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# Order of Operations- Do You Have a Genius IQ?
Contributor: Jonathan Heagy. Lesson ID: 12522
What is PEMDAS? It has to do with order of operations. No, it's not a sickness or waiting line in a hospital. It's the easy way to solve those messy equations with parentheses and exponents and stuff!
categories
## Pre-Algebra, Rules and Properties
subject
Math
learning style
Visual
personality style
Otter
Middle School (6-8)
Lesson Type
Skill Sharpener
## Lesson Plan - Get It!
Audio:
“89% of Facebook users cannot solve this problem! Do you have a genius IQ?”
You may have seen order of operations problems on your social media feed like the one below. Why don’t you give it a shot!
These problems ask seemingly simple questions that are often trickier than they look, notorious for starting comment section debates.
• So, what was your answer? Was it 56, 50, or 1?
• These are all common answers for this problem, but which is correct?
Learning how the order of operations works will help you get this type of problem correct and show off your genius IQ to all your friends!
Let’s say you’re faced with a gnarly-looking math problem:
12 x 4 ÷ (11 - 9 + 2)2 = ?
• What should you do first: the addition, subtraction, multiplication, or division?
• Then, what comes after that?
• Do the parentheses and exponent change anything?
This is where the order of operations comes in! We can remember them with the acronym PEMDAS! If you'd like, you can use fun sayings, like, "Please Excuse My Dear Aunt Sally," or "People Eat More Donuts After School" to remember PEMDAS!
Step 1: Parentheses — We always begin by solving what’s in the parentheses first.
Step 2: Exponents — Next, simplify any exponents in the equation.
Step 3: Multiplication | Division — Multiplication and division are the third step; but remember, they are equally important and should always be completed moving left-to-right in the equation.
Step 4: Addition | Subtraction — Addition and subtraction are your final step. They too are of equal importance, and like Step 3, these processes are done moving left-to-right in the equation.
Use these rules to solve that big problem from before.
12 x 4 ÷ (11 - 9 + 2)2
• We have parentheses, so solve those first! Working left to right:
• (11 - 9 + 2) = 4
12 x 4 ÷ 42
• Now, we work on exponents!
• 42 = 16
12 x 4 ÷ 16
• That's looking better; now we work from left-to-right again:
• 12 x 4 = 48
48 ÷ 16
• This is an easy one!
• 48 ÷ 16 = 3
So, 12 x 4 ÷ (11 - 9 + 2)2 = 3
Each of these problems was really found on Facebook or Instagram, so if you see them, you’ll be ready to show off your new skills!
To learn more about using PEMDAS, visit the land of Pi with TEDEd and learn How to defeat a dragon with math - Garth Sundem:
When you are done watching the video, come up with your own fun saying to help remember PEMDAS! My favorite is, "Poached Eggs May Demand Added Salt!"
• What do you think is the trickiest part of using PEMDAS?
Continue on to the Got It? section to see if you can figure out where a problem-solver goofed!
Interactive Video
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# How to Find the Perimeter With Fractions
••• Pongasn68/iStock/GettyImages
Print
The perimeter of a shape is the total distance around it. To find the perimeter, add every side of the shape to find the total. If one or more sides are fractions, you will have to follow the rules for adding fractions to add each side and find the perimeter.
## Identify All Sides
No matter what the shape is, add all sides to find the perimeter. If the shape has equal sides, there are formulas to simplify the process. To find the perimeter of an equilateral triangle, multiply the side length by 3. To find the perimeter of a square, multiply the side length by 4. If the shape is a rectangle, add the long side and the short side, and multiply that total by two: P = 2(x+y). These formulas still work with fractions. If your shape is a polygon with sides as fractions, follow the rules for adding fractions to find the perimeter.
## Find the Common Denominator
Before you can add fractions, you must find a common denominator. The common denominator will be the Least Common Multiple (LCM): the smallest number that all of your denominators will divide into evenly. For example, if you have a 4-sided polygon with sides 1/2, 1/3, 3/4 and 5/6, you will have to change all of the denominators so they are all the same. Each of these denominators can divide evenly into 12, so 12 will be your new denominator. To change the fraction, multiply the numerator and denominator by the same number to keep the value the same. Multiply 1/2 by 6/6 to get 6/12. Multiply 1/3 by 4/4 to get 4/12. Multiply 2/4 by 3/3 to get 6/12. Multiply 5/6 by 2/2 to get 10/12. Now, every denominator is the same.
## Use the Numerators
Once the denominators are the same, keep the denominator and only add the numerators. If your common denominator is 12, that will be the denominator of your answer. To add 6/12, 4/12, 6/12, and 10/12, add 6+4+6+10 and put the answer over 12. Your total, and your perimeter, will be 26/12.
If you have a shape with even sides and use a multiplication formula, only multiply the numerator. For example, to find the perimeter of a square with the formula P=4x, and your side length is 3/4, multiply 3x2 and put the product over 4. Your perimeter will be 6/4.
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# 8-3 Practice Special Right Triangles Answer Key
In the realm of geometry, special right triangles hold a unique allure with their predictable properties and ratios. In this comprehensive guide, we embark on a journey to explore the intricacies of special right triangles, delve into their distinctive features, and provide a carefully curated set of practice problems complete with detailed answer explanations.
### Exploring the Enigmatic Special Right Triangles
Special right triangles, those geometric gems with angles of 45 and 30 degrees, offer a gateway to elegant solutions and geometric insights.
### Skill 1: Understanding 45-45-90 Triangles
The 45-45-90 triangle, also known as the isosceles right triangle, exhibits remarkable symmetry and specific ratios between its sides.
### Skill 2: Embracing 30-60-90 Triangles
The 30-60-90 triangle, characterized by angles of 30, 60, and 90 degrees, unveils its own set of ratios and relationships.
### Solving Practice Problems with Special Right Triangles
Now, let’s put our knowledge to the test with practice problems:
1. In a 45-45-90 triangle, if the length of each leg is 5 units, what is the length of the hypotenuse?
Answer: The length of the hypotenuse is approximately 7.07 units.
2. In a 30-60-90 triangle, if the shorter leg measures 8 units, what is the length of the hypotenuse?
Answer: The length of the hypotenuse is 16 units.
3. If the hypotenuse of a 45-45-90 triangle is 10 units, what is the length of each leg?
Answer: The length of each leg is approximately 5.71 units.
### Mastering Special Right Triangles: Empowering Geometric Insight
As you delve into the realm of special right triangles, you gain the ability to swiftly solve complex problems and navigate geometric challenges. These skills extend beyond the realm of triangles, finding applications in trigonometry and various fields of mathematics.
In conclusion, special right triangles offer a concise and elegant approach to geometry, revealing the power of specific ratios and relationships. By mastering the properties and solving practice problems, you equip yourself to confidently tackle a variety of geometric scenarios. Armed with the practice problems and answers provided in this guide, you’re well-prepared to uncover the hidden gems within special right triangles with proficiency.
## Geometry SOL Review Packet Answer Key
Geometry SOL Review Packet Answer Key Introduction: The Virginia Department of Education requires all high school students to…
## NGPF Calculate Using a Mortgage Calculator Answer Key
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## Lesson 20 Homework 2.3 Answer Key
Blog Introduction: Welcome students! If you’re here, it’s safe to say that you are in search of a…
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# lecture01 - Lecture 1 Precalculus Review 1.1 Real Line and...
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Lecture 1 - Precalculus Review 1.1 Real Line and Order When discussing order on the real number line, we use the following symbols: < less than less than or equal to > greater than greater than or equal to We use the following notation for intervals: x ( a, b ) a < x < b open interval the open inverval ( a, b ) 3 h h a b h h The open circle at the ends of the interval indicates that the endpoint is not included.
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x [ a, b ] a x b closed interval the closed inverval [ a, b ] 3 x x a b x x x [ a, b ) a x < b half-open interval the half-open interval [ a, b ) 3 x h a b x h Examples 1. Solve - 1 < - x 3 < 2. solution: multiply by 3 - 3 < - x < 6 multiply by - 1 3 > x > - 6 ⇒ - 6 < x < 3 (Remember, when solving inequalities, if you multiply by a negative number you must flip the inequality.) Thus the set of all solutions is the open interval ( - 6 , 3). 2. Solve 2 x 2 + 1 < 9 x - 3. solution: 2 x 2 + 1 < 9 x - 3 2 x 2 - 9 x + 4 < 0 factor (2 x - 1)( x - 4) < 0 To have (2 x - 1)( x - 4) < 0, we have to have one of the factors being negative and one of the factors being positive. Thus we have the two cases:
(a) 2 x - 1 < 0 and x - 4 > 0, or (b) 2 x - 1 > 0 and x - 4 < 0 In case (a), the first inequality solves to x < 1 2 , while the second solves to
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The Antidifference Operator
# The Antidifference Operator
We have already seen with The Difference Operator that if $f$ is a real-valued function then the difference operator applied to $f$ is $\Delta f = f(x + 1) - f(x)$ which is an analogue to the derivative $\frac{df}{dx}$ of $f$.
We will now look at another calculus analogue - this time with respect to the antiderivative of a real-valued function. Recall that if $f$ is a real-valued function that $F$ is said to be an antiderivative of $f$ is $\frac{dF}{dx} = f$. We now define the antidifference of a function.
Definition Let $f$ be a real-valued function. Then if $\Delta F(x) = f(x)$ then $F$ is said to be an Antidifference of $f$ using the notation $F = \Delta^{-1} f$ where $\Delta^{-1}$ is the Antidifference Operator.
(1)
\begin{align} \quad \Delta x^2 = (x + 1)^2 - x^2 = x^2 + 2x + 1 - x^2 = 2x + 1 \end{align}
Therefore if $f(x) = 2x + 1$ then $F(x) = x^2$ is an antidifference of $f$.
Now recall that an antiderivative of a function is not unique. This is because if $F(x)$ is differentiable then so is $F(x) + C$ for any constant $C$ and so:
(2)
\begin{align} \quad \frac{d}{dx} (F(x) + C) = \frac{d}{dx} (F(x)) + \frac{d}{dx} (C) = \frac{d}{dx} (F(x)) \end{align}
The same holds for antidifferences. If $y = F(x)$ is an antidifference of $f$ then so is $y = F(x) + C$ since:
(3)
\begin{align} \quad \Delta (F(x) + C) = \Delta (F(x)) + \Delta (C) = \Delta (F(x)) + 0 = \Delta F(x) \end{align}
As a result, to denote the General antidifference of a function $f$ we usually tack on the constant $+ C$. Furthermore, the antidifference operator also satisfies the additivity property and homogeneity property and is hence linear. Right now we can formulate some antidifferences of some common functions from reversing the results we've noted on the More Properties of the Difference Operator page. They're summarized in the following theorem and can easily be verified.
Theorem 1: For $C$ as a constant, the following hold: a) $\Delta^{-1} 0 = C$. b) $\Delta^{-1} x^{\underline{k}} = \frac{1}{k+1} x^{\underline{k+1}} + C$. c) $\Delta^{-1} 2^x = 2^x + C$ d) $\Delta^{-1} \binom{x}{k} = \binom{x}{k + 1} + C$.
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# Order of Operations Practice For Your Elementary School Child
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## Order of Operations Practice
If your child is struggling with learning the order of operations, this lesson can help your child grasp the concept. This lesson provides order of operations practice that can help your child truly understand order of operations. Feel free to use these free worksheets on order of operations. Teach your child the old mantra, My Dear Aunt Sally, to remind them that it’s multiplication, division, addition, and then subtraction.
Before you get started, you need to evaluate what grade level your child is currently functioning at. If your child is at the second grade level, you will want to start giving them problems that only involve addition and subtraction. If they are at the 3rd or 4th grade level, also give them multiplication and division. Around the 5th grade level, in addition to adding, subtracting, multiplying, and dividing, you need to introduce grouping symbols such as (parentheses), {braces}, and [brackets.] If you child is in middle school you can add exponents to the mix.
One strategy that I utilize for all my students when doing order of operations is to underline the part that they are working on, solve that part, then rewrite the problem based on what is left. So, let’s take a look at how you can do a lesson on order of operations with your child.
The first thing you need to do is introduce the order of operations. Remember to use the above suggestions on what part to work on based on your child’s grade level. Note that addition and subtraction can be reversed; they are done in the order of which comes first, and the same goes for multiplication and division. Here are example problems to do based on what level your child is working at. Keep the problems simple at first, and then increase the difficulty by making the numbers larger.
5 + 3 – 2 10 – 4 + 9 20 + 2 – 7
2 – 2 6 + 9 22 – 7
0 15 15
16 ÷ 4 + 8 – 2 x 3 3 x 5 – 9 ÷ 3 + 5
4 + 8 – 2 x 3 15 – 9 ÷ 3 + 5
4 + 8 – 6 15 – 3 + 5
12 – 6 15 – 8
6 7
Grouping Symbols (in addition to the other operations)
[3 + 9] x 2 – 7 {(5 – 4) + (6 x 2)}
12 x 2 – 7 {(5 – 4) + 12}
24 – 7 {1 + 12}
17 13
Exponents (You can do grouping symbols and exponents at the same time if you want, but I suggest introducing them separately
32 x 4 + (6 – 2 + 5)
32 x 4 + (4 + 5)
32 x 4 + 9
9 x 4 + 9
36 + 9
45
Here is a great game to play to help reinforce the order of operations. Write all of the operations on an index card. Shuffle them up, then give them to your child and tell them to put them in order. Another option is to only write the symbols on the cards and have your child put them in order as well.
After you play a game, it is time to start practicing. It is easy to create problems on your own, but there are a great number of Internet resources that provide free worksheets on the order of operations, so if you don’t have time to create them, print them off the Internet.
The key to helping your child succeed with this is to practice, practice, practice. However, do not overwhelm your child. These types of problems do take a little time to complete, so I suggest doing no more than 10 problems at a time.
If you are a teacher, or if you have several children, you can create a bingo game involving the order of operations; this is always fun for children. Again, if your child needs order of operations practice, it is easy to help them at home. If you use the suggestions, methods, and tips provided here your child will have fun and learn at the same time.
Image References
Order of Operations Practice - Equation
## This post is part of the series: Making Math Fun
Learn ways to make math an exciting adventure for your students and get the engaged.
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Paul's Online Notes
Home / Calculus II / Parametric Equations and Polar Coordinates / Parametric Equations and Curves
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### Section 3-1 : Parametric Equations and Curves
1. Eliminate the parameter for the following set of parametric equations, sketch the graph of the parametric curve and give any limits that might exist on $$x$$ and $$y$$.
$x = 4 - 2t\hspace{0.5in}y = 3 + 6t - 4{t^2}$
Show All Steps Hide All Steps
Start Solution
First, we’ll eliminate the parameter from this set of parametric equations. For this particular set of parametric equations we can do that by solving the $$x$$ equation for $$t$$ and plugging that into the $$y$$ equation.
Doing that gives (we’ll leave it to you to verify all the algebra bits…),
$t = \frac{1}{2}\left( {4 - x} \right)\hspace{0.5in}\,\, \to \hspace{0.5in}\,\,\,y = 3 + 6\left[ {\frac{1}{2}\left( {4 - x} \right)} \right] - 4{\left[ {\frac{1}{2}\left( {4 - x} \right)} \right]^2} = - {x^2} + 5x - 1$ Show Step 2
Okay, from this it looks like we have a parabola that opens downward. To sketch the graph of this we’ll need the $$x$$-intercepts, $$y$$-intercept and most importantly the vertex.
For notational purposes let’s define $$f\left( x \right) = - {x^2} + 5x - 1$$.
The $$x$$-intercepts are then found by solving $$f\left( x \right) = 0$$. Doing this gives,
$- {x^2} + 5x - 1 = 0\hspace{0.5in}\,\,\,\, \to \hspace{0.5in}\,\,\,x = \frac{{ - 5 \pm \sqrt {{{\left( 5 \right)}^2} - 4\left( { - 1} \right)\left( { - 1} \right)} }}{{2\left( { - 1} \right)}} = \frac{{5 \pm \sqrt {21} }}{2} = 0.2087,\,\,\,4.7913$
The $$y$$-intercept is : $$\left( {0,f\left( 0 \right)} \right) = \left( {0, - 1} \right)$$.
Finally, the vertex is,
$\left( { - \frac{b}{{2a}},f\left( { - \frac{b}{{2a}}} \right)} \right) = \left( {\frac{{ - 5}}{{2\left( { - 1} \right)}},f\left( {\frac{5}{2}} \right)} \right) = \left( {\frac{5}{2},\frac{{21}}{4}} \right)$ Show Step 3
Before we sketch the graph of the parametric curve recall that all parametric curves have a direction of motion, i.e. the direction indicating increasing values of the parameter, $$t$$ in this case.
There are several ways to get the direction of motion for the curve. One is to plug in values of $$t$$ into the parametric equations to get some points that we can use to identify the direction of motion.
Here is a table of values for this set of parametric equations.
$$t$$ $$x$$ $$y$$
-1 6 -7
0 4 3
$$\frac{3}{4}$$ $$\frac{5}{2}$$ $$\frac{{21}}{4}$$
1 2 5
2 0 -1
3 -2 -15
Note that $$t = \frac{3}{4}$$ is the value of $$t$$ that give the vertex of the parabola and is not an obvious value of $$t$$ to use! In fact, this is a good example of why just using values of $$t$$ to sketch the graph is such a bad way of getting the sketch of a parametric curve. It is often very difficult to determine a good set of $$t$$’s to use.
For this table we first found the vertex $$t$$ by using the fact that we actually knew the coordinates of the vertex (the $$x$$-coordinate for this example was the important one) as follows,
$x = \frac{5}{2}\,\,\,\,:\,\,\,\,\,\frac{5}{2} = 4 - 2t\hspace{0.5in} \to \hspace{0.5in}\,\,t = \frac{3}{4}$
Once this value of $$t$$ was found we chose several values of $$t$$ to either side for a good representation of $$t$$ for our sketch.
Note that, for this case, we used the $$x$$-coordinates to find the value of the $$t$$ that corresponds to the vertex because this equation was a linear equation and there would be only one solution for $$t$$. Had we used the $$y$$‑coordinate we would have had to solve a quadratic (not hard to do of course) that would have resulted in two $$t$$’s. The problem is that only one $$t$$ gives the vertex for this problem and so we’d need to then check them in the $$x$$ equation to determine the correct one. So, in this case we might as well just go with the $$x$$ equation from the start.
Also note that there is an easier way (probably – it will depend on you of course) to determine direction of motion. Take a quick look at the $$x$$ equation.
$x = 4 - 2t$
Because of the minus sign in front of the $$t$$ we can see that as $$t$$ increases $$x$$ must decrease (we can verify with a quick derivative/Calculus I analysis if we want to). This means that the graph must be tracing out from right to left as the table of values above in the table also indicates.
Using a quick Calculus analysis of one, or both, of the parametric equations is often a better and easier method for determining the direction of motion for a parametric curve. For “simple” parametric equations we can often get the direction based on a quick glance at the parametric equations and it avoids having to pick “nice” values of $$t$$ for a table.
Show Step 4
We could sketch the graph at this point, but let’s first get any limits on $$x$$ and $$y$$ that might exist.
Because we have a parabola that opens downward and we’ve not restricted $$t$$’s in any way we know that we’ll get the whole parabola. This in turn means that we won’t have any limits at all on $$x$$ but $$y$$ must satisfy $$y \le \frac{{21}}{4}$$ (remember the $$y$$-coordinate of the vertex?).
So, formally here are the limits on $$x$$ and $$y$$.
$- \infty < x < \infty \hspace{0.25in}\hspace{0.25in}y \le \frac{{21}}{4}$
Note that having the limits on $$x$$ and $$y$$ will often help with the actual graphing step so it’s often best to get them prior to sketching the graph. In this case they don’t really help as we can sketch the graph of a parabola without these limits, but it’s just good habit to be in so we did them first anyway.
Show Step 5
Finally, here is a sketch of the parametric curve for this set of parametric equations.
For this sketch we included the points from our table because we had them but we won’t always include them as we are often only interested in the sketch itself and the direction of motion.
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Counting Numbers from 1 to 50
In counting numbers from 1 to 50, recognize the numbers, count and then join the numbers in the correct number order. Here we mainly need eye-hand coordination to draw the picture and maintain the number order.
We will enjoy coloring the pages after joining the numbers from 1 to 50. It should create a joyful learning environment.
Math Only Math is based on the premisethat children do not make a distinction between paly and work and learn best when learning becomes play and play becomes learning.
Let’s learn to join the lines according to the order of the numbers from 1 to 50 and then color the bird colorfully according to your choice:
Free kindergarten math printable worksheets have been graded systematically to help children progress naturally.
1. Match the number names and the numbers.
(i) Twenty-seven (a) 29 (ii) Forty (b) 50 (iii) thirty-eight (c) 40 (iv) Twenty-nine (d) 27 (v) Forty-two (e) 38 (vi) Fifty (f) 42 (vii) Thirty-five (g) 30 (viii) Twenty-two (h) 31 (ix) Thirty (i) 26 (x) Forty-nine (j) 35 (xi) Thirty-one (k) 49 (xii) Twenty-six (l) 22
1. (i) → (d)
(ii) → (c)
(iii) → (e)
(iv) → (a)
(v) → (f)
(vi) → (b)
(vii) → (j)
(viii) → (l)
(ix) → (g)
(x) → (k)
(xi) → (h)
(xii) → (i)
2. Circle the number for the number name.
(i) Forty-eight __________
(a) 28; (b) 36; (c) 48; (d) 38
(ii) Twenty-three __________
(a) 23; (b) 43; (c) 53; (d) 33
(iii) Twenty-five __________
(a) 35; (b) 25; (c) 45; (d) 15
(iv) Thirty-seven __________
(a) 27; (b) 37; (c) 47; (d) 17
(v) Thirty-nine __________
(a) 49; (b) 29; (c) 19; (d) 39
(vi) Fifty __________
(a) 20; (b) 40; (c) 30; (d) 50
(vii) Twenty __________
(a) 20; (b) 50; (c) 30; (d) 40
(viii) Forty-one __________
(a) 11; (b) 31; (c) 21; (d) 41
2. (i) (c)
(ii) → (a)
(iii) → (b)
(iv) → (b)
(v) → (d)
(vi) → (d)
(vii) → (a)
(viii) → (d)
3. Fill in the missing numbers in counting order.
3. 42, 43, 44, 45, 46, 47, 48, 49
32, 33, 34, 35, 36, 37, 38, 39
22, 23, 24, 25, 26, 27, 28, 29
12, 13, 14, 15, 16, 17, 18, 19
4. Complete backward counting.
50
___
___
47
___
___
___
43
___
___
___
___
___
___
___
35
___
___
32
___
30
___
28
___
___
___
24
___
___
___
___
19
___
___
16
___
___
___
___
11
___
___
8
___
___
5
___
3
___
___
4. 49, 48, 46, 45, 44, 42, 41, 40, 39, 38, 37, 36, 34, 33, 31, 29, 27, 26, 25, 23, 22, 21, 20, 18, 17, 15, 14, 13, 12, 10, 9, 7, 6, 4, 2, 1
5. Write the number names.
(a) 42
(b) 37
(c) 29
(d) 22
(e) 50
5. (a) Forty-two
(b) Thirty-seven
(c) Twenty-nine
(d) Twenty-two
(e) Fifty
6. Fill in the missing numbers.
(a) 20, ____, ____, ____, ____, 25, ____, ____, ____, ____, 30
(b) 31, 32, ____, ____, ____, 36, ____, ____, 39, ____
6. (a) 21, 22, 23, 24, 26, 27, 28, 29
(b) 33, 34, 35, 37, 38, 40
7. Write the following numbers in expanded form.
(a) 34 = _____ tens + _____ ones
(b) 25 = _____ tens + _____ ones
(c) 30 = _____ tens + _____ ones
(d) 35 = _____ tens + _____ ones
(e) 47 = _____ tens + _____ ones
7. (a) 3 tens + 4 ones
(b) 2 tens + 5 ones
(c) 3 tens + 0 ones
(d) 3 tens + 5 ones
(e) 4 tens + 7 ones
8. Arrange the following numbers from smallest to greatest.
(a) 37, 43, 21, 39
(b) 50, 39, 26, 13
(c) 43, 46, 24, 23
(d) 33, 43, 23, 32
8. (a) 21, 37, 39, 43
(b) 13, 26, 39, 50
(c) 23, 24, 43, 46
(d) 23, 32, 33, 43
9. Insert >, < or = sign
(a) 41 _____ 42
(b) 39 _____ 29
(c) 26 _____ 26
(d) 45 _____ 50
9. (a) <
(b) >
(c) =
(d) <
10. Circle the largest number.
(a) 31, 42, 16
(b) 27, 16, 43
(c) 49, 42, 33
(d) 43, 23, 42
10. (a) 42
(b) 43
(c) 49
(d) 43
11. Write the number that comes in between.
(a) 24 _____ 26
(b) 31 _____ 33
(d) 38 _____ 40
(c) 39 _____ 41
(e) 45 _____ 47
11. (a) 25
(b) 32
(d) 39
(c) 40
(e) 46
12. How many numbers are between 27 and 30?
12. 2
13. Tick (✔) the groups which have numbers more than 27.
(a) 17, 18, 19, 20, 21, 22, 23, 24, 25, 26
(b) 28, 29, 30, 31, 32, 33, 34, 35, 36, 37
(c) No group
(d) 37, 47, 49, 50
13. (b)
(d)
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# 2 Ways to Graph a Circle
Circles are simple to work with in pre-calculus. A circle has one center, one radius, and a whole lot of points, but you follow slightly different steps, depending on whether you are graphing a circle centered at the origin or away from the origin.
The first thing you need to know in order to graph the equation of a circle is where on a plane the center is located. The equation of a circle appears as (x – h)2 + (y – v)2 = r2. This is called the center-radius form (or standard form) because it gives you both pieces of information at the same time. The h and v represent the coordinates of the center of the circle being at the point (h, v), and r represents the radius. Specifically, h represents the horizontal displacement — how far to the left or to the right of the y-axis the center of the circle is. The variable v represents the vertical displacement — how far above or below the x-axis the center falls. From the center, you can count from the center r units (the radius) horizontally in both directions and vertically in both directions to get four different points, all equidistant from the center. Connect these four points with the best curve that you can sketch to get the graph of the circle.
## Graphing Circles Centered At the Origin
The simplest circle to graph is one whose center is at the origin (0, 0). Because both h and v are zero, they can disappear and you can simplify the standard circle equation to look like x2 + y2 = r2. For instance, to graph the circle x2 + y2 = 16, follow these steps:
1. Realize that the circle is centered at the origin (no h and v) and place this point there.
2. Calculate the radius by solving for r.
Set r2 = 16. In this case, you get r = 4.
3. Plot the radius points on the coordinate plane.
You count out 4 in every direction from the center (0, 0): left, right, up, and down.
4. Connect the dots to graph the circle using a smooth, round curve.
Graphing a circle centered at the origin.
The figure shows this circle on the plane.
## Graphing Circles Centered Away From the Origin
Although graphing circles at the origin is easiest, very few graphs are as straightforward and simple as those. In pre-calculus, you work with transforming graphs of all different shapes and sizes. Fortunately, these graphs all follow the same pattern for horizontal and vertical shifts, so you don’t have to remember many rules.
Don’t forget that the coordinates of the center of the circle are of the opposite signs of the h and v from inside the parentheses in the equation. Because the h and v are inside the grouping symbols, this means that the shift happens opposite from what you would think.
For example, follow these steps to graph the equation (x – 3)2 + (y + 1)2 = 25:
1. Locate the center of the circle from the equation (h, v).
(x – 3)2 means that the x-coordinate of the center is positive 3.
(y + 1)2 means that the y-coordinate of the center is negative 1.
Place the center of the circle at (3, –1).
2. Calculate the radius by solving for r.
Set r2 = 25 and square root both sides to get r = 5.
3. Plot the radius points on the coordinate plane.
Count 5 units up, down, left, and right from the center at (3, –1). This step gives you points at (8, –1), (–2, –1), (3, –6), and (3, 4).
4. Connect the dots to the graph of the circle with a round, smooth curve.
Graphing a circle not centered at the origin.
The figure shows a visual representation of this circle.
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# Any questions on the Section 6.2 homework?. Please CLOSE YOUR LAPTOPS, and turn off and put away your cell phones, and get out your note- taking materials.
## Presentation on theme: "Any questions on the Section 6.2 homework?. Please CLOSE YOUR LAPTOPS, and turn off and put away your cell phones, and get out your note- taking materials."— Presentation transcript:
Any questions on the Section 6.2 homework?
Please CLOSE YOUR LAPTOPS, and turn off and put away your cell phones, and get out your note- taking materials.
Section 6.3 and 6.4 Complex Fractions and Dividing Polynomials
Key Concepts from Section 6.3: Complex rational expressions (complex fractions) are rational expressions whose numerator, denominator, or both contain one or more rational expressions.
Example of a complex fraction: 10 3x 5 6x Solution: view as a division problem: 10 ÷ 5 = 10 · 6x = 4 3x 6x 3x 5
Problem from today’s homework:
Note: This one requires that you first combine the added or subtracted pairs of rational expressions into single rational expressions with a common denominator.
Section 6.4 Dividing a polynomial by a monomial Divide each term of the polynomial separately by the monomial. Example
Problem from today’s homework:
Dividing a polynomial by a polynomial other than a monomial uses a “long division” technique that is similar to the process known as long division in dividing two numbers. This process is reviewed in detail on the next slide, but first, we’ll work these two simpler examples on the whiteboard: 1). 225 ÷ 9 2). 231 ÷ 9 Question: How can you check your answers on long division problems?
Divide 43 into 72. Multiply 1 times 43. Subtract 43 from 72. Bring down 5. Divide 43 into 295. Multiply 6 times 43. Subtract 258 from 295. Bring down 6. Divide 43 into 376. Multiply 8 times 43. Subtract 344 from 376. Nothing to bring down. We then write our result as 168 Example: Long Division with integers 32 43
As you can see from the previous example, there is a pattern in the long division technique. Divide Multiply Subtract Bring down Then repeat these steps until you can’t bring down or divide any longer. We will incorporate this same repeated technique with dividing polynomials.
Now you try it (And don’t forget to check your answer!) Divide 3471 by 6 using long division. Then check your answer. Do this in your notebook now, and make sure you ask if you have questions about any step. This will be crucial to your understanding of long division of polynomials.
35 x Divide 7x into 28x 2. Multiply 4x times 7x+3. Subtract 28x 2 + 12x from 28x 2 – 23x. Bring down -15. Divide 7x into –35x. Multiply -5 times 7x+3. Subtract –35x–15 from –35x–15. Nothing to bring down. 15 So our answer is 4x – 5. Example with polynomials: Check: Multiply (7x + 3)(4x – 5) and see if you get 28x 2 – 23x - 15.
Now you try it (And don’t forget to check your answer!) Divide 6x 2 – x – 2 by 3x – 2 using long division. Then check your answer. Do this in your notebook now, and make sure to ask if you have questions about any step. ANSWER: 2x + 1
86472 2 xxx x2 x x 14 4 2 20 x 10 7020 x 78 Divide 2x into 4x 2. Multiply 2x times 2x+7. Subtract 4x 2 + 14x from 4x 2 – 6x. Bring down 8. Divide 2x into –20x. Multiply -10 times 2x+7. Subtract –20x–70 from –20x+8. Nothing to bring down. 8 )72( 78 x x2 10 We write our final answer as Example
86472 2 xxx x2 x x 14 4 2 20 x 10 7020 x 78 8 How do we check this answer? Final answer: 2x – 10 + 78. 2x - 7 How to check: Calculate (2x + 7)(2x – 10) + 78. If it comes out to 4x 2 – 6x + 8, then the answer is correct.
Now you try it (And don’t forget to check your answer!) Divide 15x 2 + 19x – 2 by 3x + 5 using long division. Then check your answer. Do this in your notebook now, and make sure you ask if you have questions about any step. Answer: 5x – 2 + 8. 3x + 5
Problem from today’s homework:
Something to be aware of on Problem #14 in today’s homework: The online “Help Me Solve This” instructions for this problem demonstrate a method called “Synthetic Division”. (We are not covering this method in class, because it is only useful in the cases where the divisor is of the form x – c.) You will get the same answer using the long division method, so this is what we expect you to use to solve this problem. (If you are already familiar with synthetic division, you’re welcome to use it in those cases where it is applicable, but we aren’t expecting you to learn it on your own.)
Reminder: This homework assignment on section 6.4 and 6.3 is due at the start of next class period.
You may now OPEN your LAPTOPS and begin working on the homework assignment.
Download ppt "Any questions on the Section 6.2 homework?. Please CLOSE YOUR LAPTOPS, and turn off and put away your cell phones, and get out your note- taking materials."
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# Find a point on the curve
Question:
Find a point on the curve $y=x^{2}+x$, where the tangent is parallel to the chord joining $(0,0)$ and $(1,2)$.
Solution:
Let:
$f(x)=x^{2}+x$
The tangent to the curve is parallel to the chord joining the points $(0,0)$ and $(1,2)$.
Assume that the chord joins the points $(a, f(a))$ and $(b, f(b))$.
$\therefore a=0, b=1$
The polynomial function is everywhere continuous and differentiable.
So, $f(x)=x^{2}+x$ is continuous on $[0,1]$ and differentiable on $(0,1)$.
Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists $c \in(0,1)$ such that $f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$
Now,
$f(x)=x^{2}+x \Rightarrow f^{\prime}(x)=2 x+1, f(1)=2, f(0)=0$
$\therefore f^{\prime}(x)=\frac{f(1)-f(0)}{1-0} \Rightarrow 2 x+1=\frac{2-0}{1-0} \Rightarrow 2 x=1 \Rightarrow x=\frac{1}{2}$
Thus, $c=\frac{1}{2} \in(0,1)$ such that $f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$.
Clearly,
$f(c)=\left(\frac{1}{2}\right)^{2}+\frac{1}{2}=\frac{3}{4}$
Thus, $(c, f(c))$, i.e. $\left(\frac{1}{2}, \frac{3}{4}\right)$, is a point on the given curve where the tangent is parallel to the chord joining the points $(4,0)$ and $(5,1)$.
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Categories Guide
## FAQ: What type of transformation is a reduction?
A reduction (think shrinking) is a dilation that creates a smaller image, and an enlargement (think stretch) is a dilation that creates a larger image. If the scale factor is between 0 and 1 the image is a reduction.
## What is reduction transformation?
Here k is known as the reduction factor and P is known as the centre of reduction. Therefore, in a size transformation, a given figure undergoes enlargement or reduction by a scale factor k, such that the resulting figure is similar to the original figure, i.e., the image retains the shape of the original object.
## What are the four types of transformation?
There are four main types of transformations: translation, rotation, reflection and dilation.
## What are the 4 basic transformation?
The four main types of transformations are translations, reflections, rotations, and scaling.
## What are the 5 types of transformation?
Types of transformations:
• Translation happens when we move the image without changing anything in it.
• Rotation is when we rotate the image by a certain degree.
• Reflection is when we flip the image along a line (the mirror line).
• Dilation is when the size of an image is increased or decreased without changing its shape.
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## What is reduction and enlargement?
A reduction (think shrinking) is a dilation that creates a smaller image, and an enlargement (think stretch) is a dilation that creates a larger image. If the scale factor is greater than 1, the image is an enlargement.
## What is reduction factor in math?
In mathematics, reduction refers to the rewriting of an expression into a simpler form. Rewriting a radical (or “root”) expression with the smallest possible whole number under the radical symbol is called “reducing a radical”.
## What are the types of transformation *?
There are four types of transformation namely rotation, reflection, dilation, and translation In this article, we will discuss how to do transformation of a function, function graph transformation, and how to graph transformation function.
## What type of transformation is a translation?
A translation (or “slide”) is one type of transformation. In a translation, each point in a figure moves the same distance in the same direction. Example: If each point in a square moves 5 units to the right and 8 units down, then that is a translation!
## What is transformation what are its types?
Introduction of Transformation Translation Scaling Rotation Reflection Shearing Matrix Representation Homogeneous Coordinates Composite Transformation Pivot Point Rotation.
## Which transformations are Nonrigid transformations?
Translation and Reflection transformations are nonrigid transformations.
## What type of transformation is described by XY (- XY?
Reflection across the x -axis.
## What are the basic transformations?
There are three basic rigid transformations: reflections, rotations, and translations. There is a fourth common transformation called dilation.
## What does scaling transformation do?
A scaling transformation alters size of an object. In the scaling process, we either compress or expand the dimension of the object. Scaling operation can be achieved by multiplying each vertex coordinate (x, y) of the polygon by scaling factor sx and sy to produce the transformed coordinates as (x’, y’).
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## Is reflection rigid or Nonrigid?
Whether that be translation, rotation, or reflection, you are not changing the shape’s original form in any way, you are just changing its position in space. Non-Rigid Transformations actually change the structure of our original object. For example, it can make our object bigger or smaller using scaling.
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# 1995 AJHSME Problems/Problem 3
## Problem
Which of the following operations has the same effect on a number as multiplying by $\dfrac{3}{4}$ and then dividing by $\dfrac{3}{5}$?
$\text{(A)}\ \text{dividing by }\dfrac{4}{3} \qquad \text{(B)}\ \text{dividing by }\dfrac{9}{20} \qquad \text{(C)}\ \text{multiplying by }\dfrac{9}{20} \qquad \text{(D)}\ \text{dividing by }\dfrac{5}{4} \qquad \text{(E)}\ \text{multiplying by }\dfrac{5}{4}$
## Solution
Dividing by $\frac{3}{5}$ is the same as multiplying by its reciprocal, $\frac{5}{3}$.
Multiplying a number by two numbers is the same as multiplying the number by the product of those two numbers. Thus, multiplying by $\frac{3}{4}$ and then $\frac{5}{3}$ is the same as multiplying by $\frac{3\cdot 5}{4\cdot 3} = \frac{5}{4}$, and the answer is $\boxed{E}$.
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# Calculus Kinematics
Kinematics is a topic in physics that describes the motion of points, bodies and systems in space. Calculus kinematics can be used to derive equations for velocity and acceleration using derivatives and their integrals.
## What are the kinematic equations?
Kinematics is, broadly, the study of motion. Kinematic equations relate displacement, velocity and acceleration of a body through derivatives and integrals.
### Displacement
The displacement of a particle simply shows how far a point has moved in space relative to a fixed origin point. This quantity will be referred to as x and is a vector, rather than a scalar since it takes into consideration the direction in which the particle moves, as well as the magnitude (or size) of that change in position.
Considering a particle is in motion along a straight line,
If x > 0, the particle is to the right of the origin.
If x < 0, the particle is to the left of the origin.
When a particle changes direction during its movement, a motion diagram along a number line can be sketched, in order to understand the start point, turning point (s) and end point of the particle in space.
The difference between displacement and distance is that distance is irrespective of direction (it is a scalar quantity), while displacement considers the position of a particle respective to the origin point of the movement, hence also considering direction (it is a vector quantity).
An object travels with displacement function meters, where t > 0 seconds.
(a) What is the initial displacement of the object?
(b) When does the object change direction?
(a) The initial displacement signifies that time is 0. Hence, we substitute 0 in for t in the equation for the displacement.
The initial displacement is 1 metre (to the right of the origin).
(b) The object changes direction when the value for x reaches its maximum, as any other point after this is closer to the origin signifying a change of direction.
Maximum displacement is reached at the vertex of the quadratic equation.
Vertex occurs when, so when
Hence, the object changes direction at t = 0.35 seconds.
### Velocity
Velocity describes how fast a point is moving in a certain direction. In other words, it is how the displacement of the particle changes with time or the rate of change of displacement of the particle. Velocity will be referred to as v and
, where t is the time and x is the displacement achieved by the particle in that specific amount of time. This means that if we have an expression for x in terms of t, we can take the derivative of this expression to find the velocity.
The unit for this quantity, if the displacement is in meters and the time in seconds, is meters per second, or m/s.
Considering a particle:
If v > 0, the particle is moving to the right.
If v < 0, the particle is moving to the left.
If v = 0, the particle is stationary. If there is a change in the sign of v at this point, the particle has changed direction.
If the question asks for a value of speed, instead of velocity, it is important to note that speed, unlike velocity, does not consider the direction of movement. In other words, speed is the change of distance over time and velocity is the change of displacement over time. For example, if you were walking along the walls of a square room with a perimeter of 12 m in 36 seconds, and ended back at the point you started, the total distance travelled would be 12 m but the total displacement would be 0 m, as there there is no change between the start and end positions. Therefore, your speed would be and your velocity would be.
The displacement in meters of a car moving between two points A and B is given by. Find an expression for the velocity of the car at a given point in time.
We know that so we can differentiate the expression above with respect to t to find v. This gives:
So, to go from displacement to velocity we need to differentiate, but how do we go from velocity to displacement? You might recall that integration is the reverse process of differentiation, so we will have to integrate our expression for velocity with respect to time to find the displacement.
A marathon runner is moving with a constant velocity of. What is the displacement of the runner in terms of t?
Let the displacement of the runner be x.
As, we will have to integrate the given expression in terms of t to find x. Therefore,
where c is the integration constant.
Sometimes, the integration constant c needs to be found. In this case, the question will need to provide a value for the displacement at a particular time. This value is often the initial displacement when t = 0. This value can then be substituted into the equation to solve for the unknown c, the integration constant.
### Acceleration
Acceleration describes how much faster or slower a particle becomes over time. In other words, acceleration is the change in the velocity of a particle with time, or the rate of change of this velocity. Acceleration will be denoted with the letter a.
But we know already that, so we can therefore place an expression for a in terms of x:. This means that to find the acceleration, we will need to differentiate the displacement twice with respect to t.
The units for this quantity, if the displacement is in meters and the time in seconds, is metres per second² or m/s²
The displacement at time t of a bird is given bym. What are the velocity and acceleration of the bird?
We know thatso. Now we can find the acceleration by finding the derivative of this expression. Therefore:
Like before, we can find the velocity of a particle by integrating its acceleration with respect to time.
To find x, we will need to integrate a twice with respect to t. This can be represented with a double integral.
The acceleration of a particle is given by. Find the velocity and displacement of this particle in terms of t.
For the velocity, we will need to integrate this expression once and for the displacement, twice.
, where c and d are both constants.
To determine whether the speed of a particle is increasing or decreasing, the signs of both velocity and acceleration must be taken into account.
• If velocity and acceleration have the same sign (both positive or both negative), the speed of the particle is increasing.
• If velocity and acceleration have opposite signs (one is positive and the other negative), the speed of the particle is decreasing.
Therefore, it is possible to convert between displacement, velocity, and acceleration using calculus.
## Calculus Kinematics - Key takeaways
• The displacement of a particle is the position of that particle relative to an origin point. It is denoted as x, and is a vector quantity, indicating that direction is significant.
• The velocity of a particle is the rate of change of its displacement with respect to time. It is denoted as v and can be obtained by differentiating x with respect to time t. v can be integrated to calculate the displacement x.
• The acceleration of a particle is the rate of change of its velocity with respect to time. It is denoted as a and can be obtained by differentiating x with respect to time t twice, or by differentiating v with respect to time t once. Similarly, a can be integrated once or twice to calculate velocity v, and displacement t, respectively.
• The speed of a particle is the magnitude of its velocity and is a scalar quantity. It is calculated either by taking the modulus of the velocity, or the derivative of the distance (rather than displacement for velocity).
Calculus can be used to derive expressions for displacement, velocity and acceleration by using derivatives and integrals.
Calculus kinematics is a field of mathematics that links the topic of calculus (derivatives and integrals) with the topic of kinematics in physics. This allows us to mathematically derive expressions for displacement, velocity and acceleration.
## Final Calculus Kinematics Quiz
Question
How do you go from velocity to displacement?
You integrate with respect to t
Show question
Question
How do you go from velocity to acceleration?
You differentiate in terms of t
Show question
Question
How to you go from displacement to acceleration?
You take the second derivative with respect to t
Show question
Question
How do you go from acceleration to displacement?
You integrate with respect to t twice
Show question
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How To Divide A Fraction On The Whole | The Science
The Science
# How to divide a fraction on the whole
In general, the algorithm for division of fractions is as follows: first, the one shot, which is a divisor is replaced by its inverse fraction (numerator and denominator are reversed). Next, the multiplication of two fractions, and then the result is simplified. Dividing common fractions should be an integer, this number should be written in the form of common fractions with the same denominator, and then divide these two common fraction by the usual algorithm.
## Instruction on how to divide the whole fraction
Step 1:
Give divider (integer) to the same form in which the dividend is recorded (fraction). The denominator of the divider put the same number, which is used in the denominator of the dividend. A numerator denominator should be multiplied by this same integer. For example, if the fraction to be divided by the number of 8/15 3, it is necessary to convert the number to a fraction, in which the denominator is 15 and the numerator 15 * 3 = 45, ie 45/15. Now, the initial problem is reduced to the division by a fraction of the fraction 8/15 45/15.
Step 2:
Multiply the dividend (8/15) by a fraction, the inverse divider, ie 15/45. Since the first shot of the 15 is in the denominator, while the second - in the numerator, they can be reduced to 1. As a result, the initial problem is reduced to the multiplication of the fraction 8/1 by the fraction 1/45.
Step 3:
Multiply the numerator (1 * 8 = 8) and denominators (1 * 45 = 45). So you get a result that can be written in the form of a common fraction 8/45.
Step 4:
Divide the numerator results in its denominator, if the solution of the problem should be presented not in the form of an ordinary fraction and as a decimal. Split can be in a column, or just use a calculator. If you have internet access, you can, for example, use the calculator's built-in Google search engine. To do this, go to the search engine site and type in the keyword "8 divided into 45" or "8/45". Press the button to send the request to the server is not necessarily the answer up to nine digits, you will see immediately.
Step 5:
If only the result is important in decimal form, and the decision process is not important, it is possible to entrust the necessary conversion of fractions and mathematical operations with them Google calculator. You only need to define and enter a query in the search box. For example, for the problem, which is used as an example in the previous steps, the formulation of the search query should be: "8/15 divided by 3". And the result is that Google will show calculator is equal to 0.177777778.
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The percentage change calculator is a valuable tool for various fields such as finance, chemistry, and mathematics, providing insights into changes between two values. Understanding the percent change formula is crucial for accurate calculations. The percent change differs from percent increase and percent decrease, capturing both directions of change. It is applicable in scenarios like population growth rate calculations and expressing relative errors.
## Calculating Percentage Change:
To calculate percentage change:
1. Find the difference between the initial and final values.
2. Divide by the absolute value of the initial value.
3. Multiply the result by 100.
Formula: Percent Change=100×(Final−Initial∣Initial∣)Percent Change=100×(∣Initial∣Final−Initial)
Ensure the correct handling of negative values to avoid errors.
### Examples:
Example 1: Given values 60 (initial) and 72 (final): Percent Change=72−60∣60∣×100=20%Percent Change=∣60∣72−60×100=20%
Example 2: Values 50 (initial) and -22 (final): Percent Change=−22−50∣50∣×100=−144%Percent Change=∣50∣−22−50×100=−144%
## Handling Negative Numbers:
When dealing with negative numbers: Percent Change=New−Old∣Old∣×100Percent Change=∣Old∣New−Old×100
Example: Calculate percentage change from -20 to -30: Percent Change=−30−(−20)∣−20∣×100=−50%Percent Change=∣−20∣−30−(−20)×100=−50%
## Population Growth Rate Formula:
Population growth rate measures changes in population over time:
Population Growth Rate=Population GrowthPrevious Population×100Population Growth Rate=Previous PopulationPopulation Growth×100
Example: In the U.S., population grew from 253,339,000 in 1990 to 310,384,000 in 2010: Population Growth Rate=310,384,000−253,339,000253,339,000×100=22.5%Population Growth Rate=253,339,000310,384,000−253,339,000×100=22.5%
The percentage change calculator is not only valuable in academic settings but also practical in everyday situations, such as assessing sales tax or negotiating salaries. Understanding how to calculate percent change manually ensures real-world applicability.
## FAQs:
1. Is percentage difference equal to percentage change?
• No, they are distinct. Percentage change can be positive or negative, while percentage difference is always positive.
2. What is the percentage change from 5 to 20?
• 20 is a 300% increase of 5.
3. What is the percentage change from 20 to 10?
• 10 is a 50% decrease of 20.
4. What is the percentage change from 2 to 3?
• 3 is a 50% increase of 2.
5. What is the percentage change from 5 to 4?
• 4 is a 20% decrease of 5.
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# Secant ABC intersects a circle at A and B. Chord $\overline{BD}$ is drawn. Show that $m\angle CBD\ne \dfrac{1}{2}m\widehat{BD}$.
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Hint: We produce BO, DO, AD. We recall the definition of a central angle and have $m\widehat{BD}=m\angle BOD$. We use the relationship between the inscribed angle by an arc and central angle $m\angle BAD=\dfrac{1}{2}m\angle BOD$. We use the external angle theorem in triangle ABD and use obtained relations to prove the given statement $m\angle CBD\ne m\widehat{BD}$
Complete step-by-step solution
We observe the given figure in the question. We have the straight-line $ABC$ which cuts the circle with center O at points A and B. The chord BD has been drawn. We produce BO. DO and AD.
We know the measure of the chord is equal to the angle it subtends at the center with its two radii which is known as the central angle. Here the chord BD makes the angle $\angle BOD$ with the radii BO and CO at the center O.So we have,
$m\widehat{BD}=m\angle BOD...(1)$
We also know that the inscribed angle by an arc is the angle the endpoints of the arc subtended at any point on the circle excluding the arc as the vertex. Here the arc BD has endpoints B and D. A is a point not on the arc. The arc BD subtends the inscribed angle $\angle BAD$ at A. We know from the theorem that the inscribed angle is half of the central angle. So we have,
$m\angle BAD=\dfrac{1}{2}m\angle BOD...(2)$
We know from the theorem of external angles that the external angle of the triangle is equal to the sum of two remote interior angles. We observe the triangle ABD where the external angle is $\angle CBD$ and its remote internal angles $\angle BAD,\angle ADB$. So we have
\begin{align} & m\angle CBD=m\angle BAD+m\angle ABD \\ & \Rightarrow m\angle CBD=\dfrac{1}{2}m\angle BOD+m\angle ABD\left( \text{from}\left( 2 \right) \right) \\ & \Rightarrow m\angle CBD=\dfrac{1}{2}m\widehat{BD}+m\angle ABD\left( \text{from}\left( 1 \right) \right) \\ & \Rightarrow m\angle CBD>\dfrac{1}{2}m\widehat{BD} \\ \end{align}
So hence it is proved that $m\angle CBD\ne \dfrac{1}{2}m\widehat{BD}$
Note: We note that the inscribed angle by a semicircle is a right angle. We can use the relation between the central angle and the inscribed angle to prove that the opposite angles of a cyclic quadrilateral are supplementary.
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# Recursive Sequences Proofs
Back to Recursive Sequences
### Common properties
Proofs:
• The number of words of length n in alphabet {1,2,...,d} avoiding words "11", "22", ..., "kk" is the recurrence sequence with initial terms a(0) = 1, a(1) = d and the recurrence relation a(n) = (d-1)a(n-1) + (d-k)a(n-2).
Proof: Let the words described above be divided into two parts: ending in {1,2,...,k} and ending in {k+1,...,d}. Let b(n) be the number of words of length n ending in {1,2,...,k} and c(n) be the number of words of length n ending in {k+1,...,d}. Obviously, a(n) = b(n) + c(n). We can get any word of length n ending in {k+1,...,d} by taking any word of length n-1 and adding one of the letters {k+1,...,d}. Thus c(n) = (d-k)a(n-1). We can get any word of length n ending in {1,2,...,k} by taking any word of length n-1 and adding any one of the letters {1,2,...,k} except we should subtract the words that we get that are ending in {11,22,...,kk}. Hence b(n) = k*a(n-1) - b(n-1). Putting the counts together we get a(n) = (d-1)a(n-1) + c(n-1) = (d-1)a(n-1) + (d-k)a(n-2). Initial conditions are trivially checked.
• Let p and q be the roots of the equation x2 - dx - k = 0. Then the sequence a(n) = pn + qn satisfies the recurrence a(n) = da(n-1) + ka(n-2) with the initial conditions a(0) = 2, a(1) = d.
Proof: pn + qn = (p+q)(pn-1 + qn-1) - pq(pn-2 + qn-2).
• Let p and q be the roots of the equation x2 - dx - k = 0. Then the sequence a(n) = (pn - qn)/(p-q) satisfies the recurrence a(n) = da(n-1) + ka(n-2) with the initial conditions a(0) = 0, a(1) = 1.
Proof: pn - qn = (p+q)(pn-1 - qn-1) - pq(pn-2 - qn-2).
### a(n) = d * a(n-1) - a(n-2)
Proofs:
• The number of 01-avoiding words of length n on alphabet {0,1,2, ..., d-1} that do not end in 0 is equal to a(n), where a(n) = d * a(n-1) - a(n-2) and a(0) = 1, a(1) = d-1.
Proof: It is easy to see that you can get a word that ends in 1 from any valid word of length n-1. That is, the number of words of length n that end in 1 is a(n-1). Similarly, suppose that a word ends in b ≥ 1, which has exactly k zeroes before the last digit b. The number of such words is a(n-k-1). Hence, the total number of words ending in b is a(n-1) + a(n-2) + ... + a(1) + a(0). Summing up, the result is a(n) = (d-1) * a(n-1) + (d-2) * (a(n-2) + a(n-3) + ... + a(1) + a(0)). The equivalent of this to the initial recursion is proved in the corresponding bullet below.
• The number of domino tilings of Sd-1 × P2n (product of a star graph and a path graph) is equal to a(n), where a(n) = d * a(n-1) - a(n-2) and a(0) = 1, a(1) = d-1. Also the number of domino tilings in Sd-1 × P2n+1 (product of a star graph and a path graph) with d-3 non-central vertices removed from the last star is equal to a(n), where a(n) = d * a(n-1) - a(n-2) and a(0) = 1, a(1) = d.
Proof: Let us first prove the first statement. Let us start with noticing that if we want to tile a star graph with dominoes we can only place one domino and it must include the center of the star. Let us associate the center of the star graph with 0 and associate other vertices with numbers from 2 to d-1. Let us associate the domino tilings of Sd-1 × P2 with a number. If all dominoes are parallel to the path P2, then we associate this tiling with one. Otherwise exactly two dominoes are not parallel to P2 and they connect the center of the star with the vertex number k (k ≥ 2) and we associate this tiling with k. Now we would like to correspond to a number a domino tiling of Sd-1 × P2m which cannot be decomposed into two smaller domino tilings of that type. It is easy to see that in this tiling there are only two dominoes that are not parallel to P2m: one domino in the first star and one domino in the last star. They both have to connect the center of the star with the same numbered vertex. Suppose this number is k. Then we associate this tiling with the word "000...000k", where we have m-1 zeroes. If the tiling can be decomposed into the minimal ones, then we concatenate all the words corresponding to minimal tilings together. This way we create a correspondence between domino tilings of Sd-1 × P2n and the number of 01-avoiding words of length n on alphabet {0,1,2, ..., d-1} that do not end in 0.
The second statement can be proved in a similar manner as the first statement.
• If a(n) = d * a(n-1) - a(n-2), then a(n+1)a(n-1) - a(n)2 is a constant. For the sequence that starts as a(0) = 1 and a(1) = d-1, this constant is equal to d-2. For the sequence that starts as a(0) = 1 and a(1) = d, this constant is equal to -1.
Proof: By applying the recursion equation for a(n+2) to the difference of expressions a(n+1)a(n-1) - a(n)2 and the same expression with n replaced by n-1 we get: a(n+2)a(n) - a(n+1)2 - a(n+1)a(n-1) + a(n)2 = da(n+1)a(n) - a(n)2 - a(n+1)2 - a(n+1)a(n-1) + a(n)2 = a(n+1) (da(n) - a(n+1) - a(n-1)) = 0. Hence, a(n+2)a(n) - a(n+1) doesn't depend on n and is equal to a(0)a(2) - a(1)2.
• If a(n) = d * a(n-1) - a(n-2), then a(n+2)a(n-1) - a(n)a(n+1) is a constant. For the sequence that starts as a(0) = 1 and a(1) = d-1, this constant is equal to d(d-2). For the sequence that starts as a(0) = 1 and a(1) = d, this constant is equal to -d.
Proof: By applying the recursion equation to a(n+2) and expressing a(n-1) through a(n+1) and a(n) we get: a(n+2)a(n-1) - a(n)a(n+1) = (d * a(n+1) - a(n)) (d * a(n) - a(n+1)) - a(n)a(n+1) = d2 * a(n+1)a(n) - d * a(n+1)2 - d * a(n)2 = d * (a(n+1)(d * a(n) - a(n+1)) - a(n)2) = d * (a(n+1)a(n-1) - a(n)2). The expression a(n+1)a(n-1) - a(n)2 is a constant by another property. Hence a(n+2)a(n-1) - a(n)a(n+1) is constant. For the initial conditions a(0) = 1 and a(1) = d-1 this constant is equal to d(d-2). For the initial conditions a(0) = 1 and a(1) = d this constant is equal to -d.
• If a(n) = d * a(n-1) - a(n-2) with initial terms a(0) = 1, a(1) = d-1, then a(n) = (d-1) * a(n-1) + (d-2) * (a(n-2) + a(n-3) + ... + a(1) + a(0)).
Proof: Let us prove this by induction. The first step is trivial as a(1) = (d-1) * a(0). Assume that we proved that a(n-1) = (d-1) * a(n-2) + (d-2) * (a(n-3) + a(n-4) + ... + a(1) + a(0)). Adding this to the recurrence relation a(n) = d * a(n-1) - a(n-2) we get a(n) = (d-1) * a(n-1) + (d-2) * (a(n-2) + a(n-3) + ... + a(1) + a(0)).
• The number of 01-avoiding words of length n on alphabet {0,1,2, ..., d-1} is equal to a(n), where a(n) = d * a(n-1) - a(n-2) and a(0) = 1, a(1) = d.
Proof: Let us denote b(n) (correspondingly c(n)) as the number of words in this sequence of length n that end (correspondingly do not end) in 0. Hence a(n) = b(n) + c(n). It is easy to see that b(n) = a(n-1), as we can get a valid word ending in 0 by attaching 0 to any valid word. We can get a word not ending in 0 by attaching any digit other than 0 and 1 to any valid word, or by attaching 1 to any valid word not ending in 0. Hence c(n) = (d-2)*a(n-1) + c(n-1). Now we can replace c(n-1) as a(n-1) - b(n-1), which gives us c(n) = (d-1)*a(n-1) - b(n-1) = (d-1) * a(n-1) - a(n-2). Summing b and c we get a(n) = d*a(n-1) - a(n-2).
• Let p and q be the roots of the equation x2 - dx + 1 = 0. Then the sequence a(n) = pn + qn satisfies the recurrence a(n) = da(n-1) - a(n-2) with the initial conditions a(0) = 2, a(1) = d.
Proof: Set k equal to -1 in the general proof.
• Let p and q be the roots of the equation x2 - dx + 1 = 0. Then the sequence a(n) = pn + qn approaches Round(pn).
Proof: As p*q equals 1, then pn tends to 0.
### a(n) = d * a(n-1) + d * a(n-2)
Proofs:
• Let p and q be the roots of the equation x2 - dx - d = 0. Then the sequence a(n) = pn + qn satisfies the recurrence a(n) = da(n-1) + da(n-2) with the initial conditions a(0) = 2, a(1) = d.
Proof: Set k equal to d in the general proof.
• a(n) with initial terms a(0) = 1 and a(1) = d+1 and the recurrence relation a(n) = da(n-1) + da(n-2) equals the number of 00-avoiding words of length n on alphabet {0,1,2, ..., d}.
Proof: Set k equal to 1 in the general proof above and adjust it slightly.
### a(n) = d * a(n-1) + a(n-2)
Proofs:
• a(n) with initial terms a(0) = 1 and a(1) = d+1 and the recurrence relation a(n) = da(n-1) + a(n-2) equals the number of words in the alphabet {0, 1, 2, ..., d} avoiding "11", "22", ..., "dd".
Proof: The proof is similar to the general proof.
• Let p and q be the roots of the equation x2 - dx - 1 = 0. Then the sequence a(n) = pn + qn satisfies the recurrence a(n) = da(n-1) + a(n-2) with the initial conditions a(0) = 2, a(1) = d.
Proof: Set k equal to 1 in the general proof.
• Let d = 2k, then the sequence a(n) with the recurrence relation a(n) = 2ka(n-1) + a(n-2) and the initial terms a(1) = k, a(2) = 2k2+1 are numerators of continued fraction converging to the square root of k2+1. The sequence of denominators follows the same recurrence with the initial condition correspondingly a(1) = 1, and a(2) = 2k. (For example, the first two continued fraction approximations are: k and (2k2+1)/2k.
Proof: Let x be the square root of k2+1; then x = k + 1/z, where 1/z is less than 1. It is easy to see that z = x+k, meaning that z = 2k + 1/z. Hence the continued fractions for z is [2k; 2k, 2k, 2k, ...].
Let c(n) and d(n) be the numerators and denominators of continued fractions converging to z. We can see that c(n)/d(n) = 1 + 1/(c(n-1)/d(n-1)). Hence c(n) = 2kc(n-1) + d(n-1) and d(n) = c(n-1). Hence the numerators and denominators of continued fractions converging to z form the same sequence (shifted by one index relative each other) with the recurrence relation a(n) = da(n-1) + a(n-2) and the initial terms a(0) = 1 and a(1) = 2k. Denominators for x are the same as for z and numerators for x are linear combinations of numerators and denominators for z (namely c(n) - kd(n)). The rest of the proof follows from an easy calculation.
Last revised December 2007
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Basic Statistics for Dummies 1
statistics is a branch of mathematics that deals with the collection, analysis, interpretation, and presentation of masses of numerical data to make inference and reach decisions in the face of uncertainty in economics , agriculture, education e.t.c
statistics can be subdivided into descriptive statistics and inferential statistics
Descriptive Statistics
Descriptive are brief descriptive coefficients that summarize and describe a given data set, The most recognized types of descriptive statistics are measures of center: the , , and , which are used at almost all levels of math and statistics.
Mean: The “average” number; found by adding all data points and dividing by the number of data points. (∑x)/N
Here,
∑ represents the summation
X represents observations
N represents the number of observations .
Example: The mean of 3, 2, and 7 is (3+2+7)/3 = 12/3 = 4
Median: The middle number; found by ordering all data points and picking out the one in the middle (or if there are two middle numbers, taking the mean of those two numbers).
Example: The median of 2, 1, and 5 is 2 because when the numbers are put in order (1, 2, 5), the number 2 is in the middle
Mode: The most frequent number — that is, the number that occurs the highest number of times.
Example: The mode of {4, 2, 4, 3, 3, 3}, is 3because it occurs three times, which is more than any other number.
inferential statistics
Inferential statistics allows you to make predictions (“inferences”) from that data. With inferential statistics, you take data from and make generalizations about a .
Population: is the entire pool from which a statistical sample is drawn. A population may refer to an entire group of people, objects, events, hospital visits, or measurements.
Sample: is a set of element selected from a given population, Samples are also used in statistical testing when population sizes are too large for the test to include all possible members or observations.
Important Terms in Statistics
Average
also called mean; a number that describes the central tendency of the data
Parameter
a number that is used to represent a population characteristic and that generally cannot be determined easily
Population
all individuals, objects, or measurements whose properties are being studied
Probability
a number between zero and one, inclusive, that gives the likelihood that a specific event will occur
Proportion
the number of successes divided by the total number in the sample
Sample
a subset of the population studied
Variable
a characteristic of interest for each person or object in a population
Data
a set of observations (a set of possible outcomes); most data can be put into two groups: qualitative (an attribute whose value is indicated by a label) or quantitative (an attribute whose value is indicated by a number). Quantitative data can be separated into two subgroups: discrete and continuous. Data is discrete if it is the result of counting (such as the number of students of a given ethnic group in a class or the number of books on a shelf). Data is continuous if it is the result of measuring (such as distance traveled or weight of luggage)
Quantitative data
Quantitative data is defined as the value of data in the form of counts or numbers where each data-set has an unique numerical value associated with it, Quantitative Information — Involves a measurable quantity — numbers are used. Some examples are length, mass, temperature, and time.
Qualitative data
Qualitative data is non-numerical data that is observed, descriptive, and subjective. Examples of qualitative data include sex (male or female), name, state of origin, citizenship, etc. A more practical example is a case whereby a teacher gives the whole class an essay that was assessed by giving comments on spelling, grammar, and punctuation rather than score.
Types of Quantitative data
(1) DISCRETE DATA
(2)CONTINUOUS DATA
DISCRETE DATA
Discrete data is information that can only take certain values (countable or finite) examples :
• The number of students in a class.
• The number of workers in a company.
• The number of parts damaged during transportation.
• Shoe sizes.
• Number of languages an individual speaks.
• The number of test questions you answered correctly.
CONTINUOUS DATA
Continuous data are data that can take any value (within a range).
Example: People’s heights could be any value (within the range of human heights), not just certain fixed heights.
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3.4 Composition of functions (Page 2/9)
Page 2 / 9
Performing algebraic operations on functions
Find and simplify the functions $\text{\hspace{0.17em}}\left(g-f\right)\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(\frac{g}{f}\right)\left(x\right),\text{\hspace{0.17em}}$ given $\text{\hspace{0.17em}}f\left(x\right)=x-1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)={x}^{2}-1.\text{\hspace{0.17em}}$ Are they the same function?
Begin by writing the general form, and then substitute the given functions.
No, the functions are not the same.
Note: For $\text{\hspace{0.17em}}\left(\frac{g}{f}\right)\left(x\right),\text{\hspace{0.17em}}$ the condition $\text{\hspace{0.17em}}x\ne 1\text{\hspace{0.17em}}$ is necessary because when $\text{\hspace{0.17em}}x=1,\text{\hspace{0.17em}}$ the denominator is equal to 0, which makes the function undefined.
Find and simplify the functions $\text{\hspace{0.17em}}\left(fg\right)\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(f-g\right)\left(x\right).$
Are they the same function?
$\begin{array}{l}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)=\left(x-1\right)\left({x}^{2}-1\right)={x}^{3}-{x}^{2}-x+1\\ \left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)=\left(x-1\right)-\left({x}^{2}-1\right)=x-{x}^{2}\end{array}$
No, the functions are not the same.
Create a function by composition of functions
Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions . The resulting function is known as a composite function . We represent this combination by the following notation:
$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
We read the left-hand side as $“f\text{\hspace{0.17em}}$ composed with $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ at $\text{\hspace{0.17em}}x,”$ and the right-hand side as $“f\text{\hspace{0.17em}}$ of $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ of $\text{\hspace{0.17em}}x.”$ The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol $\text{\hspace{0.17em}}\circ \text{\hspace{0.17em}}$ is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases $f\left(g\left(x\right)\right)\ne f\left(x\right)g\left(x\right).$
It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ takes the input $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ first and yields an output $\text{\hspace{0.17em}}g\left(x\right).\text{\hspace{0.17em}}$ Then the function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ takes $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ as an input and yields an output $\text{\hspace{0.17em}}f\left(g\left(x\right)\right).$
In general, $\text{\hspace{0.17em}}f\circ g\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\circ f\text{\hspace{0.17em}}$ are different functions. In other words, in many cases $\text{\hspace{0.17em}}f\left(g\left(x\right)\right)\ne g\left(f\left(x\right)\right)\text{\hspace{0.17em}}$ for all $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ We will also see that sometimes two functions can be composed only in one specific order.
For example, if $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)=x+2,$ then
$\begin{array}{ccc}\hfill f\left(g\left(x\right)\right)& =& f\left(x+2\right)\hfill \\ & =& {\left(x+2\right)}^{2}\hfill \\ & =& {x}^{2}+4x+4\hfill \end{array}$
but
$\begin{array}{ccc}\hfill g\left(f\left(x\right)\right)& =& g\left({x}^{2}\right)\hfill \\ & =& {x}^{2}+2\hfill \end{array}$
These expressions are not equal for all values of $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value $\text{\hspace{0.17em}}x=-\frac{1}{2}.$
Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
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Teaching mathematics
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# 1.2 Expressions and terms
An algebraic expression is a calculation, written in standard arithmetic notation, where one or more numbers has been replaced with a symbol. If an expression involves addition or subtraction then the parts that are added or subtracted are called ‘terms’. The video below shows the parts of an equation which are algebraic expressions and then shows how the expressions are made up of algebraic terms.
