8 – 8 = 0 0 – 8 = –8 3 + 3 = 6 1 + 3 = 4 0 + 3 = 3 –1 + 3 = 2 –2 + 3 = 1
3. Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation. (a) 5m=60 (10, 5, 12, 15)
(b) n+12=20 (12, 8, 20, 0)
(c) p−5=5 (0, 10, 5, –5)
(d) q/2=7 (7, 2, 10, 14)
(e) r−4=0 (4, –4, 8, 0)
(f) x+4=2 (–2, 0, 2, 4)
(a) 5m=60
Putting the given values in L.H.S.,
5 × 10 = 50, 5 × 5 = 25
∵ L.H.S. ≠ R.H.S.
∴m=10 is not the solution.
∴m=5 is not the solution.
5 × 12 = 60, 5 × 15 = 75
∵ L.H.S. = R.H.S.
∴m=12 is a solution.
∴m=15 is not the solution.
(b) n+12=20
Putting the given values in L.H.S.,
12 + 12 = 24, 8 + 12 = 20
∵ L.H.S. ≠ R.H.S.
∴n=12 is not the solution.
∴n=8 is a solution.
20 + 12 = 32, 0 + 12 = 12
∵ L.H.S. ≠ R.H.S.
∴n=20 is not the solution.
∴n=0 is not the solution.
(c) p−5=5
Putting the given values in L.H.S.,
0 – 5 = –5, 10 – 5 = 5
∵ L.H.S. ≠ R.H.S.
∴p=0 is not the solution.
∴p=10 is a solution.
5 – 5 = 0, –5 – 5 = –10
∵ L.H.S. ≠ R.H.S.
∴p=5 is not the solution.
∴p=−5 is not the solution.
(d) q/2=7
Putting the given values in L.H.S.,
7/2, 2/2=1
∵ L.H.S. ≠ R.H.S.
∴q=7 is not the solution.
∴q=2 is not the solution.
10/2=5.14/2=7
∵ L.H.S. ≠ R.H.S.
∴ q=10 is not the solution.
∴q=14 is a solution.
(e) r−4=0r−4=0
Putting the given values in L.H.S.,
4 – 4 = 0, –4 – 4 = –8
∵ L.H.S. = R.H.S.
∴r=4 is a solution.
∴r=−4 is not the solution.
8 – 4 = 4, 0 – 4 = –4
∵ L.H.S. ≠ R.H.S.
∴r=8 is not the solution.
∴r=0 is not the solution.
(f) x+4 = 2x+4 = 2
Putting the given values in L.H.S.,
–2 + 4 = 2, 0 + 4 = 4
∵ L.H.S. = R.H.S.
∴x = −2 is a solution.
∴x=0 is not the solution.
2 + 4 = 6, 4 + 4 = 8
∵ L.H.S. ≠ R.H.S.
∴x=2 is not the solution.
∴x=4 is not the solution.
4. (a) Complete the table and by inspection of the table find the solution to the equation
m+10=16.
m 1 2 3 4 5 6 7 8 9 10 --- --- --- m+10 --- --- --- --- --- --- --- --- --- --- --- --- ---
(b) Complete the table and by inspection of the table find the solution to the equation 5t=355t=35.
t 3 4 5 6 7 8 9 10 11 --- --- --- --- --- 5t --- --- --- --- --- --- --- --- --- --- --- --- --- ---
(c)Complete the table and by inspection of the table find the solution to the equation z3=4.z3=4.
z 8 9 10 11 12 13 14 15 16 --- --- --- --- z/3 2.2/3 3 3.1/3 --- --- --- --- --- --- --- --- --- ---
(d)Complete the table and by inspection of the table find the solution to the equation m−7=3.m−7=3.
m 5 6 7 8 9 10 11 12 13 --- --- m−7 --- --- --- --- --- --- --- --- --- --- ---
(a)
m 1 2 3 4 5 6 7 8 9 10 11 12 13 m+10 11 12 13 14 15 16 17 18 19 20 21 22 23
∴ m=6 is the solution.
∵ At m=6,m+10=16
(b)
t 3 4 5 6 7 8 9 10 11 12 13 14 15 16 5t 15 20 25 30 35 40 45 50 55 60 65 70 75 80
∴ t=7 is the solution.
∵ At t=7,5t=35
(c)
z 8 9 10 11 12 13 14 15 16 17 18 19 20 z/3 2.2/3 3 3.1/3 3.2/3 4 4.1/3 4.2/3 5 5.1/3 5.2/3 6 6.1/3 6.2/3
∴ z=12 is the solution.