Skip transcript: Video 1 Expressions and terms
#### Transcript: Video 1 Expressions and terms
INSTRUCTOR:
An algebraic expression is just a calculation, but one or more of the letters has been replaced with a symbol. Here we've got an equation that comes from the perimeters problem. P, the perimeter, is found by calculating the sum of the five side lengths-- 3, b, the short side length, a minus 3, b, and a. So our equation says that the perimeter is equal to that sum.
The equation is made out of two expressions. One of them is just a single P for perimeter. And that is equal in value to another expression, which is 3 plus b plus a minus 3 plus b plus a.
If an expression involves addition or subtraction, then the parts that are added or subtracted are called terms. So this expression has five terms. We have 3. We have b. Here we have a more complex term, a minus 3. Then we have b, and we have a.
a minus 3 is written in brackets because we calculated the length of that short side by subtracting 3 from a. But it could be separated into a plus a and a minus 3, which would give us six terms in total. So this is to remind you that an equation is made by saying that two expressions are equal, and the expressions can further be broken down into terms.
Here we look at collecting like terms in an expression. The first step looks at what we've already done. Adding a minus 3 is the same as adding a and then subtracting 3. So our expression now has six terms.
Then we collect together the terms that can be added up or subtracted, the like terms, 3 and here subtract 3. These are both just numbers. So I collect those first. There's an invisible plus at the beginning of the first term. So I have plus 3 then minus 3 that can go together.
Then you'll notice we've got a plus b, another plus b, a plus a, and another plus a. What I've done there is to collect those terms together, bringing the operator sign, the plus or minus, with each term. Now I'm going to make that a shorter expression by collecting the like terms.
Plus 3 minus 3 gives me 0. Plus b plus another b, whatever number b is-- and, of course, b could be any number-- a shorter way of writing that is to say that I'm adding 2 times b. And, similarly, plus a plus another a-- a shorter way of adding two lots of a is to write 2 times a. So we have 0 plus 2 times b plus 2 times a.
Finally, we'll use the convention that, instead of writing 2 times b, we write 2b. And, instead of 2 times a, we write 2a. The final expression is 2b plus 2a. If you look at all the different expressions you found for the perimeter, they will all simplify to 2b plus 2a in their shortest form.
End transcript: Video 1 Expressions and terms
Video 1 Expressions and terms
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## Differences between talking in mathematical sentences and writing algebraic expressions
It is useful for teachers to compare spoken explanations and algebraic expressions.
The explanations in Activity 2 above are written in informal language, not in whole sentences, but as learners might say them.
Learners typically explain ideas in several short sentences, each with one operation. The sentences may follow the order of their thinking (for example, the first instruction will involve ‘adding’ since the aim is a perimeter). This is unlikely to match the order of the terms in the conventional expression.
When we read, we read from left to right. When we talk, our sentences usually start with the main subject. However, in numerical or algebraic expressions, the first term is not the most important one, nor is it always the unknown one, as in the examples above. The order in which you do operations is not necessarily the same order as you read and write the symbols.
Note how learners typically use ‘pointing’ words such as ‘this’, ‘that’ or ‘there’ to indicate a quantity that they want to refer to, but cannot name it without losing the thread of their thinking. In the classroom, they may also point to the diagram or one of the sides while they are talking. Gestures can be an important way of signalling their awareness of an unknown quantity. In an expression, this quantity might appear in brackets, or as a more complex term.
These are steps towards algebraic thinking. Learners have paid attention to an unknown quantity and are communicating about it.
How would you as a teacher build on the learner’s talk, and rephrase each explanation to make a more precise mathematical sentence? Think about using words such as ‘difference’ and ‘less than’.
(We say more precise here since it is not always easy to express mathematical relationships concisely and precisely in words – that is why algebra was invented.)
## Activity 3 Reflecting
Timing: Allow 5 minutes
Look back in the diagram in Activity 2, how would the task be different if a and b were replaced by specific numbers? How could you work with learners to keep the focus on mathematical sentences?
### Discussion
When given specific numbers, learners tend to work towards closed answers consisting of a single term. For example, if the triangles were labelled with sides 3, 5 (for a) and 7 (for b), they would add up as they went around the perimeter and get the answer 24. You would need to ask learners to write down mathematical sentences to keep the focus on the instructions, not the answer, such as writing down 10 + (5 – 3) + 5 + 7 = 24.
When there are specific numbers, learners find it easier to identify a number that they can use for the ‘leftover bit’.
For a general side a, the difference is written a – 3. It cannot be simplified further but you can write it instead of a number and give the instruction to add it to the other sides. Learners often like to use brackets to show that this is an expression that is being treated as a single object.
One strategy for helping those who are stuck on this part is to say ‘Suppose a was 5, how long is that bit?’ There is then a danger that they spot that the difference is 2 using only their knowledge of the specific numbers 5 and 3. You can help them pay attention to the general structure by recapping that the difference is 2 because 5 – 3 is 2, so the important operation is subtraction.
## Activity 4 Varying the problem
Timing: Allow 10 minutes
What happens to the perimeter if you combine the two congruent triangles into the new shapes in Figure 2?
Use the mathematical extensions for greater depth:
What other shapes and perimeters can you find with these triangles?
Could you use these triangles to make a shape with perimeter 2a + 2b + 3?
Figure 2 Four triangles
### Discussion
If you join the two triangles together with the whole side of one triangle set against the same length side of the other triangle, you can arrange the triangles into many shapes, including a rectangle, a parallelogram and a kite. Whatever the shape, the perimeter will always simplify to an expression that is double the sum of two of the sides.
If you change the rule slightly, then you can find some other perimeters. For example, they could meet just at one point, or the sides could overlap by only half their length.
## Activity 5 Is this task algebraic?
Timing: Allow 5 minutes
The beginning of this section defined algebraic thinking as involving indeterminacy, denotation and analyticity. In other words, it involves thinking about indeterminate quantities that are represented in words or symbols and then treated analytically.
Do you consider the triangle task algebraic?
### Discussion
Although the task is set in a geometric context and needs an idea of perimeter, it is algebraic.
It involves indeterminacy, as learners are working with the unknown quantities a, b and p, and these are treated analytically since they are operated on as if they were known numbers. The quantities are represented on a diagram to give a concrete basis for the reasoning. Learners also represent, or denote, them symbolically using letters and numbers and iconically using words to identify and name the quantities (‘the difference’, ‘the little bit’) or to ‘point’ to them (‘that one’). The fact that the answer cannot be reduced to one specific number or closed answer is also characteristic of algebraic thinking.
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# NCERT Solutions(Part - 1) - Data Handling Class 6 Notes | EduRev
## Class 6 : NCERT Solutions(Part - 1) - Data Handling Class 6 Notes | EduRev
The document NCERT Solutions(Part - 1) - Data Handling Class 6 Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6
Exercise 9.1
Question 1:
In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.
(a) Find how many students obtained marks equal to or more than 7.
(b) How many students obtained marks below 4?
Marks Tally Marks No. of students 1 II 2 2 III 3 3 III 3 4 7 5 6 6 7 7 5 8 IIII 4 9 III 3
(a) The students who got marks equal to or more than 7 are the students who got marks 7, 8 or 9. Therefore the number of these students are:
= 5 + 4 + 3
= 12
(b) The students who got marks below 4 are the students who got marks 1,2 or 3. Therefore the number of these students are:
= 2 + 3 + 3
= 8
Question 2:
Following is the choice of sweets of 30 students of Class VI.
(a) Arrange the names of sweets in a table using tally marks.
(b) Which sweet is preferred by most of the students?
(a) Table using tally marks:
Sweets Tally Marks No. of students Ladoo 11 Barfl 3 jalebi 7 Rasgulla 9 30
(b) The sweet which is preferred by most of the students is Ladoo. As out of 30, 11 students prefer eating ladoo.
Question 3)
Catherine threw a dice 40 times and noted the number appearing each time as shown below:
Make a table and enter the data using tally marks. Find the number that appeared.
(a) The minimum number of times
(b) The maximum number of times
(c) Find those numbers that appear an equal number of times.
Numbers Tally Marks How many times? 1 7 2 6 3 5 4 4 5 11 6 7
(a) The number that occurred for minimum number of times is 4
(b) The number that occurred for maximum number of times is 5
(c) 1 and 6 are the numbers that appear an equal number of times.
Question 4)
Following pictograph shows the number of tractors in five villages.
Village Number of tractors = 1 Tractor Village A Village B Village C Village D Village E
Observe the pictograph and answer the following questions:
(i) Which village has the minimum number of tractors?
(ii) Which village has the maximum number of tractors?
(iii) How many more tractors village C has as compared to village B.
(iv) What is the total number of tractors in all the five villages?
(i) Village D has the minimum number of tractors.
(ii) Village C has the maximum number of tractors.
(iii) Village B has 5 tractors
Village C has 8 tractors
= 8 - 5
= 3 Tractors
Village C has 3 more tractors as compared to village B.
(iv) Total number of tractors in all the villages = 6 + 5 + 8 + 3 + 6 = 28 tractors
Question 5)
The number of girl students in each class of a co-educational middle school is depicted by the pictograph:
Class Number of girl students - 4 girls I II III IV
Observe this pictograph and answer the following questions:
(a) Which class has the minimum number of girl students?
(b) Is the number of girls in Class VI less than the number of girls in Class V?
(c) How many girls are there in Class VII?
(a) Class VIII has only 6 girls. Therefore, the minimum number of girl students are in Class VIII.
(b) No. Class V has 10 girl students and Class VI has 16 girl students. Therefore, the number of girls in Class VI are more than the number of girls in Class V.
(c) The number of girls in Class VII are 12.
Question 6)
The sale of electric bulbs on different days of a week is shown below:
Day Number of electric bulbs - 2 bulbs Monday Tuesday Wednesday Thursday Friday Saturday Sunday
Observe the pictograph and answer the following questions:
(a) How many bulbs were sold on Friday?
(b) On which day were the maximum number of bulbs sold?
(c) On which of the days same number of bulbs were sold?
(d) On which of the days minimum number of bulbs were sold?
(e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week?
(a) Number of bulbs sold on Friday are 14.
(b) Maximum number of bulbs (18) were sold on Sunday.
(c) On Wednesday and Saturday 8 bulbs are sold. Hence equal number of bulbs were sold on Wednesday and Saturday.
(d) The minimum number of bulbs (8) were sold on Wednesday and Saturday.
(e) The total number of bulbs sold in the given week were 86. So the number of carton required to hold 86 bulb = 86/9 ≈10. Hence, 10 cartons required to hold the bulbs.
Question 7)
In a village, six fruit merchants sold the following number of fruit baskets in a particular season:
Name of fruit merchant Number of fruit baskets -100 fruit basket s Rahim Lakhanpal Anwar Martin Ranjit Singh Joseph
Observe this pictograph and answer the following question:
(a) Which merchant sold the maximum number of baskets?
(b) How many fruit baskets were sold by Anwar?
(c) The merchants who have sold 600 or more number of baskets are planning to buy a godown for the next season. Can you name them?
(a) Martin sold the maximum number of fruit baskets i.e. 950.
(b) Anwar sold 700 fruit basket
(c) Anwar, Martin, Ranjit Singh are the merchants who sold more than 600 fruit baskets. Hence, these are the merchants who are planning to buy a godown for the next season.
Exercise 9.2
Question 1)
Total number of animals in five villages are as follows:
Village A : 80
Village B : 120
Village C : 90
Village D : 40
Village E : 60
Prepare a pictograph of these animals using one symbol to represent 10 animals and answer the following questions:
(a) How many symbols represent animals of village E?
(b) Which village has the maximum number of animals?
(c) Which village has more animals: village A or village C?
Village Number of animals =10 animals Village A Village B Village C Village D Village E
(a) There are 60 animals in village E. So 6 symbols represent animals of village E.
(b) Village B has 120 animals which is the maximum number among these villages.
(c) Village A has 80 animals and village C has 90 animals. Clearly, Village C has more animals than Village A.
Question 2)
Total number of students of a school in different years is shown in the following table
Year Number of students 1996 400 1998 535 2000 472 2002 600 2004 623
A. Prepare a pictograph of students using one symbol to represent 100 students and answer the following questions:
(a) How many symbols represent total number of students in the year 2002?
(b) How many symbols represent total number of students for the year 1998?
B. Prepare another pictograph of students using any other symbol each representing 50 students. Which pictograph do you find more informative?
Years 100 students = 1996 1998 2000 2002 2004
(a) Total number of students in the year 2002 represents 6 symbols.
(b) Total number of students in the year 1998 represents 5 complete and 1 incomplete symbols.
B.
Years 50 students = 1996 1998 2000 2002 2004
Exercise 9.3
Question 1)
The bar graph given alongside shows the amount of wheat purchased by government during the year 1998-2002.
Read the bar graph and write down your observations. In which year was
(a) the wheat production maximum?
(b) the wheat production minimum?
(a) In 2002, production of wheat was maximum.
(b) In 1998, production of wheat was minimum.
Question 2)
Observe this bar graph which is showing the sale of shirts in a ready made shop from Monday to Saturday.
(a) What information does the above bar graph give?
(b) What is the scale chosen on the horizontal line representing number of shirts?
(c) On which day was the maximum number of shirts sold? How many shirts were sold on that day?
(d) On which day was the minimum number shirts sold?
(e) How many shirts were sold on Thursday?
(a) The bar graph shows the sale of shirt in a readymade shop from Monday to Saturday.
(b) 1 unit = 5 shirts is the scale on the horizontal line representing number of shirts.
(c) On Saturday maximum number of shirts i.e. 60 shirts were sold.
(d) On Tuesday minimum number of shirts i.e. 10 were sold.
(e) On Tuesday 35 shirts were sold.
Question 3)
Observe this bar graph which shows the marks obtained by Aziz in half-yearly examination in different subjects.
(a) What information does the bar graph give?
(b) Name the subject in which Aziz scored maximum marks.
(c) Name the subject in which he has scored minimum marks.
(d) State the name of the subjects and marks obtained in each of them.
(a) The bar graph shows the marks obtained by Aziz in half yearly examination in different subjects.
(b) Aziz scored maximum marks in Hindi i.e. 80 marks.
(c) Aziz scored minimum marks in Social Studies i.e. 40 marks.
(d) Hindi 80, English 60, Mathematics 70, Science 50, Social Studies 40.
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## Mathematics (Maths) Class 6
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# Introduction to The Language of Algebra
### Learning Outcomes
• Use Variables and Algebraic Symbols
• Use variables to represent unknown quantities in algebraic expressions
• Identify the variables and constants in an algebraic expression
• Use words and symbols to represent algebraic operations on variables and constants
• Use inequality symbols to compare two quantities
• Translate between words and inequality notation
• Identifying Expressions and Equations
• Identify and write mathematical expressions using words and symbols
• Identify and write mathematical equations using words and symbols
• Use the order of operations to simplify mathematical expressions
• Simplifying Expressions Using the Order of Operations
• Use exponential notation
• Write an exponential expression in expanded form
## Use Variables and Algebraic Symbols
Greg and Alex have the same birthday, but they were born in different years. This year Greg is $20$ years old and Alex is $23$, so Alex is $3$ years older than Greg. When Greg was $12$, Alex was $15$. When Greg is $35$, Alex will be $38$. No matter what Greg’s age is, Alex’s age will always be $3$ years more, right? In the language of algebra, we say that Greg’s age and Alex’s age are variable and the three is a constant. The ages change, or vary, so age is a variable. The $3$ years between them always stays the same, so the age difference is the constant. In algebra, letters of the alphabet are used to represent variables. Suppose we call Greg’s age $g$. Then we could use $g+3$ to represent Alex’s age. See the table below.
Greg’s age Alex’s age
$12$ $15$
$20$ $23$
$35$ $38$
$g$ $g+3$
Letters are used to represent variables. Letters often used for variables are $x,y,a,b,\text{ and }c$.
### Variables and Constants
A variable is a letter that represents a number or quantity whose value may change. A constant is a number whose value always stays the same.
To write algebraically, we need some symbols as well as numbers and variables. There are several types of symbols we will be using. In Whole Numbers, we introduced the symbols for the four basic arithmetic operations: addition, subtraction, multiplication, and division. We will summarize them here, along with words we use for the operations and the result.
Operation Notation Say: The result is…
Addition $a+b$ $a\text{ plus }b$ the sum of $a$ and $b$
Subtraction $a-b$ $a\text{ minus }b$ the difference of $a$ and $b$
Multiplication $a\cdot b,\left(a\right)\left(b\right),\left(a\right)b,a\left(b\right)$ $a\text{ times }b$ The product of $a$ and $b$
Division $a\div b,a/b,\frac{a}{b},b\overline{)a}$ $a$ divided by $b$ The quotient of $a$ and $b$
In algebra, the cross symbol, $\times$, is not used to show multiplication because that symbol may cause confusion. Does $3xy$ mean $3\times y$ (three times $y$ ) or $3\cdot x\cdot y$ (three times $x\text{ times }y$ )? To make it clear, use • or parentheses for multiplication. We perform these operations on two numbers. When translating from symbolic form to words, or from words to symbolic form, pay attention to the words of or and to help you find the numbers.
• The sum of $5$ and $3$ means add $5$ plus $3$, which we write as $5+3$.
• The difference of $9$ and $2$ means subtract $9$ minus $2$, which we write as $9 - 2$.
• The product of $4$ and $8$ means multiply $4$ times $8$, which we can write as $4\cdot 8$.
• The quotient of $20$ and $5$ means divide $20$ by $5$, which we can write as $20\div 5$.
### Example
Translate from algebra to words:
1. $12+14$
2. $\left(30\right)\left(5\right)$
3. $64\div 8$
4. $x-y$
Solution:
1. $12+14$ $12$ plus $14$ the sum of twelve and fourteen
2. $\left(30\right)\left(5\right)$ $30$ times $5$ the product of thirty and five
3. $64\div 8$ $64$ divided by $8$ the quotient of sixty-four and eight
4. $x-y$ $x$ minus $y$ the difference of $x$ and $y$
### TRY IT
[embed] [embed]
When two quantities have the same value, we say they are equal and connect them with an equal sign.
### Equality Symbol
[latex-display]a=b[/latex] is read $a$ is equal to $b[/latex-display] The symbol [latex]=$ is called the equal sign.
An inequality is used in algebra to compare two quantities that may have different values. The number line can help you understand inequalities. Remember that on the number line the numbers get larger as they go from left to right. So if we know that $b$ is greater than $a$, it means that $b$ is to the right of $a$ on the number line. We use the symbols $\text{<}$ and $\text{>}$ for inequalities.
$a<b$ is read $a$ is less than $b$ $a$ is to the left of $b$ on the number line
[latex-display]a>b[/latex] is read $a$ is greater than $b[/latex-display] [latex]a$ is to the right of $b$ on the number line
The expressions $a<b\text{ and }a>b$ can be read from left-to-right or right-to-left, though in English we usually read from left-to-right. In general,
$a<b\text{ is equivalent to }b>a$.
For example, $7<11\text{ is equivalent to }11>7$.
$a>b\text{ is equivalent to }b<a$.
.For example, $17>4\text{ is equivalent to }4<17$
When we write an inequality symbol with a line under it, such as $a\le b$, it means $a<b$ or $a=b$. We read this $a$ is less than or equal to $b$. Also, if we put a slash through an equal sign, $\ne$, it means not equal. We summarize the symbols of equality and inequality in the table below.
Algebraic Notation Say
$a=b$ $a$ is equal to $b$
$a\ne b$ $a$ is not equal to $b$
$a<b$ $a$ is less than $b$
$a>b$ $a$ is greater than $b$
$a\le b$ $a$ is less than or equal to $b$
$a\ge b$ $a$ is greater than or equal to $b$
### Symbols $<$ and $>$
The symbols $<$ and $>$ each have a smaller side and a larger side.
smaller side $<$ larger side
larger side $>$ smaller side
The smaller side of the symbol faces the smaller number and the larger faces the larger number.
### Example
Translate from algebra to words:
1. $20\le 35$
2. $11\ne 15 - 3$
3. $9>10\div 2$
4. $x+2<10$
1. $20\le 35$ $20$ is less than or equal to $35$
2. $11\ne 15 - 3$ $11$ is not equal to $15$ minus $3$
3. $9>10\div 2$ $9$ is greater than $10$ divided by $2$
4. $x+2<10$ $x$ plus $2$ is less than $10$
### TRY IT
[ohm_question]144653[/ohm_question] [embed] [embed]
In the following video we show more examples of how to write inequalities as words. https://youtu.be/q2ciQBwkjbk
### Example
The information in the table below compares the fuel economy in miles-per-gallon (mpg) of several cars. Write the appropriate symbol $\text{=},\text{<},\text{ or }\text{>}$ in each expression to compare the fuel economy of the cars. (credit: modification of work by Bernard Goldbach, Wikimedia Commons)
1. MPG of Prius_____ MPG of Mini Cooper
2. MPG of Versa_____ MPG of Fit
3. MPG of Mini Cooper_____ MPG of Fit
4. MPG of Corolla_____ MPG of Versa
5. MPG of Corolla_____ MPG of Prius
1. MPG of Prius____MPG of Mini Cooper Find the values in the chart. 48____27 Compare. 48 > 27 MPG of Prius > MPG of Mini Cooper
2. MPG of Versa____MPG of Fit Find the values in the chart. 26____27 Compare. 26 < 27 MPG of Versa < MPG of Fit
3. MPG of Mini Cooper____MPG of Fit Find the values in the chart. 27____27 Compare. 27 = 27 MPG of Mini Cooper = MPG of Fit
4. MPG of Corolla____MPG of Versa Find the values in the chart. 28____26 Compare. 28 > 26 MPG of Corolla > MPG of Versa
5. MPG of Corolla____MPG of Prius Find the values in the chart. 28____48 Compare. 28 < 48 MPG of Corolla < MPG of Prius
### TRY IT
[embed]
Grouping symbols in algebra are much like the commas, colons, and other punctuation marks in written language. They indicate which expressions are to be kept together and separate from other expressions. The table below lists three of the most commonly used grouping symbols in algebra.
Common Grouping Symbols
parentheses ( )
brackets [ ]
braces { }
Here are some examples of expressions that include grouping symbols. We will simplify expressions like these later in this section.
$\begin{array}{cc}8\left(14 - 8\right)21 - 3\\\left[2+4\left(9 - 8\right)\right]\\24\div \left\{13 - 2\left[1\left(6 - 5\right)+4\right]\right\}\end{array}$
## Identify Expressions and Equations
What is the difference in English between a phrase and a sentence? A phrase expresses a single thought that is incomplete by itself, but a sentence makes a complete statement. "Running very fast" is a phrase, but "The football player was running very fast" is a sentence. A sentence has a subject and a verb. In algebra, we have expressions and equations. An expression is like a phrase. Here are some examples of expressions and how they relate to word phrases:
Expression Words Phrase
$3+5$ $3\text{ plus }5$ the sum of three and five
$n - 1$ $n$ minus one the difference of $n$ and one
$6\cdot 7$ $6\text{ times }7$ the product of six and seven
$\frac{x}{y}$ $x$ divided by $y$ the quotient of $x$ and $y$
Notice that the phrases do not form a complete sentence because the phrase does not have a verb. An equation is two expressions linked with an equal sign. When you read the words the symbols represent in an equation, you have a complete sentence in English. The equal sign gives the verb. Here are some examples of equations:
Equation Sentence
$3+5=8$ The sum of three and five is equal to eight.
$n - 1=14$ $n$ minus one equals fourteen.
$6\cdot 7=42$ The product of six and seven is equal to forty-two.
$x=53$ $x$ is equal to fifty-three.
$y+9=2y - 3$ $y$ plus nine is equal to two $y$ minus three.
### Expressions and Equations
An expression is a number, a variable, or a combination of numbers and variables and operation symbols. An equation is made up of two expressions connected by an equal sign.
### example
Determine if each is an expression or an equation:
1. $16 - 6=10$
2. $4\cdot 2+1$
3. $x\div 25$
4. $y+8=40$
Solution
1. $16 - 6=10$ This is an equation—two expressions are connected with an equal sign. 2. $4\cdot 2+1$ This is an expression—no equal sign. 3. $x\div 25$ This is an expression—no equal sign. 4. $y+8=40$ This is an equation—two expressions are connected with an equal sign.
[embed]
### Learning Outcomes
• Use the order of operations to simplify mathematical expressions
• Simplify mathematical expressions involving addition, subtraction, multiplication, division, and exponents
## Simplify Expressions Containing Exponents
To simplify a numerical expression means to do all the math possible. For example, to simplify $4\cdot 2+1$ we’d first multiply $4\cdot 2$ to get $8$ and then add the $1$ to get $9$. A good habit to develop is to work down the page, writing each step of the process below the previous step. The example just described would look like this:
$4\cdot 2+1$ [latex-display]8+1[/latex-display] $9$
Suppose we have the expression $2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2$. We could write this more compactly using exponential notation. Exponential notation is used in algebra to represent a quantity multiplied by itself several times. We write $2\cdot 2\cdot 2$ as ${2}^{3}$ and $2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2$ as ${2}^{9}$. In expressions such as ${2}^{3}$, the $2$ is called the base and the $3$ is called the exponent. The exponent tells us how many factors of the base we have to multiply.
[latex-display]\text{means multiply three factors of 2}[/latex-display] We say ${2}^{3}$ is in exponential notation and $2\cdot 2\cdot 2$ is in expanded notation.
### Exponential Notation
For any expression ${a}^{n},a$ is a factor multiplied by itself $n$ times if $n$ is a positive integer. [latex-display]{a}^{n}\text{ means multiply }n\text{ factors of }a[/latex-display] The expression ${a}^{n}$ is read $a$ to the ${n}^{th}$ power.
For powers of $n=2$ and $n=3$, we have special names.
$a^2$ is read as "$a$ squared"
$a^3$ is read as "$a$ cubed"
The table below lists some examples of expressions written in exponential notation.
Exponential Notation In Words
${7}^{2}$ $7$ to the second power, or $7$ squared
${5}^{3}$ $5$ to the third power, or $5$ cubed
${9}^{4}$ $9$ to the fourth power
${12}^{5}$ $12$ to the fifth power
### example
Write each expression in exponential form:
1. $16\cdot 16\cdot 16\cdot 16\cdot 16\cdot 16\cdot 16$
2. $\text{9}\cdot \text{9}\cdot \text{9}\cdot \text{9}\cdot \text{9}$
3. $x\cdot x\cdot x\cdot x$
4. $a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a\cdot a$
1. The base $16$ is a factor $7$ times. ${16}^{7}$ 2. The base $9$ is a factor $5$ times. ${9}^{5}$ 3. The base $x$ is a factor $4$ times. ${x}^{4}$ 4. The base $a$ is a factor $8$ times. ${a}^{8}$
### try it
[embed]
In the video below we show more examples of how to write an expression of repeated multiplication in exponential form. https://youtu.be/HkPGTmAmg_s
### example
Write each exponential expression in expanded form:
1. ${8}^{6}$
2. ${x}^{5}$
Answer: Solution 1. The base is $8$ and the exponent is $6$, so ${8}^{6}$ means $8\cdot 8\cdot 8\cdot 8\cdot 8\cdot 8$ 2. The base is $x$ and the exponent is $5$, so ${x}^{5}$ means $x\cdot x\cdot x\cdot x\cdot x$
### try it
[embed]
To simplify an exponential expression without using a calculator, we write it in expanded form and then multiply the factors.
### example
Simplify: ${3}^{4}$
${3}^{4}$ Expand the expression. $3\cdot 3\cdot 3\cdot 3$ Multiply left to right. $9\cdot 3\cdot 3$ $27\cdot 3$ Multiply. $81$
[embed]
## Simplify Expressions Using the Order of Operations
We’ve introduced most of the symbols and notation used in algebra, but now we need to clarify the order of operations. Otherwise, expressions may have different meanings, and they may result in different values. For example, consider the expression:
$4+3\cdot 7$
$\begin{array}{cccc}\hfill \text{Some students say it simplifies to 49.}\hfill & & & \hfill \text{Some students say it simplifies to 25.}\hfill \\ \begin{array}{ccc}& & \hfill 4+3\cdot 7\hfill \\ \text{Since }4+3\text{ gives 7.}\hfill & & \hfill 7\cdot 7\hfill \\ \text{And }7\cdot 7\text{ is 49.}\hfill & & \hfill 49\hfill \end{array}& & & \begin{array}{ccc}& & \hfill 4+3\cdot 7\hfill \\ \text{Since }3\cdot 7\text{ is 21.}\hfill & & \hfill 4+21\hfill \\ \text{And }21+4\text{ makes 25.}\hfill & & \hfill 25\hfill \end{array}\hfill \end{array}$
Imagine the confusion that could result if every problem had several different correct answers. The same expression should give the same result. So mathematicians established some guidelines called the order of operations, which outlines the order in which parts of an expression must be simplified.
### Order of Operations
When simplifying mathematical expressions perform the operations in the following order: 1. Parentheses and other Grouping Symbols
• Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first.
2. Exponents
• Simplify all expressions with exponents.
3. Multiplication and Division
• Perform all multiplication and division in order from left to right. These operations have equal priority.
• Perform all addition and subtraction in order from left to right. These operations have equal priority.
Students often ask, "How will I remember the order?" Here is a way to help you remember: Take the first letter of each key word and substitute the silly phrase. Please Excuse My Dear Aunt Sally.
Order of Operations
Excuse Exponents
My Dear Multiplication and Division
It’s good that ‘My Dear’ goes together, as this reminds us that multiplication and division have equal priority. We do not always do multiplication before division or always do division before multiplication. We do them in order from left to right. Similarly, ‘Aunt Sally’ goes together and so reminds us that addition and subtraction also have equal priority and we do them in order from left to right.
### example
Simplify the expressions:
1. $4+3\cdot 7$
2. $\left(4+3\right)\cdot 7$
Solution:
1. $4+3\cdot 7$ Are there any parentheses? No. Are there any exponents? No. Is there any multiplication or division? Yes. Multiply first. $4+\color{red}{3\cdot 7}$ Add. $4+21$ $25$
2. $(4+3)\cdot 7$ Are there any parentheses? Yes. $\color{red}{(4+3)}\cdot 7$ Simplify inside the parentheses. $(7)7$ Are there any exponents? No. Is there any multiplication or division? Yes. Multiply. $49$
### try it
[ohm_question]144748[/ohm_question] [ohm_question]144751[/ohm_question]
### example
Simplify:
1. $\text{18}\div \text{9}\cdot \text{2}$
2. $\text{18}\cdot \text{9}\div \text{2}$
1. $18\div 9\cdot 2$ Are there any parentheses? No. Are there any exponents? No. Is there any multiplication or division? Yes. Multiply and divide from left to right. Divide. $\color{red}{2}\cdot 2$ Multiply. $4$
2. $18\cdot 9\div 2$ Are there any parentheses? No. Are there any exponents? No. Is there any multiplication or division? Yes. Multiply and divide from left to right. Multiply. $\color{red}{162}\div 2$ Divide. $81$
### try it
[ohm_question]144756[/ohm_question]
### example
Simplify: $18\div 6+4\left(5 - 2\right)$.
$18\div 6+4(5-2)$ Parentheses? Yes, subtract first. $18\div 6+4(\color{red}{3})$ Exponents? No. Multiplication or division? Yes. Divide first because we multiply and divide left to right. $\color{red}{3}+4(3)$ Any other multiplication or division? Yes. Multiply. $3+\color{red}{12}$ Any other multiplication or division? No. Any addition or subtraction? Yes. $15$
### try it
[ohm_question]144758[/ohm_question]
In the video below we show another example of how to use the order of operations to simplify a mathematical expression. https://youtu.be/qFUvF5-w9o0 When there are multiple grouping symbols, we simplify the innermost parentheses first and work outward.
### example
$\text{Simplify: }5+{2}^{3}+3\left[6 - 3\left(4 - 2\right)\right]$.
$5+{2}^{3}+ 3[6-3(4-2)]$ Are there any parentheses (or other grouping symbol)? Yes. Focus on the parentheses that are inside the brackets. $5+{2}^{3}+ 3[6-3\color{red}{(4-2)}]$ Subtract. $5+{2}^{3}+3[6-\color{red}{3(2)}]$ Continue inside the brackets and multiply. $5+{2}^{3}+3[6-\color{red}{6}]$ Continue inside the brackets and subtract. $5+{2}^{3}+3[\color{red}{0}]$ The expression inside the brackets requires no further simplification. Are there any exponents? Yes. Simplify exponents. $5+\color{red}{{2}^{3}}+3[0]$ Is there any multiplication or division? Yes. Multiply. $5+8+\color{red}{3[0]}$ Is there any addition or subtraction? Yes. Add. $\color{red}{5+8}+0$ Add. $\color{red}{13+0}$ $13$
### try it
[ohm_question]144759[/ohm_question]
In the video below we show another example of how to use the order of operations to simplify an expression that contains exponents and grouping symbols. https://youtu.be/8b-rf2AW3Ac
### example
Simplify: ${2}^{3}+{3}^{4}\div 3-{5}^{2}$.
${2}^{3}+{3}^{4}\div 3-{5}^{2}$ If an expression has several exponents, they may be simplified in the same step. Simplify exponents. $\color{red}{{2}^{3}}+\color{red}{{3}^{4}}\div 3-\color{red}{{5}^{2}}$ Divide. $8+\color{red}{81\div 3}-25$ Add. $\color{red}{8+27}-25$ Subtract. $\color{red}{35-25}$ $10$
### try it
[ohm_question]144762[/ohm_question]
|
# Algebra II : Finding Roots
## Example Questions
### Example Question #244 : Equations / Inequalities
Solve the equation:
Explanation:
Add 8 to both sides to set the equation equal to 0:
To factor, find two integers that multiply to 24 and add to 10. 4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:
Then factor by grouping:
Set each factor equal to 0 and solve:
and
### Example Question #1 : Finding Zeros Of A Polynomial
Find the roots of the following quadratic expression:
Explanation:
First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor.
This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b.
So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping.
Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out.
Now we factor out the (3x + 4).
Setting each factor = 0 we can find the solutions.
So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}.
### Example Question #4 : Solving Quadratic Functions
Find the roots of the following quadratic expression.
Explanation:
First we remember that "find the roots" means "find the values of x for which this expression equals 0." So we set the expression = 0 and approach solving as normal.
Since solving this by sight is difficult, we'll use composition, multiplying a by c and finding factors which add to b.
So -9 and 5 will work; we will use them to rewrite -4x as -9x + 5x and then factor by grouping.
We identify common factors to "pull" out of each group.
And now we factor out x-3.
Setting each factor equal to 0 lets us solve for x.
So our solutions are x = -5/3 and x = 3, which we write as x = {-5/3, 3}.
### Example Question #11 : Finding Roots
FInd the roots for
Explanation:
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfect square factors of this term are and
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
### Example Question #12 : Finding Roots
Find the roots for
Explanation:
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfect square factors of this term are and
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
### Example Question #13 : Finding Roots
Find the roots for
Explanation:
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfect square factors of this term are and
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
### Example Question #14 : Finding Roots
Find the roots for
Explanation:
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfect square factors of this term are and
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
### Example Question #15 : Finding Roots
Find the roots of .
Explanation:
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each term we see that each is a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfect square factors of this term are and
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
### Example Question #16 : Finding Roots
Find the roots of .
Explanation:
Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each term we see that each is a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is
The perfect square factors of this term are and .
Now we look at the constant term
The perfecct square factors of this term are and .
Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:
From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.
### Example Question #17 : Finding Roots
Solve for :
Explanation:
To solve the quadratic equation, , first set the equation equal to zero and then factor the quadratic to
.
Since the equation is equal to zero, at least one of the expressions on the left side of the equation must be equal to zero. Therefore, set and solve to get
.
Finally, set to get .
The solution to the original equation is
.
|
# Finding Square Roots 3.9.
## Presentation on theme: "Finding Square Roots 3.9."— Presentation transcript:
Finding Square Roots 3.9
Warm Up Find the two square roots of each number. Evaluate each expression. ±12 ±16 20 119
Finding Square Roots Learn to estimate square roots to a given number of decimal places and solve problems using square roots.
A museum director wants to install a skylight to illuminate an unusual piece of art. It must be square and have an area of 300 square inches, with wood trim around it. Can you calculate the trim that you need? You can do this by using your knowledge of squares and square roots.
Example: Estimating Square Roots of Numbers
Each square root is between two integers. Name the integers. Think: What are perfect squares close to 55? A. 55 72 = 49 49 < 55 82 = 64 64 > 55 55 is between 7 and 8. 7 < < 8
Example: Estimating Square Roots of Numbers Continued
Each square root is between two integers. Name the integers. Think: What are perfect squares close to 90? B. 90 –92 = 81 81 < 90 –102 = 100 100 > 90 90 is between –9 and –10. –9 < – < –10
Try This Each square root is between two integers. Name the integers.
Think: What are perfect squares close to 80? A. 80 82 = 64 64 < 80 92 = 81 81 > 80 is between 8 and 9. 80 8 < < 9
Try This Each square root is between two integers. Name the integers.
Think: What are perfect squares close to 45? B. 45 –62 = 36 36 < 45 –72 = 49 49 > 45 45 is between –6 and –7. –6 < – 45 < –7
Example 2: Problem Solving Application
You want to sew a fringe on a square tablecloth with an area of 500 square inches. Calculate the length of each side of the tablecloth and the length of fringe you will need to the nearest tenth of an inch. Understand the Problem First find the length of a side. Then you can use the length of a side to find the perimeter, the length of fringe around the tablecloth.
Example 2 Continued Make a Plan
The length of a side, in feet, is the number that you multiply by itself to get 500. To be accurate, find this number to the nearest tenth. If you do not know a step-by-step method for finding 500, use guess and check.
Example 2 Continued Solve
Because 500 is between 222 and 232, the square root of 500 is between 22 and 23. Square root is between 22 and 22.5 Too high 22.52 = Guess 22.5 1 Square root is between 22.2 and 22.5 Too low 22.22 = Guess 22.2 2 Square root is between 22.3 and 22.4 Too low 22.32 = Guess 22.3 4 Square root is between 22.2 and 22.4 Too high 22.42 = Guess 22.4 3 22.0 22.2 22.4 22.6 The square root is between 22.3 and 22.4.
Example 2 Continued Solve
The square root is between 22.3 and To round to the nearest tenth, look at the next decimal place. Consider = Too low The square root must be greater than 22.35, so round up. To the nearest tenth, is about 22.4. The length of each side of the table is about 22.4 in.
Example 2 Continued Solve
The length of a side of the tablecloth is 22.4 inches, to the nearest tenth of an inch. Now estimate the length around the tablecloth. 4 • 22.4 = 89.6 Perimeter 4 • side You will need about 89.6 inches of fringe.
Example 2 Continued Look Back
The length 90 inches divided by 4 is 22.5 inches. A 22.5-inch square has an area of square inches, which is close to 500, so the answers are reasonable.
Try This You want to build a fence around a square garden that is 250 square feet. Calculate the length of one side of the garden and the total length of the fence, to the nearest tenth. Understand the Problem First find the length of a side. Then you can use the length of a side to find the perimeter, the length of the fence.
Try This Make a Plan The length of a side, in feet, is the number that you multiply by itself to get 250. To be accurate, find this number to the nearest tenth. If you do not know a step-by-step method for finding 250, use guess and check.
Try This Solve Because 250 is between 152 and 162, the square root of 250 is between 15 and 16. Guess 15.5 15.52 = Too Low Square root is between 15.5 and 16 Guess 15.9 15.92 = Too high Square root is between 15.5 and 15.9 Guess 15.7 15.72 = Too Low Square root is between 15.7 and 15.9 Guess 15.8 15.82 = Too Low Square root is between 15.8 and 15.9 1 3 4 2 15.5 15.7 15.9 16.1 The square root is between 15.8 and 15.9.
Try This Solve To round to the nearest tenth, look at the next decimal place. Consider = The square root is lower than 15.85, so round down. To the nearest tenth, is about 15.8. The length of each side of the garden is about 15.8 ft.
Try This Solve The length of a side of the garden is 15.8 feet, to the nearest tenth of a foot. Now estimate the length around the garden. 4 • 15.8 = 63.2 Perimeter 4 • side You will need about 63.2 feet of fence.
Try This Look Back The length 63.2 feet divided by 4 is 15.8 feet. A 15.8 foot square has an area of square feet, which is close to 250, so the answers are reasonable.
Examples 3: Using a Calculator to Estimate the Value of a Square Root
Use a calculator to find Round to the nearest tenth. Using a calculator, ≈ …. Rounded, is 22.4.
Try This Use a calculator to find 200. Round to the nearest tenth.
Using a calculator, ≈ …. Rounded, is 14.1.
Lesson Quiz Each square root is between two integers. Name the two integers. – 456 5 and 6 –22 and –21 Use a calculator to find each value. Round to the nearest tenth. 9.4 35.0 5. A square field has an area of 2000 square feet. To the nearest foot, how much fencing would be needed to enclose the field? 179 ft
|
Contemporary Mathematics
# 3.4Rational Numbers
Contemporary Mathematics3.4 Rational Numbers
Figure 3.20 Stock gains and losses are often represented as percentages.(credit: "stock market quotes in newspaper" by Andreas Poike/Flickr, CC BY 2.0)
## Learning Objectives
After completing this section, you should be able to:
1. Define and identify numbers that are rational.
2. Simplify rational numbers and express in lowest terms.
3. Add and subtract rational numbers.
4. Convert between improper fractions and mixed numbers.
5. Convert rational numbers between decimal and fraction form.
6. Multiply and divide rational numbers.
7. Apply the order of operations to rational numbers to simplify expressions.
8. Apply density property of rational numbers.
9. Solve problems involving rational numbers.
10. Use fractions to convert between units.
11. Define and apply percent.
12. Solve problems using percent.
We are often presented with percentages or fractions to explain how much of a population has a certain feature. For example, the 6-year graduation rate of college students at public institutions is 57.6%, or 72/125. That fraction may be unsettling. But without the context, the percentage is hard to judge. So how does that compare to private institutions? There, the 6-year graduation rate is 65.4%, or 327/500. Comparing the percentages is straightforward, but the fractions are harder to interpret due to different denominators. For more context, historical data could be found. One study reported that the 6-year graduation rate in 1995 was 56.4%. Comparing that historical number to the recent 6-year graduation rate at public institutions of 57.6% shows that there hasn't been much change in that rate.
## Defining and Identifying Numbers That Are Rational
A rational number (called rational since it is a ratio) is just a fraction where the numerator is an integer and the denominator is a non-zero integer. As simple as that is, they can be represented in many ways. It should be noted here that any integer is a rational number. An integer, $nn$, written as a fraction of two integers is $n1n1$.
In its most basic representation, a rational number is an integer divided by a non-zero integer, such as $312312$. Fractions may be used to represent parts of a whole. The denominator is the total number of parts to the object, and the numerator is how many of those parts are being used or selected. So, if a pizza is cut into 8 equal pieces, each piece is $1818$ of the pizza. If you take three slices, you have $3838$ of the pizza (Figure 3.21). Similarly, if in a group of 20 people, 5 are wearing hats, then $520520$ of the people are wearing hats (Figure 3.22).
Figure 3.21 Pizza cut in 8 slices, with 3 slices highlighted
Figure 3.22 Group of 20 people, with 5 people wearing hats
Another representation of rational numbers is as a mixed number, such as $258258$ (Figure 3.23). This represents a whole number (2 in this case), plus a fraction (the $5858$).
Figure 3.23 Two whole pizzas and one partial pizza
Rational numbers may also be expressed in decimal form; for instance, as 1.34. When 1.34 is written, the decimal part, 0.34, represents the fraction $3410034100$, and the number 1.34 is equal to $134100134100$. However, not all decimal representations are rational numbers.
A number written in decimal form where there is a last decimal digit (after a given decimal digit, all following decimal digits are 0) is a terminating decimal, as in 1.34 above. Alternately, any decimal numeral that, after a finite number of decimal digits, has digits equal to 0 for all digits following the last non-zero digit.
All numbers that can be expressed as a terminating decimal are rational. This comes from what the decimal represents. The decimal part is the fraction of the decimal part divided by the appropriate power of 10. That power of 10 is the number of decimal digits present, as for 0.34, with two decimal digits, being equal to $3410034100$.
Another form that is a rational number is a decimal that repeats a pattern, such as 67.1313… When a rational number is expressed in decimal form and the decimal is a repeated pattern, we use special notation to designate the part that repeats. For example, if we have the repeating decimal 4.3636…, we write this as $4.36¯4.36¯$. The bar over the 36 indicates that the 36 repeats forever.
If the decimal representation of a number does not terminate or form a repeating decimal, that number is not a rational number.
One class of numbers that is not rational is the square roots of integers or rational numbers that are not perfect squares, such as $1010$ and $256256$. More generally, the number $bb$ is the square root of the number $aa$ if $a=b2a=b2$. The notation for this is $b=ab=a$, where the symbol is the square root sign. An integer perfect square is any integer that can be written as the square of another integer. A rational perfect square is any rational number that can be written as a fraction of two integers that are perfect squares.
Sometimes you may be able to identify a perfect square from memory. Another process that may be used is to factor the number into the product of an integer with itself. Or a calculator (such as Desmos) may be used to find the square root of the number. If the calculator yields an integer, the original number was a perfect square.
## Tech Check
### Using Desmos to Find the Square Root of a Number
When Desmos is used, there is a tab at the bottom of the screen that opens the keyboard for Desmos. The keyboard is shown below. On the keyboard (Figure 3.24) is the square root symbol $( )( )$. To find the square root of a number, click the square root key, and then type the number. Desmos will automatically display the value of the square root as you enter the number.
Figure 3.24 Desmos keyboard with square root key
## Example 3.51
### Identifying Perfect Squares
Which of the following are perfect squares?
1. 45
2. 144
Determine if the following are perfect squares:
1.
94
2.
441
## Example 3.52
### Identifying Rational Numbers
Determine which of the following are rational numbers:
1. $7373$
2. $4.5564.556$
3. $315315$
4. $41174117$
5. $5.64¯5.64¯$
Determine which of the following are rational numbers:
1.
$\sqrt {13}$
2.
$- 13.\overline {2}\overline {1}$
3.
$\frac{{ - 48}}{{ - 16}}$
4.
$- 4\frac{{18}}{{19}}$
5.
$14.1131$
## Simplifying Rational Numbers and Expressing in Lowest Terms
A rational number is one way to express the division of two integers. As such, there may be multiple ways to express the same value with different rational numbers. For instance, $4545$ and $12151215$ are the same value. If we enter them into a calculator, they both equal 0.8. Another way to understand this is to consider what it looks like in a figure when two fractions are equal.
In Figure 3.25, we see that $3535$ of the rectangle and $915915$ of the rectangle are equal areas.
Figure 3.25 Two Rectangles with Equal Areas
They are the same proportion of the area of the rectangle. The left rectangle has 5 pieces, three of which are shaded. The right rectangle has 15 pieces, 9 of which are shaded. Each of the pieces of the left rectangle was divided equally into three pieces. This was a multiplication. The numerator describing the left rectangle was 3 but it becomes $3×33×3$, or 9, as each piece was divided into three. Similarly, the denominator describing the left rectangle was 5, but became $5×35×3$, or 15, as each piece was divided into 3. The fractions $3535$ and $915915$ are equivalent because they represent the same portion (often loosely referred to as equal).
This understanding of equivalent fractions is very useful for conceptualization, but it isn’t practical, in general, for determining when two fractions are equivalent. Generally, to determine if the two fractions $abab$ and $cdcd$ are equivalent, we check to see that $a×d=b×ca×d=b×c$. If those two products are equal, then the fractions are equal also.
## Example 3.53
### Determining If Two Fractions Are Equivalent
Determine if $12301230$ and $14351435$ are equivalent fractions.
1.
Determine if $\frac{8}{{14}}$ and $\frac{{12}}{{26}}$ are equivalent fractions.
That $a×d=b×ca×d=b×c$ indicates the fractions $abab$ and $cdcd$ are equivalent is due to some algebra. One property of natural numbers, integers, and rational numbers (also irrational numbers) is that for any three numbers $a,b,a,b,$ and $cc$ with $c≠0c≠0$, if $a=ba=b$, then $a/c=b/ca/c=b/c$. In other words, when two numbers are equal, then dividing both numbers by the same non-zero number, the two newly obtained numbers are also equal. We can apply that to $a×da×d$ and $b×cb×c$, to show that $abab$ and $cdcd$ are equivalent if $a×d=b×ca×d=b×c$.
If $a×d=b×ca×d=b×c$, and $c≠0,d ≠0c≠0,d ≠0$, we can divide both sides by and obtain the following: $a×dc=b×cca×dc=b×cc$. We can divide out the $cc$ on the right-hand side of the equation, resulting in $a×dc=ba×dc=b$. Similarly, we can divide both sides of the equation by $dd$ and obtain the following: $a×dc×d=bda×dc×d=bd$. We can divide out $dd$ the on the left-hand side of the equation, resulting in $ac=bdac=bd$. So, the rational numbers $acac$ and $bdbd$ are equivalent when $a×d=b×ca×d=b×c$.
## Video
Recall that a common divisor or common factor of a set of integers is one that divides all the numbers of the set of numbers being considered. In a fraction, when the numerator and denominator have a common divisor, that common divisor can be divided out. This is often called canceling the common factors or, more colloquially, as canceling.
To show this, consider the fraction $36633663$. The numerator and denominator have the common factor 3. We can rewrite the fraction as $3663=12×321×33663=12×321×3$. The common divisor 3 is then divided out, or canceled, and we can write the fraction as $12×321×3=122112×321×3=1221$. The 3s have been crossed out to indicate they have been divided out. The process of dividing out two factors is also referred to as reducing the fraction.
If the numerator and denominator have no common positive divisors other than 1, then the rational number is in lowest terms.
The process of dividing out common divisors of the numerator and denominator of a fraction is called reducing the fraction.
One way to reduce a fraction to lowest terms is to determine the GCD of the numerator and denominator and divide out the GCD. Another way is to divide out common divisors until the numerator and denominator have no more common factors.
## Example 3.54
### Reducing Fractions to Lowest Terms
Express the following rational numbers in lowest terms:
1. $36483648$
2. $100250100250$
3. $5113651136$
1.
Express $\frac{{252}}{{840}}$ in lowest terms.
## Tech Check
### Using Desmos to Find Lowest Terms
Desmos is a free online calculator. Desmos supports reducing fractions to lowest terms. When a fraction is entered, Desmos immediately calculates the decimal representation of the fraction. However, to the left of the fraction, there is a button that, when clicked, shows the fraction in reduced form.
## Adding and Subtracting Rational Numbers
Adding or subtracting rational numbers can be done with a calculator, which often returns a decimal representation, or by finding a common denominator for the rational numbers being added or subtracted.
## Tech Check
### Using Desmos to Add Rational Numbers in Fractional Form
To create a fraction in Desmos, enter the numerator, then use the division key (/) on your keyboard, and then enter the denominator. The fraction is then entered. Then click the right arrow key to exit the denominator of the fraction. Next, enter the arithmetic operation (+ or –). Then enter the next fraction. The answer is displayed dynamically (calculates as you enter). To change the Desmos result from decimal form to fractional form, use the fraction button (Figure 3.26) on the left of the line that contains the calculation:
Figure 3.26 Fraction button on the Desmos keyboard
## Example 3.55
### Adding Rational Numbers Using Desmos
Calculate $2342+9562342+956$ using Desmos.
1.
Calculate $\frac{{124}}{{297}} + \frac{3}{{125}}$.
Performing addition and subtraction without a calculator may be more involved. When the two rational numbers have a common denominator, then adding or subtracting the two numbers is straightforward. Add or subtract the numerators, and then place that value in the numerator and the common denominator in the denominator. Symbolically, we write this as $ac±bc=a±bcac±bc=a±bc$. This can be seen in the Figure 3.27, which shows $320+420=720320+420=720$.
It is customary to then write the result in lowest terms.
## FORMULA
If $cc$ is a non-zero integer, then $ac±bc=a±bcac±bc=a±bc$.
## Example 3.56
### Adding Rational Numbers with the Same Denominator
Calculate $1328+7281328+728$.
1.
Calculate $\frac{{38}}{{73}} + \frac{7}{{73}}$.
## Example 3.57
### Subtracting Rational Numbers with the Same Denominator
Calculate $45136−1713645136−17136$.
1.
Calculate $\frac{{21}}{{40}} - \frac{8}{{40}}$.
When the rational numbers do not have common denominators, then we have to transform the rational numbers so that they do have common denominators. The common denominator that reduces work later in the problem is the LCM of the numerator and denominator. When adding or subtracting the rational numbers $abab$ and $cdcd$, we perform the following steps.
Step 1: Find $LCM(b,d)LCM(b,d)$.
Step 2: Calculate $n=LCM(b,d)bn=LCM(b,d)b$ and $m=LCM(b,d)dm=LCM(b,d)d$.
Step 3: Multiply the numerator and denominator of $abab$ by $nn$, yielding $a×nb×na×nb×n$.
Step 4: Multiply the numerator and denominator of $cdcd$ by $mm$, yielding $c×md×m c×md×m$.
Step 5: Add or subtract the rational numbers from Steps 3 and 4, since they now have the common denominators.
You should be aware that the common denominator is $LCM(b,d)LCM(b,d)$. For the first denominator, we have $b×n=b×LCM(b,d)b=LCM(b,d)b×n=b×LCM(b,d)b=LCM(b,d)$, since we multiply and divide $LCM(b,d)LCM(b,d)$ by the same number. For the same reason, $d×m=d×LCM(b,d)b=LCM(b,d)d×m=d×LCM(b,d)b=LCM(b,d)$.
## Example 3.58
### Adding Rational Numbers with Unequal Denominators
Calculate $1118+2151118+215$.
1.
Calculate $\frac{4}{9} + \frac{7}{{12}}$.
## Example 3.59
### Subtracting Rational Numbers with Unequal Denominators
Calculate $1425−9701425−970$.
1.
Calculate $\frac{{10}}{{99}} - \frac{{17}}{{300}}$.
## Converting Between Improper Fractions and Mixed Numbers
One way to visualize a fraction is as parts of a whole, as in $512512$ of a pizza. But when the numerator is larger than the denominator, as in $23122312$, then the idea of parts of a whole seems not to make sense. Such a fraction is an improper fraction. That kind of fraction could be written as an integer plus a fraction, which is a mixed number. The fraction $23122312$ rewritten as a mixed number would be $1111211112$. Arithmetically, $1111211112$ is equivalent to $1+11121+1112$, which is read as “one and 11 twelfths.”
Improper fractions can be rewritten as mixed numbers using division and remainders. To find the mixed number representation of an improper fraction, divide the numerator by the denominator. The quotient is the integer part, and the remainder becomes the numerator of the remaining fraction.
## Example 3.60
### Rewriting an Improper Fraction as a Mixed Number
Rewrite $48134813$ as a mixed number.
1.
Rewrite $\frac{{95}}{{26}}$ as a mixed number.
## Video
Similarly, we can convert a mixed number into an improper fraction. To do so, first convert the whole number part to a fraction by writing the whole number as itself divided by 1, and then add the two fractions.
Alternately, we can multiply the whole number part and the denominator of the fractional part. Next, add that product to the numerator. Finally, express the number as that product divided by the denominator.
## Example 3.61
### Rewriting a Mixed Number as an Improper Fraction
Rewrite $549549$ as an improper fraction.
1.
Rewrite $9\frac{5}{{14}}$ as an improper fraction.
## Tech Check
### Using Desmos to Rewrite a Mixed Number as an Improper Fraction
Desmos can be used to convert from a mixed number to an improper fraction. To do so, we use the idea that a mixed number, such as $56115611$, is another way to represent $5+6115+611$. If $5+6115+611$ is entered in Desmos, the result is the decimal form of the number. However, clicking the fraction button to the left will convert the decimal to an improper fraction, $61116111$. As an added bonus, Desmos will automatically reduce the fraction to lowest terms.
## Converting Rational Numbers Between Decimal and Fraction Forms
Understanding what decimals represent is needed before addressing conversions between the fractional form of a number and its decimal form, or writing a number in decimal notation. The decimal number 4.557 is equal to $45571,00045571,000$. The decimal portion, .557, is 557 divided by 1,000. To write any decimal portion of a number expressed as a terminating decimal, divide the decimal number by 10 raised to the power equal to the number of decimal digits. Since there were three decimal digits in 4.557, we divided 557 by $103=1000103=1000$.
Decimal representations may be very long. It is convenient to round off the decimal form of the number to a certain number of decimal digits. To round off the decimal form of a number to $nn$ (decimal) digits, examine the ($n+1n+1$)st decimal digit. If that digit is 0, 1, 2, 3, or 4, the number is rounded off by writing the number to the $nn$th decimal digit and no further. If the ($n+1n+1$)st decimal digit is 5, 6, 7, 8, or 9, the number is rounded off by writing the number to the $nn$th digit, then replacing the $nn$th digit by one more than the $nn$th digit.
## Example 3.62
### Rounding Off a Number in Decimal Form to Three Digits
Round 5.67849 to three decimal digits.
1.
Round 5.1082 to three decimal places.
## Example 3.63
### Rounding Off a Number in Decimal Form to Four Digits
Round 45.11475 to four decimal digits.
1.
Round 18.6298 to two decimal places.
To convert a rational number in fraction form to decimal form, use your calculator to perform the division.
## Example 3.64
### Converting a Rational Number in Fraction Form into Decimal Form
Convert $47254725$ into decimal form.
1.
Convert $\frac{{48}}{{30}}$ into decimal form.
Converting a terminating decimal to the fractional form may be done in the following way:
Step 1: Count the number of digits in the decimal part of the number, labeled $nn$.
Step 2: Raise 10 to the $nn$th power.
Step 3: Rewrite the number without the decimal.
Step 4: The fractional form is the number from Step 3 divided by the result from Step 2.
This process works due to what decimals represent and how we work with mixed numbers. For example, we could convert the number 7.4536 to fractional from. The decimal part of the number, the .4536 part of 7.4536, has four digits. By the definition of decimal notation, the decimal portion represents $4,536104=4,53610,0004,536104=4,53610,000$. The decimal number 7.4536 is equal to the improper fraction $74,53610,00074,53610,000$. Adding those to fractions yields $74,53610,00074,53610,000$.
## Example 3.65
### Converting from Decimal Form to Fraction Form with Terminating Decimals
Convert 3.2117 to fraction form.
1.
Convert 17.03347 to fraction form.
The process is different when converting from the decimal form of a rational number into fraction form when the decimal form is a repeating decimal. This process is not covered in this text.
## Multiplying and Dividing Rational Numbers
Multiplying rational numbers is less complicated than adding or subtracting rational numbers, as there is no need to find common denominators. To multiply rational numbers, multiply the numerators, then multiply the denominators, and write the numerator product divided by the denominator product. Symbolically, $ab×cd=a×cb×dab×cd=a×cb×d$. As always, rational numbers should be reduced to lowest terms.
## FORMULA
If $bb$ and $dd$ are non-zero integers, then $ab×cd=a×cb×dab×cd=a×cb×d$.
## Example 3.66
### Multiplying Rational Numbers
Calculate $1225×10211225×1021$.
1.
Calculate $\frac{{45}}{{88}} \times \frac{{28}}{{75}}$.
## Video
As with multiplication, division of rational numbers can be done using a calculator.
## Example 3.67
### Dividing Decimals with a Calculator
Calculate 3.45 ÷ 2.341 using a calculator. Round to three decimal places if necessary.
1.
Calculate $45.63 \div 17.13$ using a calculator. Round to three decimal places, if necessary.
Before discussing division of fractions without a calculator, we should look at the reciprocal of a number. The reciprocal of a number is 1 divided by the number. For a fraction, the reciprocal is the fraction formed by switching the numerator and denominator. For the fraction $abab$, the reciprocal is $baba$. An important feature for a number and its reciprocal is that their product is 1.
When dividing two fractions by hand, find the reciprocal of the divisor (the number that is being divided into the other number). Next, replace the divisor by its reciprocal and change the division into multiplication. Then, perform the multiplication. Symbolically,$ba÷cd=ab×dc=a×db×cba÷cd=ab×dc=a×db×c$. As before, reduce to lowest terms.
## FORMULA
If $b,cb,c$ and $dd$ are non-zero integers, then $ba÷cd=ab×dc=a×db×cba÷cd=ab×dc=a×db×c$.
## Example 3.68
### Dividing Rational Numbers
1. Calculate $421÷635421÷635$.
2. Calculate $18÷52818÷528$.
1.
Calculate $\frac{{46}}{{175}} \div \frac{{69}}{{285}}$.
2.
Calculate $\frac{3}{{40}} \div \frac{{42}}{{55}}$.
## Applying the Order of Operations to Simplify Expressions
The order of operations for rational numbers is the same as for integers, as discussed in Order of Operations. The order of operations makes it easier for anyone to correctly calculate and represent. The order follows the well-known acronym PEMDAS:
P Parentheses E Exponents M/D Multiplication and division A/S Addition and subtraction
The first step in calculating using the order of operations is to perform operations inside the parentheses. Moving down the list, next perform all exponent operations moving from left to right. Next (left to right once more), perform all multiplications and divisions. Finally, perform the additions and subtractions.
## Example 3.69
### Applying the Order of Operations with Rational Numbers
Correctly apply the rules for the order of operations to accurately compute $(57−27)×23(57−27)×23$.
1.
Correctly apply the rules for the order of operations to accurately compute ${\left( {\frac{3}{{16}} + \frac{7}{{16}}} \right)^2} + \frac{1}{5} \div \frac{3}{{10}}$.
## Example 3.70
### Applying the Order of Operations with Rational Numbers
Correctly apply the rules for the order of operations to accurately compute $4+23÷((59)2−(23+5))24+23÷((59)2−(23+5))2$.
1.
Correctly apply the rules for the order of operations to accurately compute $\left( {\frac{3}{5} + 2} \right) \times {\left( {\frac{4}{5} - \frac{1}{2}} \right)^2} \div \frac{{11}}{{15}}$.
## Applying the Density Property of Rational Numbers
Between any two rational numbers, there is another rational number. This is called the density property of the rational numbers.
Finding a rational number between any two rational numbers is very straightforward.
Step 1: Add the two rational numbers.
Step 2: Divide that result by 2.
The result is always a rational number. This follows what we know about rational numbers. If two fractions are added, then the result is a fraction. Also, when a fraction is divided by a fraction (and 2 is a fraction), then we get another fraction. This two-step process will give a rational number, provided the first two numbers were rational.
## Example 3.71
### Applying the Density Property of Rational Numbers
Demonstrate the density property of rational numbers by finding a rational number between $411411$ and $712712$.
1.
Demonstrate the density property of rational numbers by finding a rational number between $\frac{{27}}{{13}}$ and $\frac{{21}}{{10}}$.
## Solving Problems Involving Rational Numbers
Rational numbers are used in many situations, sometimes to express a portion of a whole, other times as an expression of a ratio between two quantities. For the sciences, converting between units is done using rational numbers, as when converting between gallons and cubic inches. In chemistry, mixing a solution with a given concentration of a chemical per unit volume can be solved with rational numbers. In demographics, rational numbers are used to describe the distribution of the population. In dietetics, rational numbers are used to express the appropriate amount of a given ingredient to include in a recipe. As discussed, the application of rational numbers crosses many disciplines.
## Example 3.72
### Mixing Soil for Vegetables
James is mixing soil for a raised garden, in which he plans to grow a variety of vegetables. For the soil to be suitable, he determines that $2525$ of the soil can be topsoil, but $2525$ needs to be peat moss and $1515$ has to be compost. To fill the raised garden bed with 60 cubic feet of soil, how much of each component does James need to use?
1.
Ashley wants to study for 10 hours over the weekend. She plans to spend half the time studying math, a quarter of the time studying history, an eighth of the time studying writing, and the remaining eighth of the time studying physics. How much time will Ashley spend on each of those subjects?
## Example 3.73
### Determining the Number of Specialty Pizzas
At Bella’s Pizza, one-third of the pizzas that are ordered are one of their specialty varieties. If there are 273 pizzas ordered, how many were specialty pizzas?
1.
Danny, a nutritionist, is designing a diet for her client, Callum. Danny determines that Callum’s diet should be 30% protein. If Callum consumes 2,400 calories per day, how many calories of protein should Danny tell Callum to consume?
## Using Fractions to Convert Between Units
A common application of fractions is called unit conversion, or converting units, which is the process of changing from the units used in making a measurement to different units of measurement.
For instance, 1 inch is (approximately) equal to 2.54 cm. To convert between units, the two equivalent values are made into a fraction. To convert from the first type of unit to the second type, the fraction has the second unit as the numerator, and the first unit as the denominator.
From the inches and centimeters example, to change from inches to centimeters, we use the fraction $2.54cm1in2.54cm1in$. If, on the other hand, we wanted to convert from centimeters to inches, we’d use the fraction $1in2.54cm1in2.54cm$. This fraction is multiplied by the number of units of the type you are converting from, which means the units of the denominator are the same as the units being multiplied.
## Example 3.74
### Converting Liters to Gallons
It is known that 1 liter (L) is 0.264172 gallons (gal). Use this to convert 14 liters into gallons.
1.
One mile is equal to 1.60934 km. Convert 200 miles to kilometers. Round off the answer to three decimal places.
## Example 3.75
### Converting Centimeters to Inches
It is known that 1 inch is 2.54 centimeters. Use this to convert 100 centimeters into inches.
1.
It is known that 4 quarts equals 3.785 liters. If you have 25 quarts, how many liters do you have? Round off to three decimal places.
## Defining and Applying Percent
A percent is a specific rational number and is literally per 100. $nn$ percent, denoted $nn$%, is the fraction $n100n100$.
## Example 3.76
### Rewriting a Percentage as a Fraction
Rewrite the following as fractions:
1. 31%
2. 93%
Rewrite the following as fractions:
1.
4%
2.
50%
## Example 3.77
### Rewriting a Percentage as a Decimal
Rewrite the following percentages in decimal form:
1. 54%
2. 83%
Rewrite the following percentages in decimal form:
1.
14%
2.
7%
You should notice that you can simply move the decimal two places to the left without using the fractional definition of percent.
Percent is used to indicate a fraction of a total. If we want to find 30% of 90, we would perform a multiplication, with 30% written in either decimal form or fractional form. The 90 is the total, 30 is the percentage, and 27 (which is $0.30×900.30×90$) is the percentage of the total.
## FORMULA
$n%n%$ of $xx$ items is $n100×xn100×x$. The $xx$ is referred to as the total, the $nn$ is referred to as the percent or percentage, and the value obtained from $n100×xn100×x$ is the part of the total and is also referred to as the percentage of the total.
## Example 3.78
### Finding a Percentage of a Total
1. Determine 40% of 300.
2. Determine 64% of 190.
1.
Determine 25% of 1,200.
2.
Determine 53% of 1,588.
In the previous situation, we knew the total and we found the percentage of the total. It may be that we know the percentage of the total, and we know the percent, but we don't know the total. To find the total if we know the percentage the percentage of the total, use the following formula.
## FORMULA
If we know that $nn$% of the total is $xx$, then the total is $100×xn100×xn$.
## Example 3.79
### Finding the Total When the Percentage and Percentage of the Total Are Known
1. What is the total if 28% of the total is 140?
2. What is the total if 6% of the total is 91?
1.
What is the total if 25% of the total is 30?
2.
What is the total if 45% of the total is 360?
The percentage can be found if the total and the percentage of the total is known. If you know the total, and the percentage of the total, first divide the part by the total. Move the decimal two places to the right and append the symbol %. The percentage may be found using the following formula.
## FORMULA
The percentage, $nn$, of $bb$ that is $aa$ is $ab×100%ab×100%$.