∵ At z=12,z/3=4
(d)
m 5 6 7 8 9 10 11 12 13 14 15 m−7 -2 -1 0 1 2 3 4 5 6 7 8
∴ m=10 is the solution.
∵ At m=10, m−7=3
## NCERT Solutions for Class 6 Maths Chapter 11 Algebra
Chapter 11 Algebra NCERT Solutions are prepared by Studyrankers experts which allow students to cover the entire syllabus effectively without any frustration. The branch of mathematics where letters are used along with numbers is called algebra.
• An expression consisting of variables, constants and mathematical operators is called an algebraic expression.
• An unknown quantity can be represented by a variable. Generally, a variable is any letter from the English alphabet that represents an unknown quantity.
NCERT Solutions help students cope with the pressure of the large board examination syllabus. Students can cross check their answers and also whether they learned it properly or not. You can find exercisewise NCERT Solutions from the links given below.
• Exercise 11.1 Chapter 11 Class 6 Maths NCERT Solutions
• Exercise 11.2 Chapter 11 Class 6 Maths NCERT Solutions
• Exercise 11.3 Chapter 11 Class 6 Maths NCERT Solutions
• Exercise 11.4 Chapter 11 Class 6 Maths NCERT Solutions
• Exercise 11.5 Chapter 11 Class 6 Maths NCERT Solutions
These NCERT Solutions are helpful resources that can help you not only cover the entire syllabus but also provide in depth analysis of the topics. It help in improving the student's experience and guide student in a better way.
### NCERT Solutions for Class 6 Maths Chapters:
Chapter 1 Knowing Our Numbers Chapter 2 Whole Numbers Chapter 3 Playing with Numbers Chapter 4 Basic Geometrical Ideas Chapter 5 Understanding Elementary Shapes Chapter 6 Integers Chapter 7 Fractions Chapter 8 Decimals Chapter 9 Data Handling Chapter 10 Mensuration Chapter 12 Ratio and Proportion Chapter 13 Symmetry Chapter 14 Practical Geometry
FAQ on Chapter 11 Algebra
#### How many exercises in Chapter 11 Algebra?
There are 5 exercises in Chapter 11 NCERT Solutions that provide good experience and provide opportunities to learn new things. It help in solving the difficulties that lie ahead with ease and serve as beneficial tool that can be used to recall various questions any time.
#### What is associative property of addition?
This property states that three numbers can be added in any order. If a, b and c represent any three numbers, then (a + b) + c = a + (b + c).
#### What is commutative property of addition?
This property states that two numbers can be added in any order. If a and b represent any two numbers, then a+b=b+a.
#### What is distributive property of multiplication over addition?
This property states that if a, b and c represent any three numbers, then ax (b + c) = a × b + a × c.
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# Find the length of a rectangle lot with a perimeter of 82 meters if the length is 7 meters more then the width?
May 8, 2018
24 meters
#### Explanation:
If the short side of the rectangle = s, then the long side would be s+7.
The total perimeter then would be the sum of all 4 sides, where the sids are equal in pairs, i.e.
Perimeter = s+(s+7)+s+(s+7) = 4s+14 = 82
Therefore 4s=82-14 = 68
so s=17
As this is the short side, the long side is s+7=17+7=24
The length of the rectangle, therefore, is 24 meters
May 8, 2018
The length of the rectangle lot is $\textcolor{red}{24}$ meters.
#### Explanation:
To start, I'll set up an equation to help solve this...
$2 l + 2 w = p$, meaning the length twice plus the width twice equals the perimeter. We can fill this in with the known values:
$2 l + 2 \left(l - 7\right) = 82$
$w$ can be replaced by $\left(l - 7\right)$ because the problem states that the width is $7$ meters shorter than the length.
Distribute the $2$ to the $\left(l - 7\right)$ to get:
$2 l + 2 l - 14 = 82$
And combine like terms to get:
$4 l = 96$
Finally divide both sides by $4$ to finish with:
$l = 24$, meaning the length is $\textcolor{red}{24}$ meters.
I'll check my work by plugging this info back in to the original equation to see if it works:
$2 l + 2 w = p$
$2 l + 2 w = 82$
$2 \left(24\right) + 2 \left(24 - 7\right) = 82$
$2 \left(24\right) + 2 \left(17\right) = 82$
$\textcolor{g r e e n}{48 + 34 = 82}$
The equation works, meaning the length is $24$ m and the width is $17$ m.
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# How do you differentiate f(x) = (3x-2)^4 using the chain rule?