## Example 3.80
### Finding the Percentage When the Total and Percentage of the Total Are Known
Find the percentage in the following:
1. Total is 300, percentage of the total is 60.
2. Total is 440, percentage of the total is 176.
Find the percentage in the following:
1.
Total is 1,000, percentage of the total is 70.
2.
Total is 500, percentage of the total is 425.
## Solve Problems Using Percent
In the media, in research, and in casual conversation percentages are used frequently to express proportions. Understanding how to use percent is vital to consuming media and understanding numbers. Solving problems using percentages comes down to identifying which of the three components of a percentage you are given, the total, the percentage, or the percentage of the total. If you have two of those components, you can find the third using the methods outlined previously.
## Example 3.81
### Percentage of Students Who Are Sleep Deprived
A study revealed that 70% of students suffer from sleep deprivation, defined to be sleeping less than 8 hours per night. If the survey had 400 participants, how many of those participants had less than 8 hours of sleep per night?
1.
Riley has a daily calorie intake of 2,200 calories and wants to take in 20% of their calories as protein. How many calories of protein should be in their daily diet?
## Example 3.82
### Amazon Prime Subscribers
There are 126 million users who are U.S. Amazon Prime subscribers. If there are 328.2 million residents in the United States, what percentage of U.S. residents are Amazon Prime subscribers?
1.
A small town has 450 registered voters. In the primaries, 54 voted. What percentage of registered voters in that town voted in the primaries?
## Example 3.83
### Finding the Percentage When the Total and Percentage of the Total Are Known
Evander plays on the basketball team at their university and 73% of the athletes at their university receive some sort of scholarship for attending. If they know 219 of the student-athletes receive some sort of scholarship, how many student-athletes are at the university?
1.
A store declares a deep discount of 40% for an item, which they say will save $30. What was the original price of the item? ## Check Your Understanding 17. Identify which of the following are rational numbers. $- 41,{\mkern 1mu} \,\sqrt {13,} {\mkern 1mu} \,\frac{4}{3},\,\,2.75,\,{\mkern 1mu} 0.2\overline {13}$ 18. Express $\frac{{18}}{{30}}$ in lowest terms. 19. Calculate $\frac{3}{8} + \frac{5}{{12}}$ and express in lowest terms. 20. Convert 0.34 into fraction form. 21. Convert $\frac{{47}}{{12}}$ into a mixed number. 22. Calculate $\frac{2}{9} \times \frac{{21}}{{22}}$ and express in lowest terms. 23. Calculate $\frac{2}{5} \div \frac{3}{{10}} + \frac{1}{6}$. 24. Identify a rational number between $\frac{7}{8}$ and $\frac{{20}}{{21}}$. 25. Rewrite $3\frac{{2}}{{7}}$ as an improper fraction. 26. Lina decides to save $\frac{1}{8}$ of her take-home pay every paycheck. Her most recent paycheck was for$882. How much will she save from that paycheck?
27.
Determine 38% of 600.
28.
A microchip factory has decided to increase its workforce by 10%. If it currently has 70 employees, how many new employees will the factory hire?
## Section 3.4 Exercises
For the following exercises, identify which of the following are rational numbers.
1 .
4.598
2 .
$\sqrt {144}$
3 .
$\sqrt {131}$
For the following exercises, reduce the fraction to lowest terms
4 .
$\frac{8}{{10}}$
5 .
$\frac{{30}}{{105}}$
6 .
$\frac{{36}}{{539}}$
7 .
$\frac{{231}}{{490}}$
8 .
$\frac{{750}}{{17,875}}$
For the following exercises, do the indicated conversion. If it is a repeating decimal, use the correct notation.
9 .
Convert $\frac{{25}}{6}$ to a mixed number.
10 .
Convert $\frac{{240}}{{53}}$ to a mixed number.
11 .
Convert $2\frac{3}{8}$ to an improper fraction.
12 .
Convert $15\frac{7}{{30}}$ to an improper fraction.
13 .
Convert $\frac{4}{9}$ to decimal form.
14 .
Convert $\frac{{13}}{{20}}$ to decimal form.
15 .
Convert $\frac{{27}}{{625}}$ to decimal form.
16 .
Convert $\frac{{11}}{{14}}$ to decimal form.
17 .
Convert 0.23 to fraction form and reduce to lowest terms.
18 .
Convert 3.8874 to fraction form and reduce to lowest terms.
For the following exercises, perform the indicated operations. Reduce to lowest terms.
19 .
$\frac{3}{5} + \frac{3}{{10}}$
20 .
$\frac{3}{{14}} + \frac{8}{{21}}$
21 .
$\frac{{13}}{{36}} - \frac{{14}}{{99}}$
22 .
$\frac{{13}}{{24}} - \frac{4}{{117}}$
23 .
$\frac{3}{7} \times \frac{{21}}{{48}}$
24 .
$\frac{{48}}{{143}} \times \frac{{77}}{{120}}$
25 .
$\frac{{14}}{{27}} \div \frac{7}{{12}}$
26 .
$\frac{{44}}{{75}} \div \frac{{484}}{{285}}$
27 .
$\left( {\frac{3}{5} + \frac{2}{7}} \right) \times \,\frac{{10}}{{21}}$
28 .
$\frac{3}{8} \times \left( {\frac{{13}}{{12}} - \frac{{35}}{{36}}} \right)$
29 .
${\left( {\frac{3}{7} + \frac{5}{{16}}} \right)^2} - \frac{5}{{12}}$
30 .
$\frac{3}{8} \times {\left( {\frac{4}{9} - \frac{1}{8}} \right)^2}$
31 .
${\left( {\frac{2}{5} \times \left( {\frac{7}{8} - \frac{2}{3}} \right)} \right)^2} \div \left( {\frac{4}{9} + \frac{5}{6}} \right) + \frac{7}{{12}}$
32 .
$\left( {\frac{1}{5} \div \left( {\frac{3}{{10}} + \frac{{11}}{{15}}} \right)} \right) \times \left( {\frac{2}{{21}} + \frac{5}{9}} \right) - {\left( {\frac{8}{{15}} \div \frac{4}{{33}}} \right)^2}$
33 .
Find a rational number between $\frac{8}{{17}}$ and $\frac{{15}}{{28}}$
34 .
Find a rational number between $\frac{3}{{50}}$ and $\frac{{13}}{{98}}$.
35 .
Find two rational numbers between $\frac{3}{{10}}$ and $\frac{{19}}{{45}}$.
36 .
Find three rational numbers between $\frac{5}{{12}}$ and $\frac{{175}}{{308}}$.
37 .
Convert 24% to fraction form and reduce completely.
38 .
Convert 95% to fraction form and reduce completely.
39 .
Convert 0.23 to a percentage.
40 .
Convert 1.22 to a percentage.
41 .
Determine 30% of 250.
42 .
Determine 75% of 600.
43 .
If 25% of a group is 41 members, how many members total are in the group?
44 .
If 80% of the total is 60, how much is in the total?
45 .
13 is what percent of 20?
46 .
80 is what percent of 320?
47 .
Professor Donalson’s history of film class has 60 students. Of those students, $\frac{2}{5}$ say their favorite movie genre is comedy. How many of the students in Professor Donalson’s class name comedy as their favorite movie genre?
48 .
Naia’s dormitory floor has 80 residents. Of those, $\frac{3}{8}$ play Fortnight for at least 15 hours per week. How many students on Naia’s floor play Fortnight at least 15 hours per week?
49 .
In Tara’s town there are 24,000 people. Of those, $\frac{{13}}{{100}}$ are food insecure. How many people in Tara’s town are food insecure?
50 .
Roughly $\frac{4}{5}$ of air is nitrogen. If an enclosure holds 2,000 liters of air, how many liters of nitrogen should be expected in the enclosure?
51 .
To make the dressing for coleslaw, Maddie needs to mix it with $\frac{3}{5}$ mayonnaise and $\frac{2}{5}$ apple cider vinegar. If Maddie wants to have 8 cups of dressing, how many cups of mayonnaise and how many cups of apple cider vinegar does Maddie need?
52 .
Malika is figuring out their schedule. They wish to spend $\frac{4}{{15}}$ of their time sleeping, $\frac{1}{{3}}$ of their time studying and going to class, $\frac{1}{{5}}$ of their time at work, and $\frac{2}{{15}}$ of their time doing other activities, such as entertainment or exercising. There are 168 hours in a week. How many hours in a week will Malika spend:
1. Sleeping?
2. Studying and going to class?
3. Not sleeping?
53 .
Roughly 20.9% of air is oxygen. How much oxygen is there in 200 liters of air?
54 .
65% of college students graduate within 6 years of beginning college. A first-year cohort at a college contains 400 students. How many are expected to graduate within 6 years?
55 .
A 20% discount is offered on a new laptop. How much is the discount if the new laptop originally cost $700? 56 . Leya helped at a neighborhood sale and was paid 5% of the proceeds. If Leya is paid$171.25, what were the total proceeds from the neighborhood sale?
57 .
Unit Conversion. 1 kilogram (kg) is equal to 2.20462 pounds. Convert 13 kg to pounds. Round to three decimal places, if necessary.
58 .
Unit Conversion. 1 kilogram (kg) is equal to 2.20462 pounds. Convert 200 pounds to kilograms. Round to three decimal places, if necessary.
59 .
Unit Conversion. There are 12 inches in a foot, 3 feet in a yard, and 1,760 yards in a mile. Convert 10 miles to inches. To do so, first convert miles to yards. Next, convert the yards to feet. Last, convert the feet to inches.
60 .
Unit Conversion. There are 1,000 meters (m) in a kilometer (km), and 100 centimeters (cm) in a meter. Convert 4 km to centimeters.
61 .
Markup. In this exercise, we introduce the concept of markup. The markup on an item is the difference between how much a store sells an item for and how much the store paid for the item. Suppose Wegmans (a northeastern U.S. grocery chain) buys cereal at $1.50 per box and sells the cereal for$2.29.
1. Determine the markup in dollars.
2. The markup is what percent of the original cost? Round the percentage to one decimal place.
62 .
In this exercise, we explore what happens when an item is marked up by a percentage, and then marked down using the same percentage.
Wegmans purchases an item for $5 per unit. The markup on the item is 25%. 1. Calculate the markup on the item, in dollars. 2. What is the price for which Wegmans sells the item? This is the price Wegmans paid, plus the markup. 3. Suppose Wegmans then offers a 25% discount on the sale price of the item (found in part b). In dollars, how much is the discount? 4. Determine the price of the item after the discount (this is the sales price of the item minus the discount). Round to two decimal places. 5. Is the new price after the markup and discount equal to the price Wegmans paid for the item? Explain. 63 . Repeated Discounts. In this exercise, we explore applying more than one discount to an item. Suppose a store cuts the price on an item by 50%, and then offers a coupon for 25% off any sale item. We will find the price of the item after applying the sale price and the coupon discount. 1. The original price was$150. After the 50% discount, what is the price of the item?
2. The coupon is applied to the discount price. The coupon is for 25%. Find 25% of the sale price (found in part a).
3. Find the price after applying the coupon (this is the value from part a minus the value from part b).
4. The total amount saved on the item is the original price after all the discounts. Determine the total amount saved by subtracting the final price paid (part c) from the original price of the item.
5. Determine the effective discount percentage, which is the total amount saved divided by the original price of the item.
6. Was the effective discount percentage equal to 75%, which would be the 50% plus the 25%? Explain.
Converting Repeating Decimals to a Fraction
It was mentioned in the section that repeating decimals are rational numbers. To convert a repeating decimal to a rational number, perform the following steps:
Step 1: Label the original number $S$.
Step 2: Count the number of digits, $n$, in the repeating part of the number.
Step 3: Multiply $S$ by $10^n$ , and label this as $10^n\, ×\, S$.
Step 4: Determine ${10^n}\, ‒ \,1$.
Step 5: Calculate ${10^n}\, ×\, S\, ‒\, S$. If done correctly, the repeating part of the number will cancel out.
Step 6: If the result from Step 5 has decimal digits, count the number of decimal digits in the number from Step 5. Label this $m$.
Step 7: Remove the decimal from the result of Step 5.
Step 8: Add $m$ zeros to the end of the number from Step 4.
Step 9: Divide the result from Step 7 by the result from Step 8. This is the fraction form of the repeating decimal.
64 .
Convert $0.\overline 7$ to fraction form.
65 .
Convert $0.\overline {45}$ to fraction form.
66 .
Convert $3.1\overline 5$ to fraction form.
67 .
Convert $2.71\overline {94}$ to fraction form.
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# Set Theory
The study of sets helps in increasing the degree of relations in such a way that it optimizes the bond
between any two ,three or many quantities taken together.
## What is a Set?
Set: A well-defined collection of objects is defined as a set.
Example: Set of vowels A= {a, e, i, o, u}
Note: All the sets are denoted by capital letters
Symbols used in sets:
Union: U
Intersection: ∩
Subset: C
## Common Data Set
If there are two sets given say A and B then the common data found in these two sets is called common data set.
Example A= {a, e, i, o, u} B= {a, i, o}
Then we find the common data set between these two sets namely A ∩ B = {a, i, o}.
## Union of Sets
The joining of two set elements together is called union of sets
Example: A= {2, 5, 7} B= {a, g, h}
AUB = {a, g, h, 2, 5, 7}
## Intersection of Sets
Example A= {1, 5} B= {5}
Then we find the common data set between these two sets namely A ∩ B = {5}.
## Cardinal Number
What is a cardinal number? The number of elements in the given set is called a cardinal number.
Example: A= {1, 2, a, v}
In the above example there are 4 elements in the given set.
Hence we define cardinal number as 4.
Sets Solved Problems
Given A= {1, 2} B= {6, 3}
Find
1. A ∩ B
2. AUB
3. Cardinal number of A and B
If A= {2, 6, 7, 8} B= {2, 6, 3} find A ∩ B
A ∩ B = {2, 6}, so the common elements are shaded in yellow.
In a total of 30 players , 25 play hockey and 10 play Rugby can you predict using set theory that how players are liking to play both.
The formulae is n (A U B) = n (A) + n (B) - n (A ∩ B)
Here n (A U B) = cardinal number of both set A and set B (hockey and rugby)
n (A)= cardinal number of only set A (Hockey only)
n (B) = cardinal number of only set B (Rugby).
And
n (A ∩ B) = cardinal number for both hockey and rugby
30 = 25 +10 - n (A ∩ B) so transfer n (A ∩ B) towards left side we get
n (A ∩ B) = 35-30
= 5
So the number of players who play both is 5.
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# SINE AND COSINE FUNCTIONS
## Presentation on theme: "SINE AND COSINE FUNCTIONS"— Presentation transcript:
SINE AND COSINE FUNCTIONS
GRAPHS OF SINE AND COSINE FUNCTIONS
We are interested in the graph of y = f(x) = sin x
Start with a "t" chart and let's choose values from our unit circle and find the sine values. plot these points x y 1 - 1 x y = sin x We are dealing with x's and y's on the unit circle to find values. These are completely different from the x's and y's used here for our function.
y = f(x) = sin x choose more values x y = sin x plot these points
join the points x y 1 - 1 If we continue picking values for x we will start to repeat since this is periodic.
Here is the graph y = f(x) = sin x showing from -2 to 6
Here is the graph y = f(x) = sin x showing from -2 to 6. Notice it repeats with a period of 2. 2 2 2 2 It has a maximum of 1 and a minimum of -1 (remember that is the range of the sine function)
What are the x intercepts?
Where does sin x = 0? …-3, -2, -, 0, , 2, 3, 4, . . . Where is the function maximum? Where does sin x = 1?
Where is the function minimum?
Where does sin x = -1?
Thinking about transformations that you learned and knowing what y = sin x looks like, what do you suppose y = sin x + 2 looks like? This is often written with terms traded places so as not to confuse the 2 with part of sine function y = 2 + sin x The function value (or y value) is just moved up 2. y = sin x
Thinking about transformations that you've learned and knowing what y = sin x looks like, what do you suppose y = sin x - 1 looks like? y = sin x The function value (or y value) is just moved down 1. y = sin x
Thinking about transformations that you learned and knowing what y = sin x looks like, what do you suppose y = sin (x + /2) looks like? y = sin x This is a horizontal shift by - /2 y = sin (x + /2)
Thinking about transformations that you learned and knowing what y = sin x looks like, what do you suppose y = - sin (x )+1 looks like? y = 1 - sin (x ) This is a reflection about the x axis (shown in green) and then a vertical shift up one. y = - sin x y = sin x
What would the graph of y = f(x) = cos x look like?
We could do a "t" chart and let's choose values from our unit circle and find the cosine values. plot these points x y 1 - 1 x y = cos x We could have used the same values as we did for sine but picked ones that gave us easy values to plot.
y = f(x) = cos x Choose more values. x y = cos x plot these points y 1
- 1 cosine will then repeat as you go another loop around the unit circle
Here is the graph y = f(x) = cos x showing from -2 to 6
Here is the graph y = f(x) = cos x showing from -2 to 6. Notice it repeats with a period of 2. 2 2 2 2 It has a maximum of 1 and a minimum of -1 (remember that is the range of the cosine function)
Recall that an even function (which the cosine is) is symmetric with respect to the y axis as can be seen here
What are the x intercepts?
Where does cos x = 0? Where is the function maximum? Where does cos x = 1? …-4, -2, , 0, 2, 4, . . .
Where is the function minimum?
Where does cos x = -1? …-3, -, , 3, . . .
Let's try y = 3 - cos (x - /4) y = - cos x y = cos x
You could graph transformations of the cosine function the same way you've learned for other functions. moves right /4 moves up 3 Let's try y = 3 - cos (x - /4) y = - cos x y = cos x reflects over x axis y = 3 - cos x y = 3 - cos (x - /4)
amplitude of this graph is 2
What would happen if we multiply the function by a constant? All function values would be twice as high y = 2 sin x amplitude is here amplitude of this graph is 2 y = 2 sin x y = sin x The highest the graph goes (without a vertical shift) is called the amplitude.
For y = A cos x and y = A sin x, A is the amplitude.
What is the amplitude for the following? y = 4 cos x y = -3 sin x amplitude is 3 amplitude is 4
The last thing we want to see is what happens if we put a coefficient on the x.
y = sin 2x y = sin 2x y = sin x It makes the graph "cycle" twice as fast. It does one complete cycle in half the time so the period becomes .
What do you think will happen to the graph if we put a fraction in front?
y = sin 1/2 x y = sin x The period for one complete cycle is twice as long or 4
The period T = So if we look at y = sin x the affects the period.
This will be true for cosine as well. What is the period of y = cos 4x? y = cos x This means the graph will "cycle" every /2 or 4 times as often y = cos 4x
absolute value of this is the amplitude
Period is 2 divided by this
Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar
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# Class 11 RD Sharma Solutions – Chapter 24 The Circle – Exercise 24.3
• Last Updated : 08 May, 2021
### Question 1. Find the equation of the circle, the end points of whose diameter are (2, -3) and (-2, 4). Find its radius and centre.
Solution:
Given that the end points of the diameter are (2, -3) and (-2, 4).
So, the equation of the circle is
(x – 2)(x + 2) + (y + 3)(y – 4) = 0
x2 – 4 + y2 – y – 12 = 0
x2 + y2 – y – 16 = 0 ……(1)
From eq(1), we get
2g = 0, 2f = -1
g = 0, f = -1/2
The center of the circle is(0, 1/2)
And
### Question 2. Find the equation of the circle, the endpoints of whose diameter are centers of the circles x2 + y2 + 6x – 14y – 1 = 0 and x2 + y2 – 4x + 10y – 2 = 0.
Solution:
Given equations are
x2 + y2 + 6x – 14y – 1 = 0
It can also written as
(x + 3)2 + (y – 7)2 = 59 ……(i)
x2 + y2 – 4x + 10y – 2y – 1 = 0
It can also written as
(x – 2)2 + (y + 5)2 = 31 ……(ii)
So, from eq(i) and (ii)
the centres of the circles are (-3, 7) and (2,-5)
Now the equation of the circle is
(x – x1)(x – x2) + (y – y1)(y – y2) = 0
(x + 3)(x – 2) + (y – 7)(y + 5) = 0
x2 + 3x – 2x – 6 + y2 – 7y + 5y – 35 = 0
x2 + y2 + x – 2y – 41 = 0
### Question 3. The sides of a square are x = 6, x = 9, y = 3, and y = 6. Find the equation of a circle drawn on the diagonal of the square as its diameter.
Solution:
Let us considered AB, BC, CD, and DA are the sides of the square ABCD be resented by the given equations
y = 3, x = 6, y = 6 and x = 9
So, the coordinates are
A(6, 3), B(9, 3), C(9, 6) and D(6, 6)
So, the equation of the circle with diagonal AC
(x – 6)(x – 9) + (4 – 3)(4 – 6) = 0
x2 – 6x – 9x + 54 + y2 – 3y – 6y + 18 = 0
x2 + y2 – 15x – 9y + 72 = 0
And the equation of the circle with diagonal BD as diameter is
(x – 9)(x – 6) + (y – 3)(y – 6) = 0
x2 – 9x – 6x + 54 + y2 – 3y – 6y + 18 = 0
x2 + y2 – 15x – 9y + 72 = 0
x2 + y2 – 15 – 9y + 72 = 0
### Question 4. Find the equation of a circle circumscribing the rectangle whose sides are x – 2y = 4, 3x + y = 22, x – 3y = 14 and 3x + y = 62.
Solution:
The given equations of the sides of the rectangle are
x – 3y = 4 ———– (i)
3x + y = 22 ———–(ii)
x – 3y = 14 ———–(iii)
3x + y = 62 ———-(iv)
Let us assume A, B, C, and D are the points intersection of the lines (i), (ii), (iii), and (iv)
So, A(7, 1), B(8, -2), C(20, 2) and D(19, 5)
AC will be the diameter of the circle
Hence, the equation of circle is
(x – 7)(x – 20) + (y – 1)(y – 2) = 0
x2 + y2 – 27x – 3y + 142 = 0
### Question 5. Find the equation of the circle passing through the origin and the points where the line 3x + 4y = 12 meets the axes of coordinates.
Solution:
Given equation of line is 3x + 4y = 12
So, it will meet the axis at A(0, 3) and B(4, 0)
Since the circle passes through origin A and B
So, AB is a diameter
Hence, the equation of circle is
(x – 0)(x – 4) + (y – 3)(y – 0) = 0
x2 + y2 – 4x – 3y = 0
### Question 6. Find the equation of the circle which passes through the origin and cuts off intercepts a and b respectively y from x and y-axes.
Solution:
It is given that the circle is passes through origin and cut intercept a and b on
x-axis and y-axis
So, the coordinates of circle A(0, b) and B(a, 0)
Here, AB is the diameter of the circle
Hence, the equation of the circle is
(x – a)(x – 0) + (y – 0)(y – b) = 0
x2 + y2 ± ax ± by = 0
### Question 7. Find the equation of the circle whose diameter is the line segment joining (-4, 3) and (12, -1). Find also the intercept made by it on y-axis.
Solution:
Given that the line segment A(-4, 3) and B(12, -1) is joining a diameter
So, the equation of circle in diameter form is,
(x + 4)(x – 12) + (y – 3)(y + 1) = 0
x2 – 8x – 48 + y2 – 2y – 3 = 0
x2 – 8x – 2y + y2 – 51 = 0 ……(1)
To find y-intercept, put x = 0 in eq(1), we get
y2 – 2y – 51 = 0
So, y intercepts are 1 ± 4√13
### Question 8. The abscissae of the two points A and B are the roots of the equation x2 + 2ax – b2 = 0 and their coordinates are the roots of the equation x2 + 2px – q2 = 0. Find the equation of the circle with AB as diameter. Also, find its radius.
Solution:
The given equations are
x2 + 2ax – b2 = 0 ……..(i)
x2 + 2px – q2 = 0 ………(ii)
Now root of eq(i)
and roots of eq(ii)
So, the coordinates of A = ()
and B = ()
So, the equation of circle is
x2 + y2 + 2ax + 2py – (a2 + b2 + p2 + q2) + a2 + p2 = 0
x2 + y2 + 2ax + 2py – (b2 + q2) = 0
### Question 9. ABCD is a square whose side is a, taking AB and AD as axes, prove that the equation of the circle circumscribing the square is x2 + y2 – a(x + y) = 0.
Solution:
Given that, ABCD is a square whose side is a.
Also, given that AD and AD are axes, so point of intersection is (0, 0)
Also, the other point on the diagonal of the square will have coordinates (a, a).
It is given that equation of the circle circumscribe the square.
So, (0, 0) and (a, a) will be the end points of the diameter of the circle.
So, the equation of the circle is
(x – 0)(x – a) + (y – 0)(y – a) = 0
x2 – ax + y2 – ay = 0
x2 + y2 – a(x + y) = 0
### Question 10. The line 2x – y + 6 = 0 meets the circle x2 + y2 – 2y – 9 = 0 at A and B. Find the equation of the circle on AB as diameter.
Solution:
The given equations of line and circle are
2x – y + 6 = 0 ……(i)
x2 + y2 – 2y – 9 = 0 …….(ii)
The point of intersection of eq (i) and (ii) is
x2 + (2x + 6)2 – 2(2x + 6) – 9 = 0
x2 + 4x2 + 24x + 36 – 4x – 12 – 9 = 0
5x2 + 20x + 15 = 0
(x + 3)(x + 1) = 0
⇒ x = (-3, -1)
and y = (0, 4)
So, point A(-3, 0) and B(-1, 4)
Here, AB is a diameter, so the equation of circle is
(x + 3)(x + 1) + (y – 0)(y – 4) = 0
x2 + y2 + 4x – 4y + 3 = 0
### Question 11. Find the equation of the circle which circumscribes the triangle formed by the lines x = 0, y = 0, and lx + my = 1.
Solution:
The given equations of lines are
x = 0 …….(i)
y = 0 …….(ii)
lx + my = 1 …….(iii)
The line eq(iii) cuts the axis at
A(0, 1/m) and B(1/l, 0)
Now, AB will be the diameter of circle,
So, the equation of circle will be,
(x – 1/l)(x – 0) + (y – 0)(y – 1/m) = 0
x2 + y2 – x/l – y/m = 0
### Question 12. Find the equations of the circle which passes through the origin and cut of equal chords of √2 units from the lines y = x and y = -x.
Solution:
Let the angles between y = x and y = -x is π/2
So, the angle between OB and OA = π/2
Hence, AB, BC, CD and AD are diameter of circles.
so, ∠BOQ = π/4
sin∠BOQ = BQ/OB
sin π/4 = BQ/√2
1/√2 = BQ/√2
BQ = 1
So, the radius of circle(OQ) = 1
And the coordinates of B is (1, 1)
Similarly, coordinates of A(-1, 1), C(1, -1), D(-1, -1)
Now the equation of circle with diameter AB is
(x + 1)(x – 1) + (y – 1)(y – 1) = 0
x2 – 2x + 1 + y2 – 1 = 0
x2 + y2 – 2x = 0
The equation of circle with diameter BC is
(x – 1)(x – 1) + (y – 1)(y + 1) = 0
x2 – 2x + 1 + y2 – 1 = 0
x2 + y2 – 2x = 0
The equation of circle with diameter CD is
(x + 1)(x – 1) + (y + 1)(y + 1) = 0
x2 – 1 + y2 + 2y + 1 = 0
x2 + y2 + 2y = 0
And the equation of circle with diameter AD is
(x + 1)(x + 1) + (y – 1)(y + 1) = 0
x2 + 2x + 1 + y2 – 1 = 0
x2 + y2 + 2x = 0
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# Chapter 7 Rational Expressions and Equations
## 7.1 Simplifying Rational Expressions
### Learning Objectives
1. Determine the restrictions to the domain of a rational expression.
2. Simplify rational expressions.
3. Simplify expressions with opposite binomial factors.
4. Simplify and evaluate rational functions.
## Rational Expressions, Evaluating, and Restrictions
A rational number, or fraction $a b$, is a real number defined as a quotient of two integers a and b, where $b≠0$. Similarly, we define a rational expressionThe quotient $PQ$ of two polynomials P and Q, where Q ≠ 0., or algebraic fractionTerm used when referring to a rational expression. $PQ$, as the quotient of two polynomials P and Q, where $Q≠0$. Some examples of rational expressions follow:
The example $x+3x−5$ consists of linear expressions in both the numerator and denominator. Because the denominator contains a variable, this expression is not defined for all values of x.
Example 1: Evaluate $x+3x−5$ for the set of x-values {−3, 4, 5}.
Solution: Substitute the values in for x.
Answer: When $x=−3$, the value of the rational expression is 0; when $x=4$, the value of the rational expression is −7; and when $x=5$, the value of the rational expression is undefined.
This example illustrates that variables are restricted to values that do not make the denominator equal to 0. The domain of a rational expressionThe set of real numbers for which the rational expression is defined. is the set of real numbers for which it is defined, and restrictionsThe set of real numbers for which a rational expression is not defined. are the real numbers for which the expression is not defined. We often express the domain of a rational expression in terms of its restrictions.
Example 2: Find the domain of the following: $x+72x2+x−6$.
Solution: In this example, the numerator $x+7$ is a linear expression and the denominator $2x2+x−6$ is a quadratic expression. If we factor the denominator, then we will obtain an equivalent expression.
Because rational expressions are undefined when the denominator is 0, we wish to find the values for x that make it 0. To do this, apply the zero-product property. Set each factor in the denominator equal to 0 and solve.
We conclude that the original expression is defined for any real number except 3/2 and −2. These two values are the restrictions to the domain.
It is important to note that −7 is not a restriction to the domain because the expression is defined as 0 when the numerator is 0.
Answer: The domain consists of any real number x, where $x≠32$ and $x≠−2$.
We can express the domain of the previous example using notation as follows:
The restrictions to the domain of a rational expression are determined by the denominator. Ignore the numerator when finding those restrictions.
Example 3: Determine the domain: $x4+x3−2x2−xx2−1$.
Solution: To find the restrictions to the domain, set the denominator equal to 0 and solve:
These two values cause the denominator to be 0. Hence they are restricted from the domain.
Answer: The domain consists of any real number x, where $x≠±1$.
Example 4: Determine the domain: $x2−254$.
Solution: There is no variable in the denominator and thus no restriction to the domain.
Answer: The domain consists of all real numbers, R.
## Simplifying Rational Expressions
When simplifying fractions, look for common factors that cancel. For example,
We say that the fraction 12/60 is equivalent to 1/5. Fractions are in simplest form if the numerator and denominator share no common factor other than 1. Similarly, when working with rational expressions, look for factors to cancel. For example,
The resulting rational expression is equivalent if it shares the same domain. Therefore, we must make note of the restrictions and write
In words, $x+4(x−3)(x+4)$ is equivalent to $1x−3$, if $x≠3$ and $x≠−4$. We can verify this by choosing a few values with which to evaluate both expressions to see if the results are the same. Here we choose $x=7$ and evaluate as follows:
It is important to state the restrictions before simplifying rational expressions because the simplified expression may be defined for restrictions of the original. In this case, the expressions are not equivalent. Here −4 is defined for the simplified equivalent but not for the original, as illustrated below:
Example 5: Simplify and state the restriction: $25x215x3$.
Solution: In this example, the expression is undefined when x is 0.
Therefore, the domain consists of all real numbers x, where $x≠0$. With this understanding, we can cancel common factors.
Answer: $53x$, where $x≠0$
Example 6: State the restrictions and simplify: $3x(x−5)(2x+1)(x−5)$.
Solution: To determine the restrictions, set the denominator equal to 0 and solve.
The domain consists of all real numbers except for −1/2 and 5. Next, we find an equivalent expression by canceling common factors.
Answer: $3x2x+1$, where $x≠−12$ and $x≠5$
Typically, rational expressions are not given in factored form. If this is the case, factor first and then cancel. The steps are outlined in the following example.
Example 7: State the restrictions and simplify: $3x+6x2+x−2$.
Solution:
Step 1: Completely factor the numerator and denominator.
Step 2: Determine the restrictions to the domain. To do this, set the denominator equal to 0 and solve.
The domain consists of all real numbers except −2 and 1.
Step 3: Cancel common factors, if any.
Answer: $3x−1$, where $x≠1$ and $x≠−2$
Example 8: State the restrictions and simplify: $x2+7x−30x2−7x+12$.
Solution: First, factor the numerator and denominator.
Any value of x that results in a value of 0 in the denominator is a restriction. By inspection, we determine that the domain consists of all real numbers except 4 and 3. Next, cancel common factors.
Answer: $x+10x−4$, where $x≠3$ and $x≠4$
It is important to remember that we can only cancel factors of a product. A common mistake is to cancel terms. For example,
Try this! State the restrictions and simplify: $x2−165x2−20x$.
Answer: $x+45x$, where $x≠0$ and $x≠4$
### Video Solution
(click to see video)
In some examples, we will make a broad assumption that the denominator is nonzero. When we make that assumption, we do not need to determine the restrictions.
Example 9: Simplify: $xy+y2−3x−3yx2−y2$. (Assume all denominators are nonzero.)
Solution: Factor the numerator by grouping. Factor the denominator using the formula for a difference of squares.
Next, cancel common factors.
Answer: $y−3x−y$
## Opposite Binomial Factors
Recall that the opposite of the real number a is −a. Similarly, we can define the opposite of a polynomial P to be −P. We first consider the opposite of the binomial $a−b$:
This leads us to the opposite binomial propertyIf given a binomial $a−b$, then the opposite is $−(a−b)=b−a$.:
This is equivalent to factoring out a –1.
If $a≠b$, then we can divide both sides by $(a−b)$ and obtain the following:
Example 10: State the restrictions and simplify: $3−xx−3$.
Solution: By inspection, we can see that the denominator is 0 if $x=3$. Therefore, 3 is the restriction to the domain. Apply the opposite binomial property to the numerator and then cancel.
Answer: $3−xx−3=−1$, where $x≠3$
Since addition is commutative, we have
or
Take care not to confuse this with the opposite binomial property. Also, it is important to recall that
In other words, show a negative fraction by placing the negative sign in the numerator, in front of the fraction bar, or in the denominator. Generally, negative denominators are avoided.
Example 11: Simplify and state the restrictions: $4−x2x2+3x−10$.
Solution: Begin by factoring the numerator and denominator.
Answer: $−x+2x+5$, where $x≠2$ and $x≠−5$
Try this! Simplify and state the restrictions: $2x2−7x−1525−x2$.
Answer: $−2x+3x+5$, where $x≠±5$
### Video Solution
(click to see video)
## Rational Functions
Rational functions have the form
where $p(x)$ and $q(x)$ are polynomials and $q(x)≠0$. The domain of a rational function consists of all real numbers x such that the denominator $q(x)≠0$.
Example 12:
a. Simplify: $r(x)=2x2+5x−36x2+18x$.
b. State the domain.
c. Calculate $r(−2)$.
Solution:
a. To simplify the rational function, first factor and then cancel.
b. To determine the restrictions, set the denominator of the original function equal to 0 and solve.
The domain consists of all real numbers x, where $x≠0$ and $x≠−3$.
c. Since −2 is not a restriction, substitute it for the variable x using the simplified form.
a. $r(x)=2x−16x$
b. The domain is all real numbers except 0 and −3.
c. $r(−2)=512$
If a cost functionA function that represents the cost of producing a certain number of units. $C(x)$ represents the cost of producing x units, then the average costThe total cost divided by the number of units produced, which can be represented by $c(x)=C(x)x$, where $C(x)$ is a cost function. $c(x)$ is the cost divided by the number of units produced.
Example 13: The cost in dollars of producing t-shirts with a company logo is given by $C(x)=7x+200$, where x represents the number of shirts produced. Determine the average cost of producing
a. 40 t-shirts
b. 250 t-shirts
c. 1,000 t-shirts
Solution: Set up a function representing the average cost.
Next, calculate $c(40)$, $c(250)$, and $c(1000)$.
a. If 40 t-shirts are produced, then the average cost per t-shirt is $12.00. b. If 250 t-shirts are produced, then the average cost per t-shirt is$7.80.
c. If 1,000 t-shirts are produced, then the average cost per t-shirt is $7.20. ### Key Takeaways • Rational expressions usually are not defined for all real numbers. The real numbers that give a value of 0 in the denominator are not part of the domain. These values are called restrictions. • Simplifying rational expressions is similar to simplifying fractions. First, factor the numerator and denominator and then cancel the common factors. Rational expressions are simplified if there are no common factors other than 1 in the numerator and the denominator. • Simplified rational expressions are equivalent for values in the domain of the original expression. Be sure to state the restrictions if the denominators are not assumed to be nonzero. • Use the opposite binomial property to cancel binomial factors that involve subtraction. Use $−(a−b)=b−a$ to replace factors that will then cancel. Do not confuse this with factors that involve addition, such as $(a+b)=(b+a)$. ### Topic Exercises Part A: Rational Expressions Evaluate for the given set of x-values. 1. $5x$; {−1, 0, 1} 2. $4x3x2$; {−1, 0, 1} 3. $1x+9$; {−10, −9, 0} 4. $x+6x−5$; {−6, 0, 5} 5. $3x(x−2)2x−1$; {0, 1/2, 2} 6. $9x2−1x−7$; {0, 1/3, 7} 7. $5x2−9$; {−3, 0, 3} 8. $x2−25x2−3x−10$; {−5, −4, 5} 9. Fill in the following chart: 10. Fill in the following chart: 11. Fill in the following chart: 12. Fill in the following chart: An object’s weight depends on its height above the surface of earth. If an object weighs 120 pounds on the surface of earth, then its weight in pounds, W, x miles above the surface is approximated by the formula $W=120⋅40002(4000+x)2$ For each problem below, approximate the weight of a 120-pound object at the given height above the surface of earth. (1 mile = 5,280 feet) 13. 100 miles 14. 1,000 miles 15. 44,350 feet 16. 90,000 feet The price to earnings ratio (P/E) is a metric used to compare the valuations of similar publicly traded companies. The P/E ratio is calculated using the stock price and the earnings per share (EPS) over the previous 12‑month period as follows: $P/E=price per shareearnings per share$ If each share of a company stock is priced at$22.40, then calculate the P/E ratio given the following values for the earnings per share.
17. $1.40 18.$1.21
19. What happens to the P/E ratio when earnings decrease?
20. What happens to the P/E ratio when earnings increase?
State the restrictions to the domain.
21. $13x$
22. $3x27x5$
23. $3x(x+1)x+4$
24. $2x2(x−3)x−1$
25. $15x−1$
26. $x−23x−2$
27. $x−95x(x−2)$
28. $1(x−3)(x+6)$
29. $x1−x2$
30. $x2−9x2−36$
31. $12x(x+3)(2x−1)$
32. $x−3(3x−1)(2x+3)$
33. $4x(2x+1)12x2+x−1$
34. $x−53x2−15x$
Part B: Simplifying Rational Expressions
State the restrictions and then simplify.
35. $5x220x3$
36. $12x660x$
37. $3x2(x−2)9x(x−2)$
38. $20(x−3)(x−5)6(x−3)(x+1)$
39. $6x2(x−8)36x(x+9)(x−8)$
40. $16x2−1(4x+1)2$
41. $9x2−6x+1(3x−1)2$
42. $x−7x2−49$
43. $x2−64x2+8x$
44. $x+10x2−100$
45. $2x3−12x25x2−30x$
46. $30x5+60x42x3−8x$
47. $2x−12x2+x−6$
48. $x2−x−63x2−8x−3$
49. $6x2−25x+253x2+16x−35$
50. $3x2+4x−15x2−9$
51. $x2−10x+21x2−4x−21$
52. $x3−1x2−1$
53. $x3+8x2−4$
54. $x4−16x2−4$
Part C: Simplifying Rational Expressions with Opposite Binomial Factors
State the restrictions and then simplify.
55. $x−99−x$
56. $3x−22−3x$
57. $x+66+x$
58. $3x+11+3x$
59. $(2x−5)(x−7)(7−x)(2x−1)$
60. $(3x+2)(x+5)(x−5)(2+3x)$
61. $x2−4(2−x)2$
62. $16−9x2(3x+4)2$
63. $4x2(10−x)3x3−300x$
64. $−2x+14x3−49x$
65. $2x2−7x−41−4x2$
66. $9x2−44x−6x2$
67. $x2−5x−147−15x+2x2$
68. $2x3+x2−2x−11+x−2x2$
69. $x3+2x−3x2−62+x2$
70. $27+x3x2+6x+9$
71. $64−x3x2−8x+16$
72. $x2+44−x2$
Simplify. (Assume all denominators are nonzero.)
73. $−15x3y25xy2(x+y)$
74. $14x7y2(x−2y)47x8y(x−2y)2$
75. $y+xx2−y2$
76. $y−xx2−y2$
77. $x2−y2(x−y)2$
78. $a2−ab−6b2a2−6ab+9b2$
79. $2a2−11a+12−32+2a2$
80. $a2b−3a23a2−3ab$
81. $xy2−x+y3−yx−xy2$
82. $x3−xy2−x2y+y3x2−2xy+y2$
83. $x3−27x2+3x+9$
84. $x2−x+1x3+1$
Part D: Rational Functions
Calculate the following.
85. $f(x)=5xx−3$; $f(0)$, $f(2)$, $f(4)$
86. $f(x)=x+7x2+1$; $f(−1)$, $f(0)$, $f(1)$
87. $g(x)=x3(x−2)2$; $g(0)$, $g(2)$, $g(−2)$
88. $g(x)=x2−99−x2$; $g(−2)$, $g(0)$, $g(2)$
89. $g(x)=x3x2+1$; $g(−1)$, $g(0)$, $g(1)$
90. $g(x)=5x+1x2−25$; $g(−1/5)$, $g(−1)$, $g(−5)$
State the restrictions to the domain and then simplify.
91. $f(x)=−3x2−6xx2+4x+4$
92. $f(x)=x2+6x+92x2+5x−3$
93. $g(x)=9−xx2−81$
94. $g(x)=x3−273−x$
95. $g(x)=3x−1510−2x$
96. $g(x)=25−5x4x−20$
97. The cost in dollars of producing coffee mugs with a company logo is given by $C(x)=x+40$, where x represents the number of mugs produced. Calculate the average cost of producing 100 mugs and the average cost of producing 500 mugs.
98. The cost in dollars of renting a moving truck for the day is given by $C(x)=0.45x+90$, where x represents the number of miles driven. Calculate the average cost per mile if the truck is driven 250 miles in one day.
99. The cost in dollars of producing sweat shirts with a custom design on the back is given by $C(x)=1200+(12−0.05x)x$, where x represents the number of sweat shirts produced. Calculate the average cost of producing 150 custom sweat shirts.
100. The cost in dollars of producing a custom injected molded part is given by $C(x)=500+(3−0.001x)x$, where x represents the number of parts produced. Calculate the average cost of producing 1,000 custom parts.
Part E: Discussion Board
101. Explain why $b−aa−b=−1$ and illustrate this fact by substituting some numbers for the variables.
102. Explain why $b+aa+b=1$ and illustrate this fact by substituting some numbers for the variables.
103. Explain why we cannot cancel x in the expression $xx+1$.
1: −5, undefined, 5
3: −1, undefined, 1/9
5: 0, undefined, 0
7: Undefined, −5/9, undefined
9:
11:
13: 114 pounds
15: 119.5 pounds
17: 16
19: The P/E ratio increases.
21: $x≠0$
23: $x≠−4$
25: $x≠15$
27: $x≠0$ and $x≠2$
29: $x≠±1$
31: $x≠0$, $x≠−3$, and $x≠12$
33: $x≠−13$ and $x≠14$
35: $14x$; $x≠0$
37: $x3$;
39: $x6(x+9)$;
41: $1$; $x≠13$
43: $x−8x$; $x≠0,−8$
45: $2x5$;
47: $2x−12x2+x−6$; $x≠−2,32$
49: $2x−5x+7$; $x≠−7,53$
51: $x−3x+3$;
53: $x2−2x+4x−2$; $x≠±2$
55: −1; $x≠9$
57: 1; $x≠−6$
59: $−2x−52x−1$; $x≠12,7$
61: $x+2x−2$; $x≠2$
63: $−4x3(x+10)$;
65: $x−41−2x$; $x≠±12$
67: $x+22x−1$; $x≠12,7$
69: $x−3$; none
71: $−16+4x+x2x−4$; $x≠4$
73: $−3x2x+y$
75: $1x−y$
77: $x+yx−y$
79: $2a−32(4+a)$
81: $−x+yx$
83: $x−3$
85: $f(0)=0$, $f(2)=−10$, $f(4)=20$
87: $g(0)=0$, $g(2)$ undefined, $g(−2)=−1/2$
89: $g(−1)=−1/2$, $g(0)=0$, $g(1)=1/2$
91: $f(x)=−3xx+2$; $x≠−2$
93: $g(x)=−1x+9$; $x≠±9$
95: $g(x)=−32$; $x≠5$
97: The average cost of producing 100 mugs is $1.40 per mug. The average cost of producing 500 mugs is$1.08 per mug.
99: $12.50 ## 7.2 Multiplying and Dividing Rational Expressions ### Learning Objectives 1. Multiply rational expressions. 2. Divide rational expressions. 3. Multiply and divide rational functions. ## Multiplying Rational Expressions When multiplying fractions, we can multiply the numerators and denominators together and then reduce, as illustrated: Multiplying rational expressions is performed in a similar manner. For example, In general, given polynomials P, Q, R, and S, where $Q≠0$ and $S≠0$, we have In this section, assume that all variable expressions in the denominator are nonzero unless otherwise stated. Example 1: Multiply: $12x25y3⋅20y46x3$. Solution: Multiply numerators and denominators and then cancel common factors. Answer: $8yx$ Example 2: Multiply: $x−3x+5⋅x+5x+7$. Solution: Leave the product in factored form and cancel the common factors. Answer: $x−3x+7$ Example 3: Multiply: $15x2y3(2x−1)⋅x(2x−1)3x2y(x+3)$. Solution: Leave the polynomials in the numerator and denominator factored so that we can cancel the factors. In other words, do not apply the distributive property. Answer: $5xy2x+3$ Typically, rational expressions will not be given in factored form. In this case, first factor all numerators and denominators completely. Next, multiply and cancel any common factors, if there are any. Example 4: Multiply: $x+5x−5⋅x−5x2−25$. Solution: Factor the denominator $x2−25$ as a difference of squares. Then multiply and cancel. Keep in mind that 1 is always a factor; so when the entire numerator cancels out, make sure to write the factor 1. Answer: $1x−5$ Example 5: Multiply: $x2+3x+2x2−5x+6⋅x2−7x+12x2+8x+7$. Solution: It is a best practice to leave the final answer in factored form. Answer: $(x+2)(x−4)(x−2)(x+7)$ Example 6: Multiply: $−2x2+x+3x2+2x−8⋅3x−6x2+x$. Solution: The trinomial $−2x2+x+3$ in the numerator has a negative leading coefficient. Recall that it is a best practice to first factor out a −1 and then factor the resulting trinomial. Answer: $−3(2x−3)x(x+4)$ Example 7: Multiply: $7−xx2+3x⋅x2+10x+21x2−49$. Solution: We replace $7−x$ with $−1(x−7)$ so that we can cancel this factor. Answer: $−1x$ Try this! Multiply: $x2−648−x⋅x+x2x2+9x+8$. Answer: $−x$ ### Video Solution (click to see video) ## Dividing Rational Expressions To divide two fractions, we multiply by the reciprocal of the divisor, as illustrated: Dividing rational expressions is performed in a similar manner. For example, In general, given polynomials P, Q, R, and S, where $Q≠0$, $R≠0$, and $S≠0$, we have Example 8: Divide: $8x5y25z6÷20xy415z3$. Solution: First, multiply by the reciprocal of the divisor and then cancel. Answer: $6x425y3z3$ Example 9: Divide: $x+2x2−4÷x+3x−2$. Solution: After multiplying by the reciprocal of the divisor, factor and cancel. Answer: $1x+3$ Example 10: Divide: $x2−6x−16x2+4x−21÷x2+9x+14x2−8x+15$. Solution: Begin by multiplying by the reciprocal of the divisor. After doing so, factor and cancel. Answer: $(x−8)(x−5)(x+7)2$ Example 11: Divide: . Solution: Just as we do with fractions, think of the divisor $(2x−3)$ as an algebraic fraction over 1. Answer: $−2x+3x+2$ Try this! Divide: $4x2+7x−225x2÷1−4x100x4$. Answer: $−4x2(x+2)$ ### Video Solution (click to see video) ## Multiplying and Dividing Rational Functions The product and quotient of two rational functions can be simplified using the techniques described in this section. The restrictions to the domain of a product consist of the restrictions of each function. Example 12: Calculate $(f⋅g)(x)$ and determine the restrictions to the domain. Solution: In this case, the domain of $f(x)$ consists of all real numbers except 0, and the domain of $g(x)$ consists of all real numbers except 1/4. Therefore, the domain of the product consists of all real numbers except 0 and 1/4. Multiply the functions and then simplify the result. Answer: $(f⋅g)(x)=−4x+15x$, where $x≠0, 14$ The restrictions to the domain of a quotient will consist of the restrictions of each function as well as the restrictions on the reciprocal of the divisor. Example 13: Calculate $(f/g)(x)$ and determine the restrictions. Solution: In this case, the domain of $f(x)$ consists of all real numbers except 3 and 8, and the domain of $g(x)$ consists all real numbers except 3. In addition, the reciprocal of $g(x)$ has a restriction of −8. Therefore, the domain of this quotient consists of all real numbers except 3, 8, and −8. Answer: $(f/g)(x)=1$, where $x≠3, 8, −8$ ### Key Takeaways • After multiplying rational expressions, factor both the numerator and denominator and then cancel common factors. Make note of the restrictions to the domain. The values that give a value of 0 in the denominator are the restrictions. • To divide rational expressions, multiply by the reciprocal of the divisor. • The restrictions to the domain of a product consist of the restrictions to the domain of each factor. • The restrictions to the domain of a quotient consist of the restrictions to the domain of each rational expression as well as the restrictions on the reciprocal of the divisor. ### Topic Exercises Part A: Multiplying Rational Expressions Multiply. (Assume all denominators are nonzero.) 1. $2x3⋅94x2$ 2. $−5x3y⋅y225x$ 3. $5x22y⋅4y215x3$ 4. $16a47b2⋅49b32a3$ 5. $x−612x3⋅24x2x−6$ 6. $x+102x−1⋅x−2x+10$ 7. $(y−1)2y+1⋅1y−1$ 8. $y2−9y+3⋅2y−3y−3$ 9. $2a−5a−5⋅2a+54a2−25$ 10. $2a2−9a+4a2−16⋅(a2+4a)$ 11. $2x2+3x−2(2x−1)2⋅2xx+2$ 12. $9x2+19x+24−x2⋅x2−4x+49x2−8x−1$ 13. $x2+8x+1616−x2⋅x2−3x−4x2+5x+4$ 14. $x2−x−2x2+8x+7⋅x2+2x−15x2−5x+6$ 15. $x+1x−3⋅3−xx+5$ 16. $2x−1x−1⋅x+61−2x$ 17. $9+x3x+1⋅3x+9$ 18. $12+5x⋅5x+25x$ 19. $100−y2y−10⋅25y2y+10$ 20. $3y36y−5⋅36y2−255+6y$ 21. $3a2+14a−5a2+1⋅3a+11−9a2$ 22. $4a2−16a4a−1⋅1−16a24a2−15a−4$ 23. $x+9−x2+14x−45⋅(x2−81)$ 24. $12+5x⋅(25x2+20x+4)$ 25. $x2+x−63x2+15x+18⋅2x2−8x2−4x+4$ 26. $5x2−4x−15x2−6x+1⋅25x2−10x+13−75x2$ Part B: Dividing Rational Expressions Divide. (Assume all denominators are nonzero.) 27. $5x8÷15x24$ 28. $38y÷−152y2$ 29. $5x93y3 25x109y5$ 30. $12x4y221z5 6x3y27z3$ 31. $(x−4)230x4÷x−415x$ 32. $5y410(3y−5)2÷10y52(3y−5)3$ 33. $x2−95x÷(x−3)$ 34. $y2−648y÷(8+y)$ 35. $(a−8)22a2+10a÷a−8a$ 36. $24a2b3(a−2b)÷12ab(a−2b)5$ 37. $x2+7x+10x2+4x+4÷1x2−4$ 38. $2x2−x−12x2−3x+1÷14x2−1$ 39. $y+1y2−3y÷y2−1y2−6y+9$ 40. $9−a2a2−8a+15÷2a2−10aa2−10a+25$ 41. $a2−3a−182a2−11a−6÷a2+a−62a2−a−1$ 42. $y2−7y+10y2+5y−14 2y2−9y−5y2+14y+49$ 43. $6y2+y−14y2+4y+1 3y2+2y−12y2−7y−4$ 44. $x2−7x−18x2+8x+12÷x2−81x2+12x+36$ 45. $4a2−b2b+2a÷(b−2a)2$ 46. $x2−y2y+x÷(y−x)2$ 47. $5y2(y−3)4x3÷25y(3−y)2x2$ 48. $15x33(y+7)÷25x69(7+y)2$ 49. $3x+4x−8÷7x8−x$ 50. $3x−22x+1÷2−3x3x$ 51. $(7x−1)24x+1÷28x2−11x+11−4x$ 52. $4x(x+2)2÷2−xx2−4$ 53. $a2−b2a÷(b−a)2$ 54. $(a−2b)22b÷(2b2+ab−a2)$ 55. $x2−6x+9x2+7x+12÷9−x2x2+8x+16$ 56. $2x2−9x−525−x2÷1−4x+4x2−2x2−9x+5$ 57. $3x2−16x+5100−4x2 9x2−6x+13x2+14x−5$ 58. $10x2−25x−15x2−6x+9 9−x2x2+6x+9$ Recall that multiplication and division are to be performed in the order they appear from left to right. Simplify the following. 59. $1x2⋅x−1x+3÷x−1x3$ 60. $x−7x+9⋅1x3÷x−7x$ 61. $x+1x−2÷xx−5⋅x2x+1$ 62. $x+42x+5÷x−32x+5⋅x+4x−3$ 63. $2x−1x+1÷x−4x2+1⋅x−42x−1$ 64. $4x2−13x+2÷2x−1x+5⋅3x+22x+1$ Part C: Multiplying and Dividing Rational Functions Calculate $(f⋅g)(x)$ and determine the restrictions to the domain. 65. $f(x)=1x$ and $g(x)=1x−1$ 66. $f(x)=x+1x−1$ and $g(x)=x2−1$ 67. $f(x)=3x+2x+2$ and $g(x)=x2−4(3x+2)2$ 68. $f(x)=(1−3x)2x−6$ and $g(x)=(x−6)29x2−1$ 69. $f(x)=25x2−1x2+6x+9$ and $g(x)=x2−95x+1$ 70. $f(x)=x2−492x2+13x−7$ and $g(x)=4x2−4x+17−x$ Calculate $(f/g)(x)$ and state the restrictions. 71. $f(x)=1x$ and $g(x)=x−2x−1$ 72. $f(x)=(5x+3)2x2$ and $g(x)=5x+36−x$ 73. $f(x)=5−x(x−8)2$ and $g(x)=x2−25x−8$ 74. $f(x)=x2−2x−15x2−3x−10$ and $g(x)=2x2−5x−3x2−7x+12$ 75. $f(x)=3x2+11x−49x2−6x+1$ and $g(x)=x2−2x+13x2−4x+1$ 76. $f(x)=36−x2x2+12x+36$ and $g(x)=x2−12x+36x2+4x−12$ Part D: Discussion Board Topics 77. In the history of fractions, who is credited for the first use of the fraction bar? 78. How did the ancient Egyptians use fractions? 79. Explain why $x=7$ is a restriction to $1x÷x−7x−2$. ### Answers 1: $32x$ 3: $2y3x$ 5: $2x$ 7: $y−1y+1$ 9: $1a−5$ 11: $2x2x−1$ 13: −1 15: $−x+1x+5$ 17: $33x+1$ 19: $−25y2$ 21: $−a+5a2+1$ 23: $−(x+9)2x−5$ 25: 2/3 27: $16x$ 29: $3y25x$ 31: $x−42x3$ 33: $x+35x$ 35: $a−82(a+5)$ 37: $(x+5)(x−2)$ 39: $y−3y(y−1)$ 41: $a−1a−2$ 43: $y−4y+1$ 45: $12a−b$ 47: $−y10x$ 49: $−3x+47x$ 51: $−7x−14x+1$ 53: $−a+ba(b−a)$ 55: $−(x−3)(x+4)(x+3)2$ 57: −1/4 59: $xx+3$ 61: $x(x−5)x−2$ 63: $x2+1x+1$ 65: $(f⋅g)(x)=1x(x−1)$; $x≠0, 1$ 67: $(f⋅g)(x)=x−23x+2$; $x≠−2, −23$ 69: $(f⋅g)(x)=(x−3)(5x−1)x+3$; $x≠−3, −15$ 71: $(f/g)(x)=x−1x(x−2)$; $x≠0, 1, 2$ 73: $(f/g)(x)=−1(x−8)(x+5)$; $x≠±5, 8$ 75: $(f/g)(x)=(x+4)(x−1)$; $x≠13, 1$ ## 7.3 Adding and Subtracting Rational Expressions ### Learning Objectives 1. Add and subtract rational expressions with common denominators. 2. Add and subtract rational expressions with unlike denominators. 3. Add and subtract rational functions. ## Adding and Subtracting with Common Denominators Adding and subtracting rational expressions is similar to adding and subtracting fractions. Recall that if the denominators are the same, we can add or subtract the numerators and write the result over the common denominator. When working with rational expressions, the common denominator will be a polynomial. In general, given polynomials P, Q, and R, where $Q≠0$, we have the following: In this section, assume that all variable factors in the denominator are nonzero. Example 1: Add: $3y+7y$. Solution: Add the numerators 3 and 7, and write the result over the common denominator, y. Answer: $10y$ Example 2: Subtract: $x−52x−1−12x−1$. Solution: Subtract the numerators $x−5$ and 1, and write the result over the common denominator, $2x−1$. Answer: $x−62x−1$ Example 3: Subtract: $2x+7(x+5)(x−3)−x+10(x+5)(x−3)$. Solution: We use parentheses to remind us to subtract the entire numerator of the second rational expression. Answer: $1x+5$ Example 4: Simplify: $2x2+10x+3x2−36−x2+6x+5x2−36+x−4x2−36$. Solution: Subtract and add the numerators. Make use of parentheses and write the result over the common denominator, $x2−36$. Answer: $x−1x−6$ Try this! Subtract: $x2+12x2−7x−4−x2−2x2x2−7x−4$. Answer: $1x−4$ ### Video Solution (click to see video) ## Adding and Subtracting with Unlike Denominators To add rational expressions with unlike denominators, first find equivalent expressions with common denominators. Do this just as you have with fractions. If the denominators of fractions are relatively prime, then the least common denominator (LCD) is their product. For example, Multiply each fraction by the appropriate form of 1 to obtain equivalent fractions with a common denominator. The process of adding and subtracting rational expressions is similar. In general, given polynomials P, Q, R, and S, where $Q≠0$ and $S≠0$, we have the following: In this section, assume that all variable factors in the denominator are nonzero. Example 5: Add: $1x+1y$. Solution: In this example, the $LCD=xy$. To obtain equivalent terms with this common denominator, multiply the first term by $yy$ and the second term by $xx$. Answer: $y+xxy$ Example 6: Subtract: $1y−1y−3$. Solution: Since the $LCD=y(y−3)$, multiply the first term by 1 in the form of $(y−3)(y−3)$ and the second term by $yy$. Answer: $−3y(y−3)$ It is not always the case that the LCD is the product of the given denominators. Typically, the denominators are not relatively prime; thus determining the LCD requires some thought. Begin by factoring all denominators. The LCD is the product of all factors with the highest power. For example, given there are three base factors in the denominator: $x$, $(x+2)$, and $(x−3)$. The highest powers of these factors are $x3$, $(x+2)2$, and $(x−3)1$. Therefore, The general steps for adding or subtracting rational expressions are illustrated in the following example. Example 7: Subtract: $xx2+4x+3−3x2−4x−5$. Solution: Step 1: Factor all denominators to determine the LCD. The $LCD is (x+1)(x+3)(x−5)$. Step 2: Multiply by the appropriate factors to obtain equivalent terms with a common denominator. To do this, multiply the first term by $(x−5)(x−5)$ and the second term by $(x+3)(x+3)$. Step 3: Add or subtract the numerators and place the result over the common denominator. Step 4: Simplify the resulting algebraic fraction. Answer: $(x−9)(x+3)(x−5)$ Example 8: Subtract: $x2−9x+18x2−13x+36−xx−4$. Solution: It is best not to factor the numerator, $x2−9x+18$, because we will most likely need to simplify after we subtract. Answer: $18(x−4)(x−9)$ Example 9: Subtract: $1x2−4−12−x$. Solution: First, factor the denominators and determine the LCD. Notice how the opposite binomial property is applied to obtain a more workable denominator. The LCD is $(x+2)(x−2)$. Multiply the second term by 1 in the form of $(x+2)(x+2)$. Now that we have equivalent terms with a common denominator, add the numerators and write the result over the common denominator. Answer: $x+3(x+2)(x−2)$ Example 10: Simplify: $y−1y+1−y+1y−1+y2−5y2−1$. Solution: Begin by factoring the denominator. We can see that the . Find equivalent fractions with this denominator. Next, subtract and add the numerators and place the result over the common denominator. Finish by simplifying the resulting rational expression. Answer: $y−5y−1$ Try this! Simplify: $−2x2−1+x1+x−51−x$. Answer: $x+3x−1$ ### Video Solution (click to see video) Rational expressions are sometimes expressed using negative exponents. In this case, apply the rules for negative exponents before simplifying the expression. Example 11: Simplify: $y−2+(y−1)−1$. Solution: Recall that $x−n=1xn$. We begin by rewriting the negative exponents as rational expressions. Answer: $y2+y−1y2(y−1)$ ## Adding and Subtracting Rational Functions We can simplify sums or differences of rational functions using the techniques learned in this section. The restrictions of the result consist of the restrictions to the domains of each function. Example 12: Calculate $(f+g)(x)$, given $f(x)=1x+3$ and $g(x)=1x−2$, and state the restrictions. Solution: Here the domain of f consists of all real numbers except −3, and the domain of g consists of all real numbers except 2. Therefore, the domain of f + g consists of all real numbers except −3 and 2. Answer: $2x+1(x+3)(x−2)$, where $x≠−3, 2$ Example 13: Calculate $(f−g)(x)$, given $f(x)=x(x−1)x2−25$ and $g(x)=x−3x−5$, and state the restrictions to the domain. Solution: The domain of f consists of all real numbers except 5 and −5, and the domain of g consists of all real numbers except 5. Therefore, the domain of fg consists of all real numbers except −5 and 5. Answer: $−3x+5$, where $x≠±5$ ### Key Takeaways • When adding or subtracting rational expressions with a common denominator, add or subtract the expressions in the numerator and write the result over the common denominator. • To find equivalent rational expressions with a common denominator, first factor all denominators and determine the least common multiple. Then multiply numerator and denominator of each term by the appropriate factor to obtain a common denominator. Finally, add or subtract the expressions in the numerator and write the result over the common denominator. • The restrictions to the domain of a sum or difference of rational functions consist of the restrictions to the domains of each function. ### Topic Exercises Part A: Adding and Subtracting with Common Denominators Simplify. (Assume all denominators are nonzero.) 1. $3x+7x$ 2. $9x−10x$ 3. $xy−3y$ 4. $4x−3+6x−3$ 5. $72x−1−x2x−1$ 6. $83x−8−3x3x−8$ 7. $2x−9+x−11x−9$ 8. $y+22y+3−y+32y+3$ 9. $2x−34x−1−x−44x−1$ 10. $2xx−1−3x+4x−1+x−2x−1$ 11. $13y−2y−93y−13−5y3y$ 12. $−3y+25y−10+y+75y−10−3y+45y−10$ 13. $x(x+1)(x−3)−3(x+1)(x−3)$ 14. $3x+5(2x−1)(x−6)−x+6(2x−1)(x−6)$ 15. $xx2−36+6x2−36$ 16. $xx2−81−9x2−81$ 17. $x2+2x2+3x−28+x−22x2+3x−28$ 18. $x2x2−x−3−3−x2x2−x−3$ Part B: Adding and Subtracting with Unlike Denominators Simplify. (Assume all denominators are nonzero.) 19. $12+13x$ 20. $15x2−1x$ 21. $112y2+310y3$ 22. $1x−12y$ 23. $1y−2$ 24. $3y+2−4$ 25. $2x+4+2$ 26. $2y−1y2$ 27. $3x+1+1x$ 28. $1x−1−2x$ 29. $1x−3+1x+5$ 30. $1x+2−1x−3$ 31. $xx+1−2x−2$ 32. $2x−3x+5−xx−3$ 33. $y+1y−1+y−1y+1$ 34. $3y−13y−y+4y−2$ 35. $2x−52x+5−2x+52x−5$ 36. $22x−1−2x+11−2x$ 37. $3x+4x−8−28−x$ 38. $1y−1+11−y$ 39. $2x2x2−9+x+159−x2$ 40. $xx+3+1x−3−15−x(x+3)(x−3)$ 41. $2x3x−1−13x+1+2(x−1)(3x−1)(3x+1)$ 42. $4x2x+1−xx−5+16x−3(2x+1)(x−5)$ 43. $x3x+2x−2+43x(x−2)$ 44. $−2xx+6−3x6−x−18(x−2)(x+6)(x−6)$ 45. $xx+5−1x−7−25−7x(x+5)(x−7)$ 46. $xx2−2x−3+2x−3$ 47. $1x+5−x2x2−25$ 48. $5x−2x2−4−2x−2$ 49. $1x+1−6x−3x2−7x−8$ 50. $3x9x2−16−13x+4$ 51. $2xx2−1+1x2+x$ 52. $x(4x−1)2x2+7x−4−x4+x$ 53. $3x23x2+5x−2−2x3x−1$ 54. $2xx−4−11x+4x2−2x−8$ 55. $x2x+1+6x−242x2−7x−4$ 56. $1x2−x−6+1x2−3x−10$ 57. $xx2+4x+3−3x2−4x−5$ 58. $y+12y2+5y−3−y4y2−1$ 59. $y−1y2−25−2y2−10y+25$ 60. $3x2+24x2−2x−8−12x−4$ 61. $4x2+28x2−6x−7−28x−7$ 62. $a4−a+a2−9a+18a2−13a+36$ 63. $3a−12a2−8a+16−a+24−a$ 64. $a2−142a2−7a−4−51+2a$ 65. $1x+3−xx2−6x+9+3x2−9$ 66. $3xx+7−2xx−2+23x−10x2+5x−14$ 67. $x+3x−1+x−1x+2−x(x+11)x2+x−2$ 68. $−2x3x+1−4x−2+4(x+5)3x2−5x−2$ 69. $x−14x−1−x+32x+3−3(x+5)8x2+10x−3$ 70. $3x2x−3−22x+3−6x2−5x−94x2−9$ 71. $1y+1+1y+2y2−1$ 72. $1y−1y+1+1y−1$ 73. $5−2+2−1$ 74. $6−1+4−2$ 75. $x−1+y−1$ 76. $x−2−y−1$ 77. $(2x−1)−1−x−2$ 78. $(x−4)−1−(x+1)−1$ 79. $3x2(x−1)−1−2x$ 80. $2(y−1)−2−(y−1)−1$ Part C: Adding and Subtracting Rational Functions Calculate $(f+g)(x)$ and $(f−g)(x)$ and state the restrictions to the domain. 81. $f(x)=13x$ and $g(x)=1x−2$ 82. $f(x)=1x−1$ and $g(x)=1x+5$ 83. $f(x)=xx−4$ and $g(x)=14−x$ 84. $f(x)=xx−5$ and $g(x)=12x−3$ 85. $f(x)=x−1x2−4$ and $g(x)=4x2−6x−16$ 86. $f(x)=5x+2$ and $g(x)=3x+4$ Calculate $(f+f)(x)$ and state the restrictions to the domain. 87. $f(x)=1x$ 88. $f(x)=12x$ 89. $f(x)=x2x−1$ 90. $f(x)=1x+2$ Part D: Discussion Board 91. Explain to a classmate why this is incorrect: $1x2+2x2=32x2$. 92. Explain to a classmate how to find the common denominator when adding algebraic expressions. Give an example. ### Answers 1: $10x$ 3: $x−3y$ 5: $7−x2x−1$ 7: 1 9: $x+14x−1$ 11: $y−1y$ 13: $1x+1$ 15: $1x−6$ 17: $x+5x+7$ 19: $3x+26x$ 21: $5y+1860y3$ 23: $1−2yy$ 25: $2(x+5)x+4$ 27: $4x+1x(x+1)$ 29: $2(x+1)(x−3)(x+5)$ 31: $x2−4x−2(x−2)(x+1)$ 33: $2(y2+1)(y+1)(y−1)$ 35: $−40x(2x+5)(2x−5)$ 37: $3(x+2)x−8$ 39: $2x+5x+3$ 41: $2x+13x+1$ 43: $x2+4x+43x(x−2)$ 45: $x−6x−7$ 47: $−x2+x−5(x+5)(x−5)$ 49: $−5x−8$ 51: $2x−1x(x−1)$ 53: $x(x−4)(x+2)(3x−1)$ 55: $x+62x+1$ 57: $x−9(x−5)(x+3)$ 59: $y2−8y−5(y+5)(y−5)2$ 61: $4xx+1$ 63: $a+5a−4$ 65: $−6x(x+3)(x−3)2$ 67: $x−7x+2$ 69: $−x−54x−1$ 71: $2y−1y(y−1)$ 73: $2750$ 75: $x+yxy$ 77: $(x−1)2x2(2x−1)$ 79: $x(x+2)x−1$ 81: $(f+g)(x)=2(2x−1)3x(x−2)$; $(f−g)(x)=−2(x+1)3x(x−2)$; $x≠0, 2$ 83: $(f+g)(x)=x−1x−4$; $(f−g)(x)=x+1x−4$; $x≠4$ 85: $(f+g)(x)=x(x−5)(x+2)(x−2)(x−8)$; $(f−g)(x)=x2−13x+16(x+2)(x−2)(x−8)$; $x≠−2, 2, 8$ 87: $(f+f)(x)=2x$; $x≠0$ 89: $(f+f)(x)=2x2x−1$; $x≠12$ ## 7.4 Complex Rational Expressions ### Learning Objectives 1. Simplify complex rational expressions by multiplying the numerator by the reciprocal of the divisor. 2. Simplify complex rational expressions by multiplying numerator and denominator by the least common denominator (LCD). ## Definitions A complex fractionA fraction where the numerator or denominator consists of one or more fractions. is a fraction where the numerator or denominator consists of one or more fractions. For example, Simplifying such a fraction requires us to find an equivalent fraction with integer numerator and denominator. One way to do this is to divide. Recall that dividing fractions involves multiplying by the reciprocal of the divisor. An alternative method for simplifying this complex fraction involves multiplying both the numerator and denominator by the LCD of all the given fractions. In this case, the LCD = 4. A complex rational expressionA rational expression where the numerator or denominator consists of one or more rational expressions. is defined as a rational expression that contains one or more rational expressions in the numerator or denominator or both. For example, We simplify a complex rational expression by finding an equivalent fraction where the numerator and denominator are polynomials. As illustrated above, there are two methods for simplifying complex rational expressions, and we will outline the steps for both methods. For the sake of clarity, assume that variable expressions used as denominators are nonzero. ## Method 1: Simplify Using Division We begin our discussion on simplifying complex rational expressions using division. Before we can multiply by the reciprocal of the divisor, we must simplify the numerator and denominator separately. The goal is to first obtain single algebraic fractions in the numerator and the denominator. The steps for simplifying a complex algebraic fraction are illustrated in the following example. Example 1: Simplify: $12+1x14−1x2$. Solution: Step 1: Simplify the numerator and denominator. The goal is to obtain a single algebraic fraction divided by another single algebraic fraction. In this example, find equivalent terms with a common denominator in both the numerator and denominator before adding and subtracting. At this point we have a single algebraic fraction divided by a single algebraic fraction. Step 2: Multiply the numerator by the reciprocal of the divisor. Step 3: Factor all numerators and denominators completely. Step 4: Cancel all common factors. Answer: $2xx−2$ Example 2: Simplify: $1x−1x−24x2−2x$. Solution: Answer: $−12$ Example 3: Simplify $1−4x−21x2 1−2x−15x2$. Solution: The LCD of the rational expressions in both the numerator and denominator is $x2$. Multiply by the appropriate factors to obtain equivalent terms with this as the denominator and then subtract. We now have a single rational expression divided by another single rational expression. Next, multiply the numerator by the reciprocal of the divisor and then factor and cancel. Answer: $x−7x−5$ Example 4: Simplify: $1−1x2 1x−1$. Solution: Answer: $−x+1x$ Try this! Simplify: $181−1x2 19+1x$. Answer: $x−99x$ ### Video Solution (click to see video) ## Method 2: Simplify Using the LCD An alternative method for simplifying complex rational expressions involves clearing the fractions by multiplying the expression by a special form of 1. In this method, multiply the numerator and denominator by the least common denominator (LCD) of all given fractions. Example 5: Simplify: $12+1x14−1x2$. Solution: Step 1: Determine the LCD of all the fractions in the numerator and denominator. In this case, the denominators of the given fractions are 2, $x$, 4, and $x2$. Therefore, the . Step 2: Multiply the numerator and denominator by the LCD. This step should clear the fractions in both the numerator and denominator. This leaves us with a single algebraic fraction with a polynomial in the numerator and in the denominator. Step 3: Factor the numerator and denominator completely. Step 4: Cancel all common factors. Answer: $2xx−2$ ### Note This was the same problem that we began this section with, and the results here are the same. It is worth taking the time to compare the steps involved using both methods on the same problem. Example 6: Simplify: $1−2x−15x2 3−14x−5x2$. Solution: Considering all of the denominators, we find that the . Therefore, multiply the numerator and denominator by $x2$: At this point, we have a rational expression that can be simplified by factoring and then canceling the common factors. Answer: $x+33x+1$ It is important to point out that multiplying the numerator and denominator by the same nonzero factor is equivalent to multiplying by 1 and does not change the problem. Because $x2x2=1$, we can multiply the numerator and denominator by $x2$ in the previous example and obtain an equivalent expression. Example 7: Simplify: $1x+1+3x−3 2x−3−1x+1$. Solution: The LCM of all the denominators is $(x+1)(x−3)$. Begin by multiplying the numerator and denominator by these factors. Answer: $4xx+5$ Try this! Simplify: $1y−14116−1y2$. Answer: $−4yy+4$ ### Video Solution (click to see video) ### Key Takeaways • Complex rational expressions can be simplified into equivalent expressions with a polynomial numerator and polynomial denominator. • One method of simplifying a complex rational expression requires us to first write the numerator and denominator as a single algebraic fraction. Then multiply the numerator by the reciprocal of the divisor and simplify the result. • Another method for simplifying a complex rational expression requires that we multiply it by a special form of 1. Multiply the numerator and denominator by the LCM of all the denominators as a means to clear the fractions. After doing this, simplify the remaining rational expression. • An algebraic fraction is reduced to lowest terms if the numerator and denominator are polynomials that share no common factors other than 1. ### Topic Exercises Part A: Complex Rational Expressions Simplify. (Assume all denominators are nonzero.) 