Nov 24, 2015
$f ' \left(x\right) = 12 {\left(3 x - 2\right)}^{3}$
#### Explanation:
According to the Chain Rule:
$f ' \left(x\right) = 4 {\left(3 x - 2\right)}^{3} \cdot \frac{d}{\mathrm{dx}} \left[3 x - 2\right]$
$f ' \left(x\right) = 4 {\left(3 x - 2\right)}^{3} \cdot 3$
$f ' \left(x\right) = 12 {\left(3 x - 2\right)}^{3}$
Nov 24, 2015
$\frac{\mathrm{dy}}{\mathrm{dx}} = 12 {\left(3 x - 2\right)}^{3}$
#### Explanation:
Given -
$y = {\left(3 x - 2\right)}^{4}$
Let $U = 3 x - 2$
Then-
$y = {U}^{4}$
$\frac{\mathrm{dy}}{\mathrm{dU}} = 4 {U}^{3}$
$\frac{\mathrm{dU}}{\mathrm{dx}} = 3$
$\frac{\mathrm{dy}}{d} = \frac{\mathrm{dy}}{\mathrm{dU}} . \frac{\mathrm{dU}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {\left(U\right)}^{3} \left(3\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {\left(3 x - 2\right)}^{3} \left(3\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 12 {\left(3 x - 2\right)}^{3}$
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# Combinations and Permutations: The Basics
## Combinations and Permutations: The Basics
The Math section of the SAT contains problems pertaining to topics known as combinations and permutations. You may be asked the number of combinations possible in a given situation. For example, let’s say you walk into a restaurant that serves ice cream. You have a choice of five different flavors: chocolate, vanilla, strawberry, mint, and blueberry. You also have a choice of three different toppings: hot fudge, caramel, and strawberry. How many different ways can you choose one scoop of ice cream with one topping?
One way to solve this question (although there are faster ways) is to write the list of all the possible options, systematically. Thus, we can choose chocolate with hot fudge, chocolate with caramel, and chocolate with strawberry. We’ve now exhausted all the options for chocolate ice cream, so now we move to vanilla. We now have vanilla with hot fudge, vanilla with caramel, and vanilla with strawberry. We’re now finished with all the options involving vanilla. Continuing through the list in order, we come up with the following:
strawberry with hot fudge
strawberry with caramel
strawberry with strawberry
mint with hot fudge
mint with caramel
mint with strawberry
blueberry with hot fudge
blueberry with caramel
blueberry with strawberry
If you count all the possible options, you’ll see that there are 15 combinations.
Here’s another way to think about this problem: each of the five flavors had three different options for toppings. Five threes means “five times three,” which is 15. So another way to solve combinations problems is to multiply the number of options in each category by each other. Use this method instead of listing all the possible options. It’s faster.
Multiplying works no matter how many categories there are. For example, if you have three different hats, four different shirts, and five different pairs of pants, how many different outfits can you make? The process is easy: 3 X 4 X 5, which equals 60.
Permutations are similar to combinations. Permutations deal with the number of ways that things (or people) can be arranged. For example: let’s say you have five different soccer trophies, and you want to arrange them in a row on your desk. What are all the possibilities of doing so?
Pretend that there are five different blank spaces on a desk. Imagine the first blank space—let’s say it’s the space that’s furthermost left. How many different trophies can you put there? Five. Now let’s pretend you select a trophy and put it in the space. How many options do you have to put a trophy in the blank space next to the first space? Four—because that’s how many trophies you have remaining. Pretend to select a trophy again and put it in that second blank space. How many options do you have to put a trophy in the next blank space? You have three trophies left. Pretend to select a trophy again and put it in that third blank space. Now how many options do you have to put a trophy in the next blank space? Two. Pretend to select a trophy again and put it in that fourth blank space. You have one blank space left. Now how many options do you have to put a trophy in the last blank space? Only one. Now, multiply all those numbers. Why? Because this is a special type of combinations problem; therefore, we follow the abovementioned rule. So 5 X 4 X 3 X 2 X 1 = 120. There are 120 ways that you can arrange five trophies in a row.
You can perform a problem like this on your calculator by typing all those numbers, but there’s an even easier way. If your calculator sports a button with an exclamation mark, you may use that key. That key is called a factorial. Type “5,” hit the factorial key, then press the enter key. You should get the same answer, because 5! means that you multiply all the integers starting with 5 and (always, by definition) ending with 1. (Another example: 7! = 7 X 6 X 5 X 4 X 3 X 2 X 1.)
These simple rules will help you solve the most straightforward math problems involving combinations and permutations.