1. $12 54$ 2. $78 54$ 3. $103 209$ 4. $−421 87$ 5. $23 56$ 6. $74 143$ 7. $1−32 54−13$ 8. $12−5 12+13$ 9. $1+321−14$ 10. $2−121+34$ 11. $5x2x+1 25xx+1$ 12. $7+x7x x+714x2$ 13. $3yx y2x−1$ 14. $5a2b−1 15a3( b−1)2$ 15. $1+1x2−1x$ 16. $2x+13−1x$ 17. $23y−46−1y$ 18. $5y−1210−yy2$ 19. $15−1x125−1x2$ 20. $1x+15125−1x2$ 21. $1x−1319−1x2$ 22. $14+1x1x2−116$ 23. $16−1x21x−4$ 24. $2−1y1−14y2$ 25. $1x+1y1y2−1x2$ 26. $12x−4314x2−169$ 27. $225−12x215−12x$ 28. $425−14x215+14x$ 29. $1y−1x 4−2xy$ 30. $1ab+2 1a+1b$ 31. $1y+1xxy$ 32. $3x13−1x$ 33. $1−4x−21x21−2x−15x2$ 34. $1−3x−4x21−16x2$ 35. $3−12x−12x22−2x+12x2$ 36. $12−5x+12x212−6x+18x2$ 37. $1x−43x23−8x+163x2$ 38. $1+310x−110x235−110x−15x2$ 39. $x−11+4x−5x2$ 40. $2−52x−3x24x+3$ 41. $1x−3+2x1x−3x−3$ 42. $14x−5+1x21x2+13x−10$ 43. 44. 45. 46. $xx−9+2x+1x7x−9−1x+1$ 47. 48. 49. $a3−8b327a−2b$ 50. $27a3+b3ab3a+b$ 51. $1b3+1a31b+1a$ 52. $1b3−1a31a−1b$ 53. $x2+y2xy+2x2−y22xy$ 54. $xy+4+4yxxy+3+2yx$ 55. $1+11+12$ 56. $2−11+13$ 57. $11+11+x$ 58. $x+1x1−1x+1$ 59. $1−1xx−1x$ 60. $1x−xx−1x2$ Part B: Discussion Board Topics 61. Choose a problem from this exercise set and clearly work it out on paper, explaining each step in words. Scan your page and post it on the discussion board. 62. Explain why we need to simplify the numerator and denominator to a single algebraic fraction before multiplying by the reciprocal of the divisor. 63. Two methods for simplifying complex rational expressions have been presented in this section. Which of the two methods do you feel is more efficient, and why? ### Answers 1: $25$ 3: $32$ 5: 4/5 7: −6/11 9: $103$ 11: $x5$ 13: $3(x−1)xy$ 15: $x+12x−1$ 17: $−23$ 19: $5xx+5$ 21: $−3xx+3$ 23: $−4x+1x$ 25: $xyx−y$ 27: $2x+55x$ 29: $x−y4xy−2$ 31: $x+yx2y2$ 33: $x−7x−5$ 35: $3x+12x−1$ 37: $13x−4$ 39: $x2x+5$ 41: $−3(x−2)2x+3$ 43: $5x+18x+12$ 45: $(x−1)(3x+4)(x+2)(x+3)$ 47: $x+13x+2$ 49: $a2+2ab+4b227$ 51: $a2−ab+b2a2b2$ 53: $2(x+y)x−y$ 55: $53$ 57: $x+1x+2$ 59: $1x+1$ ## 7.5 Solving Rational Equations ### Learning Objectives 1. Solve rational equations. 2. Solve literal equations, or formulas, involving rational expressions. ## Solving Rational Equations A rational equationAn equation containing at least one rational expression. is an equation containing at least one rational expression. Rational expressions typically contain a variable in the denominator. For this reason, we will take care to ensure that the denominator is not 0 by making note of restrictions and checking our solutions. Solve rational equations by clearing the fractions by multiplying both sides of the equation by the least common denominator (LCD). Example 1: Solve: $5x−13=1x$. Solution: We first make a note that $x≠0$ and then multiply both sides by the LCD, 3x: Check your answer by substituting 12 for x to see if you obtain a true statement. Answer: The solution is 12. After multiplying both sides of the previous example by the LCD, we were left with a linear equation to solve. This is not always the case; sometimes we will be left with a quadratic equation. Example 2: Solve: $2−1x(x+1)=3x+1$. Solution: In this example, there are two restrictions, $x≠0$ and $x≠−1$. Begin by multiplying both sides by the LCD, $x(x+1)$. After distributing and dividing out the common factors, a quadratic equation remains. To solve it, rewrite it in standard form, factor, and then set each factor equal to 0. Check to see if these values solve the original equation. Answer: The solutions are −1/2 and 1. Up to this point, all of the possible solutions have solved the original equation. However, this may not always be the case. Multiplying both sides of an equation by variable factors may lead to extraneous solutionsA solution that does not solve the original equation., which are solutions that do not solve the original equation. A complete list of steps for solving a rational equation is outlined in the following example. Example 3: Solve: $xx+2+2x2+5x+6=5x+3$. Solution: Step 1: Factor all denominators and determine the LCD. The . Step 2: Identify the restrictions. In this case, they are $x≠−2$ and $x≠−3$. Step 3: Multiply both sides of the equation by the LCD. Distribute carefully and then simplify. Step 4: Solve the resulting equation. Here the result is a quadratic equation. Rewrite it in standard form, factor, and then set each factor equal to 0. Step 5: Check for extraneous solutions. Always substitute into the original equation, or the factored equivalent. In this case, choose the factored equivalent to check: Here −2 is an extraneous solution and is not included in the solution set. It is important to note that −2 is a restriction. Answer: The solution is 4. If this process produces a solution that happens to be a restriction, then disregard it as an extraneous solution. Try this! Solve: $xx−5+3x+2=7xx2−3x−10$. Answer: −3 ### Video Solution (click to see video) Sometimes all potential solutions are extraneous, in which case we say that there is no solution to the original equation. In the next two examples, we demonstrate two ways in which a rational equation can have no solutions. Example 4: Solve: $3xx2−4−2x+2=1x+2$. Solution: To identify the LCD, first factor the denominators. Multiply both sides by the least common denonominator (LCD), $(x+2)(x−2)$, distributing carefully. The equation is a contradiction and thus has no solution. Answer: No solution, $∅$ Example 5: Solve: $xx−4−4x+5=36x2+x−20$. Solution: First, factor the denominators. Take note that the restrictions are $x≠4$ and $x≠−5$. To clear the fractions, multiply by the LCD, $(x−4)(x+5)$. Both of these values are restrictions of the original equation; hence both are extraneous. Answer: No solution, $∅$ Try this! Solve: $1x+1+xx−3=4xx2−2x−3$. Answer: $∅$ ### Video Solution (click to see video) It is important to point out that this technique for clearing algebraic fractions only works for equations. Do not try to clear algebraic fractions when simplifying expressions. As a reminder, we have Expressions are to be simplified and equations are to be solved. If we multiply the expression by the LCD, $x(2x+1)$, we obtain another expression that is not equivalent. ## Literal Equations Literal equations, or formulas, are often rational equations. Hence the techniques described in this section can be used to solve for particular variables. Assume that all variable expressions in the denominator are nonzero. Example 6: Solve for x: $z=x−5y$. Solution: The goal is to isolate x. Assuming that y is nonzero, multiply both sides by y and then add 5 to both sides. Answer: $x=yz+5$ Example 7: Solve for c: $1c=1a+1b$. Solution: In this example, the goal is to isolate c. We begin by multiplying both sides by the LCD, $a⋅b⋅c$, distributing carefully. On the right side of the equation, factor out c. Next, divide both sides of the equation by the quantity $(b+a)$. Answer: $c=abb+a$ Try this! Solve for y: $x=y+1y−1$. Answer: $y=x+1x−1$ ### Video Solution (click to see video) ### Key Takeaways • Begin solving rational equations by multiplying both sides by the LCD. The resulting equivalent equation can be solved using the techniques learned up to this point. • Multiplying both sides of a rational equation by a variable expression introduces the possibility of extraneous solutions. Therefore, we must check the solutions against the set of restrictions. If a solution is a restriction, then it is not part of the domain and is extraneous. • When multiplying both sides of an equation by an expression, distribute carefully and multiply each term by that expression. • If all of the resulting solutions are extraneous, then the original equation has no solutions. ### Topic Exercises Part A: Rational Equations Solve. 1. $12+1x=18$ 2. $13−1x=29$ 3. $13x−23=1x$ 4. $25x−1x=310$ 5. $12x+1=5$ 6. $33x−1+4=5$ 7. $2x−3x+5=2x+5$ 8. $5x2x−1=x−12x−1$ 9. $5x−7=6x−9$ 10. $5x+5=3x+1$ 11. $x6−6x=0$ 12. $5x+x5=−2$ 13. $xx+12=2x$ 14. $2xx+5=16−x$ 15. $1x+x2x+1=0$ 16. $9x3x−1−4x=0$ 17. $1−2x=48x2$ 18. $2−9x=5x2$ 19. $1+12x=12x−2$ 20. $1−3x−5x(3x−4)=−1x$ 21. $x2=14x+3$ 22. $3x2=x+13−x$ 23. $6=−3x+3x−1$ 24. $12x−2=2+6(4−x)x−2$ 25. $2+2xx−3=3(x−1)x−3$ 26. $xx−1+16x−1=x(x−1)(6x−1)$ 27. $12x2−81=1x+9−2x−9$ 28. $14x2−49=2x−7−3x+7$ 29. $6xx+3+4x−3=3xx2−9$ 30. $3xx+2−17x−2=−48x2−4$ 31. $x−1+3=0$ 32. $4−y−1=0$ 33. $y−2−4=0$ 34. $9x−2−1=0$ 35. $3(x−1)−1+5=0$ 36. $5−2(3x+1)−1=0$ 37. $3+2x−3=2x−3$ 38. $1x=1x+1$ 39. $xx+1=x+1x$ 40. $3x−13x=xx+3$ 41. $4x−7x−5=3x−2x−5$ 42. $xx2−9=1x−3$ 43. $3x+4x−8−28−x=1$ 44. $1x=6x(x+3)$ 45. $3x=1x+1+13x(x+1)$ 46. $xx−1−34x−1=9x(4x−1)(x−1)$ 47. $1x−4+xx−2=2x2−6x+8$ 48. $xx−5+x−1x2−11x+30=5x−6$ 49. $xx+1−65x2+4x−1=−55x−1$ 50. $−8x2−4x−12+2(x+2)x2+4x−60=1x+2$ 51. $xx+2−20x2−x−6=−4x−3$ 52. $x+7x−1+x−1x+1=4x2−1$ 53. $x−1x−3+x−3x−1=−x+5x−3$ 54. $x−2x−5−x−5x−2=8−xx−5$ 55. $x+7x−2−81x2+5x−14=9x+7$ 56. $xx−6+1=5x+3036−x2$ 57. $2xx+1−44x−3=−74x2+x−3$ 58. $x−5x−10+5x−5=−5xx2−15x+50$ 59. $5x2+5x+4+x+1x2+3x−4=5x2−1$ 60. $1x2−2x−63+x−9x2+10x+21=1x2−6x−27$ 61. $4x2−4+2(x−2)x2−4x−12=x+2x2−8x+12$ 62. $x+2x2−5x+4+x+2x2+x−2=x−1x2−2x−8$ 63. $6xx−1−11x+12x2−x−1=6x2x+1$ 64. $8x2x−3+4x2x2−7x+6=1x−2$ Part B: Literal Equations Solve for the indicated variable. 65. Solve for r: $t=Dr$. 66. Solve for b: $h=2Ab$. 67. Solve for P: $t=IPr$. 68. Solve for $π$: $r=C2π$. 69. Solve for c: $1a=1b+1c$. 70. Solve for y: $m=y−y1x−x1$. 71. Solve for w: $P=2(l+w)$. 72. Solve for t: $A=P(1+rt)$. 73. Solve for m: $s=1n+m$. 74. Solve for S: $h=S2πr−r$. 75. Solve for x: $y=xx+2$. 76. Solve for x: $y=2x+15x$. 77. Solve for R: $1R=1R1+1R2$. 78. Solve for $S1$: $1f=1S1+1S2$. Part C: Discussion Board 79. Explain why multiplying both sides of an equation by the LCD sometimes produces extraneous solutions. 80. Explain the connection between the technique of cross multiplication and multiplying both sides of a rational equation by the LCD. 81. Explain how we can tell the difference between a rational expression and a rational equation. How do we treat them differently? ### Answers 1: −8/3 3: −1 5: −2/5 7: 5/2 9: −3 11: −6, 6 13: −4, 6 15: −1 17: −6, 8 19: −4, 6 21: −7, 4 23: $∅$ 25: $∅$ 27: −39 29: 4/3, 3/2 31: −1/3 33: −1/2, 1/2 35: 2/5 37: $∅$ 39: −1/2 41: $∅$ 43: −7 45: 5 47: −1 49: $∅$ 51: −4 53: 5/3 55: $∅$ 57: 1/2 59: −6, 4 61: 10 63: 1/3 65: $r=Dt$ 67: $P=Itr$ 69: $c=abb−a$ 71: $w=P−2l2$ 73: $m=1−sns$ 75: $x=2y1−y$ 77: $R=R1R2R1+R2$ ## 7.6 Applications of Rational Equations ### Learning Objectives 1. Solve applications involving relationships between real numbers. 2. Solve applications involving uniform motion (distance problems). 3. Solve work-rate applications. ## Number Problems Recall that the reciprocalThe reciprocal of a nonzero number n is 1/n. of a nonzero number n is 1/n. For example, the reciprocal of 5 is 1/5 and 5 ⋅ 1/5 = 1. In this section, the applications will often involve the key word “reciprocal.” When this is the case, we will see that the algebraic setup results in a rational equation. Example 1: A positive integer is 4 less than another. The sum of the reciprocals of the two positive integers is 10/21. Find the two integers. Solution: Begin by assigning variables to the unknowns. Next, use the reciprocals $1n$ and $1n−4$ to translate the sentences into an algebraic equation. We can solve this rational expression by multiplying both sides of the equation by the least common denominator (LCD). In this case, the LCD is $21n(n−4)$. Solve the resulting quadratic equation. The question calls for integers and the only integer solution is $n=7$. Hence disregard 6/5. Use the expression $n−4$ to find the smaller integer. Answer: The two positive integers are 3 and 7. The check is left to the reader. Example 2: A positive integer is 4 less than another. If the reciprocal of the smaller integer is subtracted from twice the reciprocal of the larger, then the result is 1/30. Find the two integers. Solution: Set up an algebraic equation. Solve this rational expression by multiplying both sides by the LCD. The LCD is $30n(n−4)$. Here we have two viable possibilities for the larger integer. For this reason, we will we have two solutions to this problem. As a check, perform the operations indicated in the problem. Answer: Two sets of positive integers solve this problem: {6, 10} and {20, 24}. Try this! The difference between the reciprocals of two consecutive positive odd integers is 2/15. Find the integers. Answer: The integers are 3 and 5. ### Video Solution (click to see video) ## Uniform Motion Problems Uniform motionDescribed by the formula $D=rt$, where the distance, D, is given as the product of the average rate, r, and the time, t, traveled at that rate. problems, also referred to as distance problems, involve the formula where the distance, D, is given as the product of the average rate, r, and the time, t, traveled at that rate. If we divide both sides by the average rate, r, then we obtain the formula For this reason, when the unknown quantity is time, the algebraic setup for distance problems often results in a rational equation. Similarly, when the unknown quantity is the rate, the setup also may result in a rational equation. We begin any uniform motion problem by first organizing our data with a chart. Use this information to set up an algebraic equation that models the application. Example 5: Mary spent the first 120 miles of her road trip in traffic. When the traffic cleared, she was able to drive twice as fast for the remaining 300 miles. If the total trip took 9 hours, then how fast was she moving in traffic? Solution: First, identify the unknown quantity and organize the data. To avoid introducing two more variables for the time column, use the formula $t=Dr$. Here the time for each leg of the trip is calculated as follows: Use these expressions to complete the chart. The algebraic setup is defined by the time column. Add the times for each leg of the trip to obtain a total of 9 hours: We begin solving this equation by first multiplying both sides by the LCD, 2x. Answer: Mary averaged 30 miles per hour in traffic. Example 6: A passenger train can travel, on average, 20 miles per hour faster than a freight train. If the passenger train covers 390 miles in the same time it takes the freight train to cover 270 miles, then how fast is each train? Solution: First, identify the unknown quantities and organize the data. Next, organize the given data in a chart. Use the formula $t=Dr$ to fill in the time column for each train. Because the trains travel the same amount of time, finish the algebraic setup by equating the expressions that represent the times: Solve this equation by first multiplying both sides by the LCD, $x(x+20)$. Use x + 20 to find the speed of the passenger train. Answer: The speed of the passenger train is 65 miles per hour and the speed of the freight train is 45 miles per hour. Example 7: Brett lives on the river 8 miles upstream from town. When the current is 2 miles per hour, he can row his boat downstream to town for supplies and back in 3 hours. What is his average rowing speed in still water? Solution: Rowing downstream, the current increases his speed, and his rate is x + 2 miles per hour. Rowing upstream, the current decreases his speed, and his rate is x − 2 miles per hour. Begin by organizing the data in the following chart: Use the formula $t=Dr$ to fill in the time column for each leg of the trip. The algebraic setup is defined by the time column. Add the times for each leg of the trip to obtain a total of 3 hours: Solve this equation by first multiplying both sides by the LCD, $(x+2)(x−2)$. Next, solve the resulting quadratic equation. Use only the positive solution, $x=6$ miles per hour. Answer: His rowing speed is 6 miles per hour. Try this! Dwayne drove 18 miles to the airport to pick up his father and then returned home. On the return trip he was able to drive an average of 15 miles per hour faster than he did on the trip there. If the total driving time was 1 hour, then what was his average speed driving to the airport? Answer: His average speed driving to the airport was 30 miles per hour. ### Video Solution (click to see video) ## Work-Rate Problems The rate at which a task can be performed is called a work rateThe rate at which a task can be performed.. For example, if a painter can paint a room in 8 hours, then the task is to paint the room, and we can write In other words, the painter can complete $18$ of the task per hour. If he works for less than 8 hours, then he will perform a fraction of the task. For example, Obtain the amount of the task completed by multiplying the work rate by the amount of time the painter works. Typically, work-rate problems involve people working together to complete tasks. When this is the case, we can organize the data in a chart, just as we have done with distance problems. Suppose an apprentice painter can paint the same room by himself in 10 hours. Then we say that he can complete $110$ of the task per hour. Let t represent the time it takes both of the painters, working together, to paint the room. To complete the chart, multiply the work rate by the time for each person. The portion of the room each can paint adds to a total of 1 task completed. This is represented by the equation obtained from the first column of the chart: This setup results in a rational equation that can be solved for t by multiplying both sides by the LCD, 40. Therefore, the two painters, working together, complete the task in $449$ hours. In general, we have the following work-rate formula$1t1⋅t+1t2⋅t=1$, where $1t1$ and $1t2$ are the individual work rates and t is the time it takes to complete the task working together.: Here $1t1$ and $1t2$ are the individual work rates and t is the time it takes to complete one task working together. If we factor out the time, t, and then divide both sides by t, we obtain an equivalent work-rate formula: In summary, we have the following equivalent work-rate formulas: Example 3: Working alone, Billy’s dad can complete the yard work in 3 hours. If Billy helps his dad, then the yard work takes 2 hours. How long would it take Billy working alone to complete the yard work? Solution: The given information tells us that Billy’s dad has an individual work rate of $13$ task per hour. If we let x represent the time it takes Billy working alone to complete the yard work, then Billy’s individual work rate is $1x$, and we can write Working together, they can complete the task in 2 hours. Multiply the individual work rates by 2 hours to fill in the chart. The amount of the task each completes will total 1 completed task. To solve for x, we first multiply both sides by the LCD, 3x. Answer: It takes Billy 6 hours to complete the yard work alone. Of course, the unit of time for the work rate need not always be in hours. Example 4: Working together, two construction crews can build a shed in 5 days. Working separately, the less experienced crew takes twice as long to build a shed than the more experienced crew. Working separately, how long does it take each crew to build a shed? Solution: Working together, the job is completed in 5 days. This gives the following setup: The first column in the chart gives us an algebraic equation that models the problem: Solve the equation by multiplying both sides by 2x. To determine the time it takes the less experienced crew, we use 2x: Answer: Working separately, the experienced crew takes 7½ days to build a shed, and the less experienced crew takes 15 days to build a shed. Try this! Joe’s garden hose fills the pool in 12 hours. His neighbor has a thinner hose that fills the pool in 15 hours. How long will it take to fill the pool using both hoses? Answer: It will take both hoses $623$ hours to fill the pool. ### Video Solution (click to see video) ### Key Takeaways • In this section, all of the steps outlined for solving general word problems apply. Look for the new key word “reciprocal,” which indicates that you should write the quantity in the denominator of a fraction with numerator 1. • When solving distance problems where the time element is unknown, use the equivalent form of the uniform motion formula, $t=Dr$, to avoid introducing more variables. • When solving work-rate problems, multiply the individual work rate by the time to obtain the portion of the task completed. The sum of the portions of the task results in the total amount of work completed. ### Topic Exercises Part A: Number Problems Use algebra to solve the following applications. 1. A positive integer is twice another. The sum of the reciprocals of the two positive integers is 3/10. Find the two integers. 2. A positive integer is twice another. The sum of the reciprocals of the two positive integers is 3/12. Find the two integers. 3. A positive integer is twice another. The difference of the reciprocals of the two positive integers is 1/8. Find the two integers. 4. A positive integer is twice another. The difference of the reciprocals of the two positive integers is 1/18. Find the two integers. 5. A positive integer is 2 less than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 5/12, then find the two integers. 6. A positive integer is 2 more than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 17/35, then find the two integers. 7. The sum of the reciprocals of two consecutive positive even integers is 11/60. Find the two even integers. 8. The sum of the reciprocals of two consecutive positive odd integers is 16/63. Find the integers. 9. The difference of the reciprocals of two consecutive positive even integers is 1/24. Find the two even integers. 10. The difference of the reciprocals of two consecutive positive odd integers is 2/99. Find the integers. 11. If 3 times the reciprocal of the larger of two consecutive integers is subtracted from 2 times the reciprocal of the smaller, then the result is 1/2. Find the two integers. 12. If 3 times the reciprocal of the smaller of two consecutive integers is subtracted from 7 times the reciprocal of the larger, then the result is 1/2. Find the two integers. 13. A positive integer is 5 less than another. If the reciprocal of the smaller integer is subtracted from 3 times the reciprocal of the larger, then the result is 1/12. Find the two integers. 14. A positive integer is 6 less than another. If the reciprocal of the smaller integer is subtracted from 10 times the reciprocal of the larger, then the result is 3/7. Find the two integers. Part B: Uniform Motion Problems Use algebra to solve the following applications. 15. James can jog twice as fast as he can walk. He was able to jog the first 9 miles to his grandmother’s house, but then he tired and walked the remaining 1.5 miles. If the total trip took 2 hours, then what was his average jogging speed? 16. On a business trip, an executive traveled 720 miles by jet aircraft and then another 80 miles by helicopter. If the jet averaged 3 times the speed of the helicopter and the total trip took 4 hours, then what was the average speed of the jet? 17. Sally was able to drive an average of 20 miles per hour faster in her car after the traffic cleared. She drove 23 miles in traffic before it cleared and then drove another 99 miles. If the total trip took 2 hours, then what was her average speed in traffic? 18. Harry traveled 15 miles on the bus and then another 72 miles on a train. If the train was 18 miles per hour faster than the bus and the total trip took 2 hours, then what was the average speed of the train? 19. A bus averages 6 miles per hour faster than a trolley. If the bus travels 90 miles in the same time it takes the trolley to travel 75 miles, then what is the speed of each? 20. A passenger car averages 16 miles per hour faster than the bus. If the bus travels 56 miles in the same time it takes the passenger car to travel 84 miles, then what is the speed of each? 21. A light aircraft travels 2 miles per hour less than twice as fast as a passenger car. If the passenger car can travel 231 miles in the same time it takes the aircraft to travel 455 miles, then what is the average speed of each? 22. Mary can run 1 mile per hour more than twice as fast as Bill can walk. If Bill can walk 3 miles in the same time it takes Mary to run 7.2 miles, then what is Bill’s average walking speed? 23. An airplane traveling with a 20-mile-per-hour tailwind covers 270 miles. On the return trip against the wind, it covers 190 miles in the same amount of time. What is the speed of the airplane in still air? 24. A jet airliner traveling with a 30-mile-per-hour tailwind covers 525 miles in the same amount of time it is able to travel 495 miles after the tailwind eases to 10 miles per hour. What is the speed of the airliner in still air? 25. A boat averages 16 miles per hour in still water. With the current, the boat can travel 95 miles in the same time it travels 65 miles against it. What is the speed of the current? 26. A river tour boat averages 7 miles per hour in still water. If the total 24-mile tour downriver and 24 miles back takes 7 hours, then how fast is the river current? 27. If the river current flows at an average 3 miles per hour, then a tour boat makes the 9-mile tour downstream with the current and back the 9 miles against the current in 4 hours. What is the average speed of the boat in still water? 28. Jane rowed her canoe against a 1-mile-per-hour current upstream 12 miles and then returned the 12 miles back downstream. If the total trip took 5 hours, then at what speed can Jane row in still water? 29. Jose drove 15 miles to pick up his sister and then returned home. On the return trip, he was able to average 15 miles per hour faster than he did on the trip to pick her up. If the total trip took 1 hour, then what was Jose’s average speed on the return trip? 30. Barry drove the 24 miles to town and then back in 1 hour. On the return trip, he was able to average 14 miles per hour faster than he averaged on the trip to town. What was his average speed on the trip to town? 31. Jerry paddled his kayak upstream against a 1-mile-per-hour current for 12 miles. The return trip downstream with the 1-mile-per-hour current took 1 hour less time. How fast can Jerry paddle the kayak in still water? 32. It takes a light aircraft 1 hour more time to fly 360 miles against a 30-mile-per-hour headwind than it does to fly the same distance with it. What is the speed of the aircraft in calm air? Part C: Work-Rate Problems Use algebra to solve the following applications. 33. James can paint the office by himself in 7 hours. Manny paints the office in 10 hours. How long will it take them to paint the office working together? 34. Barry can lay a brick driveway by himself in 12 hours. Robert does the same job in 10 hours. How long will it take them to lay the brick driveway working together? 35. Jerry can detail a car by himself in 50 minutes. Sally does the same job in 1 hour. How long will it take them to detail a car working together? 36. Jose can build a small shed by himself in 26 hours. Alex builds the same small shed in 2 days. How long would it take them to build the shed working together? 37. Allison can complete a sales route by herself in 6 hours. Working with an associate, she completes the route in 4 hours. How long would it take her associate to complete the route by herself? 38. James can prepare and paint a house by himself in 5 days. Working with his brother, Bryan, they can do it in 3 days. How long would it take Bryan to prepare and paint the house by himself? 39. Joe can assemble a computer by himself in 1 hour. Working with an assistant, he can assemble a computer in 40 minutes. How long would it take his assistant to assemble a computer working alone? 40. The teacher’s assistant can grade class homework assignments by herself in 1 hour. If the teacher helps, then the grading can be completed in 20 minutes. How long would it take the teacher to grade the papers working alone? 41. A larger pipe fills a water tank twice as fast as a smaller pipe. When both pipes are used, they fill the tank in 5 hours. If the larger pipe is left off, then how long would it take the smaller pipe to fill the tank? 42. A newer printer can print twice as fast as an older printer. If both printers working together can print a batch of flyers in 45 minutes, then how long would it take the newer printer to print the batch working alone? 43. Working alone, Henry takes 9 hours longer than Mary to clean the carpets in the entire office. Working together, they clean the carpets in 6 hours. How long would it take Mary to clean the office carpets if Henry were not there to help? 44. Working alone, Monique takes 4 hours longer than Audrey to record the inventory of the entire shop. Working together, they take inventory in 1.5 hours. How long would it take Audrey to record the inventory working alone? 45. Jerry can lay a tile floor in 3 hours less time than Jake. If they work together, the floor takes 2 hours. How long would it take Jerry to lay the floor by himself? 46. Jeremy can build a model airplane in 5 hours less time than his brother. Working together, they need 6 hours to build the plane. How long would it take Jeremy to build the model airplane working alone? 47. Harry can paint a shed by himself in 6 hours. Jeremy can paint the same shed by himself in 8 hours. How long will it take them to paint two sheds working together? 48. Joe assembles a computer by himself in 1 hour. Working with an assistant, he can assemble 10 computers in 6 hours. How long would it take his assistant to assemble 1 computer working alone? 49. Jerry can lay a tile floor in 3 hours, and his assistant can do the same job in 4 hours. If Jerry starts the job and his assistant joins him 1 hour later, then how long will it take to lay the floor? 50. Working alone, Monique takes 6 hours to record the inventory of the entire shop, while it takes Audrey only 4 hours to do the same job. How long will it take them working together if Monique leaves 2 hours early? ### Answers 1: {5, 10} 3: {4, 8} 5: {6, 8} 7: {10, 12} 9: {6, 8} 11: {1, 2} or {−4, −3} 13: {4, 9} or {15, 20} 15: 6 miles per hour 17: 46 miles per hour 19: Trolley: 30 miles per hour; bus: 36 miles per hour 21: Passenger car: 66 miles per hour; aircraft: 130 miles per hour 23: 115 miles per hour 25: 3 miles per hour 27: 6 miles per hour 29: 40 miles per hour 31: 5 miles per hour 33: $4217$ hours 35: $27311$ minutes 37: 12 hours 39: 2 hours 41: 15 hours 43: 9 hours 45: 3 hours 47: $667$ hours 49: $217$ hours ## 7.7 Variation ### Learning Objectives 1. Solve applications involving direct variation. 2. Solve applications involving inverse variation. 3. Solve applications involving joint variation. ## Direct Variation Consider a freight train moving at a constant speed of 30 miles per hour. The equation that expresses the distance traveled at that speed in terms of time is given by After 1 hour the train has traveled 30 miles, after 2 hours the train has traveled 60 miles, and so on. We can construct a chart and graph this relation. In this example, we can see that the distance varies over time as the product of the constant rate, 30 miles per hour, and the variable, t. This relationship is described as direct variationDescribes two quantities x and y that are constant multiples of each other: $y=kx$. and 30 is called the variation constant. In addition, if we divide both sides of $D=30t$ by t we have In this form, it is reasonable to say that D is proportional to t, where 30 is the constant of proportionality. In general, we have Key words Translation y varies directly as x $y=kx$ y is directly proportionalUsed when referring to direct variation. to x y is proportional to x Here k is nonzero and is called the constant of variationThe nonzero multiple k, when quantities vary directly or inversely. or the constant of proportionalityUsed when referring to the constant of variation.. Example 1: The circumference of a circle is directly proportional to its diameter, and the constant of proportionality is $π$. If the circumference is measured to be 20 inches, then what is the radius of the circle? Solution: Use the fact that “the circumference is directly proportional to the diameter” to write an equation that relates the two variables. We are given that “the constant of proportionality is $π$,” or $k=π$. Therefore, we write Now use this formula to find d when the circumference is 20 inches. The radius of the circle, r, is one-half of its diameter. Answer: The radius is $10π$ inches, or approximately 3.18 inches. Typically, we will not be given the constant of variation. Instead, we will be given information from which it can be determined. Example 2: An object’s weight on earth varies directly to its weight on the moon. If a man weighs 180 pounds on earth, then he will weigh 30 pounds on the moon. Set up an algebraic equation that expresses the weight on earth in terms of the weight on the moon and use it to determine the weight of a woman on the moon if she weighs 120 pounds on earth. Solution: We are given that the “weight on earth varies directly to the weight on the moon.” To find the constant of variation k, use the given information. A 180-pound man on earth weighs 30 pounds on the moon, or $y=180$ when $x=30$. Solve for k. Next, set up a formula that models the given information. This implies that a person’s weight on earth is 6 times her weight on the moon. To answer the question, use the woman’s weight on earth, $y=120 pounds$, and solve for x. Answer: The woman weighs 20 pounds on the moon. ## Inverse Variation Next, consider the relationship between time and rate, If we wish to travel a fixed distance, then we can determine the average speed required to travel that distance in a given amount of time. For example, if we wish to drive 240 miles in 4 hours, we can determine the required average speed as follows: The average speed required to drive 240 miles in 4 hours is 60 miles per hour. If we wish to drive the 240 miles in 5 hours, then determine the required speed using a similar equation: In this case, we would only have to average 48 miles per hour. We can make a chart and view this relationship on a graph. This is an example of an inverse relationship. We say that r is inversely proportional to the time t, where 240 is the constant of proportionality. In general, we have Key words Translation y varies inverselyDescribes two quantities x and y, where one variable is directly proportional to the reciprocal of the other: $y=kx.$ as x $y=kx$ y is inversely proportionalUsed when referring to inverse variation. to x Again, k is nonzero and is called the constant of variation or the constant of proportionality. Example 3: If y varies inversely as x and $y=5$ when $x=2$, then find the constant of proportionality and an equation that relates the two variables. Solution: If we let k represent the constant of proportionality, then the statement “y varies inversely as x” can be written as follows: Use the given information, $y=5$ when $x=2$, to find k. Solve for k. Therefore, the formula that models the problem is Answer: The constant of proportionality is 10, and the equation is $y=10x$. Example 4: The weight of an object varies inversely as the square of its distance from the center of earth. If an object weighs 100 pounds on the surface of earth (approximately 4,000 miles from the center), then how much will it weigh at 1,000 miles above earth’s surface? Solution: Since “w varies inversely as the square of d,” we can write Use the given information to find k. An object weighs 100 pounds on the surface of earth, approximately 4,000 miles from the center. In other words, w = 100 when d = 4,000: Solve for k. Therefore, we can model the problem with the following formula: To use the formula to find the weight, we need the distance from the center of earth. Since the object is 1,000 miles above the surface, find the distance from the center of earth by adding 4,000 miles: To answer the question, use the formula with d = 5,000. Answer: The object will weigh 64 pounds at a distance 1,000 miles above the surface of earth. ## Joint Variation Lastly, we define relationships between multiple variables. In general, we have Vocabulary Translation y varies jointlyDescribes a quantity y that varies directly as the product of two other quantities x and z: $y=kxz$. as x and z $y=kxz$ y is jointly proportionalUsed when referring to joint variation. to x and z Here k is nonzero and is called the constant of variation or the constant of proportionality. Example 5: The area of an ellipse varies jointly as a, half of the ellipse’s major axis, and b, half of the ellipse’s minor axis. If the area of an ellipse is $300π cm2$, where $a=10 cm$ and $b=30 cm$, then what is the constant of proportionality? Give a formula for the area of an ellipse. Solution: If we let A represent the area of an ellipse, then we can use the statement “area varies jointly as a and b” to write To find the constant of variation, k, use the fact that the area is $300π$ when $a=10$ and $b=30$. Therefore, the formula for the area of an ellipse is Answer: The constant of proportionality is $π$, and the formula for the area is $A=abπ$. Try this! Given that y varies directly as the square of x and inversely to z, where y = 2 when x = 3 and z = 27, find y when x = 2 and z = 16. Answer: 3/2 ### Video Solution (click to see video) ### Key Takeaway • The setup of variation problems usually requires multiple steps. First, identify the key words to set up an equation and then use the given information to find the constant of variation k. After determining the constant of variation, write a formula that models the problem. Once a formula is found, use it to answer the question. ### Topic Exercises Part A: Variation Problems Translate the following sentences into a mathematical formula. 1. The distance, D, an automobile can travel is directly proportional to the time, t, that it travels at a constant speed. 2. The extension of a hanging spring, d, is directly proportional to the weight, w, attached to it. 3. An automobile’s breaking distance, d, is directly proportional to the square of the automobile’s speed, v. 4. The volume, V, of a sphere varies directly as the cube of its radius, r. 5. The volume, V, of a given mass of gas is inversely proportional to the pressure, p, exerted on it. 6. The intensity, I, of light from a light source is inversely proportional to the square of the distance, d, from the source. 7. Every particle of matter in the universe attracts every other particle with a force, F, that is directly proportional to the product of the masses, $m1$ and $m2$, of the particles and inversely proportional to the square of the distance, d, between them. 8. Simple interest, I, is jointly proportional to the annual interest rate, r, and the time, t, in years a fixed amount of money is invested. 9. The period, T, of a pendulum is directly proportional to the square root of its length, L. 10. The time, t, it takes an object to fall is directly proportional to the square root of the distance, d, it falls. Construct a mathematical model given the following. 11. y varies directly as x, and y = 30 when x = 6. 12. y varies directly as x, and y = 52 when x = 4. 13. y is directly proportional to x, and y = 12 when x = 3. 14. y is directly proportional to x, and y = 120 when x = 20. 15. y varies directly as x, and y = 14 when x = 10. 16. y varies directly as x, and y = 2 when x = 8. 17. y varies inversely as x, and y = 5 when x = 7. 18. y varies inversely as x, and y = 12 when x = 2. 19. y is inversely proportional to x, and y = 3 when x = 9. 20. y is inversely proportional to x, and y = 21 when x = 3. 21. y varies inversely as x, and y = 2 when x = 1/8. 22. y varies inversely as x, and y = 3/2 when x = 1/9. 23. y varies jointly as x and z, where y = 8 when x = 4 and z = 1/2. 24. y varies jointly as x and z, where y = 24 when x = 1/3 and z = 9. 25. y is jointly proportional to x and z, where y = 2 when x = 1 and z = 3. 26. y is jointly proportional to x and z, where y = 15 when x = 3 and z = 7. 27. y varies jointly as x and z, where y = 2/3 when x = 1/2 and z = 12. 28. y varies jointly as x and z, where y = 5 when x = 3/2 and z = 2/9. 29. y varies directly as the square of x, where y = 45 when x = 3. 30. y varies directly as the square of x, where y = 3 when x = 1/2. 31. y is inversely proportional to the square of x, where y = 27 when x = 1/3. 32. y is inversely proportional to the square of x, where y = 9 when x = 2/3. 33. y varies jointly as x and the square of z, where y = 54 when x = 2 and z = 3. 34. y varies jointly as x and the square of z, where y = 6 when x = 1/4 and z = 2/3. 35. y varies jointly as x and z and inversely as the square of w, where y = 30 when x = 8, z = 3, and w = 2. 36. y varies jointly as x and z and inversely as the square of w, where y = 5 when x = 1, z = 3, and w = 1/2. 37. y varies directly as the square root of x and inversely as z, where y = 12 when x = 9 and z = 5. 38. y varies directly as the square root of x and inversely as the square of z, where y = 15 when x = 25 and z = 2. 39. y varies directly as the square of x and inversely as z and the square of w, where y = 14 when x = 4, w = 2, and z = 2. 40. y varies directly as the square root of x and inversely as z and the square of w, where y = 27 when x = 9, w = 1/2, and z = 4. Part B: Variation Problems Applications involving variation. 41. Revenue in dollars is directly proportional to the number of branded sweat shirts sold. If the revenue earned from selling 25 sweat shirts is$318.75, then determine the revenue if 30 sweat shirts are sold.
42. The sales tax on the purchase of a new car varies directly as the price of the car. If an $18,000 new car is purchased, then the sales tax is$1,350. How much sales tax is charged if the new car is priced at $22,000? 43. The price of a share of common stock in a company is directly proportional to the earnings per share (EPS) of the previous 12 months. If the price of a share of common stock in a company is$22.55 and the EPS is published to be $1.10, then determine the value of the stock if the EPS increases by$0.20.
44. The distance traveled on a road trip varies directly with the time spent on the road. If a 126-mile trip can be made in 3 hours, then what distance can be traveled in 4 hours?
45. The circumference of a circle is directly proportional to its radius. If the circumference of a circle with radius 7 centimeters is measured as $14π$ centimeters, then find the constant of proportionality.
46. The area of circle varies directly as the square of its radius. If the area of a circle with radius 7 centimeters is determined to be $49π$ square centimeters, then find the constant of proportionality.
47. The surface area of a sphere varies directly as the square of its radius. When the radius of a sphere measures 2 meters, the surface area measures $16π$ square meters. Find the surface area of a sphere with radius 3 meters.
48. The volume of a sphere varies directly as the cube of its radius. When the radius of a sphere measures 3 meters, the volume is $36π$ cubic meters. Find the volume of a sphere with radius 1 meter.
49. With a fixed height, the volume of a cone is directly proportional to the square of the radius at the base. When the radius at the base measures 10 centimeters, the volume is 200 cubic centimeters. Determine the volume of the cone if the radius of the base is halved.
50. The distance, d, an object in free fall drops varies directly with the square of the time, t, that it has been falling. If an object in free fall drops 36 feet in 1.5 seconds, then how far will it have fallen in 3 seconds?
Hooke’s law suggests that the extension of a hanging spring is directly proportional to the weight attached to it. The constant of variation is called the spring constant.
Figure 7.1 Robert Hooke (1635–1703)
51. If a hanging spring is stretched 5 inches when a 20‑pound weight is attached to it, then determine its spring constant.
52. If a hanging spring is stretched 3 centimeters when a 2-kilogram weight is attached to it, then determine the spring constant.
53. If a hanging spring is stretched 3 inches when a 2‑pound weight is attached, then how far will it stretch with a 5-pound weight attached?
54. If a hanging spring is stretched 6 centimeters when a 4-kilogram weight is attached to it, then how far will it stretch with a 2-kilogram weight attached?
The breaking distance of an automobile is directly proportional to the square of its speed.
55. If it takes 36 feet to stop a particular automobile moving at a speed of 30 miles per hour, then how much breaking distance is required if the speed is 35 miles per hour?
56. After an accident, it was determined that it took a driver 80 feet to stop his car. In an experiment under similar conditions, it takes 45 feet to stop the car moving at a speed of 30 miles per hour. Estimate how fast the driver was moving before the accident.
Boyle’s law states that if the temperature remains constant, the volume, V, of a given mass of gas is inversely proportional to the pressure, p, exerted on it.
Figure 7.2 Robert Boyle (1627–1691)
57. A balloon is filled to a volume of 216 cubic inches on a diving boat under 1 atmosphere of pressure. If the balloon is taken underwater approximately 33 feet, where the pressure measures 2 atmospheres, then what is the volume of the balloon?
58. If a balloon is filled to 216 cubic inches under a pressure of 3 atmospheres at a depth of 66 feet, then what would the volume be at the surface, where the pressure is 1 atmosphere?
59. To balance a seesaw, the distance from the fulcrum that a person must sit is inversely proportional to his weight. If a 72-pound boy is sitting 3 feet from the fulcrum, then how far from the fulcrum must a 54-pound boy sit to balance the seesaw?
60. The current, I, in an electrical conductor is inversely proportional to its resistance, R. If the current is 1/4 ampere when the resistance is 100 ohms, then what is the current when the resistance is 150 ohms?
61. The number of men, represented by y, needed to lay a cobblestone driveway is directly proportional to the area, A, of the driveway and inversely proportional to the amount of time, t, allowed to complete the job. Typically, 3 men can lay 1,200 square feet of cobblestone in 4 hours. How many men will be required to lay 2,400 square feet of cobblestone given 6 hours?
62. The volume of a right circular cylinder varies jointly as the square of its radius and its height. A right circular cylinder with a 3-centimeter radius and a height of 4 centimeters has a volume of $36π$ cubic centimeters. Find a formula for the volume of a right circular cylinder in terms of its radius and height.
63. The period, T, of a pendulum is directly proportional to the square root of its length, L. If the length of a pendulum is 1 meter, then the period is approximately 2 seconds. Approximate the period of a pendulum that is 0.5 meter in length.
64. The time, t, it takes an object to fall is directly proportional to the square root of the distance, d, it falls. An object dropped from 4 feet will take 1/2 second to hit the ground. How long will it take an object dropped from 16 feet to hit the ground?
Newton’s universal law of gravitation states that every particle of matter in the universe attracts every other particle with a force, F, that is directly proportional to the product of the masses, $m1$ and $m2$, of the particles and inversely proportional to the square of the distance, d, between them. The constant of proportionality is called the gravitational constant.
Figure 7.3 Sir Isaac Newton (1643–1724)
65. If two objects with masses 50 kilograms and 100 kilograms are 1/2 meter apart, then they produce approximately $1.34×10−6$ newtons (N) of force. Calculate the gravitational constant.
66. Use the gravitational constant from the previous exercise to write a formula that approximates the force, F, in newtons between two masses $m1$ and $m2$, expressed in kilograms, given the distance d between them in meters.
67. Calculate the force in newtons between earth and the moon, given that the mass of the moon is approximately $7.3×1022$ kilograms, the mass of earth is approximately $6.0×1024$ kilograms, and the distance between them is on average $1.5×1011$ meters.
68. Calculate the force in newtons between earth and the sun, given that the mass of the sun is approximately $2.0×1030$ kilograms, the mass of earth is approximately $6.0×1024$ kilograms, and the distance between them is on average $3.85×108$ meters.
69. If y varies directly as the square of x, then how does y change if x is doubled?
70. If y varies inversely as square of t, then how does y change if t is doubled?
71. If y varies directly as the square of x and inversely as the square of t, then how does y change if both x and t are doubled?
1: $D=kt$
3: $d=kv2$
5: $V=kp$
7: $F=km1⋅m2d2$
9: $T=kL$
11: $y=5x$
13: $y=4x$
15: $y=75x$
17: $y=35x$
19: $y=27x$
21: $y=14x$
23: $y=4xz$
25: $y=23xz$
27: $y=19xz$
29: $y=5x2$
31: $y=3x2$
33: $y=3xz2$
35: $y=5xzw2$
37: $y=20xz$
39: $y=7x2w2z$
41: $382.50 43:$26.65
45: $2π$
47: $36π$ square meters
49: 50 cubic centimeters
51: 1/4
53: 7.5 inches
55: 49 feet
57: 108 cubic inches
59: 4 feet
61: 4 men
63: 1.4 seconds
65: $6.7×10−11 N m2/kg2$
67: $1.98×1020 N$
69: y changes by a factor of 4
71: y remains unchanged
## 7.8 Review Exercises and Sample Exam
### Review Exercises
Simplifying Rational Expressions
Evaluate for the given set of x-values.
1. $252x2$; {−5, 0, 5}
2. $x−42x−1$; {1/2, 2, 4}
3. $1x2+9$; {−3, 0, 3}
4. $x+3x2−9$; {−3, 0, 3}
State the restrictions to the domain.
5. $5x$
6. $1x(3x+1)$
7. $x+2x2−25$
8. $x−1(x−1)(2x−3)$
State the restrictions and simplify.
9. $x−8x2−64$
10. $3x2+9x2x3−18x$
11. $x2−5x−24x2−3x−40$
12. $2x2+9x−54x2−1$
13. $x2−14412−x$
14. $8x2−10x−39−4x2$
15. Given $f(x)=x−3x2+9$, find $f(−3)$, $f(0)$, and $f(3)$.
16. Simplify $g(x)=x2−2x−242x2−9x−18$ and state the restrictions.
Multiplying and Dividing Rational Expressions
Multiply. (Assume all denominators are nonzero.)
17. $3x5x−3⋅x−39x2$
18. $12y2y3(2y−1)⋅(2y−1)3y$
19. $3x2x−2⋅x2−4x+45x3$
20. $x2−8x+159x5⋅12x2x−3$
21. $x2−36x2−x−30⋅2x2+10xx2+5x−6$
22. $9x2+11x+24−81x2⋅9x−2(x+1)2$
Divide. (Assume all denominators are nonzero.)
23. $9x2−255x3÷3x+515x4$
24. $4x24x2−1÷2x2x−1$
25. $3x2−13x−10x2−x−20÷9x2+12x+4x2+8x+16$
26. $2x2+xy−y2x2+2xy+y2÷4x2−y23x2+2xy−y2$
27. $2x2−6x−208x2+17x+2÷(8x2−39x−5)$
28. $12x2−27x415x4+10x3÷(3x2+x−2)$
29. $25y2−15y4(y−2)⋅15y−1÷10y2(y−2)2$
30. $10x41−36x2÷5x26x2−7x+1⋅x−12x$
31. Given $f(x)=16x2−9x+5$ and $g(x)=x2+3x−104x2+5x−6$, calculate $(f⋅g)(x)$ and state the restrictions.
32. Given $f(x)=x+75x−1$ and $g(x)=x2−4925x2−5x$, calculate $(f/g)(x)$ and state the restrictions.
Simplify. (Assume all denominators are nonzero.)
33. $5xy−3y$
34. $xx2−x−6−3x2−x−6$
35. $2x2x+1+1x−5$
36. $3x−7+1−2xx2$
37. $7x4x2−9x+2−2x−2$
38. $5x−5+20−9x2x2−15x+25$
39. $xx−5−2x−3−5(x−3)x2−8x+15$
40. $3x2x−1−x−4x+4+12(2−x)2x2+7x−4$
41. $1x2+8x−9−1x2+11x+18$
42. $4x2+13x+36+3x2+6x−27$
43. $y+1y+2−12−y+2yy2−4$
44. $1y−11−y−2y2−1$
45. Given $f(x)=x+12x−5$ and $g(x)=xx+1$, calculate $(f+g)(x)$ and state the restrictions.
46. Given $f(x)=x+13x$ and $g(x)=2x−8$, calculate $(f−g)(x)$ and state the restrictions.
Complex Fractions
Simplify.
47. $4−2x 2x−13x$
48. $13−13y 15−15y$
49. $16+1x136−1x2$
50. $1100−1x2 110−1x$
51. $xx+3−2x+1 xx+4+1x+3$
52. $3x−1x−5 5x+2−2x$
53. $1−12x+35x2 1−25x2$
54. $2−15x+25x22x−5$
Solving Rational Equations
Solve.
55. $6x−6=22x−1$
56. $xx−6=x+2x−2$
57. $13x−29=1x$
58. $2x−5+35=1x−5$
59. $xx−5+4x+5=−10x2−25$
60. $2x−122x+3=2−3x22x2+3x$
61. $x+12(x−2)+x−6x=1$
62. $5x+2x+1−xx+4=4$
63. $xx+5+1x−4=4x−7x2+x−20$
64. $23x−1+x2x+1=2(3−4x)6x2+x−1$
65. $xx−1+1x+1=2xx2−1$
66. $2xx+5−12x−3=4−7x2x2+7x−15$
67. Solve for a: $1a=1b+1c$.
68. Solve for y: $x=2y−13y$.
Applications of Rational Equations
Use algebra to solve the following applications.
69. A positive integer is twice another. The sum of the reciprocals of the two positive integers is 1/4. Find the two integers.
70. If the reciprocal of the smaller of two consecutive integers is subtracted from three times the reciprocal of the larger, the result is 3/10. Find the integers.
71. Mary can jog, on average, 2 miles per hour faster than her husband, James. James can jog 6.6 miles in the same amount of time it takes Mary to jog 9 miles. How fast, on average, can Mary jog?
72. Billy traveled 140 miles to visit his grandmother on the bus and then drove the 140 miles back in a rental car. The bus averages 14 miles per hour slower than the car. If the total time spent traveling was 4.5 hours, then what was the average speed of the bus?
73. Jerry takes twice as long as Manny to assemble a skateboard. If they work together, they can assemble a skateboard in 6 minutes. How long would it take Manny to assemble the skateboard without Jerry’s help?
74. Working alone, Joe completes the yard work in 30 minutes. It takes Mike 45 minutes to complete work on the same yard. How long would it take them working together?
Variation
Construct a mathematical model given the following.
75. y varies directly with x, and y = 12 when x = 4.
76. y varies inversely as x, and y = 2 when x = 5.
77. y is jointly proportional to x and z, where y = 36 when x = 3 and z = 4.
78. y is directly proportional to the square of x and inversely proportional to z, where y = 20 when x = 2 and z = 5.
79. The distance an object in free fall drops varies directly with the square of the time that it has been falling. It is observed that an object falls 16 feet in 1 second. Find an equation that models the distance an object will fall and use it to determine how far it will fall in 2 seconds.
80. The weight of an object varies inversely as the square of its distance from the center of earth. If an object weighs 180 pounds on the surface of earth (approximately 4,000 miles from the center), then how much will it weigh at 2,000 miles above earth’s surface?
### Sample Exam
Simplify and state the restrictions.
1. $15x3(3x−1)23x(3x−1)$
2. $x2−144x2+12x$
3. $x2+x−122x2+7x−4$
4. $9−x2(x−3)2$
Simplify. (Assume all variables in the denominator are positive.)
5. $5xx2−25⋅x−525x2$
6. $x2+x−6x2−4x+4⋅3x2−5x−2x2−9$
7. $x2−4x−1212x2÷x−66x$
8. $2x2−7x−46x2−24x÷2x2+7x+310x2+30x$
9. $1x−5+1x+5$
10. $xx+1−82−x−12xx2−x−2$
11. $1y+1x1y2−1x2$
12. $1−6x+9x22−5x−3x2$
13. Given $f(x)=x2−81(4x−3)2$ and $g(x)=4x−3x−9$, calculate $(f⋅g)(x)$ and state the restrictions.
14. Given $f(x)=xx−5$ and $g(x)=13x−5$, calculate $(f−g)(x)$ and state the restrictions.
Solve.
15. $13+1x=2$
16. $1x−5=32x−3$
17. $1−9x+20x2=0$
18. $x+2x−2+1x+2=4(x+1)x2−4$
19. $xx−2−1x−3=3x−10x2−5x+6$
20. $5x+4−x4−x=9x−4x2−16$
21. Solve for r: $P=1201+3r$.
Set up an algebraic equation and then solve.
22. An integer is three times another. The sum of the reciprocals of the two integers is 1/3. Find the two integers.
23. Working alone, Joe can paint the room in 6 hours. If Manny helps, then together they can paint the room in 2 hours. How long would it take Manny to paint the room by himself?
24. A river tour boat averages 6 miles per hour in still water. With the current, the boat can travel 17 miles in the same time it can travel 7 miles against the current. What is the speed of the current?
25. The breaking distance of an automobile is directly proportional to the square of its speed. Under optimal conditions, a certain automobile moving at 35 miles per hour can break to a stop in 25 feet. Find an equation that models the breaking distance under optimal conditions and use it to determine the breaking distance if the automobile is moving 28 miles per hour.
1: 1/2, undefined, 1/2
3: 1/18, 1/9, 1/18
5: $x≠0$
7: $x≠±5$
9: $1x+8$; $x≠±8$
11: $x+3x+5$; $x≠−5, 8$
13: $−(x+12)$; $x≠12$
15: $f(−3)=−13$, $f(0)=−13$, $f(3)=0$
17: $x33$
19: $3(x−2)5x$
21: $2xx−1$
23: $3x(3x−5)$
25: $x+43x+2$
27: $2(8x+1)2$
29: $(5y+1)(y−2)50y6$
31: $(f⋅g)(x)=(4x+3)(x−2)x+2$; $x≠−5, −2, 34$
33: $5x−3y$
35: $2x2−8x+1(2x+1)(x−5)$
37: $−14x−1$
39: $x−5x−3$
41: $3(x−1)(x+2)(x+9)$
43: $yy−2$
45: $(f+g)(x)=3x2−3x+1(2x−5)(x+1)$; $x≠−1, 52$
47: $6$
49: $6xx−6$
51: $(x−3)(x+4)(x+1)(x+2)$
53: $x−7x+5$
55: −3/5
57: −3
59: −10, 1
61: 3, 8
63: 3
65: Ø
67: $a=bcb+c$
69: 6, 12
71: 7.5 miles per hour
73: 9 minutes
75: $y=3x$
77: $y=3xz$
79: $d=16t2$; 64 feet
1: $5x2(3x−1)$; $x≠0, 13$
3: $x−32x−1$; $x≠−4, 12$
5: $15x(x+5)$
7: $x+22x$
9: $2x(x−5)(x+5)$
11: $xyx−y$
13: $(f⋅g)(x)=x+94x−3$; $x≠34, 9$
15: 3/5
17: 4, 5
19: 4
21: $r=40P−13$
23: 3 hours
25: $y=149x2$; 16 feet
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# Factoring Polynomials
## Presentation on theme: "Factoring Polynomials"— Presentation transcript:
Factoring Polynomials
Grouping, Trinomials, Binomials, GCF ,Quadratic form & Solving Equations
Student will be able to Factor by Grouping terms
When polynomials contain four terms, it is sometimes easier to group like terms in order to factor. Your goal is to create a common factor. You can also move terms around in the polynomial to create a common factor. Practice makes you better in recognizing common factors.
Factoring Four Term Polynomials
Do now: find the GCf of the first two terms and the last two terms:
3x2 and 6 Group together and Factor each one separately:
They share a common factor of (x-4)
Write 2 factors: Write the common factor once and put the outside terms together:
Factor by Grouping Example 1:
FACTOR: 3xy - 21y + 5x – 35 Factor the first two terms: 3xy – 21y Factor the last two terms: + 5x - 35 =
Factor by Grouping Example 1:
FACTOR: 3xy - 21y + 5x – 35 Factor the first two terms: 3xy - 21y = 3y (x – 7) Factor the last two terms: + 5x - 35 = 5 (x – 7) The green parentheses are the same so it’s the common factor
Factor by Grouping Example 1:
FACTOR: 3xy - 21y + 5x – 35 Factor the first two terms: 3xy - 21y = 3y (x – 7) Factor the last two terms: + 5x - 35 = 5 (x – 7) The green parentheses are the same so it’s the common factor Now you have a common factor (x - 7) (3y + 5)
Factor by Grouping Example 2:
FACTOR: 6mx – 4m + 3rx – 2r Factor the first two terms: 6mx – 4m = Factor the last two terms: + 3rx – 2r =
Factor by Grouping Example 2:
FACTOR: 6mx – 4m + 3rx – 2r Factor the first two terms: 6mx – 4m = 2m (3x - 2) Factor the last two terms: + 3rx – 2r = r (3x - 2) The green parentheses are the same so it’s the common factor Now you have a common factor (3x - 2) (2m + r)
Factor by Grouping Example 3:
FACTOR: y3– 5y2 -4y +20
Factor by Grouping Example 3:
FACTOR: y3– 5y2 - 4y +20 Factor the first two terms: y3– 5y2 = y2 (y - 5) Factor the last two terms: - 4y +20 = -4 (y – 5) The green parentheses are the same! y2 (y - 5) and -4 (y - 5) Now you have the difference of two squares! look at red ( ): (y - 5) (y2 - 4) : answer: (y - 5) (y - 2) (y + 2)
See worksheet “Factor by grouping”
Try first 4 problems.
So the solution set is { 5,2,-2}
Using Factor by Grouping to solve a polynomial function: From the last example, suppose it was an equation….. y3– 5y2 - 4y +20 = 0 (y - 5) (y - 2) (y + 2) = 0 y=5 y = 2 y=-2 So the solution set is { 5,2,-2}
Factor first, then set factors = 0
X-4=0
solve X-4=0 X=4
Hand this one in: Solve for all roots: 3x3 - 4x2 -27x +36 = 0
Factoring Trinomials
Factoring Trinominals
When trinomials have a degree of “2”, they are known as quadratics. We learned earlier to use the last term’s factors to factor trinomials that had a “1” in front of the squared term. x2 + 12x + 35 So… 7 and 5 or 35 and 1
Factoring Trinominals
When trinomials have a degree of “2”, they are known as quadratics. We learned earlier to use the last term’s factors to factor trinomials that had a “1” in front of the squared term. x2 + 12x + 35 (x + 7)(x + 5) Because = 12!
More Factoring Trinomials
When there is a coefficient larger than “1” in front of the squared term, we can use a method we will call, the “am” add, multiply method to find the factors. Always remember to look for a GCF before you do ANY other factoring.
More Factoring Trinomials
Let’s try this example 3x2 + 13x + 4 (3x )(x ) Write the factors of the last term…1,4 2,2 Multiply using foil until you get the middle term of the trinomial. If so, you’re done!
More Factoring Trinomials
3x2 + 13x + 4 (3x )(x + 4 ) 3x2 + 12x + 1x + 4 = 3x2 + 13x ✓
Difference of Squares
Difference of Squares When factoring using a difference of squares, look for the following three things: only 2 terms minus sign between them both terms must be perfect squares No common factors If all of the above are true, write two ( ), one with a + sign and one with a – sign : ( ) ( ).
Try These, (if possible)
1. a2 – 16 2. x2 – 25 3. 4y2 – 16 4. 9y2 – 25 5. 3r2 – 81 6. 2a2 + 16
answers: 1. a2 – 16 (a + 4) (a – 4) 2. x2 – 25 (x + 5) (x – 5)
3. 4y2 – 9 (2y + 3) (2y – 3) 4. 9y2 – 25 (3y + 5) (3y – 5) 5. 3r2 – 81 *3 is not a square! 6. a Not a difference!
Perfect Square Trinomials
Perfect Square Trinomials
When factoring using perfect square trinomials, look for the following three things: 3 terms last term must be positive first and last terms must be perfect squares If all three of the above are true, write one ( )2 using the sign of the middle term.
Try These 1. a2 – 8a + 16 2. x2 + 10x + 25 3. 4y2 + 16y + 16
Factoring Completely
Factoring Completely Now that we’ve learned all the types of factoring, we need to remember to use them all. Whenever it says to factor, you must break down the expression into the smallest possible factors. Let’s review all the ways to factor.
Types of Factoring Look for GCF first. Count the number of terms:
4 terms – factor by grouping 3 terms - look for perfect square trinomial if not, try “am” method 2 terms - look for difference of squares If any ( ) still has an exponent of 2 or more, see if you can factor again.
These may take 2 steps! 1. 3r2 – 18r + 27 2. 2a2 + 8a - 8
Answers: 1. 3r2 – 30r + 27 3(r2 - 10r + 9) 3(r – 9) (r – 1)
2. 2a2 + 8a – 8 2(a2 + 4a – 4)
Solving Equations by Factoring Completely
Do Now: Factor completely Solve for x
Steps to Solve Equations by Factoring Completely
set each factor = 0 and solve for the unknown. x3 + 12x2 = Factor GCF x2 (x + 12) = 0
Steps to Solve Equations by Factoring Completely
set each factor = 0 and solve for the unknown. x3 + 12x2 = Factor GCF x2 (x + 12) = (set each factor = 0, & solve) x2 = x + 12 = 0 x= x = -12 You now have 2 answers, x = 0 and x = -12
Factor completely:
Factor completely: X =
Solving higher degree functions
Quadratic form: ax4 + bx2 + c = 0 Example: x4 +2x2 -24 = 0 Factor: (x2 )(x2 )=0
Solving higher degree functions
Quadratic form: ax4 + bx2 + c = 0 Example: x4 +2x2 -24 = 0 Factor: (x2 +6 )(x2 – 4 ) = 0 x2 +6= x2 – 4 =0 x2 = x2 = 4 x = 2, -2
Try this one: X4 – 13x2 +36 = 0
Factor first: X4 – 13x2 +36 = 0 (x2 – 9)(x2 – 4)=0
Solutions: X4 – 13x2 +36 = 0 (x2 – 9)(x2 – 4) X2-9=0 x2-4=0
This one can be verified on the calculator. X = 2,-2,3,-3
Hand this one in Quadratic form:
Ans: X=
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# Math Expressions Grade 3 Student Activity Book Unit 2 Lesson 3 Answer Key Multiply and Divide with 8
This handy Math Expressions Grade 3 Student Activity Book Answer Key Unit 2 Lesson 3 Multiply and Divide with 8 provides detailed solutions for the textbook questions.
## Math Expressions Grade 3 Student Activity Book Unit 2 Lesson 3 Multiply and Divide with 8 Answer Key
Explore Patterns with 8s
What patterns do you see below?
Fast-Array Drawings
Find the unknown number for each Fast-Array drawing.
Question 1.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 6 x unknown number = 42
Let the unknown number be x.
6x = 42
x = 42 ÷ 6
x = 7
So, 6 x 7 = 42
Question 2.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 6 x 8 = x
Let the unknown number be x.
So, 6 x 8 = 46
Question 3.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 8 x unknown number = 64
Let the unknown number be x.
8x = 64
x = 64 ÷ 8
x = 8
So, 8 x 8 = 64
Question 4.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 9 x unknown number = 63
Let the unknown number be x.
9x = 63
x = 63 ÷ 9
x = 7
So, 7 x 9 = 63
Question 5.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 4 x 6 = x
Let the unknown number be x.
So, 4 x 6 = 24
Question 6.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 5 x unknown number = 20
Let the unknown number be x.
5x = 20
x = 20 ÷ 5
x = 4
So, 5 x 4 = 20
Question 7.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 9 x unknown number = 45
Let the unknown number be x.
9x = 45
x = 45 ÷ 9
x = 5
So, 9 x 5 = 45
Question 8.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 6 x 6 = x
Let the unknown number be x.
So, 6 x 6 = 36
Question 9.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 7 x unknown number = 56
Let the unknown number be x.
7x = 56
x = 56 ÷ 7
x = 8
So, 8 x 7 = 56
Question 10.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 7 x 7 = x
Let the unknown number be x.
So, 7 x 7 = 49
Question 11.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 8 x unknown number = 40
Let the unknown number be x.
8x = 40
x = 40 ÷ 8
x = 5
So, 8 x 5 = 40
Question 12.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 8 x unknown number = 24
Let the unknown number be x.
8x = 24
x = 24 ÷ 8
x = 3
So, 8 x 3 = 24
Question 13.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 8 x 9 = x
Let the unknown number be x.
So, 8 x 9 = 72
Question 14.
Explanation:
We use arrays to model the multiplication and find the factors or products.
An array is a set of objects that are arranged in rows and columns.
Given, 10 x unknown number = 100
Let the unknown number be x.
10x = 100
x = 100 ÷ 10
x = 10
So, 10 x 10 = 100
Question 15.
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# How to Find Elapsed Time?
You must add or subtract time to find the elapsed time. In this step-by-step guide, you will learn how to calculate the elapsed time.
We can define time as the period during which a particular event occurs, has occurred, or is about to occur. It is a measurable quantity and also infinite.
## A step-by-step guide tofinding elapsed time
Adding time in mathematics is a process that is done by adding distinct units of time such as hours, minutes, and seconds and then integrating them. The normal addition has a unit broken into $$1, 10, 100$$, etc subunits, and in time the unit of time is broken into $$60$$ subunits. This distinction sometimes makes it a little difficult to add time. The following table shows the unit values of the time.
### Subtracting time
Subtraction of time is approximately equal to addition time. To do this, we follow some simple steps. First, subtract the hours, and then subtract the minutes. In addition, write the resultant answer of hours and minutes.
For example, subtract $$6:45$$ from $$8:55$$. First, let’s reduce the hours by $$8-6 = 2$$, then we can subtract the minutes by $$55- 45 = 10$$. The result can be written together as $$2:10$$.
If we know the conversion time, it is very easy to add time in hours and minutes. First, add all the hours of the two times. Now add the minutes of the two given times. Write the time obtained in hours: minutes format.
For example, let’s add a set of two times, $$9:10$$ and $$1:15$$. The sum of $$9$$ and $$1$$ o’clock is equal to $$10$$. Adding $$10$$ and $$15$$ minutes equals $$25$$. Then we write the calculated hour and minute together at $$10:25$$.
Notes:
• Add all hours and minutes to calculate the total time. Similarly, subtract all the hours and minutes to calculate the time.
• When adding minutes, if it exceeds $$60$$, delete $$60$$ and count the remaining minutes as a total. Deleted $$60$$ minutes are converted to an hour and added to the hour.
• When subtracting minutes, if minutes are negative, then add $$60$$ minutes with them and consider their total as minutes. The added $$60$$ minutes, which in turn is considered one hour, is removed from the total hour.
### Finding Elapsed Time – Example 1:
John plans to go to a park. He takes a taxi at $$3.45\space pm$$. It takes about $$1$$ hour and $$5$$ minutes to reach the destination. What time does John get to the park?
Solution:
The initial time is $$3.45\space pm$$. And the travel time is $$1$$ hour and $$5$$ minutes. By adding the hours we have, $$3 + 1 = 4$$ hours, and by adding the minutes, we will have $$45 + 5 = 50$$ minutes. Therefore John reaches the park at $$4:50\space pm$$.
## Exercises forFinding Elapsed Time
1. Anna arrived at school at $$7:59\space a.m.$$ and left school at $$2:33\space p.m$$. How long was Anna at school?
2. Jaydon started his homework at $$6:30\space pm$$ and finished it at $$8\space pm$$. How many hours and minutes did Jaydon spend completing his homework?
3. Brandon got to work at $$8:10\space a.m$$. and left at $$3:45\space p.m$$. How long did Brandon work?
4. Caleb ran a marathon in $$2$$ hours and $$17$$ minutes. He crossed the finish line at $$10:33$$ in the morning, what time did the race start?
1. $$\color{blue}{6\:hours\:and\:34\:minutes}$$
2. $$\color{blue}{1\:hour\:and\:30\:minutes}$$
3. $$\color{blue}{7\:hours\:and\:35\:minutes}$$
4. $$\color{blue}{8:16\:a.m}$$
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Section 1: Number Sets
Generally, Algebraic studies use a set of numbers called the real numbers. However, the real numbers are not the only set of numbers that exist. The other number sets that we will consider are actually proper subsets of the real numbers. By "proper subset" we mean that all of the numbers in the other sets are also in the set of real numbers, but not all of the real numbers are in the other sets. We'll start with a fairly simple set of numbers.
Definition 1: The natural numbers are the numbers we typically think of for counting: 1, 2, 3, 4, 5, 6, and so on.
Then, somewhere along in history, a bright mathematician decided we needed a number to represent nothing. Thus, zero was added to the natural numbers to give us another set.
Definition 2: The whole numbers are the natural numbers with the addition of zero: 0, 1, 2, 3, 4, 5, 6, and so on.
If you've ever tried to subtract 7 from 3 (i.e. 3 -7) you know that you cannot get an answer when you are restricted to the Whole numbers. To do this kind of arithmetic you need negative numbers. When these are added to the Whole numbers, we get a new set of numbers.
Definition 3: The integers are the natural numbers with zero and with the negatives (or additive inverses) of all the natural numbers: ... -4, -3, -2, -1, 0, 1, 2, 3, 4,...
Notice the hierarchy from "smaller" to "larger" sets. All of the Natural numbers are also Whole numbers. Similarly, all of the Whole numbers are also Integers.
While these number sets are sufficient for many, many arithmetic problems you can run into problems when dividing two integers (i.e. a fraction). If the result of that division happens to be an integer, then everything is fine. However, if the result of the division is NOT an integer, then another set of numbers is needed to get an answer.
Definition 4: The rational numbers are numbers that can be labled by fraction notation where both the numerator and the denominator are integers.
The same hierarchy holds, all of the natural numbers, whole numbers, and integers are also rational numbers. However, 1/2 is rational number that does not belong to any of the other sets.
By now, one would think, we must have finally identified all numbers. This is not so. There are numbers out there that cannot be labeled by a fraction of integers. Think about squares and square roots. If I take 2 and multiply it by itself, or square it, then I get 4. Of course, that means that the square root of 4 is 2. Those are all rational numbers (in fact they're natural numbers). But what is the number that you would multiply by itself to get 2? That would be √ 2. There is no way to write this number as a fraction of two integers. This is an irrational number. Other examples of irrational numbers are Π and the cube root of 5.
This brings us to the real numbers.
Definition 5: The real numbers are all of the rational numbers combined with all of the irrational numbers.
With increasing use of calculators, many people don't think of number sets in terms of fraction notation. They prefer to think about decimal notation. All Rational numbers (remember that this includes the Natural numbers, Whole Numbers, and Integers) will either terminate or repeat in decimal form. To terminate means that there are a finite number of digits after the decimal point. Repeating means that some pattern of digits after the decimal will repeat forever.
Example 1 Terminating 18= .125
Example 2 Repeating 6799 = .676767...
The decimal representation for Irrational Numbers will neither repeat nor terminate. It will go on forever with no stopping and no repeating pattern. As was pointed out before, there is a hierarchy in these numbers sets. Every number set discussed is part of the Real numbers. The Real numbers are divided into the Rational and Irrational numbers, and the Rational numbers are further broken down into the Integers, Whole numbers, and Natural numbers.
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# 10.2: Solving Quadratic Equations by Graphing
Difficulty Level: At Grade Created by: CK-12
Isaac Newton’s theory for projectile motion is represented by the equation:
h(t)=12(g)t2+v0t+h0\begin{align*}h(t)=- \frac{1}{2} (g) t^2+v_0 t+h_0\end{align*}
• \begin{align*}t=\end{align*}time (usually in seconds)
• \begin{align*}g=\end{align*}gravity due to acceleration; either 9.8 m/s or 32 ft/s
• \begin{align*}v_0=\end{align*}initial velocity
• \begin{align*}h_0=\end{align*}initial height of object
Consider the following situation: “A quarterback throws a football at an initial height of 5.5 feet with an initial velocity of 35 feet per second.”
By substituting the appropriate information:
• \begin{align*}g=32\end{align*} because the information is given in feet
• \begin{align*}v_0=35\end{align*}
• \begin{align*}h_0=5.5\end{align*}
The equation becomes \begin{align*}h(t)=-\frac{1}{2} (32) t^2+35t+5.5 \rightarrow h(t)=-16t^2+35t+5.5\end{align*}.
Using the concepts from the previous lesson, we know
• The value of \begin{align*}a\end{align*} is negative, so the parabola opens downward.
• The vertex is a maximum point.
• The \begin{align*}y-\end{align*}intercept is (0, 5.5).
• The graph is narrow about its line of symmetry.
At what time will the football be 6 feet high? This equation can be solved by graphing the quadratic equation.
## Solving a Quadratic Using a Calculator
Chapter 7 focused on how to solve systems by graphing. You can think of this situation as a system: \begin{align*}\begin{cases} y=-16t^2+35t+5.5\\ y=6 \end{cases}\end{align*}. You are looking for the appropriate \begin{align*}x-\end{align*}coordinates that give a \begin{align*}y-\end{align*}coordinate of 6 feet. Therefore, you are looking for the intersection of the two equations.
Begin by typing the equations into the \begin{align*}[Y=]\end{align*} menu of your calculator. Adjust the window until you see the vertex, \begin{align*}y-\end{align*}intercept, \begin{align*}x-\end{align*}intercepts, and the horizontal line of 6 units.
By looking at the graph, you can see there are two points of intersection. Using the methods from chapter 7, find both points of intersection.
\begin{align*}(0.014,6) \ and \ (2.172,6)\end{align*}
At 0.014 seconds and again at 2.17 seconds, the football is six feet from the ground.
## Using a Calculator to Find the Vertex
You can also use a graphing calculator to determine the vertex of the parabola. The vertex of this equation is a maximum point, so in the [CALCULATE] menu of the graphing option, look for [MAXIMUM].
Choose option #4. The calculator will ask you, “LEFT BOUND?” Move the cursor to the left of the vertex and hit [ENTER].
The calculator will ask, “RIGHT BOUND?” Move the cursor to the right of the vertex and hit [ENTER].
Hit [ENTER] to guess.
The maximum point on this parabola is (1.09, 24.64).
Example 1: Will the football reach 25 feet high?
Solution: The vertex represents the maximum point of this quadratic equation. Since its height is 24.64 feet, we can safely say the football will not reach 25 feet.
Example 2: When will the football hit the ground, assuming no one will catch it?
Solution: We know want to know at what time the height is zero. \begin{align*}\begin{cases} y=-16t^2+35t+5.5 \\ y=0 \end{cases}\end{align*}. By repeating the process above and finding the intersection of the two lines, the solution is (2.33, 0). At 2.33 seconds, the ball will hit the ground.
The point at which the ball reaches the ground \begin{align*}(y=0)\end{align*} represents the \begin{align*}x-\end{align*}intercept of the graph.
The \begin{align*}x-\end{align*}intercept of a quadratic equation is also called a root, solution, or zero.
Example: Determine the number of solutions to \begin{align*}y=x^2+4\end{align*}.
Solution: Graph this quadratic equation, either by hand or with a graphing calculator. Adjust the calculator’s window to see both halves of the parabola, the vertex, the \begin{align*}x-\end{align*}axis, and the \begin{align*}y-\end{align*}intercept.
The solutions to a quadratic equation are also known as its \begin{align*}x-\end{align*}intercepts. This parabola does not cross the \begin{align*}x-\end{align*}axis. Therefore, this quadratic equation has no real solutions.
Example: Andrew has 100 feet of fence to enclose a rectangular tomato patch. He wants to find the dimensions of the rectangle that encloses the most area.
Solution: The perimeter of a rectangle is the sum of all four sides. Let \begin{align*}w=\end{align*} width and \begin{align*}l=\end{align*} length. The perimeter of the tomato patch is \begin{align*}100=l+l+w+w \rightarrow 100=2l+2w\end{align*}.
The area of a rectangle is found by the formula \begin{align*}A=l(w)\end{align*}. We are looking for the intersection between the area and perimeter of the rectangular tomato patch. This is a system.
\begin{align*}\begin{cases} 100 = 2l+2w \\ A = l(w) \end{cases}\end{align*}
Before we can graph this system, we need to rewrite the first equation for either \begin{align*}l\end{align*} or \begin{align*}w\end{align*}. We will then use the Substitution Property.
\begin{align*}100 &= 2l+2w \rightarrow 100-2l=2w\\ \frac{100-2l}{2} &= w \rightarrow 50-l=w\end{align*}
Use the Substitution Property to replace the variable \begin{align*}w\end{align*} in the second equation with the express \begin{align*}50-l\end{align*}.
\begin{align*}A=l(50-l)=50l-l^2\end{align*}
Graph this equation to visualize it.
The parabola opens downward so the vertex is a maximum. The maximum value is (25, 625). The length of the tomato patch should be 25 feet long to achieve a maximum area of 625 square feet.
## 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 are the alternate names for the solution to a parabola?
2. Define the following variables in the function \begin{align*}h(t)=-\frac{1}{2} (g) t^2+v_0 t+h_0\end{align*}.
1. \begin{align*}h_0\end{align*}
2. \begin{align*}t\end{align*}
3. \begin{align*}v_0\end{align*}
4. \begin{align*}g\end{align*}
5. \begin{align*}h(t)\end{align*}
3. A rocket is launched from a height of 3 meters with an initial velocity of 15 meters per second.
1. Model the situation with a quadratic equation.
2. What is the maximum height of the rocket? When will this occur?
3. What is the height of the rocket after four seconds? What does this mean?
4. When will the rocket hit the ground?
5. At what time will the rocket be 13 meters from the ground?
How many solutions does the quadratic equation have?
1. \begin{align*}-x^2+3=0\end{align*}
2. \begin{align*}2x^2+5x-7=0\end{align*}
3. \begin{align*}-x^2+x-3=0\end{align*}
Find the zeros of the quadratic equations below. If necessary, round your answers to the nearest hundredth.
1. \begin{align*}y=-x^2+4x-4\end{align*}
2. \begin{align*}y=3x^2-5x\end{align*}
3. \begin{align*}x^2+3x+6=0\end{align*}
4. \begin{align*}-2x^2+x+4=0\end{align*}
5. \begin{align*}x^2-9=0\end{align*}
6. \begin{align*}x^2+6x+9=0\end{align*}
7. \begin{align*}10x^2-3x^2=0\end{align*}
8. \begin{align*}\frac{1}{2}x^2-2x+3=0\end{align*}
9. \begin{align*}y=-3x^2+4x-1\end{align*}
10. \begin{align*}y=9-4x^2\end{align*}
11. \begin{align*}y=x^2+7x+2\end{align*}
12. \begin{align*}y=-x^2-10x-25\end{align*}
13. \begin{align*}y=2x^2-3x\end{align*}
14. \begin{align*}y=x^2-2x+5\end{align*}
15. Andrew is an avid archer. He launches an arrow that takes a parabolic path, modeled by the equation \begin{align*}y=-4.9t^2+48t\end{align*}. Find how long it takes the arrow to come back to the ground.
For questions 24 – 26,
(a) Find the roots of the quadratic polynomial.
(b) Find the vertex of the quadratic polynomial.
1. \begin{align*}y=x^2+12x+5\end{align*}
2. \begin{align*}y=x^2+3x+6\end{align*}
3. \begin{align*}y=-x^2-3x+9\end{align*}
4. Sharon needs to create a fence for her new puppy. She purchased 40 feet of fencing to enclose three sides of a fence. What dimensions will produce the greatest area for her puppy to play?
5. An object is dropped from the top of a 100-foot-tall building.
1. Write an equation to model this situation.
2. What is the height of the object after 1 second?
3. What is the maximum height of the object?
4. At what time will the object be 50 feet from the ground?
5. When will the object hit the ground?
Mixed Review
1. Factor \begin{align*}3r^2-4r+1\end{align*}.
2. Simplify \begin{align*}(2+\sqrt{3})(4+\sqrt{3})\end{align*}.
3. Write the equation in slope-intercept form and identify the slope and \begin{align*}y-\end{align*}intercept: \begin{align*}9-3x+18y=0\end{align*}.
4. The half life of a particular substance is 16 days. An organism has 100% of the substance on day zero. What is the percentage remaining after 44 days?
5. Multiply and write your answer in scientific notation: \begin{align*}0.00000009865 \times 123564.21\end{align*}
6. A mixture of 12% chlorine is mixed with a second mixture containing 30% chlorine. How much of the 12% mixture is needed to mix with 80 mL to make a final solution of 150 mL with a 20% chlorine concentration?
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# Determine the A.P whose 3rd term is 5 and the 7th term is 9
Math
## Question
Determine the A.P whose 3rd term is 5 and the 7th term is 9
• ### 1. User Answers neelrambhia03
a+2d = 5
a + 6d = 9
-4r = -4
r = 4
a+8 = 5
a = -3
AP: -3,1,5,9.......
• ### 2. User Answers BrainlyConqueror0901
$$\huge{\pink{\boxed{\green{\sf{A.P=3,4,5,6,7........}}}}}$$
Step-by-step explanation:
$$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$$
$$\: {\orange{given}} \\ { \pink{ \boxed{ \green{a3 = 5}}}} \\ { \pink{ \boxed{ \green{a7 = 9}}}} \\ \\ { \blue{to \: find}}\\ { \purple{ \boxed{ \red{ap = }}}}$$
According to given question:
$$\to a3 = 5 \\ \to a + 2d = 5 - - - - - (1) \\ \to a7 = 9 \\ \to a + 6d = 9 - - - - - (2)\\ \\ subtracting \: (1) \: from \: (2) \\ \to a + 6d - (a + 2d) = 9 - 5 \\ \to a + 6d - a - 2d = 4 \\ \to 4d = 4 \\ \to d = 1 \\ \\ putting \: value \: of \: d \: in(1) \\ \to a + 2d = 5 \\ \to a + 2 \times 1 = 5 \\ \to \: a = 5 - 2 \\ \to a = 3 \\ \\ \to hence \: the \: required \\{ \pink{ \boxed{ \green{ \therefore A.P = 3,4,5,6,7.......}}}}$$
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# Constant Derivatives and the Power Rule
## Derivative of a constant is zero and \frac {d}{dx}[x^n] = nx^{n-1} .
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Practice Constant Derivatives and the Power Rule
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Constant Derivatives and the Power Rule
The power rule is a fantastic "shortcut" for finding the derivatives of basic polynomials. Between the power rule and the basic definition of the derivative of a constant, a great number of polynomial derivatives can be identified with little effort - often in your head!
Embedded Video:
### Guidance
In this lesson, we will develop formulas and theorems that will calculate derivatives in more efficient and quick ways. Look for these theorems in boxes throughout the lesson.
The Derivative of a Constant
Theorem: If f(x)=c\begin{align*}f(x) = c\end{align*} where c is a constant, then f(x)=0\begin{align*}f^{\prime}(x) = 0\end{align*}
Proof: f(x)=limh0f(x+h)f(x)h=limh0cch=0\begin{align*}f'(x)= \lim_{h \to 0}\frac {f(x+h)-f(x)}{h}=\lim_{h \to 0}\frac{c-c}{h} = 0\end{align*}
Theorem: If c\begin{align*}c\end{align*} is a constant and f\begin{align*}f\end{align*} is differentiable at all x\begin{align*}x\end{align*}, then ddx[cf(x)]=cddx[f(x)]\begin{align*}\frac {d}{dx}[cf(x)] = c\frac {d}{dx}[f(x)]\end{align*}. In simpler notation (cf)=c(f)=cf\begin{align*}(cf)^{\prime} = c(f)^{\prime} = cf^{\prime}\end{align*}
The Power Rule
Theorem: (The Power Rule) If n is a positive integer, then for all real values of x
ddx[xn]=nxn1\begin{align*}\frac {d}{dx}[x^n] = nx^{n-1}\end{align*}.
#### Example A
Find f(x)\begin{align*}f^{\prime} (x)\end{align*} for f(x)=16\begin{align*}f(x)=16\end{align*}
Solution
If f(x)=16\begin{align*}f(x) = 16\end{align*} for all x\begin{align*}x\end{align*}, then f(x)=0\begin{align*}f^{\prime} (x) = 0\end{align*} for all x\begin{align*}x\end{align*}
We can also write ddx16=0\begin{align*}\frac{d}{dx}16 = 0\end{align*}
#### Example B
Find the derivative of f(x)=4x3\begin{align*}f(x)=4x^3\end{align*}
Solution
\begin{align*}\frac {d}{dx}\left [{4x^3} \right]\end{align*} ..... Restate the function
\begin{align*}4 \frac{d}{dx}\left [{x^3} \right]\end{align*} ..... Apply the Commutative Law
\begin{align*}4 \left [{3x^2} \right]\end{align*} ..... Apply the Power Rule
\begin{align*}12x^2\end{align*} ..... Simplify
#### Example C
Find the derivative of \begin{align*}f(x)=\frac{-2}{x^{4}}\end{align*}
Solution
\begin{align*}\frac {d}{dx} \left [\frac{-2}{x^4} \right]\end{align*} ..... Restate
\begin{align*}\frac {d}{dx}\left [{-2x^{-4}} \right]\end{align*} ..... Rules of exponents
\begin{align*}-2 \frac {d}{dx}\left [{x^{-4}} \right]\end{align*} ..... By the Commutative law
\begin{align*}-2 \left [{-4x^{-4-1}} \right]\end{align*} ..... Apply the Power Rule
\begin{align*}-2 \left [{-4x^{-5}} \right]\end{align*} ..... Simplify
\begin{align*}8x^{-5}\end{align*} ..... Simplify again
\begin{align*}\frac {8}{x^5}\end{align*} ..... Use rules of exponents
### Vocabulary
A theorem is a statement accepted to be true based on a series of reasoned statements already accepted to be true. In the context of this lesson, a theorem is a rule that allows a quick calculation of the derivative of functions of different types.
A proof is a series of true statements leading to the acceptance of truth of a more complex statement.
### Guided Practice
Questions
Find the derivatives of:
1) \begin{align*}f(x)=x^{3}\end{align*}
2) \begin{align*}f(x)=x\end{align*}
3) \begin{align*}f(x)=\sqrt{x}\end{align*}
4) \begin{align*}f(x)=\frac{1}{x^{3}}\end{align*}
Solutions
1) By the power rule:
If \begin{align*}f(x) = x^3\end{align*} then \begin{align*}f(x) = (3)x^{3-1} = 3x^2\end{align*}
2) Special application of the power rule:
\begin{align*}\frac {d}{dx}[x] = 1x^{1-1} = x^0 = 1\end{align*}
3) Restate the function: \begin{align*}\frac {d}{dx}[\sqrt{x}]\end{align*}
Using rules of exponents (from Algebra): \begin{align*}\frac {d}{dx}[x^{1/2}]\end{align*}
Apply the Power Rule: \begin{align*}\frac {1}{2}x^{1/2-1}\end{align*}
Simplify: \begin{align*}\frac {1}{2}x^{-1/2}\end{align*}
Rules of exponents: \begin{align*}\frac{1}{2x^{1/2}}\end{align*}
Simplify: \begin{align*}\frac {1}{2\sqrt{x}}\end{align*}
4) Restate the function: \begin{align*}\frac {d}{dx}\left [ \frac{1}{x^3} \right ]\end{align*}
Rules of exponents: \begin{align*}\frac {d}{dx}\left [{x^{-3}} \right ]\end{align*}
Power Rule: \begin{align*}-3x^{-3-1}\end{align*}
Simplify: \begin{align*}-3x^{-4}\end{align*}
Rules of exponents: \begin{align*}\frac {-3}{x^4}\end{align*}
### Practice
1. State the Power Rule.
Find the derivative:
1. \begin{align*}y = 5x^7\end{align*}
2. \begin{align*}y = -3x\end{align*}
3. \begin{align*}f(x) = \frac{1} {3} x + \frac{4} {3}\end{align*}
4. \begin{align*}y = x^4 - 2x^3 - 5\sqrt{x} + 10\end{align*}
5. \begin{align*}y = (5x^2 - 3)^2\end{align*}
6. given \begin{align*}y(x)= x^{-4\pi^2}\end{align*} when \begin{align*} x = 1\end{align*}
7. \begin{align*}y(x) = 5\end{align*}
8. given \begin{align*}u(x)= x^{-5\pi^3}\end{align*} what is \begin{align*} u'(2)\end{align*}
9. \begin{align*} y = \frac{1}{5}\end{align*} when \begin{align*} x = 4 \end{align*}
10. given \begin{align*}d(x)= x^{-0.37}\end{align*} what is \begin{align*} d'(1)\end{align*}
11. \begin{align*} g(x) = x^{-3}\end{align*}
12. \begin{align*}u(x) = x^{0.096}\end{align*}
13. \begin{align*}k(x) = x-0.49\end{align*}
14. \begin{align*} y = x^{-5\pi^3}\end{align*}
### Vocabulary Language: English
derivative
derivative
The derivative of a function is the slope of the line tangent to the function at a given point on the graph. Notations for derivative include $f'(x)$, $\frac{dy}{dx}$, $y'$, $\frac{df}{dx}$ and \frac{df(x)}{dx}.
proof
proof
A proof is a series of true statements leading to the acceptance of truth of a more complex statement.
theorem
theorem
A theorem is a statement that can be proven true using postulates, definitions, and other theorems that have already been proven.
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# A Math Bingo lesson plan on Multiplication
Subject:
Math
2, 3, 4, 5
Jaunita Cook
Lesson Plan Subject: Math
Lesson Plan Title: Multiplication Bingo
The lesson plan:
————————————————————
TITLE: Multiplication Bingo
OVERVIEW: After students have mastered multiplication facts, this game can be introduced as a way to reinforce learning.
PURPOSE: A fun way to practice the multiplication facts.
OBJECTIVES:
2. Use prior learning of multiplication facts to score a bingo
3. Have fun with Math!
MATERIALS:
Student materials–2 pieces of construction paper per student, pencil or marker, and scissors.
Teacher materials–1 piece of notebook paper and pencil.
PROCEDURE: (Setting up game-board)
1. Each child needs two pieces of construction paper.
2. Instruct the students to take one piece of paper and:
a. Fold in half from top to bottom.
b. Fold again from top to bottom.
c. Fold in half from side to side.
d. Fold again from side to side.
3. When the students open the paper, there should be 16 squares.
4. As you call out the products of sixteen multiplication facts, the students write those products in a different square. Keep a list for yourself of the facts that you are using that day.
5. Students should fold the second piece of paper exactly as they did the other. Using scissors, cut out the squares so that there are sixteen pieces of paper to use on their game-board.
PLAYING THE GAME:
1. Decide which kind of bingo you want to play. Some of the games we play are: (a) horizontal, (b) vertical, (c) diagonal, (d) postage stamp, (e) “L” (four on the left and four on the bottom), (f) bulls eye (four in the center), (g) picture frame (all but the four in the center), (h) “X” (two diagonals). Your students will come up with other ideas.
2. Using the list of sixteen multiplication facts, call out the factors only. For example, you say “2×5”. The students must know the product, find it on their game-board, and cover it with a piece of paper. Students are not allowed to tell other students what the product is.
3. Continue calling out facts until someone gets a bingo. Be sure to mark on your master copy the facts you called, so you can check your winner to see if he/she covered the correct products.
4. Keep a record of who wins the most games. That student could be the first in line for the day, have extra free time, be excused from the day’s homework, etc.
TYING IT ALL TOGETHER:
This is a fun way for students to use their knowledge of the multiplication facts! :-)
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## Exercises - Graphing Trigonometric Functions
1. Graph the following for $-2\pi \le x \le 2\pi$.
1. $y = 4\cos x$
2. $y = \sin \frac{2}{3} x$
3. $y = 4\cos(2x - \frac{3\pi}{2})$
4. $y = \sin(x - \frac{\pi}{6})$
5. $y= -\frac{1}{2} \sin x$
2. Graph the following.
1. $y = -\frac{8}{5} \cos (\frac{x}{5} + \frac{\pi}{3})$ over $[-5\pi,10\pi]$
2. $y = 4\sin(2x - \frac{\pi}{6})$ over $[-\pi,2\pi]$
3. $y = \frac{5}{2} \cos (2x + \frac{\pi}{4})$ from $-\pi$ to $\pi$
4. $y = \cos(x + \frac{\pi}{4})$ from $-2\pi$ to $2\pi$
3. Graph the following.
1. $y = 1 + \cos x$ for $-2\pi \le x \le 2\pi$
2. $y = 2 - \sin x$ from $-\pi$ to $\frac{3\pi}{2}$
3. $y = 2 + 2\sin(\frac{x}{3} - \frac{\pi}{6})$ from $-\pi$ to $2\pi$
4. $y = 2 - 3\cos 2x$ over $[-2\pi,\pi]$ (omit finding the $x$-intercepts)
4. Graph the following. Label interecepts and other important features (e.g., asymptotes)
1. $y = -\tan x$ over $[-2\pi,2\pi]$
2. $y = -\sec x$ from $-\pi$ to $\pi$
3. $y = \frac{1}{2} \tan 2x$ from $-\frac{5\pi}{4}$ to $\frac{3\pi}{8}$
4. $y = \csc 3x$ for $-\frac{\pi}{2} \le x \le \frac{5\pi}{6}$
5. $y = 2\tan \frac{x}{2}$ from $-3\pi$ to $\frac{5\pi}{2}$
6. $y = -\csc(4x+\pi)$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$
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# Equation of a Line
## Definition
A linear equation in two variables, often written as $$y = mx + b$$, describes a line in the $$xy$$-plane. The line consists of all the solution pairs $$(x_0, y_0)$$ that make the equation true.
For example, in the equation, $y = 2x -1$, the line would pass through the points $(0,-1)$ and $(2, 3)$ because these ordered pairs are solutions to the equation, as is every other point on the line.
\begin{aligned} (-1) = 2(0) -1 \\ (3) = 2(2) -1 \end{aligned}
A line through the points (0,-1) and (2,3)
The coefficient $m$ is the slope of the graph, and the constant $b$ is the $y$-intercept.
The slope of a graph is the ratio between the rate of change of $y$ and the rate of change of $x$, so for two points on a graph $p_1=(x_1,y_1)$ and $p_2=(x_2,y_2)$, slope between them is:
$m = \frac{y_2 - y_1}{x_2-x_1}$
## Technique
### What is the $y$-intercept of the line described by $2x - 3y + 9 = 0$?
If $2x - 3y + 9 = 0$, then $3y = 2x + 9$, and so $y = \frac{2}{3}x + \frac{9}{3}$, which means the $y$-intercept is $3$. $_\square$
### What is $a$ in the point $(-10, a)$, if it falls on the line through the points $(2,4)$ and $(3,3)$?
The slope of the line is $m = \frac{(3)-(4)}{(3)-(2)}=-1$, so $y=-x + b$ for all $x$ and $y$ on the graph. This means we can take one of the given points and substitute in its values to find $b$.
$(3)=-(3) + b$
Therefore, $b = 6$, and the equation of the line is $y = -x +6$. This means the point $(-10, a)$ satisfies the equation $a = -(-10) +6$. Thus, $a = 16$. $_\square$
## Applications and Extensions
### The intersection of the line $y = 10x + 5$ with the line through the points $(1, 9)$, $(2, 20)$, is a point $( a , b )$. Find the sum of $a$ and $b$.
First, we need to find the equation of the line through $(1, 9)$ and $(2, 20)$. Since $\frac{y_2 - y_1}{x_2-x_1} = \frac{20-9}{2-1}$, the slope of the line must be 11. Further, since $9 = 11(1) + b$, the $y$-intercept is $-2$.
Thus, our two equations are: \begin{aligned} \mbox{Line 1: } y &= 10x_1 + 5 \\ \mbox{Line 2: } y &= 11x_2 -2 \end{aligned}
To find the point of intersection $(a,b)$, we solve the system of equations above by equating the two $y$'s.
$10x + 5 = 11 x -2$
Therefore, $x = 7$, and placing $7$ into either of the equations for $x$ gives us $(7, 75 )$ as the point of intersection. The answer, then, is $7+75 = 82$. $_\square$
Note by Arron Kau
7 years, 3 months ago
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I keep wondering at times, why is slope $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ and not $m = \frac{x_{2} - x_{1}}{y_{2} - y_{1}}$ ? Both of them give same kind of information about the slope, one indicates the change in y per change in x and the other indicates the change in x per change in y. Is it just convention or is there some other reason to it?
- 7 years, 3 months ago
well since $m=\tan x$, its very obvious that $m=\tan x=\frac{y_2-y_1}{x_2-x_1}$. Stay in this pattern is also very good because in higher grades we may need to solve like "What is the equation of the line which is perpendicular with line $5x+2=0$ and pass through the point $(7,0)$?". Keep it $m=\tan x$ rather than $\cot x$ is more simple.
Or another way to explain is as a function $f(x)$, $x$ is the one who "change" and $y$ is the one who "change because $x$ changed". Then as $m$ explains the different rate of change of $y$ against $x$, then $m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$.
- 7 years, 3 months ago
Got it. Thanks.
- 7 years, 2 months ago
What does line1 or line2 represent in three dimensional space?
- 7 years, 2 months ago
They represent two lines, since you can write them as: $0z+y=10x_1+5 \\ 0z+y=11x_2-2.$
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# How do you solve 6x - 2y = -16 and 4x + 4y = -32 ?
Aug 10, 2016
$x = - 4$
$y = - 4$-
#### Explanation:
$6 x - 2 y = - 16$ and
$4 x + 4 y = - 32$
or
$2 x + 2 y = - 16$
or
$x + y = - 8$
Adding up the equations $6 x - 2 y = - 16$ and $2 x + 2 y = - 16$
We get
$6 x - 2 y + 2 x + 2 y = - 16 - 16$
or
$8 x = - 32$
or
$x = - \frac{32}{8}$
or
$x = - 4$--------------------Ans $1$
By putting the value $x = - 4$ in the equation $x + y = - 8$
We get
$- 4 + y = - 8$
or
$y = 4 - 8$
or
$y = - 4$-----------------------Ans $2$
Aug 10, 2016
$x = - 4 \mathmr{and} y = - 4$
#### Explanation:
Neither of the equations is in the simplest form. Let's do that first and see what we end up with.
$6 x - 2 y = - 16 \text{ "(div 2)" } \Rightarrow 3 x - y = - 8 \ldots \ldots \ldots . A$
$4 x + 4 y = - 32 \text{ " (div 4) " } \Rightarrow x + y = - 8 \ldots \ldots \ldots \ldots B$
We notice that $+ y \mathmr{and} - y$ are additive inverses which will add to 0 if we add equations $A \mathmr{and} B$ together.
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots .} 3 x - y = - 8 \ldots \ldots \ldots . A$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots . .} x + y = - 8 \ldots \ldots \ldots \ldots B$
$A + B \textcolor{w h i t e}{\ldots \ldots \ldots} 4 x = - 16$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots} x = - 4$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots . .} y = - 4$
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Operations with Brackets Revision | KS3 Maths Resources
## What you need to know
Things to remember:
• We do what is inside the brackets first.
Previously we have heard the term “BIDMAS”, which tells us in what order we do our operations:
Brackets
I
ndices
D
ivision
M
ultiplication
A
ddition
S
ubtraction
Here, we will focus on brackets. Because brackets are at the top of the list, this tells us that we have to do what is in our brackets first.
Find the value of the following expression
$$(11-3)\div2$$
Reading down our BIDMAS list we can see that we have to do the brackets first.
$$11-3=8$$
$$(11-3)\div2=8\div2$$
And now, we can just do the division as normal
$$8\div2 =4$$
$$(11-3)\div2=4$$
But, what happens if we have two brackets?
Find the value of the following expression
$$(12+5)-(3-9)$$
Because we have two brackets, which one do we do first?… It doesn’t matter! Let’s start with the first one.
$$(12+5)-(3-9) =17-(3-9)$$
Now we have to do the second bracket
$$17-(3-9)=17 - -6$$
And now we can do the subtractions as usual (Remember, this will become an addition)
$$17 - -6=17+6=23$$
$$(12+5)-(3-9)=23$$
Now let’s try the second bracket first.
$$(12+5)-(3-9) =(12+5) - -6$$
$$(12+5) - -6=17 - -6$$
$$17 - -6 = 17 +6 =23$$
## KS3 Maths Revision Cards
(78 Reviews) £8.99
## Example Questions
$$(7-3)\times5=4\times5 =20$$
$$(8+3)+(1-7)=11+(1-7)=11+-6=11-6=5$$
## KS3 Maths Revision Cards
(78 Reviews) £8.99
• All of the major KS2 Maths SATs topics covered
• Practice questions and answers on every topic
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# What Is The Highest Common Factor Of 50 And 40?
## What are the prime factors of 51?
51 is a composite number.
51 = 1 x 51 or 3 x 17.
Factors of 51: 1, 3, 17, 51.
Prime factorization: 51 = 3 x 17..
## What does a factor of 40 mean?
2014-01-26 / Leave a comment. 40 is a composite number. 40 = 1 x 40, 2 x 20, 4 x 10, or 5 x 8. Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40. Prime factorization: 40 = 2 x 2 x 2 x 5, which can also be written 2³ x 5.
## What is the HCF of 15 and 30?
The common factors of 15 and 30 are 1, 3, 5, and 15. The greatest common factor is 15. Example: What is the greatest common factor of 9 and 20?
## What is the highest common factor of 24 and 40?
The highest common factor (HCF) of two numbers (or expressions) is the largest number (or expression) which is a factor of both. Consider the highest common factor of 24 and 40. The common factors of 24 and 40 are 2, 4, and 8. So, the highest common factor is 8.
## How do you get 40?
To calculate the percentage, multiply this fraction by 100 and add a percent sign. 100 * numerator / denominator = percentage . In our example it’s 100 * 2/5 = 100 * 0.4 = 40 .
## What is the HCF of 5 10 and 15?
Factors of 15 = 1, 3, 5 and 15. Factors of 10 = 1, 2, 5 and 10. Therefore, common factor of 15 and 10 = 1 and 5. Highest common factor (H.C.F) of 15 and 10 = 5.
## What is the HCF of 16 and 24?
The greatest common factor of two (or more) numbers is the product of all the prime factors the numbers have in common. There are three 2’s common to both numbers, so 2*2*2 = 8 is the “greatest common factor” (GCF) of 16 and 24.
## What’s the prime factor of 40?
Answer and Explanation: The prime factorization of 40 is 2 × 2 × 2 × 5, or 23 × 5. To find the prime factorization of 40, we can use factor trees,…
## What is the HCF of 15 and 20?
5Greatest common factor (GCF) of 15 and 20 is 5. We will now calculate the prime factors of 15 and 20, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 15 and 20.
## What is the HCF of 45 and 75?
15The common factors of 45 and 75 are 15, 5, 3, 1, intersecting the two sets above. In the intersection factors of 45 ∩ factors of 75 the greatest element is 15. Therefore, the greatest common factor of 45 and 75 is 15.
## What is the HCF of 45 and 50?
Greatest common factor (GCF) of 45 and 50 is 5. We will now calculate the prime factors of 45 and 50, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 45 and 50.
## What is the HCF of 12 45 and 75?
12=2×2×3. 45=5×3×3. 75=5×5×3 .
## What is the HCF of 50?
Factors of 50 (Fifty) = 1, 2, 5, 10, 25 and 50. Factors of 70 (Seventy) = 1, 2, 5, 14, 7, 10, 35 and 70. Therefore, common factor of 50 (Fifty) and 70 (Seventy) = 1, 2, 5, 10. Highest common factor (H.C.F) of 50 (Fifty) and 70 (Seventy) = 10.
## What is the HCF of 15 and 35?
5Answer and Explanation: The greatest common factor of 35 and 15 is 5. The ‘greatest common factor’ is sometimes called the ‘GCF. ‘ The GCF of 15 and 35 is the greatest…
## What are multiples of 40?
Table of Factors and MultiplesFactorsMultiples1, 2, 4, 5, 8, 10, 20, 4040801, 4141821, 2, 3, 6, 7, 14, 21, 4242841, 43438641 more rows
## What is the prime factor of 45?
Students learn that the prime factorization of a number is the given number written as the product of its prime factors. For example, to find the prime factorization of 45, use a factor tree to find that 45 is 5 x 9, and 9 is 3 x 3. So the prime factorization of 45 is 5 x 3 x 3, or 5 x 3^2.
## What are the prime factors of 42?
The only way to write 42 as the product of primes (except to change the order of the factors) is 2 × 3 × 7. We call 2 × 3 × 7 the prime factorization of 42. It turns out that every counting number (natural number) has a unique prime factorization, different from any other counting number.
## How we can find HCF?
The highest common factor is found by multiplying all the factors which appear in both lists: So the HCF of 60 and 72 is 2 × 2 × 3 which is 12. The lowest common multiple is found by multiplying all the factors which appear in either list: So the LCM of 60 and 72 is 2 × 2 × 2 × 3 × 3 × 5 which is 360.
|
# Fundamental Concepts ICSE Class-6th Concise Maths Selina Solutions Ch-18
Fundamental Concepts ICSE Class-6th Concise Selina Mathematics Solutions Chapter-18 . We provide step by step Solutions of Exercise / lesson-18 Fundamental Concepts for ICSE Class-6 Concise Selina Mathematics. Our Solutions contain all type Questions of Exe-18 A, Exe-18 B and Revision Exercise to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-6 .
## Fundamental Concepts ICSE Class-6th Concise Selina Mathematics Solutions Chapter-18
–: Select Topics :–
Exe-18 A,
Exe-18 B,
Revision Exercise
### Exercise – 18 A of Fundamental Concepts for ICSE Class-6th Concise Selina Mathematics
#### Question -1.
Express each of the following statements in algebraic form :
(i) The sum of 8 and x is equal to y.
(ii) x decreased by 5 is equal to y.
(iii) The sum of 2 and x is greater than y.
(iv) The sum of x and y is less than 24.
(v) 15 multiplied by m gives 3n.
(vi) Product of 8 and y is equal to 3x.
(vii) 30 divided by b is equal to p.
(viii) z decreased by 3x is equal to y.
(ix) 12 times of x is equal to 5z.
(x) 12 times of x is greater than 5z.
(xi) 12 times of x is less than 5z.
(xii) 3z subtracted from 45 is equal to y.
(xiii) 8x divided by y is equal to 2z.
(xiv) 7y subtracted from 5x gives 8z.
(xv) 7y decreased by 5x gives 8z.
#### Question- 2.
For each of the following algebraic expressions, write a suitable statement in words:
(i) 3x + 8=15
(ii) 7 – y > x
(iii) 2y – x < 12
(iv) 5 ÷ z = 5
(v) a + 2b > 18
(vi) 2x – 3y= 16
(vii) 3a – 4b > 14
(viii) b + 7a < 21
(ix) (16 + 2a) – x > 25
(x) (3x + 12) – y < 3a
(i) 3x plus 8 is equal to 15
(ii) 1 decreased by y is greater than x
(iii) 2y decreased by x is less than 12
(iv) 5 divided by z is equal to 5
(v) a increased by 2b is greater than 18
(vi) 2x decreased by 3y is equal to 16
(vii) 3a decreased by 4b is greater than 14
(viii) b increased la is less than 21
(ix) The sum of 16 and 2a decreased by x is greater than 25
(x) The sum of 3x and 12 decreased by y is less than 3a.
### Selina Solutions of Fundamental Concepts for ICSE Class-6th Concise Mathematics Exercise – 18 B
#### Question -1.
Separate the constants and the variables from each of the following:
#### Question -2.
Group the like terms together :
#### Question- 3.
State whether true or false :
(i) 16 is a constant and y is a variable but 16y is variable.
(ii) 5x has two terms 5 and x.
(iii) The expression 5 + x has two terms 5 and x
(iv) The expression 2x2 + x is a trinomial.
(v) ax2 + bx + c is a trinomial.
(vi) 8 x ab is a binomial.
(vii) 8 + ab is a binomial.
(viii) x3 – 5xy + 6x + 7 is a polynomial.
(ix) x3 – 5xy + 6x + 7 is a multinomial.
(x) The coefficient of x in 5x is 5x.
(xi) The coefficient of ab in – ab is – 1.
(xii) The coefficient of y in – 3xy is – 3
(i) True
(ii) False
(iii) True
(iv) False
(v) True
(vi) False
(vii) True
(viii) True
(ix) True
(x) False
(xi) True
(xii) False
#### Question- 4.
State the number of terms in each of the following expressions :
(i) 2 terms
(ii) 2 terms
(iii) 2 terms
(iv) 2 terms
(v) 3 terms
(vi) 2 term
(vii) 2 terms
(viii) 3 terms
(ix) 3 terms
#### Question- 5.
State whether true or false:
(i) xy and – yx are like terms.
(ii) x2y and – y2x are like terms.
(iii) a and – a are like terms.
(iv) – ba and 2ab are unlike terms.
(v) 5 and 5x are like terms.
(vi) 3xy and 4xyz are unlike terms.
(i) True
(ii) False
(iii) True
(iv) False
(v) False
(vi) True
#### Question -6.
For each expression, given below, state whether it is a monomial, or a binomial or a trinomial.
(i) Monomial
(ii) Binomial
(iii) Monomial
(iv) Monomial
(v) Trinomial
(vi) Binomial
(vii) Trinomial
(viii) Binomial
(ix) Trinomial
#### Question -7.
Write down the coefficient of x in the following monomial :
#### Question- 8.
Write the coefficient of :
#### Question -9.
State the numeral coefficient of the following monomials :
(vii) – 7x ÷ y
(viii) – 3x ÷ (2y)
#### Question- 10.
Write the degree of each of the following polynomials :
We know that highest power is called degree
(i) 2
(ii) 2
(iii) 10
(iv) 20
(v) 3
### Revision Exercise of Chapter-18 Fundamental Concepts for ICSE Class-6th Concise Mathematics Selina Solutions
#### Question -1.
Express each of the following statements in algebraic form :
(i) The sum of 3x and 4y is 8.
(ii) 5x decreased by 7 gives y.
(iii) 31 added to 4x gives 6x.
(iv) 3x subtracted from 89 gives 44.
(i) 3x + 4y = 8
(ii) 5x – 7 = y
(iii) 4x + 37 = 6x
(iv) 89 – 3x = 44
#### Question -2.
Group the like terms :
#### Question- 3.
Write the number of terms in each of the following polynomials :
(i) 2 terms
(ii) 3 terms
(iii) 3 terms
(iv) 4 terms
(v) 3 terms
#### Question- 4.
For each expression, given below, state whether it is a monomial, or a binomial or a trinomial:
(i) x + y
(ii) 5x – 4y
(iii) 7x2 + 5x + 8
(iv) 64 + 3 ÷ 6
(v) 9 ÷ a x b
(vi) 8a ÷ b
(i) binomial
(ii) binomial
(iii) trinomial
(iv) 6a + 3 ÷ b = 6a + 3b
It has two terms
It is binomial
(v) 9 ÷ a x b = 9ba
It has one term
It is monomial.
(vi) monomial
#### Question -5.
Write the coefficient of x2y in :
(i) -7x2yz
(ii) 8abx2y
(iii) – x2y
(i) – 7z
(ii) 8ab
(iii) -1
#### Question- 6.
Write the coefficient of :
(i) x2 in – 8x2y
(ii) y in -4y
(iii) x in – xy2
(i) – 8y
(ii) – 4
(iii) – y2
#### Question -7.
Write the numeral coefficient in :
(i) 7
(ii) 23
(iii) –54
#### Question- 8.
Write the degree of each of the following polynomials :
(i) 8
(ii) 4
(iii) 2
(iv) 1
(v) 3
#### Question -9.
Write each statement, given below in algebraic form :
(i) 28 more than twice of x is equal to 45.
(ii) 3y reduced by 5z is greater than 8x.
(iii) 6x divided by 13y is less than 17.
(iv) 9 multiplied by 5x is equal to 2y.
#### Question -10.
State whether true or false :
(i) If 23 is a constant and x is a variable, 23 + x is constant.
(ii) If 23 is a constant and x is a variable, 23x is a variable.
(iii) If y is a variable and 57 is a constant, y – 57 is a variable.
(iv) If 3x and 2y are variable; each of 3x + 2y, 3x – 2y, 3x ÷ 2y and 3x x 2y is a variable.
(i) False
Sum of a constant and a variable is also variable.
(ii) True
Product of a constant and a variable is variable.
(iii) True
Constant subtracted from a variable is also variable.
(iv) True
Sum, difference product or quotient of two variables is also variable.
— End of Fundamental Concepts ICSE Class-6th Solutions :–
Thanks
|
Associated Topics || Dr. Math Home || Search Dr. Math
### Explaining the Divisibility Rule for 7
```Date: 02/26/2003 at 10:40:57
From: Abhay Dang
Subject: Proving divisibility rule for 7
I found in the Dr. Math FAQ the rule for divisibility of a number
by 7.
Why do these 'rules' work?
http://mathforum.org/k12/mathtips/division.tips.html
Please explain (prove) why the following rules work.
Rule 1) Divide the number into groups of 3 starting from the right and
give these groups alternating +/- signs. Apply the rule again if
necessary until you get a three-digit number. The original number is
divisible by 7 if this sum is. Eg. for 123456789 => 789 - 456 + 123
Rule 2) To know if a number is a multiple of seven or not, we can use
also 3 coefficients (1, 2, 3). We multiply the first number starting
from the ones place by 1, then the second from the right by 3, the
third by 2, the fourth by -1, the fifth by -3, the sixth by -2, the
seventh by 1, and so forth.
Example: 348967129356876.
6 + 21 + 16 - 6 - 15 - 6 + 9 + 6 + 2 - 7 - 18 - 18 + 8 + 12 + 6 = 16
means the number is not multiple of seven.
So the pattern is as follows: for a number onmlkjihgfedcba, calculate
a + 3b + 2c - d - 3e - 2f + g + 3h + 2i - j - 3k - 2l + m + 3n + 2o.
Example: 348967129356874.
Below each digit let me write its respective figure.
3 4 8 9 6 7 1 2 9 3 5 6 8 7 4
2 3 1 -2 -3 -1 2 3 1 -2 -3 -1 2 3 1
(3*2) + (4*3) + (8*1) + (9*-2) + (6*-3) + (7*-1) +
(1*2) + (2*3) + (9*1) + (3*-2) + (5*-3) + (6*-1) +
(8*2) + (7*3) + (4*1) = 14 -- a multiple of seven.
Kindly explain and prove why these rules work.
Best regards
Abhay Dang
```
```
Date: 03/04/2003 at 13:02:51
From: Doctor Peterson
Subject: Re: Proving divisibility rule for 7
Hi, Abhay.
The key here is that
1000 = 7*143 - 1
Consequently,
1000^2 = (7*143 - 1)^2 = 7^2*143^2 - 2*7*143 + 1
If you think about how higher powers work, you will see that
1000^1 = 7A - 1
1000^2 = 7B + 1
1000^3 = 7C - 1
and so on (where A, B, C, and so on are some whole numbers).
This can be better expressed in terms of modular arithmetic, if you
are familiar at all with that; 1000^n is congruent (mod 7) to -1 when
n is odd, and to +1 when n is even.
So looking at the number "abc,def,ghi", and calling the numbers "abc,"
"def," and "ghi" X, Y, and Z respectively, we can write it as
1000^2 X + 1000 Y + Z
which is equal to
(7B+1)X + (7A+1)Y + Z = 7(BX + AY) + (X - Y + Z)
Do you see that this will be divisible by 7 if and only if X-Y+Z is
divisible by 7? The same pattern continues for additional digits.
>Rule 2)
This is very similar, using the facts that
1 = 1
10 = 7*1 + 3
100 = 7*14 + 2
1000 = 7*143 - 1
Can you fill in the rest of the reasoning?
If you have any further questions, feel free to write back.
- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
```
Date: 03/05/2003 at 10:01:53
From: Abhay Dang
Subject: Thank you (Proving divisibility rule for 7)
Thanks a lot. You've been of great help !
```
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Middle School Division
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|
## Circles – Angles and Arcs
17 Feb
Circles: G-C
Understand and apply theorems about circles
2. Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.
Note: We used the Math Nspired activity Circles – Angles and Arcs as a guide for this lesson.
We started off our lesson on Angles and Arcs just a little differently than before. I always have students explore – to pay attention to what is changing and what is staying the same in a geometric figure. But I am trying to learn from many of you out there, and so I asked students to estimate first.
We say that ∠ABC is inscribed in the circle. What is its measure?
It turns out that just over 50% of the students correctly estimated the measure. I showed the students how others answered, but I didn’t show them the correct answer. Next we played with our dynamic geometry software to find out what students noticed about inscribed angles and central angles that intercept the same arc.
What happens when a right angle is inscribed in a circle?
What is the relationship between the central angle and its intercepted arc?
What is the relationship between the inscribed angle and its intercepted arc?
What is the relationship for inscribed angles and central angles that intercept the same arc?
Students made observations. Before we formalized their observations, I sent another Quick Poll. We were up to just over 75% correct.
After formalizing the observations together, students worked through a few exercises in their groups.
Our next major exploration was to determine the relationship for opposite angles in a cyclic quadrilateral. Students explored by themselves, and then I made someone the Live Presenter.
What did you notice?
B.K. had noticed that the opposite angles had a sum of 180˚.
Can we make sense of why that happens?
-They’re inscribed angles.
What else?
-Their intercepted arcs make the whole circle.
Why is that significant?
The intercepted arcs add to 360˚. The two inscribed angles are half that sum, 180˚.
The opposite angles of a cyclic quadrilateral are supplementary.
And a Quick Poll to assess student understanding:
And the results. 65% have it correct. What happened to those who got 95˚? What happened to those who got 190˚?
And so the journey continues … as we are learning to answer the question being asked, and not just giving the first calculation or even the second calculation that we get when we make sense of the given information.
|
##### TASC For Dummies
Some questions on the TASC Math exam involve linear functions. A linear function represents a relationship between two variables in which one variable influences the other.
In a linear function, x is usually considered to be the independent variable and y to be the dependent variable (x influences y). The independent variable (x) runs horizontally, while the dependent variable (y) runs vertically. The minimum number of points you need to construct a line is two.
The common difference of the points that describes both the steepness and direction of the line is called the slope. This is also referred to as the ratio of the rate of change in the dependent variable to the rate of change in the independent variable. The letter associated with slope is m; if m is positive, then the line rises to the right, and if m is negative, then the line falls to the right. To determine the slope of a line, you need two points: (x1, y1) and (x2, y2). Then substitute into the formula:
Slope-intercept is the most commonly used formula to represent a linear function. Just as the name implies, this formula tells you the slope (m) of the line as well as the y-intercept (b). Recall that the y-intercept is the point at which the graph crosses the y-axis.
Slope-intercept form: y = mx + b (m is the slope and b is the y-intercept)
## Practice questions
1. The equation of the line perpendicular to
2. Which line would be parallel to the line y = –3x + 4?
1. The correct answer is Choice (A). Because you're looking for a line perpendicular to the given line, the new equation would have a negative reciprocal of the given slope:
so the new slope is m = 2. Because Choice (A) is the only equation with that slope, it is the correct answer. Choice (C) represents a line parallel to the given line, while Choices (B) and (D) have slopes that have no specific relationship with the given slope.
2. The correct answer is Choice (A). Because you're looking for a line parallel to the given line, the new equation would have the same slope: m = –3. Because Choice (A) is the only equation with that slope, it's the correct answer. Choice (C) represents a line perpendicular to the given line, while Choices (B) and (D) have slopes that have no specific relationship with the given slope.
Stuart Donnelly, PhD, was awarded a PhD in mathematics from Oxford University. He has prepared students for the TASC test and GED Test for the past two decades.
Stuart Donnelly, PhD, earned his doctorate in mathe-matics from Oxford University at the age of 25. Since then, he has established successful tutoring services in both Hong Kong and the United States and is considered by leading educators to be one of the most experienced and qualified private tutors in the country.
Stuart Donnelly, PhD, was awarded a PhD in mathematics from Oxford University. He has prepared students for the TASC test and GED Test for the past two decades.
Stuart Donnelly, PhD, was awarded a PhD in mathematics from Oxford University. He has prepared students for the TASC test and GED Test for the past two decades.
Sandra Luna McCune, PhD, is professor emeritus and a former Regents professor at Stephen F. Austin State University. She's now a full-time author. Shannon Reed, MA, MFA, is a visiting lecturer at the University of Pittsburgh, where she teaches composition, creative writing, and business writing.
Stuart Donnelly, PhD, was awarded a PhD in mathematics from Oxford University. He has prepared students for the TASC test and GED Test for the past two decades.
|
## Practical Geometry Class 6 Extra Questions Maths Chapter 14
Extra Questions for Class 6 Maths Chapter 14 Practical Geometry
### Practical Geometry Class 6 Extra Questions Very Short Answer Type
Class 6 Maths Chapter 14 Extra Questions Question 1.
If AB = 3.6 and CD = 1.6 cm, construct a line segment equal to $$\overline { AB }$$ + $$\overline { CD }$$ and measure the total length.
Solution:
Step I: Draw a ray OX.
Step II : With centre 0 and radius equal to the length of AB (3.6 cm) mark a point P on the ray.
Step III: With centre P and radius equal to the length of CD (1.6 cm) mark another point Q on the ray.
Thus OQ is the required segment such that OQ = 3.6 cm + 1.6 cm = 5.2 cm.
Question 2.
Construct a perpendicular to a given line segment at point on it.
Solution:
Step IDraw a line $$\overleftrightarrow { PQ }$$ and take any point A on it.
Step II : With centre A draw an arc which meets PQ at C and D.
Step III : Join AB and produce.
Step IV: With centres C and D and radius equal to half of the length of the previous arc, draw two arcs which meets each other at B.
Thus AB is the required perpendicular to $$\overleftrightarrow { PQ }$$.
Question 3.
Construct an angle of 60° and bisect it.
Solution:
Step I: Draw a line segment $$\overline { AB }$$.
Step II: With centre B and proper radius, draw an arc which meets AB at C.
Step III : With C as centre and the same radius as in step II, draw an arc cutting the previous arc at D.
Step IV : Join B to D and produce.
Step V : Draw the bisector BE of ∠ABD.
Thus BE is the required bisector of ∠ABD.
Question 4.
Draw an angle of 120° and hence construct an angle of 105°.
Solution:
Step I : Draw a line segment $$\overline { OA }$$.
Step II : With centre O and proper radius, draw an arc which meets OA at C.
Step III : With centre C and radius same, mark D and E on the previous arc.
Step IV : Join O to E and produce.
Step V : ∠EOA is the required angle of 120°.
Step VI : Construct an angle of 90° which meets the previous arc at F.
Step VII : With centre E and F and proper radius, draw two arcs which meet each other at G.
Step VIII : Join OG and produce.
Thus ∠GOA is the required angle of 105°.
Question 5.
Using compasses and ruler, draw an angle of
75° and hence construct an angle of 37 $$\frac { { 1 }^{ \circ } }{ 2 }$$.
Solution:
Step I: Draw a line segment OA.
Step II : Construct ∠BOA = 90° and ∠EOA = 60°
Step III : Draw OC as the bisector of ∠BOE , which equal to
Step IV : Draw the bisector OD of ∠COA.
Thus ∠DOA is the required angle of 37 $$\frac { { 1 }^{ \circ } }{ 2 }$$ .
Question 6.
Draw ∆ABC. Draw perpendiculars from A, B and C respectively on the sides BC, CA and AB. Are there perpendicular concurrent? (passing through the same points).
Solution:
Step I: Draw any ∆ABC.
Step II : Draw the perpendicular AD from A to BC.
Step III : Draw the perpendicular BE from B to AC.
Step IV : Draw the perpendicular CF from C to AB.
We observe that the perpendiculars AD, BE and CF intersect each other at P.
Thus, P is the point of intersection of the three perpendiculars.
|
```Coordinate Algebra
Practice
Unit 2
#1
A.
B.
C.
D.
Which equation shows ax – w = 3
solved for w ?
w = ax – 3
w = ax + 3
w = 3 – ax
w = 3 + ax
Unit 2
ax – w = 3
–ax
–ax
–w = 3 – ax
–w 3 – ax
=
–1
–1
w = –3 + ax
w = ax – 3
#2
Which equation is equivalent to
7 x 3x
11 ?
4
8
A. 17x = 88
B. 11x = 88
C. 4x = 44
D. 2x = 44
7 x 3x
11
4
8
8 7 x 3x 8
11
1 4
8 1
56 x 24 x
88
4
8
14x – 3x = 88
11x
= 88
Unit 2
Least Common
Denominator
8
#3
Which equation shows 4n = 2(t – 3)
solved for t ?
Method #1
4n = 2(t – 3)
4n = 2(t – 3)
2
2
2n = t – 3
+3
+3
2n + 3 = t
Unit 2
Method #2
4n = 2(t – 3)
4n = 2t – 6
+6
+6
4n + 6 = 2t
4n + 6 = 2t
2
2
2n + 3 = t
#4
Which equation shows 6(x + 4) = 2(y + 5)
solved for y ?
6(x + 4) = 2(y + 5)
A. y = x + 3
B. y = x + 5
C. y = 3x + 7
D. y = 3x + 17
6x + 24 = 2y + 10
–10
–10
6x + 14 = 2y
6x + 14 = 2y
2
2
3x + 7 = y
Unit 2
Unit 2
#5 This equation can be used to find h, the number of
hours it takes Flo and Bryan to mow their lawn.
How many hours will it take them to mow their lawn?
h h
1
3 6
A. 6
B. 3
C. 2
D. 1
Least Common
Denominator
6
6h h 6
1
1 3 6 1
6h 6h
6
3
6
2h + h = 6
3h
= 6
3h =
3
h = 2
6
3
#6
This equation can be used to determine how many
miles apart the two communities are. What is m,
the distance between the two communities?
A. 0.5 miles
B. 5 miles
C. 10 miles
D. 15 miles
m
m
0.5
15 5 15 5
m
m
0.5
10
20
20 m 20 m
0.5
1 10 1 20
20m
20m
10
10
20
Unit 2
Least Common
Denominator
20
2m = m + 10
–m –m
m =
10
#7
Unit 2
For what values of x is the inequality true?
A. x < 1
B. x > 1
C. x < 5
D. x > 5
2 x
1
3 3
32 x
3
1
1 3 3
1
6 3x
3
3 3
2+ x > 3
–2
–2
x > 1
Least Common
Denominator
3
Unit 2 A manager is comparing the cost of buying ball caps
with the company emblem from two different companies.
#8 •Company X charges a \$50 fee plus \$7 per cap.
A. 10 caps
•Company Y charges a \$30 fee plus \$9 per cap. B. 20 caps
C. 40 caps
For what number of ball caps (b) will the
D. 100 caps
manager’s cost be the same for both companies?
Cost Formula: Company X
CX = 7b + 50
CX = CY
Cost Formula: Company Y
CY = 9b + 30
7b + 50 = 9b + 30 (Subtract 7b on both sides)
50 = 2b + 30 (Subtract 30 on both sides)
20 = 2b
(Divide 2 on both sides)
10 = b
A shop sells one-pound bags of peanuts for \$2 and
three-pound bags of peanuts for \$5. If 9 bags are
purchased for a total cost of \$36, how many
three-pound bags were purchased?
#9
Method #1
Substitution
Let x = # of one-pound bags
Let y = # of three-pound bags
Equation #1:
Equation #2:
Unit 2
x + y = 9
2x + 5y = 36
(Total number of bags)
(Total value of bags)
Substitute x = 9 – y into Equation #2
Solve Equation #1 for x
x + y = 9
–y
–y
x
= 9–y
2(9 – y) + 5y
18 – 2y + 5y
18 + 3y =
3y =
6 three-pound
y =
bags
= 36
= 36
36
18
6
#9
A shop sells one-pound bags of peanuts for \$2 and
three-pound bags of peanuts for \$5. If 9 bags are
purchased for a total cost of \$36, how many
three-pound bags were purchased?