By |2017-05-22T07:10:38+00:00August 8th, 2013|SAT, SAT Tips, Test Prep|0 Comments
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Sum Of Partial Factorials
Sum Of Partial Factorials
(December 23, 2006)
This page shows a general formula called sum of partial factorials that is used to generate many identities such that each of identity is true for all positive integers n.
General Formula
, where .
The general formula (I) can be rewritten in a product form as follows:
, where.
It is known that
or
or so on, where .
Now we want to find a general formula which is of the form
or
for all positive integers n.
Indeed, the general formula (I) or (II) is used to derive our desire. Let consider some values of m.
• m = 2, we obtain a formula for all positive integers n, namely
, which can be proved by using the method of Mathematical induction.
It can easily be checked, for instance, n = 3, both sides are equal to 20.
.
• m = 3, we get the formula
.
It can easily be checked, for instance, n = 2, both sides are equal to 30.
.
• m = 4, we obtain
.
We see that for each positive m, (I) generates a general formula in which it is true for all positive integers n.
• m = 1
or
.
• m = 2
or
.
• m = 3
or
.
• m = 4
or
.
• m = 5
or
.
General Formula
or its summation notation,
, where .
We express it in terms of factorial form
or its summation notation,
, where m, .
If we treat m as real x, we get the extended formula in terms of x as follows
or its summation notation,
, where .
In addition, we get another interested formula that is based on the above results, namely
or its expansion form,
, where
Example
Let n = 4, we obtain a beautiful identity
for all real x.
Here are the lists of identities for different values of n. These identities are true for all real x.
• n = 2
or its product notation,
.
• n = 3
or its product notation,
.
• n = 4
or
.
•
or its product notation,
.
Other finite series:
Number proceeds from unity. (Aristotle)
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# NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers| PDF Download
On this page, you will find Chapter 9 Rational Numbers Class 7 Maths NCERT Solutions which will help you in attaining good marks in the examinations and complete homework on time. You can download PDF of Chapter 9 Rational Numbers NCERT Solutions which will give good experience and provide opportunities to learn new things. Chapter 9 NCERT Solutions will serve as beneficial tool that can be used to recall various questions any time.
Class 7 Maths NCERT Solutions will be useful in understanding the concepts of the chapter properly. These NCERT Solutions are helpful resources that can help you not only cover the entire syllabus but also inculcate correct learning habits among students.
Exercise 9.1
1. List five rational numbers between:
(i) -1 and 0
(ii) -2 and - 1
(iii) -4/5 and -2/3
(iv) -1/2 and 2/3
(i) -1 and 0
Let us write -1 and 0 as rational numbers with denominator 6.
Therefore, five rational numbers between -1 and 0 would be
(ii) -2 and -1
Let us write -2 and -1 as rational numbers with denominator 6.
Therefore, Five rational numbers between -2 and -1 would be
(iii) -4/5 and -2/3
Let us write -4/5 and -2/3 as rational numbers with the same denominators.
Therefore, five rational numbers between -4/5 and -2/3 would be
(iv) -1/2 and 2/3
Let us write -1/2 and 2/3 as rational numbers with the same denominators.
Therefore, five rational numbers between -1/2 and 2/3 would be
2. Write four more rational numbers in each of the following patterns:
Therefore, four equivalent rational numbers are
(ii)
Therefore, the next four rational numbers of this pattern would be
(iii)
Therefore, the next four rational numbers of this pattern would be
(iv)
Therefore, the next four rational numbers of tins pattern would be
3. Give four rational numbers equivalent to:
(i) -2/7
(ii) 5/-3
(iii) 4/9
(iv) -2/7
(i)
Therefore, four equivalent rational numbers are
(ii) 5/-3
Therefore, four equivalent rational numbers are
(iii) 4/9
Therefore, four equivalent rational numbers are
4. Draw the number line and represent the following rational numbers on it:
(i) 3/4
(ii) -5/8
(iii) -7/4
(iv) 7/8
(i) 3/4
(ii) -5/8
(iii) -7/4
(iv) 7/8
5. The points P, Q, R, S, T, U, A and B on the number line are such that, TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S.
Therefore,
Similarly
Thus, the rational numbers represented P, Q, R and S are respectively.
6. Which of the following pairs represent the same rational numbers:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
7. Rewrite the following rational numbers in the simplest form:
(i) -8/6
(ii) 25/45
(iii) -44/72
(iv) -8/10
8. Fill in the boxes with the correct symbol out of <, > and =:
9. Which is greater in each of the following:
10. Write the following rational numbers in ascending order:
Exercise 9.2
1. Find the sum:
2. Find:
3. Find the product:
4. Find the value of:
## NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers
In Chapter 7 Maths NCERT Solutions, we will deal with rational numbers which is defined as a number that can be expressed in the form p/q , where p and q are integers and q ≠ 0. All numbers, including whole numbers, integers, fractions and decimal numbers, can be written in the numerator-denominator form.