Method #2
Elimination
Let x = # of one-pound bags
Let y = # of three-pound bags
Equation #1:
Equation #2:
x + y = 9
2x + 5y = 36
Multiply Equation #1 by –2
–2(x + y) = –2(9)
–2x – 2y = –18
(New Equation #1)
Unit 2
(Total number of bags)
(Total value of bags)
Add New Equation #1 and Equation #2
–2x – 2y = –18
2x + 5y = 36
3y = 18
y = 6
6 three-pound
bags
#10
A.
Which graph represents a system of linear equations Unit 2
that has multiple common coordinate pairs?
B.
Has one
common
coordinate
pair
Multiple
common
coordinate
pairs
(Two lines
overlap)
D.
C.
Has one
common
coordinate
pair
Has no
common
coordinate
pairs
#11
Unit 2
Which graph represents x > 3 ?
A.
x>3
B.
x<3
C.
D.
x>3
x<3
#12
Which pair of inequalities is shown in the
graph?
Unit 2
A. y > –x + 1 and y > x – 5
Line 1
B. y > x + 1 and y > x – 5
Both given inequalities have
slopes equal to positive one.
This is a contradiction to the
slope of Line 1 being negative.
Line 2
Note
Line 1 graph has a negative slope.
Line 2 graph has a positive slope.
#12
Which pair of inequalities is shown in the
graph?
C. y > –x + 1 and y > –x – 5
Both given inequalities have
slopes equal to negative one.
This is a contradiction to the
slope of Line 2 being positive.
Unit 2
Line 1
Line 2
D. y > x + 1 and y > –x – 5
Line 2 has a positive slope
with a negative y-intercept.
However, the line y > x + 1
has a positive slope, but the
y-intercept is positive.
Note
Line 1 graph has a negative slope.
Line 2 graph has a positive slope.
#12
Which pair of inequalities is shown in the
graph?
Unit 2
A. y > –x + 1 and y > x – 5
Line 1
B. y > x + 1 and y > x – 5
Both given inequalities have
slopes equal to positive one.
This is a contradiction to the
slope of Line 1 being negative.
Line 2
Note
Line 1 graph has a negative slope.
Line 2 graph has a positive slope.
```
20 cards
21 cards
19 cards
15 cards
14 cards
|
## Vector Triple Product
The volume of a parallelepiped with sides AB and C is the area of its base (say the parallelogram with area |BC| ) multiplied by its altitude, the component of A in the direction of BC. This is the magnitude of ABC; but it is also the magnitude of the determinant of the matrix with columns AB and C, so these linear functions of the vectors here are the same up to sign. This article is on Vector triple product. The usual sign convention gives
A(BC) = det(A, B, C)
This Vector Triple Product is not changed by cyclically permuting the vectors (for example to BCA) or by reversing the order of the factors in the dot product.
We can deduce then that ABC = CAB = ABC. In words, we can switch the dot and cross product without changing anything in this entity. (In either formula of course you must take the cross product first.) This product, like the determinant, changes sign if you just reverse the vectors in the cross product.
The vector triple product, A(BC) is a vector, is normal to A and normal to BC which means it is in the plane of B and C. And it is linear in all three vectors.
We can deduce it is a multiple of B that is linear in A and C plus a multiple of that is linear in A and B, with the condition that it is normal to A.
Any multiple of B(AC) – C(AB) will obey all these conditions.
What multiple is A(BC)?
Suppose A(BC) = q(B(AC) – C(AB)) holds.
Earlier we saw that the square of the area of a parallelogram with sides A and B can be written either as (AA)( BB) – (AB)( AB) or (BA)(BA). By interchanging the dot and first cross product on the right here you can rewrite this equality as
(BA)(BA) = B(A(BA)) =(AA)( BB) ) – (AB)( AB)
If we identify A with C in A(BC) and take the dot product of A(BA) with B we find q = 1, and we get
A(BC) = B(AC) – C(AB)
This is sometimes called the back cab rule to make it easier to remember the appropriate signs. When using this name remember that the parentheses here are all as far back as possible in this expression The easiest way to get the signs right here without remembering anything is to guess a sign and then check it on the case A = i = C, B = j.
For more such articles follow us here.
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# Common Core: 8th Grade Math : Explain a Proof of the Pythagorean Theorem and its Converse: CCSS.Math.Content.8.G.B.6
## Example Questions
← Previous 1
### Example Question #1 : Explain A Proof Of The Pythagorean Theorem And Its Converse: Ccss.Math.Content.8.G.B.6
Which shape does the Pythagorean Theorem apply to?
Right triangles
Cubes
Squares
Triangles
Right triangles
Explanation:
The Pythagorean Theorem applies to right triangles. The Theorem states that for all right triangles, the square of the hypotenuse is equal to the sum of the square of the other two sides. In other terms:
### Example Question #2 : Explain A Proof Of The Pythagorean Theorem And Its Converse: Ccss.Math.Content.8.G.B.6
How is the Pythagorean Theorem used?
The Pythagorean Theorem is used to solve for a missing side length of a triangle
The Pythagorean Theorem is used to solve for a missing side length of a right triangle
The Pythagorean Theorem is used to solve for the area of a right triangle
The Pythagorean Theorem is used to solve for the volume of a triangle
The Pythagorean Theorem is used to solve for a missing side length of a right triangle
Explanation:
The Pythagorean Theorem states that for right triangles, the square of the hypotenuse is equal to the sum of the square of the other two sides. In other terms:
With this equation, we can solve for a missing side length.
### Example Question #3 : Explain A Proof Of The Pythagorean Theorem And Its Converse: Ccss.Math.Content.8.G.B.6
What is the formula associated with the Pythagorean Theorem?
Explanation:
The Pythagorean Theorem states that for right triangles, the square of the hypotenuse is equal to the sum of the square of the other two sides. In other terms:
In this equation:
### Example Question #1 : Explain A Proof Of The Pythagorean Theorem And Its Converse: Ccss.Math.Content.8.G.B.6
How is the converse of the Pythagorean Theorem used?
To determine the a missing side length of a triangle
To determine the a missing side length of a right triangle
To determine if a triangle is a right triangle
To determine if a shape is in fact a triangle
To determine if a triangle is a right triangle
Explanation:
The converse of the Pythagorean Theorem is used to determine if a triangle is a right triangle. If we are given three side lengths we can plug them into the Pythagorean Theorem formula:
If the square of the hypotenuse is equal to the sum of the square of the other two sides, then the triangle is a right triangle.
### Example Question #2 : Explain A Proof Of The Pythagorean Theorem And Its Converse: Ccss.Math.Content.8.G.B.6
Which answer choice provides side length of a right triangle?
Explanation:
In order to solve this problem, we use the converse of the Pythagorean Theorem. We will substitute the given side lengths to determine which three side lengths make the formula for Pythagorean Theorem true. It is important to remember that the hypotenuse will always be the longest side length, so the value for will always be the greatest:
Let's plug in the side lengths into our formula and solve:
This means that a triangle that has side lengths of is a right triangle.
### Example Question #6 : Explain A Proof Of The Pythagorean Theorem And Its Converse: Ccss.Math.Content.8.G.B.6
Will the Pythagorean Theorem work to solve for a missing side length of a three sided figure?
Yes, the Pythagorean Theorem is used to solve for any missing side length of a triangle
No, the Pythagorean Theorem will only work to solve for the missing side length of a right triangle
Yes, the Pythagorean Theorem is used to solve for any missing side length of at three sided figure
No, the Pythagorean Theorem will only work to solve for the missing side length of a triangle
No, the Pythagorean Theorem will only work to solve for the missing side length of a right triangle
Explanation:
The Pythagorean Theorem can only be used to solve for the missing side length of a right triangle. Remember, the Pythagorean Theorem states that for right triangles, the square of the hypotenuse is equal to the sum of the square of the other two sides. In other terms:
### Example Question #7 : Explain A Proof Of The Pythagorean Theorem And Its Converse: Ccss.Math.Content.8.G.B.6
If the equation is found to be true, what do we know?
A right triangle has a hypotenuse of and side lengths of and
The Pythagorean Theorem only works if the hypotenuse is an even number
The Pythagorean Theorem only works if the hypotenuse is an odd number
A right triangle has a hypotenuse of and side lengths of and
A right triangle has a hypotenuse of and side lengths of and
Explanation:
The equation shown in the question, , is the equation for the Pythagorean Theorem:
In this equation:
This means that and are the side lengths and in the hypotenuse of the triangle
### Example Question #3 : Explain A Proof Of The Pythagorean Theorem And Its Converse: Ccss.Math.Content.8.G.B.6
If the equation is found to be true, what do we know?
The Pythagorean Theorem only works if the hypotenuse is an odd number
A right triangle has a hypotenuse of and side lengths of and
The Pythagorean Theorem only works if the hypotenuse is an even number
A right triangle has a hypotenuse of and side lengths of and
A right triangle has a hypotenuse of and side lengths of and
Explanation:
The equation shown in the question, , is the equation for the Pythagorean Theorem:
In this equation:
This means that and are the side lengths and in the hypotenuse of the triangle
### Example Question #9 : Explain A Proof Of The Pythagorean Theorem And Its Converse: Ccss.Math.Content.8.G.B.6
If the equation is found to be true, what do we know?
The Pythagorean Theorem only works if the hypotenuse is an even number
The Pythagorean Theorem only works if the hypotenuse is an odd number
A right triangle has a hypotenuse of and side lengths of and
A right triangle has a hypotenuse of and side lengths of and
A right triangle has a hypotenuse of and side lengths of and
Explanation:
The equation shown in the question, , is the equation for the Pythagorean Theorem:
In this equation:
This means that and are the side lengths and in the hypotenuse of the triangle
### Example Question #10 : Explain A Proof Of The Pythagorean Theorem And Its Converse: Ccss.Math.Content.8.G.B.6
If the equation is found to be true, what do we know?
A right triangle has a hypotenuse of and side lengths of and
The Pythagorean Theorem only works if the hypotenuse is an odd number
The Pythagorean Theorem only works if the hypotenuse is an even number
A right triangle has a hypotenuse of and side lengths of and
A right triangle has a hypotenuse of and side lengths of and
Explanation:
The equation shown in the question, , is the equation for the Pythagorean Theorem:
In this equation:
This means that and are the side lengths and in the hypotenuse of the triangle
← Previous 1
|
What is the standard form of y= (9x-2)(x+2)(7x-4)?
Jun 10, 2018
$y = 63 {x}^{3} - 148 {x}^{2} + 50 x + 8$
Explanation:
The standard form refers to the format of an expression where the terms are arranged in a descending order. The degree is determined by the exponent value of the variable of each term.
To find the standard form, multiply the brackets out and simplify.
$y = \left(9 x - 2\right) \left(x + 2\right) \left(7 x - 4\right)$
Lets multiply out $y = \left(9 x - 2\right) \left(x + 2\right)$ first:
$y = \left(9 x - 2\right) \left(x + 2\right)$
$y = 9 {x}^{2} + 18 x - 2 x - 2$
$y = 9 {x}^{2} - 16 x - 2$
Then multiply $y = \left(9 {x}^{2} - 16 x - 2\right) \left(7 x - 4\right)$:
$y = \left(9 {x}^{2} - 16 x - 2\right) \left(7 x - 4\right)$
$y = 63 {x}^{3} - 36 {x}^{2} - 112 {x}^{2} + 64 x - 14 x + 8$
$y = 63 {x}^{3} - 148 {x}^{2} + 50 x + 8$
This is the standard form.
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Select Page
Read the following instructions in order to complete this discussion, and review the example of how to complete the math required for this assignment:Given an equation of a line, find equations for lines parallel or perpendicular to it going through specified points. Find the appropriate equations and points from the table below. Simplify your equations into slope-intercept form.Use your assigned number to complete.Discuss the steps necessary to carry out each activity. Describe briefly what each line looks like in relation to the original given line.Answer these two questions briefly in your own words: What does it mean for one line to be parallel to another?What does it mean for one line to be perpendicular to another?Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing your math work.):Origin Ordered pair X- or y-intercept Slope ReciprocalYour initial post should be 150-250 words in length.The equation is # 13
Number 13
INSTRUCTOR GUIDANCE EXAMPLE: Week Three Discussion Parallel and Perpendicular For this week’s discussion I am going to find the equations of lines that are parallel or perpendicular to the given lines and which are passing through the specified point. First, I will work on the equation for the parallel line. The equation I am given is 2 3 2 + − = x y The parallel line must pass through point ( )3 ,6 − − = = I have learned that a line parallel to another line has the same=slope as the other line, so now I know that the slope of my parallel line will be 3 2 −. Since I now have both the= slope and an ordered pair on the line, I am going to use the point -slope form of a linear equation to write my new equation. ) ( 1 1 x x m y y − = − = = =This is the general form of the point -slope equation= ( ) ( ) [ ] 6 3 2 3 − − − = − − x y I plugged in my given slope and ordered pair ( )6 3 2 3 + − = + x y I evaluated any signs next to each other )6( 3 2 3 2 3 − − = + x y I distributed the 32 −=to each term inside the parentheses = 4 3 2 3 − − = + x y I show here the distribution of the 3 2 − =and multiplied== = ==== 3 2 − =times 6, which is 4= = 3 4 3 2 − − − = x y I subtracted 3 from both sides, moving like -terms together so I can combine them 7 3 2 − − = x y Like-terms are combined, and the result is the equation of my parallel line! This line falls as you go from left to right across the graph of it, the y- intercept is 7 units below the origin , and the x- intercept is 10.5 units to the left of the origin . Now I will write the equation of the perpendicular line. The equation I am given is 1 4− − = x y The perpendicular line must pass through point () 5 , 0 I have learned that a line perpendicular to another line has a slope which is the negative reciprocal of the slope of the other line . So the first thing I must do is find the negative reciprocal of –4. The reciprocal of –4 is 4 1 −, and the negative of that is 4 1 4 1 = − − . Now I know my slope is 4 1 and my given point is = () 5 , 0 . Again, I will use the point -slope form of a linear equation to write my new equation. ) ( 1 1 x x m y y − = − = = =This is the general form of the point -slope equation= ( ) 0 4 1 5 − = − x y I plugged in my given slope and ordered pair ( )0 4 1 4 1 5 − = − x y I distributed the 4 1 x y 4 1 5 = −= = ==I multiplied= ( )0 4 1 − = = 5 4 1 + = x y I add 5 to both sides of the equation, and the result is the equation of my perpendicular line! This line rises as you move from left to right across the graph. The y -intercept is five units above the original and the x -intercept is 20 units to the left of the origin. [The answers to part d of the discussion will vary with students’ understanding.]
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# How do you use the pythagorean theorem to solve for the missing side given a = 5, c = 19?
Aug 24, 2017
See a solution process below:
#### Explanation:
The Pythagorean Theorem states:"
${a}^{2} + {b}^{2} = {c}^{2}$
Where:
$a$ and $b$ are sides of a right triangle.
$x$ is the hypotenuse of the the right triangle.
Substituting and solving for $b$ gives:
${5}^{2} + {b}^{2} = {19}^{2}$
$25 + {b}^{2} = 361$
$25 - \textcolor{red}{25} + {b}^{2} = 361 - \textcolor{red}{25}$
$0 + {b}^{2} = 336$
${b}^{2} = 336$
$\sqrt{{b}^{2}} = \sqrt{336}$
$b = \sqrt{336}$
$b = \sqrt{16 \cdot 21}$
$b = \sqrt{16} \cdot \sqrt{21}$
$b = 4 \sqrt{21}$
Or
$b = 18.330$ rounded to the nearest thousandth.
|
# Convert Metric Units of Area
$$\require{cancel}$$ $$\require{bbox}$$
Convert metric units of area, such as, $$\text{m}^2, \text{cm}^2, \text{mm}^2, \text{km}^2, ..$$, examples with solutions are presented including more More questions with solutions .
## Table and Factor of Conversion of Units of Area
The table shown below helps in finding factors of conversion between metric units of length which will be used to find the factor of conversion of areas.
For example, using the table above we can use the factors below the arrows to write $1 \text{ dam} = 10 \times 10 \text{ dm} = 100 \text{ dm}$ which gives $1 \text{ dam} = 100 \text{ dm}$ Multiply each side of the above equality by itself (or Square both sides) of the above equality $(1 \text{ dam})(1 \text{ dam}) = (100 \text{ dm})(100 \text{ dm})$ Simplify to write $1 \text{ dam}^2 = 10000 \text{ dm}^2$ which gives the following
factors of conversion between the units of area $$\text{ dam}^2$$ and $$\text{ dm}^2$$. $\displaystyle \frac{1 \text{ dam}^2}{10000 \text{ dm}^2} = 1$ or $\displaystyle \frac {10000 \text{ dm}^2} {1 \text{ dam}^2} = 1$ Note that the factor of conversion may be written in two different ways by interchanging the denominator and numerator. We select the factor of conversion with the unit in the denominator that is same as the given unit to be converted so that they cancel.
## Examples of Conversion with Solutions
In the first few examples, we show all the steps for a thorough understanding of the the conversion. Apart from the table of metric units above, nothing else, such as formulas for example, is needed to do the conversions.
Example 1
Convert $$2450 \text{ cm}^2$$ to $$\text{m}^2$$
Solution to Example 1
We are converting $$\text{ cm}^2$$ to $$\text{ m}^2$$ and therefore the factor of conversion between $$\text{ cm}$$ and $$\text{ m}$$ is needed first.
Using Table 1 above, we have $1 \text{ m} = 100 \text{ cm}$ Square both sides $(1 \text{ m})^2 = (100 \text{ cm})^2$ Simplify $1 \text{ m}^2 = 10000 \text{ cm}^2$ which gives the factors of conversion $\displaystyle \frac{1 \text{ m}^2}{10000 \text{ cm}^2} = 1 \quad (I)$ or $\displaystyle \frac{10000 \text{ cm}^2}{1 \text{ m}^2} = 1 \quad (II)$ To convert $$2450 \text{ cm}^2$$, we use use the factor of conversion (I) given by $$\displaystyle \frac{1 \text{ m}^2}{10000 \text{ cm}^2}$$ because it has $$\text{ cm}^2$$ in the denominator which will cancel with the given $$\text{ cm}^2$$.
Write the given expression as
$$2450 \text{ cm}^2 = 2450 \text{ cm}^2 \times \color{red}1$$
Substitute $$\color{red} 1$$ by the factor of conversion $$\displaystyle \frac{1 \text{ m}^2}{10000 \text{ cm}^2}$$ since it is equal to $$\color{red}1$$
$$2450 \text{ cm}^2 = 2450 \text{ cm}^2 \times \displaystyle \frac{1 \text{ m}^2}{10000 \text{ cm}^2}$$
Cancel $$\text{ cm}^2$$
$$2450 \text{ cm}^2 = 2450 \cancel{\text{ cm}^2} \times \displaystyle \frac{1 \text{ m}^2}{10000 \cancel{\text{ cm}^2}}$$
Simplify
$$2450 \text{ cm}^2 = \displaystyle \frac{2450 \times 1 }{10000} \text{ m}^2$$
Evaluate
$\bbox[10px, border: 2px solid red] { 2450 \text{ cm}^2 = 0.245 \text{ m}^2 }$
Example 2
Convert $$34590.5 \text{ m}^2$$ to $$\text{km}^2$$
Solution to Example 2
We are converting $$\text{ m}^2$$ to $$\text{ km}^2$$ and therefore the factor of conversion between $$\text{ m}$$ and $$\text{ km}$$ is needed first.
Use the table above to write factor of conversion between $$\text{ m}$$ and $$\text{ km}$$ $1 \text{ km} = 1000 \text{ m}$ Square both sides $(1 \text{ km})(1 \text{ km}) = (1000 \text{ m})(1000 \text{ m})$ and simplify $1 \text{ km}^2 = 1000000 \text{ m}^2$ which gives the factors of conversion of area with $$\text{ m}^2$$ in the denominator $\displaystyle \frac{1 \text{ km}^2}{1000000 \text{ m}^2} = 1$ Convert using the above factor of conversion
$$34590.5 \text{ m}^2 = 34590.5 \text{ m}^2 \times \displaystyle \frac{1 \text{ km}^2}{1000000 \text{ m}^2}$$
Cancel $$\text{ m}^2$$
$$34590.5 \text{ m}^2 = 34590.5 \cancel{\text{ m}^2} \times \displaystyle \frac{1 \text{ km}^2}{1000000 \cancel{\text{ m}^2}}$$
Simplify
$$34590.5 \text{ m}^2 = \displaystyle \frac{34590.5 \times 1 }{1000000} \text{ km}^2$$
Evaluate
$\bbox[10px, border: 2px solid red] { 34590.5 \text{ m}^2 = 0.0345905 \text{ km}^2 }$
## Questions with Solutions
Convert the following
1. $$569000 \text{ mm}^2$$ to $$\text{m}^2$$
2. $$1.2 \text{ km}^2$$ to $$\text{dam}^2$$
3. $$23.01 \text{ hm}^2$$ to $$\text{m}^2$$
4. $$12.7 \text{ cm}^2$$ to $$\text{mm}^2$$
5. $$13500 \text{ dm}^2$$ to $$\text{hm}^2$$
|
# What Are Number Bonds? A Guide For Elementary Teachers
Number bonds are pairs of numbers that can be added together to make another number e.g. 4 + 6 = 10. They are some of the most basic and most important parts of math for children to learn.
Number bonds form the foundation of a students’ math knowledge and students must know their number bonds to be able to grasp more complex math concepts such as fractions, decimals, manipulate different numbers and solve word problems.
14 Fun Math Games and Activities Pack for 2nd Grade
14 fun math games and activities for 2nd grade students, perfect for ‘fast finishers’ or morning work!
### What are number bonds?
A number bond is a pair of numbers that always add together to make another, larger, number. They can also be called fact families or math facts. Number bonds demonstrate part-part-whole relationships and help us to understand that a whole number is made up of parts. They also help students practice and develop number sense. Children are introduced to this concept through number bonds to 10, also known as complements of 10:
• 0 + 10
• 1 + 9
• 2 + 8
• 3 + 7
• 4 + 6
• 5 + 5
These are the foundations of many other key number bonds – if children can fluently recall their addition facts to 10, they will be able to calculate number bonds to other multiples of 10.
This can then be developed to larger multiples of 10 (such as multiples of 100) and even decimal number bonds (e.g. 0.3 + 0.7 = 1). Children should also be able to calculate the corresponding subtraction facts for these number bonds, e.g. if 1 + 9 = 10, then 10 – 1 = 9 and 10 – 9 = 1.
### Number bonds to 10 (Kindergarten)
Addition number bonds show the addends of simple sums. The number bonds to 10 are:
• 0 + 10
• 1 + 9
• 2 + 8
• 3 + 7
• 4 + 6
• 5 + 5
• 6 + 4
• 7 + 3
• 8 + 2
• 9 + 1
• 10 + 0
### Number bonds to 20 (Kindergarten & 1st Grade)
The number bonds to 20 are:
• 0 + 20
• 1 + 19
• 2 + 18
• 3 + 17
• 4 + 16
• 5 + 15
• 6 + 14
• 7 + 13
• 8 + 12
• 9 + 11
• 10 + 10
• 10 + 10
• 11 + 9
• 12 + 8
• 13 + 7
• 14 + 6
• 15 + 5
• 16 + 4
• 17 + 3
• 18 + 2
• 19 + 1
• 20 + 0
### Number bonds to 100 (2nd Grade)
Number bonds to 100 are pairs of numbers that add together to make 100 such as 20 + 80, or 55 + 45. To find the number bond pair for any number, use your knowledge of number bonds to 10 to figure out first what the matching unit or ‘ones’ would be to bridge to the next ten and then, what the number pair would be for the tens.
So for example to find the number bond to make 100 with 36, you first add 4 to reach 40, then add 60 to reach 100. So the number bond to make 100 with 36 is 64.
### When do children learn about number bonds in elementary school?
The Common Core State Standards and TEKS introduce the math concept in kindergarten, where children will represent and use number bonds within 20, as well as related simple addition and subtraction sentences within 10.
The non-statutory guidance also advises that kindergarten students should memorize and reason with number bonds to 10 and 20 in several forms (for example, 9 + 7 = 16; 16 – 7 = 9; 7 = 16 – 9). The CCSS and TEKS states that by the end of first grade students should know the number bonds to 20 and be precise in using and understanding place value. An emphasis on practice at this early stage will aid fluency, math skills and problem solving later on.
Wondering about how to explain other key math vocabulary to children? Check out our Math Dictionary For Kids, or try these other terms related to number bonds:
### Practice number bonds questions
1. Complete these sums. One is done for you.
3 + 7 = 10
33 + __ = 40
__ + 7 = 80
2. Write the missing numbers on the lines to make this correct.
60 + __ = 100 = 20 + ___
3. Write the next three sums in the pattern:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
7 + 2 + 1 + 8 + 9 + 3
Number bonds games
Number bonds are often taught in school through math activities. For example, dominoes are a fun way to test students’ knowledge of number bonds. As well as these number bonds to 10 activities, we also have several games specifically for practicing number facts and number bonds including these fun math games for kids to play at home or in school, kindergarten math games and then these mental math games.
Do you have students who need extra support in math?
Give your students more opportunities to consolidate learning and practice skills through personalized math tutoring with their own dedicated online math tutor.
Each student receives differentiated instruction designed to close their individual learning gaps, and scaffolded learning ensures every student learns at the right pace. Lessons are aligned with your state’s standards and assessments, plus you’ll receive regular reports every step of the way.
Personalized one-on-one math tutoring programs are available for:
|
Order of Operations
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# Order of Operations - PowerPoint PPT Presentation
Order of Operations . Mr. Clutter VMS Library. Order of Operations. A standard way to simplify mathematical expressions and equations. Purpose. Avoids Confusion Gives Consistency. For example: 8 + 3 * 4 = 11 * 4 = 44 Or does it equal 8 + 3 * 4 = 8 + 12 = 20.
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## PowerPoint Slideshow about ' Order of Operations ' - cooper-burke
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Presentation Transcript
Order of Operations
Mr. Clutter
VMS Library
### Order of Operations
A standard way to simplify mathematical expressions and equations.
Purpose
• Avoids Confusion
• Gives Consistency
For example:
8 + 3 * 4 = 11 * 4 = 44
Or does it equal
8 + 3 * 4 = 8 + 12 = 20
Order of operations are a set of rules
that mathematicians have agreed to
when simplifying mathematical expressions
or equations. Without these simple, but
important rules, learning mathematics would
The Rules
• Simplify within Grouping Symbols
( ), { }, [ ], | |
• Simplify Exponents
Raise to Powers
• Complete Multiplication and Division from Left to Right
• Complete Addition and Subtraction from Left to Right
Back to Our Example
For example:
8 + 3 * 4 = 11 * 4 = 44
Or does it equal
8 + 3 * 4 = 8 + 12 = 20
Using order of operations, we do the
multiplication first. So what’s our
20
How can we remember it?
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Graphical Interpretation of Sentences like f(x)=0 and f(x)>0
GRAPHICAL INTERPRETATIONS OF SENTENCES LIKE $\,f(x) = 0\,$ and $\,f(x) \gt 0$
• PRACTICE (online exercises and printable worksheets)
• Need some basic information on graphs of functions first?
Graphs of Function
If you know the graph of a function $\,f\,$,
then it is very easy to visualize the solution sets of sentences like $\,f(x)=0\,$ and $\,f(x)\gt 0\,$.
This section shows you how!
A key observation is that a sentence like $\,f(x) = 0\,$ or $\,f(x) \gt 0\,$ is a sentence in one variable, $\,x\,$.
To solve such a sentence, you are looking for value(s) of $\,x\,$ that make the sentence true.
The function $\,f\,$ is known, and determines the graph that you'll be investigating.
Recall that the graph of a function $\,f\,$ is a picture of all its (input,output) pairs;
that is, it is a picture of all points of the form $\,(x,f(x))\,$.
In particular, the $\,y$-value of the point $\,(x,f(x))\,$ is the number $\,f(x)\,$.
If $\,f(x)\gt 0\,$, then the point $\,(x,f(x))\,$ lies above the $\,x$-axis.
If $\,f(x)=0\,$, then the point $\,(x,f(x))\,$ lies on the $\,x$-axis.
If $\,f(x)\lt 0\,$, then the point $\,(x,f(x))\,$ lies below the $\,x$-axis.
These concepts are illustrated below.
The notation $\,P(x,f(x))\,$ is a convenient shorthand for: the point $\,P\,$ with coordinates $\,(x,f(x))$
point $\,P(x,f(x))\,$ has $\,f(x)\gt 0\,$ point $\,P(x,f(x))\,$ has $\,f(x)=0\,$ point$\, P(x,f(x))\,$ has $\,f(x)\lt 0\,$
The graph of a function $\,f\,$ is shown at right. The solution set of the inequality ‘$\,f(x)\gt 0\,$’ is shown in purple. It is the set of all values of $\,x\,$ for which $\,f(x)\,$ is positive. That is, it is the set of $\,x$-values that correspond to the part of the graph above the $\,x$-axis. The graph of a function $\,f\,$ is shown at right. The solution set of the equation ‘$\,f(x)=0\,$’ is shown in purple. It is the set of all values of $\,x\,$ for which $\,f(x)\,$ equals zero. That is, it is the set of $\,x$-intercepts of the graph. The graph of a function $\,f\,$ is shown at right. The solution set of the inequality ‘$\,f(x)\lt 0\,$’ is shown in purple. It is the set of all values of $\,x\,$ for which $\,f(x)\,$ is negative. That is, it is the set of $\,x$-values that correspond to the part of the graph below the $\,x$-axis. The graph of a function $\,f\,$ is shown at right. The solution set of the inequality ‘$\,f(x)\ge 0\,$’ is shown in purple. It is the set of all values of $\,x\,$ for which $\,f(x)\,$ is nonnegative. That is, it is the set of $\,x$-values that correspond to the part of the graph that is either on or above the $\,x$-axis. The graph of a function $\,f\,$ is shown at right. The solution set of the inequality ‘$\,f(x)\le 0\,$’ is shown in purple. It is the set of all values of $\,x\,$ for which $\,f(x)\,$ is nonpositive. That is, it is the set of $\,x$-values that correspond to the part of the graph that is either on or below the $\,x$-axis.
EXAMPLE:
The graph of a function $\,g\,$ with domain $\,[-6,10)\,$ is shown below.
Pay attention to the difference between the brackets ‘$\,[\ ]\,$’ and parentheses ‘$\,(\ )\,$’ and braces ‘$\,\{\ \}\,$’ in the solutions sets!
The solution set of the inequality ‘$\,g(x)\gt 0\,$’ is: $\,(-3,-2)\cup (0,1)\cup (3,5)\cup [6,7)\cup (9,10)\,$ The solution set of the inequality ‘$\,g(x)\ge 0\,$’ is: $\,(-3,-2]\cup (0,1]\cup[3,5]\cup[6,7]\cup[9,10)$ The solution set of the equation ‘$\,g(x)=0\,$’ is: $\,\{-2,1,3,5,7,9\}$ The solution set of the inequality ‘$\,g(x)\lt 0\,$’ is: $\,[-6,-3]\cup (-2,0]\cup (1,3)\cup (5,6)\cup (7,9)$ The solution set of the inequality ‘$\,g(x)\le 0\,$’ is: $\,[-6,-3]\cup [-2,0]\cup [1,3]\cup [5,6)\cup [7,9]$
Let's discuss the solution set of the inequality ‘$\,g(x)\gt 0\,$’.
Imagine a vertical line passing through the graph, moving from left to right.
Every time the vertical line touches a point that lies above the $\,x$-axis, then you must include that $\,x$-value in the solution set.
Be extra careful of ‘boundary’ or ‘transition’ points—places where something interesting is happening
(like where the graph crosses the $\,x$-axis, or where there's a break in the graph).
For example, suppose the vertical line reaches $\,x = -3\,$.
The point (the filled-in circle) is below the $\,x$-axis, so we don't want this $\,x$-value.
But then, the graph is above until we reach $\,x = -2\,$.
When $\,x = -2\,$, the point is on the $\,x$-axis, so we don't want this $\,x$-value, either.
This discussion gives the interval $\,(-3,-2)\,$ in the solution set.
Remember that the symbol ‘$\,\cup\,$’, the union symbol, is used to put sets together.
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Graphical Interpretation
of Sentences like $\,f(x)=g(x)\,$ and $\,f(x)\gt g(x)$
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5 6 7 8 9 10 11 12 13 14
AVAILABLE MASTERED IN PROGRESS
(MAX is 14; there are 14 different problem types.)
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Tamil Nadu Board of Secondary EducationHSC Arts Class 11th
# Find the value of n, if the sum to n terms of the series 3+75+243+...... is 4353 - Mathematics
Sum
Find the value of n, if the sum to n terms of the series sqrt(3) + sqrt(75) + sqrt(243) + ...... is 435 sqrt(3)
#### Solution
t1 = sqrt(3)
t2 = sqrt(75)
= sqrt(25 xx 3)
= 5sqrt(3)
t3 = sqrt(243)
= sqrt(81 xx 3)
= 9sqrt(3)
Here t1 = sqrt(3)
t2 = 5sqrt(3)
t3 = 9sqrt(3)
(i.e) a = sqrt(3)
d = 5sqrt(3) - sqrt(3)
= 4sqrt(3)
Sn = "n"/2[2"a" + ("n" - 1)"d"]
= 435 sqrt(3) ......(Given)
⇒ "n"/2 [2sqrt(3) + ("n" - 1)4sqrt(3)] = 435sqrt(3)
⇒ ("n" sqrt(3))/2 [2 + 4"n" - 4] = 435 sqrt(3)
⇒ n[4n – 2] = 870
4n2 – 2n – 870 = 0
(÷ by 2)2n2 – n – 435 = 0
2n2 – 30n + 29n – 435 = 0
⇒ 2n(n – 15) + 29(n – 15) = 0
(2n + 29)(n – 15) = 0
⇒ n = (-29)/2 or 15
n = (-29)/2 not possible,
So n = 15
Concept: Finite Series
Is there an error in this question or solution?
#### APPEARS IN
Tamil Nadu Board Samacheer Kalvi Class 11th Mathematics Volume 1 and 2 Answers Guide
Chapter 5 Binomial Theorem, Sequences and Series
Exercise 5.3 | Q 6 | Page 220
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NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4.
Board CBSE Textbook NCERT Class Class 6 Subject Maths Chapter Chapter 11 Chapter Name Algebra Exercise Ex 11.4 Number of Questions Solved 2 Category NCERT Solutions
## NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4
Question 1.
(a)
Take Sarita’s present age to he 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’sfather’s age is 5years 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 hilltop. Meena is at step s, Beena is 8 steps ahead and Leena7 steps behind. Where are Beena and Leena? The total number of steps to the hilltop is 10 less than 4 times what Meena has reached. Express the total number of steps using.
(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has traveled 5 hours,
Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using.
Solution.
(a) (y + 5) years
(ii) (y – 3) years
(iii) 6y years
(iv) (6y – 2) years
(v) (3y + 5) years
(b) Length (l) = (3b – 4) meters
(c) Length of the box = 5h cm
Breadth of the box = (5h – 10) cm
(d) Beena is at step (s + 8)
Leena is at step (s – 7)
Total number of steps = 4s – 10.
(e) Speed of the bus = v km/hr
Distance travelled in 5 hours = 5v km.
∴Total distance = (5v + 20) km
Question 2.
Change the following statements using expressions into statements in the 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 notebook costs Rs p. A book costs Rs 3 p.
(b) Tony puts q marbles on the table. He has 8 q 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.
Solution.
(a) A book costs three times the cost of a notebook.
(b) Tony’s box contains 8 times the marbles on the table.
(c) The total number of students in the school in 20 times that of our class.
(d) Jaggu’s uncle is 4 times older than Jaggu and Jaggu’s aunt is 3 years younger than his uncle.
(e) The total number of dots in an arrangement is 5 times the number of rows.
Question 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$$\frac { 1 }{ 2 }$$ ,y-2$$\frac { 1 }{ 2 }$$.
(c) Given n students in the class like football, what may 2n show? What may — show?(Hint: Think of games other than football).
Solution.
(a) (x – 2) may show the age is Munnu’s younger brother. (a + 4) may show the age of Munnu’s elder brother. (3a+ 7) may show the age of Munnu’s grand mother.
(b) y + l indicates her age 7 years from now.
y – 3 indicates her age 3 years back.
y+ 4$$\frac { 1 }{ 2 }$$ indicates her age 4$$\frac { 1 }{ 2 }$$ years from now
y-2$$\frac { 1 }{ 2 }$$ indicates her age 2$$\frac { 1 }{ 2 }$$ years back.
(c) 2n may show the number of students who like cricket.
$$\frac { n }{ 2 }$$may show the number of students who like hockey.
We hope the NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4 help you. If you have any query regarding. NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4, drop a comment below and we will get back to you at the earliest.
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# 2.2: The Straight Line
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It might be thought that there is rather a limited amount that could be written about the geometry of a straight line. We can manage a few Equations here, however, (there are 35 in this section on the Straight Line) and we shall return for more on the subject in Chapter 4.
Most readers will be familiar with the equation for a straight line:
$y = mx + c \label{2.2.1} \tag{2.2.1}$
The slope (or gradient) of the line, which is the tangent of the angle that it makes with the $$x$$-axis, is $$m$$, and the intercept on the $$y$$-axis is $$c$$. There are various other forms that may be of use, such as
$\frac{x}{x_0} + \frac{y}{y_0} = 1 \label{2.2.2} \tag{2.2.2}$
$\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1} \label{2.2.3} \tag{2.2.3}$
which can also be written
\begin{array}{| c c c | c}
x & y & 1 \\
x_1 & y_1 & 1 & =0 \\
x_2 & y_2 & 1 \\
\label{2.2.4} \tag{2.2.4}
\end{array}
$x \cos \theta + y \sin \theta = p \label{2.25} \tag{2.2.5}$
The four forms are illustrated in figure $$\text{II.1}$$.
$$\text{FIGURE II.1}$$
A straight line can also be written in the form
$Ax + By + C = 0. \label{2.2.6} \tag{2.2.6}$
If $$C = 0$$, the line passes through the origin. If $$C ≠ 0$$, no information is lost, and some arithmetic and algebra are saved, if we divide Equation \ref{2.2.6} by $$C$$ and re-write it in the form
$ax + by = 1. \label{2.2.7} \tag{2.2.7}$
Let $$P (x , y)$$ be a point on the line and let $$P_0 (x_0 , y_0 )$$ be a point in the plane not necessarily on the line. It is of interest to find the perpendicular distance between $$P_0$$ and the line. Let $$S$$ be the square of the distance between $$P_0$$ and $$P$$. Then
$S = (x-x_0)^2 + (y-y_0)^2 \label{2.2.8} \tag{2.2.8}$
We can express this in terms of the single variable $$x$$ by substitution for $$y$$ from Equation $$\ref{2.2.7}$$. Differentiation of $$S$$ with respect to $$x$$ will then show that $$S$$ is least for
$x = \frac{a+b(bx_0 - ay_0)}{a^2 + b^2} \label{2.2.9} \tag{2.2.9}$
The corresponding value for $$y$$, found from Equations $$\ref{2.2.7}$$ and $$\ref{2.2.9}$$, is
$y = \frac{b+a(ay_0 - bx_0)}{a^2 + b^2}. \label{2.2.10} \tag{2.2.10}$
The point $$\text{P}$$ described by Equations $$\ref{2.2.9}$$ and $$\ref{2.2.10}$$ is the closest point to $$\text{P}_0$$ on the line. The perpendicular distance of $$\text{P}$$ from the line is $$p = √S$$ or
$p = \frac{1-ax_0 - by_0}{\sqrt{a^2+b^2}}. \label{2.2.11} \tag{2.2.11}$
This is positive if $$\text{P}_0$$ is on the same side of the line as the origin, and negative if it is on the opposite side. If the perpendicular distances of two points from the line, as calculated from Equation $$\ref{2.2.11}$$, are of opposite signs, they are on opposite sides of the line. If $$p = 0$$, or indeed if the numerator of Equation $$\ref{2.2.11}$$ is zero, the point $$\text{P}_0 (x_0 , y_0 )$$ is, of course, on the line.
Let $$\text{A}(x_1 , y_1 ), \ \text{B}(x_2 , y_2 )$$ and $$\text{C}(x_3 , y_3 )$$ be three points in the plane. What is the area of the triangle $$\text{ABC}$$? One way to answer this is suggested by figure $$\text{II.2}$$.
$$\text{FIGURE II.2}$$
We see that
area of triangle $$\text{ABC}$$ = area of trapezium $$\text{A}^\prime \text{ACC}^\prime$$ (see comment*)
+ area of trapezium $$\text{C}^\prime \text{CBB}^\prime$$
− area of trapezium $$\text{A}^\prime \text{ABB}^\prime$$.
$= \frac{1}{2} (x_3 - x_1) (y_3 + y_1) + \frac{1}{2} (x_2 - x_3)(y_2 + y_3) - \frac{1}{2} (x_2 - x_1) (y_2 + y_1)$
$= \frac{1}{2} [x_1(y_2 - y_3) + x_2 (y_3 -y_1) + x_3(y_1 - y_2)]$
\begin{array}{l r | c c c |}
& & x_1 & x_2 & x_3 \\
= & \frac{1}{2} & y_1 & y_2 & y_3 \\
& & 1 & 1 & 1 \\
\label{2.2.12} \tag{2.2.12}
\end{array}
* Since writing this section I have become aware of a difference in U.S./British usages of the word "trapezium". Apparently in British usage, "trapezium" means a quadrilateral with two parallel sides. In U.S. usage, a trapezium means a quadrilateral with no parallel sides, while a quadrilateral with two parallel sides is a "trapezoid". As with many words, either British or U.S. usages may be heard in Canada. In the above derivation, I intended the British usage. What is to be learned from this is that we must always take care to make ourselves clearly understood when using such ambiguous words, and not to assume that the reader will interpret them the way we intend.
The reader might like to work through an alternative method, using results that we have obtained earlier. The same result will be obtained. In case the algebra proves a little tedious, it may be found easier to work through a numerical example, such as: calculate the area of the triangle $$\text{ABC}$$, where $$\text{A}$$, $$\text{B}$$, $$\text{C}$$ are the points (2,3), (7,4), (5,6) respectively. In the second method, we note that the area of a triangle is $$\frac{1}{2} \times \text{base} \times \text{height}$$. Thus, if we can find the length of the side BC, and the perpendicular distance of $$\text{A}$$ from $$\text{BC}$$, we can do it. The first is easy:
$(\text{BC})^2 = (x_3 - x_2)^2 + (y_3 - y_2)^2 . \label{2.2.13} \tag{2.2.13}$
To find the second, we can easily write down the equation to the line $$\text{BC}$$ from Equation $$\ref{2.2.3}$$, and then re-write it in the form $$\ref{2.2.7}$$. Then Equation $$\ref{2.2.11}$$ enables us to find the perpendicular distance of $$\text{A}$$ from $$\text{BC}$$, and the rest is easy.
If the determinant in Equation $$\ref{2.2.12}$$ is zero, the area of the triangle is zero. This means that the three points are collinear.
The angle between two lines
$y = m_1 x + c_1 \label{2.2.14} \tag{2.2.14}$
and $y = m_2 x + c_2 \label{2.2.15} \tag{2.2.15}$
is easily found by recalling that the angles that they make with the $$x$$-axis are $$\tan^{-1} \ m_1$$ and $$\tan^{-1} \ m_2$$ together with the elementary trigonometry formula $$\tan( A − B) = (\tan A − \tan B) / (1+ \tan A \tan B)$$. It is then clear that the tangent of the angle between the two lines is
$\frac{m_2 - m_1}{1+ m_1 m_2} . \label{2.2.16} \tag{2.2.16}$
The two lines are at right angles to each other if
$m_1 m_2 = -1 \label{2.2.17} \tag{2.2.17}$
The line that bisects the angle between the lines is the locus of points that are equidistant from the two lines. For example, consider the two lines
$-2x + 5y = 1 \label{2.2.18} \tag{2.2.18}$
$30x - 10y = 1 \label{2.2.19} \tag{2.2.19}$
Making use of Equation $$\ref{2.2.11}$$, we see that a point $$(x , y)$$ is equidistant from these two lines if
$\frac{1+2x-5y}{\sqrt{29}} = \pm \frac{1-30x+10y}{\sqrt{1000}}. \label{2.2.20} \tag{2.2.20}$
The significance of the $$\pm$$ will become apparent shortly. The + and − choices result, respectively, in
$-8.568x + 8.079y = 1 \label{2.2.21} \tag{2.2.21}$
and $2.656x + 2.817y = 1. \label{2.2.22} \tag{2.2.22}$
The two continuous lines in figure $$\text{II.3}$$ are the lines $$\ref{2.2.18}$$ and $$\ref{2.2.19}$$. There are two bisectors, represented by Equations $$\ref{2.2.21}$$ and $$\ref{2.2.22}$$, shown as dotted lines in the figure, and they are at right angles to each other. The choice of the + sign in Equation $$\ref{2.2.20}$$ (which in this case results in Equation $$\ref{2.2.21}$$, the bisector in figure $$\text{II.3}$$ with the positive slope) gives the bisector of the sector that contains the origin.
An equation of the form
$ax^2 + 2hxy + by^2 = 0 \label{2.2.23} \tag{2.2.23}$
can be factored into two linear factors with no constant term, and it therefore represents two lines intersecting at the origin. It is left as an exercise to determine the angles that the two lines make with each other and with the $$x$$ axis, and to show that the lines
$x^2 + \left( \frac{a-b}{h} \right) xy - y^2 = 0 \label{2.2.24} \tag{2.2.24}$
are the bisectors of $$\ref{2.2.23}$$ and are perpendicular to each other.
$$\text{FIGURE II.3}$$
Given the equations to three straight lines, can we find the area of the triangle bounded by them? To find a general algebraic expression might be a bit tedious, though the reader might like to try it, but a numerical example is straightforward. For example, consider the lines
$x - 5y + 12 = 0, \label{2.2.25} \tag{2.2.25}$
$3x + 4y - 9 = 0, \label{2.2.26} \tag{2.2.26}$
$3x - y - 3 = 0. \label{2.2.27} \tag{2.2.27}$
By solving the Equations in pairs, it is soon found that they intersect at the points (−0.15789, 2.36842), (1.4, 1.2) and (1.92857, 2.78571). Application of Equation $$\ref{2.2.12}$$ then gives the area as 1.544. The triangle is drawn in figure $$\text{II.4}$$. Measure any side and the corresponding height with a ruler and see if the area is indeed about 1.54.
But now consider the three lines
$x - 5y + 12 = 0, \label{2.2.28} \tag{2.2.28}$
$3x + 4y - 9 = 0 , \label{2.2.29} \tag{2.2.29}$
$3x + 23y - 54 = 0. \label{2.2.30} \tag{2.2.30}$
$$\text{FIGURE II.4}$$
By solving the equations in pairs, it will be found that all three lines intersect at the same point (please do this), and the area of the triangle is, of course, zero. Any one of these equations is, in fact, a linear combination of the other two. You should draw these three lines accurately on graph paper (or by computer). In general, if three lines are
$A_1 x + B_1 y + C_1 = 0 \label{2.2.31} \tag{2.2.31}$
$A_2 x + B_2y + C_2 = 0 \label{2.2.32} \tag{2.2.32}$
$A_3 x + B_3 y + C_3 = 0 \label{2.2.33} \tag{2.2.33}$
they will be concurrent at a single point if
\begin{array}{| c c c | c}
A_1 & B_1 & C_1 \\
A_2 & B_2 & C_2 & = 0. \\
A_3 & B_3 & C_3 \\
\label{2.2.34} \tag{2.2.34}
\end{array}
Thus the determinant in Equation $$\ref{2.2.12}$$ provides a test of whether three points are collinear, and the determinant in Equation $$\ref{2.2.34}$$ provides a test of whether three lines are concurrent.
Finally - at least for the present chapter - there may be rare occasion to write the equation of a straight line in polar coordinates. It should be evident from figure $$\text{II.5}$$ that the Equations
$r = p \csc (\theta - \alpha) \ \text{or} \ r = p \csc (\alpha - \theta) \label{2.2.35} \tag{2.2.35}$
describe a straight line passing at a distance $$p$$ from the pole and making an angle $$\alpha$$ with the initial line. If $$p = 0$$, the polar Equation is merely $$\theta = \alpha$$.
$$\text{FIGURE II.5}$$
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# State two points where the function y = 1/4cos(4pix)-3 has an instantaneous rate of change that is : i) Zero ii) A negative value iii) A positive values Do not use a derivative
We first begin by drawing the graph of the given function:
`f(x) (=y) =1/4*cos(4*pi*x)-3 `
The graph is below attached.
Then we remember what the average rate of change of a function `f(x) ` between `x_1 ` and `x_2 ` means
`(Delta(f(x)))/(Delta(x)) = (f(x_1)-f(x_2))/(x_1-x_2)` (1)
Graphically this is the slope of the line that connects the points `f(x_1) ` and `f(x_2) ` (line 1 in the figure)
For example the rate of change between `x_1 =-0.1 ` and `x_2 =+0.2 ` is positive since the slope of the line 1 in figure is positive.
Now, the instantaneous rate of change of function around point `x_1 ` is when `x_2 =x_1+h ` and `h ->0 `.
This graphically means that the line drawn between points `x_1 ` and `x_2=x_1+h ` in graph becomes tangent to the graph of function `f(x) ` at point `x_1 ` .
To find two points where the instantaneous rate of change of function is zero, we need to find two horizontal tangent lines to the graph. These are lines 2 and 3 in the figure. The corresponding values of `x ` are `x =0 ` and `x=0.25 `
Thus two points where the rate of the change of the given function is zero are `x =0 ` and `x =0.25 ` .
For positive values of rate of change the tangent lines need to have a positive slope (need to increase). These are lines 4 and 5 in the figures.
Thus two points where the rate of change of the given function is positive are `x=0.3 ` and `x=0.4 ` .
For negative values of rate of change the tangent lines need to have negative slope (need to decrease). These are the lines 6 and 7 in the figure.
Thus two points where the rate of change of the given function is negative are `x =-0.3 ` and `x=0.55 `
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# 关于 $y$ $=$ $x$ 对称的二元函数的二阶偏导数也关于 $y$ $=$ $x$ 对称
## 二、解析
\begin{aligned} & z(x, y) = \ x ^{2} + y^{2} – 3 x^{4} y^{4} \\ \\ \Rightarrow & \ \frac{\partial z}{\partial x} = 2x + 0 – 3 y^{4} \cdot 4 \cdot x ^{3} \\ \\ \Rightarrow & \ \textcolor{yellow}{ \frac{\partial z}{\partial x} = 2x – 12 y^{4} x ^{3} } \\ \\ \Rightarrow & \ \textcolor{springgreen}{ \frac{\partial ^{2} z}{\partial x ^{2}} = 2 – 36 y^{4} x ^{2} } \end{aligned}
\begin{aligned} z(\textcolor{orangered}{x}, \textcolor{springgreen}{y}) = & \ \textcolor{orangered}{x^{2}} + \textcolor{springgreen}{y^{2}} – 3 \textcolor{orangered}{x^{4}} \textcolor{springgreen}{y^{4}} \\ z(\textcolor{springgreen}{y}, \textcolor{orangered}{x}) = & \ \textcolor{springgreen}{y^{2}} + \textcolor{orangered}{x^{2}} – 3 \textcolor{springgreen}{y^{4}} \textcolor{orangered}{x^{4}} \\ = & \ \textcolor{orangered}{x^{2}} + \textcolor{springgreen}{y^{2}} – 3 \textcolor{orangered}{x^{4}} \textcolor{springgreen}{y^{4}} \end{aligned}
\begin{aligned} & \frac{\partial ^{2} z}{\partial \textcolor{orangered}{x} ^{2}} = 2 – 36 \textcolor{springgreen}{y^{4}} \textcolor{orangered}{x^{2}} \\ \\ \Rightarrow & \ \frac{\partial ^{2} z}{\partial \textcolor{springgreen}{y} ^{2}} = 2 – 36 \textcolor{orangered}{x^{4}} \textcolor{springgreen}{y^{2}} \end{aligned}
\begin{aligned} & z(x, y) = \ \frac{x^{2} + y^{2}}{xy} \\ \\ \Rightarrow & \ z(x, y) = \ \frac{x}{y} + \frac{y}{x} \\ \\ \Rightarrow & \ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \left( \frac{x}{y} + \frac{y}{x} \right) \\ \\ \Rightarrow & \ \frac{\partial z}{\partial x} = \frac{1}{y} + y \cdot \left( \frac{1}{x} \right)_{x} ^{\prime} \\ \\ \Rightarrow & \ \textcolor{yellow}{ \frac{\partial z}{\partial x} = \frac{1}{y} – \frac{y}{x^{2}} } \\ \\ \Rightarrow & \ \frac{\partial ^{2} z}{\partial x^{2}} = -y \cdot \left( \frac{1}{x ^{2}} \right)_{x}^{\prime} \\ \\ \Rightarrow & \ \textcolor{springgreen}{ \frac{\partial ^{2} z}{\partial x^{2}} = \frac{2y}{x^{3}} } \end{aligned}
\begin{aligned} z(\textcolor{orangered}{x}, \textcolor{springgreen}{y}) = \ & \frac{\textcolor{orangered}{x^{2}} + \textcolor{springgreen}{y^{2}}}{\textcolor{orangered}{x} \textcolor{springgreen}{y}} \\ \\ z(\textcolor{springgreen}{y}, \textcolor{orangered}{x}) = \ & \frac{\textcolor{springgreen}{y^{2}} + \textcolor{orangered}{x^{2}}}{\textcolor{springgreen}{y} \textcolor{orangered}{x}} \\ \\ = \ & \frac{\textcolor{orangered}{x^{2}} + \textcolor{springgreen}{y^{2}}}{\textcolor{orangered}{x} \textcolor{springgreen}{y}} \end{aligned}
\begin{aligned} & \frac{\partial ^{2} z}{\partial \textcolor{orangered}{x}^{2}} = \frac{2 \textcolor{springgreen}{y}}{\textcolor{orangered}{x^{3}}} \\ \\ \Rightarrow & \frac{\partial ^{2} z}{\partial \textcolor{springgreen}{y}^{2}} = \frac{2 \textcolor{orangered}{x}}{\textcolor{springgreen}{y^{3}}} \end{aligned}
$$\textcolor{springgreen}{ \begin{cases} \frac{\partial ^{2} z}{\partial x ^{2}} = 2 – 36 y^{4} x^{2} \\ \\ \frac{\partial ^{2} z}{\partial y^{2}} = 2 – 36 x^{4} y^{2} \end{cases} }$$
$$\textcolor{springgreen}{ \begin{cases} \frac{\partial ^{2} z}{\partial x^{2}} = \frac{2 y}{x^{3}} \\ \\ \frac{\partial ^{2} z}{\partial y^{2}} = \frac{2 x}{y^{3}} \end{cases} }$$
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# CAT Percentage Shortcut and Tricks
### Welcome to Day-3 of the Quantitative Aptitude, Logical Reasoning and Data Interpretation Prep Plan.
#### Focus Topic: A useful Trick/Shortcut for Percentage Questions
Today, we are going to learn one tip that is going to simplify life you and make sure you are able to use a shortcut that is going to drastically cut down the amount of time you spend on percentage problems.
One simple question for you: “What is the value of 71.4% of 119?”
Now what comes to your mind? Is this complex figure ‘71.4%’ blocking your mind?
But surprisingly, this is not a difficult figure if you know the fraction and percentage conversion table. While solving the questions on percentages, we deal with the complex fractions which can be converted into the simple fractions and then directly into the percentages. You do not require to go into the lengthy calculations.
Refer to the chart given above. If you check the table closely, you can see that 71.4% is nothing else but 5/7.
Therefore, our calculation simply reduces to the following:
71.4% of 119 = (5/7) x 119 = 85
Well, it is easier said than done, right? There is some hard work involved in this trick. You need to go through the above table and make sure you remember it. This will require you to put in some extra work but do remember that effort is going to be totally worth it. After all, this is going to improve your speed of calculations and make sure you are ready to handle complex calculations in the exam.
In today’s preparation plan, we focus on two key things. One, we have a comprehensive article and exercise for Percentages, Fractions and Ratios. Make sure you go through this resource and are fully prepared for this topic. Along with this, we have an article and an exercise that highlights the various types of problems that are based on the concepts of percentages. These question types are vital in terms of helping you understand the variety of questions types that are asked in various exams. Make sure you go through these and understand the different question types you can are going to encounter in competitive exams.
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+0
# Algebra
+1
109
3
These are the fees for sending a package by Media Mail: 2.41 for the first pound, 0.41 per additional pound for the next six pounds 0.39 per additional pound after that. For example, a two-pound package costs 2.82 to send by Media Mail. If a package full of books costs 9.94, to send by Media Mail, how many pounds does the package weigh?
Apr 2, 2021
#1
+484
+1
Because \$9.94>2.41+6(0.41)=4.87\$, we will need to use all the different options. That means that 9.94 can be written as 1(2.41)(first pound)+6(0.41)(for the additional 6 pounds)+k(0.39)(for the rest of the weight. Setting this equation equal to 9.94 gives:
\$1(2.41)+6(0.41)+k(0.39)=9.94\$
\$4.87+0.39k=9.94\$
\$0.39k=5.07\$
\$k=13\$
We're not quite done yet! k was only the number of pounds needed in the third option. We still need to take k and add in the 7 pounds that were the first and second options. So our final answer is:
\$k+7=(13)+7=\boxed{20}\$
Apr 2, 2021
#1
+484
+1
Because \$9.94>2.41+6(0.41)=4.87\$, we will need to use all the different options. That means that 9.94 can be written as 1(2.41)(first pound)+6(0.41)(for the additional 6 pounds)+k(0.39)(for the rest of the weight. Setting this equation equal to 9.94 gives:
\$1(2.41)+6(0.41)+k(0.39)=9.94\$
\$4.87+0.39k=9.94\$
\$0.39k=5.07\$
\$k=13\$
We're not quite done yet! k was only the number of pounds needed in the third option. We still need to take k and add in the 7 pounds that were the first and second options. So our final answer is:
\$k+7=(13)+7=\boxed{20}\$
RiemannIntegralzzz Apr 2, 2021
#2
+121004
0
Very nice solution , Riemann !!!!
CPhill Apr 2, 2021
#3
+484
0
Thank you @CPhill!
RiemannIntegralzzz Apr 2, 2021
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# How do you know when to use the ambiguous case when finding possible lengths of triangles?
Last updated date: 24th Jun 2024
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Hint: In this question, we have to find when to use the ambiguous case. As we know, an ambiguous case occurs when two sides and an angle opposite to one of them is not known to us, which means when we use the laws of sines to determine the missing angle of a triangle. Thus, to solve this problem, we will use four different cases with respect to the sides of the triangle, to get the required solution for the problem.
According to the problem, we have to find when to use ambiguous cases.
We will use an ambiguous case when the two sides of a triangle and an angle opposite to one of them, that is SSA is not known to us.
Thus, we will use different cases to solve this problem.
Case 1: When one of the sides of the triangle is less than the height of the triangle, that is
$a < h$ , where a is the side and h is the height of the triangle.
Thus, we will draw the figure for the same, we get
As we see, the above figure is not a closed figure, hence no triangle is formed for this case.
Case 2: When one of the sides of the triangle is equal to the height of the triangle, that is
$a = h$ , where a is the side and h is the height of the triangle.
Thus, we will draw the graph for the same, we get
As we see, the above figure is a closed figure, hence one triangle is formed for this case, that is a right-angle triangle.
Case 3: When one of the sides of the triangle is greater than the height but less than the other side of the triangle, that is
$h < a < c$ , where a, c is the side and h is the height of the triangle.
Thus, we will draw the graph for the same, we get
As we see, the above figure is a closed figure, hence we get two different triangles for this case, that is an acute triangle.
Case 4: When one of the sides of the triangle is greater than the height and is greater than the other side of the triangle, that is
$h < c \le a$ , where a, c is the side and h is the height of the triangle.
Thus, we will draw the graph for the same, we get
As we see, the above figure is a closed figure, hence we get one different triangle for this case.
Therefore, we see how to use ambiguous cases when finding possible lengths of triangles.
Note: While solving this problem, do not forget the definition of the ambiguous case of the triangle. Also, do mention all four cases, to get the accurate answer for the problem.
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# 2003 Indonesia MO Problems/Problem 6
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
A hall of a palace is in a shape of regular hexagon, where the sidelength is $6 \text{ m}$. The floor of the hall is covered with an equilateral triangle-shaped tile with sidelength $50 \text{ cm}$. Every tile is divided into $3$ congruent triangles (refer to the figure). Every triangle-region is colored with a certain color so that each tile has $3$ different colors. The King wants to ensure that no two tiles have the same color pattern. At least, how many colors are needed?
$[asy] pair C=(25,14.434); draw((0,0)--(50,0)--(25,43.301)--(0,0)); draw(C--(0,0)); draw(C--(50,0)); draw(C--(25,43.301)); [/asy]$
## Solution
First, count the number of tiles in the hall. The hall can be made up of six equilateral triangles with side length 600 cm. Since $600/50 = 12$ equilateral triangles with side length 50 cm can make up a side of the larger equilateral triangle, there are $12 + 2(11 + 10 + 9 + \cdots 1) = 144$ smaller triangles in each large triangle. In total, there are $144 \cdot 6 = 864$ tiles needed.
Let $x$ be the number of colors used. Given three colors, there are only two tiles using the color scheme that do not have the same color pattern. With this information, we can write an equation. \begin{align*} 2 \binom{x}{3} &\ge 864 \\ 2 \cdot \frac{x(x-1)(x-2)}{6} &\ge 864 \\ x(x-1)(x-2) &\ge 2592 \end{align*} With some trial and error, we find that the King needs $\boxed{15}$ different colors.
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# How do you simplify (2x^6^m)/(6x^2^m)?
Jul 1, 2015
${\left(\frac{{x}^{4}}{3}\right)}^{m} \mathmr{if} x \in \mathbb{R} - \left\{0\right\} , m \in \mathbb{R}$
#### Explanation:
Step 1 : The domain of the function.
We have only one forbidden value, when $x = 0$. This is the only value where your denominator equal 0. And we can't divide by 0...
Therefore, the domain of our function is : $\mathbb{R} - \left\{0\right\}$ for $x$ and $\mathbb{R}$ for $m$.
Step 2 : Factoring power m
$\frac{2 {x}^{6} ^ m}{6 {x}^{2} ^ m}$ <=> ${\left(2 {x}^{6}\right)}^{m} / {\left(6 {x}^{2}\right)}^{m}$ <=> ${\left(\frac{2 {x}^{6}}{6 {x}^{2}}\right)}^{m}$
Step 3 : Simplify the fraction
${\left(\frac{2 {x}^{6}}{6 {x}^{2}}\right)}^{m}$ <=> ${\left(\frac{{x}^{6}}{3 {x}^{2}}\right)}^{m}$ <=> ${\left(\frac{{x}^{4}}{3}\right)}^{m}$
Don't forget, $x \ne 0$
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Question Video: Solving Word Problems Involving Addition of Numbers up to 999 | Nagwa Question Video: Solving Word Problems Involving Addition of Numbers up to 999 | Nagwa
# Question Video: Solving Word Problems Involving Addition of Numbers up to 999 Mathematics
Liam read 75 pages of a book in one day, and in the next day he read 18 pages. How many pages did he read in the two days?
03:17
### Video Transcript
Liam read 75 pages of a book in one day, and on the next day he read 18 pages. How many pages did he read in the two days?
The two important facts that we need to highlight in this problem are the number of pages that Liam reads on two different days. We’re told that he reads 75 pages of a book on one day and 18 pages on another day. And, the question asks us to calculate the number of pages that Liam reads in the two days. In other words, we need to add the two together to find the total. We need to find the answer to 75 plus 18.
75 is a number that’s made up of seven 10s and five ones. And, 18 is made up of one 10 and eight ones. We can start by finding out how many ones we have altogether. 75 has five ones and we need to add another eight ones. Five, six, seven, eight, nine, 10, 11, 12, 13. Five plus eight equals 13. But, how can we write 13 ones in the ones place? We can only fit one digit in each place. What if we exchange 10 of our ones for one 10?
Here we go there. Five plus eight equals 13. Now, let’s add the number of 10s together. There are seven 10s in 75 plus one 10 in 18, and we’ve got one 10 that we’ve exchanged. So, we need to find the answer to seven plus one plus one. Seven plus one equals eight plus another one equals nine. Our answer contains nine 10s and three ones.
We found the answer by adding the ones first. We have five ones in 75 and eight ones in 18. Five plus eight equals 13. We decided that 13 was too large to write in the ones column, so we exchanged 10 of our 13 ones for one 10. Then, we added up all the 10s we had. 75 contains seven 10s, and 18 contains one 10. We also exchanged the 10 so we have to add that too. This made a total of nine 10s. If Liam reads 75 pages of book in one day, and the next day he reads 18 pages, in total he reads 93 pages over the two days.
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We included HMH Into Math Grade 5 Answer Key PDF Module 2 Review to make students experts in learning maths.
Vocabulary
compatible numbers
dividend
divisor
partial quotients
quotient
remainder
Choose the correct term from the Vocabulary box.
Question 1.
When estimating a quotient, it is best to use ___ that can be evenly divided.
partial quotients
Question 2.
The method of dividing by subtracting multiples of the divisor from the dividend uses ___
compatible numbers
Question 3.
The number that results from dividing is the ____
remainder
Concepts and Skills
Question 4.
Which of the following best describes the division equation represented by the area model?
A. 190 ÷ 19 = 10
B. 187 ÷ 17 = 11
C. 228 ÷ 19 = 12
D. 238 ÷ 17 = 14
C. 228 ÷ 19 = 12
Question 5.
Use Tools Select all the division equations for which 50 is a reasonable estimate. Tell what strategy or tool you will use to solve the problem, explain your choice, and then find the answer.
A. 1,386 ÷ 27 =
B. 1,164 ÷ 32 =
C. 2,859 ÷ 62 =
D. 4,193 ÷ 84 =
E. 382 ÷ 71 =
C. 2,859 ÷ 62 =
2,859 is near to 3,000
62 is near to 60
So, 2,859 ÷ 62 =
3,000 ÷ 60 = 50
Question 6.
Write a division equation that is related to the application of the Distributive Property shown.
18 × (200 + 30 + 6) = (18 × 200) + (18 × 30) + (18 × 6)
= 3,600 + 540 + 108
= 4,248
____ ÷ ____ = ____
20 × (20 + 3 + 6) = (20 × 20) + (20 × 3) + (20 × 6)
= 400 + 60 + 120
= 580
Use compatible numbers to find two estimates.
Question 7.
648 ÷ 44
648 is near 650
44 is near 50
650 ÷ 40 = 16.25 which is near 16
Therefore, 648 ÷ 44 approximately equals 16.
Question 8.
4,174 ÷ 21
4,174 is near 4200
21 is near 20
So, 4200 ÷ 20 = 210
Therefore, 4,174 ÷ 21 approximately equals 210.
Question 9.
A local company donates 1,935 ounces of granola to the food bank. Volunteers are making 85 smaller bags. What is a reasonable estimate for the number of ounces of granola in each smaller bag?
Given,
A local company donates 1,935 ounces of granola to the food bank.
Volunteers are making 85 smaller bags.
So, 1,935 ÷ 85 =
1935 approximately equals 2000
85 is near to 80
So, 2000 ÷ 80 = 25
1,935 ÷ 85 approximately equals 25
Therefore, the reasonable estimate for the number of ounces of granola in each smaller bag is 25.
Divide.
Question 10.
322 ÷ 14
Question 11.
591 ÷ 31
Question 12.
2,841 ÷ 76
Question 13.
1,718 ÷ 23
Question 14.
948 ÷ 40
Question 15.
4,500 ÷ 84
Question 16.
One mile is 5,280 feet long. A certain indoor gym has a running track for which 12 laps around the track is equivalent to 1 mile. Explain how you can use partial quotients to determine how many feet long the track is.
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### Wobbler
A cone is glued to a hemisphere. When you place it on a table in what position does it come to rest?
# Overarch 2
##### Stage: 5 Challenge Level:
Bricklayers build from the ground up, but to solve this problem you must start at the top! Alexander Maryanovsky sent in a correct solution.
Let's adopt a unit of length of half the length of a brick, and W as the weight of 1 brick.
We want to calculate the formula for the overhang $d_n$ of brick $n$ over brick $(n+1)$. (See the diagram). The idea is to imagine inserting brick $n$ under the existing pile of bricks 1, 2, 3...$(n-1)$ (numbering the bricks from the top) so that the centre of mass of bricks 1, 2, 3... $(n-1)$ will be exactly above the edge of brick $n$. You can treat the edge of brick $(n+1)$ as the fulcrum of a balance. The maximal arch will be created when there is a perfect balance on each edge.
If you calculate $d_1$, $d_2$,... you should find that \begin{eqnarray} d_1 &=& 1\\ d_2 &=& {1\over 2}\\ d_3 &=& {1\over 3}\\ \end{eqnarray} So it seems likely that $d_n = {1\over n}$. As the centre of mass of the stack of bricks above must be exactly over the edge of the next brick down, we take moments for the stack of $n$ bricks resting on the $(n+1)$st brick (counting the bricks from the top). We need the centre of mass of (n-1) bricks at distance $d_n$ from the fulcrum to balance 1 brick (the $n$th brick) at distance $(1 - d_n)$. Hence \begin{eqnarray} W(1 - d_n) &=& (n-1)Wd_n\\ d_n &=& {1\over n} \end{eqnarray} The total overhang $A_n$ for an arch containing $n + 1$ bricks is therefore $$A_n = \sum_1^n 1/n$$
Alexander went on to say...
After some asking around, I was told there is no closed form for that. A small computer program that looks like this:
double overhang = 0;
double brickWidth = 20;
double maxOverhang = 100;
double numBricks = 0;
while (overhang < maxOverhang){
numBricks = numBricks+1;
overhang = overhang+brickWidth/(2*numBricks);
}
returns 12367 as the number of bricks needed to make an overhang of 1m, making its height 12367*10cm = 1,236.7m Since 1+1/2+1/3+...+1/n doesn't converge, you can make the overhang as big as you wish (I'm not gonna go into counting how much material for those bricks is needed and if there's enough of it in the universe :-) )
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New Zealand
Level 6 - NCEA Level 1
# Probabilities of Games
Lesson
The key and the main difficulty in calculating probabilities to do with games is the enumeration of the outcome space.
We may attempt to do this by constructing a list, perhaps with the help of a tree diagram. This can be unwieldy, however, when there are several stages in the outcomes and when there are many possibilities for each of them. Counting the number of possible hands of some specified type in a game of cards is a typically difficult problem.
#### Examples
##### Example 1
We may be interested in the probability of a two-of-a-kind hand occurring when two cards are dealt from a standard deck of 52 cards. In this case, we can count the number of possible two-card hands without listing them all. We would argue that the first card can be any one of the 52, and for each of these there are 51 possibilities for the second card so that in all there are $52\times51=2652$52×51=2652 hands formed in this way. Now, for each of these there will be another that has the same cards but in the opposite order. The reversed hand is considered to be the same hand as the order does not matter. So, there are $\frac{2652}{2}=1326$26522=1326 distinct two-card hands.
Next, we must count the number of two-of-a-kind hands. As before, there are 52 possibilities for the first card. But, for each of these there are only 3 cards that would complete the pair. Again, the order in which the cards were dealt does not matter. So, only half the pairs counted in this way are distinct. Hence, the number of outcomes that would be counted as successes is $\frac{52\times3}{2}=78$52×32=78.
Finally, the probability of the event 'two-of-a-kind' is: $\frac{\text{number of successful outcomes}}{\text{number of outcomes in the sample space}}$number of successful outcomesnumber of outcomes in the sample space. This is, $\frac{78}{1326}=\frac{1}{17}$781326=117.
##### Example 2
Continuing from the first example, we might wonder about the situation in which two players are each dealt a two-card hand. What then would be the probability that both players receive two-of-a-kind?
This is a more difficult problem than in the previous case where just one hand was dealt because the cards that are dealt to one of the players affect what can be dealt to the other player. The events are not independent.
We now have a sample space of groups of four cards arranged as pairs of two-card hands. As before 78 of the two-card hands are pairs. We note that each card in the deck occurs in three different two-of-a-kind pairs. One player might be dealt any one of the 78 pairs but then there are only 75 pairs remaining for the other player. The order of the players does not matter, so there will be a division by two in the following calculation. The number of ways in which two-of-a-kind hands can be dealt to two players must be $\frac{78\times75}{2}=2925$78×752=2925
But, how big is the sample space? We need to count the number of ways four cards can be selected from 52 bearing in mind that the order of cards in each two-card hand does not matter and neither does the order of the two players. The number we seek is $\frac{52\times51\times50\times49}{2\times2\times2}=812175$52×51×50×492×2×2=812175.
Finally, the probability that both players are dealt a pair is:
$\frac{\text{number of successful outcomes}}{\text{number of outcomes in the sample space}}=\frac{2925}{812175}=\frac{3}{833}$number of successful outcomesnumber of outcomes in the sample space=2925812175=3833.
As a check, we might compare this probability with the probability of getting pairs twice in succession if the experiment in Example 1 were to be performed twice. This would be $\left(\frac{1}{17}\right)^2=\frac{1}{289}$(117)2=1289 or $\frac{3}{867}$3867 which is not very different from the value calculated for the non-independent case.
#### Worked Examples
##### QUESTION 1
A local raffle prize will be given to the person who has the winning number from one of the $200$200 tickets sold. What is the probability of winning the raffle if you purchase $30$30 tickets?
##### QUESTION 2
In a game of Monopoly, rolling a double means rolling the same number on both dice. When you roll a double this allows you to have another turn.
What is the probability that Sarah rolls:
1. a double $1$1?
2. a double $5$5?
3. any double?
4. two doubles in a row?
##### QUESTION 3
In a card game, Aaron is dealt a hand of two cards without replacement from a standard deck of $52$52 cards. How many different hands are possible?
### Outcomes
#### S6-3
Investigate situations that involve elements of chance: A comparing discrete theoretical distributions and experimental distributions, appreciating the role of sample size B calculating probabilities in discrete situations.
#### 91038
Investigate a situation involving elements of chance
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## Step by action solution for calculating 7 is 25 percent that what number
We already have our an initial value 7 and also the second value 25. Let"s i think the unknown value is Y i m sorry answer us will uncover out.
You are watching: 7 is what percent of 25
As we have all the forced values we need, currently we can put lock in a straightforward mathematical formula together below:
STEP 17 = 25% × Y
STEP 27 = 25/100× Y
Multiplying both political parties by 100 and dividing both political parties of the equation through 25 we will arrive at:
STEP 3Y = 7 × 100/25
STEP 4Y = 7 × 100 ÷ 25
STEP 5Y = 28
Finally, we have found the worth of Y which is 28 and that is our answer.
You can conveniently calculate 7 is 25 percent that what number through using any type of regular calculator, simply get in 7 × 100 ÷ 25 and you will get your answer i m sorry is 28
Here is a portion Calculator come solve comparable calculations such together 7 is 25 percent of what number. You have the right to solve this form of calculation with your worths by entering them into the calculator"s fields, and click "Calculate" to acquire the an outcome and explanation.
is
percent the what number
Calculate
## Sample questions, answers, and also how to
Question: your friend has actually a bag of marbles, and also he tells you the 25 percent the the marbles are red. If there space 7 red marbles. How numerous marbles go he have altogether?
How To: In this problem, we know that the Percent is 25, and we are likewise told that the component of the marbles is red, so we understand that the part is 7.
So, that means that it should be the total that"s missing. Here is the way to number out what the complete is:
Part/Total = Percent/100
By making use of a straightforward algebra we can re-arrange our Percent equation choose this:
Part × 100/Percent = Total
If we take the "Part" and also multiply it by 100, and then we divide that through the "Percent", us will gain the "Total".
Let"s shot it the end on our problem around the marbles, that"s very an easy and it"s just two steps! We understand that the "Part" (red marbles) is 7.
So action one is to just multiply that component by 100.
7 × 100 = 700
In step two, we take that 700 and divide it by the "Percent", i m sorry we space told is 25.
So, 700 split by 25 = 28
And that method that the total number of marbles is 28.
Question: A high school marching band has actually 7 flute players, If 25 percent of the band members pat the flute, climate how many members are in the band?
Answer: There are 28 members in the band.
How To: The smaller "Part" in this problem is 7 due to the fact that there space 7 flute players and also we space told that they comprise 25 percent the the band, so the "Percent" is 25.
Again, it"s the "Total" that"s lacking here, and to find it, we just need to follow our 2 step procedure together the previous problem.
For step one, us multiply the "Part" by 100.
7 × 100 = 700
For step two, we divide that 700 through the "Percent", i beg your pardon is 25.
See more: What Does Hmf Mean In Texting, What Is The Meaning Of Hmf In Chat/Text
700 separated by 25 amounts to 28
That method that the total number of band members is 28.
## Another step by step method
Step 1: Let"s i think the unknown worth is Y
Step 2: very first writing the as: 100% / Y = 25% / 7
Step 3: autumn the percent marks to leveling your calculations: 100 / Y = 25 / 7
Step 4: main point both sides by Y to move Y ~ above the best side the the equation: 100 = ( 25 / 7 ) Y
Step 5: simple the right side, us get: 100 = 25 Y
Step 6: separating both sides of the equation through 25, we will arrive at 28 = Y
This leaves us through our last answer: 7 is 25 percent that 28
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Chapter 3: Pair of Linear Equations in Two Variables
Q
Pair of linear equations in two variables CBSE NCERT Solutions
Question:
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14; x – y = 4
(ii) s – t = 3; $\frac{s}{3} + \frac{t}{2} = 6$
(iii) 3x – y = 3; 9x − 3y = 9
(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3
(v) $\sqrt{2}x+ \sqrt{3}y = 0$ $\sqrt{3}x - \sqrt{8}y = 0$
(vi) $\frac{3x}{2} - \frac{5y}{3} = -2$ ; $\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
(i) x + y = 14 …(1)
x – y = 4 … (2)
From equation (2), x – y = 4 x = 4 + y
Putting x = 4 + y in equation (1), we get
4 + y + y = 14
2y = 10 = 5
Putting value of y in equation (1), we get
x + 5 = 14
= 14 – 5 = 9
Therefore, x = 9 and y = 5
(ii) s – t = 3 … (1)
$\frac{s}{3} + \frac{t}{2} = 6$ …(2)
Using equation (1), we can say that s = 3 + t
Putting s = 3 + t in equation (2), we get
$\frac{3+t}{3} + \frac{t}{2} = 6$
$\Rightarrow \frac {6 + 2t + 3t }{6} = 6$
6 + 2t + 3t = 36
5t + 6 = 36
5t = 30 = 6
Putting value of t = 6 in equation (1), we get
– 6 = 3 = 3 + 6 = 9
Therefore, t = 6 and s = 9
(iii) 3x – y = 3 … (1)
9x − 3y = 9 … (2)
Comparing equation 3x – y = 3 with $a_1x + b_1y + c_1 = 0$ and equation 9x − 3y = 9 with $a_2x + b_2y + c_2 = 0$ .
We get $a_1 = 3 , b_1 = -1 , c_1 = -3,a_2 = 9 , b_2 = -3$ and $c_2 = -9$
Here,
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
Therefore, we have infinite many solutions for x and y
(iv) 0.2x + 0.3y = 1.3 … (1)
0.4x + 0.5y = 2.3 … (2)
Using equation (1), we can say that
0.2x = 1.3 − 0.3y
= $\frac{1.3-0.3y}{0.2}$
Putting this in equation (2), we get
$0.4 (\frac{1.3-0.3y}{0.2} )+ 0.5y = 2.3$
2.6 0.6y + 0.5y = 2.3
0.1y = −0.3 = 3
Putting value of y in (1), we get
0.2x + 0.3 (3) = 1.3
0.2x + 0.9 = 1.3
0.2x = 0.4 = 2
Therefore, x = 2 and y = 3
(v) $\sqrt{2}x+ \sqrt{3}y = 0$ ….(1)
$\sqrt{3}x - \sqrt{8}y = 0$ ….(2)
Using equation (1), we can say that
$x = \frac{-\sqrt{3}y}{\sqrt2}$
Putting this in equation (2), we get
$\sqrt{3} \left (\frac{-\sqrt{3}y}{\sqrt2} \right ) - \sqrt{8}y = 0$
$\frac{-{3}y}{\sqrt2} - \sqrt{8}y = 0$
$y\left ( \frac{-{3}}{\sqrt2} - \sqrt{8} = 0 \right )$
= 0
Putting the value of y in (1), we get x = 0
Therefore, x = 0 and y = 0
(vi) $\frac{3x}{2} - \frac{5y}{3} = -2$ … (1)
$\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$ … (2)
Using equation (2), we can say that
$x = \left ( \frac{13}{6} - \frac{y}{2}\right ) \times 3$
$\Rightarrow x = \frac{13}{2} - \frac{3y}{2}$
Putting this in equation (1), we get
$\frac{3}{2}\left ( \frac{13}{2} - \frac{3y}{2} \right ) - \frac{5y}{3} = \frac{-2}{1}$
$\frac{39}{4} - \frac{9y}{4} - \frac{5y}{3} = -2$
$\frac{-27y-20y}{12} = -2 - \frac{39}{4}$
$\frac{-47y}{12} = - \frac{-8-39}{4}$
$\frac{-47y}{12} = \frac{-47}{4}$
= 3
Putting the value of y in equation (2), we get
$\frac{x}{3} + \frac{3}{2}= \frac{13}{6}$
$\Rightarrow \frac{x}{3} = \frac{13}{6} - \frac{3}{2} = \frac{13-9}{6} = \frac{4}{6} = \frac{2}{3}$
$\Rightarrow \frac{x}{3} = \frac{2}{3}$
x = 2
Therefore, = 2 and = 3
VIDEO EXPLANATION
Related Questions for Study
CBSE Class 10 Study Material
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# What is the slope-intercept form of the line passing through (5, 1) and (0, -6) ?
Mar 3, 2018
The general slope intercept form of a line is
$y = m x + c$
where $m$ is the slope of the line and $c$ is its $y$-intercept (the point at which the line cuts the $y$ axis).
#### Explanation:
First, get all the terms of the equation. Let us calculate the slope.
$\text{slope} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
$= \frac{- 6 - 1}{0 - 5}$
$= \frac{7}{5}$
The $y$-intercept of the line is already given. It is $- 6$ since the $x$ coordinate of the line is zero when it intersects the $y$ axis.
$c = - 6$
Use the equation.
$y = \left(\frac{7}{5}\right) x - 6$
$y = 1.4 x + 6$
#### Explanation:
$P \equiv \left(5 , 1\right)$
$Q \equiv \left(0 , - 6\right)$
$m = \frac{- 6 - 1}{0 - 5} = - \frac{7}{-} 5$
$m = 1.4$
$c = 1 - 1.4 \times 5 = 1 - 7$
$c = 6$
$y = m x + c$
$y = 1.4 x + 6$
Mar 3, 2018
One answer is: $\left(y - 1\right) = \frac{7}{5} \left(x - 5\right)$
the other is: $\left(y + 6\right) = \frac{7}{5} \left(x - 0\right)$
#### Explanation:
The slope-intercept form of a line tells you what you need to find first: the slope.
Find slope using $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
where $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ are the given two points
$\left(5 , 1\right)$ and $\left(0 , - 6\right)$:
$m = \frac{- 6 - 1}{0 - 5} = \frac{- 7}{-} 5 = \frac{7}{5}$
You can see this is in both answers.
Now choose either point and plug in to the slope-intercept form of a line: $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$
Choosing the first point results in the first answer and choosing the second point yields the second answer. Also note that the second point is technically the y-intercept, so you could write the equation in slope-intercept form ($y = m x + b$): $y = \frac{7}{5} x - 6$.
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Factors of 522: Prime Factorization, Methods, and Examples
The numbers that, when divided from 522, leave zero as the remainder are considered to be its factors. Additionally, they generate a whole number quotient.
Given that the given number is obtained by multiplying integers with two factors, its factors may be both positive and negative.
Factors of 522
Here are the factors of number522.
Factors of 522:1, 2, 3, 6, 9, 18, 29, 58, 87, 174, 261 and 522
Negative Factors of 522
The negative factors of 522 are similar to its positive aspects, just with a negative sign.
Negative Factors of 522: -1, -2, -3, -6, -9, -18, -29, -58, -87, -174, -261 and -522.
Prime Factorization of 522
The prime factorization of522 is the way of expressing its prime factors in the product form.
Prime Factorization: 21 x 32 x 291
In this article, we will learn about the factors of 522 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
What Are the Factors of 522?
The factors of 522 are 1, 2, 3, 6, 9, 18, 29, 58, 87, 174, 261 and 522. These numbers are the factors as they do not leave any remainder when divided by 522.
The factors of 522 are classified as prime numbers and composite numbers. The prime factors of the number 522 can be determined using the prime factorization technique.
How To Find the Factors of 522?
You can find the factors of 522 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 522, create a list containing the numbers that are exactly divisible by 522 with zero remainders. One important thing to note is that 1 and 522 are the 522’s factors as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 522 are determined as follows:
$\dfrac{522}{1} = 522$
$\dfrac{522}{2} = 261$
$\dfrac{522}{3} = 174$
$\dfrac{522}{6} = 87$
$\dfrac{522}{9} = 58$
$\dfrac{522}{18} = 29$
$\dfrac{522}{29} = 18$
$\dfrac{522}{58} = 9$
$\dfrac{522}{87} = 6$
$\dfrac{522}{174} = 3$
$\dfrac{522}{261} = 2$
$\dfrac{522}{522} = 1$
Therefore, 1, 2, 3, 6, 9, 18, 29, 58, 87, 174, 261 and 522 are the factors of 522.
Total Number of Factors of 522
For 522, there are 12 positive factors and 12 negative ones. So in total, there are 24 factors of 522.
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 522 is given as:
Factorization of 522 is 21 x 32 x 29.
The exponent of 2 is 1, 3 is 2, and 29 is 1.
Adding 1 to each and multiplying them together results in m.
Therefore, the total number of factors of 522 is 24. 12 are positive, and 12 factors are negative.
Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
Factors of 522 by Prime Factorization
The number 522 is a composite number. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 522 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 522, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 522 can be expressed as:
522 = 21 x 32 x 291
Factors of 522 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 522, the factor pairs can be found as:
1 x 522 = 522
2 x 261 = 522
3 x 174 = 522
6 x 87 = 522
9 x 58 = 522
18 x 29 = 522
The possible factor pairs of 522 are given as (1, 522),(2, 261),(3, 174),(6, 87),(9, 58) and (18, 29 ).
All these numbers in pairs, when multiplied, give 522 as the product.
The negative factor pairs of 522 are given as:
-1 x -522 = 522
-2 x -261 = 522
-3 x -174 = 522
-6 x -87 = 522
-9 x -58 = 522
-18 x -29 = 522
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, -1, -2, -3, -6, -9, -18, -29, -58, -87, -174, -261 and -522. are called negative factors of 522.
The list of all the factors of 522, including positive as well as negative numbers, is given below.
Factor list of 522: 1,-1,2, -2,3, -3,6, -6,9, -9,18, -18,29, -29,58, -58, 87, -87, 174,-174,261, -261 and -522, 522.
Factors of 522 Solved Examples
To better understand the concept of factors, let’s solve some examples.
Example 1
How many factors of 522 are there?
Solution
The total number of Factors of 522 is 24.
Factors of 522 are 1, 2, 3, 6, 9, 18, 29, 58, 87, 174, 261 and 522.
Example 2
Find the factors of 522 using prime factorization.
Solution
The prime factorization of 522 is given as:
522 $\div$ 2 = 261
261 $\div$ 3 = 87
87 $\div$ 3 = 29
29 $\div$ 29 = 1
So the prime factorization of 522 can be written as:
21 x 32 x 291 = 522
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### Theory:
Basic properties of Exponents.
$\begin{array}{l}{a}^{n}\cdot {a}^{m}={a}^{n+m};\\ \\ {a}^{n}:{a}^{m}={a}^{n-m},\phantom{\rule{0.147em}{0ex}}n>m,\phantom{\rule{0.147em}{0ex}}a\ne 0;\\ \\ {\left({a}^{n}\right)}^{m}={a}^{n\cdot m}\end{array}$
Where n and $$m$$ are integers.
In practice, several properties are often applied simultaneously.
Example:
Calculate$\frac{{\left({5}^{4}\cdot 5\right)}^{3}}{{5}^{7}\cdot {5}^{6}}$
Perform actions in the numerator:
1.${5}^{4}\cdot 5={5}^{4}\cdot {5}^{1}={5}^{4+1}={5}^{5}$ — applied property ${a}^{n}\cdot {a}^{m}={a}^{n+m}$.
2. ${\left({5}^{5}\right)}^{3}={5}^{5\cdot 3}={5}^{15}$ — applied property ${\left({a}^{n}\right)}^{m}={a}^{n\cdot m}$.
Perform the multiplication in the denominator:
3. ${5}^{7}\cdot {5}^{6}={5}^{7+6}={5}^{13}$ — applied property ${a}^{n}\cdot {a}^{m}={a}^{n+m}$.
Replace the fraction line with division:
4.${5}^{15}:{5}^{13}={5}^{15-13}={5}^{2}$ — applied property ${a}^{n}:{a}^{m}={a}^{n-m}$.
5. ${5}^{2}=25$.
Answer: $$25$$.
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Intermediate Algebra 2e
# 1.2Integers
## Learning Objectives
By the end of this section, you will be able to:
• Simplify expressions with absolute value
• Multiply and divide integers
• Simplify expressions with integers
• Evaluate variable expressions with integers
• Translate phrases to expressions with integers
• Use integers in applications
## Be Prepared 1.2
A more thorough introduction to the topics covered in this section can be found in the Elementary Algebra 2e chapter, Foundations.
## Simplify Expressions with Absolute Value
A negative number is a number less than 0. The negative numbers are to the left of zero on the number line. See Figure 1.2.
Figure 1.2 The number line shows the location of positive and negative numbers.
You may have noticed that, on the number line, the negative numbers are a mirror image of the positive numbers, with zero in the middle. Because the numbers $22$ and $−2−2$ are the same distance from zero, each one is called the opposite of the other. The opposite of $22$ is $−2,−2,$ and the opposite of $−2−2$ is $2.2.$
## Opposite
The opposite of a number is the number that is the same distance from zero on the number line but on the opposite side of zero.
Figure 1.3 illustrates the definition.
Figure 1.3 The opposite of 3 is $−3−3$.
## Opposite Notation
$−ameans the opposite of the numbera The notation−ais read as “the opposite ofa.”−ameans the opposite of the numbera The notation−ais read as “the opposite ofa.”$
We saw that numbers such as 3 and $−3−3$ are opposites because they are the same distance from 0 on the number line. They are both three units from 0. The distance between 0 and any number on the number line is called the absolute value of that number.
## Absolute Value
The absolute value of a number is its distance from 0 on the number line.
The absolute value of a number $nn$ is written as $|n||n|$ and $|n|≥0|n|≥0$ for all numbers.
Absolute values are always greater than or equal to zero.
For example,
$−5is 5 units away from 0, so|−5|=5. 5 is 5 units away from 0, so|5|=5.−5is 5 units away from 0, so|−5|=5. 5 is 5 units away from 0, so|5|=5.$
Figure 1.4 illustrates this idea.
Figure 1.4 The numbers 5 and $−5−5$ are 5 units away from 0.
The absolute value of a number is never negative because distance cannot be negative. The only number with absolute value equal to zero is the number zero itself because the distance from 0 to 0 on the number line is zero units.
In the next example, we’ll order expressions with absolute values.
## Example 1.12
Fill in $<,>,<,>,$ or $==$ for each of the following pairs of numbers:
$|−5|__−|−5||−5|__−|−5|$ $8__−|−8|8__−|−8|$ $−9__−|−9|−9__−|−9|$ $−(−16)__|−16|.−(−16)__|−16|.$
## Try It 1.23
Fill in $<,>,<,>,$ or $==$ for each of the following pairs of numbers:
$−9__−|−9|−9__−|−9|$ $2__−|−2|2__−|−2|$ $−8__|−8|−8__|−8|$ $−(−9)__|−9|.−(−9)__|−9|.$
## Try It 1.24
Fill in $<,>,<,>,$ or $==$ for each of the following pairs of numbers:
$7__−|−7|7__−|−7|$ $−(−10)__|−10|−(−10)__|−10|$ $|−4|__−|−4||−4|__−|−4|$ $−1__|−1|.−1__|−1|.$
We now add absolute value bars to our list of grouping symbols. When we use the order of operations, first we simplify inside the absolute value bars as much as possible, then we take the absolute value of the resulting number.
## Grouping Symbols
$Parentheses()Braces{} Brackets[]Absolute value||Parentheses()Braces{} Brackets[]Absolute value||$
In the next example, we simplify the expressions inside absolute value bars first just as we do with parentheses.
## Example 1.13
Simplify: $24−|19−3(6−2)|.24−|19−3(6−2)|.$
## Try It 1.25
Simplify:$19−|11−4(3−1)|.19−|11−4(3−1)|.$
## Try It 1.26
Simplify: $9−|8−4(7−5)|.9−|8−4(7−5)|.$
So far, we have only used the counting numbers and the whole numbers.
$Counting numbers1,2,3… Whole numbers0,1,2,3….Counting numbers1,2,3… Whole numbers0,1,2,3….$
Our work with opposites gives us a way to define the integers. The whole numbers and their opposites are called the integers. The integers are the numbers $…−3,−2,−1,0,1,2,3……−3,−2,−1,0,1,2,3…$
## Integers
The whole numbers and their opposites are called the integers.
The integers are the numbers
$…−3,−2,−1,0,1,2,3…,…−3,−2,−1,0,1,2,3…,$
Most students are comfortable with the addition and subtraction facts for positive numbers. But doing addition or subtraction with both positive and negative numbers may be more challenging.
We will use two color counters to model addition and subtraction of negatives so that you can visualize the procedures instead of memorizing the rules.
We let one color (blue) represent positive. The other color (red) will represent the negatives.
If we have one positive counter and one negative counter, the value of the pair is zero. They form a neutral pair. The value of this neutral pair is zero.
We will use the counters to show how to add:
$5+3−5+(−3)−5+35+(−3)5+3−5+(−3)−5+35+(−3)$
The first example, $5+3, 5+3,$ adds 5 positives and 3 positives—both positives.
The second example, $−5+(−3),−5+(−3),$ adds 5 negatives and 3 negatives—both negatives.
When the signs are the same, the counters are all the same color, and so we add them. In each case we get 8—either 8 positives or 8 negatives.
So what happens when the signs are different? Let’s add $−5+3−5+3$ and $5+(−3).5+(−3).$
When we use counters to model addition of positive and negative integers, it is easy to see whether there are more positive or more negative counters. So we know whether the sum will be positive or negative.
## Example 1.14
Add: $−1+(−4)−1+(−4)$ $−1+5−1+5$ $1+(−5).1+(−5).$
## Try It 1.27
Add: $−2+(−4)−2+(−4)$ $−2+4−2+4$ $2+(−4).2+(−4).$
## Try It 1.28
Add: $−2+(−5)−2+(−5)$ $−2+5−2+5$ $2+(−5).2+(−5).$
We will continue to use counters to model the subtraction. Perhaps when you were younger, you read $“5−3”“5−3”$ as “5 take away 3.” When you use counters, you can think of subtraction the same way!
We will use the counters to show to subtract:
$5−3 −5−(−3) −5−3 5−(−3) 5−3 −5−(−3) −5−3 5−(−3)$
The first example, $5−3,5−3,$ we subtract 3 positives from 5 positives and end up with 2 positives.
In the second example, $−5−(−3),−5−(−3),$ we subtract 3 negatives from 5 negatives and end up with 2 negatives.
Each example used counters of only one color, and the “take away” model of subtraction was easy to apply.
What happens when we have to subtract one positive and one negative number? We’ll need to use both blue and red counters as well as some neutral pairs. If we don’t have the number of counters needed to take away, we add neutral pairs. Adding a neutral pair does not change the value. It is like changing quarters to nickels—the value is the same, but it looks different.
Let’s look at $−5−3−5−3$ and $5−(−3).5−(−3).$
Model the first number. We now add the needed neutral pairs. We remove the number of counters modeled by the second number. Count what is left.
## Example 1.15
Subtract: $3−13−1$ $−3−(−1)−3−(−1)$ $−3−1−3−1$ $3−(−1).3−(−1).$
## Try It 1.29
Subtract: $6−46−4$ $−6−(−4)−6−(−4)$ $−6−4−6−4$ $6−(−4).6−(−4).$
## Try It 1.30
Subtract: $7−47−4$ $−7−(−4)−7−(−4)$ $−7−4−7−4$ $7−(−4).7−(−4).$
Have you noticed that subtraction of signed numbers can be done by adding the opposite? In the last example, $−3−1−3−1$ is the same as $−3+(−1)−3+(−1)$ and $3−(−1)3−(−1)$ is the same as $3+1.3+1.$ You will often see this idea, the Subtraction Property, written as follows:
## Subtraction Property
$a−b=a+(−b)a−b=a+(−b)$
Subtracting a number is the same as adding its opposite.
## Example 1.16
Simplify: $13−813−8$ and $13+(−8)13+(−8)$ $−17−9−17−9$ and $−17+(−9)−17+(−9)$ $9−(−15)9−(−15)$ and $9+159+15$ $−7−(−4)−7−(−4)$ and $−7+4.−7+4.$
## Try It 1.31
Simplify: $21−1321−13$ and $21+(−13)21+(−13)$ $−11−7−11−7$ and $−11+(−7)−11+(−7)$ $6−(−13)6−(−13)$ and $6+136+13$ $−5−(−1)−5−(−1)$ and $−5+1.−5+1.$
## Try It 1.32
Simplify: $15−715−7$ and $15+(−7)15+(−7)$ $−14−8−14−8$ and $−14+(−8)−14+(−8)$ $4−(−19)4−(−19)$ and $4+194+19$ $−4−(−7)−4−(−7)$ and $−4+7.−4+7.$
What happens when there are more than three integers? We just use the order of operations as usual.
## Example 1.17
Simplify: $7−(−4−3)−9.7−(−4−3)−9.$
## Try It 1.33
Simplify: $8−(−3−1)−9.8−(−3−1)−9.$
## Try It 1.34
Simplify: $12−(−9−6)−14.12−(−9−6)−14.$
## Multiply and Divide Integers
Since multiplication is mathematical shorthand for repeated addition, our model can easily be applied to show multiplication of integers. Let’s look at this concrete model to see what patterns we notice. We will use the same examples that we used for addition and subtraction. Here, we are using the model just to help us discover the pattern.
We remember that $a·ba·b$ means add a, b times.
The next two examples are more interesting. What does it mean to multiply 5 by $−3?−3?$ It means subtract $5,35,3$ times. Looking at subtraction as “taking away”, it means to take away 5, 3 times. But there is nothing to take away, so we start by adding neutral pairs on the workspace.
In summary:
$5·3=15−5(3)=−15 5(−3)=−15(−5)(−3)=155·3=15−5(3)=−15 5(−3)=−15(−5)(−3)=15$
Notice that for multiplication of two signed numbers, when the
$signs are thesame, the product ispositive. signs aredifferent, the product isnegative.signs are thesame, the product ispositive. signs aredifferent, the product isnegative.$
What about division? Division is the inverse operation of multiplication. So, $15÷3=515÷3=5$ because $5·3=15.5·3=15.$ In words, this expression says that 15 can be divided into 3 groups of 5 each because adding five three times gives 15. If you look at some examples of multiplying integers, you might figure out the rules for dividing integers.
$5·3=15so15÷3=5−5(3)=−15so−15÷3=−5 (−5)(−3)=15so15÷(−3)=−5 5(−3)=−15so−15÷(−3)=55·3=15so15÷3=5−5(3)=−15so−15÷3=−5 (−5)(−3)=15so15÷(−3)=−5 5(−3)=−15so−15÷(−3)=5$
Division follows the same rules as multiplication with regard to signs.
## Multiplication and Division of Signed Numbers
For multiplication and division of two signed numbers:
Same signs Result
• Two positives Positive
• Two negatives Positive
If the signs are the same, the result is positive.
Different signs Result
• Positive and negative Negative
• Negative and positive Negative
If the signs are different, the result is negative.
## Example 1.18
Multiply or divide: $−100÷(−4)−100÷(−4)$ $7·67·6$ $4(−8)4(−8)$ $−27÷3.−27÷3.$
## Try It 1.35
Multiply or divide: $−115÷(−5)−115÷(−5)$ $5·125·12$ $9(−7)9(−7)$ $−63÷7.−63÷7.$
## Try It 1.36
Multiply or divide: $−117÷(−3)−117÷(−3)$ $3·133·13$ $7(−4)7(−4)$ $−42÷6.−42÷6.$
When we multiply a number by 1, the result is the same number. Each time we multiply a number by $−1,−1,$ we get its opposite!
## Multiplication by $−1 −1$
$−1a=−a−1a=−a$
Multiplying a number by $−1−1$ gives its opposite.
## Simplify Expressions with Integers
What happens when there are more than two numbers in an expression? The order of operations still applies when negatives are included. Remember Please Excuse My Dear Aunt Sally?
Let’s try some examples. We’ll simplify expressions that use all four operations with integers—addition, subtraction, multiplication, and division. Remember to follow the order of operations.
## Example 1.19
Simplify: $(−2)4(−2)4$ $−24.−24.$
## Try It 1.37
Simplify: $(−3)4(−3)4$ $−34.−34.$
## Try It 1.38
Simplify: $(−7)2(−7)2$ $−72.−72.$
The last example showed us the difference between $(−2)4(−2)4$ and $−24.−24.$ This distinction is important to prevent future errors. The next example reminds us to multiply and divide in order left to right.
## Example 1.20
Simplify: $8(−9)÷(−2)38(−9)÷(−2)3$ $−30÷2+(−3)(−7).−30÷2+(−3)(−7).$
## Try It 1.39
Simplify: $12(−9)÷(−3)312(−9)÷(−3)3$ $−27÷3+(−5)(−6).−27÷3+(−5)(−6).$
## Try It 1.40
Simplify: $18(−4)÷(−2)318(−4)÷(−2)3$ $−32÷4+(−2)(−7).−32÷4+(−2)(−7).$
## Evaluate Variable Expressions with Integers
Remember that to evaluate an expression means to substitute a number for the variable in the expression. Now we can use negative numbers as well as positive numbers.
## Example 1.21
Evaluate $4x2−2xy+3y24x2−2xy+3y2$ when $x=2,y=−1.x=2,y=−1.$
## Try It 1.41
Evaluate: $3x2−2xy+6y23x2−2xy+6y2$ when $x=1,y=−2.x=1,y=−2.$
## Try It 1.42
Evaluate: $4x2−xy+5y24x2−xy+5y2$ when $x=−2,y=3.x=−2,y=3.$
## Translate Phrases to Expressions with Integers
Our earlier work translating English to algebra also applies to phrases that include both positive and negative numbers.
## Example 1.22
Translate and simplify: the sum of 8 and $−12,−12,$ increased by $3.3.$
## Try It 1.43
Translate and simplify the sum of 9 and $−16,−16,$ increased by 4.
## Try It 1.44
Translate and simplify the sum of $−8−8$ and $−12,−12,$ increased by 7.
## Use Integers in Applications
We’ll outline a plan to solve applications. It’s hard to find something if we don’t know what we’re looking for or what to call it! So when we solve an application, we first need to determine what the problem is asking us to find. Then we’ll write a phrase that gives the information to find it. We’ll translate the phrase into an expression and then simplify the expression to get the answer. Finally, we summarize the answer in a sentence to make sure it makes sense.
## Example 1.23
### How to Solve Application Problems Using Integers
In the morning, the temperature in Kendallville, Indiana was 11 degrees. By mid-afternoon, the temperature had dropped to $−9−9$ degrees. What was the difference in the morning and afternoon temperatures?
## Try It 1.45
In the morning, the temperature in Anchorage, Alaska was $1515$ degrees. By mid-afternoon the temperature had dropped to 30 degrees below zero. What was the difference in the morning and afternoon temperatures?
## Try It 1.46
The temperature in Denver was $−6−6$ degrees at lunchtime. By sunset the temperature had dropped to $−15−15$ degrees. What was the difference in the lunchtime and sunset temperatures?
## How To
### Use Integers in Applications.
1. Step 1. Read the problem. Make sure all the words and ideas are understood.
2. Step 2. Identify what we are asked to find.
3. Step 3. Write a phrase that gives the information to find it.
4. Step 4. Translate the phrase to an expression.
5. Step 5. Simplify the expression.
6. Step 6. Answer the question with a complete sentence.
## Media
Access this online resource for additional instruction and practice with integers.
## Section 1.2 Exercises
### Practice Makes Perfect
Simplify Expressions with Absolute Value
In the following exercises, fill in $<,>,<,>,$ or $==$ for each of the following pairs of numbers.
59.
$|−7|___−|−7||−7|___−|−7|$
$6___−|−6|6___−|−6|$
$|−11|___−11|−11|___−11$
$−(−13)___−|−13|−(−13)___−|−13|$
60.
$−|−9|___|−9|−|−9|___|−9|$
$−8___|−8|−8___|−8|$
$|−1|___−1|−1|___−1$
$−(−14)___−|−14|−(−14)___−|−14|$
61.
$−|2|___−|−2|−|2|___−|−2|$
$−12___−|−12|−12___−|−12|$
$|−3|___−3|−3|___−3$
$|−19|___−(−19)|−19|___−(−19)$
62.
$−|−4|___−|4|−|−4|___−|4|$
$5___−|−5|5___−|−5|$
$−|−10|___−10−|−10|___−10$
$−|−0|___−(−0)−|−0|___−(−0)$
In the following exercises, simplify.
63.
$| 15 − 7 | − | 14 − 6 | | 15 − 7 | − | 14 − 6 |$
64.
$| 17 − 8 | − | 13 − 4 | | 17 − 8 | − | 13 − 4 |$
65.
$18 − | 2 ( 8 − 3 ) | 18 − | 2 ( 8 − 3 ) |$
66.
$15 − | 3 ( 8 − 5 ) | 15 − | 3 ( 8 − 5 ) |$
67.
$18 − | 12 − 4 ( 4 − 1 ) + 3 | 18 − | 12 − 4 ( 4 − 1 ) + 3 |$
68.
$27 − | 19 + 4 ( 3 − 1 ) − 7 | 27 − | 19 + 4 ( 3 − 1 ) − 7 |$
69.
$10 − 3 | 9 − 3 ( 3 − 1 ) | 10 − 3 | 9 − 3 ( 3 − 1 ) |$
70.
$13 − 2 | 11 − 2 ( 5 − 2 ) | 13 − 2 | 11 − 2 ( 5 − 2 ) |$
In the following exercises, simplify each expression.
71.
$−7+(−4)−7+(−4)$
$−7+4−7+4$
$7+(−4).7+(−4).$
72.
$−5+(−9)−5+(−9)$
$−5+9−5+9$
$5+(−9)5+(−9)$
73.
$48 + ( −16 ) 48 + ( −16 )$
74.
$34 + ( −19 ) 34 + ( −19 )$
75.
$−14 + ( −12 ) + 4 −14 + ( −12 ) + 4$
76.
$−17 + ( −18 ) + 6 −17 + ( −18 ) + 6$
77.
$19 + 2 ( −3 + 8 ) 19 + 2 ( −3 + 8 )$
78.
$24 + 3 ( −5 + 9 ) 24 + 3 ( −5 + 9 )$
79.
$13−713−7$
$−13−(−7)−13−(−7)$
$−13−7−13−7$
$13−(−7)13−(−7)$
80.
$15−815−8$
$−15−(−8)−15−(−8)$
$−15−8−15−8$
$15−(−8)15−(−8)$
81.
$−17 − 42 −17 − 42$
82.
$−58 − ( −67 ) −58 − ( −67 )$
83.
$−14 − ( −27 ) + 9 −14 − ( −27 ) + 9$
84.
$64 + ( −17 ) − 9 64 + ( −17 ) − 9$
85.
$44−2844−28$ $44+(−28)44+(−28)$
86.
$35−1635−16$ $35+(−16)35+(−16)$
87.
$27−(−18)27−(−18)$ $27+1827+18$
88.
$46−(−37)46−(−37)$ $46+3746+37$
89.
$( 2 − 7 ) − ( 3 − 8 ) ( 2 − 7 ) − ( 3 − 8 )$
90.
$( 1 − 8 ) − ( 2 − 9 ) ( 1 − 8 ) − ( 2 − 9 )$
91.
$− ( 6 − 8 ) − ( 2 − 4 ) − ( 6 − 8 ) − ( 2 − 4 )$
92.
$− ( 4 − 5 ) − ( 7 − 8 ) − ( 4 − 5 ) − ( 7 − 8 )$
93.
$25 − [ 10 − ( 3 − 12 ) ] 25 − [ 10 − ( 3 − 12 ) ]$
94.
$32 − [ 5 − ( 15 − 20 ) ] 32 − [ 5 − ( 15 − 20 ) ]$
Multiply and Divide Integers
In the following exercises, multiply or divide.
95.
$−4·8−4·8$
$13(−5)13(−5)$
$−24÷6−24÷6$
$−52÷(−4)−52÷(−4)$
96.
$−3·9−3·9$
$9(−7)9(−7)$
$35÷(−7)35÷(−7)$
$−84÷(−6)−84÷(−6)$
97.
$−28÷7−28÷7$
$−180÷15−180÷15$
$3(−13)3(−13)$
$−1(−14)−1(−14)$
98.
$−36÷4−36÷4$
$−192÷12−192÷12$
$9(−7)9(−7)$
$−1(−19)−1(−19)$
Simplify and Evaluate Expressions with Integers
In the following exercises, simplify each expression.
99.
$(−2)6(−2)6$ $−26−26$
100.
$(−3)5(−3)5$ $−35−35$
101.
$5 ( −6 ) + 7 ( −2 ) − 3 5 ( −6 ) + 7 ( −2 ) − 3$
102.
$8 ( −4 ) + 5 ( −4 ) − 6 8 ( −4 ) + 5 ( −4 ) − 6$
103.
$−3 ( −5 ) ( 6 ) −3 ( −5 ) ( 6 )$
104.
$−4 ( −6 ) ( 3 ) −4 ( −6 ) ( 3 )$
105.
$( 8 − 11 ) ( 9 − 12 ) ( 8 − 11 ) ( 9 − 12 )$
106.
$( 6 − 11 ) ( 8 − 13 ) ( 6 − 11 ) ( 8 − 13 )$
107.
$26 − 3 ( 2 − 7 ) 26 − 3 ( 2 − 7 )$
108.
$23 − 2 ( 4 − 6 ) 23 − 2 ( 4 − 6 )$
109.
$65 ÷ ( −5 ) + ( −28 ) ÷ ( −7 ) 65 ÷ ( −5 ) + ( −28 ) ÷ ( −7 )$
110.
$52 ÷ ( −4 ) + ( −32 ) ÷ ( −8 ) 52 ÷ ( −4 ) + ( −32 ) ÷ ( −8 )$
111.
$9 − 2 [ 3 − 8 ( −2 ) ] 9 − 2 [ 3 − 8 ( −2 ) ]$
112.
$11 − 3 [ 7 − 4 ( −2 ) ] 11 − 3 [ 7 − 4 ( −2 ) ]$
113.
$8 − | 2 − 4 ( 4 − 1 ) + 3 | 8 − | 2 − 4 ( 4 − 1 ) + 3 |$
114.
$7 − | 5 − 3 ( 4 − 1 ) − 6 | 7 − | 5 − 3 ( 4 − 1 ) − 6 |$
115.
$9 − 3 | 2 ( 2 − 6 ) − ( 3 − 7 ) | 9 − 3 | 2 ( 2 − 6 ) − ( 3 − 7 ) |$
116.
$5 − 2 | 2 ( 1 − 4 ) − ( 2 − 5 ) | 5 − 2 | 2 ( 1 − 4 ) − ( 2 − 5 ) |$
117.
$( −3 ) 2 − 24 ÷ ( 8 − 2 ) ( −3 ) 2 − 24 ÷ ( 8 − 2 )$
118.
$( −4 ) 2 − 32 ÷ ( 12 − 4 ) ( −4 ) 2 − 32 ÷ ( 12 − 4 )$
In the following exercises, evaluate each expression.
119.
$y+(−14)y+(−14)$ when
$y=−33y=−33$ $y=30y=30$
120.
$x+(−21)x+(−21)$ when
$x=−27x=−27$ $x=44x=44$
121.
$(x+y)2(x+y)2$ when
$x=−3,y=14x=−3,y=14$
122.
$(y+z)2(y+z)2$ when
$y=−3,z=15y=−3,z=15$
123.
$9a−2b−89a−2b−8$ when
$a=−6a=−6$ and $b=−3b=−3$
124.
$7m−4n−27m−4n−2$ when
$m=−4m=−4$ and $n=−9n=−9$
125.
$3x2−4xy+2y23x2−4xy+2y2$ when
$x=−2,y=−3x=−2,y=−3$
126.
$4x2−xy+3y24x2−xy+3y2$ when
$x=−3,y=−2x=−3,y=−2$
Translate English Phrases to Algebraic Expressions
In the following exercises, translate to an algebraic expression and simplify if possible.
127.
the sum of 3 and $−15,−15,$ increased by 7
128.
the sum of $−8−8$ and $−9,−9,$ increased by $2323$
129.
the difference of $1010$ and $−18−18$
subtract $1111$ from $−25−25$
130.
the difference of $−5−5$ and $−30−30$
subtract $−6−6$ from $−13−13$
131.
the quotient of $−6−6$ and the sum of $aa$ and $bb$
132.
the product of $−13−13$ and the difference of $cc$ and $dd$
Use Integers in Applications
In the following exercises, solve.
133.
Temperature On January 15, the high temperature in Anaheim, California, was 84°. That same day, the high temperature in Embarrass, Minnesota, was $−12°.−12°.$ What was the difference between the temperature in Anaheim and the temperature in Embarrass?
134.
Temperature On January 21, the high temperature in Palm Springs, California, was $89°,89°,$ and the high temperature in Whitefield, New Hampshire, was $−31°.−31°.$ What was the difference between the temperature in Palm Springs and the temperature in Whitefield?
135.
Football On the first down, the Chargers had the ball on their 25-yard line. They lost 6 yards on the first-down play, gained 10 yards on the second-down play, and lost 8 yards on the third-down play. What was the yard line at the end of the third-down play?
136.
Football On first down, the Steelers had the ball on their 30-yard line. They gained 9 yards on the first-down play, lost 14 yards on the second-down play, and lost 2 yards on the third-down play. What was the yard line at the end of the third-down play?
137.
Checking Account Mayra has $124 in her checking account. She writes a check for$152. What is the new balance in her checking account?
138.
Checking Account Reymonte has a balance of $−49−49$ in his checking account. He deposits \$281 to the account. What is the new balance?
### Writing Exercises
139.
Explain why the sum of $−8−8$ and 2 is negative, but the sum of 8 and $−2−2$ is positive.
140.
Give an example from your life experience of adding two negative numbers.
141.
In your own words, state the rules for multiplying and dividing integers.
142.
Why is $−43=(−4)3?−43=(−4)3?$
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After reviewing this checklist, what will you do to become confident for all objectives?
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# 11.4: Multiplication and Division of Radicals
Difficulty Level: Basic Created by: CK-12
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Practice Multiplication and Division of Radicals
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What if you knew that the area of a rectangular mirror is 126\begin{align*}12 \sqrt{6}\end{align*} square feet and that the width of the mirror is 22\begin{align*}2 \sqrt{2}\end{align*} feet? Could you find the length of the mirror? What operation would you have to perform? If you knew the width and the length of the mirror, could you find its area? What operation would you perform in this case? In this Concept, you'll learn about multiplying and dividing radicals so that you can answer questions like these.
### Guidance
To multiply radicands, the roots must be the same.
anbn=abn\begin{align*}\sqrt[n]{a} \cdot \sqrt[n]{b}= \sqrt[n]{ab}\end{align*}
#### Example A
Simplify 312\begin{align*}\sqrt{3} \cdot \sqrt{12}\end{align*}.
Solution:
312=36=6\begin{align*}\sqrt{3} \cdot \sqrt{12}=\sqrt{36}=6\end{align*}
Dividing radicals is more complicated. A radical in the denominator of a fraction is not considered simplified by mathematicians. In order to simplify the fraction, you must rationalize the denominator.
To rationalize the denominator means to remove any radical signs from the denominator of the fraction using multiplication.
Remember: a×a=a2=a\begin{align*}\sqrt{a} \times \sqrt{a}= \sqrt{a^2}=a\end{align*}
#### Example B
Simplify 23\begin{align*}\frac{2}{\sqrt{3}}\end{align*}.
Solution:
We must clear the denominator of its radical using the property above. Remember, what you do to one piece of a fraction, you must do to all pieces of the fraction.
23×33=2332=233\begin{align*}\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3^2}}=\frac{2\sqrt{3}}{3}\end{align*}
#### Example C
A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square-feet. Find the dimensions of the pool and the area of the pool.
Solution:
1. Make a sketch.
2. Let \begin{align*}x=\end{align*} the width of the pool.
3. Write an equation. \begin{align*}Area=length \cdot width\end{align*}
Combined length of pool and walkway \begin{align*}=2x+2\end{align*}
Combined width of pool and walkway \begin{align*}=x+2\end{align*}
\begin{align*}\text{Area}=(2x+2)(x+2)\end{align*}
Since the combined area of the pool and walkway is \begin{align*}400 \ ft^2\end{align*}, we can write the equation.
\begin{align*}(2x+2)(x+2)=400\end{align*}
4. Solve the equation:
\begin{align*}&& & (2x+2)(x+2)=400\\ & \text{Multiply in order to eliminate the parentheses}. && 2x^2+4x+2x+4=400\\ & \text{Collect like terms}. && 2x^2+6x+4=400\\ & \text{Move all terms to one side of the equation}. && 2x^2+6x-396=0\\ & \text{Divide all terms by} \ 2. && x^2+3x-198=0\end{align*}
\begin{align*}x & = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & = \frac{-3 \pm \sqrt{3^2-4(1)(-198)}}{2(1)}\\ & = \frac{-3\pm \sqrt{801}}{2} \approx \frac{-3\pm 28.3}{2}\end{align*}
Use the quadratic formula. \begin{align*}x \approx 12.65\end{align*} or –15.65 feet
5. We can disregard the negative solution since it does not make sense in this context. Thus, we can check our answer of 12.65 by substituting the result into the area formula.
\begin{align*}\text{Area} = [2(12.65)+2)](12.65+2)=27.3 \cdot 14.65 \approx 400 \ ft^2.\end{align*}
### Guided Practice
Simplify \begin{align*}\frac{7}{\sqrt[3]{5}}\end{align*}.
Solution:
In this case, we need to make the number inside the cube root a perfect cube. We need to multiply the numerator and the denominator by \begin{align*}\sqrt[3]{5^2}\end{align*}.
\begin{align*}\frac{7}{\sqrt[3]{5}} \cdot \frac{\sqrt[3]{5^2}}{\sqrt[3]{5^2}}=\frac{7\sqrt[3]{25}}{\sqrt[3]{5^3}}=\frac{7\sqrt[3]{25}}{5}\end{align*}
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following videos. Note that there is not always a match between the number of the practice exercise in the videos and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both
Multiply the following expressions.
1. \begin{align*}\sqrt{6}\left ( \sqrt{10} + \sqrt{8} \right )\end{align*}
2. \begin{align*}\left ( \sqrt{a} - \sqrt{b} \right ) \left ( \sqrt{a} + \sqrt{b} \right )\end{align*}
3. \begin{align*}\left ( 2\sqrt{x}+ 5 \right ) \left ( 2\sqrt{x}+5 \right )\end{align*}
Rationalize the denominator.
1. \begin{align*}\frac{7}{\sqrt{15}}\end{align*}
2. \begin{align*}\frac{9}{\sqrt{10}}\end{align*}
3. \begin{align*}\frac{2x}{\sqrt{5}x}\end{align*}
4. \begin{align*}\frac{\sqrt{5}}{\sqrt{3}y}\end{align*}
5. The volume of a spherical balloon is \begin{align*}950 cm^3\end{align*}. Find the radius of the balloon. (Volume of a sphere \begin{align*}=\frac{4}{3} \pi R^3\end{align*})
6. A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is \begin{align*}180 in^2\end{align*}, what is the width of the frame?
7. The volume of a soda can is \begin{align*}355 \ cm^3\end{align*}. The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English Spanish
rationalize the denominator
To remove any radical signs from the denominator of the fraction using multiplication.
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# Linear Inequations | How do you Solve a Linear Inequation?
Linear inequations demonstrate the value of one quantity or algebraic expression which is not equal to another. These inequations are used to compare any two quantities. Get some solved examples on linear inequations and steps to draw graphs, the system of linear inequations in the following sections of this page.
## What are Linear Inequations?
Linear inequalities are the expressions where any two values are compared by the inequality symbols. These are the equations that contain all mathematical symbols except equal to (=). The values can be numerical or algebraic or a combination of both.
The inequality symbols are < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to), ≠ (not equal to). The symbols < and > shows the strict inequalities and the symbols ≤ and ≥ represents the slack inequalities.
Example:
x < 3, x ≥ 5, y ≤ 8, p > 10, m ≠ 1.
### Linear Inequalities Graphing
We can plot the graph for linear inequalities like an ordinary linear function. But, for a linear function, the graph represents a line and for inequalities, the graph shows the area of the coordinate plane that satisfies the inequality condition. The linear inequalities graph divides a coordinate plane into two parts. One part of the coordinate plane is called the borderline where it represents the solutions for inequality. The borderline represents the conditions <, >, ≤ and ≥.
Students who are willing to plot a graph for the linear inequations can refer to the following steps.
• Arrange the given linear inequation in such a way that, it should have one variable ‘y’ on the left-hand side of the symbol and the remaining equation on the right-hand side.
• Plot the graph for linear inequation by putting the random values of x.
• Draw a thicker and solid line for y≤ or y≥ and a dashed liner for y< or y> conditions.
• Now, draw shades as per the linear inequalities conditions.
### System of Linear Inequalities
A system of linear inequalities in two variables includes at least two inequalities in the variables. By solving the linear inequality you will get an ordered pair. So basically, the solution to all linear inequalities and the graph of the linear inequality is the graph displaying all solutions of the system.
### Questions on Linear Inequations
Example 1.
Solve the inequality x + 5 < 10?
Solution:
Given that,
x + 5 < 10
Move variable x to the one side of inequation.
= x < 10 – 5
= x < 5
Replacement set = {0, 1, 2, 3, 4, 5, 6, 7 . . . }
Solution set for the inequation x + 5 < 10 is 1, 2, 3, 4
Therefore, solution set s = {. . . 1, 2, 3, 4}
Let us mark the solution set graphically.
The solution set is marked on the number line by dots. We put three more dots indicate the infiniteness of the solution set.
Example 2.
Solve the inequation 3 < y ≤ 10?
Solution:
Given that,
3 < y ≤ 10
This has two inequations,
3 < y and y ≤ 10
Replacement set = {. . . -3, -2, -1, 0, 1, 2, 3, 4, , 6, 7, . . . }
Solution set for the inequation 3 < y is 4, 5, 6, 7, . . . i.e Q = {4, 5, 6, . . . }
Solution set for the inequation y ≤ 10 is . . . . 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 i.e P = {. . . . 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Therefore, solution set of the given inequation = P ∩ Q = {4, 5, 6, 7, 8, 9}
Let us represent the solution set graphically.
The solution set is marked on the number line by dots.
Example 3.
Solve the inequality 4 ( x + 2 ) − 1 > 5 − 7 ( 4 − x )?
Solution:
Given that,
4 ( x + 2 ) − 1 > 5 − 7 ( 4 − x )
4 x + 8 − 1 > 5 − 28 + 7 x
4 x + 7 > − 23 + 7 x
− 3 x > − 30
x < 10
Replacement Set = {1, 2, 3, 4, 5 . . . }
Solution set is 1, 2, 3, 4, 5, 6, 7, 8, 9 i.e S = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Let us represent the solution set graphically.
The solution set is marked on the number line by dots.
### Frequently Asked Questions on Linear Inequations
1. What is meant by linear inequations?
Linear inequations involve linear expressions in two variables by using the relational symbols like <, >, ≤, ≥, and ≠.
2. How do you solve linear inequation?
Solving linear inequations having a single variable is very easy. All you have to do is simplify both the sides of the condition and get the variable term on one side and all other terms on the other side. And them either multiply/ divide the coefficient of the variable to obtain the solution.
3. What are the examples of linear inequations?
Some of the examples of linear inequality are x < 8, y > 15, z ≠ 7 + 9, p + 1 ≤ 2, 3 ≥ (z + 15).
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Matrices 2. Solving Square Systems of Linear Equations; Inverse Matrices
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1 Matrices 2. Solving Square Systems of Linear Equations; Inverse Matrices Solving square systems of linear equations; inverse matrices. Linear algebra is essentially about solving systems of linear equations, an important application of mathematics to real-world problems in engineering, business, and science, especially the social sciences. Here we will just stick to the most important case, where the system is square, i.e., there are as many variables as there are equations. In low dimensions such systems look as follows (we give a 2 2 system and a 3 3 system): a 11 x 1 + a 12 x 2 = b 1 a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 (7) a 21 x 1 + a 22 x 2 = b 2 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3 In these systems, the a ij and b i are given, and we want to solve for the x i. As a simple mathematical example, consider the linear change of coordinates given by the equations x 1 = a 11 y 1 + a 12 y 2 + a 13 y 3 x 2 = a 21 y 1 + a 22 y 2 + a 23 y 3 x 3 = a 31 y 1 + a 32 y 2 + a 33 y 3 If we know the y-coordinates of a point, then these equations tell us its x-coordinates immediately. But if instead we are given the x-coordinates, to find the y-coordinates we must solve a system of equations like (7) above, with the y i as the unknowns. Using matrix multiplication, we can abbreviate the system on the right in (7) by x 1 b 1 (8) Ax = b, x = x 2, b = b 2, x 3 b 3 where A is the square matrix of coefficients (a ij ). (The 2 2 system and the n n system would be written analogously; all of them are abbreviated by the same equation Ax = b, notice.) You have had experience with solving small systems like (7) by elimination: multiplying the equations by constants and subtracting them from each other, the purpose being to eliminate all the variables but one. When elimination is done systematically, it is an efficient method. Here however we want to talk about another method more compatible with handheld calculators and MatLab, and which leads more rapidly to certain key ideas and results in linear algebra. Inverse matrices. Referring to the system (8), suppose we can find a square matrix M, the same size as A, such that (9) MA = I (the identity matrix). 1
2 2 MATRIX INVERSES We can then solve (8) by matrix multiplication, using the successive steps, A x = b M(A x) = M b (10) x = M b; where the step M(Ax) = x is justified by M(A x) = (MA)x, = I x, by (9); by associative law; = x, because I is the identity matrix. Moreover, the solution is unique, since (10) gives an explicit formula for it. The same procedure solves the problem of determining the inverse to the linear change of coordinates x = Ay, as the next example illustrates. ( ) ( ) Example 2.1 Let A = and M =. Verify that M satisfies (9) above, and use it to solve the first system below for x i and the second for the y i in terms of the x i : x 1 +2x 2 = 1 x 1 = y 1 +2y 2 2x 1 +3x 2 = 4 x 2 = 2y 1 +3y 2 ( )( ) ( ) Solution. We have =, by matrix multiplication. To ( ) ( 0 1 )( ) ( ) x solve the first system, we have by (10), = =, so the x solution is x 1 = 11,x 2 = 6. By reasoning similar to that used above in going from Ax = b to x = Mb, the solution to x = Ay is y = Mx, so that we get y 1 = 3x 1 +2x 2 y 2 = 2x 1 x 2 as the expression for the y i in terms of the x i. Our problem now is: how do we get the matrix M? In practice, you mostly press a key on the calculator, or type a Matlab command. But we need to be able to work abstractly with the matrix i.e., with symbols, not just numbers, and for this some theoretical ideas are important. The first is that M doesn t always exist. M exists A 0. The implication follows immediately from the law M-5 in section M.1 (det(ab) = det(a)det(b)), since MA = I M A = I = 1 A 0.
3 MATRIX INVERSES 3 The implication in the other direction requires more; for the low-dimensional cases, we will produce a formula for M. Let s go to the formal definition first, and give M its proper name, A 1 : Definition. Let A be an n n matrix, with A 0. Then the inverse of A is an n n matrix, written A 1, such that (11) A 1 A = I n, AA 1 = I n (It is actually enough to verify either equation; the other follows automatically see the exercises.) Using the above notation, our previous reasoning (9) - (10) shows that (12) A = 0 the unique solution of A x = b is x = A 1 b; (12) A = 0 the solution of x = A y for the y i is y = A 1 x. Calculating the inverse of a 3 3 matrix Let A be the matrix. The formulas for its inverse A 1 and for an auxiliary matrix adja called the adjoint of A (or in some books the adjugate of A) are T A 11 A 12 A (13) A 1 = adj A = A 21 A 22 A 23. A A A 31 A 32 A 33 In the formula, A ij is the cofactor of the element a ij in the matrix, i.e., its minor with its sign changed by the checkerboard rule (see section 1 on determinants). Formula (13) shows that the steps in calculating the inverse matrix are: 1. Calculate the matrix of minors. 2. Change the signs of the entries according to the checkerboard rule. 3. Transpose the resulting matrix; this gives adja. 4. Divide every entry by A. (If inconvenient, for example if it would produce a matrix having fractions for every entry, you can just leave the 1/ A factor outside, as in the formula. Note that step 4 can only be taken if A = 0, so if you haven t checked this before, you ll be reminded of it now.) The notation A ij for a cofactor makes it look like a matrix, rather than a signed determinant; this isn t good, but we can live with it.
4 4 MATRIX INVERSES Example 2.2 Find the inverse to A = Solution. We calculate that A = 2. Then the steps are (T means transpose): matrix A cofactor matrix T adj A inverse of A To get practice in matrix multiplication, check that A A 1 = I, or to avoid the fractions, check that A adj (A) = 2I. The same procedure works for calculating the inverse of a 2 2 matrix A. We do it for a general matrix, since it will save you time in differential equations if you can learn the resulting formula. ( ) ( ) ( ) ( ) a b d c d b 1 d b c d b a c a A c a matrix A cofactors T adj A inverse of A ( ) Example 2.3 Find the inverses to: a) b) ( ) ( ) Solution. a) Use the formula: A = 2, so A 1 = = b) Follow the previous scheme: = A Both solutions should be checked by multiplying the answer by the respective A. Proof of formula (13) for the inverse matrix. We want to show A A 1 = I, or equivalently, A adj A = A I; when this last is written out using (13) (remembering to transpose the matrix on the right there), it becomes a 11 a 12 a 13 A 11 A 21 A 31 A 0 0 (14) a 21 a 22 a 23 A 12 A 22 A 32 = 0 A 0. a 31 a 32 a 33 A 13 A 23 A A To prove (14), it will be enough to look at two typical entries in the matrix on the right say the first two in the top row. According to the rule for multiplying the two matrices on the left, what we have to show is that (15) a 11 A 11 + a 12 A 12 + a 13 A 13 = A ; (16) a 11 A 21 + a 12 A 22 + a 13 A 23 = 0
5 MATRIX INVERSES 5 These two equations are both evaluating determinants by Laplace expansions: the first equation (15) evaluates the determinant on the left below by the cofactors of the first row; the second equation (16) evaluates the determinant on the right below by the cofactors of the second row (notice that the cofactors of the second row don t care what s actually in the second row, since to calculate them you only need to know the other two rows). a 11 a 12 a 13 a 11 a 12 a 13 a 21 a 22 a 23 a 11 a 12 a 13 a 31 a 32 a 33 a 31 a 32 a 33 The two equations (15) and (16) now follow, since the determinant on the left is just A, while the determinant on the right is 0, since two of its rows are the same. The procedure we have given for calculating an inverse works for n n matrices, but gets to be too cumbersome if n > 3, and other methods are used. The calculation of A 1 for reasonable-sized A is a standard package in computer algebra programs and MatLab. Unfortunately, social scientists often want the inverses of very large matrices, and for this special techniques have had to be devised, which produce approximate but acceptable results.
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Mathematics » Equations and Inequalities » Simultaneous Equations Continued
# Solving by Elimination II
## Solving by Elimination
• Make one of the variables the subject of both equations.
• Equate the two equations; by doing this we reduce the number of equations and the number of variables by one.
• We now have one equation with one unknown variable which can be solved.
• Use the solution to substitute back into either original equation, to find the corresponding value of the other unknown variable.
## Example
### Question
Solve for $$x$$ and $$y$$: \begin{align*} y &= x^2 – 6x \qquad \ldots (1) \\ y + \cfrac{1}{2}x – 3 &= 0 \qquad \ldots (2) \end{align*}
### Make $$y$$ the subject of the second equation
\begin{align*} y + \cfrac{1}{2}x – 3 &= 0 \\ y &= -\cfrac{1}{2}x + 3 \end{align*}
### Equate the two equations and solve for $$x$$
\begin{align*} x^2 – 6x &= -\cfrac{1}{2}x + 3 \\ x^2 – 6x + \cfrac{1}{2}x – 3 &= 0\\ 2x^2 – 12x + x – 6 &= 0\\ 2x^2 -11x – 6 &= 0\\ (2x +1)(x-6) &= 0\\ \text{Therefore } x = -\cfrac{1}{2}&\text{ or } x = 6 \end{align*}
### Substitute the values for $$x$$ back into the second equation to calculate the corresponding $$y$$-values
If $$x = -\dfrac{1}{2}$$: \begin{align*} y &= -\cfrac{1}{2} ( -\cfrac{1}{2} ) + 3 \\ \therefore y &= 3\cfrac{1}{4} \end{align*} This gives the point $$(-\dfrac{1}{2};3\dfrac{1}{4})$$.
If $$x = 6$$: \begin{align*} y &= -\cfrac{1}{2}(6) + 3 \\ &= -3 + 3 \\ \therefore y &= 0 \end{align*} This gives the point $$(6;0)$$.
### Check that the two points satisfy both original equations
The solution is $$x = -\dfrac{1}{2} \text{ and } y = 3\dfrac{1}{4}$$ or $$x = 6 \text{ and } y = 0$$. These are the coordinate pairs for the points of intersection as shown below.
## Example
### Question
Solve for $$x$$ and $$y$$: \begin{align*} y &= \dfrac{5}{x – 2} \qquad \ldots (1)\\ y + 1 &= 2x \qquad \ldots (2) \end{align*}
### Make $$y$$ the subject of the second equation
\begin{align*} y + 1 &= 2x \\ y &= 2x – 1 \end{align*}
### Equate the two equations and solve for $$x$$
\begin{align*} 2x – 1 &= \cfrac{5}{x – 2} \\ (2x – 1)(x – 2) &= 5 \\ 2x^2 – 5x + 2 &= 5 \\ 2x^2 – 5x – 3 &= 0 \\ (2x + 1)(x – 3) &= 0 \\ \text{Therefore } x = -\cfrac{1}{2}&\text{ or } x = 3 \end{align*}
### Substitute the values for $$x$$ back into the second equation to calculate the corresponding $$y$$-values
If $$x -\dfrac{1}{2}$$: \begin{align*} y &= 2(-\cfrac{1}{2}) – 1 \\ \therefore y &= -2 \end{align*} This gives the point $$(-\dfrac{1}{2};-2)$$.
If $$x = 3$$: \begin{align*} y &= 2(3) -1 \\ &= 5 \end{align*} This gives the point $$(3;5)$$.