• The rational number zero is neither negative nor positive. Positive rational numbers are represented to the right of 0. Negative rational numbers are represented to the left of 0.
• A rational number is said to be in its standard form if its numerator and denominator have no common factor other than 1, and its denominator is a positive integer.
You can get exercisewise NCERT Solutions just by clicking on the links given below which will help you in improving experience. It can be used to enrich knowledge and make lessons for learners more exciting.
You will get NCERT Solutions which are prepared by Studyrankers experts which will set a good foundation for your future goals. It is very challenging to socre good marks in tests that is why we have prepared NCERT Solutions.
### NCERT Solutions for Class 7 Maths Chapters:
Chapter 1 Integers Chapter 2 Fractions and Decimals Chapter 3 Data Handling Chapter 4 Simple Equations Chapter 5 Lines and Angles Chapter 6 The Triangle and its Properties Chapter 7 Congruence of Triangles Chapter 8 Comparing Quantities Chapter 10 Practical Geometry Chapter 11 Perimeter and Area Chapter 12 Algebraic Expressions Chapter 13 Exponents and Powers Chapter 14 Symmetry Chapter 15 Visualising Solid Shapes
FAQ on Chapter 9 Rational Numbers
#### How many exercises are there in Chapter 9 Rational Numbers Class 7 Maths NCERT Solutions?
Chapter 9 Rational Numbers consists of only two exercises which will help you in improving the marks in the examinations and have edge over your classmates. It will help you in identify, analyze, and then rectify the mistakes.
#### Arrange -7/8, -5/6, -3/4 in the ascending order.
-7/8 < -5/6 < -3/4.
#### How to do addition of rational numbers?
• If the denominators of the given rational numbers are the same, then the denominator of their sum will also be the same. The numerator of the sum of two rational numbers with the same denominator is the sum of the numerators of the given numbers.
• To add rational numbers with different denominators, we convert them into equivalent rational numbers with the same denominator.
• Two rational numbers whose sum is zero are called additive inverses of each other.
#### How to do multiplication of rational numbers?
• To multiply two rational numbers, we simply multiply their numerators and denominators with their correct signs.
• Two rational numbers whose product is 1 are called reciprocals of each other.
• A rational number and its reciprocal will always have the same sign.
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How do you translate 6 less than or equal to x cubed into an algebraic expression?
Last updated date: 02nd Aug 2024
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Hint: We are given a statement as 6 less than or equal to x cube. We are asked to translate this into an algebraic expression. To do so we will learn what algebraic expression refers to. Then we will use the algebraic tool like $\le ,\ge ,<,>,+,-,\div ,\times ,etc.$ to simplify or change the statement written into mathematical form.
We are given a statement as 6 less than or equal to x. We have to write it into an algebraic expression. Before we solve, we should know that algebra is the equation or expression we will build up from integer, constant variables and the various algebraic tools. So, as we have to find the algebraic expression it means we have to translate the statement “6 less than x cube” into a mathematical expression. Before, we move another step we will learn how to express the word used in the given statement into the mathematical term. Now, less than is represented by the symbol ‘<’. Here, a < b. It means a is less than b. For symbol ‘>’, for example, a > b, it means a is greater than b. Now, we have that less than or equal to, so the symbol $\le$ means less than or equal if $a\le b$ means a is less than or equal to b. There is another word cube, the cube of any term means the product of any term with itself three times. Cube of a means $a\times a\times a.$ We can denote a cube as $a\times a\times a$ or ${{a}^{3}}.$ In a simple way, say we denote as ${{a}^{3}}.$ Now, we are given that x is cubed. So, it is denoted as ${{x}^{3}}.$ Now, we will work on our problem. We have that 6 less than or equal to x cubed. For less than or equal we use $\le$ and for x cube we use ${{x}^{3}}.$ So, our equation will be expressed as $6\le {{x}^{3}}.$
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# Unit 0: Observation, Measurement and Calculations Cartoon courtesy of NearingZero.net.
## Presentation on theme: "Unit 0: Observation, Measurement and Calculations Cartoon courtesy of NearingZero.net."— Presentation transcript:
Unit 0: Observation, Measurement and Calculations Cartoon courtesy of NearingZero.net
Steps in the Scientific Method 1.Observations - quantitative - qualitative 2.Formulating hypotheses - possible explanation for the observation 3.Performing experiments - gathering new information to decide whether the hypothesis is valid whether the hypothesis is valid
Outcomes Over the Long-Term Theory (Model) - A set of tested hypotheses that give an overall explanation of some natural phenomenon. overall explanation of some natural phenomenon. Natural Law - The same observation applies to many different systems different systems - Example - Law of Conservation of Mass
Law vs. Theory A law summarizes what happens A law summarizes what happens A theory (model) is an attempt to explain why it happens.
In science, we deal with some very LARGE numbers: 1 mole = 602000000000000000000000 In science, we deal with some very SMALL numbers: Mass of an electron = 0.000000000000000000000000000000091 kg Scientific Notation
Imagine the difficulty of calculating the mass of 1 mole of electrons! 0.000000000000000000000000000000091 kg x 602000000000000000000000 x 602000000000000000000000 ???????????????????????????????????
Scientific Notation: A method of representing very large or very small numbers in the form: M x 10 n M x 10 n M is a number between 1 and 10 n is an integer
2 500 000 000 Step #1: Insert an understood decimal point. Step #2: Decide where the decimal must end up so that one number is to its left up so that one number is to its left Step #3: Count how many places you bounce the decimal point the decimal point 1234567 8 9 Step #4: Re-write in the form M x 10 n
2.5 x 10 9 The exponent is the number of places we moved the decimal.
0.0000579 Step #2: Decide where the decimal must end up so that one number is to its left up so that one number is to its left Step #3: Count how many places you bounce the decimal point the decimal point Step #4: Re-write in the form M x 10 n 12345
5.79 x 10 -5 The exponent is negative because the number we started with was less than 1.
PERFORMING CALCULATIONS IN SCIENTIFIC NOTATION ADDITION AND SUBTRACTION
Review: Scientific notation expresses a number in the form: M x 10 n 1 M 10 n is an integer
4 x 10 6 + 3 x 10 6 IF the exponents are the same, we simply add or subtract the numbers in front and bring the exponent down unchanged. 7 x 10 6
4 x 10 6 - 3 x 10 6 The same holds true for subtraction in scientific notation. 1 x 10 6
4 x 10 6 + 3 x 10 5 If the exponents are NOT the same, we must move a decimal to make them the same.
4.00 x 10 6 + 3.00 x 10 5 +.30 x 10 6 4.30 x 10 6 Move the decimal on the smaller number! 4.00 x 10 6
A Problem for you… 2.37 x 10 -6 + 3.48 x 10 -4
2.37 x 10 -6 + 3.48 x 10 -4 Solution… 002.37 x 10 -6
+ 3.48 x 10 -4 Solution… 0.0237 x 10 -4 3.5037 x 10 -4
Nature of Measurement Part 1 - number Part 2 - scale (unit) Examples: 20 grams 6.63 x 10 -34 Joule seconds Measurement - quantitative observation consisting of 2 parts consisting of 2 parts
The Fundamental SI Units (le Système International, SI)
SI Units
SI Prefixes Common to Chemistry PrefixUnit Abbr.Exponent Kilok10 3 Decid10 -1 Centic10 -2 Millim10 -3 Micro 10 -6
Uncertainty in Measurement A digit that must be estimated is called uncertain. A measurement always has some degree of uncertainty.
Why Is there Uncertainty? Measurements are performed with instruments No instrument can read to an infinite number of decimal places Which of these balances has the greatest uncertainty in measurement?
Precision and Accuracy Accuracy refers to the agreement of a particular value with the true value. Precision refers to the degree of agreement among several measurements made in the same manner. Neither accurate nor precise Precise but not accurate Precise AND accurate
Types of Error Random Error (Indeterminate Error) - measurement has an equal probability of being high or low. Systematic Error (Determinate Error) - Occurs in the same direction each time (high or low), often resulting from poor technique or incorrect calibration.
Rules for Counting Significant Figures - Details Nonzero integers always count as significant figures. 3456 has 4 sig figs.
Rules for Counting Significant Figures - Details Zeros - Leading zeros do not count as significant figures. 0.0486 has 3 sig figs.
Rules for Counting Significant Figures - Details Zeros - Captive zeros always count as significant figures. 16.07 has 4 sig figs.
Rules for Counting Significant Figures - Details Zeros Trailing zeros are significant only if the number contains a decimal point. 9.300 has 4 sig figs.