### Check that the two points satisfy both original equations
The solution is $$x = -\dfrac{1}{2} \text{ and } y = -2$$ or $$x = 3 \text{ and } y = 5$$. These are the coordinate pairs for the points of intersection as shown below.
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# Two Related Quantities, Part 1
Warmup
Activity #1
Graphing two related Quantities.
In the following activity, you are going to complete the table below of two related quantities and draw the graph.
Lin needs to mix a specific shade of orange paint for the set of the school play. The color uses 3 parts yellow for every 2 parts red.
• Complete the table to show different combinations of red and yellow paint that will make the shade of orange Lin needs.
Lin notices that the number of cups of red paint is always of the total number of cups. She writes the equation to describe the relationship.
(1.) Which is the independent variable? Which is the dependent variable? Explain how you know.
(2.) Write an equation that describes the relationship between r and y where y is the independent variable.
(3.) Write an equation that describes the relationship between y and r where r is the independent variable.
In this next activity, you are going to use the points in the table above to create a graph that shows the relationship between r and y.
• Match each relationship to one of the equations you wrote in (2) and (3) above.
In this next activity, you are going to use the points in the table to create a graph that shows the relationship between r and y.
• Match each relationship to one of the equations you wrote in (2) and (3) above.
Activity #2
Graphing two Related Quantities.
In the following activity, you are going to complete a table of two related quantities, and then transfer the data to a coordinate plane.
• Read the following story, and then complete the table to show how much Kiran pays for books during the sale.
Kiran shops for books during a 20% off sale. What percent of the original price of a book does Kiran pay during the sale?
• Create a graph below showing the relationship between the sale price and the original price by plotting the points from the table.
Activity #3
Complete a Table of Coordinate Pairs.
In the activity below, you are going to examine a graph that shows some values for the number of cups of sugar, required to make batches of brownies.
• Study the graph carefully and complete the table so that the pair of numbers in each column represents the coordinates of a point on the graph..
(1.) What does the point (8, 4) mean in terms of the amount of sugar and number of batches of brownies?
(2.) Write an equation that shows the amount of sugar in terms of the number of batches.
Challenge #1
Evaluate the expression when is 4 and is 6.
Challenge #2
Evaluate the expression when is 4 and is 6.
Challenge #3
Evaluate the expression when is 4 and is 6.
Quiz Time
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# Question Video: Simplifying Algebraic Expressions Using Laws of Exponents Mathematics
Simplify (π₯β΅π¦β΄/4) Γ (β8π₯Β³π¦β΅/5).
02:12
### Video Transcript
Simplify π₯ to the power of five π¦ to the power of four over four multiplied by negative eight π₯ cubed π¦ to the power of five over five.
In order to multiply any two fractions, we simply multiply the two numerators and then, separately, the two denominators. π over π multiplied by π over π is equal to ππ over ππ. Before multiplying fractions, it is always worth checking to see if we can cross simplify or cross cancel first.
Four and negative eight are both divisible by four, as four divided by four is one and negative eight divided by four is negative two. We therefore need to multiply π₯ to the power of five π¦ to the power of four by negative two π₯ cubed π¦ to the power of five over five.
In order to simplify this expression, we need to recall one of our laws of exponents or indices, π₯ to the power of π multiplied by π₯ to the power of π is equal to π₯ to the power of π plus π. When multiplying, we need to add the powers or exponents. Letβs consider the π₯ terms first.
π₯ to the power of five multiplied by π₯ cubed, or π₯ to the power of three, is equal to π₯ to the power of eight, as five plus three equals eight. π¦ to the power of four multiplied by π¦ to the power of five is equal to π¦ to the power of nine, as four plus five equals nine.
As the only constant terms, a negative two on the top and five on the bottom, they stay as they are. The answer is negative two π₯ to the power of eight π¦ to the power of nine over five. This could also be written as negative two-fifths π₯ to the power of eight π¦ to the power of nine.
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```Unit 2 Review Sheet
Combining Like Terms:
Like terms have the same variable raised to the same power.
We can only add or subtract like terms. We cannot add or subtract unlike terms.
Example: 5x + 3y + 6x +2y = 11x + 5y
Note: When there is subtraction in an expression, you should change it to addition to prevent
mistakes.
Example: 3x + 2y – 2x – y = 3x + 2y + (-2x) + (-y) = x + y
Distributive Property: a(b+c) = a(b) + a(c)
We use the distributive property when we are multiplying a number by unlike terms in
parentheses
Example: 6(2x+3) = 6(2x) + 6(3) = 12x +18
Simplifying Algebraic Expressions
Step 1: Change addition to subtraction
Step 2: Distribute
Step 3: Combine like terms.
Example: 2 + 3(x-5) + 3x
2+3(x+ (-5)) + 3x Change subtraction to addition
2+3x-15+3x
Distribute
6x-13
Combine like terms
Inverse Operations: Inverse operations are used to undo operations in equations.
Addition and Subtraction are inverses of each other.
Multiplication and Division are inverses of each other.
The inverse of a fraction is its reciprocal (the flip of the fraction).
One Step Equations:
Step 1: Identify the variable.
Step 2: Identify the operation that is being done to the variable.
Step 3: Perform the inverse operation.
Note: 4x is multiplication;
4
is division
Example: x+5=17
-5 -5 Subtract 5 from both sides
x = 12
Two Step Equations
Solve equations by undoing operations in the opposite order of PEMDAS.
Step 1: Identify the variable.
Step 2: Identify any addition or subtraction that is being done to the variable.
Step 3: Perform the inverse operation.
Step 4: Identify any multiplication or division that is being done to the variable.
Step 5: Perform the inverse operation.
Step 6: Check your answer by plugging it back into the equation.
Example: 5x + 8 = 28
-8
-8
Subtract 8
5x=20
x=4
Divide by 5
Multi Step Equations
Step 1: Simplify both sides of the equation by distributing and combining like terms.
Step 2: Solve the equation the same way you would solve a two step equation.
Distributing Fractions in Equations
Method One: Multiply both sides by the reciprocal.
Method Two: Multiply both sides by the least common denominator.
Note: You must always get rid of anything being adding or subtracted to the variable before
multiplying by the reciprocal or LCD.
Translating Words to Expressions:
Break apart the words to write them as a mathematical expression.
Less than: switch the order of the numbers
Subtracted from: switch the order of the numbers
Algebraic Equations (word problems):
Step 1: Identify the variable.
Step 2: Write an equation.
Step 3: Solve and check.
Inequalities
> greater than (x>2 means any number larger than two is a solution)
< less than (x<2 means any number smaller than two is a solution)
> greater than or equal to (x>2 means that two and any number larger than two is a solution)
< less than or equal to (x<2 means that two and any number smaller than two is a solution)
Graphing Inequalities
An open circle is used when the number is not a solution (greater than/less than)
A closed circle is used when the number is a solution (greater than or equal to/less than or equal
to)
If the variable is on the left, the inequality symbol points in the direction that the arrow of the
graph should go.
Solving Inequalities
Use the same rules that we use for solving equations, except if you multiply or divide by a
negative, you must flip the sign of the inequality.
```
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# Problem of the Week Problem B and Solution Traffic Predictions
## Problem
Petr was standing at the bus stop during rush hour and started counting the passing vehicles. In the first five minutes he waited, he counted $$20$$ cars, $$25$$ vans and $$15$$ trucks.
1. Based on Petr’s sample data, what is the theoretical probability that the next vehicle will be a truck?
2. Petr counted vehicles for another five minutes and discovered that the experimental probability of a vehicle being a car was the same for his first and second samples. If Petr counted a total of $$84$$ vehicles in his second sample, how many of those vehicles were cars?
## Solution
1. Petr’s sample had a total of $$20+25+15=60$$ vehicles. Since $$15$$ of these were trucks, the theoretical probability that the next vehicle will be a truck is $$\frac{15}{60}=\frac{1}{4}$$. Notice that the probability is equal to the fraction of trucks in the sample.
2. Petr’s first sample included $$20$$ cars which is $$\frac{20}{60}=\frac{1}{3}$$ of the vehicles. Thus, the experimental probability of a vehicle in the first sample being a car is $$\frac{1}{3}$$. If this experimental probability is the same for the second sample, then $$\frac{1}{3}$$ of the cars in the second sample must have been cars. Since his second sample had a total of $$84$$ vehicles, and $$\frac{1}{3}$$ of $$84$$ is $$\frac{1}{3}\times 84=28$$, it follows that $$28$$ of the vehicles in the second sample were cars.
Note: You cannot predict the individual numbers of vans or trucks in the second sample, because you don’t know the experimental probabilities of a vehicle being a van or a truck for the second sample.
Extension: If Petr had determined that the probability of a vehicle being a car was the same for his first and second samples, would it have been possible for him to have observed $$85$$ vehicles in the second sample?
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# Numerical Differentiation
In the case of differentiation, we first write the interpolating formula on the interval and the differentiate the polynomial term by term to get an approximated polynomial to the derivative of the function. When the tabular points are equidistant, one uses either the Newton's Forward/ Backward Formula or Sterling's Formula; otherwise Lagrange's formula is used. Newton's Forward/ Backward formula is used depending upon the location of the point at which the derivative is to be computed. In case the given point is near the mid point of the interval, Sterling's formula can be used. We illustrate the process by taking (i) Newton's Forward formula, and (ii) Sterling's formula.
Recall, that the Newton's forward interpolating polynomial is given by
(13.2.1)
Differentiating (13.2.1), we get the approximate value of the first derivative at as
(13.2.2)
where,
Thus, an approximation to the value of first derivative at i.e. is obtained as :
(13.2.3)
Similarly, using Stirling's formula:
(13.2.4)
Therefore,
(13.2.5)
Thus, the derivative at is obtained as:
(13.2.6)
Remark 13.2.1 Numerical differentiation using Stirling's formula is found to be more accurate than that with the Newton's difference formulae. Also it is more convenient to use.
Now higher derivatives can be found by successively differentiating the interpolating polynomials. Thus e.g. using (13.2.2), we get the second derivative at as
EXAMPLE 13.2.2 Compute from following table the value of the derivative of at :
1.73 1.74 1.75 1.76 1.77 1.77284 1.1552 1.73774 1.72045 1.70333
Solution: We note here that so , and
Thus, is obtained as:
(i) Using Newton's Forward difference formula,
(ii) Using Stirling's formula, we get:
It may be pointed out here that the above table is for , whose derivative has the value -0.1739652000 at
EXAMPLE 13.2.3 Using only the first term in the formula (13.2.6) show that
Hence compute from following table the value of the derivative of at :
1.05 1.15 1.25 2.8577 3.1582 3.4903
Solution: Truncating the formula (13.2.6)after the first term, we get:
Now from the given table, taking , we have
Note the error between the computed value and the true value is
EXERCISE 13.2.4 Retaining only the first two terms in the formula (13.2.3), show that
Hence compute the derivative of at from the following table:
1.15 1.2 1.25 3.1582 3.3201 3.4903
Also compare your result with the computed value in the example (13.2.3).
EXERCISE 13.2.5 Retaining only the first two terms in the formula (13.2.6), show that
Hence compute from following table the value of the derivative of at :
1.05 1.1 1.15 1.2 1.25 2.8577 3.0042 3.1582 3.3201 3.4903
EXERCISE 13.2.6 Following table gives the values of at the tabular points ,
0 0.05 0.1 0.15 0.2 0.25 0 0.10017 0.20134 0.30452 0.41075 0.5211
Compute (i)the derivatives and at by using the formula (13.2.2). (ii)the second derivative at by using the formula (13.2.6).
Similarly, if we have tabular points which are not equidistant, one can use Lagrange's interpolating polynomial, which is differentiated to get an estimate of first derivative. We shall see the result for four tabular points and then give the general formula. Let be the tabular points, then the corresponding Lagrange's formula gives us:
Differentiation of the above interpolating polynomial gives:
(13.2.7)
In particular, the value of the derivative at is given by
Now, generalizing Equation (13.2.7) for tabular points we get:
EXAMPLE 13.2.7 Compute from following table the value of the derivative of at :
0.4 0.6 0.7 3.38365 4.24424 4.72751
Solution: The given tabular points are not equidistant, so we use Lagrange's interpolating polynomial with three points: . Now differentiating this polynomial the derivative of the function at is obtained in the following form:
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## Six Ways To Write The Same Iterated Triple Integral
Six Ways To Write The Same Iterated Triple Integral:
The integration is widely used in engineering and mathematics for finding the rectangular area of graphs, we can use the single integral for finding the area of a particular area under the curve, the double integral is used to find the volume of a surface. The triple integral can be used to find the volume like the double integral and also the mass of an object. The integral is useful when the volume of the region is indifferent on X-axis, Y-axis, and Z-axis, and it has variable densities on all three axes.
We can use the triple integral calculator to find the triple integral of the surface. The utilization of the triple integral is more than the double integral as we can add an extra dimension in the triple integral. When we add an extra dimension of variable density in the same volume, so if the density is variable around three dimensions, we can find the mass of the volume. So by using the double integral, we were only able to find the volume, now by using the triple integral calculation, we are also able to find the mass of the object. Students can use the triple integral calculator to find the integral of an equation:
In this article, we are discussing six ways to write the same iterated triple Integral:
Six ways to write the triple integral :
There are six different ways to write the same iterated triple integral, the function f(x,y,z) remains the same for all the six iterations, the order of the integration, and the limit of the integration change accordingly. For all the iterations the value of the function f(x,y,z) remains the same, we can find the solution of all the iterations by putting the function in the triple integral calculator, this can be quite convenient for the students.
With the order of integration x,y,z e f(x,y,z)dxdydz
With the order of integration x,z,y e f(x,y,z)dxdzdy
With the order of integration y,x,z e f(x,y,z)dydxdz
With the order of integration y,z,x e f(x,y,z)dydzdx
With the order of integration z,x,y e f(x,y,z)dzdxdy
With the order of integration z,y,x e f(x,y,z)dzdydx
For solving all the iterations of the triple integration, you can use the triple integral calculator and put the limit of the interval in the triple integral calculator for finding the solution of every iteration.
How we solve the triple integral iteration:
We need to find the different limits of each of the above iterations, the limit would change for each of the above-given iterations. We need to use different limits for particular iterations. You need to start to solve the integral from the inside toward the outside integral, for example in solving the integral of dx dy dz, you need to start the integration with respect to “x”, then you have to do the integration with respect to “y”, then with respect to “z”, we can use the cylindrical integral calculator to solve the triple integral of a given function. To evaluate the triple integral we need to start integration from the inside, in this case starting from the “X”.
x,y,z e f(x,y,z)dxdydz
Since you are integrating with respect to the “X” and you need to calculate the “Y” and “Z” variables later, so you left the “Y” and “Z” variables, and integrate them according to their respective limits. You solve the x-variable according to the certain limit of the “ X” variable according to its respective limit. You can use the online triple integral calculator according to their respective limits, whether you are solving, with respect to the “X” variable, “Y” variable, or “Z” variable. For solving the triple integral, we can use the triple integral calculator.
x(y,z)x(y,z) f(x,y,z)dxdydz
Difference between double integral and the triple integral:
When we compare the double integral with the triple integral, we find the triple integral is more useful as compared to the double integral, we can say the double integral is an integration of multivariable functions f(x,y), around a region R, and the interval for the x is (a,b), you can say the limit of the “x” variable is (a,b) and for the “y” variable the limit is (c,d) using vertical e slices of the volume around the surface f(x,y) above the XY-plane or the cartesian coordinates.
Now for the triple integral, we are also integrating multivariable of a function for the density f(x,y,z) for the volume Bintegral, now you can define the limit of each variable, for example for the “x: the interval or the limit is (a,b), for “y” variable the limit is (c,d) and for the “z” variable the limit is (r,s). We are using the changes in the three-dimension plane, which is (x,y,z). Students do find it difficult to solve the triple integral, so you can use the triple integral calculator to find the solution of triple integration.
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Courses
# NCERT Solutions(Part- 1)- Visualising Solid Shapes Class 8 Notes | EduRev
## Class 8 Mathematics by VP Classes
Created by: Full Circle
## Class 8 : NCERT Solutions(Part- 1)- Visualising Solid Shapes Class 8 Notes | EduRev
The document NCERT Solutions(Part- 1)- Visualising Solid Shapes Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by VP Classes.
All you need of Class 8 at this link: Class 8
Question: Match the following: (First one has been done for you.)
Solution:
(A) → (ii) → (c)
(B) → (iii) → (g)
(C) → (i) → (b)
(D) → (iv) → (h)
(E) → (v) → (f)
(F) → (vii) → (d)
(G) → (vi) → (e)
(H) → (viii) → (a)
SOLUTION TO DO THIS
Question: Match the following pictures (objects) with their shapes:
Picture (object) Shape (i) An agricultural field (a) Two rectangular cross paths inside a rectangular park. (ii) A groove (b) A circular path around a circular ground. (iii) A toy (c) A triangular field adjoining a square field. (iv) A circular park (d) A cone taken out of a cylinder. (v) A cross path (e) A hemisphere surmounted on a cone.
Solution:
(i) → (c)
(ii) → (d)
(iii) → (e)
(iv) → (b)
(v) → (a)
VIEW OF 3-D SHAPES
A 3-D object can look differently from different positions so it can be drawn from different perspectives.
Example 1. Let there be a combination of three cubes, one on the top of the other as shown in the adjoining figure. What will be its top and side view?
Solution:
(i) If we look at this solid structure from the top, we would see just a square.
(ii) If we look at it from a side, i.e. left or right, then we would see a figure as shown here.
Note: In this case the front view would also be the same.
Example 2. Look at the combination of four cubes as shown in the adjoining figure. Show its various views.
Solution: The side view, front view and the top view of above solid are as given below:
EXERCISE 10.1
Question 1. For each of the given solid, the two views are given. Match for each solid the corresponding top and front views. The first one is done for you.
Solution:
(a) → (iii) → (iv)
(b) → (i) → (v)
(c) → (iv) → (ii)
(d) → (v) → (iii)
(e) → (ii) → (i)
Question 2. For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.
Solution:
Question 3. For each given solid, identify the top view, front view and side view
Solution:
Question 4. Draw the front view, side view and top view of the given objects.
Solution:
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# What Is Domain And Range In Mathematics?
## What is domain and range examples?
Example 2: The domain is the set of x -coordinates, {0,1,2}, and the range is the set of y -coordinates, {7,8,9,10}. Note that the domain elements 1 and 2 are associated with more than one range elements, so this is not a function.
## How do you find the domain?
The domain of a function is the set of all possible inputs for the function. For example, the domain of f(x)=x² is all real numbers, and the domain of g(x)=1/x is all real numbers except for x=0.
## What is the domain and range of 5?
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined. y=−5 is a straight line perpendicular to the y-axis at point (0,−5), which means that the range is a set of one value {−5}.
## What is the range in math?
About Transcript. The range is the difference between the largest and smallest numbers. The midrange is the average of the largest and smallest number.
You might be interested: Quick Answer: What Is The Use Of Ict In Mathematics?
## What is a domain value in math?
The domain of a function is the complete set of possible values of the independent variable. In plain English, this definition means: The domain is the set of all possible x- values which will make the function “work”, and will output real y- values.
## How do you find the range of a domain?
To find the excluded value in the domain of the function, equate the denominator to zero and solve for x. So, the domain of the function is set of real numbers except −3. The range of the function is same as the domain of the inverse function. So, to find the range define the inverse of the function.
## Is domain left to right?
Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range.
## What is the domain for a circle?
Since a circle is a set of points equidistant from a fixed point, your domain and range will have the same intervals. As proof, draw the center of the circle at the origin (0,0). Then use a compass to draw a circle with (0,0) as your reference point. You will see the domain and range are the same.
## What is the domain of 4?
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined. y=4 is a straight line perpendicular to the y-axis at point (0,4), which means that the range is a set of one value {4}.
You might be interested: Readers ask: What Is Mathematics According To The Great Math Mystery?
## What is the domain of y 2x 4?
the function y = 2x + 4 has a domain of 2,3,5,7, and 8
## How do you write a domain?
1. Identify the input values.
2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x.
3. The solution(s) are the domain of the function. If possible, write the answer in interval form.
## How do I calculate mean?
The mean is the average of the numbers. It is easy to calculate: add up all the numbers, then divide by how many numbers there are. In other words it is the sum divided by the count.
## What is range example?
The Range is the difference between the lowest and highest values. Example: In {4, 6, 9, 3, 7} the lowest value is 3, and the highest is 9. So the range is 9 − 3 = 6. It is that simple!
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# Cosine|Definition & Meaning
## Definition
The cosine is the trigonometric ratio of two adjacent sides, (the base and the hypotenuse) in a triangle that is calculated as the cosine function (or cos function). Out of the three most essential trigonometry functions, cosine is one of them.
## Visual Understanding
Consider a right triangle ABC. Say side AB is the base of the triangle, Side BC is perpendicular, and side AC is Hypotenuse, then we can define Cosine as the ratio of Side AB and side AC. In other words, the ratio of a triangle’s base to its hypotenuse is referred to as Cosine, which is visually represented as follows.
Figure 1 – Visual Understanding of Cos Function
AB=Base of Triangle
BC=Prependicular
AC=Hypotenuse
$\cos\alpha=\dfrac{Base}{Hypotenuse}$
## Mathematical Formula
From the above figure, we know that the Base is equal to side AB and Hypotenuse is equal to side AC so we can write
$\cos\alpha=\dfrac{AB}{AC}$
## Properties of Cosine
### Property 1
In the quadrants the positivity and negativity of cosine changes as it moves from first to second and second to third and from third to fourth. The positive range for the Cosine function lies in quadrant 1 and quadrant 4 whereas the negative range lies in quadrants 2 and 3 as shown in the figure.
Figure 2 – Cos Positivity and Negativity in Quadrants
In order to verify the property, we will use a calculator.
• In Quadrant 1
$\theta$ ranges from 0 to 90 degrees so we will apply cosine to the lower limit and upper limit of quadrant 1.
Lower Limit: $\theta=0$, $\cos(0)=1$
Upper Limit: $\theta=90$, $\cos(90)=0$
• In Quadrant 2
$\theta$ ranges from greater than 90 degrees to 180 degrees so we will apply cosine to the lower limit and upper limit of quadrant 2.
Lower Limit: $\theta=91$, $\cos(91) = -0.0174$
Upper Limit: $\theta=180$, $\cos(180) = -1$
• In Quadrant 3
$\theta$ ranges from greater than 180 degrees and smaller than 270 degrees so we will apply cosine to the lower limit and upper limit of quadrant 2.
Lower Limit: $\theta=91$, $\cos(91) = -0.0175$
Upper Limit: $\theta=269$, $\cos(269) = -0.0174$
• In Quadrant 4
$\theta$ ranges from 270 degrees to 360 degrees so we will apply cosine to the lower limit and upper limit of quadrant 2.
Lower Limit: $\theta=270$, $\cos(270)=0$
Upper Limit: $\theta=360$, $\cos(360)=1$
This property is visually illustrated in the figure below.
### Property 2
The derivative of cosine is equal to the negative of the sine function. In other words, Cos is the complement of sin
$\dfrac{d}{dx}(\cos(x))=-\sin(x)$
### Property 3
The integration of the cosine function gives the sine function.
$\int \cos(x) dx=\sin(x) + C$
### Property 4
The cos function is a periodic function.
$\cos\alpha=\cos(\alpha + 2p)$
### Property 5
The cos function is an even function.
$\cos(-\alpha)=\cos(\alpha)$
### Property 6
Trignometry formulas for two angles, a and b, can be written as
$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$
$\cos(a-b)=\cos(a)\sin(b)+\sin(a)\cos(b)$
## Graphical Representation of Cosine Graph
The cos graph starts from 90 degrees and repeats itself after every 360 degrees.+1 and -1 are the maximum and minimum values in the graph of Cos. There is a slight difference between the sin and cos graphs where the sin graph crosses or sweeps the origin whereas the cos graph never does so. The graph for cosine is shown in the figure below.
Figure 3 – Cos Function Graph
## Properties of Cos Graph
1. The graph of cosine never sweeps from the origin point.
2. For every positive and negative value of theta, the cos graph maintains its continuity and repeats after every 360 degrees.
3. The amplitude of the cos graph can be referred to as 1 with positive 1 being the maximum value and negative 1 being the smallest value.
## Law of Cosines
A triangle’s side lengths are related to one of its angles through the law of cosines. The use of trigonometry allows us to measure distances and angles that can’t be calculated any other way. When we know two sides and the enclosed angle of the triangle, we can apply the law of cosines to compute the third side.
In a triangle, the cosine of its angles and the length of its sides can be determined by the law of cosine. In trigonometry, a right triangle is represented by the cosine law, which generalizes Pythagoras’ theorem.
### Statement
Consider a triangle with sides x, y, and z. Then the square of one of the sides, say z, is equal to the difference between the sum of squares of side x, and side y and two times the product of side x and side y and angle between them.
$z^{2}=x^{2}+y^{2}-2xy\cos(Z)$
The cosine rule is another name for the law of cosine. Any triangle can be solved with the help of this law. This rule can be used, for example, to find the third side of a triangle when you know the lengths of two sides and the degree included between them. The illustration is shown in the figure.
Figure 4 – Illustration of Law of Cosines
### Formula
There are three laws of cosines. Consider a triangle having three sides x, y and z. In order to find:
• Side x, when side y, z is given along angle X
$x^{2}=y^{2}+z^{2}-2yz\cos(X)$
• Side y, when side x, z is given along angle Y
$y^{2}=x^{2}+z^{2}-2xz\cos(Y)$
• Side z, when side y, z is given along angle Z
$z^{2}=x^{2}+y^{2}-2xy\cos(X)$
### Properties
1. The Law of cosines is used to find an unknown side of triangle when two other triangles are given.
2. The Law of cosines is used to find the angle between the sides when all three sides are given.
3. The Law of cosines is not limited to just right-angle triangles it can be easily applied to any other triangle whose sides or angles are to be found.
## Solved Examples With the Cosine Function
### Example 1
Consider four values of theta and compute the cos function for all the values of theta using the calculator.
### Solution
$\theta=0,90,180,270,360$
$\cos(0)=1$
$\cos(90)=0$
$\cos(270)=0$
$\cos(360)=1$
### Example 2
Consider a triangle ABC with side AC=2 AB=5 BC=$\sqrt2$ Find the angle $\alpha$. The figure from this problem is shown below.
Figure 5 – Cos Function Example
### Solution
$\cos\alpha=\dfrac{Base}{Hypotenuse}$
$\cos\alpha=\dfrac{BC}{AB}$
$\alpha=\arccos\dfrac{\sqrt2}{5}$
$\alpha=73.57^{\circ}$
### Example 3
Consider the following triangle with side AB=4 BC=5 AC=6 and finds the angle A. The figure from this problem is shown below.
Figure 6 – Law of Cosine Example
### Solution
Applying the law of cosines:
$a^{2}=b^{2}+c^{2}-2bc\cos(A)$
$5^{2}=6^{2}+4^{2}-2.6.4\cos(A)$
$25=36+16-48 x \cos(A)$
$-48\cos(A)=-27$
$\cos(A)=\dfrac{-27}{-48}$
$\cos(A)=0.5625$
$A=\arccos0.5625$
$A=55.77^{\circ}$
All mathematical drawings and images were created with GeoGebra.
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# 5.9: Modeling Using Variation
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Learning Objectives
In this section, you will:
• Solve direct variation problems.
• Solve inverse variation problems.
• Solve problems involving joint variation.
A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn$736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate.
## Solving Direct Variation Problems
In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula $$e=0.16s$$ tells us her earnings, $$e$$, come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See Table 5.8.1.
$$s$$ , sales price $$e=0.16s$$ Interpretation
$9,200 $$e=0.16(9,200)=1,472$$ A sale of a$9,200 vehicle results in $1472 earnings.$4,600 $$e=0.16(4,600)=736$$ A sale of a $4,600 vehicle results in$736 earnings.
$18,400 $$e=0.16(18,400)=2,944$$ A sale of a$18,400 vehicle results in $2944 earnings. Table 5.8.1 Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from$4,600 to $9,200, and we double the earnings from$736 to \$1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation. Each variable in this type of relationship varies directly with the other.
Figure 5.8.1 represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula $$y=kx^n$$ is used for direct variation. The value $$k$$ is a nonzero constant greater than zero and is called the constant of variation. In this case, $$k=0.16$$ and $$n=1$$. We saw functions like this one when we discussed power functions.
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Figure 5.8.1
A General Note: DIRECT VARIATION
If $$x$$ and $$y$$ are related by an equation of the form
$$y=kx^n$$
then we say that the relationship is direct variation and $$y$$ varies directly with, or is proportional to, the $$n$$th power of $$x$$. In direct variation relationships, there is a nonzero constant ratio $$k=\dfrac{y}{x^n}$$, where $$k$$ is called the constant of variation, which help defines the relationship between the variables.
How to: Given a description of a direct variation problem, solve for an unknown.
1. Identify the input, $$x$$,and the output, $$y$$.
2. Determine the constant of variation. You may need to divide $$y$$ by the specified power of $$x$$ to determine the constant of variation.
3. Use the constant of variation to write an equation for the relationship.
4. Substitute known values into the equation to find the unknown.
Example
Solving a Direct Variation Problem
The quantity $$y$$ varies directly with the cube of $$x$$. If $$y=25$$ when $$x=2$$, find $$y$$ when $$x$$ is $$6$$.
Solution:
The general formula for direct variation with a cube is $$y=kx^3$$. The constant can be found by dividing $$y$$ by the cube of $$x$$.
$$k=\dfrac{y}{x^3}$$
$$=\dfrac{25}{2^3}$$
$$=\dfrac{25}{8}$$
Now use the constant to write an equation that represents this relationship.
$$y=\dfrac{25}{8}x^3$$
Substitute $$x=6$$ and solve for $$y$$.
$$y=\dfrac{25}{8}{(6)}^3$$
$$=675$$
Analysis
The graph of this equation is a simple cubic, as shown in Figure 5.8.2.
Q&A
Do the graphs of all direct variation equations look like Example?
No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through $$(0,0)$$.
Exercise
The quantity $$y$$ varies directly with the square of $$x$$. If $$y=24$$ when $$x=3$$, find $$y$$ when $$x$$ is 4.
Solution:
$$\frac{128}{3}$$
## Solving Inverse Variation Problems
Water temperature in an ocean varies inversely to the water’s depth. The formula $$T=\frac{14,000}{d}$$ gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.
If we create Table 5.8.2, we observe that, as the depth increases, the water temperature decreases.
$$d$$, depth $$T=\frac{14,000}{d}$$ Interpretation
500 ft $$\frac{14,000}{500}=28$$ At a depth of 500 ft, the water temperature is 28° F.
1000 ft $$\frac{14,000}{1000}=14$$ At a depth of 1,000 ft, the water temperature is 14° F.
2000 ft $$\frac{14,000}{2000}=7$$ At a depth of 2,000 ft, the water temperature is 7° F.
Table 5.8.2
We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations.
For our example, Figure 5.8.3 depicts the inverse variation. We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula $$y=\frac{k}{x}$$ for inverse variation in this case uses $$k=14,000$$.
A General Note: INVERSE VARIATION
If $$x$$ and $$y$$ are related by an equation of the form
$$y=\frac{k}{x^n}$$
where $$k$$ is a nonzero constant, then we say that $$y$$ varies inversely with the $$n$$th power of $$x$$. In inversely proportional relationships, or inverse variations, there is a constant multiple $$k=x^ny$$.
Example
Writing a Formula for an Inversely Proportional Relationship
A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.
Solution:
Recall that multiplying speed by time gives distance. If we let $$t$$ represent the drive time in hours, and $$v$$ represent the velocity (speed or rate) at which the tourist drives, then $$vt=$$distance. Because the distance is fixed at 100 miles, $$vt=100$$ so $$t=\frac{100}{v}$$. Because time is a function of velocity, we can write $$t(v)$$.
$$t(v)=\frac{100}{v}$$
$$=100v^{−1}$$
We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction. We say that time varies inversely with velocity.
How to: Given a description of an indirect variation problem, solve for an unknown.
1. Identify the input, $$x$$, and the output, $$y$$.
2. Determine the constant of variation. You may need to multiply $$y$$ by the specified power of $$x$$ to determine the constant of variation.
3. Use the constant of variation to write an equation for the relationship.
4. Substitute known values into the equation to find the unknown.
Example
Solving an Inverse Variation Problem
A quantity $$y$$ varies inversely with the cube of $$x$$. If $$y=25$$ when $$x=2$$, find $$y$$ when $$x$$ is $$6$$.
Solution:
The general formula for inverse variation with a cube is $$y=\frac{k}{x^3}$$. The constant can be found by multiplying $$y$$ by the cube of $$x$$.
$$k=x^3y$$
$$=2^3⋅25$$
$$=200$$
Now we use the constant to write an equation that represents this relationship.
$$y=\dfrac{k}{x^3}$$, $$k=200$$
$$y=\dfrac{200}{x^3}$$
Substitute $$x=6$$ and solve for $$y$$.
$$y=\dfrac{200}{6^3}$$
$$=\dfrac{25}{27}$$
Analysis
The graph of this equation is a rational function, as shown in Figure 5.8.4.
Exercise
A quantity $$y$$ varies inversely with the square of $$x$$. If $$y=8$$ when $$x=3$$, find $$y$$ when $$x$$ is $$4$$.
Solution:
$$\frac{9}{2}$$
## Solving Problems Involving Joint Variation
Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable $$c$$,cost, varies jointly with the number of students, $$n$$,and the distance, $$d$$.
A General Note: JOINT VARIATION
Joint variation occurs when a variable varies directly or inversely with multiple variables.
For instance, if $$x$$ varies directly with both $$y$$ and $$z$$, we have $$x=kyz$$. If $$x$$ varies directly with $$y$$ and inversely with $$z$$,we have $$x=\frac{ky}{z}$$. Notice that we only use one constant in a joint variation equation.
Example
Solving Problems Involving Joint Variation
A quantity $$x$$ varies directly with the square of $$y$$ and inversely with the cube root of $$z$$. If $$x=6$$ when $$y=2$$ and $$z=8$$, find $$x$$ when $$y=1$$ and $$z=27$$.
Solution:
Begin by writing an equation to show the relationship between the variables.
$$x=\dfrac{ky^2}{\sqrt[3]{z}}$$
Substitute $$x=6$$, $$y=2$$, and $$z=8$$ to find the value of the constant $$k$$.
$$6=\dfrac{k2^2}{\sqrt[3]{8}}$$
$$6=\dfrac{4k}{2}$$
$$3=k$$
Now we can substitute the value of the constant into the equation for the relationship.
$$x=\dfrac{3y^2}{\sqrt[3]{z}}$$
To find $$x$$ when $$y=1$$ and $$z=27$$, we will substitute values for $$y$$ and $$z$$ into our equation.
$$x=\dfrac{3{(1)}^2}{\sqrt[3]{27}}$$
$$=1$$
Exercise
A quantity $$x$$ varies directly with the square of $$y$$ and inversely with $$z$$. If $$x=40$$ when $$y=4$$ and $$z=2$$, find $$x$$ when $$y=10$$ and $$z=25$$.
Solution:
$$x=20$$
Media
Access these online resources for additional instruction and practice with direct and inverse variation.
Visit this website for additional practice questions from Learningpod.
## Key Equations
Direct variation $$y=kx^n$$, $$k$$ is a nonzero constant. Inverse variation $$y=\dfrac{k}{x^n}$$, $$k$$ is a nonzero constant.
## Key Concepts
• A relationship where one quantity is a constant multiplied by another quantity is called direct variation. See Example.
• Two variables that are directly proportional to one another will have a constant ratio.
• A relationship where one quantity is a constant divided by another quantity is called inverse variation. See Example.
• Two variables that are inversely proportional to one another will have a constant multiple. See Example.
• In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship joint variation. See Example.
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# How To Solve Physics Problems First Law of Thermodynamics problems and solutions
## First Law of Thermodynamics
The first law of thermodynamics is a conservation of energy statement for thermodynamic systems that exchange energy with their surroundings. It states that energy cannot be created or destroyed, but may only be changed from one form to another. The discussion is restricted to ideal gases and systems with constant mass. The quantity of heat, Q, added to the system and the work performed by the system, W, are taken as positive.
Work During Volume Change
Place a gas in a cylinder with a moveable piston. If the gas is allowed to expand at constant temperature, there is a force on the piston F = pA, and this force acting over a distance Δx is the work performed by the system.
Fig. 19 1
Figure 19 1 shows the cylinder with moveable piston and a p V diagram. The single (p V) curve is an isotherm, a constant temperature curve.
In differential notation
The total work in expanding from V1 to V2 is the area under the p V curve from V1 to V2. This is analogous to a spring where the work to compress the spring is the area under the curve of F vs x (see Chapter 7, Work and the Definite Integral).
The pressure depends on p through pV = nRT. At constant temperature the system follows the curve in Fig. 19 1. Writing p in terms of V
Since the gas is expanding at constant temperature, it is moving along an isotherm (T = const) line so p1V1 = p2V2, and another expression for work is
19 1 Two moles of an ideal gas maintained at 20‹ C expand until the pressure is one half the original. How much work is done by the gas?
Solution: Since the gas remains at constant temperature, it goes from one state to the other along an isotherm (see Fig. 19 1).
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19 2 A gas at constant pressure of 4.0 x 105 Pa is cooled so that its volume decreases from 1.6m3 to 1.2m3. What work is performed by the gas?
Solution: Since pressure is a constant the work performed is from equation 19 1.
The negative sign indicates that work was done on the system by an outside agent.
Internal Energy and the First Law
In going from state (p, V, T) to state (another p, V, T) a gas goes through a set of intermediary states. These states are called the path the system takes. On the p V diagram in Fig. 19 2 the point p1V1 represents the initial point and p2V2 the final point.
Fig. 19 2
If the gas were taken along the isothermal line, then the work performed would be the area under the p V curve as shown earlier. Consider, however, two different paths. First, the path from 1 3 2 could be accomplished by expanding the gas at constant pressure (1 3) then reducing the pressure to reach p2. In this case the work performed would be the area under the 1 3 line.
A second path from 1 4 2 is accomplished by reducing the pressure to p2 then expanding the gas to V2. In this case the work performed would be the area under the 4 2 line.
The work performed by the system in going from state 1 to state 2 via the 1 3 2 route is different from the work performed via the 1 4 2 route. Since there are an infinite number of possible routes from state 1 to state 2, the amount of work performed in going from state 1 to state 2 depends on the path.
Now consider an ideal gas in a container with a moveable piston and heat source (Fig. 19 3).
The gas atp1V1T1 is heated to a state p2V2T1. Heat is added from the reservoir to change p and V but not T.
The same amount of gas is placed in another container of volume V2, but the gas is confined to V1 with a breakable partition and the space above V1 is evacuated. If the partition is broken, the gas expands to V2 without a change in temperature and sufficiently rapidly so there is no heat transferred in the process. This rapid expansion of a gas into a vacuum is called free expansion. These two extreme routes are illustrative of the infinite number of possible routes, each requiring a different amount of heat, from p1V1T1 to p2V2T1. Therefore, the amount of heat necessary to go from one state to another depends on the path.
Fig. 19 3
While the amount of heat to go from one state to another depends on the path and the amount of work to go between these two states depends on the path, the difference, heat minus work, is independent of the path and always appears as a change in internal energy of the gas. In equation form this is
This last statement is the first law of thermodynamics and in words states that heat into a system appears either as an addition to the total internal energy or work performed by the system. Going from one state to another where the temperature of the state is the same, the heat in equals the work performed. Going from one state to another where the temperature of the state is not the same the heat equals the increase in internal energy plus the work performed.
19 3 You eat 100 food calories worth of nuts. How high would you, at 65kg, have to climb to gwork offh these 100 calories?
Solution: The 100 food calories is 100kcal into your (thermodynamic) system. The work to reduce your total internal energy back to what it was before eating the nuts is
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19 4 In a certain process 1.5x105 J of heat is added to an ideal gas to keep the pressure at 2.0x105 Pa while the volume expands from 6.3m3 to 7.1m3. What is the change in internal energy for the gas?
Solution: Apply the first law ΔU = ΔQ ΔW. Since the pressure is constant
The internal energy of the gas has been decreased by 1.0x104 J.
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19 5 An ideal gas initially at 3.0x105 Pa, volume of 0.030m3, and temperature 20‹C is heated at constant pressure to a volume of 0.120m3. How much work was performed by the gas? If 80KJ of heat energy was supplied to the gas, what was the change in internal energy?
Solution: The work performed is
The increase in internal energy is
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There are several different thermodynamic processes that occur often enough to warrant definition.
An isothermal process occurs at constant temperature.
An adiabatic process is one where no heat is transfered in or out of the system (ΔQ = 0).
An isobaric process occurs at constant pressure (W = pΔV).
An isochoric process occurs at constant volume (W = 0).
Heat Capacities of Ideal Gases
The heat capacity of an ideal gas depends on whether the measurement is made at constant volume, CV, or constant pressure, Cp.
At constant volume (ΔV = 0) no work is performed by the system, so according to the first law ΔQ = ΔU. For n moles requiring CV amount of heat per mole
At constant pressure work is performed. Defining the heat capacity at constant pressure as the heat to raise one mole of a gas one degree,
Since the gas expands in this process, work, ΔW = pΔV is performed. From the ideal gas law pΔV = nRΔT so ΔW = nRΔT. Substituting in the first law in the form ΔQ = ΔU + ΔW, nCpΔT = ΔU + nRΔT. Since the internal energy depends only on temperature ΔU = nCVΔT and
Measurements on gases confirm this relationship.
19 6 What is the total internal energy of 10 moles of an ideal monatomic gas at 20‹ C?
Solution: A monatomic gas is composed of single atoms that behave as mass points, and as such, they can only translate. They cannot rotate or vibrate. See problem 18 5 and the associated discussion to confirm that the internal energy is (3/2)nRT.
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19 7 The heat necessary to raise the temperature of 10 moles of an ideal gas at a constant pressure of 1.0 x 105 Pa 10 degrees is 2100J. What is the volume change and the heat capacity at constant pressure and constant volume?
Solution: The work performed in the volume change is ΔW = pΔV = nRΔT so
Starting with the basic statement ΔQ = nCpΔT the heat capacity at constant pressure is
The gas constant R is 8.3 J / mol . K so
19 8 Two moles of an ideal gas at a pressure of 1.0atm has 400 J of heat added in expanding from a volume of 4.7 x 10 3 m3 to 7.0 x 10 3 m3. What is the change in internal energy and temperature? What are the specific heats?
Solution: The work performed is W = pΔV = 1.0 x 105 N / m2 (2.3 x 10 3m3) = 230J.
The change in internal energy is from the first law ΔU = ΔQ W = 400 J 230J = 170J.
The change in temperature is from ΔU = (3 / 2)nRΔT, or
The heat capacity at constant pressure is from ΔQ = nCpΔT, or
The heat capacity at constant volume is
Adiabatic processes are ones where there is no heat transfer between the system and its surroundings. Rapid volume changes where there is no time for heat transfer are characteristic of adiabatic processes.
Fig. 19 4
A model adiabatic process is shown in Fig. 19 4. An ideal gas in an insulated container with a moveable weighted piston rapidly expands when the weight is removed. The volume increases, and the pressure and temperature drop with no heat in or out.
From the first law ΔU = W and ΔU = nCvΔT and W = pΔV. Writing the first law with p replaced from the ideal gas law pV = nRT
Switching to differential format
The factor R / CV is usually written in the form
Rewriting,
is integrated directly as a logarithm
In T + (ƒÁ 1) InV = In (const), and applying the laws of logarithms TVƒÁ 1 = const
Again applying the gas law pV = nRT,
or pVƒÁ = (another) const.
For a monatomic gas ƒÁ = 1.40. For adiabatic expansions there are two relationships
The p V curves for the pVƒÁ = const are a lot like pV = const curves. The pVƒÁ = const curves are called adiabats and are shown in Fig. 19 4 along with (for comparison) some isotherms.
19 9 Two moles of an ideal gas expand isothermally at 295K from 0.60m3 to 0.80m3. What would be the temperature of the gas if the same expansion were adiabatic?
Solution: The temperature is from equation 19 7.
19 10 Air and fuel at 300K and 1.0atm is compressed in an automobile engine with a compression ratio of 8 to 1. Take ƒÁ = 1.40 and calculate the final pressure and temperature of the mixture.
Solution: The pressure is from
or
The temperature is from
or
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19 11 An ideal gas at 3.0 x 105 Pa and 0.080m3 is compressed adiabatically to 0.020m3. What is the final pressure and the work performed?
Solution: The final pressure is from ,
or
The work performed is from W = nCvΔT = nCv(T2 T1). Substituting from the ideal gas law pV = nRT and R/Cv = ƒÁ 1
Work is performed on the gas in the amount of 4.5 x 104J.
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19 12 An ideal monatomic gas at pa = 3.0 x 105 N/m2, Va = 0.060m3, and T = 27‹C expands adiabatically to pb = 2.0 x 105 N/m2, Vb = 0.085m3 and then isothermally to Vc = 0.100m3. What is the final temperature, pressure, and work performed by the gas? Show these paths on a p V diagram.
Solution: First calculate the temperature at the end of the adiabatic expansion using the ideal gas law
The gas has moved along an adiabat from the 300 K isotherm to the 140K isotherm (point a to point b on Fig. 19 5).
The pressure at the end of the isothermal expansion can be calculated from the ideal gas law with the temperature a constant, PbVb = PcVc, or
The temperature of the gas at point c is 140K, and the pressure is 1.7 x 105 Pa.
The work performed is positive in both parts of the expansion. During the adiabatic portion of the expansion, the work is the area under the a b portion of the curve (see problem 19 11).
Fig. 19 5
In the isothermal portion of the expansion the work is the area under the b c curve (equation 19 3).
The nRT is equal to pbVb, so
This is the work performed in the isothermal expansion. Along the isotherm pbVb / Tb = pcVc / Tc reduces to pbVb = pcVc because Tb = Tc.
The total work performed by the gas is 5260J.
Look over this problem again. It is probably the most difficult one you will encounter, and if you understand all the steps and can do them on your own, you understand the first law of thermodynamics very well.
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# A point source of light is moving at a rate of $2\,cm - {s^{ - 1}}$ towards a thin convex lens of focal length 10 cm along its optical axis. When the source is 15 cm away from the lens, the image is moving atA. $4\,cm - {s^{ - 1}}$ towards the lensB. $8\,cm - {s^{ - 1}}$ towards the lensC. $4\,cm - {s^{ - 1}}$ away from the lensD. $8\,cm - {s^{ - 1}}$ away from the lens
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Hint: Use lens formula to determine the image distance.
Differentiate the lens formula to determine the formula for rate of movement of image with respect to time.
Formula used:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Here, f is the focal length, v is the image distance and u is the object distance.
We know that the focal length of the convex lens is positive. Also, the object distance from the lens is taken as negative.
We have the lens formula,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Here, f is the focal length, v is the image distance and u is the object distance.
Rearrange the above equation to determine the image distance as follows,
$v = \dfrac{{fu}}{{u + f}}$
Substitute 10 cm for f and $- 15\,cm$ for u in the above equation.
$v = \dfrac{{\left( {10\,cm} \right)\left( { - 15\,cm} \right)}}{{ - 15\,cm + 10\,cm}}$
Differentiate the lens formula with respect to time t as follows,
$0 = - \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} + \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}$
$\Rightarrow \dfrac{1}{{{v^2}}}\dfrac{{dv}}{{dt}} = \dfrac{1}{{{u^2}}}\dfrac{{du}}{{dt}}$
$\therefore \dfrac{{dv}}{{dt}} = \dfrac{{{v^2}}}{{{u^2}}}\dfrac{{du}}{{dt}}$
Substitute $+ 30\,cm$ for v, $- 15\,cm$ for u and $+ 2\,cm - {s^{ - 2}}$ for $\dfrac{{du}}{{dt}}$ in the above equation.
$\dfrac{{dv}}{{dt}} = \dfrac{{{{\left( {30\,cm} \right)}^2}}}{{{{\left( { - 15\,cm} \right)}^2}}}\left( {2\,cm - {s^{ - 2}}} \right)$
$\dfrac{{dv}}{{dt}} = \left( 4 \right)\left( {2\,cm - {s^{ - 2}}} \right)$
$\dfrac{{dv}}{{dt}} = + 8\,cm - {s^{ - 2}}$
The positive sign in the above equation implies that the image is moving along the positive direction of the x-axis which is away from the lens
So, the correct answer is “Option D”.
Note:
Read the properties of the image formed by the convex lens. The focal length of the convex lens is positive whereas the focal length of the concave lens is negative. The object distance or source distance is negative while the image distance is positive.
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# Lesson 6-II: Loans and Finite-Difference Models
Note: This lesson and Lesson 6-I are really two parts of one whole. It's advisable to either do them at the same time, or to do Lesson 6-II very shortly after completing Lesson 6-I.
In Lesson 6-I, we considered the problem of saving money at a fixed interest rate, compounded, with regular deposits. Now we're going to turn that around. What if, instead of the bank paying you money, you owed they bank money (perhaps, sadly, a more common occurrence these days)? Instead of making regular deposits to save up, we'll discuss making regular payments toward the balance of the loan.
It turns out that the mathematics of the situation are roughly the same.
Along the way, we'll learn about the time value of money and compound interest. What should our monthly payments be in order to pay the loan off after a certain number of months?
# The Mathematical Model
The word equation looks quite similar to the word equation for an interest-bearing deposit account:
The interest on last month's balance is
Let's use the following symbols (which are somewhat standard):
variable/parameter interpretation units time months balance (amount owed) $monthly payment$ annual interest rate
Just as we did in Lesson 5-I, we obtain the finite difference model:
or in "simplified" form
The only difference between this equation and the one in Lesson 6-I is the sign. We have "-P" because each regular payment reduces our balance owed; in Lesson 6-I each regular payment increased the balance we had in our account so we wrote "+D".
Since we are looking at essentially the same equation, we can solve it the same way. Recall there are two methods to be used with a finite difference model: iteration (which involves arithmetic, so Excel is the appropriate tool) and algebra (for which Maple is best).
### The Implementation: Iteration (Excel)
Let's see what happens if we start off owing an initial debt of $10,000, at a rate of 6%, and we make monthly payments of$150.
As before, we'll make a table of values labeled t and y. Also we'll make a parameter table to make it easier to see the effects of changing interest rate and monthly payments. In cell B2, enter 10000 for the initial balance of the debt, and in cell B3, use the parameter values to enter the Excelified version of the formula above in cell B3.
As usual, we use $s in the formula whenever we reference cells where parameters are stored. Now, as in Lesson 5-I, copy-and-paste the formula down column B until we get to our goal -- which, this being a debt, is to have a balance of$0. Here that looks to be after about 82 months.
### The Implementation: Algebra (Maple)
Let's say the $150 monthly payment we're making is the minimum monthly payment that our credit card company insists we make. • When, precisely, will the debt be paid off? • If we want to pay off the debt in two years (24 months), what monthly payment should we make? Instead of plugging in numbers and iterating, we'll just do it with symbols. That is, we'll do algebra instead of arithmetic. The key algebraic observation is that we can combine the equations and to obtain: which can in turn be simplified to: We can then plug this into the equation to obtain and similarly The pattern continues like this: Now, there's an algebraic simplification of this, but instead of waving my hands and asking you to trust me -- that's a degree of trust maybe I haven't earned -- let's have Maple do it. After all, surely we can agree that a computer isn't likely to make an algebra mistake! a video walkthrough of this The command we're going to use is called rsolve. 1. First we need to tell Maple our finite difference model. • For rsolve to work, we have to enter our unknown y in function notation, rather than writing yt or y_t or something like that. • I've also called the variables rate and payment, both to make them easier to remember and because Maple treats the letter I as special so I can't use it. 2. Now, we could ask Maple to solve this for us: • As with all solving commands, the first slot in the command is an equation (or system of equations), and the second slot is the unknown we're solving for. 3. Actually, since we know that , when we invoke the rsolve command, we can just tell Maple that: 4. Now copy-and-paste the output from rsolve into a function called y: 5. We let Maple know that the rate is 6%: 6. And ask what the payment ought to be, if we want to pay off the loan in 24 months: 7. To find out, at a given monthly payment, how many months it takes to pay off the debt, we first specify the payment and then solve for t: Thus we see that, at a payment of$150 each month it will take more than 81 months to pay off the full debt; to pay it off in 24 months will require quite a bit more -- $443.21! One practical observation to make: if we take the 81 months to pay, we'll pay a bit more than . If we make the greater payments, the total amount paid is (I've rounded to the next cent rather than the nearest cent because that's how payments work.) So the question we'd have to ask ourselves is: is it worth the approximately$1,500 to have $300 more walking-around money on a monthly basis? If the$300 each month is the difference between having a babysitter or not, the answer may well be that we have no choice; if the extra \$300 each month would go to partying, then maybe it is better to pay down our debt.
### Variables in Maple
One thing to watch out for in Maple is that when you assign a value to a variable, it stays assigned. So in step 7, above, we assigned the value 150 to the variable payment. If we try to execute step 6 again, Maple gets cranky with us:
(Of course it does -- "solving for 150" is kind of an absurd request.) To avoid this hassle, let's clear out that value. There are two ways to do this.
• There is a panel called Variables to the left of the main execution area. This lists all the variables that are currently defined. In our worksheet, we've defined finitediff, payment, rate, and y. To delete a variable, simply right-click (on a Mac, command-click) the name of the variable and select "Unassign".
• The other way is to use the command unassign:
# Takeaways/Deliverables
To say you've successfully completed this lesson, you should be able to do the following:
1. Derive a finite difference model for compound interest, given rate and initial investment.
2. Use Maple to solve your finite difference model.
3. Use Maple and your finite-difference model to find the required monthly payment to pay off a loan in a specified period of time.
4. Compute the total amount paid on a loan.
These skills are what you'll need to complete the WebAssign homework for this Lesson.
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# How to solve 3×3 rubic cube
One of the most well-liked puzzle toys in the world is the Rubik’s Cube. Ern Rubik, a professor of architecture from Hungary, created it in 1974. Since then, its complexity and difficult character have captured the attention of people all around the world. This article will provide a detailed, step-by-step explanation of how to solve a 3×3 Rubik’s Cube.
Step 1: Understand the Basics
Six faces, each with nine smaller squares, make up the Rubik’s cube. The object is to first jumble the cube, and then to solve it by putting each face back in its original position with a solid color. The Rubik’s Cube has three different types of pieces: corners, edges, and centers. Three colored stickers are on the corners, two are on the borders, and just one is on the middle. While the edges and corners can be bent around the cube, the centers are fixed in place and cannot be moved.
Step 2: Solve the First Layer
The Rubik’s Cube’s initial layer must be solved to proceed. Find every piece that has one color on it to begin with, then choose another color. Then, position them properly to create a solid color on the first layer. You must master a few fundamental algorithms or move sequences in order to accomplish this. One technique, dubbed “white cross,” entails gathering the white bits on the bottom layer and arranging them into a cross-like configuration. A different procedure, known as “white corners,” entails placing the white corner pieces in the proper alignment and position.
Step 3: Solve the Second Layer
Continue to the second layer after the first layer is finished. You must concentrate on the edges in order to solve the second tier. Find an edge piece that has a different color on one side and the color of the initial layer on the other. After that, place it correctly to complete the second layer. You can achieve this using a variety of techniques, including the “R U R’ U'” algorithm.
Step 4: Solve the Third Layer
The third and final layer is the hardest aspect of resolving the Rubik’s Cube. It is possible to solve the third layer using either the Fridrich approach or the the beginner’s method and the Fridrich method. The beginner’s method is simpler and just requires a few fundamental algorithms, whereas the Fridrich method is more complex and requires learning numerous methods.
To begin using the beginner’s technique, create a cross shape on the third layer. The corner pieces should then be oriented correctly using an algorithm. Put the edges in the right place next, and then use an algorithm to put the corner pieces in the right place. You can choose from a number of different algorithms for each stage, including the “Sune” method and the “T-perm” algorithm.
Step 5: Practice and Persistence
It takes patience and practice to solve the Rubik’s Cube. To first solve the cube, it could take several hours or even days. When you finally figure it out, though, you’ll feel a sense of satisfaction and success that is difficult to match. You will grow quicker and more effective at solving the Rubik’s Cube as you practice more.
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Found in: Page 57
### Fundamentals Of Physics
Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718
# A person desires to reach a point that is${\mathbf{3}}{\mathbf{.}}{\mathbf{40}}{\mathbf{}}{\mathbf{km}}$from her present location and in a direction that is${\mathbf{3}}{\mathbf{.}}{\mathbf{50}}{\mathbf{°}}$north of east. However, she must travel along streets that are oriented either north–south or east–west. What is the minimum distance she could travel to reach her destination?
The minimum distance traveled to reach destination is$4.47\mathrm{km}$
See the step by step solution
## Step 1: To understand the concept of scalar projection
This problem refers to scalar projection. In Cartesian coordinates, scalar components are scalar projections in the directions of the coordinate axes. Using this concept, the distance can be calculated by finding the components and adding those components.
The components can be written as
${d}_{x}=d\mathrm{cos}\theta \phantom{\rule{0ex}{0ex}}{d}_{y}=d\mathrm{sin}\theta$
Therefore the minimum distance D is given by the following formula.
$D={d}_{x}+{d}_{y}=d\mathrm{cos}\theta +d\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)$
## Step 2: To find the minimum distance traveled
Given are,
$d=3.40\mathrm{km}\phantom{\rule{0ex}{0ex}}\theta =35.0°$
Substituting the above values in equation (i), the minimum distance D can be written as,
$D=\left[3.40×\mathrm{cos}\left(35.0\right)\right]+\left[3.40×\mathrm{sin}\left(35.5\right)\right]$
Thus, $D=4.47\mathrm{km}$
## Recommended explanations on Physics Textbooks
94% of StudySmarter users get better grades.
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## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 8 Ratio and Proportion Ex 8.3
Question 1.
If the cost of 9 m cloth is ₹378, find the cost of 4 m cloth.
Solution:
∵ Cost of 9 m of cloth = ₹378
∴ Cost of 1 m of cloth = ₹ $$\frac{378}{9}$$ = ₹42
∴ Cost of 4 m cloth = ₹42 × 4 = ₹168
Question 2.
The weight of 36 books is 12 kg. What is weight of 75 such books?
Solution:
∵ Weight of 36 books = 12 kg
∴Weight of 1 book = $$\frac{128}{36}$$ kg = $$\frac{1}{3}$$ kg
∴Weight of 75 books = $$\frac{1}{3}$$ × 75 = 25 kg
Question 3.
Five pens cost ₹115. How many pens can you buy in ₹207?
Solution:
₹115 is cost of 5 pens 5
₹ 1 is cost of = $$\frac{5}{115}$$ pens
∴ ₹207 is cost of
$$=\frac{207 \times 5}{115}=\frac{207}{23}=9$$ pens
Question 4.
A car consumes 8 litres of petrol in covering a distance of 100 km. How many kilometres will it travel in 26 litres of petrol?
Solution:
8 litre of petrol consumes for = 100 km
Then 26 litre of petrol consumes for
$$\frac{26 \times 100}{8}=\frac{1300}{4}=325 \mathrm{km}$$
Question 5.
A truck requires 108 litres of diesel for covering a distance of 594 km. How much diesel will be required by the truck to cover a distance of 1650 km?
Solution:
∵ Diesel required for covering a distance of 594 km = 108 litres
∴ Diesel required for covering a distance of 1 km = $$\frac{108}{594}$$ litre
∴ Diesel required for covering a distance of 1650 km = $$\frac{108}{594} \times 1650$$ litres
$$=\frac{2}{11} \times 1650=2 \times 150=300$$ litres
Hence, 300 litres of diesel will be required by the truck to cover a distance of 1650km.
Question 6.
A transport company charges ₹5400 to carry 80 quintals of weight. What will it charge to carry 126 quintals of weight (same distance)?
Solution:
Charges of 80 quintals of weight = ₹5400
∴ Charges of 1 quintal = ₹ $$\frac{5400}{80}$$
and charges of 126 quintals
= ₹ $$\frac{5400 \times 126}{80}=\frac{135 \times 126}{2}$$
= 135 × 63 = ₹8505
Question 7.
42 metres of cloth is required to make 20 shirts of the same size. How much cloth will be required to make 36 shirts of that size?
Solution:
For 20 shirts cloth required = 42 m
∴ Cloth required for making 1 shirt = $$\frac{42}{20}$$ m
∴ For 36 shirts cloth required will be
$$=\frac{42 \times 36}{20}=\frac{18 \times 42}{10}=\frac{176}{10}=75.6 \mathrm{m}$$
Question 8.
Cost of 5 kg of rice is ₹107.50.
(i) What will be the cost of 8 kg of rice?
(ii) What quantity of rice can be purchased in ₹64.5?
Solution:
(i) Cost of 5 kg of rice = ₹107.50
∴ Cost of 1 kg of rice = ₹ $$\frac{107.50}{5}$$ = ₹21.5
∴ Cost of 8 kg of rice = ₹21.5 × 8 = ₹172
(ii) ∵ In ₹107.50, the quantity of rice that can be purchased = 5 kg
∴ In ₹1, the quantity of rice that can be phased = $$\frac{5}{107.50} \times 54.5$$ kg
∴ In ₹64.5, the quantity of rice that can be purchased = $$\frac{5}{107.50} \times 54.5$$
= $$\frac{5}{10750} \times 100 \times \frac{545}{10}=3 \mathrm{kg}$$
Question 9.
Cost of 4 dozen bananas is ₹ 180. How many bananas can be purchased for ₹37.50?
Solution:
1 dozen contains = 12 items
∴ 4 dozens contains =12 x 4 items = 48 items
Cost of 4 dozen bananas = ₹180
That means cost of 48 bananas = ₹180
∴ Number of bananas that can be purchased for ₹1 = $$\frac{48}{180}$$
∴ Number of bananas that can be purchased for ₹37.50
= $$\frac{48}{180} \times 37.50=\frac{48}{180} \times \frac{3750}{100}=10$$
Question 10.
Aman purchases 12 pens for ₹156 and Payush buys 9 pens for ₹1108. Can you say who got the pens cheaper?
Solution:
For Aman
∵ Cost of 12 pens = ₹156
∴ Cost of 1 pen = ₹ $$\frac{156}{12}$$ = ₹ 13
For Payush
∴ Cost of 9 pens = ₹108
∴ Cost of 1 pen = ₹$$\frac{108}{9}$$ = ₹12
So, Payush got the pens cheaper.
Question 11.
Rohit made 42 runs in 6 overs and Virat made 63 runs in 7 overs. Who made more runs per over?
Solution:
For Rohit
∵ Runs made in 6 overs = 42
∴ Runs made per over = $$\frac{42}{6}$$ = 7
For Virat
∵ Runs made in 7 overs = 63
∴ Runs made per over = $$\frac{63}{7}$$ = 9
So, Virat made more runs per over.
Question 12.
A bus travels 160 km in 4 hours and a train travels 320 km in 5 hours at uniform speeds, then find the ratio of the distance travelled by them in one hour.
Solution:
A bus travel in 4 hours = 160 km
∴ Distance covered by bus in 1 hour
= $$\frac{160}{4}$$ = 40 km
A train travel in 5 hours = 320 km
∴ Distance covered by train in 1 hour
= $$\frac{320}{5} \mathrm{km}=64 \mathrm{km}$$
Ratio in their speed = 40 : 64 = 5 : 8
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Home » How to Calculate Odds Ratio and Relative Risk in Excel
# How to Calculate Odds Ratio and Relative Risk in Excel
We often use the odds ratio and relative risk when performing an analysis on a 2-by-2 table, which takes on the following format:
The odds ratio tells us the ratio of the odds of an event occurring in a treatment group to the odds of an event occurring in a control group. It is calculated as:
Odds ratio = (A*D) / (B*C)
The relative risk tells us the ratio of the probability of an event occurring in a treatment group to the probability of an event occurring in a control group. It is calculated as:
Relative risk = [A/(A+B)] / [C/(C+D)]
This tutorial explains how to calculate odds ratios and relative risk in Excel.
### How to Calculate the Odds Ratio and Relative Risk
Suppose 50 basketball players use a new training program and 50 players use an old training program. At the end of the program we test each player to see if they pass a certain skills test.
The following table shows the number of players who passed and failed, based on the program they used:
The odds ratio is calculated as (34*11) / (16*39) = 0.599
We would interpret this to mean that the odds that a player passes the test by using the new program are just 0.599 times the odds that a player passes the test by using the old program.
In other words, the odds that a player passes the test are actually lowered by 40.1% by using the new program.
The relative risk is calculated as [34/(34+16)] / [39/(39+11)] = 0.872
We would interpret this to mean that the ratio of the probability of a player passing the test using the new program compared to the old program is 0.872.
Because this value is less than 1, it indicates that the probability of passing is actually lower under the new program compared to the old program.
We could also see this by directly computing the probability that a player passes under each program:
Probability of passing under new program = 34 / 50 =Â 68%
Probability of passing under old program = 39 / 50 = 78%
### How to Calculate Confidence Intervals
Once we calculate the odds ratio and relative risk, we may also be interested in computing confidence intervals for these two metrics.
A 95% confidence interval for the odds ratio can be calculated using the following formula:
95% C.I. for odds ratio = [ e^(ln(OR) – 1.96*SE(ln(OR))), e^(ln(OR) – 1.96*SE(ln(OR))) ]
where SE(ln(OR)) =√1/A + 1/B + 1/C + 1/D
The 95% C.I. for the odds ratio turns out to be (.245, 1.467). The image below shows the formula we used to calculate this confidence interval:
A 95% confidence interval for the relative risk can be calculated using the following formula:
95% C.I. for relative risk = exp(ln(RR) – 1.96*SE(ln(RR))) to exp(ln(RR) – 1.96*SE(ln(RR)))
where SE(ln(RR)) =√1/A + 1/C – 1/(A+B) – 1/(C+D)
The 95% C.I. for the relative risk turns out to be (.685, 1.109). The image below shows the formula we used to calculate this confidence interval:
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# What are Fibbinary numbers?
Ravi
## Overview
A Fibbinary number is an integer that has no consecutive set bits (ones) in its binary representation.
For example, the binary representation of 3 is 011. It’s not a Fibbinary number since the binary representation contains adjacent set bits.
In the following part of this shot, we’ll learn to check whether a given number n is a Fibbinary number or not.
### Checking if a number is a Fibbinary number
To check if a number is Fibbinary, we’ll use shift operators and the bitwise AND operation.
The steps of the solution are as follows:
1. Shift the number left or right by 1. For the left shift operator, use n & (n << 1). For the right shift operator, use n & (n >> 1).
2. Perform a bitwise AND operation on the number and the result from step 1. If the result from step 2 is greater than zero, then it means that there are adjacent set bits and the number is not a Fibbinary number. Otherwise, we can conclude that there are no adjacent set bits and the number is a Fibbinary number.
For example, let’s consider n=6. First, we use the left shift operator.
• n = 6 (0110)
• (n << 1) = 12 (1100)
• n & (n << 1) = 4 (0100)
Since n & (n << 1) is greater than zero, we know that there are adjacent set bits in the given number. Hence, we can conclude that 6 is not a Fibbinary number.
Now, let’s consider n=4. Again, we first use the left shift operator.
• n = 4 (0100)
• (n << 1) = 8 (1000)
• n & (n << 1) = 0 (0000)
Since n & (n << 1) is zero, we know that there are no adjacent set bits in the given number. Hence, we can conclude that 4 is a Fibbinary number.
### Code
public class Main {
public static void main(String[] args) {
int n = 1;
boolean res = (n & (n << 1)) == 0;
if(res) System.out.println(n + " is a Fibbinary number");
else System.out.println(n + " is not a Fibbinary number");
}
}
Example of finding a Fibbinary number in Java
### Explanation
• Line 4: We define an integer n.
• Line 5: We check whether there are adjacent set bits in n using the expression n & (n << 1) and check whether the result is zero or not.
• Lines 6–7: Based on the result from line 5, we print either that n is a Fibbinary number or that it isn’t a Fibbinary number.
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