Rules for Counting Significant Figures - Details Exact numbers have an infinite number of significant figures. 1 inch = 2.54 cm, exactly
Sig Fig Practice #1 How many significant figures in each of the following? 1.0070 m 5 sig figs 17.10 kg 4 sig figs 100,890 L 5 sig figs 3.29 x 10 3 s 3 sig figs 0.0054 cm 2 sig figs 3,200,000 2 sig figs
Rules for Significant Figures in Mathematical Operations Multiplication and Division: # sig figs in the result equals the number in the least precise measurement used in the calculation. 6.38 x 2.0 = 12.76 13 (2 sig figs)
Sig Fig Practice #2 3.24 m x 7.0 m CalculationCalculator says:Answer 22.68 m 2 23 m 2 100.0 g ÷ 23.7 cm 3 4.219409283 g/cm 3 4.22 g/cm 3 0.02 cm x 2.371 cm 0.04742 cm 2 0.05 cm 2 710 m ÷ 3.0 s 236.6666667 m/s240 m/s 1818.2 lb x 3.23 ft5872.786 lb·ft 5870 lb·ft 1.030 g ÷ 2.87 mL 2.9561 g/mL2.96 g/mL
Rules for Significant Figures in Mathematical Operations Addition and Subtraction: The number of decimal places in the result equals the number of decimal places in the least precise measurement. 6.8 + 11.934 = 18.734 18.7 (3 sig figs)
Sig Fig Practice #3 3.24 m + 7.0 m CalculationCalculator says:Answer 10.24 m 10.2 m 100.0 g - 23.73 g 76.27 g 76.3 g 0.02 cm + 2.371 cm 2.391 cm 2.39 cm 713.1 L - 3.872 L 709.228 L709.2 L 1818.2 lb + 3.37 lb1821.57 lb 1821.6 lb 2.030 mL - 1.870 mL 0.16 mL 0.160 mL
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# Using Algebra for Hanging Pictures
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## Introduction: Using Algebra for Hanging Pictures
When traveling in Mexico last year, we bought some odd film stills of a movie called "El Profesor Erotico" (1976). At the time, I thought that it was an old campy Mexican TV show. Further research on the internet showed that it was a B-movie of the Argentinian variety*. We had purchased 4 photographs, got them framed and wanted to hang them on a wall.
In the past, I've done all sorts of division, taping and remeasuring work out how exactly to hang sets of pictures. After thinking about how negative and positive space are a matter of perception, I realized the solution to this would be to use simple 8th grade algebra.
This technique will save you tons of hassle of measuring, remeasuring, chicken scratch division and putting blue tape all over the place.
* I have no idea what they are doing with the blow-up doll, but all the film stills are equally puzzling and funny.
## Step 1: Sketch Your Space
Each frame is identical. The plan was to hang them in a row, on the same wall with an equal amount of white space between each picture. On the left side of the wall is a door frame and the right edge is another door. These edges will like be margins on a page with an equal amount of wall space as the center.
Draw a simple diagram of the space. Take your time and refer back to this when you are doing the measurements. I drew mine exceedingly quickly. The doorways are the large rectangles that drift into an imaginary floor and the pictures are the tiny squares.
## Step 2: Make Your Equation
This is a simple x and y equation. In this case x = the width of the picture frame and y = the space between the picture frames. Also, measure the total expanse of the wall.In my case, the total wall space = 126 3/4" which is 126.75 if we express in a decimal (we'll convert back to a fraction when we are done).
Our equations is: 5y + 4x = 126.75
We have 4 picture frames and 5 negative spaces of equal length.
## Step 3: Measure Your Picture Frames
Measure your frame and this becomes your x value. In this case, the picture frame = 14.5 inches wide.
The equation is: 5y + (4 * 14.5) = 126.75
## Step 4: Solve for Y
What we want to know is what y =. Remember how torturous algebra was? We're smarter now and have the full use of the web. It's pretty easy. Use your sheet as a scratch pad.
5y + (4 * 14.5) = 126.75
5y + 58 = 126.75
5y = 68.75
y = 13.75
## Step 5: Convert the Decimal to Fractions
Most rulers are in inches or millimeters and show fractional values. In this case, 13.75 easily converts to 13 3/4.
It's helpful to refer to a decimal to fraction conversion table if you have decimal values you can't convert so easily in your head to fractional measurements.
## Step 6: Now Hang Your Pictures
There are other Instructables on how to hang the pictures themselves such as this one. My Instructable will provides a calculation technique for the horizontal space. I'm a fan of using the blue tape at this point and never marking the wall with a pencil. Mark the blue tape itself. Leave no trace.
## Step 7: Apply the Technique to Other Situations
Here is another hanging situation for an art show I did several years ago.
Each picture is 30" wide. I had 150 inches total wall space, and I wanted to have the middle sections be twice as wide as the wall space on the outside.
Here the equation is 3x + 6y = 150
You can use the same mickey-mouse algebra to solve this problem. Next time someone says "Math is Hard", remember it's time to liberate your toys.
Enjoy!
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## 8 Discussions
Much easier formula: if you are hanging 3 frames, divide the length of the wall by 4 to get the centre points (from which they'll be hanging).
If you have 4 frames divide by 5, etc.
All have to be same size though
Thank you! Followed your formula and worked a treat.
I am 35 years old. This is the first time in my life I've come across any sort of practical application for algebra I learned in school.
As a math teacher, I am glad that now there is a tendency to focus on applicable aspects of algebra. Students even seem to be more intested in the subject. I've heard of a contest among students, the task was to count skittles that could fit ito 1 jar.
Here is also a bunch of interactive tests that I use during my classes.
As jmarsh2 mentioned. I have a good teaching tool now for my kids, brains both of them always have a logical excuse to argue with my logic. Got them now on why its important to do the easy math when told to.. thx.
Note:
correction to formula:
X=the number of picture frames
Not the size!!!!
@Lrcooper54 Scott has it exactly right. X is not the number of pictures - it is the width of the pictures. Did you even read the article? I hope you didn't screw anyone up. Grrrr!
you could develop the formula so that (n+1)y +(n)x = w where w is the width of the wall and n ifs the number of items to be hung.
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# Word Problems About Money Practice Problems
based on 1 rating
By McGraw-Hill Professional
Updated on Sep 27, 2011
## Money Word Problems Practice Problems
### Set 1: Introduction to Money Word Problems
To review Word Problems about Money, go to Word Problems About Money Help
#### Practice
1. A vending machine has \$19.75 in dimes and quarters. There are 100 coins in all. How many dimes and quarters are in the machine?
2. Ann has \$2.25 in coins. She has the same number of quarters as dimes. She has half as many nickels as quarters. How many of each coin does she have?
3. Sue has twice as many quarters as nickels and half as many dimes as nickels. If she has a total of \$4.80, how many of each coin does she have?
#### Solutions
1. Let x represent the number of dimes. Then 100 – x is the number of quarters. There is 0.10 x dollars in dimes and 0.25(100 – x ) dollars in quarters.
There are 35 dimes and 100 – x = 100 – 35 = 65 quarters.
2. Let x represent the number of quarters. There are as many dimes as quarters, so x also represents the number of dimes. There are half as many nickels as dimes, so (or 0.50 x , as a decimal number) is the number of nickels.
There are 6 quarters, 6 dimes, 0.50 x = 0.50(6) = 3 nickels.
3. As both the number of quarters and dimes are being compared to the number of nickels, let x represent the number of nickels. Then 2 x represents the number of quarters and (or 0.50 x ) is the number of dimes.
There are 8 nickels, 0.50 x = 0.50(8) = 4 dimes, and 2 x = 2(8) = 16 quarters.
### Set 2: Money Word Problems - One Quanitty with Different Investments and Interest Rates
To review money problems involving one quantity divided into two investments, go to Word Problems About Money Help
#### Practice
1. A businessman invested \$50,000 into two funds which yielded profits of and . If the total profit was \$8520, how much was invested in each fund?
2. A college student deposited \$3500 into two savings accounts, one with an annual yield of and the other with an annual yield of . If he earned \$171.75 total interest the first year, how much was deposited in each account?
3. A banker plans to lend \$48,000 at a simple interest rate of 16% and the remainder at 19%. How should she allot the loans in order to obtain a return of ?
#### Solutions
1. Let x represent the amount invested at . Then 50,000 – x represents the amount invested at 18%. The profit from the account is 0.165 x , and the profit from the 18% investment is 0.18(50,000 – x ). The sum of the profits is \$8520.
The amount invested at is \$32,000, and the amount invested at 18% is 50,000 – x = 50,000 – 32,000 = \$18,000.
2. Let x represent the amount deposited at . Then the amount deposited at is 3500 – x . The interest earned at is 0.0475 x ; the interest earned at is 0.0525(3500 – x ). The sum of these two quantities is 171.75.
\$2400 was deposited in the account, and 3500 – x = 3500 – 2400 = \$1100 was deposited in the account.
3. Let x represent the amount to be loaned at 16%, so 48,000 – x represents the amount to be loaned at 19%. The total amount of return should be of 48,000 which is 0.185(48,000) = 8880.
\$8000 should be loaned at 16%, and 48,000 – x = 48,000 – 8000 = \$40,000 should be loaned at 19%.
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